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Schaum Outlines - Linear Algebra Fourth Edition

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Commercial problem-solving textbook (McGraw-Hill, 2009) by Seymour Lipschutz and Marc Lars Lipson, kept as a downloaded reference in the archive's linear algebra folder. It has 13 chapters covering vectors, matrices, linear systems, vector spaces, linear mappings, inner products, determinants, eigenvalues, canonical forms, dual spaces, bilinear forms, and operators on inner product spaces. Appendices cover multilinear products, algebraic structures, polynomials, and the Moore-Penrose inverse.

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SCHAUM’S outlines Problem / een A’|Solved ————————— Fourth Edition 612fully solved problems ™Concise explanations ofallcourse concepts *Information onalgebraic systems, polynomials, and matrix applications USE WITH THESE COURSES Beginning Linear Algebra *Linear Algebra *Advanced Linear Algebra Advanced Physics *Advanced Engineering *Quantitative Analysis Seymour Lipschutz, Ph.D. ¢Marc Lipson, Ph.D. SCHAUM’S outlines Linear Algebra Fourth Edition Seymour Lipschutz, Ph.D. Temple University Marc Lars Lipson, Ph.D. University of Virginia Schaum’s Outline Series New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto SCHAUM’S outlines Copyright © 2009, 2001, 1991, 1968 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior writ-ten permission of the publisher. ISBN: 978-0-07-154353-8MHID: 0-07-154353-8The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-154352-1, MHID: 0-07-154352-X.All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book,they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. To contact a representative please e-mail us at [email protected]. TERMS OF USEThis is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or anypart of it without McGraw-Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY , ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CANBE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY , EXPRESS ORIMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, er ror or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under nocircumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from theuse of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claimor cause whatsoever whether such claim or cause arises in contract, tort or otherwise. Preface Linear algebra has in recent years become an essential part of the mathematical background required by mathematicians and mathematics teachers, engineers, computer scientists, physicists, economists, and statisticians, among others. This requirement reﬂects the importance and wide applications of the subjectmatter. This book is designed for use as a textbook for a formal course in linear algebra or as a supplement to all current standard texts. It aims to present an introduction to linear algebra which will be found helpful to all readers regardless of their ﬁelds of speciﬁcation. More material has been included than can be covered in mostﬁrst courses. This has been done to make the book more ﬂexible, to provide a useful book of reference, and tostimulate further interest in the subject. Each chapter begins with clear statements of pertinent deﬁnitions, principles, and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, and to provide the repetition of basicprinciples so vital to effective learning. Numerous proofs, especially those of all essential theorems, are included among the solved problems. The supplementary problems serve as a complete review of the material of each chapter. The ﬁrst three chapters treat vectors in Euclidean space, matrix algebra, and systems of linear equations. These chapters provide the motivation and basic computational tools for the abstract investigations of vectorspaces and linear mappings which follow. After chapters on inner product spaces and orthogonality and on determinants, there is a detailed discussion of eigenvalues and eigenvectors giving conditions for representing a linear operator by a diagonal matrix. This naturally leads to the study of various canonical forms,speciﬁcally, the triangular, Jordan, and rational canonical forms. Later chapters cover linear functions and the dual space V*, and bilinear, quadratic, and Hermitian forms. The last chapter treats linear operators on inner product spaces. The main changes in the fourth edition have been in the appendices. First of all, we have expanded Appendix A on the tensor and exterior products of vector spaces where we have now included proofs on theexistence and uniqueness of such products. We also added appendices covering algebraic structures, includingmodules, and polynomials over a ﬁeld. Appendix D, ‘‘Odds and Ends,’’ includes the Moore–Penrose generalized inverse which appears in various applications, such as statistics. There are also many additional solved and supplementary problems. Finally, we wish to thank the staff of the McGraw-Hill Schaum’s Outline Series, especially Charles Wall, for their unfailing cooperation. S EYMOUR LIPSCHUTZ MARCLARSLIPSON iii This page intentionally left blank Contents CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1 1.1 Introduction 1.2 Vectors in Rn1.3 Vector Addition and Scalar Multi- plication 1.4 Dot (Inner) Product 1.5 Located Vectors, Hyperplanes, Lines, Curves in Rn1.6 Vectors in R3(Spatial Vectors), ijkNotation 1.7 Complex Numbers 1.8 Vectors in Cn CHAPTER 2 Algebra of Matrices 27 2.1 Introduction 2.2 Matrices 2.3 Matrix Addition and Scalar Multiplica- tion 2.4 Summation Symbol 2.5 Matrix Multiplication 2.6 Transpose of aMatrix 2.7 Square Matrices 2.8 Powers of Matrices, Polynomials in Matrices 2.9 Invertible (Nonsingular) Matrices 2.10 Special Types ofSquare Matrices 2.11 Complex Matrices 2.12 Block Matrices CHAPTER 3 Systems of Linear Equations 57 3.1 Introduction 3.2 Basic Deﬁnitions, Solutions 3.3 Equivalent Systems,Elementary Operations 3.4 Small Square Systems of Linear Equations 3.5Systems in Triangular and Echelon Forms 3.6 Gaussian Elimination 3.7Echelon Matrices, Row Canonical Form, Row Equivalence 3.8 GaussianElimination, Matrix Formulation 3.9 Matrix Equation of a System of Linear Equations 3.10 Systems of Linear Equations and Linear Combinations of Vectors 3.11 Homogeneous Systems of Linear Equations 3.12 ElementaryMatrices 3.13 LUDecomposition CHAPTER 4 Vector Spaces 112 4.1 Introduction 4.2 Vector Spaces 4.3 Examples of Vector Spaces 4.4 Linear Combinations, Spanning Sets 4.5 Subspaces 4.6 Linear Spans, RowSpace of a Matrix 4.7 Linear Dependence and Independence 4.8 Basis andDimension 4.9 Application to Matrices, Rank of a Matrix 4.10 Sums andDirect Sums 4.11 Coordinates CHAPTER 5 Linear Mappings 164 5.1 Introduction 5.2 Mappings, Functions 5.3 Linear Mappings (Linear Transformations) 5.4 Kernel and Image of a Linear Mapping 5.5 Singularand Nonsingular Linear Mappings, Isomorphisms 5.6 Operations withLinear Mappings 5.7 Algebra A(V) of Linear Operators CHAPTER 6 Linear Mappings and Matrices 195 6.1 Introduction 6.2 Matrix Representation of a Linear Operator 6.3 Change of Basis 6.4 Similarity 6.5 Matrices and General Linear Mappings CHAPTER 7 Inner Product Spaces, Orthogonality 226 7.1 Introduction 7.2 Inner Product Spaces 7.3 Examples of Inner ProductSpaces 7.4 Cauchy–Schwarz Inequality, Applications 7.5 Orthogonal-ity 7.6 Orthogonal Sets and Bases 7.7 Gram–Schmidt OrthogonalizationProcess 7.8 Orthogonal and Positive Deﬁnite Matrices 7.9 Complex InnerProduct Spaces 7.10 Normed Vector Spaces (Optional) v CHAPTER 8 Determinants 264 8.1 Introduction 8.2 Determinants of Orders 1 and 2 8.3 Determinants of Order 3 8.4 Permutations 8.5 Determinants of Arbitrary Order 8.6 Proper- ties of Determinants 8.7 Minors and Cofactors 8.8 Evaluation of Determi-nants 8.9 Classical Adjoint 8.10 Applications to Linear Equations,Cramer’s Rule 8.11 Submatrices, Minors, Principal Minors 8.12 BlockMatrices and Determinants 8.13 Determinants and Volume 8.14 Determi-nant of a Linear Operator 8.15 Multilinearity and Determinants CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 292 9.1 Introduction 9.2 Polynomials of Matrices 9.3 Characteristic Polyno-mial, Cayley–Hamilton Theorem 9.4 Diagonalization, Eigenvalues andEigenvectors 9.5 Computing Eigenvalues and Eigenvectors, DiagonalizingMatrices 9.6 Diagonalizing Real Symmetric Matrices and QuadraticForms 9.7 Minimal Polynomial 9.8 Characteristic and Minimal Polyno- mials of Block Matrices CHAPTER 10 Canonical Forms 325 10.1 Introduction 10.2 Triangular Form 10.3 Invariance 10.4 InvariantDirect-Sum Decompositions 10.5 Primary Decomposition 10.6 Nilpotent Operators 10.7 Jordan Canonical Form 10.8 Cyclic Subspaces 10.9Rational Canonical Form 10.10 Quotient Spaces CHAPTER 11 Linear Functionals and the Dual Space 349 11.1 Introduction 11.2 Linear Functionals and the Dual Space 11.3 DualBasis 11.4 Second Dual Space 11.5 Annihilators 11.6 Transpose of aLinear Mapping CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 359 12.1 Introduction 12.2 Bilinear Forms 12.3 Bilinear Forms andMatrices 12.4 Alternating Bilinear Forms 12.5 Symmetric BilinearForms, Quadratic Forms 12.6 Real Symmetric Bilinear Forms, Law ofInertia 12.7 Hermitian Forms CHAPTER 13 Linear Operators on Inner Product Spaces 377 13.1 Introduction 13.2 Adjoint Operators 13.3 Analogy Between A(V) and C, Special Linear Operators 13.4 Self-Adjoint Operators 13.5 Orthogonal and Unitary Operators 13.6 Orthogonal and Unitary Matrices 13.7 Changeof Orthonormal Basis 13.8 Positive Deﬁnite and Positive Operators 13.9Diagonalization and Canonical Forms in Inner Product Spaces 13.10Spectral Theorem APPENDIX A Multilinear Products 396 APPENDIX B Algebraic Structures 403 APPENDIX C Polynomials over a Field 411 APPENDIX D Odds and Ends 415 List of Symbols 420 Index 421vi Contents CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.1 Introduction There are two ways to motivate the notion of a vector: one is by means of lists of numbers and subscripts, and the other is by means of certain objects in physics. We discuss these two ways below. Here we assume the reader is familiar with the elementary properties of the ﬁeld of real numbers, denoted by R. On the other hand, we will review properties of the ﬁeld of complex numbers, denoted by C. In the context of vectors, the elements of our number ﬁelds are called scalars . Although we will restrict ourselves in this chapter to vectors whose elements come from Rand then from C, many of our operations also apply to vectors whose entries come from some arbitrary ﬁeld K. Lists of Numbers Suppose the weights (in pounds) of eight students are listed as follows: 156;125;145;134;178;145;162;193 One can denote all the values in the list using only one symbol, say w, but with different subscripts; that is, w1;w2;w3;w4;w5;w6;w7;w8 Observe that each subscript denotes the position of the value in the list. For example, w1¼156;the first number ;w2¼125;the second number ;... Such a list of values, w¼ðw1;w2;w3;...;w8Þ is called a linear array orvector . Vectors in Physics Many physical quantities, such as temperature and speed, possess only ‘‘magnitude.’ ’ These quantities can be represented by real numbers and are called scalars . On the other hand, there are also quantities, such as force and velocity, that possess both ‘‘magnitude’’ and ‘‘direction.’ ’ These quantities, which canbe represented by arrows having appropriate lengths and directions and emanating from some givenreference point O, are called vectors . Now we assume the reader is familiar with the space R 3where all the points in space are represented by ordered triples of real numbers. Suppose the origin of the axes in R3is chosen as the reference point O for the vectors discussed above. Then every vector is uniquely determined by the coordinates of itsendpoint, and vice versa. There are two important operations, vector addition and scalar multiplication, associated with vectors in physics. The deﬁnition of these operations and the relationship between these operations and theendpoints of the vectors are as follows. 1 CHAPTER 1 (i)Vector Addition: The resultant uþvof two vectors uandvis obtained by the parallelogram law ; that is, uþvis the diagonal of the parallelogram formed by uandv. Furthermore, ifða;b;cÞand ða0;b0;c0Þare the endpoints of the vectors uandv, thenðaþa0;bþb0;cþc0Þis the endpoint of the vector uþv. These properties are pictured in Fig. 1-1(a). (ii)Scalar Multiplication: The product kuof a vector uby a real number kis obtained by multiplying the magnitude of ubykand retaining the same direction if k>0 or the opposite direction if k<0. Also, ifða;b;cÞis the endpoint of the vector u, thenðka;kb;kcÞis the endpoint of the vector ku. These properties are pictured in Fig. 1-1(b). Mathematically, we identify the vector uwith itsða;b;cÞand write u¼ða;b;cÞ. Moreover, we call the ordered triple ða;b;cÞof real numbers a point or vector depending upon its interpretation. We generalize this notion and call an n-tupleða1;a2;...;anÞof real numbers a vector. However, special notation may be used for the vectors in R3called spatial vectors (Section 1.6). 1.2 Vectors in Rn The set of all n-tuples of real numbers, denoted by Rn,i sc a l l e d n-space . A particular n-tuple in Rn,s a y u¼ða1;a2;...;anÞ is called a point orvector . The numbers aiare called the coordinates ,components ,entries ,o relements ofu. Moreover, when discussing the space Rn, we use the term scalar for the elements of R. Two vectors, uandv, are equal , written u¼v, if they have the same number of components and if the corresponding components are equal. Although the vectors ð1;2;3Þandð2;3;1Þcontain the same three numbers, these vectors are not equal because corresponding entries are not equal. The vectorð0;0;...;0Þwhose entries are all 0 is called the zero vector and is usually denoted by 0. EXAMPLE 1.1 (a) The following are vectors: ð2;/C05Þ;ð7;9Þ;ð0;0;0Þ;ð3;4;5Þ The ﬁrst two vectors belong to R2, whereas the last two belong to R3. The third is the zero vector in R3. (b) Find x;y;zsuch thatðx/C0y;xþy;z/C01Þ¼ð 4;2;3Þ. By deﬁnition of equality of vectors, corresponding entries must be equal. Thus, x/C0y¼4; xþy¼2; z/C01¼3 Solving the above system of equations yields x¼3,y¼/C01,z¼4.Figure 1-12 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors Column Vectors Sometimes a vector in n-space Rnis written vertically rather than horizontally. Such a vector is called a column vector , and, in this context, the horizontally written vectors in Example 1.1 are called row vectors . For example, the following are column vectors with 2 ;2;3, and 3 components, respectively: 1 2/C20/C21 ;3 /C04/C20/C21 ;1 5 /C062 43 5;1:5 2 3 /C0152 643 75 We also note that any operation deﬁned for row vectors is deﬁned analogously for column vectors. 1.3 Vector Addition and Scalar Multiplication Consider two vectors uand vinRn, say u¼ða1;a2;...;anÞ and v¼ðb1;b2;...;bnÞ Their sum,w r i t t e n uþv, is the vector obtained by adding corresponding components from uandv.T h a ti s , uþv¼ða1þb1;a2þb2;...;anþbnÞ The scalar product or, simply, product , of the vector uby a real number k, written ku, is the vector obtained by multiplying each component of ubyk. That is, ku¼kða1;a2;...;anÞ¼ð ka1;ka2;...;kanÞ Observe that uþvand kuare also vectors in Rn. The sum of vectors with different numbers of components is not deﬁned. Negatives and subtraction are deﬁned in Rnas follows: /C0u¼ð/C0 1Þu and u/C0v¼uþð/C0 vÞ The vector/C0uis called the negative ofu, and u/C0vis called the difference ofuand v. Now suppose we are given vectors u1;u2;...;uminRnand scalars k1;k2;...;kminR. We can multiply the vectors by the corresponding scalars and then add the resultant scalar products to form thevector v¼k 1u1þk2u2þk3u3þ/C1/C1/C1þ kmum Such a vector vis called a linear combination of the vectors u1;u2;...;um. EXAMPLE 1.2 (a) Let u¼ð2;4;/C05Þand v¼ð1;/C06;9Þ. Then uþv¼ð2þ1;4þð/C0 5Þ;/C05þ9Þ¼ð 3;/C01;4Þ 7u¼ð7ð2Þ;7ð4Þ;7ð/C05ÞÞ¼ð 14;28;/C035Þ /C0v¼ð/C0 1Þð1;/C06;9Þ¼ð/C0 1;6;/C09Þ 3u/C05v¼ð6;12;/C015Þþð/C0 5;30;/C045Þ¼ð 1;42;/C060Þ (b) The zero vector 0 ¼ð0;0;...;0ÞinRnis similar to the scalar 0 in that, for any vector u¼ða1;a2;...;anÞ. uþ0¼ða1þ0;a2þ0;...;anþ0Þ¼ð a1;a2;...;anÞ¼u (c) Let u¼2 3 /C042 43 5and v¼3 /C01 /C022 43 5. Then 2 u/C03v¼4 6 /C082 43 5þ/C09 3 62 43 5¼/C05 9 /C022 43 5.CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 3 Basic properties of vectors under the operations of vector addition and scalar multiplication are described in the following theorem. THEOREM 1.1: For any vectors u;v;winRnand any scalars k;k0inR, (i)ðuþvÞþw¼uþðvþwÞ, (v) kðuþvÞ¼kuþkv, (ii) uþ0¼u; (vi)ðkþk0Þu¼kuþk0u, (iii) uþð/C0 uÞ¼0; (vii) (kk’)u=k(k’u) ; (iv) uþv¼vþu, (viii) 1 u¼u. We postpone the proof of Theorem 1.1 until Chapter 2, where it appears in the context of matrices (Problem 2.3). Suppose uandvare vectors in Rnfor which u¼kvfor some nonzero scalar kinR. Then uis called a multiple ofv. Also, uis said to be in the same oropposite direction asvaccording to whether k>0o r k<0. 1.4 Dot (Inner) Product Consider arbitrary vectors uand vinRn; say, u¼ða1;a2;...;anÞ and v¼ðb1;b2;...;bnÞ Thedot product orinner product orscalar product ofuand vis denoted and deﬁned by u/C1v¼a1b1þa2b2þ/C1/C1/C1þ anbn That is, u/C1vis obtained by multiplying corresponding components and adding the resulting products. The vectors uand vare said to be orthogonal (orperpendicular ) if their dot product is zero—that is, if u/C1v¼0. EXAMPLE 1.3 (a) Let u¼ð1;/C02;3Þ,v¼ð4;5;/C01Þ,w¼ð2;7;4Þ. Then, u/C1v¼1ð4Þ/C02ð5Þþ3ð/C01Þ¼4/C010/C03¼/C09 u/C1w¼2/C014þ12¼0; v/C1w¼8þ35/C04¼39 Thus, uandware orthogonal. (b) Let u¼2 3 /C042 43 5and v¼3 /C01 /C022 43 5. Then u/C1v¼6/C03þ8¼11. (c) Suppose u¼ð1;2;3;4Þand v¼ð6;k;/C08;2Þ. Find kso that uand vare orthogonal. First obtain u/C1v¼6þ2k/C024þ8¼/C010þ2k. Then set u/C1v¼0 and solve for k: /C010þ2k¼0o r2 k¼10 or k¼5 Basic properties of the dot product in Rn(proved in Problem 1.13) follow. THEOREM 1.2: For any vectors u;v;winRnand any scalar kinR: (i)ðuþvÞ/C1w¼u/C1wþv/C1w; (iii) u/C1v¼v/C1u, (ii)ðkuÞ/C1v¼kðu/C1vÞ, (iv) u/C1u/C210;andu/C1u¼0 iff u¼0. Note that (ii) says that we can ‘‘take kout’ ’ from the ﬁrst position in an inner product. By (iii) and (ii), u/C1ðkvÞ¼ð kvÞ/C1u¼kðv/C1uÞ¼kðu/C1vÞ4 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors That is, we can also ‘‘take kout’ ’ from the second position in an inner product. The space Rnwith the above operations of vector addition, scalar multiplication, and dot product is usually called Euclidean n-space . Norm (Length) of a Vector Thenorm orlength of a vector uinRn, denoted bykuk, is deﬁned to be the nonnegative square root of u/C1u. In particular, if u¼ða1;a2;...;anÞ, then kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃu/C1up¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 1þa2 2þ/C1/C1/C1þ a2nq That is,kukis the square root of the sum of the squares of the components of u. Thus,kuk/C210, and kuk¼0 if and only if u¼0. A vector uis called a unitvector ifkuk¼1 or, equivalently, if u/C1u¼1. For any nonzero vector vin Rn, the vector ^v¼1 kvkv¼v kvk is the unique unit vector in the same direction as v. The process of ﬁnding ^vfrom vis called normalizing v. EXAMPLE 1.4 (a) Suppose u¼ð1;/C02;/C04;5;3Þ. To ﬁndkuk, we can ﬁrst ﬁndkuk2¼u/C1uby squaring each component of uand adding, as follows: kuk2¼12þð/C0 2Þ2þð/C0 4Þ2þ52þ32¼1þ4þ16þ25þ9¼55 Thenkuk¼ﬃﬃﬃﬃﬃ 55p . (b) Let v¼ð1;/C03;4;2Þandw¼ð1 2;/C01 6;56;16Þ. Then kvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ9þ16þ4p ¼ﬃﬃﬃﬃﬃ 30p andkwk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9 36þ1 36þ25 36þ1 36r ¼ﬃﬃﬃﬃﬃ 36 36r ¼ﬃﬃﬃ 1p ¼1 Thus wis a unit vector, but vis not a unit vector. However, we can normalize vas follows: ^v¼v kvk¼1ﬃﬃﬃﬃﬃ 30p ;/C03ﬃﬃﬃﬃﬃ 30p ;4ﬃﬃﬃﬃﬃ 30p ;2ﬃﬃﬃﬃﬃ 30p/C18/C19 This is the unique unit vector in the same direction as v. The following formula (proved in Problem 1.14) is known as the Schwarz inequality or Cauchy– Schwarz inequality. It is used in many branches of mathematics. THEOREM 1.3 (Schwarz): For any vectors u;vinRn,ju/C1vj/C20k ukkvk. Using the above inequality, we also prove (Problem 1.15) the following result known as the ‘‘triangle inequality’’ or Minkowski’s inequality. THEOREM 1.4 (Minkowski): For any vectors u;vinRn,kuþvk/C20k ukþk vk. Distance, Angles, Projections Thedistance between vectors u¼ða1;a2;...;anÞand v¼ðb1;b2;...;bnÞinRnis denoted and deﬁned by dðu;vÞ¼k u/C0vk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ða1/C0b1Þ2þða2/C0b2Þ2þ/C1/C1/C1þð an/C0bnÞ2q One can show that this deﬁnition agrees with the usual notion of distance in the Euclidean plane R2or space R3.CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 5 Theangle ybetween nonzero vectors u;vinRnis deﬁned by cosy¼u/C1v kukkvk This deﬁnition is well deﬁned, because, by the Schwarz inequality (Theorem 1.3), /C01/C20u/C1v kukkvk/C201 Note that if u/C1v¼0, then y¼90/C14(ory¼p=2). This then agrees with our previous deﬁnition of orthogonality. Theprojection of a vector uonto a nonzero vector vis the vector denoted and deﬁned by projðu;vÞ¼u/C1v kvk2v¼u/C1v v/C1vv We show below that this agrees with the usual notion of vector projection in physics. EXAMPLE 1.5 (a) Suppose u¼ð1;/C02;3Þand v¼ð2;4;5Þ. Then dðu;vÞ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð1/C02Þ2þð/C0 2/C04Þ2þð3/C05Þ2q ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ36þ4p ¼ﬃﬃﬃﬃﬃ 41p To ﬁnd cos y, where yis the angle between uand v, we ﬁrst ﬁnd u/C1v¼2/C08þ15¼9;kuk2¼1þ4þ9¼14;kvk2¼4þ16þ25¼45 Then cosy¼u/C1v kukkvk¼9ﬃﬃﬃﬃﬃ 14pﬃﬃﬃﬃﬃ 45p Also, projðu;vÞ¼u/C1v kvk2v¼9 45ð2;4;5Þ¼1 5ð2;4;5Þ¼2 5;4 5;1/C18/C19 (b) Consider the vectors uand vin Fig. 1-2(a) (with respective endpoints AandB). The (perpendicular) projection ofuonto vis the vector u* with magnitude ku*k¼k ukcosy¼kuku/C1v kukvk¼u/C1v kvk To obtain u*, we multiply its magnitude by the unit vector in the direction of v, obtaining u*¼ku*kv kvk¼u/C1v kvkv kvk¼u/C1v kvk2v This is the same as the above deﬁnition of proj ðu;vÞ. Figure 1-2z y x0u ()bB bbb(, , )123 u=B–AA aaa( ,,)123P b a b aba(– , – , –)1122 3 3 0u ()aProjection of onto u* uA u*BC θ6 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.5 Located Vectors, Hyperplanes, Lines, Curves in Rn This section distinguishes between an n-tuple PðaiÞ/C17Pða1;a2;...;anÞviewed as a point in Rnand an n-tuple u¼½c1;c2;...;cn/C138viewed as a vector (arrow) from the origin Oto the point Cðc1;c2;...;cnÞ. Located Vectors Any pair of points AðaiÞandBðbiÞinRndeﬁnes the located vector ordirected line segment from AtoB, written AB/C131!. We identify AB/C131!with the vector u¼B/C0A¼½b1/C0a1;b2/C0a2;...;bn/C0an/C138 because AB/C131!and uhave the same magnitude and direction. This is pictured in Fig. 1-2(b) for the points Aða1;a2;a3Þand Bðb1;b2;b3ÞinR3and the vector u¼B/C0Awhich has the endpoint Pðb1/C0a1,b2/C0a2,b3/C0a3Þ. Hyperplanes Ahyperplane H inRnis the set of points ðx1;x2;...;xnÞthat satisfy a linear equation a1x1þa2x2þ/C1/C1/C1þ anxn¼b where the vector u¼½a1;a2;...;an/C138of coefﬁcients is not zero. Thus a hyperplane HinR2is a line, and a hyperplane HinR3is a plane. We show below, as pictured in Fig. 1-3(a) for R3, that uis orthogonal to any directed line segment PQ/C131!, where PðpiÞandQðqiÞare points in H:[For this reason, we say that uis normal toHand that Hisnormal tou:] Because PðpiÞandQðqiÞbelong to H;they satisfy the above hyperplane equation—that is, a1p1þa2p2þ/C1/C1/C1þ anpn¼band a1q1þa2q2þ/C1/C1/C1þ anqn¼b v¼PQ/C131!¼Q/C0P¼½q1/C0p1;q2/C0p2;...;qn/C0pn/C138 Let Then u/C1v¼a1ðq1/C0p1Þþa2ðq2/C0p2Þþ/C1/C1/C1þ anðqn/C0pnÞ ¼ða1q1þa2q2þ/C1/C1/C1þ anqnÞ/C0ð a1p1þa2p2þ/C1/C1/C1þ anpnÞ¼b/C0b¼0 Thus v¼PQ/C131!is orthogonal to u;as claimed.Figure 1-3 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 7 Lines in Rn The line L inRnpassing through the point Pðb1;b2;...;bnÞand in the direction of a nonzero vector u¼½a1;a2;...;an/C138consists of the points Xðx1;x2;...;xnÞthat satisfy X¼Pþtu orx1¼a1tþb1 x2¼a2tþb2 :::::::::::::::::::: xn¼antþbnorLðtÞ¼ð aitþbiÞ8 >>< >>: where the parameter t takes on all real values. Such a line LinR3is pictured in Fig. 1-3(b). EXAMPLE 1.6 (a) Let Hbe the plane in R3corresponding to the linear equation 2 x/C05yþ7z¼4. Observe that Pð1;1;1Þand Qð5;4;2Þare solutions of the equation. Thus PandQand the directed line segment v¼PQ/C131!¼Q/C0P¼½5/C01;4/C01;2/C01/C138¼½4;3;1/C138 lie on the plane H. The vector u¼½2;/C05;7/C138is normal to H, and, as expected, u/C1v¼½2;/C05;7/C138/C1½4;3;1/C138¼8/C015þ7¼0 That is, uis orthogonal to v. (b) Find an equation of the hyperplane HinR4that passes through the point Pð1;3;/C04;2Þand is normal to the vector u¼½4;/C02;5;6/C138. The coefﬁcients of the unknowns of an equation of Hare the components of the normal vector u; hence, the equation of Hmust be of the form 4x1/C02x2þ5x3þ6x4¼k Substituting Pinto this equation, we obtain 4ð1Þ/C02ð3Þþ5ð/C04Þþ6ð2Þ¼k or 4/C06/C020þ12¼k or k¼/C010 Thus, 4 x1/C02x2þ5x3þ6x4¼/C010 is the equation of H. (c) Find the parametric representation of the line LinR4passing through the point Pð1;2;3;/C04Þand in the direction of u¼½5;6;/C07;8/C138. Also, ﬁnd the point QonLwhen t¼1. Substitution in the above equation for Lyields the following parametric representation: x1¼5tþ1; x2¼6tþ2; x3¼/C07tþ3; x4¼8t/C04 or, equivalently, LðtÞ¼ð 5tþ1;6tþ2;/C07tþ3;8t/C04Þ Note that t¼0 yields the point PonL. Substitution of t¼1 yields the point Qð6;8;/C04;4ÞonL. Curves in Rn LetDbe an interval (ﬁnite or inﬁnite) on the real line R. A continuous function F:D!Rnis acurve in Rn. Thus, to each point t2Dthere is assigned the following point in Rn: FðtÞ¼½ F1ðtÞ;F2ðtÞ;...;FnðtÞ/C138 Moreover, the derivative (if it exists) of FðtÞyields the vector VðtÞ¼dFðtÞ dt¼dF1ðtÞ dt;dF2ðtÞ dt;...;dFnðtÞ dt/C20/C218 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors which is tangent to the curve. Normalizing VðtÞyields TðtÞ¼VðtÞ kVðtÞk Thus, TðtÞis the unit tangent vector to the curve. (Unit vectors with geometrical signiﬁcance are often presented in bold type.) EXAMPLE 1.7 Consider the curve FðtÞ¼½ sint;cost;t/C138inR3. Taking the derivative of FðtÞ[or each component of FðtÞ] yields VðtÞ¼½ cost;/C0sint;1/C138 which is a vector tangent to the curve. We normalize VðtÞ. First we obtain kVðtÞk2¼cos2tþsin2tþ1¼1þ1¼2 Then the unit tangent vection TðtÞto the curve follows: TðtÞ¼VðtÞ kVðtÞk¼costﬃﬃﬃ 2p;/C0sintﬃﬃﬃ 2p ;1ﬃﬃﬃ 2p/C20/C21 1.6 Vectors in R3(Spatial Vectors), ijk Notation Vectors in R3, called spatial vectors , appear in many applications, especially in physics. In fact, a special notation is frequently used for such vectors as follows: i¼½1;0;0/C138denotes the unit vector in the xdirection : j¼½0;1;0/C138denotes the unit vector in the ydirection : k¼½0;0;1/C138denotes the unit vector in the zdirection : Then any vector u¼½a;b;c/C138inR3can be expressed uniquely in the form u¼½a;b;c/C138¼aiþbjþcj Because the vectors i;j;kare unit vectors and are mutually orthogonal, we obtain the following dot products: i/C1i¼1;j/C1j¼1;k/C1k¼1 and i/C1j¼0;i/C1k¼0;j/C1k¼0 Furthermore, the vector operations discussed above may be expressed in the ijknotation as follows. Suppose u¼a1iþa2jþa3k and v¼b1iþb2jþb3k Then uþv¼ða1þb1Þiþða2þb2Þjþða3þb3Þk and cu¼ca1iþca2jþca3k where cis a scalar. Also, u/C1v¼a1b1þa2b2þa3b3 andkuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃu/C1up¼a2 1þa2 2þa2 3 EXAMPLE 1.8 Suppose u¼3iþ5j/C02kand v¼4i/C08jþ7k. (a) To ﬁnd uþv, add corresponding components, obtaining uþv¼7i/C03jþ5k (b) To ﬁnd 3 u/C02v, ﬁrst multiply by the scalars and then add: 3u/C02v¼ð9iþ13j/C06kÞþð/C0 8iþ16j/C014kÞ¼iþ29j/C020kCHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 9 (c) To ﬁnd u/C1v, multiply corresponding components and then add: u/C1v¼12/C040/C014¼/C042 (d) To ﬁndkuk, take the square root of the sum of the squares of the components: kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9þ25þ4p ¼ﬃﬃﬃﬃﬃ 38p Cross Product There is a special operation for vectors uandvinR3that is not deﬁned in Rnforn6¼3. This operation is called the cross product and is denoted by u/C2v. One way to easily remember the formula for u/C2vis to use the determinant (of order two) and its negative, which are denoted and deﬁned as follows: ab cd/C12/C12/C12/C12/C12/C12/C12/C12¼ad/C0bc and/C0ab cd/C12/C12/C12/C12/C12/C12/C12/C12¼bc/C0ad Here aanddare called the diagonal elements and bandcare the nondiagonal elements. Thus, the determinant is the product adof the diagonal elements minus the product bcof the nondiagonal elements, but vice versa for the negative of the determinant. Now suppose u¼a 1iþa2jþa3kand v¼b1iþb2jþb3k. Then u/C2v¼ða2b3/C0a3b2Þiþða3b1/C0a1b3Þjþða1b2/C0a2b1Þk ¼a1a2a3 b1b2b3/C12/C12/C12/C12/C12/C12/C12/C12i/C0a 1a2a3 b1b2b3/C12/C12/C12/C12/C12/C12/C12/C12jþa 1a2a3 b1b2b3/C12/C12/C12/C12/C12/C12/C12/C12i That is, the three components of u/C2vare obtained from the array a 1a2a3 b1b2b3/C20/C21 (which contain the components of uabove the component of v) as follows: (1) Cover the ﬁrst column and take the determinant. (2) Cover the second column and take the negative of the determinant.(3) Cover the third column and take the determinant. Note that u/C2vis a vector; hence, u/C2vis also called the vector product orouter product ofu and v. EXAMPLE 1.9 Find u/C2vwhere: (a) u¼4iþ3jþ6k,v¼2iþ5j/C03k, (b) u¼½2;/C01;5/C138,v¼½3;7;6/C138. (a) Use43 6 25/C03/C20/C21 to get u/C2v¼ð /C0 9/C030Þiþð12þ12Þjþð20/C06Þk¼/C0 39iþ24jþ14k (b) Use2/C015 37 6/C20/C21 to get u/C2v¼½ /C0 6/C035;15/C012;14þ3/C138¼½ /C0 41;3;17/C138 Remark: The cross products of the vectors i;j;kare as follows: i/C2j¼k; j/C2k¼i; k/C2i¼j j/C2i¼/C0k; k/C2j¼/C0i; i/C2k¼/C0j Thus, if we view the triple ði;j;kÞas a cyclic permutation, where ifollows kand hence kprecedes i, then the product of two of them in the given direction is the third one, but the product of two of them in the opposite direction is the negative of the third one. Two important properties of the cross product are contained in the following theorem.10 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors THEOREM 1.5: Letu;v;wbe vectors in R3. (a) The vector u/C2vis orthogonal to both uand v. (b) The absolute value of the ‘‘triple product’ ’ u/C1v/C2w represents the volume of the parallelopiped formed by the vectors u;v,w. [See Fig. 1-4(a).] We note that the vectors u;v,u/C2vform a right-handed system, and that the following formula gives the magnitude of u/C2v: ku/C2vk¼k ukkvksiny where yis the angle between uand v. 1.7 Complex Numbers The set of complex numbers is denoted by C. Formally, a complex number is an ordered pair ða;bÞof real numbers where equality, addition, and multiplication are deﬁned as follows: ða;bÞ¼ð c;dÞif and only if a¼candb¼d ða;bÞþð c;dÞ¼ð aþc;bþdÞ ða;bÞ/C1ðc;dÞ¼ð ac/C0bd;adþbcÞ We identify the real number awith the complex number ða;0Þ; that is, a$ða;0Þ This is possible because the operations of addition and multiplication of real numbers are preserved under the correspondence; that is, ða;0Þþð b;0Þ¼ð aþb;0Þ andða;0Þ/C1ðb;0Þ¼ð ab;0Þ Thus we view Ras a subset of C, and replaceða;0Þbyawhenever convenient and possible. We note that the set Cof complex numbers with the above operations of addition and multiplication is afield of numbers, like the set Rof real numbers and the set Qofrational numbers .Figure 1-4 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 11 The complex number ð0;1Þis denoted by i. It has the important property that i2¼ii¼ð0;1Þð0;1Þ¼ð/C0 1;0Þ¼/C0 1o r i¼ﬃﬃﬃﬃﬃﬃﬃ /C01p Accordingly, any complex number z¼ða;bÞcan be written in the form z¼ða;bÞ¼ð a;0Þþð 0;bÞ¼ð a;0Þþð b;0Þ/C1ð0;1Þ¼aþbi The above notation z¼aþbi, where a/C17Rezandb/C17Imzare called, respectively, the real and imaginary parts ofz, is more convenient than ða;bÞ. In fact, the sum and product of complex numbers z¼aþbiandw¼cþdican be derived by simply using the commutative and distributive laws and i2¼/C01: zþw¼ðaþbiÞþð cþdiÞ¼aþcþbiþdi¼ðaþbÞþð cþdÞi zw¼ðaþbiÞðcþdiÞ¼acþbciþadiþbdi2¼ðac/C0bdÞþð bcþadÞi We also deﬁne the negative ofzand subtraction in Cby /C0z¼/C01z and w/C0z¼wþð/C0 zÞ Warning: The letter irepresentingﬃﬃﬃﬃﬃﬃﬃ /C01p has no relationship whatsoever to the vector i¼½1;0;0/C138in Section 1.6. Complex Conjugate, Absolute Value Consider a complex number z¼aþbi. The conjugate ofzis denoted and deﬁned by /C22z¼aþbi¼a/C0bi Then z/C22z¼ðaþbiÞða/C0biÞ¼a2/C0b2i2¼a2þb2. Note that zis real if and only if /C22z¼z. Theabsolute value ofz, denoted byjzj, is deﬁned to be the nonnegative square root of z/C22z. Namely, jzj¼ﬃﬃﬃﬃ z/C22zp ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2þb2p Note thatjzjis equal to the norm of the vector ða;bÞinR2. Suppose z6¼0. Then the inverse z/C01ofzand division in Cofwbyzare given, respectively, by z/C01¼/C22z z/C22z¼a a2þb2/C0b a2þb2i andw z/C0w/C22z z/C22z¼wz/C01 EXAMPLE 1.10 Suppose z¼2þ3iandw¼5/C02i. Then zþw¼ð2þ3iÞþð 5/C02iÞ¼2þ5þ3i/C02i¼7þi zw¼ð2þ3iÞð5/C02iÞ¼10þ15i/C04i/C06i2¼16þ11i /C22z¼2þ3i¼2/C03i and /C22w¼5/C02i¼5þ2i w z¼5/C02i 2þ3i¼ð5/C02iÞð2/C03iÞ ð2þ3iÞð2/C03iÞ¼4/C019i 13¼4 13/C019 13i jzj¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4þ9p ¼ﬃﬃﬃﬃﬃ 13p andjwj¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 25þ4p ¼ﬃﬃﬃﬃﬃ 29p Complex Plane Recall that the real numbers Rcan be represented by points on a line. Analogously, the complex numbers Ccan be represented by points in the plane. Speciﬁcally, we let the point ða;bÞin the plane represent the complex number aþbias shown in Fig. 1-4(b). In such a case, jzjis the distance from the origin Oto the point z. The plane with this representation is called the complex plane , just like the line representing Ris called the real line .12 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.8 Vectors in Cn The set of all n-tuples of complex numbers, denoted by Cn, is called complex n-space . Just as in the real case, the elements of Cnare called points orvectors , the elements of Care called scalars , and vector addition in Cnand scalar multiplication on Cnare given by ½z1;z2;...;zn/C138þ½w1;w2;...;wn/C138¼½z1þw1;z2þw2;...;znþwn/C138 z½z1;z2;...;zn/C138¼½zz1;zz2;...;zzn/C138 where the zi,wi, and zbelong to C. EXAMPLE 1.11 Consider vectors u¼½2þ3i;4/C0i;3/C138and v¼½3/C02i;5i;4/C06i/C138inC3. Then uþv¼½2þ3i;4/C0i;3/C138þ½3/C02i;5i;4/C06i/C138¼½ 5þi;4þ4i;7/C06i/C138 ð5/C02iÞu¼½ ð 5/C02iÞð2þ3iÞ;ð5/C02iÞð4/C0iÞ;ð5/C02iÞð3Þ/C138 ¼ ½ 16þ11i;18/C013i;15/C06i/C138 Dot (Inner) Product in Cn Consider vectors u¼½z1;z2;...;zn/C138andv¼½w1;w2;...;wn/C138inCn. The dotorinner product ofuandvis denoted and deﬁned by u/C1v¼z1/C22w1þz2/C22w2þ/C1/C1/C1þ zn/C22wn This deﬁnition reduces to the real case because /C22wi¼wiwhen wiis real. The norm of uis deﬁned by kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃu/C1up¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ z1/C22z1þz2/C22z2þ/C1/C1/C1þ zn/C22znp ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ jz1j2þjz2j2þ/C1/C1/C1þj vnj2q We emphasize that u/C1uand sokukare real and positive when u6¼0 and 0 when u¼0. EXAMPLE 1.12 Consider vectors u¼½2þ3i;4/C0i;3þ5i/C138and v¼½3/C04i;5i;4/C02i/C138inC3. Then u/C1v¼ð2þ3iÞð3/C04iÞþð 4/C0iÞð5iÞþð 3þ5iÞð4/C02iÞ ¼ð2þ3iÞð3þ4iÞþð 4/C0iÞð/C05iÞþð 3þ5iÞð4þ2iÞ ¼ð/C0 6þ13iÞþð/C0 5/C020iÞþð 2þ26iÞ¼/C0 9þ19i u/C1u¼j2þ3ij2þj4/C0ij2þj3þ5ij2¼4þ9þ16þ1þ9þ25¼64 kuk¼ﬃﬃﬃﬃﬃ 64p ¼8 The space Cnwith the above operations of vector addition, scalar multiplication, and dot product, is called complex Euclidean n-space . Theorem 1.2 for Rnalso holds for Cnif we replace u/C1v¼v/C1uby u/C1v¼u/C1v On the other hand, the Schwarz inequality (Theorem 1.3) and Minkowski’s inequality (Theorem 1.4) are true for Cnwith no changes. SOLVED PROBLEMS Vectors in Rn 1.1. Determine which of the following vectors are equal: u1¼ð1;2;3Þ; u2¼ð2;3;1Þ; u3¼ð1;3;2Þ; u4¼ð2;3;1Þ Vectors are equal only when corresponding entries are equal; hence, only u2¼u4.CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 13 1.2. Letu¼ð2;/C07;1Þ,v¼ð/C0 3;0;4Þ,w¼ð0;5;/C08Þ. Find: (a) 3 u/C04v, (b) 2 uþ3v/C05w. First perform the scalar multiplication and then the vector addition. (a) 3 u/C04v¼3ð2;/C07;1Þ/C04ð/C03;0;4Þ¼ð 6;/C021;3Þþð 12;0;/C016Þ¼ð 18;/C021;/C013Þ (b) 2 uþ3v/C05w¼ð4;/C014;2Þþð/C0 9;0;12Þþð 0;/C025;40Þ¼ð/C0 5;/C039;54Þ 1.3. Letu¼5 3 /C042 43 5;v¼/C01 5 22 43 5;w¼3 /C01 /C022 43 5. Find: (a) 5 u/C02v, (b)/C02uþ4v/C03w. First perform the scalar multiplication and then the vector addition: (a) 5 u/C02v¼55 3 /C042 43 5/C02/C01 5 22 43 5¼25 15 /C0202 43 5þ2 /C010 /C042 43 5¼27 5 /C0242 43 5 (b)/C02uþ4v/C03w¼/C010 /C06 82 43 5þ/C04 20 82 43 5þ/C09 3 62 43 5¼/C023 17 222 43 5 1.4. Find xandy, where: (a)ðx;3Þ¼ð 2;xþyÞ, (b)ð4;yÞ¼xð2;3Þ. (a) Because the vectors are equal, set the corresponding entries equal to each other, yielding x¼2; 3¼xþy Solve the linear equations, obtaining x¼2;y¼1: (b) First multiply by the scalar xto obtainð4;yÞ¼ð 2x;3xÞ. Then set corresponding entries equal to each other to obtain 4¼2x; y¼3x Solve the equations to yield x¼2,y¼6. 1.5. Write the vector v¼ð1;/C02;5Þas a linear combination of the vectors u1¼ð1;1;1Þ,u2¼ð1;2;3Þ, u3¼ð2;/C01;1Þ. We want to express vin the form v¼xu1þyu2þzu3with x;y;zas yet unknown. First we have 1 /C02 52 43 5¼x1 112 43 5þy1 232 43 5þz2 /C01 12 43 5¼xþyþ2z xþ2y/C0z xþ3yþz2 43 5 (It is more convenient to write vectors as columns than as rows when forming linear combinations.) Set corresponding entries equal to each other to obtain xþyþ2z¼1 xþ2y/C0z¼/C02 xþ3yþz¼5orxþyþ2z¼1 y/C03z¼/C03 2y/C0z¼4orxþyþ2z¼1 y/C03z¼/C03 5z¼10 This unique solution of the triangular system is x¼/C06,y¼3,z¼2. Thus, v¼/C06u 1þ3u2þ2u3.14 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.6. Write v¼ð2;/C05;3Þas a linear combination of u1¼ð1;/C03;2Þ;u2¼ð2;/C04;/C01Þ;u3¼ð1;/C05;7Þ: Find the equivalent system of linear equations and then solve. First, 2 /C05 32 43 5¼x1 /C03 22 43 5þy2 /C04 /C012 43 5þz1 /C05 72 43 5¼xþ2yþz /C03x/C04y/C05z 2x/C0yþ7z2 43 5 Set the corresponding entries equal to each other to obtain xþ2yþz¼2 /C03x/C04y/C05z¼/C05 2x/C0yþ7z¼3orxþ2yþz¼2 2y/C02z¼1 /C05yþ5z¼/C01orxþ2yþz¼2 2y/C02z¼1 0¼3 The third equation, 0 xþ0yþ0z¼3, indicates that the system has no solution. Thus, vcannot be written as a linear combination of the vectors u1,u2,u3. Dot (Inner) Product, Orthogonality, Norm in Rn 1.7. Find u/C1vwhere: (a)u¼ð2;/C05;6Þand v¼ð8;2;/C03Þ, (b)u¼ð4;2;/C03;5;/C01Þand v¼ð2;6;/C01;/C04;8Þ. Multiply the corresponding components and add: (a)u/C1v¼2ð8Þ/C05ð2Þþ6ð/C03Þ¼16/C010/C018¼/C012 (b)u/C1v¼8þ12þ3/C020/C08¼/C05 1.8. Letu¼ð5;4;1Þ,v¼ð3;/C04;1Þ,w¼ð1;/C02;3Þ. Which pair of vectors, if any, are perpendicular (orthogonal)? Find the dot product of each pair of vectors: u/C1v¼15/C016þ1¼0; v/C1w¼3þ8þ3¼14; u/C1w¼5/C08þ3¼0 Thus, uand vare orthogonal, uandware orthogonal, but vandware not. 1.9. Find kso that uand vare orthogonal, where: (a)u¼ð1;k;/C03Þand v¼ð2;/C05;4Þ, (b)u¼ð2;3k;/C04;1;5Þand v¼ð6;/C01;3;7;2kÞ. Compute u/C1v, set u/C1vequal to 0, and then solve for k: (a)u/C1v¼1ð2Þþkð/C05Þ/C03ð4Þ¼/C0 5k/C010. Then/C05k/C010¼0, or k¼/C02. (b)u/C1v¼12/C03k/C012þ7þ10k¼7kþ7. Then 7 kþ7¼0, or k¼/C01. 1.10. Findkuk, where: (a) u¼ð3;/C012;/C04Þ, (b) u¼ð2;/C03;8;/C07Þ. First ﬁndkuk2¼u/C1uby squaring the entries and adding. Then kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ kuk2q . (a)kuk2¼ð3Þ2þð/C0 12Þ2þð/C0 4Þ2¼9þ144þ16¼169. Thenkuk¼ﬃﬃﬃﬃﬃﬃﬃﬃ 169p ¼13. (b)kuk2¼4þ9þ64þ49¼126. Thenkuk¼ﬃﬃﬃﬃﬃﬃﬃﬃ 126p .CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 15 1.11. Recall that normalizing a nonzero vector vmeans ﬁnding the unique unit vector ^vin the same direction as v, where ^v¼1 kvkv Normalize: (a) u¼ð3;/C04Þ, (b) v¼ð4;/C02;/C03;8Þ, (c) w¼ð1 2,23,/C01 4). (a) First ﬁndkuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9þ16p ¼ﬃﬃﬃﬃﬃ 25p ¼5. Then divide each entry of uby 5, obtaining ^u¼ð3 5,/C04 5). (b) Herekvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16þ4þ9þ64p ¼ﬃﬃﬃﬃﬃ 93p . Then ^v¼4ﬃﬃﬃﬃﬃ 93p ;/C02ﬃﬃﬃﬃﬃ 93p ;/C03ﬃﬃﬃﬃﬃ 93p ;8ﬃﬃﬃﬃﬃ 93p/C18/C19 (c) Note that wand any positive multiple of wwill have the same normalized form. Hence, ﬁrst multiply w by 12 to ‘‘clear fractions’’—that is, ﬁrst ﬁnd w0¼12w¼ð6;8;/C03Þ. Then kw0k¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 36þ64þ9p ¼ﬃﬃﬃﬃﬃﬃﬃﬃ 109p and ^w¼bw0¼6ﬃﬃﬃﬃﬃﬃﬃﬃ 109p ;8ﬃﬃﬃﬃﬃﬃﬃﬃ 109p ;/C03ﬃﬃﬃﬃﬃﬃﬃﬃ 109p/C18/C19 1.12. Letu¼ð1;/C03;4Þand v¼ð3;4;7Þ. Find: (a) cos y, where yis the angle between uand v; (b) projðu;vÞ, the projection of uonto v; (c)dðu;vÞ, the distance between uand v. First ﬁnd u/C1v¼3/C012þ28¼19,kuk2¼1þ9þ16¼26,kvk2¼9þ16þ49¼74. Then (a) cos y¼u/C1v kukkvk¼19ﬃﬃﬃﬃﬃ 26pﬃﬃﬃﬃﬃ 74p , (b) projðu;vÞ¼u/C1v kvk2v¼19 74ð3;4;7Þ¼57 74;76 74;133 74/C18/C19 ¼57 74;38 37;133 74/C18/C19 ; (c)dðu;vÞ¼k u/C0vk¼kð/C0 2;/C07/C03Þk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ4þ49þ9p¼ﬃﬃﬃﬃﬃ 62p : 1.13. Prove Theorem 1.2: For any u;v;winRnandkinR: (i)ðuþvÞ/C1w¼u/C1wþv/C1w, (ii)ðkuÞ/C1v¼kðu/C1vÞ, (iii) u/C1v¼v/C1u, (iv) u/C1u/C210, and u/C1u¼0 iff u¼0. Letu¼ðu1;u2;...;unÞ,v¼ðv1;v2;...;vnÞ,w¼ðw1;w2;...;wnÞ. (i) Because uþv¼ðu1þv1;u2þv2;...;unþvnÞ, ðuþvÞ/C1w¼ðu1þv1Þw1þðu2þv2Þw2þ/C1/C1/C1þð unþvnÞwn ¼u1w1þv1w1þu2w2þ/C1/C1/C1þ unwnþvnwn ¼ðu1w1þu2w2þ/C1/C1/C1þ unwnÞþð v1w1þv2w2þ/C1/C1/C1þ vnwnÞ ¼u/C1wþv/C1w (ii) Because ku¼ðku1;ku2;...;kunÞ, ðkuÞ/C1v¼ku1v1þku2v2þ/C1/C1/C1þ kunvn¼kðu1v1þu2v2þ/C1/C1/C1þ unvnÞ¼kðu/C1vÞ (iii) u/C1v¼u1v1þu2v2þ/C1/C1/C1þ unvn¼v1u1þv2u2þ/C1/C1/C1þ vnun¼v/C1u (iv) Because u2 iis nonnegative for each i, and because the sum of nonnegative real numbers is nonnegative, u/C1u¼u2 1þu2 2þ/C1/C1/C1þ u2 n/C210 Furthermore, u/C1u¼0 iff ui¼0 for each i, that is, iff u¼0.16 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.14. Prove Theorem 1.3 (Schwarz): ju/C1vj/C20k ukkvk. For any real number t, and using Theorem 1.2, we have 0/C20ðtuþvÞ/C1ðtuþvÞ¼t2ðu/C1uÞþ2tðu/C1vÞþð v/C1vÞ¼k uk2t2þ2ðu/C1vÞtþkvk2 Leta¼kuk2,b¼2ðu/C1vÞ,c¼kvk2. Then, for every value of t,at2þbtþc/C210. This means that the quadratic polynomial cannot have two real roots. This implies that the discriminant D¼b2/C04ac/C200 or, equivalently, b2/C204ac. Thus, 4ðu/C1vÞ2/C204kuk2kvk2 Dividing by 4 gives us our result. 1.15. Prove Theorem 1.4 (Minkowski): kuþvk/C20k ukþk vk. By the Schwarz inequality and other properties of the dot product, kuþvk2¼ðuþvÞ/C1ðuþvÞ¼ð u/C1uÞþ2ðu/C1vÞþð v/C1vÞ/C20k uk2þ2kukkvkþk vk2¼ðk ukþk vkÞ2 Taking the square root of both sides yields the desired inequality. Points, Lines, Hyperplanes in Rn Here we distinguish between an n-tuple Pða1;a2;...;anÞviewed as a point in Rnand an n-tuple u¼½c1;c2;...;cn/C138viewed as a vector (arrow) from the origin Oto the point Cðc1;c2;...;cnÞ. 1.16. Find the vector uidentiﬁed with the directed line segment PQ/C131!for the points: (a) Pð1;/C02;4ÞandQð6;1;/C05ÞinR3, (b) Pð2;3;/C06;5ÞandQð7;1;4;/C08ÞinR4. (a)u¼PQ/C131!¼Q/C0P¼½6/C01;1/C0ð/C0 2Þ;/C05/C04/C138¼½5;3;/C09/C138 (b)u¼PQ/C131!¼Q/C0P¼½7/C02;1/C03;4þ6;/C08/C05/C138¼½5;/C02;10;/C013/C138 1.17. Find an equation of the hyperplane HinR4that passes through Pð3;/C04;1;/C02Þand is normal to u¼½2;5;/C06;/C03/C138. The coefﬁcients of the unknowns of an equation of Hare the components of the normal vector u. Thus, an equation of His of the form 2 x1þ5x2/C06x3/C03x4¼k. Substitute Pinto this equation to obtain k¼/C026. Thus, an equation of His 2x1þ5x2/C06x3/C03x4¼/C026. 1.18. Find an equation of the plane HinR3that contains Pð1;/C03;/C04Þand is parallel to the plane H0 determined by the equation 3 x/C06yþ5z¼2. The planes HandH0are parallel if and only if their normal directions are parallel or antiparallel (opposite direction). Hence, an equation of His of the form 3 x/C06yþ5z¼k. Substitute Pinto this equation to obtain k¼1. Then an equation of His 3x/C06yþ5z¼1. 1.19. Find a parametric representation of the line LinR4passing through Pð4;/C02;3;1Þin the direction ofu¼½2;5;/C07;8/C138. Here Lconsists of the points XðxiÞthat satisfy X¼Pþtu or xi¼aitþbi or LðtÞ¼ð aitþbiÞ where the parameter ttakes on all real values. Thus we obtain x1¼4þ2t;x2¼/C02þ2t;x3¼3/C07t;x4¼1þ8torLðtÞ¼ð 4þ2t;/C02þ2t;3/C07t;1þ8tÞCHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 17 1.20. LetCbe the curve FðtÞ¼ð t2;3t/C02;t3;t2þ5ÞinR4, where 0/C20t/C204. (a) Find the point PonCcorresponding to t¼2. (b) Find the initial point Qand terminal point Q0ofC. (c) Find the unit tangent vector Tto the curve Cwhen t¼2. (a) Substitute t¼2 into FðtÞto get P¼fð2Þ¼ð 4;4;8;9Þ. (b) The parameter tranges from t¼0t o t¼4. Hence, Q¼fð0Þ¼ð 0;/C02;0;5Þand Q0¼Fð4Þ¼ð 16;10;64;21Þ. (c) Take the derivative of FðtÞ—that is, of each component of FðtÞ—to obtain a vector Vthat is tangent to the curve: VðtÞ¼dFðtÞ dt¼½2t;3;3t2;2t/C138 Now ﬁnd Vwhen t¼2; that is, substitute t¼2 in the equation for VðtÞto obtain V¼Vð2Þ¼½ 4;3;12;4/C138. Then normalize Vto obtain the desired unit tangent vector T. We have kVk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16þ9þ144þ16p ¼ﬃﬃﬃﬃﬃﬃﬃﬃ 185p and T¼4ﬃﬃﬃﬃﬃﬃﬃﬃ 185p ;3ﬃﬃﬃﬃﬃﬃﬃﬃ 185p ;12ﬃﬃﬃﬃﬃﬃﬃﬃ 185p ;4ﬃﬃﬃﬃﬃﬃﬃﬃ 185p/C20/C21 Spatial Vectors (Vectors in R3),ijkNotation, Cross Product 1.21. Letu¼2i/C03jþ4k,v¼3iþj/C02k,w¼iþ5jþ3k. Find: (a)uþv, (b) 2 u/C03vþ4w, (c) u/C1vandu/C1w, (d)kukandkvk. Treat the coefﬁcients of i,j,kjust like the components of a vector in R3. (a) Add corresponding coefﬁcients to get uþv¼5i/C02j/C02k. (b) First perform the scalar multiplication and then the vector addition: 2u/C03vþ4w¼ð4i/C06jþ8kÞþð/C0 9iþ3jþ6kÞþð 4iþ20jþ12kÞ ¼/C0iþ17jþ26k (c) Multiply corresponding coefﬁcients and then add: u/C1v¼6/C03/C08¼/C05 and u/C1w¼2/C015þ12¼/C01 (d) The norm is the square root of the sum of the squares of the coefﬁcients: kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4þ9þ16p ¼ﬃﬃﬃﬃﬃ 29p andkvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9þ1þ4p ¼ﬃﬃﬃﬃﬃ 14p 1.22. Find the (parametric) equation of the line L: (a) through the points Pð1;3;2ÞandQð2;5;/C06Þ; (b) containing the point Pð1;/C02;4Þand perpendicular to the plane Hgiven by the equation 3xþ5yþ7z¼15: (a) First ﬁnd v¼PQ/C131!¼Q/C0P¼½1;2;/C08/C138¼iþ2j/C08k. Then LðtÞ¼ð tþ1;2tþ3;/C08tþ2Þ¼ð tþ1Þiþð2tþ3Þjþð/C0 8tþ2Þk (b) Because Lis perpendicular to H, the line Lis in the same direction as the normal vector N¼3iþ5jþ7ktoH. Thus, LðtÞ¼ð 3tþ1;5t/C02;7tþ4Þ¼ð 3tþ1Þiþð5t/C02Þjþð7tþ4Þk 1.23. LetSbe the surface xy2þ2yz¼16 in R3. (a) Find the normal vector Nðx;y;zÞto the surface S. (b) Find the tangent plane HtoSat the point Pð1;2;3Þ.18 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors (a) The formula for the normal vector to a surface Fðx;y;zÞ¼0i s Nðx;y;zÞ¼FxiþFyjþFzk where Fx,Fy,Fzare the partial derivatives. Using Fðx;y;zÞ¼xy2þ2yz/C016, we obtain Fx¼y2; Fy¼2xyþ2z; Fz¼2y Thus, Nðx;y;zÞ¼y2iþð2xyþ2zÞjþ2yk. (b) The normal to the surface Sat the point Pis NðPÞ¼Nð1;2;3Þ¼4iþ10jþ4k Hence, N¼2iþ5jþ2kis also normal to SatP. Thus an equation of Hhas the form 2 xþ5yþ2z¼c. Substitute Pin this equation to obtain c¼18. Thus the tangent plane HtoSatPis 2xþ5yþ2z¼18. 1.24. Evaluate the following determinants and negative of determinants of order two: (a) (i)34 59/C12/C12/C12/C12/C12/C12/C12/C12, (ii)2/C01 43/C12/C12/C12/C12/C12/C12/C12/C12, (iii)4/C05 3/C02/C12/C12/C12/C12/C12/C12/C12/C12 (b) (i)/C036 42/C12/C12/C12/C12/C12/C12/C12/C12, (ii)/C07/C05 32/C12/C12/C12/C12/C12/C12/C12/C12, (iii)/C04/C01 8/C03/C12/C12/C12/C12/C12/C12/C12/C12 Useab cd/C12/C12/C12/C12/C12/C12/C12/C12¼ad/C0bcand/C0ab cd/C12/C12/C12/C12/C12/C12/C12/C12¼bc/C0ad. Thus, (a) (i) 27/C020¼7, (ii) 6þ4¼10, (iii)/C08þ15¼7: (b) (i) 24/C06¼18, (ii)/C015/C014¼/C029, (iii)/C08þ12¼4: 1.25. Let u¼2i/C03jþ4k,v¼3iþj/C02k,w¼iþ5jþ3k. Find: (a) u/C2v,(b)u/C2w (a) Use2/C034 31/C02/C20/C21 to get u/C2v¼ð6/C04Þiþð12þ4Þjþð2þ9Þk¼2iþ16jþ11k: (b) Use2/C034 15 3/C20/C21 to get u/C2w¼ð/C0 9/C020Þiþð4/C06Þjþð10þ3Þk¼/C029i/C02jþ13k: 1.26. Find u/C2v, where: (a) u¼ð1;2;3Þ,v¼ð4;5;6Þ; (b) u¼ð/C0 4;7;3Þ,v¼ð6;/C05;2Þ. (a) Use123 456/C20/C21 to get u/C2v¼½12/C015;12/C06;5/C08/C138¼½/C0 3;6;/C03/C138: (b) Use/C047 3 6/C052/C20/C21 to get u/C2v¼½14þ15;18þ8;20/C042/C138¼½29;26;/C022/C138: 1.27. Find a unit vector uorthogonal to v¼½1;3;4/C138andw¼½2;/C06;/C05/C138. First ﬁnd v/C2w, which is orthogonal to vandw. The array134 2/C06/C05/C20/C21 gives v/C2w¼½/C0 15þ24;8þ5;/C06/C061/C138¼½9;13;/C012/C138: Normalize v/C2wto get u¼½9=ﬃﬃﬃﬃﬃﬃﬃﬃ 394p ,1 3=ﬃﬃﬃﬃﬃﬃﬃﬃ 394p ,/C012=ﬃﬃﬃﬃﬃﬃﬃﬃ 394p /C138: 1.28. Let u¼ða1;a2;a3Þand v¼ðb1;b2;b3Þsou/C2v¼ða2b3/C0a3b2;a3b1/C0a1b3;a1b2/C0a2b1Þ. Prove: (a)u/C2vis orthogonal to uand v[Theorem 1.5(a)]. (b)ku/C2vk2¼ðu/C1uÞðv/C1vÞ/C0ð u/C1vÞ2(Lagrange’s identity).CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 19 (a) We have u/C1ðu/C2vÞ¼a1ða2b3/C0a3b2Þþa2ða3b1/C0a1b3Þþa3ða1b2/C0a2b1Þ ¼a1a2b3/C0a1a3b2þa2a3b1/C0a1a2b3þa1a3b2/C0a2a3b1¼0 Thus, u/C2vis orthogonal to u. Similarly, u/C2vis orthogonal to v. (b) We have ku/C2vk2¼ða2b3/C0a3b2Þ2þða3b1/C0a1b3Þ2þða1b2/C0a2b1Þ2ð1Þ ðu/C1uÞðv/C1vÞ/C0ð u/C1vÞ2¼ða2 1þa2 2þa2 3Þðb2 1þb2 2þb2 3Þ/C0ð a1b1þa2b2þa3b3Þ2ð2Þ Expansion of the right-hand sides of (1) and (2) establishes the identity. Complex Numbers, Vectors in Cn 1.29. Suppose z¼5þ3iandw¼2/C04i. Find: (a) zþw, (b) z/C0w, (c) zw. Use the ordinary rules of algebra together with i2¼/C01 to obtain a result in the standard form aþbi. (a)zþw¼ð5þ3iÞþð 2/C04iÞ¼7/C0i (b)z/C0w¼ð5þ3iÞ/C0ð 2/C04iÞ¼5þ3i/C02þ4i¼3þ7i (c)zw¼ð5þ3iÞð2/C04iÞ¼10/C014i/C012i2¼10/C014iþ12¼22/C014i 1.30. Simplify: (a)ð5þ3iÞð2/C07iÞ, (b)ð4/C03iÞ2, (c)ð1þ2iÞ3. (a)ð5þ3iÞð2/C07iÞ¼10þ6i/C035i/C021i2¼31/C029i (b)ð4/C03iÞ2¼16/C024iþ9i2¼7/C024i (c)ð1þ2iÞ3¼1þ6iþ12i2þ8i3¼1þ6i/C012/C08i¼/C011/C02i 1.31. Simplify: (a) i0;i3;i4, (b) i5;i6;i7;i8, (c) i39;i174,i252,i317: (a)i0¼1,i3¼i2ðiÞ¼ð/C0 1ÞðiÞ¼/C0 i;i4¼ði2Þði2Þ¼ð/C0 1Þð/C01Þ¼1 (b)i5¼ði4ÞðiÞ¼ð 1ÞðiÞ¼i,i6¼ði4Þði2Þ¼ð 1Þði2Þ¼i2¼/C01,i7¼i3¼/C0i,i8¼i4¼1 (c) Using i4¼1 and in¼i4qþr¼ði4Þqir¼1qir¼ir, divide the exponent nby 4 to obtain the remainder r: i39¼i4ð9Þþ3¼ði4Þ9i3¼19i3¼i3¼/C0i; i174¼i2¼/C01; i252¼i0¼1; i317¼i1¼i 1.32. Find the complex conjugate of each of the following: (a) 6þ4i,7/C05i,4þi,/C03/C0i, (b) 6,/C03, 4i,/C09i. (a)6þ4i¼6/C04i,7/C05i¼7þ5i,4þi¼4/C0i,/C03/C0i¼/C03þi (b) /C226¼6,/C03¼/C03,4i¼/C04i,/C09i¼9i (Note that the conjugate of a real number is the original number, but the conjugate of a pure imaginary number is the negative of the original number.) 1.33. Find z/C22zandjzjwhen z¼3þ4i. Forz¼aþbi, use z/C22z¼a2þb2andz¼ﬃﬃﬃﬃz/C22zp¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2þb2p . z/C22z¼9þ16¼25;jzj¼ﬃﬃﬃﬃﬃ 25p ¼5 1.34. Simpify2/C07i 5þ3i: To simplify a fraction z=wof complex numbers, multiply both numerator and denominator by /C22w, the conjugate of the denominator: 2/C07i 5þ3i¼ð2/C07iÞð5/C03iÞ ð5þ3iÞð5/C03iÞ¼/C011/C041i 34¼/C011 34/C041 34i20 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.35. Prove: For any complex numbers z,w2C, (i) zþw¼/C22zþ/C22w, (ii) zw¼/C22z/C22w, (iii) /C22z¼z. Suppose z¼aþbiandw¼cþdiwhere a;b;c;d2R. (i) zþw¼ðaþbiÞþð cþdiÞ¼ðaþcÞþð bþdÞi ¼ðaþcÞ/C0ð bþdÞi¼aþc/C0bi/C0di ¼ða/C0biÞþð c/C0diÞ¼ /C22zþ/C22w (ii) zw¼ðaþbiÞðcþdiÞ¼ðac/C0bdÞþð adþbcÞi ¼ðac/C0bdÞ/C0ð adþbcÞi¼ða/C0biÞðc/C0diÞ¼ /C22z/C22w (iii) /C22z¼aþbi¼a/C0bi¼a/C0ð/C0 bÞi¼aþbi¼z 1.36. Prove: For any complex numbers z;w2C,jzwj¼jzjjwj. By (ii) of Problem 1.35, jzwj2¼ðzwÞðzwÞ¼ð zwÞð/C22z/C22wÞ¼ð z/C22zÞðw/C22wÞ¼j zj2jwj2 The square root of both sides gives us the desired result. 1.37. Prove: For any complex numbers z;w2C,jzþwj/C20jzjþjwj. Suppose z¼aþbiandw¼cþdiwhere a;b;c;d2R. Consider the vectors u¼ða;bÞand v¼ðc;dÞin R2. Note that jzj¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2þb2p ¼kuk;jwj¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ c2þd2p ¼kvk and jzþwj¼jð aþcÞþð bþdÞij¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðaþcÞ2þðbþdÞ2q ¼kð aþc;bþdÞk¼k uþvk By Minkowski’s inequality (Problem 1.15), kuþvk/C20k ukþk vk, and so jzþwj¼k uþvk/C20k ukþk vk¼j zjþjwj 1.38. Find the dot products u/C1vand v/C1uwhere: (a) u¼ð1/C02i;3þiÞ,v¼ð4þ2i;5/C06iÞ; (b) u¼ð3/C02i;4i;1þ6iÞ,v¼ð5þi;2/C03i;7þ2iÞ. Recall that conjugates of the second vector appear in the dot product ðz1;...;znÞ/C1ðw1;...;wnÞ¼z1/C22w1þ/C1/C1/C1þ zn/C22wn (a)u/C1v¼ð1/C02iÞð4þ2iÞþð 3þiÞð5/C06iÞ ¼ð1/C02iÞð4/C02iÞþð 3þiÞð5þ6iÞ¼/C0 10iþ9þ23i¼9þ13i v/C1u¼ð4þ2iÞð1/C02iÞþð 5/C06iÞð3þiÞ ¼ð4þ2iÞð1þ2iÞþð 5/C06iÞð3/C0iÞ¼ 10iþ9/C023i¼9/C013i (b)u/C1v¼ð3/C02iÞð5þiÞþð 4iÞð2/C03iÞþð 1þ6iÞð7þ2iÞ ¼ð3/C02iÞð5/C0iÞþð 4iÞð2þ3iÞþð 1þ6iÞð7/C02iÞ¼ 20þ35i v/C1u¼ð5þiÞð3/C02iÞþð 2/C03iÞð4iÞþð 7þ2iÞð1þ6iÞ ¼ð5þiÞð3þ2iÞþð 2/C03iÞð/C04iÞþð 7þ2iÞð1/C06iÞ¼ 20/C035i In both cases, v/C1u¼u/C1v. This holds true in general, as seen in Problem 1.40. 1.39. Letu¼ð7/C02i;2þ5iÞand v¼ð1þi;/C03/C06iÞ. Find: (a) uþv, (b) 2 iu, (c)ð3/C0iÞv, (d) u/C1v, (e)kukandkvk. (a)uþv¼ð7/C02iþ1þi;2þ5i/C03/C06iÞ¼ð 8/C0i;/C01/C0iÞ (b) 2 iu¼ð14i/C04i2;4iþ10i2Þ¼ð 4þ14i;/C010þ4iÞ (c)ð3/C0iÞv¼ð3þ3i/C0i/C0i2;/C09/C018iþ3iþ6i2Þ¼ð 4þ2i;/C015/C015iÞCHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 21 (d)u/C1v¼ð7/C02iÞð1þiÞþð 2þ5iÞð/C03/C06iÞ ¼ð7/C02iÞð1/C0iÞþð 2þ5iÞð/C03þ6iÞ¼ 5/C09i/C036/C03i¼/C0 31/C012i (e)kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 72þð/C0 2Þ2þ22þ52q ¼ﬃﬃﬃﬃﬃ 82p andkvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 12þ12þð/C0 3Þ2þð/C0 6Þ2q ¼ﬃﬃﬃﬃﬃ 47p 1.40. Prove: For any vectors u;v2Cnand any scalar z2C, (i) u/C1v¼v/C1u, (ii)ðzuÞ/C1v¼zðu/C1vÞ, (iii)u/C1ðzvÞ¼ /C22zðu/C1vÞ. Suppose u¼ðz1;z2;...;znÞand v¼ðw1;w2;...;wnÞ. (i) Using the properties of the conjugate, v/C1u¼w1/C22z1þw2/C22z2þ/C1/C1/C1þ wn/C22zn¼w1/C22z1þw2/C22z2þ/C1/C1/C1þ wn/C22zn ¼/C22w1z1þ/C22w2z2þ/C1/C1/C1þ /C22wnzn¼z1/C22w1þz2/C22w2þ/C1/C1/C1þ zn/C22wn¼u/C1v (ii) Because zu¼ðzz1;zz2;...;zznÞ, ðzuÞ/C1v¼zz1/C22w1þzz2/C22w2þ/C1/C1/C1þ zzn/C22wn¼zðz1/C22w1þz2/C22w2þ/C1/C1/C1þ zn/C22wnÞ¼zðu/C1vÞ (Compare with Theorem 1.2 on vectors in Rn.) (iii) Using (i) and (ii), u/C1ðzvÞ¼ðzvÞ/C1u¼zðv/C1uÞ¼ /C22zðv/C1uÞ¼ /C22zðu/C1vÞ SUPPLEMENTARY PROBLEMS Vectors in Rn 1.41. Letu¼ð1;/C02;4Þ,v¼ð3;5;1Þ,w¼ð2;1;/C03Þ. Find: (a) 3 u/C02v; (b) 5 uþ3v/C04w; (c) u/C1v,u/C1w,v/C1w; (d)kuk,kvk; (e) cos y, where yis the angle between uand v;( f ) dðu;vÞ; (g) projðu;vÞ. 1.42. Repeat Problem 1.41 for vectors u¼1 3 /C042 43 5,v¼2 1 52 43 5,w¼3 /C02 62 43 5. 1.43. Letu¼ð2;/C05;4;6;/C03Þand v¼ð5;/C02;1;/C07;/C04Þ. Find: (a) 4 u/C03v; (b) 5 uþ2v; (c) u/C1v; (d)kukandkvk; (e) projðu;vÞ; (f) dðu;vÞ. 1.44. Normalize each vector: (a) u¼ð5;/C07Þ; (b) v¼ð1;2;/C02;4Þ; (c) w¼1 2;/C01 3;3 4/C18/C19 . 1.45. Letu¼ð1;2;/C02Þ,v¼ð3;/C012;4Þ, and k¼/C03. (a) Findkuk,kvk,kuþvk,kkuk: (b) Verify thatkkuk¼j kjkukandkuþvk/C20k ukþk vk. 1.46. Find xandywhere: (a)ðx;yþ1Þ¼ð y/C02;6Þ; (b) xð2;yÞ¼yð1;/C02Þ. 1.47. Find x;y;zwhereðx;yþ1;yþzÞ¼ð 2xþy;4;3zÞ.22 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.48. Write v¼ð2;5Þas a linear combination of u1andu2, where: (a) u1¼ð1;2Þandu2¼ð3;5Þ; (b) u1¼ð3;/C04Þandu2¼ð2;/C03Þ. 1.49. Write v¼9 /C03 162 43 5as a linear combination of u1¼1 332 43 5,u 2¼2 5 /C012 43 5,u3¼4 /C02 32 43 5. 1.50. Find kso that uand vare orthogonal, where: (a) u¼ð3;k;/C02Þ,v¼ð6;/C04;/C03Þ; (b) u¼ð5;k;/C04;2Þ,v¼ð1;/C03;2;2kÞ; (c) u¼ð1;7;kþ2;/C02Þ,v¼ð3;k;/C03;kÞ. Located Vectors, Hyperplanes, Lines in Rn 1.51. Find the vector videntiﬁed with the directed line segment PQ! for the points: (a) Pð2;3;/C07ÞandQð1;/C06;/C05ÞinR3; (b) Pð1;/C08;/C04;6ÞandQð3;/C05;2;/C04ÞinR4. 1.52. Find an equation of the hyperplane HinR4that: (a) contains Pð1;2;/C03;2Þand is normal to u¼½2;3;/C05;6/C138; (b) contains Pð3;/C01;2;5Þand is parallel to 2 x1/C03x2þ5x3/C07x4¼4. 1.53. Find a parametric representation of the line in R4that: (a) passes through the points Pð1;2;1;2ÞandQð3;/C05;7;/C09Þ; (b) passes through Pð1;1;3;3Þand is perpendicular to the hyperplane 2 x1þ4x2þ6x3/C08x4¼5. Spatial Vectors (Vectors in R3),ijkNotation 1.54. Given u¼3i/C04jþ2k,v¼2iþ5j/C03k,w¼4iþ7jþ2k. Find: (a) 2 u/C03v; (b) 3 uþ4v/C02w; (c) u/C1v,u/C1w,v/C1w; (d)kuk,kvk,kwk. 1.55. Find the equation of the plane H: (a) with normal N¼3i/C04jþ5kand containing the point Pð1;2;/C03Þ; (b) parallel to 4 xþ3y/C02z¼11 and containing the point Qð2;/C01;3Þ. 1.56. Find the (parametric) equation of the line L: (a) through the point Pð2;5;/C03Þand in the direction of v¼4i/C05jþ7k; (b) perpendicular to the plane 2 x/C03yþ7z¼4 and containing Pð1;/C05;7Þ. 1.57. Consider the following curve CinR3where 0/C20t/C205: FðtÞ¼t3i/C0t2jþð2t/C03Þk (a) Find the point PonCcorresponding to t¼2. (b) Find the initial point Qand the terminal point Q0. (c) Find the unit tangent vector Tto the curve Cwhen t¼2. 1.58. Consider a moving body Bwhose position at time tis given by RðtÞ¼t2iþt3jþ3tk. [Then VðtÞ¼dRðtÞ=dtand AðtÞ¼dVðtÞ=dtdenote, respectively, the velocity and acceleration of B.] When t¼1, ﬁnd for the body B: (a) position; (b) velocity v; (c) speed s; (d) acceleration a.CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 23 1.59. Find a normal vector Nand the tangent plane Hto each surface at the given point: (a) surface x2yþ3yz¼20 and point Pð1;3;2Þ; (b) surface x2þ3y2/C05z2¼160 and point Pð3;/C02;1Þ: Cross Product 1.60. Evaluate the following determinants and negative of determinants of order two: (a)25 36/C12/C12/C12/C12/C12/C12/C12/C12;3/C06 1/C04/C12/C12/C12/C12/C12/C12/C12/C12;/C04/C02 7/C03/C12/C12/C12/C12/C12/C12/C12/C12 (b)/C064 75/C12/C12/C12/C12/C12/C12/C12/C12;/C01/C03 24/C12/C12/C12/C12/C12/C12/C12/C12;/C08/C03 /C06/C02/C12/C12/C12/C12/C12/C12/C12/C12 1.61. Given u¼3i/C04jþ2k,v¼2iþ5j/C03k,w¼4iþ7jþ2k, ﬁnd: (a) u/C2v, (b) u/C2w, (c) v/C2w. 1.62. Given u¼½2;1;3/C138,v¼½4;/C02;2/C138,w¼½1;1;5/C138, ﬁnd: (a) u/C2v, (b) u/C2w, (c) v/C2w. 1.63. Find the volume Vof the parallelopiped formed by the vectors u;v;wappearing in: (a) Problem 1.60 (b) Problem 1.61. 1.64. Find a unit vector uorthogonal to: (a) v¼½1;2;3/C138andw¼½1;/C01;2/C138; (b) v¼3i/C0jþ2kandw¼4i/C02j/C0k. 1.65. Prove the following properties of the cross product: (a) u/C2v¼/C0ð v/C2uÞ (d) u/C2ðvþwÞ¼ð u/C2vÞþð u/C2wÞ (b) u/C2u¼0 for any vector u (e)ðvþwÞ/C2u¼ðv/C2uÞþð w/C2uÞ (c)ðkuÞ/C2v¼kðu/C2vÞ¼u/C2ðkvÞ (f)ðu/C2vÞ/C2w¼ðu/C1wÞv/C0ðv/C1wÞu Complex Numbers 1.66. Simplify: (a)ð4/C07iÞð9þ2iÞ; (b)ð3/C05iÞ2; (c)1 4/C07i; (d)9þ2i 3/C05i; (e)ð1/C0iÞ3. 1.67. Simplify: (a)1 2i; (b)2þ3i 7/C03i; (c) i15;i25;i34; (d)1 3/C0i/C18/C192 . 1.68. Letz¼2/C05iandw¼7þ3i. Find: (a) vþw; (b) zw; (c) z=w; (d) /C22z;/C22w; (e)jzj,jwj. 1.69. Show that for complex numbers zandw: (a) Re z¼1 2ðzþ/C22zÞ, (b) Im z¼1 2ðz/C0/C22z), (c) zw¼0 implies z¼0o r w¼0. Vectors in Cn 1.70. Letu¼ð1þ7i;2/C06iÞand v¼ð5/C02i;3/C04iÞ. Find: (a)uþv(b)ð3þiÞu(c) 2 iuþð4þ7iÞv(d) u/C1v(e)kukandkvk.24 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 1.71. Prove: For any vectors u;v;winCn: (a)ðuþvÞ/C1w¼u/C1wþv/C1w, (b) w/C1ðuþvÞ¼w/C1uþw/C1v. 1.72. Prove that the norm in Cnsatisﬁes the following laws: ½N1/C138For any vector u,kuk/C210; andkuk¼0 if and only if u¼0. ½N2/C138For any vector uand complex number z,kzuk¼j zjkuk. ½N3/C138For any vectors uand v,kuþvk/C20k ukþk vk. ANSWERS TO SUPPLEMENTARY PROBLEMS 1.41. (a)ð/C03;/C016;4Þ; (b) (6,1,35); (c) /C03;12;8; (d)ﬃﬃﬃﬃﬃ 21p ,ﬃﬃﬃﬃﬃ 35p ,ﬃﬃﬃﬃﬃ 14p ; (e)/C03=ﬃﬃﬃﬃﬃ 21pﬃﬃﬃﬃﬃ 35p ;( f )ﬃﬃﬃﬃﬃ 62p ; (g)/C03 35ð3;5;1Þ¼ð/C09 35,/C015 35,/C03 35) 1.42. (Column vectors) (a) ð/C01;7;/C022Þ; (b)ð/C01;26;/C029Þ; (c)/C015;/C027;34; (d)ﬃﬃﬃﬃﬃ 26p ,ﬃﬃﬃﬃﬃ 30p ; (e)/C015=ðﬃﬃﬃﬃﬃ 26pﬃﬃﬃﬃﬃ 30p Þ;( f )ﬃﬃﬃﬃﬃ 86p ; (g)/C015 30v¼ð/C0 1;/C01 2;/C05 2Þ 1.43. (a)ð/C013;/C014;13;45;0Þ; (b)ð20;/C029;22;16;/C023Þ; (c)/C06; (d)ﬃﬃﬃﬃﬃ 90p ;ﬃﬃﬃﬃﬃ 95p ; (e)/C06 95v;( f )ﬃﬃﬃﬃﬃﬃﬃﬃ 167p 1.44. (a)ð5=ﬃﬃﬃﬃﬃ 76p ;9=ﬃﬃﬃﬃﬃ 76p Þ; (b)ð1 5;25;/C02 5;45Þ; (c)ð6=ﬃﬃﬃﬃﬃﬃﬃﬃ 133p ;/C04ﬃﬃﬃﬃﬃﬃﬃﬃ 133p ;9ﬃﬃﬃﬃﬃﬃﬃﬃ 133p Þ 1.45. (a) 3 ;13;ﬃﬃﬃﬃﬃﬃﬃﬃ 120p ;9 1.46. (a) x¼/C03;y¼5; (b) x¼0;y¼0, and x¼1;y¼2 1.47. x¼/C03;y¼3;z¼3 2 1.48. (a) v¼5u1/C0u2; (b) v¼16u1/C023u2 1.49. v¼3u1/C0u2þ2u3 1.50. (a) 6; (b) 3; (c)3 2 1.51. (a) v¼½/C0 1;/C09;2/C138; (b) [2 ;3;6;/C010] 1.52. (a) 2 x1þ3x2/C05x3þ6x4¼35; (b) 2 x1/C03x2þ5x3/C07x4¼/C016 1.53. (a)½2tþ1;/C07tþ2;6tþ1;/C011tþ2/C138; (b)½2tþ1;4tþ1;6tþ3;/C08tþ3/C138 1.54. (a)/C023jþ13k; (b) 9 i/C06j/C010k; (c)/C020;/C012;37; (d)ﬃﬃﬃﬃﬃ 29p ;ﬃﬃﬃﬃﬃ 38p ;ﬃﬃﬃﬃﬃ 69p 1.55. (a) 3 x/C04yþ5z¼/C020; (b) 4 xþ3y/C02z¼/C01 1.56. (a)½4tþ2;/C05tþ5;7t/C03/C138; (b)½2tþ1;/C03t/C05;7tþ7/C138 1.57. (a) P¼Fð2Þ¼8i/C04jþk; (b) Q¼Fð0Þ¼/C0 3k,Q0¼Fð5Þ¼125i/C025jþ7k; (c)T¼ð6i/C02jþkÞ=ﬃﬃﬃﬃﬃ 41p 1.58. (a)iþjþ2k; (b) 2 iþ3jþ2k; (c)ﬃﬃﬃﬃﬃ 17p ; (d) 2 iþ6j 1.59. (a)N¼6iþ7jþ9k,6xþ7yþ9z¼45; (b) N¼6i/C012j/C010k,3x/C06y/C05z¼16CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors 25 1.60. (a)/C03;/C06;26; (b)/C02;/C010;34 1.61. (a) 2 iþ13jþ23k; (b)/C022iþ2jþ37k; (c) 31 i/C016j/C06k 1.62. (a)½5;8;/C06/C138; (b)½2;/C07;1/C138; (c)½/C07;/C018;5/C138 1.63. (a) 143; (b) 17 1.64. (a)ð7;1;/C03Þ=ﬃﬃﬃﬃﬃ 59p ; (b)ð5iþ11j/C02kÞ=ﬃﬃﬃﬃﬃﬃﬃﬃ 150p 1.66. (a) 50/C055i; (b)/C016/C030i; (c)1 65ð4þ7iÞ; (d)1 2ð1þ3iÞ; (e)/C02/C02i 1.67. (a)/C01 2i; (b)1 58ð5þ27iÞ; (c)/C01;i;/C01; (d)1 50ð4þ3iÞ 1.68. (a) 9/C02i; (b) 29/C029i; (c)1 61ð/C01/C041iÞ; (d) 2þ5i,7/C03i; (e)ﬃﬃﬃﬃﬃ 29p ,ﬃﬃﬃﬃﬃ 58p 1.69. (c) Hint: Ifzw¼0, thenjzwj¼jzjjwj¼j0j¼0 1.70. (a)ð6þ5i,5/C010iÞ; (b)ð/C04þ22i,1 2/C016iÞ; (c)ð/C08/C041i,/C04/C033iÞ; (d) 12þ2i; (e)ﬃﬃﬃﬃﬃ 90p ,ﬃﬃﬃﬃﬃ 54p26 CHAPTER 1 Vectors in Rnand Cn, Spatial Vectors Algebra of Matrices 2.1 Introduction This chapter investigates matrices and algebraic operations deﬁned on them. These matrices may be viewed as rectangular arrays of elements where each entry depends on two subscripts (as compared with vectors, where each entry depended on only one subscript). Systems of linear equations and theirsolutions (Chapter 3) may be efﬁciently investigated using the language of matrices. Furthermore, certainabstract objects introduced in later chapters, such as ‘‘change of basis,’’ ‘‘linear transformations,’’ and ‘‘quadratic forms,’’ can be represented by these matrices (rectangular arrays). On the other hand, theabstract treatment of linear algebra presented later on will give us new insight into the structure of thesematrices. The entries in our matrices will come from some arbitrary, but ﬁxed, ﬁeld K. The elements of Kare called numbers orscalars . Nothing essential is lost if the reader assumes that Kis the real ﬁeld R. 2.2 Matrices Amatrix A over a ﬁeld K or, simply, a matrix A (when Kis implicit) is a rectangular array of scalars usually presented in the following form: A¼a11a12 ... a1n a21a22 ... a2n /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 am1am2... amn2 6643 775 Therows of such a matrix Aare the mhorizontal lists of scalars: ða11;a12;...;a1nÞ;ða21;a22;...;a2nÞ; ...;ðam1;am2;...;amnÞ and the columns ofAare the nvertical lists of scalars: a11 a21 ... am12 6643 775;a12 a22 ... am22 6643 775; ...;a1n a2n ... amn2 6643 775 Note that the element aij, called the ij-entry orij-element , appears in row iand column j. We frequently denote such a matrix by simply writing A¼½aij/C138. A matrix with mrows and ncolumns is called an mb yn matrix, written m/C2n. The pair of numbers m andnis called the sizeof the matrix. Two matrices AandBareequal , written A¼B, if they have the same size and if corresponding elements are equal. Thus, the equality of two m/C2nmatrices is equivalent to a system of mnequalities, one for each corresponding pair of elements. A matrix with only one row is called a row matrix orrow vector , and a matrix with only one column is called a column matrix orcolumn vector . A matrix whose entries are all zero is called a zero matrix and will usually be denoted by 0. 27CHAPTER 2 Matrices whose entries are all real numbers are called real matrices and are said to be matrices over R. Analogously, matrices whose entries are all complex numbers are called complex matrices and are said to bematrices over C. This text will be mainly concerned with such real and complex matrices. EXAMPLE 2.1 (a) The rectangular array A¼1/C045 03/C02/C20/C21 is a 2/C23 matrix. Its rows are ð1;/C04;5Þandð0;3;/C02Þ, and its columns are 1 0/C20/C21 ;/C04 3/C20/C21 ;5 /C02/C20/C21 (b) The 2/C24 zero matrix is the matrix 0 ¼0000 0000/C20/C21 . (c) Find x;y;z;tsuch that xþy2zþt x/C0yz/C0t/C20/C21 ¼37 15/C20/C21 By deﬁnition of equality of matrices, the four corresponding entries must be equal. Thus, xþy¼3; x/C0y¼1; 2zþt¼7; z/C0t¼5 Solving the above system of equations yields x¼2,y¼1,z¼4,t¼/C01. 2.3 Matrix Addition and Scalar Multiplication LetA¼½aij/C138andB¼½bij/C138be two matrices with the same size, say m/C2nmatrices. The sumofAandB, written AþB, is the matrix obtained by adding corresponding elements from AandB. That is, AþB¼a11þb11 a12þb12 ... a1nþb1n a21þb21 a22þb22 ... a2nþb2n /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 am1þbm1am2þbm2... amnþbmn2 6643 775 The product of the matrix Aby a scalar k, written k/C1Aor simply kA, is the matrix obtained by multiplying each element of Abyk. That is, kA¼ka11ka12 ... ka1n ka21ka22 ... ka2n /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 /C1/C1/C1 kam1kam2... kamn2 6643 775 Observe that AþBandkAare also m/C2nmatrices. We also deﬁne /C0A¼ð/C0 1ÞA and A/C0B¼Aþð/C0 BÞ The matrix/C0Ais called the negative of the matrix A, and the matrix A/C0Bis called the difference ofA andB. The sum of matrices with different sizes is not deﬁned.28 CHAPTER 2 Algebra of Matrices EXAMPLE 2.2 LetA¼1/C023 04 5/C20/C21 andB¼468 1/C03/C07/C20/C21 . Then AþB¼1þ4/C02þ63þ8 0þ14þð/C0 3Þ5þð/C0 7Þ"# ¼54 1 1 11/C02"# 3A¼3ð1Þ3ð/C02Þ3ð3Þ 3ð0Þ 3ð4Þ3ð5Þ"# ¼3/C069 01 2 1 5"# 2A/C03B¼2/C046 08 1 0"# þ/C012/C018/C024 /C039 2 1"# ¼/C010/C022/C018 /C031 73 1"# The matrix 2 A/C03Bis called a linear combination ofAandB. Basic properties of matrices under the operations of matrix addition and scalar multiplication follow. THEOREM 2.1: Consider any matrices A;B;C(with the same size) and any scalars kandk0. Then (i)ðAþBÞþC¼AþðBþCÞ, (v) kðAþBÞ¼kAþkB, (ii) Aþ0¼0þA¼A, (vi) ðkþk0ÞA¼kAþk0A, (iii) Aþð/C0 AÞ¼ð/C0 AÞþA¼0;(vii)ðkk0ÞA¼kðk0AÞ, (iv) AþB¼BþA, (viii) 1 /C1A¼A. Note ﬁrst that the 0 in (ii) and (iii) refers to the zero matrix. Also, by (i) and (iv), any sum of matrices A1þA2þ/C1/C1/C1þ An requires no parentheses, and the sum does not depend on the order of the matrices. Furthermore, using (vi) and (viii), we also have AþA¼2A; AþAþA¼3A; ... and so on. The proof of Theorem 2.1 reduces to showing that the ij-entries on both sides of each matrix equation are equal. (See Problem 2.3.) Observe the similarity between Theorem 2.1 for matrices and Theorem 1.1 for vectors. In fact, the above operations for matrices may be viewed as generalizations of the corresponding operations forvectors. 2.4 Summation Symbol Before we deﬁne matrix multiplication, it will be instructive to ﬁrst introduce the summation symbol S (the Greek capital letter sigma). Suppose fðkÞis an algebraic expression involving the letter k. Then the expression Pn k¼1fðkÞ or equivalentlyPn k¼1fðkÞ has the following meaning. First we set k¼1i n fðkÞ, obtaining fð1Þ Then we set k¼2i n fðkÞ, obtaining fð2Þ, and add this to fð1Þ, obtaining fð1Þþfð2ÞCHAPTER 2 Algebra of Matrices 29 Then we set k¼3i n fðkÞ, obtaining fð3Þ, and add this to the previous sum, obtaining fð1Þþfð2Þþfð3Þ We continue this process until we obtain the sum fð1Þþfð2Þþ/C1/C1/C1þ fðnÞ Observe that at each step we increase the value of kby 1 until we reach n. The letter kis called the index , and 1 and nare called, respectively, the lower andupper limits. Other letters frequently used as indices areiandj. We also generalize our deﬁnition by allowing the sum to range from any integer n1to any integer n2. That is, we deﬁne Pn2 k¼n1fðkÞ¼fðn1Þþfðn1þ1Þþfðn1þ2Þþ/C1/C1/C1þ fðn2Þ EXAMPLE 2.3 (a)P5 k¼1xk¼x1þx2þx3þx4þx5andPn i¼1aibi¼a1b1þa2b2þ/C1/C1/C1þ anbn (b)P5 j¼2j2¼22þ32þ42þ52¼54 andPn i¼0aixi¼a0þa1xþa2x2þ/C1/C1/C1þ anxn (c)Pp k¼1aikbkj¼ai1b1jþai2b2jþai3b3jþ/C1/C1/C1þ aipbpj 2.5 Matrix Multiplication The product of matrices AandB, written AB, is somewhat complicated. For this reason, we ﬁrst begin with a special case. The product ABof a row matrix A¼½ai/C138and a column matrix B¼½bi/C138with the same number of elements is deﬁned to be the scalar (or 1 /C21 matrix) obtained by multiplying corresponding entries and adding; that is, AB¼½a1;a2;...;an/C138b1 b2 ... bn2 6643 775¼a 1b1þa2b2þ/C1/C1/C1þ anbn¼Pn k¼1akbk We emphasize that ABis a scalar (or a 1/C21 matrix). The product ABis not deﬁned when AandBhave different numbers of elements. EXAMPLE 2.4 (a)½7;/C04;5/C1383 2 /C012 43 5¼7ð3Þþð/C0 4Þð2Þþ5ð/C01Þ¼21/C08/C05¼8 (b)½6;/C01;8;3/C1384 /C09 /C02 52 6643 775¼24þ9/C016þ15¼32 We are now ready to deﬁne matrix multiplication in general.30 CHAPTER 2 Algebra of Matrices DEFINITION: Suppose A¼½aik/C138andB¼½bkj/C138are matrices such that the number of columns of Ais equal to the number of rows of B; say, Ais an m/C2pmatrix and Bis ap/C2nmatrix. Then the product ABis the m/C2nmatrix whose ij-entry is obtained by multiplying the ith row of Aby the jth column of B. That is, a11 ... a1p : ... : ai1 ... aip : ... : am1... amp2 666643 77775b11 ... b1j... b1n : ... : ... : : ... : ... : : ... : ... : bp1... bpj ... bpn2 666643 77775¼c11 ... c1n : ... : :cij : : ... : cm1... cmn2 666643 77775 where cij¼ai1b1jþai2b2jþ/C1/C1/C1þ aipbpj¼Pp k¼1aikbkj The product ABis not deﬁned if Ais an m/C2pmatrix and Bis aq/C2nmatrix, where p6¼q. EXAMPLE 2.5 (a) Find ABwhere A¼13 2/C01/C20/C21 andB¼20/C04 5/C026/C20/C21 . Because Ais 2/C22 and Bis 2/C23, the product ABis deﬁned and ABis a 2/C23 matrix. To obtain the ﬁrst row of the product matrix AB, multiply the ﬁrst row [1, 3] of Aby each column of B, 2 5/C20/C21 ;0 /C02/C20/C21 ;/C04 6/C20/C21 respectively. That is, AB¼2þ15 0/C06/C04þ18/C20/C21 ¼17/C061 4/C20/C21 To obtain the second row of AB, multiply the second row ½2;/C01/C138ofAby each column of B. Thus, AB¼17/C061 4 4/C050þ2/C08/C06/C20/C21 ¼17/C061 4 /C012/C014/C20/C21 (b) Suppose A¼12 34/C20/C21 andB¼56 0/C02/C20/C21 . Then AB¼5þ06/C04 15þ01 8/C08/C20/C21 ¼52 15 10/C20/C21 and BA¼5þ18 10þ24 0/C060/C08/C20/C21 ¼23 34 /C06/C08/C20/C21 The above example shows that matrix multiplication is not commutative—that is, in general, AB6¼BA. However, matrix multiplication does satisfy the following properties. THEOREM 2.2: LetA;B;Cbe matrices. Then, whenever the products and sums are deﬁned, (i)ðABÞC¼AðBCÞ(associative law), (ii) AðBþCÞ¼ABþAC(left distributive law), (iii)ðBþCÞA¼BAþCA(right distributive law), (iv) kðABÞ¼ð kAÞB¼AðkBÞ, where kis a scalar. We note that 0 A¼0 and B0¼0, where 0 is the zero matrix.CHAPTER 2 Algebra of Matrices 31 2.6 Transpose of a Matrix Thetranspose of a matrix A, written AT, is the matrix obtained by writing the columns of A, in order, as rows. For example, 123 456/C20/C21T ¼14 25 362 43 5 and½1;/C03;/C05/C138T¼1 /C03 /C052 43 5 In other words, if A¼½aij/C138is an m/C2nmatrix, then AT¼½bij/C138is the n/C2mmatrix where bij¼aji. Observe that the tranpose of a row vector is a column vector. Similarly, the transpose of a column vector is a row vector. The next theorem lists basic properties of the transpose operation. THEOREM 2.3: LetAandBbe matrices and let kbe a scalar. Then, whenever the sum and product are deﬁned, (i)ðAþBÞT¼ATþBT, (iii)ðkAÞT¼kAT, (ii)ðATÞT¼A; (iv)ðABÞT¼BTAT. We emphasize that, by (iv), the transpose of a product is the product of the transposes, but in the reverse order. 2.7 Square Matrices Asquare matrix is a matrix with the same number of rows as columns. An n/C2nsquare matrix is said to be of order n and is sometimes called an n-square matrix . Recall that not every two matrices can be added or multiplied. However, if we only consider square matrices of some given order n, then this inconvenience disappears. Speciﬁcally, the operations of addition, multiplication, scalar multiplication, and transpose can be performed on any n/C2nmatrices, and the result is again an n/C2nmatrix. EXAMPLE 2.6 The following are square matrices of order 3: A¼123 /C04/C04/C04 5672 43 5 and B¼2/C051 03/C02 12/C042 43 5 The following are also matrices of order 3: AþB¼3/C034 /C04/C01/C06 6832 643 75; 2A¼246 /C08/C08/C08 10 12 142 643 75; AT¼1/C045 2/C046 3/C0472 643 75 AB¼57/C015 /C012 0 20 17 7/C0352 643 75; BA¼27 30 33 /C022/C024/C026 /C027/C030/C0332 643 75 Diagonal and Trace LetA¼½aij/C138be an n-square matrix. The diagonal ormain diagonal ofAconsists of the elements with the same subscripts—that is, a11;a22;a33; ...;ann32 CHAPTER 2 Algebra of Matrices Thetrace ofA, written trðAÞ, is the sum of the diagonal elements. Namely, trðAÞ¼a11þa22þa33þ/C1/C1/C1þ ann The following theorem applies. THEOREM 2.4: Suppose A¼½aij/C138andB¼½bij/C138aren-square matrices and kis a scalar. Then (i) trðAþBÞ¼trðAÞþtrðBÞ, (iii) trðATÞ¼trðAÞ, (ii) trðkAÞ¼ktrðAÞ, (iv) tr ðABÞ¼trðBAÞ. EXAMPLE 2.7 LetAandBbe the matrices AandBin Example 2.6. Then diagonal of A¼f1;/C04;7g and trðAÞ¼1/C04þ7¼4 diagonal of B¼f2;3;/C04g and trðBÞ¼2þ3/C04¼1 Moreover, trðAþBÞ¼3/C01þ3¼5; trð2AÞ¼2/C08þ14¼8;trðATÞ¼1/C04þ7¼4 trðABÞ¼5þ0/C035¼/C030; trðBAÞ¼27/C024/C033¼/C030 As expected from Theorem 2.4, trðAþBÞ¼trðAÞþtrðBÞ; trðATÞ¼trðAÞ; trð2AÞ¼2t rðAÞ Furthermore, although AB6¼BA, the traces are equal. Identity Matrix, Scalar Matrices Then-square identity orunitmatrix, denoted by In, or simply I, is the n-square matrix with 1’s on the diagonal and 0’s elsewhere. The identity matrix Iis similar to the scalar 1 in that, for any n-square matrix A, AI¼IA¼A More generally, if Bis an m/C2nmatrix, then BIn¼ImB¼B. For any scalar k, the matrix kIthat contains k’s on the diagonal and 0’s elsewhere is called the scalar matrix corresponding to the scalar k. Observe that ðkIÞA¼kðIAÞ¼kA That is, multiplying a matrix Aby the scalar matrix kIis equivalent to multiplying Aby the scalar k. EXAMPLE 2.8 The following are the identity matrices of orders 3 and 4 and the corresponding scalar matrices for k¼5: 100 0100012 43 5;1 1 1 12 6643 775;500 0500052 43 5;5 5 5 52 6643 775 Remark 1: It is common practice to omit blocks or patterns of 0’s when there is no ambiguity, as in the above second and fourth matrices. Remark 2: TheKronecker delta function d ijis deﬁned by dij¼0i f i6¼j 1i f i¼j/C26 Thus, the identity matrix may be deﬁned by I¼½dij/C138.CHAPTER 2 Algebra of Matrices 33 2.8 Powers of Matrices, Polynomials in Matrices LetAbe an n-square matrix over a ﬁeld K.Powers ofAare deﬁned as follows: A2¼AA; A3¼A2A; ...; Anþ1¼AnA; ...; and A0¼I Polynomials in the matrix Aare also deﬁned. Speciﬁcally, for any polynomial fðxÞ¼a0þa1xþa2x2þ/C1/C1/C1þ anxn where the aiare scalars in K,fðAÞis deﬁned to be the following matrix: fðAÞ¼a0Iþa1Aþa2A2þ/C1/C1/C1þ anAn [Note that fðAÞis obtained from fðxÞby substituting the matrix Afor the variable xand substituting the scalar matrix a0Ifor the scalar a0.] If fðAÞis the zero matrix, then Ais called a zero orroot offðxÞ. EXAMPLE 2.9 Suppose A¼12 3/C04/C20/C21 . Then A2¼12 3/C04/C20/C21 12 3/C04/C20/C21 ¼7/C06 /C092 2/C20/C21 and A3¼A2A¼7/C06 /C092 2/C20/C21 12 3/C04/C20/C21 ¼/C011 38 57/C0106/C20/C21 Suppose fðxÞ¼2x2/C03xþ5 and gðxÞ¼x2þ3x/C010. Then fðAÞ¼27/C06 /C092 2/C20/C21 /C0312 3/C04/C20/C21 þ510 01/C20/C21 ¼16/C018 /C027 61/C20/C21 gðAÞ¼7/C06 /C092 2/C20/C21 þ312 3/C04/C20/C21 /C01010 01/C20/C21 ¼00 00/C20/C21 Thus, Ais a zero of the polynomial gðxÞ. 2.9 Invertible (Nonsingular) Matrices A square matrix Ais said to be invertible ornonsingular if there exists a matrix Bsuch that AB¼BA¼I where Iis the identity matrix. Such a matrix Bis unique. That is, if AB1¼B1A¼IandAB2¼B2A¼I, then B1¼B1I¼B1ðAB2Þ¼ð B1AÞB2¼IB2¼B2 We call such a matrix Btheinverse ofAand denote it by A/C01. Observe that the above relation is symmetric; that is, if Bis the inverse of A, then Ais the inverse of B. EXAMPLE 2.10 Suppose that A¼25 13/C20/C21 andB¼3/C05 /C012/C20/C21 . Then AB¼6/C05/C010þ10 3/C03/C05þ6/C20/C21 ¼10 01/C20/C21 and BA¼6/C051 5/C015 /C02þ2/C05þ6/C20/C21 ¼10 01/C20/C21 Thus, AandBare inverses. It is known (Theorem 3.16) that AB¼Iif and only if BA¼I. Thus, it is necessary to test only one product to determine whether or not two given matrices are inverses. (See Problem 2.17.) Now suppose AandBare invertible. Then ABis invertible andðABÞ/C01¼B/C01A/C01. More generally, if A1;A2;...;Akare invertible, then their product is invertible and ðA1A2...AkÞ/C01¼A/C01 k...A/C01 2A/C01 1 the product of the inverses in the reverse order.34 CHAPTER 2 Algebra of Matrices Inverse of a 2/C22 Matrix LetAbe an arbitrary 2/C22 matrix, say A¼ab cd/C20/C21 . We want to derive a formula for A/C01, the inverse ofA. Speciﬁcally, we seek 22¼4 scalars, say x1,y1,x2,y2, such that ab cd/C20/C21 x1x2 y1y2/C20/C21 ¼10 01/C20/C21 orax1þby1ax2þby2 cx1þdy1cx2þdy2/C20/C21 ¼10 01/C20/C21 Setting the four entries equal to the corresponding entries in the identity matrix yields four equations, which can be partitioned into two 2 /C22 systems as follows: ax1þby1¼1; ax2þby2¼0 cx1þdy1¼0; cx2þdy2¼1 Suppose we letjAj¼ab/C0bc(called the determinant ofA). AssumingjAj6¼0, we can solve uniquely for the above unknowns x1,y1,x2,y2, obtaining x1¼d jAj; y1¼/C0c jAj; x2¼/C0b jAj; y2¼a jAj Accordingly, A/C01¼ab cd/C20/C21/C01 ¼d=jAj/C0 b=jAj /C0c=jAj a=jAj/C20/C21 ¼1 jAjd/C0b /C0ca/C20/C21 In other words, when jAj6¼0, the inverse of a 2 /C22 matrix Amay be obtained from Aas follows: (1) Interchange the two elements on the diagonal. (2) Take the negatives of the other two elements. (3) Multiply the resulting matrix by 1 =jAjor, equivalently, divide each element by jAj. In casejAj¼0, the matrix Ais not invertible. EXAMPLE 2.11 Find the inverse of A¼23 45/C20/C21 andB¼13 26/C20/C21 . First evaluatejAj¼2ð5Þ/C03ð4Þ¼10/C012¼/C02. BecausejAj6¼0, the matrix Ais invertible and A/C01¼1 /C025/C03 /C042/C20/C21 ¼/C05 232 2/C01/C20/C21 Now evaluatejBj¼1ð6Þ/C03ð2Þ¼6/C06¼0. BecausejBj¼0, the matrix Bhas no inverse. Remark: The above property that a matrix is invertible if and only if Ahas a nonzero determinant is true for square matrices of any order. (See Chapter 8.) Inverse of an n/C2nMatrix Suppose Ais an arbitrary n-square matrix. Finding its inverse A/C01reduces, as above, to ﬁnding the solution of a collection of n/C2nsystems of linear equations. The solution of such systems and an efﬁcient way of solving such a collection of systems is treated in Chapter 3. 2.10 Special Types of Square Matrices This section describes a number of special kinds of square matrices. Diagonal and Triangular Matrices A square matrix D¼½dij/C138isdiagonal if its nondiagonal entries are all zero. Such a matrix is sometimes denoted by D¼diagðd11;d22;...;dnnÞCHAPTER 2 Algebra of Matrices 35 where some or all the diimay be zero. For example, 30 0 0/C070 00 22 43 5;40 0/C05/C20/C21 ;6 0 /C09 82 6643 775 are diagonal matrices, which may be represented, respectively, by diagð3;/C07;2Þ; diagð4;/C05Þ; diagð6;0;/C09;8Þ (Observe that patterns of 0’s in the third matrix have been omitted.) A square matrix A¼½aij/C138isupper triangular or simply triangular if all entries below the (main) diagonal are equal to 0—that is, if aij¼0 for i>j. Generic upper triangular matrices of orders 2, 3, 4 are as follows: a11a12 0a22/C20/C21 ;b11b12b13 b22b23 b332 43 5;c11c12c13c14 c22c23c24 c33c34 c442 6643 775 (As with diagonal matrices, it is common practice to omit patterns of 0’s.) The following theorem applies. THEOREM 2.5: Suppose A¼½aij/C138andB¼½bij/C138aren/C2n(upper) triangular matrices. Then (i) AþB,kA,ABare triangular with respective diagonals: ða11þb11;...;annþbnnÞ;ðka11;...;kannÞ;ða11b11;...;annbnnÞ (ii) For any polynomial fðxÞ, the matrix fðAÞis triangular with diagonal ðfða11Þ;fða22Þ;...;fðannÞÞ (iii) Ais invertible if and only if each diagonal element aii6¼0, and when A/C01exists it is also triangular. Alower triangular matrix is a square matrix whose entries above the diagonal are all zero. We note that Theorem 2.5 is true if we replace ‘‘triangular’’ by either ‘‘lower triangular’’ or ‘‘diagonal.’’ Remark: A nonempty collection Aof matrices is called an algebra (of matrices) if Ais closed under the operations of matrix addition, scalar multiplication, and matrix multiplication. Clearly, the square matrices with a given order form an algebra of matrices, but so do the scalar, diagonal, triangular, and lower triangular matrices. Special Real Square Matrices: Symmetric, Orthogonal, Normal [Optional until Chapter 12] Suppose now Ais a square matrix with real entries—that is, a real square matrix. The relationship between Aand its transpose ATyields important kinds of matrices. (a) Symmetric Matrices A matrix Aissymmetric ifAT¼A. Equivalently, A¼½aij/C138is symmetric if symmetric elements (mirror elements with respect to the diagonal) are equal—that is, if each aij¼aji. A matrix Aisskew-symmetric ifAT¼/C0Aor, equivalently, if each aij¼/C0aji. Clearly, the diagonal elements of such a matrix must be zero, because aii¼/C0aiiimplies aii¼0. (Note that a matrix Amust be square if AT¼AorAT¼/C0A.)36 CHAPTER 2 Algebra of Matrices EXAMPLE 2.12 LetA¼2/C035 /C0367 57/C082 43 5;B¼03/C04 /C0305 4/C0502 43 5;C¼100 001/C20/C21 : (a) By inspection, the symmetric elements in Aare equal, or AT¼A. Thus, Ais symmetric. (b) The diagonal elements of Bare 0 and symmetric elements are negatives of each other, or BT¼/C0B. Thus, Bis skew-symmetric. (c) Because Cis not square, Cis neither symmetric nor skew-symmetric. (b) Orthogonal Matrices A real matrix Aisorthogonal ifAT¼A/C01—that is, if AAT¼ATA¼I. Thus, Amust necessarily be square and invertible. EXAMPLE 2.13 LetA¼1 989/C049 49/C049/C079 8 919 492 643 75. Multiplying AbyATyields I; that is, AAT¼I. This means ATA¼I, as well. Thus, AT¼A/C01; that is, Ais orthogonal. Now suppose Ais a real orthogonal 3 /C23 matrix with rows u1¼ða1;a2;a3Þ; u2¼ðb1;b2;b3Þ; u3¼ðc1;c2;c3Þ Because Ais orthogonal, we must have AAT¼I. Namely, AAT¼a1a2a3 b1b2b3 c1c2c32 43 5a1b1c1 a2b2c2 a3b3c32 43 5¼100 010 0012 43 5¼I Multiplying AbyATand setting each entry equal to the corresponding entry in Iyields the following nine equations: a2 1þa2 2þa2 3¼1; a1b1þa2b2þa3b3¼0; a1c1þa2c2þa3c3¼0 b1a1þb2a2þb3a3¼0; b2 1þb2 2þb2 3¼1; b1c1þb2c2þb3c3¼0 c1a1þc2a2þc3a3¼0; c1b1þc2b2þc3b3¼0; c2 1þc2 2þc2 3¼1 Accordingly, u1/C1u1¼1,u2/C1u2¼1,u3/C1u3¼1, and ui/C1uj¼0 for i6¼j. Thus, the rows u1,u2,u3are unit vectors and are orthogonal to each other. Generally speaking, vectors u1,u2;...;uminRnare said to form an orthonormal set of vectors if the vectors are unit vectors and are orthogonal to each other; that is, ui/C1uj¼0i f i6¼j 1i f i¼j/C26 In other words, ui/C1uj¼dijwhere dijis the Kronecker delta function : We have shown that the condition AAT¼Iimplies that the rows of Aform an orthonormal set of vectors. The condition ATA¼Isimilarly implies that the columns of Aalso form an orthonormal set of vectors. Furthermore, because each step is reversible, the converse is true. The above results for 3 /C23 matrices are true in general. That is, the following theorem holds. THEOREM 2.6: LetAbe a real matrix. Then the following are equivalent: (a)Ais orthogonal. (b) The rows of Aform an orthonormal set. (c) The columns of Aform an orthonormal set. Forn¼2, we have the following result (proved in Problem 2.28).CHAPTER 2 Algebra of Matrices 37 THEOREM 2.7: LetAbe a real 2/C22orthogonal matrix. Then, for some real number y, A¼cosysiny /C0sinycosy/C20/C21 or A¼cosy siny siny/C0cosy/C20/C21 (c) Normal Matrices A real matrix Aisnormal if itcommutes with its transpose AT—that is, if AAT¼ATA.I fAis symmetric, orthogonal, or skew-symmetric, then Ais normal. There are also other normal matrices. EXAMPLE 2.14 LetA¼6/C03 36/C20/C21 . Then AAT¼6/C03 36/C20/C21 63 /C036/C20/C21 ¼45 0 04 5/C20/C21 and ATA¼63 /C036/C20/C21 6/C03 36/C20/C21 ¼45 0 04 5/C20/C21 Because AAT¼ATA, the matrix Ais normal. 2.11 Complex Matrices LetAbe a complex matrix—that is, a matrix with complex entries. Recall (Section 1.7) that if z¼aþbi is a complex number, then /C22z¼a/C0biis its conjugate. The conjugate of a complex matrix A, written /C22A,i s the matrix obtained from Aby taking the conjugate of each entry in A. That is, if A¼½aij/C138, then /C22A¼½bij/C138, where bij¼/C22aij. (We denote this fact by writing /C22A¼½/C22aij/C138.) The two operations of transpose and conjugation commute for any complex matrix A, and the special notation AHis used for the conjugate transpose of A. That is, AH¼ð /C22AÞT¼ðATÞ Note that if Ais real, then AH¼AT. [Some texts use A* instead of AH:] EXAMPLE 2.15 LetA¼2þ8i5/C03i4/C07i 6i 1/C04i3þ2i/C20/C21 . Then AH¼2/C08i/C06i 5þ3i1þ4i 4þ7i3/C02i2 43 5. Special Complex Matrices: Hermitian, Unitary, Normal [Optional until Chapter 12] Consider a complex matrix A. The relationship between Aand its conjugate transpose AHyields important kinds of complex matrices (which are analogous to the kinds of real matrices described above). A complex matrix Ais said to be Hermitian orskew-Hermitian according as to whether AH¼A or AH¼/C0A: Clearly, A¼½aij/C138is Hermitian if and only if symmetric elements are conjugate—that is, if each aij¼/C22aji—in which case each diagonal element aiimust be real. Similarly, if Ais skew-symmetric, then each diagonal element aii¼0. (Note that Amust be square if AH¼AorAH¼/C0A.) A complex matrix Aisunitary ifAHA/C01¼A/C01AH¼I—that is, if AH¼A/C01: Thus, Amust necessarily be square and invertible. We note that a complex matrix Ais unitary if and only if its rows (columns) form an orthonormal set relative to the dot product of complex vectors. A complex matrix Ais said to be normal if it commutes with AH—that is, if AAH¼AHA38 CHAPTER 2 Algebra of Matrices (Thus, Amust be a square matrix.) This deﬁnition reduces to that for real matrices when Ais real. EXAMPLE 2.16 Consider the following complex matrices: A¼31/C02i4þ7i 1þ2i/C04/C02i 4/C07i 2i 52 43 5 B¼1 21/C0i/C01þi i 11þi 1þi/C01þi 02 43 5 C¼2þ3i 1 i 1þ2i/C20/C21 (a) By inspection, the diagonal elements of Aare real, and the symmetric elements 1 /C02iand 1þ2iare conjugate, 4þ7iand 4/C07iare conjugate, and /C02iand 2 iare conjugate. Thus, Ais Hermitian. (b) Multiplying BbyBHyields I; that is, BBH¼I. This implies BHB¼I, as well. Thus, BH¼B/C01, which means Bis unitary. (c) To show Cis normal, we evaluate CCHandCHC: CCH¼2þ3i 1 i 1þ2i/C20/C21 2/C03i/C0i 11/C02i/C20/C21 ¼14 4/C04i 4þ4i 6/C20/C21 and similarly CHC¼14 4/C04i 4þ4i 6/C20/C21 . Because CCH¼CHC, the complex matrix Cis normal. We note that when a matrix Ais real, Hermitian is the same as symmetric, and unitary is the same as orthogonal. 2.12 Block Matrices Using a system of horizontal and vertical (dashed) lines, we can partition a matrix Ainto submatrices called blocks (orcells)o fA. Clearly a given matrix may be divided into blocks in different ways. For example, 1/C020 13 235 7 /C02 314 5946/C031 82 6643 775;1/C020 13 235 7 /C02 314 5946/C031 82 6643 775;1/C020 13 235 7 /C02 314 5946/C031 82 6643 775 The convenience of the partition of matrices, say AandB, into blocks is that the result of operations on A andBcan be obtained by carrying out the computation with the blocks, just as if they were the actual elements of the matrices. This is illustrated below, where the notation A¼½A ij/C138will be used for a block matrix Awith blocks Aij. Suppose that A¼½Aij/C138andB¼½Bij/C138are block matrices with the same numbers of row and column blocks, and suppose that corresponding blocks have the same size. Then adding the corresponding blocksofAandBalso adds the corresponding elements of AandB, and multiplying each block of Aby a scalar kmultiplies each element of Abyk. Thus, AþB¼A 11þB11 A12þB12 ... A1nþB1n A21þB21 A22þB22 ... A2nþB2n ... ... ... ... Am1þBm1Am2þBm2... AmnþBmn2 66643 7775 and kA¼kA11kA12 ... kA1n kA21kA22 ... kA2n ... ... ... ... kAm1kAm2... kAmn2 6643 775CHAPTER 2 Algebra of Matrices 39 The case of matrix multiplication is less obvious, but still true. That is, suppose that U¼½Uik/C138and V¼½Vkj/C138are block matrices such that the number of columns of each block Uikis equal to the number of rows of each block Vkj. (Thus, each product UikVkjis deﬁned.) Then UV¼W11W12 ... W1n W21W22 ... W2n ... ... ... ... Wm1Wm2... Wmn2 6643 775; where Wij¼Ui1V1jþUi2V2jþ/C1/C1/C1þ UipVpj The proof of the above formula for UVis straightforward but detailed and lengthy. It is left as an exercise (Problem 2.85). Square Block Matrices LetMbe a block matrix. Then Mis called a square block matrix if (i)Mis a square matrix. (ii) The blocks form a square matrix. (iii) The diagonal blocks are also square matrices. The latter two conditions will occur if and only if there are the same number of horizontal and vertical lines and they are placed symmetrically. Consider the following two block matrices: A¼12345 11111 9876544444 353532 666643 77775and B¼12345 11111 98765 44444 353532 666643 77775 The block matrix Ais not a square block matrix, because the second and third diagonal blocks are not square. On the other hand, the block matrix Bis a square block matrix. Block Diagonal Matrices LetM¼½Aij/C138be a square block matrix such that the nondiagonal blocks are all zero matrices; that is, Aij¼0 when i6¼j. Then Mis called a block diagonal matrix . We sometimes denote such a block diagonal matrix by writing M¼diagðA11;A22;...;ArrÞ or M¼A11/C8A22/C8/C1/C1/C1/C8 Arr The importance of block diagonal matrices is that the algebra of the block matrix is frequently reduced to the algebra of the individual blocks. Speciﬁcally, suppose fðxÞis a polynomial and Mis the above block diagonal matrix. Then fðMÞis a block diagonal matrix, and fðMÞ¼diagðfðA11Þ;fðA22Þ;...;fðArrÞÞ Also, Mis invertible if and only if each Aiiis invertible, and, in such a case, M/C01is a block diagonal matrix, and M/C01¼diagðA/C01 11;A/C01 22;...;A/C01 rrÞ Analogously, a square block matrix is called a block upper triangular matrix if the blocks below the diagonal are zero matrices and a block lower triangular matrix if the blocks above the diagonal are zero matrices.40 CHAPTER 2 Algebra of Matrices EXAMPLE 2.17 Determine which of the following square block matrices are upper diagonal, lower diagonal, or diagonal: A¼120 345 0062 43 5; B¼1000 2340 5060 07892 6643 775; C¼100 023 0452 43 5; D¼120 345 0672 43 5 (a)Ais upper triangular because the block below the diagonal is a zero block. (b)Bis lower triangular because all blocks above the diagonal are zero blocks. (c)Cis diagonal because the blocks above and below the diagonal are zero blocks. (d)Dis neither upper triangular nor lower triangular. Also, no other partitioning of Dwill make it into either a block upper triangular matrix or a block lower triangular matrix. SOLVED PROBLEMS Matrix Addition and Scalar Multiplication 2.1 Given A¼1/C023 45/C06/C20/C21 andB¼302 /C0718/C20/C21 , ﬁnd: (a)AþB, (b) 2 A/C03B. (a) Add the corresponding elements: AþB¼1þ3/C02þ03þ2 4/C075þ1/C06þ8/C20/C21 ¼4/C025 /C036 2/C20/C21 (b) First perform the scalar multiplication and then a matrix addition: 2A/C03B¼2/C046 81 0/C012/C20/C21 þ/C090/C06 21/C03/C024/C20/C21 ¼/C07/C040 29 7/C036/C20/C21 (Note that we multiply Bby/C03 and then add, rather than multiplying Bby 3 and subtracting. This usually prevents errors.) 2.2. Find x;y;z;twhere 3xy zt/C20/C21 ¼x6 /C012 t/C20/C21 þ4 xþy zþt 3/C20/C21 : Write each side as a single equation: 3x 3y 3z 3t/C20/C21 ¼xþ4xþyþ6 zþt/C012 tþ3/C20/C21 Set corresponding entries equal to each other to obtain the following system of four equations: 3x¼xþ4; 3y¼xþyþ6; 3z¼zþt/C01; 3t¼2tþ3 or 2 x¼4; 2y¼6þx; 2z¼t/C01; t¼3 The solution is x¼2,y¼4,z¼1,t¼3. 2.3. Prove Theorem 2.1 (i) and (v): (i) ðAþBÞþC¼AþðBþCÞ, (v) kðAþBÞ¼kAþkB. Suppose A¼½aij/C138,B¼½bij/C138,C¼½cij/C138. The proof reduces to showing that corresponding ij-entries in each side of each matrix equation are equal. [We prove only (i) and (v), because the other parts of Theorem 2.1 are proved similarly.]CHAPTER 2 Algebra of Matrices 41 (i) The ij-entry of AþBisaijþbij; hence, the ij-entry ofðAþBÞþCisðaijþbijÞþcij. On the other hand, theij-entry of BþCisbijþcij; hence, the ij-entry of AþðBþCÞisaijþðbijþcijÞ. However, for scalars in K, ðaijþbijÞþcij¼aijþðbijþcijÞ Thus,ðAþBÞþCandAþðBþCÞhave identical ij-entries. Therefore, ðAþBÞþC¼AþðBþCÞ. (v) The ij-entry of AþBisaijþbij; hence, kðaijþbijÞis the ij-entry of kðAþBÞ. On the other hand, the ij- entries of kAandkBarekaijandkbij, respectively. Thus, kaijþkbijis the ij-entry of kAþkB. However, for scalars in K, kðaijþbijÞ¼kaijþkbij Thus, kðAþBÞandkAþkBhave identical ij-entries. Therefore, kðAþBÞ¼kAþkB. Matrix Multiplication 2.4. Calculate: (a)½8;/C04;5/C1383 2 /C012 43 5, (b)½6;/C01;7;5/C1384 /C09 /C03 22 6643 775, (c)½3;8;/C02;4/C1385 /C01 62 43 5 (a) Multiply the corresponding entries and add: ½8;/C04;5/C1383 2 /C012 43 5¼8ð3Þþð/C0 4Þð2Þþ5ð/C01Þ¼24/C08/C05¼11 (b) Multiply the corresponding entries and add: ½6;/C01;7;5/C1384 /C09 /C03 22 66643 7775¼24þ9/C021þ10¼22 (c) The product is not deﬁned when the row matrix and the column matrix have different numbers of elements. 2.5. Letðr/C2sÞdenote an r/C2smatrix. Find the sizes of those matrix products that are deﬁned: (a)ð2/C23Þð3/C24Þ; (c)ð1/C22Þð3/C21Þ; (e)ð4/C24Þð3/C23Þ (b)ð4/C21Þð1/C22Þ, (d)ð5/C22Þð2/C23Þ, (f)ð2/C22Þð2/C24Þ In each case, the product is deﬁned if the inner numbers are equal, and then the product will have the size of the outer numbers in the given order. (a) 2/C24, (c) not deﬁned, (e) not deﬁned (b) 4/C22, (d) 5/C23, (f) 2 /C24 2.6. LetA¼13 2/C01/C20/C21 andB¼20/C04 3/C026/C20/C21 . Find: (a) AB, (b) BA. (a) Because Ais a 2/C22 matrix and Ba2/C23 matrix, the product ABis deﬁned and is a 2 /C23 matrix. To obtain the entries in the ﬁrst row of AB, multiply the ﬁrst row ½1;3/C138ofAby the columns 2 3/C20/C21 ;0 /C02/C20/C21 ;/C04 6/C20/C21 ofB, respectively, as follows: AB¼13 2/C01/C20/C21 20/C04 3/C026/C20/C21 ¼2þ90/C06/C04þ18/C20/C21 ¼11/C061 4/C20/C2142 CHAPTER 2 Algebra of Matrices To obtain the entries in the second row of AB, multiply the second row ½2;/C01/C138ofAby the columns of B: AB¼13 2/C01/C20/C2120/C04 3/C026/C20/C21 ¼11/C061 4 4/C030þ2/C08/C06/C20/C21 Thus, AB¼11/C061 4 12/C014/C20/C21 : (b) The size of Bis 2/C23 and that of Ais 2/C22. The inner numbers 3 and 2 are not equal; hence, the product BAis not deﬁned. 2.7. Find AB, where A¼23/C01 4/C025/C20/C21 andB¼2/C010 6 13/C051 41/C0222 43 5. Because Ais a 2/C23 matrix and Ba3/C24 matrix, the product ABis deﬁned and is a 2 /C24 matrix. Multiply the rows of Aby the columns of Bto obtain AB¼4þ3/C04/C02þ9/C010/C015þ21 2þ3/C02 8/C02þ20/C04/C06þ50þ10/C010 24/C02þ10/C20/C21 ¼36/C013 13 26/C050 3 2/C20/C21 : 2.8. Find: (a)16 /C035/C20/C21 2 /C07/C20/C21 , (b)2 /C07/C20/C21 16 /C035/C20/C21 , (c)½2;/C07/C13816 /C035/C20/C21 . (a) The ﬁrst factor is 2 /C22 and the second is 2 /C21, so the product is deﬁned as a 2 /C21 matrix: 16 /C035/C20/C21 2 /C07/C20/C21 ¼2/C042 /C06/C035/C20/C21 ¼/C040 /C041/C20/C21 (b) The product is not deﬁned, because the ﬁrst factor is 2 /C21 and the second factor is 2 /C22. (c) The ﬁrst factor is 1 /C22 and the second factor is 2 /C22, so the product is deﬁned as a 1 /C22 (row) matrix: ½2;/C07/C13816 /C035/C20/C21 ¼½2þ21;12/C035/C138¼½23;/C023/C138 2.9. Clearly, 0 A¼0 and A0¼0, where the 0’s are zero matrices (with possibly different sizes). Find matrices AandBwith no zero entries such that AB¼0. LetA¼12 24/C20/C21 andB¼62 /C03/C01/C20/C21 . Then AB¼00 00/C20/C21 . 2.10. Prove Theorem 2.2(i): ðABÞC¼AðBCÞ. Let A¼½aij/C138,B¼½bjk/C138,C¼½ckl/C138, and let AB¼S¼½sik/C138,BC¼T¼½tjl/C138. Then sik¼Pm j¼1aijbjk and tjl¼Pn k¼1bjkckl Multiplying S¼ABbyC, the il-entry ofðABÞCis si1c1lþsi2c2lþ/C1/C1/C1þ sincnl¼Pn k¼1sikckl¼Pn k¼1Pm j¼1ðaijbjkÞckl On the other hand, multiplying AbyT¼BC, the il-entry of AðBCÞis ai1t1lþai2t2lþ/C1/C1/C1þ aintnl¼Pm j¼1aijtjl¼Pm j¼1Pn k¼1aijðbjkcklÞ The above sums are equal; that is, corresponding elements in ðABÞCand AðBCÞare equal. Thus, ðABÞC¼AðBCÞ.CHAPTER 2 Algebra of Matrices 43 2.11. Prove Theorem 2.2(ii): AðBþCÞ¼ABþAC. LetA¼½aij/C138,B¼½bjk/C138,C¼½cjk/C138, and let D¼BþC¼½djk/C138,E¼AB¼½eik/C138,F¼AC¼½fik/C138. Then djk¼bjkþcjk; eik¼Pm j¼1aijbjk; fik¼Pm j¼1aijcjk Thus, the ik-entry of the matrix ABþACis eikþfik¼Pm j¼1aijbjkþPm j¼1aijcjk¼Pm j¼1aijðbjkþcjkÞ On the other hand, the ik-entry of the matrix AD¼AðBþCÞis ai1d1kþai2d2kþ/C1/C1/C1þ aimdmk¼Pm j¼1aijdjk¼Pm j¼1aijðbjkþcjkÞ Thus, AðBþCÞ¼ABþAC, because the corresponding elements are equal. Transpose 2.12. Find the transpose of each matrix: A¼1/C023 78/C09/C20/C21 ; B¼123 245 3562 43 5; C¼½1;/C03;5;/C07/C138; D¼2 /C04 62 43 5 Rewrite the rows of each matrix as columns to obtain the transpose of the matrix: AT¼17 /C028 3/C092 43 5; BT¼123 2453562 43 5; C T¼1 /C03 5 /C072 6643 775; DT¼½2;/C04;6/C138 (Note that BT¼B; such a matrix is said to be symmetric . Note also that the transpose of the row vector Cis a column vector, and the transpose of the column vector Dis a row vector.) 2.13. Prove Theorem 2.3(iv): ðABÞT¼BTAT. LetA¼½aik/C138andB¼½bkj/C138. Then the ij-entry of ABis ai1b1jþai2b2jþ/C1/C1/C1þ aimbmj This is the ji-entry (reverse order) of ðABÞT. Now column jofBbecomes row jofBT, and row iofAbecomes column iofAT. Thus, the ij-entry of BTATis ½b1j;b2j;...;bmj/C138½ai1;ai2;...;aim/C138T¼b1jai1þb2jai2þ/C1/C1/C1þ bmjaim Thus,ðABÞT¼BTATon because the corresponding entries are equal. Square Matrices 2.14. Find the diagonal and trace of each matrix: (a) A¼13 6 2/C058 4/C0292 43 5, (b) B¼24 8 3/C079 /C050 22 43 5, (c) C¼12/C03 4/C056/C20/C21 . (a) The diagonal of Aconsists of the elements from the upper left corner of Ato the lower right corner of Aor, in other words, the elements a11,a22,a33. Thus, the diagonal of Aconsists of the numbers 1 ;/C05, and 9. The trace of Ais the sum of the diagonal elements. Thus, trðAÞ¼1/C05þ9¼5 (b) The diagonal of Bconsists of the numbers 2 ;/C07, and 2. Hence, trðBÞ¼2/C07þ2¼/C03 (c) The diagonal and trace are only deﬁned for square matrices.44 CHAPTER 2 Algebra of Matrices 2.15. LetA¼12 4/C03/C20/C21 , and let fðxÞ¼2x3/C04xþ5 and gðxÞ¼x2þ2xþ11. Find (a)A2, (b) A3, (c) fðAÞ, (d) gðAÞ. (a) A2¼AA¼12 4/C03/C20/C21 12 4/C03/C20/C21 ¼1þ82/C06 4/C012 8þ9/C20/C21 ¼9/C04 /C081 7/C20/C21 (b) A3¼AA2¼12 4/C03/C20/C21 9/C04 /C081 7/C20/C21 ¼9/C016/C04þ34 36þ24/C016/C051/C20/C21 ¼/C073 0 60/C067/C20/C21 (c) First substitute Aforxand 5 Ifor the constant in fðxÞ, obtaining fðAÞ¼2A3/C04Aþ5I¼2/C073 0 60/C067/C20/C21 /C0412 4/C03/C20/C21 þ510 01/C20/C21 Now perform the scalar multiplication and then the matrix addition: fðAÞ¼/C014 60 120/C0134/C20/C21 þ/C04/C08 /C016 12/C20/C21 þ50 05/C20/C21 ¼/C013 52 104/C0117/C20/C21 (d) Substitute Aforxand 11 Ifor the constant in gðxÞ, and then calculate as follows: gðAÞ¼A2þ2A/C011I¼9/C04 /C081 7/C20/C21 þ212 4/C03/C20/C21 /C01110 01/C20/C21 ¼9/C04 /C081 7/C20/C21 þ24 8/C06/C20/C21 þ/C011 0 0/C011/C20/C21 ¼00 00/C20/C21 Because gðAÞis the zero matrix, Ais a root of the polynomial gðxÞ. 2.16. Let A¼13 4/C03/C20/C21 . (a) Find a nonzero column vector u¼x y/C20/C21 such that Au¼3u. (b) Describe all such vectors. (a) First set up the matrix equation Au¼3u, and then write each side as a single matrix (column vector) as follows: 13 4/C03/C20/C21 x y/C20/C21 ¼3x y/C20/C21 ; and thenxþ3y 4x/C03y/C20/C21 ¼3x 3y/C20/C21 Set the corresponding elements equal to each other to obtain a system of equations: xþ3y¼3x 4x/C03y¼3yor2x/C03y¼0 4x/C06y¼0or 2 x/C03y¼0 The system reduces to one nondegenerate linear equation in two unknowns, and so has an inﬁnite number of solutions. To obtain a nonzero solution, let, say, y¼2; then x¼3. Thus, u¼ð3;2ÞTis a desired nonzero vector. (b) To ﬁnd the general solution, set y¼a, where ais a parameter. Substitute y¼ainto 2 x/C03y¼0 to obtain x¼3 2a. Thus, u¼ð3 2a;aÞTrepresents all such solutions. Invertible Matrices, Inverses 2.17. Show that A¼10 2 2/C013 41 82 43 5andB¼/C011 2 2 /C0401 6/C01/C012 43 5are inverses. Compute the product AB, obtaining AB¼/C011þ0þ12 2þ0/C022þ0/C02 /C022þ4þ18 4þ0/C034/C01/C03 /C044/C04þ48 8þ0/C088þ1/C082 43 5¼100 010 0012 43 5¼I Because AB¼I, we can conclude (Theorem 3.16) that BA¼I. Accordingly, AandBare inverses.CHAPTER 2 Algebra of Matrices 45 2.18. Find the inverse, if possible, of each matrix: (a) A¼53 42/C20/C21 ; (b) B¼2/C03 13/C20/C21 ; (c)/C026 3/C09/C20/C21 : Use the formula for the inverse of a 2 /C22 matrix appearing in Section 2.9. (a) First ﬁndjAj¼5ð2Þ/C03ð4Þ¼10/C012¼/C02. Next interchange the diagonal elements, take the negatives of the nondiagonal elements, and multiply by 1 =jAj: A/C01¼/C01 22/C03 /C045/C20/C21 ¼/C013 2 2/C05 2"# (b) First ﬁndjBj¼2ð3Þ/C0ð/C0 3Þð1Þ¼6þ3¼9. Next interchange the diagonal elements, take the negatives of the nondiagonal elements, and multiply by 1 =jBj: B/C01¼1 933 /C012/C20/C21 ¼1 313 /C01 929"# (c) First ﬁndjCj¼/C0 2ð/C09Þ/C06ð3Þ¼18/C018¼0. BecausejCj¼0;Chas no inverse. 2.19. LetA¼111 0121242 6643 775. Find A /C01¼x1x2x3 y1y2y3 z1z2z32 43 5. Multiplying AbyA/C01and setting the nine entries equal to the nine entries of the identity matrix Iyields the following three systems of three equations in three of the unknowns: x1þy1þz1¼1 x2þy2þz2¼0 x3þy3þz3¼0 y1þ2z1¼0 y2þ2z2¼1 y3þ2z3¼0 x1þ2y1þ4z1¼0 x2þ2y2þ4z2¼0 x3þ2y3þ4z3¼1 [Note that Ais the coefﬁcient matrix for all three systems.] Solving the three systems for the nine unknowns yields x1¼0;y1¼2;z1¼/C01; x2¼/C02;y2¼3;z2¼/C01; x3¼1;y3¼/C02;z3¼1 Thus ; A/C01¼0/C021 23/C02 /C01/C0112 643 75 (Remark: Chapter 3 gives an efﬁcient way to solve the three systems.) 2.20. LetAandBbe invertible matrices (with the same size). Show that ABis also invertible and ðABÞ/C01¼B/C01A/C01. [Thus, by induction, ðA1A2...AmÞ/C01¼A/C01 m...A/C01 2A/C01 1.] Using the associativity of matrix multiplication, we get ðABÞðB/C01A/C01Þ¼AðBB/C01ÞA/C01¼AIA/C01¼AA/C01¼I ðB/C01A/C01ÞðABÞ¼B/C01ðA/C01AÞB¼A/C01IB¼B/C01B¼I Thus,ðABÞ/C01¼B/C01A/C01.46 CHAPTER 2 Algebra of Matrices Diagonal and Triangular Matrices 2.21. Write out the diagonal matrices A¼diagð4;/C03;7Þ,B¼diagð2;/C06Þ,C¼diagð3;/C08;0;5Þ. Put the given scalars on the diagonal and 0’s elsewhere: A¼40 0 0/C030 00 72 43 5; B¼20 0/C06/C20/C21 ; C¼3 /C08 0 52 6643 775 2.22. LetA¼diagð2;3;5ÞandB¼diagð7;0;/C04Þ. Find (a)AB,A2,B2; (b) fðAÞ, where fðxÞ¼x2þ3x/C02; (c) A/C01andB/C01. (a) The product matrix ABis a diagonal matrix obtained by multiplying corresponding diagonal entries; hence, AB¼diagð2ð7Þ;3ð0Þ;5ð/C04ÞÞ¼ diagð14;0;/C020Þ Thus, the squares A2andB2are obtained by squaring each diagonal entry; hence, A2¼diagð22;32;52Þ¼diagð4;9;25Þ and B2¼diagð49;0;16Þ (b)fðAÞis a diagonal matrix obtained by evaluating fðxÞat each diagonal entry. We have fð2Þ¼4þ6/C02¼8; fð3Þ¼9þ9/C02¼16; fð5Þ¼25þ15/C02¼38 Thus, fðAÞ¼diagð8;16;38Þ. (c) The inverse of a diagonal matrix is a diagonal matrix obtained by taking the inverse (reciprocal) of each diagonal entry. Thus, A/C01¼diagð1 2;13;15Þ, but Bh a sn oi n v e r s eb e c a u s et h e r ei sa0o nt h e diagonal. 2.23. Find a 2/C22 matrix Asuch that A2is diagonal but not A. LetA¼12 3/C01/C20/C21 . Then A2¼70 07/C20/C21 , which is diagonal. 2.24. Find an upper triangular matrix Asuch that A3¼8/C057 02 7/C20/C21 . SetA¼xy 0z/C20/C21 . Then x3¼8, so x¼2; and z3¼27, so z¼3. Next calculate A3using x¼2 and y¼3: A2¼2y 03/C20/C21 2y 03/C20/C21 ¼45 y 09/C20/C21 and A3¼2y 03/C20/C21 45 y 09/C20/C21 ¼81 9 y 02 7/C20/C21 Thus, 19 y¼/C057, or y¼/C03. Accordingly, A¼2/C03 03/C20/C21 . 2.25. LetA¼½aij/C138andB¼½bij/C138be upper triangular matrices. Prove that ABis upper triangular with diagonal a11b11,a22b22;...;annbnn. LetAB¼½cij/C138. Then cij¼Pn k¼1aikbkjandcii¼Pn k¼1aikbki. Suppose i>j. Then, for any k, either i>kor k>j, so that either aik¼0o r bkj¼0. Thus, cij¼0, and ABis upper triangular. Suppose i¼j. Then, for k<i, we have aik¼0; and, for k>i, we have bki¼0. Hence, cii¼aiibii, as claimed. [This proves one part of Theorem 2.5(i); the statements for AþBandkAare left as exercises.]CHAPTER 2 Algebra of Matrices 47 Special Real Matrices: Symmetric and Orthogonal 2.26. Determine whether or not each of the following matrices is symmetric —that is, AT¼A—or skew-symmetric —that is, AT¼/C0A: (a) A¼5/C071 /C0782 12/C042 43 5; (b) B¼04/C03 /C0405 3/C0502 43 5; (c) C¼000 000/C20/C21 (a) By inspection, the symmetric elements (mirror images in the diagonal) are /C07 and/C07, 1 and 1, 2 and 2. Thus, Ais symmetric, because symmetric elements are equal. (b) By inspection, the diagonal elements are all 0, and the symmetric elements, 4 and /C04,/C03 and 3, and 5 and /C05, are negatives of each other. Hence, Bis skew-symmetric. (c) Because Cis not square, Cis neither symmetric nor skew-symmetric. 2.27. Suppose B¼4 xþ2 2x/C03xþ1/C20/C21 is symmetric. Find xandB. Set the symmetric elements xþ2 and 2 x/C03 equal to each other, obtaining 2 x/C03¼xþ2o r x¼5. Hence, B¼47 76/C20/C21 . 2.28. LetAbe an arbitrary 2/C22 (real) orthogonal matrix . (a) Prove: Ifða;bÞis the ﬁrst row of A, then a2þb2¼1 and A¼ab /C0ba/C20/C21 or A¼ab b/C0a/C20/C21 : (b) Prove Theorem 2.7: For some real number y, A¼cosysiny /C0sinycosy/C20/C21 or A¼cosy siny siny/C0cosy/C20/C21 (a) Supposeðx;yÞis the second row of A. Because the rows of Aform an orthonormal set, we get a2þb2¼1; x2þy2¼1; axþby¼0 Similarly, the columns form an orthogonal set, so a2þx2¼1; b2þy2¼1; abþxy¼0 Therefore, x2¼1/C0a2¼b2, whence x¼/C6b: Case (i): x¼b. Then bðaþyÞ¼0, so y¼/C0a. Case (ii): x¼/C0b. Then bðy/C0aÞ¼0, so y¼a. This means, as claimed, A¼ab /C0ba/C20/C21 or A¼ab b/C0a/C20/C21 (b) Because a2þb2¼1, we have/C01/C20a/C201. Let a¼cosy. Then b2¼1/C0cos2y,s ob¼siny. This proves the theorem. 2.29. Find a 2/C22 orthogonal matrix Awhose ﬁrst row is a (positive) multiple of ð3;4Þ. Normalizeð3;4Þto getð3 5;45Þ. Then, by Problem 2.28, A¼3 545 /C04 535"# or A¼3 545 4 5/C035"# : 2.30. Find a 3/C23 orthogonal matrix Pwhose ﬁrst two rows are multiples of u1¼ð1;1;1Þand u2¼ð0;/C01;1Þ, respectively. (Note that, as required, u1andu2are orthogonal.)48 CHAPTER 2 Algebra of Matrices First ﬁnd a nonzero vector u3orthogonal to u1andu2; say (cross product) u3¼u1/C2u2¼ð2;/C01;/C01Þ.L e t Abe the matrix whose rows are u1;u2;u3;a n dl e t Pbe the matrix obtained from Aby normalizing the rows of A. Thus, A¼111 0/C011 2/C01/C012 643 75 and P¼1=ﬃﬃﬃ 3p 1=ﬃﬃﬃ 3p 1=ﬃﬃﬃ 3p 0/C01=ﬃﬃﬃ 2p 1=ﬃﬃﬃ 2p 2=ﬃﬃﬃ 6p /C01=ﬃﬃﬃ 6p /C01=ﬃﬃﬃ 6p2 66643 7775 Complex Matrices: Hermitian and Unitary Matrices 2.31. Find AHwhere (a) A¼3/C05i2þ4i 6þ7i1þ8i/C20/C21 , (b) A¼2/C03i5þ8i /C043/C07i /C06/C0i 5i2 43 5 Recall that AH¼/C22AT, the conjugate tranpose of A. Thus, (a) AH¼3þ5i6/C07i 2/C04i1/C08i/C20/C21 , (b) AH¼2þ3i/C04/C06þi 5/C08i3þ7i/C05i/C20/C21 2.32. Show that A¼1 3/C023i23i /C02 3i/C01 3/C023i"# is unitary. The rows of Aform an orthonormal set: 1 3/C02 3i;2 3i/C18/C19 /C11 3/C02 3i;2 3i/C18/C19 ¼1 9þ4 9/C18/C19 þ4 9¼1 1 3/C02 3i;2 3i/C18/C19 /C1/C02 3i;/C01 3/C02 3i/C18/C19 ¼2 9iþ4 9/C18/C19 þ/C02 9i/C04 9/C18/C19 ¼0 /C02 3i;/C01 3/C02 3i/C18/C19 /C1/C02 3i;/C01 3/C02 3i/C18/C19 ¼4 9þ1 9þ4 9/C18/C19 ¼1 Thus, Ais unitary. 2.33. Prove the complex analogue of Theorem 2.6: Let Abe a complex matrix. Then the following are equivalent: (i) Ais unitary. (ii) The rows of Aform an orthonormal set. (iii) The columns of A form an orthonormal set. (The proof is almost identical to the proof on page 37 for the case when Ais a 3/C23r e a lm a t r i x . ) First recall that the vectors u1;u2;...;uninCnform an orthonormal set if they are unit vectors and are orthogonal to each other, where the dot product in Cnis deﬁned by ða1;a2;...;anÞ/C1ðb1;b2;...;bnÞ¼a1/C22b1þa2/C22b2þ/C1/C1/C1þ an/C22bn Suppose Ais unitary, and R1;R2;...;Rnare its rows. Then /C22RT 1;/C22RT 2;...;/C22RT nare the columns of AH. Let AAH¼½cij/C138. By matrix multiplication, cij¼Ri/C22RT j¼Ri/C1Rj. Because Ais unitary, we have AAH¼I. Multi- plying AbyAHand setting each entry cijequal to the corresponding entry in Iyields the following n2 equations: R1/C1R1¼1;R2/C1R2¼1; ...;Rn/C1Rn¼1; and Ri/C1Rj¼0;fori6¼j Thus, the rows of Aare unit vectors and are orthogonal to each other; hence, they form an orthonormal set of vectors. The condition ATA¼Isimilarly shows that the columns of Aalso form an orthonormal set of vectors. Furthermore, because each step is reversible, the converse is true. This proves the theorem. Block Matrices 2.34. Consider the following block matrices (which are partitions of the same matrix): (a)1/C0201 3 23 5 7/C02 31 4 592 43 5, (b)1/C0201 3 2 357/C02 3 145 92 43 5CHAPTER 2 Algebra of Matrices 49 Find the size of each block matrix and also the size of each block. (a) The block matrix has two rows of matrices and three columns of matrices; hence, its size is 2 /C23. The block sizes are 2/C22, 2/C22, and 2/C21 for the ﬁrst row; and 1 /C22, 1/C22, and 1/C21 for the second row. (b) The size of the block matrix is 3 /C22; and the block sizes are 1 /C23 and 1/C22 for each of the three rows. 2.35. Compute ABusing block multiplication, where A¼121 340 0022 43 5 and B¼1231 4561 00012 43 5 Here A¼EF 01/C22G/C20/C21 andB¼RS 01/C23T/C20/C21 , where E;F;G;R;S;Tare the given blocks, and 01/C22and 01/C23 are zero matrices of the indicated sites. Hence, AB¼ER ESþFT 01/C23 GT/C20/C21 ¼ ½00 0/C13891 21 5 19 26 33/C20/C21 23 7/C20/C21 þ1 0/C20/C212 643 75¼91 21 54 19 26 33 7 000 22 43 5 2.36. LetM¼diagðA;B;CÞ, where A¼12 34/C20/C21 ,B¼½5/C138,C¼13 57/C20/C21 . Find M2. Because Mis block diagonal, square each block: A2¼71 0 15 22/C20/C21 ; B2¼½25/C138; C2¼16 24 40 64/C20/C21 ; so M2¼71 0 15 22 25 16 2440 642 666643 77775 Miscellaneous Problem 2.37. LetfðxÞandgðxÞbe polynomials and let Abe a square matrix. Prove (a)ðfþgÞðAÞ¼fðAÞþgðAÞ, (b)ðf/C1gÞðAÞ¼fðAÞgðAÞ, (c)fðAÞgðAÞ¼gðAÞfðAÞ. Suppose fðxÞ¼Pr i¼1aixiandgðxÞ¼Ps j¼1bjxj. (a) We can assume r¼s¼nby adding powers of xwith 0 as their coefﬁcients. Then fðxÞþgðxÞ¼Pn i¼1ðaiþbiÞxi Hence, ðfþgÞðAÞ¼Pn i¼1ðaiþbiÞAi¼Pn i¼1aiAiþPn i¼1biAi¼fðAÞþgðAÞ (b) We have fðxÞgðxÞ¼P i;jaibjxiþj. Then fðAÞgðAÞ¼ P iaiAi! P jbjAj ! ¼P i;jaibjAiþj¼ðfgÞðAÞ (c) Using fðxÞgðxÞ¼gðxÞfðxÞ, we have fðAÞgðAÞ¼ð fgÞðAÞ¼ð gfÞðAÞ¼gðAÞfðAÞ50 CHAPTER 2 Algebra of Matrices SUPPLEMENTARY PROBLEMS Algebra of Matrices Problems 2.38–2.41 refer to the following matrices: A¼12 3/C04/C20/C21 ;B¼50 /C067/C20/C21 ;C¼1/C034 26/C05/C20/C21 ;D¼37/C01 4/C089/C20/C21 2.38. Find (a) 5 A/C02B, (b) 2 Aþ3B, (c) 2 C/C03D. 2.39. Find (a) ABandðABÞC, (b) BCandAðBCÞ. [Note thatðABÞC¼AðBCÞ.] 2.40. Find (a) A2andA3, (b) ADandBD, (c) CD. 2.41. Find (a) AT, (b) BT, (c)ðABÞT, (d) ATBT. [Note that ATBT6¼ðABÞT.] Problems 2.42 and 2.43 refer to the following matrices: A¼1/C012 03 4/C20/C21 ; B¼40/C03 /C01/C023/C20/C21 ; C¼2/C030 1 5/C01/C042 /C0100 32 43 5; D¼2 /C01 32 43 5: 2.42. Find (a) 3 A/C04B, (b) AC, (c) BC, (d) AD, (e) BD,(f)CD. 2.43. Find (a) AT, (b) ATB, (c) ATC. 2.44. LetA¼12 36/C20/C21 . Find a 2/C23 matrix Bwith distinct nonzero entries such that AB¼0. 2.45 Lete1¼½1;0;0/C138,e2¼½0;1;0/C138,e3¼½0;0;1/C138, and A¼a1a2a3a4 b1b2b3b4 c1c2c3c42 43 5. Find e1A,e2A,e3A. 2.46. Letei¼½0;...;0;1;0;...;0/C138, where 1 is the ith entry. Show (a)eiA¼Ai,ith row of A. (c) If eiA¼eiB, for each i, then A¼B. (b)BeT j¼Bj,jth column of B. (d) If AeT j¼BeT j, for each j, then A¼B. 2.47. Prove Theorem 2.2(iii) and (iv): (iii) ðBþCÞA¼BAþCA, (iv) kðABÞ¼ð kAÞB¼AðkBÞ. 2.48. Prove Theorem 2.3: (i) ðAþBÞT¼ATþBT, (ii)ðATÞT¼A, (iii)ðkAÞT¼kAT. 2.49. Show (a) If Ahas a zero row, then ABhas a zero row. (b) If Bhas a zero column, then ABhas a zero column. Square Matrices, Inverses 2.50. Find the diagonal and trace of each of the following matrices: (a)A¼2/C058 3/C06/C07 40/C012 43 5, (b) B¼13/C04 6172/C05/C012 43 5, (c) C¼43/C06 2/C050/C20/C21 Problems 2.51–2.53 refer to A¼2/C05 31/C20/C21 ,B¼4/C02 1/C06/C20/C21 ,C¼6/C04 3/C02/C20/C21 . 2.51. Find (a) A 2andA3, (b) fðAÞandgðAÞ, where fðxÞ¼x3/C02x2/C05; gðxÞ¼x2/C03xþ17:CHAPTER 2 Algebra of Matrices 51 2.52. Find (a) B2andB3, (b) fðBÞandgðBÞ, where fðxÞ¼x2þ2x/C022; gðxÞ¼x2/C03x/C06: 2.53. Find a nonzero column vector usuch that Cu¼4u. 2.54. Find the inverse of each of the following matrices (if it exists): A¼74 53/C20/C21 ; B¼23 45/C20/C21 ; C¼4/C06 /C023/C20/C21 ; D¼5/C02 6/C03/C20/C21 2.55. Find the inverses of A¼112 125 1372 43 5andB¼1/C011 01/C01 13/C022 43 5.[Hint: See Problem 2.19.] 2.56. Suppose Ais invertible. Show that if AB¼AC, then B¼C. Give an example of a nonzero matrix Asuch that AB¼ACbutB6¼C. 2.57. Find 2/C22 invertible matrices AandBsuch that AþB6¼0 and AþBis not invertible. 2.58. Show (a) Ais invertible if and only if ATis invertible. (b) The operations of inversion and transpose commute; that is, ðATÞ/C01¼ðA/C01ÞT. (c) If Ahas a zero row or zero column, then Ais not invertible. Diagonal and triangular matrices 2.59. LetA¼diagð1;2;/C03ÞandB¼diagð2;/C05;0Þ. Find (a)AB,A2,B2; (b) fðAÞ, where fðxÞ¼x2þ4x/C03; (c) A/C01andB/C01. 2.60. LetA¼12 01/C20/C21 andB¼110 011 0012 43 5. (a) Find An. (b) Find Bn. 2.61. Find all real triangular matrices Asuch that A2¼B, where (a) B¼42 1 02 5/C20/C21 , (b) B¼14 0/C09/C20/C21 . 2.62. LetA¼52 0k/C20/C21 . Find all numbers kfor which Ais a root of the polynomial: (a)fðxÞ¼x2/C07xþ10, (b) gðxÞ¼x2/C025, (c) hðxÞ¼x2/C04. 2.63. LetB¼10 26 27/C20/C21 :Find a matrix Asuch that A3¼B. 2.64. LetB¼185 095 0042 43 5. Find a triangular matrix Awith positive diagonal entries such that A2¼B. 2.65. Using only the elements 0 and 1, ﬁnd the number of 3 /C23 matrices that are (a) diagonal, (b) upper triangular, (c) nonsingular and upper triangular. Generalize to n/C2nmatrices. 2.66. LetDk¼kI, the scalar matrix belonging to the scalar k. Show (a)DkA¼kA, (b) BDk¼kB, (c) DkþDk0¼Dkþk0, (d) DkDk0¼Dkk0 2.67. Suppose AB¼C, where AandCare upper triangular. (a) Find 2/C22 nonzero matrices A;B;C, where Bis not upper triangular. (b) Suppose Ais also invertible. Show that Bmust also be upper triangular.52 CHAPTER 2 Algebra of Matrices Special Types of Real Matrices 2.68. Find x;y;zsuch that Ais symmetric, where (a)A¼2x3 45 y z172 43 5, (b) A¼7/C062 x yz/C02 x/C0252 43 5. 2.69. Suppose Ais a square matrix. Show (a) AþATis symmetric, (b) A/C0ATis skew-symmetric, (c)A¼BþC, where Bis symmetric and Cis skew-symmetric. 2.70. Write A¼45 13/C20/C21 as the sum of a symmetric matrix Band a skew-symmetric matrix C. 2.71. Suppose AandBare symmetric. Show that the following are also symmetric: (a)AþB; (b) kA, for any scalar k; (c) A2; (d)An, for n>0; (e) fðAÞ, for any polynomial fðxÞ. 2.72. Find a 2/C22 orthogonal matrix Pwhose ﬁrst row is a multiple of (a)ð3;/C04Þ, (b)ð1;2Þ. 2.73. Find a 3/C23 orthogonal matrix Pwhose ﬁrst two rows are multiples of (a)ð1;2;3Þandð0;/C02;3Þ, (b)ð1;3;1Þandð1;0;/C01Þ. 2.74. Suppose AandBare orthogonal matrices. Show that AT,A/C01,ABare also orthogonal. 2.75. Which of the following matrices are normal? A¼3/C04 43/C20/C21 ,B¼1/C02 23/C20/C21 ,C¼111 011 0012 43 5. Complex Matrices 2.76. Find real numbers x;y;zsuch that Ais Hermitian, where A¼3 xþ2iy i 3/C02i 01þzi yi 1/C0xi/C012 43 5: 2.77. Suppose Ais a complex matrix. Show that AAHandAHAare Hermitian. 2.78. LetAbe a square matrix. Show that (a) AþAHis Hermitian, (b) A/C0AHis skew-Hermitian, (c) A¼BþC, where Bis Hermitian and Cis skew-Hermitian. 2.79. Determine which of the following matrices are unitary: A¼i=2/C0ﬃﬃﬃ 3p =2ﬃﬃﬃ 3p =2/C0i=2/C20/C21 ; B¼1 21þi1/C0i 1/C0i1þi/C20/C21 ; C¼1 21/C0i/C01þi i 11þi 1þi/C01þi 02 43 5 2.80. Suppose AandBare unitary. Show that AH,A/C01,ABare unitary. 2.81. Determine which of the following matrices are normal: A¼3þ4i 1 i 2þ3i/C20/C21 and B¼10 1/C0ii/C20/C21 .CHAPTER 2 Algebra of Matrices 53 Block Matrices 2.82. LetU¼12000 34000 00512 003412 6643 775andV¼3/C0200 2400 0012 002/C03 00/C0412 666643 77775. (a) Find UVusing block multiplication. (b) Are UandVblock diagonal matrices? (c) Is UVblock diagonal? 2.83. Partition each of the following matrices so that it becomes a square block matrix with as many diagonal blocks as possible: A¼100 002 0032 43 5; B¼12000 30000 00400 00500 000062 666643 77775; C¼010 000 2002 43 5 2.84. Find M 2andM3for (a) M¼2000 0140 021000032 6643 775, (b) M¼1100 2300 001200452 6643 775. 2.85. For each matrix Min Problem 2.84, ﬁnd fðMÞwhere fðxÞ¼x 2þ4x/C05. 2.86. Suppose U¼½Uik/C138andV¼½Vkj/C138are block matrices for which UVis deﬁned and the number of columns of each block Uikis equal to the number of rows of each block Vkj. Show that UV¼½Wij/C138, where Wij¼P kUikVkj. 2.87. Suppose MandNare block diagonal matrices where corresponding blocks have the same size, sayM¼diagðAiÞandN¼diagðBiÞ. Show (i)MþN¼diagðAiþBiÞ, (iii) MN¼diagðAiBiÞ, (ii)kM¼diagðkAiÞ, (iv) fðMÞ¼diagðfðAiÞÞfor any polynomial fðxÞ. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: A¼½R1;R2; .../C138denotes a matrix Awith rows R1;R2;.... 2.38. (a)½/C05;10;27;/C034/C138, (b)½17;4;/C012;13/C138, (c)½/C07;/C027;11;/C08;36;/C037/C138 2.39. (a)½/C07;14;39;/C028/C138,½21;105;/C098;/C017;/C0285;296/C138 (b)½5;/C015;20;8;60;/C059/C138,½21;105;/C098;/C017;/C0285;296/C138 2.40. (a)½7;/C06;/C09;22/C138,½/C011;38;57;/C0106/C138; (b)½11;/C09;17;/C07;53;/C039/C138,½15;35;/C05;10;/C098;69/C138; (c) not deﬁned 2.41. (a)½1;3;2;/C04/C138, (b)½5;/C06;0;7/C138, (c)½/C07;39;14;/C028/C138;(d)½5;15;10;/C040/C138 2.42. (a)½/C013;/C03;18;4;17;0/C138, (b)½/C05;/C02;4;5;11;/C03;/C012;18/C138, (c)½11;/C012;0;/C05;/C015;5;8;4/C138, (d)½9;9/C138, (e)½/C01;9/C138, (f ) not deﬁned54 CHAPTER 2 Algebra of Matrices 2.43. (a)½1;0;/C01;3;2;4/C138, (b)½4;0;/C03;/C07;/C06;12;4;/C08;6], (c) not deﬁned 2.44.½2;4;6;/C01;/C02;/C03/C138 2.45.½a1;a2;a3;a4/C138,½b1;b2;b3;b4/C138,½c1;c2;c3;c4/C138 2.50. (a) 2 ;/C06;/C01;trðAÞ¼/C0 5, (b) 1 ;1;/C01;trðBÞ¼1, (c) not deﬁned 2.51. (a)½/C011;/C015; 9 ;/C014/C138,½/C067;40;/C024;/C059/C138, (b)½/C050;70;/C042;/C036/C138,gðAÞ¼0 2.52. (a)½14;4;/C02;34/C138,½60;/C052;26;/C0200/C138, (b) fðBÞ¼0,½/C04;10;/C05;46/C138 2.53. u¼½2a;a/C138T 2.54.½3;/C04;/C05;7/C138,½/C05 2;32;2 ;/C01/C138, not deﬁned,½1;/C02 3;2;/C05 3/C138 2.55.½1;1;/C01;2;/C05;3;/C01;2;/C01/C138,½1;1;0;/C01;/C03;1;/C01;/C04;1/C138 2.56. A¼½1;2;1;2/C138,B¼½0;0;1;1/C138,C¼½2;2;0;0/C138 2.57. A¼½1;2;0;3/C138;B¼½4;3;3;0/C138 2.58. (c) Hint: Use Problem 2.48 2.59. (a) AB¼diagð2;/C010;0Þ,A2¼diagð1;4;9Þ,B2¼diagð4;25;0Þ; (b) fðAÞ¼diagð2;9;/C06Þ; (c) A/C01¼diagð1;1 2;/C01 3Þ,C/C01does not exist 2.60. (a)½1;2n;0;1/C138, (b)½1;n;1 2nðn/C01Þ;0;1;n;0;0;1/C138 2.61. (a)½2;3;0;5/C138,½/C02;/C03;0;/C05/C138,½2;/C07;0;/C05/C138,½/C02;7;0;5/C138, (b) none 2.62. (a) k¼2, (b) k¼/C05, (c) none 2.63.½1;0;2;3/C138 2.64.½1;2;1;0;3;1;0;0;2/C138 2.65. All entries below the diagonal must be 0 to be upper triangular, and all diagonal entries must be 1 to be nonsingular. (a) 8ð2nÞ, (b) 26ð2nðnþ1Þ=2Þ, (c) 23ð2nðn/C01Þ=2Þ 2.67. (a) A¼½1;1;0;0/C138,B¼½1;2;3;4/C138,C¼½4;6;0;0/C138 2.68. (a) x¼4,y¼1,z¼3; (b) x¼0,y¼/C06,zany real number 2.69. (c) Hint: LetB¼1 2ðAþATÞandC¼1 2ðA/C0ATÞ: 2.70. B¼½4;3;3;3/C138,C¼½0;2;/C02;0/C138 2.72. (a)½3 5,/C04 5;45,35], (b)½1=ﬃﬃﬃ 5p ,2=ﬃﬃﬃ 5p ;2=ﬃﬃﬃ 5p ,/C01=ﬃﬃﬃ 5p /C138 2.73. (a)½1=ﬃﬃﬃﬃﬃ 14p ,2=ﬃﬃﬃﬃﬃ 14p ,3=ﬃﬃﬃﬃﬃ 14p ;0 ;/C02=ﬃﬃﬃﬃﬃ 13p ,3=ﬃﬃﬃﬃﬃ 13p ;1 2 =ﬃﬃﬃﬃﬃﬃﬃﬃ 157p ,/C03=ﬃﬃﬃﬃﬃﬃﬃﬃ 157p ,/C02=ﬃﬃﬃﬃﬃﬃﬃﬃ 157p /C138 (b)½1=ﬃﬃﬃﬃﬃ 11p ,3=ﬃﬃﬃﬃﬃ 11p ,1=ﬃﬃﬃﬃﬃ 11p ;1=ﬃﬃﬃ 2p ,0;/C01=ﬃﬃﬃ 2p ;3=ﬃﬃﬃﬃﬃ 22p ,/C02=ﬃﬃﬃﬃﬃ 22p ,3=ﬃﬃﬃﬃﬃ 22p /C138 2.75. A;CCHAPTER 2 Algebra of Matrices 55 2.76. x¼3,y¼0,z¼3 2.78. (c) Hint: LetB¼1 2ðAþAHÞandC¼1 2ðA/C0AHÞ. 2.79. A;B;C 2.81. A 2.82. (a) UV¼diagð½7;6;17;10/C138;½/C01;9;7;/C05/C138); (b) no; (c) yes 2.83. A: line between ﬁrst and second rows (columns); B: line between second and third rows (columns) and between fourth and ﬁfth rows (columns); C:Citself—no further partitioning of Cis possible. 2.84. (a) M2¼diagð½4/C138,½9;8;4;9/C138,½9/C138Þ, M3¼diagð½8/C138;½25;44;22;25/C138,½27/C138Þ (b) M2¼diagð½3;4;8;11/C138,½9;12;24;33/C138Þ M3¼diagð½11;15;30;41/C138,½57;78;156;213/C138Þ 2.85. (a) diagð½7/C138,½8;24;12;8/C138,½16/C138Þ, (b) diagð½2;8;16;181],½8;20; 40 ;48/C138Þ56 CHAPTER 2 Algebra of Matrices Systems of Linear Equations 3.1 Introduction Systems of linear equations play an important and motivating role in the subject of linear algebra. In fact, many problems in linear algebra reduce to ﬁnding the solution of a system of linear equations. Thus, the techniques introduced in this chapter will be applicable to abstract ideas introduced later. On the otherhand, some of the abstract results will give us new insights into the structure and properties of systems oflinear equations. All our systems of linear equations involve scalars as both coefﬁcients and constants, and such scalars may come from any number ﬁeld K. There is almost no loss in generality if the reader assumes that all our scalars are real numbers—that is, that they come from the real ﬁeld R. 3.2 Basic Deﬁnitions, Solutions This section gives basic deﬁnitions connected with the solutions of systems of linear equations. The actual algorithms for ﬁnding such solutions will be treated later. Linear Equation and Solutions Alinear equation in unknowns x1;x2;...;xnis an equation that can be put in the standard form a1x1þa2x2þ/C1/C1/C1þ anxn¼b ð3:1Þ where a1;a2;...;an, and bare constants. The constant akis called the coefﬁcient ofxk, and bis called the constant term of the equation. A solution of the linear equation (3.1) is a list of values for the unknowns or, equivalently, a vector uin Kn, say x1¼k1;x2¼k2; ...;xn¼kn or u¼ðk1;k2;...;knÞ such that the following statement (obtained by substituting kiforxiin the equation) is true: a1k1þa2k2þ/C1/C1/C1þ ankn¼b In such a case we say that u satisﬁes the equation. Remark: Equation (3.1) implicitly assumes there is an ordering of the unknowns. In order to avoid subscripts, we will usually use x;yfor two unknowns; x;y;zfor three unknowns; and x;y;z;tfor four unknowns; they will be ordered as shown. 57CHAPTER 3 EXAMPLE 3.1 Consider the following linear equation in three unknowns x;y;z: xþ2y/C03z¼6 We note that x¼5;y¼2;z¼1, or, equivalently, the vector u¼ð5;2;1Þis a solution of the equation. That is, 5þ2ð2Þ/C03ð1Þ¼6o r5þ4/C03¼6o r 6¼6 On the other hand, w¼ð1;2;3Þis not a solution, because on substitution, we do not get a true statement: 1þ2ð2Þ/C03ð3Þ¼6o r1þ4/C09¼6o r/C04¼6 System of Linear Equations A system of linear equations is a list of linear equations with the same unknowns. In particular, a system ofmlinear equations L1;L2;...;Lminnunknowns x1;x2;...;xncan be put in the standard form a11x1þa12x2þ/C1/C1/C1þ a1nxn¼b1 a21x1þa22x2þ/C1/C1/C1þ a2nxn¼b2ð3:2Þ ::::::::::::::::::::::::::::::::::::::::::::::::::: am1x1þam2x2þ/C1/C1/C1þ amnxn¼bm where the aijandbiare constants. The number aijis the coefﬁcient of the unknown xjin the equation Li, and the number biis the constant of the equation Li. The system (3.2) is called an m/C2n(read: mbyn) system. It is called a square system ifm¼n—that is, if the number mof equations is equal to the number nof unknowns. The system (3.2) is said to be homogeneous if all the constant terms are zero—that is, if b1¼0, b2¼0;...;bm¼0. Otherwise the system is said to be nonhomogeneous . Asolution (or a particular solution ) of the system (3.2) is a list of values for the unknowns or, equivalently, a vector uinKn, which is a solution of each of the equations in the system. The set of all solutions of the system is called the solution set or the general solution of the system. EXAMPLE 3.2 Consider the following system of linear equations: x1þx2þ4x3þ3x4¼5 2x1þ3x2þx3/C02x4¼1 x1þ2x2/C05x3þ4x4¼3 It is a 3/C24 system because it has three equations in four unknowns. Determine whether (a) u¼ð/C0 8;6;1;1Þand (b)v¼ð/C0 10;5;1;2Þare solutions of the system. (a) Substitute the values of uin each equation, obtaining /C08þ6þ4ð1Þþ3ð1Þ¼5o r/C08þ6þ4þ3¼5o r5¼5 2ð/C08Þþ3ð6Þþ1/C02ð1Þ¼1o r/C016þ18þ1/C02¼1o r1¼1 /C08þ2ð6Þ/C05ð1Þþ4ð1Þ¼3o r/C08þ12/C05þ4¼3o r3¼3 Yes, uis a solution of the system because it is a solution of each equation. (b) Substitute the values of vinto each successive equation, obtaining /C010þ5þ4ð1Þþ3ð2Þ¼5o r/C010þ5þ4þ6¼5o r 5¼5 2ð/C010Þþ3ð5Þþ1/C02ð2Þ¼1o r/C020þ15þ1/C04¼1o r/C08¼1 No, vis not a solution of the system, because it is not a solution of the second equation. (We do not need to substitute vinto the third equation.)58 CHAPTER 3 Systems of Linear Equations The system (3.2) of linear equations is said to be consistent if it has one or more solutions, and it is said to be inconsistent if it has no solution. If the ﬁeld Kof scalars is inﬁnite, such as when Kis the real ﬁeld Ror the complex ﬁeld C, then we have the following important result. THEOREM 3.1: Suppose the ﬁeld Kis inﬁnite. Then any system lof linear equations has (i) a unique solution, (ii) no solution, or (iii) an inﬁnite number of solutions. This situation is pictured in Fig. 3-1. The three cases have a geometrical description when the system lconsists of two equations in two unknowns (Section 3.4). Augmented and Coefﬁcient Matrices of a System Consider again the general system (3.2) of mequations in nunknowns. Such a system has associated with it the following two matrices: M¼a11a12 ... a1nb1 a21a22 ... a2nb2 ::::::::::::::::::::::::::::::::::::::: am1am2... amn bn2 6643 775and A¼a11a12 ... a1n a21a22 ... a2n ::::::::::::::::::::::::::::::: am1am2... amn2 6643 775 The ﬁrst matrix Mis called the augmented matrix of the system, and the second matrix Ais called the coefﬁcient matrix . The coefﬁcient matrix Ais simply the matrix of coefﬁcients, which is the augmented matrix Mwithout the last column of constants. Some texts write M¼½A;B/C138to emphasize the two parts of M, where B denotes the column vector of constants. The augmented matrix Mand the coefﬁcient matrix Aof the system in Example 3.2 are as follows: M¼11 4 35 23 1/C021 12/C054 32 43 5 and A¼1 143 23 1/C02 12/C0542 43 5 As expected, Aconsists of all the columns of Mexcept the last, which is the column of constants. Clearly, a system of linear equations is completely determined by its augmented matrix M, and vice versa. Speciﬁcally, each row of Mcorresponds to an equation of the system, and each column of M corresponds to the coefﬁcients of an unknown, except for the last column, which corresponds to theconstants of the system. Degenerate Linear Equations A linear equation is said to be degenerate if all the coefﬁcients are zero—that is, if it has the form 0x1þ0x2þ/C1/C1/C1þ 0xn¼b ð3:3Þ Figure 3-1CHAPTER 3 Systems of Linear Equations 59 The solution of such an equation depends only on the value of the constant b. Speciﬁcally, (i) If b6¼0, then the equation has no solution. (ii) If b¼0, then every vector u¼ðk1;k2;...;knÞinKnis a solution. The following theorem applies. THEOREM 3.2: Letlbe a system of linear equations that contains a degenerate equation L, say with constant b. (i) If b6¼0, then the system lhas no solution. (ii) If b¼0, then Lmay be deleted from the system without changing the solution set of the system. Part (i) comes from the fact that the degenerate equation has no solution, so the system has no solution. Part (ii) comes from the fact that every element in Knis a solution of the degenerate equation. Leading Unknown in a Nondegenerate Linear Equation Now let Lbe a nondegenerate linear equation. This means one or more of the coefﬁcients of Lare not zero. By the leading unknown ofL, we mean the ﬁrst unknown in Lwith a nonzero coefﬁcient. For example, x3andyare the leading unknowns, respectively, in the equations 0x1þ0x2þ5x3þ6x4þ0x5þ8x6¼7 and 0 xþ2y/C04z¼5 We frequently omit terms with zero coefﬁcients, so the above equations would be written as 5x3þ6x4þ8x6¼7 and 2 y/C04z¼5 In such a case, the leading unknown appears ﬁrst. 3.3 Equivalent Systems, Elementary Operations Consider the system (3.2) of mlinear equations in nunknowns. Let Lbe the linear equation obtained by multiplying the mequations by constants c1;c2;...;cm, respectively, and then adding the resulting equations. Speciﬁcally, let Lbe the following linear equation: ðc1a11þ/C1/C1/C1þ cmam1Þx1þ/C1/C1/C1þð c1a1nþ/C1/C1/C1þ cmamnÞxn¼c1b1þ/C1/C1/C1þ cmbm Then Lis called a linear combination of the equations in the system. One can easily show (Problem 3.43) that any solution of the system (3.2) is also a solution of the linear combination L. EXAMPLE 3.3 LetL1,L2,L3denote, respectively, the three equations in Example 3.2. Let Lbe the equation obtained by multiplying L1,L2,L3by 3 ;/C02;4, respectively, and then adding. Namely, 3L1: 3x1þ3x2þ12x3þ9x4¼15 /C02L2:/C04x1/C06x2/C02x3þ4x4¼/C02 4L1: 4x1þ8x2/C020x3þ16x4¼12 ðSumÞL: 3x1þ5x2/C010x3þ29x4¼2560 CHAPTER 3 Systems of Linear Equations Then Lis a linear combination of L1,L2,L3. As expected, the solution u¼ð/C0 8;6;1;1Þof the system is also a solution of L. That is, substituting uinL, we obtain a true statement: 3ð/C08Þþ5ð6Þ/C010ð1Þþ29ð1Þ¼25 or/C024þ30/C010þ29¼25 or 9¼9 The following theorem holds. THEOREM 3.3: Two systems of linear equations have the same solutions if and only if each equation in each system is a linear combination of the equations in the other system. Two systems of linear equations are said to be equivalent if they have the same solutions. The next subsection shows one way to obtain equivalent systems of linear equations. Elementary Operations The following operations on a system of linear equations L1;L2;...;Lmare called elementary operations . ½E1/C138Interchange two of the equations. We indicate that the equations LiandLjare interchanged by writing: ‘‘Interchange LiandLj’’ or ‘‘ Li !Lj’’ ½E2/C138Replace an equation by a nonzero multiple of itself. We indicate that equation Liis replaced by kLi (where k6¼0) by writing ‘‘Replace LibykLi’’ or ‘‘ kLi!Li’’ ½E3/C138Replace an equation by the sum of a multiple of another equation and itself. We indicate that equation Ljis replaced by the sum of kLiandLjby writing ‘‘Replace LjbykLiþLj’’ or ‘‘ kLiþLj!Lj’’ The arrow!in½E2/C138and½E3/C138may be read as ‘‘replaces.’’ The main property of the above elementary operations is contained in the following theorem (proved in Problem 3.45). THEOREM 3.4: Suppose a system of mof linear equations is obtained from a system lof linear equations by a ﬁnite sequence of elementary operations. Then mandlhave the same solutions. Remark: Sometimes (say to avoid fractions when all the given scalars are integers) we may apply ½E2/C138and½E3/C138in one step; that is, we may apply the following operation: ½E/C138Replace equation Ljby the sum of kLiandk0Lj(where k06¼0), written ‘‘Replace LjbykLiþk0Lj’’ or ‘‘ kLiþk0Lj!Lj’’ We emphasize that in operations ½E3/C138and [E], only equation Ljis changed. Gaussian elimination, our main method for ﬁnding the solution of a given system of linear equations, consists of using the above operations to transform a given system into an equivalentsystem whose solution can be easily obtained. The details of Gaussian elimination are discussed in subsequent sections. 3.4 Small Square Systems of Linear Equations This section considers the special case of one equation in one unknown, and two equations in two unknowns. These simple systems are treated separately because their solution sets can be describedgeometrically, and their properties motivate the general case.CHAPTER 3 Systems of Linear Equations 61 Linear Equation in One Unknown The following simple basic result is proved in Problem 3.5. THEOREM 3.5: Consider the linear equation ax¼b. (i) If a6¼0, then x¼b=ais a unique solution of ax¼b. (ii) If a¼0, but b6¼0, then ax¼bhas no solution. (iii) If a¼0 and b¼0, then every scalar kis a solution of ax¼b. EXAMPLE 3.4 Solve (a) 4 x/C01¼xþ6, (b) 2 x/C05/C0x¼xþ3, (c) 4þx/C03¼2xþ1/C0x. (a) Rewrite the equation in standard form obtaining 3 x¼7. Then x¼7 3is the unique solution [Theorem 3.5(i)]. (b) Rewrite the equation in standard form, obtaining 0 x¼8. The equation has no solution [Theorem 3.5(ii)]. (c) Rewrite the equation in standard form, obtaining 0 x¼0. Then every scalar kis a solution [Theorem 3.5(iii)]. System of Two Linear Equations in Two Unknowns (2 /C22 System) Consider a system of two nondegenerate linear equations in two unknowns xandy, which can be put in the standard form A1xþB1y¼C1 A2xþB2y¼C2ð3:4Þ Because the equations are nondegenerate, A1andB1are not both zero, and A2andB2are not both zero. The general solution of the system (3.4) belongs to one of three types as indicated in Fig. 3-1. If Ris the ﬁeld of scalars, then the graph of each equation is a line in the plane R2and the three types may be described geometrically as pictured in Fig. 3-2. Speciﬁcally, (1) The system has exactly one solution . Here the two lines intersect in one point [Fig. 3-2(a)]. This occurs when the lines have distinct slopes or, equivalently, when the coefﬁcients of xandyare not proportional: A1 A26¼B1 B2or;equivalently ; A1B2/C0A2B16¼0 For example, in Fig. 3-2(a), 1 =36¼/C01=2. y L1x L20 –3 3 –33 Lx y Lxy1 2:– = – 1 :3 +2 =1 26 (a)y (b)L1x L20 3 –33 Lx y Lxy1 2:+ 3 = 3 :2 +6 =– 86 –3y (c)L L 12and x 0 3 –33 Lx y Lx y1 2:+ 2 = 4 :2 +4 =86 –3 Figure 3-262 CHAPTER 3 Systems of Linear Equations (2) The system has no solution . Here the two lines are parallel [Fig. 3-2(b)]. This occurs when the lines have the same slopes butdifferent yintercepts, or when A1 A2¼B1 B26¼C1 C2 For example, in Fig. 3-2( b), 1=2¼3=66¼/C03=8. (3) The system has an inﬁnite number of solutions . Here the two lines coincide [Fig. 3-2(c)]. This occurs when the lines have the same slopes and same yintercepts, or when the coefﬁcients and constants are proportional, A1 A2¼B1 B2¼C1 C2 For example, in Fig. 3-2(c), 1 =2¼2=4¼4=8. Remark: The following expression and its value is called a determinant of order two : A1B1 A2B2/C12/C12/C12/C12/C12/C12/C12/C12¼A1B2/C0A2B1 Determinants will be studied in Chapter 8. Thus, the system (3.4) has a unique solution if and only if the determinant of its coefﬁcients is not zero. (We show later that this statement is true for any square systemof linear equations.) Elimination Algorithm The solution to system (3.4) can be obtained by the process of elimination, whereby we reduce the systemto a single equation in only one unknown. Assuming the system has a unique solution, this eliminationalgorithm has two parts. ALGORITHM 3.1: The input consists of two nondegenerate linear equations L1and L2in two unknowns with a unique solution. Part A. (Forward Elimination) Multiply each equation by a constant so that the resulting coefﬁcients of one unknown are negatives of each other, and then add the two equations to obtain a new equation Lthat has only one unknown. Part B. (Back-Substitution) Solve for the unknown in the new equation L(which contains only one unknown), substitute this value of the unknown into one of the original equations, and then solve to obtain the value of the other unknown. Part A of Algorithm 3.1 can be applied to any system even if the system does not have a unique solution. In such a case, the new equation Lwill be degenerate and Part B will not apply. EXAMPLE 3.5 (Unique Case). Solve the system L1:2x/C03y¼/C08 L2:3xþ4y¼5 The unknown xis eliminated from the equations by forming the new equation L¼/C03L1þ2L2. That is, we multiply L1by/C03 and L2by 2 and add the resulting equations as follows: /C03L1:/C06xþ9y¼24 2L2: 6xþ8y¼10 Addition : 17 y¼34CHAPTER 3 Systems of Linear Equations 63 We now solve the new equation for y, obtaining y¼2. We substitute y¼2 into one of the original equations, say L1, and solve for the other unknown x, obtaining 2x/C03ð2Þ¼/C0 8o r2 x/C06¼8o r2 x¼/C02o r x¼/C01 Thus, x¼/C01,y¼2, or the pair u¼ð/C0 1;2Þis the unique solution of the system. The unique solution is expected, because 2 =36¼/C03=4. [Geometrically, the lines corresponding to the equations intersect at the point ð/C01;2Þ.] EXAMPLE 3.6 (Nonunique Cases) (a) Solve the system L1: x/C03y¼4 L2:/C02xþ6y¼5 We eliminated xfrom the equations by multiplying L1by 2 and adding it to L2—that is, by forming the new equation L¼2L1þL2. This yields the degenerate equation 0xþ0y¼13 which has a nonzero constant b¼13. Thus, this equation and the system have no solution. This is expected, because 1 =ð/C02Þ¼/C0 3=66¼4=5. (Geometrically, the lines corresponding to the equations are parallel.) (b) Solve the system L1: x/C03y¼4 L2:/C02xþ6y¼/C08 We eliminated xfrom the equations by multiplying L1by 2 and adding it to L2—that is, by forming the new equation L¼2L1þL2. This yields the degenerate equation 0xþ0y¼0 where the constant term is also zero. Thus, the system has an inﬁnite number of solutions, which correspond to the solutions of either equation. This is expected, because 1 =ð/C02Þ¼/C0 3=6¼4=ð/C08Þ. (Geometrically, the lines corresponding to the equations coincide.) To ﬁnd the general solution, let y¼a, and substitute into L1to obtain x/C03a¼4o r x¼3aþ4 Thus, the general solution of the system is x¼3aþ4;y¼a or u¼ð3aþ4;aÞ where a(called a parameter ) is any scalar. 3.5 Systems in Triangular and Echelon Forms The main method for solving systems of linear equations, Gaussian elimination, is treated in Section 3.6. Here we consider two simple types of systems of linear equations: systems in triangular form and themore general systems in echelon form. Triangular Form Consider the following system of linear equations, which is in triangular form : 2x1/C03x2þ5x3/C02x4¼9 5x2/C0x3þ3x4¼1 7x3/C0x4¼3 2x4¼864 CHAPTER 3 Systems of Linear Equations That is, the ﬁrst unknown x1is the leading unknown in the ﬁrst equation, the second unknown x2is the leading unknown in the second equation, and so on. Thus, in particular, the system is square and eachleading unknown is directly to the right of the leading unknown in the preceding equation. Such a triangular system always has a unique solution, which may be obtained by back-substitution . That is, (1) First solve the last equation for the last unknown to get x 4¼4. (2) Then substitute this value x4¼4 in the next-to-last equation, and solve for the next-to-last unknown x3as follows: 7x3/C04¼3o r7 x3¼7o r x3¼1 (3) Now substitute x3¼1 and x4¼4 in the second equation, and solve for the second unknown x2as follows: 5x2/C01þ12¼1o r5 x2þ11¼1o r5 x2¼/C010 or x2¼/C02 (4) Finally, substitute x2¼/C02,x3¼1,x4¼4 in the ﬁrst equation, and solve for the ﬁrst unknown x1as follows: 2x1þ6þ5/C08¼9o r2 x1þ3¼9o r2 x1¼6o r x1¼3 Thus, x1¼3,x2¼/C02,x3¼1,x4¼4, or, equivalently, the vector u¼ð3;/C02;1;4Þis the unique solution of the system. Remark: There is an alternative form for back-substitution (which will be used when solving a system using the matrix format). Namely, after ﬁrst ﬁnding the value of the last unknown, we substitutethis value for the last unknown in all the preceding equations before solving for the next-to-lastunknown. This yields a triangular system with one less equation and one less unknown. For example, inthe above triangular system, we substitute x 4¼4 in all the preceding equations to obtain the triangular system 2x1/C03x2þ5x3¼17 5x2/C0x3¼/C01 7x3¼7 We then repeat the process using the new last equation. And so on. Echelon Form, Pivot and Free Variables The following system of linear equations is said to be in echelon form : 2x1þ6x2/C0x3þ4x4/C02x5¼15 x3þ2x4þ2x5¼5 3x4/C09x5¼6 That is, no equation is degenerate and the leading unknown in each equation other than the ﬁrst is to the right of the leading unknown in the preceding equation. The leading unknowns in the system, x1,x3,x4, are called pivot variables, and the other unknowns, x2andx5, are called freevariables. Generally speaking, an echelon system or a system in echelon form has the following form: a11x1þa12x2þa13x3þa14x4þ/C1/C1/C1þ a1nxn¼b1 a2j2xj2þa2;j2þ1xj2þ1þ/C1/C1/C1þ a2nxn¼b2 :::::::::::::::::::::::::::::::::::::::::::::: arjrxjrþ/C1/C1/C1þ arnxn¼brð3:5Þ where 1 <j2</C1/C1/C1<jranda11,a2j2;...;arjrare not zero. The pivot variables are x1,xj2;...;xjr. Note thatr/C20n. The solution set of any echelon system is described in the following theorem (proved in Problem 3.10).CHAPTER 3 Systems of Linear Equations 65 THEOREM 3.6: Consider a system of linear equations in echelon form, say with requations in n unknowns. There are two cases: (i) r¼n. That is, there are as many equations as unknowns (triangular form). Then the system has a unique solution. (ii) r<n. That is, there are more unknowns than equations. Then we can arbitrarily assign values to the n/C0rfree variables and solve uniquely for the rpivot variables, obtaining a solution of the system. Suppose an echelon system contains more unknowns than equations. Assuming the ﬁeld Kis inﬁnite, the system has an inﬁnite number of solutions, because each of the n/C0rfree variables may be assigned any scalar. The general solution of a system with free variables may be described in either of two equivalent ways, which we illustrate using the above echelon system where there are r¼3 equations and n¼5 unknowns. One description is called the ‘‘Parametric Form’’ of the solution, and the other description is called the‘‘Free-Variable Form.’’ Parametric Form Assign arbitrary values, called parameters , to the free variables x2andx5, say x2¼aandx5¼b, and then use back-substitution to obtain values for the pivot variables x1,x3,x5in terms of the parameters a andb. Speciﬁcally, (1) Substitute x5¼bin the last equation, and solve for x4: 3x4/C09b¼6o r3 x4¼6þ9b or x4¼2þ3b (2) Substitute x4¼2þ3bandx5¼binto the second equation, and solve for x3: x3þ2ð2þ3bÞþ2b¼5o r x3þ4þ8b¼5o r x3¼1/C08b (3) Substitute x2¼a,x3¼1/C08b,x4¼2þ3b,x5¼binto the ﬁrst equation, and solve for x1: 2x1þ6a/C0ð1/C08bÞþ4ð2þ3bÞ/C02b¼15 or x1¼4/C03a/C09b Accordingly, the general solution in parametric form is x1¼4/C03a/C09b; x2¼a; x3¼1/C08b; x4¼2þ3b; x5¼b or, equivalently, v¼ð4/C03a/C09b;a;1/C08b;2þ3b;bÞwhere aandbare arbitrary numbers. Free-Variable Form Use back-substitution to solve for the pivot variables x1,x3,x4directly in terms of the free variables x2 and x5. That is, the last equation gives x4¼2þ3x5. Substitution in the second equation yields x3¼1/C08x5, and then substitution in the ﬁrst equation yields x1¼4/C03x2/C09x5. Accordingly, x1¼4/C03x2/C09x5;x2¼free variable ;x3¼1/C08x5; x4¼2þ3x5; x5¼free variable or, equivalently, v¼ð4/C03x2/C09x5;x2;1/C08x5;2þ3x5;x5Þ is the free-variable form for the general solution of the system. We emphasize that there is no difference between the above two forms of the general solution, and the use of one or the other to represent the general solution is simply a matter of taste. Remark: A particular solution of the above system can be found by assigning any values to the free variables and then solving for the pivot variables by back-substitution. For example, setting x2¼1 and x5¼1, we obtain x4¼2þ3¼5; x3¼1/C08¼/C07; x1¼4/C03/C09¼/C08 Thus, u¼ð/C0 8;1;7;5;1Þis the particular solution corresponding to x2¼1 and x5¼1.66 CHAPTER 3 Systems of Linear Equations 3.6 Gaussian Elimination The main method for solving the general system (3.2) of linear equations is called Gaussian elimination . It essentially consists of two parts: Part A. (Forward Elimination) Step-by-step reduction of the system yielding either a degenerate equation with no solution (which indicates the system has no solution) or an equivalent simplersystem in triangular or echelon form. Part B. (Backward Elimination) Step-by-step back-substitution to ﬁnd the solution of the simpler system. Part B has already been investigated in Section 3.4. Accordingly, we need only give the algorithm for Part A, which is as follows. ALGORITHM 3.2 for (Part A): Input: Them/C2nsystem (3.2) of linear equations. ELIMINATION STEP: Find the ﬁrst unknown in the system with a nonzero coefﬁcient (which now must be x1). (a) Arrange so that a116¼0. That is, if necessary, interchange equations so that the ﬁrst unknown x1 appears with a nonzero coefﬁcient in the ﬁrst equation. (b) Use a11as a pivot to eliminate x1from all equations except the ﬁrst equation. That is, for i>1: (1) Set m¼/C0ai1=a11; (2) Replace LibymL1þLi The system now has the following form: a11x1þa12x2þa13x3þ/C1/C1/C1þ a1nxn¼b1 a2j2xj2þ/C1/C1/C1þ a2nxn¼b2 ::::::::::::::::::::::::::::::::::::::: amj2xj2þ/C1/C1/C1þ amnxn¼bn where x1does not appear in any equation except the ﬁrst, a116¼0, and xj2denotes the ﬁrst unknown with a nonzero coefﬁcient in any equation other than the ﬁrst. (c) Examine each new equation L. (1) If Lhas the form 0 x1þ0x2þ/C1/C1/C1þ 0xn¼bwith b6¼0, then STOP The system is inconsistent and has no solution. (2) If Lhas the form 0 x1þ0x2þ/C1/C1/C1þ 0xn¼0o ri f Lis a multiple of another equation, then delete Lfrom the system. RECURSION STEP: Repeat the Elimination Step with each new ‘‘smaller’’ subsystem formed by all the equations excluding the ﬁrst equation. OUTPUT: Finally, the system is reduced to triangular or echelon form, or a degenerate equation withno solution is obtained indicating an inconsistent system. The next remarks refer to the Elimination Step in Algorithm 3.2. (1) The following number min (b) is called the multiplier : m¼/C0ai1 a11¼/C0coefficient to be deleted pivot (2) One could alternatively apply the following operation in (b): Replace Liby/C0ai1L1þa11Li This would avoid fractions if all the scalars were originally integers.CHAPTER 3 Systems of Linear Equations 67 Gaussian Elimination Example Here we illustrate in detail Gaussian elimination using the following system of linear equations: L1: x/C03y/C02z¼6 L2: 2x/C04y/C03z¼8 L3:/C03xþ6yþ8z¼/C05 Part A. We use the coefﬁcient 1 of xin the ﬁrst equation L1as the pivot in order to eliminate xfrom the second equation L2and from the third equation L3. This is accomplished as follows: (1) Multiply L1by the multiplier m¼/C02 and add it to L2; that is, ‘‘Replace L2by/C02L1þL2.’’ (2) Multiply L1by the multiplier m¼3 and add it to L3; that is, ‘‘Replace L3by 3L1þL3.’’ These steps yield ð/C02ÞL1:/C02xþ6yþ4z¼/C012 L2: 2x/C04y/C03z¼ 8 New L2: 2yþz¼/C043L1: 3x/C09y/C06z¼18 L3:/C03xþ6yþ8z¼/C05 New L3:/C03yþ2z¼13 Thus, the original system is replaced by the following system: L1: x/C03y/C02z¼6 L2: 2yþz¼/C04 L3:/C03yþ2z¼13 (Note that the equations L2andL3form a subsystem with one less equation and one less unknown than the original system.) Next we use the coefﬁcient 2 of yin the (new) second equation L2as the pivot in order to eliminate y from the (new) third equation L3. This is accomplished as follows: (3) Multiply L2by the multiplier m¼3 2and add it to L3; that is, ‘‘Replace L3by3 2L2þL3:’’ (Alternately, ‘‘Replace L3by 3L2þ2L3,’’ which will avoid fractions.) This step yields 3 2L2: 3yþ3 2z¼/C06 L3:/C03yþ2z¼13 New L3:7 2z¼7or3L2: 6yþ3z¼/C012 2L3:/C06yþ4z¼26 New L3: 7z¼14 Thus, our system is replaced by the following system: L1: x/C03y/C02z¼6 L2: 2yþz¼/C04 L3: 7z¼14ðor7 2z¼7Þ The system is now in triangular form, so Part A is completed. Part B. The values for the unknowns are obtained in reverse order, z;y;x, by back-substitution. Speciﬁcally, (1) Solve for zinL3to get z¼2. (2) Substitute z¼2i n L2, and solve for yto get y¼/C03. (3) Substitute y¼/C03 and z¼2i n L1, and solve for xto get x¼1. Thus, the solution of the triangular system and hence the original system is as follows: x¼1;y¼/C03;z¼2o r ;equivalently ; u¼ð1;/C03;2Þ:68 CHAPTER 3 Systems of Linear Equations Condensed Format The Gaussian elimination algorithm involves rewriting systems of linear equations. Sometimes we can avoid excessive recopying of some of the equations by adopting a ‘‘condensed format.’’ This format forthe solution of the above system follows: Number Equation Operation ð1Þ x/C03y/C02z¼6 ð2Þ 2x/C04y/C03z¼8 ð3Þ/C0 3xþ6yþ8z¼/C05 ð2 0Þ 2yþz¼/C04 Replace L2by/C02L1þL2 ð30Þ/C0 3yþ2z¼13 Replace L3by 3L1þL3 ð300Þ 7z¼14 Replace L3by 3L2þ2L3 That is, ﬁrst we write down the number of each of the original equations. As we apply the Gaussian elimination algorithm to the system, we only write down the new equations, and we label each new equation using the same number as the original correspondin g equation, but with an added prime. (After each new equation, we will indicate, for instructional purp oses, the elementary operation that yielded the new equation.) The system in triangular form consists of equations (1), ð20Þ, andð300Þ, the numbers with the largest number of primes. Applying back-substitution to these equations again yields x¼1,y¼/C03,z¼2. Remark: If two equations need to be interchanged, say to obtain a nonzero coefﬁcient as a pivot, then this is easily accomplished in the format by simply renumbering the two equations rather thanchanging their positions. EXAMPLE 3.7 Solve the following system: xþ2y/C03z¼1 2xþ5y/C08z¼4 3xþ8y/C013z¼7 We solve the system by Gaussian elimination. Part A. (Forward Elimination) We use the coefﬁcient 1 of xin the ﬁrst equation L1as the pivot in order to eliminate xfrom the second equation L2and from the third equation L3. This is accomplished as follows: (1) Multiply L1by the multiplier m¼/C02 and add it to L2; that is, ‘‘Replace L2by/C02L1þL2.’’ (2) Multiply L1by the multiplier m¼/C03 and add it to L3; that is, ‘‘Replace L3by/C03L1þL3.’’ The two steps yield xþ2y/C03z¼1 y/C02z¼2 2y/C04z¼4orxþ2y/C03z¼1 y/C02z¼2 (The third equation is deleted, because it is a multiple of the second equation.) The system is now in echelon form with free variable z. Part B. (Backward Elimination) To obtain the general solution, let the free variable z¼a, and solve for xandy by back-substitution. Substitute z¼ain the second equation to obtain y¼2þ2a. Then substitute z¼aand y¼2þ2ainto the ﬁrst equation to obtain xþ2ð2þ2aÞ/C03a¼1o r xþ4þ4a/C03a¼1o r x¼/C03/C0a Thus, the following is the general solution where ais a parameter: x¼/C03/C0a;y¼2þ2a;z¼a or u¼ð/C0 3/C0a;2þ2a;aÞCHAPTER 3 Systems of Linear Equations 69 EXAMPLE 3.8 Solve the following system: x1þ3x2/C02x3þ5x4¼4 2x1þ8x2/C0 x3þ9x4¼9 3x1þ5x2/C012x3þ17x4¼7 We use Gaussian elimination. Part A. (Forward Elimination) We use the coefﬁcient 1 of x1in the ﬁrst equation L1as the pivot in order to eliminate x1from the second equation L2and from the third equation L3. This is accomplished by the following operations: (1) ‘‘Replace L2by/C02L1þL2’’ and (2) ‘‘Replace L3by/C03L1þL3’’ These yield: x1þ3x2/C02x3þ5x4¼4 2x2þ3x3/C0x4¼1 /C04x2/C06x3þ2x4¼/C05 We now use the coefﬁcient 2 of x2in the second equation L2as the pivot and the multiplier m¼2 in order to eliminate x2from the third equation L3. This is accomplished by the operation ‘‘Replace L3by 2L2þL3,’’ which then yields the degenerate equation 0x1þ0x2þ0x3þ0x4¼/C03 This equation and, hence, the original system have no solution: DO NOT CONTINUE Remark 1: As in the above examples, Part A of Gaussian elimination tells us whether or not the system has a solution—that is, whether or not the system is consistent. Accordingly, Part B need never beapplied when a system has no solution. Remark 2: If a system of linear equations has more than four unknowns and four equations, then it may be more convenient to use the matrix format for solving the system. This matrix format is discussed later. 3.7 Echelon Matrices, Row Canonical Form, Row Equivalence One way to solve a system of linear equations is by working with its augmented matrix Mrather than the system itself. This section introduces the necessary matrix concepts for such a discussion. Theseconcepts, such as echelon matrices and elementary row operations, are also of independent interest. Echelon Matrices A matrix Ais called an echelon matrix , or is said to be in echelon form , if the following two conditions hold (where a leading nonzero element of a row of Ais the ﬁrst nonzero element in the row): (1) All zero rows, if any, are at the bottom of the matrix. (2) Each leading nonzero entry in a row is to the right of the leading nonzero entry in the preceding row. That is, A¼½aij/C138is an echelon matrix if there exist nonzero entries a1j1;a2j2;...;arjr; where j1<j2</C1/C1/C1<jr70 CHAPTER 3 Systems of Linear Equations with the property that aij¼0 forðiÞi/C20r;j<ji ðiiÞi>r/C26 The entries a1j1,a2j2;...;arjr, which are the leading nonzero elements in their respective rows, are called thepivots of the echelon matrix. EXAMPLE 3.9 The following is an echelon matrix whose pivots have been circled: A¼02345907 00034125 00000572 00000086000000002 666643 77775 Observe that the pivots are in columns C2;C4;C6;C7, and each is to the right of the one above. Using the above notation, the pivots are a1j1¼2; a2j2¼3; a3j3¼5; a4j4¼8 where j1¼2,j2¼4,j3¼6,j4¼7. Here r¼4. Row Canonical Form A matrix Ais said to be in row canonical form (orrow-reduced echelon form ) if it is an echelon matrix— that is, if it satisﬁes the above properties (1) and (2), and if it satisﬁes the following additional two properties: (3) Each pivot (leading nonzero entry) is equal to 1. (4) Each pivot is the only nonzero entry in its column. The major difference between an echelon matrix and a matrix in row canonical form is that in an echelon matrix there must be zeros below the pivots [Properties (1) and (2)], but in a matrix in rowcanonical form, each pivot must also equal 1 [Property (3)] and there must also be zeros above the pivots[Property (4)]. The zero matrix 0 of any size and the identity matrix Iof any size are important special examples of matrices in row canonical form. EXAMPLE 3.10 The following are echelon matrices whose pivots have been circled: 2320 45 /C06 0001/C032 0 0 0 0 00 62 0 0 0 00 002 6643 775;123 001 0002 43 5;01300 4 00010/C03 00001 22 43 5 The third matrix is also an example of a matrix in row canonical form. The second matrix is not in row canonical form, because it does not satisfy property (4); that is, there is a nonzero entry above the second pivot in the third column. The ﬁrst matrix is not in row canonical form, because it satisﬁes neither property (3) nor property (4); that is, some pivots are not equal to 1 and there are nonzero entries above the pivots.CHAPTER 3 Systems of Linear Equations 71 Elementary Row Operations Suppose Ais a matrix with rows R1;R2;...;Rm. The following operations on Aare called elementary row operations . ½E1/C138(Row Interchange): Interchange rows RiandRj. This may be written as ‘‘Interchange RiandRj’’ or ‘‘ Ri ! Rj’’ ½E2/C138(Row Scaling): Replace row Riby a nonzero multiple kRiof itself. This may be written as ‘‘Replace RibykRiðk6¼0Þ’’ or ‘‘ kRi!Ri’’ ½E3/C138(Row Addition): Replace row Rjby the sum of a multiple kRiof a row Riand itself. This may be written as ‘‘Replace RjbykRiþRj’’ or ‘‘ kRiþRj!Rj’’ The arrow!in E2and E3may be read as ‘‘replaces.’’ Sometimes (say to avoid fractions when all the given scalars are integers) we may apply ½E2/C138and½E3/C138 in one step; that is, we may apply the following operation: ½E/C138Replace Rjby the sum of a multiple kRiof a row Riand a nonzero multiple k0Rjof itself. This may be written as ‘‘Replace RjbykRiþk0Rjðk06¼0Þ’’ or ‘‘ kRiþk0Rj!Rj’’ We emphasize that in operations ½E3/C138and½E/C138only row Rjis changed. Row Equivalence, Rank of a Matrix A matrix Ais said to be row equivalent to a matrix B, written A/C24B ifBcan be obtained from Aby a sequence of elementary row operations. In the case that Bis also an echelon matrix, Bis called an echelon form ofA. The following are two basic results on row equivalence. THEOREM 3.7: Suppose A¼½aij/C138andB¼½bij/C138are row equivalent echelon matrices with respective pivot entries a1j1;a2j2;...arjrand b1k1;b2k2;...bsks Then AandBhave the same number of nonzero rows—that is, r¼s—and the pivot entries are in the same positions—that is, j1¼k1,j2¼k2; ...;jr¼kr. THEOREM 3.8: Every matrix Ais row equivalent to a unique matrix in row canonical form. The proofs of the above theorems will be postponed to Chapter 4. The unique matrix in Theorem 3.8 is called the row canonical form ofA. Using the above theorems, we can now give our ﬁrst deﬁnition of the rank of a matrix. DEFINITION: Therank of a matrix A, written rankðAÞ, is equal to the number of pivots in an echelon form of A. The rank is a very important property of a matrix and, depending on the context in which the matrix is used, it will be deﬁned in many different ways. Of course, all the deﬁnitions lead to thesame number. The next section gives the matrix format of Gaussian elimination, which ﬁnds an echelon form of any matrix A(and hence the rank of A), and also ﬁnds the row canonical form of A.72 CHAPTER 3 Systems of Linear Equations One can show that row equivalence is an equivalence relation . That is, (1)A/C24Afor any matrix A. (2) If A/C24B, then B/C24A. (3) If A/C24BandB/C24C, then A/C24C. Property (2) comes from the fact that each elementary row operation has an inverse operation of the same type. Namely, (i) ‘‘Interchange RiandRj’’ is its own inverse. (ii) ‘‘Replace RibykRi’’ and ‘‘Replace Ribyð1=kÞRi’’ are inverses. (iii) ‘‘Replace RjbykRiþRj’’ and ‘‘Replace Rjby/C0kRiþRj’’ are inverses. There is a similar result for operation [E] (Problem 3.73). 3.8 Gaussian Elimination, Matrix Formulation This section gives two matrix algorithms that accomplish the following: (1) Algorithm 3.3 transforms any matrix Ainto an echelon form. (2) Algorithm 3.4 transforms the echelon matrix into its row canonical form. These algorithms, which use the elementary row operations, are simply restatements of Gaussian elimination as applied to matrices rather than to linear equations. (The term ‘‘row reduce’’ or simply ‘‘reduce’’ will mean to transform a matrix by the elementary row operations.) ALGORITHM 3.3 (Forward Elimination): The input is any matrix A. (The algorithm puts 0’s below each pivot, working from the ‘‘top-down.’’) The output isan echelon form of A. Step 1. Find the ﬁrst column with a nonzero entry. Let j 1denote this column. (a) Arrange so that a1j16¼0. That is, if necessary, interchange rows so that a nonzero entry appears in the ﬁrst row in column j1. (b) Use a1j1as a pivot to obtain 0’s below a1j1. Speciﬁcally, for i>1: ð1ÞSetm¼/C0aij1=a1j1;ð2ÞReplace RibymR1þRi [That is, apply the operation /C0ðaij1=a1j1ÞR1þRi!Ri:] Step 2. Repeat Step 1 with the submatrix formed by all the rows excluding the ﬁrst row. Here we let j2 denote the ﬁrst column in the subsystem with a nonzero entry. Hence, at the end of Step 2, we have a2j26¼0. Steps 3 to r.Continue the above process until a submatrix has only zero rows. We emphasize that at the end of the algorithm, the pivots will be a1j1;a2j2;...;arjr where rdenotes the number of nonzero rows in the ﬁnal echelon matrix. Remark 1: The following number min Step 1(b) is called the multiplier : m¼/C0aij1 a1j1¼/C0entry to be deleted pivotCHAPTER 3 Systems of Linear Equations 73 Remark 2: One could replace the operation in Step 1(b) by the following which would avoid fractions if all the scalars were originally integers. Replace Riby/C0aij1R1þa1j1Ri: ALGORITHM 3.4 (Backward Elimination): The input is a matrix A¼½aij/C138in echelon form with pivot entries a1j1;a2j2; ...;arjr The output is the row canonical form of A. Step 1. (a) (Use row scaling so the last pivot equals 1.) Multiply the last nonzero row Rrby 1 =arjr. (b) (Use arjr¼1 to obtain 0’s above the pivot.) For i¼r/C01;r/C02; ...;2;1: ð1ÞSetm¼/C0aijr;ð2ÞReplace RibymRrþRi (That is, apply the operations /C0aijrRrþRi!Ri.) Steps 2 to r/C01.Repeat Step 1 for rows Rr/C01,Rr/C02;...;R2. Step r.(Use row scaling so the ﬁrst pivot equals 1.) Multiply R1by 1 =a1j1. There is an alternative form of Algorithm 3.4, which we describe here in words. The formal description of this algorithm is left to the reader as a supplementary problem. ALTERNATIVE ALGORITHM 3.4 Puts 0’s above the pivots row by row from the bottom up (rather than column by column from right to left). The alternative algorithm, when applied to an augmented matrix Mof a system of linear equations, is essentially the same as solving for the pivot unknowns one after the other from the bottom up. Remark: We emphasize that Gaussian elimination is a two-stage process. Speciﬁcally, Stage A (Algorithm 3.3). Puts 0’s below each pivot, working from the top row R1down. Stage B (Algorithm 3.4). Puts 0’s above each pivot, working from the bottom row Rrup. There is another algorithm, called Gauss–Jordan , that also row reduces a matrix to its row canonical form. The difference is that Gauss–Jordan puts 0’s both below and above each pivot as it works its way from the top row R1down. Although Gauss–Jordan may be easier to state and understand, it is much less efﬁcient than the two-stage Gaussian elimination algorithm. EXAMPLE 3.11 Consider the matrix A¼12/C031 2 24/C0461 0 36/C0691 32 43 5. (a) Use Algorithm 3.3 to reduce Ato an echelon form. (b) Use Algorithm 3.4 to further reduce Ato its row canonical form. (a) First use a11¼1 as a pivot to obtain 0’s below a11; that is, apply the operations ‘‘Replace R2by/C02R1þR2’’ and ‘‘Replace R3by/C03R1þR3.’’ Then use a23¼2 as a pivot to obtain 0 below a23; that is, apply the operation ‘‘Replace R3by/C03 2R2þR3.’’ This yields A/C2412/C0312 00 246 00 3672 43 5/C2412/C031 2 00 24 6 00 00/C022 43 5 The matrix is now in echelon form.74 CHAPTER 3 Systems of Linear Equations (b) Multiply R3by/C01 2so the pivot entry a35¼1, and then use a35¼1 as a pivot to obtain 0’s above it by the operations ‘‘Replace R2by/C06R3þR2’’ and then ‘‘Replace R1by/C02R3þR1.’’ This yields A/C2412/C0312 00 246 00 0012 43 5/C2412/C0310 00 240 00 0012 43 5: Multiply R2by1 2so the pivot entry a23¼1, and then use a23¼1 as a pivot to obtain 0’s above it by the operation ‘‘Replace R1by 3R2þR1.’’ This yields A/C2412/C0310 00 120 00 0012 43 5/C2412070 00120 000012 43 5: The last matrix is the row canonical form of A. Application to Systems of Linear Equations One way to solve a system of linear equations is by working with its augmented matrix Mrather than the equations themselves. Speciﬁcally, we reduce Mto echelon form (which tells us whether the system has a solution), and then further reduce Mto its row canonical form (which essentially gives the solution of the original system of linear equations). The justiﬁcation for this process comes from the following facts: (1) Any elementary row operation on the augmented matrix Mof the system is equivalent to applying the corresponding operation on the system itself. (2) The system has a solution if and only if the echelon form of the augmented matrix Mdoes not have a row of the formð0;0;...;0;bÞwith b6¼0. (3) In the row canonical form of the augmented matrix M(excluding zero rows), the coefﬁcient of each basic variable is a pivot entry equal to 1, and it is the only nonzero entry in its respective column;hence, the free-variable form of the solution of the system of linear equations is obtained by simplytransferring the free variables to the other side. This process is illustrated below. EXAMPLE 3.12 Solve each of the following systems: (a)x1þx2/C02x3þ4x4¼5 2x1þ2x2/C03x3þx4¼3 3x1þ3x2/C04x3/C02x4¼1 (b)x1þx2/C02x3þ3x4¼4 2x1þ3x2þ3x3/C0x4¼3 5x1þ7x2þ4x3þx4¼5 (c)xþ2yþz¼3 2xþ5y/C0z¼/C04 3x/C02y/C0z¼5 (a) Reduce its augmented matrix Mto echelon form and then to row canonical form as follows: M¼11/C024 5 22/C031 3 33/C04/C0212 43 5/C2411/C0245 00 1/C07/C07 00 2/C014/C0142 43 5/C24110/C010/C09 001/C07/C07 000 0 02 43 5 Rewrite the row canonical form in terms of a system of linear equations to obtain the free variable form of the solution. That is, x1þx2/C010x4¼/C09 x3/C07x4¼/C07orx1¼/C09/C0x2þ10x4 x3¼/C07þ7x4 (The zero row is omitted in the solution.) Observe that x1andx3are the pivot variables, and x2andx4are the free variables.CHAPTER 3 Systems of Linear Equations 75 (b) First reduce its augmented matrix Mto echelon form as follows: M¼11/C023 4 23 3/C013 57 4 152 43 5/C2411/C0234 01 7/C07/C05 02 1 4/C014/C0152 43 5/C2411/C0234 01 7/C07/C05 0 000 /C052 43 5 There is no need to continue to ﬁnd the row canonical form of M, because the echelon form already tells us that the system has no solution. Speciﬁcally, the third row of the echelon matrix corresponds to the degenerate equation 0x1þ0x2þ0x3þ0x4¼/C05 which has no solution. Thus, the system has no solution. (c) Reduce its augmented matrix Mto echelon form and then to row canonical form as follows: M¼1213 25/C01/C04 3/C02/C0152 643 75/C24121 3 01/C03/C010 0/C08/C04/C042 643 75/C2412 1 3 01/C03/C010 00/C028/C0842 643 75 /C2412 1 3 01/C03/C010 00 1 32 643 75/C24120 0 010/C01 001 32 643 75/C24100 2 010/C01 001 32 643 75 Thus, the system has the unique solution x¼2,y¼/C01,z¼3, or, equivalently, the vector u¼ð2;/C01;3Þ.W e note that the echelon form of Malready indicated that the solution was unique, because it corresponded to a triangular system. Application to Existence and Uniqueness Theorems This subsection gives theoretical conditions for the existence and uniqueness of a solution of a system of linear equations using the notion of the rank of a matrix. THEOREM 3.9: Consider a system of linear equations in nunknowns with augmented matrix M¼½A;B/C138. Then, (a) The system has a solution if and only if rank ðAÞ¼rankðMÞ. (b) The solution is unique if and only if rank ðAÞ¼rankðMÞ¼n. Proof of (a). The system has a solution if and only if an echelon form of M¼½A;B/C138does not have a row of the form ð0;0;...;0;bÞ; with b6¼0 If an echelon form of Mdoes have such a row, then bis a pivot of Mbut not of A, and hence, rankðMÞ>rankðAÞ. Otherwise, the echelon forms of Aand Mhave the same pivots, and hence, rankðAÞ¼rankðMÞ. This proves (a). Proof of (b). The system has a unique solution if and only if an echelon form has no free variable. This means there is a pivot for each unknown. Accordingly, n¼rankðAÞ¼rankðMÞ. This proves (b). The above proof uses the fact (Problem 3.74) that an echelon form of the augmented matrix M¼½A;B/C138also automatically yields an echelon form of A.76 CHAPTER 3 Systems of Linear Equations 3.9 Matrix Equation of a System of Linear Equations The general system (3.2) of mlinear equations in nunknowns is equivalent to the matrix equation a11a12 ... a1n a21a22 ... a2n ::::::::::::::::::::::::::::::: am1am2... amn2 6643 775x1 x2 x3 ... xn2 666643 77775¼b1 b2 ... bm2 6643 775or AX¼B where A¼½aij/C138is the coefﬁcient matrix, X¼½xj/C138is the column vector of unknowns, and B¼½bi/C138is the column vector of constants. (Some texts write Ax¼brather than AX¼B, in order to emphasize that x andbare simply column vectors.) The statement that the system of linear equations and the matrix equation are equivalent means that any vector solution of the system is a solution of the matrix equation, and vice versa. EXAMPLE 3.13 The following system of linear equations and matrix equation are equivalent: x1þ2x2/C04x3þ7x4¼4 3x1/C05x2þ6x3/C08x4¼8 4x1/C03x2/C02x3þ6x4¼11and12/C047 3/C056/C08 4/C03/C0262 43 5x1 x2 x3 x42 6643 775¼4 8 112 43 5 We note that x1¼3,x2¼1,x3¼2,x4¼1, or, in other words, the vector u¼½3;1;2;1/C138is a solution of the system. Thus, the (column) vector uis also a solution of the matrix equation. The matrix form AX¼Bof a system of linear equations is notationally very convenient when discussing and proving properties of systems of linear equations. This is illustrated with our ﬁrst theorem (described in Fig. 3-1), which we restate for easy reference. THEOREM 3.1: Suppose the ﬁeld Kis inﬁnite. Then the system AX¼Bhas: (a) a unique solution, (b) no solution, or (c) an inﬁnite number of solutions. Proof. It sufﬁces to show that if AX¼Bhas more than one solution, then it has inﬁnitely many. Suppose uand vare distinct solutions of AX¼B; that is, Au¼BandAv¼B. Then, for any k2K, A½uþkðu/C0vÞ/C138¼ AuþkðAu/C0AvÞ¼BþkðB/C0BÞ¼B Thus, for each k2K, the vector uþkðu/C0vÞis a solution of AX¼B. Because all such solutions are distinct (Problem 3.47), AX¼Bhas an inﬁnite number of solutions. Observe that the above theorem is true when Kis the real ﬁeld R(or the complex ﬁeld C). Section 3.3 shows that the theorem has a geometrical description when the system consists of two equations in twounknowns, where each equation represents a line in R 2. The theorem also has a geometrical description when the system consists of three nondegenerate equations in three unknowns, where the three equations correspond to planes H1,H2,H3inR3. That is, (a)Unique solution: Here the three planes intersect in exactly one point. (b)No solution: Here the planes may intersect pairwise but with no common point of intersection, or two of the planes may be parallel. (c)Inﬁnite number of solutions: Here the three planes may intersect in a line (one free variable), or they may coincide (two free variables). These three cases are pictured in Fig. 3-3. Matrix Equation of a Square System of Linear Equations A system AX¼Bof linear equations is square if and only if the matrix Aof coefﬁcients is square. In such a case, we have the following important result.CHAPTER 3 Systems of Linear Equations 77 THEOREM 3.10: A square system AX¼Bof linear equations has a unique solution if and only if the matrix Ais invertible. In such a case, A/C01Bis the unique solution of the system. We only prove here that if Ais invertible, then A/C01Bis a unique solution. If Ais invertible, then AðA/C01BÞ¼ð AA/C01ÞB¼IB¼B and hence, A/C01Bis a solution. Now suppose vis any solution, so Av¼B. Then v¼Iv¼ðA/C01AÞv¼A/C01ðAvÞ¼A/C01B Thus, the solution A/C01Bis unique. EXAMPLE 3.14 Consider the following system of linear equations, whose coefﬁcient matrix Aand inverse A/C01are also given: xþ2yþ3z¼1 xþ3yþ6z¼3 2xþ6yþ13z¼5; A¼12 3 13 6261 32 43 5; A /C01¼3/C083 /C017/C03 0/C0212 43 5 By Theorem 3.10, the unique solution of the system is A/C01B¼3/C083 /C017/C03 0/C0212 43 51 352 43 5¼/C06 5 /C012 43 5 That is, x¼/C06,y¼5,z¼/C01. Remark: We emphasize that Theorem 3.10 does not usually help us to ﬁnd the solution of a square system. That is, ﬁnding the inverse of a coefﬁcient matrix Ais not usually any easier than solving the system directly. Thus, unless we are given the inverse of a coefﬁcient matrix A, as in Example 3.14, we usually solve a square system by Gaussian elimination (or some iterative method whose discussionlies beyond the scope of this text).( ) Unique solutionaH2H3 H1H1H2H3 ( ) Infinite number of solutionscH3 HH12and(i) (ii) (iii)HH H12 3, , and (i) ( ) No solutionsbH3 H2 H1 (ii) (iii) (i )vH1H2H3 H2H3 H1H3 Figure 3-378 CHAPTER 3 Systems of Linear Equations 3.10 Systems of Linear Equations and Linear Combinations of Vectors The general system (3.2) of linear equations may be rewritten as the following vector equation: x1a11 a21 ... am12 6643 775þx2a12 a22 ... am22 6643 775þ/C1/C1/C1þ xna1n a2n ... amn2 6643 775¼b1 b2 ... bm2 6643 775 Recall that a vector vinKnis said to be a linear combination of vectors u1;u2;...;uminKnif there exist scalars a1;a2;...;aminKsuch that v¼a1u1þa2u2þ/C1/C1/C1þ amum Accordingly, the general system (3.2) of linear equations and the above equivalent vector equation have a solution if and only if the column vector of constants is a linear combination of the columns of thecoefﬁcient matrix. We state this observation formally. THEOREM 3.11: A system AX¼Bof linear equations has a solution if and only if Bis a linear combination of the columns of the coefﬁcient matrix A. Thus, the answer to the problem of expressing a given vector vinKnas a linear combination of vectors u1;u2;...;uminKnreduces to solving a system of linear equations. Linear Combination Example Suppose we want to write the vector v¼ð1;/C02;5Þas a linear combination of the vectors u1¼ð1;1;1Þ; u2¼ð1;2;3Þ; u3¼ð2;/C01;1Þ First we write v¼xu1þyu2þzu3with unknowns x;y;z, and then we ﬁnd the equivalent system of linear equations which we solve. Speciﬁcally, we ﬁrst write 1 /C02 52 43 5¼x1 112 43 5þy1 232 43 5þz2 /C01 12 43 5 ð*Þ Then 1 /C02 52 43 5¼x x x2 43 5þy 2y 3y2 43 5þ2z /C0z z2 43 5¼xþyþ2z xþ2y/C0z xþ3yþz2 43 5 Setting corresponding entries equal to each other yields the following equivalent system: xþyþ2z¼1 xþ2y/C0z¼/C02 xþ3yþz¼5ð**Þ For notational convenience, we have written the vectors in R nas columns, because it is then easier to ﬁnd the equivalent system of linear equations. In fact, one can easily go from the vector equation (*) directlyto the system (**). Now we solve the equivalent system of linear equations by reducing the system to echelon form. This yields xþyþ2z¼1 y/C03z¼/C03 2y/C0z¼4and thenxþyþ2z¼1 y/C03z¼/C03 5z¼10 Back-substitution yields the solution x¼/C06,y¼3,z¼2. Thus, v¼/C06u 1þ3u2þ2u3.CHAPTER 3 Systems of Linear Equations 79 EXAMPLE 3.15 (a) Write the vector v¼ð4;9;19Þas a linear combination of u1¼ð1;/C02;3Þ; u2¼ð3;/C07;10Þ; u3¼ð2;1;9Þ: Find the equivalent system of linear equations by writing v¼xu1þyu2þzu3, and reduce the system to an echelon form. We have xþ3yþ2z¼4 /C02x/C07yþz¼9 3xþ10yþ9z¼19orxþ3yþ2z¼4 /C0yþ5z¼17 yþ3z¼7orxþ3yþ2z¼4 /C0yþ5z¼17 8z¼24 Back-substitution yields the solution x¼4,y¼/C02,z¼3. Thus, vis a linear combination of u1;u2;u3. Speciﬁcally, v¼4u1/C02u2þ3u3. (b) Write the vector v¼ð2;3;/C05Þas a linear combination of u1¼ð1;2;/C03Þ; u2¼ð2;3;/C04Þ; u3¼ð1;3;/C05Þ Find the equivalent system of linear equations by writing v¼xu1þyu2þzu3, and reduce the system to an echelon form. We have xþ2yþz¼2 2xþ3yþ3z¼3 /C03x/C04y/C05z¼/C05orxþ2yþz¼2 /C0yþz¼/C01 2y/C02z¼1orxþ2yþz¼2 /C05yþ5z¼/C01 0¼3 The system has no solution. Thus, it is impossible to write vas a linear combination of u1;u2;u3. Linear Combinations of Orthogonal Vectors, Fourier Coefﬁcients Recall ﬁrst (Section 1.4) that the dot (inner) product u/C1vof vectors u¼ða1;...;anÞandv¼ðb1;...;bnÞ inRnis deﬁned by u/C1v¼a1b1þa2b2þ/C1/C1/C1þ anbn Furthermore, vectors uand vare said to be orthogonal if their dot product u/C1v¼0. Suppose that u1;u2;...;uninRnarennonzero pairwise orthogonal vectors. This means ðiÞui/C1uj¼0 for i6¼j andðiiÞui/C1ui6¼0 for each i Then, for any vector vinRn, there is an easy way to write vas a linear combination of u1;u2;...;un, which is illustrated in the next example. EXAMPLE 3.16 Consider the following three vectors in R3: u1¼ð1;1;1Þ; u2¼ð1;/C03;2Þ; u3¼ð5;/C01;/C04Þ These vectors are pairwise orthogonal; that is, u1/C1u2¼1/C03þ2¼0; u1/C1u3¼5/C01/C04¼0; u2/C1u3¼5þ3/C08¼0 Suppose we want to write v¼ð4;14;/C09Þas a linear combination of u1;u2;u3. Method 1. Find the equivalent system of linear equations as in Example 3.14 and then solve, obtaining v¼3u1/C04u2þu3. Method 2. (This method uses the fact that the vectors u1;u2;u3are mutually orthogonal, and hence, the arithmetic is much simpler.) Set vas a linear combination of u1;u2;u3using unknown scalars x;y;zas follows: ð4;14;/C09Þ¼xð1;1;1Þþyð1;/C03;2Þþzð5;/C01;/C04Þð *Þ80 CHAPTER 3 Systems of Linear Equations Take the dot product of (*) with respect to u1to get ð4;14;/C09Þ/C1ð1;1;1Þ¼xð1;1;1Þ/C1ð1;1;1Þ or 9¼3x or x¼3 (The last two terms drop out, because u1is orthogonal to u2and to u3.) Next take the dot product of (*) with respect tou2to obtain ð4;14;/C09Þ/C1ð1;/C03;2Þ¼yð1;/C03;2Þ/C1ð1;/C03;2Þ or/C056¼14y or y¼/C04 Finally, take the dot product of (*) with respect to u3to get ð4;14;/C09Þ/C1ð5;/C01;/C04Þ¼zð5;/C01;/C04Þ/C1ð5;/C01;/C04Þ or 42¼42z or z¼1 Thus, v¼3u1/C04u2þu3. The procedure in Method 2 in Example 3.16 is valid in general. Namely, THEOREM 3.12: Suppose u1;u2;...;unare nonzero mutually orthogonal vectors in Rn. Then, for any vector vinRn, v¼v/C1u1 u1/C1u1u1þv/C1u2 u2/C1u2u2þ/C1/C1/C1þv/C1un un/C1unun We emphasize that there must be nsuch orthogonal vectors uiinRnfor the formula to be used. Note also that each ui/C1ui6¼0, because each uiis a nonzero vector. Remark: The following scalar ki(appearing in Theorem 3.12) is called the Fourier coefﬁcient ofv with respect to ui: ki¼v/C1ui ui/C1ui¼v/C1ui kuik2 It is analogous to a coefﬁcient in the celebrated Fourier series of a function. 3.11 Homogeneous Systems of Linear Equations A system of linear equations is said to be homogeneous if all the constant terms are zero. Thus, a homogeneous system has the form AX¼0. Clearly, such a system always has the zero vector 0¼ð0;0;...;0Þas a solution, called the zero ortrivial solution. Accordingly, we are usually interested in whether or not the system has a nonzero solution. Because a homogeneous system AX¼0 has at least the zero solution, it can always be put in an echelon form, say a11x1þa12x2þa13x3þa14x4þ/C1/C1/C1þ a1nxn¼0 a2j2xj2þa2;j2þ1xj2þ1þ/C1/C1/C1þ a2nxn¼0 :::::::::::::::::::::::::::::::::::::::::::: arjrxjrþ/C1/C1/C1þ arnxn¼0 Here rdenotes the number of equations in echelon form and ndenotes the number of unknowns. Thus, the echelon system has n/C0rfree variables. The question of nonzero solutions reduces to the following two cases: (i)r¼n. The system has only the zero solution. (ii)r<n. The system has a nonzero solution. Accordingly, if we begin with fewer equations than unknowns, then, in echelon form, r<n, and the system has a nonzero solution. This proves the following important result. THEOREM 3.13: A homogeneous system AX¼0with more unknowns than equations has a nonzero solution.CHAPTER 3 Systems of Linear Equations 81 EXAMPLE 3.17 Determine whether or not each of the following homogeneous systems has a nonzero solution: (a)xþy/C0z¼0 2x/C03yþz¼0 x/C04yþ2z¼0 (b)xþy/C0z¼0 2xþ4y/C0z¼0 3xþ2yþ2z¼0 (c)x1þ2x2/C03x3þ4x4¼0 2x1/C03x2þ5x3/C07x4¼0 5x1þ6x2/C09x3þ8x4¼0 (a) Reduce the system to echelon form as follows: xþy/C0z¼0 /C05yþ3z¼0 /C05yþ3z¼0and thenxþy/C0z¼0 /C05yþ3z¼0 The system has a nonzero solution, because there are only two equations in the three unknowns in echelon form. Here zis a free variable. Let us, say, set z¼5. Then, by back-substitution, y¼3 and x¼2. Thus, the vector u¼ð2;3;5Þis a particular nonzero solution. (b) Reduce the system to echelon form as follows: xþy/C0z¼0 2yþz¼0 /C0yþ5z¼0and thenxþy/C0z¼0 2yþz¼0 11z¼0 In echelon form, there are three equations in three unknowns. Thus, the system has only the zero solution. (c) The system must have a nonzero solution (Theorem 3.13), because there are four unknowns but only three equations. (Here we do not need to reduce the system to echelon form.) Basis for the General Solution of a Homogeneous System LetWdenote the general solution of a homogeneous system AX¼0. A list of nonzero solution vectors u1;u2;...;usof the system is said to be a basis forWif each solution vector w2Wcan be expressed uniquely as a linear combination of the vectors u1;u2;...;us; that is, there exist unique scalars a1;a2;...;assuch that w¼a1u1þa2u2þ/C1/C1/C1þ asus The number sof such basis vectors is equal to the number of free variables. This number sis called the dimension ofW, written as dim W¼s. When W¼f0g—that is, the system has only the zero solution— we deﬁne dim W¼0. The following theorem, proved in Chapter 5, page 171, tells us how to ﬁnd such a basis. THEOREM 3.14: LetWbe the general solution of a homogeneous system AX¼0, and suppose that the echelon form of the homogeneous system has sfree variables. Let u1;u2;...;us be the solutions obtained by setting one of the free variables equal to 1 (or any nonzero constant) and the remaining free variables equal to 0. Then dimW¼s, and the vectors u1;u2;...;usform a basis of W. We emphasize that the general solution Wmay have many bases, and that Theorem 3.12 only gives us one such basis. EXAMPLE 3.18 Find the dimension and a basis for the general solution Wof the homogeneous system x1þ2x2/C03x3þ2x4/C04x5¼0 2x1þ4x2/C05x3þx4/C06x5¼0 5x1þ10x2/C013x3þ4x4/C016x5¼082 CHAPTER 3 Systems of Linear Equations First reduce the system to echelon form. Apply the following operations: ‘‘Replace L2by/C02L1þL2’’ and ‘‘Replace L3by/C05L1þL3’’ and then ‘‘Replace L3by/C02L2þL3’’ These operations yield x1þ2x2/C03x3þ2x4/C04x5¼0 x3/C03x4þ2x5¼0 2x3/C06x4þ4x5¼0andx1þ2x2/C03x3þ2x4/C04x5¼0 x3/C03x4þ2x5¼0 The system in echelon form has three free variables, x2;x4;x5; hence, dim W¼3. Three solution vectors that form a basis for Ware obtained as follows: (1) Set x2¼1,x4¼0,x5¼0. Back-substitution yields the solution u1¼ð/C0 2;1;0;0;0Þ. (2) Set x2¼0,x4¼1,x5¼0. Back-substitution yields the solution u2¼ð7;0;3;1;0Þ. (3) Set x2¼0,x4¼0,x5¼1. Back-substitution yields the solution u3¼ð/C0 2;0;/C02;0;1Þ. The vectors u1¼ð/C0 2;1;0;0;0Þ,u2¼ð7;0;3;1;0Þ,u3¼ð/C0 2;0;/C02;0;1Þform a basis for W. Remark: Any solution of the system in Example 3.18 can be written in the form au1þbu2þcu3¼að/C02;1;0;0;0Þþbð7;0;3;1;0Þþcð/C02;0;/C02;0;1Þ ¼ð/C0 2aþ7b/C02c;a;3b/C02c;b;cÞ or x1¼/C02aþ7b/C02c; x2¼a; x3¼3b/C02c; x4¼b; x5¼c where a;b;care arbitrary constants. Observe that this representation is nothing more than the parametric form of the general solution under the choice of parameters x2¼a,x4¼b,x5¼c. Nonhomogeneous and Associated Homogeneous Systems LetAX¼Bbe a nonhomogeneous system of linear equations. Then AX¼0 is called the associated homogeneous system . For example, xþ2y/C04z¼7 3x/C05yþ6z¼8andxþ2y/C04z¼0 3x/C05yþ6z¼0 show a nonhomogeneous system and its associated homogeneous system. The relationship between the solution Uof a nonhomogeneous system AX¼Band the solution Wof its associated homogeneous system AX¼0 is contained in the following theorem. THEOREM 3.15: Let v0be a particular solution of AX¼Band let Wbe the general solution of AX¼0. Then the following is the general solution of AX¼B: U¼v0þW¼fv0þw:w2Wg That is, U¼v0þWis obtained by adding v0to each element in W. We note that this theorem has a geometrical interpretation in R3. Speciﬁcally, suppose Wis a line through the origin O. Then, as pictured in Fig. 3-4, U¼v0þWis the line parallel to Wobtained by adding v0to each element of W. Similarly, whenever Wis a plane through the origin O, then U¼v0þWis a plane parallel to W.CHAPTER 3 Systems of Linear Equations 83 3.12 Elementary Matrices Letedenote an elementary row operation and let eðAÞdenote the results of applying the operation eto a matrix A. Now let Ebe the matrix obtained by applying eto the identity matrix I; that is, E¼eðIÞ Then Eis called the elementary matrix corresponding to the elementary row operation e. Note that Eis always a square matrix. EXAMPLE 3.19 Consider the following three elementary row operations: ð1ÞInterchange R2andR3:ð2ÞReplace R2by/C06R2:ð3ÞReplace R3by/C04R1þR3: The 3/C23 elementary matrices corresponding to the above elementary row operations are as follows: E1¼100 001 0102 43 5; E2¼10 0 0/C060 00 12 43 5; E3¼100 010 /C04012 43 5 The following theorem, proved in Problem 3.34, holds. THEOREM 3.16: Letebe an elementary row operation and let Ebe the corresponding m/C2m elementary matrix. Then eðAÞ¼EA where Ais any m/C2nmatrix. In other words, the result of applying an elementary row operation eto a matrix Acan be obtained by premultiplying Aby the corresponding elementary matrix E. Now suppose e0is the inverse of an elementary row operation e, and let E0andEbe the corresponding matrices. We note (Problem 3.33) that Eis invertible and E0is its inverse. This means, in particular, that any product P¼Ek...E2E1 of elementary matrices is invertible. Figure 3-484 CHAPTER 3 Systems of Linear Equations Applications of Elementary Matrices Using Theorem 3.16, we are able to prove (Problem 3.35) the following important properties of matrices. THEOREM 3.17: LetAbe a square matrix. Then the following are equivalent: (a) Ais invertible (nonsingular). (b) Ais row equivalent to the identity matrix I. (c) Ais a product of elementary matrices. Recall that square matrices AandBare inverses if AB¼BA¼I. The next theorem (proved in Problem 3.36) demonstrates that we need only show that one of the products is true, say AB¼I, to prove that matrices are inverses. THEOREM 3.18: Suppose AB¼I. Then BA¼I, and hence, B¼A/C01. Row equivalence can also be deﬁned in terms of matrix multiplication. Speciﬁcally, we will prove (Problem 3.37) the following. THEOREM 3.19: Bis row equivalent to Aif and only if there exists a nonsingular matrix Psuch that B¼PA. Application to Finding the Inverse of an n/C2nMatrix The following algorithm ﬁnds the inverse of a matrix. ALGORITHM 3.5: The input is a square matrix A. The output is the inverse of Aor that the inverse does not exist. Step 1. Form the n/C22n(block) matrix M¼½A;I/C138, where Ais the left half of Mand the identity matrix Iis the right half of M. Step 2. Row reduce Mto echelon form. If the process generates a zero row in the Ahalf of M, then STOP Ahas no inverse. (Otherwise Ais in triangular form.) Step 3. Further row reduce Mto its row canonical form M/C24½I;B/C138 where the identity matrix Ihas replaced Ain the left half of M. Step 4. SetA/C01¼B, the matrix that is now in the right half of M. The justiﬁcation for the above algorithm is as follows. Suppose Ais invertible and, say, the sequence of elementary row operations e1;e2;...;eqapplied to M¼½A;I/C138reduces the left half of M, which is A,t o the identity matrix I. Let Eibe the elementary matrix corresponding to the operation ei. Then, by applying Theorem 3.16. we get Eq...E2E1A¼I orðEq...E2E1IÞA¼I; so A/C01¼Eq...E2E1I That is, A/C01can be obtained by applying the elementary row operations e1;e2;...;eqto the identity matrix I, which appears in the right half of M. Thus, B¼A/C01, as claimed. EXAMPLE 3.20 Find the inverse of the matrix A¼10 2 2/C013 41 82 43 5.CHAPTER 3 Systems of Linear Equations 85 First form the (block) matrix M¼½A;I/C138and row reduce Mto an echelon form: M¼1 02100 2/C013010 4 180012 43 5/C241021 0 0 0/C01/C01/C0210 010/C04012 43 5/C241021 0 0 0/C01/C01/C0210 00/C01/C06112 43 5 In echelon form, the left half of Mis in triangular form; hence, Ahas an inverse. Next we further row reduce Mto its row canonical form: M/C2410 0/C011 2 2 0/C010 4 0 /C01 00 1 6 /C01/C012 43 5/C24100/C011 2 2 010/C0401 001 6/C01/C012 43 5 The identity matrix is now in the left half of the ﬁnal matrix; hence, the right half is A/C01. In other words, A/C01¼/C011 2 2 /C0401 6/C01/C012 43 5 Elementary Column Operations Now let Abe a matrix with columns C1;C2;...;Cn. The following operations on A, analogous to the elementary row operations, are called elementary column operations : ½F1/C138(Column Interchange): Interchange columns CiandCj. ½F2/C138(Column Scaling): Replace CibykCi(where k6¼0). ½F3/C138(Column Addition): Replace CjbykCiþCj. We may indicate each of the column operations by writing, respectively, ð1ÞCi$Cj;ð2ÞkCi!Ci;ð3ÞðkCiþCjÞ!Cj Moreover, each column operation has an inverse operation of the same type, just like the corresponding row operation. Now let fdenote an elementary column operation, and let Fbe the matrix obtained by applying fto the identity matrix I; that is, F¼fðIÞ Then Fis called the elementary matrix corresponding to the elementary column operation f. Note that F is always a square matrix. EXAMPLE 3.21 Consider the following elementary column operations: ð1ÞInterchange C1andC3;ð2ÞReplace C3by/C02C3;ð3ÞReplace C3by/C03C2þC3 The corresponding three 3 /C23 elementary matrices are as follows: F1¼001 010 1002 43 5; F2¼10 0 01 0 00/C022 43 5; F3¼10 0 01/C03 00 12 43 5 The following theorem is analogous to Theorem 3.16 for the elementary row operations. THEOREM 3.20: For any matrix A;fðAÞ¼AF. That is, the result of applying an elementary column operation fon a matrix Acan be obtained by postmultiplying Aby the corresponding elementary matrix F.86 CHAPTER 3 Systems of Linear Equations Matrix Equivalence A matrix Bisequivalent to a matrix AifBcan be obtained from Aby a sequence of row and column operations. Alternatively, Bis equivalent to A, if there exist nonsingular matrices PandQsuch that B¼PAQ . Just like row equivalence, equivalence of matrices is an equivalence relation. The main result of this subsection (proved in Problem 3.38) is as follows. THEOREM 3.21: Every m/C2nmatrix Ais equivalent to a unique block matrix of the form Ir0 00/C20/C21 where Iris the r-square identity matrix. The following deﬁnition applies. DEFINITION: The nonnegative integer rin Theorem 3.18 is called the rank ofA, written rankðAÞ. Note that this deﬁnition agrees with the previous deﬁnition of the rank of a matrix. 3.13 LUDECOMPOSITION Suppose Ais a nonsingular matrix that can be brought into (upper) triangular form Uusing only row- addition operations; that is, suppose Acan be triangularized by the following algorithm, which we write using computer notation. ALGORITHM 3.6: The input is a matrix Aand the output is a triangular matrix U. Step 1. Repeat for i¼1;2;...;n/C01: Step 2. Repeat for j¼iþ1,iþ2;...;n (a) Set mij:¼/C0aij=aii. (b) Set Rj:¼mijRiþRj [End of Step 2 inner loop.] [End of Step 1 outer loop.] The numbers mijare called multipliers . Sometimes we keep track of these multipliers by means of the following lower triangular matrix L: L¼100 ... 00 /C0m21 10 ... 00 /C0m31/C0m32 1 ... 00 /C0mn1/C0mn2/C0mn3.../C0mn;n/C0112 666643 77775 That is, Lhas 1’s on the diagonal, 0’s above the diagonal, and the negative of the multiplier mijas its ij-entry below the diagonal. The above matrix Land the triangular matrix Uobtained in Algorithm 3.6 give us the classical LU factorization of such a matrix A. Namely, THEOREM 3.22: LetAbe a nonsingular matrix that can be brought into triangular form Uusing only row-addition operations. Then A¼LU, where Lis the above lower triangular matrix with 1’s on the diagonal, and Uis an upper triangular matrix with no 0’s on the diagonal..........................................................CHAPTER 3 Systems of Linear Equations 87 EXAMPLE 3.22 Suppose A¼12/C03 /C03/C041 3 21/C052 43 5.W en o t et h a t Amay be reduced to triangular form by the operations ‘‘Replace R2by 3R1þR2’’;‘‘Replace R3by/C02R1þR3’’;and then ‘‘Replace R3by3 2R2þR3’’ That is, A/C2412/C03 024 0/C0312 43 5/C2412/C03 02 4 00 72 43 5 This gives us the classical factorization A¼LU, where L¼10 0 /C031 0 2/C03 212 643 75 and U¼12/C03 02 400 72 643 75 We emphasize: (1) The entries/C03;2;/C0 3 2inLare the negatives of the multipliers in the above elementary row operations. (2)Uis the triangular form of A. Application to Systems of Linear Equations Consider a computer algorithm M. Let CðnÞdenote the running time of the algorithm as a function of the size nof the input data. [The function CðnÞis sometimes called the time complexity or simply the complexity of the algorithm M.] Frequently, CðnÞsimply counts the number of multiplications and divisions executed by M, but does not count the number of additions and subtractions because they take much less time to execute. Now consider a square system of linear equations AX¼B, where A¼½aij/C138; X¼½x1;...;xn/C138T; B¼½b1;...;bn/C138T and suppose Ahas an LUfactorization. Then the system can be brought into triangular form (in order to apply back-substitution) by applying Algorithm 3.6 to the augmented matrix M¼½A;B/C138of the system. The time complexity of Algorithm 3.6 and back-substitution are, respectively, CðnÞ/C251 2n3and CðnÞ/C251 2n2 where nis the number of equations. On the other hand, suppose we already have the factorization A¼LU. Then, to triangularize the system, we need only apply the row operations in the algorithm (retained by the matrix L) to the column vector B. In this case, the time complexity is CðnÞ/C251 2n2 Of course, to obtain the factorization A¼LUrequires the original algorithm where CðnÞ/C251 2n3. Thus, nothing may be gained by ﬁrst ﬁnding the LUfactorization when a single system is involved. However, there are situations, illustrated below, where the LUfactorization is useful. Suppose, for a given matrix A, we need to solve the system AX¼B88 CHAPTER 3 Systems of Linear Equations repeatedly for a sequence of different constant vectors, say B1;B2;...;Bk. Also, suppose some of the Bi depend upon the solution of the system obtained while using preceding vectors Bj. In such a case, it is more efﬁcient to ﬁrst ﬁnd the LUfactorization of A, and then to use this factorization to solve the system for each new B. EXAMPLE 3.23 Consider the following system of linear equations: xþ2yþz¼k1 2xþ3yþ3z¼k2 /C03xþ10yþ2z¼k3orAX¼B;where A¼12 1 23 3 /C031 022 43 5and B¼k1 k2 k32 43 5 Suppose we want to solve the system three times where Bis equal, say, to B1;B2;B3. Furthermore, suppose B1¼½1;1;1/C138T, and suppose Bjþ1¼BjþXjðforj¼1;2Þ where Xjis the solution of AX¼Bj. Here it is more efﬁcient to ﬁrst obtain the LUfactorization of Aand then use the LUfactorization to solve the system for each of the B’s. (This is done in Problem 3.42.) SOLVED PROBLEMS Linear Equations, Solutions, 2 /C22 Systems 3.1. Determine whether each of the following equations is linear: (a) 5 xþ7y/C08yz¼16, (b) xþpyþez¼log 5, (c) 3 xþky/C08z¼16 (a) No, because the product yzof two unknowns is of second degree. (b) Yes, because p;e, and log 5 are constants. (c) As it stands, there are four unknowns: x;y;z;k. Because of the term kyit is not a linear equation. However, assuming kis a constant, the equation is linear in the unknowns x;y;z. 3.2. Determine whether the following vectors are solutions of x1þ2x2/C04x3þ3x4¼15: (a)u¼ð3;2;1;4Þand (b) v¼ð1;2;4;5Þ: (a) Substitute to obtain 3 þ2ð2Þ/C04ð1Þþ3ð4Þ¼15, or 15¼15; yes, it is a solution. (b) Substitute to obtain 1 þ2ð2Þ/C04ð4Þþ3ð5Þ¼15, or 4¼15; no, it is not a solution. 3.3. Solve (a) ex¼p, (b) 3 x/C04/C0x¼2xþ3, (c) 7þ2x/C04¼3xþ3/C0x (a) Because e6¼0, multiply by 1 =eto obtain x¼p=e. (b) Rewrite in standard form, obtaining 0 x¼7. The equation has no solution. (c) Rewrite in standard form, obtaining 0 x¼0. Every scalar kis a solution. 3.4. Prove Theorem 3.4: Consider the equation ax¼b. (i) If a6¼0, then x¼b=ais a unique solution of ax¼b. (ii) If a¼0 but b6¼0, then ax¼bhas no solution. (iii) If a¼0 and b¼0, then every scalar kis a solution of ax¼b. Suppose a6¼0. Then the scalar b=aexists. Substituting b=ainax¼byields aðb=aÞ¼b,o rb¼b; hence, b=ais a solution. On the other hand, suppose x0is a solution to ax¼b, so that ax0¼b. Multiplying both sides by 1 =ayields x0¼b=a. Hence, b=ais the unique solution of ax¼b. Thus, (i) is proved. On the other hand, suppose a¼0. Then, for any scalar k, we have ak¼0k¼0. If b6¼0, then ak6¼b. Accordingly, kis not a solution of ax¼b, and so (ii) is proved. If b¼0, then ak¼b. That is, any scalar kis a solution of ax¼b, and so (iii) is proved.CHAPTER 3 Systems of Linear Equations 89 3.5. Solve each of the following systems: (a)2x/C05y¼11 3xþ4y¼5(b)2x/C03y¼8 /C06xþ9y¼6(c)2x/C03y¼ 8 /C04xþ6y¼/C016 (a) Eliminate xfrom the equations by forming the new equation L¼/C03L1þ2L2. This yields the equation 23y¼/C023; and so y¼/C01 Substitute y¼/C01 in one of the original equations, say L1, to get 2x/C05ð/C01Þ¼11 or 2 xþ5¼11 or 2 x¼6o r x¼3 Thus, x¼3,y¼/C01 or the pair u¼ð3;/C01Þis the unique solution of the system. (b) Eliminate xfrom the equations by forming the new equation L¼3L1þL2. This yields the equation 0xþ0y¼30 This is a degenerate equation with a nonzero constant; hence, this equation and the system have no solution. (Geometrically, the lines corresponding to the equations are parallel.) (c) Eliminate xfrom the equations by forming the new equation L¼2L1þL2. This yields the equation 0xþ0y¼0 This is a degenerate equation where the constant term is also zero. Thus, the system has an inﬁnite number of solutions, which correspond to the solution of either equation. (Geometrically, the linescorresponding to the equations coincide.) To ﬁnd the general solution, set y¼aand substitute in L 1to obtain 2x/C03a¼8o r2 x¼3aþ8o r x¼3 2aþ4 Thus, the general solution is x¼3 2aþ4;y¼a or u¼3 2aþ4;a/C0/C1 where ais any scalar. 3.6. Consider the system xþay¼4 axþ9y¼b (a) For which values of adoes the system have a unique solution? (b) Find those pairs of values ( a;b) for which the system has more than one solution. (a) Eliminate xfrom the equations by forming the new equation L¼/C0aL1þL2. This yields the equation ð9/C0a2Þy¼b/C04a ð1Þ The system has a unique solution if and only if the coefﬁcient of yin (1) is not zero—that is, if 9/C0a26¼0o ri f a6¼/C63. (b) The system has more than one solution if both sides of (1) are zero. The left-hand side is zero when a¼/C63. When a¼3, the right-hand side is zero when b/C012¼0o r b¼12. When a¼/C03, the right- hand side is zero when bþ12/C00o r b¼/C012. Thus, (3 ;12) andð/C03;/C012Þare the pairs for which the system has more than one solution. Systems in Triangular and Echelon Form 3.7. Determine the pivot and free variables in each of the following systems: 2x1/C03x2/C06x3/C05x4þ2x5¼7 x3þ3x4/C07x5¼6 x4/C02x5¼1 (a)2x/C06yþ7z¼1 4yþ3z¼8 2z¼4 (b)xþ2y/C03z¼2 2xþ3yþz¼4 3xþ4yþ5z¼8 (c) (a) In echelon form, the leading unknowns are the pivot variables, and the others are the free variables. Here x1,x3,x4are the pivot variables, and x2andx5are the free variables.90 CHAPTER 3 Systems of Linear Equations (b) The leading unknowns are x;y;z, so they are the pivot variables. There are no free variables (as in any triangular system). (c) The notion of pivot and free variables applies only to a system in echelon form. 3.8. Solve the triangular system in Problem 3.7(b). Because it is a triangular system, solve by back-substitution. (i) The last equation gives z¼2. (ii) Substitute z¼2 in the second equation to get 4 yþ6¼8o r y¼1 2. (iii) Substitute z¼2 and y¼1 2in the ﬁrst equation to get 2x/C061 2/C18/C19 þ7ð2Þ¼1o r2 xþ11¼1o r x¼/C05 Thus, x¼/C05,y¼1 2,z¼2o r u¼ð/C0 5;1 2;2Þis the unique solution to the system. 3.9. Solve the echelon system in Problem 3.7(a). Assign parameters to the free variables, say x2¼aandx5¼b, and solve for the pivot variables by back- substitution. (i) Substitute x5¼bin the last equation to get x4/C02b¼1o r x4¼2bþ1. (ii) Substitute x5¼bandx4¼2bþ1 in the second equation to get x3þ3ð2bþ1Þ/C07b¼6o r x3/C0bþ3¼6o r x3¼bþ3 (iii) Substitute x5¼b,x4¼2bþ1,x3¼bþ3,x2¼ain the ﬁrst equation to get 2x1/C03a/C06ðbþ3Þ/C05ð2bþ1Þþ2b¼7o r2 x1/C03a/C014b/C023¼7 or x1¼3 2aþ7bþ15 Thus, x1¼3 2aþ7bþ15; x2¼a; x3¼bþ3; x4¼2bþ1;x5¼b or u¼3 2aþ7bþ15;a;bþ3;2bþ1;b/C18/C19 is the parametric form of the general solution. Alternatively, solving for the pivot variable x1;x3;x4in terms of the free variables x2andx5yields the following free-variable form of the general solution: x1¼3 2x2þ7x5þ15; x3¼x5þ3; x4¼2x5þ1 3.10. Prove Theorem 3.6. Consider the system (3.4) of linear equations in echelon form with requations andnunknowns. (i) If r¼n, then the system has a unique solution. (ii) If r<n, then we can arbitrarily assign values to the n/C0rfree variable and solve uniquely for therpivot variables, obtaining a solution of the system. (i) Suppose r¼n. Then we have a square system AX¼Bwhere the matrix Aof coefﬁcients is (upper) triangular with nonzero diagonal elements. Thus, Ais invertible. By Theorem 3.10, the system has a unique solution. (ii) Assigning values to the n/C0rfree variables yields a triangular system in the pivot variables, which, by (i), has a unique solution.CHAPTER 3 Systems of Linear Equations 91 Gaussian Elimination 3.11. Solve each of the following systems: xþ2y/C04z¼/C0 4 2xþ5y/C09z¼/C010 3x/C02yþ3z¼11 (a)xþ2y/C03z¼/C01 /C03xþy/C02z¼/C07 5xþ3y/C04z¼2 (b)xþ2y/C03z¼1 2xþ5y/C08z¼4 3xþ8y/C013z¼7 (c) Reduce each system to triangular or echelon form using Gaussian elimination: (a) Apply ‘‘Replace L2by/C02L1þL2’’ and ‘‘Replace L3by/C03L1þL3’’ to eliminate xfrom the second and third equations, and then apply ‘‘Replace L3by 8L2þL3’’ to eliminate yfrom the third equation. These operations yield xþ2y/C04z¼/C04 y/C0 z¼/C02 /C08yþ15z¼23and thenxþ2y/C04z¼/C04 y/C0z¼/C02 7z¼7 The system is in triangular form. Solve by back-substitution to obtain the unique solution u¼ð2;/C01;1Þ. (b) Eliminate xfrom the second and third equations by the operations ‘‘Replace L2by 3 L1þL2’’ and ‘‘Replace L3by/C05L1þL3.’’ This gives the equivalent system xþ2y/C03z¼/C0 1 7y/C011z¼/C010 /C07yþ11z¼ 7 The operation ‘‘Replace L3byL2þL3’’ yields the following degenerate equation with a nonzero constant: 0xþ0yþ0z¼/C03 This equation and hence the system have no solution. (c) Eliminate xfrom the second and third equations by the operations ‘‘Replace L2by/C02L1þL2’’ and ‘‘Replace L3by/C03L1þL3.’’ This yields the new system xþ2y/C03z¼1 y/C02z¼2 2y/C04z¼4orxþ2y/C03z¼1 y/C02z¼2 (The third equation is deleted, because it is a multiple of the second equation.) The system is in echelon form with pivot variables xandyand free variable z. To ﬁnd the parametric form of the general solution, set z¼aand solve for xandyby back- substitution. Substitute z¼ain the second equation to get y¼2þ2a. Then substitute z¼aand y¼2þ2ain the ﬁrst equation to get xþ2ð2þ2aÞ/C03a¼1o r xþ4þa¼1o r x¼/C03/C0a Thus, the general solution is x¼/C03/C0a;y¼2þ2a;z¼a or u¼ð/C0 3/C0a;2þ2a;aÞ where ais a parameter. 3.12. Solve each of the following systems: x1/C03x2þ2x3/C0x4þ2x5¼2 3x1/C09x2þ7x3/C0x4þ3x5¼7 2x1/C06x2þ7x3þ4x4/C05x5¼7 (a)x1þ2x2/C03x3þ4x4¼2 2x1þ5x2/C02x3þx4¼1 5x1þ12x2/C07x3þ6x4¼3 (b) Reduce each system to echelon form using Gaussian elimination:92 CHAPTER 3 Systems of Linear Equations (a) Apply ‘‘Replace L2by/C03L1þL2’’ and ‘‘Replace L3by/C02L1þL3’’ to eliminate xfrom the second and third equations. This yields x1/C03x2þ2x3/C0x4þ2x5¼2 x3þ2x4/C03x5¼1 3x3þ6x4/C09x5¼3orx1/C03x2þ2x3/C0x4þ2x5¼2 x3þ2x4/C03x5¼1 (We delete L3, because it is a multiple of L2.) The system is in echelon form with pivot variables x1and x3and free variables x2;x4;x5. To ﬁnd the parametric form of the general solution, set x2¼a,x4¼b,x5¼c, where a;b;care parameters. Back-substitution yields x3¼1/C02bþ3candx1¼3aþ5b/C08c. The general solution is x1¼3aþ5b/C08c;x2¼a;x3¼1/C02bþ3c;x4¼b;x5¼c or, equivalently, u¼ð3aþ5b/C08c;a;1/C02bþ3c;b;cÞ. (b) Eliminate x1from the second and third equations by the operations ‘‘Replace L2by/C02L1þL2’’ and ‘‘Replace L3by/C05L1þL3.’’ This yields the system x1þ2x2/C03x3þ4x4¼2 x2þ4x3/C07x4¼/C03 2x2þ8x3/C014x4¼/C07 The operation ‘‘Replace L3by/C02L2þL3’’ yields the degenerate equation 0 ¼/C01. Thus, the system has no solution (even though the system has more unknowns than equations). 3.13. Solve using the condensed format: 2yþ3z¼3 xþyþz¼4 4xþ8y/C03z¼35 The condensed format follows: Number Equation Operation ð2Þð 1=Þ 2yþ3z¼3 L1$L2 ð1Þð 2=Þ xþyþ z¼4 L1$L2 ð3Þ 4xþ8y/C03z¼35 ð30Þ 4y/C07z¼19 Replace L3by/C04L1þL3 ð300Þ/C0 13z¼13 Replace L3by/C02L2þL3 Here (1), (2), and (300) form a triangular system. (We emphasize that the interchange of L1andL2is accomplished by simply renumbering L1andL2as above.) Using back-substitution with the triangular system yields z¼/C01 from L3,y¼3 from L2, and x¼2 from L1. Thus, the unique solution of the system is x¼2,y¼3,z¼/C01 or the triple u¼ð2;3;/C01Þ. 3.14. Consider the system xþ2yþz¼3 ayþ5z¼10 2xþ7yþaz¼b (a) Find those values of afor which the system has a unique solution. (b) Find those pairs of values ða;bÞfor which the system has more than one solution. Reduce the system to echelon form. That is, eliminate xfrom the third equation by the operation ‘‘Replace L3by/C02L1þL3’’ and then eliminate yfrom the third equation by the operationCHAPTER 3 Systems of Linear Equations 93 ‘‘Replace L3by/C03L2þaL3.’’ This yields xþ2yþz¼3 ayþ5z¼10 3yþða/C02Þz¼b/C06and thenxþ2yþz¼3 ayþ5z¼10 ða2/C02a/C015Þz¼ab/C06a/C030 Examine the last equation ða2/C02a/C015Þz¼ab/C06a/C030. (a) The system has a unique solution if and only if the coefﬁcient of zis not zero; that is, if a2/C02a/C015¼ða/C05Þðaþ3Þ6¼0o r a6¼5 and a6¼/C03: (b) The system has more than one solution if both sides are zero. The left-hand side is zero when a¼5o r a¼/C03. When a¼5, the right-hand side is zero when 5 b/C060¼0, or b¼12. When a¼/C03, the right- hand side is zero when /C03b/C012¼0, or b¼/C04. Thus,ð5;12Þandð/C03;/C04Þare the pairs for which the system has more than one solution. Echelon Matrices, Row Equivalence, Row Canonical Form 3.15. Row reduce each of the following matrices to echelon form: (a) A¼12/C030 24/C022 36/C0432 43 5; (b) B¼/C041/C06 12/C05 63/C042 43 5 (a) Use a11¼1 as a pivot to obtain 0’s below a11; that is, apply the row operations ‘‘Replace R2by /C02R1þR2’’ and ‘‘Replace R3by/C03R1þR3:’’ Then use a23¼4 as a pivot to obtain a 0 below a23; that is, apply the row operation ‘‘Replace R3by/C05R2þ4R3.’’ These operations yield A/C2412/C030 00 42 00 532 43 5/C2412/C030 00 42 00 022 43 5 The matrix is now in echelon form. (b) Hand calculations are usually simpler if the pivot element equals 1. Therefore, ﬁrst interchange R1andR2. Next apply the operations ‘‘Replace R2by 4R1þR2’’ and ‘‘Replace R3by/C06R1þR3’’; and then apply the operation ‘‘Replace R3byR2þR3.’’ These operations yield B/C2412/C05 /C041/C06 63/C042 43 5/C2412/C05 09/C026 0/C092 62 43 5/C2412/C05 09/C026 00 02 43 5 The matrix is now in echelon form. 3.16. Describe the pivoting row-reduction algorithm. Also describe the advantages, if any, of using this pivoting algorithm. The row-reduction algorithm becomes a pivoting algorithm if the entry in column jof greatest absolute value is chosen as the pivot a1j1and if one uses the row operation ð/C0aij1=a1j1ÞR1þRi!Ri The main advantage of the pivoting algorithm is that the above row operation involves division by the (current) pivot a1j1, and, on the computer, roundoff errors may be substantially reduced when one divides by a number as large in absolute value as possible. 3.17. LetA¼2/C022 1 /C036 0/C01 1/C071 0 22 43 5. Reduce Ato echelon form using the pivoting algorithm.94 CHAPTER 3 Systems of Linear Equations First interchange R1andR2so that/C03 can be used as the pivot, and then apply the operations ‘‘Replace R2 by2 3R1þR2’’ and ‘‘Replace R3by1 3R1þR3.’’ These operations yield A/C24/C036 0/C01 2/C022 1 1/C071 0 22 43 5/C24/C036 0/C01 02 21 3 0/C051 05 32 643 75 Now interchange R2andR3so that/C05 can be used as the pivot, and then apply the operation ‘‘Replace R3by 2 5R2þR3.’’ We obtain A/C24/C036 0/C01 0/C051 05 3 02 2132 43 5/C24/C036 0/C01 0/C051 05 3 00 612 43 5 The matrix has been brought to echelon form using partial pivoting. 3.18. Reduce each of the following matrices to row canonical form: (a) A¼22/C0164 44 11 01 3 88/C012 62 32 43 5; (b) B¼5/C096 02 3 00 72 43 5 (a) First reduce Ato echelon form by applying the operations ‘‘Replace R2by/C02R1þR2’’ and ‘‘Replace R3 by/C04R1þR3,’’ and then applying the operation ‘‘Replace R3by/C0R2þR3.’’ These operations yield A/C2422/C016 4 00 3/C025 0 032 72 43 5/C2422/C016 4 00 3/C025 00 0 422 43 5 Now use back-substitution on the echelon matrix to obtain the row canonical form of A. Speciﬁcally, ﬁrst multiply R3by1 4to obtain the pivot a34¼1, and then apply the operations ‘‘Replace R2by 2R3þR2’’ and ‘‘Replace R1by/C06R3þR1.’’ These operations yield A/C2422/C016 4 00 3/C025 0 0011 22 43 5/C2422/C0101 00 306 00 011 22 43 5 Now multiply R2by1 3, making the pivot a23¼1, and then apply ‘‘Replace R1byR2þR1,’’ yielding A/C2422/C0101 00 102 00 011 22 43 5/C2422003 00102 00011 22 43 5 Finally, multiply R1by1 2, so the pivot a11¼1. Thus, we obtain the following row canonical form of A: A/C2411003 2 00102 00011 22 43 5 (b) Because Bis in echelon form, use back-substitution to obtain B/C245/C096 02 3 00 12 643 75/C245/C090 02 0 00 12 643 75/C245/C090 01 0 00 12 643 75/C24500 010 0012 643 75/C24100 010 0012 643 75 The last matrix, which is the identity matrix I, is the row canonical form of B. (This is expected, because Bis invertible, and so its row canonical form must be I.) 3.19. Describe the Gauss–Jordan elimination algorithm, which also row reduces an arbitrary matrix Ato its row canonical form.CHAPTER 3 Systems of Linear Equations 95 The Gauss–Jordan algorithm is similar in some ways to the Gaussian elimination algorithm, except that here each pivot is used to place 0’s both below and above the pivot, not just below the pivot, before working with the next pivot. Also, one variation of the algorithm ﬁrst normalizes each row—that is, obtains a unit pivot—before it is used to produce 0’s in the other rows, rather than normalizing the rows at the end of thealgorithm. 3.20. LetA¼1/C023 12 11 4/C013 25 9/C0282 43 5. Use Gauss–Jordan to ﬁnd the row canonical form of A. Usea11¼1 as a pivot to obtain 0’s below a11by applying the operations ‘‘Replace R2by/C0R1þR2’’ and ‘‘Replace R3by/C02R1þR3.’’ This yields A/C241/C023 12 03 1/C021 09 3/C0442 43 5 Multiply R2by1 3to make the pivot a22¼1, and then produce 0’s below and above a22by applying the operations ‘‘Replace R3by/C09R2þR3’’ and ‘‘Replace R1by 2R2þR1.’’ These operations yield A/C241/C023 12 011 3/C023 13 09 3/C0442 6643 775/C2410 11 3/C013 83 0113/C023 13 00 0 212 6643 775 Finally, multiply R3by1 2to make the pivot a34¼1, and then produce 0’s above a34by applying the operations ‘‘Replace R2by2 3R3þR2’’ and ‘‘Replace R1by1 3R3þR1.’’ These operations yield A/C241011 3/C01383 0113/C02313 00 0 11 22 6643 775/C241011 3017 6 0113023 00 011 22 6643 775 which is the row canonical form of A. Systems of Linear Equations in Matrix Form 3.21. Find the augmented matrix Mand the coefﬁcient matrix Aof the following system: xþ2y/C03z¼4 3y/C04zþ7x¼5 6zþ8x/C09y¼1 First align the unknowns in the system, and then use the aligned system to obtain MandA. We have xþ2y/C03z¼4 7xþ3y/C04z¼5 8x/C09yþ6z¼1; then M¼12/C034 73/C045 8/C096 12 43 5 and A¼12/C03 73/C04 8/C0962 43 5 3.22. Solve each of the following systems using its augmented matrix M: xþ2y/C0z¼3 xþ3yþz¼5 3xþ8yþ4z¼17 (a)x/C02yþ4z¼2 2x/C03yþ5z¼3 3x/C04yþ6z¼7 (b)xþyþ3z¼1 2xþ3y/C0z¼3 5xþ7yþz¼7 (c) (a) Reduce the augmented matrix Mto echelon form as follows: M¼12/C013 13 1 5 38 41 72 43 5/C2412/C013 01 22 02 782 43 5/C2412/C013 01 22 00 342 43 596 CHAPTER 3 Systems of Linear Equations Now write down the corresponding triangular system xþ2y/C0z¼3 yþ2z¼2 3z¼4 and solve by back-substitution to obtain the unique solution x¼17 3;y¼/C02 3;z¼4 3or u¼ð17 3;/C02 3;43Þ Alternately, reduce the echelon form of Mto row canonical form, obtaining M/C2412/C013 01 2200 1 4 32 6643 775/C2412013 3 010/C02 3 0014 32 6643 775/C2410017 3 010/C02 3 0014 32 6643 775 This also corresponds to the above solution. (b) First reduce the augmented matrix Mto echelon form as follows: M¼1/C0242 2/C0353 3/C04672 43 5/C241/C0242 01/C03/C01 02/C0612 43 5/C241/C0242 01/C03/C01 00032 43 5 The third row corresponds to the degenerate equation 0 xþ0yþ0z¼3, which has no solution. Thus, ‘‘DO NOT CONTINUE.’’ The original system also has no solution. (Note that the echelon form indicates whether or not the system has a solution.) (c) Reduce the augmented matrix Mto echelon form and then to row canonical form: M¼11 31 23/C013 57 172 43 5/C2411 31 01/C071 02/C014 22 43 5/C241 0 10 0 01/C071/C20/C21 (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) Write down the system corresponding to the row canonical form of Mand then transfer the free variables to the other side to obtain the free-variable form of the solution: xþ10z¼0 y/C07z¼1andx¼/C010z y¼1þ7z Here zis the only free variable. The parametric solution, using z¼a, is as follows: x¼/C010a;y¼1þ7a;z¼a or u¼ð/C0 10a;1þ7a;aÞ 3.23. Solve the following system using its augmented matrix M: x1þ2x2/C03x3/C02x4þ4x5¼1 2x1þ5x2/C08x3/C0x4þ6x5¼4 x1þ4x2/C07x3þ5x4þ2x5¼8 Reduce the augmented matrix Mto echelon form and then to row canonical form: M¼12/C03/C0241 25/C08/C0164 14/C075 2 82 643 75/C2412/C03/C024 1 01/C023/C022 02/C047/C0272 643 75/C2412/C03/C024 1 01/C023/C022 0 0012 32 643 75 /C2412/C03 087 01/C020/C08/C07 00 01 2 32 643 75/C2410 10 2 4 2 1 01/C020/C08/C07 00 01 2 32 643 75 Write down the system corresponding to the row canonical form of Mand then transfer the free variables to the other side to obtain the free-variable form of the solution: x1þx3þ 24x5¼21 x2/C02x3/C0 8x5¼/C07 x4þ2x5¼3andx1¼21/C0x3/C024x5 x2¼/C07þ2x3þ8x5 x4¼3/C02x5CHAPTER 3 Systems of Linear Equations 97 Here x1;x2;x4are the pivot variables and x3andx5are the free variables. Recall that the parametric form of the solution can be obtained from the free-variable form of the solution by simply setting the free variables equal to parameters, say x3¼a,x5¼b. This process yields x1¼21/C0a/C024b;x2¼/C07þ2aþ8b;x3¼a;x4¼3/C02b;x5¼b or u¼ð21/C0a/C024b;/C07þ2aþ8b;a;3/C02b;bÞ which is another form of the solution. Linear Combinations, Homogeneous Systems 3.24. Write vas a linear combination of u1;u2;u3, where (a) v¼ð3;10;7Þandu1¼ð1;3;/C02Þ;u2¼ð1;4;2Þ;u3¼ð2;8;1Þ; (b) v¼ð2;7;10Þandu1¼ð1;2;3Þ,u2¼ð1;3;5Þ,u3¼ð1;5;9Þ; (c)v¼ð1;5;4Þandu1¼ð1;3;/C02Þ,u2¼ð2;7;/C01Þ,u3¼ð1;6;7Þ. Find the equivalent system of linear equations by writing v¼xu1þyu2þzu3. Alternatively, use the augmented matrix Mof the equivalent system, where M¼½u1;u2;u3;v/C138. (Here u1;u2;u3;vare the columns ofM.) (a) The vector equation v¼xu1þyu2þzu3for the given vectors is as follows: 3 10 72 43 5¼x1 3 /C022 43 5þy1 4 22 43 5þz2 8 12 43 5¼xþyþ2z 3xþ4yþ8z /C02xþ2yþz2 43 5 Form the equivalent system of linear equations by setting corresponding entries equal to each other, and then reduce the system to echelon form: xþyþ2z¼3 3xþ4yþ8z¼10 /C02xþ2yþz¼7orxþyþ2z¼3 yþ2z¼1 4yþ5z¼13orxþyþ2z¼3 yþ2z¼1 /C03z¼9 The system is in triangular form. Back-substitution yields the unique solution x¼2,y¼7,z¼/C03. Thus, v¼2u1þ7u2/C03u3. Alternatively, form the augmented matrix M¼[u1;u2;u3;v] of the equivalent system, and reduce Mto echelon form: M¼112 3 3481 0 /C0221 72 43 5/C24112 3 012 1 0451 32 43 5/C2411 23 01 21 00/C0392 43 5 The last matrix corresponds to a triangular system that has a unique solution. Back-substitution yields the solution x¼2,y¼7,z¼/C03. Thus, v¼2u1þ7u2/C03u3. (b) Form the augmented matrix M¼½u1;u2;u3;v/C138of the equivalent system, and reduce Mto the echelon form: M¼111 2 235 7 3591 02 43 5/C241112 0133 02642 43 5/C24111 2 013 3 000/C022 43 5 The third row corresponds to the degenerate equation 0 xþ0yþ0z¼/C02, which has no solution. Thus, the system also has no solution, and vcannot be written as a linear combination of u1;u2;u3. (c) Form the augmented matrix M¼½u1;u2;u3;v/C138of the equivalent system, and reduce Mto echelon form: M¼12 1 1 37 6 5 /C02/C01742 43 5/C241211 0132 03962 43 5/C241211 0132 00002 43 598 CHAPTER 3 Systems of Linear Equations The last matrix corresponds to the following system with free variable z: xþ2yþz¼1 yþ3z¼2 Thus, vcan be written as a linear combination of u1;u2;u3in many ways. For example, let the free variable z¼1, and, by back-substitution, we get y¼/C02 and x¼2. Thus, v¼2u1/C02u2þu3. 3.25. Letu1¼ð1;2;4Þ,u2¼ð2;/C03;1Þ,u3¼ð2;1;/C01ÞinR3. Show that u1;u2;u3are orthogonal, and write vas a linear combination of u1;u2;u3, where (a) v¼ð7;16;6Þ, (b) v¼ð3;5;2Þ. Take the dot product of pairs of vectors to get u1/C1u2¼2/C06þ4¼0;u1/C1u3¼2þ2/C04¼0;u2/C1u3¼4/C03/C01¼0 Thus, the three vectors in R3are orthogonal, and hence Fourier coefﬁcients can be used. That is, v¼xu1þyu2þzu3, where x¼v/C1u1 u1/C1u1; y¼v/C1u2 u2/C1u2; z¼v/C1u3 u3/C1u3 (a) We have x¼7þ32þ24 1þ4þ16¼63 21¼3; y¼14/C048þ6 4þ9þ1¼/C028 14¼/C02; z¼14þ16/C06 4þ1þ1¼24 6¼4 Thus, v¼3u1/C02u2þ4u3. (b) We have x¼3þ10þ8 1þ4þ16¼21 21¼1; y¼6/C015þ2 4þ9þ1¼/C07 14¼/C01 2; z¼6þ5/C02 4þ1þ1¼9 6¼3 2 Thus, v¼u1/C01 2u2þ32u3. 3.26. Find the dimension and a basis for the general solution Wof each of the following homogeneous systems: 2x1þ4x2/C05x3þ3x4¼0 3x1þ6x2/C07x3þ4x4¼0 5x1þ10x2/C011x3þ6x4¼0 (a)x/C02y/C03z¼0 2xþyþ3z¼0 3x/C04y/C02z¼0 (b) (a) Reduce the system to echelon form using the operations ‘‘Replace L2by/C03L1þ2L2,’’ ‘‘Replace L3by /C05L1þ2L3,’’ and then ‘‘Replace L3by/C02L2þL3.’’ These operations yield 2x1þ4x2/C05x3þ3x4¼0 x3/C0x4¼0 3x3/C03x4¼0and2x1þ4x2/C05x3þ3x4¼0 x3/C0x4¼0 The system in echelon form has two free variables, x2andx4, so dim W¼2. A basis½u1;u2/C138forWmay be obtained as follows:(1) Set x 2¼1,x4¼0. Back-substitution yields x3¼0, and then x1¼/C02. Thus, u1¼ð/C0 2;1;0;0Þ. (2) Set x2¼0,x4¼1. Back-substitution yields x3¼1, and then x1¼1. Thus, u2¼ð1;0;1;1Þ. (b) Reduce the system to echelon form, obtaining x/C02y/C03z¼0 5yþ9z¼0 2yþ7z¼0andx/C02y/C03z¼0 5yþ9z¼0 17z¼0 There are no free variables (the system is in triangular form). Hence, dim W¼0, and Whas no basis. Speciﬁcally, Wconsists only of the zero solution; that is, W¼f0g. 3.27. Find the dimension and a basis for the general solution Wof the following homogeneous system using matrix notation: x1þ2x2þ3x3/C02x4þ4x5¼0 2x1þ4x2þ8x3þx4þ9x5¼0 3x1þ6x2þ13x3þ4x4þ14x5¼0 Show how the basis gives the parametric form of the general solution of the system. When a system is homogeneous, we represent the system by its coefﬁcient matrix Arather than by itsCHAPTER 3 Systems of Linear Equations 99 augmented matrix M, because the last column of the augmented matrix Mis a zero column, and it will remain a zero column during any row-reduction process. Reduce the coefﬁcient matrix Ato echelon form, obtaining A¼12 3/C024 24 8 1 9 361 3 41 42 43 5/C24123/C024 002 51 004 1 022 43 5/C24123/C024 002 51/C20/C21 (The third row of the second matrix is deleted, because it is a multiple of the second row and will result in a zero row.) We can now proceed in one of two ways. (a) Write down the corresponding homogeneous system in echelon form: x1þ2x2þ3x3/C02x4þ4x5¼0 2x3þ5x4þx5¼0 The system in echelon form has three free variables, x2;x4;x5, so dim W¼3. A basis½u1;u2;u3/C138forW may be obtained as follows: (1) Set x2¼1,x4¼0,x5¼0. Back-substitution yields x3¼0, and then x1¼/C02. Thus, u1¼ð/C0 2;1;0;0;0Þ. (2) Set x2¼0,x4¼1,x5¼0. Back-substitution yields x3¼/C05 2, and then x1¼19 2. Thus, u2¼ð19 2;0;/C05 2;1;0Þ. (3) Set x2¼0,x4¼0,x5¼1. Back-substitution yields x3¼/C01 2, and then x1¼/C05 2. Thus, u3¼ð/C05 2,0 ,/C01 2;0;1Þ. [One could avoid fractions in the basis by choosing x4¼2 in (2) and x5¼2 in (3), which yields multiples of u2andu3.] The parametric form of the general solution is obtained from the following linear combination of the basis vectors using parameters a;b;c: au1þbu2þcu3¼ð/C0 2aþ19 2b/C05 2c;a;/C05 2b/C01 2c;b;cÞ (b) Reduce the echelon form of Ato row canonical form: A/C24123/C024 0015 212"# /C24123/C019 252 0015 212"# Write down the corresponding free-variable solution: x1¼/C02x2þ19 2x4/C05 2x5 x3¼/C05 2x4/C01 2x5 Using these equations for the pivot variables x1andx3, repeat the above process to obtain a basis ½u1;u2;u3/C138 forW. That is, set x2¼1,x4¼0,x5¼0 to get u1; set x2¼0,x4¼1,x5¼0 to get u2; and set x2¼0, x4¼0,x5¼1 to get u3. 3.28. Prove Theorem 3.15. Let v0be a particular solution of AX¼B, and let Wbe the general solution ofAX¼0. Then U¼v0þW¼fv0þw:w2Wgis the general solution of AX¼B. Letwbe a solution of AX¼0. Then Aðv0þwÞ¼Av0þAw¼Bþ0¼B Thus, the sum v0þwis a solution of AX¼B. On the other hand, suppose vis also a solution of AX¼B. Then Aðv/C0v0Þ¼Av/C0Av0¼B/C0B¼0 Therefore, v/C0v0belongs to W. Because v¼v0þðv/C0v0Þ, we ﬁnd that any solution of AX¼Bcan be obtained by adding a solution of AX¼0 to a solution of AX¼B. Thus, the theorem is proved.100 CHAPTER 3 Systems of Linear Equations Elementary Matrices, Applications 3.29. Lete1;e2;e3denote, respectively, the elementary row operations ‘‘Interchange rows R1andR2;’’ ‘‘Replace R3by 7R3;’’ ‘‘Replace R2by/C03R1þR2’’ Find the corresponding three-square elementary matrices E1;E2;E3. Apply each operation to the 3 /C23 identity matrix I3to obtain E1¼010 100 0012 43 5; E2¼100 010 0072 43 5; E3¼100 /C0310 0012 43 5 3.30. Consider the elementary row operations in Problem 3.29. (a) Describe the inverse operations e/C01 1,e/C01 2,e/C01 3. (b) Find the corresponding three-square elementary matrices E0 1,E0 2,E0 3. (c) What is the relationship between the matrices E0 1,E0 2,E0 3and the matrices E1,E2,E3? (a) The inverses of e1,e2,e3are, respectively, ‘‘Interchange rows R1andR2;’’ ‘‘Replace R3by1 7R3;’’ ‘‘Replace R2by 3R1þR2:’’ (b) Apply each inverse operation to the 3 /C23 identity matrix I3to obtain E0 1¼010 1000012 43 5; E 0 2¼100 010 001 72 43 5; E0 3¼100 3100012 43 5 (c) The matrices E0 1,E0 2,E0 3are, respectively, the inverses of the matrices E1,E2,E3. 3.31. Write each of the following matrices as a product of elementary matrices: (a) A¼1/C03 /C024/C20/C21 ; (b) B¼123 014 0012 43 5; (c) C¼11 2 23 8 /C03/C0122 43 5 The following three steps write a matrix Mas a product of elementary matrices: Step 1. Row reduce Mto the identity matrix I, keeping track of the elementary row operations. Step 2. Write down the inverse row operations. Step 3. Write Mas the product of the elementary matrices corresponding to the inverse operations. This gives the desired result. If a zero row appears in Step 1, then Mis not row equivalent to the identity matrix I, and Mcannot be written as a product of elementary matrices. (a) (1) We have A¼1/C03 /C024/C20/C21 /C241/C03 0/C02/C20/C21 /C241/C03 01/C20/C21 /C2410 01/C20/C21 ¼I where the row operations are, respectively, ‘‘Replace R2by 2R1þR2;’’ ‘‘Replace R2by/C01 2R2;’’ ‘‘Replace R1by 3R2þR1’’ (2) Inverse operations: ‘‘Replace R2by/C02R1þR2;’’ ‘‘Replace R2by/C02R2;’’ ‘‘Replace R1by/C03R2þR1’’ (3) A¼10 /C021/C20/C21 10 0/C02/C20/C21 1/C03 01/C20/C21CHAPTER 3 Systems of Linear Equations 101 (b) (1) We have B¼123 014 0012 43 5/C24120 010 0012 43 5/C24100 010 0012 43 5¼I where the row operations are, respectively, ‘‘Replace R2by/C04R3þR2;’’ ‘‘Replace R1by/C03R3þR1;’’ ‘‘Replace R1by/C02R2þR1’’ (2) Inverse operations: ‘‘Replace R2by 4R3þR2;’’ ‘‘Replace R1by 3R3þR1;’’ ‘‘Replace R1by 2R2þR1’’ (3) B¼100 014 0012 43 5103 010 0012 43 5120 010 0012 43 5 (c) (1) First row reduce Cto echelon form. We have C¼11 2 23 8 /C03/C0122 43 5/C24112 014 0282 43 5/C24112 014 0002 43 5 In echelon form, Chas a zero row. ‘‘STOP.’’ The matrix Ccannot be row reduced to the identity matrix I, and Ccannot be written as a product of elementary matrices. (We note, in particular, that Chas no inverse.) 3.32. Find the inverse of (a) A¼12/C04 /C01/C015 27/C032 43 5;(b) B¼13/C04 15/C01 31 3/C062 43 5. (a) Form the matrix M¼[A;I] and row reduce Mto echelon form: M¼12/C04100 /C01/C01 5010 27/C030012 643 75/C2412/C04 100 01 1 110 03 5/C02012 643 75 /C2412/C0410 0 0 1111 0 00 2/C05/C0312 643 75 In echelon form, the left half of Mis in triangular form; hence, Ahas an inverse. Further reduce Mto row canonical form: M/C24120/C09/C062 0107 252/C012 001/C05 2/C032 122 6643 775/C24100/C016/C011 3 0107 252/C012 001/C05 2/C032 122 6643 775 The ﬁnal matrix has the form ½I;A/C01/C138; that is, A/C01is the right half of the last matrix. Thus, A/C01¼/C016/C011 3 7 252/C012 /C05 2/C032 122 6643 775 (b) Form the matrix M¼½B;I/C138and row reduce Mto echelon form: M¼13/C04100 15/C01010 31 3/C060012 43 5/C2413/C041 0 0 02 3/C0110 04 6/C03012 43 5/C2413/C0410 0 02 3/C011 0 00 0/C01/C0212 43 5 In echelon form, Mhas a zero row in its left half; that is, Bis not row reducible to triangular form. Accordingly, Bhas no inverse.102 CHAPTER 3 Systems of Linear Equations 3.33. Show that every elementary matrix Eis invertible, and its inverse is an elementary matrix. LetEbe the elementary matrix corresponding to the elementary operation e; that is, eðIÞ¼E. Let e0be the inverse operation of eand let E0be the corresponding elementary matrix; that is, e0ðIÞ¼E0. Then I¼e0ðeðIÞÞ¼ e0ðEÞ¼E0E and I¼eðe0ðIÞÞ¼ eðE0Þ¼EE0 Therefore, E0is the inverse of E. 3.34. Prove Theorem 3.16: Let ebe an elementary row operation and let Ebe the corresponding m-square elementary matrix; that is, E¼eðIÞ. Then eðAÞ¼EA, where Ais any m/C2nmatrix. LetRibe the row iofA; we denote this by writing A¼½R1;...;Rm/C138.I fBis a matrix for which ABis deﬁned then AB¼½R1B;...;RmB/C138. We also let ei¼ð0;...;0;^1;0;...;0Þ; ^¼i Here ^¼imeans 1 is the ith entry. One can show (Problem 2.45) that eiA¼Ri. We also note that I¼½e1;e2;...;em/C138is the identity matrix. (i) Let ebe the elementary row operation ‘‘Interchange rows RiandRj.’’ Then, for ^¼iand ^^¼j, E¼eðIÞ¼½ e1;...;bej;...;bbei;...;em/C138 and eðAÞ¼½ R1;...;bRj;...;bbRi;...;Rm/C138 Thus, EA¼½e1A;...;cejA;...;cceiA;...;emA/C138¼½R1;...;bRj;...;bbRi;...;Rm/C138¼eðAÞ (ii) Let ebe the elementary row operation ‘‘Replace RibykRiðk6¼0Þ.’’ Then, for ^¼i, E¼eðIÞ¼½ e1;...;bkei;...;em/C138 and eðAÞ¼½ R1;...;ckRi;...;Rm/C138 Thus, EA¼½e1A;...;dkeiA;...;emA/C138¼½R1;...;ckRi;...;Rm/C138¼eðAÞ (iii) Let ebe the elementary row operation ‘‘Replace RibykRjþRi.’’ Then, for ^¼i, E¼eðIÞ¼½ e1;...;dkejþei;...;em/C138 and eðAÞ¼½ R1;...;dkRjþRi;...;Rm/C138 UsingðkejþeiÞA¼kðejAÞþeiA¼kRjþRi, we have EA¼½e1A; ...;ðkejþeiÞA; ...;emA/C138 ¼½R1; ...;dkRjþRi; ...;Rm/C138¼eðAÞ 3.35. Prove Theorem 3.17: Let Abe a square matrix. Then the following are equivalent: (a)Ais invertible (nonsingular). (b)Ais row equivalent to the identity matrix I. (c)Ais a product of elementary matrices. Suppose Ais invertible and suppose Ais row equivalent to matrix Bin row canonical form. Then there exist elementary matrices E1;E2;...;Essuch that Es...E2E1A¼B. Because Ais invertible and each elementary matrix is invertible, Bis also invertible. But if B6¼I, then Bhas a zero row; whence Bis not invertible. Thus, B¼I, and (a) implies (b).CHAPTER 3 Systems of Linear Equations 103 If (b) holds, then there exist elementary matrices E1;E2;...;Essuch that Es...E2E1A¼I. Hence, A¼ðEs...E2E1Þ/C01¼E/C01 1E/C01 2...;E/C01 s. But the E/C01 iare also elementary matrices. Thus (b) implies (c). If (c) holds, then A¼E1E2...Es. The Eiare invertible matrices; hence, their product Ais also invertible. Thus, (c) implies (a). Accordingly, the theorem is proved. 3.36. Prove Theorem 3.18: If AB¼I, then BA¼I, and hence B¼A/C01. Suppose Ais not invertible. Then Ais not row equivalent to the identity matrix I,a n ds o Ais row equivalent to a matrix with a zero row. In other words, there exist elementary matrices E1;...;Essuch that Es...E2E1Ahas a zero row. Hence, Es...E2E1AB¼Es...E2E1, an invertible matrix, also has a zero row. But invertible matrices cannot have zero rows; hence Ais invertible, with inverse A/C01.T h e n also, B¼IB¼ðA/C01AÞB¼A/C01ðABÞ¼A/C01I¼A/C01 3.37. Prove Theorem 3.19: Bis row equivalent to A(written B/C24AÞif and only if there exists a nonsingular matrix Psuch that B¼PA. IfB/C24A, then B¼esð...ðe2ðe1ðAÞÞÞ...Þ¼Es...E2E1A¼PAwhere P¼Es...E2E1is nonsingular. Conversely, suppose B¼PA, where Pis nonsingular. By Theorem 3.17, Pis a product of elementary matrices, and so Bcan be obtained from Aby a sequence of elementary row operations; that is, B/C24A. Thus, the theorem is proved. 3.38. Prove Theorem 3.21: Every m/C2nmatrix Ais equivalent to a unique block matrix of the form Ir0 00/C20/C21 , where Iris the r/C2ridentity matrix. The proof is constructive, in the form of an algorithm. Step 1. Row reduce Ato row canonical form, with leading nonzero entries a1j1,a2j2;...;arjr. Step 2. Interchange C1andC1j1, interchange C2andC2j2;..., and interchange CrandCjr. This gives a matrix in the formIrB 00/C20/C21 , with leading nonzero entries a11;a22;...;arr. Step 3. Use column operations, with the aiias pivots, to replace each entry in Bwith a zero; that is, for i¼1;2;...;randj¼rþ1,rþ2;...;n, apply the operation /C0bijCiþCj!Cj. The ﬁnal matrix has the desired formIr0 00/C20/C21 . Lu Factorization 3.39. Find the LU factorization of (a) A¼1/C035 2/C047 /C01/C0212 43 5;(b) B¼14/C03 281 /C05/C0972 43 5: (a) Reduce Ato triangular form by the following operations: ‘‘Replace R2by/C02R1þR2;’’ ‘‘Replace R3byR1þR3;’’ and then ‘‘Replace R3by5 2R2þR3’’ These operations yield the following, where the triangular form is U: A/C241/C035 02/C03 0/C0562 43 5/C241/C035 02/C03 00/C03 22 43 5¼U and L¼10 0 21 0 /C01/C05 212 43 5 The entries 2 ;/C01;/C05 2inLare the negatives of the multipliers /C02;1;5 2in the above row operations. (As a check, multiply LandUto verify A¼LU.)104 CHAPTER 3 Systems of Linear Equations (b) Reduce Bto triangular form by ﬁrst applying the operations ‘‘Replace R2by/C02R1þR2’’ and ‘‘Replace R3by 5R1þR3.’’ These operations yield B/C2414/C03 00 7 01 1/C082 43 5: Observe that the second diagonal entry is 0. Thus, Bcannot be brought into triangular form without row interchange operations. Accordingly, Bis not LU-factorable. (There does exist a PLU factorization of such a matrix B, where Pis a permutation matrix, but such a factorization lies beyond the scope of this text.) 3.40. Find the LDU factorization of the matrix Ain Problem 3.39. The A¼LDU factorization refers to the situation where Lis a lower triangular matrix with 1’s on the diagonal (as in the LUfactorization of A),Dis a diagonal matrix, and Uis an upper triangular matrix with 1’s on the diagonal. Thus, simply factor out the diagonal entries in the matrix Uin the above LUfactorization of A to obtain DandL. That is, L¼10 0 21 0 /C01/C05 212 43 5; D¼10 0 02 0 00/C03 22 43 5; U¼1/C035 01/C03 0012 43 5 3.41. Find the LUfactorization of the matrix A¼12 1 23 3 /C03/C010 22 43 5. Reduce Ato triangular form by the following operations: ð1Þ‘‘Replace R2by/C02R1þR2;’’ð2Þ‘‘Replace R3by 3R1þR3;’’ð3Þ‘‘Replace R3by/C04R2þR3’’ These operations yield the following, where the triangular form is U: A/C2412 1 0/C011 0/C0452 43 5/C2412 1 0/C011 00 12 43 5¼U and L¼100 210 /C03412 43 5 The entries 2 ;/C03;4i n Lare the negatives of the multipliers /C02;3;/C04 in the above row operations. (As a check, multiply LandUto verify A¼LU.) 3.42. LetAbe the matrix in Problem 3.41. Find X1;X2;X3, where Xiis the solution of AX¼Bifor (a) B1¼ð1;1;1Þ, (b) B2¼B1þX1, (c) B3¼B2þX2. (a) Find L/C01B1by applying the row operations (1), (2), and then (3) in Problem 3.41 to B1: B1¼1 1 12 43 5/C0/C0/C0/C0/C0!ð1Þandð2Þ1 /C01 42 43 5/C0/C0/C0/C0/C0!ð3Þ1 /C01 82 43 5 Solve UX¼BforB¼ð1;/C01;8Þby back-substitution to obtain X1¼ð/C0 25;9;8Þ. (b) First ﬁnd B2¼B1þX1¼ð1;1;1Þþð/C0 25;9;8Þ¼ð/C0 24;10;9Þ. Then as above B2¼½/C0 24;10;9/C138T/C0/C0/C0/C0/C0!ð1Þandð2Þ½/C024;58;/C063/C138T/C0/C0/C0/C0/C0!ð3Þ½/C024;58;/C0295/C138T Solve UX¼BforB¼ð/C0 24;58;/C0295Þby back-substitution to obtain X2¼ð943;/C0353;/C0295Þ. (c) First ﬁnd B3¼B2þX2¼ð/C0 24;10;9Þþð 943;/C0353;/C0295Þ¼ð 919;/C0343;/C0286Þ. Then, as above B3¼½943;/C0353;/C0295/C138T/C0/C0/C0/C0/C0!ð1Þandð2Þ½919;/C02181 ;2671/C138T/C0/C0/C0/C0/C0!ð3Þ½919;/C02181 ;11 395/C138T Solve UX¼BforB¼ð919;/C02181 ;11 395Þby back-substitution to obtain X3¼ð/C0 37 628 ;13 576 ;11 395Þ.CHAPTER 3 Systems of Linear Equations 105 Miscellaneous Problems 3.43. LetLbe a linear combination of the mequations in nunknowns in the system (3.2). Say Lis the equation ðc1a11þ/C1/C1/C1þ cmam1Þx1þ/C1/C1/C1þð c1a1nþ/C1/C1/C1þ cmamnÞxn¼c1b1þ/C1/C1/C1þ cmbmð1Þ Show that any solution of the system (3.2) is also a solution of L. Letu¼ðk1;...;knÞbe a solution of (3.2). Then ai1k1þai2k2þ/C1/C1/C1þ ainkn¼biði¼1;2;...;mÞð 2Þ Substituting uin the left-hand side of (1) and using (2), we get ðc1a11þ/C1/C1/C1þ cmam1Þk1þ/C1/C1/C1þð c1a1nþ/C1/C1/C1þ cmamnÞkn ¼c1ða11k1þ/C1/C1/C1þ a1nknÞþ/C1/C1/C1þ cmðam1k1þ/C1/C1/C1þ amnknÞ ¼c1b1þ/C1/C1/C1þ cmbm This is the right-hand side of (1); hence, uis a solution of (1). 3.44. Suppose a system mof linear equations is obtained from a system lby applying an elementary operation (page 64). Show that mandlhave the same solutions. Each equation Linmis a linear combination of equations in l. Hence, by Problem 3.43, any solution oflwill also be a solution of m. On the other hand, each elementary operation has an inverse elementary operation, so lcan be obtained from mby an elementary operation. This means that any solution of mis a solution of l. Thus,landmhave the same solutions. 3.45. Prove Theorem 3.4: Suppose a system mof linear equations is obtained from a system lby a sequence of elementary operations. Then mandlhave the same solutions. Each step of the sequence does not change the solution set (Problem 3.44). Thus, the original system l and the ﬁnal system m(and any system in between) have the same solutions. 3.46. A system lof linear equations is said to be consistent if no linear combination of its equations is a degenerate equation Lwith a nonzero constant. Show that lis consistent if and only if lis reducible to echelon form. Supposelis reducible to echelon form. Then lhas a solution, which must also be a solution of every linear combination of its equations. Thus, L, which has no solution, cannot be a linear combination of the equations in l. Thus,lis consistent. On the other hand, suppose lis not reducible to echelon form. Then, in the reduction process, it must yield a degenerate equation Lwith a nonzero constant, which is a linear combination of the equations in l. Therefore, lis not consistent; that is, lis inconsistent. 3.47. Suppose uandvare distinct vectors. Show that, for distinct scalars k, the vectors uþkðu/C0vÞare distinct. Suppose uþk1ðu/C0vÞ¼uþk2ðu/C0vÞ:We need only show that k1¼k2. We have k1ðu/C0vÞ¼k2ðu/C0vÞ; and soðk1/C0k2Þðu/C0vÞ¼0 Because uand vare distinct, u/C0v6¼0. Hence, k1/C0k2¼0, and so k1¼k2. 3.48. Suppose ABis deﬁned. Prove (a) Suppose Ahas a zero row. Then ABhas a zero row. (b) Suppose Bhas a zero column. Then ABhas a zero column.106 CHAPTER 3 Systems of Linear Equations (a) Let Ribe the zero row of A, and C1;...;Cnthe columns of B. Then the ith row of ABis ðRiC1;RiC2;...;RiCnÞ¼ð 0;0;0;...;0Þ (b)BThas a zero row, and so BTAT¼ðABÞThas a zero row. Hence, ABhas a zero column. SUPPLEMENTARY PROBLEMS Linear Equations, 2 /C22 Systems 3.49. Determine whether each of the following systems is linear: (a) 3 x/C04yþ2yz¼8, (b) exþ3y¼p, (c) 2 x/C03yþkz¼4 3.50. Solve (a) px¼2, (b) 3 xþ2¼5xþ7/C02x, (c) 6 xþ2/C04x¼5þ2x/C03 3.51. Solve each of the following systems: (a) 2 xþ3y¼1 5xþ7y¼3(b) 4 x/C02y¼5 /C06xþ3y¼1(c) 2 x/C04¼3y 5y/C0x¼5(d) 2 x/C04y¼10 3x/C06y¼15 3.52. Consider each of the following systems in unknowns xandy: (a) x/C0ay¼1 ax/C04y¼b(b) axþ3y¼2 12xþay¼b(c) xþay¼3 2xþ5y¼b For which values of adoes each system have a unique solution, and for which pairs of values ða;bÞdoes each system have more than one solution? General Systems of Linear Equations 3.53. Solve (a) xþyþ2z¼4 2xþ3yþ6z¼10 3xþ6yþ10z¼17(b) x/C02yþ3z¼2 2x/C03yþ8z¼7 3x/C04yþ13z¼8(c) xþ2yþ3z¼3 2xþ3yþ8z¼4 5xþ8yþ19z¼11 3.54. Solve (a) x/C02y¼5 2xþ3y¼3 3xþ2y¼7(b) xþ2y/C03zþ2t¼2 2xþ5y/C08zþ6t¼5 3xþ4y/C05zþ2t¼4(c) xþ2yþ4z/C05t¼3 3x/C0yþ5zþ2t¼4 5x/C04yþ6zþ9t¼2 3.55. Solve (a) 2 x/C0y/C04z¼2 4x/C02y/C06z¼5 6x/C03y/C08z¼8(b) xþ2y/C0zþ3t¼3 2xþ4yþ4zþ3t¼9 3xþ6y/C0zþ8t¼10 3.56. Consider each of the following systems in unknowns x;y;z: (a) x/C02y¼1 x/C0yþaz¼2 ayþ9z¼b(b) xþ2yþ2z¼1 xþayþ3z¼3 xþ11yþaz¼b(c) xþyþaz¼1 xþayþz¼4 axþyþz¼b For which values of adoes the system have a unique solution, and for which pairs of values ða;bÞdoes the system have more than one solution? The value of bdoes not have any effect on whether the system has a unique solution. Why?CHAPTER 3 Systems of Linear Equations 107 Linear Combinations, Homogeneous Systems 3.57. Write vas a linear combination of u1;u2;u3, where (a) v¼ð4;/C09;2Þ,u1¼ð1;2;/C01Þ,u2¼ð1;4;2Þ,u3¼ð1;/C03;2Þ; (b) v¼ð1;3;2Þ,u1¼ð1;2;1Þ,u2¼ð2;6;5Þ,u3¼ð1;7;8Þ; (c) v¼ð1;4;6Þ,u1¼ð1;1;2Þ,u2¼ð2;3;5Þ,u3¼ð3;5;8Þ. 3.58. Letu1¼ð1;1;2Þ,u2¼ð1;3;/C02Þ,u3¼ð4;/C02;/C01ÞinR3. Show that u1;u2;u3are orthogonal, and write v as a linear combination of u1;u2;u3, where (a) v¼ð5;/C05;9Þ, (b) v¼ð1;/C03;3Þ, (c) v¼ð1;1;1Þ. (Hint: Use Fourier coefﬁcients.) 3.59. Find the dimension and a basis of the general solution Wof each of the following homogeneous systems: (a) x/C0yþ2z¼0 2xþyþz¼0 5xþyþ4z¼0(b) xþ2y/C03z¼0 2xþ5yþ2z¼0 3x/C0y/C04z¼0(c) xþ2yþ3zþt¼0 2xþ4yþ7zþ4t¼0 3xþ6yþ10zþ5t¼0 3.60. Find the dimension and a basis of the general solution Wof each of the following systems: (a) x1þ3x2þ2x3/C0x4/C0x5¼0 2x1þ6x2þ5x3þx4/C0x5¼0 5x1þ15x2þ12x3þx4/C03x5¼0(b) 2 x1/C04x2þ3x3/C0x4þ2x5¼0 3x1/C06x2þ5x3/C02x4þ4x5¼0 5x1/C010x2þ7x3/C03x4þ18x5¼0 Echelon Matrices, Row Canonical Form 3.61. Reduce each of the following matrices to echelon form and then to row canonical form: (a)11 2 24 9 151 22 43 5; (b)12/C012 1 24 1/C025 36 3/C0772 43 5; (c)242/C025 1 3 6 220 4 482 6/C0572 43 5 3.62. Reduce each of the following matrices to echelon form and then to row canonical form: (a)1212 1 2 2435 5 7 36491 01 1 1243 6 92 6643 775; (b)012 3 0381 2004 6 0271 02 6643 775; (c)13 13 28 5 1 017 7 1 1 31 171 52 6643 775 3.63. Using only 0’s and 1’s, list all possible 2 /C22 matrices in row canonical form. 3.64. Using only 0’s and 1’s, ﬁnd the number nof possible 3/C23 matrices in row canonical form. Elementary Matrices, Applications 3.65. Lete1;e2;e3denote, respectively, the following elementary row operations: ‘‘Interchange R2andR3;’’ ‘‘Replace R2by 3R2;’’ ‘‘Replace R1by 2R3þR1’’ (a) Find the corresponding elementary matrices E1;E2;E3. (b) Find the inverse operations e/C01 1,e/C01 2,e/C01 3; their corresponding elementary matrices E0 1,E0 2,E0 3; and the relationship between them and E1;E2;E3. (c) Describe the corresponding elementary column operations f1;f2;f3. (d) Find elementary matrices F1;F2;F3corresponding to f1;f2;f3, and the relationship between them and E1;E2;E3.108 CHAPTER 3 Systems of Linear Equations 3.66. Express each of the following matrices as a product of elementary matrices: A¼12 34/C20/C21 ; B¼3/C06 /C024/C20/C21 ; C¼26 /C03/C07/C20/C21 ; D¼120 0133872 43 5 3.67. Find the inverse of each of the following matrices (if it exists): A¼1/C02/C01 2/C031 3/C0442 43 5; B¼12 3 26 1 31 0/C012 43 5; C¼13/C02 28/C03 17 12 43 5; D¼21/C01 52/C03 02 12 43 5 3.68. Find the inverse of each of the following n/C2nmatrices: (a) Ahas 1’s on the diagonal and superdiagonal (entries directly above the diagonal) and 0’s elsewhere. (b) Bhas 1’s on and above the diagonal, and 0’s below the diagonal. Lu Factorization 3.69. Find the LUfactorization of each of the following matrices: (a)1/C01/C01 3/C04/C02 2/C03/C022 43 5, (b)13/C01 25 1 34 22 43 5, (c)236 479 3542 43 5, (d)12 3 24 7 371 02 43 5 3.70. LetAbe the matrix in Problem 3.69(a). Find X1;X2;X3;X4, where (a) X1is the solution of AX¼B1, where B1¼ð1;1;1ÞT. (b) For k>1,Xkis the solution of AX¼Bk, where Bk¼Bk/C01þXk/C01. 3.71. LetBbe the matrix in Problem 3.69(b). Find the LDU factorization of B. Miscellaneous Problems 3.72. Consider the following systems in unknowns xandy: ðaÞaxþby¼1 cxþdy¼0ðbÞaxþby¼0 cxþdy¼1 Suppose D¼ad/C0bc6¼0. Show that each system has the unique solution: (a) x¼d=D,y¼/C0c=D, (b) x¼/C0b=D,y¼a=D. 3.73. Find the inverse of the row operation ‘‘Replace RibykRjþk0Riðk06¼0Þ.’’ 3.74. Prove that deleting the last column of an echelon form (respectively, the row canonical form) of an augmented matrix M¼½A;B/C138yields an echelon form (respectively, the row canonical form) of A. 3.75. Letebe an elementary row operation and Eits elementary matrix, and let fbe the corresponding elementary column operation and Fits elementary matrix. Prove (a) fðAÞ¼ð eðATÞÞT, (b) F¼ET, (c) fðAÞ¼AF. 3.76. Matrix Aisequivalent to matrix B, written A/C25B, if there exist nonsingular matrices PandQsuch that B¼PAQ . Prove that/C25is an equivalence relation; that is, (a) A/C25A, (b) If A/C25B, then B/C25A, (c) If A/C25BandB/C25C, then A/C25C.CHAPTER 3 Systems of Linear Equations 109 ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: A¼½R1;R2; .../C138denotes the matrix Awith rows R1;R2;.... The elements in each row are separated by commas (which may be omitted with single digits), the rows are separated by semicolons, and 0 denotes a zero row. For example, A¼½1;2;3;4;5;/C06;7;/C08;0/C138¼12 34 5/C067/C08 00 002 43 5 3.49. (a) no, (b) yes, (c) linear in x;y;z, not linear in x;y;z;k 3.50. (a) x¼2=p, (b) no solution, (c) every scalar kis a solution 3.51. (a)ð2;/C01Þ, (b) no solution, (c) ð5;2Þ, (d)ð5/C02a;aÞ 3.52. (a) a6¼/C62;ð2;2Þ;ð/C02;/C02Þ, (b) a6¼/C66;ð6;4Þ;ð/C06;/C04Þ, (c) a6¼5 2;ð5 2;6Þ 3.53. (a)ð2;1;1 2Þ, (b) no solution, (c) u¼ð/C0 7a/C01;2aþ2;aÞ. 3.54. (a)ð3;/C01Þ, (b) u¼ð/C0 aþ2b;1þ2a/C02b;a;bÞ, (c) no solution 3.55. (a) u¼ð1 2aþ2;a;1 2Þ, (b) u¼ð1 2ð7/C05b/C04aÞ;a;1 2ð1þbÞ;bÞ 3.56. (a) a6¼/C63;ð3;3Þ;ð/C03;/C03Þ, (b) a6¼5 and a6¼/C01;ð5;7Þ;ð/C01;/C05Þ, (c) a6¼1 and a6¼/C02;ð/C02;5Þ 3.57. (a) 2 ;/C01;3, (b) 6 ;/C03;1, (c) not possible 3.58. (a) 3 ;/C02;1, (b)2 3;/C01;1 3, (c)23;17;1 21 3.59. (a) dim W¼1;u1¼ð/C0 1;1;1Þ, (b) dim W¼0, no basis, (c) dim W¼2;u1¼ð/C0 2;1;0;0Þ;u2¼ð5;0;/C02;1Þ 3.60. (a) dim W¼3;u1¼ð/C0 3;1;0;0;0Þ,u2¼ð7;0;/C03;1;0Þ,u3¼ð3;0;/C01;0;1Þ, (b) dim W¼2,u1¼ð2;1;0;0;0Þ,u2¼ð5;0;/C05;/C03;1Þ 3.61. (a)½1;0;/C01 2;0;1;5 2;0/C138, (b)½1;2;0;0;2;0;0;1;0;5;0;0;0;1;2/C138, (c)½1;2;0;4;/C05;3;0;0;1;/C05;15 2;/C05 2;0/C138 3.62. (a)½1;2;0;0;/C04;/C02;0;0;1;0;1;2;0;0;0;1;2;1;0/C138, (b)½0;1;0;0;0;0;1;0;0;0;0;1;0/C138, (c)½1;0;0;4;0;1;0;/C01;0;0;1;2;0/C138 3.63. 5:½1;0;0;1/C138,½1;1;0;0/C138,½1;0;0;0/C138,½0;1;0;0/C138;0 3.64. 16 3.65. (a)½1;0;0;0;0;1;0;1;0/C138,½1;0;0;0;3;0;0;0;1/C138,½1;0;2;0;1;0;0;0;1/C138, (b) R2$R3;1 3R2!R2;/C02R3þR1!R1; each E0 i¼E/C01 i, (c) C2$C3;3C2!C2;2C3þC1!C1, (d) each Fi¼ET i. 3.66. A¼½1;0;3;1/C138½1;0;0;/C02/C138½1;2;0;1/C138, Bis not invertible, C¼½1;0;/C03 2;1/C138½1;0;0;2/C138½1;6;0;1/C138½2;0;0;1/C138, D¼½100 ;010 ;301/C138½100 ;010 ;021/C138½100 ;013 ;001/C138½120 ;010 ;001/C138 3.67. A/C01¼½/C0 8;12;/C05;/C05;7;/C03;1;/C02;1/C138, Bhas no inverse, C/C01¼½29 2;/C017 2;72;/C05 2;32;/C01 2;3;/C02;1/C138; D/C01¼½8;/C03;/C01;/C05;2;1;10;/C04;/C01/C138110 CHAPTER 3 Systems of Linear Equations 3.68. A/C01¼½1;/C01;1;/C01;...; 0;1;/C01;1;/C01;...; 0;0;1;/C01;1;/C01;1;...; ...; ...; 0;...0;1/C138 B/C01has 1’s on diagonal, /C01’s on superdiagonal, and 0’s elsewhere. 3.69. (a)½100 ;310 ;211/C138½1;/C01;/C01;0;/C01;1;0;0;/C01/C138, (b)½100 ;210 ;351/C138½1;3;/C01;0;/C01;3;0;0;/C010/C138, (c)½100 ;210 ;3 2;12;1/C138½2;3;6;0;1;/C03;0;0;/C07 2/C138, (d) There is no LUdecomposition. 3.70. X1¼½1;1;/C01/C138T;B2¼½2;2;0/C138T,X2¼½6;4;0/C138T,B3¼½8;6;0/C138T,X3¼½22;16;/C02/C138T, B4¼½30;22;/C02/C138T,X4¼½86;62;/C06/C138T 3.71. B¼½100 ;210 ;351/C138diagð1;/C01;/C010Þ½1;3;/C01;0;1;3;0;0;1/C138 3.73. Replace Riby/C0kRjþð1=k0ÞRi. 3.75. (c) fðAÞ¼ð eðATÞÞT¼ðEATÞT¼ðATÞTET¼AF 3.76. (a) A¼IAI:(b) If A¼PBQ , then B¼P/C01AQ/C01. (c) If A¼PBQ andB¼P0CQ0, then A¼ðPP0ÞCðQ0QÞ.CHAPTER 3 Systems of Linear Equations 111 Vector Spaces 4.1 Introduction This chapter introduces the underlying structure of linear algebra, that of a ﬁnite-dimensional vector space. The deﬁnition of a vector space V, whose elements are called vectors , involves an arbitrary ﬁeld K, whose elements are called scalars . The following notation will be used (unless otherwise stated or implied): V the given vector space u;v;w vectors in V K the given number field a;b;c;ork scalars in K Almost nothing essential is lost if the reader assumes that Kis the real ﬁeld Ror the complex ﬁeld C. The reader might suspect that the real line Rhas ‘‘dimension’’ one, the cartesian plane R2has ‘‘dimension’’ two, and the space R3has ‘‘dimension’’ three. This chapter formalizes the notion of ‘‘dimension,’’ and this deﬁnition will agree with the reader’s intuition. Throughout this text, we will use the following set notation: a2A Element abelongs to set A a;b2A Elements aandbbelong to A 8x2A For every xinA 9x2A There exists an xinA A/C18BA is a subset of B A\B Intersection of AandB A[B Union of AandB ; Empty set 4.2 Vector Spaces The following deﬁnes the notion of a vector space Vwhere Kis the ﬁeld of scalars. DEFINITION: LetVbe a nonempty set with two operations: (i) Vector Addition: This assigns to any u;v2Vasum uþvinV. (ii) Scalar Multiplication: This assigns to any u2V,k2Kaproduct ku2V. Then Vis called a vector space (over the ﬁeld K) if the following axioms hold for any vectors u;v;w2V: 112 CHAPTER 4 [A1]ðuþvÞþw¼uþðvþwÞ [A2] There is a vector in V, denoted by 0 and called the zero vector , such that, for any u2V; uþ0¼0þu¼u [A3] For each u2V;there is a vector in V, denoted by/C0u, and called the negative ofu, such that uþð/C0 uÞ¼ð/C0 uÞþu¼0. [A4]uþv¼vþu. [M1]kðuþvÞ¼kuþkv, for any scalar k2K: [M2]ðaþbÞu¼auþbu;for any scalars a;b2K. [M3]ðabÞu¼aðbuÞ;for any scalars a;b2K. [M4]1u¼u, for the unit scalar 1 2K. The above axioms naturally split into two sets (as indicated by the labeling of the axioms). The ﬁrst four are concerned only with the additive structure of Vand can be summarized by saying Vis a commutative group under addition. This means (a) Any sum v1þv2þ/C1/C1/C1þ vmof vectors requires no parentheses and does not depend on the order of the summands. (b) The zero vector 0 is unique, and the negative /C0uof a vector uis unique. (c) (Cancellation Law) If uþw¼vþw, then u¼v. Also, subtraction inVis deﬁned by u/C0v¼uþð/C0 vÞ, where/C0vis the unique negative of v. On the other hand, the remaining four axioms are concerned with the ‘‘action’’ of the ﬁeld Kof scalars on the vector space V. Using these additional axioms, we prove (Problem 4.2) the following simple properties of a vector space. THEOREM 4.1: LetVbe a vector space over a ﬁeld K. (i) For any scalar k2Kand 02V;k0¼0. (ii) For 02Kand any vector u2V;0u¼0. (iii) If ku¼0, where k2Kandu2V, then k¼0o r u¼0. (iv) For any k2Kand any u2V;ð/C0kÞu¼kð/C0uÞ¼/C0 ku. 4.3 Examples of Vector Spaces This section lists important examples of vector spaces that will be used throughout the text. Space Kn LetKbe an arbitrary ﬁeld. The notation Knis frequently used to denote the set of all n-tuples of elements inK. Here Knis a vector space over Kusing the following operations: (i)Vector Addition:ða1;a2;...;anÞþð b1;b2;...;bnÞ¼ð a1þb1;a2þb2;...;anþbnÞ (ii)Scalar Multiplication: kða1;a2;...;anÞ¼ð ka1;ka2;...;kanÞ The zero vector in Knis the n-tuple of zeros, 0¼ð0;0;...;0Þ and the negative of a vector is deﬁned by /C0ða1;a2;...;anÞ¼ð/C0 a1;/C0a2;...;/C0anÞ Observe that these are the same as the operations deﬁned for Rnin Chapter 1. The proof that Knis a vector space is identical to the proof of Theorem 1.1, which we now regard as stating that Rnwith the operations deﬁned there is a vector space over R.CHAPTER 4 Vector Spaces 113 Polynomial Space PðtÞ LetPðtÞdenote the set of all polynomials of the form pðtÞ¼a0þa1tþa2t2þ/C1/C1/C1þ astsðs¼1;2;...Þ where the coefﬁcients aibelong to a ﬁeld K.T h e n PðtÞis a vector space over Kusing the following operations: (i)Vector Addition: Here pðtÞþqðtÞinPðtÞis the usual operation of addition of polynomials. (ii)Scalar Multiplication: Here kpðtÞinPðtÞis the usual operation of the product of a scalar kand a polynomial pðtÞ. The zero polynomial 0 is the zero vector in PðtÞ. Polynomial Space PnðtÞ LetPnðtÞdenote the set of all polynomials pðtÞover a ﬁeld K, where the degree of pðtÞis less than or equal to n; that is, pðtÞ¼a0þa1tþa2t2þ/C1/C1/C1þ asts where s/C20n. Then PnðtÞis a vector space over Kwith respect to the usual operations of addition of polynomials and of multiplication of a polynomial by a constant (just like the vector space PðtÞabove). We include the zero polynomial 0 as an element of PnðtÞ, even though its degree is undeﬁned. Matrix Space Mm;n The notation Mm;n, or simply M;will be used to denote the set of all m/C2nmatrices with entries in a ﬁeld K. Then Mm;nis a vector space over Kwith respect to the usual operations of matrix addition and scalar multiplication of matrices, as indicated by Theorem 2.1. Function Space FðXÞ LetXbe a nonempty set and let Kbe an arbitrary ﬁeld. Let FðXÞdenote the set of all functions of Xinto K. [Note that FðXÞis nonempty, because Xis nonempty.] Then FðXÞis a vector space over Kwith respect to the following operations: (i)Vector Addition: The sum of two functions fandginFðXÞis the function fþginFðXÞdeﬁned by ðfþgÞðxÞ¼fðxÞþgðxÞ8 x2X (ii) Scalar Multiplication: The product of a scalar k2Kand a function finFðXÞis the function kfin FðXÞdeﬁned by ðkfÞðxÞ¼kfðxÞ8 x2X The zero vector in FðXÞis the zero function 0, which maps every x2Xinto the zero element 0 2K; 0ðxÞ¼08x2X Also, for any function finFðXÞ, negative of fis the function/C0finFðXÞdeﬁned by ð/C0fÞðxÞ¼/C0 fðxÞ8 x2X Fields and Subﬁelds Suppose a ﬁeld Eis an extension of a ﬁeld K; that is, suppose Eis a ﬁeld that contains Kas a subﬁeld. Then Emay be viewed as a vector space over Kusing the following operations: (i)Vector Addition: Here uþvinEis the usual addition in E. (ii)Scalar Multiplication: Here kuinE, where k2Kandu2E, is the usual product of kanduas elements of E. That is, the eight axioms of a vector space are satisﬁed by Eand its subﬁeld Kwith respect to the above two operations.114 CHAPTER 4 Vector Spaces 4.4 Linear Combinations, Spanning Sets LetVbe a vector space over a ﬁeld K. A vector vinVis alinear combination of vectors u1;u2;...;umin Vif there exist scalars a1;a2;...;aminKsuch that v¼a1u1þa2u2þ/C1/C1/C1þ amum Alternatively, vis a linear combination of u1;u2;...;umif there is a solution to the vector equation v¼x1u1þx2u2þ/C1/C1/C1þ xmum where x1;x2;...;xmare unknown scalars. EXAMPLE 4.1 (Linear Combinations in Rn) Suppose we want to express v¼ð3;7;/C04ÞinR3as a linear combination of the vectors u1¼ð1;2;3Þ; u2¼ð2;3;7Þ; u3¼ð3;5;6Þ We seek scalars x,y,zsuch that v¼xu1þyu2þzu3; that is, 3 3 /C042 43 5¼x1 2 32 43 5þy2 3 72 43 5þz3 5 62 43 5 orxþ2yþ3z¼3 2xþ3yþ5z¼7 3xþ7yþ6z¼/C04 (For notational convenience, we have written the vectors in R3as columns, because it is then easier to ﬁnd the equivalent system of linear equations.) Reducing the system to echelon form yields xþ2yþ3z¼ 3 /C0y/C0z¼ 1 y/C03z¼/C013and thenxþ2yþ3z¼ 3 /C0y/C0z¼ 1 /C04z¼/C012 Back-substitution yields the solution x¼2,y¼/C04,z¼3. Thus, v¼2u1/C04u2þ3u3. Remark: Generally speaking, the question of expressing a given vector vinKnas a linear combination of vectors u1;u2;...;uminKnis equivalent to solving a system AX¼Bof linear equations, where vis the column Bof constants, and the u’s are the columns of the coefﬁcient matrix A. Such a system may have a unique solution (as above), many solutions, or no solution. The last case—nosolution—means that vcannot be written as a linear combination of the u’s. EXAMPLE 4.2 (Linear combinations in PðtÞ) Suppose we want to express the polynomial v¼3t2þ5t/C05a sa linear combination of the polynomials p1¼t2þ2tþ1; p2¼2t2þ5tþ4; p3¼t2þ3tþ6 We seek scalars x,y,zsuch that v¼xp1þyp2þzp3; that is, 3t2þ5t/C05¼xðt2þ2tþ1Þþyð2t2þ5tþ4Þþzðt2þ3tþ6Þð *Þ There are two ways to proceed from here. (1) Expand the right-hand side of (*) obtaining: 3t2þ5t/C05¼xt2þ2xtþxþ2yt2þ5ytþ4yþzt2þ3ztþ6z ¼ðxþ2yþzÞt2þð2xþ5yþ3zÞtþðxþ4yþ6zÞ Set coefﬁcients of the same powers of tequal to each other, and reduce the system to echelon form: xþ2yþz¼3 2xþ5yþ3z¼5 xþ4yþ6z¼/C05orxþ2yþz¼3 yþz¼/C01 2yþ5z¼/C08orxþ2yþz¼3 yþz¼/C01 3z¼/C06CHAPTER 4 Vector Spaces 115 The system is in triangular form and has a solution. Back-substitution yields the solution x¼3,y¼1,z¼/C02. Thus, v¼3p1þp2/C02p3 (2) The equation (*) is actually an identity in the variable t; that is, the equation holds for any value oft. We can obtain three equations in the unknowns x,y,zby setting tequal to any three values. For example, Sett¼0i nð1Þto obtain : xþ4yþ6z¼/C05 Sett¼1i nð1Þto obtain : 4xþ11yþ10z¼3 Sett¼/C01i nð1Þto obtain : yþ4z¼/C07 Reducing this system to echelon form and solving by back-substitution again yields the solution x¼3,y¼1, z¼/C02. Thus (again), v¼3p1þp2/C02p3. Spanning Sets LetVbe a vector space over K. Vectors u1;u2;...;uminVare said to span V or to form a spanning set of Vif every vinVis a linear combination of the vectors u1;u2;...;um—that is, if there exist scalars a1;a2;...;aminKsuch that v¼a1u1þa2u2þ/C1/C1/C1þ amum The following remarks follow directly from the deﬁnition. Remark 1: Suppose u1;u2;...;umspan V. Then, for any vector w, the set w;u1;u2;...;umalso spans V. Remark 2: Suppose u1;u2;...;umspan Vand suppose ukis a linear combination of some of the other u’s. Then the u’s without ukalso span V. Remark 3: Suppose u1;u2;...;umspan Vand suppose one of the u’s is the zero vector. Then the u’s without the zero vector also span V. EXAMPLE 4.3 Consider the vector space V¼R3. (a) We claim that the following vectors form a spanning set of R3: e1¼ð1;0;0Þ; e2¼ð0;1;0Þ; e3¼ð0;0;1Þ Speciﬁcally, if v¼ða;b;cÞis any vector in R3, then v¼ae1þbe2þce3 For example, v¼ð5;/C06;2Þ¼/C0 5e1/C06e2þ2e3. (b) We claim that the following vectors also form a spanning set of R3: w1¼ð1;1;1Þ; w2¼ð1;1;0Þ; w3¼ð1;0;0Þ Speciﬁcally, if v¼ða;b;cÞis any vector in R3, then (Problem 4.62) v¼ða;b;cÞ¼cw1þðb/C0cÞw2þða/C0bÞw3 For example, v¼ð5;/C06;2Þ¼2w1/C08w2þ11w3. (c) One can show (Problem 3.24) that v¼ð2;7;8Þcannot be written as a linear combination of the vectors u1¼ð1;2;3Þ; u2¼ð1;3;5Þ; u3¼ð1;5;9Þ Accordingly, u1,u2,u3do not span R3.116 CHAPTER 4 Vector Spaces EXAMPLE 4.4 Consider the vector space V¼PnðtÞconsisting of all polynomials of degree /C20n. (a) Clearly every polynomial in PnðtÞcan be expressed as a linear combination of the nþ1 polynomials 1;t;t2;t3; ...;tn Thus, these powers of t(where 1¼t0) form a spanning set for PnðtÞ. (b) One can also show that, for any scalar c, the following nþ1 powers of t/C0c, 1;t/C0c;ðt/C0cÞ2;ðt/C0cÞ3; ...;ðt/C0cÞn (whereðt/C0cÞ0¼1), also form a spanning set for PnðtÞ. EXAMPLE 4.5 Consider the vector space M¼M2;2consisting of all 2 /C22 matrices, and consider the following four matrices in M: E11¼10 00/C20/C21 ; E12¼01 00/C20/C21 ; E21¼00 10/C20/C21 ; E22¼00 01/C20/C21 Then clearly any matrix AinMcan be written as a linear combination of the four matrices. For example, A¼5/C06 78/C20/C21 ¼5E11/C06E12þ7E21þ8E22 Accordingly, the four matrices E11,E12,E21,E22span M. 4.5 Subspaces This section introduces the important notion of a subspace. DEFINITION: LetVbe a vector space over a ﬁeld Kand let Wbe a subset of V. Then Wis asubspace ofVifWis itself a vector space over Kwith respect to the operations of vector addition and scalar multiplication on V. The way in which one shows that any set Wis a vector space is to show that Wsatisﬁes the eight axioms of a vector space. However, if Wis a subset of a vector space V, then some of the axioms automatically hold in W, because they already hold in V. Simple criteria for identifying subspaces follow. THEOREM 4.2: Suppose Wis a subset of a vector space V. Then Wis a subspace of Vif the following two conditions hold: (a) The zero vector 0 belongs to W. (b) For every u;v2W;k2K: (i) The sum uþv2W. (ii) The multiple ku2W. Property (i) in (b) states that Wisclosed under vector addition , and property (ii) in (b) states that Wis closed under scalar multiplication . Both properties may be combined into the following equivalent single statement: (b0) For every u;v2W;a;b2K, the linear combination auþbv2W. Now let Vbe any vector space. Then Vautomatically contains two subspaces: the set {0} consisting of the zero vector alone and the whole space Vitself. These are sometimes called the trivial subspaces of V. Examples of nontrivial subspaces follow. EXAMPLE 4.6 Consider the vector space V¼R3. (a) Let Uconsist of all vectors in R3whose entries are equal; that is, U¼fða;b;cÞ:a¼b¼cg For example, (1, 1, 1), ( 73,73,73), (7, 7, 7), ( 72,72,72) are vectors in U. Geometrically, Uis the line through the origin Oand the point (1, 1, 1) as shown in Fig. 4-1(a). Clearly 0 ¼ð0;0;0Þbelongs to U, becauseCHAPTER 4 Vector Spaces 117 all entries in 0 are equal. Further, suppose uandvare arbitrary vectors in U, say, u¼ða;a;aÞandv¼ðb;b;bÞ. Then, for any scalar k2R, the following are also vectors in U: uþv¼ðaþb;aþb;aþbÞ and ku¼ðka;ka;kaÞ Thus, Uis a subspace of R3. (b) Let Wbe any plane in R3passing through the origin, as pictured in Fig. 4-1(b). Then 0 ¼ð0;0;0Þbelongs to W, because we assumed Wpasses through, the origin O. Further, suppose uand vare vectors in W. Then uand v may be viewed as arrows in the plane Wemanating from the origin O, as in Fig. 4-1(b). The sum uþvand any multiple kuofualso lie in the plane W. Thus, Wis a subspace of R3. EXAMPLE 4.7 (a) Let V¼Mn;n, the vector space of n/C2nmatrices. Let W1be the subset of all (upper) triangular matrices and let W2be the subset of all symmetric matrices. Then W1is a subspace of V, because W1contains the zero matrix 0 andW1is closed under matrix addition and scalar multiplication; that is, the sum and scalar multiple of such triangular matrices are also triangular. Similarly, W2is a subspace of V. (b) Let V¼PðtÞ, the vector space PðtÞof polynomials. Then the space PnðtÞof polynomials of degree at most n may be viewed as a subspace of PðtÞ. Let QðtÞbe the collection of polynomials with only even powers of t. For example, the following are polynomials in QðtÞ: p1¼3þ4t2/C05t6and p2¼6/C07t4þ9t6þ3t12 (We assume that any constant k¼kt0is an even power of t.) Then QðtÞis a subspace of PðtÞ. (c) Let Vbe the vector space of real-valued functions. Then the collection W1of continuous functions and the collection W2of differentiable functions are subspaces of V. Intersection of Subspaces LetUandWbe subspaces of a vector space V. We show that the intersection U\Wis also a subspace of V. Clearly, 02Uand 02W, because UandWare subspaces; whence 0 2U\W. Now suppose uandv belong to the intersection U\W. Then u;v2Uandu;v2W. Further, because UandWare subspaces, for any scalars a;b2K, auþbv2U and auþbv2W Thus, auþbv2U\W. Therefore, U\Wis a subspace of V. The above result generalizes as follows. THEOREM 4.3: The intersection of any number of subspaces of a vector space Vis a subspace of V.Figure 4-1118 CHAPTER 4 Vector Spaces Solution Space of a Homogeneous System Consider a system AX¼Bof linear equations in nunknowns. Then every solution umay be viewed as a vector in Kn. Thus, the solution set of such a system is a subset of Kn. Now suppose the system is homogeneous; that is, suppose the system has the form AX¼0. Let Wbe its solution set. Because A0¼0, the zero vector 0 2W. Moreover, suppose uand vbelong to W. Then uand vare solutions of AX¼0, or, in other words, Au¼0 and Av¼0. Therefore, for any scalars aandb, we have AðauþbvÞ¼aAuþbAv¼a0þb0¼0þ0¼0 Thus, auþbvbelongs to W, because it is a solution of AX¼0. Accordingly, Wis a subspace of Kn. We state the above result formally. THEOREM 4.4: The solution set Wof a homogeneous system AX¼0innunknowns is a subspace ofKn. We emphasize that the solution set of a nonhomogeneous system AX¼Bis not a subspace of Kn.I n fact, the zero vector 0 does not belong to its solution set. 4.6 Linear Spans, Row Space of a Matrix Suppose u1;u2;...;umare any vectors in a vector space V. Recall (Section 4.4) that any vector of the form a1u1þa2u2þ/C1/C1/C1þ amum, where the aiare scalars, is called a linear combination ofu1;u2;...;um. The collection of all such linear combinations, denoted by spanðu1;u2;...;umÞ or spanðuiÞ is called the linear span ofu1;u2;...;um. Clearly the zero vector 0 belongs to span ðuiÞ, because 0¼0u1þ0u2þ/C1/C1/C1þ 0um Furthermore, suppose vand v0belong to spanðuiÞ, say, v¼a1u1þa2u2þ/C1/C1/C1þ amum and v0¼b1u1þb2u2þ/C1/C1/C1þ bmum Then, vþv0¼ða1þb1Þu1þða2þb2Þu2þ/C1/C1/C1þð amþbmÞum and, for any scalar k2K, kv¼ka1u1þka2u2þ/C1/C1/C1þ kamum Thus, vþv0andkvalso belong to span ðuiÞ. Accordingly, span ðuiÞis a subspace of V. More generally, for any subset SofV, spanðSÞconsists of all linear combinations of vectors in Sor, when S¼f, span( S)¼f0g. Thus, in particular, Sis a spanning set (Section 4.4) of span ðSÞ. The following theorem, which was partially proved above, holds. THEOREM 4.5: LetSbe a subset of a vector space V. (i) Then spanðSÞis a subspace of Vthat contains S. (ii) If Wis a subspace of Vcontaining S, then spanðSÞ/C18W. Condition (ii) in theorem 4.5 may be interpreted as saying that span ðSÞis the ‘‘smallest’’ subspace of Vcontaining S. EXAMPLE 4.8 Consider the vector space V¼R3. (a) Let ube any nonzero vector in R3. Then spanðuÞconsists of all scalar multiples of u. Geometrically, span ðuÞis the line through the origin Oand the endpoint of u, as shown in Fig. 4-2(a).CHAPTER 4 Vector Spaces 119 (b) Let uandvbe vectors in R3that are not multiples of each other. Then span ðu;vÞis the plane through the origin Oand the endpoints of uand vas shown in Fig. 4-2(b). (c) Consider the vectors e1¼ð1;0;0Þ,e2¼ð0;1;0Þ,e3¼ð0;0;1ÞinR3. Recall [Example 4.1(a)] that every vector inR3is a linear combination of e1,e2,e3. That is, e1,e2,e3form a spanning set of R3. Accordingly, spanðe1;e2;e3Þ¼R3. Row Space of a Matrix LetA¼½aij/C138be an arbitrary m/C2nmatrix over a ﬁeld K. The rows of A, R1¼ða11;a12;...;a1nÞ; R2¼ða21;a22;...;a2nÞ; ...; Rm¼ðam1;am2;...;amnÞ may be viewed as vectors in Kn; hence, they span a subspace of Kncalled the row space ofAand denoted by rowsp(A). That is, rowspðAÞ¼spanðR1;R2;...;RmÞ Analagously, the columns of Amay be viewed as vectors in Kmcalled the column space ofAand denoted by colsp(A). Observe that colsp ðAÞ¼rowspðATÞ. Recall that matrices AandBare row equivalent, written A/C24B,i fBcan be obtained from Aby a sequence of elementary row operations. Now suppose Mis the matrix obtained by applying one of the following elementary row operations on a matrix A: ð1ÞInterchange RiandRj;ð2ÞReplace RibykRi;ð3ÞReplace RjbykRiþRj Then each row of Mis a row of Aor a linear combination of rows of A. Hence, the row space of Mis contained in the row space of A. On the other hand, we can apply the inverse elementary row operation on Mto obtain A; hence, the row space of Ais contained in the row space of M. Accordingly, AandMhave the same row space. This will be true each time we apply an elementary row operation. Thus, we have proved the following theorem. THEOREM 4.6: Row equivalent matrices have the same row space. We are now able to prove (Problems 4.45–4.47) basic results on row equivalence (which ﬁrst appeared as Theorems 3.7 and 3.8 in Chapter 3). THEOREM 4.7: Suppose A¼½aij/C138andB¼½bij/C138are row equivalent echelon matrices with respective pivot entries a1j1;a2j2;...;arjrand b1k1;b2k2;...;bsks Then AandBhave the same number of nonzero rows—that is, r¼s—and their pivot entries are in the same positions—that is, j1¼k1;j2¼k2;...;jr¼kr. THEOREM 4.8: Suppose AandBare row canonical matrices. Then AandBhave the same row space if and only if they have the same nonzero rows.0 (a)u Figure 4-20 (b)u120 CHAPTER 4 Vector Spaces COROLLARY 4.9: Every matrix Ais row equivalent to a unique matrix in row canonical form. We apply the above results in the next example. EXAMPLE 4.9 Consider the following two sets of vectors in R4: u1¼ð1;2;/C01;3Þ; u2¼ð2;4;1;/C02Þ; u3¼ð3;6;3;/C07Þ w1¼ð1;2;/C04;11Þ; w2¼ð2;4;/C05;14Þ LetU¼spanðuiÞandW¼spanðwiÞ. There are two ways to show that U¼W. (a) Show that each uiis a linear combination of w1andw2, and show that each wiis a linear combination of u1,u2, u3. Observe that we have to show that six systems of linear equations are consistent. (b) Form the matrix Awhose rows are u1,u2,u3and row reduce Ato row canonical form, and form the matrix B whose rows are w1andw2and row reduce Bto row canonical form: A¼12/C013 24 1/C02 36 3/C072 643 75/C2412/C013 00 3/C08 00 6/C0162 643 75/C241201 3 001/C08 3 000 02 643 75 B¼12/C041 1 24/C051 4/C20/C21 /C2412/C041 1 00 3/C08/C20/C21 /C241201 3 001/C08 3"# Because the nonzero rows of the matrices in row canonical form are identical, the row spaces of AandBare equal. Therefore, U¼W. Clearly, the method in (b) is more efﬁcient than the method in (a). 4.7 Linear Dependence and Independence LetVbe a vector space over a ﬁeld K. The following deﬁnes the notion of linear dependence and independence of vectors over K. (One usually suppresses mentioning Kwhen the ﬁeld is understood.) This concept plays an essential role in the theory of linear algebra and in mathematics in general. DEFINITION: We say that the vectors v1;v2;...;vminVarelinearly dependent if there exist scalars a1;a2;...;aminK, not all of them 0, such that a1v1þa2v2þ/C1/C1/C1þ amvm¼0 Otherwise, we say that the vectors are linearly independent . The above deﬁnition may be restated as follows. Consider the vector equation x1v1þx2v2þ/C1/C1/C1þ xmvm¼0 ð*Þ where the x’s are unknown scalars. This equation always has the zero solution x1¼0; x2¼0;...;xm¼0. Suppose this is the only solution; that is, suppose we can show: x1v1þx2v2þ/C1/C1/C1þ xmvm¼0 implies x1¼0;x2¼0; ...;xm¼0 Then the vectors v1;v2;...;vmare linearly independent, On the other hand, suppose the equation (*) has a nonzero solution; then the vectors are linearly dependent. A set S¼fv1;v2;...;vmgof vectors in Vis linearly dependent or independent according to whether the vectors v1;v2;...;vmare linearly dependent or independent. An inﬁnite set Sof vectors is linearly dependent or independent according to whether there do or do not exist vectors v1;v2;...;vkinSthat are linearly dependent. Warning: The set S¼fv1;v2;...;vmgabove represents a listor, in other words, a ﬁnite sequence of vectors where the vectors are ordered and repetition is permitted.CHAPTER 4 Vector Spaces 121 The following remarks follow directly from the above deﬁnition. Remark 1: Suppose 0 is one of the vectors v1;v2;...;vm, say v1¼0. Then the vectors must be linearly dependent, because we have the following linear combination where the coefﬁcient of v16¼0: 1v1þ0v2þ/C1/C1/C1þ 0vm¼1/C10þ0þ/C1/C1/C1þ 0¼0 Remark 2: Suppose vis a nonzero vector. Then v, by itself, is linearly independent, because kv¼0; v6¼0 implies k¼0 Remark 3: Suppose two of the vectors v1;v2;...;vmare equal or one is a scalar multiple of the other, say v1¼kv2. Then the vectors must be linearly dependent, because we have the following linear combination where the coefﬁcient of v16¼0: v1/C0kv2þ0v3þ/C1/C1/C1þ 0vm¼0 Remark 4: Two vectors v1andv2are linearly dependent if and only if one of them is a multiple of the other. Remark 5: If the setfv1;...;vmgis linearly independent, then any rearrangement of the vectors fvi1;vi2;...;vimgis also linearly independent. Remark 6: If a set Sof vectors is linearly independent, then any subset of Sis linearly independent. Alternatively, if Scontains a linearly dependent subset, then Sis linearly dependent. EXAMPLE 4.10 (a) Let u¼ð1;1;0Þ,v¼ð1;3;2Þ,w¼ð4;9;5Þ. Then u,v,ware linearly dependent, because 3uþ5v/C02w¼3ð1;1;0Þþ5ð1;3;2Þ/C02ð4;9;5Þ¼ð 0;0;0Þ¼0 (b) We show that the vectors u¼ð1;2;3Þ,v¼ð2;5;7Þ,w¼ð1;3;5Þare linearly independent. We form the vector equation xuþyvþzw¼0, where x,y,zare unknown scalars. This yields x1 232 43 5þy2 572 43 5þz1 352 43 5¼0 002 43 5 orxþ2yþz¼0 2xþ5yþ3z¼0 3xþ7yþ5z¼0orxþ2yþz¼0 yþz¼0 2z¼0 Back-substitution yields x¼0,y¼0,z¼0. We have shown that xuþyvþzw¼0 implies x¼0;y¼0;z¼0 Accordingly, u,v,ware linearly independent. (c) Let Vbe the vector space of functions from RintoR. We show that the functions fðtÞ¼sint,gðtÞ¼et, hðtÞ¼t2are linearly independent. We form the vector (function) equation xfþygþzh¼0, where x,y,zare unknown scalars. This function equation means that, for every value of t, xsintþyetþzt2¼0 Thus, in this equation, we choose appropriate values of tto easily get x¼0,y¼0,z¼0. For example, ðiÞSubstitute t¼0 ðiiÞSubstitute t¼p ðiiiÞSubstitute t¼p=2to obtain xð0Þþyð1Þþzð0Þ¼0 to obtain xð0Þþ0ðepÞþzðp2Þ¼0 to obtain xð1Þþ0ðep=2Þþ0ðp2=4Þ¼0or or ory¼0 z¼0 x¼0 We have shown xfþygþzf¼0 implies x¼0;y¼0;z¼0 Accordingly, u,v,ware linearly independent.122 CHAPTER 4 Vector Spaces Linear Dependence in R3 Linear dependence in the vector space V¼R3can be described geometrically as follows: (a) Any two vectors uandvinR3are linearly dependent if and only if they lie on the same line through the origin O, as shown in Fig. 4-3(a). (b) Any three vectors u,v,winR3are linearly dependent if and only if they lie on the same plane through the origin O, as shown in Fig. 4-3(b). Later, we will be able to show that any four or more vectors in R3are automatically linearly dependent. Linear Dependence and Linear Combinations The notions of linear dependence and linear combinations are closely related. Speciﬁcally, for more than one vector, we show that the vectors v1;v2;...;vmare linearly dependent if and only if one of them is a linear combination of the others. Suppose, say, viis a linear combination of the others, vi¼a1v1þ/C1/C1/C1þ ai/C01vi/C01þaiþ1viþ1þ/C1/C1/C1þ amvm Then by adding/C0vito both sides, we obtain a1v1þ/C1/C1/C1þ ai/C01vi/C01/C0viþaiþ1viþ1þ/C1/C1/C1þ amvm¼0 where the coefﬁcient of viis not 0. Hence, the vectors are linearly dependent. Conversely, suppose the vectors are linearly dependent, say, b1v1þ/C1/C1/C1þ bjvjþ/C1/C1/C1þ bmvm¼0; where bj6¼0 Then we can solve for vjobtaining vj¼b/C01 jb1v1/C0/C1/C1/C1/C0 b/C01 jbj/C01vj/C01/C0b/C01 jbjþ1vjþ1/C0/C1/C1/C1/C0 b/C01 jbmvm and so vjis a linear combination of the other vectors. We now state a slightly stronger statement than the one above. This result has many important consequences. LEMMA 4.10: Suppose two or more nonzero vectors v1;v2;...;vmare linearly dependent. Then one of the vectors is a linear combination of the preceding vectors; that is, there existsk>1such that v k¼c1v1þc2v2þ/C1/C1/C1þ ck/C01vk/C01Figure 4-3CHAPTER 4 Vector Spaces 123 Linear Dependence and Echelon Matrices Consider the following echelon matrix A, whose pivots have been circled: A¼0/C13234567 00/C1343234 0000/C13789 00000/C1367 00000002 666643 77775 Observe that the rows R 2,R3,R4have 0’s in the second column below the nonzero pivot in R1, and hence any linear combination of R2,R3,R4must have 0 as its second entry. Thus, R1cannot be a linear combination of the rows below it. Similarly, the rows R3andR4have 0’s in the third column below the nonzero pivot in R2, and hence R2cannot be a linear combination of the rows below it. Finally, R3cannot be a multiple of R4, because R4has a 0 in the ﬁfth column below the nonzero pivot in R3. Viewing the nonzero rows from the bottom up, R4,R3,R2,R1, no row is a linear combination of the preceding rows. Thus, the rows are linearly independent by Lemma 4.10. The argument used with the above echelon matrix Acan be used for the nonzero rows of any echelon matrix. Thus, we have the following very useful result. THEOREM 4.11: The nonzero rows of a matrix in echelon form are linearly independent. 4.8 Basis and Dimension First we state two equivalent ways to deﬁne a basis of a vector space V. (The equivalence is proved in Problem 4.28.) DEFINITION A: A set S¼fu1;u2;...;ungof vectors is a basis ofVif it has the following two properties: (1) Sis linearly independent. (2) Sspans V. DEFINITION B: A set S¼fu1;u2;...;ungof vectors is a basis ofVif every v2Vcan be written uniquely as a linear combination of the basis vectors. The following is a fundamental result in linear algebra. THEOREM 4.12: LetVbe a vector space such that one basis has melements and another basis has n elements. Then m¼n. A vector space Vis said to be of ﬁnite dimension n orn-dimensional , written dimV¼n ifVhas a basis with nelements. Theorem 4.12 tells us that all bases of Vhave the same number of elements, so this deﬁnition is well deﬁned. The vector space {0} is deﬁned to have dimension 0.Suppose a vector space Vdoes not have a ﬁnite basis. Then Vis said to be of inﬁnite dimension or to beinﬁnite-dimensional . The above fundamental Theorem 4.12 is a consequence of the following ‘‘replacement lemma’’ (proved in Problem 4.35). LEMMA 4.13: Supposefv1;v2;...;vngspans V, and supposefw1;w2;...;wmgis linearly indepen- dent. Then m/C20n, and Vis spanned by a set of the form fw1;w2;...;wm;vi1;vi2;...;vin/C0mg Thus, in particular, nþ1 or more vectors in Vare linearly dependent. Observe in the above lemma that we have replaced mof the vectors in the spanning set of Vby the m independent vectors and still retained a spanning set.124 CHAPTER 4 Vector Spaces Examples of Bases This subsection presents important examples of bases of some of the main vector spaces appearing in this text. (a) Vector space Kn:Consider the following nvectors in Kn: e1¼ð1;0;0;0;...;0;0Þ;e2¼ð0;1;0;0;...;0;0Þ;...;en¼ð0;0;0;0;...;0;1Þ These vectors are linearly independent. (For example, they form a matrix in echelon form.) Furthermore, any vector u¼ða1;a2;...;anÞinKncan be written as a linear combination of the above vectors. Speciﬁcally, v¼a1e1þa2e2þ/C1/C1/C1þ anen Accordingly, the vectors form a basis of Kncalled the usual orstandard basis of Kn. Thus (as one might expect), Knhas dimension n. In particular, any other basis of Knhasnelements. (b) Vector space M ¼Mr;sof all r/C2smatrices: The following six matrices form a basis of the vector space M2;3of all 2/C23 matrices over K: 100 000/C20/C21 ;010 000/C20/C21 ;001 000/C20/C21 ;000 100/C20/C21 ;000 010/C20/C21 ;000 001/C20/C21 More generally, in the vector space M¼Mr;sof all r/C2smatrices, let Eijbe the matrix with ij-entry 1 and 0’s elsewhere. Then all such matrices form a basis of Mr;scalled the usual orstandard basis of Mr;s. Accordingly, dim Mr;s¼rs. (c) Vector space PnðtÞof all polynomials of degree /C20n:The set S¼f1;t;t2;t3;...;tngofnþ1 polynomials is a basis of PnðtÞ. Speciﬁcally, any polynomial fðtÞof degree/C20ncan be expessed as a linear combination of these powers of t, and one can show that these polynomials are linearly independent. Therefore, dim PnðtÞ¼nþ1. (d) Vector space P ðtÞof all polynomials: Consider any ﬁnite set S¼ff1ðtÞ;f2ðtÞ;...;fmðtÞgof polynomials in PðtÞ, and let mdenote the largest of the degrees of the polynomials. Then any polynomial gðtÞof degree exceeding mcannot be expressed as a linear combination of the elements of S. Thus, Scannot be a basis of PðtÞ. This means that the dimension of PðtÞis inﬁnite. We note that the inﬁnite set S0¼f1;t;t2;t3;...g, consisting of all the powers of t, spans PðtÞand is linearly independent. Accordingly, S0is an inﬁnite basis of PðtÞ. Theorems on Bases The following three theorems (proved in Problems 4.37, 4.38, and 4.39) will be used frequently. THEOREM 4.14: LetVbe a vector space of ﬁnite dimension n. Then: (i) Any nþ1 or more vectors in Vare linearly dependent. (ii) Any linearly independent set S¼fu1;u2;...;ungwith nelements is a basis ofV. (iii) Any spanning set T¼fv1;v2;...;vngofVwith nelements is a basis of V. THEOREM 4.15: Suppose Sspans a vector space V. Then: (i) Any maximum number of linearly independent vectors in Sform a basis of V. (ii) Suppose one deletes from Severy vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V.CHAPTER 4 Vector Spaces 125 THEOREM 4.16: LetVbe a vector space of ﬁnite dimension and let S¼fu1;u2;...;urgbe a set of linearly independent vectors in V. Then Sis part of a basis of V; that is, Smay be extended to a basis of V. EXAMPLE 4.11 (a) The following four vectors in R4form a matrix in echelon form: ð1;1;1;1Þ;ð0;1;1;1Þ;ð0;0;1;1Þ;ð0;0;0;1Þ Thus, the vectors are linearly independent, and, because dim R4¼4, the four vectors form a basis of R4. (b) The following nþ1 polynomials in PnðtÞare of increasing degree: 1;t/C01;ðt/C01Þ2;...;ðt/C01Þn Therefore, no polynomial is a linear combination of preceding polynomials; hence, the polynomials are linear independent. Furthermore, they form a basis of PnðtÞ, because dim PnðtÞ¼nþ1. (c) Consider any four vectors in R3, say ð257;/C0132;58Þ;ð43;0;/C017Þ;ð521;/C0317;94Þ;ð328;/C0512;/C0731Þ By Theorem 4.14(i), the four vectors must be linearly dependent, because they come from the three-dimensional vector space R3. Dimension and Subspaces The following theorem (proved in Problem 4.40) gives the basic relationship between the dimension of a vector space and the dimension of a subspace. THEOREM 4.17: LetWbe a subspace of an n-dimensional vector space V. Then dimW/C20n.I n particular, if dimW¼n, then W¼V. EXAMPLE 4.12 LetWbe a subspace of the real space R3. Note that dim R3¼3. Theorem 4.17 tells us that the dimension of Wcan only be 0, 1, 2, or 3. The following cases apply: (a) If dim W¼0, then W¼f0g, a point. (b) If dim W¼1, then Wis a line through the origin 0. (c) If dim W¼2, then Wis a plane through the origin 0. (d) If dim W¼3, then Wis the entire space R3. 4.9 Application to Matrices, Rank of a Matrix LetAbe any m/C2nmatrix over a ﬁeld K. Recall that the rows of Amay be viewed as vectors in Knand that the row space of A, written rowsp(A), is the subspace of Knspanned by the rows of A. The following deﬁnition applies. DEFINITION: Therank of a matrix A, written rank( A), is equal to the maximum number of linearly independent rows of Aor, equivalently, the dimension of the row space of A. Recall, on the other hand, that the columns of an m/C2nmatrix Amay be viewed as vectors in Kmand that the column space of A, written colsp(A), is the subspace of Kmspanned by the columns of A. Although mmay not be equal to n—that is, the rows and columns of Amay belong to different vector spaces—we have the following fundamental result. THEOREM 4.18: The maximum number of linearly independent rows of any matrix Ais equal to the maximum number of linearly independent columns of A. Thus, the dimension of the row space of Ais equal to the dimension of the column space of A. Accordingly, one could restate the above deﬁnition of the rank of Ausing columns instead of rows.126 CHAPTER 4 Vector Spaces Basis-Finding Problems This subsection shows how an echelon form of any matrix Agives us the solution to certain problems about Aitself. Speciﬁcally, let AandBbe the following matrices, where the echelon matrix B(whose pivots are circled) is an echelon form of A: A¼1 213 12 2 556 45 37 61 16 9 1 5 10 8 9 926 81 191 22 666643 77775and B¼/C13121312 0/C1313121 000/C13112 000000 0000002 666643 77775 We solve the following four problems about the matrix A, where C 1;C2;...;C6denote its columns: (a) Find a basis of the row space of A. (b) Find each column CkofAthat is a linear combination of preceding columns of A. (c) Find a basis of the column space of A. (d) Find the rank of A. (a) We are given that AandBare row equivalent, so they have the same row space. Moreover, Bis in echelon form, so its nonzero rows are linearly independent and hence form a basis of the row space ofB. Thus, they also form a basis of the row space of A. That is, basis of rowspðAÞ:ð1;2;1;3;1;2Þ;ð0;1;3;1;2;1Þ;ð0;0;0;1;1;2Þ (b) Let Mk¼½C1;C2;...;Ck/C138, the submatrix of Aconsisting of the ﬁrst kcolumns of A. Then Mk/C01and Mkare, respectively, the coefﬁcient matrix and augmented matrix of the vector equation x1C1þx2C2þ/C1/C1/C1þ xk/C01Ck/C01¼Ck Theorem 3.9 tells us that the system has a solution, or, equivalently, Ckis a linear combination of the preceding columns of Aif and only if rank ðMkÞ¼rankðMk/C01Þ, where rankðMkÞmeans the number of pivots in an echelon form of Mk. Now the ﬁrst kcolumn of the echelon matrix Bis also an echelon form of Mk. Accordingly, rankðM2Þ¼rankðM3Þ¼2 and rank ðM4Þ¼rankðM5Þ¼rankðM6Þ¼3 Thus, C3,C5,C6are each a linear combination of the preceding columns of A. (c) The fact that the remaining columns C1,C2,C4are not linear combinations of their respective preceding columns also tells us that they are linearly independent. Thus, they form a basis of thecolumn space of A. That is, basis of colspðAÞ:½1;2;3;1;2/C138T;½2;5;7;5;6/C138T;½3;6;11;8;11/C138T Observe that C1,C2,C4may also be characterized as those columns of Athat contain the pivots in any echelon form of A. (d) Here we see that three possible deﬁnitions of the rank of Ayield the same value. (i) There are three pivots in B, which is an echelon form of A. (ii) The three pivots in Bcorrespond to the nonzero rows of B, which form a basis of the row space of A. (iii) The three pivots in Bcorrespond to the columns of A, which form a basis of the column space ofA. Thus, rankðAÞ¼3.CHAPTER 4 Vector Spaces 127 Application to Finding a Basis for W¼spanðu1;u2;...;urÞ Frequently, we are given a list S¼fu1;u2;...;urgof vectors in Knand we want to ﬁnd a basis for the subspace WofKnspanned by the given vectors—that is, a basis of W¼spanðSÞ¼spanðu1;u2;...;urÞ The following two algorithms, which are essentially described in the above subsection, ﬁnd such a basis (and hence the dimension) of W. Algorithm 4.1 (Row space algorithm) Step 1. Form the matrix Mwhose rows are the given vectors. Step 2. Row reduce Mto echelon form. Step 3. Output the nonzero rows of the echelon matrix. Sometimes we want to ﬁnd a basis that only comes from the original given vectors. The next algorithm accomplishes this task. Algorithm 4.2 (Casting-out algorithm) Step 1. Form the matrix Mwhose columns are the given vectors. Step 2. Row reduce Mto echelon form. Step 3. For each column Ckin the echelon matrix without a pivot, delete (cast out) the vector ukfrom the list Sof given vectors. Step 4. Output the remaining vectors in S(which correspond to columns with pivots). We emphasize that in the ﬁrst algorithm we form a matrix whose rows are the given vectors, whereas in the second algorithm we form a matrix whose columns are the given vectors. EXAMPLE 4.13 LetWbe the subspace of R5spanned by the following vectors: u1¼ð1;2;1;3;2Þ; u2¼ð1;3;3;5;3Þ; u3¼ð3;8;7;13;8Þ u4¼ð1;4;6;9;7Þ; u5¼ð5;13;13;25;19Þ Find a basis of Wconsisting of the original given vectors, and ﬁnd dim W. Form the matrix Mwhose columns are the given vectors, and reduce Mto echelon form: M¼11 31 5 23 841 3 13 761 3351 392 5 23 871 92 666643 77775/C2411315 01223 0001200000 000002 666643 77775 The pivots in the echelon matrix appear in columns C1,C2,C4. Accordingly, we ‘‘cast out’’ the vectors u3andu5 from the original ﬁve vectors. The remaining vectors u1,u2,u4, which correspond to the columns in the echelon matrix with pivots, form a basis of W. Thus, in particular, dim W¼3. Remark: The justiﬁcation of the casting-out algorithm is essentially described above, but we repeat it again here for emphasis. The fact that column C3in the echelon matrix in Example 4.13 does not have a pivot means that the vector equation xu1þyu2¼u3 has a solution, and hence u3is a linear combination of u1andu2. Similarly, the fact that C5does not have a pivot means that u5is a linear combination of the preceding vectors. We have deleted each vector in the original spanning set that is a linear combination of preceding vectors. Thus, the remaining vectors arelinearly independent and form a basis of W.128 CHAPTER 4 Vector Spaces Application to Homogeneous Systems of Linear Equations Consider again a homogeneous system AX¼0 of linear equations over Kwith nunknowns. By Theorem 4.4, the solution set Wof such a system is a subspace of Kn, and hence Whas a dimension. The following theorem, whose proof is postponed until Chapter 5, holds. THEOREM 4.19: The dimension of the solution space Wof a homogeneous system AX¼0isn/C0r, where nis the number of unknowns and ris the rank of the coefﬁcient matrix A. In the case where the system AX¼0 is in echelon form, it has precisely n/C0rfree variables, say xi1;xi2;...;xin/C0r. Let vjbe the solution obtained by setting xij¼1 (or any nonzero constant) and the remaining free variables equal to 0. We show (Problem 4.50) that the solutions v1;v2;...;vn/C0rare linearly independent; hence, they form a basis of the solution space W. We have already used the above process to ﬁnd a basis of the solution space Wof a homogeneous system AX¼0 in Section 3.11. Problem 4.48 gives three other examples. 4.10 Sums and Direct Sums LetUandWbe subsets of a vector space V. The sum of UandW, written UþW, consists of all sums uþwwhere u2Uandw2W. That is, UþW¼fv:v¼uþw;where u2Uandw2Wg Now suppose UandWare subspaces of V. Then one can easily show (Problem 4.53) that UþWis a subspace of V. Recall that U\Wis also a subspace of V. The following theorem (proved in Problem 4.58) relates the dimensions of these subspaces. THEOREM 4.20: Suppose UandWare ﬁnite-dimensional subspaces of a vector space V. Then UþWhas ﬁnite dimension and dimðUþWÞ¼dimUþdimW/C0dimðU\WÞ EXAMPLE 4.14 LetV¼M2;2, the vector space of 2 /C22 matrices. Let Uconsist of those matrices whose second row is zero, and let Wconsist of those matrices whose second column is zero. Then U¼ab 00/C20/C21/C26/C27 ;W¼a0 c0/C20/C21/C26/C27 and UþW¼ab c0/C20/C21/C26/C27 ;U\W¼a0 00/C20/C21/C26/C27 That is, UþWconsists of those matrices whose lower right entry is 0, and U\Wconsists of those matrices whose second row and second column are zero. Note that dim U¼2, dim W¼2, dimðU\WÞ¼1. Also, dimðUþWÞ¼3, which is expected from Theorem 4.20. That is, dimðUþWÞ¼dimUþdimV/C0dimðU\WÞ¼2þ2/C01¼3 Direct Sums The vector space Vis said to be the direct sum of its subspaces UandW, denoted by V¼U/C8W if every v2Vcan be written in one and only one way as v¼uþwwhere u2Uandw2W. The following theorem (proved in Problem 4.59) characterizes such a decomposition. THEOREM 4.21: The vector space Vis the direct sum of its subspaces UandWif and only if: (i)V¼UþW, (ii) U\W¼f0g.CHAPTER 4 Vector Spaces 129 EXAMPLE 4.15 Consider the vector space V¼R3: (a) Let Ube the xy-plane and let Wbe the yz-plane; that is, U¼fða;b;0Þ:a;b2Rg and W¼fð 0;b;cÞ:b;c2Rg Then R3¼UþW, because every vector in R3is the sum of a vector in Uand a vector in W. However, R3is not the direct sum of UandW, because such sums are not unique. For example, ð3;5;7Þ¼ð 3;1;0Þþð 0;4;7Þ and alsoð3;5;7Þ¼ð 3;/C04;0Þþð 0;9;7Þ (b) Let Ube the xy-plane and let Wbe the z-axis; that is, U¼fða;b;0Þ:a;b2Rg and W¼fð 0;0;cÞ:c2Rg Now any vectorða;b;cÞ2R3can be written as the sum of a vector in Uand a vector in Vin one and only one way: ða;b;cÞ¼ð a;b;0Þþð 0;0;cÞ Accordingly, R3is the direct sum of UandW; that is, R3¼U/C8W. General Direct Sums The notion of a direct sum is extended to more than one factor in the obvious way. That is, Vis the direct sum of subspaces W1;W2;...;Wr, written V¼W1/C8W2/C8/C1/C1/C1/C8 Wr if every vector v2Vcan be written in one and only one way as v¼w1þw2þ/C1/C1/C1þ wr where w12W1;w22W2;...;wr2Wr. The following theorems hold. THEOREM 4.22: Suppose V¼W1/C8W2/C8/C1/C1/C1/C8 Wr. Also, for each k, suppose Skis a linearly independent subset of Wk.T h e n (a) The union S¼S kSkis linearly independent in V. (b) If each Skis a basis of Wk, thenS kSkis a basis of V. (c) dim V¼dimW1þdimW2þ/C1/C1/C1þ dimWr. THEOREM 4.23: Suppose V¼W1þW2þ/C1/C1/C1þ WranddimV¼P kdimWk. Then V¼W1/C8W2/C8/C1/C1/C1/C8 Wr: 4.11 Coordinates LetVbe an n-dimensional vector space over Kwith basis S¼fu1;u2;...;ung. Then any vector v2V can be expressed uniquely as a linear combination of the basis vectors in S, say v¼a1u1þa2u2þ/C1/C1/C1þ anun These nscalars a1;a2;...;anare called the coordinates ofvrelative to the basis S, and they form a vector [a1;a2;...;an]i nKncalled the coordinate vector ofvrelative to S. We denote this vector by ½v/C138S,o r simply½v/C138;when Sis understood. Thus, ½v/C138S¼½a1;a2;...;an/C138 For notational convenience, brackets ½.../C138, rather than parentheses ð...Þ, are used to denote the coordinate vector.130 CHAPTER 4 Vector Spaces Remark: The above nscalars a1;a2;...;analso form the coordinate column vector ½a1;a2;...;an/C138Tofvrelative to S. The choice of the column vector rather than the row vector to represent vdepends on the context in which it is used. The use of such column vectors will become clear later in Chapter 6. EXAMPLE 4.16 Consider the vector space P2ðtÞof polynomials of degree /C202. The polynomials p1¼tþ1; p2¼t/C01; p3¼ðt/C01Þ2¼t2/C02tþ1 form a basis SofP2ðtÞ. The coordinate vector [ v]o f v¼2t2/C05tþ9 relative to Sis obtained as follows. Setv¼xp1þyp2þzp3using unknown scalars x,y,z, and simplify: 2t2/C05tþ9¼xðtþ1Þþyðt/C01Þþzðt2/C02tþ1Þ ¼xtþxþyt/C0yþzt2/C02ztþz ¼zt2þðxþy/C02zÞtþðx/C0yþzÞ Then set the coefﬁcients of the same powers of tequal to each other to obtain the system z¼2; xþy/C02z¼/C05; x/C0yþz¼9 The solution of the system is x¼3,y¼/C04,z¼2. Thus, v¼3p1/C04p2þ2p3;and hence ;½v/C138¼½3;/C04;2/C138 EXAMPLE 4.17 Consider real space R3. The following vectors form a basis SofR3: u1¼ð1;/C01;0Þ; u2¼ð1;1;0Þ; u3¼ð0;1;1Þ The coordinates of v¼ð5;3;4Þrelative to the basis Sare obtained as follows. Setv¼xv1þyv2þzv3; that is, set vas a linear combination of the basis vectors using unknown scalars x,y,z. This yields 5 3 42 43 5¼x1 /C01 02 43 5þy1 1 02 43 5þz0 1 12 43 5 The equivalent system of linear equations is as follows: xþy¼5;/C0xþyþz¼3; z¼4 The solution of the system is x¼3,y¼2,z¼4. Thus, v¼3u1þ2u2þ4u3; and so½v/C138s¼½3;2;4/C138 Remark 1: There is a geometrical interpretation of the coordinates of a vector vrelative to a basis Sfor the real space Rn, which we illustrate using the basis SofR3in Example 4.17. First consider the space R3with the usual x,y,zaxes. Then the basis vectors determine a new coordinate system of R3, say with x0,y0,z0axes, as shown in Fig. 4-4. That is, (1) The x0-axis is in the direction of u1with unit lengthku1k. (2) The y0-axis is in the direction of u2with unit lengthku2k. (3) The z0-axis is in the direction of u3with unit lengthku3k. Then each vector v¼ða;b;cÞor, equivalently, the point Pða;b;cÞinR3will have new coordinates with respect to the new x0,y0,z0axes. These new coordinates are precisely ½v/C138S, the coordinates of vwith respect to the basis S. Thus, as shown in Example 4.17, the coordinates of the point Pð5;3;4Þwith the new axes form the vector [3, 2, 4]. Remark 2: Consider the usual basis E¼fe1;e2;...;engofKndeﬁned by e1¼ð1;0;0;...;0;0Þ; e2¼ð0;1;0;...;0;0Þ; ...;en¼ð0;0;0;...;0;1ÞCHAPTER 4 Vector Spaces 131 Letv¼ða1;a2;...;anÞbe any vector in Kn. Then one can easily show that v¼a1e1þa2e2þ/C1/C1/C1þ anen; and so½v/C138E¼½a1;a2;...;an/C138 That is, the coordinate vector ½v/C138Eof any vector vrelative to the usual basis EofKnis identical to the original vector v. Isomorphism of VandKn LetVbe a vector space of dimension nover K, and suppose S¼fu1;u2;...;ungis a basis of V. Then each vector v2Vcorresponds to a unique n-tuple½v/C138SinKn. On the other hand, each n-tuple [c1;c2;...;cn]i n Kncorresponds to a unique vector c1u1þc2u2þ/C1/C1/C1þ cnuninV. Thus, the basis S induces a one-to-one correspondence between VandKn. Furthermore, suppose v¼a1u1þa2u2þ/C1/C1/C1þ anun and w¼b1u1þb2u2þ/C1/C1/C1þ bnun Then vþw¼ða1þb1Þu1þða2þb2Þu2þ/C1/C1/C1þð anþbnÞun kv¼ðka1Þu1þðka2Þu2þ/C1/C1/C1þð kanÞun where kis a scalar. Accordingly, ½vþw/C138S¼½a1þb1; ...;anþbn/C138¼½a1;...;an/C138þ½b1;...;bn/C138¼½ v/C138Sþ½w/C138S ½kv/C138S¼½ka1;ka2;...;kan/C138¼k½a1;a2;...;an/C138¼k½v/C138S Thus, the above one-to-one correspondence between VandKnpreserves the vector space operations of vector addition and scalar multiplication. We then say that VandKnare isomorphic, written VﬃKn We state this result formally. Figure 4-4132 CHAPTER 4 Vector Spaces THEOREM 4.24: LetVbe an n-dimensional vector space over a ﬁeld K. Then Vand Knare isomorphic. The next example gives a practical application of the above result. EXAMPLE 4.18 Suppose we want to determine whether or not the following matrices in V¼M2;3are linearly dependent: A¼12/C03 40 1/C20/C21 ; B¼13/C04 65 4/C20/C21 ; C¼38/C011 16 10 9/C20/C21 The coordinate vectors of the matrices in the usual basis of M2;3are as follows: ½A/C138¼½1;2;/C03;4;0;1/C138;½B/C138¼½1;3;/C04;6;5;4/C138;½C/C138¼½3;8;/C011;16;10;9/C138 Form the matrix Mwhose rows are the above coordinate vectors and reduce Mto an echelon form: M¼12/C0340 1 13/C0465 4 38/C011 16 10 92 43 5/C2412/C034 01 01/C012 53 02/C02 4 10 62 43 5/C2412/C03401 01/C01253 00 00002 43 5 Because the echelon matrix has only two nonzero rows, the coordinate vectors [ A], [B], [C] span a subspace of dimension 2 and so are linearly dependent. Accordingly, the original matrices A,B,Care linearly dependent. SOLVED PROBLEMS Vector Spaces, Linear Combinations 4.1. Suppose uand vbelong to a vector space V. Simplify each of the following expressions: (a) E1¼3ð2u/C04vÞþ5uþ7v, (c) E3¼2uvþ3ð2uþ4vÞ (b) E2¼3u/C06ð3u/C05vÞþ7u, (d) E4¼5u/C03 vþ5u Multiply out and collect terms: (a) E1¼6u/C012vþ5uþ7v¼11u/C05v (b) E2¼3u/C018uþ30vþ7u¼/C08uþ30v (c) E3is not deﬁned because the product uvof vectors is not deﬁned. (d) E4is not deﬁned because division by a vector is not deﬁned. 4.2. Prove Theorem 4.1: Let Vbe a vector space over a ﬁeld K. (i)k0¼0. (ii) 0 u¼0. (iii) If ku¼0, then k¼0o r u¼0. (iv)ð/C0kÞu¼kð/C0uÞ¼/C0 ku. (i) By Axiom [A 2] with u¼0, we have 0þ0¼0. Hence, by Axiom [M 1], we have k0¼kð0þ0Þ¼k0þk0 Adding/C0k0 to both sides gives the desired result. (ii) For scalars, 0 þ0¼0. Hence, by Axiom [M 2], we have 0u¼ð0þ0Þu¼0uþ0u Adding/C00uto both sides gives the desired result. (iii) Suppose ku¼0 and k6¼0. Then there exists a scalar k/C01such that k/C01k¼1. Thus, u¼1u¼ðk/C01kÞu¼k/C01ðkuÞ¼k/C010¼0 (iv) Using uþð/C0 uÞ¼0 and kþð/C0 kÞ¼0 yields 0¼k0¼k½uþð/C0 uÞ/C138¼ kuþkð/C0uÞ and 0¼0u¼½kþð/C0 kÞ/C138u¼kuþð/C0 kÞu Adding/C0kuto both sides of the ﬁrst equation gives /C0ku¼kð/C0uÞ;and adding/C0kuto both sides of the second equation gives /C0ku¼ð/C0 kÞu. Thus,ð/C0kÞu¼kð/C0uÞ¼/C0 ku.CHAPTER 4 Vector Spaces 133 4.3. Show that (a) kðu/C0vÞ¼ku/C0kv, (b) uþu¼2u. (a) Using the deﬁnition of subtraction, that u/C0v¼uþð/C0 vÞ, and Theorem 4.1(iv), that kð/C0vÞ¼/C0 kv,w e have kðu/C0vÞ¼k½uþð/C0 vÞ/C138¼ kuþkð/C0vÞ¼kuþð/C0 kvÞ¼ku/C0kv (b) Using Axiom [M 4] and then Axiom [M 2], we have uþu¼1uþ1u¼ð1þ1Þu¼2u 4.4. Express v¼ð1;/C02;5ÞinR3as a linear combination of the vectors u1¼ð1;1;1Þ; u2¼ð1;2;3Þ; u3¼ð2;/C01;1Þ We seek scalars x,y,z, as yet unknown, such that v¼xu1þyu2þzu3. Thus, we require 1 /C02 52 43 5¼x1 1 12 43 5þy1 2 32 43 5þz2 /C01 12 43 5 orxþyþ2z¼1 xþ2y/C0z¼/C02 xþ3yþz¼5 (For notational convenience, we write the vectors in R3as columns, because it is then easier to ﬁnd the equivalent system of linear equations.) Reducing the system to echelon form yields the triangular system xþyþ2z¼1; y/C03z¼/C03; 5z¼10 The system is consistent and has a solution. Solving by back-substitution yields the solution x¼/C06,y¼3, z¼2. Thus, v¼/C06u1þ3u2þ2u3. Alternatively, write down the augmented matrix Mof the equivalent system of linear equations, where u1,u2,u3are the ﬁrst three columns of Mand vis the last column, and then reduce Mto echelon form: M¼1 121 12/C01/C02 1 3152 43 5/C241 121 01/C03/C03 02/C0142 43 5/C241 121 01/C03/C03 00 5 1 02 43 5 The last matrix corresponds to a triangular system, which has a solution. Solving the triangular system by back-substitution yields the solution x¼/C06,y¼3,z¼2. Thus, v¼/C06u1þ3u2þ2u3. 4.5. Express v¼ð2;/C05;3ÞinR3as a linear combination of the vectors u1¼ð1;/C03;2Þ;u2¼ð2;/C04;/C01Þ;u3¼ð1;/C05;7Þ We seek scalars x,y,z, as yet unknown, such that v¼xu1þyu2þzu3. Thus, we require 2 /C05 32 43 5¼x1 /C03 22 43 5þy2 /C04 /C012 43 5þz1 /C05 72 43 5 orxþ2yþz¼2 /C03x/C04y/C05z¼/C05 2x/C0yþ7z¼3 Reducing the system to echelon form yields the system xþ2yþz¼2; 2y/C02z¼1; 0¼3 The system is inconsistent and so has no solution. Thus, vcannot be written as a linear combination of u1,u2,u3. 4.6. Express the polynomial v¼t2þ4t/C03i nPðtÞas a linear combination of the polynomials p1¼t2/C02tþ5; p2¼2t2/C03t; p3¼tþ1 Setvas a linear combination of p1,p2,p3using unknowns x,y,zto obtain t2þ4t/C03¼xðt2/C02tþ5Þþyð2t2/C03tÞþzðtþ1Þð *Þ We can proceed in two ways.134 CHAPTER 4 Vector Spaces Method 1. Expand the right side of (*) and express it in terms of powers of tas follows: t2þ4t/C03¼xt2/C02xtþ5xþ2yt2/C03ytþztþz ¼ðxþ2yÞt2þð/C0 2x/C03yþzÞtþð5xþ3zÞ Set coefﬁcients of the same powers of tequal to each other, and reduce the system to echelon form. This yields xþ2y¼1 /C02x/C03yþz¼4 5xþ3z¼/C03orxþ2y¼1 yþz¼6 /C010yþ3z¼/C08orxþ2y¼1 yþz¼6 13z¼52 The system is consistent and has a solution. Solving by back-substitution yields the solution x¼/C03,y¼2, z¼4. Thus, v¼/C03p1þ2p2þ4p2. Method 2. The equation (*) is an identity in t; that is, the equation holds for any value of t. Thus, we can settequal to any numbers to obtain equations in the unknowns. (a) Set t¼0 in (*) to obtain the equation /C03¼5xþz. (b) Set t¼1 in (*) to obtain the equation 2 ¼4x/C0yþ2z. (c) Set t¼/C01 in (*) to obtain the equation /C06¼8xþ5y. Solve the system of the three equations to again obtain the solution x¼/C03,y¼2,z¼4. Thus, v¼/C03p1þ2p2þ4p3. 4.7. Express Mas a linear combination of the matrices A,B,C, where M¼47 79/C20/C21 ; and A¼11 11/C20/C21 ; B¼12 34/C20/C21 ; C¼11 45/C20/C21 SetMas a linear combination of A,B,Cusing unknown scalars x,y,z; that is, set M¼xAþyBþzC. This yields 47 79/C20/C21 ¼x11 11/C20/C21 þy12 34/C20/C21 þz11 45/C20/C21 ¼xþyþzxþ2yþz xþ3yþ4zxþ4yþ5z/C20/C21 Form the equivalent system of equations by setting corresponding entries equal to each other: xþyþz¼4; xþ2yþz¼7; xþ3yþ4z¼7; xþ4yþ5z¼9 Reducing the system to echelon form yields xþyþz¼4; y¼3; 3z¼/C03; 4z¼/C04 The last equation drops out. Solving the system by back-substitution yields z¼/C01,y¼3,x¼2. Thus, M¼2Aþ3B/C0C. Subspaces 4.8. Prove Theorem 4.2: Wis a subspace of Vif the following two conditions hold: (a) 02W. (b) If u;v2W, then uþv,ku2W. By (a), Wis nonempty, and, by (b), the operations of vector addition and scalar multiplication are well deﬁned for W. Axioms [A 1], [A 4], [M 1], [M 2], [M 3], [M 4] hold in Wbecause the vectors in Wbelong to V. Thus, we need only show that [A 2] and [A 3] also hold in W. Now [A 2] holds because the zero vector in V belongs to Wby (a). Finally, if v2W, thenð/C01Þv¼/C0 v2W, and vþð/C0 vÞ¼0. Thus [A 3] holds. 4.9. LetV¼R3. Show that Wis not a subspace of V, where (a) W¼fð a;b;cÞ:a/C210g, (b) W¼fð a;b;cÞ:a2þb2þc2/C201g. In each case, show that Theorem 4.2 does not hold.CHAPTER 4 Vector Spaces 135 (a) Wconsists of those vectors whose ﬁrst entry is nonnegative. Thus, v¼ð1;2;3Þbelongs to W. Let k¼/C03. Then kv¼ð/C0 3;/C06;/C09Þdoes not belong to W, because/C03 is negative. Thus, Wis not a subspace of V. (b) Wconsists of vectors whose length does not exceed 1. Hence, u¼ð1;0;0Þand v¼ð0;1;0Þbelong to W, but uþv¼ð1;1;0Þdoes not belong to W, because 12þ12þ02¼2>1. Thus, Wis not a subspace of V. 4.10. LetV¼PðtÞ, the vector space of real polynomials. Determine whether or not Wis a subspace of V, where (a) Wconsists of all polynomials with integral coefﬁcients. (b) Wconsists of all polynomials with degree /C216 and the zero polynomial. (c) Wconsists of all polynomials with only even powers of t. (a) No, because scalar multiples of polynomials in Wdo not always belong to W. For example, fðtÞ¼3þ6tþ7t22W but1 2fðtÞ¼3 2þ3tþ7 2t262W (b and c) Yes. In each case, Wcontains the zero polynomial, and sums and scalar multiples of polynomials inWbelong to W. 4.11. LetVbe the vector space of functions f:R!R. Show that Wis a subspace of V, where (a) W¼ffðxÞ:fð1Þ¼0g, all functions whose value at 1 is 0. (b) W¼ffðxÞ:fð3Þ¼fð1Þg, all functions assigning the same value to 3 and 1. (c) W¼ffðtÞ:fð/C0xÞ¼/C0 fðxÞg, all odd functions . Let ^0 denote the zero function, so ^0ðxÞ¼0 for every value of x. (a) ^02W, because ^0ð1Þ¼0. Suppose f;g2W. Then fð1Þ¼0 and gð1Þ¼0. Also, for scalars aandb,w e have ðafþbgÞð1Þ¼afð1Þþbgð1Þ¼a0þb0¼0 Thus, afþbg2W, and hence Wis a subspace. (b) ^02W, because ^0ð3Þ¼0¼^0ð1Þ. Suppose f;g2W. Then fð3Þ¼fð1Þandgð3Þ¼gð1Þ. Thus, for any scalars aandb, we have ðafþbgÞð3Þ¼afð3Þþbgð3Þ¼afð1Þþbgð1Þ¼ð afþbgÞð1Þ Thus, afþbg2W, and hence Wis a subspace. (c) ^02W, because ^0ð/C0xÞ¼0¼/C00¼/C0 ^0ðxÞ. Suppose f;g2W.T h e n fð/C0xÞ¼/C0 fðxÞandgð/C0xÞ¼/C0 gðxÞ. Also, for scalars aandb, ðafþbgÞð/C0xÞ¼afð/C0xÞþbgð/C0xÞ¼/C0 afðxÞ/C0bgðxÞ¼/C0ð afþbgÞðxÞ Thus, abþgf2W, and hence Wis a subspace of V. 4.12. Prove Theorem 4.3: The intersection of any number of subspaces of Vis a subspace of V. LetfWi:i2Igbe a collection of subspaces of Vand let W¼\ð Wi:i2IÞ. Because each Wiis a subspace of V, we have 02Wi, for every i2I. Hence, 02W. Suppose u;v2W. Then u;v2Wi, for every i2I. Because each Wiis a subspace, auþbv2Wi, for every i2I. Hence, auþbv2W. Thus, Wis a subspace of V. Linear Spans 4.13. Show that the vectors u1¼ð1;1;1Þ,u2¼ð1;2;3Þ,u3¼ð1;5;8Þspan R3. We need to show that an arbitrary vector v¼ða;b;cÞinR3is a linear combination of u1,u2,u3. Set v¼xu1þyu2þzu3; that is, set ða;b;cÞ¼xð1;1;1Þþyð1;2;3Þþzð1;5;8Þ¼ð xþyþz;xþ2yþ5z;xþ3yþ8zÞ136 CHAPTER 4 Vector Spaces Form the equivalent system and reduce it to echelon form: xþyþz¼a xþ2yþ5z¼b xþ3yþ8z¼corxþyþz¼a yþ4z¼b/C0a 2yþ7c¼c/C0aorxþyþz¼a yþ4z¼b/C0a /C0z¼c/C02bþa The above system is in echelon form and is consistent; in fact, x¼/C0aþ5b/C03c;y¼3a/C07bþ4c;z¼aþ2b/C0c is a solution. Thus, u1,u2,u3span R3. 4.14. Find conditions on a,b,cso that v¼ða;b;cÞinR3belongs to W¼spanðu1;u2;u3Þ;where u1¼ð1;2;0Þ;u2¼ð/C0 1;1;2Þ;u3¼ð3;0;/C04Þ Setvas a linear combination of u1,u2,u3using unknowns x,y,z; that is, set v¼xu1þyu2þzu3:This yields ða;b;cÞ¼xð1;2;0Þþyð/C01;1;2Þþzð3;0;/C04Þ¼ð x/C0yþ3z;2xþy;2y/C04zÞ Form the equivalent system of linear equations and reduce it to echelon form: x/C0yþ3z¼a 2xþy¼b 2y/C04z¼corx/C0yþ3z¼a 3y/C06z¼b/C02a 2y/C04z¼corx/C0yþ3z¼a 3y/C06z¼b/C02a 0¼4a/C02bþ3c The vector v¼ða;b;cÞbelongs to Wif and only if the system is consistent, and it is consistent if and only if 4a/C02bþ3c¼0. Note, in particular, that u1,u2,u3do not span the whole space R3. 4.15. Show that the vector space V¼PðtÞof real polynomials cannot be spanned by a ﬁnite number of polynomials. Any ﬁnite set Sof polynomials contains a polynomial of maximum degree, say m. Then the linear span span(S) of Scannot contain a polynomial of degree greater than m. Thus, spanðSÞ6¼V, for any ﬁnite set S. 4.16. Prove Theorem 4.5: Let Sbe a subset of V. (i) Then span(S) is a subspace of Vcontaining S. (ii) If Wis a subspace of Vcontaining S, then spanðSÞ/C18W. (i) Suppose Sis empty. By deﬁnition, span ðSÞ¼f 0g. Hence spanðSÞ¼f 0gis a subspace of Vand S/C18spanðSÞ. Suppose Sis not empty and v2S. Then v¼1v2spanðSÞ; hence, S/C18spanðSÞ. Also 0¼0v2spanðSÞ. Now suppose u;w2spanðSÞ, say u¼a1u1þ/C1/C1/C1þ arur¼P iaiui and w¼b1w1þ/C1/C1/C1þ bsws¼P jbjwj where ui,wj2Sandai;bj2K. Then uþv¼P iaiuiþP jbjwj and ku¼kP iaiui/C18/C19 ¼P ikaiui belong to span(S) because each is a linear combination of vectors in S. Thus, span(S) is a subspace of V. (ii) Suppose u1;u2;...;ur2S. Then all the uibelong to W. Thus, all multiples a1u1;a2u2;...;arur2W, and so the sum a1u1þa2u2þ/C1/C1/C1þ arur2W. That is, Wcontains all linear combinations of elements inS, or, in other words, span ðSÞ/C18W, as claimed. Linear Dependence 4.17. Determine whether or not uand vare linearly dependent, where (a) u¼ð1;2Þ,v¼ð3;/C05Þ, (c) u¼ð1;2;/C03Þ,v¼ð4;5;/C06Þ (b) u¼ð1;/C03Þ,v¼ð/C0 2;6Þ, (d) u¼ð2;4;/C08Þ,v¼ð3;6;/C012Þ Two vectors uand vare linearly dependent if and only if one is a multiple of the other. (a) No. (b) Yes; for v¼/C02u. (c) No. (d) Yes, for v¼3 2u.CHAPTER 4 Vector Spaces 137 4.18. Determine whether or not uand vare linearly dependent, where (a) u¼2t2þ4t/C03,v¼4t2þ8t/C06, (b) u¼2t2/C03tþ4,v¼4t2/C03tþ2, (c) u¼13/C04 50/C01/C20/C21 ;v¼/C04/C012 16 /C020 0 4/C20/C21 , (d) u¼111 222/C20/C21 ;v¼222 333/C20/C21 Two vectors uand vare linearly dependent if and only if one is a multiple of the other. (a) Yes; for v¼2u. (b) No. (c) Yes, for v¼/C04u. (d) No. 4.19. Determine whether or not the vectors u¼ð1;1;2Þ,v¼ð2;3;1Þ,w¼ð4;5;5ÞinR3are linearly dependent. Method 1. Set a linear combination of u,v,wequal to the zero vector using unknowns x,y,zto obtain the equivalent homogeneous system of linear equations and then reduce the system to echelon form. This yields x1 1 12 43 5þy2 3 12 43 5þz4 5 52 43 5¼0 0 02 43 5 orxþ2yþ4z¼0 xþ3yþ5z¼0 2xþyþ5z¼0orxþ2yþ4z¼0 yþz¼0 The echelon system has only two nonzero equations in three unknowns; hence, it has a free variable and a nonzero solution. Thus, u,v,ware linearly dependent. Method 2. Form the matrix Awhose columns are u,v,wand reduce to echelon form: A¼124 135 2152 43 5/C24124 011 0/C03/C032 43 5/C24124 011 0002 43 5 The third column does not have a pivot; hence, the third vector wis a linear combination of the ﬁrst two vectors uand v. Thus, the vectors are linearly dependent. (Observe that the matrix Ais also the coefﬁcient matrix in Method 1. In other words, this method is essentially the same as the ﬁrst method.) Method 3. Form the matrix Bwhose rows are u,v,w, and reduce to echelon form: B¼112 231 4552 43 5/C2401 2 01/C03 01/C032 43 5/C2411 2 01/C03 00 02 43 5 Because the echelon matrix has only two nonzero rows, the three vectors are linearly dependent. (The three given vectors span a space of dimension 2.) 4.20. Determine whether or not each of the following lists of vectors in R3is linearly dependent: (a) u1¼ð1;2;5Þ,u2¼ð1;3;1Þ,u3¼ð2;5;7Þ,u4¼ð3;1;4Þ, (b) u¼ð1;2;5Þ,v¼ð2;5;1Þ,w¼ð1;5;2Þ, (c) u¼ð1;2;3Þ,v¼ð0;0;0Þ,w¼ð1;5;6Þ. (a) Yes, because any four vectors in R3are linearly dependent. (b) Use Method 2 above; that is, form the matrix Awhose columns are the given vectors, and reduce the matrix to echelon form: A¼121 255 5122 43 5/C24121 013 0/C09/C032 43 5/C2412 1 01 3 002 42 43 5 Every column has a pivot entry; hence, no vector is a linear combination of the previous vectors. Thus, the vectors are linearly independent. (c) Because 0¼ð0;0;0Þis one of the vectors, the vectors are linearly dependent.138 CHAPTER 4 Vector Spaces 4.21. Show that the functions fðtÞ¼sint,gðtÞcost,hðtÞ¼tfrom RintoRare linearly independent. Set a linear combination of the functions equal to the zero function 0using unknown scalars x,y,z; that is, set xfþygþzh¼0. Then show x¼0,y¼0,z¼0. We emphasize that xfþygþzh¼0means that, for every value of t, we have xfðtÞþygðtÞþzhðtÞ¼0. Thus, in the equation xsintþycostþzt¼0: ðiÞSett¼0 ðiiÞSett¼p=2 ðiiiÞSett¼pto obtain to obtainto obtainxð0Þþyð1Þþzð0Þ¼0 xð1Þþyð0Þþzp=2¼0 xð0Þþyð/C01ÞþzðpÞ¼0or or ory¼0: xþpz=2¼0: /C0yþpz¼0: The three equations have only the zero solution; that is, x¼0,y¼0,z¼0. Thus, f,g,hare linearly independent. 4.22. Suppose the vectors u,v,ware linearly independent. Show that the vectors uþv,u/C0v, u/C02vþware also linearly independent. Suppose xðuþvÞþyðu/C0vÞþzðu/C02vþwÞ¼0. Then xuþxvþyu/C0yvþzu/C02zvþzw¼0 or ðxþyþzÞuþðx/C0y/C02zÞvþzw¼0 Because u,v,ware linearly independent, the coefﬁcients in the above equation are each 0; hence, xþyþz¼0; x/C0y/C02z¼0; z¼0 The only solution to the above homogeneous system is x¼0,y¼0,z¼0. Thus, uþv,u/C0v,u/C02vþw are linearly independent. 4.23. Show that the vectors u¼ð1þi;2iÞandw¼ð1;1þiÞinC2are linearly dependent over the complex ﬁeld Cbut linearly independent over the real ﬁeld R. Recall that two vectors are linearly dependent (over a ﬁeld K) if and only if one of them is a multiple of the other (by an element in K). Because ð1þiÞw¼ð1þiÞð1;1þiÞ¼ð 1þi;2iÞ¼u uandware linearly dependent over C. On the other hand, uandware linearly independent over R, as no real multiple of wcan equal u. Speciﬁcally, when kis real, the ﬁrst component of kw¼ðk;kþkiÞmust be real, and it can never equal the ﬁrst component 1 þiofu, which is complex. Basis and Dimension 4.24. Determine whether or not each of the following form a basis of R3: (a) (1, 1, 1), (1, 0, 1); (c) (1, 1, 1), (1, 2, 3), ð2;/C01;1Þ; (b) (1, 2, 3), (1, 3, 5), (1, 0, 1), (2, 3, 0); (d) (1, 1, 2), (1, 2, 5), (5, 3, 4). (a and b) No, because a basis of R3must contain exactly three elements because dim R3¼3. (c) The three vectors form a basis if and only if they are linearly independent. Thus, form the matrix whose rows are the given vectors, and row reduce the matrix to echelon form: 11 1 12 3 2/C0112 43 5/C24111 012 0/C03/C012 43 5/C24111 012 0052 43 5 The echelon matrix has no zero rows; hence, the three vectors are linearly independent, and so they do form a basis of R3.CHAPTER 4 Vector Spaces 139 (d) Form the matrix whose rows are the given vectors, and row reduce the matrix to echelon form: 112 125 5342 43 5/C24112 013 0/C02/C062 43 5/C24112 013 0002 43 5 The echelon matrix has a zero row; hence, the three vectors are linearly dependent, and so they do not form a basis of R3. 4.25. Determine whether (1, 1, 1, 1), (1, 2, 3, 2), (2, 5, 6, 4), (2, 6, 8, 5) form a basis of R4. If not, ﬁnd the dimension of the subspace they span. Form the matrix whose rows are the given vectors, and row reduce to echelon form: B¼1111 1232 2564 26852 6643 775/C241111 0121 0342 04632 6643 775/C241 111 0 121 00/C02/C01 00/C02/C012 6643 775/C241111 0121 0021 00002 6643 775 The echelon matrix has a zero row. Hence, the four vectors are linearly dependent and do not form a basis of R 4. Because the echelon matrix has three nonzero rows, the four vectors span a subspace of dimension 3. 4.26. Extendfu1¼ð1;1;1;1Þ;u2¼ð2;2;3;4Þgto a basis of R4. First form the matrix with rows u1andu2, and reduce to echelon form: 1111 2234/C20/C21 /C241111 0012/C20/C21 Then w1¼ð1;1;1;1Þandw2¼ð0;0;1;2Þspan the same set of vectors as spanned by u1andu2. Let u3¼ð0;1;0;0Þandu4¼ð0;0;0;1Þ. Then w1,u3,w2,u4form a matrix in echelon form. Thus, they are linearly independent, and they form a basis of R4. Hence, u1,u2,u3,u4also form a basis of R4. 4.27. Consider the complex ﬁeld C, which contains the real ﬁeld R, which contains the rational ﬁeld Q. (Thus, Cis a vector space over R, and Ris a vector space over Q.) (a) Show thatf1;igis a basis of Cover R; hence, Cis a vector space of dimension 2 over R. (b) Show that Ris a vector space of inﬁnite dimension over Q. (a) For any v2C, we have v¼aþbi¼að1ÞþbðiÞ, where a;b2R. Hence,f1;igspans Cover R. Furthermore, if xð1ÞþyðiÞ¼0o r xþyi¼0, where x,y2R, then x¼0 and y¼0. Hence,f1;igis linearly independent over R. Thus,f1;igis a basis for Cover R. (b) It can be shown that pis a transcendental number; that is, pis not a root of any polynomial over Q. Thus, for any n, the nþ1 real numbers 1 ;p;p2;...;pnare linearly independent over Q.Rcannot be of dimension nover Q. Accordingly, Ris of inﬁnite dimension over Q. 4.28. Suppose S¼fu1;u2;...;ungis a subset of V. Show that the following Deﬁnitions A and B of a basis of Vare equivalent: (A) Sis linearly independent and spans V. (B) Every v2Vis a unique linear combination of vectors in S. Suppose (A) holds. Because Sspans V, the vector vis a linear combination of the ui, say u¼a1u1þa2u2þ/C1/C1/C1þ anun and u¼b1u1þb2u2þ/C1/C1/C1þ bnun Subtracting, we get 0¼v/C0v¼ða1/C0b1Þu1þða2/C0b2Þu2þ/C1/C1/C1þð an/C0bnÞun140 CHAPTER 4 Vector Spaces But the uiare linearly independent. Hence, the coefﬁcients in the above relation are each 0: a1/C0b1¼0; a2/C0b2¼0; ...; an/C0bn¼0 Therefore, a1¼b1;a2¼b2;...;an¼bn. Hence, the representation of vas a linear combination of the uiis unique. Thus, (A) implies (B). Suppose (B) holds. Then Sspans V. Suppose 0¼c1u1þc2u2þ/C1/C1/C1þ cnun However, we do have 0¼0u1þ0u2þ/C1/C1/C1þ 0un By hypothesis, the representation of 0 as a linear combination of the uiis unique. Hence, each ci¼0 and the uiare linearly independent. Thus, (B) implies (A). Dimension and Subspaces 4.29. Find a basis and dimension of the subspace WofR3where (a) W¼fð a;b;cÞ:aþbþc¼0g, (b) W¼fð a;b;cÞ:ða¼b¼cÞg (a) Note that W6¼R3, because, for example, ð1;2;3Þ62W. Thus, dim W<3. Note that u1¼ð1;0;/C01Þ andu2¼ð0;1;/C01Þare two independent vectors in W. Thus, dim W¼2, and so u1andu2form a basis ofW. (b) The vector u¼ð1;1;1Þ2W. Any vector w2Whas the form w¼ðk;k;kÞ. Hence, w¼ku. Thus, u spans Wand dim W¼1. 4.30. LetWbe the subspace of R4spanned by the vectors u1¼ð1;/C02;5;/C03Þ; u2¼ð2;3;1;/C04Þ; u3¼ð3;8;/C03;/C05Þ (a) Find a basis and dimension of W. (b) Extend the basis of Wto a basis of R4. (a) Apply Algorithm 4.1, the row space algorithm. Form the matrix whose rows are the given vectors, and reduce it to echelon form: A¼1/C025/C03 231/C04 38/C03/C052 43 5/C241/C025/C03 07/C092 01 4/C018 42 43 5/C241/C025/C03 07/C092 00002 43 5 The nonzero rows ð1;/C02;5;/C03Þandð0;7;/C09;2Þof the echelon matrix form a basis of the row space ofAand hence of W. Thus, in particular, dim W¼2. (b) We seek four linearly independent vectors, which include the above two vectors. The four vectors ð1;/C02;5;/C03Þ,ð0;7;/C09;2Þ, (0, 0, 1, 0), and (0, 0, 0, 1) are linearly independent (because they form an echelon matrix), and so they form a basis of R4, which is an extension of the basis of W. 4.31. Let Wbe the subspace of R5spanned by u1¼ð1;2;/C01;3;4Þ,u2¼ð2;4;/C02;6;8Þ, u3¼ð1;3;2;2;6Þ,u4¼ð1;4;5;1;8Þ,u5¼ð2;7;3;3;9Þ. Find a subset of the vectors that form a basis of W. Here we use Algorithm 4.2, the casting-out algorithm. Form the matrix Mwhose columns (not rows) are the given vectors, and reduce it to echelon form: M¼1 2112 2 4347 /C01/C02253 3 6213 4 86892 666643 77775/C241 2112 0 0123 0 0365 00/C01/C02/C03 0 02412 666643 77775/C241211 2 0012 3 0000/C04 0000 0 0000 02 666643 77775 The pivot positions are in columns C1,C3,C5. Hence, the corresponding vectors u1,u3,u5form a basis of W, and dim W¼3.CHAPTER 4 Vector Spaces 141 4.32. LetVbe the vector space of 2 /C22 matrices over K. Let Wbe the subspace of symmetric matrices. Show that dim W¼3, by ﬁnding a basis of W. Recall that a matrix A¼½aij/C138is symmetric if AT¼A, or, equivalently, each aij¼aji. Thus, A¼ab bd/C20/C21 denotes an arbitrary 2 /C22 symmetric matrix. Setting (i) a¼1,b¼0,d¼0; (ii) a¼0,b¼1,d¼0; (iii)a¼0,b¼0,d¼1, we obtain the respective matrices: E1¼10 00/C20/C21 ; E2¼01 10/C20/C21 ; E3¼00 01/C20/C21 We claim that S¼fE1;E2;E3gis a basis of W; that is, (a) Sspans Wand (b) Sis linearly independent. (a) The above matrix A¼ab bd/C20/C21 ¼aE1þbE2þdE3. Thus, Sspans W. (b) Suppose xE1þyE2þzE3¼0, where x,y,zare unknown scalars. That is, suppose x10 00/C20/C21 þy01 10/C20/C21 þz00 01/C20/C21 ¼00 00/C20/C21 orxy yz/C20/C21 ¼00 00/C20/C21 Setting corresponding entries equal to each other yields x¼0,y¼0,z¼0. Thus, Sis linearly independent. Therefore, Sis a basis of W, as claimed. Theorems on Linear Dependence, Basis, and Dimension 4.33. Prove Lemma 4.10: Suppose two or more nonzero vectors v1;v2;...;vmare linearly dependent. Then one of them is a linear combination of the preceding vectors. Because the viare linearly dependent, there exist scalars a1;...;am, not all 0, such that a1v1þ/C1/C1/C1þ amvm¼0. Let kbe the largest integer such that ak6¼0. Then a1v1þ/C1/C1/C1þ akvkþ0vkþ1þ/C1/C1/C1þ 0vm¼0o r a1v1þ/C1/C1/C1þ akvk¼0 Suppose k¼1; then a1v1¼0,a16¼0, and so v1¼0. But the viare nonzero vectors. Hence, k>1 and vk¼/C0a/C01 ka1v1/C0/C1/C1/C1/C0 a/C01 kak/C01vk/C01 That is, vkis a linear combination of the preceding vectors. 4.34. Suppose S¼fv1;v2;...;vmgspans a vector space V. (a) If w2V, thenfw;v1;...;vmgis linearly dependent and spans V. (b) If viis a linear combination of v1;...;vi/C01, then Swithout vispans V. (a) The vector wis a linear combination of the vi, becausefvigspans V. Accordingly,fw;v1;...;vmgis linearly dependent. Clearly, wwith the vispan V, as the viby themselves span V; that is,fw;v1;...;vmg spans V. (b) Suppose vi¼k1v1þ/C1/C1/C1þ ki/C01vi/C01. Let u2V. Becausefvigspans V,uis a linear combination of the vj’s, say u¼a1v1þ/C1/C1/C1þ amvm:Substituting for vi, we obtain u¼a1v1þ/C1/C1/C1þ ai/C01vi/C01þaiðk1v1þ/C1/C1/C1þ ki/C01vi/C01Þþaiþ1viþ1þ/C1/C1/C1þ amvm ¼ða1þaik1Þv1þ/C1/C1/C1þð ai/C01þaiki/C01Þvi/C01þaiþ1viþ1þ/C1/C1/C1þ amvm Thus,fv1;...;vi/C01;viþ1;...;vmgspans V. In other words, we can delete vifrom the spanning set and still retain a spanning set. 4.35. Prove Lemma 4.13: Suppose fv1;v2;...;vngspans V, and supposefw1;w2;...;wmgis linearly independent. Then m/C20n, and Vis spanned by a set of the form fw1;w2;...;wm;vi1;vi2;...;vin/C0mg Thus, any nþ1 or more vectors in Vare linearly dependent.142 CHAPTER 4 Vector Spaces It sufﬁces to prove the lemma in the case that the viare all not 0. (Prove!) Because fvigspans V,w e have by Problem 4.34 that fw1;v1;...;vngð 1Þ is linearly dependent and also spans V. By Lemma 4.10, one of the vectors in (1) is a linear combination of the preceding vectors. This vector cannot be w1, so it must be one of the v’s, say vj:Thus by Problem 4.34, we can delete vjfrom the spanning set (1) and obtain the spanning set fw1;v1;...;vj/C01;vjþ1;...;vngð 2Þ Now we repeat the argument with the vector w2. That is, because (2) spans V, the set fw1;w2;v1;...;vj/C01;vjþ1;...;vngð 3Þ is linearly dependent and also spans V. Again by Lemma 4.10, one of the vectors in (3) is a linear combination of the preceding vectors. We emphasize that this vector cannot be w1orw2, because fw1;...;wmgis independent; hence, it must be one of the v’s, say vk. Thus, by Problem 4.34, we can delete vkfrom the spanning set (3) and obtain the spanning set fw1;w2;v1;...;vj/C01;vjþ1;...;vk/C01;vkþ1;...;vng We repeat the argument with w3, and so forth. At each step, we are able to add one of the w’s and delete one of the v’s in the spanning set. If m/C20n, then we ﬁnally obtain a spanning set of the required form: fw1;...;wm;vi1;...;vin/C0mg Finally, we show that m>nis not possible. Otherwise, after nof the above steps, we obtain the spanning setfw1;...;wng. This implies that wnþ1is a linear combination of w1;...;wn, which contradicts the hypothesis that fwigis linearly independent. 4.36. Prove Theorem 4.12: Every basis of a vector space Vhas the same number of elements. Supposefu1;u2;...;ungis a basis of V, and supposefv1;v2;...gis another basis of V. Becausefuig spans V, the basisfv1;v2;...gmust contain nor less vectors, or else it is linearly dependent by Problem 4.35—Lemma 4.13. On the other hand, if the basis fv1;v2;...gcontains less than nelements, thenfu1;u2;...;ungis linearly dependent by Problem 4.35. Thus, the basis fv1;v2;...gcontains exactly n vectors, and so the theorem is true. 4.37. Prove Theorem 4.14: Let Vbe a vector space of ﬁnite dimension n. Then (i) Any nþ1 or more vectors must be linearly dependent. (ii) Any linearly independent set S¼fu1;u2;...ungwith nelements is a basis of V. (iii) Any spanning set T¼fv1;v2;...;vngofVwith nelements is a basis of V. Suppose B¼fw1;w2;...;wngis a basis of V. (i) Because Bspans V, any nþ1 or more vectors are linearly dependent by Lemma 4.13. (ii) By Lemma 4.13, elements from Bcan be adjoined to Sto form a spanning set of Vwith nelements. Because Salready has nelements, Sitself is a spanning set of V. Thus, Sis a basis of V. (iii) Suppose Tis linearly dependent. Then some viis a linear combination of the preceding vectors. By Problem 4.34, Vis spanned by the vectors in Twithout viand there are n/C01 of them. By Lemma 4.13, the independent set Bcannot have more than n/C01 elements. This contradicts the fact that Bhas nelements. Thus, Tis linearly independent, and hence Tis a basis of V. 4.38. Prove Theorem 4.15: Suppose Sspans a vector space V. Then (i) Any maximum number of linearly independent vectors in Sform a basis of V. (ii) Suppose one deletes from Severy vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V. (i) Supposefv1;...;vmgis a maximum linearly independent subset of S, and suppose w2S. Accord- ingly,fv1;...;vm;wgis linearly dependent. No vkcan be a linear combination of preceding vectors.CHAPTER 4 Vector Spaces 143 Hence, wis a linear combination of the vi. Thus, w2spanðviÞ, and hence S/C18spanðviÞ. This leads to V¼spanðSÞ/C18spanðviÞ/C18V Thus,fvigspans V, and, as it is linearly independent, it is a basis of V. (ii) The remaining vectors form a maximum linearly independent subset of S; hence, by (i), it is a basis ofV. 4.39. Prove Theorem 4.16: Let Vbe a vector space of ﬁnite dimension and let S¼fu1;u2;...;urgbe a set of linearly independent vectors in V. Then Sis part of a basis of V; that is, Smay be extended to a basis of V. Suppose B¼fw1;w2;...;wngis a basis of V. Then Bspans V, and hence Vis spanned by S[B¼fu1;u2;...;ur;w1;w2;...;wng By Theorem 4.15, we can delete from S[Beach vector that is a linear combination of preceding vectors to obtain a basis B0forV. Because Sis linearly independent, no ukis a linear combination of preceding vectors. Thus, B0contains every vector in S, and Sis part of the basis B0forV. 4.40. Prove Theorem 4.17: Let Wbe a subspace of an n-dimensional vector space V. Then dim W/C20n. In particular, if dim W¼n, then W¼V. Because Vis of dimension n, any nþ1 or more vectors are linearly dependent. Furthermore, because a basis of Wconsists of linearly independent vectors, it cannot contain more than nelements. Accordingly, dimW/C20n. In particular, iffw1;...;wngis a basis of W, then, because it is an independent set with nelements, it is also a basis of V. Thus, W¼Vwhen dim W¼n. Rank of a Matrix, Row and Column Spaces 4.41. Find the rank and basis of the row space of each of the following matrices: (a) A¼12 0/C01 26/C03/C03 31 0/C06/C052 43 5, (b) B¼13 1/C02/C03 14 3/C01/C04 23/C04/C07/C03 38 1/C07/C082 6643 775. (a) Row reduce Ato echelon form: A/C2412 0/C01 02/C03/C01 04/C06/C022 43 5/C2412 0/C01 02/C03/C01 0 0002 43 5 The two nonzero rows ð1;2;0;/C01Þandð0;2;/C03;/C01Þof the echelon form of Aform a basis for rowsp(A). In particular, rank ðAÞ¼2. (b) Row reduce Bto echelon form: B/C24131/C02/C03 0121 /C01 0/C03/C06/C033 0/C01/C02/C0112 6643 775/C24131/C02/C03 012 1/C01 0 0 000 0 0 0002 6643 775 The two nonzero rows ð1;3;1;/C02;/C03Þandð0;1;2;1;/C01Þof the echelon form of Bform a basis for rowsp(B). In particular, rank ðBÞ¼2. 4.42. Show that U¼W, where UandWare the following subspaces of R3: U¼spanðu1;u2;u3Þ¼spanð1;1;/C01Þ;ð2;3;/C01Þ;ð3;1;/C05Þg W¼spanðw1;w2;w3Þ¼spanð1;/C01;/C03Þ;ð3;/C02;/C08Þ;ð2;1;/C03Þg144 CHAPTER 4 Vector Spaces Form the matrix Awhose rows are the ui, and row reduce Ato row canonical form: A¼11/C01 23/C01 31/C052 43 5/C2411/C01 0110/C02/C022 43 5/C2410/C02 01 100 02 43 5 Next form the matrix Bwhose rows are the w j, and row reduce Bto row canonical form: B¼1/C01/C03 3/C02/C08 21/C032 43 5/C241/C01/C03 011 0332 43 5/C2410/C02 01 1 00 02 43 5 Because AandBhave the same row canonical form, the row spaces of AandBare equal, and so U¼W. 4.43. LetA¼121 2 3 1 243 7 7 4 122 5 5 6 3 6 6 15 14 152 6643 775. (a) Find rankðMkÞ, for k¼1;2;...;6, where Mkis the submatrix of Aconsisting of the ﬁrst k columns C1;C2;...;CkofA. (b) Which columns Ckþ1are linear combinations of preceding columns C1;...;Ck? (c) Find columns of Athat form a basis for the column space of A. (d) Express column C4as a linear combination of the columns in part (c). (a) Row reduce Ato echelon form: A/C2412123 1 00131 2 00132 5 003951 22 6643 775/C24121231 001312 000013 0000002 6643 775 Observe that this simultaneously reduces all the matrices Mkto echelon form; for example, the ﬁrst four columns of the echelon form of Aare an echelon form of M4. We know that rank ðMkÞis equal to the number of pivots or, equivalently, the number of nonzero rows in an echelon form of Mk. Thus, rankðM1Þ¼rankðM2Þ¼1; rankðM3Þ¼rankðM4Þ¼2 rankðM5Þ¼rankðM6Þ¼3 (b) The vector equation x1C1þx2C2þ/C1/C1/C1þ xkCk¼Ckþ1yields the system with coefﬁcient matrix Mk and augmented Mkþ1. Thus, Ckþ1is a linear combination of C1;...;Ckif and only if rankðMkÞ¼rankðMkþ1Þor, equivalently, if Ckþ1does not contain a pivot. Thus, each of C2,C4,C6 is a linear combination of preceding columns. (c) In the echelon form of A, the pivots are in the ﬁrst, third, and ﬁfth columns. Thus, columns C1,C3,C5 ofAform a basis for the columns space of A. Alternatively, deleting columns C2,C4,C6from the spanning set of columns (they are linear combinations of other columns), we obtain, again, C1,C3,C5. (d) The echelon matrix tells us that C4is a linear combination of columns C1andC3. The augmented matrix Mof the vector equation C4¼xC1þyC2consists of the columns C1,C3,C4ofAwhich, when reduced to echelon form, yields the matrix (omitting zero rows) 112 013/C20/C21 orxþy¼2 y¼3or x¼/C01;y¼3 Thus, C4¼/C0C1þ3C3¼/C0C1þ3C3þ0C5. 4.44. Suppose u¼ða1;a2;...;anÞis a linear combination of the rows R1;R2;...;Rmof a matrix B¼½bij/C138, say u¼k1R1þk2R2þ/C1/C1/C1þ kmRm:Prove that ai¼k1b1iþk2b2iþ/C1/C1/C1þ kmbmi; i¼1;2;...;n where b1i;b2i;...;bmiare the entries in the ith column of B.CHAPTER 4 Vector Spaces 145 We are given that u¼k1R1þk2R2þ/C1/C1/C1þ kmRm. Hence, ða1;a2;...;anÞ¼k1ðb11;...;b1nÞþ/C1/C1/C1þ kmðbm1;...;bmnÞ ¼ðk1b11þ/C1/C1/C1þ kmbm1;...;k1b1nþ/C1/C1/C1þ kmbmnÞ Setting corresponding components equal to each other, we obtain the desired result. 4.45. Prove Theorem 4.7: Suppose A¼½aij/C138andB¼½bij/C138are row equivalent echelon matrices with respective pivot entries a1j1;a2j2;...;arjrand b1k1;b2k2;...;bsks (pictured in Fig. 4-5). Then AandBhave the same number of nonzero rows—that is, r¼s—and their pivot entries are in the same positions; that is, j1¼k1;j2¼k2;...;jr¼kr. Clearly A¼0 if and only if B¼0, and so we need only prove the theorem when r/C211 and s/C211. We ﬁrst show that j1¼k1. Suppose j1<k1. Then the j1th column of Bis zero. Because the ﬁrst row R*o fAis in the row space of B, we have R*¼c1R1þc1R2þ/C1/C1/C1þ cmRm, where the Riare the rows of B. Because the j1th column of Bis zero, we have a1j1¼c10þc20þ/C1/C1/C1þ cm0¼0 But this contradicts the fact that the pivot entry a1j16¼0. Hence, j1/C21k1and, similarly, k1/C21j1. Thus j1¼k1. Now let A0be the submatrix of Aobtained by deleting the ﬁrst row of A, and let B0be the submatrix of B obtained by deleting the ﬁrst row of B. We prove that A0andB0have the same row space. The theorem will then follow by induction, because A0andB0are also echelon matrices. LetR¼ða1;a2;...;anÞbe any row of A0and let R1;...;Rmbe the rows of B. Because Ris in the row space of B, there exist scalars d1;...;dmsuch that R¼d1R1þd2R2þ/C1/C1/C1þ dmRm. Because Ais in echelon form and Ris not the ﬁrst row of A,t h e j1th entry of Ris zero: ai¼0f o r i¼j1¼k1. Furthermore, because Bis in echelon form, all the entries in the k1th column of Bare 0 except the ﬁrst: b1k16¼0, but b2k1¼0;...;bmk1¼0. Thus, 0¼ak1¼d1b1k1þd20þ/C1/C1/C1þ dm0¼d1b1k1 Now b1k16¼0 and so d1¼0. Thus, Ris a linear combination of R2;...;Rmand so is in the row space of B0. Because Rwas any row of A0, the row space of A0is contained in the row space of B0. Similarly, the row space of B0is contained in the row space of A0. Thus, A0andB0have the same row space, and so the theorem is proved. 4.46. Prove Theorem 4.8: Suppose AandBare row canonical matrices. Then AandBhave the same row space if and only if they have the same nonzero rows. Obviously, if AandBhave the same nonzero rows, then they have the same row space. Thus we only have to prove the converse. Suppose AandBhave the same row space, and suppose R6¼0 is the ith row of A. Then there exist scalars c1;...;cssuch that R¼c1R1þc2R2þ/C1/C1/C1þ csRs ð1Þ where the Riare the nonzero rows of B. The theorem is proved if we show that R¼Ri; that is, that ci¼1 but ck¼0 for k6¼i.A¼a1j1/C3/C3/C3/C3/C3/C3 a2j2/C3/C3/C3/C3 :::::::::::::::::::::::::::::::::::::: arjr/C3/C32 6643 775; b¼b1k1/C3/C3/C3/C3/C3/C3 b2k2/C3/C3/C3/C3 :::::::::::::::::::::::::::::::::::::: bsks/C3/C32 6643 775 Figure 4-5146 CHAPTER 4 Vector Spaces Letaij, be the pivot entry in R—that is, the ﬁrst nonzero entry of R. By (1) and Problem 4.44, aiji¼c1b1jiþc2b2jiþ/C1/C1/C1þ csbsjið2Þ But, by Problem 4.45, bijiis a pivot entry of B, and, as Bis row reduced, it is the only nonzero entry in the jth column of B. Thus, from (2), we obtain aiji¼cibiji. However, aiji¼1 and biji¼1, because AandBare row reduced; hence, ci¼1. Now suppose k6¼i, and bkjkis the pivot entry in Rk. By (1) and Problem 4.44, aijk¼c1b1jkþc2b2jkþ/C1/C1/C1þ csbsjkð3Þ Because Bis row reduced, bkjkis the only nonzero entry in the jth column of B. Hence, by (3), aijk¼ckbkjk. Furthermore, by Problem 4.45, akjkis a pivot entry of A, and because Ais row reduced, aijk¼0. Thus, ckbkjk¼0, and as bkjk¼1,ck¼0. Accordingly R¼Ri;and the theorem is proved. 4.47. Prove Corollary 4.9: Every matrix Ais row equivalent to a unique matrix in row canonical form. Suppose Ais row equivalent to matrices A1andA2, where A1andA2are in row canonical form. Then rowspðAÞ¼rowspðA1Þand rowspðAÞ¼rowspðA2Þ. Hence, rowspðA1Þ¼rowspðA2Þ. Because A1andA2are in row canonical form, A1¼A2by Theorem 4.8. Thus, the corollary is proved. 4.48. Suppose RBandABare deﬁned, where Ris a row vector and AandBare matrices. Prove (a) RBis a linear combination of the rows of B. (b) The row space of ABis contained in the row space of B. (c) The column space of ABis contained in the column space of A. (d) If Cis a column vector and ACis deﬁned, then ACis a linear combination of the columns ofA: (e) rankðABÞ/C20rankðBÞand rankðABÞ/C20rankðAÞ. (a) Suppose R¼ða1;a2;...;amÞandB¼½bij/C138. Let B1;...;Bmdenote the rows of BandB1;...;Bnits columns. Then RB¼ðRB1;RB2;...;RBnÞ ¼ða1b11þa2b21þ/C1/C1/C1þ ambm1; ...;a1b1nþa2b2nþ/C1/C1/C1þ ambmnÞ ¼a1ðb11;b12;...;b1nÞþa2ðb21;b22;...;b2nÞþ/C1/C1/C1þ amðbm1;bm2;...;bmnÞ ¼a1B1þa2B2þ/C1/C1/C1þ amBm Thus, RBis a linear combination of the rows of B, as claimed. (b) The rows of ABareRiB, where Riis the ith row of A. Thus, by part (a), each row of ABis in the row space of B. Thus, rowspðABÞ/C18rowspðBÞ, as claimed. (c) Using part (b), we have colsp ðABÞ¼rowspðABÞT¼rowspðBTATÞ/C18rowspðATÞ¼colspðAÞ: (d) Follows from ðcÞwhere Creplaces B: (e) The row space of ABis contained in the row space of B; hence, rankðABÞ/C20rankðBÞ. Furthermore, the column space of ABis contained in the column space of A; hence, rankðABÞ/C20rankðAÞ. 4.49. LetAbe an n-square matrix. Show that Ais invertible if and only if rank ðAÞ¼n. Note that the rows of the n-square identity matrix Inare linearly independent, because Inis in echelon form; hence, rankðInÞ¼n. Now if Ais invertible, then Ais row equivalent to In; hence, rankðAÞ¼n. But if Ais not invertible, then Ais row equivalent to a matrix with a zero row; hence, rank ðAÞ<n; that is, Ais invertible if and only if rank ðAÞ¼n.CHAPTER 4 Vector Spaces 147 Applications to Linear Equations 4.50. Find the dimension and a basis of the solution space Wof each homogeneous system: xþ2yþ2z/C0sþ3t¼0 xþ2yþ3zþsþt¼0 3xþ6yþ8zþsþ5t¼0 (a)xþ2yþz/C02t¼0 2xþ4yþ4z/C03t¼0 3xþ6yþ7z/C04t¼0 (b)xþyþ2z¼0 2xþ3yþ3z¼0 xþ3yþ5z¼0 (c) (a) Reduce the system to echelon form: xþ2yþ2z/C0sþ3t¼0 zþ2s/C02t¼0 2zþ4s/C04t¼0orxþ2yþ2z/C0sþ3t¼0 zþ2s/C02t¼0 The system in echelon form has two (nonzero) equations in ﬁve unknowns. Hence, the system has 5/C02¼3 free variables, which are y,s,t. Thus, dim W¼3. We obtain a basis for W: ð1ÞSety¼1;s¼0;t¼0 to obtain the solution v1¼ð/C0 2;1;0;0;0Þ: ð2ÞSety¼0;s¼1;t¼0 to obtain the solution v2¼ð5;0;/C02;1;0Þ: ð3ÞSety¼0;s¼0;t¼1 to obtain the solution v3¼ð/C0 7;0;2;0;1Þ: The setfv1;v2;v3gis a basis of the solution space W. (b) (Here we use the matrix format of our homogeneous system.) Reduce the coefﬁcient matrix Ato echelon form: A¼121/C02 244/C03 367/C042 43 5/C24121/C02 002 1 004 22 43 5/C24121/C02 002 1 000 02 43 5 This corresponds to the system xþ2yþ2z/C02t¼0 2zþt¼0 The free variables are yandt, and dim W¼2. (i) Set y¼1,z¼0 to obtain the solution u1¼ð/C0 2;1;0;0Þ. (ii) Set y¼0,z¼2 to obtain the solution u2¼ð6;0;/C01;2Þ. Thenfu1;u2gis a basis of W. (c) Reduce the coefﬁcient matrix Ato echelon form: A¼112 233 1352 43 5/C2411 2 01/C01 02 32 43 5/C2411 2 01/C01 00 52 43 5 This corresponds to a triangular system with no free variables. Thus, 0 is the only solution; that is, W¼f0g. Hence, dim W¼0. 4.51. Find a homogeneous system whose solution set Wis spanned by fu1;u2;u3g¼fð 1;/C02;0;3Þ;ð1;/C01;/C01;4Þ;ð1;0;/C02;5Þg Letv¼ðx;y;z;tÞ. Then v2Wif and only if vis a linear combination of the vectors u1,u2,u3that span W. Thus, form the matrix Mwhose ﬁrst columns are u1,u2,u3and whose last column is v, and then row reduce Mto echelon form. This yields M¼111 x /C02/C010 y 0/C01/C02z 345 t2 6643 775/C24111 x 012 2 xþy 0/C01/C02 z 012/C03xþt2 6643 775/C24111 x 012 2 xþy 000 2 xþyþz 000/C05x/C0yþt2 6643 775148 CHAPTER 4 Vector Spaces Then vis a linear combination of u1,u2,u3if rankðMÞ¼rankðAÞ, where Ais the submatrix without column v. Thus, set the last two entries in the fourth column on the right equal to zero to obtain the required homogeneous system: 2xþyþz¼0 5xþy/C0t¼0 4.52. Letxi1;xi2;...;xikbe the free variables of a homogeneous system of linear equations with n unknowns. Let vjbe the solution for which xij¼1, and all other free variables equal 0. Show that the solutions v1;v2;...;vkare linearly independent. LetAbe the matrix whose rows are the vi. We interchange column 1 and column i1, then column 2 and column i2;...;then column kand column ik, and we obtain the k/C2nmatrix B¼½I;C/C138¼100 ... 00 c1;kþ1... c1n 010 ... 00 c2;kþ1... c2n ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 000 ... 01 ck;kþ1... ckn2 6643 775 The above matrix Bis in echelon form, and so its rows are independent; hence, rank ðBÞ¼k. Because Aand Bare column equivalent, they have the same rank—rank ðAÞ¼k. But Ahaskrows; hence, these rows (i.e., thev i) are linearly independent, as claimed. Sums, Direct Sums, Intersections 4.53. LetUandWbe subspaces of a vector space V. Show that (a) UþVis a subspace of V. (b) UandWare contained in UþW. (c) UþWis the smallest subspace containing UandW; that is, UþW¼spanðU;WÞ. (d) WþW¼W. (a) Because UandWare subspaces, 02Uand 02W. Hence, 0¼0þ0 belongs to UþW. Now suppose v;v02UþW. Then v¼uþwand v0¼u0þv0, where u;u02Uandw;w02W. Then avþbv0¼ðauþbu0Þþð awþbw0Þ2UþW Thus, UþWis a subspace of V. (b) Let u2U. Because Wis a subspace, 02W. Hence, u¼uþ0 belongs to UþW. Thus, U/C18UþW. Similarly, W/C18UþW. (c) Because UþWis a subspace of Vcontaining UandW, it must also contain the linear span of Uand W. That is, spanðU;WÞ/C18UþW. On the other hand, if v2UþW, then v¼uþw¼1uþ1w, where u2Uandw2W. Thus, vis a linear combination of elements in U[W, and so v2spanðU;WÞ. Hence, UþW/C18spanðU;WÞ. The two inclusion relations give the desired result. (d) Because Wis a subspace of V, we have that Wis closed under vector addition; hence, WþW/C18W.B y part (a), W/C18WþW. Hence, WþW¼W. 4.54. Consider the following subspaces of R5: U¼spanðu1;u2;u3Þ¼spanfð1;3;/C02;2;3Þ;ð1;4;/C03;4;2Þ;ð2;3;/C01;/C02;9Þg W¼spanðw1;w2;w3Þ¼spanfð1;3;0;2;1Þ;ð1;5;/C06;6;3Þ;ð2;5;3;2;1Þg Find a basis and the dimension of (a) UþW, (b) U\W.CHAPTER 4 Vector Spaces 149 (a) UþWis the space spanned by all six vectors. Hence, form the matrix whose rows are the given six vectors, and then row reduce to echelon form: 13/C022 3 14/C034 2 23/C01/C029 1 302 115/C066 3 2 532 12 66666643 7777775/C2413/C0223 01/C012/C01 0/C033/C063 0020 /C02 02/C0440 0/C017/C02/C052 66666643 7777775/C2413/C022 3 01/C012/C01 00 10/C01 00 00 0 00 00 0 00 00 02 66666643 7777775 The following three nonzero rows of the echelon matrix form a basis of U\W: ð1;3;/C02;2;2;3Þ;ð0;1;/C01;2;/C01Þ;ð0;0;1;0;/C01Þ Thus, dimðUþWÞ¼3. (b) Let v¼ðx;y;z;s;tÞdenote an arbitrary element in R 5. First ﬁnd, say as in Problem 4.49, homogeneous systems whose solution sets are UandW, respectively. LetMbe the matrix whose columns are the uiand v, and reduce Mto echelon form: M¼112 x 343 y /C02/C03/C01z 24/C02s 329 t2 666643 77775/C2411 2 x 01/C03/C03xþy 00 0/C0xþyþz 00 0 4 x/C02yþs 00 0/C06xþyþt2 666643 77775 Set the last three entries in the last column equal to zero to obtain the following homogeneous system whose solution set is U: /C0xþyþz¼0; 4x/C02yþs¼0;/C06xþyþt¼0 Now let M 0be the matrix whose columns are the wiand v, and reduce M0to echelon form: M0¼11 2 x 35 5 y 0/C063 z 26 2 s 13 1 t2 666643 77775/C2411 2 x 02/C01/C03xþy 00 0/C09xþ3yþz 00 0 4 x/C02yþs 00 0 2 x/C0yþt2 666643 77775 Again set the last three entries in the last column equal to zero to obtain the following homogeneous system whose solution set is W: /C09þ3þz¼0; 4x/C02yþs¼0; 2x/C0yþt¼0 Combine both of the above systems to obtain a homogeneous system, whose solution space is U\W, and reduce the system to echelon form, yielding /C0xþyþz¼0 2yþ4zþs¼0 8zþ5sþ2t¼0 s/C02t¼0 There is one free variable, which is t; hence, dimðU\WÞ¼1. Setting t¼2, we obtain the solution u¼ð1;4;/C03;4;2Þ, which forms our required basis of U\W. 4.55. Suppose UandWare distinct four-dimensional subspaces of a vector space V, where dim V¼6. Find the possible dimensions of U\W. Because UandWare distinct, UþWproperly contains UandW; consequently, dim ðUþWÞ>4. But dimðUþWÞcannot be greater than 6, as dim V¼6. Hence, we have two possibilities: (a) dimðUþWÞ¼5 or (b) dimðUþWÞ¼6. By Theorem 4.20, dimðU\WÞ¼dimUþdimW/C0dimðUþWÞ¼8/C0dimðUþWÞ Thus (a) dimðU\WÞ¼3 or (b) dimðU\WÞ¼2.150 CHAPTER 4 Vector Spaces 4.56. LetUandWbe the following subspaces of R3: U¼fð a;b;cÞ:a¼b¼cg and W¼fð 0;b;cÞg (Note that Wis the yz-plane.) Show that R3¼U/C8W. First we show that U\W¼f0g. Suppose v¼ða;b;cÞ2U\W. Then a¼b¼canda¼0. Hence, a¼0,b¼0,c¼0. Thus, v¼0¼ð0;0;0Þ. Next we show that R3¼UþW. For, if v¼ða;b;cÞ2R3, then v¼ða;a;aÞþð 0;b/C0a;c/C0aÞ whereða;a;aÞ2Uandð0;b/C0a;c/C0aÞ2W Both conditions U\W¼f0gandUþW¼R3imply that R3¼U/C8W. 4.57. Suppose that UandWare subspaces of a vector space Vand that S¼fuigspans UandS0¼fwjg spans W. Show that S[S0spans UþW. (Accordingly, by induction, if Sispans Wi, for i¼1;2;...;n, then S1[...[Snspans W1þ/C1/C1/C1þ Wn.) Let v2UþW. Then v¼uþw, where u2Uand w2W. Because Sspans U,uis a linear combination of ui, and as S0spans W,wis a linear combination of wj; say u¼a1ui1þa2ui2þ/C1/C1/C1þ aruirand v¼b1wj1þb2wj2þ/C1/C1/C1þ bswjs where ai;bj2K. Then v¼uþw¼a1ui1þa2ui2þ/C1/C1/C1þ aruirþb1wj1þb2wj2þ/C1/C1/C1þ bswjs Accordingly, S[S0¼fui;wjgspans UþW. 4.58. Prove Theorem 4.20: Suppose UandVare ﬁnite-dimensional subspaces of a vector space V. Then UþWhas ﬁnite dimension and dimðUþWÞ¼dimUþdimW/C0dimðU\WÞ Observe that U\Wis a subspace of both Uand W. Suppose dim U¼m, dim W¼n, dimðU\WÞ¼r. Supposefv1;...;vrgis a basis of U\W. By Theorem 4.16, we can extend fvigto a basis of Uand to a basis of W; say fv1;...;vr;u1;...;um/C0rg andfv1;...;vr;w1;...;wn/C0rg are bases of UandW, respectively. Let B¼fv1;...;vr;u1;...;um/C0r;w1;...;wn/C0rg Note that Bhas exactly mþn/C0relements. Thus, the theorem is proved if we can show that Bis a basis ofUþW. Becausefvi;ujgspans Uandfvi;wkgspans W, the union B¼fvi;uj;wkgspans UþW. Thus, it sufﬁces to show that Bis independent. Suppose a1v1þ/C1/C1/C1þ arvrþb1u1þ/C1/C1/C1þ bm/C0rum/C0rþc1w1þ/C1/C1/C1þ cn/C0rwn/C0r¼0ð1Þ where ai,bj,ckare scalars. Let v¼a1v1þ/C1/C1/C1þ arvrþb1u1þ/C1/C1/C1þ bm/C0rum/C0r ð2Þ By (1), we also have v¼/C0c1w1/C0/C1/C1/C1/C0 cn/C0rwn/C0r ð3Þ Becausefvi;ujg/C18U,v2Uby (2); and asfwkg/C18W,v2Wby (3). Accordingly, v2U\W. Nowfvigis a basis of U\W, and so there exist scalars d1;...;drfor which v¼d1v1þ/C1/C1/C1þ drvr. Thus, by (3), we have d1v1þ/C1/C1/C1þ drvrþc1w1þ/C1/C1/C1þ cn/C0rwn/C0r¼0 Butfvi;wkgis a basis of W, and so is independent. Hence, the above equation forces c1¼0;...;cn/C0r¼0. Substituting this into (1), we obtain a1v1þ/C1/C1/C1þ arvrþb1u1þ/C1/C1/C1þ bm/C0rum/C0r¼0 Butfvi;ujgis a basis of U, and so is independent. Hence, the above equation forces a1¼ 0;...;ar¼0;b1¼0;...;bm/C0r¼0. Because (1) implies that the ai,bj,ckare all 0, B¼fvi;uj;wkgis independent, and the theorem is proved.CHAPTER 4 Vector Spaces 151 4.59. Prove Theorem 4.21: V¼U/C8Wif and only if (i) V¼UþW, (ii) U\W¼f0g. Suppose V¼U/C8W. Then any v2Vcan be uniquely written in the form v¼uþw, where u2Uand w2W. Thus, in particular, V¼UþW. Now suppose v2U\W. Then ð1Þv¼vþ0;where v2U;02W;ð2Þv¼0þv;where 02U;v2W: Thus, v¼0þ0¼0 and U\W¼f0g. On the other hand, suppose V¼UþWandU\W¼f0g. Let v2V. Because V¼UþW, there exist u2Uandw2Wsuch that v¼uþw. We need to show that such a sum is unique. Suppose also that v¼u0þw0, where u02Uandw02W. Then uþw¼u0þw0; and so u/C0u0¼w0/C0w Butu/C0u02Uandw0/C0w2W; hence, by U\W¼f0g, u/C0u0¼0;w0/C0w¼0; and so u¼u0;w¼w0 Thus, such a sum for v2Vis unique, and V¼U/C8W. 4.60. Prove Theorem 4.22 (for two factors): Suppose V¼U/C8W. Also, suppose S¼fu1;...;umgand S0¼fw1;...;wngare linearly independent subsets of UandW, respectively. Then (a) The union S[S0is linearly independent in V. (b) If SandS0are bases of UandW, respectively, then S[S0is a basis of V. (c) dim V¼dimUþdimW. (a) Suppose a1u1þ/C1/C1/C1þ amumþb1w1þ/C1/C1/C1þ bnwn¼0, where ai,bjare scalars. Then ða1u1þ/C1/C1/C1þ amumÞþð b1w1þ/C1/C1/C1þ bnwnÞ¼0¼0þ0 where 0 ;a1u1þ/C1/C1/C1þ amum2Uand 0 ;b1w1þ/C1/C1/C1þ bnwn2W. Because such a sum for 0 is unique, this leads to a1u1þ/C1/C1/C1þ amum¼0 and b1w1þ/C1/C1/C1þ bnwn¼0 Because S1is linearly independent, each ai¼0, and because S2is linearly independent, each bj¼0. Thus, S¼S1[S2is linearly independent. (b) By part (a), S¼S1[S2is linearly independent, and, by Problem 4.55, S¼S1[S2spans V¼UþW. Thus, S¼S1[S2is a basis of V. (c) This follows directly from part (b). Coordinates 4.61. Relative to the basis S¼fu1;u2g¼fð 1;1Þ;ð2;3ÞgofR2, ﬁnd the coordinate vector of v, where (a)v¼ð4;/C03Þ, (b) v¼ða;bÞ. In each case, set v¼xu1þyu2¼xð1;1Þþyð2;3Þ¼ð xþ2y;xþ3yÞ and then solve for xandy. (a) We have ð4;/C03Þ¼ð xþ2y;xþ3yÞ orxþ2y¼ 4 xþ3y¼/C03 The solution is x¼18,y¼/C07. Hence,½v/C138¼½18;/C07/C138. (b) We have ða;bÞ¼ð xþ2y;xþ3yÞ orxþ2y¼a xþ3y¼b The solution is x¼3a/C02b,y¼/C0aþb. Hence,½v/C138¼½3a/C02b;aþb/C138.152 CHAPTER 4 Vector Spaces 4.62. Find the coordinate vector of v¼ða;b;cÞinR3relative to (a) the usual basis E¼fð 1;0;0Þ;ð0;1;0Þ;ð0;0;1Þg, (b) the basis S¼fu1;u2;u3g¼fð 1;1;1Þ;ð1;1;0Þ;ð1;0;0Þg. (a) Relative to the usual basis E, the coordinates of ½v/C138Eare the same as v. That is,½v/C138E¼½a;b;c/C138. (b) Set vas a linear combination of u1,u2,u3using unknown scalars x,y,z. This yields a b c2 43 5¼x1 1 12 43 5þy1 1 02 43 5þz1 0 02 43 5 orxþyþz¼a xþy¼b x¼c Solving the system yields x¼c,y¼b/C0c,z¼a/C0b. Thus,½v/C138S¼½c;b/C0c;a/C0b/C138. 4.63. Consider the vector space P3ðtÞof polynomials of degree /C203. (a) Show that S¼fð t/C01Þ3;ðt/C01Þ2;t/C01;1gis a basis of P3ðtÞ. (b) Find the coordinate vector ½v/C138ofv¼3t3/C04t2þ2t/C05 relative to S. (a) The degree of ðt/C01Þkisk; writing the polynomials of Sin reverse order, we see that no polynomial is a linear combination of preceding polynomials. Thus, the polynomials are linearly independent, and, because dim P3ðtÞ¼4, they form a basis of P3ðtÞ. (b) Set vas a linear combination of the basis vectors using unknown scalars x,y,z,s. We have v¼3t3þ4t2þ2t/C05¼xðt/C01Þ3þyðt/C01Þ2þzðt/C01Þþsð1Þ ¼xðt3/C03t2þ3t/C01Þþyðt2/C02tþ1Þþzðt/C01Þþsð1Þ ¼xt3/C03xt2þ3xt/C0xþyt2/C02ytþyþzt/C0zþs ¼xt3þð/C0 3xþyÞt2þð3x/C02yþzÞtþð/C0 xþy/C0zþsÞ Then set coefﬁcients of the same powers of tequal to each other to obtain x¼3;/C03xþy¼4; 3x/C02yþz¼2;/C0xþy/C0zþs¼/C05 Solving the system yields x¼3,y¼13,z¼19,s¼4. Thus,½v/C138¼½3;13;19;4/C138. 4.64. Find the coordinate vector of A¼23 4/C07/C20/C21 in the real vector space M¼M2;2relative to (a) the basis S¼11 11/C20/C21 ;1/C01 10/C20/C21 ;1/C01 00/C20/C21 ;10 00/C20/C21 /C26/C27 , (b) the usual basis E¼10 00/C20/C21 ;01 00/C20/C21 ;00 10/C20/C21 ;00 01/C20/C21 /C26/C27 (a) Set Aas a linear combination of the basis vectors using unknown scalars x,y,z,tas follows: A¼23 4/C07/C20/C21 ¼x11 11/C20/C21 þy1/C01 10/C20/C21 þz1/C01 00/C20/C21 þt10 00/C20/C21 ¼xþzþtx/C0y/C0z xþyx/C20/C21 Set corresponding entries equal to each other to obtain the system xþzþt¼2; x/C0y/C0z¼3; xþy¼4; x¼/C07 Solving the system yields x¼/C07,y¼11,z¼/C021,t¼30. Thus,½A/C138S¼½/C0 7;11;/C021;30/C138. (Note that the coordinate vector of Ais a vector in R4, because dim M¼4.) (b) Expressing Aas a linear combination of the basis matrices yields 23 4/C07/C20/C21 ¼x10 00/C20/C21 þy01 00/C20/C21 þz00 10/C20/C21 þt00 01/C20/C21 ¼xy zt/C20/C21 Thus, x¼2,y¼3,z¼4,t¼/C07. Hence,½A/C138¼½2;3;4;/C07/C138, whose components are the elements of A written row by row.CHAPTER 4 Vector Spaces 153 Remark: This result is true in general; that is, if Ais any m/C2nmatrix in M¼Mm;n, then the coordinates of Arelative to the usual basis of Mare the elements of Awritten row by row. 4.65. In the space M¼M2;3, determine whether or not the following matrices are linearly dependent: A¼123 405/C20/C21 ; B¼24 7 10 1 13/C20/C21 ; C¼12 5 821 1/C20/C21 If the matrices are linearly dependent, ﬁnd the dimension and a basis of the subspace WofM spanned by the matrices. The coordinate vectors of the above matrices relative to the usual basis of Mare as follows: ½A/C138¼½1;2;3;4;0;5/C138;½B/C138¼½2;4;7;10;1;13/C138;½C/C138¼½1;2;5;8;2;11/C138 Form the matrix Mwhose rows are the above coordinate vectors, and reduce Mto echelon form: M¼1 2 34 05 2471 011 3 125 821 12 43 5/C24123405 001213 0000002 43 5 Because the echelon matrix has only two nonzero rows, the coordinate vectors ½A/C138,½B/C138,½C/C138span a space of dimension two, and so they are linearly dependent. Thus, A,B,Care linearly dependent. Furthermore, dimW¼2, and the matrices w1¼123 405/C20/C21 and w2¼001 213/C20/C21 corresponding to the nonzero rows of the echelon matrix form a basis of W. Miscellaneous Problems 4.66. Consider a ﬁnite sequence of vectors S¼fv1;v2;...;vng. Let Tbe the sequence of vectors obtained from Sby one of the following ‘‘elementary operations’’: (i) interchange two vectors, (ii) multiply a vector by a nonzero scalar, (iii) add a multiple of one vector to another. Show that S andTspan the same space W. Also show that Tis independent if and only if Sis independent. Observe that, for each operation, the vectors in Tare linear combinations of vectors in S. On the other hand, each operation has an inverse of the same type (Prove!); hence, the vectors in Sare linear combinations of vectors in T. Thus SandTspan the same space W. Also, Tis independent if and only if dim W¼n, and this is true if and only if Sis also independent. 4.67. LetA¼½aij/C138andB¼½bij/C138be row equivalent m/C2nmatrices over a ﬁeld K, and let v1;...;vnbe any vectors in a vector space Vover K. Let u1¼a11v1þa12v2þ/C1/C1/C1þ a1nvn u2¼a21v1þa22v2þ/C1/C1/C1þ a2nvn um¼am1v1þam2v2þ/C1/C1/C1þ amnvnw1¼b11v1þb12v2þ/C1/C1/C1þ b1nvn w2¼b21v1þb22v2þ/C1/C1/C1þ b2nvn ::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::: wm¼bm1v1þbm2v2þ/C1/C1/C1þ bmnvn Show thatfuigandfwigspan the same space. Applying an ‘‘elementary operation’’ of Problem 4.66 to fuigis equivalent to applying an elementary row operation to the matrix A. Because AandBare row equivalent, Bcan be obtained from Aby a sequence of elementary row operations; hence, fwigcan be obtained from fuigby the corresponding sequence of operations. Accordingly, fuigandfwigspan the same space. 4.68. Letv1;...;vnbelong to a vector space Vover K, and let P¼½aij/C138be an n-square matrix over K.L e t w1¼a11v1þa12v2þ/C1/C1/C1þ a1nvn; ...; wn¼an1v1þan2v2þ/C1/C1/C1þ annvn (a) Suppose Pis invertible. Show that fwigandfvigspan the same space; hence, fwigis independent if and only if fvigis independent. (b) Suppose Pis not invertible. Show that fwigis dependent. (c) Supposefwigis independent. Show that Pis invertible.154 CHAPTER 4 Vector Spaces (a) Because Pis invertible, it is row equivalent to the identity matrix I. Hence, by Problem 4.67, fwigand fvigspan the same space. Thus, one is independent if and only if the other is. (b) Because Pis not invertible, it is row equivalent to a matrix with a zero row. This means that fwigspans a space that has a spanning set of less than nelements. Thus,fwigis dependent. (c) This is the contrapositive of the statement of (b), and so it follows from (b). 4.69. Suppose that A1;A2;...are linearly independent sets of vectors, and that A1/C18A2/C18.... Show that the union A¼A1[A2[...is also linearly independent. Suppose Ais linearly dependent. Then there exist vectors v1;...;vn2Aand scalars a1;...;an2K, not all of them 0, such that a1v1þa2v2þ/C1/C1/C1þ anvn¼0 ð1Þ Because A¼[Aiand the vi2A, there exist sets Ai1;...;Ainsuch that v12Ai1; v22Ai2; ...; vn2Ain Letkbe the maximum index of the sets Aij:k¼maxði1;...;inÞ. It follows then, as A1/C18A2/C18...;that each Aijis contained in Ak. Hence, v1;v2;...;vn2Ak, and so, by (1), Akis linearly dependent, which contradicts our hypothesis. Thus, Ais linearly independent. 4.70. LetKbe a subﬁeld of a ﬁeld L, and let Lbe a subﬁeld of a ﬁeld E. (Thus, K/C18L/C18E, and Kis a subﬁeld of E.) Suppose Eis of dimension nover L, and Lis of dimension mover K. Show that Eis of dimension mnover K. Supposefv1;...;vngis a basis of Eover Landfa1;...;amgis a basis of Lover K. We claim that faivj:i¼1;...;m;j¼1;...;ngis a basis of Eover K. Note thatfaivjgcontains mnelements. Letwbe any arbitrary element in E. Becausefv1;...;vngspans Eover L,wis a linear combination of theviwith coefﬁcients in L: w¼b1v1þb2v2þ/C1/C1/C1þ bnvn; bi2L ð1Þ Becausefa1;...;amgspans Lover K, each bi2Lis a linear combination of the ajwith coefﬁcients in K: b1¼k11a1þk12a2þ/C1/C1/C1þ k1mam b2¼k21a1þk22a2þ/C1/C1/C1þ k2mam :::::::::::::::::::::::::::::::::::::::::::::::::: bn¼kn1a1þkn2a2þ/C1/C1/C1þ kmnam where kij2K. Substituting in (1), we obtain w¼ðk11a1þ/C1/C1/C1þ k1mamÞv1þðk21a1þ/C1/C1/C1þ k2mamÞv2þ/C1/C1/C1þð kn1a1þ/C1/C1/C1þ knmamÞvn ¼k11a1v1þ/C1/C1/C1þ k1mamv1þk21a1v2þ/C1/C1/C1þ k2mamv2þ/C1/C1/C1þ kn1a1vnþ/C1/C1/C1þ knmamvn ¼P i;jkjiðaivjÞ where kji2K. Thus, wis a linear combination of the aivjwith coefﬁcients in K; hence,faivjgspans Eover K. The proof is complete if we show that faivjgis linearly independent over K. Suppose, for scalars xji2K;we haveP i;jxjiðaivjÞ¼0; that is, ðx11a1v1þx12a2v1þ/C1/C1/C1þ x1mamv1Þþ/C1/C1/C1þð xn1a1vnþxn2a2vnþ/C1/C1/C1þ xnmamvmÞ¼0 or ðx11a1þx12a2þ/C1/C1/C1þ x1mamÞv1þ/C1/C1/C1þð xn1a1þxn2a2þ/C1/C1/C1þ xnmamÞvn¼0 Becausefv1;...;vngis linearly independent over Land the above coefﬁcients of the vibelong to L, each coefﬁcient must be 0: x11a1þx12a2þ/C1/C1/C1þ x1mam¼0; ...; xn1a1þxn2a2þ/C1/C1/C1þ xnmam¼0CHAPTER 4 Vector Spaces 155 Butfa1;...;amgis linearly independent over K; hence, because the xji2K, x11¼0;x12¼0;...;x1m¼0;...;xn1¼0;xn2¼0;...;xnm¼0 Accordingly,faivjgis linearly independent over K, and the theorem is proved. SUPPLEMENTARY PROBLEMS Vector Spaces 4.71. Suppose uand vbelong to a vector space V. Simplify each of the following expressions: (a) E1¼4ð5u/C06vÞþ2ð3uþvÞ, (c) E3¼6ð3uþ2vÞþ5u/C07v, (b) E2¼5ð2u/C03vÞþ4ð7vþ8Þ, (d) E4¼3ð5uþ2=vÞ: 4.72. LetVbe the set of ordered pairs ( a;b) of real numbers with addition in Vand scalar multiplication on V deﬁned by ða;bÞþð c;dÞ¼ð aþc;bþdÞ and kða;bÞ¼ð ka;0Þ Show that Vsatisﬁes all the axioms of a vector space except [M 4]—that is, except 1 u¼u. Hence, [M 4]i s not a consequence of the other axioms. 4.73. Show that Axiom [A 4] of a vector space V(that uþv¼vþu) can be derived from the other axioms for V. 4.74. LetVbe the set of ordered pairs ( a;b) of real numbers. Show that Vis not a vector space over Rwith addition and scalar multiplication deﬁned by (i)ða;bÞþð c;dÞ¼ð aþd;bþcÞandkða;bÞ¼ð ka;kbÞ, (ii)ða;bÞþð c;dÞ¼ð aþc;bþdÞandkða;bÞ¼ð a;bÞ, (iii)ða;bÞþð c;dÞ¼ð 0;0Þandkða;bÞ¼ð ka;kbÞ, (iv)ða;bÞþð c;dÞ¼ð ac;bdÞandkða;bÞ¼ð ka;kbÞ. 4.75. LetVbe the set of inﬁnite sequences ( a1;a2;...) in a ﬁeld K. Show that Vis a vector space over Kwith addition and scalar multiplication deﬁned by ða1;a2;...Þþð b1;b2;...Þ¼ð a1þb1;a2þb2;...Þand kða1;a2;...Þ¼ð ka1;ka2;...Þ 4.76. LetUandWbe vector spaces over a ﬁeld K. Let Vbe the set of ordered pairs ( u;w) where u2Uand w2W. Show that Vis a vector space over Kwith addition in Vand scalar multiplication on Vdeﬁned by ðu;wÞþð u0;w0Þ¼ð uþu0;wþw0Þ and kðu;wÞ¼ð ku;kwÞ (This space Vis called the external direct product ofUandW.) Subspaces 4.77. Determine whether or not Wis a subspace of R3where Wconsists of all vectors ( a;b;c)i nR3such that (a)a¼3b, (b) a/C20b/C20c, (c) ab¼0, (d) aþbþc¼0, (e) b¼a2,(f)a¼2b¼3c. 4.78. LetVbe the vector space of n-square matrices over a ﬁeld K. Show that Wis a subspace of VifWconsists of all matrices A¼½aij/C138that are (a) symmetric ( AT¼Aoraij¼aji), (b) (upper) triangular, (c) diagonal, (d) scalar. 4.79. LetAX¼Bbe a nonhomogeneous system of linear equations in nunknowns; that is, B6¼0. Show that the solution set is not a subspace of Kn. 4.80. Suppose UandWare subspaces of Vfor which U[Wis a subspace. Show that U/C18WorW/C18U. 4.81. LetVbe the vector space of all functions from the real ﬁeld RintoR. Show that Wis a subspace of V where Wconsists of all: (a) bounded functions, (b) even functions. [Recall that f:R!Risbounded if 9M2Rsuch that8x2R, we havejfðxÞj/C20 M; and fðxÞiseven iffð/C0xÞ¼fðxÞ;8x2R.]156 CHAPTER 4 Vector Spaces 4.82. LetVbe the vector space (Problem 4.75) of inﬁnite sequences ( a1;a2;...) in a ﬁeld K. Show that Wis a subspace of VifWconsists of all sequences with (a) 0 as the ﬁrst element, (b) only a ﬁnite number of nonzero elements. Linear Combinations, Linear Spans 4.83. Consider the vectors u¼ð1;2;3Þand v¼ð2;3;1ÞinR3. (a) Write w¼ð1;3;8Þas a linear combination of uand v. (b) Write w¼ð2;4;5Þas a linear combination of uand v. (c) Find kso that w¼ð1;k;4Þis a linear combination of uand v. (d) Find conditions on a,b,cso that w¼ða;b;cÞis a linear combination of uand v. 4.84. Write the polynomial fðtÞ¼at2þbtþcas a linear combination of the polynomials p1¼ðt/C01Þ2, p2¼t/C01,p3¼1. [Thus, p1,p2,p3span the space P2ðtÞof polynomials of degree /C202.] 4.85. Find one vector in R3that spans the intersection of Uand Wwhere Uis the xy-plane—that is, U¼fð a;b;0Þg—and Wis the space spanned by the vectors (1, 1, 1) and (1, 2, 3). 4.86. Prove that span( S) is the intersection of all subspaces of Vcontaining S. 4.87. Show that spanðSÞ¼spanðS[f0gÞ. That is, by joining or deleting the zero vector from a set, we do not change the space spanned by the set. 4.88. Show that (a) If S/C18T, then spanðSÞ/C18spanðTÞ. (b) span½spanðSÞ/C138¼ spanðSÞ. Linear Dependence and Linear Independence 4.89. Determine whether the following vectors in R4are linearly dependent or independent: (a)ð1;2;/C03;1Þ,ð3;7;1;/C02Þ,ð1;3;7;/C04Þ; (b)ð1;3;1;/C02Þ,ð2;5;/C01;3Þ,ð1;3;7;/C02Þ. 4.90. Determine whether the following polynomials u,v,winPðtÞare linearly dependent or independent: (a) u¼t3/C04t2þ3tþ3,v¼t3þ2t2þ4t/C01,w¼2t3/C0t2/C03tþ5; (b) u¼t3/C05t2/C02tþ3,v¼t3/C04t2/C03tþ4,w¼2t3/C017t2/C07tþ9. 4.91. Show that the following functions f,g,hare linearly independent: (a) fðtÞ¼et,gðtÞ¼sint,hðtÞ¼t2; (b) fðtÞ¼et,gðtÞ¼e2t,hðtÞ¼t. 4.92. Show that u¼ða;bÞand v¼ðc;dÞinK2are linearly dependent if and only if ad/C0bc¼0. 4.93. Suppose u,v,ware linearly independent vectors. Prove that Sis linearly independent where (a) S¼fuþv/C02w;u/C0v/C0w;uþwg; (b) S¼fuþv/C03w;uþ3v/C0w;vþwg. 4.94. Supposefu1;...;ur;w1;...;wsgis a linearly independent subset of V. Show that spanðuiÞ\spanðwjÞ¼f 0g 4.95. Suppose v1;v2;...;vnare linearly independent. Prove that Sis linearly independent where (a) S¼fa1v1;a2v2;...;anvngand each ai6¼0. (b) S¼fv1;...;vk/C01;w;vkþ1;...;vngandw¼P ibiviandbk6¼0. 4.96. Supposeða11;...;a1nÞ;ða21;...;a2nÞ;...;ðam1;...;amnÞare linearly independent vectors in Kn, and suppose v1;v2;...;vnare linearly independent vectors in a vector space Vover K. Show that the followingCHAPTER 4 Vector Spaces 157 vectors are also linearly independent: w1¼a11v1þ/C1/C1/C1þ a1nvn; w2¼a21v1þ/C1/C1/C1þ a2nvn; ...; wm¼am1v1þ/C1/C1/C1þ amnvn Basis and Dimension 4.97. Find a subset of u1,u2,u3,u4that gives a basis for W¼spanðuiÞofR5, where (a) u1¼ð1;1;1;2;3Þ,u2¼ð1;2;/C01;/C02;1Þ,u3¼ð3;5;/C01;/C02;5Þ,u4¼ð1;2;1;/C01;4Þ (b) u1¼ð1;/C02;1;3;/C01Þ,u2¼ð/C0 2;4;/C02;/C06;2Þ,u3¼ð1;/C03;1;2;1Þ,u4¼ð3;/C07;3;8;/C01Þ (c) u1¼ð1;0;1;0;1Þ,u2¼ð1;1;2;1;0Þ,u3¼ð2;1;3;1;1Þ,u4¼ð1;2;1;1;1Þ (d) u1¼ð1;0;1;1;1Þ,u2¼ð2;1;2;0;1Þ,u3¼ð1;1;2;3;4Þ,u4¼ð4;2;5;4;6Þ 4.98. Consider the subspaces U¼fð a;b;c;dÞ:b/C02cþd¼0gandW¼fð a;b;c;dÞ:a¼d;b¼2cgofR4. Find a basis and the dimension of (a) U, (b) W, (c) U\W. 4.99. Find a basis and the dimension of the solution space Wof each of the following homogeneous systems: ðaÞxþ2y/C02zþ2s/C0t¼0 xþ2y/C0zþ3s/C02t¼0 2xþ4y/C07zþsþt¼0ðbÞxþ2y/C0zþ3s/C04t¼0 2xþ4y/C02z/C0sþ5t¼0 2xþ4y/C02zþ4s/C02t¼0 4.100. Find a homogeneous system whose solution space is spanned by the following sets of three vectors: (a)ð1;/C02;0;3;/C01Þ,ð2;/C03;2;5;/C03Þ,ð1;/C02;1;2;/C02Þ; (b) (1, 1, 2, 1, 1), (1, 2, 1, 4, 3), (3, 5, 4, 9, 7). 4.101. Determine whether each of the following is a basis of the vector space PnðtÞ: (a)f1;1þt;1þtþt2;1þtþt2þt3; ...;1þtþt2þ/C1/C1/C1þ tn/C01þtng; (b)f1þt;tþt2;t2þt3; ...;tn/C02þtn/C01;tn/C01þtng: 4.102. Find a basis and the dimension of the subspace WofPðtÞspanned by (a) u¼t3þ2t2/C02tþ1, v¼t3þ3t2/C03tþ4,w¼2t3þt2/C07t/C07, (b) u¼t3þt2/C03tþ2, v¼2t3þt2þt/C04,w¼4t3þ3t2/C05tþ2. 4.103. Find a basis and the dimension of the subspace WofV¼M2;2spanned by A¼1/C05 /C042/C20/C21 ; B¼11 /C015/C20/C21 ; C¼2/C04 /C057/C20/C21 ; D¼1/C07 /C051/C20/C21 Rank of a Matrix, Row and Column Spaces 4.104. Find the rank of each of the following matrices: (a)13/C025 4 14 13 5 14 24 3 27/C0361 32 6643 775, (b)12/C03/C02 13/C020 38/C07/C02 21/C09/C0102 6643 775, (c)11 2 45 5 58 1 /C01/C0222 6643 775 4.105. Fork¼1;2;...;5, ﬁnd the number nkof linearly independent subsets consisting of kcolumns for each of the following matrices: (a) A¼11023 12025 130272 43 5, (b) B¼12102 12304 115062 43 5158 CHAPTER 4 Vector Spaces 4.106. Let (a) A¼1 2 13 16 243 831 5 122 531 1 4861 673 22 6643 775, (b) B¼12212 1 24545 5 12344 6 367791 02 6643 775 For each matrix (where C 1;...;C6denote its columns): (i) Find its row canonical form M. (ii) Find the columns that are linear combinations of preceding columns. (iii) Find columns (excluding C6) that form a basis for the column space. (iv) Express C6as a linear combination of the basis vectors obtained in (iii). 4.107. Determine which of the following matrices have the same row space: A¼1/C02/C01 3/C045/C20/C21 ; B¼1/C012 23/C01/C20/C21 ; C¼1/C013 2/C011 0 3/C0512 43 5 4.108. Determine which of the following subspaces of R3are identical: U1¼span½ð1;1;/C01Þ;ð2;3;/C01Þ;ð3;1;/C05Þ/C138; U2¼span½ð1;/C01;/C03Þ;ð3;/C02;/C08Þ;ð2;1;/C03Þ/C138 U3¼span½ð1;1;1Þ;ð1;/C01;3Þ;ð3;/C01;7Þ/C138 4.109. Determine which of the following subspaces of R4are identical: U1¼span½ð1;2;1;4Þ;ð2;4;1;5Þ;ð3;6;2;9Þ/C138; U2¼span½ð1;2;1;2Þ;ð2;4;1;3Þ/C138; U3¼span½ð1;2;3;10Þ;ð2;4;3;11Þ/C138 4.110. Find a basis for (i) the row space and (ii) the column space of each matrix M: (a) M¼00 3 1 4 13 1 2 1 39 4 5 2 41 28872 6643 775, (b) M¼121 01 122 13 365 27 241/C0102 6643 775. 4.111. Show that if any row is deleted from a matrix in echelon (respectively, row canonical) form, then the resulting matrix is still in echelon (respectively, row canonical) form. 4.112. LetAandBbe arbitrary m/C2nmatrices. Show that rank ðAþBÞ/C20rankðAÞþrankðBÞ. 4.113. Letr¼rankðAþBÞ. Find 2/C22 matrices AandBsuch that (a)r<rankðAÞ, rank(B); (b) r¼rankðAÞ¼rankðBÞ; (c) r>rankðAÞ, rank(B). Sums, Direct Sums, Intersections 4.114. Suppose UandWare two-dimensional subspaces of K3. Show that U\W6¼f0g. 4.115. Suppose UandWare subspaces of Vsuch that dim U¼4, dim W¼5, and dim V¼7. Find the possible dimensions of U\W. 4.116. LetUandWbe subspaces of R3for which dim U¼1, dim W¼2, and U6/C18W. Show that R3¼U/C8W. 4.117. Consider the following subspaces of R5: U¼span½ð1;/C01;/C01;/C02;0Þ;ð1;/C02;/C02;0;/C03Þ;ð1;/C01;/C02;/C02;1Þ/C138 W¼span½ð1;/C02;/C03;0;/C02Þ;ð1;/C01;/C03;2;/C04Þ;ð1;/C01;/C02;2;/C05Þ/C138CHAPTER 4 Vector Spaces 159 (a) Find two homogeneous systems whose solution spaces are UandW, respectively. (b) Find a basis and the dimension of U\W. 4.118. LetU1,U2,U3be the following subspaces of R3: U1¼fð a;b;cÞ:a¼cg; U2¼fð a;b;cÞ:aþbþc¼0g; U3¼fð 0;0;cÞg Show that (a) R3¼U1þU2, (b) R3¼U2þU3, (c)R3¼U1þU3. When is the sum direct? 4.119. Suppose U,W1,W2are subspaces of a vector space V. Show that ðU\W1Þþð U\W2Þ/C18U\ðW1þW2Þ Find subspaces of R2for which equality does not hold. 4.120. Suppose W1;W2;...;Wrare subspaces of a vector space V. Show that (a) spanðW1;W2;...;WrÞ¼W1þW2þ/C1/C1/C1þ Wr. (b) If Sispans Wifori¼1;...;r, then S1[S2[/C1/C1/C1[ Srspans W1þW2þ/C1/C1/C1þ Wr. 4.121. Suppose V¼U/C8W. Show that dim V¼dimUþdimW. 4.122. LetSandTbe arbitrary nonempty subsets (not necessarily subspaces) of a vector space Vand let kbe a scalar. The sum SþTand the scalar product kSare deﬁned by SþT¼ðuþv:u2S;v2Tg; kS¼fku:u2Sg [We also write wþSforfwgþS.] Let S¼fð 1;2Þ;ð2;3Þg; T¼fð 1;4Þ;ð1;5Þ;ð2;5Þg; w¼ð1;1Þ; k¼3 Find: (a) SþT, (b) wþS, (c) kS, (d) kT, (e) kSþkT,( f ) kðSþTÞ. 4.123. Show that the above operations of SþTandkSsatisfy (a) Commutative law: SþT¼TþS. (b) Associative law: ðS1þS2ÞþS3¼S1þðS2þS3Þ. (c) Distributive law: kðSþTÞ¼kSþkT. (d) Sþf0g¼f 0gþS¼SandSþV¼VþS¼V. 4.124. LetVbe the vector space of n-square matrices. Let Ube the subspace of upper triangular matrices, and let Wbe the subspace of lower triangular matrices. Find (a) U\W, (b) UþW. 4.125. LetVbe the external direct sum of vector spaces UandWover a ﬁeld K. (See Problem 4.76.) Let ^U¼fð u;0Þ:u2Ug and ^W¼fð 0;wÞ:w2Wg Show that (a) ^Uand ^Ware subspaces of V, (b) V¼^U/C8^W. 4.126. Suppose V¼UþW. Let ^Vbe the external direct sum of UandW. Show that Vis isomorphic to ^Vunder the correspondence v¼uþw$ðu;wÞ. 4.127. Use induction to prove (a) Theorem 4.22, (b) Theorem 4.23. Coordinates 4.128. The vectors u1¼ð1;/C02Þandu2¼ð4;/C07Þform a basis SofR2. Find the coordinate vector ½v/C138ofvrelative toSwhere (a) v¼ð5;3Þ, (b) v¼ða;bÞ. 4.129. The vectors u1¼ð1;2;0Þ,u2¼ð1;3;2Þ,u3¼ð0;1;3Þform a basis SofR3. Find the coordinate vector ½v/C138 ofvrelative to Swhere (a) v¼ð2;7;/C04Þ, (b) v¼ða;b;cÞ.160 CHAPTER 4 Vector Spaces 4.130. S¼ft3þt2;t2þt;tþ1;1gis a basis of P3ðtÞ. Find the coordinate vector ½v/C138ofvrelative to S where (a) v¼2t3þt2/C04tþ2, (b) v¼at3þbt2þctþd. 4.131. LetV¼M2;2. Find the coordinate vector [ A]o fArelative to Swhere S¼11 11/C20/C21 ;1/C01 10/C20/C21 ;11 00/C20/C21 ;10 00/C20/C21 /C26/C27 andðaÞA¼3/C05 67/C20/C21 ;ðbÞA¼ab cd/C20/C21 4.132. Find the dimension and a basis of the subspace WofP3ðtÞspanned by u¼t3þ2t2/C03tþ4; v¼2t3þ5t2/C04tþ7; w¼t3þ4t2þtþ2 4.133. Find the dimension and a basis of the subspace WofM¼M2;3spanned by A¼121 312/C20/C21 ; B¼243 756/C20/C21 ; C¼123 576/C20/C21 Miscellaneous Problems 4.134. Answer true or false. If false, prove it with a counterexample. (a) If u1,u2,u3span V, then dim V¼3. (b) If Ais a 4/C28 matrix, then any six columns are linearly dependent. (c) If u1,u2,u3are linearly independent, then u1,u2,u3,ware linearly dependent. (d) If u1,u2,u3,u4are linearly independent, then dim V/C214. (e) If u1,u2,u3span V, then w,u1,u2,u3span V. (f) If u1,u2,u3,u4are linearly independent, then u1,u2,u3are linearly independent. 4.135. Answer true or false. If false, prove it with a counterexample. (a) If any column is deleted from a matrix in echelon form, then the resulting matrix is still in echelon form. (b) If any column is deleted from a matrix in row canonical form, then the resulting matrix is still in row canonical form. (c) If any column without a pivot is deleted from a matrix in row canonical form, then the resulting matrix is in row canonical form. 4.136. Determine the dimension of the vector space Wof the following n-square matrices: (a) symmetric matrices, (b) antisymmetric matrices, (d) diagonal matrices, (c) scalar matrices. 4.137. Lett1;t2;...;tnbe symbols, and let Kbe any ﬁeld. Let Vbe the following set of expressions where ai2K: a1t1þa2t2þ/C1/C1/C1þ antn Deﬁne addition in Vand scalar multiplication on Vby ða1t1þ/C1/C1/C1þ antnÞþð b1t1þ/C1/C1/C1þ bntnÞ¼ð a1þb1Þt1þ/C1/C1/C1þð anbnmÞtn kða1t1þa2t2þ/C1/C1/C1þ antnÞ¼ka1t1þka2t2þ/C1/C1/C1þ kantn Show that Vis a vector space over Kwith the above operations. Also, show that ft1;...;tngis a basis of V, where tj¼0t1þ/C1/C1/C1þ 0tj/C01þ1tjþ0tjþ1þ/C1/C1/C1þ 0tnCHAPTER 4 Vector Spaces 161 ANSWERS TO SUPPLEMENTARY PROBLEMS [Some answers, such as bases, need not be unique.] 4.71. (a) E1¼26u/C022v; (b) The sum 7 vþ8 is not deﬁned, so E2is not deﬁned; (c) E3¼23uþ5v; (d) Division by vis not deﬁned, so E4is not deﬁned. 4.77. (a) Yes; (b) No; e.g., ð1;2;3Þ2Wbut/C02ð1;2;3Þ62W; (c) No; e.g.,ð1;0;0Þ;ð0;1;0Þ2W, but not their sum; (d) Yes; (e) No; e.g.,ð1;1;1Þ2W, but 2ð1;1;1Þ62W; (f) Yes 4.79. The zero vector 0 is not a solution. 4.83. (a) w¼3u1/C0u2, (b) Impossible, (c) k¼11 5, (d) 7 a/C05bþc¼0 4.84. Using f¼xp1þyp2þzp3, we get x¼a,y¼2aþb,z¼aþbþc 4.85. v¼ð2;1;0Þ 4.89. (a) Dependent, (b) Independent 4.90. (a) Independent, (b) Dependent 4.97. (a) u1,u2,u4; (b) u1,u2,u3; (c) u1,u2,u4; (d) u1,u2,u3 4.98. (a) dim U¼3, (b) dim W¼2, (c) dimðU\WÞ¼1 4.99. (a) Basis:fð2;/C01;0;0;0Þ;ð4;0;1;/C01;0Þ;ð3;0;1;0;1Þg; dim W¼3; (b) Basis:fð2;/C01;0;0;0Þ;ð1;0;1;0;0Þg; dim W¼2 4.100. (a) 5 xþy/C0z/C0s¼0;xþy/C0z/C0t¼0; (b) 3 x/C0y/C0z¼0;2x/C03yþs¼0;x/C02yþt¼0 4.101. (a) Yes, (b) No, because dim PnðtÞ¼nþ1, but the set contains only nelements. 4.102. (a) dim W¼2, (b) dim W¼3 4.103. dimW¼2 4.104. (a) 3, (b) 2, (c) 3 4.105. (a) n1¼4;n2¼5;n3¼n4¼n5¼0; (b) n1¼4;n2¼6;n3¼3;n4¼n5¼0 4.106. (a) (i) M¼½1;2;0;1;0;3;0;0;1;2;0;1;0;0;0;0;1;2;0/C138; (ii) C2,C4,C6; (iii) C1,C3,C5; (iv) C6¼3C1þC3þ2C5. (b) (i) M¼½1;2;0;0;3;1;0;0;1;0;/C01;/C01;0;0;0;1;1;2;0/C138; (ii) C2,C5,C6; (iii) C1,C3,C4; (iv) C6¼C1/C0C3þ2C4 4.107. AandCare row equivalent to107 014/C20/C21 , but not B 4.108. U1andU2are row equivalent to10/C02 01 1/C20/C21 , but not U3 4.109. U1andU3are row equivalent to1201 0013/C20/C21 ;but not U2 4.110. (a) (i)ð1;3;1;2;1Þ,ð0;0;1;/C01;/C01Þ,ð0;0;0;4;7Þ; (ii) C1,C3,C4; (b) (i)ð1;2;1;0;1Þ,ð0;0;1;1;2Þ; (ii) C1,C3162 CHAPTER 4 Vector Spaces 4.113. (a) A¼11 00/C20/C21 ;B¼/C01/C01 00/C20/C21 ; (b) A¼10 00/C20/C21 ;B¼02 00/C20/C21 ; (c) A¼10 00/C20/C21 ;B¼00 01/C20/C21 4.115. dimðU\WÞ¼2, 3, or 4 4.117. (a) (i)3xþ4y/C0z/C0t¼0 4xþ2yþs¼0(ii)4xþ2y/C0s¼0 9xþ2yþzþt¼0; (b) Basis:fð1;/C02;/C05;0;0Þ;ð0;0;1;0;/C01Þg; dimðU\WÞ¼2 4.118. The sum is direct in (b) and (c). 4.119. InR2, let U,V,Wbe, respectively, the line y¼x, the x-axis, the y-axis. 4.122. (a)fð2;6Þ;ð2;7Þ;ð3;7Þ;ð3;8Þ;ð4;8Þg; (b)fð2;3Þ;ð3;4Þg; (c)fð3;6Þ;ð6;9Þg; (d)fð3;12Þ;ð3;15Þ;ð6;15Þg; (e and f)fð6;18Þ;ð6;21Þ;ð9;21Þ;ð9;24Þ;ð12;24Þg 4.124. (a) Diagonal matrices, (b) V 4.128. (a) [/C041;11], (b) [/C07a/C04b;2aþb] 4.129. (a) [/C011;13;/C010], (b) [ c/C03bþ7a;/C0cþ3b/C06a;c/C02bþ4a] 4.130. (a) [2 ;/C01;/C02;2], (b) [ a;b/C0c;c/C0bþa;d/C0cþb/C0a] 4.131. (a) [7 ;/C01;/C013;10], (b) [ d;c/C0d;bþc/C02d;a/C0b/C02cþ2d] 4.132. dimW¼2; basis:ft3þ2t2/C03tþ4;t2þ2t/C01g 4.133. dimW¼2; basis:f½1;2;1;3;1;2/C138;½0;0;1;1;3;2/C138g 4.134. (a) False; (1, 1), (1, 2), (2, 1) span R2; (b) True; (c) False; (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), w¼ð0;0;0;1Þ; (d) True; (e) True; (f) True 4.135. (a) True; (b) False; e.g. delete C2from103 012/C20/C21 ; (c) True 4.136. (a)1 2nðnþ1Þ, (b)1 2nðn/C01Þ, (c) n, (d) 1CHAPTER 4 Vector Spaces 163 Linear Mappings 5.1 Introduction The main subject matter of linear algebra is the study of linear mappings and their representation by means of matrices. This chapter introduces us to these linear maps and Chapter 6 shows how they can be represented by matrices. First, however, we begin with a study of mappings in general. 5.2 Mappings, Functions LetAandBbe arbitrary nonempty sets. Suppose to each element in a2Athere is assigned a unique element of B; called the image ofa. The collection fof such assignments is called a mapping (or map) from AintoB, and it is denoted by f:A!B The set Ais called the domain of the mapping, and Bis called the target set . We write fðaÞ, read ‘‘ fofa;’’ for the unique element of Bthatfassigns to a2A. One may also view a mapping f:A!Bas a computer that, for each input value a2A, produces a unique output fðaÞ2B. Remark: The term function is used synonymously with the word mapping , although some texts reserve the word ‘‘function’’ for a real-valued or complex-valued mapping. Consider a mapping f:A!B.I fA0is any subset of A, then fðA0Þdenotes the set of images of elements of A0; and if B0is any subset of B, then f/C01ðB0Þdenotes the set of elements of A;each of whose image lies in B. That is, fðA0Þ¼f fðaÞ:a2A0g and f/C01ðB0Þ¼f a2A:fðaÞ2B0g We call fðA0) the image ofA0andf/C01ðB0Þtheinverse image orpreimage ofB0. In particular, the set of all images (i.e., fðAÞ) is called the image or range off. To each mapping f:A!Bthere corresponds the subset of A/C2Bgiven byfða;fðaÞÞ:a2Ag.W e call this set the graph off. Two mappings f:A!Bandg:A!Bare deﬁned to be equal , written f¼g,i ffðaÞ¼gðaÞfor every a2A—that is, if they have the same graph. Thus, we do not distinguish between a function and its graph. The negation of f¼gis written f6¼gand is the statement: There exists an a2Afor which fðaÞ6¼gðaÞ: Sometimes the ‘‘barred’’ arrow 7!is used to denote the image of an arbitrary element x2Aunder a mapping f:A!Bby writing x7!fðxÞ This is illustrated in the following example. 164 CHAPTER 5 EXAMPLE 5.1 (a) Let f:R!Rbe the function that assigns to each real number xits square x2. We can denote this function by writing fðxÞ¼x2or x7!x2 Here the image of /C03 is 9, so we may write fð/C03Þ¼9. However, f/C01ð9Þ¼f 3;/C03g. Also, fðRÞ¼½ 0;1Þ¼f x:x/C210gis the image of f. (b) Let A¼fa;b;c;dgandB¼fx;y;z;tg. Then the following deﬁnes a mapping f:A!B: fðaÞ¼y;fðbÞ¼x;fðcÞ¼z;fðdÞ¼y or f¼fð a;yÞ;ðb;xÞ;ðc;zÞ;ðd;yÞg The ﬁrst deﬁnes the mapping explicitly, and the second deﬁnes the mapping by its graph. Here, fðfa;b;dgÞ¼f fðaÞ;fðbÞ;fðdÞg¼f y;x;yg¼f x;yg Furthermore, fðAÞ¼f x;y;zgis the image of f. EXAMPLE 5.2 LetVbe the vector space of polynomials over R, and let pðtÞ¼3t2/C05tþ2. (a) The derivative deﬁnes a mapping D:V!Vwhere, for any polynomials fðtÞ, we have DðfÞ¼df=dt. Thus, DðpÞ¼Dð3t2/C05tþ2Þ¼6t/C05 (b) The integral, say from 0 to 1, deﬁnes a mapping J:V!R. That is, for any polynomial fðtÞ, JðfÞ¼ð1 0fðtÞdt; and so JðpÞ¼ð1 0ð3t2/C05tþ2Þ¼1 2 Observe that the mapping in ( b) is from the vector space Vinto the scalar ﬁeld R, whereas the mapping in ( a) is from the vector space Vinto itself. Matrix Mappings LetAbe any m/C2nmatrix over K. Then Adetermines a mapping FA:Kn!Kmby FAðuÞ¼Au where the vectors in KnandKmare written as columns. For example, suppose A¼1/C045 23/C06/C20/C21 and u¼1 3 /C052 43 5 then FAðuÞ¼Au¼1/C045 23/C06/C20/C21 1 3 /C052 43 5¼/C036 41/C20/C21 Remark: For notational convenience, we will frequently denote the mapping FAby the letter A, the same symbol as used for the matrix. Composition of Mappings Consider two mappings f:A!Bandg:B!C, illustrated below: A/C0!fB/C0!gC Thecomposition offandg, denoted by g/C14f, is the mapping g/C14f:A!Cdeﬁned by ðg/C14fÞðaÞ/C17gðfðaÞÞCHAPTER 5 Linear Mappings 165 That is, ﬁrst we apply ftoa2A, and then we apply gtofðaÞ2Bto get gðfðaÞÞ2 C. Viewing fandg as ‘‘computers,’’ the composition means we ﬁrst input a2Ato get the output fðaÞ2Busing f, and then we input fðaÞto get the output gðfðaÞÞ2 Cusing g. Our ﬁrst theorem tells us that the composition of mappings satisﬁes the associative law. THEOREM 5.1: Letf:A!B,g:B!C,h:C!D. Then h/C14ðg/C14fÞ¼ð h/C14gÞ/C14f We prove this theorem here. Let a2A. Then ðh/C14ðg/C14fÞÞðaÞ¼hððg/C14fÞðaÞÞ¼ hðgðfðaÞÞÞ ððh/C14gÞ/C14fÞðaÞ¼ð h/C14gÞðfðaÞÞ¼ hðgðfðaÞÞÞ Thus,ðh/C14ðg/C14fÞÞðaÞ¼ðð h/C14gÞ/C14fÞðaÞfor every a2A, and so h/C14ðg/C14fÞ¼ð h/C14gÞ/C14f. One-to-One and Onto Mappings We formally introduce some special types of mappings. DEFINITION: A mapping f:A!Bis said to be one-to-one (or 1-1 or injective ) if different elements ofAhave distinct images; that is, IffðaÞ¼fða0Þ;then a¼a0: DEFINITION: A mapping f:A!Bis said to be onto (orfmaps Aonto Borsurjective ) if every b2B is the image of at least one a2A. DEFINITION: A mapping f:A!Bis said to be a one-to-one correspondence between AandB(or bijective )i ffis both one-to-one and onto. EXAMPLE 5.3 Let f:R!R,g:R!R,h:R!Rbe deﬁned by fðxÞ¼2x; gðxÞ¼x3/C0x; hðxÞ¼x2 The graphs of these functions are shown in Fig. 5-1. The function fis one-to-one. Geometrically, this means that each horizontal line does not cont ain more than one point of f. The function gis onto. Geometrically, this means that each horizontal line contains at least one point of g. The function his neither one-to-one nor onto. For example, both 2 and /C02 have the same image 4, and /C016 has no preimage. Identity and Inverse Mappings LetAbe any nonempty set. The mapping f:A!Adeﬁned by fðaÞ¼a—that is, the function that assigns to each element in Aitself—is called identity mapping . It is usually denoted by 1Aor1orI. Thus, for any a2A, we have 1AðaÞ¼a. Figure 5-1166 CHAPTER 5 Linear Mappings Now let f:A!B. We call g:B!Athe inverse of f, written f/C01,i f f/C14g¼1B and g/C14f¼1A We emphasize that fhas an inverse if and only if fis a one-to-one correspondence between AandB; that is,fis one-to-one and onto (Problem 5.7). Also, if b2B, then f/C01ðbÞ¼a, where ais the unique element ofAfor which fðaÞ¼b 5.3 Linear Mappings (Linear Transformations) We begin with a deﬁnition. DEFINITION: LetVandUbe vector spaces over the same ﬁeld K. A mapping F:V!Uis called a linear mapping orlinear transformation if it satisﬁes the following two conditions: (1) For any vectors v;w2V,FðvþwÞ¼FðvÞþFðwÞ. (2) For any scalar kand vector v2V,FðkvÞ¼kFðvÞ. Namely, F:V!Uis linear if it ‘‘preserves’’ the two basic operations of a vector space, that of vector addition and that of scalar multiplication. Substituting k¼0 into condition (2), we obtain Fð0Þ¼0. Thus, every linear mapping takes the zero vector into the zero vector. Now for any scalars a;b2Kand any vector v;w2V, we obtain FðavþbwÞ¼FðavÞþFðbwÞ¼aFðvÞþbFðwÞ More generally, for any scalars ai2Kand any vectors vi2V, we obtain the following basic property of linear mappings: Fða1v1þa2v2þ/C1/C1/C1þ amvmÞ¼a1Fðv1Þþa2Fðv2Þþ/C1/C1/C1þ amFðvmÞ Remark 1: A linear mapping F:V!Uis completely characterized by the condition FðavþbwÞ¼aFðvÞþbFðwÞð *Þ and so this condition is sometimes used as its deﬁntion. Remark 2: The term linear transformation rather than linear mapping is frequently used for linear mappings of the form F:Rn!Rm. EXAMPLE 5.4 (a) Let F:R3!R3be the ‘‘projection’’ mapping into the xy-plane; that is, Fis the mapping deﬁned by Fðx;y;zÞ¼ð x;y;0Þ. We show that Fis linear. Let v¼ða;b;cÞandw¼ða0;b0;c0Þ. Then FðvþwÞ¼Fðaþa0;bþb0;cþc0Þ¼ð aþa0;bþb0;0Þ ¼ða;b;0Þþð a0;b0;0Þ¼FðvÞþFðwÞ and, for any scalar k, FðkvÞ¼Fðka;kb;kcÞ¼ð ka;kb;0Þ¼kða;b;0Þ¼kFðvÞ Thus, Fis linear. (b) Let G:R2!R2be the ‘‘translation’’ mapping deﬁned by Gðx;yÞ¼ð xþ1;yþ2Þ. [That is, Gadds the vector (1, 2) to any vector v¼ðx;yÞinR2.] Note that Gð0Þ¼Gð0;0Þ¼ð 1;2Þ6¼0 Thus, the zero vector is not mapped into the zero vector. Hence, Gis not linear.CHAPTER 5 Linear Mappings 167 EXAMPLE 5.5 (Derivative and Integral Mappings) Consider the vector space V¼PðtÞof polynomials over the real ﬁeld R. Let uðtÞand vðtÞbe any polynomials in Vand let kbe any scalar. (a) Let D:V!Vbe the derivative mapping. One proves in calculus that dðuþvÞ dt¼du dtþdv dtanddðkuÞ dt¼kdu dt That is, DðuþvÞ¼DðuÞþDðvÞandDðkuÞ¼kDðuÞ. Thus, the derivative mapping is linear. (b) Let J:V!Rbe an integral mapping, say JðfðtÞÞ¼ð1 0fðtÞdt One also proves in calculus that, ð1 0½uðtÞþvðtÞ/C138dt¼ð1 0uðtÞdtþð1 0vðtÞdt and ð1 0kuðtÞdt¼kð1 0uðtÞdt That is, JðuþvÞ¼JðuÞþJðvÞandJðkuÞ¼kJðuÞ. Thus, the integral mapping is linear. EXAMPLE 5.6 (Zero and Identity Mappings) (a) Let F:V!Ube the mapping that assigns the zero vector 0 2Uto every vector v2V. Then, for any vectors v;w2Vand any scalar k2K, we have FðvþwÞ¼0¼0þ0¼FðvÞþFðwÞ and FðkvÞ¼0¼k0¼kFðvÞ Thus, Fis linear. We call Fthezero mapping , and we usually denote it by 0. (b) Consider the identity mapping I:V!V, which maps each v2Vinto itself. Then, for any vectors v;w2V and any scalars a;b2K, we have IðavþbwÞ¼avþbw¼aIðvÞþbIðwÞ Thus, Iis linear. Our next theorem (proved in Problem 5.13) gives us an abundance of examples of linear mappings. In particular, it tells us that a linear mapping is complete ly determined by its values on the elements of a basis. THEOREM 5.2: LetVandUbe vector spaces over a ﬁeld K. Letfv1;v2;...;vngbe a basis of Vand letu1;u2;...;unbe any vectors in U. Then there exists a unique linear mapping F:V!Usuch that Fðv1Þ¼u1;Fðv2Þ¼u2;...;FðvnÞ¼un. We emphasize that the vectors u1;u2;...;unin Theorem 5.2 are completely arbitrary; they may be linearly dependent or they may even be equal to each other. Matrices as Linear Mappings LetAbe any real m/C2nmatrix. Recall that Adetermines a mapping FA:Kn!KmbyFAðuÞ¼Au (where the vectors in KnandKmare written as columns). We show FAis linear. By matrix multiplication, FAðvþwÞ¼AðvþwÞ¼AvþAw¼FAðvÞþFAðwÞ FAðkvÞ¼AðkvÞ¼kðAvÞ¼kFAðvÞ In other words, using Ato represent the mapping, we have AðvþwÞ¼AvþAw and AðkvÞ¼kðAvÞ Thus, the matrix mapping Ais linear.168 CHAPTER 5 Linear Mappings Vector Space Isomorphism The notion of two vector spaces being isomorphic was deﬁned in Chapter 4 when we investigated the coordinates of a vector relative to a basis. We now redeﬁne this concept. DEFINITION: Two vector spaces VandUover Kareisomorphic , written VﬃU, if there exists a bijective (one-to-one and onto) linear mapping F:V!U. The mapping Fis then called an isomorphism between VandU. Consider any vector space Vof dimension nand let Sbe any basis of V. Then the mapping v7!½v/C138S which maps each vector v2Vinto its coordinate vector ½v/C138S, is an isomorphism between VandKn. 5.4 Kernel and Image of a Linear Mapping We begin by deﬁning two concepts. DEFINITION: LetF:V!Ube a linear mapping. The kernel ofF, written Ker F, is the set of elements in Vthat map into the zero vector 0 in U; that is, KerF¼fv2V:FðvÞ¼0g Theimage (orrange )o fF, written Im F, is the set of image points in U; that is, ImF¼fu2U:there exists v2Vfor which FðvÞ¼ug The following theorem is easily proved (Problem 5.22). THEOREM 5.3: LetF:V!Ube a linear mapping. Then the kernel of Fis a subspace of Vand the image of Fis a subspace of U. Now suppose that v1;v2;...;vmspan a vector space Vand that F:V!Uis linear. We show that Fðv1Þ;Fðv2Þ;...;FðvmÞspan Im F. Let u2ImF. Then there exists v2Vsuch that FðvÞ¼u. Because thevi’s span Vand v2V, there exist scalars a1;a2;...;amfor which v¼a1v1þa2v2þ/C1/C1/C1þ amvm Therefore, u¼FðvÞ¼Fða1v1þa2v2þ/C1/C1/C1þ amvmÞ¼a1Fðv1Þþa2Fðv2Þþ/C1/C1/C1þ amFðvmÞ Thus, the vectors Fðv1Þ;Fðv2Þ;...;FðvmÞspan Im F. We formally state the above result. PROPOSITION 5.4: Suppose v1;v2;...;vmspan a vector space V, and suppose F:V!Uis linear. Then Fðv1Þ;Fðv2Þ;...;FðvmÞspan ImF. EXAMPLE 5.7 (a) Let F:R3!R3be the projection of a vector vinto the xy-plane [as pictured in Fig. 5-2(a)]; that is, Fðx;y;zÞ¼ð x;y;0Þ Clearly the image of Fis the entire xy-plane—that is, points of the form ( x;y;0). Moreover, the kernel of Fis thez-axis—that is, points of the form (0 ;0;c). That is, ImF¼fð a;b;cÞ:c¼0g¼xy-plane and Ker F¼fð a;b;cÞ:a¼0;b¼0g¼z-axis (b) Let G:R3!R3be the linear mapping that rotates a vector vabout the z-axis through an angle y[as pictured in Fig. 5-2(b)]; that is, Gðx;y;zÞ¼ð xcosy/C0ysiny;xsinyþycosy;zÞCHAPTER 5 Linear Mappings 169 Observe that the distance of a vector vfrom the origin Odoes not change under the rotation, and so only the zero vector 0 is mapped into the zero vector 0. Thus, Ker G¼f0g. On the other hand, every vector uinR3is the image of a vector vinR3that can be obtained by rotating uback by an angle of y.T h u s ,I m G¼R3, the entire space. EXAMPLE 5.8 Consider the vector space V¼PðtÞof polynomials over the real ﬁeld R, and let H:V!Vbe the third-derivative operator; that is, H½fðtÞ/C138¼ d3f=dt3. [Sometimes the notation D3is used for H, where Dis the derivative operator.] We claim that KerH¼fpolynomials of degree /C202g¼P2ðtÞ and Im H¼V The ﬁrst comes from the fact that Hðat2þbtþcÞ¼0 but HðtnÞ6¼0 for n/C213. The second comes from that fact that every polynomial gðtÞinVis the third derivative of some polynomial fðtÞ(which can be obtained by taking the antiderivative of gðtÞthree times). Kernel and Image of Matrix Mappings Consider, say, a 3 /C24 matrix Aand the usual basis fe1;e2;e3;e4gofK4(written as columns): A¼a1a2a3a4 b1b2b3b4 c1c2c3c42 43 5; e1¼1 0 0 02 6643 775; e2¼1 0 0 02 6643 775; e3¼1 0 0 02 6643 775; e4¼1 0 0 02 6643 775 Recall that Amay be viewed as a linear mapping A:K4!K3, where the vectors in K4andK3are viewed as column vectors. Now the usual basis vectors span K4, so their images Ae1,Ae2,Ae3,Ae4span the image of A. But the vectors Ae1,Ae2,Ae3,Ae4are precisely the columns of A: Ae1¼½a1;b1;c1/C138T; Ae2¼½a2;b2;c2/C138T; Ae3¼½a3;b3;c3/C138T; Ae4¼½a4;b4;c4/C138T Thus, the image of Ais precisely the column space of A. On the other hand, the kernel of Aconsists of all vectors vfor which Av¼0. This means that the kernel of Ais the solution space of the homogeneous system AX¼0, called the null space ofA. We state the above results formally. PROPOSITION 5.5: LetAbe any m/C2nmatrix over a ﬁeld Kviewed as a linear map A:Kn!Km. Then KerA¼nullspðAÞ and Im A¼colspðAÞ Here colsp( A) denotes the column space of A, and nullsp( A) denotes the null space of A.Figure 5-2170 CHAPTER 5 Linear Mappings Rank and Nullity of a Linear Mapping LetF:V!Ube a linear mapping. The rank ofFis deﬁned to be the dimension of its image, and the nullity ofFis deﬁned to be the dimension of its kernel; namely, rankðFÞ¼dimðImFÞ and nullityðFÞ¼dimðKerFÞ The following important theorem (proved in Problem 5.23) holds. THEOREM 5.6 LetVbe of ﬁnite dimension, and let F:V!Ube linear. Then dimV¼dimðKerFÞþdimðImFÞ¼nullityðFÞþrankðFÞ Recall that the rank of a matrix Awas also deﬁned to be the dimension of its column space and row space. If we now view Aas a linear mapping, then both deﬁnitions correspond, because the image of Ais precisely its column space. EXAMPLE 5.9 LetF:R4!R3be the linear mapping deﬁned by Fðx;y;z;tÞ¼ð x/C0yþzþt;2x/C02yþ3zþ4t;3x/C03yþ4zþ5tÞ (a) Find a basis and the dimension of the image of F. First ﬁnd the image of the usual basis vectors of R4, Fð1;0;0;0Þ¼ð 1;2;3Þ; Fð0;0;1;0Þ¼ð 1;3;4Þ Fð0;1;0;0Þ¼ð/C0 1;/C02;/C03Þ; Fð0;0;0;1Þ¼ð 1;4;5Þ By Proposition 5.4, the image vectors span Im F. Hence, form the matrix Mwhose rows are these image vectors and row reduce to echelon form: M¼123 /C01/C02/C03 1341452 6643 775/C24123 000 0110222 6643 775/C24123 011 0000002 6643 775 Thus, (1, 2, 3) and (0, 1, 1) form a basis of Im F. Hence, dimðImFÞ¼2 and rankðFÞ¼2. (b) Find a basis and the dimension of the kernel of the map F. SetFðvÞ¼0, where v¼ðx;y;z;tÞ, Fðx;y;z;tÞ¼ð x/C0yþzþt;2x/C02yþ3zþ4t;3x/C03yþ4zþ5tÞ¼ð 0;0;0Þ Set corresponding components equal to each other to form the following homogeneous system whose solution space is Ker F: x/C0yþzþt¼0 2x/C02yþ3zþ4t¼0 3x/C03yþ4zþ5t¼0orx/C0yþzþt¼0 zþ2t¼0 zþ2t¼0orx/C0yþzþt¼0 zþ2t¼0 The free variables are yandt. Hence, dimðKerFÞ¼2 or nullityðFÞ¼2. (i) Set y¼1,t¼0 to obtain the solution ( /C01;1;0;0Þ, (ii) Set y¼0,t¼1 to obtain the solution (1 ;0;/C02;1Þ. Thus, (/C01;1;0;0) and (1 ;0;/C02;1) form a basis for Ker F. As expected from Theorem 5.6, dim ðImFÞþdimðKerFÞ¼4¼dimR4. Application to Systems of Linear Equations LetAX¼Bdenote the matrix form of a system of mlinear equations in nunknowns. Now the matrix A may be viewed as a linear mapping A:Kn!KmCHAPTER 5 Linear Mappings 171 Thus, the solution of the equation AX¼Bmay be viewed as the preimage of the vector B2Kmunder the linear mapping A. Furthermore, the solution of the associated homogeneous system AX¼0 may be viewed as the kernel of the linear mapping A. Applying Theorem 5.6 to this homogeneous system yields dimðKerAÞ¼dimKn/C0dimðImAÞ¼n/C0rank A Butnis exactly the number of unknowns in the homogeneous system AX¼0. Thus, we have proved the following theorem of Chapter 4. THEOREM 4.19: The dimension of the solution space Wof a homogenous system AX¼0of linear equations is s¼n/C0r, where nis the number of unknowns and ris the rank of the coefﬁcient matrix A. Observe that ris also the number of pivot variables in an echelon form of AX¼0, so s¼n/C0ris also the number of free variables. Furthermore, the ssolution vectors of AX¼0 described in Theorem 3.14 are linearly independent (Problem 4.52). Accordingly, because dim W¼s, they form a basis for the solution space W. Thus, we have also proved Theorem 3.14. 5.5 Singular and Nonsingular Linear Mappings, Isomorphisms LetF:V!Ube a linear mapping. Recall that Fð0Þ¼0.Fis said to be singular if the image of some nonzero vector vis 0—that is, if there exists v6¼0 such that FðvÞ¼0. Thus, F:V!Uisnonsingular if the zero vector 0 is the only vector whose image under Fis 0 or, in other words, if Ker F¼f0g. EXAMPLE 5.10 Consider the projection map F:R3!R3and the rotation map G:R3!R3appearing in Fig. 5-2. (See Example 5.7.) Because the kernel of Fis the z-axis, Fis singular. On the other hand, the kernel of G consists only of the zero vector 0. Thus, Gis nonsingular. Nonsingular linear mappings may also be characterized as those mappings that carry independent sets into independent sets. Speciﬁcally, we prove (Problem 5.28) the following theorem. THEOREM 5.7: LetF:V!Ube a nonsingular linear mapping. Then the image of any linearly independent set is linearly independent. Isomorphisms Suppose a linear mapping F:V!Uis one-to-one. Then only 0 2Vcan map into 02U, and so Fis nonsingular. The converse is also true. For suppose Fis nonsingular and FðvÞ¼FðwÞ, then Fðv/C0wÞ¼FðvÞ/C0FðwÞ¼0, and hence, v/C0w¼0o r v¼w. Thus, FðvÞ¼FðwÞimplies v¼w— that is, Fis one-to-one. We have proved the following proposition. PROPOSITION 5.8: A linear mapping F:V!Uis one-to-one if and only if Fis nonsingular. Recall that a mapping F:V!Uis called an isomorphism ifFis linear and if Fis bijective (i.e., if F is one-to-one and onto). Also, recall that a vector space Vis said to be isomorphic to a vector space U, written VﬃU, if there is an isomorphism F:V!U. The following theorem (proved in Problem 5.29) applies. THEOREM 5.9: Suppose Vhas ﬁnite dimension and dimV¼dimU. Suppose F:V!Uis linear. Then Fis an isomorphism if and only if Fis nonsingular.172 CHAPTER 5 Linear Mappings 5.6 Operations with Linear Mappings We are able to combine linear mappings in various ways to obtain new linear mappings. These operations are very important and will be used throughout the text. LetF:V!UandG:V!Ube linear mappings over a ﬁeld K. The sum FþGand the scalar product kF, where k2K, are deﬁned to be the following mappings from VintoU: ðFþGÞðvÞ/C17FðvÞþGðvÞ andðkFÞðvÞ/C17kFðvÞ We now show that if FandGare linear, then FþGandkFare also linear. Speciﬁcally, for any vectors v;w2Vand any scalars a;b2K, ðFþGÞðavþbwÞ¼FðavþbwÞþGðavþbwÞ ¼aFðvÞþbFðwÞþaGðvÞþbGðwÞ ¼a½FðvÞþGðvÞ/C138þb½FðwÞþGðwÞ/C138 ¼aðFþGÞðvÞþbðFþGÞðwÞ and ðkFÞðavþbwÞ¼kFðavþbwÞ¼k½aFðvÞþbFðwÞ/C138 ¼akFðvÞþbkFðwÞ¼aðkFÞðvÞþbðkFÞðwÞ Thus, FþGandkFare linear. The following theorem holds. THEOREM 5.10: LetVandUbe vector spaces over a ﬁeld K. Then the collection of all linear mappings from Vinto Uwith the above operations of addition and scalar multi- plication forms a vector space over K. The vector space of linear mappings in Theorem 5.10 is usually denoted by HomðV;UÞ Here Hom comes from the word ‘‘homomorphism.’’ We emphasize that the proof of Theorem 5.10 reduces to showing that Hom ðV;UÞdoes satisfy the eight axioms of a vector space. The zero element of HomðV;UÞis the zero mapping from VintoU, denoted by 0and deﬁned by 0ðvÞ¼0 for every vector v2V. Suppose VandUare of ﬁnite dimension. Then we have the following theorem. THEOREM 5.11: Suppose dimV¼manddimU¼n. Then dim½HomðV;UÞ/C138¼ mn. Composition of Linear Mappings Now suppose V,U, and Ware vector spaces over the same ﬁeld K, and suppose F:V!Uand G:U!Ware linear mappings. We picture these mappings as follows: V/C0!FU/C0!GW Recall that the composition function G/C14Fis the mapping from Vinto Wdeﬁned by ðG/C14FÞðvÞ¼GðFðvÞÞ. We show that G/C14Fis linear whenever FandGare linear. Speciﬁcally, for any vectors v;w2Vand any scalars a;b2K, we have ðG/C14FÞðavþbwÞ¼GðFðavþbwÞÞ¼ GðaFðvÞþbFðwÞÞ ¼aGðFðvÞÞþ bGðFðwÞÞ¼ aðG/C14FÞðvÞþbðG/C14FÞðwÞ Thus, G/C14Fis linear. The composition of linear mappings and the operations of addition and scalar multiplication are related as follows.CHAPTER 5 Linear Mappings 173 THEOREM 5.12: LetV,U,Wbe vector spaces over K. Suppose the following mappings are linear: F:V!U; F0:V!U and G:U!W; G0:U!W Then, for any scalar k2K: (i) G/C14ðFþF0Þ¼G/C14FþG/C14F0. (ii)ðGþG0Þ/C14F¼G/C14FþG0/C14F. (iii) kðG/C14FÞ¼ð kGÞ/C14F¼G/C14ðkFÞ. 5.7 Algebra AðVÞof Linear Operators LetVbe a vector space over a ﬁeld K. This section considers the special case of linear mappings from the vector space Vinto itself—that is, linear mappings of the form F:V!V. They are also called linear operators orlinear transformations onV. We will write AðVÞ, instead of HomðV;VÞ, for the space of all such mappings. Now AðVÞis a vector space over K(Theorem 5.8), and, if dim V¼n,t h e nd i m AðVÞ¼n2.M o r e o v e r , for any mappings F;G2AðVÞ, the composition G/C14Fexists and also belongs to AðVÞ. Thus, we have a ‘‘multiplication’’ deﬁned in AðVÞ. [We sometimes write FGinstead of G/C14Fin the space AðVÞ.] Remark: Analgebra A over a ﬁeld Kis a vector space over Kin which an operation of multiplication is deﬁned satisfying, for every F;G;H2Aand every k2K: (i) FðGþHÞ¼FGþFH, (ii)ðGþHÞF¼GFþHF, (iii) kðGFÞ¼ð kGÞF¼GðkFÞ. The algebra is said to be associative if, in addition,ðFGÞH¼FðGHÞ. The above deﬁnition of an algebra and previous theorems give us the following result. THEOREM 5.13: LetVbe a vector space over K. Then AðVÞis an associative algebra over Kwith respect to composition of mappings. If dimV¼n, then dimAðVÞ¼n2. This is why AðVÞis called the algebra of linear operators onV. Polynomials and Linear Operators Observe that the identity mapping I:V!Vbelongs to AðVÞ. Also, for any linear operator FinAðVÞ, we have FI¼IF¼F. We can also form ‘‘powers’’ of F. Namely, we deﬁne F0¼I; F2¼F/C14F; F3¼F2/C14F¼F/C14F/C14F; F4¼F3/C14F; ... Furthermore, for any polynomial pðtÞover K, say, pðtÞ¼a0þa1tþa2t2þ/C1/C1/C1þ ast2 we can form the linear operator pðFÞdeﬁned by pðFÞ¼a0Iþa1Fþa2F2þ/C1/C1/C1þ asFs (For any scalar k, the operator kIis sometimes denoted simply by k.) In particular, we say Fis azero of the polynomial pðtÞifpðFÞ¼0. EXAMPLE 5.11 LetF:K3!K3be deﬁned by Fðx;y;zÞ¼ð 0;x;yÞ. For anyða;b;cÞ2K3, ðFþIÞða;b;cÞ¼ð 0;a;bÞþð a;b;cÞ¼ð a;aþb;bþcÞ F3ða;b;cÞ¼F2ð0;a;bÞ¼Fð0;0;aÞ¼ð 0;0;0Þ Thus, F3¼0, the zero mapping in AðVÞ. This means Fis a zero of the polynomial pðtÞ¼t3.174 CHAPTER 5 Linear Mappings Square Matrices as Linear Operators LetM¼Mn;nbe the vector space of all square n/C2nmatrices over K. Then any matrix AinMdeﬁnes a linear mapping FA:Kn!KnbyFAðuÞ¼Au(where the vectors in Knare written as columns). Because the mapping is from Kninto itself, the square matrix Ais a linear operator, not simply a linear mapping. Suppose AandBare matrices in M. Then the matrix product ABis deﬁned. Furthermore, for any (column) vector uinKn, FABðuÞ¼ð ABÞu¼AðBuÞ¼AðFBðUÞÞ¼ FAðFBðuÞÞ¼ð FA/C14FBÞðuÞ In other words, the matrix product ABcorresponds to the composition of AandBas linear mappings. Similarly, the matrix sum AþBcorresponds to the sum of AandBas linear mappings, and the scalar product kAcorresponds to the scalar product of Aas a linear mapping. Invertible Operators in AðVÞ LetF:V!Vbe a linear operator. Fis said to be invertible if it has an inverse—that is, if there exists F/C01inAðVÞsuch that FF/C01¼F/C01F¼I. On the other hand, Fis invertible as a mapping if Fis both one-to-one and onto. In such a case, F/C01is also linear and F/C01is the inverse of Fas a linear operator (proved in Problem 5.15). Suppose Fis invertible. Then only 0 2Vcan map into itself, and so Fis nonsingular. The converse is not true, as seen by the following example. EXAMPLE 5.12 LetV¼PðtÞ, the vector space of polynomials over K. Let Fbe the mapping on Vthat increases by 1 the exponent of tin each term of a polynomial; that is, Fða0þa1tþa2t2þ/C1/C1/C1þ astsÞ¼a0tþa1t2þa2t3þ/C1/C1/C1þ astsþ1 Then Fis a linear mapping and Fis nonsingular. However, Fis not onto, and so Fis not invertible. The vector space V¼PðtÞin the above example has inﬁnite dimension. The situation changes signiﬁcantly when Vhas ﬁnite dimension. Namely, the following theorem applies. THEOREM 5.14: LetFbe a linear operator on a ﬁnite-dimensional vector space V. Then the following four conditions are equivalent. (i) Fis nonsingular: Ker F¼f0g. (iii) Fis an onto mapping. (ii) Fis one-to-one. (iv) Fis invertible. The proof of the above theorem mainly follows from Theorem 5.6, which tells us that dimV¼dimðKerFÞþdimðImFÞ By Proposition 5.8, (i) and (ii) are equivalent. Note that (iv) is equivalent to (ii) and (iii). Thus, to prove the theorem, we need only show that (i) and (iii) are equivalent. This we do below. (a) Suppose (i) holds. Then dim ðKerFÞ¼0, and so the above equation tells us that dim V¼dimðImFÞ. This means V¼ImFor, in other words, Fis an onto mapping. Thus, (i) implies (iii). (b) Suppose (iii) holds. Then V¼ImF, and so dim V¼dimðImFÞ. Therefore, the above equation tells us that dimðKerFÞ¼0, and so Fis nonsingular. Therefore, (iii) implies (i). Accordingly, all four conditions are equivalent. Remark: Suppose Ais a square n/C2nmatrix over K. Then Amay be viewed as a linear operator on Kn. Because Knhas ﬁnite dimension, Theorem 5.14 holds for the square matrix A. This is why the terms ‘‘nonsingular’’ and ‘‘invertible’’ are used interchangeably when applied to square matrices. EXAMPLE 5.13 LetFbe the linear operator on R2deﬁned by Fðx;yÞ¼ð 2xþy;3xþ2yÞ. (a) To show that Fis invertible, we need only show that Fis nonsingular. Set Fðx;yÞ¼ð 0;0Þto obtain the homogeneous system 2xþy¼0 and 3 xþ2y¼0CHAPTER 5 Linear Mappings 175 Solve for xandyto get x¼0,y¼0. Hence, Fis nonsingular and so invertible. (b) To ﬁnd a formula for F/C01, we set Fðx;yÞ¼ð s;tÞand so F/C01ðs;tÞ¼ð x;yÞ. We have ð2xþy;3xþ2yÞ¼ð s;tÞ or2xþy¼s 3xþ2y¼t Solve for xandyin terms of sandtto obtain x¼2s/C0t,y¼/C03sþ2t. Thus, F/C01ðs;tÞ¼ð 2s/C0t;/C03sþ2tÞ or F/C01ðx;yÞ¼ð 2x/C0y;/C03xþ2yÞ where we rewrite the formula for F/C01using xandyinstead of sandt. SOLVED PROBLEMS Mappings 5.1. State whether each diagram in Fig. 5-3 deﬁnes a mapping from A¼fa;b;cgintoB¼fx;y;zg. (a) No. There is nothing assigned to the element b2A. (b) No. Two elements, xandz, are assigned to c2A. (c) Yes. 5.2. Letf:A!Bandg:B!Cbe deﬁned by Fig. 5-4. (a) Find the composition mapping ðg/C14fÞ:A!C. (b) Find the images of the mappings f,g,g/C14f. (a) Use the deﬁnition of the composition mapping to compute ðg/C14fÞðaÞ¼gðfðaÞÞ¼ gðyÞ¼t;ðg/C14fÞðbÞ¼gðfðbÞÞ¼ gðxÞ¼s ðg/C14fÞðcÞ¼gðfðcÞÞ¼ gðyÞ¼t Observe that we arrive at the same answer if we ‘‘follow the arrows’’ in Fig. 5-4: a!y!t; b!x!s; c!y!t (b) By Fig. 5-4, the image values under the mapping farexandy, and the image values under garer,s,t. Figure 5-3 Figure 5-4176 CHAPTER 5 Linear Mappings Hence, Imf¼fx;yg and Im g¼fr;s;tg Also, by part (a), the image values under the composition mapping g/C14faretands; accordingly, Img/C14f¼fs;tg. Note that the images of gandg/C14fare different. 5.3. Consider the mapping F:R3!R2deﬁned by Fðx;y;zÞ¼ð yz;x2Þ. Find (a)Fð2;3;4Þ; (b) Fð5;/C02;7Þ; (c) F/C01ð0;0Þ, that is, all v2R3such that FðvÞ¼0. (a) Substitute in the formula for Fto get Fð2;3;4Þ¼ð 3/C14;22Þ¼ð 12;4Þ. (b)Fð5;/C02;7Þ¼ð/C0 2/C17;52Þ¼ð/C0 14;25Þ. (c) Set FðvÞ¼0, where v¼ðx;y;zÞ, and then solve for x,y,z: Fðx;y;zÞ¼ð yz;x2Þ¼ð 0;0Þ or yz¼0;x2¼0 Thus, x¼0 and either y¼0o r z¼0. In other words, x¼0,y¼0o r x¼0;z¼0—that is, the z-axis and the y-axis. 5.4. Consider the mapping F:R2!R2deﬁned by Fðx;yÞ¼ð 3y;2xÞ. Let Sbe the unit circle in R2, that is, the solution set of x2þy2¼1. (a) Describe FðSÞ. (b) Find F/C01ðSÞ. (a) Let ( a;b) be an element of FðSÞ. Then there exists ðx;yÞ2Ssuch that Fðx;yÞ¼ð a;bÞ. Hence, ð3y;2xÞ¼ð a;bÞ or 3 y¼a;2x¼b or y¼a 3;x¼b 2 Becauseðx;yÞ2S—that is, x2þy2¼1—we have b 2/C18/C192 þa 3/C16/C172 ¼1o ra2 9þb2 4¼1 Thus, FðSÞis an ellipse. (b) Let Fðx;yÞ¼ð a;bÞ, whereða;bÞ2S. Thenð3y;2xÞ¼ð a;bÞor 3y¼a,2x¼b. Becauseða;bÞ2S,w e have a2þb2¼1. Thus,ð3yÞ2þð2xÞ2¼1. Accordingly, F/C01ðSÞis the ellipse 4 x2þ9y2¼1. 5.5. Let the mappings f:A!B,g:B!C,h:C!Dbe deﬁned by Fig. 5-5. Determine whether or not each function is (a) one-to-one; (b) onto; (c) invertible (i.e., has an inverse). (a) The mapping f:A!Bis one-to-one, as each element of Ahas a different image. The mapping g:B!Cis not one-to one, because xandzboth have the same image 4. The mapping h:C!Dis one-to-one. (b) The mapping f:A!Bis not onto, because z2Bis not the image of any element of A. The mapping g:B!Cis onto, as each element of Cis the image of some element of B. The mapping h:C!Dis also onto. (c) A mapping has an inverse if and only if it is one-to-one and onto. Hence, only hhas an inverse. zyx wB g f C h 5 64 1 aD b c2 3A Figure 5-5CHAPTER 5 Linear Mappings 177 5.6. Suppose f:A!Bandg:B!C. Hence,ðg/C14fÞ:A!Cexists. Prove (a) If fandgare one-to-one, then g/C14fis one-to-one. (b) If fandgare onto mappings, then g/C14fis an onto mapping. (c) If g/C14fis one-to-one, then fis one-to-one. (d) If g/C14fis an onto mapping, then gis an onto mapping. (a) Supposeðg/C14fÞðxÞ¼ð g/C14fÞðyÞ. Then gðfðxÞÞ¼ gðfðyÞÞ. Because gis one-to-one, fðxÞ¼fðyÞ. Because fis one-to-one, x¼y. We have proven that ðg/C14fÞðxÞ¼ð g/C14fÞðyÞimplies x¼y; hence g/C14f is one-to-one. (b) Suppose c2C. Because gis onto, there exists b2Bfor which gðbÞ¼c. Because fis onto, there exists a2Afor which fðaÞ¼b. Thus,ðg/C14fÞðaÞ¼gðfðaÞÞ¼ gðbÞ¼c. Hence, g/C14fis onto. (c) Suppose fis not one-to-one. Then there exist distinct elements x;y2Afor which fðxÞ¼fðyÞ. Thus, ðg/C14fÞðxÞ¼gðfðxÞÞ¼ gðfðyÞÞ¼ð g/C14fÞðyÞ. Hence, g/C14fis not one-to-one. Therefore, if g/C14fis one-to- one, then fmust be one-to-one. (d) If a2A, thenðg/C14fÞðaÞ¼gðfðaÞÞ2 gðBÞ. Hence,ðg/C14fÞðAÞ/C18gðBÞ. Suppose gis not onto. Then gðBÞ is properly contained in Cand soðg/C14fÞðAÞis properly contained in C; thus, g/C14fis not onto. Accordingly, if g/C14fis onto, then gmust be onto. 5.7. Prove that f:A!Bhas an inverse if and only if fis one-to-one and onto. Suppose fhas an inverse—that is, there exists a function f/C01:B!Afor which f/C01/C14f¼1Aand f/C14f/C01¼1B. Because 1Ais one-to-one, fis one-to-one by Problem 5.6(c), and because 1Bis onto, fis onto by Problem 5.6( d); that is, fis both one-to-one and onto. Now suppose fis both one-to-one and onto. Then each b2Bis the image of a unique element in A, say b*. Thus, if fðaÞ¼b, then a¼b*; hence, fðb*Þ¼b. Now let gdenote the mapping from BtoAdeﬁned by b7!b*. We have (i)ðg/C14fÞðaÞ¼gðfðaÞÞ¼ gðbÞ¼b*¼afor every a2A; hence, g/C14f¼1A. (ii)ðf/C14gÞðbÞ¼fðgðbÞÞ¼ fðb*Þ¼bfor every b2B; hence, f/C14g¼1B. Accordingly, fhas an inverse. Its inverse is the mapping g. 5.8. Letf:R!Rbe deﬁned by fðxÞ¼2x/C03. Now fis one-to-one and onto; hence, fhas an inverse mapping f/C01. Find a formula for f/C01. Letybe the image of xunder the mapping f;t h a ti s , y¼fðxÞ¼2x/C03. Hence, xwill be the image of y under the inverse mapping f/C01. Thus, solve for xin terms of yin the above equation to obtain x¼1 2ðyþ3Þ. Then the formula deﬁning the inverse function is f/C01ðyÞ¼1 2ðyþ3Þ,o r ,u s i n g xinstead of y,f/C01ðxÞ¼1 2ðxþ3Þ. Linear Mappings 5.9. Suppose the mapping F:R2!R2is deﬁned by Fðx;yÞ¼ð xþy;xÞ. Show that Fis linear. We need to show that FðvþwÞ¼FðvÞþFðwÞandFðkvÞ¼kFðvÞ, where uandvare any elements of R2andkis any scalar. Let v¼ða;bÞandw¼ða0;b0Þ. Then vþw¼ðaþa0;bþb0Þ and kv¼ðka;kbÞ We have FðvÞ¼ð aþb;aÞandFðwÞ¼ð a0þb0;a0Þ. Thus, FðvþwÞ¼Fðaþa0;bþb0Þ¼ð aþa0þbþb0;aþa0Þ ¼ðaþb;aÞþð a0þb0;a0Þ¼FðvÞþFðwÞ and FðkvÞ¼Fðka;kbÞ¼ð kaþkb;kaÞ¼kðaþb;aÞ¼kFðvÞ Because v,w,kwere arbitrary, Fis linear.178 CHAPTER 5 Linear Mappings 5.10. Suppose F:R3!R2is deﬁned by Fðx;y;zÞ¼ð xþyþz;2x/C03yþ4zÞ. Show that Fis linear. We argue via matrices. Writing vectors as columns, the mapping Fmay be written in the form FðvÞ¼Av, where v¼½x;y;z/C138Tand A¼11 1 2/C034/C20/C21 Then, using properties of matrices, we have FðvþwÞ¼AðvþwÞ¼AvþAw¼FðvÞþFðwÞ FðkvÞ¼AðkvÞ¼kðAvÞ¼kFðvÞ and Thus, Fis linear. 5.11. Show that the following mappings are not linear: (a) F:R2!R2deﬁned by Fðx;yÞ¼ð xy;xÞ (b) F:R2!R3deﬁned by Fðx;yÞ¼ð xþ3;2y;xþyÞ (c) F:R3!R2deﬁned by Fðx;y;zÞ¼ðj xj;yþzÞ (a) Let v¼ð1;2Þandw¼ð3;4Þ; then vþw¼ð4;6Þ. Also, FðvÞ¼ð 1ð2Þ;1Þ¼ð 2;1Þ and FðwÞ¼ð 3ð4Þ;3Þ¼ð 12;3Þ Hence, FðvþwÞ¼ð 4ð6Þ;4Þ¼ð 24;6Þ6¼FðvÞþFðwÞ (b) Because Fð0;0Þ¼ð 3;0;0Þ6¼ð0;0;0Þ,Fcannot be linear. (c) Let v¼ð1;2;3Þandk¼/C03. Then kv¼ð/C0 3;/C06;/C09Þ. We have FðvÞ¼ð 1;5Þand kFðvÞ¼/C0 3ð1;5Þ¼ð/C0 3;/C015Þ: Thus, FðkvÞ¼Fð/C03;/C06;/C09Þ¼ð 3;/C015Þ6¼kFðvÞ Accordingly, Fis not linear. 5.12. LetVbe the vector space of n-square real matrices. Let Mbe an arbitrary but ﬁxed matrix in V. LetF:V!Vbe deﬁned by FðAÞ¼AMþMA, where Ais any matrix in V. Show that Fis linear. For any matrices AandBinVand any scalar k, we have FðAþBÞ¼ð AþBÞMþMðAþBÞ¼AMþBMþMAþMB ¼ðAMþMAÞ¼ð BMþMBÞ¼FðAÞþFðBÞ and FðkAÞ¼ð kAÞMþMðkAÞ¼kðAMÞþkðMAÞ¼kðAMþMAÞ¼kFðAÞ Thus, Fis linear. 5.13. Prove Theorem 5.2: Let VandUbe vector spaces over a ﬁeld K. Letfv1;v2;...;vngbe a basis of Vand let u1;u2;...;unbe any vectors in U. Then there exists a unique linear mapping F:V!U such that Fðv1Þ¼u1;Fðv2Þ¼u2;...;FðvnÞ¼un. There are three steps to the proof of the theorem: (1) Deﬁne the mapping F:V!Usuch that FðviÞ¼ui;i¼1;...;n. (2) Show that Fis linear. (3) Show that Fis unique. Step 1. Let v2V. Becausefv1;...;vngis a basis of V, there exist unique scalars a1;...;an2Kfor which v¼a1v1þa2v2þ/C1/C1/C1þ anvn. We deﬁne F:V!Uby FðvÞ¼a1u1þa2u2þ/C1/C1/C1þ anunCHAPTER 5 Linear Mappings 179 (Because the aiare unique, the mapping Fis well deﬁned.) Now, for i¼1;...;n, vi¼0v1þ/C1/C1/C1þ 1viþ/C1/C1/C1þ 0vn Hence, FðviÞ¼0u1þ/C1/C1/C1þ 1uiþ/C1/C1/C1þ 0un¼ui Thus, the ﬁrst step of the proof is complete. Step 2. Suppose v¼a1v1þa2v2þ/C1/C1/C1þ anvnandw¼b1v1þb2v2þ/C1/C1/C1þ bnvn. Then vþw¼ða1þb1Þv1þða2þb2Þv2þ/C1/C1/C1þð anþbnÞvn and, for any k2K,kv¼ka1v1þka2v2þ/C1/C1/C1þ kanvn. By deﬁnition of the mapping F, FðvÞ¼a1u1þa2u2þ/C1/C1/C1þ anvn and FðwÞ¼b1u1þb2u2þ/C1/C1/C1þ bnun Hence, FðvþwÞ¼ð a1þb1Þu1þða2þb2Þu2þ/C1/C1/C1þð anþbnÞun ¼ða1u1þa2u2þ/C1/C1/C1þ anunÞþð b1u1þb2u2þ/C1/C1/C1þ bnunÞ ¼FðvÞþFðwÞ and FðkvÞ¼kða1u1þa2u2þ/C1/C1/C1þ anunÞ¼kFðvÞ Thus, Fis linear. Step 3. Suppose G:V!Uis linear and Gðv1Þ¼ui;i¼1;...;n. Let v¼a1v1þa2v2þ/C1/C1/C1þ anvn Then GðvÞ¼Gða1v1þa2v2þ/C1/C1/C1þ anvnÞ¼a1Gðv1Þþa2Gðv2Þþ/C1/C1/C1þ anGðvnÞ ¼a1u1þa2u2þ/C1/C1/C1þ anun¼FðvÞ Because GðvÞ¼FðvÞfor every v2V;G¼F. Thus, Fis unique and the theorem is proved. 5.14. LetF:R2!R2be the linear mapping for which Fð1;2Þ¼ð 2;3ÞandFð0;1Þ¼ð 1;4Þ. [Note that fð1;2Þ;ð0;1Þgis a basis of R2, so such a linear map Fexists and is unique by Theorem 5.2.] Find a formula for F; that is, ﬁnd Fða;bÞ. Writeða;bÞas a linear combination of (1, 2) and (0, 1) using unknowns xandy, ða;bÞ¼xð1;2Þþyð0;1Þ¼ð x;2xþyÞ; so a¼x;b¼2xþy Solve for xandyin terms of aandbto get x¼a,y¼/C02aþb. Then Fða;bÞ¼xFð1;2ÞþyFð0;1Þ¼að2;3Þþð/C0 2aþbÞð1;4Þ¼ð b;/C05aþ4bÞ 5.15. Suppose a linear mapping F:V!Uis one-to-one and onto. Show that the inverse mapping F/C01:U!Vis also linear. Suppose u;u02U. Because Fis one-to-one and onto, there exist unique vectors v;v02Vfor which FðvÞ¼uandFðv0Þ¼u0. Because Fis linear, we also have Fðvþv0Þ¼FðvÞþFðv0Þ¼uþu0and FðkvÞ¼kFðvÞ¼ku By deﬁnition of the inverse mapping, F/C01ðuÞ¼v;F/C01ðu0Þ¼v0;F/C01ðuþu0Þ¼vþv0;F/C01ðkuÞ¼kv: Then F/C01ðuþu0Þ¼vþv0¼F/C01ðuÞþF/C01ðu0Þ and F/C01ðkuÞ¼kv¼kF/C01ðuÞ Thus, F/C01is linear.180 CHAPTER 5 Linear Mappings Kernel and Image of Linear Mappings 5.16. LetF:R4!R3be the linear mapping deﬁned by Fðx;y;z;tÞ¼ð x/C0yþzþt;xþ2z/C0t;xþyþ3z/C03tÞ Find a basis and the dimension of (a) the image of F;(b) the kernel of F. (a) Find the images of the usual basis of R4: Fð1;0;0;0Þ¼ð 1;1;1Þ; Fð0;0;1;0Þ¼ð 1;2;3Þ Fð0;1;0;0Þ¼ð/C0 1;0;1Þ; Fð0;0;0;1Þ¼ð 1;/C01;/C03Þ By Proposition 5.4, the image vectors span Im F. Hence, form the matrix whose rows are these image vectors, and row reduce to echelon form: 111 /C0101 123 1/C01/C032 66643 7775/C24111 012 012 0/C02/C042 66643 7775/C24111 012 000 0002 66643 7775 Thus, (1, 1, 1) and (0, 1, 2) form a basis for Im F; hence, dimðImFÞ¼2. (b) Set FðvÞ¼0, where v¼ðx;y;z;tÞ; that is, set Fðx;y;z;tÞ¼ð x/C0yþzþt;xþ2z/C0t;xþyþ3z/C03tÞ¼ð 0;0;0Þ Set corresponding entries equal to each other to form the following homogeneous system whose solution space is Ker F: x/C0yþzþt¼0 xþ2z/C0t¼0 xþyþ3z/C03t¼0orx/C0yþzþt¼0 yþz/C02t¼0 2yþ2z/C04t¼0orx/C0yþzþt¼0 yþz/C02t¼0 The free variables are zandt. Hence, dimðKerFÞ¼2. (i) Set z¼/C01,t¼0 to obtain the solution (2 ;1;/C01;0). (ii) Set z¼0,t¼1 to obtain the solution (1, 2, 0, 1). Thus, (2 ;1;/C01;0) and (1, 2, 0, 1) form a basis of Ker F. [As expected, dimðImFÞþdimðKerFÞ¼2þ2¼4¼dimR 4, the domain of F.] 5.17. LetG:R3!R3be the linear mapping deﬁned by Gðx;y;zÞ¼ð xþ2y/C0z;yþz;xþy/C02zÞ Find a basis and the dimension of (a) the image of G, (b) the kernel of G. (a) Find the images of the usual basis of R3: Gð1;0;0Þ¼ð 1;0;1Þ; Gð0;1;0Þ¼ð 2;1;1Þ; Gð0;0;1Þ¼ð/C0 1;1;/C02Þ By Proposition 5.4, the image vectors span Im G. Hence, form the matrix Mwhose rows are these image vectors, and row reduce to echelon form: M¼10 1 21 1 /C011/C022 43 5/C2410 1 01/C01 01/C012 43 5/C2410 1 01/C01 00 02 43 5 Thus, (1, 0, 1) and (0 ;1;/C01) form a basis for Im G; hence, dimðImGÞ¼2. (b) Set GðvÞ¼0, where v¼ðx;y;zÞ; that is, Gðx;y;zÞ¼ð xþ2y/C0z;yþz;xþy/C02zÞ¼ð 0;0;0ÞCHAPTER 5 Linear Mappings 181 Set corresponding entries equal to each other to form the following homogeneous system whose solution space is Ker G: xþ2y/C0z¼0 yþz¼0 xþy/C02z¼0orxþ2y/C0z¼0 yþz¼0 /C0y/C0z¼0orxþ2y/C0z¼0 yþz¼0 The only free variable is z; hence, dimðKerGÞ¼1. Set z¼1; then y¼/C01 and x¼3. Thus, (3 ;/C01;1) forms a basis of Ker G. [As expected, dim ðImGÞþdimðKerGÞ¼2þ1¼3¼dimR3, the domain ofG.] 5.18. Consider the matrix mapping A:R4!R3, where A¼12 3 1 13 5/C02 381 3/C032 43 5. Find a basis and the dimension of (a) the image of A, (b) the kernel of A. (a) The column space of Ais equal to Im A. Now reduce ATto echelon form: AT¼113 238 35 1 31/C02/C032 6643 775/C24113 012 0240/C03/C062 6643 775/C24113 012 0000002 6643 775 Thus,fð1;1;3Þ;ð0;1;2Þgis a basis of Im A, and dimðImAÞ¼2. (b) Here Ker Ais the solution space of the homogeneous system AX¼0, where X¼fx;y;z;tÞ T. Thus, reduce the matrix Aof coefﬁcients to echelon form: 123 1 012/C03 024/C062 43 5/C24123 1 012/C03 000 02 43 5 orxþ2yþ3zþt¼0 yþ2z/C03t¼0 The free variables are zandt. Thus, dimðKerAÞ¼2. (i) Set z¼1,t¼0 to get the solution (1 ;/C02;1;0). (ii) Set z¼0,t¼1 to get the solution ( /C07;3;0;1). Thus, (1 ;/C02;1;0) and (/C07;3;0;1) form a basis for Ker A. 5.19. Find a linear map F:R3!R4whose image is spanned by (1 ;2;0;/C04) and (2 ;0;/C01;/C03). Form a 4/C23 matrix whose columns consist only of the given vectors, say A¼122 200 0/C01/C01 /C04/C03/C032 6643 775 Recall that Adetermines a linear map A:R3!R4whose image is spanned by the columns of A. Thus, A satisﬁes the required condition. 5.20. Suppose f:V!Uis linear with kernel W, and that fðvÞ¼u. Show that the ‘‘coset’’ vþW¼fvþw:w2Wgis the preimage of u; that is, f/C01ðuÞ¼vþW. We must prove that (i) f/C01ðuÞ/C18vþWand (ii) vþW/C18f/C01ðuÞ. We ﬁrst prove (i). Suppose v02f/C01ðuÞ. Then fðv0Þ¼u, and so fðv0/C0vÞ¼fðv0Þ/C0fðvÞ¼u/C0u¼0 that is, v0/C0v2W. Thus, v0¼vþðv0/C0vÞ2vþW, and hence f/C01ðuÞ/C18vþW.182 CHAPTER 5 Linear Mappings Now we prove (ii). Suppose v02vþW. Then v0¼vþw, where w2W. Because Wis the kernel of f; we have fðwÞ¼0. Accordingly, fðv0Þ¼fðvþwÞþfðvÞþfðwÞ¼fðvÞþ0¼fðvÞ¼u Thus, v02f/C01ðuÞ, and so vþW/C18f/C01ðuÞ. Both inclusions imply f/C01ðuÞ¼vþW. 5.21. Suppose F:V!UandG:U!Ware linear. Prove (a) rankðG/C14FÞ/C20rankðGÞ, (b) rankðG/C14FÞ/C20rankðFÞ. (a) Because FðVÞ/C18U, we also have GðFðVÞÞ/C18 GðUÞ, and so dim½GðFðVÞÞ/C138/C20 dim½GðUÞ/C138. Then rankðG/C14FÞ¼dim½ðG/C14FÞðVÞ/C138¼ dim½GðFðVÞÞ/C138/C20 dim½GðUÞ/C138¼ rankðGÞ. (b) We have dim½GðFðVÞÞ/C138/C20 dim½FðVÞ/C138. Hence, rankðG/C14FÞ¼dim½ðG/C14FÞðVÞ/C138¼ dim½GðFðVÞÞ/C138/C20 dim½FðVÞ/C138¼ rankðFÞ 5.22. Prove Theorem 5.3: Let F:V!Ube linear. Then, (a) Im Fis a subspace of U, (b) Ker Fis a subspace of V. (a) Because Fð0Þ¼0;we have 02ImF. Now suppose u;u02ImFanda;b2K. Because uandu0 belong to the image of F, there exist vectors v;v02Vsuch that FðvÞ¼uandFðv0Þ¼u0. Then Fðavþbv0Þ¼aFðvÞþbFðv0Þ¼auþbu02ImF Thus, the image of Fis a subspace of U. (b) Because Fð0Þ¼0;we have 02KerF. Now suppose v;w2KerFanda;b2K. Because vandw belong to the kernel of F,FðvÞ¼0 and FðwÞ¼0. Thus, FðavþbwÞ¼aFðvÞþbFðwÞ¼a0þb0¼0þ0¼0; and so avþbw2KerF Thus, the kernel of Fis a subspace of V. 5.23. Prove Theorem 5.6: Suppose Vhas ﬁnite dimension and F:V!Uis linear. Then dimV¼dimðKerFÞþdimðImFÞ¼nullityðFÞþrankðFÞ Suppose dimðKerFÞ¼randfw1;...;wrgis a basis of Ker F, and suppose dim ðImFÞ¼sand fu1;...;usgis a basis of Im F. (By Proposition 5.4, Im Fhas ﬁnite dimension.) Because every uj2ImF, there exist vectors v1;...;vsinVsuch that Fðv1Þ¼u1;...;FðvsÞ¼us. We claim that the set B¼fw1;...;wr;v1;...;vsg is a basis of V; that is, (i) Bspans V, and (ii) Bis linearly independent. Once we prove (i) and (ii), then dimV¼rþs¼dimðKerFÞþdimðImFÞ. (i) B spans V . Let v2V. Then FðvÞ2ImF. Because the ujspan Im F, there exist scalars a1;...;assuch thatFðvÞ¼a1u1þ/C1/C1/C1þ asus. Set ^v¼a1v1þ/C1/C1/C1þ asvs/C0v. Then Fð^vÞ¼Fða1v1þ/C1/C1/C1þ asvs/C0vÞ¼a1Fðv1Þþ/C1/C1/C1þ asFðvsÞ/C0FðvÞ ¼a1u1þ/C1/C1/C1þ asus/C0FðvÞ¼0 Thus, ^v2KerF. Because the wispan Ker F, there exist scalars b1;...;br, such that ^v¼b1w1þ/C1/C1/C1þ brwr¼a1v1þ/C1/C1/C1þ asvs/C0v Accordingly, v¼a1v1þ/C1/C1/C1þ asvs/C0b1w1/C0/C1/C1/C1/C0 brwr Thus, Bspans V.CHAPTER 5 Linear Mappings 183 (ii) B is linearly independent . Suppose x1w1þ/C1/C1/C1þ xrwrþy1v1þ/C1/C1/C1þ ysvs¼0 ð1Þ where xi;yj2K. Then 0¼Fð0Þ¼Fðx1w1þ/C1/C1/C1þ xrwrþy1v1þ/C1/C1/C1þ ysvsÞ ¼x1Fðw1Þþ/C1/C1/C1þ xrFðwrÞþy1Fðv1Þþ/C1/C1/C1þ ysFðvsÞð 2Þ But FðwiÞ¼0, since wi2KerF, and FðvjÞ¼uj. Substituting into (2), we will obtain y1u1þ/C1/C1/C1þ ysus¼0. Since the ujare linearly independent, each yj¼0. Substitution into (1) gives x1w1þ/C1/C1/C1þ xrwr¼0. Since the wiare linearly independent, each xi¼0. Thus Bis linearly independent. Singular and Nonsingular Linear Maps, Isomorphisms 5.24. Determine whether or not each of the following linear maps is nonsingular. If not, ﬁnd a nonzero vector vwhose image is 0. (a)F:R2!R2deﬁned by Fðx;yÞ¼ð x/C0y;x/C02yÞ. (b)G:R2!R2deﬁned by Gðx;yÞ¼ð 2x/C04y;3x/C06yÞ. (a) Find Ker Fby setting FðvÞ¼0, where v¼ðx;yÞ, ðx/C0y;x/C02yÞ¼ð 0;0Þ orx/C0y¼0 x/C02y¼0orx/C0y¼0 /C0y¼0 The only solution is x¼0,y¼0. Hence, Fis nonsingular. (b) Set Gðx;yÞ¼ð 0;0Þto ﬁnd Ker G: ð2x/C04y;3x/C06yÞ¼ð 0;0Þ or2x/C04y¼0 3x/C06y¼0or x/C02y¼0 The system has nonzero solutions, because yis a free variable. Hence, Gis singular. Let y¼1 to obtain the solution v¼ð2;1Þ, which is a nonzero vector, such that GðvÞ¼0. 5.25. The linear map F:R2!R2deﬁned by Fðx;yÞ¼ð x/C0y;x/C02yÞis nonsingular by the previous Problem 5.24. Find a formula for F/C01. SetFðx;yÞ¼ð a;bÞ, so that F/C01ða;bÞ¼ð x;yÞ. We have ðx/C0y;x/C02yÞ¼ð a;bÞ orx/C0y¼a x/C02y¼borx/C0y¼a y¼a/C0b Solve for xandyin terms of aandbto get x¼2a/C0b,y¼a/C0b. Thus, F/C01ða;bÞ¼ð 2a/C0b;a/C0bÞ or F/C01ðx;yÞ¼ð 2x/C0y;x/C0yÞ (The second equation is obtained by replacing aandbbyxandy, respectively.) 5.26. LetG:R2!R3be deﬁned by Gðx;yÞ¼ð xþy;x/C02y;3xþyÞ. (a) Show that Gis nonsingular. (b) Find a formula for G/C01. (a) Set Gðx;yÞ¼ð 0;0;0Þto ﬁnd Ker G. We have ðxþy;x/C02y;3xþyÞ¼ð 0;0;0Þ or xþy¼0;x/C02y¼0;3xþy¼0 The only solution is x¼0,y¼0; hence, Gis nonsingular. (b) Although Gis nonsingular, it is not invertible, because R2andR3have different dimensions. (Thus, Theorem 5.9 does not apply.) Accordingly, G/C01does not exist.184 CHAPTER 5 Linear Mappings 5.27. Suppose that F:V!Uis linear and that Vis of ﬁnite dimension. Show that Vand the image of Fhave the same dimension if and only if Fis nonsingular. Determine all nonsingular linear mappings T:R4!R3. By Theorem 5.6, dim V¼dimðImFÞþdimðKerFÞ. Hence, Vand Im Fhave the same dimension if and only if dimðKerFÞ¼0 or Ker F¼f0g(i.e., if and only if Fis nonsingular). Because dim R3is less than dim R4, we have that dim ðImTÞis less than the dimension of the domain R4ofT. Accordingly no linear mapping T:R4!R3can be nonsingular. 5.28. Prove Theorem 5.7: Let F:V!Ube a nonsingular linear mapping. Then the image of any linearly independent set is linearly independent. Suppose v1;v2;...;vnare linearly independent vectors in V. We claim that Fðv1Þ;Fðv2Þ;...;FðvnÞare also linearly independent. Suppose a1Fðv1Þþa2Fðv2Þþ/C1/C1/C1þ anFðvnÞ¼0, where ai2K. Because Fis linear, Fða1v1þa2v2þ/C1/C1/C1þ anvnÞ¼0. Hence, a1v1þa2v2þ/C1/C1/C1þ anvn2KerF But Fis nonsingular—that is, Ker F¼f0g. Hence, a1v1þa2v2þ/C1/C1/C1þ anvn¼0. Because the viare linearly independent, all the aiare 0. Accordingly, the FðviÞare linearly independent. Thus, the theorem is proved. 5.29. Prove Theorem 5.9: Suppose Vhas ﬁnite dimension and dim V¼dimU. Suppose F:V!Uis linear. Then Fis an isomorphism if and only if Fis nonsingular. IfFis an isomorphism, then only 0 maps to 0; hence, Fis nonsingular. Conversely, suppose Fis nonsingular. Then dim ðKerFÞ¼0. By Theorem 5.6, dim V¼dimðKerFÞþdimðImFÞ. Thus, dimU¼dimV¼dimðImFÞ Because Uhas ﬁnite dimension, Im F¼U. This means Fmaps Vonto U. Thus, Fis one-to-one and onto; that is, Fis an isomorphism. Operations with Linear Maps 5.30. Deﬁne F:R3!R2and G:R3!R2byFðx;y;zÞ¼ð 2x;yþzÞand Gðx;y;zÞ¼ð x/C0z;yÞ. Find formulas deﬁning the maps: (a) FþG, (b) 3 F, (c) 2 F/C05G. (a)ðFþGÞðx;y;zÞ¼Fðx;y;zÞþGðx;y;zÞ¼ð 2x;yþzÞþð x/C0z;yÞ¼ð 3x/C0z;2yþzÞ (b)ð3FÞðx;y;zÞ¼3Fðx;y;zÞ¼3ð2x;yþzÞ¼ð 6x;3yþ3zÞ (c)ð2F/C05GÞðx;y;zÞ¼2Fðx;y;zÞ/C05Gðx;y;zÞ¼2ð2x;yþzÞ/C05ðx/C0z;yÞ ¼ð4x;2yþ2zÞþð/C0 5xþ5z;/C05yÞ¼ð/C0 xþ5z;/C03yþ2zÞ 5.31. LetF:R3!R2andG:R2!R2be deﬁned by Fðx;y;zÞ¼ð 2x;yþzÞandGðx;yÞ¼ð y;xÞ. Derive formulas deﬁning the mappings: (a) G/C14F, (b) F/C14G. (a)ðG/C14FÞðx;y;zÞ¼GðFðx;y;zÞÞ¼ Gð2x;yþzÞ¼ð yþz;2xÞ (b) The mapping F/C14Gis not deﬁned, because the image of Gis not contained in the domain of F. 5.32. Prove: (a) The zero mapping 0, deﬁned by 0ðvÞ¼02Ufor every v2V, is the zero element of HomðV;UÞ. (b) The negative of F2HomðV;UÞis the mappingð/C01ÞF, that is,/C0F¼ð/C0 1ÞF. LetF2HomðV;UÞ. Then, for every v2V: ðFþ0ÞðvÞ¼FðvÞþ0ðvÞ¼FðvÞþ0¼FðvÞ ðaÞ BecauseðFþ0ÞðvÞ¼FðvÞfor every v2V, we have Fþ0¼F. Similarly, 0þF¼F: ðFþð/C0 1ÞFÞðvÞ¼FðvÞþð/C0 1ÞFðvÞ¼FðvÞ/C0FðvÞ¼0¼0ðvÞ ðbÞ Thus, Fþð/C0 1ÞF¼0:Similarlyð/C01ÞFþF¼0:Hence,/C0F¼ð/C0 1ÞF:CHAPTER 5 Linear Mappings 185 5.33. Suppose F1;F2;...;Fnare linear maps from VintoU. Show that, for any scalars a1;a2;...;an, and for any v2V, ða1F1þa2F2þ/C1/C1/C1þ anFnÞðvÞ¼a1F1ðvÞþa2F2ðvÞþ/C1/C1/C1þ anFnðvÞ The mapping a1F1is deﬁned byða1F1ÞðvÞ¼a1FðvÞ. Hence, the theorem holds for n¼1. Accordingly, by induction, ða1F1þa2F2þ/C1/C1/C1þ anFnÞðvÞ¼ð a1F1ÞðvÞþð a2F2þ/C1/C1/C1þ anFnÞðvÞ ¼a1F1ðvÞþa2F2ðvÞþ/C1/C1/C1þ anFnðvÞ 5.34. Consider linear mappings F:R3!R2,G:R3!R2,H:R3!R2deﬁned by Fðx;y;zÞ¼ð xþyþz;xþyÞ; Gðx;y;zÞ¼ð 2xþz;xþyÞ; Hðx;y;zÞ¼ð 2y;xÞ Show that F,G,Hare linearly independent [as elements of Hom ðR3;R2Þ]. Suppose, for scalars a;b;c2K, aFþbGþcH¼0 ð1Þ (Here 0is the zero mapping.) For e1¼ð1;0;0Þ2R3, we have 0ðe1Þ¼ð 0;0Þand ðaFþbGþcHÞðe1Þ¼aFð1;0;0ÞþbGð1;0;0ÞþcHð1;0;0Þ ¼að1;1Þþbð2;1Þþcð0;1Þ¼ð aþ2b;aþbþcÞ Thus by (1),ðaþ2b;aþbþcÞ¼ð 0;0Þand so aþ2b¼0 and aþbþc¼0 ð2Þ Similarly for e2¼ð0;1;0Þ2R3, we have 0ðe2Þ¼ð 0;0Þand ðaFþbGþcHÞðe2Þ¼aFð0;1;0ÞþbGð0;1;0ÞþcHð0;1;0Þ ¼að1;1Þþbð0;1Þþcð2;0Þ¼ð aþ2c;aþbÞ Thus,aþ2c¼0 and aþb¼0 ð3Þ Using (2) and (3), we obtain a¼0; b¼0; c¼0 ð4Þ Because (1) implies (4), the mappings F,G,Hare linearly independent. 5.35. Letkbe a nonzero scalar. Show that a linear map Tis singular if and only if kTis singular. Hence, Tis singular if and only if /C0Tis singular. Suppose Tis singular. Then TðvÞ¼0 for some vector v6¼0. Hence, ðkTÞðvÞ¼kTðvÞ¼k0¼0 and so kTis singular. Now suppose kTis singular. ThenðkTÞðwÞ¼0 for some vector w6¼0. Hence, TðkwÞ¼kTðwÞ¼ð kTÞðwÞ¼0 Butk6¼0 and w6¼0 implies kw6¼0. Thus, Tis also singular. 5.36. Find the dimension dof: (a) HomðR3;R4Þ, (b) HomðR5;R3Þ, (c) HomðP3ðtÞ;R2Þ,(d) HomðM2;3;R4Þ. Use dim½HomðV;UÞ/C138¼ mn, where dim V¼mand dim U¼n. (a)d¼3ð4Þ¼12. (c) Because dim P3ðtÞ¼4,d¼4ð2Þ¼8. (b)d¼5ð3Þ¼15. (d) Because dim M2;3¼6,d¼6ð4Þ¼24.186 CHAPTER 5 Linear Mappings 5.37. Prove Theorem 5.11. Suppose dim V¼mand dim U¼n. Then dim½HomðV;UÞ/C138¼ mn. Supposefv1;...;vmgis a basis of Vandfu1;...;ungis a basis of U. By Theorem 5.2, a linear mapping in HomðV;UÞis uniquely determined by arbitrarily assigning elements of Uto the basis elements viofV.W e deﬁne Fij2HomðV;UÞ; i¼1;...;m;j¼1;...;n to be the linear mapping for which FijðviÞ¼uj, and FijðvkÞ¼0 for k6¼i. That is, Fijmaps viintoujand the other v’s into 0. Observe that fFijgcontains exactly mnelements; hence, the theorem is proved if we show that it is a basis of Hom ðV;UÞ. Proof thatfFijggenerates HomðV;UÞ. Consider an arbitrary function F2HomðV;UÞ. Suppose Fðv1Þ¼w1;Fðv2Þ¼w2;...;FðvmÞ¼wm. Because wk2U, it is a linear combination of the u’s; say, wk¼ak1u1þak2u2þ/C1/C1/C1þ aknun; k¼1;...;m;aij2K ð1Þ Consider the linear mapping G¼Pm i¼1Pn j¼1aijFij. Because Gis a linear combination of the Fij, the proof thatfFijggenerates HomðV;UÞis complete if we show that F¼G. We now compute GðvkÞ;k¼1;...;m. Because FijðvkÞ¼0 for k6¼iandFkiðvkÞ¼ui; GðvkÞ¼Pm i¼1Pn j¼1aijFijðvkÞ¼Pn j¼1akjFkjðvkÞ¼Pn j¼1akjuj ¼ak1u1þak2u2þ/C1/C1/C1þ aknun Thus, by (1), GðvkÞ¼wkfor each k. But FðvkÞ¼wkfor each k. Accordingly, by Theorem 5.2, F¼G; hence,fFijggenerates HomðV;UÞ. Proof thatfFijgis linearly independent . Suppose, for scalars cij2K, Pm i¼1Pn j¼1cijFij¼0 For vk;k¼1;...;m, 0¼0ðvkÞ¼Pm i¼1Pn j¼1cijFijðvkÞ¼Pn j¼1ckjFkjðvkÞ¼Pn j¼1ckjuj ¼ck1u1þck2u2þ/C1/C1/C1þ cknun But the uiare linearly independent; hence, for k¼1;...;m, we have ck1¼0;ck2¼0;...;ckn¼0. In other words, all the cij¼0, and sofFijgis linearly independent. 5.38. Prove Theorem 5.12: (i) G/C14ðFþF0Þ¼G/C14FþG/C14F0. (ii)ðGþG0Þ/C14F¼G/C14FþG0/C14F. (iii)kðG/C14FÞ¼ð kGÞ/C14F¼G/C14ðkFÞ. (i) For every v2V, ðG/C14ðFþF0ÞÞðvÞ¼GððFþF0ÞðvÞÞ¼ GðFðvÞþF0ðvÞÞ ¼GðFðvÞÞþ GðF0ðvÞÞ¼ð G/C14FÞðvÞþð G/C14F0ÞðvÞ¼ð G/C14FþG/C14F0ÞðvÞ Thus, G/C14ðFþF0Þ¼G/C14FþG/C14F0. (ii) For every v2V, ððGþG0Þ/C14FÞðvÞ¼ð GþG0ÞðFðvÞÞ¼ GðFðvÞÞþ G0ðFðvÞÞ ¼ðG/C14FÞðvÞþð G0/C14FÞðvÞ¼ð G/C14FþG0/C14FÞðvÞ Thus,ðGþG0Þ/C14F¼G/C14FþG0/C14F.CHAPTER 5 Linear Mappings 187 (iii) For every v2V, ðkðG/C14FÞÞðvÞ¼kðG/C14FÞðvÞ¼kðGðFðvÞÞÞ¼ð kGÞðFðvÞÞ¼ð kG/C14FÞðvÞ and ðkðG/C14FÞÞðvÞ¼kðG/C14FÞðvÞ¼kðGðFðvÞÞÞ¼ GðkFðvÞÞ¼ GððkFÞðvÞÞ¼ð G/C14kFÞðvÞ Accordingly, kðG/C14FÞ¼ð kGÞ/C14F¼G/C14ðkFÞ. (We emphasize that two mappings are shown to be equal by showing that each of them assigns the same image to each point in the domain.) Algebra of Linear Maps 5.39. LetFandGbe the linear operators on R2deﬁned by Fðx;yÞ¼ð y;xÞandGðx;yÞ¼ð 0;xÞ. Find formulas deﬁning the following operators:(a)FþG, (b) 2 F/C03G, (c) FG, (d) GF, (e) F 2,( f ) G2. (a)ðFþGÞðx;yÞ¼Fðx;yÞþGðx;yÞ¼ð y;xÞþð 0;xÞ¼ð y;2xÞ. (b)ð2F/C03GÞðx;yÞ¼2Fðx;yÞ/C03Gðx;yÞ¼2ðy;xÞ/C03ð0;xÞ¼ð 2y;/C0xÞ. (c)ðFGÞðx;yÞ¼FðGðx;yÞÞ¼ Fð0;xÞ¼ð x;0Þ. (d)ðGFÞðx;yÞ¼GðFðx;yÞÞ¼ Gðy;xÞ¼ð 0;yÞ. (e)F2ðx;yÞ¼FðFðx;yÞÞ¼ Fðy;xÞ¼ð x;yÞ. (Note that F2¼I, the identity mapping.) (f)G2ðx;yÞ¼GðGðx;yÞÞ¼ Gð0;xÞ¼ð 0;0Þ. (Note that G2¼0, the zero mapping.) 5.40. Consider the linear operator TonR3deﬁned by Tðx;y;zÞ¼ð 2x;4x/C0y;2xþ3y/C0zÞ. (a) Show that Tis invertible. Find formulas for (b) T/C01, (c) T2,(d)T/C02. (a) Let W¼KerT. We need only show that Tis nonsingular (i.e., that W¼f0g). Set Tðx;y;zÞ¼ð 0;0;0Þ, which yields Tðx;y;zÞ¼ð 2x;4x/C0y;2xþ3y/C0zÞ¼ð 0;0;0Þ Thus, Wis the solution space of the homogeneous system 2x¼0; 4x/C0y¼0; 2xþ3y/C0z¼0 which has only the trivial solution (0, 0, 0). Thus, W¼f0g. Hence, Tis nonsingular, and so Tis invertible. (b) Set Tðx;y;zÞ¼ð r;s;tÞ[and so T/C01ðr;s;tÞ¼ð x;y;zÞ]. We have ð2x;4x/C0y;2xþ3y/C0zÞ¼ð r;s;tÞ or 2 x¼r;4x/C0y¼s;2xþ3y/C0z¼t Solve for x,y,zin terms of r,s,tto get x¼1 2r,y¼2r/C0s,z¼7r/C03s/C0t. Thus, T/C01ðr;s;tÞ¼ð1 2r;2r/C0s;7r/C03s/C0tÞ or T/C01ðx;y;zÞ¼ð1 2x;2x/C0y;7x/C03y/C0zÞ (c) Apply Ttwice to get T2ðx;y;zÞ¼Tð2x;4x/C0y;2xþ3y/C0zÞ ¼½4x;4ð2xÞ/C0ð 4x/C0yÞ;2ð2xÞþ3ð4x/C0yÞ/C0ð 2xþ3y/C0zÞ/C138 ¼ð4x;4xþy;14x/C06yþzÞ (d) Apply T/C01twice to get T/C02ðx;y;zÞ¼T/C02ð1 2x;2x/C0y;7x/C03y/C0zÞ ¼½1 4x;2ð1 2xÞ/C0ð 2x/C0yÞ;7ð1 2xÞ/C03ð2x/C0yÞ/C0ð 7x/C03y/C0zÞ/C138 ¼ð1 4x;/C0xþy;/C019 2xþ6yþzÞ188 CHAPTER 5 Linear Mappings 5.41. LetVbe of ﬁnite dimension and let Tbe a linear operator on Vfor which TR¼I, for some operator RonV. (We call Raright inverse ofT.) (a) Show that Tis invertible. (b) Show that R¼T/C01. (c) Give an example showing that the above need not hold if Vis of inﬁnite dimension. (a) Let dim V¼n. By Theorem 5.14, Tis invertible if and only if Tis onto; hence, Tis invertible if and only if rankðTÞ¼n. We have n¼rankðIÞ¼rankðTRÞ/C20rankðTÞ/C20n. Hence, rankðTÞ¼nandTis invertible. (b)TT/C01¼T/C01T¼I. Then R¼IR¼ðT/C01TÞR¼T/C01ðTRÞ¼T/C01I¼T/C01. (c) Let Vbe the space of polynomials in tover K; say, pðtÞ¼a0þa1tþa2t2þ/C1/C1/C1þ asts. Let TandRbe the operators on Vdeﬁned by TðpðtÞÞ¼ 0þa1þa2tþ/C1/C1/C1þ asts/C01and RðpðtÞÞ¼ a0tþa1t2þ/C1/C1/C1þ astsþ1 We have ðTRÞðpðtÞÞ¼ TðRðpðtÞÞÞ¼ Tða0tþa1t2þ/C1/C1/C1þ astsþ1Þ¼a0þa1tþ/C1/C1/C1þ asts¼pðtÞ and so TR¼I, the identity mapping. On the other hand, if k2Kandk6¼0, then ðRTÞðkÞ¼RðTðkÞÞ¼ Rð0Þ¼06¼k Accordingly, RT6¼I. 5.42. LetFandGbe linear operators on R2deﬁned by Fðx;yÞ¼ð 0;xÞandGðx;yÞ¼ð x;0Þ. Show that (a)GF¼0, the zero mapping, but FG6¼0. (b) G2¼G. (a)ðGFÞðx;yÞ¼GðFðx;yÞÞ¼ Gð0;xÞ¼ð 0;0Þ. Because GFassigns 0¼ð0;0Þto every vector ( x;y)i nR2, it is the zero mapping; that is, GF¼0. On the other hand, ðFGÞðx;yÞ¼FðGðx;yÞÞ¼ Fðx;0Þ¼ð 0;xÞ. For example,ðFGÞð2;3Þ¼ð 0;2Þ. Thus, FG6¼0, as it does not assign 0 ¼ð0;0Þto every vector in R2. (b) For any vector ( x;y)i nR2,w eh a v e G2ðx;yÞ¼GðGðx;yÞÞ¼ Gðx;0Þ¼ð x;0Þ¼Gðx;yÞ. Hence, G2¼G. 5.43. Find the dimension of (a) AðR4Þ, (b) AðP2ðtÞÞ, (c) AðM2;3). Use dim½AðVÞ/C138¼ n2where dim V¼n. Hence, (a) dim½AðR4Þ/C138¼ 42¼16, (b) dim½AðP2ðtÞÞ/C138¼ 32¼9, (c) dim½AðM2;3Þ/C138¼ 62¼36. 5.44. LetEbe a linear operator on Vfor which E2¼E. (Such an operator is called a projection .) Let U be the image of E, and let Wbe the kernel. Prove (a) If u2U, then EðuÞ¼u(i.e., Eis the identity mapping on U). (b) If E6¼I, then Eis singular—that is, EðvÞ¼0 for some v6¼0. (c) V¼U/C8W. (a) If u2U, the image of E, then EðvÞ¼ufor some v2V. Hence, using E2¼E, we have u¼EðvÞ¼E2ðvÞ¼EðEðvÞÞ¼ EðuÞ (b) If E6¼I, then for some v2V,EðvÞ¼u, where v6¼u. By (i), EðuÞ¼u. Thus, Eðv/C0uÞ¼EðvÞ/C0EðuÞ¼u/C0u¼0; where v/C0u6¼0 (c) We ﬁrst show that V¼UþW. Let v2V. Set u¼EðvÞandw¼v/C0EðvÞ. Then v¼EðvÞþv/C0EðvÞ¼uþw By deﬂnition, u¼EðvÞ2U, the image of E. We now show that w2W, the kernel of E, EðwÞ¼Eðv/C0EðvÞÞ¼ EðvÞ/C0E2ðvÞ¼EðvÞ/C0EðvÞ¼0 and thus w2W. Hence, V¼UþW. We next show that U\W¼f0g. Let v2U\W. Because v2U,EðvÞ¼vby part (a). Because v2W,EðvÞ¼0. Thus, v¼EðvÞ¼0 and so U\W¼f0g. The above two properties imply that V¼U/C8W.CHAPTER 5 Linear Mappings 189 SUPPLEMENTARY PROBLEMS Mappings 5.45. Determine the number of different mappings from ðaÞf1;2gintof1;2;3g;ðbÞf1;2;...;rgintof1;2;...;sg: 5.46. Letf:R!Randg:R!Rbe deﬁned by fðxÞ¼x2þ3xþ1 and gðxÞ¼2x/C03. Find formulas deﬁning the composition mappings: (a) f/C14g; (b) g/C14f; (c) g/C14g;(d)f/C14f. 5.47. For each mappings f:R!Rﬁnd a formula for its inverse: (a) fðxÞ¼3x/C07, (b) fðxÞ¼x3þ2. 5.48. For any mapping f:A!B, show that 1B/C14f¼f¼f/C141A. Linear Mappings 5.49. Show that the following mappings are linear: (a) F:R3!R2deﬁned by Fðx;y;zÞ¼ð xþ2y/C03z;4x/C05yþ6zÞ. (b) F:R2!R2deﬁned by Fðx;yÞ¼ð axþby;cxþdyÞ, where a,b,c,dbelong to R. 5.50. Show that the following mappings are not linear: (a) F:R2!R2deﬁned by Fðx;yÞ¼ð x2;y2Þ. (b) F:R3!R2deﬁned by Fðx;y;zÞ¼ð xþ1;yþzÞ. (c) F:R2!R2deﬁned by Fðx;yÞ¼ð xy;yÞ. (d) F:R3!R2deﬁned by Fðx;y;zÞ¼ðj xj;yþzÞ. 5.51. Find Fða;bÞ, where the linear map F:R2!R2is deﬁned by Fð1;2Þ¼ð 3;/C01ÞandFð0;1Þ¼ð 2;1Þ. 5.52. Find a 2/C22 matrix Athat maps (a)ð1;3ÞTandð1;4ÞTintoð/C02;5ÞTandð3;/C01ÞT, respectively. (b)ð2;/C04ÞTandð/C01;2ÞTintoð1;1ÞTandð1;3ÞT, respectively. 5.53. Find a 2/C22 singular matrix Bthat mapsð1;1ÞTintoð1;3ÞT. 5.54. LetVbe the vector space of real n-square matrices, and let Mbe a ﬁxed nonzero matrix in V. Show that the ﬁrst two of the following mappings T:V!Vare linear, but the third is not: (a)TðAÞ¼MA, (b) TðAÞ¼AMþMA, (c) TðAÞ¼MþA. 5.55. Give an example of a nonlinear map F:R2!R2such that F/C01ð0Þ¼f 0gbutFis not one-to-one. 5.56. LetF:R2!R2be deﬁned by Fðx;yÞ¼ð 3xþ5y;2xþ3yÞ, and let Sbe the unit circle in R2.(Sconsists of all points satisfying x2þy2¼1.) Find (a) the image FðSÞ, (b) the preimage F/C01ðSÞ. 5.57. Consider the linear map G:R3!R3deﬁned by Gðx;y;zÞ¼ð xþyþz;y/C02z;y/C03zÞand the unit sphere S2inR3, which consists of the points satisfying x2þy2þz2¼1. Find (a) GðS2Þ, (b) G/C01ðS2Þ. 5.58. Let Hbe the plane xþ2y/C03z¼4i n R3and let Gbe the linear map in Problem 5.57. Find (a)GðHÞ, (b) G/C01ðHÞ. 5.59. LetWbe a subspace of V. The inclusion map, denoted by i:W,!V, is deﬁned by iðwÞ¼wfor every w2W. Show that the inclusion map is linear. 5.60. Suppose F:V!Uis linear. Show that Fð/C0vÞ¼/C0 FðvÞ. Kernel and Image of Linear Mappings 5.61. For each linear map Fﬁnd a basis and the dimension of the kernel and the image of F: (a) F:R3!R3deﬁned by Fðx;y;zÞ¼ð xþ2y/C03z;2xþ5y/C04z;xþ4yþzÞ, (b) F:R4!R3deﬁned by Fðx;y;z;tÞ¼ð xþ2yþ3zþ2t;2xþ4yþ7zþ5t;xþ2yþ6zþ5tÞ.190 CHAPTER 5 Linear Mappings 5.62. For each linear map G, ﬁnd a basis and the dimension of the kernel and the image of G: (a) G:R3!R2deﬁned by Gðx;y;zÞ¼ð xþyþz;2xþ2yþ2zÞ, (b) G:R3!R2deﬁned by Gðx;y;zÞ¼ð xþy;yþzÞ, (c) G:R5!R3deﬁned by Gðx;y;z;s;tÞ¼ð xþ2yþ2zþsþt;xþ2yþ3zþ2s/C0t;3xþ6yþ8zþ5s/C0tÞ: 5.63. Each of the following matrices determines a linear map from R4intoR3: (a) A¼12 01 2/C012/C01 1/C032/C022 43 5, (b) B¼10 2/C01 23/C011 /C020/C0532 43 5. Find a basis as well as the dimension of the kernel and the image of each linear map. 5.64. Find a linear mapping F:R3!R3whose image is spanned by (1, 2, 3) and (4, 5, 6). 5.65. Find a linear mapping G:R4!R3whose kernel is spanned by (1, 2, 3, 4) and (0, 1, 1, 1). 5.66. LetV¼P10ðtÞ, the vector space of polynomials of degree /C2010. Consider the linear map D4:V!V, where D4denotes the fourth derivative d4ðfÞ=dt4. Find a basis and the dimension of (a) the image of D4; (b) the kernel of D4. 5.67. Suppose F:V!Uis linear. Show that (a) the image of any subspace of Vis a subspace of U; (b) the preimage of any subspace of Uis a subspace of V. 5.68. Show that if F:V!Uis onto, then dim U/C20dimV. Determine all linear maps F:R3!R4that are onto. 5.69. Consider the zero mapping 0:V!Udeﬁned by 0ðvÞ¼0;8v2V. Find the kernel and the image of 0. Operations with linear Mappings 5.70. LetF:R3!R2andG:R3!R2be deﬁned by Fðx;y;zÞ¼ð y;xþzÞandGðx;y;zÞ¼ð 2z;x/C0yÞ. Find formulas deﬁning the mappings FþGand 3 F/C02G. 5.71. LetH:R2!R2be deﬁned by Hðx;yÞ¼ð y;2xÞ. Using the maps FandGin Problem 5.70, ﬁnd formulas deﬁning the mappings: (a) H/C14FandH/C14G, (b) F/C14HandG/C14H, (c) H/C14ðFþGÞandH/C14FþH/C14G. 5.72. Show that the following mappings F,G,Hare linearly independent: (a) F;G;H2HomðR2;R2Þdeﬁned by Fðx;yÞ¼ð x;2yÞ,Gðx;yÞ¼ð y;xþyÞ,Hðx;yÞ¼ð 0;xÞ, (b) F;G;H2HomðR3;RÞdeﬁned by Fðx;y;zÞ¼xþyþz,Gðx;y;zÞ¼yþz,Hðx;y;zÞ¼x/C0z. 5.73. ForF;G2HomðV;UÞ, show that rankðFþGÞ/C20rankðFÞþrankðGÞ. (Here Vhas ﬁnite dimension.) 5.74. LetF:V!UandG:U!Vbe linear. Show that if FandGare nonsingular, then G/C14Fis nonsingular. Give an example where G/C14Fis nonsingular but Gis not. [Hint: Let dim V<dimU:/C138 5.75. Find the dimension dof (a) HomðR2;R8Þ, (b) HomðP4ðtÞ;R3Þ, (c) HomðM2;4;P2ðtÞÞ. 5.76. Determine whether or not each of the following linear maps is nonsingular. If not, ﬁnd a nonzero vector v whose image is 0; otherwise ﬁnd a formula for the inverse map: (a) F:R3!R3deﬁned by Fðx;y;zÞ¼ð xþyþz;2xþ3yþ5z;xþ3yþ7zÞ, (b) G:R3!P2ðtÞdeﬁned by Gðx;y;zÞ¼ð xþyÞt2þðxþ2yþ2zÞtþyþz, (c) H:R2!P2ðtÞdeﬁned by Hðx;yÞ¼ð xþ2yÞt2þðx/C0yÞtþxþy. 5.77. When can dim½HomðV;UÞ/C138¼ dimV?CHAPTER 5 Linear Mappings 191 Algebra of Linear Operators 5.78. LetFandGbe the linear operators on R2deﬁned by Fðx;yÞ¼ð xþy;0ÞandGðx;yÞ¼ð/C0 y;xÞ. Find formulas deﬁning the linear operators: (a) FþG, (b) 5 F/C03G, (c) FG,(d)GF,(e)F2,(f)G2. 5.79. Show that each linear operator TonR2is nonsingular and ﬁnd a formula for T/C01, where (a)Tðx;yÞ¼ð xþ2y;2xþ3yÞ, (b) Tðx;yÞ¼ð 2x/C03y;3x/C04yÞ. 5.80. Show that each of the following linear operators TonR3is nonsingular and ﬁnd a formula for T/C01, where (a)Tðx;y;zÞ¼ð x/C03y/C02z;y/C04z;zÞ; (b) Tðx;y;zÞ¼ð xþz;x/C0y;yÞ. 5.81. Find the dimension of AðVÞ, where (a) V¼R7, (b) V¼P5ðtÞ, (c) V¼M3;4. 5.82. Which of the following integers can be the dimension of an algebra AðVÞof linear maps: 5, 9, 12, 25, 28, 36, 45, 64, 88, 100? 5.83. LetTbe the linear operator on R2deﬁned by Tðx;yÞ¼ð xþ2y;3xþ4yÞ. Find a formula for fðTÞ, where (a)fðtÞ¼t2þ2t/C03, (b) fðtÞ¼t2/C05t/C02. Miscellaneous Problems 5.84. Suppose F:V!Uis linear and kis a nonzero scalar. Prove that the maps FandkFhave the same kernel and the same image. 5.85. Suppose FandGare linear operators on Vand that Fis nonsingular. Assume that Vhas ﬁnite dimension. Show that rankðFGÞ¼rankðGFÞ¼rankðGÞ. 5.86. Suppose Vhas ﬁnite dimension. Suppose Tis a linear operator on Vsuch that rankðT2Þ¼rankðTÞ. Show that Ker T\ImT¼f0g. 5.87. Suppose V¼U/C8W. Let E1andE2be the linear operators on Vdeﬁned by E1ðvÞ¼u,E2ðvÞ¼w, where v¼uþw,u2U,w2W. Show that (a) E2 1¼E1and E2 2¼E2(i.e., that E1and E2are projections); (b)E1þE2¼I, the identity mapping; (c) E1E2¼0andE2E1¼0. 5.88. LetE1andE2be linear operators on Vsatisfying parts (a), (b), (c) of Problem 5.88. Prove V¼ImE1/C8ImE2 5.89. Letvandwbe elements of a real vector space V. The line segment L from vtovþwis deﬁned to be the set of vectors vþtwfor 0/C20t/C201. (See Fig. 5.6.) (a) Show that the line segment Lbetween vectors vanduconsists of the points: (i)ð1/C0tÞvþtufor 0/C20t/C201, (ii) t1vþt2ufort1þt2¼1,t1/C210,t2/C210. (b) Let F:V!Ube linear. Show that the image FðLÞof a line segment LinVis a line segment in U. Figure 5-6192 CHAPTER 5 Linear Mappings 5.90. LetF:V!Ube linear and let Wbe a subspace of V. The restriction ofFtoWis the map FjW:W!U deﬁned by FjWðvÞ¼FðvÞfor every vinW. Prove the following: (a)FjWis linear; (b) KerðFjWÞ¼ð KerFÞ\W; (c) ImðFjWÞ¼FðWÞ. 5.91. A subset Xof a vector space Vis said to be convex if the line segment Lbetween any two points (vectors) P;Q2Xis contained in X. (a) Show that the intersection of convex sets is convex; (b) suppose F:V!U is linear and Xis convex. Show that FðXÞis convex. ANSWERS TO SUPPLEMENTARY PROBLEMS 5.45.ðaÞ32¼9;ðbÞsr 5.46. (a)ðf/C14gÞðxÞ¼4x2þ1, (b)ðg/C14fÞðxÞ¼2x2þ6x/C01, (c)ðg/C14gÞðxÞ¼4x/C09, (d)ðf/C14fÞðxÞ¼x4þ6x3þ14x2þ15xþ5 5.47. (a)f/C01ðxÞ¼1 3ðxþ7Þ, (b) f/C01ðxÞ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ x/C023p 5.49. Fðx;y;zÞ¼Aðx;y;zÞT, where (a) A¼12/C03 4/C056/C20/C21 , (b) A¼ab cd/C20/C21 5.50. (a)u¼ð2;2Þ,k¼3; then FðkuÞ¼ð 36;36ÞbutkFðuÞ¼ð 12;12Þ; (b) Fð0Þ6¼0; (c)u¼ð1;2Þ,v¼ð3;4Þ; then FðuþvÞ¼ð 24;6ÞbutFðuÞþFðvÞ¼ð 14;6Þ; (d)u¼ð1;2;3Þ,k¼/C02; then FðkuÞ¼ð 2;/C010ÞbutkFðuÞ¼ð/C0 2;/C010Þ. 5.51. Fða;bÞ¼ð/C0 aþ2b;/C03aþbÞ 5.52. (a)A¼/C017 5 23/C06/C20/C21 ; (b) None. (2 ;/C04) and (/C01;2) are linearly dependent but not (1, 1) and (1, 3). 5.53. B¼10 30/C20/C21 [Hint: Sendð0;1ÞTintoð0;0ÞT.] 5.55. Fðx;yÞ¼ð x2;y2Þ 5.56. (a) 13 x2/C042xyþ34y2¼1, (b) 13 x2þ42xyþ34y2¼1 5.57. (a)x2/C08xyþ26y2þ6xz/C038yzþ14z2¼1, (b) x2þ2xyþ3y2þ2xz/C08yzþ14z2¼1 5.58. (a)x/C0yþ2z¼4, (b) xþ6z¼4 5.61. (a) dimðKerFÞ¼1,fð7;/C02;1Þg; dimðImFÞ¼2,fð1;2;1Þ;ð0;1;2Þg; (b) dimðKerFÞ¼2,fð/C02;1;0;0Þ;ð1;0;/C01;1Þg; dimðImFÞ¼2,fð1;2;1Þ;ð0;1;3Þg 5.62. (a) dimðKerGÞ¼2,fð1;0;/C01Þ;ð1;/C01;0Þg; dimðImGÞ¼1,fð1;2Þg; (b) dimðKerGÞ¼1,fð1;/C01;1Þg;I m G¼R2,fð1;0Þ;ð0;1Þg; (c) dimðKerGÞ¼3,fð/C02;1;0;0;0Þ;ð1;0;/C01;1;0Þ;ð/C05;0;2;0;1Þg; dimðImGÞ¼2, fð1;1;3Þ;ð0;1;2Þg 5.63. (a) dimðKerAÞ¼2,fð4;/C02;/C05;0Þ;ð1;/C03;0;5Þg; dimðImAÞ¼2,fð1;2;1Þ;ð0;1;1Þg; (b) dimðKerBÞ¼1,fð/C01;2 3;1;1Þg;I m B¼R3 5.64. Fðx;y;zÞ¼ð xþ4y;2xþ5y;3xþ6yÞCHAPTER 5 Linear Mappings 193 5.65. Fðx;y;z;tÞ¼ð xþy/C0z;2xþy/C0t;0Þ 5.66. (a)f1;t;t2;...;t6g, (b)f1;t;t2;t3g 5.68. None, because dim R4>dimR3: 5.69. Ker0¼V,I m0¼f0g 5.70.ðFþGÞðx;y;zÞ¼ð yþ2z;2x/C0yþzÞ,ð3F/C02GÞðx;y;zÞ¼ð 3y/C04z;xþ2yþ3zÞ 5.71. (a)ðH/C14FÞðx;y;zÞ¼ð xþz;2yÞ,ðH/C14GÞðx;y;zÞ¼ð x/C0y;4zÞ; (b) not deﬁned; (c)ðH/C14ðFþGÞÞðx;y;zÞ¼ð H/C14FþH/C14GÞðx;y;zÞ¼ð 2x/C0yþz;2yþ4zÞ 5.74. Fðx;yÞ¼ð x;y;yÞ;Gðx;y;zÞ¼ð x;yÞ 5.75. (a) 16, (b) 15, (c) 24 5.76. (a)v¼ð2;/C03;1Þ; (b) G/C01ðat2þbtþcÞ¼ð b/C02c;a/C0bþ2c;/C0aþb/C0cÞ; (c)His nonsingular, but not invertible, because dim P2ðtÞ>dimR2. 5.77. dimU¼1; that is, U¼K. 5.78. (a)ðFþGÞðx;yÞ¼ð x;xÞ; (b)ð5F/C03GÞðx;yÞ¼ð 5xþ8y;/C03xÞ; (c)ðFGÞðx;yÞ¼ð x/C0y;0Þ; (d)ðGFÞðx;yÞ¼ð 0;xþyÞ;(e)F2ðx;yÞ¼ð xþy;0Þ(note that F2¼F); (f)G2ðx;yÞ¼ð/C0 x;/C0yÞ. [Note that G2þI¼0; hence, Gis a zero of fðtÞ¼t2þ1.] 5.79. (a)T/C01ðx;yÞ¼ð/C0 3xþ2y;2x/C0yÞ, (b) T/C01ðx;yÞ¼ð/C0 4xþ3y;/C03xþ2yÞ 5.80. (a)T/C01ðx;y;zÞ¼ð xþ3yþ14z;y/C04z;zÞ, (b) T/C01ðx;y;zÞ¼ð yþz;y;x/C0y/C0zÞ 5.81. (a) 49, (b) 36, (c) 144 5.82. Squares: 9, 25, 36, 64, 100 5.83. (a)Tðx;yÞ¼ð 6xþ14y;21xþ27yÞ; (b) Tðx;yÞ¼ð 0;0Þ—that is, fðTÞ¼0194 CHAPTER 5 Linear Mappings Linear Mappings and Matrices 6.1 Introduction Consider a basis S¼fu1;u2;...;ungof a vector space Vover a ﬁeld K. For any vector v2V, suppose v¼a1u1þa2u2þ/C1/C1/C1þ anun Then the coordinate vector of vrelative to the basis S, which we assume to be a column vector (unless otherwise stated or implied), is denoted and deﬁned by ½v/C138S¼½a1;a2;...;an/C138T Recall (Section 4.11) that the mapping v7!½v/C138S, determined by the basis S, is an isomorphism between V andKn. This chapter shows that there is also an isomorphism, determined by the basis S, between the algebra AðVÞof linear operators on Vand the algebra Mofn-square matrices over K. Thus, every linear mapping F:V!Vwill correspond to an n-square matrix½F/C138Sdetermined by the basis S. We will also show how our matrix representation changes when we choose another basis. 6.2 Matrix Representation of a Linear Operator Let Tbe a linear operator (transformation) from a vector space Vinto itself, and suppose S¼fu1;u2;...;ungis a basis of V. Now Tðu1Þ,Tðu2Þ;...;TðunÞare vectors in V, and so each is a linear combination of the vectors in the basis S; say, Tðu1Þ¼a11u1þa12u2þ/C1/C1/C1þ a1nun Tðu2Þ¼a21u1þa22u2þ/C1/C1/C1þ a2nun :::::::::::::::::::::::::::::::::::::::::::::::::::::: TðunÞ¼an1u1þan2u2þ/C1/C1/C1þ annun The following deﬁnition applies. DEFINITION: The transpose of the above matrix of coefﬁcients, denoted by mSðTÞor½T/C138S, is called thematrix representation ofTrelative to the basis S, or simply the matrix of Tin the basis S. (The subscript Smay be omitted if the basis Sis understood.) Using the coordinate (column) vector notation, the matrix representation of Tmay be written in the form mSðTÞ¼½ T/C138S¼½Tðu1Þ/C138S;½Tðu2Þ/C138S;...;½Tðu1Þ/C138S/C2/C3 That is, the columns of mðTÞare the coordinate vectors of Tðu1Þ,Tðu2Þ;...;TðunÞ, respectively. CHAPTER 6 195 EXAMPLE 6.1 LetF:R2!R2be the linear operator deﬁned by Fðx;yÞ¼ð 2xþ3y;4x/C05yÞ. (a) Find the matrix representation of Frelative to the basis S¼fu1;u2g¼fð 1;2Þ;ð2;5Þg. (1) First ﬁnd Fðu1Þ, and then write it as a linear combination of the basis vectors u1andu2. (For notational convenience, we use column vectors.) We have Fðu1Þ¼F1 2/C20/C21/C18/C19 ¼8 /C06/C20/C21 ¼x1 2/C20/C21 þy2 5/C20/C21 andxþ2y¼8 2xþ5y¼/C06 Solve the system to obtain x¼52,y¼/C022. Hence, Fðu1Þ¼52u1/C022u2. (2) Next ﬁnd Fðu2Þ, and then write it as a linear combination of u1andu2: Fðu2Þ¼F2 5/C20/C21/C18/C19 ¼19 /C017/C20/C21 ¼x1 2/C20/C21 þy2 5/C20/C21 andxþ2y¼19 2xþ5y¼/C017 Solve the system to get x¼129, y¼/C055. Thus, Fðu2Þ¼129u1/C055u2. Now write the coordinates of Fðu1ÞandFðu2Þas columns to obtain the matrix ½F/C138S¼52 129 /C022/C055/C20/C21 (b) Find the matrix representation of Frelative to the (usual) basis E¼fe1;e2g¼fð 1;0Þ;ð0;1Þg. Find Fðe1Þand write it as a linear combination of the usual basis vectors e1ande2, and then ﬁnd Fðe2Þand write it as a linear combination of e1ande2. We have Fðe1Þ¼Fð1;0Þ¼ð 2;2Þ¼ 2e1þ4e2 Fðe2Þ¼Fð0;1Þ¼ð 3;/C05Þ¼3e1/C05e2and so½F/C138E¼23 4/C05/C20/C21 Note that the coordinates of Fðe1ÞandFðe2Þform the columns, not the rows, of ½F/C138E. Also, note that the arithmetic is much simpler using the usual basis of R2. EXAMPLE 6.2 LetVbe the vector space of functions with basis S¼fsint;cost;e3tg, and let D:V!V be the differential operator deﬁned by DðfðtÞÞ¼ dðfðtÞÞ=dt. We compute the matrix representing Din the basis S: DðsintÞ¼ cost¼0ðsintÞþ1ðcostÞþ0ðe3tÞ DðcostÞ¼/C0 sint¼/C01ðsintÞþ0ðcostÞþ0ðe3tÞ Dðe3tÞ¼ 3e3t¼0ðsintÞþ0ðcostÞþ3ðe3tÞ and so ½D/C138¼0/C010 10 000 32 643 75 Note that the coordinates of DðsintÞ,DðcostÞ,Dðe3tÞform the columns, not the rows, of ½D/C138. Matrix Mappings and Their Matrix Representation Consider the following matrix A, which may be viewed as a linear operator on R2, and basis SofR2: A¼3/C02 4/C05/C20/C21 and S¼fu1;u2g¼1 2/C20/C21 ;2 5/C20/C21/C26/C27 (We write vectors as columns, because our map is a matrix.) We ﬁnd the matrix representation of A relative to the basis S.196 CHAPTER 6 Linear Mappings and Matrices (1) First we write Aðu1Þas a linear combination of u1andu2. We have Aðu1Þ¼3/C02 4/C05/C20/C21 1 2/C20/C21 ¼/C01 /C06/C20/C21 ¼x1 2/C20/C21 þy2 5/C20/C21 and soxþ2y¼/C01 2xþ5y¼/C06 Solving the system yields x¼7,y¼/C04. Thus, Aðu1Þ¼7u1/C04u2. (2) Next we write Aðu2Þas a linear combination of u1andu2. We have Aðu2Þ¼3/C02 4/C05/C20/C21 2 5/C20/C21 ¼/C04 /C07/C20/C21 ¼x1 2/C20/C21 þy2 5/C20/C21 and soxþ2y¼/C04 2xþ5y¼/C07 Solving the system yields x¼/C06,y¼1. Thus, Aðu2Þ¼/C0 6u1þu2. Writing the coordinates of Aðu1ÞandAðu2Þas columns gives us the following matrix representation of A: ½A/C138S¼7/C06 /C041/C20/C21 Remark: Suppose we want to ﬁnd the matrix representation of Arelative to the usual basis E¼fe1;e2g¼f½ 1;0/C138T;½0;1/C138TgofR2:We have Aðe1Þ¼3/C02 4/C05/C20/C21 1 0/C20/C21 ¼3 4/C20/C21 ¼3e1þ4e2 Aðe2Þ¼3/C02 4/C05/C20/C21 0 1/C20/C21 ¼/C02 /C05/C20/C21 ¼/C02e1/C05e2and so½A/C138E¼3/C02 4/C05/C20/C21 Note that½A/C138Eis the original matrix A. This result is true in general: The matrix representation of any n/C2nsquare matrix Aover a field Krelative to the usual basis EofKnis the matrix Aitself; that is ; ½A/C138E¼A Algorithm for Finding Matrix Representations Next follows an algorithm for ﬁnding matrix representations. The ﬁrst Step 0 is optional. It may be useful to use it in Step 1(b), which is repeated for each basis vector. ALGORITHM 6.1: The input is a linear operator Ton a vector space Vand a basis S¼fu1;u2;...;ungofV. The output is the matrix representation ½T/C138S. Step 0. Find a formula for the coordinates of an arbitrary vector vrelative to the basis S. Step 1. Repeat for each basis vector ukinS: (a) Find TðukÞ. (b) Write TðukÞas a linear combination of the basis vectors u1;u2;...;un. Step 2. Form the matrix½T/C138Swhose columns are the coordinate vectors in Step 1(b). EXAMPLE 6.3 LetF:R2!R2be deﬁned by Fðx;yÞ¼ð 2xþ3y;4x/C05yÞ. Find the matrix representa- tion½F/C138SofFrelative to the basis S¼fu1;u2g¼fð 1;/C02Þ;ð2;/C05Þg. (Step 0) First ﬁnd the coordinates of ða;bÞ2R2relative to the basis S. We have a b/C20/C21 ¼x1 /C02/C20/C21 þy2 /C05/C20/C21 orxþ2y¼a /C02x/C05y¼borxþ2y¼a /C0y¼2aþbCHAPTER 6 Linear Mappings and Matrices 197 Solving for xandyin terms of aandbyields x¼5aþ2b,y¼/C02a/C0b. Thus, ða;bÞ¼ð 5aþ2bÞu1þð/C0 2a/C0bÞu2 (Step 1) Now we ﬁnd Fðu1Þand write it as a linear combination of u1andu2using the above formula for ða;bÞ, and then we repeat the process for Fðu2Þ. We have Fðu1Þ¼Fð1;/C02Þ¼ð/C0 4;14Þ¼8u1/C06u2 Fðu2Þ¼Fð2;/C05Þ¼ð/C0 11;33Þ¼11u1/C011u2 (Step 2) Finally, we write the coordinates of Fðu1ÞandFðu2Þas columns to obtain the required matrix: ½F/C138S¼81 1 /C06/C011/C20/C21 Properties of Matrix Representations This subsection gives the main properties of the matrix representations of linear operators Ton a vector space V. We emphasize that we are always given a particular basis SofV. Our ﬁrst theorem, proved in Problem 6.9, tells us that the ‘‘action’’ of a linear operator Ton a vector v is preserved by its matrix representation. THEOREM 6.1: LetT:V!Vbe a linear operator, and let Sbe a (ﬁnite) basis of V. Then, for any vector vinV,½T/C138S½v/C138S¼½TðvÞ/C138S. EXAMPLE 6.4 Consider the linear operator FonR2and the basis Sof Example 6.3; that is, Fðx;yÞ¼ð 2xþ3y;4x/C05yÞ and S¼fu1;u2g¼fð 1;/C02Þ;ð2;/C05Þg Let v¼ð5;/C07Þ; and so FðvÞ¼ð/C0 11;55Þ Using the formula from Example 6.3, we get ½v/C138¼½11;/C03/C138Tand½FðvÞ/C138¼½ 55;/C033/C138T We verify Theorem 6.1 for this vector v(where½F/C138is obtained from Example 6.3): ½F/C138½v/C138¼81 1 /C06/C011/C20/C21 11 /C03/C20/C21 ¼55 /C033/C20/C21 ¼½FðvÞ/C138 Given a basis Sof a vector space V, we have associated a matrix ½T/C138to each linear operator Tin the algebra AðVÞof linear operators on V. Theorem 6.1 tells us that the ‘‘action’’ of an individual linear operator Tis preserved by this representation. The next two theorems (proved in Problems 6.10 and 6.11) tell us that the three basic operations in AðVÞwith these operators—namely (i) addition, (ii) scalar multiplication, and (iii) composition—are also preserved. THEOREM 6.2: LetVbe an n-dimensional vector space over K, letSbe a basis of V, and let Mbe the algebra of n/C2nmatrices over K. Then the mapping m:AðVÞ!M defined by mðTÞ¼½ T/C138S is a vector space isomorphism. That is, for any F;G2AðVÞand any k2K, (i) mðFþGÞ¼mðFÞþmðGÞor½FþG/C138¼½F/C138þ½G/C138 (ii) mðkFÞ¼kmðFÞor½kF/C138¼k½F/C138 (iii) mis bijective (one-to-one and onto).198 CHAPTER 6 Linear Mappings and Matrices THEOREM 6.3: For any linear operators F;G2AðVÞ, mðG/C14FÞ¼mðGÞmðFÞor½G/C14F/C138¼½G/C138½F/C138 (Here G/C14Fdenotes the composition of the maps GandF.) 6.3 Change of Basis LetVbe an n-dimensional vector space over a ﬁeld K. We have shown that once we have selected a basis SofV, every vector v2Vcan be represented by means of an n-tuple½v/C138SinKn, and every linear operator TinAðVÞcan be represented by an n/C2nmatrix over K. We ask the following natural question: How do our representations change if we select another basis? In order to answer this question, we ﬁrst need a deﬁnition. DEFINITION: LetS¼fu1;u2;...;ungbe a basis of a vector space V;and let S0¼fv1;v2;...;vng be another basis. (For reference, we will call Sthe ‘‘old’’ basis and S0the ‘‘new’’ basis.) Because Sis a basis, each vector in the ‘‘new’’ basis S0can be written uniquely as a linear combination of the vectors in S; say, v1¼a11u1þa12u2þ/C1/C1/C1þ a1nun v2¼a21u1þa22u2þ/C1/C1/C1þ a2nun ::::::::::::::::::::::::::::::::::::::::::::::::: vn¼an1u1þan2u2þ/C1/C1/C1þ annun LetPbe the transpose of the above matrix of coefﬁcients; that is, let P¼½pij/C138, where pij¼aji. Then Pis called the change-of-basis matrix (ortransition matrix ) from the ‘‘old’’ basis Sto the ‘‘new’’ basis S0. The following remarks are in order. Remark 1: The above change-of-basis matrix Pmay also be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the ‘‘new’’ basis vectors virelative to the ‘‘old’’ basis S; namely, P¼½ v1/C138S;½v2/C138S;...;½vn/C138S/C2/C3 Remark 2: Analogously, there is a change-of-basis matrix Qfrom the ‘‘new’’ basis S0to the ‘‘old’’ basis S. Similarly, Qmay be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the ‘‘old’’ basis vectors uirelative to the ‘‘new’’ basis S0; namely, Q¼½u1/C138S0;½u2/C138S0;...;½un/C138S0/C2/C3 Remark 3: Because the vectors v1;v2;...;vnin the new basis S0are linearly independent, the matrix Pis invertible (Problem 6.18). Similarly, Qis invertible. In fact, we have the following proposition (proved in Problem 6.18). PROPOSITION 6.4: LetPandQbe the above change-of-basis matrices. Then Q¼P/C01. Now suppose S¼fu1;u2;...;ungis a basis of a vector space V, and suppose P¼½pij/C138is any nonsingular matrix. Then the nvectors vi¼p1iuiþp2iu2þ/C1/C1/C1þ pniun; i¼1;2;...;n corresponding to the columns of P, are linearly independent [Problem 6.21(a)]. Thus, they form another basis S0ofV. Moreover, Pwill be the change-of-basis matrix from Sto the new basis S0.CHAPTER 6 Linear Mappings and Matrices 199 EXAMPLE 6.5 Consider the following two bases of R2: S¼fu1;u2g¼fð 1;2Þ;ð3;5Þg and S0¼fv1;v2g¼fð 1;/C01Þ;ð1;/C02Þg (a) Find the change-of-basis matrix Pfrom Sto the ‘‘new’’ basis S0. Write each of the new basis vectors of S0as a linear combination of the original basis vectors u1andu2of S. We have 1 /C01/C20/C21 ¼x1 2/C20/C21 þy3 5/C20/C21 orxþ3y¼1 2xþ5y¼/C01yielding x¼/C08;y¼3 1 /C01/C20/C21 ¼x1 2/C20/C21 þy3 5/C20/C21 orxþ3y¼1 2xþ5y¼/C01yielding x¼/C011;y¼4 Thus, v1¼/C0 8u1þ3u2 v2¼/C011u1þ4u2and hence ; P¼/C08/C011 34/C20/C21 : Note that the coordinates of v1and v2are the columns, not rows, of the change-of-basis matrix P. (b) Find the change-of-basis matrix Qfrom the ‘‘new’’ basis S0back to the ‘‘old’’ basis S. Here we write each of the ‘‘old’’ basis vectors u1andu2ofS0as a linear combination of the ‘‘new’’ basis vectors v1and v2ofS0. This yields u1¼4v1/C03v2 u2¼11v1/C08v2and hence ; Q¼41 1 /C03/C08/C20/C21 As expected from Proposition 6.4, Q¼P/C01. (In fact, we could have obtained Qby simply ﬁnding P/C01.) EXAMPLE 6.6 Consider the following two bases of R3: E¼fe1;e2;e3g¼fð 1;0;0Þ;ð0;1;0Þ;ð0;0;1Þg and S¼fu1;u2;u3g¼fð 1;0;1Þ;ð2;1;2Þ;ð1;2;2Þg (a) Find the change-of-basis matrix Pfrom the basis Eto the basis S. Because Eis the usual basis, we can immediately write each basis element of Sas a linear combination of the basis elements of E. Speciﬁcally, u1¼ð1;0;1Þ¼ e1þ e3 u2¼ð2;1;2Þ¼2e1þe2þ2e3 u3¼ð1;2;2Þ¼ e1þ2e2þ2e3and hence ; P¼121 012 1222 643 75 Again, the coordinates of u1;u2;u3appear as the columns in P. Observe that Pis simply the matrix whose columns are the basis vectors of S. This is true only because the original basis was the usual basis E. (b) Find the change-of-basis matrix Qfrom the basis Sto the basis E. The deﬁnition of the change-of-basis matrix Qtells us to write each of the (usual) basis vectors in Eas a linear combination of the basis elements of S. This yields e1¼ð1;0;0Þ¼/C0 2u1þ2u2/C0u3 e2¼ð0;1;0Þ¼/C0 2u1þu2 e3¼ð0;0;1Þ¼ 3u1/C02u2þu3and hence ; Q¼/C02/C023 21/C02 /C01012 643 75 We emphasize that to ﬁnd Q, we need to solve three 3 /C23 systems of linear equations—one 3 /C23 system for each of e1;e2;e3.200 CHAPTER 6 Linear Mappings and Matrices Alternatively, we can ﬁnd Q¼P/C01by forming the matrix M¼½P;I/C138and row reducing Mto row canonical form: M¼121100 012010 1220012 643 75/C24100/C02/C023 0 1 021 /C02 001/C01012 643 75¼½I;P/C01/C138 thus; Q¼P/C01¼/C02/C023 21/C02 /C01012 643 75 (Here we have used the fact that Qis the inverse of P.) The result in Example 6.6(a) is true in general. We state this result formally, because it occurs often. PROPOSITION 6.5: The change-of-basis matrix from the usual basis EofKnto any basis SofKnis the matrix Pwhose columns are, respectively, the basis vectors of S. Applications of Change-of-Basis Matrix First we show how a change of basis affects the coordinates of a vector in a vector space V. The following theorem is proved in Problem 6.22. THEOREM 6.6: LetPbe the change-of-basis matrix from a basis Sto a basis S0in a vector space V. Then, for any vector v2V, we have P½v/C138S0¼½v/C138S and hence ; P/C01½v/C138S¼½v/C138S0 Namely, if we multiply the coordinates of vin the original basis SbyP/C01, we get the coordinates of v in the new basis S0. Remark 1: Although Pis called the change-of-basis matrix from the old basis Sto the new basis S0, we emphasize that P/C01transforms the coordinates of vin the original basis Sinto the coordinates of v in the new basis S0. Remark 2: Because of the above theorem, many texts call Q¼P/C01, not P, the transition matrix from the old basis Sto the new basis S0. Some texts also refer to Qas the change-of-coordinates matrix. We now give the proof of the above theorem for the special case that dim V¼3. Suppose Pis the change-of-basis matrix from the basis S¼fu1;u2;u3gto the basis S0¼fv1;v2;v3g; say, v1¼a1u1þa2u2þa3a3 v2¼b1u1þb2u2þb3u3 v3¼c1u1þc2u2þc3u3and hence ; P¼a1b1c1 a2b2c2 a3b3c32 43 5 Now suppose v2Vand, say, v¼k1v1þk2v2þk3v3. Then, substituting for v1;v2;v3from above, we obtain v¼k1ða1u1þa2u2þa3u3Þþk2ðb1u1þb2u2þb3u3Þþk3ðc1u1þc2u2þc3u3Þ ¼ða1k1þb1k2þc1k3Þu1þða2k1þb2k2þc2k3Þu2þða3k1þb3k2þc3k3Þu3CHAPTER 6 Linear Mappings and Matrices 201 Thus, ½v/C138S0¼k1 k2 k32 43 5 and½v/C138S¼a1k1þb1k2þc1k3 a2k1þb2k2þc2k3 a3k1þb3k2þc3k32 43 5 Accordingly, P½v/C138S0¼a1b1c1 a2b2c2 a3b3c32 43 5k1 k2 k32 43 5¼a1k1þb1k2þc1k3 a2k1þb2k2þc2k3 a3k1þb3k2þc3k32 43 5¼½v/C138S Finally, multiplying the equation ½v/C138S¼P½v/C138S,b y P/C01, we get P/C01½v/C138S¼P/C01P½v/C138S0¼I½v/C138S0¼½v/C138S0 The next theorem (proved in Problem 6.26) shows how a change of basis affects the matrix representation of a linear operator. THEOREM 6.7: LetPbe the change-of-basis matrix from a basis Sto a basis S0in a vector space V. Then, for any linear operator TonV, ½T/C138S0¼P/C01½T/C138SP That is, if AandBare the matrix representations of Trelative, respectively, to Sand S0, then B¼P/C01AP EXAMPLE 6.7 Consider the following two bases of R3: E¼fe1;e2;e3g¼fð 1;0;0Þ;ð0;1;0Þ;ð0;0;1Þg and S¼fu1;u2;u3g¼fð 1;0;1Þ;ð2;1;2Þ;ð1;2;2Þg The change-of-basis matrix Pfrom EtoSand its inverse P/C01were obtained in Example 6.6. (a) Write v¼ð1;3;5Þas a linear combination of u1;u2;u3, or, equivalently, ﬁnd ½v/C138S. One way to do this is to directly solve the vector equation v¼xu1þyu2þzu3; that is, 1 3 52 43 5¼x1 0 12 43 5þy2 1 22 43 5þz1 2 22 43 5 orxþ2yþz¼1 yþ2z¼3 xþ2yþ2z¼5 The solution is x¼7,y¼/C05,z¼4, so v¼7u1/C05u2þ4u3. On the other hand, we know that ½v/C138E¼½1;3;5/C138T, because Eis the usual basis, and we already know P/C01. Therefore, by Theorem 6.6, ½v/C138S¼P/C01½v/C138E¼/C02/C023 21/C02 /C01012 43 51 3 52 43 5¼7 /C05 42 43 5 Thus, again, v¼7u1/C05u2þ4u3. (b) Let A¼13/C02 2/C041 3/C0122 43 5, which may be viewed as a linear operator on R3. Find the matrix Bthat represents A relative to the basis S.202 CHAPTER 6 Linear Mappings and Matrices The deﬁnition of the matrix representation of Arelative to the basis Stells us to write each of Aðu1Þ,Aðu2Þ, Aðu3Þas a linear combination of the basis vectors u1;u2;u3ofS. This yields Aðu1Þ¼ð/C0 1;3;5Þ¼11u1/C05u2þ6u3 Aðu2Þ¼ð 1;2;9Þ¼ 21u1/C014u2þ8u3 Aðu3Þ¼ð 3;/C04;5Þ¼17u1/C08e2þ2u3and hence ;B¼11 21 17 /C05/C014/C08 68 22 643 75 We emphasize that to ﬁnd B, we need to solve three 3 /C23 systems of linear equations—one 3 /C23 system for each of Aðu1Þ,Aðu2Þ,Aðu3Þ. On the other hand, because we know PandP/C01, we can use Theorem 6.7. That is, B¼P/C01AP¼/C02/C023 21/C02 /C01012 43 513/C02 2/C041 3/C0122 43 5121 012 1222 43 5¼11 21 17 /C05/C014/C08 68 22 43 5 This, as expected, gives the same result. 6.4 Similarity Suppose AandBare square matrices for which there exists an invertible matrix Psuch that B¼P/C01AP; then Bis said to be similar toA,o rBis said to be obtained from Aby a similarity transformation .W e show (Problem 6.29) that similarity of matrices is an equivalence relation. By Theorem 6.7 and the above remark, we have the following basic result. THEOREM 6.8: Two matrices represent the same linear operator if and only if the matrices are similar. That is, all the matrix representations of a linear operator Tform an equivalence class of similar matrices. A linear operator Tis said to be diagonalizable if there exists a basis SofVsuch that Tis represented by a diagonal matrix; the basis Sis then said to diagonalize T . The preceding theorem gives us the following result. THEOREM 6.9: LetAbe the matrix representation of a linear operator T.T h e n Tis diagonalizable if and only if there exists an invertible matrix Psuch that P/C01APis a diagonal matrix. That is, Tis diagonalizable if and only if its matrix representation can be diagonalized by a similarity transformation. We emphasize that not every operator is diagonalizable. However, we will show (Chapter 10) that every linear operator can be represented by certain ‘‘standard’’ matrices called its normal orcanonical forms. Such a discussion will require some theory of ﬁelds, polynomials, and determinants. Functions and Similar Matrices Suppose fis a function on square matrices that assigns the same value to similar matrices; that is, fðAÞ¼fðBÞwhenever Ais similar to B. Then finduces a function, also denoted by f, on linear operators Tin the following natural way. We deﬁne fðTÞ¼fð½T/C138SÞ where Sis any basis. By Theorem 6.8, the function is well deﬁned. The determinant (Chapter 8) is perhaps the most important example of such a function. The trace (Section 2.7) is another important example of such a function.CHAPTER 6 Linear Mappings and Matrices 203 EXAMPLE 6.8 Consider the following linear operator Fand bases EandSofR2: Fðx;yÞ¼ð 2xþ3y;4x/C05yÞ; E¼fð 1;0Þ;ð0;1Þg; S¼fð 1;2Þ;ð2;5Þg By Example 6.1, the matrix representations of Frelative to the bases EandSare, respectively, A¼23 4/C05/C20/C21 and B¼52 129 /C022/C055/C20/C21 Using matrix A, we have (i) Determinant of F¼detðAÞ¼/C0 10/C012¼/C022; (ii) Trace of F¼trðAÞ¼2/C05¼/C03: On the other hand, using matrix B, we have (i) Determinant of F¼detðBÞ¼/C0 2860þ2838¼/C022; (ii) Trace of F¼trðBÞ¼52/C055¼/C03. As expected, both matrices yield the same result. 6.5 Matrices and General Linear Mappings Last, we consider the general case of linear mappings from one vector space into another. Suppose Vand Uare vector spaces over the same ﬁeld Kand, say, dim V¼mand dim U¼n. Furthermore, suppose S¼fv1;v2;...;vmg and S0¼fu1;u2;...;ung are arbitrary but ﬁxed bases, respectively, of VandU. Suppose F:V!Uis a linear mapping. Then the vectors Fðv1Þ,Fðv2Þ;...;FðvmÞbelong to U, and so each is a linear combination of the basis vectors in S0; say, Fðv1Þ¼a11u1þa12u2þ/C1/C1/C1þ a1nun Fðv2Þ¼a21u1þa22u2þ/C1/C1/C1þ a2nun ::::::::::::::::::::::::::::::::::::::::::::::::::::::: FðvmÞ¼am1u1þam2u2þ/C1/C1/C1þ amnun DEFINITION: The transpose of the above matrix of coefﬁcients, denoted by mS;S0ðFÞor½F/C138S;S0,i s called the matrix representation ofFrelative to the bases SandS0. [We will use the simple notation mðFÞand½F/C138when the bases are understood.] The following theorem is analogous to Theorem 6.1 for linear operators (Problem 6.67). THEOREM 6.10: For any vector v2V,½F/C138S;S0½v/C138S¼½FðvÞ/C138S0. That is, multiplying the coordinates of vin the basis SofVby½F/C138, we obtain the coordinates of FðvÞ in the basis S0ofU. Recall that for any vector spaces VandU, the collection of all linear mappings from VintoUis a vector space and is denoted by Hom ðV;UÞ. The following theorem is analogous to Theorem 6.2 for linear operators, where now we let M¼Mm;ndenote the vector space of all m/C2nmatrices (Problem 6.67). THEOREM 6.11: The mapping m:HomðV;UÞ!Mdeﬁned by mðFÞ¼½ F/C138is a vector space isomorphism. That is, for any F;G2HomðV;UÞand any scalar k, (i) mðFþGÞ¼mðFÞþmðGÞor½FþG/C138¼½F/C138þ½G/C138 (ii) mðkFÞ¼kmðFÞor½kF/C138¼k½F/C138 (iii) mis bijective (one-to-one and onto).204 CHAPTER 6 Linear Mappings and Matrices Our next theorem is analogous to Theorem 6.3 for linear operators (Problem 6.67). THEOREM 6.12: LetS;S0;S00be bases of vector spaces V;U;W, respectively. Let F:V!Uand G/C14U!Wbe linear mappings. Then ½G/C14F/C138S;S00¼½G/C138S0;S00½F/C138S;S0 That is, relative to the appropriate bases, the matrix representation of the composition of two mappings is the matrix product of the matrix representations of the individual mappings. Next we show how the matrix representation of a linear mapping F:V!Uis affected when new bases are selected (Problem 6.67). THEOREM 6.13: LetPbe the change-of-basis matrix from a basis eto a basis e0inV, and let Qbe the change-of-basis matrix from a basis fto a basis f0inU. Then, for any linear map F:V!U, ½F/C138e0;f0¼Q/C01½F/C138e;fP In other words, if Ais the matrix representation of a linear mapping Frelative to the bases eandf, andBis the matrix representation of Frelative to the bases e0andf0, then B¼Q/C01AP Our last theorem, proved in Problem 6.36, shows that any linear mapping from one vector space V into another vector space Ucan be represented by a very simple matrix. We note that this theorem is analogous to Theorem 3.18 for m/C2nmatrices. THEOREM 6.14: LetF:V!Ube linear and, say, rankðFÞ¼r. Then there exist bases of VandU such that the matrix representation of Fhas the form A¼Ir0 00/C20/C21 where Iris the r-square identity matrix. The above matrix Ais called the normal orcanonical form of the linear map F. SOLVED PROBLEMS Matrix Representation of Linear Operators 6.1. Consider the linear mapping F:R2!R2deﬁned by Fðx;yÞ¼ð 3xþ4y;2x/C05yÞand the following bases of R2: E¼fe1;e2g¼fð 1;0Þ;ð0;1Þg and S¼fu1;u2g¼fð 1;2Þ;ð2;3Þg (a) Find the matrix Arepresenting Frelative to the basis E. (b) Find the matrix Brepresenting Frelative to the basis S. (a) Because Eis the usual basis, the rows of Aare simply the coefﬁcients in the components of Fðx;yÞ; that is, usingða;bÞ¼ae1þbe2, we have Fðe1Þ¼Fð1;0Þ¼ð 3;2Þ¼ 3e1þ2e2 Fðe2Þ¼Fð0;1Þ¼ð 4;/C05Þ¼4e1/C05e2and so A¼34 2/C05/C20/C21 Note that the coefﬁcients of the basis vectors are written as columns in the matrix representation.CHAPTER 6 Linear Mappings and Matrices 205 (b) First ﬁnd Fðu1Þand write it as a linear combination of the basis vectors u1andu2. We have Fðu1Þ¼Fð1;2Þ¼ð 11;/C08Þ¼xð1;2Þþyð2;3Þ; and soxþ2y¼11 2xþ3y¼/C08 Solve the system to obtain x¼/C049,y¼30. Therefore, Fðu1Þ¼/C0 49u1þ30u2 Next ﬁnd Fðu2Þand write it as a linear combination of the basis vectors u1andu2. We have Fðu2Þ¼Fð2;3Þ¼ð 18;/C011Þ¼xð1;2Þþyð2;3Þ; and soxþ2y¼18 2xþ3y¼/C011 Solve for xandyto obtain x¼/C076,y¼47. Hence, Fðu2Þ¼/C0 76u1þ47u2 Write the coefﬁcients of u1andu2as columns to obtain B¼/C049/C076 30 47/C20/C21 (b0) Alternatively, one can ﬁrst ﬁnd the coordinates of an arbitrary vector ða;bÞinR2relative to the basis S. We have ða;bÞ¼xð1;2Þþyð2;3Þ¼ð xþ2y;2xþ3yÞ; and soxþ2y¼a 2xþ3y¼b Solve for xandyin terms of aandbto get x¼/C03aþ2b,y¼2a/C0b. Thus, ða;bÞ¼ð/C0 3aþ2bÞu1þð2a/C0bÞu2 Then use the formula for ða;bÞto ﬁnd the coordinates of Fðu1ÞandFðu2Þrelative to S: Fðu1Þ¼Fð1;2Þ¼ð 11;/C08Þ¼/C0 49u1þ30u2 Fðu2Þ¼Fð2;3Þ¼ð 18;/C011Þ¼/C0 76u1þ47u2and so B¼/C049/C076 30 47/C20/C21 6.2. Consider the following linear operator GonR2and basis S: Gðx;yÞ¼ð 2x/C07y;4xþ3yÞ and S¼fu1;u2g¼fð 1;3Þ;ð2;5Þg (a) Find the matrix representation ½G/C138SofGrelative to S. (b) Verify½G/C138S½v/C138S¼½GðvÞ/C138Sfor the vector v¼ð4;/C03ÞinR2. First ﬁnd the coordinates of an arbitrary vector v¼ða;bÞinR2relative to the basis S.W e have a b/C20/C21 ¼x1 3/C20/C21 þy2 5/C20/C21 ; and soxþ2y¼a 3xþ5y¼b Solve for xandyin terms of aandbto get x¼/C05aþ2b,y¼3a/C0b. Thus, ða;bÞ¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2; and so½v/C138¼½/C0 5aþ2b;3a/C0b/C138T (a) Using the formula for ða;bÞandGðx;yÞ¼ð 2x/C07y;4xþ3yÞ, we have Gðu1Þ¼Gð1;3Þ¼ð/C0 19;13Þ¼121u1/C070u2 Gðu2Þ¼Gð2;5Þ¼ð/C0 31;23Þ¼201u1/C0116u2and so½G/C138S¼121 201 /C070/C0116/C20/C21 (We emphasize that the coefﬁcients of u1andu2are written as columns, not rows, in the matrix representation.) (b) Use the formula ða;bÞ¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2to get v¼ð4;/C03Þ¼/C0 26u1þ15u2 GðvÞ¼Gð4;/C03Þ¼ð 20;7Þ¼/C0 131u1þ80u2 Then ½v/C138S¼½/C0 26;15/C138Tand½GðvÞ/C138S¼½/C0 131;80/C138T206 CHAPTER 6 Linear Mappings and Matrices Accordingly, ½G/C138S½v/C138S¼121 201 /C070/C0116/C20/C21 /C026 15/C20/C21 ¼/C0131 80/C20/C21 ¼½GðvÞ/C138S (This is expected from Theorem 6.1.) 6.3. Consider the following 2 /C22 matrix Aand basis SofR2: A¼24 56/C20/C21 and S¼fu1;u2g¼1 /C02/C20/C21 ;3 /C07/C20/C21/C26/C27 The matrix Adeﬁnes a linear operator on R2. Find the matrix Bthat represents the mapping A relative to the basis S. First ﬁnd the coordinates of an arbitrary vector ða;bÞTwith respect to the basis S. We have a b/C20/C21 ¼x1 /C02/C20/C21 þy3 /C07/C20/C21 orxþ3y¼a /C02x/C07y¼b Solve for xandyin terms of aandbto obtain x¼7aþ3b,y¼/C02a/C0b. Thus, ða;bÞT¼ð7aþ3bÞu1þð/C0 2a/C0bÞu2 Then use the formula for ða;bÞTto ﬁnd the coordinates of Au1andAu2relative to the basis S: Au1¼24 56/C20/C211 /C02/C20/C21 ¼/C06 /C07/C20/C21 ¼/C063u1þ19u2 Au2¼24 56/C20/C213 /C07/C20/C21 ¼/C022 /C027/C20/C21 ¼/C0235u1þ71u2 Writing the coordinates as columns yields B¼/C063/C0235 19 71/C20/C21 6.4. Find the matrix representation of each of the following linear operators FonR3relative to the usual basis E¼fe1;e2;e3gofR3; that is, ﬁnd½F/C138¼½F/C138E: (a)Fdeﬁned by Fðx;y;zÞ¼ð xþ2y/C03z;4x/C05y/C06z;7xþ8yþ9z). (b)Fdeﬁned by the 3/C23 matrix A¼111 2345552 43 5. (c)Fdeﬁned by Fðe 1Þ¼ð 1;3;5Þ;Fðe2Þ¼ð 2;4;6Þ,Fðe3Þ¼ð 7;7;7Þ. (Theorem 5.2 states that a linear map is completely deﬁned by its action on the vectors in a basis.) (a) Because Eis the usual basis, simply write the coefﬁcients of the components of Fðx;y;zÞas rows: ½F/C138¼12/C03 4/C05/C06 7892 43 5 (b) Because Eis the usual basis, ½F/C138¼A, the matrix Aitself. (c) Here Fðe1Þ¼ð 1;3;5Þ¼ e1þ3e2þ5e3 Fðe2Þ¼ð 2;4;6Þ¼2e1þ4e2þ6e3 Fðe3Þ¼ð 7;7;7Þ¼7e1þ7e2þ7e3and so½F/C138¼127 347 5672 43 5 That is, the columns of ½F/C138are the images of the usual basis vectors. 6.5. LetGbe the linear operator on R3deﬁned by Gðx;y;zÞ¼ð 2yþz;x/C04y;3xÞ. (a) Find the matrix representation of Grelative to the basis S¼fw1;w2;w3g¼fð 1;1;1Þ;ð1;1;0Þ;ð1;0;0Þg (b) Verify that½G/C138½v/C138¼½GðvÞ/C138for any vector vinR3.CHAPTER 6 Linear Mappings and Matrices 207 First ﬁnd the coordinates of an arbitrary vector ða;b;cÞ2R3with respect to the basis S. Writeða;b;cÞas a linear combination of w1;w2;w3using unknown scalars x;y, and z: ða;b;cÞ¼xð1;1;1Þþyð1;1;0Þþzð1;0;0Þ¼ð xþyþz;xþy;xÞ Set corresponding components equal to each other to obtain the system of equations xþyþz¼a; xþy¼b; x¼c Solve the system for x;y,zin terms of a;b,cto ﬁnd x¼c,y¼b/C0c,z¼a/C0b. Thus, ða;b;cÞ¼cw1þðb/C0cÞw2þða/C0bÞw3, or equivalently, ½ða;b;cÞ/C138¼½ c;b/C0c;a/C0b/C138T (a) Because Gðx;y;zÞ¼ð 2yþz;x/C04y;3xÞ, Gðw1Þ¼Gð1;1;1Þ¼ð 3;/C03;3Þ¼3w1/C06x2þ6x3 Gðw2Þ¼Gð1;1;0Þ¼ð 2;/C03;3Þ¼3w1/C06w2þ5w3 Gðw3Þ¼Gð1;0;0Þ¼ð 0;1;3Þ¼3w1/C02w2/C0w3 Write the coordinates Gðw1Þ,Gðw2Þ,Gðw3Þas columns to get ½G/C138¼333 /C06/C06/C02 65/C012 43 5 (b) Write GðvÞas a linear combination of w1;w2;w3, where v¼ða;b;cÞis an arbitrary vector in R3, GðvÞ¼Gða;b;cÞ¼ð 2bþc;a/C04b;3aÞ¼3aw1þð/C0 2a/C04bÞw2þð/C0 aþ6bþcÞw3 or equivalently, ½GðvÞ/C138¼½ 3a;/C02a/C04b;/C0aþ6bþc/C138T Accordingly, ½G/C138½v/C138¼333 /C06/C06/C02 65/C012 43 5c b/C0c a/C0b2 43 5¼3a /C02a/C04b /C0aþ6bþc2 43 5¼½GðvÞ/C138 6.6. Consider the following 3 /C23 matrix Aand basis SofR3: A¼1/C021 3/C010 14/C022 43 5 and S¼fu1;u2;u3g¼1 1 12 43 5;0 1 12 43 5;1 2 32 43 58 < :9 = ; The matrix Adeﬁnes a linear operator on R3. Find the matrix Bthat represents the mapping A relative to the basis S. (Recall that Arepresents itself relative to the usual basis of R3.) First ﬁnd the coordinates of an arbitrary vector ða;b;cÞinR3with respect to the basis S. We have a b c2 43 5¼x1 1 12 43 5þy0 1 12 43 5þz1 2 32 43 5 orxþ z¼a xþyþ2z¼b xþyþ3z¼c Solve for x;y;zin terms of a;b;cto get x¼aþb/C0c;y¼/C0aþ2b/C0c;z¼c/C0b thus;ða;b;cÞT¼ðaþb/C0cÞu1þð/C0 aþ2b/C0cÞu2þðc/C0bÞu3208 CHAPTER 6 Linear Mappings and Matrices Then use the formula for ða;b;cÞTto ﬁnd the coordinates of Au1,Au2,Au3relative to the basis S: Aðu1Þ¼Að1;1;1ÞT¼ð0;2;3ÞT¼/C0u1þu2þu3 Aðu2Þ¼Að1;1;0ÞT¼ð/C0 1;/C01;2ÞT¼/C04u1/C03u2þ3u3 Aðu3Þ¼Að1;2;3ÞT¼ð0;1;3ÞT¼/C02u1/C0u2þ2u3so B¼/C01/C04/C02 1/C03/C01 1322 43 5 6.7. For each of the following linear transformations (operators) LonR2, ﬁnd the matrix Athat represents L(relative to the usual basis of R2): (a)Lis deﬁned by Lð1;0Þ¼ð 2;4ÞandLð0;1Þ¼ð 5;8Þ. (b)Lis the rotation in R2counterclockwise by 90/C14. (c)Lis the reﬂection in R2about the line y¼/C0x. (a) Becausefð1;0Þ;ð0;1Þgis the usual basis of R2, write their images under Las columns to get A¼25 48/C20/C21 (b) Under the rotation L, we have Lð1;0Þ¼ð 0;1ÞandLð0;1Þ¼ð/C0 1;0Þ. Thus, A¼0/C01 10/C20/C21 (c) Under the reﬂection L, we have Lð1;0Þ¼ð 0;/C01ÞandLð0;1Þ¼ð/C0 1;0Þ. Thus, A¼0/C01 /C010/C20/C21 6.8. The set S¼fe3t,te3t,t2e3tgis a basis of a vector space Vof functions f:R!R. Let Dbe the differential operator on V; that is, DðfÞ¼df=dt. Find the matrix representation of Drelative to the basis S. Find the image of each basis function: Dðe3tÞ¼ 3e3t Dðte3tÞ¼ e3tþ3te3t Dðt2e3tÞ¼2te3tþ3t2e3t¼3ðe3tÞþ0ðte3tÞþ0ðt2e3tÞ ¼1ðe3tÞþ3ðte3tÞþ0ðt2e3tÞ ¼0ðe3tÞþ2ðte3tÞþ3ðt2e3tÞand thus ;½D/C138¼310 032 0032 43 5 6.9. Prove Theorem 6.1: Let T:V!Vbe a linear operator, and let Sbe a (ﬁnite) basis of V. Then, for any vector vinV,½T/C138S½v/C138S¼½TðvÞ/C138S. Suppose S¼fu1;u2;...;ung, and suppose, for i¼1;...;n, TðuiÞ¼ai1u1þai2u2þ/C1/C1/C1þ ainun¼Pn j¼1aijuj Then½T/C138Sis the n-square matrix whose jth row is ða1j;a2j;...;anjÞð 1Þ Now suppose v¼k1u1þk2u2þ/C1/C1/C1þ knun¼Pn i¼1kiui Writing a column vector as the transpose of a row vector, we have ½v/C138S¼½k1;k2;...;kn/C138Tð2ÞCHAPTER 6 Linear Mappings and Matrices 209 Furthermore, using the linearity of T, TðvÞ¼TPn i¼1kiui/C18/C19 ¼Pn i¼1kiTðuiÞ¼Pn i¼1ki/C18Pn j¼1aijuj/C19 ¼Pn j¼1Pn i¼1aijki/C18/C19 uj¼Pn j¼1ða1jk1þa2jk2þ/C1/C1/C1þ anjknÞuj Thus,½TðvÞ/C138Sis the column vector whose jth entry is a1jk1þa2jk2þ/C1/C1/C1þ anjkn ð3Þ On the other hand, the jth entry of½T/C138S½v/C138Sis obtained by multiplying the jth row of½T/C138Sby½v/C138S—that is (1) by (2). But the product of (1) and (2) is (3). Hence, ½T/C138S½v/C138Sand½TðvÞ/C138Shave the same entries. Thus, ½T/C138S½v/C138S¼½TðvÞ/C138S. 6.10. Prove Theorem 6.2: Let S¼fu1;u2;...;ungbe a basis for Vover K, and let Mbe the algebra of n-square matrices over K. Then the mapping m:AðVÞ!Mdeﬁned by mðTÞ¼½ T/C138Sis a vector space isomorphism. That is, for any F;G2AðVÞand any k2K, we have (i)½FþG/C138¼½F/C138þ½G/C138, (ii)½kF/C138¼k½F/C138, (iii) mis one-to-one and onto. (i) Suppose, for i¼1;...;n, FðuiÞ¼Pn j¼1aijuj and GðuiÞ¼Pn j¼1bijuj Consider the matrices A¼½aij/C138andB¼½bij/C138. Then½F/C138¼ATand½G/C138¼BT. We have, for i¼1;...;n, ðFþGÞðuiÞ¼FðuiÞþGðuiÞ¼Pn j¼1ðaijþbijÞuj Because AþBis the matrixðaijþbijÞ, we have ½FþG/C138¼ð AþBÞT¼ATþBT¼½F/C138þ½G/C138 (ii) Also, for i¼1;...;n; ðkFÞðuiÞ¼kFðuiÞ¼kPn j¼1aijuj¼Pn j¼1ðkaijÞuj Because kAis the matrixðkaijÞ, we have ½kF/C138¼ð kAÞT¼kAT¼k½F/C138 (iii) Finally, mis one-to-one, because a linear mapping is completely determined by its values on a basis. Also, mis onto, because matrix A¼½aij/C138inMis the image of the linear operator, FðuiÞ¼Pn j¼1aijuj; i¼1;...;n Thus, the theorem is proved. 6.11. Prove Theorem 6.3: For any linear operators G;F2AðVÞ,½G/C14F/C138¼½G/C138½F/C138. Using the notation in Problem 6.10, we have ðG/C14FÞðuiÞ¼GðFðuiÞÞ¼ G/C18Pn j¼1aijuj/C19 ¼Pn j¼1aijGðujÞ ¼Pn j¼1aijPn k¼1bjkuk/C18/C19 ¼Pn k¼1/C18Pn j¼1aijbjk/C19 uk Recall that ABis the matrix AB¼½cik/C138, where cik¼Pn j¼1aijbjk. Accordingly, ½G/C14F/C138¼ð ABÞT¼BTAT¼½G/C138½F/C138 The theorem is proved.210 CHAPTER 6 Linear Mappings and Matrices 6.12. LetAbe the matrix representation of a linear operator T. Prove that, for any polynomial fðtÞ,w e have that fðAÞis the matrix representation of fðTÞ. [Thus, fðTÞ¼0 if and only if fðAÞ¼0.] Letfbe the mapping that sends an operator Tinto its matrix representation A. We need to prove that fðfðTÞÞ¼ fðAÞ. Suppose fðtÞ¼antnþ/C1/C1/C1þ a1tþa0. The proof is by induction on n, the degree of fðtÞ. Suppose n¼0. Recall that fðI0Þ¼I, where I0is the identity mapping and Iis the identity matrix. Thus, fðfðTÞÞ¼ fða0I0Þ¼a0fðI0Þ¼a0I¼fðAÞ and so the theorem holds for n¼0. Now assume the theorem holds for polynomials of degree less than n. Then, because fis an algebra isomorphism, fðfðTÞÞ¼ fðanTnþan/C01Tn/C01þ/C1/C1/C1þ a1Tþa0I0Þ ¼anfðTÞfðTn/C01Þþfðan/C01Tn/C01þ/C1/C1/C1þ a1Tþa0I0Þ ¼anAAn/C01þðan/C01An/C01þ/C1/C1/C1þ a1Aþa0IÞ¼fðAÞ and the theorem is proved. Change of Basis The coordinate vector ½v/C138Sin this section will always denote a column vector; that is, ½v/C138S¼½a1;a2;...;an/C138T 6.13. Consider the following bases of R2: E¼fe1;e2g¼fð 1;0Þ;ð0;1Þg and S¼fu1;u2g¼fð 1;3Þ;ð1;4Þg (a) Find the change-of-basis matrix Pfrom the usual basis EtoS. (b) Find the change-of-basis matrix Qfrom Sback to E. (c) Find the coordinate vector ½v/C138ofv¼ð5;/C03Þrelative to S. (a) Because Eis the usual basis, simply write the basis vectors in Sas columns: P¼11 34/C20/C21 (b)Method 1. Use the deﬁnition of the change-of-basis matrix. That is, express each vector in Eas a linear combination of the vectors in S. We do this by ﬁrst ﬁnding the coordinates of an arbitrary vector v¼ða;bÞrelative to S. We have ða;bÞ¼xð1;3Þþyð1;4Þ¼ð xþy;3xþ4yÞ orxþy¼a 3xþ4y¼b Solve for xandyto obtain x¼4a/C0b,y¼/C03aþb. Thus, v¼ð4a/C0bÞu1þð/C0 3aþbÞu2 and½v/C138S¼½ða;bÞ/C138S¼½4a/C0b;/C03aþb/C138T Using the above formula for ½v/C138Sand writing the coordinates of the eias columns yields e1¼ð1;0Þ¼ 4u1/C03u2 e2¼ð0;1Þ¼/C0 u1þu2and Q¼4/C01 /C031/C20/C21 Method 2. Because Q¼P/C01;ﬁndP/C01, say by using the formula for the inverse of a 2 /C22 matrix. Thus, P/C01¼4/C01 /C031/C20/C21 (c)Method 1. Write vas a linear combination of the vectors in S, say by using the above formula for v¼ða;bÞ. We have v¼ð5;/C03Þ¼23u1/C018u2;and so½v/C138S¼½23;/C018/C138T. Method 2. Use, from Theorem 6.6, the fact that ½v/C138S¼P/C01½v/C138Eand the fact that½v/C138E¼½5;/C03/C138T: ½v/C138S¼P/C01½v/C138E¼4/C01 /C031/C20/C21 5 /C03/C20/C21 ¼23 /C018/C20/C21CHAPTER 6 Linear Mappings and Matrices 211 6.14. The vectors u1¼ð1;2;0Þ,u2¼ð1;3;2Þ,u3¼ð0;1;3Þform a basis SofR3. Find (a) The change-of-basis matrix Pfrom the usual basis E¼fe1;e2;e3gtoS. (b) The change-of-basis matrix Qfrom Sback to E. (a) Because Eis the usual basis, simply write the basis vectors of Sas columns: P¼110 231 0232 43 5 (b)Method 1. Express each basis vector of Eas a linear combination of the basis vectors of Sby ﬁrst ﬁnding the coordinates of an arbitrary vector v¼ða;b;cÞrelative to the basis S. We have a b c2 43 5¼x1 202 43 5þy1 322 43 5þz0 132 43 5 orxþy¼a 2xþ3yþz¼b 2yþ3z¼c Solve for x;y;zto get x¼7a/C03bþc,y¼/C06aþ3b/C0c,z¼4a/C02bþc. Thus, v¼ða;b;cÞ¼ð 7a/C03bþcÞu 1þð/C0 6aþ3b/C0cÞu2þð4a/C02bþcÞu3 or½v/C138S¼½ða;b;cÞ/C138S¼½7a/C03bþc;/C06aþ3b/C0c;4a/C02bþc/C138T Using the above formula for ½v/C138Sand then writing the coordinates of the eias columns yields e1¼ð1;0;0Þ¼ 7u1/C06u2þ4u3 e2¼ð0;1;0Þ¼/C0 3u1þ3u2/C02u3 e3¼ð0;0;1Þ¼ u1/C0u2þu3and Q¼7/C031 /C063/C01 4/C0212 43 5 Method 2. Find P/C01by row reducing M¼½P;I/C138to the form½I;P/C01/C138: M¼110100 2310100230012 643 75/C24110 100 011/C0210 023 0012 643 75 /C241 1 010 0 011/C021 0 001 4/C0212 643 75/C24100 7/C031 010/C063/C01 001 4/C0212 643 75¼½I;P /C01/C138 Thus, Q¼P/C01¼7/C031 /C063/C01 4/C0212 43 5. 6.15. Suppose the x-axis and y-axis in the plane R2are rotated counterclockwise 45/C14so that the new x0-axis and y0-axis are along the line y¼xand the line y¼/C0x, respectively. (a) Find the change-of-basis matrix P. (b) Find the coordinates of the point Að5;6Þunder the given rotation. (a) The unit vectors in the direction of the new x0- and y0-axes are u1¼ð1 2ﬃﬃﬃ 2p ;1 2ﬃﬃﬃ 2p Þ and u2¼ð/C01 2ﬃﬃﬃ 2p ;1 2ﬃﬃﬃ 2p Þ (The unit vectors in the direction of the original xandyaxes are the usual basis of R2.) Thus, write the coordinates of u1andu2as columns to obtain P¼1 2ﬃﬃﬃ 2p /C01 2ﬃﬃﬃ 2p 1 2ﬃﬃﬃ 2p1 2ﬃﬃﬃ 2p"# (b) Multiply the coordinates of the point by P/C01: 1 2ﬃﬃﬃ 2p1 2ﬃﬃﬃ 2p /C01 2ﬃﬃﬃ 2p1 2ﬃﬃﬃ 2p"# 5 6/C20/C21 ¼11 2ﬃﬃﬃ 2p 1 2ﬃﬃﬃ 2p"# (Because Pis orthogonal, P/C01is simply the transpose of P.)212 CHAPTER 6 Linear Mappings and Matrices 6.16. The vectors u1¼ð1;1;0Þ,u2¼ð0;1;1Þ,u3¼ð1;2;2Þform a basis SofR3. Find the coordinates of an arbitrary vector v¼ða;b;cÞrelative to the basis S. Method 1. Express vas a linear combination of u1;u2;u3using unknowns x;y;z. We have ða;b;cÞ¼xð1;1;0Þþyð0;1;1Þþzð1;2;2Þ¼ð xþz;xþyþ2z;yþ2zÞ this yields the system xþ z¼a xþyþ2z¼b yþ2z¼corxþ z¼a yþz¼/C0aþb yþ2z¼corxþ z¼a yþz¼/C0aþb z¼a/C0bþc Solving by back-substitution yields x¼b/C0c,y¼/C02aþ2b/C0c,z¼a/C0bþc. Thus, ½v/C138S¼½b/C0c;/C02aþ2b/C0c;a/C0bþc/C138T Method 2. Find P/C01by row reducing M¼½P;I/C138to the form½I;P/C01/C138, where Pis the change-of-basis matrix from the usual basis EtoSor, in other words, the matrix whose columns are the basis vectors of S. We have M¼101100 1120100120012 643 75/C24101 100 011/C0110 012 0012 643 75 /C241 0 110 0 011/C011 0 001 1/C0112 643 75/C241 0 001 /C01 010/C022/C01 001 1/C0112 643 75¼½I;P /C01/C138 Thus ; P/C01¼01/C01 /C022/C01 1/C0112 643 75and½v/C138S¼P/C01½v/C138E¼01/C01 /C022/C01 1/C0112 643 75a b c2 643 75¼b/C0c /C02aþ2b/C0c a/C0bþc2 643 75 6.17. Consider the following bases of R2: S¼fu1;u2g¼fð 1;/C02Þ;ð3;/C04Þg and S0¼fv1;v2g¼fð 1;3Þ;ð3;8Þg (a) Find the coordinates of v¼ða;bÞrelative to the basis S. (b) Find the change-of-basis matrix Pfrom StoS0. (c) Find the coordinates of v¼ða;bÞrelative to the basis S0. (d) Find the change-of-basis matrix Qfrom S0back to S. (e) Verify Q¼P/C01. (f ) Show that, for any vector v¼ða;bÞinR2,P/C01½v/C138S¼½v/C138S0. (See Theorem 6.6.) (a) Let v¼xu1þyu2for unknowns xandy; that is, a b/C20/C21 ¼x1 /C02/C20/C21 þy3 /C04/C20/C21 orxþ3y¼a /C02x/C04y¼borxþ3y¼a 2y¼2aþb Solve for xandyin terms of aandbto get x¼/C02a/C03 2band y¼aþ1 2b. Thus, ða;bÞ¼ð/C0 2a/C03 2Þu1þðaþ1 2bÞu2 or½ða;bÞ/C138S¼½/C0 2a/C03 2b;aþ1 2b/C138T (b) Use part ( a) to write each of the basis vectors v1andv2ofS0as a linear combination of the basis vectors u1andu2ofS; that is, v1¼ð1;3Þ¼ð/C0 2/C09 2Þu1þð1þ3 2Þu2¼/C013 2u1þ52u2 v2¼ð3;8Þ¼ð/C0 6/C012Þu1þð3þ4Þu2¼/C018u1þ7u2CHAPTER 6 Linear Mappings and Matrices 213 Then Pis the matrix whose columns are the coordinates of v1and v2relative to the basis S; that is, P¼/C013 2/C018 5 27"# (c) Let v¼xv1þyv2for unknown scalars xandy: a b/C20/C21 ¼x1 3/C20/C21 þy3 8/C20/C21 orxþ3y¼a 3xþ8y¼borxþ3y¼a /C0y¼b/C03a Solve for xandyto get x¼/C08aþ3bandy¼3a/C0b. Thus, ða;bÞ¼ð/C0 8aþ3bÞv1þð3a/C0bÞv2 or½ða;bÞ/C138S0¼½/C0 8aþ3b;3a/C0b/C138T (d) Use part ( c) to express each of the basis vectors u1andu2ofSas a linear combination of the basis vectors v1and v2ofS0: u1¼ð1;/C02Þ¼ð/C0 8/C06Þv1þð3þ2Þv2¼/C014v1þ5v2 u2¼ð3;/C04Þ¼ð/C0 24/C012Þv1þð9þ4Þv2¼/C036v1þ13v2 Write the coordinates of u1andu2relative to S0as columns to obtain Q¼/C014/C036 51 3/C20/C21 . (e) QP¼/C014/C036 51 3/C20/C21/C013 2/C018 5 27"# ¼10 01/C20/C21 ¼I (f ) Use parts (a), (c), and (d) to obtain P/C01½v/C138S¼Q½v/C138S¼/C014/C036 51 3/C20/C21/C02a/C03 2b aþ1 2b"# ¼/C08aþ3b 3a/C0b/C20/C21 ¼½v/C138S0 6.18. Suppose Pis the change-of-basis matrix from a basis fuigto a basisfwig, and suppose Qis the change-of-basis matrix from the basis fwigback tofuig. Prove that Pis invertible and that Q¼P/C01. Suppose, for i¼1;2;...;n, that wi¼ai1u1þai2u2þ...þainun¼Pn j¼1aijuj ð1Þ and, for j¼1;2;...;n, uj¼bj1w1þbj2w2þ/C1/C1/C1þ bjnwn¼Pn k¼1bjkwk ð2Þ LetA¼½aij/C138andB¼½bjk/C138. Then P¼ATandQ¼BT. Substituting (2) into (1) yields wi¼Pn j¼1aij/C18Pn k¼1bjkwk/C19 ¼Pn k¼1/C18Pn j¼1aijbjk/C19 wk Becausefwigis a basis,Paijbjk¼dik, where dikis the Kronecker delta; that is, dik¼1i fi¼kbutdik¼0 ifi6¼k. Suppose AB¼½cik/C138. Then cik¼dik. Accordingly, AB¼I, and so QP¼BTAT¼ðABÞT¼IT¼I Thus, Q¼P/C01. 6.19. Consider a ﬁnite sequence of vectors S¼fu1;u2;...;ung. Let S0be the sequence of vectors obtained from Sby one of the following ‘‘elementary operations’’: (1) Interchange two vectors. (2) Multiply a vector by a nonzero scalar. (3) Add a multiple of one vector to another vector. Show that SandS0span the same subspace W. Also, show that S0is linearly independent if and only if Sis linearly independent.214 CHAPTER 6 Linear Mappings and Matrices Observe that, for each operation, the vectors S0are linear combinations of vectors in S. Also, because each operation has an inverse of the same type, each vector in Sis a linear combination of vectors in S0. Thus, SandS0span the same subspace W. Moreover, S0is linearly independent if and only if dim W¼n, and this is true if and only if Sis linearly independent. 6.20. LetA¼½aij/C138andB¼½bij/C138be row equivalent m/C2nmatrices over a ﬁeld K, and let v1;v2;...;vn be any vectors in a vector space Vover K. For i¼1;2;...;m, let uiandwibe deﬁned by ui¼ai1v1þai2v2þ/C1/C1/C1þ ainvn and wi¼bi1v1þbi2v2þ/C1/C1/C1þ binvn Show thatfuigandfwigspan the same subspace of V. Applying an ‘‘elementary operation’’ of Problem 6.19 to fuigis equivalent to applying an elementary row operation to the matrix A. Because AandBare row equivalent, Bcan be obtained from Aby a sequence of elementary row operations. Hence, fwigcan be obtained from fuigby the corresponding sequence of operations. Accordingly, fuigandfwigspan the same space. 6.21. Suppose u1;u2;...;unbelong to a vector space Vover a ﬁeld K, and suppose P¼½aij/C138is an n-square matrix over K. For i¼1;2;...;n, let vi¼ai1u1þai2u2þ/C1/C1/C1þ ainun. (a) Suppose Pis invertible. Show that fuigandfvigspan the same subspace of V. Hence,fuigis linearly independent if and only if fvigis linearly independent. (b) Suppose Pis singular (not invertible). Show that fvigis linearly dependent. (c) Supposefvigis linearly independent. Show that Pis invertible. (a) Because Pis invertible, it is row equivalent to the identity matrix I. Hence, by Problem 6.19, fvigand fuigspan the same subspace of V. Thus, one is linearly independent if and only if the other is linearly independent. (b) Because Pis not invertible, it is row equivalent to a matrix with a zero row. This means fvigspans a substance that has a spanning set with less than nelements. Thus,fvigis linearly dependent. (c) This is the contrapositive of the statement of part (b), and so it follows from part (b). 6.22. Prove Theorem 6.6: Let Pbe the change-of-basis matrix from a basis Sto a basis S0in a vector space V. Then, for any vector v2V, we have P½v/C138S0¼½v/C138S, and hence, P/C01½v/C138S¼½v/C138S0. Suppose S¼fu1;...;ungandS0¼fw1;...;wng, and suppose, for i¼1;...;n, wi¼ai1u1þai2u2þ/C1/C1/C1þ ainun¼Pn j¼1aijuj Then Pis the n-square matrix whose jth row is ða1j;a2j;...;anjÞð 1Þ Also suppose v¼k1w1þk2w2þ/C1/C1/C1þ knwn¼Pn i¼1kiwi. Then ½v/C138S0¼½k1;k2;...;kn/C138Tð2Þ Substituting for wiin the equation for v, we obtain v¼Pn i¼1kiwi¼Pn i¼1ki/C18Pn j¼1aijuj/C19 ¼Pn j¼1/C18Pn i¼1aijki/C19 uj ¼Pn j¼1ða1jk1þa2jk2þ/C1/C1/C1þ anjknÞuj Accordingly,½v/C138Sis the column vector whose jth entry is a1jk1þa2jk2þ/C1/C1/C1þ anjkn ð3Þ On the other hand, the jth entry of P½v/C138S0is obtained by multiplying the jth row of Pby½v/C138S0—that is, (1) by (2). However, the product of (1) and (2) is (3). Hence, P½v/C138S0and½v/C138Shave the same entries. Thus, P½v/C138S0¼½v/C138S0, as claimed. Furthermore, multiplying the above by P/C01gives P/C01½v/C138S¼P/C01P½v/C138S0¼½v/C138S0.CHAPTER 6 Linear Mappings and Matrices 215 Linear Operators and Change of Basis 6.23. Consider the linear transformation FonR2deﬁned by Fðx;yÞ¼ð 5x/C0y;2xþyÞand the following bases of R2: E¼fe1;e2g¼fð 1;0Þ;ð0;1Þg and S¼fu1;u2g¼fð 1;4Þ;ð2;7Þg (a) Find the change-of-basis matrix Pfrom EtoSand the change-of-basis matrix Qfrom Sback toE. (b) Find the matrix Athat represents Fin the basis E. (c) Find the matrix Bthat represents Fin the basis S. (a) Because Eis the usual basis, simply write the vectors in Sas columns to obtain the change-of-basis matrix P. Recall, also, that Q¼P/C01. Thus, P¼12 47/C20/C21 and Q¼P/C01¼/C072 4/C01/C20/C21 (b) Write the coefﬁcients of xandyinFðx;yÞ¼ð 5x/C0y;2xþyÞas rows to get A¼5/C01 21/C20/C21 (c)Method 1. Find the coordinates of Fðu1ÞandFðu2Þrelative to the basis S. This may be done by ﬁrst ﬁnding the coordinates of an arbitrary vector ða;bÞinR2relative to the basis S. We have ða;bÞ¼xð1;4Þþyð2;7Þ¼ð xþ2y;4xþ7yÞ; and soxþ2y¼a 4xþ7y¼b Solve for xandyin terms of aandbto get x¼/C07aþ2b,y¼4a/C0b. Then ða;bÞ¼ð/C0 7aþ2bÞu1þð4a/C0bÞu2 Now use the formula for ða;bÞto obtain Fðu1Þ¼Fð1;4Þ¼ð 1;6Þ¼ 5u1/C02u2 Fðu2Þ¼Fð2;7Þ¼ð 3;11Þ¼ u1þu2and so B¼51 /C021/C20/C21 Method 2. By Theorem 6.7, B¼P/C01AP. Thus, B¼P/C01AP¼/C072 4/C01/C20/C21 5/C01 21/C20/C21 12 47/C20/C21 ¼51 /C021/C20/C21 6.24. LetA¼23 4/C01/C20/C21 . Find the matrix Bthat represents the linear operator Arelative to the basis S¼fu1;u2g¼f½ 1;3/C138T;½2;5/C138Tg. [Recall Adeﬁnes a linear operator A:R2!R2relative to the usual basis EofR2]. Method 1. Find the coordinates of Aðu1ÞandAðu2Þrelative to the basis Sby ﬁrst ﬁnding the coordinates of an arbitrary vector ½a;b/C138TinR2relative to the basis S. By Problem 6.2, ½a;b/C138T¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2 Using the formula for ½a;b/C138T, we obtain Aðu1Þ¼23 4/C01/C20/C211 3/C20/C21 ¼11 1/C20/C21 ¼/C053u1þ32u2 and Aðu2Þ¼23 4/C01/C20/C212 5/C20/C21 ¼19 3/C20/C21 ¼/C089u1þ54u2 Thus ; B¼/C053/C089 32 54/C20/C21 Method 2. Use B¼P/C01AP, where Pis the change-of-basis matrix from the usual basis EtoS. Thus, simply write the vectors in S(as columns) to obtain the change-of-basis matrix Pand then use the formula216 CHAPTER 6 Linear Mappings and Matrices forP/C01. This gives P¼12 35/C20/C21 and P/C01¼/C052 3/C01/C20/C21 Then B¼P/C01AP¼12 35/C20/C2123 4/C01/C20/C21/C052 3/C01/C20/C21 ¼/C053/C089 32 54/C20/C21 6.25. LetA¼131 25/C04 1/C0222 43 5:Find the matrix Bthat represents the linear operator Arelative to the basis S¼fu1;u2;u3g¼f½ 1;1;0/C138T;½0;1;1/C138T;½1;2;2/C138Tg [Recall Athat deﬁnes a linear operator A:R3!R3relative to the usual basis EofR3.] Method 1. Find the coordinates of Aðu1Þ,Aðu2Þ,Aðu3Þrelative to the basis Sby ﬁrst ﬁnding the coordinates of an arbitrary vector v¼ða;b;cÞinR3relative to the basis S. By Problem 6.16, ½v/C138S¼ðb/C0cÞu1þð/C0 2aþ2b/C0cÞu2þða/C0bþcÞu3 Using this formula for ½a;b;c/C138T, we obtain Aðu1Þ¼½ 4;7;/C01/C138T¼8u1þ7u2/C05u3; Aðu2Þ¼½ 4;1;0/C138T¼u1/C06u2þ3u3 Aðu3Þ¼½ 9;4;1/C138T¼3u1/C011u2þ6u3 Writing the coefﬁcients of u1;u2;u3as columns yields B¼81 3 7/C06/C011 /C053 62 43 5 Method 2. UseB¼P/C01AP, where Pis the change-of-basis matrix from the usual basis EtoS. The matrix P(whose columns are simply the vectors in S) and P/C01appear in Problem 6.16. Thus, B¼P/C01AP¼01/C01 /C022/C01 1/C0112 43 5131 25/C04 1/C0222 43 5101 112 0122 43 5¼81 3 7/C06/C011 /C053 62 43 5 6.26. Prove Theorem 6.7: Let Pbe the change-of-basis matrix from a basis Sto a basis S0in a vector space V. Then, for any linear operator TonV,½T/C138S0¼P/C01½T/C138SP. Letvbe a vector in V. Then, by Theorem 6.6, P½v/C138S0¼½v/C138S. Therefore, P/C01½T/C138SP½v/C138S0¼P/C01½T/C138S½v/C138S¼P/C01½TðvÞ/C138S¼½TðvÞ/C138S0 But½T/C138S0½v/C138S0¼½TðvÞ/C138S0. Hence, P/C01½T/C138SP½v/C138S0¼½T/C138S0½v/C138S0 Because the mapping v7!½v/C138S0is onto Kn, we have P/C01½T/C138SPX¼½T/C138S0Xfor every X2Kn. Thus, P/C01½T/C138SP¼½T/C138S0, as claimed. Similarity of Matrices 6.27. LetA¼4/C02 36/C20/C21 andP¼12 34/C20/C21 . (a) Find B¼P/C01AP. (b) Verify tr ðBÞ¼trðAÞ: (c) Verify detðBÞ¼detðAÞ: (a) First ﬁnd P/C01using the formula for the inverse of a 2 /C22 matrix. We have P/C01¼/C021 3 2/C012"#CHAPTER 6 Linear Mappings and Matrices 217 Then B¼P/C01AP¼/C021 3 2/C012/C20/C21 4/C02 36/C20/C21 12 34/C20/C21 ¼25 30 /C027 2/C015/C20/C21 (b) trðAÞ¼4þ6¼10 and trðBÞ¼25/C015¼10. Hence, trðBÞ¼trðAÞ. (c) detðAÞ¼24þ6¼30 and detðBÞ¼/C0 375þ405¼30. Hence, detðBÞ¼detðAÞ. 6.28. Find the trace of each of the linear transformations FonR3in Problem 6.4. Find the trace (sum of the diagonal elements) of any matrix representation of Fsuch as the matrix representation½F/C138¼½F/C138EofFrelative to the usual basis Egiven in Problem 6.4. (a) trðFÞ¼trð½F/C138Þ¼ 1/C05þ9¼5. (b) trðFÞ¼trð½F/C138Þ¼ 1þ3þ5¼9. (c) trðFÞ¼trð½F/C138Þ¼ 1þ4þ7¼12. 6.29. Write A/C25BifAis similar to B—that is, if there exists an invertible matrix Psuch that A¼P/C01BP. Prove that/C25is an equivalence relation (on square matrices); that is, (a) A/C25A, for every A. (b) If A/C25B, then B/C25A. (c) If A/C25BandB/C25C, then A/C25C. (a) The identity matrix Iis invertible, and I/C01¼I. Because A¼I/C01AI, we have A/C25A. (b) Because A/C25B, there exists an invertible matrix Psuch that A¼P/C01BP. Hence, B¼PAP/C01¼ðP/C01Þ/C01APandP/C01is also invertible. Thus, B/C25A. (c) Because A/C25B, there exists an invertible matrix Psuch that A¼P/C01BP, and as B/C25C, there exists an invertible matrix Qsuch that B¼Q/C01CQ. Thus, A¼P/C01BP¼P/C01ðQ/C01CQÞP¼ðP/C01Q/C01ÞCðQPÞ¼ð QPÞ/C01CðQPÞ andQPis also invertible. Thus, A/C25C. 6.30. Suppose Bis similar to A, say B¼P/C01AP. Prove (a)Bn¼P/C01AnP, and so Bnis similar to An. (b)fðBÞ¼P/C01fðAÞP, for any polynomial fðxÞ, and so fðBÞis similar to fðAÞ: (c)Bis a root of a polynomial gðxÞif and only if Ais a root of gðxÞ. (a) The proof is by induction on n. The result holds for n¼1 by hypothesis. Suppose n>1 and the result holds for n/C01. Then Bn¼BBn/C01¼ðP/C01APÞðP/C01An/C01PÞ¼P/C01AnP (b) Suppose fðxÞ¼anxnþ/C1/C1/C1þ a1xþa0. Using the left and right distributive laws and part (a), we have P/C01fðAÞP¼P/C01ðanAnþ/C1/C1/C1þ a1Aþa0IÞP ¼P/C01ðanAnÞPþ/C1/C1/C1þ P/C01ða1AÞPþP/C01ða0IÞP ¼anðP/C01AnPÞþ/C1/C1/C1þ a1ðP/C01APÞþa0ðP/C01IPÞ ¼anBnþ/C1/C1/C1þ a1Bþa0I¼fðBÞ (c) By part (b), gðBÞ¼0 if and only if P/C01gðAÞP¼0 if and only if gðAÞ¼P0P/C01¼0. Matrix Representations of General Linear Mappings 6.31. LetF:R3!R2be the linear map deﬁned by Fðx;y;zÞ¼ð 3xþ2y/C04z;x/C05yþ3zÞ. (a) Find the matrix of Fin the following bases of R3andR2: S¼fw1;w2;w3g¼fð 1;1;1Þ;ð1;1;0Þ;ð1;0;0Þg and S0¼fu1;u2g¼fð 1;3Þ;ð2;5Þg218 CHAPTER 6 Linear Mappings and Matrices (b) Verify Theorem 6.10: The action of Fis preserved by its matrix representation; that is, for any vinR3, we have½F/C138S;S0½v/C138S¼½FðvÞ/C138S0. (a) From Problem 6.2, ða;bÞ¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2. Thus, Fðw1Þ¼Fð1;1;1Þ¼ð 1;/C01Þ¼/C0 7u1þ4u2 Fðw2Þ¼Fð1;1;0Þ¼ð 5;/C04Þ¼/C0 33u1þ19u2 Fðw3Þ¼Fð1;0;0Þ¼ð 3;1Þ¼/C0 13u1þ8u2 Write the coordinates of Fðw1Þ,Fðw2Þ;Fðw3Þas columns to get ½F/C138S;S0¼/C07/C033 13 41 9 8/C20/C21 (b) If v¼ðx;y;zÞ, then, by Problem 6.5, v¼zw1þðy/C0zÞw2þðx/C0yÞw3. Also, FðvÞ¼ð 3xþ2y/C04z;x/C05yþ3zÞ¼ð/C0 13x/C020yþ26zÞu1þð8xþ11y/C015zÞu2 Hence ;½v/C138S¼ðz;y/C0z;x/C0yÞTand½FðvÞ/C138S0¼/C013x/C020yþ26z 8xþ11y/C015z/C20/C21 Thus,½F/C138S;S0½v/C138S¼/C07/C033/C013 41 9 8/C20/C21 z y/C0x x/C0y2 43 5¼/C013x/C020yþ26z 8xþ11y/C015z/C20/C21 ¼½FðvÞ/C138S0 6.32. LetF:Rn!Rmbe the linear mapping deﬁned as follows: Fðx1;x2;...;xnÞ¼ð a11x1þ/C1/C1/C1þ a1nxn,a21x1þ/C1/C1/C1þ a2nxn;...;am1x1þ/C1/C1/C1þ amnxnÞ (a) Show that the rows of the matrix ½F/C138representing Frelative to the usual bases of RnandRm are the coefﬁcients of the xiin the components of Fðx1;...;xnÞ. (b) Find the matrix representation of each of the following linear mappings relative to the usual basis of Rn: (i) F:R2!R3deﬁned by Fðx;yÞ¼ð 3x/C0y;2xþ4y;5x/C06yÞ. (ii) F:R4!R2deﬁned by Fðx;y;s;tÞ¼ð 3x/C04yþ2s/C05t;5xþ7y/C0s/C02tÞ. (iii) F:R3!R4deﬁned by Fðx;y;zÞ¼ð 2xþ3y/C08z;xþyþz;4x/C05z;6yÞ. (a) We have Fð1;0;...;0Þ¼ð a11;a21;...;am1Þ Fð0;1;...;0Þ¼ð a12;a22;...;am2Þ ::::::::::::::::::::::::::::::::::::::::::::::::::::: Fð0;0;...;1Þ¼ð a1n;a2n;...;amnÞand thus ;½F/C138¼a11a12 ... a1n a21a22 ... a2n ::::::::::::::::::::::::::::::::: am1am2... amn2 6643 775 (b) By part (a), we need only look at the coefﬁcients of the unknown x;y;...inFðx;y;...Þ. Thus, ðiÞ½F/C138¼3/C01 24 5/C062 43 5;ðiiÞ½F/C138¼3/C042/C05 57/C01/C02/C20/C21 ;ðiiiÞ½F/C138¼23/C08 11 1 40/C05 06 02 6643 775 6.33. LetA¼25/C03 1/C047/C20/C21 . Recall that Adetermines a mapping F:R3!R2deﬁned by FðvÞ¼Av, where vectors are written as columns. Find the matrix ½F/C138that represents the mapping relative to the following bases of R3andR2: (a) The usual bases of R3and of R2. (b)S¼fw1;w2;w3g¼fð 1;1;1Þ;ð1;1;0Þ;ð1;0;0ÞgandS0¼fu1;u2g¼fð 1;3Þ;ð2;5Þg. (a) Relative to the usual bases, ½F/C138is the matrix A.CHAPTER 6 Linear Mappings and Matrices 219 (b) From Problem 9.2, ða;bÞ¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2. Thus, Fðw1Þ¼25/C03 1/C047/C20/C21 1 112 643 75¼4 4/C20/C21 ¼/C012u 1þ8u2 Fðw2Þ¼25/C03 1/C047/C20/C21 1 1 02 643 75¼7 /C03/C20/C21 ¼/C041u1þ24u2 Fðw3Þ¼25/C03 1/C047/C20/C21 1 0 02 643 75¼2 1/C20/C21 ¼/C08u1þ5u2 Writing the coefﬁcients of Fðw1Þ,Fðw2Þ,Fðw3Þas columns yields ½F/C138¼/C012/C041/C08 82 45/C20/C21 . 6.34. Consider the linear transformation TonR2deﬁned by Tðx;yÞ¼ð 2x/C03y;xþ4yÞand the following bases of R2: E¼fe1;e2g¼fð 1;0Þ;ð0;1Þg and S¼fu1;u2g¼fð 1;3Þ;ð2;5Þg (a) Find the matrix Arepresenting Trelative to the bases EandS. (b) Find the matrix Brepresenting Trelative to the bases SandE. (We can view Tas a linear mapping from one space into another, each having its own basis.) (a) From Problem 6.2, ða;bÞ¼ð/C0 5aþ2bÞu1þð3a/C0bÞu2. Hence, Tðe1Þ¼Tð1;0Þ¼ð 2;1Þ¼ /C0 8u1þ5u2 Tðe2Þ¼Tð0;1Þ¼ð/C0 3;4Þ¼ 23u1/C013u2and so A¼/C082 3 5/C013/C20/C21 (b) We have Tðu1Þ¼Tð1;3Þ¼ð/C0 7;13Þ¼/C0 7e1þ13e2 Tðu2Þ¼Tð2;5Þ¼ð/C0 11;22Þ¼/C0 11e1þ22e2and so B¼/C07/C011 13 22/C20/C21 6.35. How are the matrices AandBin Problem 6.34 related? By Theorem 6.12, the matrices AandBare equivalent to each other; that is, there exist nonsingular matrices PandQsuch that B¼Q/C01AP, where Pis the change-of-basis matrix from StoE, and Qis the change-of-basis matrix from EtoS. Thus, P¼12 35/C20/C21 ; Q¼/C052 3/C01/C20/C21 ; Q/C01¼12 35/C20/C21 and Q/C01AP¼12 35/C20/C21/C08/C023 5/C013/C20/C2112 35/C20/C21 ¼/C07/C011 13 22/C20/C21 ¼B 6.36. Prove Theorem 6.14: Let F:V!Ube linear and, say, rank ðFÞ¼r. Then there exist bases Vand ofUsuch that the matrix representation of Fhas the following form, where Iris the r-square identity matrix: A¼Ir0 00/C20/C21 Suppose dim V¼mand dim U¼n. Let Wbe the kernel of FandU0the image of F. We are given that rankðFÞ¼r. Hence, the dimension of the kernel of Fism/C0r. Letfw1;...;wm/C0rgbe a basis of the kernel ofFand extend this to a basis of V: fv1;...;vr;w1;...;wm/C0rg Set u1¼Fðv1Þ;u2¼Fðv2Þ;...;ur¼FðvrÞ220 CHAPTER 6 Linear Mappings and Matrices Thenfu1;...;urgis a basis of U0, the image of F. Extend this to a basis of U, say fu1;...;ur;urþ1;...;ung Observe that Fðv1Þ¼ u1¼1u1þ0u2þ/C1/C1/C1þ 0urþ0urþ1þ/C1/C1/C1þ 0un Fðv2Þ¼ u2¼0u1þ1u2þ/C1/C1/C1þ 0urþ0urþ1þ/C1/C1/C1þ 0un :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: FðvrÞ¼ ur¼0u1þ0u2þ/C1/C1/C1þ 1urþ0urþ1þ/C1/C1/C1þ 0un Fðw1Þ¼ 0¼0u1þ0u2þ/C1/C1/C1þ 0urþ0urþ1þ/C1/C1/C1þ 0un :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Fðwm/C0rÞ¼0¼0u1þ0u2þ/C1/C1/C1þ 0urþ0urþ1þ/C1/C1/C1þ 0un Thus, the matrix of Fin the above bases has the required form. SUPPLEMENTARY PROBLEMS Matrices and Linear Operators 6.37. LetF:R2!R2be deﬁned by Fðx;yÞ¼ð 4xþ5y;2x/C0yÞ. (a) Find the matrix Arepresenting Fin the usual basis E. (b) Find the matrix Brepresenting Fin the basis S¼fu1;u2g¼fð 1;4Þ;ð2;9Þg. (c) Find Psuch that B¼P/C01AP. (d) For v¼ða;bÞ, ﬁnd½v/C138Sand½FðvÞ/C138S. Verify that½F/C138S½v/C138S¼½FðvÞ/C138S. 6.38. LetA:R2!R2be deﬁned by the matrix A¼5/C01 24/C20/C21 . (a) Find the matrix Brepresenting Arelative to the basis S¼fu1;u2g¼fð 1;3Þ;ð2;8Þg. (Recall that A represents the mapping Arelative to the usual basis E.) (b) For v¼ða;bÞ, ﬁnd½v/C138Sand½AðvÞ/C138S. 6.39. For each linear transformation LonR2, ﬁnd the matrix Arepresenting L(relative to the usual basis of R2): (a) Lis the rotation in R2counterclockwise by 45/C14. (b) Lis the reﬂection in R2about the line y¼x. (c) Lis deﬁned by Lð1;0Þ¼ð 3;5ÞandLð0;1Þ¼ð 7;/C02Þ. (d) Lis deﬁned by Lð1;1Þ¼ð 3;7ÞandLð1;2Þ¼ð 5;/C04Þ. 6.40. Find the matrix representing each linear transformation TonR3relative to the usual basis of R3: (a) Tðx;y;zÞ¼ð x;y;0Þ. (b) Tðx;y;zÞ¼ð z;yþz;xþyþzÞ. (c) Tðx;y;zÞ¼ð 2x/C07y/C04z;3xþyþ4z;6x/C08yþzÞ. 6.41. Repeat Problem 6.40 using the basis S¼fu1;u2;u3g¼fð 1;1;0Þ;ð1;2;3Þ;ð1;3;5Þg. 6.42. LetLbe the linear transformation on R3deﬁned by Lð1;0;0Þ¼ð 1;1;1Þ; Lð0;1;0Þ¼ð 1;3;5Þ; Lð0;0;1Þ¼ð 2;2;2Þ (a) Find the matrix Arepresenting Lrelative to the usual basis of R3. (b) Find the matrix Brepresenting Lrelative to the basis Sin Problem 6.41. 6.43. LetDdenote the differential operator; that is, DðfðtÞÞ¼ df=dt. Each of the following sets is a basis of a vector space Vof functions. Find the matrix representing Din each basis: (a)fet;e2t;te2tg. (b)f1;t;sin 3t;cos 3 tg. (c)fe5t;te5t;t2e5tg.CHAPTER 6 Linear Mappings and Matrices 221 6.44. LetDdenote the differential operator on the vector space Vof functions with basis S¼fsiny, cos yg. (a) Find the matrix A¼½D/C138S. (b) Use Ato show that Dis a zero of fðtÞ¼t2þ1. 6.45. LetVbe the vector space of 2 /C22 matrices. Consider the following matrix Mand usual basis EofV: M¼ab cd/C20/C21 and E¼10 00/C20/C21 ;01 00/C20/C21 ;00 10/C20/C21 ;00 01/C20/C21 /C26/C27 Find the matrix representing each of the following linear operators TonVrelative to E: (a) TðAÞ¼MA. (b) TðAÞ¼AM. (c) TðAÞ¼MA/C0AM. 6.46. Let1Vand0Vdenote the identity and zero operators, respectively, on a vector space V. Show that, for any basis SofV, (a)½1V/C138S¼I, the identity matrix. (b) ½0V/C138S¼0, the zero matrix. Change of Basis 6.47. Find the change-of-basis matrix Pfrom the usual basis EofR2to a basis S, the change-of-basis matrix Q from Sback to E, and the coordinates of v¼ða;bÞrelative to S, for the following bases S: (a) S¼fð 1;2Þ;ð3;5Þg. (c) S¼fð 2;5Þ;ð3;7Þg. (b) S¼fð 1;/C03Þ;ð3;/C08Þg. (d) S¼fð 2;3Þ;ð4;5Þg. 6.48. Consider the bases S¼fð 1;2Þ;ð2;3ÞgandS0¼fð 1;3Þ;ð1;4ÞgofR2. Find the change-of-basis matrix: (a) Pfrom StoS0. (b) Qfrom S0back to S. 6.49. Suppose that the x-axis and y-axis in the plane R2are rotated counterclockwise 30/C14to yield new x0-axis and y0-axis for the plane. Find (a) The unit vectors in the direction of the new x0-axis and y0-axis. (b) The change-of-basis matrix Pfor the new coordinate system. (c) The new coordinates of the points Að1;3Þ,Bð2;/C05Þ,Cða;bÞ. 6.50. Find the change-of-basis matrix Pfrom the usual basis EofR3to a basis S, the change-of-basis matrix Q from Sback to E, and the coordinates of v¼ða;b;cÞrelative to S, where Sconsists of the vectors: (a) u1¼ð1;1;0Þ;u2¼ð0;1;2Þ;u3¼ð0;1;1Þ. (b) u1¼ð1;0;1Þ;u2¼ð1;1;2Þ;u3¼ð1;2;4Þ. (c) u1¼ð1;2;1Þ;u2¼ð1;3;4Þ;u3¼ð2;5;6Þ. 6.51. Suppose S1;S2;S3are bases of V. Let PandQbe the change-of-basis matrices, respectively, from S1toS2 and from S2toS3. Prove that PQis the change-of-basis matrix from S1toS3. Linear Operators and Change of Basis 6.52. Consider the linear operator FonR2deﬁned by Fðx;yÞ¼ð 5xþy;3x/C02yÞand the following bases of R2: S¼fð 1;2Þ;ð2;3Þg and S0¼fð 1;3Þ;ð1;4Þg (a) Find the matrix Arepresenting Frelative to the basis S. (b) Find the matrix Brepresenting Frelative to the basis S0. (c) Find the change-of-basis matrix Pfrom StoS0. (d) How are AandBrelated? 6.53. LetA:R2!R2be deﬁned by the matrix A¼1/C01 32/C20/C21 . Find the matrix Bthat represents the linear operator Arelative to each of the following bases: (a) S¼fð 1;3ÞT;ð2;5ÞTg. (b) S¼fð 1;3ÞT;ð2;4ÞTg.222 CHAPTER 6 Linear Mappings and Matrices 6.54. LetF:R2!R2be deﬁned by Fðx;yÞ¼ð x/C03y;2x/C04yÞ. Find the matrix Athat represents Frelative to each of the following bases: (a) S¼fð 2;5Þ;ð3;7Þg. (b) S¼fð 2;3Þ;ð4;5Þg. 6.55. LetA:R3!R3be deﬁned by the matrix A¼131 274 1432 43 5. Find the matrix Bthat represents the linear operator Arelative to the basis S¼fð 1;1;1ÞT;ð0;1;1ÞT;ð1;2;3ÞTg. Similarity of Matrices 6.56. LetA¼11 2/C03/C20/C21 andP¼1/C02 3/C05/C20/C21 . (a) Find B¼P/C01AP. (b) Verify that tr ðBÞ¼trðAÞ: (c) Verify that det ðBÞ¼detðAÞ. 6.57. Find the trace and determinant of each of the following linear maps on R2: (a) Fðx;yÞ¼ð 2x/C03y;5xþ4yÞ. (b) Gðx;yÞ¼ð axþby;cxþdyÞ. 6.58. Find the trace and determinant of each of the following linear maps on R3: (a) Fðx;y;zÞ¼ð xþ3y;3x/C02z;x/C04y/C03zÞ. (b) Gðx;y;zÞ¼ð yþ3z;2x/C04z;5xþ7yÞ. 6.59. Suppose S¼fu1;u2gis a basis of V, and T:V!Vis deﬁned by Tðu1Þ¼3u1/C02u2andTðu2Þ¼u1þ4u2. Suppose S0¼fw1;w2gis a basis of Vfor which w1¼u1þu2andw2¼2u1þ3u2. (a) Find the matrices AandBrepresenting Trelative to the bases SandS0, respectively. (b) Find the matrix Psuch that B¼P/C01AP. 6.60. LetAbe a 2/C22 matrix such that only Ais similar to itself. Show that Ais a scalar matrix, that is, that A¼a0 0a/C20/C21 . 6.61. Show that all matrices similar to an invertible matrix are invertible. More generally, show that similar matrices have the same rank. Matrix Representation of General Linear Mappings 6.62. Find the matrix representation of each of the following linear maps relative to the usual basis for Rn: (a) F:R3!R2deﬁned by Fðx;y;zÞ¼ð 2x/C04yþ9z;5xþ3y/C02zÞ. (b) F:R2!R4deﬁned by Fðx;yÞ¼ð 3xþ4y;5x/C02y;xþ7y;4xÞ: (c) F:R4!Rdeﬁned by Fðx1;x2;x3;x4Þ¼2x1þx2/C07x3/C0x4. 6.63. LetG:R3!R2be deﬁned by Gðx;y;zÞ¼ð 2xþ3y/C0z;4x/C0yþ2zÞ. (a) Find the matrix Arepresenting Grelative to the bases S¼fð 1;1;0Þ;ð1;2;3Þ;ð1;3;5Þg and S0¼fð 1;2Þ;ð2;3Þg (b) For any v¼ða;b;cÞinR3, ﬁnd½v/C138Sand½GðvÞ/C138S0. (c) Verify that A½v/C138S¼½GðvÞ/C138S0. 6.64. LetH:R2!R2be deﬁned by Hðx;yÞ¼ð 2xþ7y;x/C03yÞand consider the following bases of R2: S¼fð 1;1Þ;ð1;2Þg and S0¼fð 1;4Þ;ð1;5Þg (a) Find the matrix Arepresenting Hrelative to the bases SandS0. (b) Find the matrix Brepresenting Hrelative to the bases S0andS.CHAPTER 6 Linear Mappings and Matrices 223 6.65. LetF:R3!R2be deﬁned by Fðx;y;zÞ¼ð 2xþy/C0z;3x/C02yþ4zÞ. (a) Find the matrix Arepresenting Frelative to the bases S¼fð 1;1;1Þ;ð1;1;0Þ;ð1;0;0Þg and S0¼ð1;3Þ;ð1;4Þg (b) Verify that, for any v¼ða;b;cÞinR3,A½v/C138S¼½FðvÞ/C138S0. 6.66. LetSandS0be bases of V,a n dl e t 1Vbe the identity mapping on V. Show that the matrix Arepresenting 1Vrelative to the bases SandS0is the inverse of the change-of-basis matrix Pfrom StoS0;t h a ti s , A¼P/C01. 6.67. Prove (a) Theorem 6.10, (b) Theorem 6.11, (c) Theorem 6.12, (d) Theorem 6.13. [ Hint: See the proofs of the analogous Theorems 6.1 (Problem 6.9), 6.2 (Problem 6.10), 6.3 (Problem 6.11), and 6.7 (Problem 6.26).] Miscellaneous Problems 6.68. Suppose F:V!Vis linear. A subspace WofVis said to be invariant under FifFðWÞ/C18W. Suppose Wis invariant under Fand dim W¼r. Show that Fhas a block triangular matrix representation M¼AB 0C/C20/C21 where Ais an r/C2rsubmatrix. 6.69. Suppose V¼UþW, and suppose UandVare each invariant under a linear operator F:V!V. Also, suppose dim U¼rand dim W¼S. Show that Fhas a block diagonal matrix representation M¼A0 0B/C20/C21 where AandBarer/C2rands/C2ssubmatrices. 6.70. Two linear operators FandGonVare said to be similar if there exists an invertible linear operator TonV such that G¼T/C01/C14F/C14T. Prove (a) FandGare similar if and only if, for any basis SofV,½F/C138Sand½G/C138Sare similar matrices. (b) If Fis diagonalizable (similar to a diagonal matrix), then any similar matrix Gis also diagonalizable. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M¼½R1;R2;. . ./C138represents a matrix Mwith rows R1;R2;...: 6.37. (a) A¼½4;5;2;/C01/C138; (b) B¼½220;487 ;/C098;/C0217/C138; (c) P¼½1;2;4;9/C138; (d)½v/C138S¼½9a/C02b;/C04aþb/C138Tand½FðvÞ/C138S¼½32aþ47b;/C014a/C021b/C138T 6.38. (a) B¼½/C0 6;/C028;4;15/C138; (b)½v/C138S¼½4a/C0b;/C03 2aþ1 2b/C138Tand½AðvÞ/C138S¼½18a/C08b;1 2ð/C013aþ7bÞ/C138 6.39. (a)½ﬃﬃﬃ 2p ;/C0ﬃﬃﬃ 2p ;ﬃﬃﬃ 2p ;ﬃﬃﬃ 2p /C138; (b)½0;1;1;0/C138; (c)½3;7;5;/C02/C138; (d)½1;2;18;/C011/C138 6.40. (a)½1;0;0;0;1;0;0;0;0/C138; (b)½0;0;1;0;1;1;1;1;1/C138; (c)½2;/C07;/C04;3;1;4;6;/C08;1/C138 6.41. (a)½1;3;5;0;/C05;/C010;0;3;6/C138; (b)½0;1;2;/C01;2;3;1;0;0/C138; (c)½15;65;104 ;/C049;/C0219;/C0351 ;29;130;208/C138 6.42. (a)½1;1;2;1;3;2;1;5;2/C138; (b)½0;2;14;22;0;/C05;/C08/C138 6.43. (a)½1;0;0;0;2;1;0;0;2/C138; (b)½0;1;0;0;0;0;0;0;/C03;0;0;3;0/C138; (c)½5;1;0;0;5;2;0;0;5/C138224 CHAPTER 6 Linear Mappings and Matrices 6.44. (a) A¼½0;/C01;1;0/C138; (b) A2þI¼0 6.45. (a)½a;0;b;0;0;a;0;b;c;0;d;0;0;c;0;d/C138; (b)½a;c;0;0;b;d;0;0;0;0;a;c;0;0;b;d/C138; (c)½0;/C0c;b;0;/C0b;a/C0d;0;b;c;0;d/C0a;/C0c;0;c;/C0b;0/C138 6.47. (a)½1;3;2;5/C138;½/C05;3;2;/C01/C138;½v/C138¼½/C0 5aþ3b;2a/C0b/C138T; (b)½1;3;/C03;/C08/C138;½/C08;/C03;3;1/C138;½v/C138¼½/C0 8a/C03b;3aþb/C138T; (c)½2;3;5;7/C138;½/C07;3;5;/C02/C138;½v/C138¼½/C0 7aþ3b;5a/C02b/C138T; (d)½2;4;3;5/C138;½/C05 2;2;3 2;/C01/C138;½v/C138¼½/C05 2aþ2b;3 2a/C0b/C138T 6.48. (a) P¼½3;5;/C01;/C02/C138; (b) Q¼½2;5;/C01;/C03/C138 6.49. Here K¼ﬃﬃﬃ 3p : (a)1 2ðK;1Þ;1 2ð/C01;KÞ; ðbÞP¼1 2½K;/C01;1;K/C138; ðcÞ1 2½Kþ3;3K/C01/C138T;1 2½2K/C05;/C05K/C02/C138T;1 2½aKþb;bK/C0a/C138T 6.50. Pis the matrix whose columns are u1;u2;u3;Q¼P/C01;½v/C138¼Q½a;b;c/C138T: (a) Q¼½1;0;0;1;/C01;1;/C02;2;/C01/C138;½v/C138¼½a;a/C0bþc;/C02aþ2b/C0c/C138T; (b) Q¼½0;/C02;1;2;3;/C02;/C01;/C01;1/C138;½v/C138¼½/C0 2bþc;2aþ3b/C02c;/C0a/C0bþc/C138T; (c) Q¼½/C0 2;2;/C01;/C07;4;/C01;5;/C03;1/C138;½v/C138¼½/C0 2aþ2b/C0c;/C07aþ4b/C0c;5a/C03bþc/C138T 6.52. (a)½/C023;/C039;15;26/C138; (b)½35;41;/C027;/C032/C138; (c)½3;5;/C01;/C02/C138; (d) B¼P/C01AP 6.53. (a)½28;47;/C015;/C025/C138; (b)½13;18;/C015 2;/C010/C138 6.54. (a)½43;60;/C033;/C046/C138; (b)1 2½3;7;/C05;/C09/C138 6.55.½10;8;20;13;11;28;/C05;/C04;/C010/C138 6.56. (a)½/C034;57;/C019;32/C138; (b) trðBÞ¼trðAÞ¼/C0 2; (c) detðBÞ¼detðAÞ¼/C0 5 6.57. (a) trðFÞ¼6;detðFÞ¼23; (b) trðGÞ¼aþd;detðGÞ¼ad/C0bc 6.58. (a) trðFÞ¼/C0 2;detðFÞ¼13; (b) trðGÞ¼0;detðGÞ¼22 6.59. (a) A¼½3;1;/C02;4/C138;B¼½8;11;/C02;/C01/C138; (b) P¼½1;2;1;3/C138 6.62. (a)½2;/C04;9;5;3;/C02/C138; (b)½3;5;1;4;4;/C02;7;0/C138; (c)½2;1;/C07;/C01/C138 6.63. (a)½/C09;1;4;7;2;1/C138; (b)½v/C138S¼½/C0 aþ2b/C0c;5a/C05bþ2c;/C03aþ3b/C0c/C138T, and ½GðvÞ/C138S0¼½2a/C011bþ7c;7b/C04c/C138T 6.64. (a) A¼½47;85;/C038;/C069/C138; (b) B¼½71;88;/C041;/C051/C138 6.65. A¼½3;11;5;/C01;/C08;/C03/C138CHAPTER 6 Linear Mappings and Matrices 225 Inner Product Spaces, Orthogonality 7.1 Introduction The deﬁnition of a vector space Vinvolves an arbitrary ﬁeld K. Here we ﬁrst restrict Kto be the real ﬁeld R, in which case Vis called a real vector space ; in the last sections of this chapter, we extend our results to the case where Kis the complex ﬁeld C, in which case Vis called a complex vector space . Also, we adopt the previous notation that u;v;w are vectors in V a;b;c;k are scalars in K Furthermore, the vector spaces Vin this chapter have ﬁnite dimension unless otherwise stated or implied. Recall that the concepts of ‘‘length’’ and ‘‘orthogonality’’ did not appear in the investigation of arbitrary vector spaces V(although they did appear in Section 1.4 on the spaces RnandCn). Here we place an additional structure on a vector space Vto obtain an inner product space, and in this context these concepts are deﬁned. 7.2 Inner Product Spaces We begin with a deﬁnition. DEFINITION: LetVbe a real vector space. Suppose to each pair of vectors u;v2Vthere is assigned a real number, denoted by hu;vi. This function is called a ( real)inner product onVif it satisﬁes the following axioms: ½I1/C138(Linear Property ):hau1þbu2;vi¼ahu1;viþbhu2;vi. ½I2/C138(Symmetric Property ):hu;vi¼h v;ui. ½I3/C138(Positive Deﬁnite Property ):hu;ui/C210.; andhu;ui¼0 if and only if u¼0. The vector space Vwith an inner product is called a ( real)inner product space . Axiom½I1/C138states that an inner product function is linear in the ﬁrst position. Using ½I1/C138and the symmetry axiom½I2/C138, we obtain hu;cv1þdv2i¼h cv1þdv2;ui¼chv1;uiþdhv2;ui¼chu;v1iþdhu;v2i That is, the inner product function is also linear in its second position. Combining these two properties and using induction yields the following general formula: /C28P iaiui;P jbjvj/C29 ¼P iP jaibjhui;vji CHAPTER 7 226 That is, an inner product of linear combinations of vectors is equal to a linear combination of the inner products of the vectors. EXAMPLE 7.1 LetVbe a real inner product space. Then, by linearity, h3u1/C04u2;2v1/C05v2þ6v3i¼6hu1;v1i/C015hu1;v2iþ18hu1;v3i /C08hu2;v1iþ20hu2;v2i/C024hu2;v3i h2u/C05v;4uþ6vi¼8hu;uiþ12hu;vi/C020hv;ui/C030hv;vi ¼8hu;ui/C08hv;ui/C030hv;vi Observe that in the last equation we have used the symmetry property that hu;vi¼h v;ui. Remark: Axiom½I1/C138by itself implies h0;0i¼h 0v;0i¼0hv;0i¼0:Thus,½I1/C138,½I2/C138,½I3/C138are equivalent to½I1/C138,½I2/C138, and the following axiom: ½I0 3/C138If u6¼0;thenhu;uiis positive : That is, a function satisfying ½I1/C138,½I2/C138,½I0 3/C138is an inner product. Norm of a Vector By the third axiom ½I3/C138of an inner product, hu;uiis nonnegative for any vector u. Thus, its positive square root exists. We use the notation kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hu;uip This nonnegative number is called the norm orlength ofu. The relationkuk2¼hu;uiwill be used frequently. Remark: Ifkuk¼1 or, equivalently, if hu;ui¼1, then uis called a unit vector and it is said to be normalized . Every nonzero vector vinVcan be multiplied by the reciprocal of its length to obtain the unit vector ^v¼1 kvkv which is a positive multiple of v. This process is called normalizing v. 7.3 Examples of Inner Product Spaces This section lists the main examples of inner product spaces used in this text. Euclidean n-Space Rn Consider the vector space Rn. The dot product orscalar product inRnis deﬁned by u/C1v¼a1b1þa2b2þ/C1/C1/C1þ anbn where u¼ðaiÞand v¼ðbiÞ. This function deﬁnes an inner product on Rn. The normkukof the vector u¼ðaiÞin this space is as follows: kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃu/C1up¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2 1þa2 2þ/C1/C1/C1þ a2nq On the other hand, by the Pythagorean theorem, the distance from the origin O in R3to a point Pða;b;cÞis given byﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2þb2þc2p . This is precisely the same as the above-deﬁned norm of the vector v¼ða;b;cÞinR3. Because the Pythagorean theorem is a consequence of the axioms ofCHAPTER 7 Inner Product Spaces, Orthogonality 227 Euclidean geometry, the vector space Rnwith the above inner product and norm is called Euclidean n-space . Although there are many ways to deﬁne an inner product on Rn, we shall assume this inner product unless otherwise stated or implied. It is called the usual (orstandard )inner product onRn. Remark: Frequently the vectors in Rnwill be represented by column vectors—that is, by n/C21 column matrices. In such a case, the formula hu;vi¼uTv deﬁnes the usual inner product on Rn. EXAMPLE 7.2 Let u¼ð1;3;/C04;2Þ,v¼ð4;/C02;2;1Þ,w¼ð5;/C01;/C02;6ÞinR4. (a) Showh3u/C02v;wi¼3hu;wi/C02hv;wi: By deﬁnition, hu;wi¼5/C03þ8þ12¼22 andhv;wi¼20þ2/C04þ6¼24 Note that 3 u/C02v¼ð/C0 5;13;/C016;4Þ. Thus, h3u/C02v;wi¼/C0 25/C013þ32þ24¼18 As expected, 3hu;wi/C02hv;wi¼3ð22Þ/C02ð24Þ¼18¼h3u/C02v;wi. (b) Normalize uand v: By deﬁnition, kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ9þ16þ4p ¼ﬃﬃﬃﬃﬃ 30p andkvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16þ4þ4þ1p ¼5 We normalize uand vto obtain the following unit vectors in the directions of uand v, respectively: ^u¼1 kuku¼1ﬃﬃﬃﬃﬃ 30p ;3ﬃﬃﬃﬃﬃ 30p ;/C04ﬃﬃﬃﬃﬃ 30p ;2ﬃﬃﬃﬃﬃ 30p/C18/C19 and ^v¼1 kvkv¼4 5;/C02 5;2 5;1 5/C18/C19 Function Space C½a;b/C138and Polynomial Space P ðtÞ The notation C½a;b/C138is used to denote the vector space of all continuous functions on the closed interval ½a;b/C138—that is, where a/C20t/C20b. The following deﬁnes an inner product on C½a;b/C138, where fðtÞandgðtÞ are functions in C½a;b/C138: hf;gi¼ðb afðtÞgðtÞdt It is called the usual inner product onC½a;b/C138. The vector space PðtÞof all polynomials is a subspace of C½a;b/C138for any interval½a;b/C138, and hence, the above is also an inner product on PðtÞ. EXAMPLE 7.3 Consider fðtÞ¼3t/C05 and gðtÞ¼t2in the polynomial space PðtÞwith inner product hf;gi¼ð1 0fðtÞgðtÞdt: (a) Findhf;gi. We have fðtÞgðtÞ¼3t3/C05t2. Hence, hf;gi¼ð1 0ð3t3/C05t2Þdt¼3 4t4/C053t3/C12/C12/C12/C121 0¼3 4/C053¼/C01112228 CHAPTER 7 Inner Product Spaces, Orthogonality (b) Findkfkandkgk. We have½fðtÞ/C1382¼fðtÞfðtÞ¼9t2/C030tþ25 and½gðtÞ/C1382¼t4. Then kfk2¼hf;fi¼ð1 0ð9t2/C030tþ25Þdt¼3t3/C015t2þ25t/C12/C12/C12/C121 0¼13 kgk2¼hg;gi¼ð1 0t4dt¼1 5t5/C12/C12/C12/C121 0¼1 5 Therefore,kfk¼ﬃﬃﬃﬃﬃ 13p andkgk¼ﬃﬃ 1 5q ¼1 5ﬃﬃﬃ 5p . Matrix Space M ¼Mm;n LetM¼Mm;n, the vector space of all real m/C2nmatrices. An inner product is deﬁned on Mby hA;Bi¼trðBTAÞ where, as usual, tr ðÞis the trace—the sum of the diagonal elements. If A¼½aij/C138andB¼½bij/C138, then hA;Bi¼trðBTAÞ¼Pm i¼1Pn j¼1aijbij andkAk2¼hA;Ai¼Pm i¼1Pn j¼1a2 ij That is,hA;Biis the sum of the products of the corresponding entries in AandBand, in particular,hA;Ai is the sum of the squares of the entries of A. Hilbert Space LetVbe the vector space of all inﬁnite sequences of real numbers ða1;a2;a3;...Þsatisfying P1 i¼1a2 i¼a2 1þa2 2þ/C1/C1/C1 <1 that is, the sum converges. Addition and scalar multiplication are deﬁned in Vcomponentwise; that is, if u¼ða1;a2;...Þ and v¼ðb1;b2;...Þ then uþv¼ða1þb1;a2þb2;...Þ and ku¼ðka1;ka2;...Þ An inner product is deﬁned in vby hu;vi¼a1b1þa2b2þ/C1/C1/C1 The above sum converges absolutely for any pair of points in V. Hence, the inner product is well deﬁned. This inner product space is called l2-space orHilbert space . 7.4 Cauchy–Schwarz Inequality, Applications The following formula (proved in Problem 7.8) is called the Cauchy–Schwarz inequality or Schwarz inequality. It is used in many branches of mathematics. THEOREM 7.1: (Cauchy–Schwarz) For any vectors uand vin an inner product space V, hu;vi2/C20hu;uihv;vi orjhu;vij/C20k ukkvk Next we examine this inequality in speciﬁc cases. EXAMPLE 7.4 (a) Consider any real numbers a1;...;an,b1;...;bn. Then, by the Cauchy–Schwarz inequality, ða1b1þa2b2þ/C1/C1/C1þ anbnÞ2/C20ða2 1þ/C1/C1/C1þ a2 nÞðb2 1þ/C1/C1/C1þ b2 nÞ That is,ðu/C1vÞ2/C20kuk2kvk2, where u¼ðaiÞand v¼ðbiÞ.CHAPTER 7 Inner Product Spaces, Orthogonality 229 (b) Let fandgbe continuous functions on the unit interval ½0;1/C138. Then, by the Cauchy–Schwarz inequality, ð1 0fðtÞgðtÞdt/C20/C21 2 /C20ð1 0f2ðtÞdtð1 0g2ðtÞdt That is,ðhf;giÞ2/C20kfk2kvk2. Here Vis the inner product space C½0;1/C138. The next theorem (proved in Problem 7.9) gives the basic properties of a norm. The proof of the third property requires the Cauchy–Schwarz inequality. THEOREM 7.2: LetVbe an inner product space. Then the norm in Vsatisﬁes the following properties: ½N1/C138k vk/C210; andkvk¼0 if and only if v¼0. ½N2/C138kkvk¼j kjkvk. ½N3/C138kuþvk/C20k ukþk vk. The property½N3/C138is called the triangle inequality , because if we view uþvas the side of the triangle formed with sides uand v(as shown in Fig. 7-1), then ½N3/C138states that the length of one side of a triangle cannot be greater than the sum of the lengths of the other two sides. Angle Between Vectors For any nonzero vectors uand vin an inner product space V, the angle between u and vis deﬁned to be the angle ysuch that 0/C20y/C20pand cosy¼hu;vi kukkvk By the Cauchy–Schwartz inequality, /C01/C20cosy/C201, and so the angle exists and is unique. EXAMPLE 7.5 (a) Consider vectors u¼ð2;3;5Þand v¼ð1;/C04;3ÞinR3. Then hu;vi¼2/C012þ15¼5;kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4þ9þ25p ¼ﬃﬃﬃﬃﬃ 38p ;kvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ16þ9p ¼ﬃﬃﬃﬃﬃ 26p Then the angle ybetween uand vis given by cosy¼5ﬃﬃﬃﬃﬃ 38pﬃﬃﬃﬃﬃ 26p Note that yis an acute angle, because cos yis positive. (b) Let fðtÞ¼3t/C05 and gðtÞ¼t2in the polynomial space PðtÞwith inner product hf;gi¼Ð1 0fðtÞgðtÞdt.B y Example 7.3, hf;gi¼/C011 12;kfk¼ﬃﬃﬃﬃﬃ 13p ;kgk¼1 5ﬃﬃﬃ 5p Then the ‘‘angle’’ ybetween fandgis given by cosy¼/C011 12 ðﬃﬃﬃﬃﬃ 13p Þ1 5ﬃﬃﬃ 5p/C0/C1¼/C055 12ﬃﬃﬃﬃﬃ 13pﬃﬃﬃ 5p Note that yis an obtuse angle, because cos yis negative. Figure 7-1230 CHAPTER 7 Inner Product Spaces, Orthogonality 7.5 Orthogonality LetVbe an inner product space. The vectors u;v2Vare said to be orthogonal anduis said to be orthogonal tovif hu;vi¼0 The relation is clearly symmetric—if uis orthogonal to v, thenhv;ui¼0, and so vis orthogonal to u.W e note that 02Vis orthogonal to every v2V, because h0;vi¼h 0v;vi¼0hv;vi¼0 Conversely, if uis orthogonal to every v2V, thenhu;ui¼0 and hence u¼0b y½I3/C138:Observe that uand vare orthogonal if and only if cos y¼0, where yis the angle between uand v. Also, this is true if and only if uand vare ‘‘perpendicular’’—that is, y¼p=2 (or y¼90/C14). EXAMPLE 7.6 (a) Consider the vectors u¼ð1;1;1Þ,v¼ð1;2;/C03Þ,w¼ð1;/C04;3ÞinR3. Then hu;vi¼1þ2/C03¼0;hu;wi¼1/C04þ3¼0;hv;wi¼1/C08/C09¼/C016 Thus, uis orthogonal to vandw, but vandware not orthogonal. (b) Consider the functions sin tand cos tin the vector space C½/C0p;p/C138of continuous functions on the closed interval ½/C0p;p/C138. Then hsint;costi¼ðp /C0psintcostd t¼1 2sin2tjp /C0p¼0/C00¼0 Thus, sin tand cos tare orthogonal functions in the vector space C½/C0p;p/C138. Remark: A vector w¼ðx1;x2;...;xnÞis orthogonal to u¼ða1;a2;...;anÞin Rnif hu;wi¼a1x1þa2x2þ/C1/C1/C1þ anxn¼0 That is, wis orthogonal to uifwsatisﬁes a homogeneous equation whose coefﬁcients are the elements ofu. EXAMPLE 7.7 Find a nonzero vector wthat is orthogonal to u1¼ð1;2;1Þandu2¼ð2;5;4Þin R3. Letw¼ðx;y;zÞ. Then we wanthu1;wi¼0 andhu2;wi¼0. This yields the homogeneous system xþ2yþz¼0 2xþ5yþ4z¼0orxþ2yþz¼0 yþ2z¼0 Here zis the only free variable in the echelon system. Set z¼1 to obtain y¼/C02 and x¼3. Thus, w¼ð3;/C02;1Þis a desired nonzero vector orthogonal to u1andu2. Any multiple of wwill also be orthogonal to u1andu2. Normalizing w, we obtain the following unit vector orthogonal to u1andu2: ^w¼w kwk¼3ﬃﬃﬃﬃﬃ 14p ;/C02ﬃﬃﬃﬃﬃ 14p ;1ﬃﬃﬃﬃﬃ 14p/C18/C19 Orthogonal Complements LetSbe a subset of an inner product space V. The orthogonal complement of S, denoted by S?(read ‘‘ S perp’’) consists of those vectors in Vthat are orthogonal to every vector u2S; that is, S?¼fv2V:hv;ui¼0 for every u2SgCHAPTER 7 Inner Product Spaces, Orthogonality 231 In particular, for a given vector uinV, we have u?¼fv2V:hv;ui¼0g that is, u?consists of all vectors in Vthat are orthogonal to the given vector u. We show that S?is a subspace of V. Clearly 02S?, because 0 is orthogonal to every vector in V. Now suppose v,w2S?. Then, for any scalars aandband any vector u2S, we have havþbw;ui¼ahv;uiþbhw;ui¼a/C10þb/C10¼0 Thus, avþbw2S?, and therefore S?is a subspace of V. We state this result formally. PROPOSITION 7.3: LetSbe a subset of a vector space V. Then S?is a subspace of V. Remark 1: Suppose uis a nonzero vector in R3. Then there is a geometrical description of u?. Speciﬁcally, u?is the plane in R3through the origin Oand perpendicular to the vector u. This is shown in Fig. 7-2. Remark 2: LetWbe the solution space of an m/C2nhomogeneous system AX¼0, where A¼½aij/C138 andX¼½xi/C138. Recall that Wmay be viewed as the kernel of the linear mapping A:Rn!Rm. Now we can give another interpretation of Wusing the notion of orthogonality. Speciﬁcally, each solution vector w¼ðx1;x2;...;xnÞis orthogonal to each row of A; hence, Wis the orthogonal complement of the row space of A. EXAMPLE 7.8 Find a basis for the subspace u?ofR3, where u¼ð1;3;/C04Þ. Note that u?consists of all vectors w¼ðx;y;zÞsuch thathu;wi¼0, or xþ3y/C04z¼0. The free variables areyandz. (1) Set y¼1,z¼0 to obtain the solution w1¼ð/C0 3;1;0Þ. (2) Set y¼0,z¼1 to obtain the solution w1¼ð4;0;1Þ. The vectors w1andw2form a basis for the solution space of the equation, and hence a basis for u?. Suppose Wis a subspace of V. Then both WandW?are subspaces of V. The next theorem, whose proof (Problem 7.28) requires results of later sections, is a basic result in linear algebra. THEOREM 7.4: LetWbe a subspace of V. Then Vis the direct sum of Wand W?; that is, V¼W/C8W?. Figure 7-2232 CHAPTER 7 Inner Product Spaces, Orthogonality 7.6 Orthogonal Sets and Bases Consider a set S¼fu1;u2;...;urgof nonzero vectors in an inner product space V.Sis called orthogonal if each pair of vectors in Sare orthogonal, and Sis called orthonormal ifSis orthogonal and each vector inShas unit length. That is, (i)Orthogonal:hui;uji¼0 for i6¼j (ii)Orthonormal:hui;uji¼0 for i6¼j 1 for i¼j/C26 Normalizing an orthogonal set Srefers to the process of multiplying each vector in Sby the reciprocal of its length in order to transform Sinto an orthonormal set of vectors. The following theorems apply. THEOREM 7.5: Suppose Sis an orthogonal set of nonzero vectors. Then Sis linearly independent. THEOREM 7.6: (Pythagoras) Suppose fu1;u2;...;urgis an orthogonal set of vectors. Then ku1þu2þ/C1/C1/C1þ urk2¼ku1k2þku2k2þ/C1/C1/C1þk urk2 These theorems are proved in Problems 7.15 and 7.16, respectively. Here we prove the Pythagorean theorem in the special and familiar case for two vectors. Speciﬁcally, suppose hu;vi¼0. Then kuþvk2¼huþv;uþvi¼h u;uiþ2hu;viþh v;vi¼h u;uiþh v;vi¼k uk2þkvk2 which gives our result. EXAMPLE 7.9 (a) Let E¼fe1;e2;e3g¼fð 1;0;0Þ;ð0;1;0Þ;ð0;0;1Þgbe the usual basis of Euclidean space R3. It is clear that he1;e2i¼h e1;e3i¼h e2;e3i¼0 andhe1;e1i¼h e2;e2i¼h e3;e3i¼1 Namely, Eis an orthonormal basis of R3. More generally, the usual basis of Rnis orthonormal for every n. (b) Let V¼C½/C0p;p/C138be the vector space of continuous functions on the interval /C0p/C20t/C20pwith inner product deﬁned byhf;gi¼Ðp /C0pfðtÞgðtÞdt. Then the following is a classical example of an orthogonal set in V: f1;cost;cos 2 t;cos 3 t;...;sint;sin 2t;sin 3t;...g This orthogonal set plays a fundamental role in the theory of Fourier series. Orthogonal Basis and Linear Combinations, Fourier Coefﬁcients LetSconsist of the following three vectors in R3: u1¼ð1;2;1Þ; u2¼ð2;1;/C04Þ; u3¼ð3;/C02;1Þ The reader can verify that the vectors are orthogonal; hence, they are linearly independent. Thus, Sis an orthogonal basis of R3. Suppose we want to write v¼ð7;1;9Þas a linear combination of u1;u2;u3. First we set vas a linear combination of u1;u2;u3using unknowns x1;x2;x3as follows: v¼x1u1þx2u2þx3u3 orð7;1;9Þ¼x1ð1;2;1Þþx2ð2;1;/C04Þþx3ð3;/C02;1Þð *Þ We can proceed in two ways. METHOD 1: Expandð*Þ(as in Chapter 3) to obtain the system x1þ2x2þ3x3¼7; 2x1þx2/C02x3¼1; x1/C04x2þx3¼7 Solve the system by Gaussian elimination to obtain x1¼3,x2¼/C01,x3¼2. Thus, v¼3u1/C0u2þ2u3.CHAPTER 7 Inner Product Spaces, Orthogonality 233 METHOD 2: (This method uses the fact that the basis vectors are orthogonal, and the arithmetic is much simpler.) If we take the inner product of each side of ð*Þwith respect to ui, we get hv;uii¼h x1u2þx2u2þx3u3;uii orhv;uii¼xihui;uii or xi¼hv;uii hui;uii Here two terms drop out, because u1;u2;u3are orthogonal. Accordingly, x1¼hv;u1i hu1;u1i¼7þ2þ9 1þ4þ1¼18 6¼3; x2¼hv;u2i hu2;u2i¼14þ1/C036 4þ1þ16¼/C021 21¼/C01 x3¼hv;u3i hu3;u3i¼21/C02þ9 9þ4þ1¼28 14¼2 Thus, again, we get v¼3u1/C0u2þ2u3. The procedure in Method 2 is true in general. Namely, we have the following theorem (proved in Problem 7.17). THEOREM 7.7: Letfu1;u2;...;ungbe an orthogonal basis of V. Then, for any v2V, v¼hv;u1i hu1;u1iu1þhv;u2i hu2;u2iu2þ/C1/C1/C1þhv;uni hun;uniun Remark: The scalar ki/C17hv;uii hui;uiiis called the Fourier coefﬁcient ofvwith respect to ui, because it is analogous to a coefﬁcient in the Fourier series of a function. This scalar also has a geometric interpretation, which is discussed below. Projections LetVbe an inner product space. Suppose wis a given nonzero vector in V, and suppose vis another vector. We seek the ‘‘projection of valong w,’’ which, as indicated in Fig. 7-3(a), will be the multiple cw ofwsuch that v0¼v/C0cwis orthogonal to w. This means hv/C0cw;wi¼0o rhv;wi/C0chw;wi¼0o r c¼hv;wi hw;wi Accordingly, the projection of valong w is denoted and deﬁned by projðv;wÞ¼cw¼hv;wi hw;wiw Such a scalar cis unique, and it is called the Fourier coefﬁcient ofvwith respect to wor the component of valong w. The above notion is generalized as follows (see Problem 7.25). Figure 7-3 234 CHAPTER 7 Inner Product Spaces, Orthogonality THEOREM 7.8: Suppose w1;w2;...;wrform an orthogonal set of nonzero vectors in V. Let vbe any vector in V. Deﬁne v0¼v/C0ðc1w1þc2w2þ/C1/C1/C1þ crwrÞ where c1¼hv;w1i hw1;w1i; c2¼hv;w2i hw2;w2i; ...; cr¼hv;wri hwr;wri Then v0is orthogonal to w1;w2;...;wr. Note that each ciin the above theorem is the component (Fourier coefﬁcient) of valong the given wi. Remark: The notion of the projection of a vector v2Valong a subspace WofVis deﬁned as follows. By Theorem 7.4, V¼W/C8W?. Hence, vmay be expressed uniquely in the form v¼wþw0; where w2W and w02W? We deﬁne wto be the projection of valong W , and denote it by proj ðv;WÞ, as pictured in Fig. 7-2(b). In particular, if W¼spanðw1;w2;...;wrÞ, where the wiform an orthogonal set, then projðv;WÞ¼c1w1þc2w2þ/C1/C1/C1þ crwr Here ciis the component of valong wi, as above. 7.7 Gram–Schmidt Orthogonalization Process Supposefv1;v2;...;vngis a basis of an inner product space V. One can use this basis to construct an orthogonal basisfw1;w2;...;wngofVas follows. Set w1¼v1 w2¼v2/C0hv2;w1i hw1;w1iw1 w3¼v3/C0hv3;w1i hw1;w1iw1/C0hv3;w2i hw2;w2iw2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: wn¼vn/C0hvn;w1i hw1;w1iw1/C0hvn;w2i hw2;w2iw2/C0/C1/C1/C1/C0hvn;wn/C01i hwn/C01;wn/C01iwn/C01 In other words, for k¼2;3;...;n, we deﬁne wk¼vk/C0ck1w1/C0ck2w2/C0/C1/C1/C1/C0 ck;k/C01wk/C01 where cki¼hvk;wii=hwi;wiiis the component of vkalong wi. By Theorem 7.8, each wkis orthogonal to the preceeding w’s. Thus, w1;w2;...;wnform an orthogonal basis for Vas claimed. Normalizing each wi will then yield an orthonormal basis for V. The above construction is known as the Gram–Schmidt orthogonalization process . The following remarks are in order. Remark 1: Each vector wkis a linear combination of vkand the preceding w’s. Hence, one can easily show, by induction, that each wkis a linear combination of v1;v2;...;vn. Remark 2: Because taking multiples of vectors does not affect orthogonality, it may be simpler in hand calculations to clear fractions in any new wk, by multiplying wkby an appropriate scalar, before obtaining the next wkþ1.CHAPTER 7 Inner Product Spaces, Orthogonality 235 Remark 3: Suppose u1;u2;...;urare linearly independent, and so they form a basis for U¼spanðuiÞ. Applying the Gram–Schmidt orthogonalization process to the u’s yields an orthogonal basis for U. The following theorems (proved in Problems 7.26 and 7.27) use the above algorithm and remarks. THEOREM 7.9: Letfv1;v2;...;vngbe any basis of an inner product space V. Then there exists an orthonormal basis fu1;u2;...;ungofVsuch that the change-of-basis matrix from fvigtofuigis triangular; that is, for k¼1;...;n, uk¼ak1v1þak2v2þ/C1/C1/C1þ akkvk THEOREM 7.10: Suppose S¼fw1;w2;...;wrgis an orthogonal basis for a subspace Wof a vector space V. Then one may extend Sto an orthogonal basis for V; that is, one may ﬁnd vectors wrþ1;...;wnsuch thatfw1;w2;...;wngis an orthogonal basis for V. EXAMPLE 7.10 Apply the Gram–Schmidt orthogonalization process to ﬁnd an orthogonal basis and then an orthonormal basis for the subspace UofR4spanned by v1¼ð1;1;1;1Þ; v2¼ð1;2;4;5Þ; v3¼ð1;/C03;/C04;/C02Þ (1) First set w1¼v1¼ð1;1;1;1Þ. (2) Compute v2/C0hv2;w1i hw1;w1iw1¼v2/C012 4w1¼ð/C0 2;/C01;1;2Þ Setw2¼ð/C0 2;/C01;1;2Þ. (3) Compute v3/C0hv3;w1i hw1;w1iw1/C0hv3;w2i hw2;w2iw2¼v3/C0ð/C08Þ 4w1/C0ð/C07Þ 10w2¼8 5;/C017 10;/C013 10;75/C0/C1 Clear fractions to obtain w3¼ð/C0 6;/C017;/C013;14Þ. Thus, w1;w2;w3form an orthogonal basis for U. Normalize these vectors to obtain an orthonormal basis fu1;u2;u3gofU. We havekw1k2¼4,kw2k2¼10,kw3k2¼910, so u1¼1 2ð1;1;1;1Þ; u2¼1ﬃﬃﬃﬃﬃ 10pð/C02;/C01;1;2Þ; u3¼1ﬃﬃﬃﬃﬃﬃﬃﬃ 910pð16;/C017;/C013;14Þ EXAMPLE 7.11 Let Vbe the vector space of polynomials fðtÞwith inner product hf;gi¼Ð1 /C01fðtÞgðtÞdt. Apply the Gram–Schmidt orthogonalization process to f1;t;t2;t3gto ﬁnd an orthogonal basisff0;f1;f2;f3gwith integer coefﬁcients for P3ðtÞ. Here we use the fact that, for rþs¼n, htr;tsi¼ð1 /C01tndt¼tnþ1 nþ1/C12/C12/C12/C121 /C01¼2=ðnþ1Þwhen nis even 0 when nis odd/C26 (1) First set f0¼1. (2) Compute t¼ht;1i h1;1ið1Þ¼t/C00¼t. Set f1¼t. (3) Compute t2/C0ht2;1i h1;1ið1Þ/C0ht2;ti ht;tiðtÞ¼t2/C02 3 2ð1Þþ0ðtÞ¼t2/C01 3 Multiply by 3 to obtain f2¼3t2¼1.236 CHAPTER 7 Inner Product Spaces, Orthogonality (4) Compute t3/C0ht3;1i h1;1ið1Þ/C0ht3;ti ht;tiðtÞ/C0ht3;3t2/C01i h3t2/C01;3t2/C01ið3t2/C01Þ ¼t3/C00ð1Þ/C02 5 23ðtÞ/C00ð3t2/C01Þ¼t3/C03 5t Multiply by 5 to obtain f3¼5t3/C03t. Thus,f1;t;3t2/C01;5t3/C03tgis the required orthogonal basis. Remark: Normalizing the polynomials in Example 7.11 so that pð1Þ¼1 yields the polynomials 1;t;1 2ð3t2/C01Þ;1 2ð5t3/C03tÞ These are the ﬁrst four Legendre polynomials , which appear in the study of differential equations. 7.8 Orthogonal and Positive Deﬁnite Matrices This section discusses two types of matrices that are closely related to real inner product spaces V. Here vectors in Rnwill be represented by column vectors. Thus, hu;vi¼uTvdenotes the inner product in Euclidean space Rn. Orthogonal Matrices A real matrix Pisorthogonal ifPis nonsingular and P/C01¼PT, or, in other words, if PPT¼PTP¼I. First we recall (Theorem 2.6) an important characterization of such matrices. THEOREM 7.11: LetPbe a real matrix. Then the following are equivalent: (a) Pis orthogonal; (b) the rows of Pform an orthonormal set; (c) the columns of Pform an orthonormal set. (This theorem is true only using the usual inner product on Rn. It is not true if Rnis given any other inner product.) EXAMPLE 7.12 (a) Let P¼1=ﬃﬃﬃ 3p 1=ﬃﬃﬃ 3p 1=ﬃﬃﬃ 3p 01 =ﬃﬃﬃ 2p 1=ﬃﬃﬃ 2p 2=ﬃﬃﬃ 6p /C01=ﬃﬃﬃ 6p /C01=ﬃﬃﬃ 6p2 43 5:The rows of Pare orthogonal to each other and are unit vectors. Thus Pis an orthogonal matrix. (b) Let Pbe a 2/C22 orthogonal matrix. Then, for some real number y, we have P¼cosysiny /C0sinycosy/C20/C21 or P¼cosy siny siny/C0cosy/C20/C21 The following two theorems (proved in Problems 7.37 and 7.38) show important relationships between orthogonal matrices and orthonormal bases of a real inner product space V. THEOREM 7.12: Suppose E¼feigandE0¼fe0 igare orthonormal bases of V. Let Pbe the change- of-basis matrix from the basis Eto the basis E0. Then Pis orthogonal. THEOREM 7.13: Letfe1;...;engbe an orthonormal basis of an inner product space V. Let P¼½aij/C138 be an orthogonal matrix. Then the following nvectors form an orthonormal basis forV: e0 i¼a1ie1þa2ie2þ/C1/C1/C1þ anien; i¼1;2;...;nCHAPTER 7 Inner Product Spaces, Orthogonality 237 Positive Deﬁnite Matrices LetAbe a real symmetric matrix; that is, AT¼A. Then Ais said to be positive deﬁnite if, for every nonzero vector uinRn, hu;Aui¼uTAu>0 Algorithms to decide whether or not a matrix Ais positive deﬁnite will be given in Chapter 12. However, for 2/C22 matrices, we have simple criteria that we state formally in the following theorem (proved in Problem 7.43). THEOREM 7.14: A2/C22real symmetric matrix A¼ab cd/C20/C21 ¼ab bd/C20/C21 is positive deﬁnite if and only if the diagonal entries aand dare positive and the determinant jAj¼ad/C0bc¼ad/C0b2is positive. EXAMPLE 7.13 Consider the following symmetric matrices: A¼13 34/C20/C21 ; B¼1/C02 /C02/C03/C20/C21 ; C¼1/C02 /C025/C20/C21 Ais not positive deﬁnite, because jAj¼4/C09¼/C05 is negative. Bis not positive deﬁnite, because the diagonal entry/C03 is negative. However, Cis positive deﬁnite, because the diagonal entries 1 and 5 are positive, and the determinantjCj¼5/C04¼1 is also positive. The following theorem (proved in Problem 7.44) holds. THEOREM 7.15: LetAbe a real positive deﬁnite matrix. Then the function hu;vi¼uTAvis an inner product on Rn. Matrix Representation of an Inner Product (Optional) Theorem 7.15 says that every positive deﬁnite matrix Adetermines an inner product on Rn. This subsection may be viewed as giving the converse of this result. LetVbe a real inner product space with basis S¼fu1;u2;...;ung. The matrix A¼½aij/C138; where aij¼hui;uji is called the matrix representation of the inner product on V relative to the basis S . Observe that Ais symmetric, because the inner product is symmetric; that is, hui;uji¼h uj;uii. Also, A depends on both the inner product on Vand the basis SforV. Moreover, if Sis an orthogonal basis, then Ais diagonal, and if Sis an orthonormal basis, then Ais the identity matrix. EXAMPLE 7.14 The vectors u1¼ð1;1;0Þ,u2¼ð1;2;3Þ,u3¼ð1;3;5Þform a basis Sfor Euclidean space R3. Find the matrix Athat represents the inner product in R3relative to this basis S. First compute each hui;ujito obtain hu1;u1i¼1þ1þ0¼2; hu2;u2i¼1þ4þ9¼14;hu1;u2i¼1þ2þ0¼3; hu2;u3i¼1þ6þ15¼22;hu1;u3i¼1þ3þ0¼4 hu3;u3i¼1þ9þ25¼35 Then A¼234 31 42 2 42 23 52 43 5. As expected, Ais symmetric. The following theorems (proved in Problems 7.45 and 7.46, respectively) hold. THEOREM 7.16: LetAbe the matrix representation of an inner product relative to basis SforV. Then, for any vectors u;v2V, we have hu;vi¼½ u/C138TA½v/C138 where½u/C138and½v/C138denote the (column) coordinate vectors relative to the basis S.238 CHAPTER 7 Inner Product Spaces, Orthogonality THEOREM 7.17: LetAbe the matrix representation of any inner product on V. Then Ais a positive deﬁnite matrix. 7.9 Complex Inner Product Spaces This section considers vector spaces over the complex ﬁeld C. First we recall some properties of the complex numbers (Section 1.7), especially the relations between a complex number z¼aþbi;where a;b2R;and its complex conjugate /C22z¼a/C0bi: z/C22z¼a2þb2;jzj¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ a2þb2p ; z1þz2¼z1þz2 z1z2¼z1z2; z/C22/C22¼z Also, zis real if and only if /C22z¼z. The following deﬁnition applies. DEFINITION: LetVbe a vector space over C. Suppose to each pair of vectors, u;v2Vthere is assigned a complex number, denoted by hu;vi. This function is called a ( complex )inner product onVif it satisﬁes the following axioms: ½I1*/C138(Linear Property )hau1þbu2;vi¼ahu1;viþbhu2;vi ½I2*/C138(Conjugate Symmetric Property )hu;vi¼hv;ui ½I3*/C138(Positive Deﬁnite Property )hu;ui/C210; andhu;ui¼0 if and only if u¼0. The vector space Vover Cwith an inner product is called a ( complex )inner product space . Observe that a complex inner product differs from the real case only in the second axiom ½I2*/C138: Axiom½I1*/C138(Linear Property) is equivalent to the two conditions: ðaÞhu1þu2;vi¼h u1;viþh u2;vi;ðbÞhku;vi¼khu;vi On the other hand, applying ½I1*/C138and½I2*/C138, we obtain hu;kvi¼hkv;ui¼khv;ui¼ /C22khv;ui¼ /C22khu;vi That is, we must take the conjugate of a complex number when it is taken out of the second position of a complex inner product. In fact (Problem 7.47), the inner product is conjugate linear in the second position; that is, hu;av1þbv2i¼ /C22ahu;v1iþ /C22bhu;v2i Combining linear in the ﬁrst position and conjugate linear in the second position, we obtain, by induction, P iaiui;P jbjvj*+ ¼P i;jaibjhui;vji The following remarks are in order. Remark 1: Axiom½I1*/C138by itself implies that h0;0i¼h 0v;0i¼0hv;0i¼0. Accordingly,½I1*/C138,½I2*/C138, and½I3*/C138are equivalent to½I1*/C138,½I2*/C138, and the following axiom: ½I3*0/C138Ifu6¼0;thenhu;ui>0: That is, a function satisfying ½I1/C138,½I2*/C138, and½I3*0/C138is a (complex) inner product on V. Remark 2: By½I2*/C138;hu;ui¼hu;ui. Thus,hu;uimust be real. By½I3*/C138;hu;uimust be nonnegative, and hence, its positive real square root exists. As with real inner product spaces, we deﬁne kuk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hu;uip to be the norm or length of u. Remark 3: In addition to the norm, we deﬁne the notions of orthogonality, orthogonal comple- ment, and orthogonal and orthonormal sets as before. In fact, the deﬁnitions of distance and Fouriercoefﬁcient and projections are the same as in the real case.CHAPTER 7 Inner Product Spaces, Orthogonality 239 EXAMPLE 7.15 (Complex Euclidean Space Cn). Let V¼Cn, and let u¼ðziÞandv¼ðwiÞbe vectors in Cn. Then hu;vi¼P kzkwk¼z1w1þz2w2þ/C1/C1/C1þ znwn is an inner product on V, called the usual orstandard inner product onCn.Vwith this inner product is called Complex Euclidean Space. We assume this inner product on Cnunless otherwise stated or implied. Assuming uand vare column vectors, the above inner product may be deﬁned by hu;vi¼uT/C22v where, as with matrices, /C22vmeans the conjugate of each element of v.I fuand vare real, we have wi¼wi. In this case, the inner product reduced to the analogous one on Rn. EXAMPLE 7.16 (a) Let Vbe the vector space of complex continuous functions on the (real) interval a/C20t/C20b. Then the following is the usual inner product onV: hf;gi¼ðb afðtÞgðtÞdt (b) Let Ube the vector space of m/C2nmatrices over C. Suppose A¼ðzijÞandB¼ðwijÞare elements of U. Then the following is the usual inner product on U: hA;Bi¼trðBHAÞ¼Pm i¼1Pn j¼1/C22wijzij As usual, BH¼/C22BT; that is, BHis the conjugate transpose of B. The following is a list of theorems for complex inner product spaces that are analogous to those for the real case. Here a Hermitian matrix A(i.e., one where AH¼/C22AT¼AÞplays the same role that a symmetric matrix A(i.e., one where AT¼A) plays in the real case. (Theorem 7.18 is proved in Problem 7.50.) THEOREM 7.18: (Cauchy–Schwarz) Let Vbe a complex inner product space. Then jhu;vij/C20k ukkvk THEOREM 7.19: LetWbe a subspace of a complex inner product space V. Then V¼W/C8W?. THEOREM 7.20: Supposefu1;u2;...;ungis a basis for a complex inner product space V. Then, for any v2V, v¼hv;u1i hu1;u1iu1þhv;u2i hu2;u2iu2þ/C1/C1/C1þhv;uni hun;uniun THEOREM 7.21: Supposefu1;u2;...;ungis a basis for a complex inner product space V. Let A¼½aij/C138be the complex matrix deﬁned by aij¼hui;uji. Then, for any u;v2V, hu;vi¼½ u/C138TA½v/C138 where½u/C138and½v/C138are the coordinate column vectors in the given basis fuig. (Remark : This matrix Ais said to represent the inner product on V.) THEOREM 7.22: LetAbe a Hermitian matrix (i.e., AH¼/C22AT¼AÞsuch that XTA/C22Xis real and positive for every nonzero vector X2Cn. Thenhu;vi¼uTA/C22vis an inner product onCn. THEOREM 7.23: LetAbe the matrix that represents an inner product on V. Then Ais Hermitian, and XTAXis real and positive for any nonzero vector in Cn.240 CHAPTER 7 Inner Product Spaces, Orthogonality 7.10 Normed Vector Spaces (Optional) We begin with a deﬁnition. DEFINITION: LetVbe a real or complex vector space. Suppose to each v2Vthere is assigned a real number, denoted by kvk. This functionk/C1k is called a norm onVif it satisﬁes the following axioms: ½N1/C138k vk/C210; andkvk¼0 if and only if v¼0. ½N2/C138kkvk¼j kjkvk. ½N3/C138kuþvk/C20k ukþk vk. A vector space Vwith a norm is called a normed vector space . Suppose Vis a normed vector space. The distance between two vectors uand vinVis denoted and deﬁned by dðu;vÞ¼k u/C0vk The following theorem (proved in Problem 7.56) is the main reason why dðu;vÞis called the distance between uand v. THEOREM 7.24: LetVbe a normed vector space. Then the function dðu;vÞ¼k u/C0vksatisﬁes the following three axioms of a metric space: ½M1/C138dðu;vÞ/C210; and dðu;vÞ¼0 if and only if u¼v. ½M2/C138dðu;vÞ¼dðv;uÞ. ½M3/C138dðu;vÞ/C20dðu;wÞþdðw;vÞ. Normed Vector Spaces and Inner Product Spaces Suppose Vis an inner product space. Recall that the norm of a vector vinVis deﬁned by kvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv;vip One can prove (Theorem 7.2) that this norm satisﬁes ½N1/C138,½N2/C138, and½N3/C138. Thus, every inner product space Vis a normed vector space. On the other hand, there may be norms on a vector space Vthat do not come from an inner product on V, as shown below. Norms on Rnand Cn The following deﬁne three important norms on RnandCn: kða1;...;anÞk1¼maxðjaijÞ kða1;...;anÞk1¼ja1jþja2jþ/C1/C1/C1þj anj kða1;...;anÞk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ja1j2þja2j2þ/C1/C1/C1þj anj2q (Note that subscripts are used to distinguish between the three norms.) The norms k/C1k1,k/C1k1, andk/C1k2 are called the inﬁnity-norm ,one-norm , and two-norm , respectively. Observe that k/C1k2is the norm on Rn (respectively, Cn) induced by the usual inner product on Rn(respectively, Cn). We will let d1,d1,d2 denote the corresponding distance functions. EXAMPLE 7.17 Consider vectors u¼ð1;/C05;3Þand v¼ð4;2;/C03ÞinR3. (a) The inﬁnity norm chooses the maximum of the absolute values of the components. Hence, kuk1¼5 andkvk1¼4CHAPTER 7 Inner Product Spaces, Orthogonality 241 (b) The one-norm adds the absolute values of the components. Thus, kuk1¼1þ5þ3¼9 andkvk1¼4þ2þ3¼9 (c) The two-norm is equal to the square root of the sum of the squares of the components (i.e., the norm induced by the usual inner product on R3). Thus, kuk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ25þ9p ¼ﬃﬃﬃﬃﬃ 35p andkvk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 16þ4þ9p ¼ﬃﬃﬃﬃﬃ 29p (d) Because u/C0v¼ð1/C04;/C05/C02;3þ3Þ¼ð/C0 3;/C07;6Þ, we have d1ðu;vÞ¼7; d1ðu;vÞ¼3þ7þ6¼16; d2ðu;vÞ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9þ49þ36p ¼ﬃﬃﬃﬃﬃ 94p EXAMPLE 7.18 Consider the Cartesian plane R2shown in Fig. 7-4. (a) Let D1be the set of points u¼ðx;yÞinR2such thatkuk2¼1. Then D1consists of the points ðx;yÞsuch that kuk2 2¼x2þy2¼1. Thus, D1is the unit circle, as shown in Fig. 7-4. (b) Let D2be the set of points u¼ðx;yÞinR2such thatkuk1¼1. Then D1consists of the points ðx;yÞsuch that kuk1¼jxjþjyj¼1. Thus, D2is the diamond inside the unit circle, as shown in Fig. 7-4. (c) Let D3be the set of points u¼ðx;yÞinR2such thatkuk1¼1. Then D3consists of the points ðx;yÞsuch that kuk1¼maxðjxj,jyjÞ¼ 1. Thus, D3is the square circumscribing the unit circle, as shown in Fig. 7-4. Norms on C½a;b/C138 Consider the vector space V¼C½a;b/C138of real continuous functions on the interval a/C20t/C20b. Recall that the following deﬁnes an inner product on V: hf;gi¼ðb afðtÞgðtÞdt Accordingly, the above inner product deﬁnes the following norm on V¼C½a;b/C138(which is analogous to thek/C1k2norm on Rn): kfk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃðb a½fðtÞ/C1382dts Figure 7-4242 CHAPTER 7 Inner Product Spaces, Orthogonality The following deﬁne the other norms on V¼C½a;b/C138: kfk1¼ðb ajfðtÞjdt andkfk1¼maxðjfðtÞjÞ There are geometrical descriptions of these two norms and their corresponding distance functions, which are described below. The ﬁrst norm is pictured in Fig. 7-5. Here kfk1¼area between the function jfjand the t-axis d1ðf;gÞ¼area between the functions fandg This norm is analogous to the norm k/C1k1onRn. The second norm is pictured in Fig. 7-6. Here kfk1¼maximum distance between fand the t-axis d1ðf;gÞ¼maximum distance between fandg This norm is analogous to the norms k/C1k1onRn. SOLVED PROBLEMS Inner Products 7.1. Expand: (a)h5u1þ8u2;6v1/C07v2i, (b)h3uþ5v;4u/C06vi, (c)k2u/C03vk2 Use linearity in both positions and, when possible, symmetry, hu;vi¼h v;ui.Figure 7-5 Figure 7-6 CHAPTER 7 Inner Product Spaces, Orthogonality 243 (a) Take the inner product of each term on the left with each term on the right: h5u1þ8u2;6v1/C07v2i¼h 5u1;6v1iþh 5u1;/C07v2iþh 8u2;6v1iþh 8u2;/C07v2i ¼30hu1;v1i/C035hu1;v2iþ48hu2;v1i/C056hu2;v2i [Remark: Observe the similarity between the above expansion and the expansion (5 a–8b)(6c–7d)i n ordinary algebra.] (b)h3uþ5v;4u/C06vi¼12hu;ui/C018hu;viþ20hv;ui/C030hv;vi ¼12hu;uiþ2hu;vi/C030hv;vi (c)k2u/C03vk2¼h2u/C03v;2u/C03vi¼4hu;ui/C06hu;vi/C06hv;uiþ9hv;vi ¼4kuk2/C012ðu;vÞþ9kvk2 7.2. Consider vectors u¼ð1;2;4Þ;v¼ð2;/C03;5Þ;w¼ð4;2;/C03ÞinR3. Find (a) u/C1v, (b) u/C1w;(c) v/C1w, (d)ðuþvÞ/C1w, (e)kuk,( f )kvk. (a) Multiply corresponding components and add to get u/C1v¼2/C06þ20¼16: (b)u/C1w¼4þ4/C012¼/C04. (c) v/C1w¼8/C06/C015¼/C013. (d) First ﬁnd uþv¼ð3;/C01;9Þ. ThenðuþvÞ/C1w¼12/C02/C027¼/C017. Alternatively, using ½I1/C138, ðuþvÞ/C1w¼u/C1wþv/C1w¼/C04/C013¼/C017. (e) First ﬁndkuk2by squaring the components of uand adding: kuk2¼12þ22þ42¼1þ4þ16¼21; and sokuk¼ﬃﬃﬃﬃﬃ 21p (f )kvk2¼4þ9þ25¼38, and sokvk¼ﬃﬃﬃﬃﬃ 38p . 7.3. Verify that the following deﬁnes an inner product in R2: hu;vi¼x1y1/C0x1y2/C0x2y1þ3x2y2; where u¼ðx1;x2Þ;v¼ðy1;y2Þ We argue via matrices. We can write hu;viin matrix notation as follows: hu;vi¼uTAv¼½x1;x2/C1381/C01 /C013/C20/C21 y1 y2/C20/C21 Because Ais real and symmetric, we need only show that Ais positive deﬁnite. The diagonal elements 1 and 3 are positive, and the determinant kAk¼3/C01¼2 is positive. Thus, by Theorem 7.14, Ais positive deﬁnite. Accordingly, by Theorem 7.15, hu;viis an inner product. 7.4. Consider the vectors u¼ð1;5Þand v¼ð3;4ÞinR2. Find (a)hu;viwith respect to the usual inner product in R2. (b)hu;viwith respect to the inner product in R2in Problem 7.3. (c)kvkusing the usual inner product in R2. (d)kvkusing the inner product in R2in Problem 7.3. (a)hu;vi¼3þ20¼23. (b)hu;vi¼1/C13/C01/C14/C05/C13þ3/C15/C14¼3/C04/C015þ60¼44. (c)kvk2¼hv;vi¼hð 3;4Þ;ð3;4Þi¼ 9þ16¼25; hence,jvk¼5. (d)kvk2¼hv;vi¼hð 3;4Þ;ð3;4Þi¼ 9/C012/C012þ48¼33; hence,kvk¼ﬃﬃﬃﬃﬃ 33p . 7.5. Consider the following polynomials in PðtÞwith the inner product hf;gi¼Ð1 0fðtÞgðtÞdt: fðtÞ¼tþ2; gðtÞ¼3t/C02; hðtÞ¼t2/C02t/C03 (a) Findhf;giandhf;hi. (b) Findkfkandkgk. (c) Normalize fandg.244 CHAPTER 7 Inner Product Spaces, Orthogonality (a) Integrate as follows: hf;gi¼ð1 0ðtþ2Þð3t/C02Þdt¼ð1 0ð3t2þ4t/C04Þdt¼/C18 t3þ2t2/C04t/C19/C12/C12/C12/C121 0¼/C01 hf;hi¼ð1 0ðtþ2Þðt2/C02t/C03Þdt¼t4 4/C07t2 2/C06t/C18/C19 /C12/C12/C12/C121 0¼/C037 4 (b)hf;fi¼Ð1 0ðtþ2Þðtþ2Þdt¼19 3; hence,kfk¼ﬃﬃﬃﬃ 19 3q ¼1 3ﬃﬃﬃﬃﬃ 57p hg;gi¼ð1 0ð3t/C02Þð3t/C02Þ¼1; hence ;kgk¼ﬃﬃﬃ 1p ¼1 (c) Becausekfk¼1 3ﬃﬃﬃﬃﬃ 57p andgis already a unit vector, we have ^f¼1 kfkf¼3ﬃﬃﬃﬃﬃ 57pðtþ2Þ and ^g¼g¼3t/C02 7.6. Find cos ywhere yis the angle between: (a)u¼ð1;3;/C05;4Þand v¼ð2;/C03;4;1ÞinR4, (b)A¼987 654/C20/C21 andB¼123 456/C20/C21 , wherehA;Bi¼trðBTAÞ: Use cos y¼hu;vi kukkvk (a) Compute: hu;vi¼2/C09/C020þ4¼/C023;kuk2¼1þ9þ25þ16¼51;kvk2¼4þ9þ16þ1¼30 Thus ; cosy¼/C023ﬃﬃﬃﬃﬃ 51pﬃﬃﬃﬃﬃ 30p¼/C023 3ﬃﬃﬃﬃﬃﬃﬃﬃ 170p (b) UsehA;Bi¼trðBTAÞ¼Pm i¼1Pn j¼1aijbij, the sum of the products of corresponding entries. hA;Bi¼9þ16þ21þ24þ25þ24¼119 UsekAk2¼hA;Ai¼Pm i¼1Pn j¼1a2 ij;the sum of the squares of all the elements of A. kAk2¼hA;Ai¼92þ82þ72þ62þ52þ42¼271; kBk2¼hB;Bi¼12þ22þ32þ42þ52þ62¼91;and so and sokAk¼ﬃﬃﬃﬃﬃﬃﬃﬃ 271p kBk¼ﬃﬃﬃﬃﬃ 91p Thus ; cosy¼119ﬃﬃﬃﬃﬃﬃﬃﬃ 271pﬃﬃﬃﬃﬃ 91p 7.7. Verify each of the following: (a) Parallelogram Law (Fig. 7-7): kuþvk2þku/C0vk2¼2kuk2þ2kvk2. (b) Polar form for hu;vi(which shows the inner product can be obtained from the norm function): hu;vi¼1 4ðkuþvk2/C0ku/C0vk2Þ: Expand as follows to obtain kuþvk2¼huþv;uþvi¼k uk2þ2hu;viþk vk2ð1Þ ku/C0vk2¼hu/C0v;u/C0vi¼k uk2/C02hu;viþk vk2ð2Þ Add (1) and (2) to get the Parallelogram Law (a). Subtract (2) from (1) to obtain kuþvk2/C0ku/C0vk2¼4hu;vi Divide by 4 to obtain the (real) polar form (b).CHAPTER 7 Inner Product Spaces, Orthogonality 245 7.8. Prove Theorem 7.1 (Cauchy–Schwarz): For uand vin a real inner product space V; hu;ui2/C20hu;uihv;viorjhu;vij/C20k ukkvk: For any real number t, htuþv;tuþvi¼t2hu;uiþ2thu;viþh v;vi¼t2kuk2þ2thu;viþk vk2 Leta¼kuk2,b¼2hu;vÞ,c¼kvk2. Becausektuþvk2/C210, we have at2þbtþc/C210 for every value of t. This means that the quadratic polynomial cannot have two real roots, which implies that b2/C04ac/C200o r b2/C204ac. Thus, 4hu;vi2/C204kuk2kvk2 Dividing by 4 gives our result. 7.9. Prove Theorem 7.2: The norm in an inner product space Vsatisﬁes (a)½N1/C138kvk/C210; andkvk¼0 if and only if v¼0. (b)½N2/C138kkvk¼j kjkvk. (c)½N3/C138kuþvk/C20k ukþk vk. (a) If v6¼0, thenhv;vi>0, and hence,kvk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv;vip >0. If v¼0, thenh0;0i¼0. Consequently, k0k¼ﬃﬃﬃ 0p ¼0. Thus,½N1/C138is true. (b) We havekkvk2¼hkv;kvi¼k2hv;vi¼k2kvk2. Taking the square root of both sides gives ½N2/C138. (c) Using the Cauchy–Schwarz inequality, we obtain kuþvk2¼huþv;uþvi¼h u;uiþh u;viþh u;viþh v;vi /C20kuk2þ2kukkvkþk vk2¼ðk ukþk vkÞ2 Taking the square root of both sides yields ½N3/C138. Orthogonality, Orthonormal Complements, Orthogonal Sets 7.10. Find kso that u¼ð1;2;k;3Þand v¼ð3;k;7;/C05ÞinR4are orthogonal. First ﬁnd hu;vi¼ð 1;2;k;3Þ/C1ð3;k;7;/C05Þ¼3þ2kþ7k/C015¼9k/C012 Then sethu;vi¼9k/C012¼0 to obtain k¼4 3. 7.11. LetWbe the subspace of R5spanned by u¼ð1;2;3;/C01;2Þand v¼ð2;4;7;2;/C01Þ. Find a basis of the orthogonal complement W?ofW. We seek all vectors w¼ðx;y;z;s;tÞsuch that hw;ui¼ xþ2yþ3z/C0sþ2t¼0 hw;vi¼2xþ4yþ7zþ2s/C0t¼0 Eliminating xfrom the second equation, we ﬁnd the equivalent system xþ2yþ3z/C0sþ2t¼0 zþ4s/C05t¼0 Figure 7-7246 CHAPTER 7 Inner Product Spaces, Orthogonality The free variables are y;s, and t. Therefore, (1) Set y¼/C01,s¼0,t¼0 to obtain the solution w1¼ð2;/C01;0;0;0Þ. (2) Set y¼0,s¼1,t¼0 to ﬁnd the solution w2¼ð13;0;/C04;1;0Þ. (3) Set y¼0,s¼0,t¼1 to obtain the solution w3¼ð/C0 17;0;5;0;1Þ. The setfw1;w2;w3gis a basis of W?. 7.12. Letw¼ð1;2;3;1Þbe a vector in R4. Find an orthogonal basis for w?. Find a nonzero solution of xþ2yþ3zþt¼0, say v1¼ð0;0;1;/C03Þ. Now ﬁnd a nonzero solution of the system xþ2yþ3zþt¼0; z/C03t¼0 sayv2¼ð0;/C05;3;1Þ. Last, ﬁnd a nonzero solution of the system xþ2yþ3zþt¼0;/C05yþ3zþt¼0; z/C03t¼0 sayv3¼ð/C0 14;2;3;1Þ. Thus, v1,v2,v3form an orthogonal basis for w?. 7.13. Let S consist of the following vectors in R4: u1¼ð1;1;0;/C01Þ;u2¼ð1;2;1;3Þ;u3¼ð1;1;/C09;2Þ;u4¼ð16;/C013;1;3Þ (a) Show that Sis orthogonal and a basis of R4. (b) Find the coordinates of an arbitrary vector v¼ða;b;c;dÞinR4relative to the basis S. (a) Compute u1/C1u2¼1þ2þ0/C03¼0; u2/C1u3¼1þ2/C09þ6¼0;u1/C1u3¼1þ1þ0/C02¼0; u2/C1u4¼16/C026þ1þ9¼0;u1/C1u4¼16/C013þ0/C03¼0 u3/C1u4¼16/C013/C09þ6¼0 Thus, Sis orthogonal, and Sis linearly independent. Accordingly, Sis a basis for R4because any four linearly independent vectors form a basis of R4. (b) Because Sis orthogonal, we need only ﬁnd the Fourier coefﬁcients of vwith respect to the basis vectors, as in Theorem 7.7. Thus, k1¼hv;u1i hu1;u1i¼aþb/C0d 3; k2¼hv;u2i hu2;u2i¼aþ2bþcþ3d 15;k3¼hv;u3i hu3;u3i¼aþb/C09cþ2d 87 k4¼hv;u4i hu4;u4i¼16a/C013bþcþ3d 435 are the coordinates of vwith respect to the basis S. 7.14. Suppose S,S1,S2are the subsets of V. Prove the following: (a)S/C18S??. (b) If S1/C18S2, then S? 2/C18S? 1. (c)S?¼spanðSÞ?. (a) Let w2S. Thenhw;vi¼0 for every v2S?; hence, w2S??. Accordingly, S/C18S??. (b) Let w2S? 2. Thenhw;vi¼0 for every v2S2. Because S1/C18S2,hw;vi¼0 for every v¼S1. Thus, w2S? 1, and hence, S? 2/C18S? 1. (c) Because S/C18spanðSÞ, part (b) gives us span ðSÞ?/C18S?. Suppose u2S?and v2spanðSÞ. Then there exist w1;w2;...;wkinSsuch that v¼a1w1þa2w2þ/C1/C1/C1þ akwk. Then, using u2S?, we have hu;vi¼h u;a1w1þa2w2þ/C1/C1/C1þ akwki¼a1hu;w1iþa2hu;w2iþ/C1/C1/C1þ akhu;wki ¼a1ð0Þþa2ð0Þþ/C1/C1/C1þ akð0Þ¼0 Thus, u2spanðSÞ?. Accordingly, S?/C18spanðSÞ?. Both inclusions give S?¼spanðSÞ?. 7.15. Prove Theorem 7.5: Suppose Sis an orthogonal set of nonzero vectors. Then Sis linearly independent.CHAPTER 7 Inner Product Spaces, Orthogonality 247 Suppose S¼fu1;u2;...;urgand suppose a1u1þa2u2þ/C1/C1/C1þ arur¼0 ð1Þ Taking the inner product of (1) with u1, we get 0¼h0;u1i¼h a1u1þa2u2þ/C1/C1/C1þ arur;u1i ¼a1hu1;u1iþa2hu2;u1iþ/C1/C1/C1þ arhur;u1i ¼a1hu1;u1iþa2/C10þ/C1/C1/C1þ ar/C10¼a1hu1;u1i Because u16¼0, we havehu1;u1i6¼0. Thus, a1¼0. Similarly, for i¼2;...;r, taking the inner product of (1) with ui, 0¼h0;uii¼h a1u1þ/C1/C1/C1þ arur;uii ¼a1hu1;uiiþ/C1/C1/C1þ aihui;uiiþ/C1/C1/C1þ arhur;uii¼aihui;uii Buthui;uii6¼0, and hence, every ai¼0. Thus, Sis linearly independent. 7.16. Prove Theorem 7.6 (Pythagoras): Suppose fu1;u2;...;urgis an orthogonal set of vectors. Then ku1þu2þ/C1/C1/C1þ urk2¼ku1k2þku2k2þ/C1/C1/C1þk urk2 Expanding the inner product, we have ku1þu2þ/C1/C1/C1þ urk2¼hu1þu2þ/C1/C1/C1þ ur;u1þu2þ/C1/C1/C1þ uri ¼hu1;u1iþh u2;u2iþ/C1/C1/C1þh ur;uriþP i6¼jhui;uji The theorem follows from the fact that hui;uii¼k uik2andhui;uji¼0 for i6¼j. 7.17. Prove Theorem 7.7: Let fu1;u2;...;ungbe an orthogonal basis of V. Then for any v2V, v¼hv;u1i hu1;u1iu1þhv;u2i hu2;u2iu2þ/C1/C1/C1þhv;uni hun;uniun Suppose v¼k1u1þk2u2þ/C1/C1/C1þ knun. Taking the inner product of both sides with u1yields hv;u1i¼h k1u2þk2u2þ/C1/C1/C1þ knun;u1i ¼k1hu1;u1iþk2hu2;u1iþ/C1/C1/C1þ knhun;u1i ¼k1hu1;u1iþk2/C10þ/C1/C1/C1þ kn/C10¼k1hu1;u1i Thus, k1¼hv;u1i hu1;u1i. Similarly, for i¼2;...;n, hv;uii¼h k1uiþk2u2þ/C1/C1/C1þ knun;uii ¼k1hu1;uiiþk2hu2;uiiþ/C1/C1/C1þ knhun;uii ¼k1/C10þ/C1/C1/C1þ kihui;uiiþ/C1/C1/C1þ kn/C10¼kihui;uii Thus, ki¼hv;uii hu1;uii. Substituting for kiin the equation v¼k1u1þ/C1/C1/C1þ knun, we obtain the desired result. 7.18. Suppose E¼fe1;e2;...;engis an orthonormal basis of V. Prove (a) For any u2V, we have u¼hu;e1ie1þhu;e2ie2þ/C1/C1/C1þh u;enien. (b)ha1e1þ/C1/C1/C1þ anen;b1e1þ/C1/C1/C1þ bneni¼a1b1þa2b2þ/C1/C1/C1þ anbn. (c) For any u;v2V, we havehu;vi¼h u;e1ihv;e1iþ/C1/C1/C1þh u;enihv;eni. (a) Suppose u¼k1e1þk2e2þ/C1/C1/C1þ knen. Taking the inner product of uwith e1, hu;e1i¼h k1e1þk2e2þ/C1/C1/C1þ knen;e1i ¼k1he1;e1iþk2he2;e1iþ/C1/C1/C1þ knhen;e1i ¼k1ð1Þþk2ð0Þþ/C1/C1/C1þ knð0Þ¼k1248 CHAPTER 7 Inner Product Spaces, Orthogonality Similarly, for i¼2;...;n, hu;eii¼h k1e1þ/C1/C1/C1þ kieiþ/C1/C1/C1þ knen;eii ¼k1he1;eiiþ/C1/C1/C1þ kihei;eiiþ/C1/C1/C1þ knhen;eii ¼k1ð0Þþ/C1/C1/C1þ kið1Þþ/C1/C1/C1þ knð0Þ¼ki Substitutinghu;eiiforkiin the equation u¼k1e1þ/C1/C1/C1þ knen, we obtain the desired result. (b) We have Pn i¼1aiei;Pn j¼1bjej*+ ¼Pn i;j¼1aibjhei;eji¼Pn i¼1aibihei;eiiþP i6¼jaibjhei;eji Buthei;eji¼0 for i6¼j, andhei;eji¼1 for i¼j. Hence, as required, Pn i¼1aiei;Pn j¼1bjej*+ ¼Pn i¼1aibi¼a1b1þa2b2þ/C1/C1/C1þ anbn (c) By part (a), we have u¼hu;e1ie1þ/C1/C1/C1þh u;enien and v¼hv;e1ie1þ/C1/C1/C1þh v;enien Thus, by part (b), hu;vi¼h u;e1ihv;e1iþh u;e2ihv;e2iþ/C1/C1/C1þh u;enihv;eni Projections, Gram–Schmidt Algorithm, Applications 7.19. Suppose w6¼0. Let vbe any vector in V. Show that c¼hv;wi hw;wi¼hv;wi kwk2 is the unique scalar such that v0¼v/C0cwis orthogonal to w. In order for v0to be orthogonal to wwe must have hv/C0cw;wi¼0o rhv;wi/C0chw;wi¼0o rhv;wi¼chw;wi Thus, chv;wi hw;wi. Conversely, suppose c¼hv;wi hw;wi. Then hv/C0cw;wi¼h v;wi/C0chw;wi¼h v;wi/C0hv;wi hw;wihw;wi¼0 7.20. Find the Fourier coefﬁcient cand the projection of v¼ð1;/C02;3;/C04Þalong w¼ð1;2;1;2ÞinR4. Computehv;wi¼1/C04þ3/C08¼/C08 andkwk2¼1þ4þ1þ4¼10. Then c¼/C08 10¼/C045 and projðv;wÞ¼cw¼ð/C04 5;/C08 5;/C04 5;/C08 5Þ 7.21. Consider the subspace UofR4spanned by the vectors: v1¼ð1;1;1;1Þ; v2¼ð1;1;2;4Þ; v3¼ð1;2;/C04;/C03Þ Find (a) an orthogonal basis of U; (b) an orthonormal basis of U. (a) Use the Gram–Schmidt algorithm. Begin by setting w1¼u¼ð1;1;1;1Þ. Next ﬁnd v2/C0hv2;w1i hw1;w1iw1¼ð1;1;2;4Þ/C08 4ð1;1;1;1Þ¼ð/C0 1;/C01;0;2Þ Setw2¼ð/C0 1;/C01;0;2Þ. Then ﬁnd v3/C0hv3;w1i hw1;w1iw1/C0hv3;w2i hw2;w2iw2¼ð1;2;/C04;/C03Þ/C0ð/C04Þ 4ð1;1;1;1Þ/C0ð/C09Þ 6ð/C01;/C01;0;2Þ ¼ð1 2;32;/C03;1Þ Clear fractions to obtain w3¼ð1;3;/C06;2Þ. Then w1;w2;w3form an orthogonal basis of U.CHAPTER 7 Inner Product Spaces, Orthogonality 249 (b) Normalize the orthogonal basis consisting of w1;w2;w3. Becausekw1k2¼4,kw2k2¼6, and kw3k2¼50, the following vectors form an orthonormal basis of U: u1¼1 2ð1;1;1;1Þ; u2¼1ﬃﬃﬃ 6pð/C01;/C01;0;2Þ; u3¼1 5ﬃﬃﬃ 2pð1;3;/C06;2Þ 7.22. Consider the vector space PðtÞwith inner product hf;gi¼Ð1 0fðtÞgðtÞdt. Apply the Gram– Schmidt algorithm to the set f1;t;t2gto obtain an orthogonal set ff0;f1;f2gwith integer coefﬁcients. First set f0¼1. Then ﬁnd t/C0ht;1i h1;1i/C11¼t/C01 2 1/C11¼t/C01 2 Clear fractions to obtain f1¼2t/C01. Then ﬁnd t2/C0ht2;1i h1;1ið1Þ/C0ht2;2t/C01i h2t/C01;2t/C01ið2t/C01Þ¼t2/C01 3 1ð1Þ/C01 6 1 3ð2t/C01Þ¼t2/C0tþ1 6 Clear fractions to obtain f2¼6t2/C06tþ1. Thus,f1;2t/C01;6t2/C06tþ1gis the required orthogonal set. 7.23. Suppose v¼ð1;3;5;7Þ. Find the projection of vonto Wor, in other words, ﬁnd w2Wthat minimizeskv/C0wk, where Wis the subspance of R4spanned by (a)u1¼ð1;1;1;1Þandu2¼ð1;/C03;4;/C02Þ, (b) v1¼ð1;1;1;1Þand v2¼ð1;2;3;2Þ. (a) Because u1andu2are orthogonal, we need only compute the Fourier coefﬁcients: c1¼hv;u1i hu1;u1i¼1þ3þ5þ7 1þ1þ1þ1¼16 4¼4 c2¼hv;u2i hu2;u2i¼1/C09þ20/C014 1þ9þ16þ4¼/C02 30¼/C01 15 Then w¼projðv;WÞ¼c1u1þc2u2¼4ð1;1;1;1Þ/C01 15ð1;/C03;4;/C02Þ¼ð59 15;63 5;5615;6215Þ: (b) Because v1and v2are not orthogonal, ﬁrst apply the Gram–Schmidt algorithm to ﬁnd an orthogonal basis for W. Set w1¼v1¼ð1;1;1;1Þ. Then ﬁnd v2/C0hv2;w1i hw1;w1iw1¼ð1;2;3;2Þ/C08 4ð1;1;1;1Þ¼ð/C0 1;0;1;0Þ Setw2¼ð/C0 1;0;1;0Þ. Now compute c1¼hv;w1i hw1;w1i¼1þ3þ5þ7 1þ1þ1þ1¼16 4¼4 c2¼hv;w2i hw2;w2i/C0/C01þ0þ5þ0 1þ0þ1þ0¼/C06 2¼/C03 Then w¼projðv;WÞ¼c1w1þc2w2¼4ð1;1;1;1Þ/C03ð/C01;0;1;0Þ¼ð 7;4;1;4Þ. 7.24. Suppose w1andw2are nonzero orthogonal vectors. Let vbe any vector in V. Find c1andc2so that v0is orthogonal to w1andw2, where v0¼v/C0c1w1/C0c2w2. Ifv0is orthogonal to w1, then 0¼hv/C0c1w1/C0c2w2;w1i¼h v;w1i/C0c1hw1;w1i/C0c2hw2;w1i ¼hv;w1i/C0c1hw1;w1i/C0c20¼hv;w1i/C0c1hw1;w1i Thus, c1¼hv;w1i=hw1;w1i. (That is, c1is the component of valong w1.) Similarly, if v0is orthogonal to w2, then 0¼hv/C0c1w1/C0c2w2;w2i¼h v;w2i/C0c2hw2;w2i Thus, c2¼hv;w2i=hw2;w2i. (That is, c2is the component of valong w2.)250 CHAPTER 7 Inner Product Spaces, Orthogonality 7.25. Prove Theorem 7.8: Suppose w1;w2;...;wrform an orthogonal set of nonzero vectors in V. Let v2V. Deﬁne v0¼v/C0ðc1w1þc2w2þ/C1/C1/C1þ crwrÞ; where ci¼hv;wii hwi;wii Then v0is orthogonal to w1;w2;...;wr. Fori¼1;2;...;rand usinghwi;wji¼0 for i6¼j, we have hv/C0c1w1/C0c2x2/C0/C1/C1/C1/C0 crwr;wii¼h v;wii/C0c1hw1;wii/C0/C1/C1/C1/C0 cihwi;wii/C0/C1/C1/C1/C0 crhwr;wii ¼hv;wii/C0c1/C10/C0/C1/C1/C1/C0 cihwi;wii/C0/C1/C1/C1/C0 cr/C10 ¼hv;wii/C0cihwi;wii¼h v;wii/C0hv;wii hwi;wiihwi;wii¼0 The theorem is proved. 7.26. Prove Theorem 7.9: Let fv1;v2;...;vngbe any basis of an inner product space V. Then there exists an orthonormal basis fu1;u2;...;ungofVsuch that the change-of-basis matrix from fvigto fuigis triangular; that is, for k¼1;2;...;n, uk¼ak1v1þak2v2þ/C1/C1/C1þ akkvk The proof uses the Gram–Schmidt algorithm and Remarks 1 and 3 of Section 7.7. That is, apply the algorithm tofvigto obtain an orthogonal basis fwi;...;wng, and then normalize fwigto obtain an orthonormal basis fuigofV. The speciﬁc algorithm guarantees that each wkis a linear combination of v1;...;vk, and hence, each ukis a linear combination of v1;...;vk. 7.27. Prove Theorem 7.10: Suppose S¼fw1;w2;...;wrg, is an orthogonal basis for a subspace WofV. Then one may extend Sto an orthogonal basis for V; that is, one may ﬁnd vectors wrþ1;...;wr such thatfw1;w2;...;wngis an orthogonal basis for V. Extend Sto a basis S0¼fw1;...;wr;vrþ1;...;vngforV. Applying the Gram–Schmidt algorithm to S0, we ﬁrst obtain w1;w2;...;wrbecause Sis orthogonal, and then we obtain vectors wrþ1;...;wn, where fw1;w2;...;wngis an orthogonal basis for V. Thus, the theorem is proved. 7.28. Prove Theorem 7.4: Let Wbe a subspace of V. Then V¼W/C8W?. By Theorem 7.9, there exists an orthogonal basis fu1;...;urgofW, and by Theorem 7.10 we can extend it to an orthogonal basis fu1;u2;...;ungofV. Hence, urþ1;...;un2W?.I fv2V, then v¼a1u1þ/C1/C1/C1þ anun;where a1u1þ/C1/C1/C1þ arur2Wandarþ1urþ1þ/C1/C1/C1þ anun2W? Accordingly, V¼WþW?. On the other hand, if w2W\W?, thenhw;wi¼0. This yields w¼0. Hence, W\W?¼f0g. The two conditions V¼WþW?andW\W?¼f0ggive the desired result V¼W/C8W?. Remark: Note that we have proved the theorem for the case that Vhas ﬁnite dimension. We remark that the theorem also holds for spaces of arbitrary dimension. 7.29. Suppose Wis a subspace of a ﬁnite-dimensional space V. Prove that W¼W??. By Theorem 7.4, V¼W/C8W?, and also V¼W?/C8W??. Hence, dimW¼dimV/C0dimW?and dim W??¼dimV/C0dimW? This yields dim W¼dimW??. But W/C18W??(see Problem 7.14). Hence, W¼W??, as required. 7.30. Prove the following: Suppose w1;w2;...;wrform an orthogonal set of nonzero vectors in V.L e t vbe any vector in Vand let cibe the component of valong wi. Then, for any scalars a1;...;ar, we have v/C0Pr k¼1ckwk/C13/C13/C13/C13/C13/C13/C13/C13/C20v/C0P r k¼1akwk/C13/C13/C13/C13/C13/C13/C13/C13 That is,Pc iwiis the closest approximation to vas a linear combination of w1;...;wr.CHAPTER 7 Inner Product Spaces, Orthogonality 251 By Theorem 7.8, v/C0Pckwkis orthogonal to every wiand hence orthogonal to any linear combination ofw1;w2;...;wr. Therefore, using the Pythagorean theorem and summing from k¼1t o r, v/C0Pakwkkk2¼v/C0PckwkþPðck/C0akÞwk kk2¼v/C0Pckwkkk2þPðck/C0akÞwkkk2 /C21v/C0Pckwkkk2 The square root of both sides gives our theorem. 7.31. Supposefe1;e2;...;ergis an orthonormal set of vectors in V. Let vbe any vector in Vand let ci be the Fourier coefﬁcient of vwith respect to ui. Prove Bessel’s inequality: Pr k¼1c2 k/C20kvk2 Note that ci¼hv;eii, becausekeik¼1. Then, usinghei;eji¼0 for i6¼jand summing from k¼1t or, we get 0/C20v/C0Pckek;v/C0Pck;ek hi ¼hv;vi/C02v;PckekiþPc2 k¼hv;vi/C0P2ckhv;ekiþPc2 k/C10 ¼hv;vi/C0P2c2 kþPc2 k¼hv;vi/C0Pc2 k This gives us our inequality. Orthogonal Matrices 7.32. Find an orthogonal matrix Pwhose ﬁrst row is u1¼ð1 3;23;23Þ. First ﬁnd a nonzero vector w2¼ðx;y;zÞthat is orthogonal to u1—that is, for which 0¼hu1;w2i¼x 3þ2y 3þ2z 3¼0o r xþ2yþ2z¼0 One such solution is w2¼ð0;1;/C01Þ. Normalize w2to obtain the second row of P: u2¼ð0;1=ﬃﬃﬃ 2p ;/C01=ﬃﬃﬃ 2p Þ Next ﬁnd a nonzero vector w3¼ðx;y;zÞthat is orthogonal to both u1andu2—that is, for which 0¼hu1;w3i¼x 3þ2y 3þ2z 3¼0o r xþ2yþ2z¼0 0¼hu2;w3i¼yﬃﬃﬃ 2p/C0yﬃﬃﬃ 2p¼0o r y/C0z¼0 Setz¼/C01 and ﬁnd the solution w3¼ð4;/C01;/C01Þ. Normalize w3and obtain the third row of P; that is, u3¼ð4=ﬃﬃﬃﬃﬃ 18p ;/C01=ﬃﬃﬃﬃﬃ 18p ;/C01=ﬃﬃﬃﬃﬃ 18p Þ: P¼1 323 23 01 =ﬃﬃﬃ 2p /C01=ﬃﬃﬃ 2p 4=3ﬃﬃﬃ 2p /C01=3ﬃﬃﬃ 2p /C01=3ﬃﬃﬃ 2p2 43 5 Thus ; We emphasize that the above matrix Pis not unique. 7.33. LetA¼11/C01 134 7/C0522 43 5. Determine whether or not: (a) the rows of Aare orthogonal; (b)Ais an orthogonal matrix; (c) the columns of Aare orthogonal. (a) Yes, because ð1;1;/C01Þ/C1ð1;3;4Þ¼1þ3/C04¼0,ð1;1/C01Þ/C1ð7;/C05;2Þ¼7/C05/C02¼0, and ð1;3;4Þ/C1ð7;/C05;2Þ¼7/C015þ8¼0. (b) No, because the rows of Aare not unit vectors, for example, ð1;1;/C01Þ2¼1þ1þ1¼3. (c) No; for example, ð1;1;7Þ/C1ð1;3;/C05Þ¼1þ3/C035¼/C0316¼0. 7.34. LetBbe the matrix obtained by normalizing each row of Ain Problem 7.33. (a) Find B. (b) Is Ban orthogonal matrix? (c) Are the columns of Borthogonal?252 CHAPTER 7 Inner Product Spaces, Orthogonality (a) We have kð1;1;/C01Þk2¼1þ1þ1¼3;kð1;3;4Þk2¼1þ9þ16¼26 kð7;/C05;2Þk2¼49þ25þ4¼78 Thus ; B¼1=ﬃﬃﬃ 3p 1=ﬃﬃﬃ 3p /C01=ﬃﬃﬃ 3p 1=ﬃﬃﬃﬃﬃ 26p 3=ﬃﬃﬃﬃﬃ 26p 4=ﬃﬃﬃﬃﬃ 26p 7=ﬃﬃﬃﬃﬃ 78p /C05=ﬃﬃﬃﬃﬃ 78p 2=ﬃﬃﬃﬃﬃ 78p2 643 75 (b) Yes, because the rows of Bare still orthogonal and are now unit vectors. (c) Yes, because the rows of Bform an orthonormal set of vectors. Then, by Theorem 7.11, the columns of Bmust automatically form an orthonormal set. 7.35. Prove each of the following: (a)Pis orthogonal if and only if PTis orthogonal. (b) If Pis orthogonal, then P/C01is orthogonal. (c) If PandQare orthogonal, then PQis orthogonal. (a) We haveðPTÞT¼P. Thus, Pis orthogonal if and only if PPT¼Iif and only if PTTPT¼Iif and only if PTis orthogonal. (b) We have PT¼P/C01, because Pis orthogonal. Thus, by part (a), P/C01is orthogonal. (c) We have PT¼P/C01and QT¼Q/C01. Thus,ðPQÞðPQÞT¼PQQTPT¼PQQ/C01P/C01¼I. Therefore, ðPQÞT¼ðPQÞ/C01, and so PQis orthogonal. 7.36. Suppose Pis an orthogonal matrix. Show that (a)hPu;Pvi¼h u;vifor any u;v2V; (b)kPuk¼k ukfor every u2V. UsePTP¼Iandhu;vi¼uTv. (a)hPu;Pvi¼ð PuÞTðPvÞ¼uTPTPv¼uTv¼hu;vi. (b) We have kPuk2¼hPu;Pui¼uTPTPu¼uTu¼hu;ui¼k uk2 Taking the square root of both sides gives our result. 7.37. Prove Theorem 7.12: Suppose E¼feigandE0¼fe0 igare orthonormal bases of V. Let Pbe the change-of-basis matrix from EtoE0. Then Pis orthogonal. Suppose e0 i¼bi1e1þbi2e2þ/C1/C1/C1þ binen; i¼1;...;n ð1Þ Using Problem 7.18(b) and the fact that E0is orthonormal, we get dij¼he0 i;e0 ji¼bi1bj1þbi2bj2þ/C1/C1/C1þ binbjn ð2Þ LetB¼½bij/C138be the matrix of the coefﬁcients in (1). (Then P¼BT.) Suppose BBT¼½cij/C138. Then cij¼bi1bj1þbi2bj2þ/C1/C1/C1þ binbjn ð3Þ By (2) and (3), we have cij¼dij. Thus, BBT¼I. Accordingly, Bis orthogonal, and hence, P¼BTis orthogonal. 7.38. Prove Theorem 7.13: Let fe1;...;engbe an orthonormal basis of an inner product space V. Let P¼½aij/C138be an orthogonal matrix. Then the following nvectors form an orthonormal basis for V: e0 i¼a1ie1þa2ie2þ/C1/C1/C1þ anien; i¼1;2;...;nCHAPTER 7 Inner Product Spaces, Orthogonality 253 Becausefeigis orthonormal, we get, by Problem 7.18(b), he0 i;e0 ji¼a1ia1jþa2ia2jþ/C1/C1/C1þ anianj¼hCi;Cji where Cidenotes the ith column of the orthogonal matrix P¼½aij/C138:Because Pis orthogonal, its columns form an orthonormal set. This implies he0 i;e0 ji¼h Ci;Cji¼dij:Thus,fe0 igis an orthonormal basis. Inner Products And Positive Deﬁnite Matrices 7.39. Which of the following symmetric matrices are positive deﬁnite? (a) A¼34 45/C20/C21 , (b) B¼8/C03 /C032/C20/C21 , (c) C¼21 1/C03/C20/C21 , (d) D¼35 59/C20/C21 Use Theorem 7.14 that a 2 /C22 real symmetric matrix is positive deﬁnite if and only if its diagonal entries are positive and if its determinant is positive. (a) No, becausejAj¼15/C016¼/C01 is negative. (b) Yes.(c) No, because the diagonal entry /C03 is negative. (d) Yes. 7.40. Find the values of kthat make each of the following matrices positive deﬁnite: (a) A¼2/C04 /C04 k/C20/C21 , (b) B¼4k k9/C20/C21 , (c) C¼k 5 5/C02/C20/C21 (a) First, kmust be positive. Also, jAj¼2k/C016 must be positive; that is, 2 k/C016>0. Hence, k>8. (b) We needjBj¼36/C0k2positive; that is, 36 /C0k2>0. Hence, k2<36 or/C06<k<6. (c)Ccan never be positive deﬁnite, because Chas a negative diagonal entry /C02. 7.41. Find the matrix Athat represents the usual inner product on R2relative to each of the following bases of R2:ðaÞf v1¼ð1;3Þ;v2¼ð2;5Þg;ðbÞfw1¼ð1;2Þ;w2¼ð4;/C02Þg: (a) Computehv1;v1i¼1þ9¼10,hv1;v2i¼2þ15¼17,hv2;v2i¼4þ25¼29. Thus, A¼10 17 17 29/C20/C21. (b) Computehw1;w1i¼1þ4¼5,hw1;w2i¼4/C04¼0,hw2;w2i¼16þ4¼20. Thus, A¼50 02 0/C20/C21 . (Because the basis vectors are orthogonal, the matrix Ais diagonal.) 7.42. Consider the vector space P2ðtÞwith inner product hf;gi¼Ð1 /C01fðtÞgðtÞdt. (a) Findhf;gi, where fðtÞ¼tþ2 and gðtÞ¼t2/C03tþ4. (b) Find the matrix Aof the inner product with respect to the basis f1;t;t2gofV. (c) Verify Theorem 7.16 by showing that hf;gi¼½ f/C138TA½g/C138with respect to the basis f1;t;t2g. (a)hf;gi¼ð1 /C01ðtþ2Þðt2/C03tþ4Þdt¼ð1 /C01ðt3/C0t2/C02tþ8Þdt¼t4 4/C0t3 3/C0t2þ8t/C18/C19 /C12/C12/C12/C121 /C01¼46 3 (b) Here we use the fact that if rþs¼n, htr;tri¼ð1 /C01tndt¼tnþ1 nþ1/C12/C12/C12/C121 /C01¼2=ðnþ1Þifnis even ; 0i f nis odd :/C26 Thenh1;1i¼2,h1;ti¼0,h1;t2i¼2 3,ht;ti¼2 3,ht;t2i¼0,ht2;t2i¼2 5. Thus, A¼202 3 02 30 2 30252 43 5254 CHAPTER 7 Inner Product Spaces, Orthogonality (c) We have½f/C138T¼ð2;1;0Þand½g/C138T¼ð4;/C03;1Þrelative to the given basis. Then ½f/C138TA½g/C138¼ð 2;1;0Þ202 3 02 30 2 30252 43 54 /C03 12 43 5¼ð4;2 3;43Þ4 /C03 12 43 5¼46 3¼hf;gi 7.43. Prove Theorem 7.14: A¼ab bc/C20/C21 is positive deﬁnite if and only if aanddare positive and jAj¼ad/C0b2is positive. Letu¼½x;y/C138T. Then fðuÞ¼uTAu¼½x;y/C138ab bd/C20/C21 x y/C20/C21 ¼ax2þ2bxyþdy2 Suppose fðuÞ>0 for every u6¼0. Then fð1;0Þ¼a>0 and fð0;1Þ¼d>0. Also, we have fðb;/C0aÞ¼aðad/C0b2Þ>0. Because a>0, we get ad/C0b2>0. Conversely, suppose a>0,b¼0,ad/C0b2>0. Completing the square gives us fðuÞ¼ax2þ2b axyþb2 a2y2/C18/C19 þdy2/C0b2 ay2¼axþby a/C18/C192 þad/C0b2 ay2 Accordingly, fðuÞ>0 for every u6¼0. 7.44. Prove Theorem 7.15: Let Abe a real positive deﬁnite matrix. Then the function hu;vi¼uTAvis an inner product on Rn. For any vectors u1;u2, and v, hu1þu2;vi¼ð u1þu2ÞTAv¼ðuT 1þuT 2ÞAv¼uT 1AvþuT 2Av¼hu1;viþh u2;vi and, for any scalar kand vectors u;v, hku;vi¼ð kuÞTAv¼kuTAv¼khu;vi Thus½I1/C138is satisﬁed. Because uTAvis a scalar,ðuTAvÞT¼uTAv. Also, AT¼Abecause Ais symmetric. Therefore, hu;vi¼uTAv¼ðuTAvÞT¼vTATuTT¼vTAu¼hv;ui Thus,½I2/C138is satisﬁed. Last, because Ais positive deﬁnite, XTAX>0 for any nonzero X2Rn. Thus, for any nonzero vector v;hv;vi¼vTAv>0. Also,h0;0i¼0TA0¼0. Thus,½I3/C138is satisﬁed. Accordingly, the function hu;vi¼Av is an inner product. 7.45. Prove Theorem 7.16: Let Abe the matrix representation of an inner product relative to a basis Sof V. Then, for any vectors u;v2V, we have hu;vi¼½ u/C138TA½v/C138 Suppose S¼fw1;w2;...;wngandA¼½kij/C138. Hence, kij¼hwi;wji. Suppose u¼a1w1þa2w2þ/C1/C1/C1þ anwn and v¼b1w1þb2w2þ/C1/C1/C1þ bnwn Then hu;vi¼Pn i¼1Pn j¼1aibjhwi;wjið 1Þ On the other hand, ½u/C138TA½v/C138¼ð a1;a2;...;anÞk11k12 ... k1n k21k22 ... k2n :::::::::::::::::::::::::::::: kn1kn2... knn2 66643 7775b1 b2 ... bn2 666643 77775 ¼Pn i¼1aiki1;Pn i¼1aiki2;...;Pn i¼1aikin/C18/C19b1 b2 ... bn2 666643 77775¼Pn j¼1Pn i¼1aibjkijð2Þ Equationsð1Þand (2) give us our result.CHAPTER 7 Inner Product Spaces, Orthogonality 255 7.46. Prove Theorem 7.17: Let Abe the matrix representation of any inner product on V. Then Ais a positive deﬁnite matrix. Becausehwi;wji¼h wj;wiifor any basis vectors wiandwj, the matrix Ais symmetric. Let Xbe any nonzero vector in Rn. Then½u/C138¼Xfor some nonzero vector u2V. Theorem 7.16 tells us that XTAX¼½u/C138TA½u/C138¼h u;ui>0. Thus, Ais positive deﬁnite. Complex Inner Product Spaces 7.47. LetVbe a complex inner product space. Verify the relation hu;av1þbv2i¼ /C22ahu;v1iþ /C22bhu;v2i Using½I2*/C138,½I1*/C138, and then½I2*/C138, we ﬁnd hu;av1þbv2i¼hav1þbv2;ui¼ahv1;uiþbhv2;ui¼ /C22ahv1;uiþ /C22bhv2;ui¼ /C22ahu;v1iþ /C22bhu;v2i 7.48. Supposehu;vi¼3þ2iin a complex inner product space V. Find (a)hð2/C04iÞu;vi; (b)hu;ð4þ3iÞvi; (c)hð3/C06iÞu;ð5/C02iÞvi: (a)hð2/C04iÞu;vi¼ð 2/C04iÞhu;vi¼ð 2/C04iÞð3þ2iÞ¼14/C08i (b)hu;ð4þ3iÞvi¼ð4þ3iÞhu;vi¼ð 4/C03iÞð3þ2iÞ¼18/C0i (c)hð3/C06iÞu;ð5/C02iÞvi¼ð 3/C06iÞð5/C02iÞhu;vi¼ð 3/C06iÞð5þ2iÞð3þ2iÞ¼129/C018i 7.49. Find the Fourier coefﬁcient (component) cand the projection cwofv¼ð3þ4i;2/C03iÞalong w¼ð5þi;2iÞinC2. Recall that c¼hv;wi=hw;wi. Compute hv;wi¼ð 3þ4iÞð5þiÞþð 2/C03iÞð2iÞ¼ð 3þ4iÞð5/C0iÞþð 2/C03iÞð/C02iÞ ¼19þ17i/C06/C04i¼13þ13i hw;wi¼25þ1þ4¼30 Thus, c¼ð13þ13iÞ=30¼13 30þ1330i:Accordingly, projðv;wÞ¼cw¼ð26 15þ3915i;/C013 15þ1 15iÞ 7.50. Prove Theorem 7.18 (Cauchy–Schwarz): Let Vbe a complex inner product space. Then jhu;vij/C20k ukkvk. Ifv¼0, the inequality reduces to 0 /C200 and hence is valid. Now suppose v6¼0. Using z/C22z¼jzj2(for any complex number z) andhv;ui¼hu;vi, we expandku/C0hu;vitvk2/C210, where tis any real value: 0/C20ku/C0hu;vitvk2¼hu/C0hu;vitv;u/C0hu;vitvi ¼hu;ui/C0hu;vithu;vi/C0h u;vÞthv;uiþh u;vihu;vit2hv;vi ¼kuk2/C02tjhu;vij2þjhu;vij2t2kvk2 Sett¼1=kvk2to ﬁnd 0/C20kuk2/C0jhu;vij2 kvk2, from whichjhu;vij2/C20kvk2kvk2. Taking the square root of both sides, we obtain the required inequality. 7.51. Find an orthogonal basis for u?inC3where u¼ð1;i;1þiÞ. Here u?consists of all vectors s¼ðx;y;zÞsuch that hw;ui¼x/C0iyþð1/C0iÞz¼0 Find one solution, say w1¼ð0;1/C0i;iÞ. Then ﬁnd a solution of the system x/C0iyþð1/C0iÞz¼0;ð1þiÞy/C0iz¼0 Here zis a free variable. Set z¼1 to obtain y¼i=ð1þiÞ¼ð 1þiÞ=2 and x¼ð3i/C03Þ2. Multiplying by 2 yields the solution w2¼ð3i/C03, 1þi, 2). The vectors w1andw2form an orthogonal basis for u?.256 CHAPTER 7 Inner Product Spaces, Orthogonality 7.52. Find an orthonormal basis of the subspace WofC3spanned by v1¼ð1;i;0Þ and v2¼ð1;2;1/C0iÞ: Apply the Gram–Schmidt algorithm. Set w1¼v1¼ð1;i;0Þ. Compute v2/C0hv2;w1i hw1;w1iw1¼ð1;2;1/C0iÞ/C01/C02i 2ð1;i;0Þ¼ð1 2þi;1/C01 2i;1/C0iÞ Multiply by 2 to clear fractions, obtaining w2¼ð1þ2i;2/C0i;2/C02iÞ. Next ﬁndkw1k¼ﬃﬃﬃ 2p and then kw2k¼ﬃﬃﬃﬃﬃ 18p . Normalizingfw1;w2g, we obtain the following orthonormal basis of W: u1¼1ﬃﬃﬃ 2p;iﬃﬃﬃ 2p;0/C18/C19 ;u2¼1þ2iﬃﬃﬃﬃﬃ 18p ;2/C0iﬃﬃﬃﬃﬃ 18p ;2/C02iﬃﬃﬃﬃﬃ 18p/C18/C19 /C26/C27 7.53. Find the matrix Pthat represents the usual inner product on C3relative to the basis f1;i;1/C0ig. Compute the following six inner products: h1;1i¼1; hi;ii¼i/C22i¼1;h1;ii¼ /C22i¼/C0i; hi;1/C0ii¼ið1/C0iÞ¼/C0 1þi;h1;1/C0ii¼1/C0i¼1þi h1/C0i;1/C0ii¼2 Then, usingðu;vÞ¼hv;ui, we obtain P¼1/C0i 1þi i 1/C01þi 1/C0i/C01/C0i 22 43 5 (As expected, Pis Hermitian; that is, PH¼P.) Normed Vector Spaces 7.54. Consider vectors u¼ð1;3;/C06;4Þand v¼ð3;/C05;1;/C02ÞinR4. Find (a)kuk1andkvj1, (b)kuk1andkvk1, (c)kuk2andkvk2, (d)d1ðu;vÞ;d1ðu;vÞ,d2ðu;vÞ. (a) The inﬁnity norm chooses the maximum of the absolute values of the components. Hence, kuk1¼6 andkvk1¼5 (b) The one-norm adds the absolute values of the components. Thus, kuk1¼1þ3þ6þ4¼14 andkvk1¼3þ5þ1þ2¼11 (c) The two-norm is equal to the square root of the sum of the squares of the components (i.e., the norm induced by the usual inner product on R3). Thus, kuk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1þ9þ36þ16p ¼ﬃﬃﬃﬃﬃ 62p andkvk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 9þ25þ1þ4p ¼ﬃﬃﬃﬃﬃ 39p (d) First ﬁnd u/C0v¼ð/C0 2;8;/C07;6Þ. Then d1ðu;vÞ¼k u/C0vk1¼8 d1ðu;vÞ¼k u/C0vk1¼2þ8þ7þ6¼23 d2ðu;vÞ¼k u/C0vk2¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4þ64þ49þ36p ¼ﬃﬃﬃﬃﬃﬃﬃﬃ 153p 7.55. Consider the function fðtÞ¼t2/C04tinC½0;3/C138. (a) Findkfk1, (b) Plot fðtÞin the plane R2, (c) Findkfk1, (d) Findkfk2. (a) We seekkfk1¼maxðjfðtÞjÞ. Because fðtÞis differentiable on ½0;3/C138,jfðtÞjhas a maximum at a critical point of fðtÞ(i.e., when the derivative f0ðtÞ¼0), or at an endpoint of ½0;3/C138. Because f0ðtÞ¼2t/C04, we set 2 t/C04¼0 and obtain t¼2 as a critical point. Compute fð2Þ¼4/C08¼/C04; fð0Þ¼0/C00¼0; fð3Þ¼9/C012¼/C03 Thus,kfk1¼jfð2Þj¼j/C0 4j¼4.CHAPTER 7 Inner Product Spaces, Orthogonality 257 (b) Compute fðtÞfor various values of tin½0;3/C138, for example, t 0123 fðtÞ0/C03/C04/C03 Plot the points in R2and then draw a continuous curve through the points, as shown in Fig. 7-8. (c) We seekkfk1¼Ð3 0jfðtÞjdt. As indicated in Fig. 7-3, fðtÞis negative in½0;3/C138; hence, jfðtÞj¼/C0ð t2/C04tÞ¼4t/C0t2 kfk1¼ð3 0ð4t/C0t2Þdt¼2t2/C0t3 3/C18/C19/C12/C12/C12/C123 0¼18/C09¼9 Thus ; (d)kfk2 2¼ð3 0fðtÞ2dt¼ð3 0ðt4/C08t3þ16t2Þdt¼t5 5/C02t4þ16t3 3/C18/C19 /C12/C12/C12/C123 0¼153 5. Thus,kfk2¼ﬃﬃﬃﬃﬃﬃﬃﬃ 153 5r . 7.56. Prove Theorem 7.24: Let Vbe a normed vector space. Then the function dðu;vÞ¼k u/C0vk satisﬁes the following three axioms of a metric space: ½M1/C138dðu;vÞ/C210; and dðu;vÞ¼0 iff u¼v. ½M2/C138dðu;vÞ¼dðv;uÞ. ½M3/C138dðu;vÞ/C20dðu;wÞþdðw;vÞ. Ifu6¼v, then u/C0v6¼0, and hence, dðu;vÞ¼k u/C0vk>0. Also, dðu;uÞ¼k u/C0uk¼k 0k¼0. Thus, ½M1/C138is satisﬁed. We also have dðu;vÞ¼k u/C0vk¼k/C0 1ðv/C0uÞk¼j/C0 1jkv/C0uk¼k v/C0uk¼dðv;uÞ and dðu;vÞ¼k u/C0vk¼kð u/C0wÞþð w/C0vÞk/C20k u/C0wkþk w/C0vk¼dðu;wÞþdðw;vÞ Thus,½M2/C138and½M3/C138are satisﬁed. SUPPLEMENTARY PROBLEMS Inner Products 7.57. Verify that the following is an inner product on R2, where u¼ðx1;x2Þand v¼ðy1;y2Þ: fðu;vÞ¼x1y1/C02x1y2/C02x2y1þ5x2y2 7.58. Find the values of kso that the following is an inner product on R2, where u¼ðx1;x2Þand v¼ðy1;y2Þ: fðu;vÞ¼x1y1/C03x1y2/C03x2y1þkx2y2 Figure 7-8258 CHAPTER 7 Inner Product Spaces, Orthogonality 7.59. Consider the vectors u¼ð1;/C03Þand v¼ð2;5ÞinR2. Find (a)hu;viwith respect to the usual inner product in R2. (b)hu;viwith respect to the inner product in R2in Problem 7.57. (c)kvkusing the usual inner product in R2. (d)kvkusing the inner product in R2in Problem 7.57. 7.60. Show that each of the following is not an inner product on R3, where u¼ðx1;x2;x3Þand v¼ðy1;y2;y3Þ: (a)hu;vi¼x1y1þx2y2;(b)hu;vi¼x1y2x3þy1x2y3. 7.61. LetVbe the vector space of m/C2nmatrices over R. Show thathA;Bi¼trðBTAÞdeﬁnes an inner product inV. 7.62. Supposejhu;vij¼k ukkvk. (That is, the Cauchy–Schwarz inequality reduces to an equality.) Show that u and vare linearly dependent. 7.63. Suppose fðu;vÞandgðu;vÞare inner products on a vector space Vover R. Prove (a) The sum fþgis an inner product on V, whereðfþgÞðu;vÞ¼fðu;vÞþgðu;vÞ. (b) The scalar product kf, for k>0, is an inner product on V, whereðkfÞðu;vÞ¼kfðu;vÞ. Orthogonality, Orthogonal Complements, Orthogonal Sets 7.64. Let Vbe the vector space of polynomials over Rof degree/C202 with inner product deﬁned by hf;gi¼Ð1 0fðtÞgðtÞdt. Find a basis of the subspace Worthogonal to hðtÞ¼2tþ1. 7.65. Find a basis of the subspace WofR4orthogonal to u1¼ð1;/C02;3;4Þandu2¼ð3;/C05;7;8Þ. 7.66. Find a basis for the subspace WofR5orthogonal to the vectors u1¼ð1;1;3;4;1Þandu2¼ð1;2;1;2;1Þ. 7.67. Letw¼ð1;/C02;/C01;3Þbe a vector in R4. Find (a) an orthogonal basis for w?;(b) an orthonormal basis for w?. 7.68. LetWbe the subspace of R4orthogonal to u1¼ð1;1;2;2Þandu2¼ð0;1;2;/C01Þ. Find (a) an orthogonal basis for W;(b) an orthonormal basis for W. (Compare with Problem 7.65.) 7.69. LetSconsist of the following vectors in R4: u1¼ð1;1;1;1Þ; u2¼ð1;1;/C01;/C01Þ; u3¼ð1;/C01;1;/C01Þ; u4¼ð1;/C01;/C01;1Þ (a) Show that Sis orthogonal and a basis of R4. (b) Write v¼ð1;3;/C05;6Þas a linear combination of u1;u2;u3;u4. (c) Find the coordinates of an arbitrary vector v¼ða;b;c;dÞinR4relative to the basis S. (d) Normalize Sto obtain an orthonormal basis of R4. 7.70. LetM¼M2;2with inner product hA;Bi¼trðBTAÞ. Show that the following is an orthonormal basis for M: 10 00/C20/C21 ;01 00/C20/C21 ;00 10/C20/C21 ;00 01/C20/C21 /C26/C27 7.71. LetM¼M2;2with inner product hA;Bi¼trðBTAÞ. Find an orthogonal basis for the orthogonal complement of (a) diagonal matrices, (b) symmetric matrices.CHAPTER 7 Inner Product Spaces, Orthogonality 259 7.72. Supposefu1;u2;...;urgis an orthogonal set of vectors. Show that fk1u1;k2u2;...;krurgis an orthogonal set for any scalars k1;k2;...;kr. 7.73. LetUandWbe subspaces of a ﬁnite-dimensional inner product space V. Show that (a)ðUþWÞ?¼U?\W?;(b)ðU\WÞ?¼U?þW?. Projections, Gram–Schmidt Algorithm, Applications 7.74. Find the Fourier coefﬁcient cand projection cwofvalong w, where (a) v¼ð2;3;/C05Þandw¼ð1;/C05;2ÞinR3: (b) v¼ð1;3;1;2Þandw¼ð1;/C02;7;4ÞinR4: (c) v¼t2andw¼tþ3i nPðtÞ;with inner product hf;gi¼Ð1 0fðtÞgðtÞdt (d) v¼12 34/C20/C21 andw¼11 55/C20/C21 inM¼M2;2;with inner product hA;Bi¼trðBTAÞ: 7.75. LetUbe the subspace of R4spanned by v1¼ð1;1;1;1Þ; v2¼ð1;/C01;2;2Þ; v3¼ð1;2;/C03;/C04Þ (a) Apply the Gram–Schmidt algorithm to ﬁnd an orthogonal and an orthonormal basis for U. (b) Find the projection of v¼ð1;2;/C03;4Þonto U. 7.76. Suppose v¼ð1;2;3;4;6Þ. Find the projection of vonto W, or, in other words, ﬁnd w2Wthat minimizes kv/C0wk, where Wis the subspace of R5spanned by (a) u1¼ð1;2;1;2;1Þandu2¼ð1;/C01;2;/C01;1Þ, (b) v1¼ð1;2;1;2;1Þand v2¼ð1;0;1;5;/C01Þ. 7.77. Consider the subspace W¼P2ðtÞofPðtÞwith inner product hf;gi¼Ð1 0fðtÞgðtÞdt. Find the projection of fðtÞ¼t3onto W.(Hint: Use the orthogonal polynomials 1 ;2t/C01, 6t2/C06tþ1 obtained in Problem 7.22.) 7.78. Consider PðtÞwith inner product hf;gi¼Ð1 /C01fðtÞgðtÞdtand the subspace W¼P3ðtÞ: (a) Find an orthogonal basis for Wby applying the Gram–Schmidt algorithm to f1;t;t2;t3g. (b) Find the projection of fðtÞ¼t5onto W. Orthogonal Matrices 7.79. Find the number and exhibit all 2 /C22 orthogonal matrices of the form1 3x yz/C20/C21 . 7.80. Find a 3/C23 orthogonal matrix Pwhose ﬁrst two rows are multiples of u¼ð1;1;1Þand v¼ð1;/C02;3Þ, respectively. 7.81. Find a symmetric orthogonal matrix Pwhose ﬁrst row is ð1 3;23;23Þ. (Compare with Problem 7.32.) 7.82. Real matrices AandBare said to be orthogonally equivalent if there exists an orthogonal matrix Psuch that B¼PTAP. Show that this relation is an equivalence relation. Positive Deﬁnite Matrices and Inner Products 7.83. Find the matrix Athat represents the usual inner product on R2relative to each of the following bases: (a)fv1¼ð1;4Þ;v2¼ð2;/C03Þg, (b)fw1¼ð1;/C03Þ;w2¼ð6;2Þg. 7.84. Consider the following inner product on R2: fðu;vÞ¼x1y1/C02x1y2/C02x2y1þ5x2y2; where u¼ðx1;x2Þ v¼ðy1;y2Þ Find the matrix Bthat represents this inner product on R2relative to each basis in Problem 7.83.260 CHAPTER 7 Inner Product Spaces, Orthogonality 7.85. Find the matrix Cthat represents the usual basis on R3relative to the basis SofR3consisting of the vectors u1¼ð1;1;1Þ,u2¼ð1;2;1Þ,u3¼ð1;/C01;3Þ. 7.86. LetV¼P2ðtÞwith inner product hf;gi¼Ð1 0fðtÞgðtÞdt. (a) Findhf;gi, where fðtÞ¼tþ2 and gðtÞ¼t2/C03tþ4. (b) Find the matrix Aof the inner product with respect to the basis f1;t;t2gofV. (c) Verify Theorem 7.16 that hf;gi¼½ f/C138TA½g/C138with respect to the basis f1;t;t2g. 7.87. Determine which of the following matrices are positive deﬁnite: (a)13 35/C20/C21 , (b)34 47/C20/C21 , (c)42 21/C20/C21 , (d)6/C07 /C079/C20/C21 . 7.88. Suppose AandBare positive deﬁnite matrices. Show that: (a)AþBis positive deﬁnite and (b) kAis positive deﬁnite for k>0. 7.89. Suppose Bis a real nonsingular matrix. Show that: (a) BTBis symmetric and (b) BTBis positive deﬁnite. Complex Inner Product Spaces 7.90. Verify that ha1u1þa2u2b1v1þb2v2i¼a1/C22b1hu1;v1iþa1/C22b2hu1;v2iþa2/C22b1hu2;v1iþa2/C22b2hu2;v2i More generally, prove that hPm i¼1aiui;Pn j¼1bjvji¼P i;jai/C22bjhui;vii. 7.91. Consider u¼ð1þi;3;4/C0iÞand v¼ð3/C04i;1þi;2iÞinC3. Find (a)hu;vi, (b)hv;ui, (c)kuk, (d)kvk, (e) dðu;vÞ. 7.92. Find the Fourier coefﬁcient cand the projection cwof (a) u¼ð3þi;5/C02iÞalong w¼ð5þi;1þiÞinC2, (b) u¼ð1/C0i;3i;1þiÞalong w¼ð1;2/C0i;3þ2iÞinC3. 7.93. Letu¼ðz1;z2Þand v¼ðw1;w2Þbelong to C2. Verify that the following is an inner product of C2: fðu;vÞ¼z1/C22w1þð1þiÞz1/C22w2þð1/C0iÞz2/C22w1þ3z2/C22w2 7.94. Find an orthogonal basis and an orthonormal basis for the subspace WofC3spanned by u1¼ð1;i;1Þand u2¼ð1þi;0;2Þ. 7.95. Letu¼ðz1;z2Þand v¼ðw1;w2Þbelong to C2. For what values of a;b;c;d2Cis the following an inner product on C2? fðu;vÞ¼az1/C22w1þbz1/C22w2þcz2/C22w1þdz2/C22w2 7.96. Prove the following form for an inner product in a complex space V: hu;vi¼1 4kuþvk2/C01 4ku/C0vk2þ1 4kuþivk2/C01 4ku/C0ivk2 [Compare with Problem 7.7(b).] 7.97. LetVbe a real inner product space. Show that (i)kuk¼k vkif and only ifhuþv;u/C0vi¼0; (ii)kuþvk2¼kuk2þkvk2if and only ifhu;vi¼0. Show by counterexamples that the above statements are not true for, say, C2. 7.98. Find the matrix Pthat represents the usual inner product on C3relative to the basis f1;1þi;1/C02ig.CHAPTER 7 Inner Product Spaces, Orthogonality 261 7.99. A complex matrix Aisunitary if it is invertible and A/C01¼AH. Alternatively, Ais unitary if its rows (columns) form an orthonormal set of vectors (relative to the usual inner product of Cn). Find a unitary matrix whose ﬁrst row is: (a) a multiple of ð1;1/C0iÞ; (b) a multiple of ð1 2;12i;1 2/C012iÞ. Normed Vector Spaces 7.100. Consider vectors u¼ð1;/C03;4;1;/C02Þand v¼ð3;1;/C02;/C03;1ÞinR5. Find (a)kuk1andkvk1, (b)kuk1andkvk1, (c)kuk2andkvk2, (d) d1ðu;vÞ;d1ðu;vÞ,d2ðu;vÞ 7.101. Repeat Problem 7.100 for u¼ð1þi;2/C04iÞand v¼ð1/C0i;2þ3iÞinC2. 7.102. Consider the functions fðtÞ¼5t/C0t2andgðtÞ¼3t/C0t2inC½0;4/C138. Find (a)d1ðf;gÞ, (b) d1ðf;gÞ, (c) d2ðf;gÞ 7.103. Prove (a)k/C1k1is a norm on Rn. (b)k/C1k1is a norm on Rn. 7.104. Prove (a)k/C1k1is a norm on C½a;b/C138. (b)k/C1k1is a norm on C½a;b/C138. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation :M¼½R1;R2; .../C138denotes a matrix Mwith rows R1;R2;:...Also, basis need not be unique. 7.58. k>9 7.59. (a)/C013, (b)/C071, (c)ﬃﬃﬃﬃﬃ 29p , (d)ﬃﬃﬃﬃﬃ 89p 7.60. Letu¼ð0;0;1Þ; thenhu;ui¼0 in both cases 7.64.f7t2/C05t;12t2/C05g 7.65.fð1;2;1;0Þ;ð4;4;0;1Þg 7.66.ð/C01;0;0;0;1Þ;ð/C06;2;0;1;0Þ;ð/C05;2;1;0;0Þ 7.67. (a) u1¼ð0;0;3;1Þ;u2¼ð0;5;/C01;3Þ;u3¼ð/C0 14;/C02;/C01;3Þ; (b) u1=ﬃﬃﬃﬃﬃ 10p ;u2=ﬃﬃﬃﬃﬃ 35p ;u3=ﬃﬃﬃﬃﬃﬃﬃﬃ 210p 7.68. (a)ð0;2;/C01;0Þ;ð/C015;1;2;5Þ, (b)ð0;2;/C01;0Þ=ﬃﬃﬃ 5p ;ð/C015;1;2;5Þ=ﬃﬃﬃﬃﬃﬃﬃﬃ 255p 7.69. (b) v¼1 4ð5u1þ3u2/C013u3þ9u4Þ, (c)½v/C138¼1 4½aþbþcþd;aþb/C0c/C0d;a/C0bþc/C0d;a/C0b/C0cþd/C138 7.71. (a)½0;1;0;0/C138;½0;0;1;0/C138, (b)½0;/C01;1;0/C138 7.74. (a) c¼/C023 30, (b) c¼1 7, (c) c¼15 148, (d) c¼19 26 7.75. (a) w1¼ð1;1;1;1Þ;w2¼ð0;/C02;1;1Þ;w3¼ð12;/C04;/C01;/C07Þ, (b) projðv;UÞ¼1 5ð/C01;12;3;6Þ 7.76. (a) projðv;WÞ¼1 8ð23;25;30;25;23Þ, (b) First ﬁnd an orthogonal basis for W; say, w1¼ð1;2;1;2;1Þandw2¼ð0;2;0;/C03;2Þ. Then projðv;WÞ¼1 17ð34;76;34;56;42Þ 7.77. projðf;WÞ¼3 2t2/C035tþ1 20262 CHAPTER 7 Inner Product Spaces, Orthogonality 7.78. (a)f1;t;3t2/C01;5t3/C03tg, projðf;WÞ¼10 9t3/C05 21t 7.79. Four:½a;b;b;/C0a/C138,½a;b;/C0b;/C0a/C138,½a;/C0b;b;a/C138,½a;/C0b;/C0b;/C0a/C138, where a¼1 3andb¼1 3ﬃﬃﬃ 8p 7.80. P¼½1=a;1=a;1=a;1=b;/C02=b;3=b;5=c;/C02=c;/C03=c/C138, where a¼ﬃﬃﬃ 3p ;b¼ﬃﬃﬃﬃﬃ 14p ;c¼ﬃﬃﬃﬃﬃ 38p 7.81.1 3½1;2;2;2;/C02;1;2;1;/C02/C138 7.83. (a)½17;/C010;/C010;13/C138, (b)½10;0;0;40/C138 7.84. (a)½65;/C068;/C068;73/C138, (b)½58;8;8;8/C138 7.85.½3;4;3;4;6;2;3;2;11/C138 7.86. (a)83 12, (b)½1;a;b;a;b;c;b;c;d/C138, where a¼1 2,b¼1 3,c¼1 4,d¼1 5 7.87. (a) No, (b) Yes, (c) No, (d) Yes 7.91. (a)/C04i, (b) 4 i, (c)ﬃﬃﬃﬃﬃ 28p , (d)ﬃﬃﬃﬃﬃ 31p , (e)ﬃﬃﬃﬃﬃ 59p 7.92. (a) c¼1 28ð19/C05iÞ, (b) c¼1 19ð3þ6iÞ 7.94.fv1¼ð1;i;1Þ=ﬃﬃﬃ 3p ;v2¼ð2i;1/C03i;3/C0iÞ=ﬃﬃﬃﬃﬃ 24p g 7.95. aanddreal and positive, c¼/C22bandad/C0bcpositive. 7.97. u¼ð1;2Þ;v¼ði;2iÞ 7.98. P¼½1;1/C0i;1þ2i; 1þi;2;/C01þ3i; 1/C02i;/C01/C03i;5/C138 7.99. (a)ð1=ﬃﬃﬃ 3p Þ½1;1/C0i; 1þi;/C01/C138, (b)½a;ai;a/C0ai; bi;b;0;a;ai;/C0a/C0ai/C138, where a¼1 2andb¼1=ﬃﬃﬃ 2p . 7.100. (a) 4 and 3, (b) 11 and 10, (c)ﬃﬃﬃﬃﬃ 31p andﬃﬃﬃﬃﬃ 24p , (d) 6 ;19;9 7.101. (a)ﬃﬃﬃﬃﬃ 20p andﬃﬃﬃﬃﬃ 13p , (b)ﬃﬃﬃ 2p þﬃﬃﬃﬃﬃ 20p andﬃﬃﬃ 2p þﬃﬃﬃﬃﬃ 13p , (c)ﬃﬃﬃﬃﬃ 22p andﬃﬃﬃﬃﬃ 15p , (d) 7 ;9;ﬃﬃﬃﬃﬃ 53p 7.102. (a) 8, (b) 16, (c) 16 =ﬃﬃﬃ 3pCHAPTER 7 Inner Product Spaces, Orthogonality 263 Determinants 8.1 Introduction Each n-square matrix A¼½aij/C138is assigned a special scalar called the determinant ofA, denoted by detðAÞ orjAjor a11a12 ... a1n a21a22 ... a2n ::::::::::::::::::::::::::::: an1an2... ann/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 We emphasize that an n/C2narray of scalars enclosed by straight lines, called a determinant of order n ,i s not a matrix but denotes the determinant of the enclosed array of scalars (i.e., the enclosed matrix). The determinant function was ﬁrst discovered during the investigation of systems of linear equations. We shall see that the determinant is an indispensable tool in investigating and obtaining properties ofsquare matrices. The deﬁnition of the determinant and most of its properties also apply in the case where the entries of a matrix come from a commutative ring. We begin with a special case of determinants of orders 1, 2, and 3. Then we deﬁne a determinant of arbitrary order. This general deﬁnition is preceded by a discussion of permutations, which is necessary forour general deﬁnition of the determinant. 8.2 Determinants of Orders 1 and 2 Determinants of orders 1 and 2 are deﬁned as follows: ja11j¼a11 anda11a12 a21a22/C12/C12/C12/C12/C12/C12/C12/C12¼a11a22/C0a12a21 Thus, the determinant of a 1 /C21 matrix A¼½a11/C138is the scalar a11; that is, detðAÞ¼j a11j¼a11. The determinant of order two may easily be remembered by using the following diagram: a11a12 a21a22/C12/C12/C12/C12/C12/C12/C12/C12 That, is, the determinant is equal to the product of the elements along the plus-labeled arrow minus the product of the elements along the minus-labeled arrow. (There is an analogous diagram for determinantsof order 3, but not for higher-order determinants.) EXAMPLE 8.1 (a) Because the determinant of order 1 is the scalar itself, we have: detð27Þ¼27; detð/C07Þ¼/C0 7; detðt/C03Þ¼t/C03 (b)53 46/C12/C12/C12/C12/C12/C12/C12/C12¼5ð6Þ/C03ð4Þ¼30/C012¼18;32 /C057/C12/C12/C12/C12/C12/C12/C12/C12¼21þ10¼31/C131/C131/C131/C131/C131/C131/C131/C131 /C131! /C131/C131/C131/C131/C131/C131/C131/C131/C131!þ/C0 CHAPTER 8 264 Application to Linear Equations Consider two linear equations in two unknowns, say a1zþb1y¼c1 a2xþb2y¼c2 LetD¼a1b2/C0a2b1, the determinant of the matrix of coefﬁcients. Then the system has a unique solution if and only if D6¼0. In such a case, the unique solution may be expressed completely in terms of determinants as follows: x¼Nx D¼b2c1/C0b1c2 a1b2/C0a2b1¼c1b1 c2b2/C12/C12/C12/C12/C12/C12/C12/C12 a1b1 a2b2/C12/C12/C12/C12/C12/C12/C12/C12; y¼Ny D¼a1c2/C0a2c1 a1b2/C0a2b1¼a1c1 a2c2/C12/C12/C12/C12/C12/C12/C12/C12 a1b1 a2b2/C12/C12/C12/C12/C12/C12/C12/C12 Here Dappears in the denominator of both quotients. The numerators NxandNyof the quotients for xand y, respectively, can be obtained by substituting the column of constant terms in place of the column of coefﬁcients of the given unknown in the matrix of coefﬁcients. On the other hand, if D¼0, then the system may have no solution or more than one solution. EXAMPLE 8.2 Solve by determinants the system4x/C03y¼15 2xþ5y¼1/C26 First ﬁnd the determinant Dof the matrix of coefﬁcients: D¼4/C03 25/C12/C12/C12/C12/C12/C12/C12/C12¼4ð5Þ/C0ð/C0 3Þð2Þ¼20þ6¼26 Because D6¼0, the system has a unique solution. To obtain the numerators NxandNy, simply replace, in the matrix of coefﬁcients, the coefﬁcients of xandy, respectively, by the constant terms, and then take their determinants: Nx¼15/C03 15/C12/C12/C12/C12/C12/C12/C12/C12¼75þ3¼78 Ny¼41 5 21/C12/C12/C12/C12/C12/C12/C12/C12¼4/C030¼/C026 Then the unique solution of the system is x¼Nx D¼78 26¼3; y¼Ny D¼/C026 26¼/C01 8.3 Determinants of Order 3 Consider an arbitrary 3 /C23 matrix A¼½aij/C138. The determinant of Ais deﬁned as follows: detðAÞ¼a11a12a13 a21a22a23 a31a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼a11a22a33þa12a23a31þa13a21a32/C0a13a22a31/C0a12a21a33/C0a11a23a32 Observe that there are six products, each product consisting of three elements of the original matrix. Three of the products are plus-labeled (keep their sign) and three of the products are minus-labeled(change their sign). The diagrams in Fig. 8-1 may help us to remember the above six products in det ðAÞ. That is, the determinant is equal to the sum of the products of the elements along the three plus-labeled arrows inCHAPTER 8 Determinants 265 Fig. 8-1 plus the sum of the negatives of the products of the elements along the three minus-labeled arrows. We emphasize that there are no such diagrammatic devices with which to remember determinantsof higher order. EXAMPLE 8.3 LetA¼211 05/C02 1/C0342 43 5andB¼321 /C045/C01 2/C0342 43 5. Find detðAÞand detðBÞ. Use the diagrams in Fig. 8-1: detðAÞ¼2ð5Þð4Þþ1ð/C02Þð1Þþ1ð/C03Þð0Þ/C01ð5Þð1Þ/C0ð/C0 3Þð/C02Þð2Þ/C04ð1Þð0Þ ¼40/C02þ0/C05/C012/C00¼21 detðBÞ¼60/C04þ12/C010/C09þ32¼81 Alternative Form for a Determinant of Order 3 The determinant of the 3 /C23 matrix A¼½aij/C138may be rewritten as follows: detðAÞ¼a11ða22a23/C0a23a32Þ/C0a12ða21a33/C0a23a31Þþa13ða21a32/C0a22a31Þ ¼a11a22a23 a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C0a12a21a23 a31a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þa13a21a22 a31a32/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 which is a linear combination of three determinants of order 2 whose coefﬁcients (with alternating signs) form the ﬁrst row of the given matrix. This linear combination may be indicated in the form a11a11a12a13 a21a22a23 a31a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C0a12a11a12a13 a21a22a23 a31a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þa13a11a12a13 a21a22a23 a31a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 Note that each 2/C22 matrix can be obtained by deleting, in the original matrix, the row and column containing its coefﬁcient. EXAMPLE 8.4 123 4/C023 05/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼1123 4/C023 05/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C02123 4/C023 05/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þ3123 4/C023 05/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 ¼1/C023 5/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C0243 0/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þ34/C02 05/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 ¼1ð2/C015Þ/C02ð/C04þ0Þþ3ð20þ0Þ¼/C0 13þ8þ60¼55 Figure 8-1266 CHAPTER 8 Determinants 8.4 Permutations A permutation sof the setf1;2;...;ngis a one-to-one mapping of the set onto itself or, equivalently, a rearrangement of the numbers 1 ;2;...;n. Such a permutation sis denoted by s¼12 ... n j1j2... jn/C18/C19 or s¼j1j2/C1/C1/C1jn; where ji¼sðiÞ The set of all such permutations is denoted by Sn, and the number of such permutations is n!.I fs2Sn; then the inverse mapping s/C012Sn; and if s;t2Sn, then the composition mapping s/C14t2Sn. Also, the identity mapping e¼s/C14s/C012Sn. (In fact, e¼123 ...n.) EXAMPLE 8.5 (a) There are 2 !¼2/C11¼2 permutations in S2; they are 12 and 21. (b) There are 3 !¼3/C12/C11¼6 permutations in S3; they are 123, 132, 213, 231, 312, 321. Sign (Parity) of a Permutation Consider an arbitrary permutation sinSn, say s¼j1j2/C1/C1/C1jn:We say sis an even or odd permutation according to whether there is an even or odd number of inversions in s.B ya n inversion inswe mean a pair of integersði;kÞsuch that i>k, but iprecedes kins. We then deﬁne the sign or parity of s, written sgns,b y sgns¼1i f sis even /C01i f sis odd/C26 EXAMPLE 8.6 (a) Find the sign of s¼35142 in S5. For each element k, we count the number of elements isuch that i>kandiprecedes kins. There are 2 numbersð3 and 5Þgreater than and preceding 1 ; 3 numbersð3;5;and 4Þgreater than and preceding 2 ; 1 numberð5Þgreater than and preceding 4 : (There are no numbers greater than and preceding either 3 or 5.) Because there are, in all, six inversions, sis even and sgn s¼1. (b) The identity permutation e¼123 ...nis even because there are no inversions in e. (c) In S2, the permutation 12 is even and 21 is odd. In S3, the permutations 123, 231, 312 are even and the permutations 132, 213, 321 are odd. (d) Let tbe the permutation that interchanges two numbers iandjand leaves the other numbers ﬁxed. That is, tðiÞ¼j; tðjÞ¼i; tðkÞ¼k;where k6¼i;j We call tatransposition .I fi<j, then there are 2ðj/C0iÞ/C01 inversions in t, and hence, the transposition t is odd. Remark: One can show that, for any n, half of the permutations in Snare even and half of them are odd. For example, 3 of the 6 permutations in S3are even, and 3 are odd. 8.5. Determinants of Arbitrary Order LetA¼½aij/C138be a square matrix of order nover a ﬁeld K. Consider a product of nelements of Asuch that one and only one element comes from each row and one and only one element comes from each column. Such a product can be written in the form a1j1a2j2/C1/C1/C1anjnCHAPTER 8 Determinants 267 that is, where the factors come from successive rows, and so the ﬁrst subscripts are in the natural order 1;2;...;n. Now because the factors come from different columns, the sequence of second subscripts forms a permutation s¼j1j2/C1/C1/C1jninSn. Conversely, each permutation in Sndetermines a product of the above form. Thus, the matrix Acontains n!such products. DEFINITION: The determinant of A¼½aij/C138, denoted by detðAÞorjAj, is the sum of all the above n! products, where each such product is multiplied by sgn s. That is, jAj¼P sðsgnsÞa1j1a2j2/C1/C1/C1anjn or jAj¼P s2SnðsgnsÞa1sð1Þa2sð2Þ/C1/C1/C1ansðnÞ The determinant of the n-square matrix Ais said to be of order n. The next example shows that the above deﬁnition agrees with the previous deﬁnition of determinants of orders 1, 2, and 3. EXAMPLE 8.7 (a) Let A¼½a11/C138be a 1/C21 matrix. Because S1has only one permutation, which is even, det ðAÞ¼a11, the number itself. (b) Let A¼½aij/C138be a 2/C22 matrix. In S2, the permutation 12 is even and the permutation 21 is odd. Hence, detðAÞ¼a11a12 a21a22/C12/C12/C12/C12/C12/C12/C12/C12¼a11a22/C0a12a21 (c) Let A¼½aij/C138be a 3/C23 matrix. In S3, the permutations 123, 231, 312 are even, and the permutations 321, 213, 132 are odd. Hence, detðAÞ¼a11a12a13 a21a22a23 a31a32a33/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼a 11a22a33þa12a23a31þa13a21a32/C0a13a22a31/C0a12a21a33/C0a11a23a32 Remark: Asnincreases, the number of terms in the determinant becomes astronomical. Accordingly, we use indirect methods to evaluate determinants rather than the deﬁnition of the determinant. In fact, we prove a number of properties about determinants that will permit us to shortenthe computation considerably. In particular, we show that a determinant of order nis equal to a linear combination of determinants of order n/C01, as in the case n¼3 above. 8.6 Properties of Determinants We now list basic properties of the determinant. THEOREM 8.1: The determinant of a matrix Aand its transpose ATare equal; that is,jAj¼jATj. By this theorem (proved in Problem 8.22), any theorem about the determinant of a matrix Athat concerns the rows of Awill have an analogous theorem concerning the columns of A. The next theorem (proved in Problem 8.24) gives certain cases for which the determinant can be obtained immediately. THEOREM 8.2: LetAbe a square matrix. (i) If Ahas a row (column) of zeros, then jAj¼0. (ii) If Ahas two identical rows (columns), then jAj¼0.268 CHAPTER 8 Determinants (iii) If Ais triangular (i.e., Ahas zeros above or below the diagonal), then jAj¼product of diagonal elements. Thus, in particular, jIj¼1, where Iis the identity matrix. The next theorem (proved in Problems 8.23 and 8.25) shows how the determinant of a matrix is affected by the elementary row and column operations. THEOREM 8.3: Suppose Bis obtained from Aby an elementary row (column) operation. (i) If two rows (columns) of Awere interchanged, then jBj¼/C0j Aj. (ii) If a row (column) of Awere multiplied by a scalar k, thenjBj¼kjAj. (iii) If a multiple of a row (column) of Awere added to another row (column) of A, thenjBj¼jAj. Major Properties of Determinants We now state two of the most important and useful theorems on determinants. THEOREM 8.4: The determinant of a product of two matrices Aand Bis the product of their determinants; that is, detðABÞ¼detðAÞdetðBÞ The above theorem says that the determinant is a multiplicative function. THEOREM 8.5: LetAbe a square matrix. Then the following are equivalent: (i) Ais invertible; that is, Ahas an inverse A/C01. (ii) AX¼0 has only the zero solution. (iii) The determinant of Ais not zero; that is, det ðAÞ6¼0. Remark: Depending on the author and the text, a nonsingular matrix Ais deﬁned to be an invertible matrix A, or a matrix Afor whichjAj6¼0, or a matrix Afor which AX¼0 has only the zero solution. The above theorem shows that all such deﬁnitions are equivalent. We will prove Theorems 8.4 and 8.5 (in Problems 8.29 and 8.28, respectively) using the theory of elementary matrices and the following lemma (proved in Problem 8.26), which is a special case ofTheorem 8.4. LEMMA 8.6: LetEbe an elementary matrix. Then, for any matrix A;jEAj¼jEjjAj. Recall that matrices AandBare similar if there exists a nonsingular matrix Psuch that B¼P/C01AP. Using the multiplicative property of the determinant (Theorem 8.4), one can easily prove (Problem 8.31)the following theorem. THEOREM 8.7: Suppose AandBare similar matrices. Then jAj¼jBj. 8.7 Minors and Cofactors Consider an n-square matrix A¼½aij/C138. Let Mijdenote theðn/C01Þ-square submatrix of Aobtained by deleting its ith row and jth column. The determinant jMijjis called the minor of the element aijofA, and we deﬁne the cofactor ofaij, denoted by Aij;to be the ‘‘signed’’ minor: Aij¼ð/C0 1ÞiþjjMijjCHAPTER 8 Determinants 269 Note that the ‘‘signs’’ ð/C01Þiþjaccompanying the minors form a chessboard pattern with þ’s on the main diagonal: þ/C0þ/C0 ... /C0þ/C0þ ... þ/C0þ/C0 ... :::::::::::::::::::::::::::::::2 6643 775 We emphasize that Mijdenotes a matrix, whereas Aijdenotes a scalar. Remark: The signð/C01Þiþjof the cofactor Aijis frequently obtained using the checkerboard pattern. Speciﬁcally, beginning with þand alternating signs: þ;/C0;þ;/C0;...; count from the main diagonal to the appropriate square. EXAMPLE 8.8 LetA¼123 456 7892 43 5. Find the following minors and cofactors: (a) jM23jand A23, (b)jM31jandA31. (a)jM23j¼123 456 789/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼12 78/C12/C12/C12/C12/C12/C12/C12/C12¼8/C014¼/C06, and so A23¼ð/C0 1Þ2þ3jM23j¼/C0ð/C0 6Þ¼6 (b)jM31j¼123 456 789/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼23 56/C12/C12/C12/C12/C12/C12/C12/C12¼12/C015¼/C03, and so A 31¼ð/C0 1Þ1þ3jM31j¼þð/C0 3Þ¼/C0 3 Laplace Expansion The following theorem (proved in Problem 8.32) holds. THEOREM 8.8: (Laplace) The determinant of a square matrix A¼½aij/C138is equal to the sum of the products obtained by multiplying the elements of any row (column) by theirrespective cofactors: jAj¼a i1Ai1þai2Ai2þ/C1/C1/C1þ ainAin¼Pn j¼1aijAij jAj¼a1jA1jþa2jA2jþ/C1/C1/C1þ anjAnj¼Pn i¼1aijAij The above formulas for jAjare called the Laplace expansions of the determinant of Aby the ith row and the jth column. Together with the elementary row (column) operations, they offer a method of simplifying the computation of jAj, as described below. 8.8 Evaluation of Determinants The following algorithm reduces the evaluation of a determinant of order nto the evaluation of a determinant of order n/C01. ALGORITHM 8.1: (Reduction of the order of a determinant) The input is a nonzero n-square matrix A¼½aij/C138with n>1. Step 1. Choose an element aij¼1 or, if lacking, aij6¼0. Step 2. Using aijas a pivot, apply elementary row (column) operations to put 0’s in all the other positions in the column (row) containing aij. Step 3. Expand the determinant by the column (row) containing aij.270 CHAPTER 8 Determinants The following remarks are in order. Remark 1: Algorithm 8.1 is usually used for determinants of order 4 or more. With determinants of order less than 4, one uses the speciﬁc formulas for the determinant. Remark 2: Gaussian elimination or, equivalently, repeated use of Algorithm 8.1 together with row interchanges can be used to transform a matrix Ainto an upper triangular matrix whose determinant is the product of its diagonal entries. However, one must keep track of the number of row interchanges, becauseeach row interchange changes the sign of the determinant. EXAMPLE 8.9 Use Algorithm 8.1 to ﬁnd the determinant of A¼5421 231/C02 /C05/C07/C039 1/C02/C0142 6643 775. Usea23¼1 as a pivot to put 0’s in the other positions of the third column; that is, apply the row operations ‘‘Replace R1by/C02R2þR1,’’ ‘‘Replace R3by 3R2þR3,’’ and ‘‘Replace R4byR2þR4.’’ By Theorem 8.3(iii), the value of the determinant does not change under these operations. Thus, jAj¼5421 231/C02 /C05/C07/C039 1/C02/C014/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼1/C020 5 23 1/C02 12 03 31 02/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 Now expand by the third column. Speciﬁcally, neglect all terms that contain 0 and use the fact that the sign of the minor M23isð/C01Þ2þ3¼/C01. Thus, jAj¼/C0120 5 231/C02 120 3 310 2/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C01/C025 12 3 31 2/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C0ð 4/C018þ5/C030/C03þ4Þ¼/C0ð/C0 38Þ¼38 8.9 Classical Adjoint LetA¼½aij/C138be an n/C2nmatrix over a ﬁeld Kand let Aijdenote the cofactor of aij. The classical adjoint ofA, denoted by adj A, is the transpose of the matrix of cofactors of A. Namely, adjA¼½Aij/C138T We say ‘‘classical adjoint’’ instead of simply ‘‘adjoint’’ because the term ‘‘adjoint’’ is currently used for an entirely different concept. EXAMPLE 8.10 LetA¼23/C04 0/C042 1/C0152 43 5. The cofactors of the nine elements of Afollow: A11¼þ/C042 /C015/C12/C12/C12/C12/C12/C12/C12/C12¼/C018; A21¼/C03/C04 /C015/C12/C12/C12/C12/C12/C12/C12/C12¼/C011; A31¼þ3/C04 /C042/C12/C12/C12/C12/C12/C12/C12/C12¼/C010;A12¼/C002 15/C12/C12/C12/C12/C12/C12/C12/C12¼2; A22¼þ2/C04 15/C12/C12/C12/C12/C12/C12/C12/C12¼14; A32¼/C02/C04 02/C12/C12/C12/C12/C12/C12/C12/C12¼/C04;A13¼þ0/C04 1/C01/C12/C12/C12/C12/C12/C12/C12/C12¼4 A23¼/C023 1/C01/C12/C12/C12/C12/C12/C12/C12/C12¼5 A33¼þ23 0/C04/C12/C12/C12/C12/C12/C12/C12/C12¼/C08CHAPTER 8 Determinants 271 The transpose of the above matrix of cofactors yields the classical adjoint of A; that is, adjA¼/C018/C011/C010 21 4/C04 45/C082 43 5 The following theorem (proved in Problem 8.34) holds. THEOREM 8.9: LetAbe any square matrix. Then AðadjAÞ¼ð adjAÞA¼jAjI where Iis the identity matrix. Thus, if jAj6¼0, A/C01¼1 jAjðadjAÞ EXAMPLE 8.11 LetAbe the matrix in Example 8.10. We have detðAÞ¼/C0 40þ6þ0/C016þ4þ0¼/C046 Thus, Adoes have an inverse, and, by Theorem 8.9, A/C01¼1 jAjðadjAÞ¼/C01 46/C018/C011/C010 21 4/C04 45/C082 643 75¼9 231146 5 23 /C01 23/C07 232 23 /C02 23/C05 464 232 643 75 8.10 Applications to Linear Equations, Cramer’s Rule Consider a system AX¼Bofnlinear equations in nunknowns. Here A¼½aij/C138is the (square) matrix of coefﬁcients and B¼½bi/C138is the column vector of constants. Let Aibe the matrix obtained from Aby replacing the ith column of Aby the column vector B. Furthermore, let D¼detðAÞ; N1¼detðA1Þ; N2¼detðA2Þ; ...; Nn¼detðAnÞ The fundamental relationship between determinants and the solution of the system AX¼Bfollows. THEOREM 8.10: The (square) system AX¼Bhas a solution if and only if D6¼0. In this case, the unique solution is given by x1¼N1 D; x2¼N2 D; ...; xn¼Nn D The above theorem (proved in Problem 8.10) is known as Cramer’s rule for solving systems of linear equations. We emphasize that the theorem only refers to a system with the same number of equations asunknowns, and that it only gives the solution when D6¼0. In fact, if D¼0, the theorem does not tell us whether or not the system has a solution. However, in the case of a homogeneous system, we have thefollowing useful result (to be proved in Problem 8.54). THEOREM 8.11: A square homogeneous system AX¼0has a nonzero solution if and only if D¼jAj¼0.272 CHAPTER 8 Determinants EXAMPLE 8.12 Solve the system using determinantsxþyþz¼5 x/C02y/C03z¼/C01 2xþy/C0z¼38 < : First compute the determinant Dof the matrix of coefﬁcients: D¼111 1/C02/C03 21/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼2/C06þ1þ4þ3þ1¼5 Because D6¼0, the system has a unique solution. To compute N x,Ny,Nz, we replace, respectively, the coefﬁcients ofx;y;zin the matrix of coefﬁcients by the constant terms. This yields Nx¼511 /C01/C02/C03 31/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼20; N y¼151 1/C01/C03 23/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C010; N z¼115 1/C02/C01 213/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼15 Thus, the unique solution of the system is x¼N x=D¼4,y¼Ny=D¼/C02, z¼Nz=D¼3; that is, the vector u¼ð4;/C02;3Þ. 8.11 Submatrices, Minors, Principal Minors LetA¼½aij/C138be a square matrix of order n. Consider any rrows and rcolumns of A. That is, consider any setI¼ði1;i2;...;irÞofrrow indices and any set J¼ðj1;j2;...;jrÞofrcolumn indices. Then IandJ deﬁne an r/C2rsubmatrix of A, denoted by AðI;JÞ, obtained by deleting the rows and columns of Awhose subscripts do not belong to IorJ, respectively. That is, AðI;JÞ¼½ ast:s2I;t2J/C138 The determinantjAðI;JÞjis called a minor ofAof order rand ð/C01Þi1þi2þ/C1/C1/C1þ irþj1þj2þ/C1/C1/C1þ jrjAðI;JÞj is the corresponding signed minor. (Note that a minor of order n/C01 is a minor in the sense of Section 8.7, and the corresponding signed minor is a cofactor.) Furthermore, if I0andJ0denote, respectively, the remaining row and column indices, then jAðI0;J0Þj denotes the complementary minor , and its sign (Problem 8.74) is the same sign as the minor. EXAMPLE 8.13 LetA¼½aij/C138be a 5-square matrix, and let I¼f1;2;4gand J¼f2;3;5g. Then I0¼f3;5gandJ0¼f1;4g, and the corresponding minor jMjand complementary minor jM0jare as follows: jMj¼jAðI;JÞj¼a12a13a15 a22a23a25 a42a43a45/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12andjM0j¼jAðI0;J0Þj¼a31a34 a51a54/C12/C12/C12/C12/C12/C12/C12/C12 Because 1þ2þ4þ2þ3þ5¼17 is odd,/C0jMjis the signed minor, and /C0jM0jis the signed complementary minor. Principal Minors A minor is principal if the row and column indices are the same, or equivalently, if the diagonal elements of the minor come from the diagonal of the matrix. We note that the sign of a principal minor is always þ1, because the sum of the row and identical column subscripts must always be even.CHAPTER 8 Determinants 273 EXAMPLE 8.14 LetA¼12/C01 35 4 /C031/C022 43 5. Find the sums C1,C2, and C3of the principal minors of Aof orders 1, 2, and 3, respectively. (a) There are three principal minors of order 1. These are j1j¼1;j5j¼5;j/C02j¼/C0 2; and so C1¼1þ5/C02¼4 Note that C1is simply the trace of A. Namely, C1¼trðAÞ: (b) There are three ways to choose two of the three diagonal elements, and each choice gives a minor of order 2. These are 12 35/C12/C12/C12/C12/C12/C12/C12/C12¼/C01;1/C01 /C03/C02/C12/C12/C12/C12/C12/C12/C12/C12¼1;54 1/C02/C12/C12/C12/C12/C12/C12/C12/C12¼/C014 (Note that these minors of order 2 are the cofactors A33,A22, and A11ofA, respectively.) Thus, C2¼/C01þ1/C014¼/C014 (c) There is only one way to choose three of the three diagonal elements. Thus, the only minor of order 3 is the determinant of Aitself. Thus, C3¼jAj¼/C0 10/C024/C03/C015/C04þ12¼/C044 8.12 Block Matrices and Determinants The following theorem (proved in Problem 8.36) is the main result of this section. THEOREM 8.12: Suppose Mis an upper (lower) triangular block matrix with the diagonal blocks A1;A2;...;An. Then detðMÞ¼detðA1ÞdetðA2Þ...detðAnÞ EXAMPLE 8.15 FindjMjwhere M¼234 78 /C0153 21 002 15003/C014 005 262 666643 77775 Note that Mis an upper triangular block matrix. Evaluate the determinant of each diagonal block: 23 /C015/C12/C12/C12/C12/C12/C12/C12/C12¼10þ3¼13;21 5 3/C014 52 6/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C012þ20þ30þ25/C016/C018¼29 ThenjMj¼13ð29Þ¼377. Remark: Suppose M¼AB CD/C20/C21 , where A;B;C;Dare square matrices. Then it is not generally true thatjMj¼jAjjDj/C0jBjjCj. (See Problem 8.68.) 8.13 Determinants and Volume Determinants are related to the notions of area and volume as follows. Let u1;u2;...;unbe vectors in Rn. LetSbe the (solid) parallelopiped determined by the vectors; that is, S¼fa1u1þa2u2þ/C1/C1/C1þ anun:0/C20ai/C201 for i¼1;...;ng (When n¼2;Sis a parallelogram.) Let VðSÞdenote the volume of S(or area of Swhen n¼2Þ. Then VðSÞ¼absolute value of det ðAÞ274 CHAPTER 8 Determinants where Ais the matrix with rows u1;u2;...;un. In general, VðSÞ¼0 if and only if the vectors u1;...;un do not form a coordinate system for Rn(i.e., if and only if the vectors are linearly dependent). EXAMPLE 8.16 Letu1¼ð1;1;0Þ,u2¼ð1;1;1Þ,u3¼ð0;2;3Þ. Find the volume VðSÞof the parallelo- piped SinR3(Fig. 8-2) determined by the three vectors. Evaluate the determinant of the matrix whose rows are u1;u2;u3: 110 111023/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼3þ0þ0/C00/C02/C03¼/C02 Hence, VðSÞ¼j/C0 2j¼2. 8.14 Determinant of a Linear Operator LetFbe a linear operator on a vector space Vwith ﬁnite dimension. Let Abe the matrix representation of Frelative to some basis SofV. Then we deﬁne the determinant of F, written detðFÞ,b y detðFÞ¼j Aj IfBwere another matrix representation of Frelative to another basis S0ofV, then AandBare similar matrices (Theorem 6.7) and jBj¼jAj(Theorem 8.7). In other words, the above deﬁnition det ðFÞis independent of the particular basis SofV. (We say that the deﬁnition is well deﬁned .) The next theorem (to be proved in Problem 8.62) follows from analogous theorems on matrices. THEOREM 8.13: LetFandGbe linear operators on a vector space V. Then (i) detðF/C14GÞ¼detðFÞdetðGÞ. (ii) Fis invertible if and only if det ðFÞ6¼0. EXAMPLE 8.17 LetFbe the following linear operator on R3and let Abe the matrix that represents F relative to the usual basis of R3: Fðx;y;zÞ¼ð 2x/C04yþz;x/C02yþ3z;5xþy/C0zÞ and A¼2/C041 1/C023 51/C012 43 5 Then detðFÞ¼j Aj¼4/C060þ1þ10/C06/C04¼/C055z y x0u3 u2 u1 Figure 8-2CHAPTER 8 Determinants 275 8.15 Multilinearity and Determinants LetVbe a vector space over a ﬁeld K. Leta¼Vn; that is,aconsists of all the n-tuples A¼ðA1;A2;...;AnÞ where the Aiare vectors in V. The following deﬁnitions apply. DEFINITION: A function D:a!Kis said to be multilinear if it is linear in each component: (i) If Ai¼BþC, then DðAÞ¼ Dð...;BþC;...Þ¼ Dð...;B;...;ÞþDð...;C;...Þ (ii) If Ai¼kB, where k2K, then DðAÞ¼ Dð...;kB;...Þ¼ kDð...;B;...Þ We also say n-linear for multilinear if there are ncomponents. DEFINITION: A function D:a!Kis said to be alternating ifDðAÞ¼0 whenever Ahas two identical elements: DðA1;A2;...;AnÞ¼0 whenever Ai¼Aj;i6¼j Now let Mdenote the set of all n-square matrices Aover a ﬁeld K. We may view Aas an n-tuple consisting of its row vectors A1;A2;...;An; that is, we may view Ain the form A¼ðA1;A2;...;AnÞ. The following theorem (proved in Problem 8.37) characterizes the determinant function. THEOREM 8.14: There exists a unique function D:M!Ksuch that (i) Dis multilinear, (ii) Dis alternating, (iii) DðIÞ¼1. This function Dis the determinant function; that is, DðAÞ¼j Aj;for any matrix A2M. SOLVED PROBLEMS Computation of Determinants 8.1. Evaluate the determinant of each of the following matrices: (a) A¼65 23/C20/C21 , (b) B¼2/C03 47/C20/C21 ;(c) C¼4/C05 /C01/C02/C20/C21 ;(d) D¼t/C056 3 tþ2/C20/C21 Use the formulaab cd/C12/C12/C12/C12/C12/C12/C12/C12¼ad/C0bc: (a)jAj¼6ð3Þ/C05ð2Þ¼18/C010¼8 (b)jBj¼14þ12¼26 (c)jCj¼/C0 8/C05¼/C013 (d)jDj¼ð t/C05Þðtþ2Þ/C018¼t2/C03t/C010/C018¼t2/C010t/C028 8.2. Evaluate the determinant of each of the following matrices: (a) A¼234 543 1212 43 5, (b) B¼1/C023 24/C01 15/C022 43 5, (c) C¼13/C05 3/C012 1/C0212 43 5276 CHAPTER 8 Determinants Use the diagram in Fig. 8-1 to obtain the six products: (a)jAj¼2ð4Þð1Þþ3ð3Þð1Þþ4ð2Þð5Þ/C01ð4Þð4Þ/C02ð3Þð2Þ/C01ð3Þð5Þ¼8þ9þ40/C016/C012/C015¼14 (b)jBj¼/C0 8þ2þ30/C012þ5/C08¼9 (c)jCj¼/C0 1þ6þ30/C05þ4/C09¼25 8.3. Compute the determinant of each of the following matrices: (a) A¼234 567 8912 43 5, (b) B¼4/C068 9 0/C027/C03 00 56 00 032 6643 775, (c) C¼1 2/C01/C01 3 34 12/C01 1/C0412 643 75: (a) One can simplify the entries by ﬁrst subtracting twice the ﬁrst row from the second row—that is, by applying the row operation ‘‘Replace R2by/C021þR2.’’ Then jAj¼234 567 891/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼23 4 10/C01 89 1/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼0/C024þ36/C00þ18/C03¼27 (b)Bis triangular, sojBj¼product of the diagonal entries ¼/C0120. (c) The arithmetic is simpler if fractions are ﬁrst eliminated. Hence, multiply the ﬁrst row R 1by 6 and the second row R2by 4. Then j24Cj¼3/C06/C02 32/C04 1/C041/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼6þ24þ24þ4/C048þ18¼28; sojCj¼28 24¼7 6 8.4. Compute the determinant of each of the following matrices: (a) A¼25/C03/C02 /C02/C032/C05 13/C022 /C01/C06432 6643 775, (b) B¼62105 211/C021 112/C023 3023 /C01 /C01/C01/C03422 666643 77775 (a) Use a31¼1 as a pivot to put 0’s in the ﬁrst column, by applying the row operations ‘‘Replace R1by /C02R3þR1,’’ ‘‘Replace R2by 2R3þR2,’’ and ‘‘Replace R4byR3þR4.’’ Then jAj¼25/C03/C02 /C02/C032/C05 13/C022 /C01/C0643/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼0/C011/C06 03/C02/C01 13/C022 0/C0325/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C011/C06 3/C02/C01 /C0325/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 ¼10þ3/C036þ36/C02/C015¼/C04 (b) First reducejBjto a determinant of order 4, and then to a determinant of order 3, for which we can use Fig. 8-1. First use c 22¼1 as a pivot to put 0’s in the second column, by applying the row operations ‘‘Replace R1by/C02R2þR1,’’ ‘‘Replace R3by/C0R2þR3,’’ and ‘‘Replace R5byR2þR5.’’ Then jBj¼20/C0143 21 1/C021 /C01 0102 3 023 /C01 10/C0223/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼2/C014 3 /C011 02 32 3/C01 1/C022 3/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼11 45 01 00 52 3/C05 /C01/C022 7/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 ¼14 5 53/C05 /C012 7/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼21þ20þ50þ15þ10/C0140¼/C034CHAPTER 8 Determinants 277 Cofactors, Classical Adjoints, Minors, Principal Minors 8.5. LetA¼21/C034 5/C047/C02 406/C03 3/C02522 6643 775: (a) Find A23, the cofactor (signed minor) of 7 in A. (b) Find the minor and the signed minor of the submatrix M¼Að2;4;2;3Þ. (c) Find the principal minor determined by the ﬁrst and third diagonal entries—that is, by M¼Að1;3;1;3Þ. (a) Take the determinant of the submatrix of Aobtained by deleting row 2 and column 3 (those which contain the 7), and multiply the determinant by ð/C01Þ2þ3: A23¼/C0214 40/C03 3/C022/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C0ð/C0 61Þ¼61 The exponent 2þ3 comes from the subscripts of A 23—that is, from the fact that 7 appears in row 2 and column 3. (b) The row subscripts are 2 and 4 and the column subscripts are 2 and 3. Hence, the minor is the determinant jMj¼a22a23 a42a43/C12/C12/C12/C12/C12/C12/C12/C12¼/C047 /C025/C12/C12/C12/C12/C12/C12/C12/C12¼/C020þ14¼/C06 and the signed minor is ð/C01Þ2þ4þ2þ3jMj¼/C0j Mj¼/C0ð/C0 6Þ¼6. (c) The principal minor is the determinant jMj¼a11a13 a31a33/C12/C12/C12/C12/C12/C12/C12/C12¼2/C03 46/C12/C12/C12/C12/C12/C12/C12/C12¼12þ12¼24 Note that now the diagonal entries of the submatrix are diagonal entries of the original matrix. Also, the sign of the principal minor is positive. 8.6. LetB¼111 234 5892 43 5. Find: (a)jBj, (b) adj B, (c) B/C01using adj B. (a)jBj¼27þ20þ16/C015/C032/C018¼/C02 (b) Take the transpose of the matrix of cofactors: adjB¼34 89/C12/C12/C12/C12/C12/C12/C12/C12/C024 59/C12/C12/C12/C12/C12/C12/C12/C1223 58/C12/C12/C12/C12/C12/C12/C12/C12 /C011 89/C12/C12/C12/C12/C12/C12/C12/C1211 59/C12/C12/C12/C12/C12/C12/C12/C12/C011 58/C12/C12/C12/C12/C12/C12/C12/C12 11 34/C12/C12/C12/C12/C12/C12/C12/C12/C011 24/C12/C12/C12/C12/C12/C12/C12/C1211 23/C12/C12/C12/C12/C12/C12/C12/C122 6666666643 777777775T ¼/C0521 /C014/C03 1/C0212 643 75T ¼/C05/C011 24/C02 1/C0312 643 75 (c) BecausejBj6¼0,B/C01¼1 jBjðadjBÞ¼1 /C02/C05/C011 24/C02 1/C0312 43 5¼5 212/C012 /C01/C021 /C01 232/C0122 643 75 8.7. LetA¼123 456 0782 43 5, and let Skdenote the sum of its principal minors of order k. Find Skfor (a)k¼1, (b) k¼2, (c) k¼3.278 CHAPTER 8 Determinants (a) The principal minors of order 1 are the diagonal elements. Thus, S1is the trace of A; that is, S1¼trðAÞ¼1þ5þ8¼14 (b) The principal minors of order 2 are the cofactors of the diagonal elements. Thus, S2¼A11þA22þA33¼56 78/C12/C12/C12/C12/C12/C12/C12/C12þ13 08/C12/C12/C12/C12/C12/C12/C12/C12þ12 45/C12/C12/C12/C12/C12/C12/C12/C12¼/C02þ8/C03¼3 (c) There is only one principal minor of order 3, the determinant of A. Then S 3¼jAj¼40þ0þ84/C00/C042/C064¼18 8.8. LetA¼13 0/C01 /C042 51 10 3/C02 3/C021 42 6643 775. Find the number Nkand sum Skof principal minors of order: (a)k¼1, (b) k¼2, (c) k¼3, (d) k¼4. Each (nonempty) subset of the diagonal (or equivalently, each nonempty subset of f1;2;3;4gÞ determines a principal minor of A, and Nk¼n k/C18/C19 ¼n! k!ðn/C0kÞ!of them are of order k. Thus ;N1¼4 1/C18/C19 ¼4; N2¼4 2/C18/C19 ¼6; N3¼4 3/C18/C19 ¼4; N4¼4 4/C18/C19 ¼1 (a)S1¼j1jþj2jþj3jþj4j¼1þ2þ3þ4¼10 (b)S2¼13 /C042/C12/C12/C12/C12/C12/C12/C12/C12þ10 13/C12/C12/C12/C12/C12/C12/C12/C12þ1/C01 34/C12/C12/C12/C12/C12/C12/C12/C12þ25 03/C12/C12/C12/C12/C12/C12/C12/C12þ21 /C024/C12/C12/C12/C12/C12/C12/C12/C12þ3/C02 14/C12/C12/C12/C12/C12/C12/C12/C12 ¼14þ3þ7þ6þ10þ14¼54 (c)S 3¼130 /C0425 103/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þ13/C01 /C0421 3/C024/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þ10/C01 13/C02 31 4/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12þ25 1 03/C02 /C021 4/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 ¼57þ65þ22þ54¼198 (d)S 4¼detðAÞ¼378 Determinants and Systems of Linear Equations 8.9. Use determinants to solve the system3yþ2x¼zþ1 3xþ2z¼8/C05y 3z/C01¼x/C02y:8 < : First arrange the equation in standard form, then compute the determinant Dof the matrix of coefﬁcients: 2xþ3y/C0z¼1 3xþ5yþ2z¼8 x/C02y/C03z¼/C01and D¼23/C01 352 1/C02/C03/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C030þ6þ6þ5þ8þ27¼22 Because D6¼0, the system has a unique solution. To compute Nx;Ny;Nz, we replace, respectively, the coefﬁcients of x;y;zin the matrix of coefﬁcients by the constant terms. Then Nx¼13/C01 852 /C01/C02/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼66; Ny¼21/C01 382 1/C01/C03/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C022; Nz¼231 358 1/C02/C01/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼44CHAPTER 8 Determinants 279 Thus, x¼Nx D¼66 22¼3; y¼Ny D¼/C022 22¼/C01; z¼Nz D¼44 22¼2 8.10. Consider the systemkxþyþz¼1 xþkyþz¼1 xþyþkz¼18 < : Use determinants to ﬁnd those values of kfor which the system has (a) a unique solution, (b) more than one solution, (c) no solution. (a) The system has a unique solution when D6¼0, where Dis the determinant of the matrix of coefﬁcients. Compute D¼k11 1k1 11 k/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼k3þ1þ1/C0k/C0k/C0k¼k3/C03kþ2¼ðk/C01Þ2ðkþ2Þ Thus, the system has a unique solution when ðk/C01Þ2ðkþ2Þ6¼0;when k6¼1 and k6¼2 (b and c) Gaussian elimination shows that the system has more than one solution when k¼1, and the system has no solution when k¼/C02. Miscellaneous Problems 8.11. Find the volume VðSÞof the parallelepiped SinR3determined by the vectors: (a)u1¼ð1;1;1Þ;u2¼ð1;3;/C04Þ;u3¼ð1;2;/C05Þ. (b)u1¼ð1;2;4Þ;u2¼ð2;1;/C03Þ;u3¼ð5;7;9Þ. VðSÞis the absolute value of the determinant of the matrix Mwhose rows are the given vectors. Thus, (a)jMj¼11 1 13/C04 12/C05/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼/C015/C04þ2/C03þ8þ5¼/C07. Hence, VðSÞ¼j/C0 7j¼7. (b)jMj¼12 4 21/C03 57 9/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12¼9/C030þ56/C020þ21/C036¼0. Thus, VðSÞ¼0, or, in other words, u 1;u2;u3 lie in a plane and are linearly dependent. 8.12. Find detðMÞwhere M¼34000 25000 0920005067 004342 666643 77775¼34000 25000 0920005067 004342 666643 77775 Mis a (lower) triangular block matrix; hence, evaluate the determinant of each diagonal block: 34 25/C12/C12/C12/C12/C12/C12/C12/C12¼15/C08¼7;j2j¼2;67 34/C12/C12/C12/C12/C12/C12/C12/C12¼24/C021¼3 Thus,jMj¼7ð2Þð3Þ¼42. 8.13. Find the determinant of F:R3!R3deﬁned by Fðx;y;zÞ¼ð xþ3y/C04z;2yþ7z;xþ5y/C03zÞ280 CHAPTER 8 Determinants The determinant of a linear operator Fis equal to the determinant of any matrix that represents F. Thus ﬁrst ﬁnd the matrix Arepresenting Fin the usual basis (whose rows, respectively, consist of the coefﬁcients ofx;y;z). Then A¼13/C04 02 7 15/C032 43 5; and so detðFÞ¼j Aj¼/C0 6þ21þ0þ8/C035/C00¼/C08 8.14. Write out g¼gðx1;x2;x3;x4Þexplicitly where gðx1;x2;...;xnÞ¼Q i<jðxi/C0xjÞ: The symbolQis used for a product of terms in the same way that the symbolPis used for a sum of terms. That is,Q i<jðxi/C0xjÞmeans the product of all terms ðxi/C0xjÞfor which i<j. Hence, g¼gðx1;...;x4Þ¼ð x1/C0x2Þðx1/C0x3Þðx1/C0x4Þðx2/C0x3Þðx2/C0x4Þðx3/C0x4Þ 8.15. LetDbe a 2-linear, alternating function. Show that DðA;BÞ¼/C0 DðB;AÞ. Because Dis alternating, DðA;AÞ¼0,DðB;BÞ¼0. Hence, DðAþB;AþBÞ¼DðA;AÞþDðA;BÞþDðB;AÞþDðB;BÞ¼DðA;BÞþDðB;AÞ However, DðAþB;AþBÞ¼0. Hence, DðA;BÞ¼/C0 DðB;AÞ, as required. Permutations 8.16. Determine the parity (sign) of the permutation s¼364152. Count the number of inversions. That is, for each element k, count the number of elements iinssuch thati>kandiprecedes kins. Namely, k¼1:3 numbersð3;6;4Þ k¼2:4 numbersð3;6;4;5Þ k¼3:0 numbersk¼4:1 numberð6Þ k¼5:1 numberð6Þ k¼6:0 numbers Because 3þ4þ0þ1þ1þ0¼9 is odd, sis an odd permutation, and sgn s¼/C01. 8.17. Lets¼24513 and t¼41352 be permutations in S5. Find (a) t/C14s, (b) s/C01. Recall that s¼24513 and t¼41352 are short ways of writing s¼12345 24513/C18/C19 or sð1Þ¼2;sð2Þ¼4;sð3Þ¼5;sð4Þ¼1;sð5Þ¼3 t¼1234 5 41352 /C24c/C18/C19 or tð1Þ¼4;tð2Þ¼1;tð3Þ¼3;tð4Þ¼5;tð5Þ¼2 (a) The effects of sand then ton 1 ;2;3;4;5 are as follows: 1!2!1; 2!4!5; 3!5!2; 4!1!4; 5!3!3 [That is, for example, ðt/C14sÞð1Þ¼tðsð1ÞÞ¼ tð2Þ¼1:/C138Thus, t/C14s¼15243. (b) By deﬁnition, s/C01ðjÞ¼kif and only if sðkÞ¼j. Hence, s/C01¼24513 12345/C18/C19 ¼12345 41523/C18/C19 or s/C01¼41523 8.18. Lets¼j1j2...jnbe any permutation in Sn. Show that, for each inversion ði;kÞwhere i>kbuti precedes kins, there is a pairði*;j*Þsuch that i*<k* and sði*Þ>sðj*Þð 1Þ and vice versa. Thus, sis even or odd according to whether there is an even or an odd number of pairs satisfying (1).CHAPTER 8 Determinants 281 Choose i* and k* so that sði*Þ¼iandsðk*Þ¼k. Then i>kif and only if sði*Þ>sðk*Þ, and i precedes kinsif and only if i*<k*. 8.19. Consider the polynomials g¼gðx1;...;xnÞandsðgÞ, deﬁned by g¼gðx1;...;xnÞ¼Q i<jðxi/C0xjÞ and sðgÞ¼Q i<jðxsðiÞ/C0xsðjÞÞ (See Problem 8.14.) Show that sðgÞ¼gwhen sis an even permutation, and sðgÞ¼/C0 gwhen sis an odd permutation. That is, sðgÞ¼ð sgnsÞg. Because sis one-to-one and onto, sðgÞ¼Q i<jðxsðiÞ/C0xsðjÞÞ¼Q i<jori>jðxi/C0xjÞ Thus, sðgÞorsðgÞ¼/C0 gaccording to whether there is an even or an odd number of terms of the form xi/C0xj, where i>j. Note that for each pair ði;jÞfor which i<j and sðiÞ>sðjÞ there is a termðxsðiÞ/C0xsðjÞÞinsðgÞfor which sðiÞ>sðjÞ. Because sis even if and only if there is an even number of pairs satisfying (1), we have sðgÞ¼gif and only if sis even. Hence, sðgÞ¼/C0 gif and only if s is odd. 8.20. Lets;t2Sn. Show that sgnðt/C14sÞ¼ð sgntÞðsgnsÞ. Thus, the product of two even or two odd permutations is even, and the product of an odd and an even permutation is odd. Using Problem 8.19, we have sgnðt/C14sÞg¼ðt/C14sÞðgÞ¼tðsðgÞÞ¼ tððsgnsÞgÞ¼ð sgntÞðsgnsÞg Accordingly, sgnðt/C14sÞ¼ð sgntÞðsgnsÞ. 8.21. Consider the permutation s¼j1j2/C1/C1/C1jn. Show that sgn s/C01¼sgnsand, for scalars aij, show that aj11aj22/C1/C1/C1ajnn¼a1k1a2k2/C1/C1/C1ankn where s/C01¼k1k2/C1/C1/C1kn. We have s/C01/C14s¼e, the identity permutation. Because eis even, s/C01andsare both even or both odd. Hence sgn s/C01¼sgns. Because s¼j1j2/C1/C1/C1jnis a permutation, aj11aj22/C1/C1/C1ajnn¼a1k1a2k2/C1/C1/C1ankn. Then k1;k2;...;knhave the property that sðk1Þ¼1; sðk2Þ¼2; ...; sðknÞ¼n Lett¼k1k2/C1/C1/C1kn. Then, for i¼1;...;n, ðs/C14tÞðiÞ¼sðtðiÞÞ¼ sðkiÞ¼i Thus, s/C14t¼e, the identity permutation. Hence, t¼s/C01. Proofs of Theorems 8.22. Prove Theorem 8.1: jATj¼jAj. IfA¼½aij/C138, then AT¼½bij/C138, with bij¼aji. Hence, jATj¼P s2SnðsgnsÞb1sð1Þb2sð2Þ/C1/C1/C1bnsðnÞ¼P s2SnðsgnsÞasð1Þ;1asð2Þ;2/C1/C1/C1asðnÞ;n Lett¼s/C01. By Problem 8.21 sgn t¼sgns, and asð1Þ;1asð2Þ;2/C1/C1/C1asðnÞ;n¼a1tð1Þa2tð2Þ/C1/C1/C1antðnÞ. Hence, jATj¼P s2SnðsgntÞa1tð1Þa2tð2Þ/C1/C1/C1antðnÞ282 CHAPTER 8 Determinants However, as sruns through all the elements of Sn;t¼s/C01also runs through all the elements of Sn. Thus, jATj¼jAj. 8.23. Prove Theorem 8.3(i): If two rows (columns) of Aare interchanged, then jBj¼/C0j Aj. We prove the theorem for the case that two columns are interchanged. Let tbe the transposition that interchanges the two numbers corresponding to the two columns of Athat are interchanged. If A¼½aij/C138and B¼½bij/C138, then bij¼aitðjÞ. Hence, for any permutation s, b1sð1Þb2sð2Þ/C1/C1/C1bnsðnÞ¼a1ðt/C14sÞð1Þa2ðt/C14sÞð2Þ/C1/C1/C1anðt/C14sÞðnÞ Thus, jBj¼P s2SnðsgnsÞb1sð1Þb2sð2Þ/C1/C1/C1bnsðnÞ¼P s2SnðsgnsÞa1ðt/C14sÞð1Þa2ðt/C14sÞð2Þ/C1/C1/C1anðt/C14sÞðnÞ Because the transposition tis an odd permutation, sgn ðt/C14sÞ¼ð sgntÞðsgnsÞ¼/C0 sgns. Accordingly, sgns¼/C0sgnðt/C14sÞ;and so jBj¼/C0P s2Sn½sgnðt/C14sÞ/C138a1ðt/C14sÞð1Þa2ðt/C14sÞð2Þ/C1/C1/C1anðt/C14sÞðnÞ But as sruns through all the elements of Sn;t/C14salso runs through all the elements of Sn:Hence,jBj¼/C0j Aj. 8.24. Prove Theorem 8.2. (i) If Ahas a row (column) of zeros, then jAj¼0. (ii) If Ahas two identical rows (columns), then jAj¼0. (iii) If Ais triangular, then jAj¼product of diagonal elements. Thus, jIj¼1. (i) Each term injAjcontains a factor from every row, and so from the row of zeros. Thus, each term of jAj is zero, and sojAj¼0. (ii) Suppose 1þ16¼0i n K. If we interchange the two identical rows of A, we still obtain the matrix A. Hence, by Problem 8.23, jAj¼/C0j Aj, and sojAj¼0. Now suppose 1þ1¼0i n K. Then sgn s¼1 for every s2Sn:Because Ahas two identical rows, we can arrange the terms of Ainto pairs of equal terms. Because each pair is 0, the determinant ofAis zero. (iii) Suppose A¼½aij/C138is lower triangular; that is, the entries above the diagonal are all zero: aij¼0 whenever i<j. Consider a term tof the determinant of A: t¼ðsgnsÞa1i1a2i2/C1/C1/C1anin; where s¼i1i2/C1/C1/C1in Suppose i16¼1. Then 1 <i1and so a1i1¼0;hence, t¼0:That is, each term for which i16¼1i s zero. Now suppose i1¼1 but i26¼2. Then 2 <i2, and so a2i2¼0; hence, t¼0. Thus, each term for which i16¼1o r i26¼2 is zero. Similarly, we obtain that each term for which i16¼1o r i26¼2o r ...orin6¼nis zero. Accordingly,jAj¼a11a22/C1/C1/C1ann¼product of diagonal elements. 8.25. Prove Theorem 8.3: Bis obtained from Aby an elementary operation. (i) If two rows (columns) of Awere interchanged, then jBj¼/C0j Aj. (ii) If a row (column) of Awere multiplied by a scalar k, thenjBj¼kjAj. (iii) If a multiple of a row (column) of Awere added to another row (column) of A;thenjBj¼jAj. (i) This result was proved in Problem 8.23. (ii) If the jth row of Ais multiplied by k, then every term in jAjis multiplied by k,a n ds ojBj¼kjAj.T h a ti s , jBj¼P sðsgnsÞa1i1a2i2/C1/C1/C1ðkajijÞ/C1/C1/C1anin¼kP sðsgnsÞa1i1a2i2/C1/C1/C1anin¼kjAjCHAPTER 8 Determinants 283 (iii) Suppose ctimes the kth row is added to the jth row of A. Using the symbol ^to denote the jth position in a determinant term, we have jBj¼P sðsgnsÞa1i1a2i2/C1/C1/C1ðcakikþajijÞ...anin ¼cP sðsgnsÞa1i1a2i2/C1/C1/C1cakik/C1/C1/C1aninþP sðsgnsÞa1i1a2i2/C1/C1/C1ajij/C1/C1/C1anin The ﬁrst sum is the determinant of a matrix whose kth and jth rows are identical. Accordingly, by Theorem 8.2(ii), the sum is zero. The second sum is the determinant of A. Thus,jBj¼c/C10þjAj¼jAj. 8.26. Prove Lemma 8.6: Let Ebe an elementary matrix. Then jEAj¼jEjjAj. Consider the elementary row operations: (i) Multiply a row by a constant k6¼0, (ii) Interchange two rows, (iii) Add a multiple of one row to another. LetE1;E2;E3be the corresponding elementary matrices That is, E1;E2;E3are obtained by applying the above operations to the identity matrix I. By Problem 8.25, jE1j¼kjIj¼k;jE2j¼/C0j Ij¼/C0 1;jE3j¼jIj¼1 Recall (Theorem 3.11) that EiAis identical to the matrix obtained by applying the corresponding operation toA. Thus, by Theorem 8.3, we obtain the following which proves our lemma: jE1Aj¼kjAj¼jE1jjAj;jE2Aj¼/C0j Aj¼jE2jjAj;jE3Aj¼jAj¼1jAj¼jE3jjAj 8.27. Suppose Bis row equivalent to a square matrix A. Prove thatjBj¼0 if and only ifjAj¼0. By Theorem 8.3, the effect of an elementary row operation is to change the sign of the determinant or to multiply the determinant by a nonzero scalar. Hence, jBj¼0 if and only ifjAj¼0. 8.28. Prove Theorem 8.5: Let Abe an n-square matrix. Then the following are equivalent: (i) Ais invertible, (ii) AX¼0 has only the zero solution, (iii) det ðAÞ6¼0. The proof is by the Gaussian algorithm. If Ais invertible, it is row equivalent to I. ButjIj6¼0. Hence, by Problem 8.27,jAj6¼0. If Ais not invertible, it is row equivalent to a matrix with a zero row. Hence, detðAÞ¼0. Thus, (i) and (iii) are equivalent. IfAX¼0 has only the solution X¼0, then Ais row equivalent to IandAis invertible. Conversely, if Ais invertible with inverse A/C01, then X¼IX¼ðA/C01AÞX¼A/C01ðAXÞ¼A/C010¼0 is the only solution of AX¼0. Thus, (i) and (ii) are equivalent. 8.29. Prove Theorem 8.4: jABj¼jAjjBj. IfAis singular, then ABis also singular, and so jABj¼0¼jAjjBj. On the other hand, if Ais nonsingular, then A¼En/C1/C1/C1E2E1, a product of elementary matrices. Then, Lemma 8.6 and induction yields jABj¼jEn/C1/C1/C1E2E1Bj¼jEnj/C1/C1/C1j E2jjE1jjBj¼jAjjBj 8.30. Suppose Pis invertible. Prove that jP/C01j¼jPj/C01. P/C01P¼I:Hence ;1¼jIj¼jP/C01Pj¼jP/C01jjPj;and sojP/C01j¼jPj/C01: 8.31. Prove Theorem 8.7: Suppose AandBare similar matrices. Then jAj¼jBj. Because AandBare similar, there exists an invertible matrix Psuch that B¼P/C01AP. Therefore, using Problem 8.30, we get jBj¼jP/C01APj¼jP/C01jjAjjPj¼jAjjP/C01jjP¼jAj. We remark that although the matrices P/C01andAmay not commute, their determinants jP/C01jandjAjdo commute, because they are scalars in the ﬁeld K. 8.32. Prove Theorem 8.8 (Laplace): Let A¼½aij/C138,a n dl e t Aijdenote the cofactor of aij. Then, for any iorj jAj¼ai1Ai1þ/C1/C1/C1þ ainAin andjAj¼a1jA1jþ/C1/C1/C1þ anjAnjd284 CHAPTER 8 Determinants BecausejAj¼jATj, we need only prove one of the expansions, say, the ﬁrst one in terms of rows of A. Each term injAjcontains one and only one entry of the ith rowðai1;ai2;...;ainÞofA. Hence, we can write jAjin the form jAj¼ai1A*i1þai2A*i2þ/C1/C1/C1þ ainA*in (Note that A*ijis a sum of terms involving no entry of the ith row of A.) Thus, the theorem is proved if we can show that A*ij¼Aij¼ð/C0 1ÞiþjjMijj where Mijis the matrix obtained by deleting the row and column containing the entry aij:(Historically, the expression A*ijwas deﬁned as the cofactor of aij, and so the theorem reduces to showing that the two deﬁnitions of the cofactor are equivalent.) First we consider the case that i¼n,j¼n. Then the sum of terms in jAjcontaining annis annA*nn¼annP sðsgnsÞa1sð1Þa2sð2Þ/C1/C1/C1an/C01;sðn/C01Þ where we sum over all permutations s2Snfor which sðnÞ¼n. However, this is equivalent (Prove!) to summing over all permutations of f1;...;n/C01g. Thus, A*nn¼jMnnj¼ð/C0 1ÞnþnjMnnj. Now we consider any iandj. We interchange the ith row with each succeeding row until it is last, and we interchange the jth column with each succeeding column until it is last. Note that the determinant jMijjis not affected, because the relative positions of the other rows and columns are not affected by these interchanges. However, the ‘‘sign’’ of jAjand of A*ijis changed n/C01 and then n/C0jtimes. Accordingly, A*ij¼ð/C0 1Þn/C0iþn/C0jjMijj¼ð/C0 1ÞiþjjMijj 8.33. LetA¼½aij/C138and let Bbe the matrix obtained from Aby replacing the ith row of Aby the row vectorðbi1;...;binÞ. Show that jBj¼bi1Ai1þbi2Ai2þ/C1/C1/C1þ binAin Furthermore, show that, for j6¼i, aj1Ai1þaj2Ai2þ/C1/C1/C1þ ajnAin¼0 and a1jA1iþa2jA2iþ/C1/C1/C1þ anjAni¼0 LetB¼½bij/C138. By Theorem 8.8, jBj¼bi1Bi1þbi2Bi2þ/C1/C1/C1þ binBin Because Bijdoes not depend on the ith row of B;we get Bij¼Aijforj¼1;...;n. Hence, jBj¼bi1Ai1þbi2Ai2þ/C1/C1/C1þ binAin Now let A0be obtained from Aby replacing the ith row of Aby the jth row of A. Because A0has two identical rows,jA0j¼0. Thus, by the above result, jA0j¼aj1Ai1þaj2Ai2þ/C1/C1/C1þ ajnAin¼0 UsingjATj¼jAj, we also obtain that a1jA1iþa2jA2iþ/C1/C1/C1þ anjAni¼0. 8.34. Prove Theorem 8.9: AðadjAÞ¼ð adjAÞA¼jAjI. LetA¼½aij/C138and let AðadjAÞ¼½ bij/C138. The ith row of Ais ðai1;ai2;...;ainÞð 1Þ Because adj Ais the transpose of the matrix of cofactors, the jth column of adj Ais the tranpose of the cofactors of the jth row of A: ðAj;Aj2;...;AjnÞTð2Þ Now bij;theijentry in AðadjAÞ, is obtained by multiplying expressions (1) and (2): bij¼ai1Aj1þai2Aj2þ/C1/C1/C1þ ainAjnCHAPTER 8 Determinants 285 By Theorem 8.8 and Problem 8.33, bij¼jAjifi¼j 0i f i6¼j/C26 Accordingly, AðadjAÞis the diagonal matrix with each diagonal element jAj. In other words, AðadjAÞ¼j AjI. Similarly,ðadjAÞA¼jAjI. 8.35. Prove Theorem 8.10 (Cramer’s rule): The (square) system AX¼Bhas a unique solution if and only if D6¼0. In this case, xi¼Ni=Dfor each i. By previous results, AX¼Bhas a unique solution if and only if Ais invertible, and Ais invertible if and only if D¼jAj6¼0. Now suppose D6¼0. By Theorem 8.9, A/C01¼ð1=DÞðadjAÞ. Multiplying AX¼BbyA/C01, we obtain X¼A/C01AX¼ð1=DÞðadjAÞB ð1Þ Note that the ith row ofð1=DÞðadjAÞisð1=DÞðA1i;A2i;...;AniÞ.I fB¼ðb1;b2;...;bnÞT, then, by (1), xi¼ð1=DÞðb1A1iþb2A2iþ/C1/C1/C1þ bnAniÞ However, as in Problem 8.33, b1A1iþb2A2iþ/C1/C1/C1þ bnAni¼Ni, the determinant of the matrix obtained by replacing the ith column of Aby the column vector B. Thus, xi¼ð1=DÞNi, as required. 8.36. Prove Theorem 8.12: Suppose Mis an upper (lower) triangular block matrix with diagonal blocks A1;A2;...;An. Then detðMÞ¼detðA1ÞdetðA2Þ/C1/C1/C1detðAnÞ We need only prove the theorem for n¼2—that is, when Mis a square matrix of the form M¼AC 0B/C20/C21 . The proof of the general theorem follows easily by induction. Suppose A¼½aij/C138isr-square, B¼½bij/C138iss-square, and M¼½mij/C138isn-square, where n¼rþs.B y deﬁnition, detðMÞ¼P s2SnðsgnsÞm1sð1Þm2sð2Þ/C1/C1/C1mnsðnÞ Ifi>randj/C20r, then mij¼0. Thus, we need only consider those permutations ssuch that sfrþ1;rþ2;...;rþsg¼f rþ1;rþ2;...;rþsg and sf1;2;...;rg¼f 1;2;...;rg Lets1ðkÞ¼sðkÞfork/C20r, and let s2ðkÞ¼sðrþkÞ/C0rfork/C20s. Then ðsgnsÞm1sð1Þm2sð2Þ/C1/C1/C1mnsðnÞ¼ðsgns1Þa1s1ð1Þa2s1ð2Þ/C1/C1/C1ars1ðrÞðsgns2Þb1s2ð1Þb2s2ð2Þ/C1/C1/C1bss2ðsÞ which implies detðMÞ¼detðAÞdetðBÞ. 8.37. Prove Theorem 8.14: There exists a unique function D:M!Ksuch that (i) Dis multilinear, (ii) Dis alternating, (iii) DðIÞ¼1. This function Dis the determinant function; that is, DðAÞ¼j Aj. LetDbe the determinant function, DðAÞ¼j Aj. We must show that Dsatisﬁes (i), (ii), and (iii), and that Dis the only function satisfying (i), (ii), and (iii). By Theorem 8.2, Dsatisﬁes (ii) and (iii). Hence, we show that it is multilinear. Suppose the ith row of A¼½aij/C138has the formðbi1þci1;bi2þci2;...;binþcinÞ. Then DðAÞ¼DðA1;...;BiþCi;...;AnÞ ¼P SnðsgnsÞa1sð1Þ/C1/C1/C1ai/C01;sði/C01ÞðbisðiÞþcisðiÞÞ/C1/C1/C1ansðnÞ ¼P SnðsgnsÞa1sð1Þ/C1/C1/C1bisðiÞ/C1/C1/C1ansðnÞþP SnðsgnsÞa1sð1Þ/C1/C1/C1cisðiÞ/C1/C1/C1ansðnÞ ¼DðA1;...;Bi;...;AnÞþDðA1;...;Ci;...;AnÞ286 CHAPTER 8 Determinants Also, by Theorem 8.3(ii), DðA1;...;kAi;...;AnÞ¼kDðA1;...;Ai;...;AnÞ Thus, Dis multilinear— Dsatisﬁes (i). We next must prove the uniqueness of D. Suppose Dsatisﬁes (i), (ii), and (iii). If fe1;...;engis the usual basis of Kn, then, by (iii), Dðe1;e2;...;enÞ¼DðIÞ¼1. Using (ii), we also have that Dðei1;ei2;...;einÞ¼sgns; where s¼i1i2/C1/C1/C1in ð1Þ Now suppose A¼½aij/C138. Observe that the kth row AkofAis Ak¼ðak1;ak2;...;aknÞ¼ak1e1þak2e2þ/C1/C1/C1þ aknen Thus, DðAÞ¼Dða11e1þ/C1/C1/C1þ a1nen;a21e1þ/C1/C1/C1þ a2nen;...;an1e1þ/C1/C1/C1þ annenÞ Using the multilinearity of D, we can write DðAÞas a sum of terms of the form DðAÞ¼PDða1i1ei1;a2i2ei2;...;anineinÞ ¼Pða1i1a2i2/C1/C1/C1aninÞDðei1;ei2;...;einÞð 2Þ where the sum is summed over all sequences i1i2...in, where ik2f1;...;ng. If two of the indices are equal, sayij¼ikbutj6¼k, then, by (ii), Dðei1;ei2;...;einÞ¼0 Accordingly, the sum in (2) need only be summed over all permutations s¼i1i2/C1/C1/C1in. Using (1), we ﬁnally have that DðAÞ¼P sða1i1a2i2/C1/C1/C1aninÞDðei1;ei2;...;einÞ ¼P sðsgnsÞa1i1a2i2/C1/C1/C1anin; where s¼i1i2/C1/C1/C1in Hence, Dis the determinant function, and so the theorem is proved. SUPPLEMENTARY PROBLEMS Computation of Determinants 8.38. Evaluate: (a)26 41/C12/C12/C12/C12/C12/C12/C12/C12, (b)51 3/C02/C12/C12/C12/C12/C12/C12/C12/C12, (c)/C028 /C05/C03/C12/C12/C12/C12/C12/C12/C12/C12, (d)49 1/C03/C12/C12/C12/C12/C12/C12/C12/C12, (e)aþba baþb/C12/C12/C12/C12/C12/C12/C12/C12 8.39. Find all tsuch that (a)t/C043 2 t/C09/C12/C12/C12/C12/C12/C12/C12/C12¼0, (b)t/C014 3 t/C02/C12/C12/C12/C12/C12/C12/C12/C12¼0 8.40. Compute the determinant of each of the following matrices: (a)211 05/C02 1/C0342 43 5, (b)3/C02/C04 25/C01 0612 43 5, (c)/C02/C014 6/C03/C02 4122 43 5, (d)76 5 12 1 3/C0212 43 5CHAPTER 8 Determinants 287 8.41. Find the determinant of each of the following matrices: (a)1223 10/C020 3/C011/C02 4/C03022 6643 775, (b)2132 301/C02 1/C0143 22/C0112 6643 775 8.42. Evaluate: (a)2/C013/C04 21/C021 33/C054 52/C014/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12, (b)2/C014/C03 /C011 02 32 3/C01 1/C022/C03/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12, (c)1/C023/C01 11/C020 204/C05 144/C06/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 8.43. Evaluate each of the following determinants: (a)12/C0131 2/C011/C023 3102 /C01 512/C034 /C023/C011/C02/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12, (b)13579 2424200123 00562 00231/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12, (c)12345 5432100651 00074 00023/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 Cofactors, Classical Adjoints, Inverses 8.44. Find detðAÞ, adj A, and A/C01, where (a) A¼110 111 0212 43 5, (b) A¼122 310 1112 43 5 8.45. Find the classical adjoint of each matrix in Problem 8.41. 8.46. LetA¼ab cd/C20/C21 . (a) Find adj A, (b) Show that adj ðadjAÞ¼A, (c) When does A¼adjA? 8.47. Show that if Ais diagonal (triangular) then adj Ais diagonal (triangular). 8.48. Suppose A¼½aij/C138is triangular. Show that (a) Ais invertible if and only if each diagonal element aii6¼0. (b) The diagonal elements of A/C01(if it exists) are a/C01 ii, the reciprocals of the diagonal elements of A. Minors, Principal Minors 8.49. LetA¼1232 10/C023 3/C0125 4/C030/C012 6643 775and B¼13/C015 2/C0314 0/C0521 305/C022 6643 775. Find the minor and the signed minor corresponding to the following submatrices: (a) Að1;4;3;4Þ, (b) Bð1;4;3;4Þ, (c) Að2;3;2;4Þ, (d) Bð2;3;2;4Þ. 8.50. Fork¼1;2;3, ﬁnd the sum Skof all principal minors of order kfor (a) A¼13 2 2/C043 5/C0212 43 5, (b) B¼15/C04 261 3/C0202 43 5, (c) C¼1/C043 21 5 4/C071 12 43 5288 CHAPTER 8 Determinants 8.51. Fork¼1;2;3;4, ﬁnd the sum Skof all principal minors of order kfor (a) A¼123/C01 1/C0205 01/C022 40/C01/C032 6643 775, (b) B¼1212 0123 1304 27452 6643 775 Determinants and Linear Equations 8.52. Solve the following systems by determinants: (a)3xþ5y¼8 4x/C02y¼1/C26 , (b)2x/C03y¼/C01 4xþ7y¼/C01/C26 , (c)ax/C02by¼c 3ax/C05by¼2cðab6¼0Þ/C26 8.53. Solve the following systems by determinants: (a)2x/C05yþ2z¼2 xþ2y/C04z¼5 3x/C04y/C06z¼18 < :, (b)2zþ3¼yþ3x x/C03z¼2yþ1 3yþz¼2/C02x8 < : 8.54. Prove Theorem 8.11: The system AX¼0 has a nonzero solution if and only if D¼jAj¼0. Permutations 8.55. Find the parity of the permutations s¼32154, t¼13524, p¼42531 in S5. 8.56. For the permutations in Problem 8.55, ﬁnd (a) t/C14s, (b) p/C14s, (c) s/C01, (d) t/C01. 8.57. Lett2Sn:Show that t/C14sruns through Snassruns through Sn;that is, Sn¼ft/C14s:s2Sng: 8.58. Lets2Snhave the property that sðnÞ¼n. Let s*2Sn/C01be deﬁned by s*ðxÞ¼sðxÞ. (a) Show that sgn s*¼sgns, (b) Show that as sruns through Sn, where sðnÞ¼n,s* runs through Sn/C01; that is, Sn/C01¼fs*:s2Sn;sðnÞ¼ng: 8.59. Consider a permutation s¼j1j2...jn. Letfeigbe the usual basis of Kn, and let Abe the matrix whose ith row is eji[i.e., A¼ðej1,ej2;...;ejnÞ]. Show thatjAj¼sgns. Determinant of Linear Operators 8.60. Find the determinant of each of the following linear transformations: (a) T:R2!R2deﬁned by Tðx;yÞ¼ð 2x/C09y;3x/C05yÞ, (b) T:R3!R3deﬁned by Tðx;y;zÞ¼ð 3x/C02z;5yþ7z;xþyþzÞ, (c) T:R3!R2deﬁned by Tðx;y;zÞ¼ð 2xþ7y/C04z;4x/C06yþ2zÞ. 8.61. LetD:V!Vbe the differential operator; that is, DðfðtÞÞ¼ df=dt. Find detðDÞifVis the vector space of functions with the following bases: (a) f1;t;...;t5g, (b)fet;e2t;e3tg, (c)fsint;costg. 8.62. Prove Theorem 8.13: Let FandGbe linear operators on a vector space V. Then (i) detðF/C14GÞ¼detðFÞdetðGÞ, (ii) Fis invertible if and only if det ðFÞ6¼0. 8.63. Prove (a) detð1VÞ¼1, where 1Vis the identity operator, (b) -det ðT/C01Þ¼detðTÞ/C01when Tis invertible.CHAPTER 8 Determinants 289 Miscellaneous Problems 8.64. Find the volume VðSÞof the parallelopiped SinR3determined by the following vectors: (a) u1¼ð1;2;/C03Þ,u2¼ð3;4;/C01Þ,u3¼ð2;/C01;5Þ, (b) u1¼ð1;1;3Þ,u2¼ð1;/C02;/C04Þ,u3¼ð4;1;5Þ. 8.65. Find the volume VðSÞof the parallelepiped SinR4determined by the following vectors: u1¼ð1;/C02;5;/C01Þ;u2¼ð2;1;/C02;1Þ;u3¼ð3;0;1/C02Þ;u4¼ð1;/C01;4;/C01Þ 8.66. LetVbe the space of 2/C22 matrices M¼ab cd/C20/C21 over R. Determine whether D:V!Ris 2-linear (with respect to the rows), where ðaÞDðMÞ¼aþd; ðbÞDðMÞ¼ad;ðcÞDðMÞ¼ac/C0bd; ðdÞDðMÞ¼ab/C0cd;ðeÞDðMÞ¼0 ðfÞDðMÞ¼1 8.67. LetAbe an n-square matrix. Prove jkAj¼knjAj. 8.68. LetA;B;C;Dbe commuting n-square matrices. Consider the 2 n-square block matrix M¼AB CD/C20/C21 . Prove thatjMj¼jAjjDj/C0jBjjCj. Show that the result may not be true if the matrices do not commute. 8.69. Suppose Ais orthogonal; that is, ATA¼I. Show that detðAÞ¼/C6 1. 8.70. LetVbe the space of m-square matrices viewed as m-tuples of row vectors. Suppose D:V!Kism-linear and alternating. Show that (a) Dð...;A;...;B;...Þ¼/C0 Dð...;B;...;A;...Þ; sign changed when two rows are interchanged. (b) If A1;A2;...;Amare linearly dependent, then DðA1;A2;...;AmÞ¼0. 8.71. LetVbe the space of m-square matrices (as above), and suppose D:V!K. Show that the following weaker statement is equivalent to Dbeing alternating: DðA1;A2;...;AnÞ¼0 whenever Ai¼Aiþ1for some i LetVbe the space of n-square matrices over K. Suppose B2Vis invertible and so det ðBÞ6¼0. Deﬁne D:V!KbyDðAÞ¼detðABÞ=detðBÞ, where A2V. Hence, DðA1;A2;...;AnÞ¼detðA1B;A2B;...;AnBÞ=detðBÞ where Aiis the ith row of A, and so AiBis the ith row of AB. Show that Dis multilinear and alternating, and thatDðIÞ¼1. (This method is used by some texts to prove that jABj¼jAjjBj.) 8.72. Show that g¼gðx1;...;xnÞ¼ð/C0 1ÞnVn/C01ðxÞwhere g¼gðxiÞis the difference product in Problem 8.19, x¼xn, and Vn/C01is the Vandermonde determinant deﬁned by Vn/C01ðxÞ/C1711 ... 11 x1 x2 ... xn/C01 x x2 1 x22 ... x2 n/C01x2 :::::::::::::::::::::::::::::::::::::::::::: xn/C01 1 xn/C01 2 ... xn/C01 n/C01xn/C012 66666664/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 8.73. LetAbe any matrix. Show that the signs of a minor A½I;J/C138and its complementary minor A½I0;J0/C138are equal.290 CHAPTER 8 Determinants 8.74. LetAbe an n-square matrix. The determinantal rank ofAis the order of the largest square submatrix of A (obtained by deleting rows and columns of A) whose determinant is not zero. Show that the determinantal rank of Ais equal to its rank—the maximum number of linearly independent rows (or columns). ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M¼½R1;R2;. . ./C138denotes a matrix with rows R1;R2;:... 8.38. (a)/C022, (b)/C013, (c) 46, (d) /C021, (e) a2þabþb2 8.39. (a) 3 ;10; (b) 5 ;/C02 8.40. (a) 21, (b) /C011, (c) 100, (d) 0 8.41. (a)/C0131, (b)/C055 8.42. (a) 33, (b) 0, (c) 45 8.43. (a)/C032, (b)/C014, (c)/C0468 8.44. (a)jAj¼/C0 2; adjA¼½/C0 1;/C01;1;/C01;1;/C01;2;/C02;0/C138, (b)jAj¼/C0 1; adjA¼½1;0;/C02;/C03;/C01;6;2;1;/C05/C138. Also, A/C01¼ðadjAÞ=jAj 8.45. (a)½/C016;/C029;/C026;/C02;/C030;/C038;/C016;29;/C08;51;/C013;/C01;/C013;1;28;/C018/C138, (b)½21;/C014;/C017;/C019;/C044;11;33;11;/C029;1;13;21; 17;7;/C019;/C018/C138 8.46. (a) adj A¼½d;/C0b;/C0c;a/C138, (c) A¼kI 8.49. (a)/C03;/C03, (b)/C023;/C023, (c) 3 ;/C03, (d) 17 ;/C017 8.50. (a)/C02;/C017;73, (b) 7 ;10;105, (c) 13 ;54;0 8.51. (a)/C06;13;62;/C0219 ; (b) 7 ;/C037;30;20 8.52. (a) x¼21 26;y¼29 26; (b) x¼/C05 13;y¼1 13; (c) x¼/C0c a;y¼/C0c b 8.53. (a) x¼5;y¼2;z¼1, (b) Because D¼0, the system cannot be solved by determinants. 8.55. (a) sgn s¼1;sgnt¼/C01;sgnp¼/C01 8.56. (a) t/C14s¼53142, (b) p/C14s¼52413, (c) s/C01¼32154, (d) t/C01¼14253 8.60. (a) detðTÞ¼17, (b) detðTÞ¼4, (c) not deﬁned 8.61. (a) 0, (b) 6, (c) 1 8.64. (a) 18, (b) 0 8.65. 17 8.66. (a) no, (b) yes, (c) yes, (d) no, (e) yes, (f ) noCHAPTER 8 Determinants 291 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 9.1 Introduction The ideas in this chapter can be discussed from two points of view. Matrix Point of View Suppose an n-square matrix Ais given. The matrix Ais said to be diagonalizable if there exists a nonsingular matrix Psuch that B¼P/C01AP is diagonal. This chapter discusses the diagonalization of a matrix A. In particular, an algorithm is given to ﬁnd the matrix Pwhen it exists. Linear Operator Point of View Suppose a linear operator T:V!Vis given. The linear operator Tis said to be diagonalizable if there exists a basis SofVsuch that the matrix representation of Trelative to the basis Sis a diagonal matrix D. This chapter discusses conditions under which the linear operator Tis diagonalizable. Equivalence of the Two Points of View The above two concepts are essentially the same. Speciﬁcally, a square matrix Amay be viewed as a linear operator Fdeﬁned by FðXÞ¼AX where Xis a column vector, and B¼P/C01APrepresents Frelative to a new coordinate system (basis) Swhose elements are the columns of P. On the other hand, any linear operator Tcan be represented by a matrix Arelative to one basis and, when a second basis is chosen, Tis represented by the matrix B¼P/C01AP where Pis the change-of-basis matrix. Most theorems will be stated in two ways: one in terms of matrices Aand again in terms of linear mappings T. Role of Underlying Field K The underlying number ﬁeld Kdid not play any special role in our previous discussions on vector spaces and linear mappings. However, the diagonalization of a matrix Aor a linear operator Twill depend on the CHAPTER 9 292 roots of a polynomial DðtÞover K, and these roots do depend on K. For example, suppose DðtÞ¼t2þ1. ThenDðtÞhas no roots if K¼R, the real ﬁeld; but DðtÞhas roots/C6iifK¼C, the complex ﬁeld. Furthermore, ﬁnding the roots of a polynomial with degree greater than two is a subject unto itself (frequently discussed in numerical analysis courses). Accordingly, our examples will usually lead to those polynomials DðtÞwhose roots can be easily determined. 9.2 Polynomials of Matrices Consider a polynomial fðtÞ¼antnþ/C1/C1/C1þ a1tþa0over a ﬁeld K. Recall (Section 2.8) that if Ais any square matrix, then we deﬁne fðAÞ¼anAnþ/C1/C1/C1þ a1Aþa0I where Iis the identity matrix. In particular, we say that Ais aroot offðtÞiffðAÞ¼0, the zero matrix. EXAMPLE 9.1 LetA¼12 34/C20/C21 . Then A2¼71 0 15 22/C20/C21 . Let fðtÞ¼2t2/C03tþ5 and gðtÞ¼t2/C05t/C02 Then fðAÞ¼2A2/C03Aþ5I¼14 20 30 44/C20/C21 þ/C03/C06 /C09/C012/C20/C21 þ50 05/C20/C21 ¼16 14 21 37/C20/C21 and gðAÞ¼A2/C05A/C02I¼71 0 15 22/C20/C21 þ/C05/C010 /C015/C020/C20/C21 þ/C020 0/C02/C20/C21 ¼00 00/C20/C21 Thus, Ais a zero of gðtÞ. The following theorem (proved in Problem 9.7) applies. THEOREM 9.1: Letfandgbe polynomials. For any square matrix Aand scalar k, (i)ðfþgÞðAÞ¼fðAÞþgðAÞ (iii)ðkfÞðAÞ¼kfðAÞ (ii)ðfgÞðAÞ¼fðAÞgðAÞ (iv) fðAÞgðAÞ¼gðAÞfðAÞ: Observe that (iv) tells us that any two polynomials in Acommute. Matrices and Linear Operators Now suppose that T:V!Vis a linear operator on a vector space V. Powers of Tare deﬁned by the composition operation: T2¼T/C14T; T3¼T2/C14T; ... Also, for any polynomial fðtÞ¼antnþ/C1/C1/C1þ a1tþa0, we deﬁne fðTÞin the same way as we did for matrices: fðTÞ¼anTnþ/C1/C1/C1þ a1Tþa0I where Iis now the identity mapping. We also say that Tis azero orroot offðtÞiffðTÞ¼0;the zero mapping. We note that the relations in Theorem 9.1 hold for linear operators as they do for matrices. Remark: Suppose Ais a matrix representation of a linear operator T. Then fðAÞis the matrix representation of fðTÞ, and, in particular, fðTÞ¼0 if and only if fðAÞ¼0.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 293 9.3 Characteristic Polynomial, Cayley–Hamilton Theorem LetA¼½aij/C138be an n-square matrix. The matrix M¼A/C0tIn, where Inis the n-square identity matrix and tis an indeterminate, may be obtained by subtracting tdown the diagonal of A. The negative of Mis the matrix tIn/C0A, and its determinant DðtÞ¼detðtIn/C0AÞ¼ð/C0 1ÞndetðA/C0tInÞ which is a polynomial in tof degree nand is called the characteristic polynomial ofA. We state an important theorem in linear algebra (proved in Problem 9.8). THEOREM 9.2: (Cayley–Hamilton) Every matrix Ais a root of its characteristic polynomial. Remark: Suppose A¼½aij/C138is a triangular matrix. Then tI/C0Ais a triangular matrix with diagonal entries t/C0aii; hence, DðtÞ¼detðtI/C0AÞ¼ð t/C0a11Þðt/C0a22Þ/C1/C1/C1ð t/C0annÞ Observe that the roots of DðtÞare the diagonal elements of A. EXAMPLE 9.2 LetA¼13 45/C20/C21 . Its characteristic polynomial is DðtÞ¼j tI/C0Aj¼t/C01/C03 /C04t/C05/C12/C12/C12/C12¼ðt/C01Þðt/C05Þ/C012¼t2/C06t/C07/C12/C12/C12/C12 As expected from the Cayley–Hamilton theorem, Ais a root of DðtÞ; that is, DðAÞ¼A2/C06A/C07I¼13 18 24 37/C20/C21 þ/C06/C018 /C024/C030/C20/C21 þ/C070 0/C07/C20/C21 ¼00 00/C20/C21 Now suppose AandBare similar matrices, say B¼P/C01AP, where Pis invertible. We show that A andBhave the same characteristic polynomial. Using tI¼P/C01tIP, we have DBðtÞ¼detðtI/C0BÞ¼detðtI/C0P/C01APÞ¼detðP/C01tIP/C0P/C01APÞ ¼det½P/C01ðtI/C0AÞP/C138¼detðP/C01ÞdetðtI/C0AÞdetðPÞ Using the fact that determinants are scalars and commute and that det ðP/C01ÞdetðPÞ¼1, we ﬁnally obtain DBðtÞ¼detðtI/C0AÞ¼DAðtÞ Thus, we have proved the following theorem. THEOREM 9.3: Similar matrices have the same characteristic polynomial. Characteristic Polynomials of Degrees 2 and 3 There are simple formulas for the characteristic polynomials of matrices of orders 2 and 3. (a) Suppose A¼a11a12 a21a22/C20/C21 . Then DðtÞ¼t2/C0ða11þa22ÞtþdetðAÞ¼t2/C0trðAÞtþdetðAÞ Here trðAÞdenotes the trace of A—that is, the sum of the diagonal elements of A. (b) Suppose A¼a11a12a13 a21a22a23 a31a32a332 43 5. Then DðtÞ¼t3/C0trðAÞt2þðA11þA22þA33Þt/C0detðAÞ (Here A11,A22,A33denote, respectively, the cofactors of a11,a22,a33.)294 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors EXAMPLE 9.3 Find the characteristic polynomial of each of the following matrices: (a)A¼53 21 0/C20/C21 , (b) B¼7/C01 62/C20/C21 , (c) C¼5/C02 4/C04/C20/C21 . (a) We have trðAÞ¼5þ10¼15 andjAj¼50/C06¼44; hence, DðtÞþt2/C015tþ44. (b) We have trðBÞ¼7þ2¼9 andjBj¼14þ6¼20; hence, DðtÞ¼t2/C09tþ20. (c) We have trðCÞ¼5/C04¼1 andjCj¼/C0 20þ8¼/C012; hence, DðtÞ¼t2/C0t/C012. EXAMPLE 9.4 Find the characteristic polynomial of A¼112 0321392 43 5. We have trðAÞ¼1þ3þ9¼13. The cofactors of the diagonal elements are as follows: A11¼32 39/C12/C12/C12/C12/C12/C12/C12/C12¼21; A22¼12 19/C12/C12/C12/C12/C12/C12/C12/C12¼7; A33¼11 03/C12/C12/C12/C12/C12/C12/C12/C12¼3 Thus, A11þA22þA33¼31. Also,jAj¼27þ2þ0/C06/C06/C00¼17. Accordingly, DðtÞ¼t3/C013t2þ31t/C017 Remark: The coefﬁcients of the characteristic polynomial DðtÞof the 3-square matrix Aare, with alternating signs, as follows: S1¼trðAÞ; S2¼A11þA22þA33; S3¼detðAÞ We note that each Skis the sum of all principal minors of Aof order k. The next theorem, whose proof lies beyond the scope of this text, tells us that this result is true in general. THEOREM 9.4: LetAbe an n-square matrix. Then its characteristic polynomial is DðtÞ¼tn/C0S1tn/C01þS2tn/C02þ/C1/C1/C1þð/C0 1ÞnSn where Skis the sum of the principal minors of order k. Characteristic Polynomial of a Linear Operator Now suppose T:V!Vis a linear operator on a vector space Vof ﬁnite dimension. We deﬁne the characteristic polynomial DðtÞofTto be the characteristic polynomial of any matrix representation of T. Recall that if AandBare matrix representations of T, then B¼P/C01AP, where Pis a change-of-basis matrix. Thus, AandBare similar, and by Theorem 9.3, AandBhave the same characteristic polynomial. Accordingly, the characteristic polynomial of Tis independent of the particular basis in which the matrix representation of Tis computed. Because fðTÞ¼0 if and only if fðAÞ¼0, where fðtÞis any polynomial and Ais any matrix representation of T, we have the following analogous theorem for linear operators. THEOREM 9.20:(Cayley–Hamilton) A linear operator Tis a zero of its characteristic polynomial.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 295 9.4 Diagonalization, Eigenvalues and Eigenvectors LetAbe any n-square matrix. Then Acan be represented by (or is similar to) a diagonal matrix D¼diagðk1;k2;...;knÞif and only if there exists a basis Sconsisting of (column) vectors u1;u2;...;un such that Au1¼k1u1 Au2¼ k2u2 :::::::::::::::::::::::::::::::::::: Aun¼ knun In such a case, Ais said to be diagonizable . Furthermore, D¼P/C01AP, where Pis the nonsingular matrix whose columns are, respectively, the basis vectors u1;u2;...;un. The above observation leads us to the following deﬁnition. DEFINITION: LetAbe any square matrix. A scalar lis called an eigenvalue ofAif there exists a nonzero (column) vector vsuch that Av¼lv Any vector satisfying this relation is called an eigenvector ofA belonging to the eigenvalue l. We note that each scalar multiple kvof an eigenvector vbelonging to lis also such an eigenvector, because AðkvÞ¼kðAvÞ¼kðlvÞ¼lðkvÞ The set Elof all such eigenvectors is a subspace of V(Problem 9.19), called the eigenspace ofl. (If dimEl¼1, then Elis called an eigenline andlis called a scaling factor .) The terms characteristic value andcharacteristic vector (orproper value andproper vector ) are sometimes used instead of eigenvalue and eigenvector. The above observation and deﬁnitions give us the following theorem. THEOREM 9.5: Ann-square matrix Ais similar to a diagonal matrix Di fa n do n l yi f Ahasnlinearly independent eigenvectors. In this case, the diagonal elements of Dare the corresponding eigenvalues and D¼P/C01AP,w h e r e Pis the matrix whose columns are the eigenvectors. Suppose a matrix Acan be diagonalized as above, say P/C01AP¼D, where Dis diagonal. Then Ahas the extremely useful diagonal factorization : A¼PDP/C01 Using this factorization, the algebra of Areduces to the algebra of the diagonal matrix D, which can be easily calculated. Speciﬁcally, suppose D¼diagðk1;k2;...;knÞ. Then Am¼ðPDP/C01Þm¼PDmP/C01¼Pdiagðkm 1;...;km nÞP/C01 More generally, for any polynomial fðtÞ, fðAÞ¼fðPDP/C01Þ¼PfðDÞP/C01¼Pdiagðfðk1Þ;fðk2Þ;...;fðknÞÞP/C01 Furthermore, if the diagonal entries of Dare nonnegative, let B¼Pdiagðﬃﬃﬃﬃﬃ k1p ;ﬃﬃﬃﬃﬃ k2p ;...;ﬃﬃﬃﬃﬃ knp ÞP/C01 Then Bis anonnegative square root ofA; that is, B2¼Aand the eigenvalues of Bare nonnegative.296 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors EXAMPLE 9.5 LetA¼31 22/C20/C21 and let v1¼1 /C02/C20/C21 and v2¼1 1/C20/C21 . Then Av1¼31 22/C20/C21 1 /C02/C20/C21 ¼1 /C02/C20/C21 ¼v1 and Av2¼31 22/C20/C21 1 1/C20/C21 ¼4 4/C20/C21 ¼4v2 Thus, v1andv2are eigenvectors of Abelonging, respectively, to the eigenvalues l1¼1 and l2¼4. Observe that v1 and v2are linearly independent and hence form a basis of R2. Accordingly, Ais diagonalizable. Furthermore, let P be the matrix whose columns are the eigenvectors v1and v2. That is, let P¼" 11 /C021# ; and so P/C01¼1 3/C013 23 13"# Then Ais similar to the diagonal matrix D¼P/C01AP¼1 3/C013 23 13"# " 31 22#" 11 /C021# ¼" 10 04# As expected, the diagonal elements 1 and 4 in Dare the eigenvalues corresponding, respectively, to the eigenvectors v1and v2, which are the columns of P. In particular, Ahas the factorization A¼PDP/C01¼" 11 /C021#" 10 04#1 3/C013 23 13"# Accordingly, A4¼" 11 /C021#" 10 0 256#1 3/C013 23 13"# ¼" 171 85 170 86# Moreover, suppose fðtÞ¼t3/C05t2þ3tþ6; hence, fð1Þ¼5 and fð4Þ¼2. Then fðAÞ¼PfðDÞP/C01¼11 /C021/C20/C2150 02/C20/C211 3/C013 23 13"# ¼3/C01 /C024/C20/C21 Last, we obtain a ‘‘positive square root’’ of A. Speciﬁcally, usingﬃﬃﬃ 1p ¼1 andﬃﬃﬃ 4p ¼2, we obtain the matrix B¼Pﬃﬃﬃﬃ Dp P/C01¼11 /C021/C20/C2110 02/C20/C211 3/C013 2 313"# ¼5 313 2 343"# where B2¼Aand where Bhas positive eigenvalues 1 and 2. Remark: Throughout this chapter, we use the following fact: IfP¼ab cd/C20/C21 ;then P/C01¼d=jPj/C0 b=jPj /C0c=jPj a=jPj/C20/C21 : That is, P/C01is obtained by interchanging the diagonal elements aanddofP, taking the negatives of the nondiagonal elements bandc, and dividing each element by the determinant jPj. Properties of Eigenvalues and Eigenvectors Example 9.5 indicates the advantages of a diagonal representation (factorization) of a square matrix. In the following theorem (proved in Problem 9.20), we list properties that help us to ﬁnd such arepresentation.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 297 THEOREM 9.6: LetAbe a square matrix. Then the following are equivalent. (i) A scalar lis an eigenvalue of A. (ii) The matrix M¼A/C0lIis singular. (iii) The scalar lis a root of the characteristic polynomial DðtÞofA. The eigenspace Elof an eigenvalue lis the solution space of the homogeneous system MX¼0, where M¼A/C0lI; that is, Mis obtained by subtracting ldown the diagonal of A. Some matrices have no eigenvalues and hence no eigenvectors. However, using Theorem 9.6 and the Fundamental Theorem of Algebra (every polynomial over the complex ﬁeld Chas a root), we obtain the following result. THEOREM 9.7: LetAbe a square matrix over the complex ﬁeld C. Then Ahas at least one eigenvalue. The following theorems will be used subsequently. (The theorem equivalent to Theorem 9.8 for linear operators is proved in Problem 9.21, and Theorem 9.9 is proved in Problem 9.22.) THEOREM 9.8: Suppose v1;v2;...;vnare nonzero eigenvectors of a matrix Abelonging to distinct eigenvalues l1;l2;...;ln. Then v1;v2;...;vnare linearly independent. THEOREM 9.9: Suppose the characteristic polynomial DðtÞof an n-square matrix Ais a product of n distinct factors, say, DðtÞ¼ð t/C0a1Þðt/C0a2Þ/C1/C1/C1ð t/C0anÞ. Then Ais similar to the diagonal matrix D¼diagða1;a2;...;anÞ. Iflis an eigenvalue of a matrix A, then the algebraic multiplicity oflis deﬁned to be the multiplicity oflas a root of the characteristic polynomial of A, and the geometric multiplicity oflis deﬁned to be the dimension of its eigenspace, dim El. The following theorem (whose equivalent for linear operators is proved in Problem 9.23) holds. THEOREM 9.10: The geometric multiplicity of an eigenvalue lof a matrix Adoes not exceed its algebraic multiplicity. Diagonalization of Linear Operators Consider a linear operator T:V!V. Then Tis said to be diagonalizable if it can be represented by a diagonal matrix D. Thus, Tis diagonalizable if and only if there exists a basis S¼fu1;u2;...;ungofV for which Tðu1Þ¼k1u1 Tðu2Þ¼ k2u2 ::::::::::::::::::::::::::::::::::::::: TðunÞ¼ knun In such a case, Tis represented by the diagonal matrix D¼diagðk1;k2;...;knÞ relative to the basis S. The above observation leads us to the following deﬁnitions and theorems, which are analogous to the deﬁnitions and theorems for matrices discussed above. DEFINITION: LetTbe a linear operator. A scalar lis called an eigenvalue ofTif there exists a nonzero vector vsuch that TðvÞ¼lv. Every vector satisfying this relation is called an eigenvector ofT belonging to the eigenvalue l.298 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors The set Elof all eigenvectors belonging to an eigenvalue lis a subspace of V, called the eigenspace ofl. (Alternatively, lis an eigenvalue of TiflI/C0Tis singular, and, in this case, Elis the kernel of lI/C0T.) The algebraic andgeometric multiplicities of an eigenvalue lof a linear operator Tare deﬁned in the same way as those of an eigenvalue of a matrix A. The following theorems apply to a linear operator Ton a vector space Vof ﬁnite dimension. THEOREM 9.50:Tcan be represented by a diagonal matrix Dif and only if there exists a basis SofV consisting of eigenvectors of T. In this case, the diagonal elements of Dare the corresponding eigenvalues. THEOREM 9.60:LetTbe a linear operator. Then the following are equivalent: (i) A scalar lis an eigenvalue of T. (ii) The linear operator lI/C0Tis singular. (iii) The scalar lis a root of the characteristic polynomial DðtÞofT. THEOREM 9.70:Suppose Vis a complex vector space. Then Thas at least one eigenvalue. THEOREM 9.80:Suppose v1;v2;...;vnare nonzero eigenvectors of a linear operator Tbelonging to distinct eigenvalues l1;l2;...;ln. Then v1;v2;...;vnare linearly independent. THEOREM 9.90:Suppose the characteristic polynomial DðtÞofTis a product of ndistinct factors, say, DðtÞ¼ð t/C0a1Þðt/C0a2Þ/C1/C1/C1ð t/C0anÞ. Then Tcan be represented by the diagonal matrix D¼diagða1;a2;...;anÞ. THEOREM 9.100:The geometric multiplicity of an eigenvalue lofTdoes not exceed its algebraic multiplicity. Remark: The following theorem reduces the investigation of the diagonalization of a linear operator Tto the diagonalization of a matrix A. THEOREM 9.11: Suppose Ais a matrix representation of T. Then Tis diagonalizable if and only if A is diagonalizable. 9.5 Computing Eigenvalues and Eigenvectors, Diagonalizing Matrices This section gives an algorithm for computing eigenvalues and eigenvectors for a given square matrix A and for determining whether or not a nonsingular matrix Pexists such that P/C01APis diagonal. ALGORITHM 9.1: (Diagonalization Algorithm) The input is an n-square matrix A. Step 1. Find the characteristic polynomial DðtÞofA. Step 2. Find the roots of DðtÞto obtain the eigenvalues of A. Step 3. Repeat (a) and (b) for each eigenvalue lofA. (a) Form the matrix M¼A/C0lIby subtracting ldown the diagonal of A. (b) Find a basis for the solution space of the homogeneous system MX¼0. (These basis vectors are linearly independent eigenvectors of Abelonging to l.)CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 299 Step 4. Consider the collection S¼fv1;v2;...;vmgof all eigenvectors obtained in Step 3. (a) If m6¼n, then Ais not diagonalizable. (b) If m¼n, then Ais diagonalizable. Speciﬁcally, let Pbe the matrix whose columns are the eigenvectors v1;v2;...;vn. Then D¼P/C01AP¼diagðl1;l2;...;lnÞ where liis the eigenvalue corresponding to the eigenvector vi. EXAMPLE 9.6 The diagonalizable algorithm is applied to A¼42 3/C01/C20/C21 . (1) The characteristic polynomial DðtÞofAis computed. We have trðAÞ¼4/C01¼/C03;jAj¼/C0 4/C06¼/C010; hence, DðtÞ¼t2/C03t/C010¼ðt/C05Þðtþ2Þ (2) Set DðtÞ¼ð t/C05Þðtþ2Þ¼0. The roots l1¼5 and l2¼/C02 are the eigenvalues of A. (3) (i) We ﬁnd an eigenvector v1ofAbelonging to the eigenvalue l1¼5. Subtract l1¼5 down the diagonal of Ato obtain the matrix M¼/C012 3/C06/C20/C21 . The eigenvectors belonging to l1¼5 form the solution of the homogeneous system MX¼0; that is, /C012 3/C06/C20/C21 x y/C20/C21 ¼0 0/C20/C21 or/C0xþ2y¼0 3x/C06y¼0or/C0xþ2y¼0 The system has only one free variable. Thus, a nonzero solution, for example, v1¼ð2;1Þ,i sa n eigenvector that spans the eigenspace of l1¼5. (ii) We ﬁnd an eigenvector v2ofAbelonging to the eigenvalue l2¼/C02. Subtract/C02 (or add 2) down the diagonal of Ato obtain the matrix M¼62 31/C20/C21 and the homogenous system6xþ2y¼0 3xþy¼0or 3 xþy¼0: The system has only one independent solution. Thus, a nonzero solution, say v2¼ð/C0 1;3Þ;is an eigenvector that spans the eigenspace of l2¼/C02: (4) Let Pbe the matrix whose columns are the eigenvectors v1and v2. Then P¼2/C01 13/C20/C21 ; and so P/C01¼3 717 /C01 727"# Accordingly, D¼P/C01APis the diagonal matrix whose diagonal entries are the corresponding eigenvalues; that is, D¼P/C01AP¼3 717 /C01727"# 42 3/C01/C20/C212/C01 13/C20/C21 ¼50 0/C02/C20/C21 EXAMPLE 9.7 Consider the matrix B¼5/C01 13/C20/C21 . We have trðBÞ¼5þ3¼8;jBj¼15þ1¼16; soDðtÞ¼t2/C08tþ16¼ðt/C04Þ2 Accordingly, l¼4 is the only eigenvalue of B.300 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Subtract l¼4 down the diagonal of Bto obtain the matrix M¼1/C01 1/C01/C20/C21 and the homogeneous systemx/C0y¼0 x/C0y¼0or x/C0y¼0 The system has only one independent solution; for example, x¼1;y¼1. Thus, v¼ð1;1Þand its multiples are the only eigenvectors of B. Accordingly, Bis not diagonalizable, because there does not exist a basis consisting of eigenvectors of B. EXAMPLE 9.8 Consider the matrix A¼3/C05 2/C03/C20/C21 . Here trðAÞ¼3/C03¼0 andjAj¼/C0 9þ10¼1. Thus, DðtÞ¼t2þ1 is the characteristic polynomial of A. We consider two cases: (a)Ais a matrix over the real ﬁeld R. Then DðtÞhas no (real) roots. Thus, Ahas no eigenvalues and no eigenvectors, and so Ais not diagonalizable. (b)Ais a matrix over the complex ﬁeld C. ThenDðtÞ¼ð t/C0iÞðtþiÞhas two roots, iand/C0i. Thus, Ahas two distinct eigenvalues iand/C0i, and hence, Ahas two independent eigenvectors. Accordingly there exists a nonsingular matrix Pover the complex ﬁeld Cfor which P/C01AP¼i0 0/C0i/C20/C21 Therefore, Ais diagonalizable (over C). 9.6 Diagonalizing Real Symmetric Matrices and Quadratic Forms There are many real matrices Athat are not diagonalizable. In fact, some real matrices may not have any (real) eigenvalues. However, if Ais a real symmetric matrix, then these problems do not exist. Namely, we have the following theorems. THEOREM 9.12: LetAbe a real symmetric matrix. Then each root lof its characteristic polynomial is real. THEOREM 9.13: LetAbe a real symmetric matrix. Suppose uand vare eigenvectors of Abelonging to distinct eigenvalues l1andl2. Then uand vare orthogonal, that; is, hu;vi¼0. The above two theorems give us the following fundamental result. THEOREM 9.14: LetAbe a real symmetric matrix. Then there exists an orthogonal matrix Psuch that D¼P/C01APis diagonal. The orthogonal matrix Pis obtained by normalizing a basis of orthogonal eigenvectors of Aas illustrated below. In such a case, we say that Ais ‘‘orthogonally diagonalizable.’’ EXAMPLE 9.9 LetA¼2/C02 /C025/C20/C21 , a real symmetric matrix. Find an orthogonal matrix Psuch that P/C01APis diagonal. First we ﬁnd the characteristic polynomial DðtÞofA. We have trðAÞ¼2þ5¼7;jAj¼10/C04¼6; soDðtÞ¼t2/C07tþ6¼ðt/C06Þðt/C01Þ Accordingly, l1¼6 and l2¼1 are the eigenvalues of A. (a) Subtracting l1¼6 down the diagonal of Ayields the matrix M¼/C04/C02 /C02/C01/C20/C21 and the homogeneous system/C04x/C02y¼0 /C02x/C0y¼0or 2 xþy¼0 A nonzero solution is u1¼ð1;/C02Þ.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 301 (b) Subtracting l2¼1 down the diagonal of Ayields the matrix M¼1/C02 /C024/C20/C21 and the homogeneous system x/C02y¼0 (The second equation drops out, because it is a multiple of the ﬁrst equation.) A nonzero solution is u2¼ð2;1Þ. As expected from Theorem 9.13, u1andu2are orthogonal. Normalizing u1andu2yields the orthonormal vectors ^u1¼ð1=ﬃﬃﬃ 5p ;/C02=ﬃﬃﬃ 5p Þ and ^u2¼ð2=ﬃﬃﬃ 5p ;1=ﬃﬃﬃ 5p Þ Finally, let Pbe the matrix whose columns are ^u1and ^u2, respectively. Then P¼1=ﬃﬃﬃ 5p 2=ﬃﬃﬃ 5p /C02=ﬃﬃﬃ 5p 1=ﬃﬃﬃ 5p/C20/C21 and P/C01AP¼60 01/C20/C21 As expected, the diagonal entries of P/C01APare the eigenvalues corresponding to the columns of P. The procedure in the above Example 9.9 is formalized in the following algorithm, which ﬁnds an orthogonal matrix Psuch that P/C01APis diagonal. ALGORITHM 9.2: (Orthogonal Diagonalization Algorithm) The input is a real symmetric matrix A. Step 1. Find the characteristic polynomial DðtÞofA. Step 2. Find the eigenvalues of A, which are the roots of DðtÞ. Step 3. For each eigenvalue lofAin Step 2, ﬁnd an orthogonal basis of its eigenspace. Step 4. Normalize all eigenvectors in Step 3, which then forms an orthonormal basis of Rn. Step 5. LetPbe the matrix whose columns are the normalized eigenvectors in Step 4. Application to Quadratic Forms Letqbe a real polynomial in variables x1;x2;...;xnsuch that every term in qhas degree two; that is, qðx1;x2;...;xnÞ¼P icix2 iþP i<jdijxixj; where ci;dij2R Then qis called a quadratic form . If there are no cross-product terms xixj(i.e., all dij¼0), then qis said to be diagonal . The above quadratic form qdetermines a real symmetric matrix A¼½aij/C138, where aii¼ciand aij¼aji¼1 2dij. Namely, qcan be written in the matrix form qðXÞ¼XTAX where X¼½x1;x2;...;xn/C138Tis the column vector of the variables. Furthermore, suppose X¼PYis a linear substitution of the variables. Then substitution in the quadratic form yields qðYÞ¼ð PYÞTAðPYÞ¼YTðPTAPÞY Thus, PTAPis the matrix representation of qin the new variables. We seek an orthogonal matrix Psuch that the orthogonal substitution X ¼PYyields a diagonal quadratic form for which PTAPis diagonal. Because Pis orthogonal, PT¼P/C01, and hence, PTAP¼P/C01AP. The above theory yields such an orthogonal matrix P.302 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors EXAMPLE 9.10 Consider the quadratic form qðx;yÞ¼2x2/C04xyþ5y2¼XTAX; where A¼2/C02 /C025/C20/C21 and X¼x y/C20/C21 By Example 9.9, P/C01AP¼60 01/C20/C21 ¼PTAP; where P¼1=ﬃﬃﬃ 5p 2=ﬃﬃﬃ 5p /C02=ﬃﬃﬃ 5p 1=ﬃﬃﬃ 5p"# LetY¼½s;t/C138T:Then matrix Pcorresponds to the following linear orthogonal substitution x¼PYof the variables x andyin terms of the variables sandt: x¼1ﬃﬃﬃ 5psþ2ﬃﬃﬃ 5pt; y¼/C02ﬃﬃﬃ 5psþ1ﬃﬃﬃ 5pt This substitution in qðx;yÞyields the diagonal quadratic form qðs;tÞ¼6s2þt2. 9.7 Minimal Polynomial LetAbe any square matrix. Let JðAÞdenote the collection of all polynomials fðtÞfor which Ais a root— that is, for which fðAÞ¼0. The set JðAÞis not empty, because the Cayley–Hamilton Theorem 9.1 tells us that the characteristic polynomial DAðtÞofAbelongs to JðAÞ. Let mðtÞdenote the monic polynomial of lowest degree in JðAÞ. (Such a polynomial mðtÞexists and is unique.) We call mðtÞtheminimal polynomial of the matrix A. Remark: A polynomial fðtÞ6¼0i smonic if its leading coefﬁcient equals one. The following theorem (proved in Problem 9.33) holds. THEOREM 9.15: The minimal polynomial mðtÞof a matrix (linear operator) Adivides every polynomial that has Aas a zero. In particular, mðtÞdivides the characteristic polynomial DðtÞofA. There is an even stronger relationship between mðtÞandDðtÞ. THEOREM 9.16: The characteristic polynomial DðtÞand the minimal polynomial mðtÞof a matrix A have the same irreducible factors. This theorem (proved in Problem 9.35) does not say that mðtÞ¼DðtÞ, only that any irreducible factor of one must divide the other. In particular, because a linear factor is irreducible, mðtÞandDðtÞhave the same linear factors. Hence, they have the same roots. Thus, we have the following theorem. THEOREM 9.17: A scalar lis an eigenvalue of the matrix Aif and only if lis a root of the minimal polynomial of A. EXAMPLE 9.11 Find the minimal polynomial mðtÞofA¼22/C05 37/C015 12/C042 43 5. First ﬁnd the characteristic polynomial DðtÞofA. We have trðAÞ¼5; A11þA22þA33¼2/C03þ8¼7; andjAj¼3 Hence, DðtÞ¼t3/C05t2þ7t/C03¼ðt/C01Þ2ðt/C03ÞCHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 303 The minimal polynomial mðtÞmust divide DðtÞ. Also, each irreducible factor of DðtÞ(i.e., t/C01 and t/C03) must also be a factor of mðtÞ. Thus, mðtÞis exactly one of the following: fðtÞ¼ð t/C03Þðt/C01Þ or gðtÞ¼ð t/C03Þðt/C01Þ2 We know, by the Cayley–Hamilton theorem, that gðAÞ¼DðAÞ¼0. Hence, we need only test fðtÞ. We have fðAÞ¼ð A/C0IÞðA/C03IÞ¼12/C05 36/C015 12/C052 43 5/C012/C05 34/C015 12/C072 43 5¼000 000 0002 43 5 Thus, fðtÞ¼mðtÞ¼ð t/C01Þðt/C03Þ¼t2/C04tþ3 is the minimal polynomial of A. EXAMPLE 9.12 (a) Consider the following two r-square matrices, where a6¼0: Jðl;rÞ¼l10 ... 00 0l1 ... 00 ::::::::::::::::::::::::::::::::: 000 ... l 1 000 ... 0 l2 666643 77775and A¼la0 ... 00 0la ... 00 ::::::::::::::::::::::::::::::::: 000 ... l a 000 ... 0 l2 666643 77775 The ﬁrst matrix, called a Jordan Block, has l’s on the diagonal, 1’s on the superdiagonal (consisting of the entries above the diagonal entries), and 0’s elsewhere. The second matrix Ahasl’s on the diagonal, a’s on the superdiagonal, and 0’s elsewhere. [Thus, Ais a generalization of Jðl;rÞ.] One can show that fðtÞ¼ð t/C0lÞr is both the characteristic and minimal polynomial of both Jðl;rÞandA. (b) Consider an arbitrary monic polynomial: fðtÞ¼tnþan/C01tn/C01þ/C1/C1/C1þ a1tþa0 LetCðfÞbe the n-square matrix with 1’s on the subdiagonal (consisting of the entries below the diagonal entries), the negatives of the coefﬁcients in the last column, and 0’s elsewhere as follows: CðfÞ¼00 ... 0/C0a0 10 ... 0/C0a1 01 ... 0/C0a2 :::::::::::::::::::::::::::::::::: 00 ... 1/C0an/C012 666643 77775 Then CðfÞis called the companion matrix of the polynomial fðtÞ. Moreover, the minimal polynomial mðtÞand the characteristic polynomial DðtÞof the companion matrix CðfÞare both equal to the original polynomial fðtÞ. Minimal Polynomial of a Linear Operator The minimal polynomial m ðtÞof a linear operator Tis deﬁned to be the monic polynomial of lowest degree for which Tis a root. However, for any polynomial fðtÞ, we have fðTÞ¼0 if and only if fðAÞ¼0 where Ais any matrix representation of T. Accordingly, TandAhave the same minimal polynomials. Thus, the above theorems on the minimal polynomial of a matrix also hold for the minimal polynomial ofa linear operator. That is, we have the following theorems. THEOREM 9.150:The minimal polynomial mðtÞof a linear operator Tdivides every polynomial that hasTas a root. In particular, mðtÞdivides the characteristic polynomial DðtÞofT.304 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors THEOREM 9.160:The characteristic and minimal polynomials of a linear operator Thave the same irreducible factors. THEOREM 9.170:A scalar lis an eigenvalue of a linear operator Tif and only if lis a root of the minimal polynomial mðtÞofT. 9.8 Characteristic and Minimal Polynomials of Block Matrices This section discusses the relationship of the characteristic polynomial and the minimal polynomial to certain (square) block matrices. Characteristic Polynomial and Block Triangular Matrices Suppose Mis a block triangular matrix, say M¼A1B 0A2/C20/C21 , where A1andA2are square matrices. Then tI/C0Mis also a block triangular matrix, with diagonal blocks tI/C0A1andtI/C0A2. Thus, jtI/C0Mj¼tI/C0A1/C0B 0 tI/C0A2/C12/C12/C12/C12/C12/C12/C12/C12¼jtI/C0A1jjtI/C0A2j That is, the characteristic polynomial of Mis the product of the characteristic polynomials of the diagonal blocks A1andA2. By induction, we obtain the following useful result. THEOREM 9.18: Suppose Mis a block triangular matrix with diagonal blocks A1;A2;...;Ar. Then the characteristic polynomial of Mis the product of the characteristic polynomials of the diagonal blocks Ai; that is, DMðtÞ¼DA1ðtÞDA2ðtÞ...DArðtÞ EXAMPLE 9.13 Consider the matrix M¼9/C0157 832/C04 0036 00/C0182 6643 775. Then Mis a block triangular matrix with diagonal blocks A¼9/C01 83/C20/C21 andB¼36 /C018/C20/C21 . Here trðAÞ¼9þ3¼12; trðBÞ¼3þ8¼11;detðAÞ¼27þ8¼35; detðBÞ¼24þ6¼30;and so and soDAðtÞ¼t2/C012tþ35¼ðt/C05Þðt/C07Þ DBðtÞ¼t2/C011tþ30¼ðt/C05Þðt/C06Þ Accordingly, the characteristic polynomial of Mis the product DMðtÞ¼DAðtÞDBðtÞ¼ð t/C05Þ2ðt/C06Þðt/C07Þ Minimal Polynomial and Block Diagonal Matrices The following theorem (proved in Problem 9.36) holds. THEOREM 9.19: Suppose Mis a block diagonal matrix with diagonal blocks A1;A2;...;Ar. Then the minimal polynomial of Mis equal to the least common multiple (LCM) of the minimal polynomials of the diagonal blocks Ai. Remark: We emphasize that this theorem applies to block diagonal matrices, whereas the analogous Theorem 9.18 on characteristic polynomials applies to block triangular matrices.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 305 EXAMPLE 9.14 Find the characteristic polynomal DðtÞand the minimal polynomial mðtÞof the block diagonal matrix: M¼25000 0200000420 00350 000072 666643 77775¼diagðA 1;A2;A3Þ;where A1¼25 02/C20/C21 ;A2¼42 35/C20/C21 ;A3¼½7/C138 ThenDðtÞis the product of the characterization polynomials D1ðtÞ,D2ðtÞ,D3ðtÞofA1;A2;A3, respectively. One can show that D1ðtÞ¼ð t/C02Þ2; D2ðtÞ¼ð t/C02Þðt/C07Þ; D3ðtÞ¼t/C07 Thus,DðtÞ¼ð t/C02Þ3ðt/C07Þ2. [As expected, deg DðtÞ¼5:/C138 The minimal polynomials m1ðtÞ,m2ðtÞ,m3ðtÞof the diagonal blocks A1;A2;A3, respectively, are equal to the characteristic polynomials; that is, m1ðtÞ¼ð t/C02Þ2; m2ðtÞ¼ð t/C02Þðt/C07Þ; m3ðtÞ¼t/C07 ButmðtÞis equal to the least common multiple of m1ðtÞ;m2ðtÞ;m3ðtÞ. Thus, mðtÞ¼ð t/C02Þ2ðt/C07Þ. SOLVED PROBLEMS Polynomials of Matrices, Characteristic Polynomials 9.1. LetA¼1/C02 45/C20/C21 . Find fðAÞ, where ðaÞfðtÞ¼t2/C03tþ7;ðbÞfðtÞ¼t2/C06tþ13 First ﬁnd A2¼1/C02 45/C20/C21 1/C02 45/C20/C21 ¼/C07/C012 24 17/C20/C21 . Then (a) fðAÞ¼A2/C03Aþ7I¼/C07/C012 24 17/C20/C21 þ/C036 /C012/C015/C20/C21 þ70 07/C20/C21 ¼/C03/C06 12 9/C20/C21 (b) fðAÞ¼A2/C06Aþ13I¼/C07/C012 24 17/C20/C21 þ/C061 2 /C024/C030/C20/C21 þ13 0 01 3/C20/C21 ¼00 00/C20/C21 [Thus, Ais a root of fðtÞ.] 9.2. Find the characteristic polynomial DðtÞof each of the following matrices: (a) A¼25 41/C20/C21 , (b) B¼7/C03 5/C02/C20/C21 , (c) C¼3/C02 9/C03/C20/C21 Use the formulaðtÞ¼t2/C0trðMÞtþjMjfor a 2/C22 matrix M: (a) trðAÞ¼2þ1¼3,jAj¼2/C020¼/C018, so DðtÞ¼t2/C03t/C018 (b) trðBÞ¼7/C02¼5,jBj¼/C0 14þ15¼1, so DðtÞ¼t2/C05tþ1 (c) trðCÞ¼3/C03¼0,jCj¼/C0 9þ18¼9, so DðtÞ¼t2þ9 9.3. Find the characteristic polynomial DðtÞof each of the following matrices: (a) A¼123 304 6452 43 5, (b) B¼16/C02 /C032 0 03/C042 43 5306 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Use the formula DðtÞ¼t3/C0trðAÞt2þðA11þA22þA33Þt/C0jAj, where Aiiis the cofactor of aiiin the 3/C23 matrix A¼½aij/C138. (a) trðAÞ¼1þ0þ5¼6, A11¼04 45/C12/C12/C12/C12/C12/C12/C12/C12¼/C016; A22¼13 65/C12/C12/C12/C12/C12/C12/C12/C12¼/C013; A33¼12 30/C12/C12/C12/C12/C12/C12/C12/C12¼/C06 A11þA22þA33¼/C035, andjAj¼48þ36/C016/C030¼38 Thus ; DðtÞ¼t3/C06t2/C035t/C038 (b) trðBÞ¼1þ2/C04¼/C01 B11¼20 3/C04/C12/C12/C12/C12/C12/C12/C12/C12¼/C08; B22¼1/C02 0/C04/C12/C12/C12/C12/C12/C12/C12/C12¼/C04; B33¼16 /C032/C12/C12/C12/C12/C12/C12/C12/C12¼20 B11þB22þB33¼8, andjBj¼/C0 8þ18/C072¼/C062 Thus ; DðtÞ¼t3þt2/C08tþ62 9.4. Find the characteristic polynomial DðtÞof each of the following matrices: (a)A¼251 1 142 2 006/C05 002 32 6643 775, (b) B¼1122 0334 0055 00062 6643 775 (a)Ais block triangular with diagonal blocks A1¼25 14/C20/C21 and A2¼6/C05 23/C20/C21 Thus ; DðtÞ¼DA1ðtÞDA2ðtÞ¼ð t2/C06tþ3Þðt2/C09tþ28Þ (b) Because Bis triangular, DðtÞ¼ð t/C01Þðt/C03Þðt/C05Þðt/C06Þ. 9.5. Find the characteristic polynomial DðtÞof each of the following linear operators: (a)F:R2!R2deﬁned by Fðx;yÞ¼ð 3xþ5y;2x/C07yÞ. (b)D:V!Vdeﬁned by DðfÞ¼df=dt, where Vis the space of functions with basis S¼fsint;costg. The characteristic polynomial DðtÞof a linear operator is equal to the characteristic polynomial of any matrix Athat represents the linear operator. (a) Find the matrix Athat represents Trelative to the usual basis of R2. We have A¼35 2/C07/C20/C21 ; soDðtÞ¼t2/C0trðAÞtþjAj¼t2þ4t/C031 (b) Find the matrix Arepresenting the differential operator Drelative to the basis S. We have DðsintÞ¼cost¼0ðsintÞþ1ðcostÞ DðcostÞ¼/C0 sint¼/C01ðsintÞþ0ðcostÞand so A¼0/C01 10/C20/C21 DðtÞ¼t2/C0trðAÞtþjAj¼t2þ1 Therefore ; 9.6. Show that a matrix Aand its transpose AThave the same characteristic polynomial. By the transpose operation, ðtI/C0AÞT¼tIT/C0AT¼tI/C0AT. Because a matrix and its transpose have the same determinant, DAðtÞ¼j tI/C0Aj¼jð tI/C0AÞTj¼jtI/C0ATj¼DATðtÞCHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 307 9.7. Prove Theorem 9.1: Let fandgbe polynomials. For any square matrix Aand scalar k, (i)ðfþgÞðAÞ¼fðAÞþgðAÞ, (iii)ðkfÞðAÞ¼kfðAÞ, (ii)ðfgÞðAÞ¼fðAÞgðAÞ, (iv) fðAÞgðAÞ¼gðAÞfðAÞ. Suppose f¼antnþ/C1/C1/C1þ a1tþa0andg¼bmtmþ/C1/C1/C1þ b1tþb0. Then, by deﬁnition, fðAÞ¼anAnþ/C1/C1/C1þ a1Aþa0I and gðAÞ¼bmAmþ/C1/C1/C1þ b1Aþb0I (i) Suppose m/C20nand let bi¼0i fi>m. Then fþg¼ðanþbnÞtnþ/C1/C1/C1þð a1þb1Þtþða0þb0Þ Hence, ðfþgÞðAÞ¼ð anþbnÞAnþ/C1/C1/C1þð a1þb1ÞAþða0þb0ÞI ¼anAnþbnAnþ/C1/C1/C1þ a1Aþb1Aþa0Iþb0I¼fðAÞþgðAÞ (ii) By deﬁnition, fg¼cnþmtnþmþ/C1/C1/C1þ c1tþc0¼Pnþm k¼0cktk, where ck¼a0bkþa1bk/C01þ/C1/C1/C1þ akb0¼Pk i¼0aibk/C0i Hence,ðfgÞðAÞ¼Pnþm k¼0ckAkand fðAÞgðAÞ¼Pn i¼0aiAi/C18/C19 /C18Pm j¼0bjAj/C19 ¼Pn i¼0Pm j¼0aibjAiþj¼Pnþm k¼0ckAk¼ðfgÞðAÞ (iii) By deﬁnition, kf¼kantnþ/C1/C1/C1þ ka1tþka0, and so ðkfÞðAÞ¼kanAnþ/C1/C1/C1þ ka1Aþka0I¼kðanAnþ/C1/C1/C1þ a1Aþa0IÞ¼kfðAÞ (iv) By (ii), gðAÞfðAÞ¼ð gfÞðAÞ¼ð fgÞðAÞ¼fðAÞgðAÞ. 9.8. Prove the Cayley–Hamilton Theorem 9.2: Every matrix Ais a root of its characterstic polynomial DðtÞ. LetAbe an arbitrary n-square matrix and let DðtÞbe its characteristic polynomial, say, DðtÞ¼j tI/C0Aj¼tnþan/C01tn/C01þ/C1/C1/C1þ a1tþa0 Now let BðtÞdenote the classical adjoint of the matrix tI/C0A. The elements of BðtÞare cofactors of the matrix tI/C0Aand hence are polynomials in tof degree not exceeding n/C01. Thus, BðtÞ¼Bn/C01tn/C01þ/C1/C1/C1þ B1tþB0 where the Biaren-square matrices over Kwhich are independent of t. By the fundamental property of the classical adjoint (Theorem 8.9), ðtI/C0AÞBðtÞ¼j tI/C0AjI,o r ðtI/C0AÞðBn/C01tn/C01þ/C1/C1/C1þ B1tþB0Þ¼ð tnþan/C01tn/C01þ/C1/C1/C1þ a1tþa0ÞI Removing the parentheses and equating corresponding powers of tyields Bn/C01¼I; Bn/C02/C0ABn/C01¼an/C01I; ...; B0/C0AB1¼a1I;/C0AB0¼a0I Multiplying the above equations by An;An/C01;...;A;I, respectively, yields AnBn/C01¼AnI; An/C01Bn/C02/C0AnBn/C01¼an/C01An/C01; ...; AB0/C0A2B1¼a1A;/C0AB0¼a0I Adding the above matrix equations yields 0 on the left-hand side and DðAÞon the right-hand side; that is, 0¼Anþan/C01An/C01þ/C1/C1/C1þ a1Aþa0I Therefore, DðAÞ¼0, which is the Cayley–Hamilton theorem.308 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Eigenvalues and Eigenvectors of 2 /C22 Matrices 9.9. LetA¼3/C04 2/C06/C20/C21 . (a) Find all eigenvalues and corresponding eigenvectors. (b) Find matrices PandDsuch that Pis nonsingular and D¼P/C01APis diagonal. (a) First ﬁnd the characteristic polynomial DðtÞofA: DðtÞ¼t2/C0trðAÞtþjAj¼t2þ3t/C010¼ðt/C02Þðtþ5Þ The roots l¼2 and l¼/C05o fDðtÞare the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l¼2 down the diagonal of Ato obtain the matrix M¼A/C02I, where the corresponding homogeneous system MX¼0 yields the eigenvectors corresponding to l¼2. We have M¼1/C04 2/C08/C20/C21 ; corresponding tox/C04y¼0 2x/C08y¼0or x/C04y¼0 The system has only one free variable, and v1¼ð4;1Þis a nonzero solution. Thus, v1¼ð4;1Þis an eigenvector belonging to (and spanning the eigenspace of) l¼2. (ii) Subtract l¼/C05 (or, equivalently, add 5) down the diagonal of Ato obtain M¼8/C04 2/C01/C20/C21 ; corresponding to8x/C04y¼0 2x/C0y¼0or 2 x/C0y¼0 The system has only one free variable, and v2¼ð1;2Þis a nonzero solution. Thus, v2¼ð1;2Þis an eigenvector belonging to l¼5. (b) Let Pbe the matrix whose columns are v1and v2. Then P¼41 12/C20/C21 and D¼P/C01AP¼20 0/C05/C20/C21 Note that Dis the diagonal matrix whose diagonal entries are the eigenvalues of Acorresponding to the eigenvectors appearing in P. Remark: Here Pis the change-of-basis matrix from the usual basis of R2to the basis S¼fv1;v2g,a n d Dis the matrix that represents (the matrix function) Arelative to the new basis S. 9.10. LetA¼22 13/C20/C21 . (a) Find all eigenvalues and corresponding eigenvectors. (b) Find a nonsingular matrix Psuch that D¼P/C01APis diagonal, and P/C01. (c) Find A6andfðAÞ, where t4/C03t3/C06t2þ7tþ3. (d) Find a ‘‘real cube root’’ of B—that is, a matrix Bsuch that B3¼AandBhas real eigenvalues. (a) First ﬁnd the characteristic polynomial DðtÞofA: DðtÞ¼t2/C0trðAÞtþjAj¼t2/C05tþ4¼ðt/C01Þðt/C04Þ The roots l¼1 and l¼4o fDðtÞare the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l¼1 down the diagonal of Ato obtain the matrix M¼A/C0lI, where the corresponding homogeneous system MX¼0 yields the eigenvectors belonging to l¼1. We have M¼12 12/C20/C21 ; corresponding toxþ2y¼0 xþ2y¼0or xþ2y¼0CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 309 The system has only one independent solution; for example, x¼2,y¼/C01. Thus, v1¼ð2;/C01Þis an eigenvector belonging to (and spanning the eigenspace of) l¼1. (ii) Subtract l¼4 down the diagonal of Ato obtain M¼/C022 1/C01/C20/C21 ; corresponding to/C02xþ2y¼0 x/C0y¼0or x/C0y¼0 The system has only one independent solution; for example, x¼1,y¼1. Thus, v2¼ð1;1Þis an eigenvector belonging to l¼4. (b) Let Pbe the matrix whose columns are v1and v2. Then P¼21 /C011/C20/C21 and D¼P/C01AP¼10 04/C20/C21 ; where P/C01¼1 3/C013 13 23"# (c) Using the diagonal factorization A¼PDP/C01, and 16¼1 and 46¼4096, we get A6¼PD6P/C01¼21 /C011"# 10 0 4096"#1 3/C013 1 323"# ¼1366 2230 1365 2731"# Also, fð1Þ¼2 and fð4Þ¼/C0 1. Hence, fðAÞ¼PfðDÞP/C01¼21 /C011"# 20 0/C01"#1 3/C013 1 323"# ¼12 /C010"# (d) Here10 0ﬃﬃﬃ 43p/C20/C21 is the real cube root of D. Hence the real cube root of Ais B¼Pﬃﬃﬃﬃ D3p P/C01¼21 /C011"# 10 0ﬃﬃﬃ 43p"#1 3/C013 1 323"# ¼1 32þﬃﬃﬃ 43p /C02þ2ﬃﬃﬃ 43p /C01þﬃﬃﬃ 43p 1þ2ﬃﬃﬃ 43p"# 9.11. Each of the following real matrices deﬁnes a linear transformation on R2: (a)A¼56 3/C02/C20/C21 , (b) B¼1/C01 2/C01/C20/C21 , (c) C¼5/C01 13/C20/C21 Find, for each matrix, all eigenvalues and a maximum set Sof linearly independent eigenvectors. Which of these linear operators are diagonalizable—that is, which can be represented by a diagonal matrix? (a) First ﬁnd DðtÞ¼t2/C03t/C028¼ðt/C07Þðtþ4Þ. The roots l¼7 and l¼/C04 are the eigenvalues of A. We ﬁnd corresponding eigenvectors. (i) Subtract l¼7 down the diagonal of Ato obtain M¼/C026 3/C09/C20/C21 ; corresponding to/C02xþ6y¼0 3x/C09y¼0or x/C03y¼0 Here v1¼ð3;1Þis a nonzero solution. (ii) Subtract l¼/C04 (or add 4) down the diagonal of Ato obtain M¼96 32/C20/C21 ; corresponding to9xþ6y¼0 3xþ2y¼0or 3 xþ2y¼0 Here v2¼ð2;/C03Þis a nonzero solution. Then S¼fv1;v2g¼fð 3;1Þ;ð2;/C03Þgis a maximal set of linearly independent eigenvectors. Because Sis ab a s i so f R2,Ais diagonalizable. Using the basis S,Ais represented by the diagonal matrix D¼diagð7;/C04Þ. (b) First ﬁnd the characteristic polynomial DðtÞ¼t2þ1. There are no real roots. Thus B, a real matrix representing a linear transformation on R2, has no eigenvalues and no eigenvectors. Hence, in particular, Bis not diagonalizable.310 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (c) First ﬁnd DðtÞ¼t2/C08tþ16¼ðt/C04Þ2. Thus, l¼4 is the only eigenvalue of C. Subtract l¼4 down the diagonal of Cto obtain M¼1/C01 1/C01/C20/C21 ; corresponding to x/C0y¼0 The homogeneous system has only one independent solution; for example, x¼1,y¼1. Thus, v¼ð1;1Þis an eigenvector of C. Furthermore, as there are no other eigenvalues, the singleton set S¼fvg¼fð 1;1Þgis a maximal set of linearly independent eigenvectors of C. Furthermore, because S is not a basis of R2,Cis not diagonalizable. 9.12. Suppose the matrix Bin Problem 9.11 represents a linear operator on complex space C2. Show that, in this case, Bis diagonalizable by ﬁnding a basis SofC2consisting of eigenvectors of B. The characteristic polynomial of Bis stillDðtÞ¼t2þ1. As a polynomial over C,DðtÞdoes factor; speciﬁcally, DðtÞ¼ð t/C0iÞðtþiÞ. Thus, l¼iandl¼/C0iare the eigenvalues of B. (i) Subtract l¼idown the diagonal of Bto obtain the homogeneous system ð1/C0iÞx/C0 y¼0 2xþð/C0 1/C0iÞy¼0orð1/C0iÞx/C0y¼0 The system has only one independent solution; for example, x¼1,y¼1/C0i. Thus, v1¼ð1;1/C0iÞis an eigenvector that spans the eigenspace of l¼i. (ii) Subtract l¼/C0i(or add i) down the diagonal of Bto obtain the homogeneous system ð1þiÞx/C0 y¼0 2xþð/C0 1þiÞy¼0orð1þiÞx/C0y¼0 The system has only one independent solution; for example, x¼1,y¼1þi. Thus, v2¼ð1;1þiÞis an eigenvector that spans the eigenspace of l¼/C0i. As a complex matrix, Bis diagonalizable. Speciﬁcally, S¼fv1;v2g¼fð 1;1/C0iÞ;ð1;1þiÞgis a basis of C2consisting of eigenvectors of B. Using this basis S,Bis represented by the diagonal matrix D¼diagði;/C0iÞ. 9.13. LetLbe the linear transformation on R2that reﬂects each point Pacross the line y¼kx, where k>0. (See Fig. 9-1.) (a) Show that v1¼ðk;1Þand v2¼ð1;/C0kÞare eigenvectors of L. (b) Show that Lis diagonalizable, and ﬁnd a diagonal representation D. (a) The vector v1¼ðk;1Þlies on the line y¼kx, and hence is left ﬁxed by L; that is, Lðv1Þ¼v1. Thus, v1 is an eigenvector of Lbelonging to the eigenvalue l1¼1. The vector v2¼ð1;/C0kÞis perpendicular to the line y¼kx, and hence, Lreﬂects v2into its negative; that is, Lðv2Þ¼/C0 v2. Thus, v2is an eigenvector of Lbelonging to the eigenvalue l2¼/C01.y x 0LP() PL()v2 v2yk= x Figure 9-1CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 311 (b) Here S¼fv1;v2gis a basis of R2consisting of eigenvectors of L. Thus, Lis diagonalizable, with the diagonal representation D¼10 0/C01/C20/C21 (relative to the basis S). Eigenvalues and Eigenvectors 9.14. LetA¼41/C01 25/C02 11 22 43 5:(a) Find all eigenvalues of A. (b) Find a maximum set Sof linearly independent eigenvectors of A. (c) Is Adiagonalizable? If yes, ﬁnd Psuch that D¼P/C01APis diagonal. (a) First ﬁnd the characteristic polynomial DðtÞofA. We have trðAÞ¼4þ5þ2¼11 andjAj¼40/C02/C02þ5þ8/C04¼45 Also, ﬁnd each cofactor AiiofaiiinA: A11¼5/C02 12/C12/C12/C12/C12/C12/C12/C12/C12¼12; A 22¼4/C01 12/C12/C12/C12/C12/C12/C12/C12/C12¼9; A 33¼41 25/C12/C12/C12/C12/C12/C12/C12/C12¼18 Hence ; DðtÞ¼t 3/C0trðAÞt2þðA11þA22þA33Þt/C0jAj¼t3/C011t2þ39t/C045 Assuming Dthas a rational root, it must be among /C61,/C63,/C65,/C69,/C615,/C645. Testing, by synthetic division, we get 31/C011þ39/C045 3/C024þ45 1/C08þ15þ0 Thus, t¼3 is a root of DðtÞ. Also, t/C03 is a factor and t2/C08tþ15 is a factor. Hence, DðtÞ¼ð t/C03Þðt2/C08tþ15Þ¼ð t/C03Þðt/C05Þðt/C03Þ¼ð t/C03Þ2ðt/C05Þ Accordingly, l¼3 and l¼5 are eigenvalues of A. (b) Find linearly independent eigenvectors for each eigenvalue of A. (i) Subtract l¼3 down the diagonal of Ato obtain the matrix M¼11/C01 22/C02 11/C012 43 5; corresponding to xþy/C0z¼0 Here u¼ð1;/C01;0Þand v¼ð1;0;1Þare linearly independent solutions. (ii) Subtract l¼5 down the diagonal of Ato obtain the matrix M¼/C011/C01 20/C02 11/C032 43 5; corresponding to/C0xþy/C0z¼0 2x/C0 2z¼0 xþy/C03z¼0orx/C0z¼0 y/C02z¼0 Only zis a free variable. Here w¼ð1;2;1Þis a solution. Thus, S¼fu;v;wg¼fð 1;/C01;0Þ;ð1;0;1Þ;ð1;2;1Þgis a maximal set of linearly independent eigenvectors of A. Remark: The vectors uand vwere chosen so that they were independent solutions of the system xþy/C0z¼0. On the other hand, wis automatically independent of uand vbecause wbelongs to a different eigenvalue of A. Thus, the three vectors are linearly independent.312 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (c)Ais diagonalizable, because it has three linearly independent eigenvectors. Let Pbe the matrix with columns u;v;w. Then P¼111 /C0102 0112 43 5 and D¼P/C01AP¼3 3 52 43 5 9.15. Repeat Problem 9.14 for the matrix B¼3/C011 7/C051 6/C0622 43 5. (a) First ﬁnd the characteristic polynomial DðtÞofB. We have trðBÞ¼0;jBj¼/C0 16; B11¼/C04; B22¼0; B33¼/C08; soP iBii¼/C012 Therefore, DðtÞ¼t3/C012tþ16¼ðt/C02Þ2ðtþ4Þ. Thus, l1¼2 and l2¼/C04 are the eigen- values of B. (b) Find a basis for the eigenspace of each eigenvalue of B. (i) Subtract l1¼2 down the diagonal of Bto obtain M¼1/C011 7/C071 6/C0602 43 5; corresponding tox/C0yþz¼0 7x/C07yþz¼0 6x/C06y¼0orx/C0yþz¼0 z¼0 The system has only one independent solution; for example, x¼1,y¼1,z¼0. Thus, u¼ð1;1;0Þforms a basis for the eigenspace of l1¼2. (ii) Subtract l2¼/C04 (or add 4) down the diagonal of Bto obtain M¼7/C011 7/C011 6/C0662 43 5; corresponding to7x/C0yþz¼0 7x/C0yþz¼0 6x/C06yþ6z¼0orx/C0yþz¼0 6y/C06z¼0 The system has only one independent solution; for example, x¼0,y¼1,z¼1. Thus, v¼ð0;1;1Þforms a basis for the eigenspace of l2¼/C04. Thus S¼fu;vgis a maximal set of linearly independent eigenvectors of B. (c) Because Bhas at most two linearly independent eigenvectors, Bis not similar to a diagonal matrix; that is,Bis not diagonalizable. 9.16. Find the algebraic and geometric multiplicities of the eigenvalue l1¼2 of the matrix Bin Problem 9.15. The algebraic multiplicity of l1¼2 is 2, because t/C02 appears with exponent 2 in DðtÞ. However, the geometric multiplicity of l1¼2 is 1, because dim El1¼1 (where El1is the eigenspace of l1). 9.17. LetT:R3!R3be deﬁned by Tðx;y;zÞ¼ð 2xþy/C02z;2xþ3y/C04z;xþy/C0zÞ. Find all eigenvalues of T, and ﬁnd a basis of each eigenspace. Is Tdiagonalizable? If so, ﬁnd the basis Sof R3that diagonalizes T;and ﬁnd its diagonal representation D. First ﬁnd the matrix Athat represents Trelative to the usual basis of R3by writing down the coefﬁcients ofx;y;zas rows, and then ﬁnd the characteristic polynomial of A(and T). We have A¼½T/C138¼21/C02 23/C04 11/C012 43 5 andtrðAÞ¼4;jAj¼2 A11¼1;A22¼0;A33¼4P iAii¼5 Therefore, DðtÞ¼t3/C04t2þ5t/C02¼ðt/C01Þ2ðt/C02Þ, and so l¼1 and l¼2 are the eigenvalues of A(and T). We next ﬁnd linearly independent eigenvectors for each eigenvalue of A.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 313 (i) Subtract l¼1 down the diagonal of Ato obtain the matrix M¼11/C02 22/C04 11/C022 43 5; corresponding to xþy/C02z¼0 Here yandzare free variables, and so there are two linearly independent eigenvectors belonging tol¼1. For example, u¼ð1;/C01;0Þand v¼ð2;0;1Þare two such eigenvectors. (ii) Subtract l¼2 down the diagonal of Ato obtain M¼01/C02 21/C04 11/C032 43 5; corresponding toy/C02z¼0 2xþy/C04z¼0 xþy/C03z¼0orxþy/C03z¼0 y/C02z¼0 Only zis a free variable. Here w¼ð1;2;1Þis a solution. Thus, Tis diagonalizable, because it has three independent eigenvectors. Speciﬁcally, choosing S¼fu;v;wg¼fð 1;/C01;0Þ;ð2;0;1Þ;ð1;2;1Þg as a basis, Tis represented by the diagonal matrix D¼diagð1;1;2Þ. 9.18. Prove the following for a linear operator (matrix) T: (a) The scalar 0 is an eigenvalue of Tif and only if Tis singular. (b) If lis an eigenvalue of T, where Tis invertible, then l/C01is an eigenvalue of T/C01. (a) We have that 0 is an eigenvalue of Tif and only if there is a vector v6¼0 such that TðvÞ¼0v—that is, if and only if Tis singular. (b) Because Tis invertible, it is nonsingular; hence, by (a), l6¼0. By deﬁnition of an eigenvalue, there exists v6¼0 such that TðvÞ¼lv. Applying T/C01to both sides, we obtain v¼T/C01ðlvÞ¼lT/C01ðvÞ; and so T/C01ðvÞ¼l/C01v Therefore, l/C01is an eigenvalue of T/C01. 9.19. Letlbe an eigenvalue of a linear operator T:V!V, and let Elconsists of all the eigenvectors belonging to l(called the eigenspace ofl). Prove that Elis a subspace of V. That is, prove (a) If u2El, then ku2Elfor any scalar k. (b) If u;v;2El, then uþv2El. (a) Because u2El, we have TðuÞ¼lu. Then TðkuÞ¼kTðuÞ¼kðluÞ¼lðkuÞ;and so ku2El: (We view the zero vector 0 2Vas an ‘‘eigenvector’’ of lin order for Elto be a subspace of V.) (b) As u;v2El, we have TðuÞ¼luandTðvÞ¼lv. Then TðuþvÞ¼TðuÞþTðvÞ¼luþlv¼lðuþvÞ;and so uþv2El 9.20. Prove Theorem 9.6: The following are equivalent: (i) The scalar lis an eigenvalue of A. (ii) The matrix lI/C0Ais singular. (iii) The scalar lis a root of the characteristic polynomial DðtÞofA. The scalar lis an eigenvalue of Aif and only if there exists a nonzero vector vsuch that Av¼lv orðlIÞv/C0Av¼0o rðlI/C0AÞv¼0 orlI/C0Ais singular. In such a case, lis a root of DðtÞ¼j tI/C0Aj. Also, vis in the eigenspace Eloflif and only if the above relations hold. Hence, vis a solution ofðlI/C0AÞX¼0. 9.21. Prove Theorem 9.80: Suppose v1;v2;...;vnare nonzero eigenvectors of Tbelonging to distinct eigenvalues l1;l2;...;ln. Then v1;v2;...;vnare linearly independent.314 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Suppose the theorem is not true. Let v1;v2;...;vsbe a minimal set of vectors for which the theorem is not true. We have s>1, because v16¼0. Also, by the minimality condition, v2;...;vsare linearly independent. Thus, v1is a linear combination of v2;...;vs, say, v1¼a2v2þa3v3þ/C1/C1/C1þ asvs ð1Þ (where some ak6¼0Þ. Applying Tto (1) and using the linearity of Tyields Tðv1Þ¼Tða2v2þa3v3þ/C1/C1/C1þ asvsÞ¼a2Tðv2Þþa3Tðv3Þþ/C1/C1/C1þ asTðvsÞð 2Þ Because vjis an eigenvector of Tbelonging to lj, we have TðvjÞ¼ljvj. Substituting in (2) yields l1v1¼a2l2v2þa3l3v3þ/C1/C1/C1þ aslsvs ð3Þ Multiplying (1) by l1yields l1v1¼a2l1v2þa3l1v3þ/C1/C1/C1þ asl1vs ð4Þ Setting the right-hand sides of (3) and (4) equal to each other, or subtracting (3) from (4) yields a2ðl1/C0l2Þv2þa3ðl1/C0l3Þv3þ/C1/C1/C1þ asðl1/C0lsÞvs¼0 ð5Þ Because v2;v3;...;vsare linearly independent, the coefﬁcients in (5) must all be zero. That is, a2ðl1/C0l2Þ¼0; a3ðl1/C0l3Þ¼0; ...; asðl1/C0lsÞ¼0 However, the liare distinct. Hence l1/C0lj6¼0 for j>1. Hence, a2¼0,a3¼0;...;as¼0. This contradicts the fact that some ak6¼0. The theorem is proved. 9.22. Prove Theorem 9.9. Suppose DðtÞ¼ð t/C0a1Þðt/C0a2Þ...ðt/C0anÞis the characteristic polynomial of an n-square matrix A, and suppose the nroots aiare distinct. Then Ais similar to the diagonal matrix D¼diagða1;a2;...;anÞ. Letv1;v2;...;vnbe (nonzero) eigenvectors corresponding to the eigenvalues ai. Then the neigenvectors viare linearly independent (Theorem 9.8), and hence form a basis of Kn. Accordingly, Ais diagonalizable (i.e., Ais similar to a diagonal matrix D), and the diagonal elements of Dare the eigenvalues ai. 9.23. Prove Theorem 9.100: The geometric multiplicity of an eigenvalue lofTdoes not exceed its algebraic multiplicity. Suppose the geometric multiplicity of lisr. Then its eigenspace Elcontains rlinearly independent eigenvectors v1;...;vr. Extend the setfvigto a basis of V, say,fvi;...;vr;w1;...;wsg. We have Tðv1Þ¼lv1; Tðv2Þ¼lv2; ...; TðvrÞ¼lvr; Tðw1Þ¼a11v1þ/C1/C1/C1þ a1rvrþb11w1þ/C1/C1/C1þ b1sws Tðw2Þ¼a21v1þ/C1/C1/C1þ a2rvrþb21w1þ/C1/C1/C1þ b2sws :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TðwsÞ¼as1v1þ/C1/C1/C1þ asrvrþbs1w1þ/C1/C1/C1þ bssws Then M¼lIrA 0B/C20/C21 is the matrix of Tin the above basis, where A¼½aij/C138TandB¼½bij/C138T: Because Mis block diagonal, the characteristic polynomial ðt/C0lÞrof the block lIrmust divide the characteristic polynomial of Mand hence of T. Thus, the algebraic multiplicity of lforTis at least r,a s required. Diagonalizing Real Symmetric Matrices and Quadratic Forms 9.24. LetA¼73 3/C01/C20/C21 . Find an orthogonal matrix Psuch that D¼P/C01APis diagonal.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 315 First ﬁnd the characteristic polynomial DðtÞofA. We have DðtÞ¼t2/C0trðAÞtþjAj¼t2/C06t/C016¼ðt/C08Þðtþ2Þ Thus, the eigenvalues of Aarel¼8 and l¼/C02. We next ﬁnd corresponding eigenvectors. Subtract l¼8 down the diagonal of Ato obtain the matrix M¼/C013 3/C09/C20/C21 ; corresponding to/C0xþ3y¼0 3x/C09y¼0or x/C03y¼0 A nonzero solution is u1¼ð3;1Þ. Subtract l¼/C02 (or add 2) down the diagonal of Ato obtain the matrix M¼93 31/C20/C21 ; corresponding to9xþ3y¼0 3xþy¼0or 3 xþy¼0 A nonzero solution is u2¼ð1;/C03Þ. As expected, because Ais symmetric, the eigenvectors u1andu2are orthogonal. Normalize u1andu2to obtain, respectively, the unit vectors ^u1¼ð3=ﬃﬃﬃﬃﬃ 10p ;1=ﬃﬃﬃﬃﬃ 10p Þ and ^u2¼ð1=ﬃﬃﬃﬃﬃ 10p ;/C03=ﬃﬃﬃﬃﬃ 10p Þ: Finally, let Pbe the matrix whose columns are the unit vectors ^u1and ^u2, respectively. Then P¼3=ﬃﬃﬃﬃﬃ 10p 1=ﬃﬃﬃﬃﬃ 10p 1=ﬃﬃﬃﬃﬃ 10p /C03=ﬃﬃﬃﬃﬃ 10p"# and D¼P/C01AP¼80 0/C02/C20/C21 As expected, the diagonal entries in Dare the eigenvalues of A. 9.25. LetB¼11/C084 /C08/C01/C02 4/C02/C042 43 5. (a) Find all eigenvalues of B. (b) Find a maximal set Sof nonzero orthogonal eigenvectors of B. (c) Find an orthogonal matrix Psuch that D¼P/C01BPis diagonal. (a) First ﬁnd the characteristic polynomial of B. We have trðBÞ¼6;jBj¼400; B11¼0; B22¼/C060; B33¼/C075; soP iBii¼/C0135 Hence,DðtÞ¼t3/C06t2/C0135t/C0400. IfDðtÞhas an integer root it must divide 400. Testing t¼/C05, by synthetic division, yields /C051/C06/C0135/C0400 /C05þ55þ400 1/C011/C080þ 0 Thus, tþ5 is a factor of DðtÞ, and t2/C011t/C080 is a factor. Thus, DðtÞ¼ð tþ5Þðt2/C011t/C080Þ¼ð tþ5Þ2ðt/C016Þ The eigenvalues of Barel¼/C05 (multiplicity 2), and l¼16 (multiplicity 1). (b) Find an orthogonal basis for each eigenspace. Subtract l¼/C05 (or, add 5) down the diagonal of Bto obtain the homogeneous system 16x/C08yþ4z¼0;/C08xþ4y/C02z¼0; 4x/C02yþz¼0 That is, 4 x/C02yþz¼0. The system has two independent solutions. One solution is v1¼ð0;1;2Þ.W e seek a second solution v2¼ða;b;cÞ, which is orthogonal to v1, such that 4a/C02bþc¼0; and also b/C02c¼0316 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors One such solution is v2¼ð/C0 5;/C08;4Þ. Subtract l¼16 down the diagonal of Bto obtain the homogeneous system /C05x/C08yþ4z¼0;/C08x/C017y/C02z¼0; 4x/C02y/C020z¼0 This system yields a nonzero solution v3¼ð4;/C02;1Þ. (As expected from Theorem 9.13, the eigenvector v3is orthogonal to v1and v2.) Then v1;v2;v3form a maximal set of nonzero orthogonal eigenvectors of B. (c) Normalize v1;v2;v3to obtain the orthonormal basis: ^v1¼v1=ﬃﬃﬃ 5p ; ^v2¼v2=ﬃﬃﬃﬃﬃﬃﬃﬃ 105p ; ^v3¼v3=ﬃﬃﬃﬃﬃ 21p Then Pis the matrix whose columns are ^v1;^v2;^v3. Thus, P¼0/C05=ﬃﬃﬃﬃﬃﬃﬃﬃ 105p 4=ﬃﬃﬃﬃﬃ 21p 1=ﬃﬃﬃ 5p /C08=ﬃﬃﬃﬃﬃﬃﬃﬃ 105p /C02=ﬃﬃﬃﬃﬃ 21p 2=ﬃﬃﬃ 5p 4=ﬃﬃﬃﬃﬃﬃﬃﬃ 105p 1=ﬃﬃﬃﬃﬃ 21p2 643 75 and D¼P/C01BP¼/C05 /C05 162 643 75 9.26. Letqðx;yÞ¼x2þ6xy/C07y2. Find an orthogonal substitution that diagonalizes q. Find the symmetric matrix Athat represents qand its characteristic polynomial DðtÞ. We have A¼13 3/C07/C20/C21 and DðtÞ¼t2þ6t/C016¼ðt/C02Þðtþ8Þ The eigenvalues of Aarel¼2 and l¼/C08. Thus, using sandtas new variables, a diagonal form of qis qðs;tÞ¼2s2/C08t2 The corresponding orthogonal substitution is obtained by ﬁnding an orthogonal set of eigenvectors of A. (i) Subtract l¼2 down the diagonal of Ato obtain the matrix M¼/C013 3/C09/C20/C21 ; corresponding to/C0xþ3y¼0 3x/C09y¼0or/C0xþ3y¼0 A nonzero solution is u1¼ð3;1Þ. (ii) Subtract l¼/C08 (or add 8) down the diagonal of Ato obtain the matrix M¼93 31/C20/C21 ; corresponding to9xþ3y¼0 3xþy¼0or 3 xþy¼0 A nonzero solution is u2¼ð/C0 1;3Þ. As expected, because Ais symmetric, the eigenvectors u1andu2are orthogonal. Now normalize u1andu2to obtain, respectively, the unit vectors ^u1¼ð3=ﬃﬃﬃﬃﬃ 10p ;1=ﬃﬃﬃﬃﬃ 10p Þ and ^u2¼ð/C0 1=ﬃﬃﬃﬃﬃ 10p ;3=ﬃﬃﬃﬃﬃ 10p Þ: Finally, let Pbe the matrix whose columns are the unit vectors ^u1and ^u2, respectively, and then ½x;y/C138T¼P½s;t/C138Tis the required orthogonal change of coordinates. That is, P¼3=ﬃﬃﬃﬃﬃ 10p /C01=ﬃﬃﬃﬃﬃ 10p 1=ﬃﬃﬃﬃﬃ 10p 3=ﬃﬃﬃﬃﬃ 10p/C12/C12/C12/C12/C12# and x¼3s/C0tﬃﬃﬃﬃﬃ 10p ; y¼sþ3tﬃﬃﬃﬃﬃ 10p One can also express sandtin terms of xandyby using P/C01¼PT. That is, s¼3xþyﬃﬃﬃﬃﬃ 10p ; t¼/C0xþ3tﬃﬃﬃﬃﬃ 10pCHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 317 Minimal Polynomial 9.27. LetA¼4/C022 6/C034 3/C0232 43 5andB¼3/C022 4/C046 2/C0352 43 5. The characteristic polynomial of both matrices is DðtÞ¼ð t/C02Þðt/C01Þ2. Find the minimal polynomial mðtÞof each matrix. The minimal polynomial mðtÞmust divide DðtÞ. Also, each factor of DðtÞ(i.e., t/C02 and t/C01) must also be a factor of mðtÞ. Thus, mðtÞmust be exactly one of the following: fðtÞ¼ð t/C02Þðt/C01Þ or gðtÞ¼ð t/C02Þðt/C01Þ2 (a) By the Cayley–Hamilton theorem, gðAÞ¼DðAÞ¼0, so we need only test fðtÞ. We have fðAÞ¼ð A/C02IÞðA/C0IÞ¼2/C022 6/C054 3/C0212 43 53/C022 6/C044 3/C0222 43 5¼000 000 0002 43 5 Thus, mðtÞ¼fðtÞ¼ð t/C02Þðt/C01Þ¼t2/C03tþ2 is the minimal polynomial of A. (b) Again gðBÞ¼DðBÞ¼0, so we need only test fðtÞ. We get fðBÞ¼ð B/C02IÞðB/C0IÞ¼1/C022 4/C066 2/C0332 43 52/C022 4/C056 2/C0342 43 5¼/C022/C02 /C044/C04 /C022/C022 43 56¼0 Thus, mðtÞ6¼fðtÞ. Accordingly, mðtÞ¼gðtÞ¼ð t/C02Þðt/C01Þ2is the minimal polynomial of B. [We emphasize that we do not need to compute gðBÞ; we know gðBÞ¼0 from the Cayley–Hamilton theorem.] 9.28. Find the minimal polynomial mðtÞof each of the following matrices: (a) A¼51 37/C20/C21 , (b) B¼123 023 0032 43 5, (c) C¼4/C01 12/C20/C21 (a) The characteristic polynomial of AisDðtÞ¼t2/C012tþ32¼ðt/C04Þðt/C08Þ. Because DðtÞhas distinct factors, the minimal polynomial mðtÞ¼DðtÞ¼t2/C012tþ32. (b) Because Bis triangular, its eigenvalues are the diagonal elements 1 ;2;3; and so its characteristic polynomial is DðtÞ¼ð t/C01Þðt/C02Þðt/C03Þ. Because DðtÞhas distinct factors, mðtÞ¼DðtÞ. (c) The characteristic polynomial of CisDðtÞ¼t2/C06tþ9¼ðt/C03Þ2. Hence the minimal polynomial of C isfðtÞ¼t/C03o r gðtÞ¼ð t/C03Þ2. However, fðCÞ6¼0; that is, C/C03I6¼0. Hence, mðtÞ¼gðtÞ¼DðtÞ¼ð t/C03Þ2: 9.29. Suppose S¼fu1;u2;...;ungis a basis of V, and suppose FandGare linear operators on Vsuch that½F/C138has 0’s on and below the diagonal, and ½G/C138hasa6¼0 on the superdiagonal and 0’s elsewhere. That is, ½F/C138¼0a21a31 ... an1 00 a32 ... an2 :::::::::::::::::::::::::::::::::::::::: 00 0 ... an;n/C01 00 0 ... 02 666643 77775;½G/C138¼0a0 ... 0 00 a ... 0 ::::::::::::::::::::::::::: 000 ... a 000 ... 02 666643 77775 Show that (a) Fn¼0, (b) Gn/C016¼0, but Gn¼0. (These conditions also hold for ½F/C138and½G/C138.) (a) We have Fðu1Þ¼0 and, for r>1,FðurÞis a linear combination of vectors preceding urinS. That is, FðurÞ¼ar1u1þar2u2þ/C1/C1/C1þ ar;r/C01ur/C01318 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Hence, F2ðurÞ¼FðFðurÞÞis a linear combination of vectors preceding ur/C01, and so on. Hence, FrðurÞ¼0 for each r. Thus, for each r,FnðurÞ¼Fn/C0rð0Þ¼0, and so Fn¼0, as claimed. (b) We have Gðu1Þ¼0 and, for each k>1,GðukÞ¼auk/C01.H e n c e , GrðukÞ¼aruk/C0rforr<k. Because a6¼0, an/C016¼0. Therefore, Gn/C01ðunÞ¼an/C01u16¼0, and so Gn/C016¼0. On the other hand, by (a), Gn¼0. 9.30. LetBbe the matrix in Example 9.12(a) that has 1’s on the diagonal, a’s on the superdiagonal, where a6¼0, and 0’s elsewhere. Show that fðtÞ¼ð t/C0lÞnis both the characteristic polynomial DðtÞand the minimum polynomial mðtÞofA. Because Ais triangular with l’s on the diagonal, DðtÞ¼fðtÞ¼ð t/C0lÞnis its characteristic polynomial. Thus, mðtÞis a power of t/C0l. By Problem 9.29, ðA/C0lIÞr/C016¼0. Hence, mðtÞ¼DðtÞ¼ð t/C0lÞn. 9.31. Find the characteristic polynomial DðtÞand minimal polynomial mðtÞof each matrix: (a)M¼41000 04100 00400 00041000042 666643 77775, (b) M 0¼27 00 02 00 00 11 00/C0242 6643 775 (a)Mis block diagonal with diagonal blocks A¼410 041 0042 43 5 and B¼41 04/C20/C21 The characteristic and minimal polynomial of AisfðtÞ¼ð t/C04Þ3and the characteristic and minimal polynomial of BisgðtÞ¼ð t/C04Þ2. Then DðtÞ¼fðtÞgðtÞ¼ð t/C04Þ5but mðtÞ¼LCM½fðtÞ;gðtÞ/C138¼ð t/C04Þ3 (where LCM means least common multiple). We emphasize that the exponent in mðtÞis the size of the largest block. (b) Here M0is block diagonal with diagonal blocks A0¼27 02/C20/C21 and B0¼11 /C024/C20/C21 The char- acteristic and minimal polynomial of A0isfðtÞ¼ð t/C02Þ2. The characteristic polynomial of B0is gðtÞ¼t2/C05tþ6¼ðt/C02Þðt/C03Þ, which has distinct factors. Hence, gðtÞis also the minimal polynomial ofB. Accordingly, DðtÞ¼fðtÞgðtÞ¼ð t/C02Þ3ðt/C03Þ but mðtÞ¼LCM½fðtÞ;gðtÞ/C138¼ð t/C02Þ2ðt/C03Þ 9.32. Find a matrix Awhose minimal polynomial is fðtÞ¼t3/C08t2þ5tþ7. Simply let A¼00/C07 10/C05 01 82 43 5, the companion matrix of fðtÞ[deﬁned in Example 9.12(b)]. 9.33. Prove Theorem 9.15: The minimal polynomial mðtÞof a matrix (linear operator) Adivides every polynomial that has Aas a zero. In particular (by the Cayley–Hamilton theorem), mðtÞdivides the characteristic polynomial DðtÞofA. Suppose fðtÞis a polynomial for which fðAÞ¼0. By the division algorithm, there exist polynomials qðtÞandrðtÞfor which fðtÞ¼mðtÞqðtÞþrðtÞandrðtÞ¼0 or deg rðtÞ<degmðtÞ. Substituting t¼Ain this equation, and using that fðAÞ¼0 and mðAÞ¼0, we obtain rðAÞ¼0. If rðtÞ6¼0, then rðtÞis a polynomial of degree less than mðtÞthat has Aas a zero. This contradicts the deﬁnition of the minimal polynomial. Thus, rðtÞ¼0, and so fðtÞ¼mðtÞqðtÞ; that is, mðtÞdivides fðtÞ.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 319 9.34. LetmðtÞbe the minimal polynomial of an n-square matrix A. Prove that the characteristic polynomial DðtÞofAdivides½mðtÞ/C138n. Suppose mðtÞ¼trþc1tr/C01þ/C1/C1/C1þ cr/C01tþcr. Deﬁne matrices Bjas follows: B0¼I B1¼Aþc1I B2¼A2þc1Aþc2I Br/C01¼Ar/C01þc1Ar/C02þ/C1/C1/C1þ cr/C01Iso so so soI¼B0 c1I¼B1/C0A¼B1/C0AB0 c2I¼B2/C0AðAþc1IÞ¼B2/C0AB1 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::: cr/C01I¼Br/C01/C0ABr/C02 Then /C0ABr/C01¼crI/C0ðArþc1Ar/C01þ/C1/C1/C1þ cr/C01AþcrIÞ¼crI/C0mðAÞ¼crI Set BðtÞ¼tr/C01B0þtr/C02B1þ/C1/C1/C1þ tBr/C02þBr/C01 Then ðtI/C0AÞBðtÞ¼ð trB0þtr/C01B1þ/C1/C1/C1þ tBr/C01Þ/C0ð tr/C01AB0þtr/C02AB1þ/C1/C1/C1þ ABr/C01Þ ¼trB0þtr/C01ðB1/C0AB0Þþtr/C02ðB2/C0AB1Þþ/C1/C1/C1þ tðBr/C01/C0ABr/C02Þ/C0ABr/C01 ¼trIþc1tr/C01Iþc2tr/C02Iþ/C1/C1/C1þ cr/C01tIþcrI¼mðtÞI Taking the determinant of both sides gives jtI/C0AjjBðtÞj¼j mðtÞIj¼½mðtÞ/C138n. BecausejBðtÞjis a poly- nomial,jtI/C0Ajdivides½mðtÞ/C138n; that is, the characteristic polynomial of Adivides½mðtÞ/C138n. 9.35. Prove Theorem 9.16: The characteristic polynomial DðtÞand the minimal polynomial mðtÞofA have the same irreducible factors. Suppose fðtÞis an irreducible polynomial. If fðtÞdivides mðtÞ, then fðtÞalso divides DðtÞ[because mðtÞ divides DðtÞ/C138. On the other hand, if fðtÞdivides DðtÞ, then by Problem 9.34, fðtÞalso divides½mðtÞ/C138n. But fðtÞ is irreducible; hence, fðtÞalso divides mðtÞ. Thus, mðtÞandDðtÞhave the same irreducible factors. 9.36. Prove Theorem 9.19: The minimal polynomial mðtÞof a block diagonal matrix Mwith diagonal blocks Aiis equal to the least common multiple (LCM) of the minimal polynomials of the diagonal blocks Ai. We prove the theorem for the case r¼2. The general theorem follows easily by induction. Suppose M¼A0 0B/C20/C21 , where AandBare square matrices. We need to show that the minimal polynomial mðtÞofM is the LCM of the minimal polynomials gðtÞandhðtÞofAandB, respectively. Because mðtÞis the minimal polynomial of M;mðMÞ¼mðAÞ 0 0 mðBÞ/C20/C21 ¼0, and mðAÞ¼0 and mðBÞ¼0. Because gðtÞis the minimal polynomial of A,gðtÞdivides mðtÞ. Similarly, hðtÞdivides mðtÞ. Thus mðtÞis a multiple of gðtÞandhðtÞ. Now let fðtÞbe another multiple of gðtÞandhðtÞ. Then fðMÞ¼fðAÞ 0 0 fðBÞ/C20/C21 ¼00 00/C20/C21 ¼0. But mðtÞis the minimal polynomial of M; hence, mðtÞdivides fðtÞ. Thus, mðtÞis the LCM of gðtÞandhðtÞ. 9.37. Suppose mðtÞ¼trþar/C01tr/C01þ/C1/C1/C1þ a1tþa0is the minimal polynomial of an n-square matrix A. Prove the following: (a)Ais nonsingular if and only if the constant term a06¼0. (b) If Ais nonsingular, then A/C01is a polynomial in Aof degree r/C01<n. (a) The following are equivalent: (i) Ais nonsingular, (ii) 0 is not a root of mðtÞ, (iii) a06¼0. Thus, the statement is true.320 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors (b) Because Ais nonsingular, a06¼0 by (a). We have mðAÞ¼Arþar/C01Ar/C01þ/C1/C1/C1þ a1Aþa0I¼0 Thus ; /C01 a0ðAr/C01þar/C01Ar/C02þ/C1/C1/C1þ a1IÞA¼I Accordingly ; A/C01¼/C01 a0ðAr/C01þar/C01Ar/C02þ/C1/C1/C1þ a1IÞ SUPPLEMENTARY PROBLEMS Polynomials of Matrices 9.38. Let A¼2/C03 51/C20/C21 and B¼12 03/C20/C21 . Find fðAÞ,gðAÞ,fðBÞ,gðBÞ, where fðtÞ¼2t2/C05tþ6 and gðtÞ¼t3/C02t2þtþ3. 9.39. LetA¼12 01/C20/C21 . Find A2,A3,An, where n>3, and A/C01. 9.40. LetB¼81 2 0 08 1 2 0082 43 5. Find a real matrix Asuch that B¼A3. 9.41. For each matrix, ﬁnd a polynomial having the following matrix as a root: (a) A¼25 1/C03/C20/C21 , (b) B¼2/C03 7/C04/C20/C21 , (c) C¼112 123 2142 43 5 9.42. LetAbe any square matrix and let fðtÞbe any polynomial. Prove (a) ðP/C01APÞn¼P/C01AnP. (b) fðP/C01APÞ¼P/C01fðAÞP. (c) fðATÞ¼½ fðAÞ/C138T. (d) If Ais symmetric, then fðAÞis symmetric. 9.43. LetM¼diag½A1;...;Ar/C138be a block diagonal matrix, and let fðtÞbe any polynomial. Show that fðMÞis block diagonal and fðMÞ¼diag½fðA1Þ;...;fðArÞ/C138: 9.44. LetMbe a block triangular matrix with diagonal blocks A1;...;Ar, and let fðtÞbe any polynomial. Show thatfðMÞis also a block triangular matrix, with diagonal blocks fðA1Þ;...;fðArÞ. Eigenvalues and Eigenvectors 9.45. For each of the following matrices, ﬁnd all eigenvalues and corresponding linearly independent eigen- vectors: (a) A¼2/C03 2/C05/C20/C21 , (b) B¼24 /C016/C20/C21 , (c) C¼1/C04 3/C07/C20/C21 When possible, ﬁnd the nonsingular matrix Pthat diagonalizes the matrix. 9.46. LetA¼2/C01 /C023/C20/C21 . (a) Find eigenvalues and corresponding eigenvectors. (b) Find a nonsingular matrix Psuch that D¼P/C01APis diagonal. (c) Find A8andfðAÞwhere fðtÞ¼t4/C05t3þ7t2/C02tþ5. (d) Find a matrix Bsuch that B2¼A.CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 321 9.47. Repeat Problem 9.46 for A¼56 /C02/C02/C20/C21 . 9.48. For each of the following matrices, ﬁnd all eigenvalues and a maximum set Sof linearly independent eigenvectors: (a) A¼1/C033 3/C053 6/C0642 43 5, (b) B¼3/C011 7/C051 6/C0622 43 5, (c) C¼12 2 12/C01 /C011 42 43 5 Which matrices can be diagonalized, and why? 9.49. For each of the following linear operators T:R2!R2, ﬁnd all eigenvalues and a basis for each eigenspace: (a) Tðx;yÞ¼ð 3xþ3y;xþ5yÞ, (b) Tðx;yÞ¼ð 3x/C013y;x/C03yÞ. 9.50. LetA¼ab cd/C20/C21 be a real matrix. Find necessary and sufﬁcient conditions on a;b;c;dso that Ais diagonalizable—that is, so that Ahas two (real) linearly independent eigenvectors. 9.51. Show that matrices AandAThave the same eigenvalues. Give an example of a 2 /C22 matrix Awhere Aand AThave different eigenvectors. 9.52. Suppose vis an eigenvector of linear operators FandG. Show that vis also an eigenvector of the linear operator kFþk0G, where kandk0are scalars. 9.53. Suppose vis an eigenvector of a linear operator Tbelonging to the eigenvalue l. Prove (a) For n>0;vis an eigenvector of Tnbelonging to ln. (b) fðlÞis an eigenvalue of fðTÞfor any polynomial fðtÞ. 9.54. Suppose l6¼0 is an eigenvalue of the composition F/C14Gof linear operators FandG. Show that lis also an eigenvalue of the composition G/C14F.[Hint: Show that GðvÞis an eigenvector of G/C14F.] 9.55. LetE:V!Vbe a projection mapping; that is, E2¼E. Show that Eis diagonalizable and, in fact, can be represented by the diagonal matrix M¼Ir0 00/C20/C21 , where ris the rank of E. Diagonalizing Real Symmetric Matrices and Quadratic Forms 9.56. For each of the following symmetric matrices A, ﬁnd an orthogonal matrix Pand a diagonal matrix Dsuch thatD¼P/C01AP: (a) A¼54 4/C01/C20/C21 , (b) A¼4/C01 /C014/C20/C21 , (c) A¼73 3/C01/C20/C21 9.57. For each of the following symmetric matrices B, ﬁnd its eigenvalues, a maximal orthogonal set Sof eigenvectors, and an orthogonal matrix Psuch that D¼P/C01BPis diagonal: (a) B¼011 101 1102 43 5, (b) B¼22 4 25 8 481 72 43 5 9.58. Using variables sandt, ﬁnd an orthogonal substitution that diagonalizes each of the following quadratic forms: (a) qðx;yÞ¼4x2þ8xy/C011y2, (b) qðx;yÞ¼2x2/C06xyþ10y2 9.59. For each of the following quadratic forms qðx;y;zÞ, ﬁnd an orthogonal substitution expressing x;y;zin terms of variables r;s;t, and ﬁnd qðr;s;tÞ: (a) qðx;y;zÞ¼5x2þ3y2þ12xz; (b) qðx;y;zÞ¼3x2/C04xyþ6y2þ2xz/C04yzþ3z2322 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 9.60. Find a real 2/C22 symmetric matrix Awith eigenvalues: (a) l¼1 and l¼4 and eigenvector u¼ð1;1Þbelonging to l¼1; (b) l¼2 and l¼3 and eigenvector u¼ð1;2Þbelonging to l¼2. In each case, ﬁnd a matrix Bfor which B2¼A. Characteristic and Minimal Polynomials 9.61. Find the characteristic and minimal polynomials of each of the following matrices: (a) A¼31/C01 24/C02 /C01/C0132 43 5, (b) B¼32/C01 38/C03 36/C012 43 5 9.62. Find the characteristic and minimal polynomials of each of the following matrices: (a) A¼25000 02000 00420 00350000072 666643 77775, (b) B¼4/C01000 12 0 0 0 00 3 1 0 00 0 3 1 00 0 0 32 666643 77775, (c) C¼32000 14000 00310 00130 000042 666643 77775 9.63. LetA¼110 020 0012 43 5andB¼200 022 0012 43 5. Show that AandBhave different characteristic polynomials (and so are not similar) but have the same minimal polynomial. Thus, nonsimilar matrices may have the same minimal polynomial. 9.64. LetAbe an n-square matrix for which A k¼0 for some k>n. Show that An¼0. 9.65. Show that a matrix Aand its transpose AThave the same minimal polynomial. 9.66. Suppose fðtÞis an irreducible monic polynomial for which fðAÞ¼0 for a matrix A. Show that fðtÞis the minimal polynomial of A. 9.67. Show that Ais a scalar matrix kIif and only if the minimal polynomial of AismðtÞ¼t/C0k. 9.68. Find a matrix Awhose minimal polynomial is (a) t3/C05t2þ6tþ8, (b) t4/C05t3/C02tþ7tþ4. 9.69. LetfðtÞandgðtÞbe monic polynomials (leading coefﬁcient one) of minimal degree for which Ais a root. Show fðtÞ¼gðtÞ:[Thus, the minimal polynomial of Ais unique.] ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M¼½R1;R2; .../C138denotes a matrix Mwith rows R1;R2;...: 9.38. fðAÞ¼½/C0 26;/C03;5;/C027/C138, gðAÞ¼½/C0 40;39;/C065;/C027/C138, fðBÞ¼½ 3;6;0;9/C138, gðBÞ¼½ 3;12;0;15/C138 9.39. A2¼½1;4;0;1/C138, A3¼½1;6;0;1/C138, An¼½1;2n;0;1/C138, A/C01¼½1;/C02;0;1/C138 9.40. LetA¼½2;a;b;0;2;c;0;0;2/C138. Set B¼A3and then a¼1,b¼/C01 2,c¼1CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors 323 9.41. FindDðtÞ: (a) t2þt/C011, (b) t2þ2tþ13, (c) t3/C07t2þ6t/C01 9.45. (a) l¼1;u¼ð3;1Þ;l¼/C04;v¼ð1;2Þ, (b) l¼4;u¼ð2;1Þ, (c) l¼/C01;u¼ð2;1Þ;l¼/C05;v¼ð2;3Þ. Only AandCcan be diagonalized; use P¼½u;v/C138. 9.46. (a) l¼1;u¼ð1;1Þ;l¼4;v¼ð1;/C02Þ, (b) P¼½u;v/C138, (c) fðAÞ¼½ 3;1;2;1/C138; A8¼½21 846 ;/C021 845 ;/C043 690 ;43 691/C138, (d) B¼4 3;/C01 3;/C02 3;53/C2/C3 9.47. (a) l¼1;u¼ð3;/C02Þ;l¼2;v¼ð2;/C01Þ, (b) P¼½u;v/C138, (c) fðAÞ¼½ 2;/C06;2;9/C138; A8¼½1021 ;1530 ;/C0510;/C0764/C138, (d) B¼½/C0 3þ4ﬃﬃﬃ 2p ;/C06þ6ﬃﬃﬃ 2p ; 2/C02ﬃﬃﬃ 2p ;4/C03ﬃﬃﬃ 2p /C138 9.48. (a) l¼/C02;u¼ð1;1;0Þ;v¼ð1;0;/C01Þ;l¼4;w¼ð1;1;2Þ, (b) l¼2;u¼ð1;1;0Þ;l¼/C04;v¼ð0;1;1Þ, (c) l¼3;u¼ð1;1;0Þ;v¼ð1;0;1Þ;l¼1;w¼ð2;/C01;1Þ. Only Aand Ccan be diagonalized; use P¼½u;v;w/C138: 9.49. (a) l¼2;u¼ð3;/C01Þ;l¼6;v¼ð1;1Þ, (b) No real eigenvalues 9.50. We need½/C0trðAÞ/C1382/C04½detðAÞ/C138/C21 0o rða/C0dÞ2þ4bc/C210. 9.51. A¼½1;1;0;1/C138 9.56. (a) P¼½2;/C01;1;2/C138=ﬃﬃﬃ 5p , D¼½7;0;0;3/C138, (b) P¼½1;1;1;/C01/C138=ﬃﬃﬃ 2p , D¼½3;0;0;5/C138, (c) P¼½3;/C01;1;3/C138=ﬃﬃﬃﬃﬃ 10p , D¼½8;0;0;2/C138 9.57. (a) l¼/C01;u¼ð1;/C01;0Þ;v¼ð1;1;/C02Þ;l¼2;w¼ð1;1;1Þ, (b) l¼1;u¼ð2;1;/C01Þ;v¼ð2;/C03;1Þ;l¼22;w¼ð1;2;4Þ; Normalize u;v;w, obtaining ^u;^v;^w, and set P¼½^u;^v;^w/C138.(Remark: u and vare not unique.) 9.58. (a) x¼ð4sþtÞ=ﬃﬃﬃﬃﬃ 17p ; y¼ð/C0 sþ4tÞ=ﬃﬃﬃﬃﬃ 17p ; qðs;tÞ¼5s2/C012t2, (b) x¼ð3s/C0tÞ=ﬃﬃﬃﬃﬃ 10p ; y¼ðsþ3tÞ=ﬃﬃﬃﬃﬃ 10p ; qðs;tÞ¼s2þ11t2 9.59. (a) x¼ð3sþ2tÞ=ﬃﬃﬃﬃﬃ 13p ; y¼r; z¼ð2s/C03tÞ=ﬃﬃﬃﬃﬃ 13p ; qðr;s;tÞ¼3r2þ9s2/C04t2, (b) x¼5KsþLt; y¼Jrþ2Ks/C02Lt; z¼2Jr/C0Ks/C0Lt, where J¼1=ﬃﬃﬃ 5p ,K¼1=ﬃﬃﬃﬃﬃ 30p , L¼1=ﬃﬃﬃ 6p ; qðr;s;tÞ¼2r2þ2s2þ8t2 9.60. (a) A¼1 2½5;/C03;/C03;5/C138;B¼1 2½3;/C01;/C01;3/C138, (b) A¼1 5½14;/C02;/C02;11/C138,B¼1 5½ﬃﬃﬃ 2p þ4ﬃﬃﬃ 3p ;2ﬃﬃﬃ 2p /C02ﬃﬃﬃ 3p ;2ﬃﬃﬃ 2p /C02ﬃﬃﬃ 3p ;4ﬃﬃﬃ 2p þﬃﬃﬃ 3p /C138 9.61. (a)DðtÞ¼mðtÞ¼ð t/C02Þ2ðt/C06Þ, (b) DðtÞ¼ð t/C02Þ2ðt/C06Þ;mðtÞ¼ð t/C02Þðt/C06Þ 9.62. (a)DðtÞ¼ð t/C02Þ3ðt/C07Þ2; mðtÞ¼ð t/C02Þ2ðt/C07Þ, (b)DðtÞ¼ð t/C03Þ5; mðtÞ¼ð t/C03Þ3, (c)DðtÞ¼ð t/C02Þ2ðt/C04Þ2ðt/C05Þ; mðtÞ¼ð t/C02Þðt/C04Þðt/C05Þ 9.68. LetAbe the companion matrix [Example 9.12(b)] with last column: (a) ½/C08;/C06;5/C138T,( b )½/C04;/C07;2;5/C138T 9.69. Hint:Ais a root of hðtÞ¼fðtÞ/C0gðtÞ, where hðtÞ/C170 or the degree of hðtÞis less than the degree of fðtÞ:324 CHAPTER 9 Diagonalization: Eigenvalues and Eigenvectors Canonical Forms 10.1 Introduction LetTbe a linear operator on a vector space of ﬁnite dimension. As seen in Chapter 6, Tmay not have a diagonal matrix representation. However, it is still possible to ‘‘simplify’’ the matrix representation of T in a number of ways. This is the main topic of this chapter. In particular, we obtain the primarydecomposition theorem, and the triangular, Jordan, and rational canonical forms. We comment that the triangular and Jordan canonical forms exist for Tif and only if the characteristic polynomial DðtÞofThas all its roots in the base ﬁeld K. This is always true if Kis the complex ﬁeld C but may not be true if Kis the real ﬁeld R. We also introduce the idea of a quotient space . This is a very powerful tool, and it will be used in the proof of the existence of the triangular and rational canonical forms. 10.2 Triangular Form LetTbe a linear operator on an n-dimensional vector space V. Suppose Tcan be represented by the triangular matrix A¼a11a12 ... a1n a22 ... a2n ... ... ann2 6643 775 Then the characteristic polynomial DðtÞofTis a product of linear factors; that is, DðtÞ¼detðtI/C0AÞ¼ð t/C0a11Þðt/C0a22Þ/C1/C1/C1ð t/C0annÞ The converse is also true and is an important theorem (proved in Problem 10.28). THEOREM 10.1: LetT:V!Vbe a linear operator whose characteristic polynomial factors into linear polynomials. Then there exists a basis of Vin which Tis represented by a triangular matrix. THEOREM 10.1: (Alternative Form) Let Abe a square matrix whose characteristic polynomial factors into linear polynomials. Then Ais similar to a triangular matrix—that is, there exists an invertible matrix Psuch that P/C01APis triangular. We say that an operator Tcan be brought into triangular form if it can be represented by a triangular matrix. Note that in this case, the eigenvalues of Tare precisely those entries appearing on the main diagonal. We give an application of this remark. CHAPTER 10 325 EXAMPLE 10.1 LetAbe a square matrix over the complex ﬁeld C. Suppose lis an eigenvalue of A2. Show thatﬃﬃﬃ lp or/C0ﬃﬃﬃ lp is an eigenvalue of A. By Theorem 10.1, AandA2are similar, respectively, to triangular matrices of the form B¼m1* ... * m2... * ... ... mn2 6643 775and B2¼m2 1* ... * m2 2... * ... ... m2 n2 6643 775 Because similar matrices have the same eigenvalues, l¼m2 ifor some i. Hence, mi¼ﬃﬃﬃ lp ormi¼/C0ﬃﬃﬃ lp is an eigenvalue of A. 10.3 Invariance LetT:V!Vbe linear. A subspace WofVis said to be invariant under T orT-invariant ifTmaps W into itself—that is, if v2Wimplies TðvÞ2W. In this case, Trestricted to Wdeﬁnes a linear operator on W; that is, Tinduces a linear operator ^T:W!Wdeﬁned by ^TðwÞ¼TðwÞfor every w2W. EXAMPLE 10.2 (a) Let T:R3!R3be the following linear operator, which rotates each vector vabout the z-axis by an angle y (shown in Fig. 10-1): Tðx;y;zÞ¼ð xcosy/C0ysiny;xsinyþycosy;zÞ Observe that each vector w¼ða;b;0Þin the xy-plane Wremains in Wunder the mapping T; hence, Wis T-invariant. Observe also that the z-axis Uis invariant under T. Furthermore, the restriction of TtoWrotates each vector about the origin O, and the restriction of TtoUis the identity mapping of U. (b) Nonzero eigenvectors of a linear operator T:V!Vmay be characterized as generators of T-invariant one-dimensional subspaces. Suppose TðvÞ¼lv,v6¼0. Then W¼fkv;k2Kg, the one-dimensional subspace generated by v, is invariant under Tbecause TðkvÞ¼kTðvÞ¼kðlvÞ¼klv2W Conversely, suppose dim U¼1 and u6¼0 spans U, and Uis invariant under T. Then TðuÞ2Uand so TðuÞis a multiple of u—that is, TðuÞ¼mu. Hence, uis an eigenvector of T. The next theorem (proved in Problem 10.3) gives us an important class of invariant subspaces. THEOREM 10.2: LetT:V!Vbe any linear operator, and let fðtÞbe any polynomial. Then the kernel of fðTÞis invariant under T. The notion of invariance is related to matrix representations (Problem 10.5) as follows. THEOREM 10.3: Suppose Wis an invariant subspace of T:V!V. Then Thas a block matrix repre- sentationAB 0C/C20/C21 ,w h e r e Ais a matrix representation of the restriction ^TofTtoW.0 Wyz xUT()v Tw()θ θv w Figure 10-1326 CHAPTER 10 Canonical Forms 10.4 Invariant Direct-Sum Decompositions A vector space Vis termed the direct sum of subspaces W1;...;Wr, written V¼W1/C8W2/C8.../C8Wr if every vector v2Vcan be written uniquely in the form v¼w1þw2þ...þwr; with wi2Wi The following theorem (proved in Problem 10.7) holds. THEOREM 10.4: Suppose W1;W2;...;Wrare subspaces of V, and suppose B1¼fw11;w12;...;w1n1g; ...; Br¼fwr1;wr2;...;wrnrg are bases of W1;W2;...;Wr, respectively. Then Vis the direct sum of the Wiif and only if the union B¼B1[...[Bris a basis of V. Now suppose T:V!Vis linear and Vis the direct sum of (nonzero) T-invariant subspaces W1;W2;...;Wr; that is, V¼W1/C8.../C8Wr and TðWiÞ/C18Wi; i¼1;...;r LetTidenote the restriction of TtoWi. Then Tis said to be decomposable into the operators TiorTis said to be the direct sum of the Ti;written T¼T1/C8.../C8Tr:Also, the subspaces W1;...;Wrare said to reduce T or to form a T-invariant direct-sum decomposition ofV. Consider the special case where two subspaces UandWreduce an operator T:V!V; say dim U¼2 and dim W¼3, and supposefu1;u2gandfw1;w2;w3gare bases of UandW, respectively. If T1andT2 denote the restrictions of TtoUandW, respectively, then T1ðu1Þ¼a11u1þa12u2 T1ðu2Þ¼a21u1þa22u2T2ðw1Þ¼b11w1þb12w2þb13w3 T2ðw2Þ¼b21w1þb22w2þb23w3 T2ðw3Þ¼b31w1þb32w2þb33w3 Accordingly, the following matrices A;B;Mare the matrix representations of T1,T2,T, respectively, A¼a11a21 a12a22/C20/C21 ; B¼b11b21b31 b12b22b32 b13b23b332 43 5; M¼A0 0B/C20/C21 The block diagonal matrix Mresults from the fact that fu1;u2;w1;w2;w3gis a basis of V(Theorem 10.4), and that TðuiÞ¼T1ðuiÞandTðwjÞ¼T2ðwjÞ. A generalization of the above argument gives us the following theorem. THEOREM 10.5: Suppose T:V!Vis linear and suppose Vis the direct sum of T-invariant subspaces, say, W1;...;Wr.I fAiis a matrix representation of the restriction of TtoWi, then Tcan be represented by the block diagonal matrix: M¼diagðA1;A2;...;ArÞ 10.5 Primary Decomposition The following theorem shows that any operator T:V!Vis decomposable into operators whose minimum polynomials are powers of irreducible polynomials. This is the ﬁrst step in obtaining a canonical form for T.CHAPTER 10 Canonical Forms 327 THEOREM 10.6: (Primary Decomposition Theorem) Let T:V!Vbe a linear operator with minimal polynomial mðtÞ¼f1ðtÞn1f2ðtÞn2/C1/C1/C1frðtÞnr where the fiðtÞare distinct monic irreducible polynomials. Then Vis the direct sum ofT-invariant subspaces W1;...;Wr, where Wiis the kernel of fiðTÞni. Moreover, fiðtÞniis the minimal polynomial of the restriction of TtoWi. The above polynomials fiðtÞniare relatively prime. Therefore, the above fundamental theorem follows (Problem 10.11) from the next two theorems (proved in Problems 10.9 and 10.10, respectively). THEOREM 10.7: Suppose T:V!Vis linear, and suppose fðtÞ¼gðtÞhðtÞare polynomials such that fðTÞ¼0andgðtÞandhðtÞare relatively prime. Then Vis the direct sum of the T-invariant subspace UandW, where U¼KergðTÞandW¼KerhðTÞ. THEOREM 10.8: In Theorem 10.7, if fðtÞis the minimal polynomial of T[and gðtÞandhðtÞare monic], then gðtÞandhðtÞare the minimal polynomials of the restrictions of TtoU andW, respectively. We will also use the primary decomposition theorem to prove the following useful characterization of diagonalizable operators (see Problem 10.12 for the proof). THEOREM 10.9: A linear operator T:V!Vis diagonalizable if and only if its minimal polynomial mðtÞis a product of distinct linear polynomials. THEOREM 10.9: (Alternative Form) A matrix Ais similar to a diagonal matrix if and only if its minimal polynomial is a product of distinct linear polynomials. EXAMPLE 10.3 Suppose A6¼Iis a square matrix for which A3¼I. Determine whether or not Ais similar to a diagonal matrix if Ais a matrix over: (i) the real ﬁeld R, (ii) the complex ﬁeld C. Because A3¼I,Ais a zero of the polynomial fðtÞ¼t3/C01¼ðt/C01Þðt2þtþ1Þ:The minimal polynomial mðtÞ ofAcannot be t/C01, because A6¼I. Hence, mðtÞ¼t2þtþ1o r mðtÞ¼t3/C01 Because neither polynomial is a product of linear polynomials over R,Ais not diagonalizable over R.O nt h e other hand, each of the polynomials is a product of distinct linear polynomials over C. Hence, Ais diagonalizable over C. 10.6 Nilpotent Operators A linear operator T:V!Vis termed nilpotent ifTn¼0for some positive integer n; we call ktheindex of nilpotency ofTifTk¼0butTk/C016¼0:Analogously, a square matrix Ais termed nilpotent if An¼0 for some positive integer n, and of index kifAk¼0 but Ak/C016¼0. Clearly the minimum polynomial of a nilpotent operator (matrix) of index kismðtÞ¼tk; hence, 0 is its only eigenvalue. EXAMPLE 10.4 The following two r-square matrices will be used throughout the chapter: N¼NðrÞ¼010 ... 00 001 ... 00 :::::::::::::::::::::::::::::::: 000 ... 01 000 ... 002 666643 77775and JðlÞ¼l10 ... 00 0l1 ... 00 :::::::::::::::::::::::::::::::: 000 ... l1 000 ... 0 l2 666643 77775328 CHAPTER 10 Canonical Forms The ﬁrst matrix N, called a Jordan nilpotent block , consists of 1’s above the diagonal (called the super- diagonal ), and 0’s elsewhere. It is a nilpotent matrix of index r. (The matrix Nof order 1 is just the 1 /C21 zero matrix [0].) The second matrix JðlÞ, called a Jordan block belonging to the eigenvalue l, consists of l’s on the diagonal, 1’s on the superdiagonal, and 0’s elsewhere. Observe that JðlÞ¼lIþN In fact, we will prove that any linear operator Tcan be decomposed into operators, each of which is the sum of a scalar operator and a nilpotent operator. The following (proved in Problem 10.16) is a fundamental result on nilpotent operators. THEOREM 10.10: LetT:V!Vbe a nilpotent operator of index k. Then Thas a block diagonal matrix representation in which each diagonal entry is a Jordan nilpotent block N. There is at least one Nof order k, and all other Nare of orders/C20k. The number of Nof each possible order is uniquely determined by T. The total number of Nof all orders is equal to the nullity of T. The proof of Theorem 10.10 shows that the number of Nof order iis equal to 2 mi/C0miþ1/C0mi/C01, where miis the nullity of Ti. 10.7 Jordan Canonical Form An operator Tcan be put into Jordan canonical form if its characteristic and minimal polynomials factor into linear polynomials. This is always true if Kis the complex ﬁeld C. In any case, we can always extend the base ﬁeld Kto a ﬁeld in which the characteristic and minimal polynomials do factor into linear factors; thus, in a broad sense, every operator has a Jordan canonical form. Analogously, every matrix issimilar to a matrix in Jordan canonical form. The following theorem (proved in Problem 10.18) describes the Jordan canonical form J of a linear operator T. THEOREM 10.11: LetT:V!Vbe a linear operator whose characteristic and minimal polynomials are, respectively, DðtÞ¼ð t/C0l1Þn1/C1/C1/C1ðt/C0lrÞnrand mðtÞ¼ð t/C0l1Þm1/C1/C1/C1ðt/C0lrÞmr where the liare distinct scalars. Then Thas a block diagonal matrix representa- tionJin which each diagonal entry is a Jordan block Jij¼JðliÞ. For each lij, the corresponding Jijhave the following properties: (i) There is at least one Jijof order mi; all other Jijare of order/C20mi. (ii) The sum of the orders of the Jijisni. (iii) The number of Jijequals the geometric multiplicity of li. (iv) The number of Jijof each possible order is uniquely determined by T. EXAMPLE 10.5 Suppose the characteristic and minimal polynomials of an operator Tare, respec- tively, DðtÞ¼ð t/C02Þ4ðt/C05Þ3and mðtÞ¼ð t/C02Þ2ðt/C05Þ3CHAPTER 10 Canonical Forms 329 Then the Jordan canonical form of Tis one of the following block diagonal matrices: diag21 02/C20/C21 ;21 02/C20/C21 ;510 051 0052 43 50 @1 A or diag21 02/C20/C21 ;½2/C138;½2/C138;510 051 0052 43 50 @1 A The ﬁrst matrix occurs if Thas two independent eigenvectors belonging to the eigenvalue 2; and the second matrix occurs if Thas three independent eigenvectors belonging to the eigenvalue 2. 10.8 Cyclic Subspaces LetTbe a linear operator on a vector space Vof ﬁnite dimension over K. Suppose v2Vandv6¼0. The set of all vectors of the form fðTÞðvÞ, where fðtÞranges over all polynomials over K,i sa T-invariant subspace of Vcalled the T-cyclic subspace of V generated by v; we denote it by Zðv;TÞand denote the restriction of TtoZðv;TÞbyTv:By Problem 10.56, we could equivalently deﬁne Zðv;TÞas the intersection of all T-invariant subspaces of Vcontaining v. Now consider the sequence v;TðvÞ;T2ðvÞ;T3ðvÞ;... of powers of Tacting on v. Let kbe the least integer such that TkðvÞis a linear combination of those vectors that precede it in the sequence, say, TkðvÞ¼/C0 ak/C01Tk/C01ðvÞ/C0/C1/C1/C1/C0 a1TðvÞ/C0a0v mvðtÞ¼tkþak/C01tk/C01þ/C1/C1/C1þ a1tþa0Then is the unique monic polynomial of lowest degree for which mvðTÞðvÞ¼0. We call mvðtÞthe T-annihilator of vand Zðv;TÞ. The following theorem (proved in Problem 10.29) holds. THEOREM 10.12: Let Zðv;TÞ,Tv,mvðtÞbe deﬁned as above. Then (i) The setfv;TðvÞ;...;Tk/C01ðvÞgis a basis of Zðv;TÞ; hence, dim Zðv;TÞ¼k. (ii) The minimal polynomial of TvismvðtÞ. (iii) The matrix representation of Tvin the above basis is just the companion matrix CðmvÞofmvðtÞ; that is, CðmvÞ¼000 ... 0/C0a0 100 ... 0/C0a1 010 ... 0/C0a2 :::::::::::::::::::::::::::::::::::::::: 000 ... 0/C0ak/C02 000 ... 1/C0ak/C012 66666643 7777775 10.9 Rational Canonical Form In this section, we present the rational canonical form for a linear operator T:V!V. We emphasize that this form exists even when the minimal polynomial cannot be factored into linear polynomials. (Recallthat this is not the case for the Jordan canonical form.)330 CHAPTER 10 Canonical Forms LEMMA 10.13: LetT:V!Vbe a linear operator whose minimal polynomial is fðtÞn, where fðtÞis a monic irreducible polynomial. Then Vis the direct sum V¼Zðv1;TÞ/C8/C1/C1/C1/C8 Zðvr;TÞ ofT-cyclic subspaces Zðvi;TÞwith corresponding T-annihilators fðtÞn1;fðtÞn2;...;fðtÞnr; n¼n1/C21n2/C21.../C21nr Any other decomposition of Vinto T-cyclic subspaces has the same number of components and the same set of T-annihilators. We emphasize that the above lemma (proved in Problem 10.31) does not say that the vectors vior other T-cyclic subspaces Zðvi;TÞare uniquely determined by T, but it does say that the set of T-annihilators is uniquely determined by T. Thus, Thas a unique block diagonal matrix representation: M¼diagðC1;C2;...;CrÞ where the Ciare companion matrices. In fact, the Ciare the companion matrices of the polynomials fðtÞni. Using the Primary Decomposition Theorem and Lemma 10.13, we obtain the following result. THEOREM 10.14: LetT:V!Vbe a linear operator with minimal polynomial mðtÞ¼f1ðtÞm1f2ðtÞm2/C1/C1/C1fsðtÞms where the fiðtÞare distinct monic irreducible polynomials. Then Thas a unique block diagonal matrix representation: M¼diagðC11;C12;...;C1r1;...;Cs1;Cs2;...;CsrsÞ where the Cijare companion matrices. In particular, the Cijare the companion matrices of the polynomials fiðtÞnij, where m1¼n11/C21n12/C21/C1/C1/C1/C21 n1r1; ...; ms¼ns1/C21ns2/C21/C1/C1/C1/C21 nsrs The above matrix representation of Tis called its rational canonical form . The polynomials fiðtÞnij are called the elementary divisors ofT. EXAMPLE 10.6 LetVbe a vector space of dimension 8 over the rational ﬁeld Q, and let Tbe a linear operator on Vwhose minimal polynomial is mðtÞ¼f1ðtÞf2ðtÞ2¼ðt4/C04t3þ6t2/C04t/C07Þðt/C03Þ2 Thus, because dim V¼8;the characteristic polynomial DðtÞ¼f1ðtÞf2ðtÞ4:Also, the rational canonical form MofT must have one block the companion matrix of f1ðtÞand one block the companion matrix of f2ðtÞ2. There are two possibilities: (a) diag½Cðt4/C04t3þ6t2/C04t/C07Þ,Cððt/C03Þ2Þ,Cððt/C03Þ2Þ/C138 (b) diag½Cðt4/C04t3þ6t2/C04t/C07Þ,Cððt/C03Þ2Þ,Cðt/C03Þ;Cðt/C03Þ/C138 That is, (a) diag000 7 100 4 010/C06 001 42 6643 775;0/C09 16/C20/C21 ;0/C09 16/C20/C210 BB@1 CCA;(b) diag000 7 100 4 010/C06 001 42 6643 775;0/C09 16/C20/C21 ;½3/C138;½3/C1380 BB@1 CCA 10.10 Quotient Spaces LetVbe a vector space over a ﬁeld Kand let Wbe a subspace of V.I fvis any vector in V, we write vþWfor the set of sums vþwwith w2W; that is, vþW¼fvþw:w2WgCHAPTER 10 Canonical Forms 331 These sets are called the cosets ofWinV. We show (Problem 10.22) that these cosets partition Vinto mutually disjoint subsets. EXAMPLE 10.7 LetWbe the subspace of R2deﬁned by W¼fð a;bÞ:a¼bg; that is, Wis the line given by the equation x/C0y¼0. We can view vþWas a translation of the line obtained by adding the vector v to each point in W. As shown in Fig. 10-2, the coset vþWis also a line, and it is parallel to W. Thus, the cosets of WinR2are precisely all the lines parallel to W. In the following theorem, we use the cosets of a subspace Wof a vector space Vto deﬁne a new vector space; it is called the quotient space ofVbyWand is denoted by V=W. THEOREM 10.15: LetWbe a subspace of a vector space over a ﬁeld K. Then the cosets of WinV form a vector space over Kwith the following operations of addition and scalar multiplication: ðiÞðuþwÞþð vþWÞ¼ð uþvÞþW;ðiiÞkðuþWÞ¼kuþW;where k2K We note that, in the proof of Theorem 10.15 (Problem 10.24), it is ﬁrst necessary to show that the operations are well deﬁned; that is, whenever uþW¼u0þWand vþW¼v0þW, then ðiÞðuþvÞþW¼ðu0þv0ÞþW andðiiÞkuþW¼ku0þW for any k2K In the case of an invariant subspace, we have the following useful result (proved in Problem 10.27). THEOREM 10.16: Suppose Wis a subspace invariant under a linear operator T:V!V. Then T induces a linear operator /C22TonV=Wdeﬁned by /C22TðvþWÞ¼TðvÞþW. Moreover, ifTis a zero of any polynomial, then so is /C22T. Thus, the minimal polynomial of /C22T divides the minimal polynomial of T. SOLVED PROBLEMS Invariant Subspaces 10.1. Suppose T:V!Vis linear. Show that each of the following is invariant under T: (a)f0g, (b) V, (c) kernel of T, (d) image of T. (a) We have Tð0Þ¼02f0g; hence,f0gis invariant under T. (b) For every v2V,TðvÞ2V; hence, Vis invariant under T. (c) Let u2KerT. Then TðuÞ¼02KerTbecause the kernel of Tis a subspace of V. Thus, Ker Tis invariant under T. (d) Because TðvÞ2ImTfor every v2V, it is certainly true when v2ImT. Hence, the image of Tis invariant under T. 10.2. SupposefWigis a collection of T-invariant subspaces of a vector space V. Show that the intersection W¼T iWiis also T-invariant. Suppose v2W; then v2Wifor every i. Because WiisT-invariant, TðvÞ2Wifor every i. Thus, TðvÞ2Wand so WisT-invariant. Figure 10-2332 CHAPTER 10 Canonical Forms 10.3. Prove Theorem 10.2: Let T:V!Vbe linear. For any polynomial fðtÞ, the kernel of fðTÞis invariant under T. Suppose v2KerfðTÞ—that is, fðTÞðvÞ¼0. We need to show that TðvÞalso belongs to the kernel of fðTÞ—that is, fðTÞðTðvÞÞ¼ð fðTÞ/C14TÞðvÞ¼0. Because fðtÞt¼tfðtÞ, we have fðTÞ/C14T¼T/C14fðTÞ. Thus, as required, ðfðTÞ/C14TÞðvÞ¼ð T/C14fðTÞÞðvÞ¼TðfðTÞðvÞÞ¼ Tð0Þ¼0 10.4. Find all invariant subspaces of A¼2/C05 1/C02/C20/C21 viewed as an operator on R2. By Problem 10.1, R2andf0gare invariant under A. Now if Ahas any other invariant subspace, it must be one-dimensional. However, the characteristic polynomial of Ais DðtÞ¼t2/C0trðAÞtþjAj¼t2þ1 Hence, Ahas no eigenvalues (in R) and so Ahas no eigenvectors. But the one-dimensional invariant subspaces correspond to the eigenvectors; thus, R2andf0gare the only subspaces invariant under A. 10.5. Prove Theorem 10.3: Suppose WisT-invariant. Then Thas a triangular block representation AB 0C/C20/C21 , where Ais the matrix representation of the restriction ^TofTtoW. We choose a basis fw1;...;wrgofWand extend it to a basis fw1;...;wr;v1;...;vsgofV. We have ^Tðw1Þ¼Tðw1Þ¼a11w1þ/C1/C1/C1þ a1rwr ^Tðw2Þ¼Tðw2Þ¼a21w1þ/C1/C1/C1þ a2rwr :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ^TðwrÞ¼TðwrÞ¼ar1w1þ/C1/C1/C1þ arrwr Tðv1Þ¼b11w1þ/C1/C1/C1þ b1rwrþc11v1þ/C1/C1/C1þ c1svs Tðv2Þ¼b21w1þ/C1/C1/C1þ b2rwrþc21v1þ/C1/C1/C1þ c2svs :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TðvsÞ¼bs1w1þ/C1/C1/C1þ bsrwrþcs1v1þ/C1/C1/C1þ cssvs But the matrix of Tin this basis is the transpose of the matrix of coefﬁcients in the above system of equations (Section 6.2). Therefore, it has the formAB 0C/C20/C21 ,w h e r e Ais the transpose of the matrix of coefﬁcients for the obvious subsystem. By the same argument, Ais the matrix of ^Trelative to the basis fwig ofW. 10.6. Let ^Tdenote the restriction of an operator Tto an invariant subspace W. Prove (a) For any polynomial fðtÞ,fð^TÞðwÞ¼fðTÞðwÞ. (b) The minimal polynomial of ^Tdivides the minimal polynomial of T. (a) If fðtÞ¼0o ri f fðtÞis a constant (i.e., of degree 1), then the result clearly holds. Assume deg f¼n>1 and that the result holds for polynomials of degree less than n. Suppose that fðtÞ¼antnþan/C01tn/C01þ/C1/C1/C1þ a1tþa0 fð^TÞðwÞ¼ð an^Tnþan/C01^Tn/C01þ/C1/C1/C1þ a0IÞðwÞ ¼ðan^Tn/C01Þð^TðwÞÞþð an/C01^Tn/C01þ/C1/C1/C1þ a0IÞðwÞ ¼ðanTn/C01ÞðTðwÞÞþð an/C01Tn/C01þ/C1/C1/C1þ a0IÞðwÞ¼fðTÞðwÞThen (b) Let mðtÞdenote the minimal polynomial of T. Then by (a), mð^TÞðwÞ¼mðTÞðwÞ¼0ðwÞ¼0 for every w2W; that is, ^Tis a zero of the polynomial mðtÞ. Hence, the minimal polynomial of ^Tdivides mðtÞ.CHAPTER 10 Canonical Forms 333 Invariant Direct-Sum Decompositions 10.7. Prove Theorem 10.4: Suppose W1;W2;...;Wrare subspaces of Vwith respective bases B1¼fw11;w12;...;w1n1g; ...; Br¼fwr1;wr2;...;wrnrg Then Vis the direct sum of the Wiif and only if the union B¼S iBiis a basis of V. Suppose Bis a basis of V. Then, for any v2V, v¼a11w11þ/C1/C1/C1þ a1n1w1n1þ/C1/C1/C1þ ar1wr1þ/C1/C1/C1þ arnrwrnr¼w1þw2þ/C1/C1/C1þ wr where wi¼ai1wi1þ/C1/C1/C1þ ainiwini2Wi. We next show that such a sum is unique. Suppose v¼w0 1þw0 2þ/C1/C1/C1þ w0 r; where w0 i2Wi Becausefwi1;...;winigis a basis of Wi,w0 i¼bi1wi1þ/C1/C1/C1þ biniwini, and so v¼b11w11þ/C1/C1/C1þ b1n1w1n1þ/C1/C1/C1þ br1wr1þ/C1/C1/C1þ brnrwrnr Because Bis a basis of V;aij¼bij, for each iand each j. Hence, wi¼w0 i, and so the sum for vis unique. Accordingly, Vis the direct sum of the Wi. Conversely, suppose Vis the direct sum of the Wi. Then for any v2V,v¼w1þ/C1/C1/C1þ wr, where wi2Wi. Becausefwijigis a basis of Wi, each wiis a linear combination of the wiji, and so vis a linear combination of the elements of B. Thus, Bspans V. We now show that Bis linearly independent. Suppose a11w11þ/C1/C1/C1þ a1n1w1n1þ/C1/C1/C1þ ar1wr1þ/C1/C1/C1þ arnrwrnr¼0 Note that ai1wi1þ/C1/C1/C1þ ainiwini2Wi. We also have that 0 ¼0þ0/C1/C1/C102Wi. Because such a sum for 0 is unique, ai1wi1þ/C1/C1/C1þ ainiwini¼0 for i¼1;...;r The independence of the bases fwijigimplies that all the a’s are 0. Thus, Bis linearly independent and is a basis of V. 10.8. Suppose T:V!Vis linear and suppose T¼T1/C8T2with respect to a T-invariant direct-sum decomposition V¼U/C8W. Show that (a)mðtÞis the least common multiple of m1ðtÞandm2ðtÞ, where mðtÞ,m1ðtÞ,m2ðtÞare the minimum polynomials of T;T1;T2, respectively. (b)DðtÞ¼D1ðtÞD2ðtÞ, where DðtÞ;D1ðtÞ,D2ðtÞare the characteristic polynomials of T;T1;T2, respectively. (a) By Problem 10.6, each of m1ðtÞandm2ðtÞdivides mðtÞ. Now suppose fðtÞis a multiple of both m1ðtÞ andm2ðtÞ, then fðT1ÞðUÞ¼0 and fðT2ÞðWÞ¼0. Let v2V, then v¼uþwwith u2Uandw2W. Now fðTÞv¼fðTÞuþfðTÞw¼fðT1ÞuþfðT2Þw¼0þ0¼0 That is, Tis a zero of fðtÞ. Hence, mðtÞdivides fðtÞ, and so mðtÞis the least common multiple of m1ðtÞ andm2ðtÞ. (b) By Theorem 10.5, Thas a matrix representation M¼A0 0B/C20/C21 ,w h e r e AandBare matrix representations ofT1andT2, respectively. Then, as required, DðtÞ¼j tI/C0Mj¼tI/C0A 0 0 tI/C0B/C12/C12/C12/C12/C12/C12/C12/C12¼jtI/C0AjjtI/C0Bj¼D1ðtÞD2ðtÞ 10.9. Prove Theorem 10.7: Suppose T:V!Vis linear, and suppose fðtÞ¼gðtÞhðtÞare polynomials such that fðTÞ¼0andgðtÞandhðtÞare relatively prime. Then Vis the direct sum of the T-invariant subspaces UandWwhere U¼KergðTÞandW¼KerhðTÞ.334 CHAPTER 10 Canonical Forms Note ﬁrst that UandWareT-invariant by Theorem 10.2. Now, because gðtÞandhðtÞare relatively prime, there exist polynomials rðtÞandsðtÞsuch that rðtÞgðtÞþsðtÞhðtÞ¼1 Hence ;for the operator T; rðTÞgðTÞþsðTÞhðTÞ¼I ð*Þ Letv2V;then;byð*Þ; v¼rðTÞgðTÞvþsðTÞhðTÞv But the ﬁrst term in this sum belongs to W¼KerhðTÞ, because hðTÞrðTÞgðTÞv¼rðTÞgðTÞhðTÞv¼rðTÞfðTÞv¼rðTÞ0v¼0 Similarly, the second term belongs to U. Hence, Vis the sum of UandW. To prove that V¼U/C8W, we must show that a sum v¼uþwwith u2U,w2W, is uniquely determined by v. Applying the operator rðTÞgðTÞtov¼uþwand using gðTÞu¼0, we obtain rðTÞgðTÞv¼rðTÞgðTÞuþrðTÞgðTÞw¼rðTÞgðTÞw Also, applyingð*Þtowalone and using hðTÞw¼0, we obtain w¼rðTÞgðTÞwþsðTÞhðTÞw¼rðTÞgðTÞw Both of the above formulas give us w¼rðTÞgðTÞv, and so wis uniquely determined by v. Similarly uis uniquely determined by v. Hence, V¼U/C8W, as required. 10.10. Prove Theorem 10.8: In Theorem 10.7 (Problem 10.9), if fðtÞis the minimal polynomial of T (and gðtÞandhðtÞare monic), then gðtÞis the minimal polynomial of the restriction T1ofTtoU andhðtÞis the minimal polynomial of the restriction T2ofTtoW. Letm1ðtÞandm2ðtÞbe the minimal polynomials of T1andT2, respectively. Note that gðT1Þ¼0 and hðT2Þ¼0 because U¼KergðTÞandW¼KerhðTÞ. Thus, m1ðtÞdivides gðtÞ and m2ðtÞdivides hðtÞð 1Þ By Problem 10.9, fðtÞis the least common multiple of m1ðtÞandm2ðtÞ. But m1ðtÞandm2ðtÞare relatively prime because gðtÞandhðtÞare relatively prime. Accordingly, fðtÞ¼m1ðtÞm2ðtÞ. We also have that fðtÞ¼gðtÞhðtÞ. These two equations together with (1) and the fact that all the polynomials are monic imply thatgðtÞ¼m1ðtÞandhðtÞ¼m2ðtÞ, as required. 10.11. Prove the Primary Decomposition Theorem 10.6: Let T:V!Vbe a linear operator with minimal polynomial mðtÞ¼f1ðtÞn1f2ðtÞn2...frðtÞnr where the fiðtÞare distinct monic irreducible polynomials. Then Vis the direct sum of T- invariant subspaces W1;...;Wrwhere Wiis the kernel of fiðTÞni. Moreover, fiðtÞniis the minimal polynomial of the restriction of TtoWi. The proof is by induction on r. The case r¼1 is trivial. Suppose that the theorem has been proved for r/C01. By Theorem 10.7, we can write Vas the direct sum of T-invariant subspaces W1andV1, where W1is the kernel of f1ðTÞn1and where V1is the kernel of f2ðTÞn2/C1/C1/C1frðTÞnr. By Theorem 10.8, the minimal polynomials of the restrictions of TtoW1andV1aref1ðtÞn1andf2ðtÞn2/C1/C1/C1frðtÞnr, respectively. Denote the restriction of TtoV1by ^T1. By the inductive hypothesis, V1is the direct sum of subspaces W2;...;Wrsuch that Wiis the kernel of fiðT1Þniand such that fiðtÞniis the minimal polynomial for the restriction of ^T1toWi. But the kernel of fiðTÞni, for i¼2;...;ris necessarily contained in V1, because fiðtÞnidivides f2ðtÞn2/C1/C1/C1frðtÞnr. Thus, the kernel of fiðTÞniis the same as the kernel of fiðT1Þni, which is Wi. Also, the restriction of TtoWiis the same as the restriction of ^T1toWi(fori¼2;...;r); hence, fiðtÞniis also the minimal polynomial for the restriction of TtoWi. Thus, V¼W1/C8W2/C8/C1/C1/C1/C8 Wris the desired decomposition of T. 10.12. Prove Theorem 10.9: A linear operator T:V!Vhas a diagonal matrix representation if and only if its minimal polynomal mðtÞis a product of distinct linear polynomials.CHAPTER 10 Canonical Forms 335 Suppose mðtÞis a product of distinct linear polynomials, say, mðtÞ¼ð t/C0l1Þðt/C0l2Þ/C1/C1/C1ð t/C0lrÞ where the liare distinct scalars. By the Primary Decomposition Theorem, Vis the direct sum of subspaces W1;...;Wr, where Wi¼KerðT/C0liIÞ. Thus, if v2Wi, thenðT/C0liIÞðvÞ¼0o r TðvÞ¼liv. In other words, every vector in Wiis an eigenvector belonging to the eigenvalue li. By Theorem 10.4, the union of bases for W1;...;Wris a basis of V. This basis consists of eigenvectors, and so Tis diagonalizable. Conversely, suppose Tis diagonalizable (i.e., Vhas a basis consisting of eigenvectors of T). Let l1;...;lsbe the distinct eigenvalues of T. Then the operator fðTÞ¼ð T/C0l1IÞðT/C0l2IÞ/C1/C1/C1ð T/C0lsIÞ maps each basis vector into 0. Thus, fðTÞ¼0, and hence, the minimal polynomial mðtÞofTdivides the polynomial fðtÞ¼ð t/C0l1Þðt/C0l2Þ/C1/C1/C1ð t/C0lsIÞ Accordingly, mðtÞis a product of distinct linear polynomials. Nilpotent Operators, Jordan Canonical Form 10.13. LetT:Vbe linear. Suppose, for v2V,TkðvÞ¼0 but Tk/C01ðvÞ6¼0. Prove (a) The set S¼fv;TðvÞ;...;Tk/C01ðvÞgis linearly independent. (b) The subspace Wgenerated by SisT-invariant. (c) The restriction ^TofTtoWis nilpotent of index k. (d) Relative to the basis fTk/C01ðvÞ;...;TðvÞ;vgofW, the matrix of Tis the k-square Jordan nilpotent block Nkof index k(see Example 10.5). (a) Suppose avþa1TðvÞþa2T2ðvÞþ/C1/C1/C1þ ak/C01Tk/C01ðvÞ¼0 ð*Þ Applying Tk/C01toð*Þand using TkðvÞ¼0, we obtain aTk/C01ðvÞ¼0; because Tk/C01ðvÞ6¼0,a¼0. Now applying Tk/C02toð*Þand using TkðvÞ¼0 and a¼0, we ﬁind a1Tk/C01ðvÞ¼0; hence, a1¼0. Next applying Tk/C03toð*Þand using TkðvÞ¼0 and a¼a1¼0, we obtain a2Tk/C01ðvÞ¼0; hence, a2¼0. Continuing this process, we ﬁnd that all the a’s are 0; hence, Sis independent. (b) Let v2W. Then v¼bvþb1TðvÞþb2T2ðvÞþ/C1/C1/C1þ bk/C01Tk/C01ðvÞ Using TkðvÞ¼0, we have TðvÞ¼bTðvÞþb1T2ðvÞþ/C1/C1/C1þ bk/C02Tk/C01ðvÞ2W Thus, WisT-invariant. (c) By hypothesis, TkðvÞ¼0. Hence, for i¼0;...;k/C01, ^TkðTiðvÞÞ¼ TkþiðvÞ¼0 That is, applying ^Tkto each generator of W, we obtain 0; hence, ^Tk¼0and so ^Tis nilpotent of index at most k. On the other hand, ^Tk/C01ðvÞ¼Tk/C01ðvÞ6¼0; hence, Tis nilpotent of index exactly k. (d) For the basis fTk/C01ðvÞ,Tk/C02ðvÞ;...;TðvÞ;vgofW, ^TðTk/C01ðvÞÞ ¼ TkðvÞ¼0 ^TðTk/C02ðvÞÞ ¼ Tk/C01ðvÞ ^TðTk/C03ðvÞÞ ¼ Tk/C02ðvÞ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ^TðTðvÞÞ ¼ T2ðvÞ ^TðvÞ¼ TðvÞ Hence, as required, the matrix of Tin this basis is the k-square Jordan nilpotent block Nk.336 CHAPTER 10 Canonical Forms 10.14. LetT:V!Vbe linear. Let U¼KerTiandW¼KerTiþ1. Show that (a)U/C18W, (b) TðWÞ/C18U. (a) Suppose u2U¼KerTi. Then TiðuÞ¼0 and so Tiþ1ðuÞ¼TðTiðuÞÞ¼ Tð0Þ¼0. Thus, u2KerTiþ1¼W. But this is true for every u2U; hence, U/C18W. (b) Similarly, if w2W¼KerTiþ1, then Tiþ1ðwÞ¼0:Thus, Tiþ1ðwÞ¼TiðTðwÞÞ¼ Tið0Þ¼0 and so TðWÞ/C18U. 10.15. LetT:Vbe linear. Let X¼KerTi/C02,Y¼KerTi/C01,Z¼KerTi. Therefore (Problem 10.14), X/C18Y/C18Z. Suppose fu1;...;urg;fu1;...;ur;v1;...;vsg;fu1;...;ur;v1;...;vs;w1;...;wtg are bases of X;Y;Z, respectively. Show that S¼fu1;...;ur;Tðw1Þ;...;TðwtÞg is contained in Yand is linearly independent. By Problem 10.14, TðZÞ/C18Y, and hence S/C18Y. Now suppose Sis linearly dependent. Then there exists a relation a1u1þ/C1/C1/C1þ arurþb1Tðw1Þþ/C1/C1/C1þ btTðwtÞ¼0 where at least one coefﬁcient is not zero. Furthermore, because fuigis independent, at least one of the bk must be nonzero. Transposing, we ﬁnd b1Tðw1Þþ/C1/C1/C1þ btTðwtÞ¼/C0 a1u1/C0/C1/C1/C1/C0 arur2X¼KerTi/C02 Hence ; Ti/C02ðb1Tðw1Þþ/C1/C1/C1þ btTðwtÞÞ¼ 0 Thus ; Ti/C01ðb1w1þ/C1/C1/C1þ btwtÞ¼0; and so b1w1þ/C1/C1/C1þ btwt2Y¼KerTi/C01 Becausefui;vjggenerates Y, we obtain a relation among the ui,vj,wkwhere one of the coefﬁcients (i.e., one of the bk) is not zero. This contradicts the fact that fui;vj;wkgis independent. Hence, Smust also be independent. 10.16. Prove Theorem 10.10: Let T:V!Vbe a nilpotent operator of index k. Then Thas a unique block diagonal matrix representation consisting of Jordan nilpotent blocks N. There is at least oneNof order k, and all other Nare of orders/C20k. The total number of Nof all orders is equal to the nullity of T. Suppose dim V¼n. Let W1¼KerT,W2¼KerT2;...;Wk¼KerTk. Let us set mi¼dimWi, for i¼1;...;k. Because Tis of index k,Wk¼VandWk/C016¼Vand so mk/C01<mk¼n. By Problem 10.14, W1/C18W2/C18/C1/C1/C1/C18 Wk¼V Thus, by induction, we can choose a basis fu1;...;ungofVsuch thatfu1;...;umigis a basis of Wi. We now choose a new basis for Vwith respect to which Thas the desired form. It will be convenient to label the members of this new basis by pairs of indices. We begin by setting vð1;kÞ¼umk/C01þ1; vð2;kÞ¼umk/C01þ2; ...; vðmk/C0mk/C01;kÞ¼umk and setting vð1;k/C01Þ¼Tvð1;kÞ; vð2;k/C01Þ¼Tvð2;kÞ; ...; vðmk/C0mk/C01;k/C01Þ¼Tvðmk/C0mk/C01;kÞ By the preceding problem, S1¼fu1...;umk/C02;vð1;k/C01Þ;...;vðmk/C0mk/C01;k/C01Þg is a linearly independent subset of Wk/C01. We extend S1to a basis of Wk/C01by adjoining new elements (if necessary), which we denote by vðmk/C0mk/C01þ1;k/C01Þ; vðmk/C0mk/C01þ2;k/C01Þ; ...; vðmk/C01/C0mk/C02;k/C01Þ Next we set vð1;k/C02Þ¼Tvð1;k/C01Þ; vð2;k/C02Þ¼Tvð2;k/C01Þ; ...; vðmk/C01/C0mk/C02;k/C02Þ¼Tvðmk/C01/C0mk/C02;k/C01ÞCHAPTER 10 Canonical Forms 337 Again by the preceding problem, S2¼fu1;...;umk/C0s;vð1;k/C02Þ;...;vðmk/C01/C0mk/C02;k/C02Þg is a linearly independent subset of Wk/C02, which we can extend to a basis of Wk/C02by adjoining elements vðmk/C01/C0mk/C02þ1;k/C02Þ; vðmk/C01/C0mk/C02þ2;k/C02Þ; ...; vðmk/C02/C0mk/C03;k/C02Þ Continuing in this manner, we get a new basis for V, which for convenient reference we arrange as follows: vð1;kÞ ...;vðmk/C0mk/C01;kÞ vð1;k/C01Þ;...;vðmk/C0mk/C01;k/C01Þ...;vðmk/C01/C0mk/C02;k/C01Þ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: vð1;2Þ; ...;vðmk/C0mk/C01;2Þ; ...;vðmk/C01/C0mk/C02;2Þ; ...;vðm2/C0m1;2Þ vð1;1Þ; ...;vðmk/C0mk/C01;1Þ; ...;vðmk/C01/C0mk/C02;1Þ; ...;vðm2/C0m1;1Þ;...;vðm1;1Þ The bottom row forms a basis of W1, the bottom two rows form a basis of W2, and so forth. But what is important for us is that Tmaps each vector into the vector immediately below it in the table or into 0 if the vector is in the bottom row. That is, Tvði;jÞ¼vði;j/C01Þforj>1 0 for j¼1/C26 Now it is clear [see Problem 10.13(d)] that Twill have the desired form if the vði;jÞare ordered lexicographically: beginning with vð1;1Þand moving up the ﬁrst column to vð1;kÞ, then jumping to vð2;1Þ and moving up the second column as far as possible. Moreover, there will be exactly mk/C0mk/C01diagonal entries of order k:Also, there will be ðmk/C01/C0mk/C02Þ/C0ð mk/C0mk/C01Þ¼ 2mk/C01/C0mk/C0mk/C02diagonal entries of order k/C01 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 2m2/C0m1/C0m3 diagonal entries of order 2 2m1/C0m2 diagonal entries of order 1 as can be read off directly from the table. In particular, because the numbers m1;...;mkare uniquely determined by T, the number of diagonal entries of each order is uniquely determined by T. Finally, the identity m1¼ðmk/C0mk/C01Þþð 2mk/C01/C0mk/C0mk/C02Þþ/C1/C1/C1þð 2m2/C0m1/C0m3Þþð 2m1/C0m2Þ shows that the nullity m1ofTis the total number of diagonal entries of T. 10.17. LetA¼01101 0011100000 00000 000002 666643 77775andB¼01100 0011100011 00000 000002 666643 77775. The reader can verify that AandB are both nilpotent of index 3; that is, A 3¼0 but A26¼0, and B3¼0 but B26¼0. Find the nilpotent matrices MAandMBin canonical form that are similar to AandB, respectively. Because AandBare nilpotent of index 3, MAandMBmust each contain a Jordan nilpotent block of order 3, and none greater then 3. Note that rank ðAÞ¼2 and rankðBÞ¼3, so nullityðAÞ¼5/C02¼3 and nullityðBÞ¼5/C03¼2. Thus, MAmust contain three diagonal blocks, which must be one of order 3 and two of order 1; and MBmust contain two diagonal blocks, which must be one of order 3 and one of order 2. Namely, MA¼01000 00100 00000 00000 000002 666643 77775and MB¼01000 00100 00000 00001 000002 666643 77775338 CHAPTER 10 Canonical Forms 10.18. Prove Theorem 10.11 on the Jordan canonical form for an operator T. By the primary decomposition theorem, Tis decomposable into operators T1;...;Tr; that is, T¼T1/C8/C1/C1/C1/C8 Tr, whereðt/C0liÞmiis the minimal polynomial of Ti. Thus, in particular, ðT1/C0l1IÞm1¼0;...;ðTr/C0lrIÞmr¼0 SetNi¼Ti/C0liI. Then, for i¼1;...;r, Ti¼NiþliI; where Nmi i¼0 That is, Tiis the sum of the scalar operator liIand a nilpotent operator Ni, which is of index mibecause ðt/C0liÞm iis the minimal polynomial of Ti. Now, by Theorem 10.10 on nilpotent operators, we can choose a basis so that Niis in canonical form. In this basis, Ti¼NiþliIis represented by a block diagonal matrix Miwhose diagonal entries are the matrices Jij. The direct sum Jof the matrices Miis in Jordan canonical form and, by Theorem 10.5, is a matrix representation of T. Last, we must show that the blocks Jijsatisfy the required properties. Property (i) follows from the fact thatNiis of index mi. Property (ii) is true because TandJhave the same characteristic polynomial. Property (iii) is true because the nullity of Ni¼Ti/C0liIis equal to the geometric multiplicity of the eigenvalue li. Property (iv) follows from the fact that the Tiand hence the Niare uniquely determined by T. 10.19. Determine all possible Jordan canonical forms Jfor a linear operator T:V!Vwhose characteristic polynomial DðtÞ¼ð t/C02Þ5and whose minimal polynomial mðtÞ¼ð t/C02Þ2. Jmust be a 5/C25 matrix, because DðtÞhas degree 5, and all diagonal elements must be 2, because 2 is the only eigenvalue. Moreover, because the exponent of t/C02i nmðtÞis 2,Jmust have one Jordan block of order 2, and the others must be of order 2 or 1. Thus, there are only two possibilities: J¼diag21 2/C20/C21 ;21 2/C20/C21 ;½2/C138/C18/C19 or J¼diag21 2/C20/C21 ;½2/C138;½2/C138;½2/C138/C18/C19 10.20. Determine all possible Jordan canonical forms for a linear operator T:V!Vwhose character- istic polynomial DðtÞ¼ð t/C02Þ3ðt/C05Þ2. In each case, ﬁnd the minimal polynomial mðtÞ. Because t/C02 has exponent 3 in DðtÞ, 2 must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) diag21 21 22 43 5;51 5/C20/C210 @1 A, (b) diag21 21 22 43 5;½5/C138;½5/C1380 @1 A, (c) diag21 2/C20/C21 ;½2/C138;51 5/C20/C21 /C18/C19 , (d) diag21 2/C20/C21 ;½2/C138;½5/C138;½5/C138/C18/C19 , (e) diag½2/C138;½2/C138;½2/C138;51 5/C20/C21 /C18/C19 , (f ) diagð½2/C138;½2/C138;½2/C138;½5/C138;½5/C138Þ The exponent in the minimal polynomial mðtÞis equal to the size of the largest block. Thus, (a) mðtÞ¼ð t/C02Þ3ðt/C05Þ2, (b) mðtÞ¼ð t/C02Þ3ðt/C05Þ, (c) mðtÞ¼ð t/C02Þ2ðt/C05Þ2, (d) mðtÞ¼ð t/C02Þ2ðt/C05Þ, (e) mðtÞ¼ð t/C02Þðt/C05Þ2,( f ) mðtÞ¼ð t/C02Þðt/C05Þ Quotient Space and Triangular Form 10.21. LetWbe a subspace of a vector space V. Show that the following are equivalent: (i) u2vþW, (ii) u/C0v2W, (iii) v2uþW. Suppose u2vþW. Then there exists w02Wsuch that u¼vþw0. Hence, u/C0v¼w02W. Conversely, suppose u/C0v2W.T h e n u/C0v¼w0where w02W. Hence, u¼vþw02vþW. Thus, (i) and (ii) are equivalent. We also have u/C0v2Wiff/C0ðu/C0vÞ¼v/C0u2Wiffv2uþW. Thus, (ii) and (iii) are also equivalent.CHAPTER 10 Canonical Forms 339 10.22. Prove the following: The cosets of WinVpartition Vinto mutually disjoint sets. That is, (a) Any two cosets uþWand vþWare either identical or disjoint. (b) Each v2Vbelongs to a coset; in fact, v2vþW. Furthermore, uþW¼vþWif and only if u/C0v2W, and soðvþwÞþW¼vþWfor any w2W. Letv2V. Because 02W, we have v¼vþ02vþW, which proves (b). Now suppose the cosets uþWand vþWare not disjoint; say, the vector xbelongs to both uþW and vþW. Then u/C0x2Wandx/C0v2W. The proof of (a) is complete if we show that uþW¼vþW. Letuþw0be any element in the coset uþW. Because u/C0x,x/C0v,w0belongs to W, ðuþw0Þ/C0v¼ðu/C0xÞþð x/C0vÞþw02W Thus, uþw02vþW, and hence the cost uþWis contained in the coset vþW. Similarly, vþWis contained in uþW, and so uþW¼vþW. The last statement follows from the fact that uþW¼vþWif and only if u2vþW, and, by Problem 10.21, this is equivalent to u/C0v2W. 10.23. LetWbe the solution space of the homogeneous equation 2 xþ3yþ4z¼0. Describe the cosets ofWinR3. Wis a plane through the origin O¼ð0;0;0Þ, and the cosets of Ware the planes parallel to W. Equivalently, the cosets of Ware the solution sets of the family of equations 2xþ3yþ4z¼k; k2R In fact, the coset vþW, where v¼ða;b;cÞ, is the solution set of the linear equation 2xþ3yþ4z¼2aþ3bþ4c or 2ðx/C0aÞþ3ðy/C0bÞþ4ðz/C0cÞ¼0 10.24. Suppose Wis a subspace of a vector space V. Show that the operations in Theorem 10.15 are well deﬁned; namely, show that if uþW¼u0þWand vþW¼v0þW, then ðaÞðuþvÞþW¼ðu0þv0ÞþW andðbÞkuþW¼ku0þW for any k2K (a) Because uþW¼u0þWand vþW¼v0þW, both u/C0u0and v/C0v0belong to W. But then ðuþvÞ/C0ð u0þv0Þ¼ð u/C0u0Þþð v/C0v0Þ2W. Hence,ðuþvÞþW¼ðu0þv0ÞþW. (b) Also, because u/C0u02Wimplies kðu/C0u0Þ2W, then ku/C0ku0¼kðu/C0u0Þ2W; accordingly, kuþW¼ku0þW. 10.25. LetVbe a vector space and Wa subspace of V. Show that the natural map Z:V!V=W, deﬁned byZðvÞ¼vþW, is linear. For any u;v2Vand any k2K, we have nðuþvÞ¼uþvþW¼uþWþvþW¼ZðuÞþZðvÞ and ZðkvÞ¼kvþW¼kðvþWÞ¼kZðvÞ Accordingly, Zis linear. 10.26. LetWbe a subspace of a vector space V. Supposefw1;...;wrgis a basis of Wand the set of cosetsf/C22v1;...;/C22vsg, where /C22vj¼vjþW, is a basis of the quotient space. Show that the set of vectors B¼fv1;...;vs,w1;...;wrgis a basis of V. Thus, dim V¼dimWþdimðV=WÞ. Suppose u2V. Becausef/C22vjgis a basis of V=W, /C22u¼uþW¼a1/C22v1þa2/C22v2þ/C1/C1/C1þ as/C22vs Hence, u¼a1v1þ/C1/C1/C1þ asvsþw, where w2W. Sincefwigis a basis of W, u¼a1v1þ/C1/C1/C1þ asvsþb1w1þ/C1/C1/C1þ brwr340 CHAPTER 10 Canonical Forms Accordingly, Bspans V. We now show that Bis linearly independent. Suppose c1v1þ/C1/C1/C1þ csvsþd1w1þ/C1/C1/C1þ drwr¼0 ð1Þ Then c1/C22v1þ/C1/C1/C1þ cs/C22vs¼/C220¼W Becausef/C22vjgis independent, the c’s are all 0. Substituting into (1), we ﬁnd d1w1þ/C1/C1/C1þ drwr¼0. Becausefwigis independent, the d’s are all 0. Thus, Bis linearly independent and therefore a basis of V. 10.27. Prove Theorem 10.16: Suppose Wis a subspace invariant under a linear operator T:V!V. Then Tinduces a linear operator /C22TonV=Wdeﬁned by /C22TðvþWÞ¼TðvÞþW. Moreover, if Tis a zero of any polynomial, then so is /C22T. Thus, the minimal polynomial of /C22Tdivides the minimal polynomial of T. We ﬁrst show that /C22Tis well deﬁned; that is, if uþW¼vþW, then /C22TðuþWÞ¼ /C22TðvþWÞ.I f uþW¼vþW, then u/C0v2W, and, as WisT-invariant, Tðu/C0vÞ¼TðuÞ/C0TðvÞ2W. Accordingly, /C22TðuþWÞ¼TðuÞþW¼TðvÞþW¼/C22TðvþWÞ as required. We next show that /C22Tis linear. We have /C22TððuþWÞþð vþWÞÞ¼ /C22TðuþvþWÞ¼TðuþvÞþW¼TðuÞþTðvÞþW ¼TðuÞþWþTðvÞþW¼/C22TðuþWÞþ /C22TðvþWÞ Furthermore, /C22TðkðuþWÞÞ¼ /C22TðkuþWÞ¼TðkuÞþW¼kTðuÞþW¼kðTðuÞþWÞ¼k^TðuþWÞ Thus, /C22Tis linear. Now, for any coset uþWinV=W, T2ðuþWÞ¼T2ðuÞþW¼TðTðuÞÞþ W¼/C22TðTðuÞþWÞ¼ /C22Tð/C22TðuþWÞÞ¼ /C22T2ðuþWÞ Hence, T2¼/C22T2. Similarly, Tn¼/C22Tnfor any n. Thus, for any polynomial fðtÞ¼antnþ/C1/C1/C1þ a0¼Paiti fðTÞðuþWÞ¼fðTÞðuÞþW¼PaiTiðuÞþW¼PaiðTiðuÞþWÞ ¼PaiTiðuþWÞ¼Pai/C22TiðuþWÞ¼ðPai/C22TiÞðuþWÞ¼fð/C22TÞðuþWÞ and so fðTÞ¼fð/C22TÞ. Accordingly, if Tis a root of fðtÞthen fðTÞ¼ /C220¼W¼fð/C22TÞ; that is, /C22Tis also a root offðtÞ. The theorem is proved. 10.28. Prove Theorem 10.1: Let T:V!Vbe a linear operator whose characteristic polynomial factors into linear polynomials. Then Vhas a basis in which Tis represented by a triangular matrix. The proof is by induction on the dimension of V. If dim V¼1, then every matrix representation of T is a 1/C21 matrix, which is triangular. Now suppose dim V¼n>1 and that the theorem holds for spaces of dimension less than n. Because the characteristic polynomial of Tfactors into linear polynomials, Thas at least one eigenvalue and so at least one nonzero eigenvector v,s a y TðvÞ¼a11v.L e t Wbe the one-dimensional subspace spanned by v. Set /C22V¼V=W. Then (Problem 10.26) dim /C22V¼dimV/C0dimW¼n/C01. Note also that Wis invariant under T. By Theorem 10.16, Tinduces a linear operator /C22Ton /C22Vwhose minimal polynomial divides the minimal polynomial of T. Because the characteristic polynomial of Tis a product of linear polynomials, so is its minimal polynomial, and hence, so are the minimal and characteristic polynomials of /C22T. Thus, /C22V and /C22Tsatisfy the hypothesis of the theorem. Hence, by induction, there exists a basis f/C22v2;...;/C22vngof /C22V such that /C22Tð/C22v2Þ¼a22/C22v2/C22Tð/C22v3Þ¼a32/C22v2þa33/C22v3 ::::::::::::::::::::::::::::::::::::::::: /C22Tð/C22vnÞ¼an2/C22vnþan3/C22v3þ/C1/C1/C1þ ann/C22vnCHAPTER 10 Canonical Forms 341 Now let v2;...;vnbe elements of Vthat belong to the cosets v2;...;vn, respectively. Then fv;v2;...;vng is a basis of V(Problem 10.26). Because /C22Tðv2Þ¼a22/C22v2, we have /C22Tð/C22v2Þ/C0a22/C22v22¼0; and so Tðv2Þ/C0a22v22W ButWis spanned by v; hence, Tðv2Þ/C0a22v2is a multiple of v, say, Tðv2Þ/C0a22v2¼a21v; and so Tðv2Þ¼a21vþa22v2 Similarly, for i¼3;...;n TðviÞ/C0ai2v2/C0ai3v3/C0/C1/C1/C1/C0 aiivi2W; and so TðviÞ¼ai1vþai2v2þ/C1/C1/C1þ aiivi Thus, TðvÞ¼a11v Tðv2Þ¼a21vþa22v2 :::::::::::::::::::::::::::::::::::::::: TðvnÞ¼an1vþan2v2þ/C1/C1/C1þ annvn and hence the matrix of Tin this basis is triangular. Cyclic Subspaces, Rational Canonical Form 10.29. Prove Theorem 10.12: Let Zðv;TÞbe a T-cyclic subspace, Tvthe restriction of TtoZðv;TÞ, and mvðtÞ¼tkþak/C01tk/C01þ/C1/C1/C1þ a0theT-annihilator of v. Then, (i) The setfv;TðvÞ;...;Tk/C01ðvÞgis a basis of Zðv;TÞ; hence, dim Zðv;TÞ¼k. (ii) The minimal polynomial of TvismvðtÞ. (iii) The matrix of Tvin the above basis is the companion matrix C¼CðmvÞofmvðtÞ[which has 1’s below the diagonal, the negative of the coefﬁcients a0;a1;...;ak/C01ofmvðtÞin the last column, and 0’s elsewhere]. (i) By deﬁnition of mvðtÞ,TkðvÞis the ﬁrst vector in the sequence v,TðvÞ,T2ðvÞ;...that, is a linear combination of those vectors that precede it in the sequence; hence, the set B¼fv;TðvÞ;...;Tk/C01ðvÞgis linearly independent. We now only have to show that Zðv;TÞ¼LðBÞ, the linear span of B. By the above, TkðvÞ2LðBÞ. We prove by induction that TnðvÞ2LðBÞfor every n. Suppose n>kand Tn/C01ðvÞ2LðBÞ—that is, Tn/C01ðvÞis a linear combination of v;...;Tk/C01ðvÞ.T h e n TnðvÞ¼TðTn/C01ðvÞÞis a linear combination of TðvÞ;...;TkðvÞ.B u t TkðvÞ2LðBÞ; hence, TnðvÞ2LðBÞfor every n. Consequently, fðTÞðvÞ2LðBÞfor any polynomial fðtÞ. Thus, Zðv;TÞ¼LðBÞ,a n ds o Bis a basis, as claimed. (ii) Suppose mðtÞ¼tsþbs/C01ts/C01þ/C1/C1/C1þ b0is the minimal polynomial of Tv. Then, because v2Zðv;TÞ, 0¼mðTvÞðvÞ¼mðTÞðvÞ¼TsðvÞþbs/C01Ts/C01ðvÞþ/C1/C1/C1þ b0v Thus, TsðvÞis a linear combination of v,TðvÞ;...;Ts/C01ðvÞ, and therefore k/C20s. However, mvðTÞ¼0and so mvðTvÞ¼0:Then mðtÞdivides mvðtÞ;and so s/C20k:Accordingly, k¼sand hence mvðtÞ¼mðtÞ. (iii)TvðvÞ¼ TðvÞ TvðTðvÞÞ ¼ T2ðvÞ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: TvðTk/C02ðvÞÞ ¼ Tk/C01ðvÞ TvðTk/C01ðvÞÞ ¼ TkðvÞ¼/C0 a0v/C0a1TðvÞ/C0a2T2ðvÞ/C0/C1/C1/C1/C0 ak/C01Tk/C01ðvÞ By deﬁnition, the matrix of Tvin this basis is the tranpose of the matrix of coefﬁcients of the above system of equations; hence, it is C, as required. 10.30. LetT:V!Vbe linear. Let Wbe a T-invariant subspace of Vand /C22Tthe induced operator on V=W. Prove (a) The T-annihilator of v2Vdivides the minimal polynomial of T. (b) The /C22T-annihilator of /C22v2V=Wdivides the minimal polynomial of T.342 CHAPTER 10 Canonical Forms (a) The T-annihilator of v2Vis the minimal polynomial of the restriction of TtoZðv;TÞ; therefore, by Problem 10.6, it divides the minimal polynomial of T. (b) The /C22T-annihilator of /C22v2V=Wdivides the minimal polynomial of /C22T, which divides the minimal polynomial of Tby Theorem 10.16. Remark: In the case where the minimum polynomial of TisfðtÞn, where fðtÞis a monic irreducible polynomial, then the T-annihilator of v2Vand the /C22T-annihilator of /C22v2V=Ware of the form fðtÞm, where m/C20n. 10.31. Prove Lemma 10.13: Let T:V!Vbe a linear operator whose minimal polynomial is fðtÞn, where fðtÞis a monic irreducible polynomial. Then Vis the direct sum of T-cyclic subspaces Zi¼Zðvi;TÞ,i¼1;...;r, with corresponding T-annihilators fðtÞn1;fðtÞn2;...;fðtÞnr; n¼n1/C21n2/C21/C1/C1/C1/C21 nr Any other decomposition of Vinto the direct sum of T-cyclic subspaces has the same number of components and the same set of T-annihilators. The proof is by induction on the dimension of V. If dim V¼1, then VisT-cyclic and the lemma holds. Now suppose dim V>1 and that the lemma holds for those vector spaces of dimension less than that of V. Because the minimal polynomial of TisfðtÞn, there exists v12Vsuch that fðTÞn/C01ðv1Þ6¼0; hence, theT-annihilator of v1isfðtÞn. Let Z1¼Zðv1;TÞand recall that Z1isT-invariant. Let /C22V¼V=Z1and let /C22T be the linear operator on /C22Vinduced by T. By Theorem 10.16, the minimal polynomial of /C22Tdivides fðtÞn; hence, the hypothesis holds for /C22Vand /C22T. Consequently, by induction, /C22Vis the direct sum of /C22T-cyclic subspaces; say, /C22V¼Zð/C22v2;/C22TÞ/C8/C1/C1/C1/C8 Zð/C22vr;/C22TÞ where the corresponding /C22T-annihilators are fðtÞn2;...;fðtÞnr,n/C21n2/C21/C1/C1/C1/C21 nr. We claim that there is a vector v2in the coset /C22v2whose T-annihilator is fðtÞn2, the /C22T-annihilator of /C22v2. Letwbe any vector in /C22v2. Then fðTÞn2ðwÞ2Z1. Hence, there exists a polynomial gðtÞfor which fðTÞn2ðwÞ¼gðTÞðv1Þð 1Þ Because fðtÞnis the minimal polynomial of T, we have, by (1), 0¼fðTÞnðwÞ¼fðTÞn/C0n2gðTÞðv1Þ ButfðtÞnis the T-annihilator of v1; hence, fðtÞndivides fðtÞn/C0n2gðtÞ, and so gðtÞ¼fðtÞn2hðtÞfor some polynomial hðtÞ. We set v2¼w/C0hðTÞðv1Þ Because w/C0v2¼hðTÞðv1Þ2Z1,v2also belongs to the coset /C22v2. Thus, the T-annihilator of v2is a multiple of the /C22T-annihilator of /C22v2. On the other hand, by (1), fðTÞn2ðv2Þ¼fðTÞnsðw/C0hðTÞðv1ÞÞ¼ fðTÞn2ðwÞ/C0gðTÞðv1Þ¼0 Consequently, the T-annihilator of v2isfðtÞn2, as claimed. Similarly, there exist vectors v3;...;vr2Vsuch that vi2viand that the T-annihilator of viisfðtÞni, the /C22T-annihilator of vi. We set Z2¼Zðv2;TÞ; ...; Zr¼Zðvr;TÞ Letddenote the degree of fðtÞ, so that fðtÞnihas degree dni. Then, because fðtÞniis both the T-annihilator ofviand the /C22T-annihilator of vi, we know that fvi;TðviÞ;...;Tdni/C01ðviÞg andf/C22vi:/C22TðviÞ;...;/C22Tdni/C01ðviÞg are bases for Zðvi;TÞand Zðvi;/C22TÞ, respectively, for i¼2;...;r. But /C22V¼Zðv2;/C22TÞ/C8/C1/C1/C1/C8 Zðvr;/C22TÞ; hence, f/C22v2;...;/C22Tdn2/C01ð/C22v2Þ;...;/C22vr;...;/C22Tdnr/C01ð/C22vrÞgCHAPTER 10 Canonical Forms 343 is a basis for /C22V. Therefore, by Problem 10.26 and the relation /C22Tið/C22vÞ¼TiðvÞ(see Problem 10.27), fv1;...;Tdn1/C01ðv1Þ;v2;...;Ten2/C01ðv2Þ;...;vr;...;Tdnr/C01ðvrÞg is a basis for V. Thus, by Theorem 10.4, V¼Zðv1;TÞ/C8/C1/C1/C1/C8 Zðvr;TÞ, as required. It remains to show that the exponents n1;...;nrare uniquely determined by T. Because d¼degree of fðtÞ; dimV¼dðn1þ/C1/C1/C1þ nrÞ and dim Zi¼dni; i¼1;...;r Also, if sis any positive integer, then (Problem 10.59) fðTÞsðZiÞis a cyclic subspace generated by fðTÞsðviÞ, and it has dimension dðni/C0sÞifni>sand dimension 0 if ni/C20s. Now any vector v2Vcan be written uniquely in the form v¼w1þ/C1/C1/C1þ wr, where wi2Zi. Hence, any vector in fðTÞsðVÞcan be written uniquely in the form fðTÞsðvÞ¼fðTÞsðw1Þþ/C1/C1/C1þ fðTÞsðwrÞ where fðTÞsðwiÞ2fðTÞsðZiÞ. Let tbe the integer, dependent on s, for which n1>s; ...; nt>s; ntþ1/C21s Then fðTÞsðVÞ¼fðTÞsðZ1Þ/C8/C1/C1/C1/C8 fðTÞsðZtÞ and so dim ½fðTÞsðVÞ/C138¼ d½ðn1/C0sÞþ/C1/C1/C1þð nt/C0sÞ/C138 ð 2Þ The numbers on the left of (2) are uniquely determined by T. Set s¼n/C01, and (2) determines the number ofniequal to n. Next set s¼n/C02, and (2) determines the number of ni(if any) equal to n/C01. We repeat the process until we set s¼0 and determine the number of niequal to 1. Thus, the niare uniquely determined by TandV, and the lemma is proved. 10.32. LetVbe a seven-dimensional vector space over R,a n dl e t T:V!Vbe a linear operator with minimal polynomial mðtÞ¼ð t2/C02tþ5Þðt/C03Þ3. Find all possible rational canonical forms M ofT. Because dim V¼7;there are only two possible characteristic polynomials, D1ðtÞ¼ð t2/C02tþ5Þ2 ðt/C03Þ3orD1ðtÞ¼ð t2/C02tþ5Þðt/C03Þ5:Moreover, the sum of the orders of the companion matrices must add up to 7. Also, one companion matrix must be Cðt2/C02tþ5Þand one must be Cððt/C03Þ3Þ¼ Cðt3/C09t2þ27t/C027Þ. Thus, Mmust be one of the following block diagonal matrices: (a) diag0/C05 12/C20/C21 ;0/C05 12/C20/C21 ;00 2 7 10/C027 01 92 43 50 @1 A; (b) diag0/C05 12/C20/C21 ;00 2 7 10/C027 01 92 43 5;0/C09 16/C20/C210 @1 A; (c) diag0/C05 12/C20/C21 ;00 2 7 10/C027 01 92 43 5;½3/C138;½3/C1380 @1 A Projections 10.33. Suppose V¼W1/C8/C1/C1/C1/C8 Wr. The projection ofVinto its subspace Wkis the mapping E:V!V deﬁned by EðvÞ¼wk, where v¼w1þ/C1/C1/C1þ wr;wi2Wi. Show that (a) Eis linear, (b) E2¼E. (a) Because the sum v¼w1þ/C1/C1/C1þ wr,wi2Wis uniquely determined by v, the mapping Eis well deﬁned. Suppose, for u2V,u¼w0 1þ/C1/C1/C1þ w0 r,w0 i2Wi. Then vþu¼ðw1þw0 1Þþ/C1/C1/C1þð wrþw0 rÞ and kv¼kw1þ/C1/C1/C1þ kwr;kwi;wiþw0 i2Wi are the unique sums corresponding to vþuandkv. Hence, EðvþuÞ¼wkþw0 k¼EðvÞþEðuÞ and EðkvÞ¼kwkþkEðvÞ and therefore Eis linear.344 CHAPTER 10 Canonical Forms (b) We have that wk¼0þ/C1/C1/C1þ 0þwkþ0þ/C1/C1/C1þ 0 is the unique sum corresponding to wk2Wk; hence, EðwkÞ¼wk. Then, for any v2V, E2ðvÞ¼EðEðvÞÞ¼ EðwkÞ¼wk¼EðvÞ Thus, E2¼E, as required. 10.34. Suppose E:V!Vis linear and E2¼E. Show that (a) EðuÞ¼ufor any u2ImE(i.e., the restriction of Eto its image is the identity mapping); (b) Vis the direct sum of the image and kernel of E:V¼ImE/C8KerE; (c) Eis the projection of Vinto Im E, its image. Thus, by the preceding problem, a linear mapping T:V!Vis a projection if and only if T2¼T; this characterization of a projection is frequently used as its deﬁnition. (a) If u2ImE, then there exists v2Vfor which EðvÞ¼u; hence, as required, EðuÞ¼EðEðvÞÞ¼ E2ðvÞ¼EðvÞ¼u (b) Let v2V. We can write vin the form v¼EðvÞþv/C0EðvÞ. Now EðvÞ2ImEand, because Eðv/C0EðvÞÞ¼ EðvÞ/C0E2ðvÞ¼EðvÞ/C0EðvÞ¼0 v/C0EðvÞ2KerE. Accordingly, V¼ImEþKerE. Now suppose w2ImE\KerE.B y( i),EðwÞ¼wbecause w2ImE. On the other hand, EðwÞ¼0 because w2KerE. Thus, w¼0, and so Im E\KerE¼f0g. These two conditions imply that Vis the direct sum of the image and kernel of E. (c) Let v2Vand suppose v¼uþw, where u2ImEandw2KerE. Note that EðuÞ¼uby (i), and EðwÞ¼0 because w2KerE. Hence, EðvÞ¼EðuþwÞ¼EðuÞþEðwÞ¼uþ0¼u That is, Eis the projection of Vinto its image. 10.35. Suppose V¼U/C8Wand suppose T:V!Vis linear. Show that UandWare both T-invariant if and only if TE¼ET, where Eis the projection of VintoU. Observe that EðvÞ2Ufor every v2V, and that (i) EðvÞ¼viffv2U, (ii) EðvÞ¼0 iff v2W. Suppose ET¼TE. Let u2U. Because EðuÞ¼u, TðuÞ¼TðEðuÞÞ¼ð TEÞðuÞ¼ð ETÞðuÞ¼EðTðuÞÞ2 U Hence, UisT-invariant. Now let w2W. Because EðwÞ¼0, EðTðwÞÞ¼ð ETÞðwÞ¼ð TEÞðwÞ¼TðEðwÞÞ¼ Tð0Þ¼0; and so TðwÞ2W Hence, Wis also T-invariant. Conversely, suppose UandWare both T-invariant. Let v2Vand suppose v¼uþw, where u2T andw2W. Then TðuÞ2UandTðwÞ2W; hence, EðTðuÞÞ¼ TðuÞandEðTðwÞÞ¼ 0. Thus, ðETÞðvÞ¼ð ETÞðuþwÞ¼ð ETÞðuÞþð ETÞðwÞ¼EðTðuÞÞþ EðTðwÞÞ¼ TðuÞ and ðTEÞðvÞ¼ð TEÞðuþwÞ¼TðEðuþwÞÞ¼ TðuÞ That is,ðETÞðvÞ¼ð TEÞðvÞfor every v2V; therefore, ET¼TE, as required. SUPPLEMENTARY PROBLEMS Invariant Subspaces 10.36. Suppose Wis invariant under T:V!V. Show that Wis invariant under fðTÞfor any polynomial fðtÞ. 10.37. Show that every subspace of Vis invariant under Iand0, the identity and zero operators.CHAPTER 10 Canonical Forms 345 10.38. LetWbe invariant under T1:V!VandT2:V!V. Prove Wis also invariant under T1þT2andT1T2. 10.39. LetT:V!Vbe linear. Prove that any eigenspace, ElisT-invariant. 10.40. LetVbe a vector space of odd dimension (greater than 1) over the real ﬁeld R. Show that any linear operator on Vhas an invariant subspace other than Vorf0g. 10.41. Determine the invariant subspace of A¼2/C04 5/C02/C20/C21 viewed as a linear operator on (a) R2, (b) C2. 10.42. Suppose dim V¼n. Show that T:V!Vhas a triangular matrix representation if and only if there exist T-invariant subspaces W1/C26W2/C26/C1/C1/C1/C26 Wn¼Vfor which dim Wk¼k,k¼1;...;n. Invariant Direct Sums 10.43. The subspaces W1;...;Wrare said to be independent ifw1þ/C1/C1/C1þ wr¼0,wi2Wi, implies that each wi¼0. Show that span ðWiÞ¼W1/C8/C1/C1/C1/C8 Wrif and only if the Wiare independent. [Here span ðWiÞ denotes the linear span of the Wi.] 10.44. Show that V¼W1/C8/C1/C1/C1/C8 Wrif and only if (i) V¼spanðWiÞand (ii) for k¼1;2;...;r, Wk\spanðW1;...;Wk/C01;Wkþ1;...;WrÞ¼f 0g. 10.45. Show that spanðWiÞ¼W1/C8/C1/C1/C1/C8 Wrif and only if dim ½spanðWiÞ/C138¼ dimW1þ/C1/C1/C1þ dimWr. 10.46. Suppose the characteristic polynomial of T:V!VisDðtÞ¼f1ðtÞn1f2ðtÞn2/C1/C1/C1frðtÞnr, where the fiðtÞare distinct monic irreducible polynomials. Let V¼W1/C8/C1/C1/C1/C8 Wrbe the primary decomposition of VintoT- invariant subspaces. Show that fiðtÞniis the characteristic polynomial of the restriction of TtoWi. Nilpotent Operators 10.47. Suppose T1andT2are nilpotent operators that commute (i.e., T1T2¼T2T1). Show that T1þT2andT1T2 are also nilpotent. 10.48. Suppose Ais a supertriangular matrix (i.e., all entries on and below the main diagonal are 0). Show that Ais nilpotent. 10.49. LetVbe the vector space of polynomials of degree /C20n. Show that the derivative operator on Vis nilpotent of index nþ1. 10.50. Show that any Jordan nilpotent block matrix Nis similar to its transpose NT(the matrix with 1’s below the diagonal and 0’s elsewhere). 10.51. Show that two nilpotent matrices of order 3 are similar if and only if they have the same index of nilpotency. Show by example that the statement is not true for nilpotent matrices of order 4. Jordan Canonical Form 10.52. Find all possible Jordan canonical forms for those matrices whose characteristic polynomial DðtÞand minimal polynomial mðtÞare as follows: (a)DðtÞ¼ð t/C02Þ4ðt/C03Þ2;mðtÞ¼ð t/C02Þ2ðt/C03Þ2, (b)DðtÞ¼ð t/C07Þ5;mðtÞ¼ð t/C07Þ2, (c)DðtÞ¼ð t/C02Þ7;mðtÞ¼ð t/C02Þ3 10.53. Show that every complex matrix is similar to its transpose. ( Hint: Use its Jordan canonical form.) 10.54. Show that all n/C2ncomplex matrices Afor which An¼IbutAk6¼Ifork<nare similar. 10.55. Suppose Ais a complex matrix with only real eigenvalues. Show that Ais similar to a matrix with only real entries.346 CHAPTER 10 Canonical Forms Cyclic Subspaces 10.56. Suppose T:V!Vis linear. Prove that Zðv;TÞis the intersection of all T-invariant subspaces containing v. 10.57. LetfðtÞandgðtÞbe the T-annihilators of uand v, respectively. Show that if fðtÞandgðtÞare relatively prime, then fðtÞgðtÞis the T-annihilator of uþv. 10.58. Prove that Zðu;TÞ¼Zðv;TÞif and only if gðTÞðuÞ¼vwhere gðtÞis relatively prime to the T-annihilator of u. 10.59. LetW¼Zðv;TÞ, and suppose the T-annihilator of visfðtÞn, where fðtÞis a monic irreducible polynomial of degree d. Show that fðTÞsðWÞis a cyclic subspace generated by fðTÞsðvÞand that it has dimension dðn/C0sÞifn>sand dimension 0 if n/C20s. Rational Canonical Form 10.60. Find all possible rational forms for a 6 /C26 matrix over Rwith minimal polynomial: (a) mðtÞ¼ð t2/C02tþ3Þðtþ1Þ2, (b) mðtÞ¼ð t/C02Þ3. 10.61. LetAbe a 4/C24 matrix with minimal polynomial mðtÞ¼ð t2þ1Þðt2/C03Þ. Find the rational canonical form forAifAis a matrix over (a) the rational ﬁeld Q, (b) the real ﬁeld R, (c) the complex ﬁeld C. 10.62. Find the rational canonical form for the four-square Jordan block with l’s on the diagonal. 10.63. Prove that the characteristic polynomial of an operator T:V!Vis a product of its elementary divisors. 10.64. Prove that two 3/C23 matrices with the same minimal and characteristic polynomials are similar. 10.65. Let CðfðtÞÞdenote the companion matrix to an arbitrary polynomial fðtÞ. Show that fðtÞis the characteristic polynomial of CðfðtÞÞ. Projections 10.66. Suppose V¼W1/C8/C1/C1/C1/C8 Wr. Let Eidenote the projection of Vinto Wi. Prove (i) EiEj¼0,i6¼j; (ii)I¼E1þ/C1/C1/C1þ Er. 10.67. LetE1;...;Erbe linear operators on Vsuch that (i)E2 i¼Ei(i.e., the Eiare projections); (ii) EiEj¼0,i6¼j; (iii) I¼E1þ/C1/C1/C1þ Er Prove that V¼ImE1/C8/C1/C1/C1/C8 ImEr. 10.68. Suppose E:V!Vis a projection (i.e., E2¼E). Prove that Ehas a matrix representation of the form Ir0 00/C20/C21 , where ris the rank of EandIris the r-square identity matrix. 10.69. Prove that any two projections of the same rank are similar. ( Hint: Use the result of Problem 10.68.) 10.70. Suppose E:V!Vis a projection. Prove (i)I/C0Eis a projection and V¼ImE/C8ImðI/C0EÞ, (ii) IþEis invertible (if 1þ16¼0). Quotient Spaces 10.71. LetWbe a subspace of V. Suppose the set of cosets fv1þW;v2þW;...;vnþWginV=Wis linearly independent. Show that the set of vectors fv1;v2;...;vnginVis also linearly independent. 10.72. LetWbe a substance of V. Suppose the set of vectors fu1;u2;...;unginVis linearly independent, and that LðuiÞ\W¼f0g. Show that the set of cosets fu1þW;...;unþWginV=Wis also linearly independent.CHAPTER 10 Canonical Forms 347 10.73. Suppose V¼U/C8Wand thatfu1;...;ungis a basis of U. Show thatfu1þW;...;unþWgis a basis of the quotient spaces V=W. (Observe that no condition is placed on the dimensionality of VorW.) 10.74. LetWbe the solution space of the linear equation a1x1þa2x2þ/C1/C1/C1þ anxn¼0; ai2K and let v¼ðb1;b2;...;bnÞ2Kn. Prove that the coset vþWofWinKnis the solution set of the linear equation a1x1þa2x2þ/C1/C1/C1þ anxn¼b; where b¼a1b1þ/C1/C1/C1þ anbn 10.75. LetVbe the vector space of polynomials over Rand let Wbe the subspace of polynomials divisible by t4 (i.e., of the form a0t4þa1t5þ/C1/C1/C1þ an/C04tn). Show that the quotient space V=Whas dimension 4. 10.76. LetUandWbe subspaces of Vsuch that W/C26U/C26V. Note that any coset uþWofWinUmay also be viewed as a coset of WinV, because u2Uimplies u2V; hence, U=Wis a subset of V=W. Prove that (i)U=Wis a subspace of V=W, (ii) dimðV=WÞ/C0dimðU=WÞ¼dimðV=UÞ. 10.77. LetUandWbe subspaces of V. Show that the cosets of U\WinVcan be obtained by intersecting each of the cosets of UinVby each of the cosets of WinV: V=ðU\WÞ¼fð vþUÞ\ð v0þWÞ:v;v02Vg 10.78. LetT:V!V0be linear with kernel Wand image U. Show that the quotient space V=Wis isomorphic to Uunder the mapping y:V=W!Udeﬁned by yðvþWÞ¼TðvÞ. Furthermore, show that T¼i/C14y/C14Z, where Z:V!V=W is the natural mapping of VintoV=W(i.e., ZðvÞ¼vþW), and i:U,!V0is the inclusion mapping (i.e., iðuÞ¼u). (See diagram.) ANSWERS TO SUPPLEMENTARY PROBLEMS 10.41. (a) R2andf0g, (b) C2;f0g;W1¼spanð2;1/C02iÞ;W2¼spanð2;1þ2iÞ 10.52. (a) diag21 2/C20/C21 ;21 2/C20/C21 ;31 3/C20/C21 /C18/C19 ; diag21 2/C20/C21 ;½2/C138:½2/C138;31 3/C20/C21 /C18/C19 ; (b) diag71 7/C20/C21 ;71 7/C20/C21 ;½7/C138/C18/C19 ; diag71 7/C20/C21 ;½7/C138;½7/C138;½7/C138/C18/C19 ; (c) Let Mkdenote a Jordan block with l¼2 and order k. Then diagðM3;M3;M1Þ, diagðM3;M2;M2Þ, diagðM3;M2;M1;M1Þ, diagðM3;M1;M1;M1;M1Þ 10.60. LetA¼0/C03 12/C20/C21 ;B¼0/C01 1/C02/C20/C21 ;C¼00 8 10/C012 01 62 43 5;D¼0/C04 14/C20/C21 . (a) diagðA;A;BÞ;diagðA;B;BÞ;diagðA;B;/C01;/C01Þ; (b) diagðC;CÞ;diagðC;D;2Þ;diagðC;2;2;2Þ 10.61. LetA¼0/C01 10/C20/C21 ;B¼03 10/C20/C21 . (a) diagðA;BÞ, (b) diagðA;ﬃﬃﬃ 3p ;/C0ﬃﬃﬃ 3p Þ, (c) diagði;/C0i;ﬃﬃﬃ 3p ;/C0ﬃﬃﬃ 3p Þ 10.62. Companion matrix with the last column ½/C0l4;4l3;/C06l2;4l/C138T 348 CHAPTER 10 Canonical Forms CHAPTER 11 Linear Functionals and the Dual Space 11.1 Introduction In this chapter, we study linear mappings from a vector space Vinto its ﬁeld Kof scalars. (Unless otherwise stated or implied, we view Kas a vector space over itself.) Naturally all the theorems and results for arbitrary mappings on Vhold for this special case. However, we treat these mappings separately because of their fundamental importance and because the special relationship of VtoKgives rise to new notions and results that do not apply in the general case. 11.2 Linear Functionals and the Dual Space LetVbe a vector space over a ﬁeld K. A mapping f:V!Kis termed a linear functional (orlinear form ) if, for every u;v2Vand every a;b;2K, fðauþbvÞ¼afðuÞþbfðvÞ In other words, a linear functional on Vis a linear mapping from VintoK. EXAMPLE 11.1 (a) Let pi:Kn!Kbe the ith projection mapping; that is, piða1;a2;...anÞ¼ai. Then piis linear and so it is a linear functional on Kn. (b) Let Vbe the vector space of polynomials in toverR. Let J:V!Rbe the integral operator deﬁned by JðpðtÞÞ¼Ð1 0pðtÞdt. Recall that Jis linear; and hence, it is a linear functional on V. (c) Let Vbe the vector space of n-square matrices over K. Let T:V!Kbe the trace mapping TðAÞ¼a11þa22þ/C1/C1/C1þ ann; where A¼½aij/C138 That is, Tassigns to a matrix Athe sum of its diagonal elements. This map is linear (Problem 11.24), and so it is a linear functional on V. By Theorem 5.10, the set of linear functionals on a vector space Vover a ﬁeld Kis also a vector space over K, with addition and scalar multiplication deﬁned by ðfþsÞðvÞ¼fðvÞþsðvÞ andðkfÞðvÞ¼kfðvÞ where fandsare linear functionals on Vandk2K. This space is called the dual space ofVand is denoted by V*. EXAMPLE 11.2 LetV¼Kn, the vector space of n-tuples, which we write as column vectors. Then the dual space V*c a n be identiﬁed with the space of row vectors . In particular, any linear functional f¼ða1;...;anÞinV* has the representation fðx1;x2;...;xnÞ¼½ a1;a2;...;an/C138½x2;x2;...;xn/C138T¼a1x1þa2x2þ/C1/C1/C1þ anxn Historically, the formal expression on the right was termed a linear form . CHAPTER 11 349 11.3 Dual Basis Suppose Vis a vector space of dimension nover K. By Theorem 5.11, the dimension of the dual space V* is also n(because Kis of dimension 1 over itself). In fact, each basis of Vdetermines a basis of V*a s follows (see Problem 11.3 for the proof). THEOREM 11.1: Supposefv1;...;vngis a basis of Vover K. Let f1;...;fn2V*be the linear functionals as deﬁned by fiðvjÞ¼dij¼1i f i¼j 0i f i6¼j/C26 Thenff1;...;fngis a basis of V*: The above basisffigis termed the basis dual tofvigor the dual basis . The above formula, which uses the Kronecker delta dij, is a short way of writing f1ðv1Þ¼1;f1ðv2Þ¼0;f1ðv3Þ¼0;...;f1ðvnÞ¼0 f2ðv1Þ¼0;f2ðv2Þ¼1;f2ðv3Þ¼0;...;f2ðvnÞ¼0 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: fnðv1Þ¼0;fnðv2Þ¼0;...;fnðvn/C01Þ¼0;fnðvnÞ¼1 By Theorem 5.2, these linear mappings fiare unique and well deﬁned. EXAMPLE 11.3 Consider the basis fv1¼ð2;1Þ;v2¼ð3;1ÞgofR2. Find the dual basis ff1;f2g. We seek linear functionals f1ðx;yÞ¼axþbyandf2ðx;yÞ¼cxþdysuch that f1ðv1Þ¼1; f1ðv2Þ¼0; f2ðv2Þ¼0; f2ðv2Þ¼1 These four conditions lead to the following two systems of linear equations: f1ðv1Þ¼f1ð2;1Þ¼2aþb¼1 f1ðv2Þ¼f1ð3;1Þ¼3aþb¼0/C27 andf2ðv1Þ¼f2ð2;1Þ¼2cþd¼0 f2ðv2Þ¼f2ð3;1Þ¼3cþd¼1/C27 The solutions yield a¼/C01,b¼3 and c¼1,d¼/C02. Hence, f1ðx;yÞ¼/C0 xþ3yandf2ðx;yÞ¼x/C02yform the dual basis. The next two theorems (proved in Problems 11.4 and 11.5, respectively) give relationships between bases and their duals. THEOREM 11.2: Letfv1;...;vngbe a basis of Vand letff1;...;fngbe the dual basis in V*. Then (i) For any vector u2V,u¼f1ðuÞv1þf2ðuÞv2þ/C1/C1/C1þ fnðuÞvn. (ii) For any linear functional s2V*,s¼sðv1Þf1þsðv2Þf2þ/C1/C1/C1þ sðvnÞfn. THEOREM 11.3: Letfv1;...;vngandfw1;...;wngbe bases of Vand letff1;...;fngand fs1;...;sngbe the bases of V*dual tofvigandfwig, respectively. Suppose Pis the change-of-basis matrix from fvigtofwig. ThenðP/C01ÞTis the change-of-basis matrix fromffigtofsig. 11.4 Second Dual Space We repeat: Every vector space Vhas a dual space V*, which consists of all the linear functionals on V. Thus, V* has a dual space V**, called the second dual ofV, which consists of all the linear functionals onV*. We now show that each v2Vdetermines a speciﬁc element ^v2V**. First, for any f2V*, we deﬁne ^vðfÞ¼fðvÞ350 CHAPTER 11 Linear Functionals and the Dual Space It remains to be shown that this map ^v:V*!Kis linear. For any scalars a;b2Kand any linear functionals f;s2V*, we have ^vðafþbsÞ¼ð afþbsÞðvÞ¼afðvÞþbsðvÞ¼a^vðfÞþb^vðsÞ That is, ^vis linear and so ^v2V**. The following theorem (proved in Problem 12.7) holds. THEOREM 11.4: IfVhas ﬁnite dimensions, then the mapping v7!^vis an isomorphism of V onto V**. The above mapping v7!^vis called the natural mapping ofVinto V**. We emphasize that this mapping is never onto V** if Vis not ﬁnite-dimensional. However, it is always linear, and moreover, it is always one-to-one. Now suppose Vdoes have ﬁnite dimension. By Theorem 11.4, the natural mapping determines an isomorphism between Vand V**. Unless otherwise stated, we will identify Vwith V** by this mapping. Accordingly, we will view Vas the space of linear functionals on V*a n dw r i t e V¼V**. We remark that ifffigis the basis of V* dual to a basisfvigofV, thenfvigis the basis of V**¼Vthat is dual toffig. 11.5 Annihilators LetWbe a subset (not necessarily a subspace) of a vector space V. A linear functional f2V* is called an annihilator ofWiffðwÞ¼0 for every w2W—that is, if fðWÞ¼f 0g. We show that the set of all such mappings, denoted by W0and called the annihilator ofW, is a subspace of V*. Clearly, 02W0: Now suppose f;s2W0. Then, for any scalars a;b;2Kand for any w2W, ðafþbsÞðwÞ¼afðwÞþbsðwÞ¼a0þb0¼0 Thus, afþbs2W0, and so W0is a subspace of V*. In the case that Wis a subspace of V, we have the following relationship between Wand its annihilator W0(see Problem 11.11 for the proof). THEOREM 11.5: Suppose Vhas ﬁnite dimension and Wis a subspace of V. Then ðiÞdimWþdimW0¼dimV andðiiÞW00¼W Here W00¼fv2V:fðvÞ¼0 for every f2W0gor, equivalently, W00¼ðW0Þ0,w h e r e W00is viewed as a subspace of Vunder the identiﬁcation of VandV**. 11.6 Transpose of a Linear Mapping LetT:V!Ube an arbitrary linear mapping from a vector space Vinto a vector space U. Now for any linear functional f2U*, the composition f/C14Tis a linear mapping from VintoK: That is, f/C14T2V*. Thus, the correspondence f7!f/C14T is a mapping from U* into V*; we denote it by Ttand call it the transpose of T. In other words, Tt:U*!V* is deﬁned by TtðfÞ¼f/C14T Thus,ðTtðfÞÞðvÞ¼fðTðvÞÞfor every v2V.CHAPTER 11 Linear Functionals and the Dual Space 351 THEOREM 11.6: The transpose mapping Ttdeﬁned above is linear. Proof. For any scalars a;b2Kand any linear functionals f;s2U*, TtðafþbsÞ¼ð afþbsÞ/C14T¼aðf/C14TÞþbðs/C14TÞ¼aTtðfÞþbTtðsÞ That is, Ttis linear, as claimed. We emphasize that if Tis a linear mapping from VintoU, then Ttis a linear mapping from U* into V*. The same ‘‘transpose’’ for the mapping Ttno doubt derives from the following theorem (proved in Problem 11.16). THEOREM 11.7: LetT:V!Ube linear, and let Abe the matrix representation of Trelative to bases fvigofVandfuigofU. Then the transpose matrix ATis the matrix representation of Tt:U*!V*relative to the bases dual to fuigandfvig. SOLVED PROBLEMS Dual Spaces and Dual Bases 11.1. Find the basisff1;f2;f3gthat is dual to the following basis of R3: fv1¼ð1;/C01;3Þ;v2¼ð0;1;/C01Þ;v3¼ð0;3;/C02Þg The linear functionals may be expressed in the form f1ðx;y;zÞ¼a1xþa2yþa3z; f2ðx;y;zÞ¼b1xþb2yþb3z; f3ðx;y;zÞ¼c1xþc2yþc3z By deﬁnition of the dual basis, fiðvjÞ¼0 for i6¼j, but fiðvjÞ¼1 for i¼j. We ﬁnd f1by setting f1ðv1Þ¼1;f1ðv2Þ¼0;f1ðv3Þ¼0:This yields f1ð1;/C01;3Þ¼a1/C0a2þ3a3¼1; f1ð0;1;/C01Þ¼a2/C0a3¼0; f1ð0;3;/C02Þ¼3a2/C02a3¼0 Solving the system of equations yields a1¼1,a2¼0,a3¼0. Thus, f1ðx;y;zÞ¼x. We ﬁnd f2by setting f2ðv1Þ¼0,f2ðv2Þ¼1,f2ðv3Þ¼0. This yields f2ð1;/C01;3Þ¼b1/C0b2þ3b3¼0; f2ð0;1;/C01Þ¼b2/C0b3¼1; f2ð0;3;/C02Þ¼3b2/C02b3¼0 Solving the system of equations yields b1¼7,b2¼/C02,a3¼/C03. Thus, f2ðx;y;zÞ¼7x/C02y/C03z. We ﬁnd f3by setting f3ðv1Þ¼0,f3ðv2Þ¼0,f3ðv3Þ¼1. This yields f3ð1;/C01;3Þ¼c1/C0c2þ3c3¼0; f3ð0;1;/C01Þ¼c2/C0c3¼0; f3ð0;3;/C02Þ¼3c2/C02c3¼1 Solving the system of equations yields c1¼/C02,c2¼1,c3¼1. Thus, f3ðx;y;zÞ¼/C0 2xþyþz. 11.2. LetV¼faþbt:a;b2Rg, the vector space of real polynomials of degree /C201. Find the basis fv1;v2gofVthat is dual to the basis ff1;f2gofV* deﬁned by f1ðfðtÞÞ¼ð1 0fðtÞdt and f2ðfðtÞÞ¼ð2 0fðtÞdt Letv1¼aþbtand v2¼cþdt. By deﬁnition of the dual basis, f1ðv1Þ¼1; f1ðv2Þ¼0 and f2ðv1Þ¼0; fiðvjÞ¼1 Thus, f1ðv1Þ¼Ð1 0ðaþbtÞdt¼aþ1 2b¼1 f2ðv1Þ¼Ð2 0ðaþbtÞdt¼2aþ2b¼0) andf1ðv2Þ¼Ð1 0ðcþdtÞdt¼cþ1 2d¼0 f2ðv2Þ¼Ð2 0ðcþdtÞdt¼2cþ2d¼1) Solving each system yields a¼2,b¼/C02 and c¼/C01 2,d¼1. Thus,fv1¼2/C02t;v2¼/C01 2þtgis the basis of Vthat is dual toff1;f2g.352 CHAPTER 11 Linear Functionals and the Dual Space 11.3. Prove Theorem 11.1: Suppose fv1;...;vngis a basis of Vover K. Let f1;...;fn2V*b e deﬁned by fiðvjÞ¼0 for i6¼j, but fiðvjÞ¼1 for i¼j. Thenff1;...;fngis a basis of V*. We ﬁrst show that ff1;...;fngspans V*. Let fbe an arbitrary element of V*, and suppose fðv1Þ¼k1; fðv2Þ¼k2; ...; fðvnÞ¼kn Sets¼k1f1þ/C1/C1/C1þ knfn. Then sðv1Þ¼ð k1f1þ/C1/C1/C1þ knfnÞðv1Þ¼k1f1ðv1Þþk2f2ðv1Þþ/C1/C1/C1þ knfnðv1Þ ¼k1/C11þk2/C10þ/C1/C1/C1þ kn/C10¼k1 Similarly, for i¼2;...;n, sðviÞ¼ð k1f1þ/C1/C1/C1þ knfnÞðviÞ¼k1f1ðviÞþ/C1/C1/C1þ kifiðviÞþ/C1/C1/C1þ knfnðviÞ¼ki Thus, fðviÞ¼sðviÞfor i¼1;...;n. Because fand sagree on the basis vectors, f¼s¼k1f1þ/C1/C1/C1þ knfn. Accordingly,ff1;...;fngspans V*. It remains to be shown that ff1;...;fngis linearly independent. Suppose a1f1þa2f2þ/C1/C1/C1þ anfn¼0 Applying both sides to v1, we obtain 0¼0ðv1Þ¼ð a1f1þ/C1/C1/C1þ anfnÞðv1Þ¼a1f1ðv1Þþa2f2ðv1Þþ/C1/C1/C1þ anfnðv1Þ ¼a1/C11þa2/C10þ/C1/C1/C1þ an/C10¼a1 Similarly, for i¼2;...;n, 0¼0ðviÞ¼ð a1f1þ/C1/C1/C1þ anfnÞðviÞ¼a1f1ðviÞþ/C1/C1/C1þ aifiðviÞþ/C1/C1/C1þ anfnðviÞ¼ai That is, a1¼0;...;an¼0. Hence,ff1;...;fngis linearly independent, and so it is a basis of V*. 11.4. Prove Theorem 11.2: Let fv1;...;vngbe a basis of Vand letff1;...;fngbe the dual basis in V*. For any u2Vand any s2V*, (i) u¼P ifiðuÞvi. (ii) s¼P ifðviÞfi. Suppose u¼a1v1þa2v2þ/C1/C1/C1þ anvn ð1Þ Then f1ðuÞ¼a1f1ðv1Þþa2f1ðv2Þþ/C1/C1/C1þ anf1ðvnÞ¼a1/C11þa2/C10þ/C1/C1/C1þ an/C10¼a1 Similarly, for i¼2;...;n, fiðuÞ¼a1fiðv1Þþ/C1/C1/C1þ aifiðviÞþ/C1/C1/C1þ anfiðvnÞ¼ai That is, f1ðuÞ¼a1,f2ðuÞ¼a2;...;fnðuÞ¼an. Substituting these results into (1), we obtain (i). Next we proveðiiÞ. Applying the linear functional sto both sides of (i), sðuÞ¼f1ðuÞsðv1Þþf2ðuÞsðv2Þþ/C1/C1/C1þ fnðuÞsðvnÞ ¼sðv1Þf1ðuÞþsðv2Þf2ðuÞþ/C1/C1/C1þ sðvnÞfnðuÞ ¼ðsðv1Þf1þsðv2Þf2þ/C1/C1/C1þ sðvnÞfnÞðuÞ Because the above holds for every u2V,s¼sðv1Þf2þsðv2Þf2þ/C1/C1/C1þ sðvnÞfn, as claimed. 11.5. Prove Theorem 11.3. Let fvigandfwigbe bases of Vand letffigandfsigbe the respective dual bases in V*. Let Pbe the change-of-basis matrix from fvigtofwig:ThenðP/C01ÞTis the change-of-basis matrix from ffigtofsig. Suppose, for i¼1;...;n, wi¼ai1v1þai2v2þ/C1/C1/C1þ ainvn and si¼bi1f1þbi2f2þ/C1/C1/C1þ ainvn Then P¼½aij/C138andQ¼½bij/C138. We seek to prove that Q¼ðP/C01ÞT. LetRidenote the ith row of Qand let Cjdenote the jth column of PT. Then Ri¼ðbi1;bi2;...;binÞ and Cj¼ðaj1;aj2;...;ajnÞTCHAPTER 11 Linear Functionals and the Dual Space 353 By deﬁnition of the dual basis, siðwjÞ¼ð bi1f1þbi2f2þ/C1/C1/C1þ binfnÞðaj1v1þaj2v2þ/C1/C1/C1þ ajnvnÞ ¼bi1aj1þbi2aj2þ/C1/C1/C1þ binajn¼RiCj¼dij where dijis the Kronecker delta. Thus, QPT¼½RiCj/C138¼½dij/C138¼I Therefore, Q¼ðPTÞ/C01¼ðP/C01ÞT, as claimed. 11.6. Suppose v2V,v6¼0, and dim V¼n. Show that there exists f2V* such that fðvÞ6¼0. We extendfvgto a basisfv;v2;...;vngofV. By Theorem 5.2, there exists a unique linear mapping f:V!Ksuch that fðvÞ¼1 and fðviÞ¼0,i¼2;...;n. Hence, fhas the desired property. 11.7. Prove Theorem 11.4: Suppose dim V¼n. Then the natural mapping v7!^vis an isomorphism of Vonto V**. We ﬁrst prove that the map v7!^vis linear—that is, for any vectors v;w2Vand any scalars a;b2K, avþbw¼a^vþb^w. For any linear functional f2V*, avþbwðfÞ¼fðavþbwÞ¼afðvÞþbfðwÞ¼a^vðfÞþb^wðfÞ¼ð a^vþb^wÞðfÞ Because avþbwðfÞ¼ð a^vþb^wÞðfÞfor every f2V*, we have avþbw¼a^vþb^w. Thus, the map v7!^vis linear. Now suppose v2V,v6¼0. Then, by Problem 11.6, there exists f2V* for which fðvÞ6¼0. Hence, ^vðfÞ¼fðvÞ6¼0, and thus ^v6¼0. Because v6¼0 implies ^v6¼0, the map v7!^vis nonsingular and hence an isomorphism (Theorem 5.64). Now dim V¼dimV*¼dimV**, because Vhas ﬁnite dimension. Accordingly, the mapping v7!^v is an isomorphism of Vonto V**. Annihilators 11.8. Show that if f2V* annihilates a subset SofV, then fannihilates the linear span LðSÞofS. Hence, S0¼½spanðSÞ/C1380. Suppose v2spanðSÞ. Then there exists w1;...;wr2Sfor which v¼a1w1þa2w2þ/C1/C1/C1þ arwr. fðvÞ¼a1fðw1Þþa2fðw2Þþ/C1/C1/C1þ arfðwrÞ¼a10þa20þ/C1/C1/C1þ ar0¼0 Because vwas an arbitrary element of span ðSÞ;fannihilates spanðSÞ, as claimed. 11.9. Find a basis of the annihilator W0of the subspace WofR4spanned by v1¼ð1;2;/C03;4Þand v2¼ð0;1;4;/C01Þ By Problem 11.8, it sufﬁces to ﬁnd a basis of the set of linear functionals fsuch that fðv1Þ¼0 and fðv2Þ¼0, where fðx1;x2;x3;x4Þ¼ax1þbx2þcx3þdx4. Thus, fð1;2;/C03;4Þ¼aþ2b/C03cþ4d¼0 and fð0;1;4;/C01Þ¼bþ4c/C0d¼0 The system of two equations in the unknowns a;b;c;dis in echelon form with free variables candd. (1) Set c¼1,d¼0 to obtain the solution a¼11,b¼/C04,c¼1,d¼0. (2) Set c¼0,d¼1 to obtain the solution a¼6,b¼/C01,c¼0,d¼1. The linear functions f1ðxiÞ¼11x1/C04x2þx3andf2ðxiÞ¼6x1/C0x2þx4form a basis of W0. 11.10. Show that (a) For any subset SofV;S/C18S00. (b) If S1/C18S2, then S0 2/C18S0 1. (a) Let v2S. Then for every linear functional f2S0,^vðfÞ¼fðvÞ¼0. Hence, ^v2ðS0Þ0. Therefore, under the identiﬁcation of VandV**,v2S00. Accordingly, S/C18S00. (b) Let f2S0 2. Then fðvÞ¼0 for every v2S2. But S1/C18S2; hence, fannihilates every element of S1 (i.e., f2S0 1). Therefore, S0 2/C18S0 1.d d d d354 CHAPTER 11 Linear Functionals and the Dual Space 11.11. Prove Theorem 11.5: Suppose Vhas ﬁnite dimension and Wis a subspace of V. Then (i) dim WþdimW0¼dimV, (ii) W00¼W. (i) Suppose dim V¼nand dim W¼r/C20n. We want to show that dim W0¼n/C0r. We choose a basis fw1;...;wrgofWand extend it to a basis of V, sayfw1;...;wr;v1;...;vn/C0rg. Consider the dual basis ff1;...;fr;s1;...;sn/C0rg By deﬁnition of the dual basis, each of the above s’s annihilates each wi; hence, s1;...;sn/C0r2W0. We claim thatfsigis a basis of W0. Nowfsjgis part of a basis of V*, and so it is linearly independent. We next show that ffjgspans W0. Let s2W0. By Theorem 11.2, s¼sðw1Þf1þ/C1/C1/C1þ sðwrÞfrþsðv1Þs1þ/C1/C1/C1þ sðvn/C0rÞsn/C0r ¼0f1þ/C1/C1/C1þ 0frþsðv1Þs1þ/C1/C1/C1þ sðvn/C0rÞsn/C0r ¼sðv1Þs1þ/C1/C1/C1þ sðvn/C0rÞsn/C0r Consequently,fs1;...;sn/C0rgspans W0and so it is a basis of W0. Accordingly, as required dimW0¼n/C0r¼dimV/C0dimW: (ii) Suppose dim V¼nand dim W¼r. Then dim V*¼nand, by (i), dim W0¼n/C0r. Thus, by (i), dimW00¼n/C0ðn/C0rÞ¼r; therefore, dim W¼dimW00. By Problem 11.10, W/C18W00. Accord- ingly, W¼W00. 11.12. LetUandWbe subspaces of V. Prove thatðUþWÞ0¼U0\W0. Letf2ðUþWÞ0. Then fannihilates UþW;and so, in particular, fannihilates UandW:That is, f2U0andf2W0;hence, f2U0\W0:Thus,ðUþWÞ0/C18U0\W0: On the other hand, suppose s2U0\W0:Then sannihilates Uand also W.I f v2UþW, then v¼uþw, where u2Uandw2W. Hence, sðvÞ¼sðuÞþsðwÞ¼0þ0¼0. Thus, sannihilates UþW; that is, s2ðUþWÞ0. Accordingly, U0þW0/C18ðUþWÞ0. The two inclusion relations together give us the desired equality. Remark: Observe that no dimension argument is employed in the proof; hence, the result holds for spaces of ﬁnite or inﬁnite dimension. Transpose of a Linear Mapping 11.13. Letfbe the linear functional on R2deﬁned by fðx;yÞ¼x/C02y. For each of the following linear operators TonR2, ﬁndðTtðfÞÞðx;yÞ: (a)Tðx;yÞ¼ð x;0Þ, (b) Tðx;yÞ¼ð y;xþyÞ, (c) Tðx;yÞ¼ð 2x/C03y;5xþ2yÞ By deﬁnition, TtðfÞ¼f/C14T; that is,ðTtðfÞÞðvÞ¼fðTðvÞÞfor every v. Hence, (a)ðTtðfÞÞðx;yÞ¼fðTðx;yÞÞ¼ fðx;0Þ¼x (b)ðTtðfÞÞðx;yÞ¼fðTðx;yÞÞ¼ fðy;xþyÞ¼y/C02ðxþyÞ¼/C0 2x/C0y (c)ðTtðfÞÞðx;yÞ¼fðTðx;yÞÞ¼ fð2x/C03y;5xþ2yÞ¼ð 2x/C03yÞ/C02ð5xþ2yÞ¼/C0 8x/C07y 11.14. LetT:V!Ube linear and let Tt:U*!V* be its transpose. Show that the kernel of Ttis the annihilator of the image of T—that is, Ker Tt¼ðImTÞ0. Suppose f2KerTt; that is, TtðfÞ¼f/C14T¼0. If u2ImT, then u¼TðvÞfor some v2V; hence, fðuÞ¼fðTðvÞÞ¼ð f/C14TÞðvÞ¼0ðvÞ¼0 We have that fðuÞ¼0 for every u2ImT; hence, f2ðImTÞ0. Thus, Ker Tt/C18ðImTÞ0. On the other hand, suppose s2ðImTÞ0; that is, sðImTÞ¼f 0g. Then, for every v2V, ðTtðsÞÞðvÞ¼ð s/C14TÞðvÞ¼sðTðvÞÞ¼ 0¼0ðvÞCHAPTER 11 Linear Functionals and the Dual Space 355 We haveðTtðsÞÞðvÞ¼0ðvÞfor every v2V; hence, TtðsÞ¼0. Thus, s2KerTt, and so ðImTÞ0/C18KerTt. The two inclusion relations together give us the required equality. 11.15. Suppose VandUhave ﬁnite dimension and T:V!Uis linear. Prove rank ðTÞ¼rankðTtÞ. Suppose dim V¼nand dim U¼m, and suppose rank ðTÞ¼r. By Theorem 11.5, dimðImTÞ0¼dimu/C0dimðImTÞ¼m/C0rankðTÞ¼m/C0r By Problem 11.14, Ker Tt¼ðImTÞ0. Hence, nullityðTtÞ¼m/C0r. It then follows that, as claimed, rankðTtÞ¼dimU*/C0nullityðTtÞ¼m/C0ðm/C0rÞ¼r¼rankðTÞ 11.16. Prove Theorem 11.7: Let T:V!Ube linear and let Abe the matrix representation of Tin the basesfvjgofVandfuigofU. Then the transpose matrix ATis the matrix representation of Tt:U*!V* in the bases dual to fuigandfvjg. Suppose, for j¼1;...;m, TðvjÞ¼aj1u1þaj2u2þ/C1/C1/C1þ ajnun ð1Þ We want to prove that, for i¼1;...;n, TtðsiÞ¼a1if1þa2if2þ/C1/C1/C1þ amifm ð2Þ wherefsigandffjgare the bases dual to fuigandfvjg, respectively. Letv2Vand suppose v¼k1v1þk2v2þ/C1/C1/C1þ kmvm. Then, by (1), TðvÞ¼k1Tðv1Þþk2Tðv2Þþ/C1/C1/C1þ kmTðvmÞ ¼k1ða11u1þ/C1/C1/C1þ a1nunÞþk2ða21u1þ/C1/C1/C1þ a2nunÞþ/C1/C1/C1þ kmðam1u1þ/C1/C1/C1þ amnunÞ ¼ðk1a11þk2a21þ/C1/C1/C1þ kmam1Þu1þ/C1/C1/C1þð k1a1nþk2a2nþ/C1/C1/C1þ kmamnÞun ¼Pn i¼1ðk1a1iþk2a2iþ/C1/C1/C1þ kmamiÞui Hence, for j¼1;...;n. ðTtðsjÞðvÞÞ¼ sjðTðvÞÞ¼ sjPn i¼1ðk1a1iþk2a2iþ/C1/C1/C1þ kmamiÞui/C18/C19 ¼k1a1jþk2a2jþ/C1/C1/C1þ kmamj ð3Þ On the other hand, for j¼1;...;n, ða1jf1þa2jf2þ/C1/C1/C1þ amjfmÞðvÞ¼ð a1jf1þa2jf2þ/C1/C1/C1þ amjfmÞðk1v1þk2v2þ/C1/C1/C1þ kmvmÞ ¼k1a1jþk2a2jþ/C1/C1/C1þ kmamj ð4Þ Because v2Vwas arbitrary, (3) and (4) imply that TtðsjÞ¼a1jf1þa2jf2þ/C1/C1/C1þ amjfm; j¼1;...;n which is (2). Thus, the theorem is proved. SUPPLEMENTARY PROBLEMS Dual Spaces and Dual Bases 11.17. Find (a) fþs, (b) 3 f, (c) 2 f/C05s, where f:R3!Rands:R3!Rare deﬁned by fðx;y;zÞ¼2x/C03yþz and sðx;y;zÞ¼4x/C02yþ3z 11.18. Find the dual basis of each of the following bases of R3: (a)fð1;0;0Þ;ð0;1;0Þ;ð0;0;1Þg, (b)fð1;/C02;3Þ;ð1;/C01;1Þ;ð2;/C04;7Þg.356 CHAPTER 11 Linear Functionals and the Dual Space 11.19. LetVbe the vector space of polynomials over Rof degree/C202. Let f1;f2;f3be the linear functionals on Vdeﬁned by f1ðfðtÞÞ¼ð1 0fðtÞdt; f2ðfðtÞÞ¼ f0ð1Þ; f3ðfðtÞÞ¼ fð0Þ Here fðtÞ¼aþbtþct22Vandf0ðtÞdenotes the derivative of fðtÞ. Find the basisff1ðtÞ;f2ðtÞ;f3ðtÞgof Vthat is dual toff1;f2;f3g. 11.20. Suppose u;v2Vand that fðuÞ¼0 implies fðvÞ¼0 for all f2V*. Show that v¼kufor some scalar k. 11.21. Suppose f;s2V* and that fðvÞ¼0 implies sðvÞ¼0 for all v2V. Show that s¼kffor some scalar k. 11.22. LetVbe the vector space of polynomials over K. For a2K, deﬁne fa:V!KbyfaðfðtÞÞ¼ fðaÞ. Show that (a) fais linear; (b) if a6¼b, then fa6¼fb. 11.23. LetVbe the vector space of polynomials of degree /C202. Let a;b;c2Kbe distinct scalars. Let fa;fb;fc be the linear functionals deﬁned by faðfðtÞÞ¼ fðaÞ,fbðfðtÞÞ¼ fðbÞ,fcðfðtÞÞ¼ fðcÞ. Show that ffa;fb;fcgis linearly independent, and ﬁnd the basis ff1ðtÞ;f2ðtÞ;f3ðtÞgofVthat is its dual. 11.24. LetVbe the vector space of square matrices of order n. Let T:V!Kbe the trace mapping; that is, TðAÞ¼a11þa22þ/C1/C1/C1þ ann, where A¼ðaijÞ. Show that Tis linear. 11.25. LetWbe a subspace of V. For any linear functional fonW, show that there is a linear functional sonV such that sðwÞ¼fðwÞfor any w2W; that is, fis the restriction of stoW. 11.26. Letfe1;...;engbe the usual basis of Kn. Show that the dual basis is fp1;...;pngwhere piis the ith projection mapping; that is, piða1;...;anÞ¼ai. 11.27. LetVbe a vector space over R. Let f1;f22V* and suppose s:V!R;deﬁned by sðvÞ¼f1ðvÞf2ðvÞ; also belongs to V*. Show that either f1¼0orf2¼0. Annihilators 11.28. LetWbe the subspace of R4spanned byð1;2;/C03;4Þ,ð1;3;/C02;6Þ,ð1;4;/C01;8Þ. Find a basis of the annihilator of W. 11.29. LetWbe the subspace of R3spanned byð1;1;0Þandð0;1;1Þ. Find a basis of the annihilator of W. 11.30. Show that, for any subset SofV;spanðSÞ¼S00, where spanðSÞis the linear span of S. 11.31. LetUandWbe subspaces of a vector space Vof ﬁnite dimension. Prove that ðU\WÞ0¼U0þW0. 11.32. Suppose V¼U/C8W. Prove that V0¼U0/C8W0. Transpose of a Linear Mapping 11.33. Letfbe the linear functional on R2deﬁned by fðx;yÞ¼3x/C02y. For each of the following linear mappings T:R3!R2, ﬁndðTtðfÞÞðx;y;zÞ: (a) Tðx;y;zÞ¼ð xþy;yþzÞ, (b) Tðx;y;zÞ¼ð xþyþz;2x/C0yÞ 11.34. Suppose T1:U!VandT2:V!Ware linear. Prove that ðT2/C14T1Þt¼Tt 1/C14Tt 2. 11.35. Suppose T:V!Uis linear and Vhas ﬁnite dimension. Prove that Im Tt¼ðKerTÞ0.CHAPTER 11 Linear Functionals and the Dual Space 357 11.36. Suppose T:V!Uis linear and u2U. Prove that u2ImTor there exists f2V* such that TtðfÞ¼0 andfðuÞ¼1. 11.37. LetVbe of ﬁnite dimension. Show that the mapping T7!Ttis an isomorphism from Hom ðV;VÞonto HomðV*;V*Þ. (Here Tis any linear operator on V.) Miscellaneous Problems 11.38. Let Vbe a vector space over R. The line segment uvjoining points u;v2Vis deﬁned by uv¼ftuþð1/C0tÞv:0/C20t/C201g. A subset SofVisconvex ifu;v2Simplies uv/C18S. Let f2V*. Deﬁne Wþ¼fv2V:fðvÞ>0g; W¼fv2V:fðvÞ¼0g; W/C0¼fv2V:fðvÞ<0g Prove that Wþ;W, and W/C0are convex. 11.39. LetVbe a vector space of ﬁnite dimension. A hyperplane H ofVmay be deﬁned as the kernel of a nonzero linear functional fonV. Show that every subspace of Vis the intersection of a ﬁnite number of hyperplanes. ANSWERS TO SUPPLEMENTARY PROBLEMS 11.17. (a) 6 x/C05yþ4z, (b) 6 x/C09yþ3z, (c)/C016xþ4y/C013z 11.18. (a) f1¼x;f2¼y;f3¼z; (b) f1¼/C03x/C05y/C02z;f2¼2xþy;f3¼xþ2yþz 11.19. f1ðtÞ¼3t/C03 2t2;f2ðtÞ¼/C01 2tþ3 4t2;f3ðtÞ¼1/C03tþ3 2t2 11.22. (b) Let fðtÞ¼t. Then faðfðtÞÞ¼ a6¼b¼fbðfðtÞÞ; and therefore, fa6¼fb 11.23. f1ðtÞ¼t2/C0ðbþcÞtþbc ða/C0bÞða/C0cÞ;f2ðtÞ¼t2/C0ðaþcÞtþac ðb/C0aÞðb/C0cÞ;f3ðtÞ¼t2/C0ðaþbÞtþab ðc/C0aÞðc/C0bÞ/C26 /C27 11.28.ff1ðx;y;z;tÞ¼5x/C0yþz;f2ðx;y;z;tÞ¼2y/C0tg 11.29.ffðx;y;zÞ¼x/C0yþzg 11.33. (a)ðTtðfÞÞðx;y;zÞ¼3xþy/C02z, (b)ðTtðfÞÞðx;y;zÞ¼/C0 xþ5yþ3z358 CHAPTER 11 Linear Functionals and the Dual Space Bilinear, Quadratic, and Hermitian Forms 12.1 Introduction This chapter generalizes the notions of linear mappings and linear functionals. Speciﬁcally, we introduce the notion of a bilinear form. These bilinear maps also give rise to quadratic and Hermitian forms. Although quadratic forms were discussed previously, this chapter is treated independently of the previous results. Although the ﬁeld Kis arbitrary, we will later specialize to the cases K¼RandK¼C. Furthermore, we may sometimes need to divide by 2. In such cases, we must assume that 1 þ16¼0, which is true when K¼RorK¼C. 12.2 Bilinear Forms LetVbe a vector space of ﬁnite dimension over a ﬁeld K.A bilinear form onVis a mapping f:V/C2V!Ksuch that, for all a;b2Kand all ui;vi2V: (i)fðau1þbu2;vÞ¼afðu1;vÞþbfðu2;vÞ, (ii)fðu;av1þbv2Þ¼afðu;v1Þþbfðu;v2Þ We express condition (i) by saying fislinear in the ﬁrst variable , and condition (ii) by saying fislinear in the second variable . EXAMPLE 12.1 (a) Let fbe the dot product on Rn; that is, for u¼ðaiÞand v¼ðbiÞ, fðu;vÞ¼u/C1v¼a1b1þa2b2þ/C1/C1/C1þ anbn Then fis a bilinear form on Rn. (In fact, any inner product on a real vector space Vis a bilinear form onV.) (b) Let fandsbe arbitrarily linear functionals on V. Let f:V/C2V!Kbe deﬁned by fðu;vÞ¼fðuÞsðvÞ. Then fis a bilinear form, because fandsare each linear. (c) Let A¼½aij/C138be any n/C2nmatrix over a ﬁeld K. Then Amay be identiﬁed with the following bilinear form Fon Kn, where X¼½xi/C138andY¼½yi/C138are column vectors of variables: fðX;YÞ¼XTAY¼P i;jaijxiyi¼a11x1y1þa12x1y2þ/C1/C1/C1þ annxnyn The above formal expression in the variables xi;yiis termed the bilinear polynomial corresponding to the matrix A. Equation (12.1) shows that, in a certain sense, every bilinear form is of this type. CHAPTER 12 359 Space of Bilinear Forms LetBðVÞdenote the set of all bilinear forms on V. A vector space structure is placed on BðVÞ, where for anyf;g2BðVÞand any k2K, we deﬁne fþgandkfas follows: ðfþgÞðu;vÞ¼fðu;vÞþgðu;vÞ andðkfÞðu;vÞ¼kfðu;vÞ The following theorem (proved in Problem 12.4) applies. THEOREM 12.1: LetVbe a vector space of dimension nover K. Letff1;...;fngbe any basis of the dual space V*. Thenffij:i;j¼1;...;ngis a basis of BðVÞ, where fijis deﬁned by fijðu;vÞ¼fiðuÞfjðvÞ. Thus, in particular, dimBðVÞ¼n2. 12.3 Bilinear Forms and Matrices Letfbe a bilinear form on Vand let S¼fu1;...;ungbe a basis of V. Suppose u;v2Vand u¼a1u1þ/C1/C1/C1þ anun and v¼b1u1þ/C1/C1/C1þ bnun Then fðu;vÞ¼fða1u1þ/C1/C1/C1þ anun;b1u1þ/C1/C1/C1þ bnunÞ¼P i;jaibjfðui;ujÞ Thus, fis completely determined by the n2values fðui;ujÞ. The matrix A¼½aij/C138where aij¼fðui;ujÞis called the matrix representation offrelative to the basis S or, simply, the ‘‘matrix of finS.’’ It ‘‘represents’’ fin the sense that, for all u;v2V, fðu;vÞ¼P i;jaibjfðui;ujÞ¼½ u/C138T SA½v/C138S ð12:1Þ [As usual,½u/C138Sdenotes the coordinate (column) vector of uin the basis S.] Change of Basis, Congruent Matrices We now ask, how does a matrix representing a bilinear form transform when a new basis is selected? The answer is given in the following theorem (proved in Problem 12.5). THEOREM 12.2: LetPbe a change-of-basis matrix from one basis Sto another basis S0.I fAis the matrix representing a bilinear form fin the original basis S, then B¼PTAPis the matrix representing fin the new basis S0. The above theorem motivates the following deﬁnition. DEFINITION: A matrix Biscongruent to a matrix A, written B’A, if there exists a nonsingular matrix Psuch that B¼PTAP. Thus, by Theorem 12.2, matrices representing the same bilinear form are congruent. We remark that congruent matrices have the same rank, because PandPTare nonsingular; hence, the following deﬁnition is well deﬁned. DEFINITION: The rank of a bilinear form fonV, written rankðfÞ, is the rank of any matrix representation of f. We say fisdegenerate ornondegenerate according to whether rankðfÞ<dimVor rankðfÞ¼dimV. 12.4 Alternating Bilinear Forms Letfbe a bilinear form on V. Then fis called (i)alternating iffðv;vÞ¼0 for every v2V; (ii)skew-symmetric iffðu;vÞ¼/C0 fðv;uÞfor every u;v2V.360 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Now suppose (i) is true. Then (ii) is true, because, for any u;v;2V, 0¼fðuþv;uþvÞ¼fðu;uÞþfðu;vÞþfðv;uÞþfðv;vÞ¼fðu;vÞþfðv;uÞ On the other hand, suppose (ii) is true and also 1 þ16¼0. Then (i) is true, because, for every v2V,w e have fðv;vÞ¼/C0 fðv;vÞ. In other words, alternating and skew-symmetric are equivalent when 1 þ16¼0. The main structure theorem of alternating bilinear forms (proved in Problem 12.23) is as follows. THEOREM 12.3: Letfbe an alternating bilinear form on V. Then there exists a basis of Vin which f is represented by a block diagonal matrix Mof the form M¼diag01 /C010/C20/C21 ;01 /C010/C20/C21 ;...;01 /C010/C20/C21 ;½0/C138;½0/C138;...½0/C138/C18/C19 Moreover, the number of nonzero blocks is uniquely determined by f[because it is equal to1 2rankðfÞ/C138. In particular, the above theorem shows that any alternating bilinear form must have even rank. 12.5 Symmetric Bilinear Forms, Quadratic Forms This section investigates the important notions of symmetric bilinear forms and quadratic forms and their representation by means of symmetric matrices. The only restriction on the ﬁeld Kis that 1þ16¼0. In Section 12.6, we will restrict Kto be the real ﬁeld R, which yields important special results. Symmetric Bilinear Forms Letfbe a bilinear form on V. Then fis said to be symmetric if, for every u;v2V, fðu;vÞ¼fðv;uÞ One can easily show that fis symmetric if and only if any matrix representation Aoffis a symmetric matrix. The main result for symmetric bilinear forms (proved in Problem 12.10) is as follows. (We emphasize that we are assuming that 1 þ16¼0.) THEOREM 12.4: Letfbe a symmetric bilinear form on V. Then Vhas a basisfv1;...;vngin which f is represented by a diagonal matrix—that is, where fðvi;vjÞ¼0fori6¼j. THEOREM 12.4: (Alternative Form) Let Abe a symmetric matrix over K. Then Ais congruent to a diagonal matrix; that is, there exists a nonsingular matrix Psuch that PTAPis diagonal. Diagonalization Algorithm Recall that a nonsingular matrix Pis a product of elementary matrices. Accordingly, one way of obtaining the diagonal form D¼PTAPis by a sequence of elementary row operations and the same sequence of elementary column operations. This same sequence of elementary row operations on theidentity matrix Iwill yield P T. This algorithm is formalized below. ALGORITHM 12.1: (Congruence Diagonalization of a Symmetric Matrix) The input is a symmetric matrix A¼½aij/C138of order n. Step 1. Form the n/C22n(block) matrix M¼½A1;I/C138, where A1¼Ais the left half of Mand the identity matrix Iis the right half of M. Step 2. Examine the entry a11. There are three cases.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 361 Case I: a116¼0. (Use a11as a pivot to put 0’s below a11inMand to the right of a11inA1:Þ Fori¼2;...;n: (a) Apply the row operation ‘‘Replace Riby/C0ai1R1þa11Ri.’’ (b) Apply the corresponding column operation ‘‘Replace Ciby/C0ai1C1þa11Ci.’’ These operations reduce the matrix Mto the form M/C24a11 0* * 0A1**/C20/C21 ð*Þ Case II: a11¼0 but akk6¼0, for some k>1. (a) Apply the row operation ‘‘Interchange R1andRk.’’ (b) Apply the corresponding column operation ‘‘Interchange C1andCk.’’ (These operations bring akkinto the ﬁrst diagonal position, which reduces the matrix to Case I.) Case III: All diagonal entries aii¼0 but some aij6¼0. (a) Apply the row operation ‘‘Replace RibyRjþRi.’’ (b) Apply the corresponding column operation ‘‘Replace CibyCjþCi.’’ (These operations bring 2 aijinto the ith diagonal position, which reduces the matrix to Case II.) Thus, Mis ﬁnally reduced to the form ð*Þ, where A2is a symmetric matrix of order less than A. Step 3. Repeat Step 2 with each new matrix Ak(by neglecting the ﬁrst row and column of the preceding matrix) until Ais diagonalized. Then Mis transformed into the form M0¼½D;Q/C138, where Dis diagonal. Step 4. SetP¼QT. Then D¼PTAP. Remark 1: We emphasize that in Step 2, the row operations will change both sides of M, but the column operations will only change the left half of M. Remark 2: The condition 1þ16¼0 is used in Case III, where we assume that 2 aij6¼0 when aij6¼0. The justiﬁcation for the above algorithm appears in Problem 12.9. EXAMPLE 12.2 LetA¼12/C03 25/C04 /C03/C0482 43 5. Apply Algorithm 9.1 to ﬁnd a nonsingular matrix Psuch thatD¼PTAPis diagonal. First form the block matrix M¼½A;I/C138; that is, let M¼½A;I/C138¼12/C03100 25/C04010 /C03/C048 0 0 12 43 5 Apply the row operations ‘‘Replace R2by/C02R1þR2’’ and ‘‘Replace R3by 3R1þR3’’ to M, and then apply the corresponding column operations ‘‘Replace C2by/C02C1þC2’’ and ‘‘Replace C3by 3C1þC3’’ to obtain 12/C031 0 0 01 2/C0210 02/C013 0 12 43 5 and then1 001 0 0 01 2/C0210 02/C013 0 12 43 5362 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Next apply the row operation ‘‘Replace R3by/C02R2þR3’’ and then the corresponding column operation ‘‘Replace C3by/C02C2þC3’’ to obtain 1 0010 0 01 2/C021 0 00/C057/C0212 43 5 and then1 0010 0 01 0/C021 0 00/C057/C0212 43 5 Now Ahas been diagonalized. Set P¼1/C027 01/C02 0012 43 5 and then D¼P/C01AP¼10 0 01 000/C052 43 5 We emphasize that Pis the transpose of the right half of the ﬁnal matrix. Quadratic Forms We begin with a deﬁnition. DEFINITION A: A mapping q:V!Kis a quadratic form ifqðvÞ¼fðv;vÞfor some symmetric bilinear form fonV. If 1þ16¼0i nK, then the bilinear form fcan be obtained from the quadratic form qby the following polar form off: fðu;vÞ¼1 2½qðuþvÞ/C0qðuÞ/C0qðvÞ/C138 Now suppose fis represented by a symmetric matrix A¼½aij/C138, and 1þ16¼0. Letting X¼½xi/C138 denote a column vector of variables, qcan be represented in the form qðXÞ¼fðX;XÞ¼XTAX¼P i;jaijxixj¼P iaiix2 iþ2P i<jaijxixj The above formal expression in the variables xiis also called a quadratic form. Namely, we have the following second deﬁnition. DEFINITION B: Aquadratic form q in variables x1;x2;...;xnis a polynomial such that every term has degree two; that is, qðx1;x2;...;xnÞ¼P icix2 iþP i<jdijxixj Using 1þ16¼0, the quadratic form qin Deﬁnition B determines a symmetric matrix A¼½aij/C138where aii¼ciandaij¼aji¼1 2dij. Thus, Deﬁnitions A and B are essentially the same. If the matrix representation Aofqis diagonal, then qhas the diagonal representation qðXÞ¼XTAX¼a11x2 1þa22x2 2þ/C1/C1/C1þ annx2 n That is, the quadratic polynomial representing qwill contain no ‘‘cross product’’ terms. Moreover, by Theorem 12.4, every quadratic form has such a representation (when 1 þ16¼0Þ. 12.6 Real Symmetric Bilinear Forms, Law of Inertia This section treats symmetric bilinear forms and quadratic forms on vector spaces Vover the real ﬁeld R. The special nature of Rpermits an independent theory. The main result (proved in Problem 12.14) is as follows. THEOREM 12.5: Letfbe a symmetric form on VoverR. Then there exists a basis of Vin which fis represented by a diagonal matrix. Every other diagonal matrix representation of fhas the same number pof positive entries and the same number nof negative entries.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 363 The above result is sometimes called the Law of Inertia orSylvester’s Theorem . The rank and signature of the symmetric bilinear form fare denoted and deﬁned by rankðfÞ¼pþn and sigðfÞ¼p/C0n These are uniquely deﬁned by Theorem 12.5. A real symmetric bilinear form fis said to be (i)positive deﬁnite ifqðvÞ¼fðv;vÞ>0 for every v6¼0, (ii)nonnegative semideﬁnite ifqðvÞ¼fðv;vÞ/C210 for every v. EXAMPLE 12.3 Letfbe the dot product on Rn. Recall that fis a symmetric bilinear form on Rn. We note thatfis also positive deﬁnite. That is, for any u¼ðaiÞ6¼0i nRn, fðu;uÞ¼a2 1þa2 2þ/C1/C1/C1þ a2 n>0 Section 12.5 and Chapter 13 tell us how to diagonalize a real quadratic form qor, equivalently, a real symmetric matrix Aby means of an orthogonal transition matrix P.I fPis merely nonsingular, then qcan be represented in diagonal form with only 1’s and /C01’s as nonzero coefﬁcients. Namely, we have the following corollary. COROLLARY 12.6: Any real quadratic form qhas a unique representation in the form qðx1;x2;...;xnÞ¼x2 1þ/C1/C1/C1þ x2 p/C0x2 pþ1/C0/C1/C1/C1/C0 x2 r where r¼pþnis the rank of the form. COROLLARY 12.6: (Alternative Form) Any real symmetric matrix Ais congruent to the unique diagonal matrix D¼diagðIp;/C0In;0Þ where r¼pþnis the rank of A. 12.7 Hermitian Forms LetVbe a vector space of ﬁnite dimension over the complex ﬁeld C.AHermitian form onVis a mapping f:V/C2V!Csuch that, for all a;b2Cand all ui;v2V, (i)fðau1þbu2;vÞ¼afðu1;vÞþbfðu2;vÞ, (ii)fðu;vÞ¼fðv;uÞ. (As usual, /C22kdenotes the complex conjugate of k2C.) Using (i) and (ii), we get fðu;av1þbv2Þ¼fðav1þbv2;uÞ¼afðv1;uÞþbfðv2;uÞ ¼^afðv1;uÞþbfðv2;uÞ¼ /C22afðu;v1Þþ /C22bfðu;v2Þ That is, ðiiiÞfðu;av1þbv2Þ¼ /C22afðu;v1Þþ /C22bfðu;v2Þ: As before, we express condition (i) by saying fis linear in the ﬁrst variable. On the other hand, we express condition (iii) by saying fis ‘‘conjugate linear’’ in the second variable. Moreover, condition (ii) tells us that fðv;vÞ¼fðv;vÞ, and hence, fðv;vÞis real for every v2V. The results of Sections 12.5 and 12.6 for symmetric forms have their analogues for Hermitian forms. Thus, the mapping q:V!R, deﬁned by qðvÞ¼fðv;vÞ, is called the Hermitian quadratic form or complex quadratic form associated with the Hermitian form f. We can obtain ffrom qby the polar form fðu;vÞ¼1 4½qðuþvÞ/C0qðu/C0vÞ/C138þ1 4½qðuþivÞ/C0qðu/C0ivÞ/C138364 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Now suppose S¼fu1;...;ungis a basis of V. The matrix H¼½hij/C138where hij¼fðui;ujÞis called the matrix representation offin the basis S. By (ii), fðui;ujÞ¼fðuj;uiÞ; hence, His Hermitian and, in particular, the diagonal entries of Hare real. Thus, any diagonal representation of fcontains only real entries. The next theorem (to be proved in Problem 12.47) is the complex analog of Theorem 12.5 on real symmetric bilinear forms. THEOREM 12.7: Letfbe a Hermitian form on VoverC. Then there exists a basis of Vin which fis represented by a diagonal matrix. Every other diagonal matrix representation of f has the same number pof positive entries and the same number nof negative entries. Again the rank andsignature of the Hermitian form fare denoted and deﬁned by rankðfÞ¼pþn and sigðfÞ¼p/C0n These are uniquely deﬁned by Theorem 12.7. Analogously, a Hermitian form fis said to be (i)positive deﬁnite ifqðvÞ¼fðv;vÞ>0 for every v6¼0, (ii)nonnegative semideﬁnite ifqðvÞ¼fðv;vÞ/C210 for every v. EXAMPLE 12.4 Letfbe the dot product on Cn; that is, for any u¼ðziÞand v¼ðwiÞinCn, fðu;vÞ¼u/C1v¼z1/C22w1þz2/C22w2þ/C1/C1/C1þ zn/C22wn Then fis a Hermitian form on Cn. Moreover, fis also positive deﬁnite, because, for any u¼ðziÞ6¼0i nCn, fðu;uÞ¼z1/C22z1þz2/C22z2þ/C1/C1/C1þ zn/C22zn¼jz1j2þjz2j2þ/C1/C1/C1þj znj2>0 SOLVED PROBLEMS Bilinear Forms 12.1. Letu¼ðx1;x2;x3Þand v¼ðy1;y2;y3Þ. Express fin matrix notation, where fðu;vÞ¼3x1y1/C02x1y3þ5x2y1þ7x2y2/C08x2y3þ4x3y2/C06x3y3 LetA¼½aij/C138, where aijis the coefﬁcient of xiyj. Then fðu;vÞ¼XTAY¼½x1;x2;x3/C13830/C02 57/C08 04/C062 43 5y1 y2 y32 43 5 12.2. LetAbe an n/C2nmatrix over K. Show that the mapping fdeﬁned by fðX;YÞ¼XTAYis a bilinear form on Kn. For any a;b2Kand any Xi;Yi2Kn, fðaX1þbX2;YÞ¼ð aX1þbX2ÞTAY¼ðaXT 1þbXT 2ÞAY ¼aXT 1AYþbXT 2AY¼afðX1;YÞþbfðX2;YÞ Hence, fis linear in the ﬁrst variable. Also, fðX;aY1þbY2Þ¼XTAðaY1þbY2Þ¼aXTAY1þbXTAY2¼afðX;Y1ÞþbfðX;Y2Þ Hence, fis linear in the second variable, and so fis a bilinear form on Kn.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 365 12.3. Letfbe the bilinear form on R2deﬁned by f½ðx1;x2Þ;ðy1;y2Þ/C138¼ 2x1y1/C03x1y2þ4x2y2 (a) Find the matrix Aoffin the basisfu1¼ð1;0Þ;u2¼ð1;1Þg. (b) Find the matrix Boffin the basisfv1¼ð2;1Þ;v2¼ð1;/C01Þg. (c) Find the change-of-basis matrix Pfrom the basisfuigto the basisfvig, and verify that B¼PTAP. (a) Set A¼½aij/C138, where aij¼fðui;ujÞ. This yields a11¼f½ð1;0Þ;ð1;0Þ/C138¼ 2/C00/C00¼2; a21¼f½ð1;1Þ;ð1;0Þ/C138¼ 2/C00þ0¼2 a12¼f½ð1;0Þ;ð1;1Þ/C138¼ 2/C03/C00¼/C01; a22¼f½ð1;1Þ;ð1;1Þ/C138¼ 2/C03þ4¼3 Thus, A¼2/C01 23/C20/C21 is the matrix of fin the basisfu1;u2g. (b) Set B¼½bij/C138, where bij¼fðvi;vjÞ. This yields b11¼f½ð2;1Þ;ð2;1Þ/C138¼ 8/C06þ4¼6; b21¼f½ð1;/C01Þ;ð2;1Þ/C138¼ 4/C03/C04¼/C03 b12¼f½ð2;1Þ;ð1;/C01Þ/C138¼ 4þ6/C04¼6; b22¼f½ð1;/C01Þ;ð1;/C01Þ/C138¼ 2þ3þ4¼9 Thus, B¼66 /C039/C20/C21 is the matrix of fin the basisfv1;v2g. (c) Writing v1and v2in terms of the uiyields v1¼u1þu2and v2¼2u1/C0u2. Then P¼12 1/C01/C20/C21 ; PT¼11 2/C01/C20/C21 PTAP¼11 2/C01/C20/C21 2/C01 23/C20/C21 12 1/C01/C20/C21 ¼66 /C039/C20/C21 ¼B and 12.4. Prove Theorem 12.1: Let Vbe an n-dimensional vector space over K. Letff1;...;fngbe any basis of the dual space V*. Thenffij:i;j¼1;...;ngis a basis of BðVÞ, where fijis deﬁned by fijðu;vÞ¼fiðuÞfjðvÞ. Thus, dim BðVÞ¼n2. Letfu1;...;ungbe the basis of Vdual toffig. We ﬁrst show that ffijgspans BðVÞ. Let f2BðVÞand suppose fðui;ujÞ¼aij:We claim that f¼P i;jaijfij. It sufﬁces to show that fðus;utÞ¼Paijfij/C0/C1 ðus;utÞ for s;t¼1;...;n We have Paijfij/C0/C1 ðus;utÞ¼Paijfijðus;utÞ¼PaijfiðusÞfjðutÞ¼Paijdisdjt¼ast¼fðus;utÞ as required. Hence, ffijgspans BðVÞ. Next, supposePaijfij¼0. Then for s;t¼1;...;n, 0¼0ðus;utÞ¼ðPaijfijÞðus;utÞ¼ars The last step follows as above. Thus, ffijgis independent, and hence is a basis of BðVÞ. 12.5. Prove Theorem 12.2. Let Pbe the change-of-basis matrix from a basis Sto a basis S0. Let Abe the matrix representing a bilinear form in the basis S. Then B¼PTAPis the matrix representing fin the basis S0. Letu;v2V. Because Pis the change-of-basis matrix from StoS0, we have P½u/C138S0¼½u/C138Sand also P½v/C138S0¼½v/C138S; hence,½u/C138T S¼½u/C138T S0PT. Thus, fðu;vÞ¼½ u/C138T SA½v/C138S¼½u/C138T S0PTAP½v/C138S0 Because uand vare arbitrary elements of V,PTAPis the matrix of fin the basis S0.366 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Symmetric Bilinear Forms, Quadratic Forms 12.6. Find the symmetric matrix that corresponds to each of the following quadratic forms: (a)qðx;y;zÞ¼3x2þ4xy/C0y2þ8xz/C06yzþz2, (b)q0ðx;y;zÞ¼3x2þxz/C02yz, (c) q00ðx;y;zÞ¼2x2/C05y2/C07z2 The symmetric matrix A¼½aij/C138that represents qðx1;...;xnÞhas the diagonal entry aiiequal to the coefﬁcient of the square term x2 iand the nondiagonal entries aijandajieach equal to half of the coefﬁcient of the cross-product term xixj. Thus, (a) A¼324 2/C01/C03 4/C0312 43 5, (b) A0¼301 2 00/C01 1 2/C0102 43 5, (c) A00¼200 0/C050 00/C072 43 5 The third matrix A00is diagonal, because the quadratic form q00is diagonal; that is, q00has no cross-product terms. 12.7. Find the quadratic form qðXÞthat corresponds to each of the following symmetric matrices: (a) A¼5/C03 /C038/C20/C21 ;(b) B¼4/C057 /C05/C068 78/C092 43 5, (c) C¼24/C015 4/C07/C068 /C01/C063 9 589 12 6643 775 The quadratic form qðXÞthat corresponds to a symmetric matrix Mis deﬁned by qðXÞ¼XTMX, where X¼½xi/C138is the column vector of unknowns. (a) Compute as follows: qðx;yÞ¼XTAX¼½x;y/C1385/C03 /C038/C20/C21x y/C20/C21 ¼½5x/C03y;/C03xþ8y/C138x y/C20/C21 ¼5x2/C03xy/C03xyþ8y2¼5x2/C06xyþ8y2 As expected, the coefﬁcient 5 of the square term x2and the coefﬁcient 8 of the square term y2are the diagonal elements of A, and the coefﬁcient /C06 of the cross-product term xyis the sum of the nondiagonal elements /C03 and/C03o f A(or twice the nondiagonal element /C03, because Ais symmetric). (b) Because Bis a three-square matrix, there are three unknowns, say x;y;zorx1;x2;x3. Then qðx;y;zÞ¼4x2/C010xy/C06y2þ14xzþ16yz/C09z2 qðx1;x2;x3Þ¼4x2 1/C010x1x2/C06x2 2þ14x1x3þ16x2x3/C09x2 3 or Here we use the fact that the coefﬁcients of the square terms x2 1;x2 2;x2 3(orx2;y2;z2) are the respective diagonal elements 4 ;/C06;/C09o f B, and the coefﬁcient of the cross-product term xixjis the sum of the nondiagonal elements bijandbji(or twice bij, because bij¼bji). (c) Because Cis a four-square matrix, there are four unknowns. Hence, qðx1;x2;x3;x4Þ¼2x2 1/C07x2 2þ3x2 3þx2 4þ8x1x2/C02x1x3 þ10x1x4/C012x2x3þ16x2x4þ18x3x4 12.8. LetA¼1/C032 /C037/C05 2/C0582 43 5. Apply Algorithm 12.1 to ﬁnd a nonsingular matrix Psuch that D¼PTAPis diagonal, and ﬁnd sig ðAÞ, the signature of A.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 367 First form the block matrix M¼½A;I/C138: M¼½A;I/C138¼1/C03 2100 /C037/C05010 2/C05 80012 43 5 Using a11¼1 as a pivot, apply the row operations ‘‘Replace R2by 3 R1þR2’’ and ‘‘Replace R3by /C02R1þR3’’ to Mand then apply the corresponding column operations ‘‘Replace C2by 3C1þC2’’ and ‘‘Replace C3by/C02C1þC3’’ to Ato obtain 1/C032 100 0/C021 310 01 4/C02012 43 5 and then10 01 0 0 0/C021 310 01 4/C02012 43 5: Next apply the row operation ‘‘Replace R3byR2þ2R3’’ and then the corresponding column operation ‘‘Replace C3byC2þ2C3’’ to obtain 10 01 0 0 0/C021 310 00 9/C01122 43 5 and then10 01 0 0 0/C02 0 310 00 1 8/C01122 43 5 Now Ahas been diagonalized and the transpose of Pis in the right half of M. Thus, set P¼13/C01 01 1 00 22 43 5 and then D¼PTAP¼10 0 0/C020 00 1 82 43 5 Note Dhasp¼2 positive and n¼1 negative diagonal elements. Thus, the signature of Ais sigðAÞ¼p/C0n¼2/C01¼1. 12.9. Justify Algorithm 12.1, which diagonalizes (under congruence) a symmetric matrix A. Consider the block matrix M¼½A;I/C138. The algorithm applies a sequence of elementary row operations and the corresponding column operations to the left side of M, which is the matrix A. This is equivalent to premultiplying Aby a sequence of elementary matrices, say, E1;E2;...;Er, and postmultiplying Aby the transposes of the Ei. Thus, when the algorithm ends, the diagonal matrix Don the left side of Mis equal to D¼Er/C1/C1/C1E2E1AET 1ET 2/C1/C1/C1ET r¼QAQT; where Q¼Er/C1/C1/C1E2E1 On the other hand, the algorithm only applies the elementary row operations to the identity matrix Ion the right side of M. Thus, when the algorithm ends, the matrix on the right side of Mis equal to Er/C1/C1/C1E2E1I¼Er/C1/C1/C1E2E1¼Q Setting P¼QT, we get D¼PTAP, which is a diagonalization of Aunder congruence. 12.10. Prove Theorem 12.4: Let fbe a symmetric bilinear form on Vover K(where 1þ16¼0). Then Vhas a basis in which fis represented by a diagonal matrix. Algorithm 12.1 shows that every symmetric matrix over Kis congruent to a diagonal matrix. This is equivalent to the statement that fhas a diagonal representation. 12.11. Letqbe the quadratic form associated with the symmetric bilinear form f. Verify the polar identity fðu;vÞ¼1 2½qðuþvÞ/C0qðuÞ/C0qðvÞ/C138. (Assume that 1þ16¼0.) We have qðuþvÞ/C0qðuÞ/C0qðvÞ¼fðuþv;uþvÞ/C0fðu;uÞ/C0fðv;vÞ ¼fðu;uÞþfðu;vÞþfðv;uÞþfðv;vÞ/C0fðu;uÞ/C0fðv;vÞ¼2fðu;vÞ If 1þ16¼0, we can divide by 2 to obtain the required identity.368 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 12.12. Consider the quadratic form qðx;yÞ¼3x2þ2xy/C0y2and the linear substitution x¼s/C03t; y¼2sþt (a) Rewrite qðx;yÞin matrix notation, and ﬁnd the matrix Arepresenting qðx;yÞ. (b) Rewrite the linear substitution using matrix notation, and ﬁnd the matrix Pcorresponding to the substitution. (c) Find qðs;tÞusing direct substitution. (d) Find qðs;tÞusing matrix notation. (a) Here qðx;yÞ¼½ x;y/C13831 1/C01/C20/C21 x y/C20/C21 . Thus, A¼31 1/C01/C20/C21 ; and qðXÞ¼XTAX, where X¼½x;y/C138T. (b) Herex y/C20/C21 ¼1/C03 21/C20/C21 s t/C20/C21 . Thus, P¼1/C03 21/C20/C21 ; and X¼x y/C20/C21 ;Y¼s t/C20/C21 andX¼PY. (c) Substitute for xandyinqto obtain qðs;tÞ¼3ðs/C03tÞ2þ2ðs/C03tÞð2sþtÞ/C0ð 2sþtÞ2 ¼3ðs2/C06stþ9t2Þþ2ð2s2/C05st/C03t2Þ/C0ð 4s2þ4stþt2Þ¼3s2/C032stþ20t2 (d) Here qðXÞ¼XTAXandX¼PY. Thus, XT¼YTPT. Therefore, qðs;tÞ¼qðYÞ¼YTPTAPY¼½s;t/C13812 /C031/C20/C2131 1/C01/C20/C211/C03 21/C20/C21s t/C20/C21 ¼½s;t/C1383/C016 /C016 20/C20/C21s t/C20/C21 ¼3s2/C032stþ20t2 [As expected, the results in parts (c) and (d) are equal.] 12.13. Consider any diagonal matrix A¼diagða1;...;anÞover K. Show that for any nonzero scalars k1;...;kn2K;Ais congruent to a diagonal matrix Dwith diagonal entries a1k2 1;...;ank2 n. Furthermore, show that (a) If K¼C, then we can choose Dso that its diagonal entries are only 1’s and 0’s. (b) If K¼R, then we can choose Dso that its diagonal entries are only 1’s, /C01’s, and 0’s. LetP¼diagðk1;...;knÞ. Then, as required, D¼PTAP¼diagðkiÞdiagðaiÞdiagðkiÞ¼diagða1k2 1;...;ank2 nÞ (a) Let P¼diagðbiÞ, where bi¼1=ﬃﬃﬃﬃaipifai6¼0 1i f ai¼0/C26 Then PTAPhas the required form. (b) Let P¼diagðbiÞ, where bi¼1=ﬃﬃﬃﬃﬃﬃﬃ jaijp ifai6¼0 1i f ai¼0/C26 Then PTAPhas the required form. Remark: We emphasize that (b) is no longer true if ‘‘congruence’’ is replaced by ‘‘Hermitian congruence.’’ 12.14. Prove Theorem 12.5: Let fbe a symmetric bilinear form on VoverR. Then there exists a basis ofVin which fis represented by a diagonal matrix. Every other diagonal matrix representation offhas the same number pof positive entries and the same number nof negative entries. By Theorem 12.4, there is a basis fu1;...;ungofVin which fis represented by a diagonal matrix with, say, ppositive and nnegative entries. Now suppose fw1;...;wngis another basis of V, in which fis represented by a diagonal matrix with p0positive and n0negative entries. We can assume without loss of generality that the positive entries in each matrix appear ﬁrst. Because rank ðfÞ¼pþn¼p0þn0,i t sufﬁces to prove that p¼p0.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 369 LetUbe the linear span of u1;...;upand let Wbe the linear span of wp0þ1;...;wn. Then fðv;vÞ>0 for every nonzero v2U, and fðv;vÞ/C200 for every nonzero v2W. Hence, U\W¼f0g. Note that dimU¼pand dim W¼n/C0p0. Thus, dimðUþWÞ¼dimUþdimW/C0dimðU\WÞ¼pþðn/C0p0Þ/C00¼p/C0p0þn But dimðUþWÞ/C20dimV¼n; hence, p/C0p0þn/C20norp/C20p0. Similarly, p0/C20pand therefore p¼p0, as required. Remark: The above theorem and proof depend only on the concept of positivity. Thus, the theorem is true for any subﬁeld Kof the real ﬁeld Rsuch as the rational ﬁeld Q. Positive Deﬁnite Real Quadratic Forms 12.15. Prove that the following deﬁnitions of a positive deﬁnite quadratic form qare equivalent: (a) The diagonal entries are all positive in any diagonal representation of q. (b)qðYÞ>0, for any nonzero vector YinRn. Suppose qðYÞ¼a1y2 1þa2y2 2þ/C1/C1/C1þ any2 n. If all the coefﬁcients are positive, then clearly qðYÞ>0 whenever Y6¼0. Thus, (a) implies (b). Conversely, suppose (a) is not true; that is, suppose some diagonal entry ak/C200. Let ek¼ð0;...;1;...0Þbe the vector whose entries are all 0 except 1 in the kth position. Then qðekÞ¼akis not positive, and so (b) is not true. That is, (b) implies (a). Accordingly, (a) and (b) are equivalent. 12.16. Determine whether each of the following quadratic forms qis positive deﬁnite: (a)qðx;y;zÞ¼x2þ2y2/C04xz/C04yzþ7z2 (b)qðx;y;zÞ¼x2þy2þ2xzþ4yzþ3z2 Diagonalize (under congruence) the symmetric matrix Acorresponding to q. (a) Apply the operations ‘‘Replace R3by 2R1þR3’’ and ‘‘Replace C3by 2C1þC3,’’ and then ‘‘Replace R3byR2þR3’’ and ‘‘Replace C3byC2þC3.’’ These yield A¼10/C02 02/C02 /C02/C0272 43 5’100 02/C02 0/C0232 43 5’100 020 0012 43 5 The diagonal representation of qonly contains positive entries, 1 ;2;1, on the diagonal. Thus, qis positive deﬁnite. (b) We have A¼101 012 1232 43 5’100 012 0222 43 5’10 0 01 0 00/C022 43 5 There is a negative entry /C02 on the diagonal representation of q. Thus, qis not positive deﬁnite. 12.17. Show that qðx;yÞ¼ax2þbxyþcy2is positive deﬁnite if and only if a>0 and the discriminant D¼b2/C04ac<0. Suppose v¼ðx;yÞ6¼0. Then either x6¼0o r y6¼0; say, y6¼0. Let t¼x=y. Then qðvÞ¼y2½aðx=yÞ2þbðx=yÞþc/C138¼y2ðat2þbtþcÞ However, the following are equivalent: (i) s¼at2þbtþcis positive for every value of t. (ii) s¼at2þbtþclies above the t-axis. (iii) a>0 and D¼b2/C04ac<0.370 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Thus, qis positive deﬁnite if and only if a>0 and D<0. [Remark :D<0 is the same as det ðAÞ>0, where Ais the symmetric matrix corresponding to q.] 12.18. Determine whether or not each of the following quadratic forms qis positive deﬁnite: (a)qðx;yÞ¼x2/C04xyþ7y2, (b) qðx;yÞ¼x2þ8xyþ5y2, (c) qðx;yÞ¼3x2þ2xyþy2 Compute the discriminant D¼b2/C04ac, and then use Problem 12.17. (a) D¼16/C028¼/C012. Because a¼1>0 and D<0;qis positive deﬁnite. (b) D¼64/C020¼44. Because D>0;qis not positive deﬁnite. (c) D¼4/C012¼/C08. Because a¼3>0 and D<0;qis positive deﬁnite. Hermitian Forms 12.19. Determine whether the following matrices are Hermitian: (a)22þ3i4/C05i 2/C03i 56þ2i 4þ5i6/C02i/C072 43 5, (b)32/C0i4þi 2/C0i 6 i 4þii 72 43 5, (c)4/C035 /C0321 51/C062 43 5 A complex matrix A¼½aij/C138is Hermitian if A*¼A—that is, if aij¼/C22aji: (a) Yes, because it is equal to its conjugate transpose.(b) No, even though it is symmetric.(c) Yes. In fact, a real matrix is Hermitian if and only if it is symmetric. 12.20. LetAbe a Hermitian matrix. Show that fis a Hermitian form on Cnwhere fis deﬁned by fðX;YÞ¼XTA/C22Y. For all a;b2Cand all X1;X2;Y2Cn, fðaX1þbX2;YÞ¼ð aX1þbX2ÞTA/C22Y¼ðaXT 1þbXT 2ÞA/C22Y ¼aXT 1A/C22YþbXT 2A/C22Y¼afðX1;YÞþbfðX2;YÞ Hence, fis linear in the ﬁrst variable. Also, fðX;YÞ¼XTA/C22Y¼ðXTA/C22YÞT¼/C22YTATX¼YTA*/C22X¼YTA/C22X¼fðY;XÞ Hence, fis a Hermitian form on Cn. Remark: We use the fact that XTA/C22Yis a scalar and so it is equal to its transpose. 12.21. Letfbe a Hermitian form on V. Let Hbe the matrix of fin a basis S¼fuigofV. Prove the following: (a)fðu;vÞ¼½ u/C138T SH½v/C138Sfor all u;v2V. (b) If Pis the change-of-basis matrix from Sto a new basis S0ofV, then B¼PTH/C22P(or B¼Q*HQ, where Q¼/C22PÞis the matrix of fin the new basis S0. Note that (b) is the complex analog of Theorem 12.2. (a) Let u;v2Vand suppose u¼a1u1þ/C1/C1/C1þ anunand v¼b1u1þ/C1/C1/C1þ bnun. Then, as required, fðu;vÞ¼fða1u1þ/C1/C1/C1þ anun;b1u1þ/C1/C1/C1þ bnunÞ ¼P i;jai/C22bjfðui;vjÞ¼½ a1;...;an/C138H½/C22b1;...;/C22bn/C138T¼½u/C138T SH½v/C138SCHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 371 (b) Because Pis the change-of-basis matrix from StoS0, we have P½u/C138S0¼½u/C138SandP½v/C138S0¼½v/C138S; hence, ½u/C138T S¼½u/C138T S0PTand½v/C138S¼/C22P½v/C138S0:Thus, by (a), fðu;vÞ¼½ u/C138T SH½v/C138S¼½u/C138T S0PTH/C22P½v/C138S0 Butuand vare arbitrary elements of V;hence, PTH/C22Pis the matrix of fin the basis S0: 12.22. LetH¼11þi 2i 1/C0i 42/C03i /C02i2þ3i 72 43 5, a Hermitian matrix. Find a nonsingular matrix Psuch that D¼PTH/C22Pis diagonal. Also, ﬁnd the signature of H. Use the modiﬁed Algorithm 12.1 that applies the same row operations but the corresponding conjugate column operations. Thus, ﬁrst form the block matrix M¼½H;I/C138: M¼11þi 2i 100 1/C0i 42/C03i010 /C02i2þ3i 70 0 12 43 5 Apply the row operations ‘‘Replace R2byð/C01þiÞR1þR2’’ and ‘‘Replace R3by 2iR1þR3’’ and then the corresponding conjugate column operations ‘‘Replace C2byð/C01/C0iÞC1þC2’’ and ‘‘Replace C3by /C02iC1þC3’’ to obtain 11þi 2i 10 0 02/C05i/C01þi10 05 i 32 i012 43 5 and then10 0 1 0 0 02/C05i/C01þi10 05 i 32 i012 43 5 Next apply the row operation ‘‘Replace R3by/C05iR2þ2R3’’ and the corresponding conjugate column operation ‘‘Replace C3by 5iC2þ2C3’’ to obtain 10 0 1 0 0 02/C05i/C01þi 10 00/C019 5þ9i/C05i22 43 5 and then10 0 1 0 0 02 0/C01þi 10 00/C038 5þ9i/C05i22 43 5 Now Hhas been diagonalized, and the transpose of the right half of MisP. Thus, set P¼1/C01þi5þ9i 01/C05i 00 22 43 5; and then D¼PTH/C22P¼10 0 02 000/C0382 43 5: Note Dhasp¼2 positive elements and n¼1 negative elements. Thus, the signature of His sigðHÞ¼2/C01¼1. Miscellaneous Problems 12.23. Prove Theorem 12.3: Let fbe an alternating form on V. Then there exists a basis of Vin which f is represented by a block diagonal matrix Mwith blocks of the form01 /C010/C20/C21 or 0. The number of nonzero blocks is uniquely determined by f[because it is equal to1 2rankðfÞ/C138. Iff¼0, then the theorem is obviously true. Also, if dim V¼1, then fðk1u;k2uÞ¼k1k2fðu;uÞ¼0 and so f¼0. Accordingly, we can assume that dim V>1 and f6¼0. Because f6¼0, there exist (nonzero) u1;u22Vsuch that fðu1;u2Þ6¼0. In fact, multiplying u1by an appropriate factor, we can assume that fðu1;u2Þ¼1a n ds o fðu2;u1Þ¼/C0 1. Now u1and u2are linearly independent; because if, say, u2¼ku1,t h e n fðu1;u2Þ¼fðu1;ku1Þ¼kfðu1;u1Þ¼0. Let U¼spanðu1;u2Þ; then,372 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms (i) The matrix representation of the restriction of ftoUin the basisfu1;u2gis01 /C010/C20/C21 , (ii) If u2U, say u¼au1þbu2, then fðu;u1Þ¼fðau1þbu2;u1Þ¼/C0 b and fðu;u2Þ¼fðau1þbu2;u2Þ¼a LetWconsists of those vectors w2Vsuch that fðw;u1Þ¼0 and fðw;u2Þ¼0:Equivalently, W¼fw2V:fðw;uÞ¼0 for every u2Ug We claim that V¼U/C8W. It is clear that U\W¼f0g, and so it remains to show that V¼UþW. Let v2V. Set u¼fðv;u2Þu1/C0fðv;u1Þu2 and w¼v/C0u ð1Þ Because uis a linear combination of u1andu2;u2U. We show next that w2W. By (1) and (ii), fðu;u1Þ¼fðv;u1Þ; hence, fðw;u1Þ¼fðv/C0u;u1Þ¼fðv;u1Þ/C0fðu;u1Þ¼0 Similarly, fðu;u2Þ¼fðv;u2Þand so fðw;u2Þþfðv/C0u;u2Þ¼fðv;u2Þ/C0fðu;u2Þ¼0 Then w2Wand so, by (1), v¼uþw, where u2W. This shows that V¼UþW; therefore, V¼U/C8W. Now the restriction of ftoWis an alternating bilinear form on W. By induction, there exists a basis u3;...;unofWin which the matrix representing frestricted to Whas the desired form. Accordingly, u1;u2;u3;...;unis a basis of Vin which the matrix representing fhas the desired form. SUPPLEMENTARY PROBLEMS Bilinear Forms 12.24. Letu¼ðx1;x2Þand v¼ðy1;y2Þ. Determine which of the following are bilinear forms on R2: (a) fðu;vÞ¼2x1y2/C03x2y1, (c) fðu;vÞ¼3x2y2, (e) fðu;vÞ¼1, (b) fðu;vÞ¼x1þy2, (d) fðu;vÞ¼x1x2þy1y2,( f ) fðu;vÞ¼0 12.25. Letfbe the bilinear form on R2deﬁned by f½ðx1;x2Þ;ðy1;y2Þ/C138¼ 3x1y1/C02x1y2þ4x2y1/C0x2y2 (a) Find the matrix Aoffin the basisfu1¼ð1;1Þ;u2¼ð1;2Þg. (b) Find the matrix Boffin the basisfv1¼ð1;/C01Þ;v2¼ð3;1Þg. (c) Find the change-of-basis matrix Pfromfuigtofvig, and verify that B¼PTAP. 12.26. LetVbe the vector space of two-square matrices over R. Let M¼12 35/C20/C21 , and let fðA;BÞ¼trðATMBÞ, where A;B2Vand ‘‘tr’’ denotes trace. (a) Show that fis a bilinear form on V. (b) Find the matrix of fin the basis 10 00/C20/C21 ;01 00/C20/C21 ;00 10/C20/C21 ;00 01/C20/C21 /C26/C27 12.27. LetBðVÞbe the set of bilinear forms on Vover K. Prove the following: (a) If f;g2BðVÞ, then fþg,kg2BðVÞfor any k2K. (b) If fandsare linear functions on V, then fðu;vÞ¼fðuÞsðvÞbelongs to BðVÞ. 12.28. Let½f/C138denote the matrix representation of a bilinear form fonVrelative to a basis fuig. Show that the mapping f7!½f/C138is an isomorphism of BðVÞonto the vector space Vofn-square matrices.CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 373 12.29. Letfbe a bilinear form on V. For any subset SofV, let S?¼fv2V:fðu;vÞ¼0 for every u2SgandS>¼fv2V:fðv;uÞ¼0 for every u2Sg Show that: (a) S>and S>are subspaces of V; (b) S1/C18S2implies S? 2/C18S? 1and S> 2/C18S> 1; (c)f0g?¼f0g>¼V. 12.30. Suppose fis a bilinear form on V. Prove that: rankðfÞ¼dimV/C0dimV?¼dimV/C0dimV>, and hence, dimV?¼dimV>. 12.31. Letfbe a bilinear form on V. For each u2V, let ^u:V!Kand ~u:V!Kbe deﬁned by ^uðxÞ¼fðx;uÞand ~uðxÞ¼fðu;xÞ. Prove the following: (a) ^uand ~uare each linear; i.e., ^u;~u2V*, (b) u7!^uandu7!~uare each linear mappings from VintoV*, (c) rankðfÞ¼rankðu7!^uÞ¼rankðu7!~uÞ. 12.32. Show that congruence of matrices (denoted by ’) is an equivalence relation; that is, (i)A’A; (ii) If A’B, then B’A; (iii) If A’BandB’C, then A’C. Symmetric Bilinear Forms, Quadratic Forms 12.33. Find the symmetric matrix Abelonging to each of the following quadratic forms: (a) qðx;y;zÞ/C02x2/C08xyþy2/C016xzþ14yzþ5z2, (c) qðx;y;zÞ¼xyþy2þ4xzþz2 (b) qðx;y;zÞ¼x2/C0xzþy2, (d) qðx;y;zÞ¼xyþyz 12.34. For each of the following symmetric matrices A, ﬁnd a nonsingular matrix Psuch that D¼PTAPis diagonal: (a) A¼102 036 2672 43 5, (b) A¼1/C021 /C0253 13/C022 43 5, (c) A¼1/C010 2 /C012 10 01 12 20 2/C012 6643 775 12.35. Letqðx;yÞ¼2x2/C06xy/C03y2and x¼sþ2t,y¼3s/C0t. (a) Rewrite qðx;yÞin matrix notation, and ﬁnd the matrix Arepresenting the quadratic form. (b) Rewrite the linear substitution using matrix notation, and ﬁnd the matrix Pcorresponding to the substitution. (c) Find qðs;tÞusing (i) direct substitution, (ii) matrix notation. 12.36. For each of the following quadratic forms qðx;y;zÞ, ﬁnd a nonsingular linear substitution expressing the variables x;y;zin terms of variables r;s;tsuch that qðr;s;tÞis diagonal: (a) qðx;y;zÞ¼x2þ6xyþ8y2/C04xzþ2yz/C09z2, (b) qðx;y;zÞ¼2x2/C03y2þ8xzþ12yzþ25z2, (c) qðx;y;zÞ¼x2þ2xyþ3y2þ4xzþ8yzþ6z2. In each case, ﬁnd the rank and signature. 12.37. Give an example of a quadratic form qðx;yÞsuch that qðuÞ¼0 and qðvÞ¼0 but qðuþvÞ6¼0. 12.38. LetSðVÞdenote all symmetric bilinear forms on V. Show that (a) SðVÞis a subspace of BðVÞ; (b) If dim V¼n, then dim SðVÞ¼1 2nðnþ1Þ. 12.39. Consider a real quadratic polynomial qðx1;...;xnÞ¼Pn i;j¼1aijxixj;where aij¼aji.374 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms (a) If a116¼0, show that the substitution x1¼y1/C01 a11ða12y2þ/C1/C1/C1þ a1nynÞ; x2¼y2; ...; xn¼yn yields the equation qðx1;...;xnÞ¼a11y2 1þq0ðy2;...;ynÞ, where q0is also a quadratic polynomial. (b) If a11¼0 but, say, a126¼0, show that the substitution x1¼y1þy2; x2¼y1/C0y2; x3¼y3; ...; xn¼yn yields the equation qðx1;...;xnÞ¼Pbijyiyj, where b116¼0, which reduces this case to case (a). Remark: This method of diagonalizing qis known as completing the square . Positive Deﬁnite Quadratic Forms 12.40. Determine whether or not each of the following quadratic forms is positive deﬁnite: (a) qðx;yÞ¼4x2þ5xyþ7y2, (c) qðx;y;zÞ¼x2þ4xyþ5y2þ6xzþ2yzþ4z2 (b) qðx;yÞ¼2x2/C03xy/C0y2; (d) qðx;y;zÞ¼x2þ2xyþ2y2þ4xzþ6yzþ7z2 12.41. Find those values of ksuch that the given quadratic form is positive deﬁnite: (a) qðx;yÞ¼2x2/C05xyþky2, (b) qðx;yÞ¼3x2/C0kxyþ12y2 (c) qðx;y;zÞ¼x2þ2xyþ2y2þ2xzþ6yzþkz2 12.42. Suppose Ais a real symmetric positive deﬁnite matrix. Show that A¼PTPfor some nonsingular matrix P. Hermitian Forms 12.43. Modify Algorithm 12.1 so that, for a given Hermitian matrix H, it ﬁnds a nonsingular matrix Pfor which D¼PTA/C22Pis diagonal. 12.44. For each Hermitian matrix H, ﬁnd a nonsingular matrix Psuch that D¼PTH/C22Pis diagonal: (a) H¼1i /C0i2/C20/C21 , (b) H¼12þ3i 2/C03i/C01/C20/C21 , (c) H¼1 i 2þi /C0i 21/C0i 2/C0i1þi 22 43 5 Find the rank and signature in each case. 12.45. LetAbe a complex nonsingular matrix. Show that H¼A*Ais Hermitian and positive deﬁnite. 12.46. We say that BisHermitian congruent toAif there exists a nonsingular matrix Psuch that B¼PTA/C22Por, equivalently, if there exists a nonsingular matrix Qsuch that B¼Q*AQ. Show that Hermitian congruence is an equivalence relation. ( Note:I fP¼/C22Q, then PTA/C22P¼Q*AQ.) 12.47. Prove Theorem 12.7: Let fbe a Hermitian form on V. Then there is a basis SofVin which fis represented by a diagonal matrix, and every such diagonal representation has the same number pof positive entries and the same number nof negative entries. Miscellaneous Problems 12.48. Letedenote an elementary row operation, and let f* denote the corresponding conjugate column operation (where each scalar kineis replaced by /C22kinf*). Show that the elementary matrix corresponding to f*i s the conjugate transpose of the elementary matrix corresponding to e. 12.49. LetVandWbe vector spaces over K. A mapping f:V/C2W!Kis called a bilinear form onVandWif (i)fðav1þbv2;wÞ¼afðv1;wÞþbfðv2;wÞ, (ii) fðv;aw1þbw2Þ¼afðv;w1Þþbfðv;w2Þ for every a;b2K;vi2V;wj2W. Prove the following:CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms 375 (a) The set BðV;WÞof bilinear forms on VandWis a subspace of the vector space of functions from V/C2WintoK. (b) Ifff1;...;fmgis a basis of V*a n dfs1;...;sngis a basis of W*, then ffij:i¼1;...;m;j¼1;...;ngis a basis of BðV;WÞ,w h e r e fijis deﬁned by fijðv;wÞ¼fiðvÞsjðwÞ. Thus, dim BðV;WÞ¼dimVdimW. [Note that if V¼W, then we obtain the space BðVÞinvestigated in this chapter.] 12.50. LetVbe a vector space over K. A mapping f:V/C2V/C2.../C2Vzﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ}|ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{mtimes !Kis called a multilinear (orm-linear ) form onViffis linear in each variable; that is, for i¼1;...;m, fð...;auþbv;...Þ¼afð...;^u;...Þþbfð...;^v;...Þ wherec...denotes the ith element, and other elements are held ﬁxed. An m-linear form fis said to be alternating iffðv1;...vmÞ¼0 whenever vi¼vjfori6¼j. Prove the following: (a) The set BmðVÞofm-linear forms on Vis a subspace of the vector space of functions from V/C2V/C2/C1/C1/C1/C2 VintoK. (b) The set AmðVÞof alternating m-linear forms on Vis a subspace of BmðVÞ. Remark 1: Ifm¼2, then we obtain the space BðVÞinvestigated in this chapter. Remark 2: IfV¼Km, then the determinant function is an alternating m-linear form on V. ANSWERS TO SUPPLEMENTARY PROBLEMS Notation: M¼½R1;R2; .../C138denotes a matrix Mwith rows R1;R2;.... 12.24. (a) yes, (b) no, (c) yes, (d) no, (e) no, (f ) yes 12.25. (a) A¼½4;1;7;3/C138, (b) B¼½0;/C04;20;32/C138, (c) P¼½3;5;/C02;/C02/C138 12.26. (b)½1;0;2;0;0;1;0;2;3;0;5;0;0;3;0;5/C138 12.33. (a)½2;/C04;/C08;/C04;1;7;/C08;7;5/C138, (b)½1;0;/C01 2;0;1;0;/C01 2;0;0/C138, (c)½0;1 2;2;1 2;1;0;2;0;1/C138, (d)½0;1 2;0;1 2;0;1;1 2;0;1 2;0;1 2;0/C138 12.34. (a) P¼½1;0;/C02;0;1;/C02;0;0;1/C138;D¼diagð1;3;/C09Þ; (b) P¼½1;2;/C011;0;1;/C05;0;0;1/C138;D¼diagð1;1;/C028Þ; (c) P¼½1;1;/C01;/C04;0;1;/C01;/C02;0;0;1;0;0;0;0;1/C138;D¼diagð1;1;0;/C09Þ 12.35. A¼½2;/C03;/C03;/C03/C138,P¼½1;2;3;/C01/C138,qðs;tÞ¼/C0 43s2/C04stþ17t2 12.36. (a) x¼r/C03s/C019t,y¼sþ7t,z¼t;qðr;s;tÞ¼r2/C0s2þ36t2; (b) x¼r/C02t;y¼sþ2t;z¼t;qðr;s;tÞ¼2r2/C03s2þ29t2; (c) x¼r/C0s/C0t;y¼s/C0t;z¼t;qðr;s;tÞ¼r2/C02s2 12.37. qðx;yÞ¼x2/C0y2,u¼ð1;1Þ,v¼ð1;/C01Þ 12.40. (a) yes, (b) no, (c) no, (d) yes 12.41. (a) k>25 8, (b)/C012<k<12, (c) k>5 12.44. (a) P¼½1;i;0;1/C138,D¼I;s¼2; (b) P¼½1;/C02þ3i;0;1/C138,D¼diagð1;/C014Þ,s¼0; (c) P¼½1;i;/C03þi;0;1;i;0;0;1/C138,D¼diagð1;1;/C04Þ;s¼1d376 CHAPTER 12 Bilinear, Quadratic, and Hermitian Forms Linear Operators on Inner Product Spaces 13.1 Introduction This chapter investigates the space AðVÞof linear operators Ton an inner product space V.( S e e Chapter 7.) Thus, the base ﬁeld Kis either the real numbers Ror the complex numbers C. In fact, different terminologies will be used for the real case and the complex case. We also use the fact that the innerproducts on real Euclidean space R nand complex Euclidean space Cnmay be deﬁned, respectively, by hu;vi¼uTv andhu;vi¼uT/C22v where uand vare column vectors. The reader should review the material in Chapter 7 and be very familiar with the notions of norm (length), orthogonality, and orthonormal bases. We also note that Chapter 7 mainly dealt with real inner product spaces, whereas here we assume that Vis a complex inner product space unless otherwise stated or implied. Lastly, we note that in Chapter 2, we used AHto denote the conjugate transpose of a complex matrix A; that is, AH¼AT. This notation is not standard. Many texts, expecially advanced texts, use A* to denote such a matrix; we will use that notation in this chapter. That is, now A*¼AT. 13.2 Adjoint Operators We begin with the following basic deﬁnition. DEFINITION: A linear operator Ton an inner product space Vis said to have an adjoint operator T * onVifhTðuÞ;vi¼h u;T*ðvÞifor every u;v2V. The following example shows that the adjoint operator has a simple description within the context of matrix mappings. EXAMPLE 13.1 (a) Let Abe a real n-square matrix viewed as a linear operator on Rn. Then, for every u;v2Rn; hAu;vi¼ð AuÞTv¼uTATv¼hu;ATvi Thus, the transpose ATofAis the adjoint of A. (b) Let Bbe a complex n-square matrix viewed as a linear operator on Cn. Then for every u;v;2Cn, hBu;vi¼ð BuÞT/C22v¼uTBT/C22v¼uTB*/C22v¼hu;B*vi Thus, the conjugate transpose B*o f Bis the adjoint of B. CHAPTER 13 377 Remark: B* may mean either the adjoint of Bas a linear operator or the conjugate transpose of B as a matrix. By Example 13.1(b), the ambiguity makes no difference, because they denote the sameobject. The following theorem (proved in Problem 13.4) is the main result in this section. THEOREM 13.1: LetTbe a linear operator on a ﬁnite-dimensional inner product space Vover K. Then (i) There exists a unique linear operator T*o n Vsuch thathTðuÞ;vi¼hu;T*ðvÞi for every u;v2V. (That is, Thas an adjoint T*.) (ii) If Ais the matrix representation Twith respect to any orthonormal basis S¼fuigofV, then the matrix representation of T* in the basis Sis the conjugate transpose A*o f A(or the transpose ATofAwhen Kis real). We emphasize that no such simple relationship exists between the matrices representing TandT*i f the basis is not orthonormal. Thus, we see one useful property of orthonormal bases. We also emphasizethat this theorem is not valid if Vhas inﬁnite dimension (Problem 13.31). The following theorem (proved in Problem 13.5) summarizes some of the properties of the adjoint. THEOREM 13.2: LetT;T1;T2be linear operators on Vand let k2K. Then (i)ðT1þT2Þ*¼T1*þT2*, (iii)ðT1T2Þ*¼T2*T1*, (ii)ðkTÞ*¼/C22kT*, (iv) ðT*Þ*¼T. Observe the similarity between the above theorem and Theorem 2.3 on properties of the transpose operation on matrices. Linear Functionals and Inner Product Spaces Recall (Chapter 11) that a linear functional fon a vector space Vis a linear mapping f:V!K. This subsection contains an important result (Theorem 13.3) that is used in the proof of the above basicTheorem 13.1. LetVbe an inner product space. Each u2Vdetermines a mapping ^u:V!Kdeﬁned by ^uðvÞ¼h v;ui Now, for any a;b2Kand any v 1;v22V, ^uðav1þbv2Þ¼h av1þbv2;ui¼ahv1;uiþbhv2;ui¼a^uðv1Þþb^uðv2Þ That is, ^uis a linear functional on V. The converse is also true for spaces of ﬁnite dimension and it is contained in the following important theorem (proved in Problem 13.3). THEOREM 13.3: Letfbe a linear functional on a ﬁnite-dimensional inner product space V. Then there exists a unique vector u2Vsuch that fðvÞ¼h v;uifor every v2V. We remark that the above theorem is not valid for spaces of inﬁnite dimension (Problem 13.24). 13.3 Analogy Between AðVÞand C, Special Linear Operators LetAðVÞdenote the algebra of all linear operators on a ﬁnite-dimensional inner product space V. The adjoint mapping T7!T*o n AðVÞis quite analogous to the conjugation mapping z7!/C22zon the complex ﬁeldC. To illustrate this analogy we identify in Table 13-1 certain classes of operators T2AðVÞwhose behavior under the adjoint map imitates the behavior under conjugation of familiar classes of complex numbers. The analogy between these operators Tand complex numbers zis reﬂected in the next theorem.378 CHAPTER 13 Linear Operators on Inner Product Spaces THEOREM 13.4: Letlbe an eigenvalue of a linear operator TonV. (i) If T*¼T/C01(i.e., Tis orthogonal or unitary), then jlj¼1. (ii) If T*¼T(i.e., Tis self-adjoint), then lis real. (iii) If T*¼/C0T(i.e., Tis skew-adjoint), then lis pure imaginary. (iv) If T¼S*Swith Snonsingular (i.e., Tis positive deﬁnite), then lis real and positive. Proof. In each case let vbe a nonzero eigenvector of Tbelonging to l; that is, TðvÞ¼lvwith v6¼0. Hence,hv;viis positive. Proof of (i). We show that l/C22lhv;vi¼h v;vi: l/C22lhv;vi¼h lv;lvi¼h TðvÞ;TðvÞi¼h v;T*TðvÞi¼h v;IðvÞi¼h v;vi Buthv;vi6¼0; hence, l/C22l¼1 and sojlj¼1. Proof of (ii). We show that lhv;vi¼ /C22lhv;vi: lhv;vi¼h lv;vi¼h TðvÞ;vi¼h v;T*ðvÞi¼h v;TðvÞi¼h v;lvi¼ /C22lhv;vi Buthv;vi6¼0; hence, l¼/C22land so lis real. Proof of (iii). We show that lhv;vi¼/C0 /C22lhv;vi: lhv;vi¼h lv;vi¼h TðvÞ;vi¼h v;T*ðvÞi¼h v;/C0TðvÞi¼h v;/C0lvi¼/C0 /C22lhv;vi Buthv;vi6¼0; hence, l¼/C0 /C22lor/C22l¼/C0l, and so lis pure imaginary. Proof of (iv). Note ﬁrst that SðvÞ6¼0 because Sis nonsingular; hence, hSðvÞ,SðvÞiis positive. We show that lhv;vi¼h SðvÞ;SðvÞi: lhv;vi¼h lv;vi¼h TðvÞ;vi¼h S*SðvÞ;vi¼h SðvÞ;SðvÞi Buthv;viandhSðvÞ;SðvÞiare positive; hence, lis positive.Table 13-1 Class of complex numbersBehavior under conjugation Class of operators in AðVÞBehavior under the adjoint map Unit circleðjzj¼1Þ /C22z¼1=z Orthogonal operators (real case) T*¼T/C01 Unitary operators (complex case) Self-adjoint operators Also called: Real axis /C22z¼z symmetric (real case) T*¼T Hermitian (complex case) Skew-adjoint operators Also called: Imaginary axis /C22z¼/C0z skew-symmetric (real case) T*¼/C0T skew-Hermitian (complex case) Positive real axis z¼/C22ww;w6¼0 Positive deﬁnite operators T¼S*S ð0;1Þ with SnonsingularCHAPTER 13 Linear Operators on Inner Product Spaces 379 Remark: Each of the above operators Tcommutes with its adjoint; that is, TT*¼T*T. Such operators are called normal operators. 13.4 Self-Adjoint Operators LetTbe a self-adjoint operator on an inner product space V; that is, suppose T*¼T (IfTis deﬁned by a matrix A, then Ais symmetric or Hermitian according as Ais real or complex.) By Theorem 13.4, the eigenvalues of Tare real. The following is another important property of T. THEOREM 13.5: LetTbe a self-adjoint operator on V. Suppose uand vare eigenvectors of T belonging to distinct eigenvalues. Then uand vare orthogonal; that is, hu;vi¼0. Proof . Suppose TðuÞ¼l1uandTðvÞ¼l2v, where l16¼l2. We show that l1hu;vi¼l2hu;vi: l1hu;vi¼h l1u;vi¼h TðuÞ;vi¼h u;T*ðvÞi¼h u;TðvÞi ¼hu;l2vi¼ /C22l2hu;vi¼l2hu;vi (The fourth equality uses the fact that T*¼T, and the last equality uses the fact that the eigenvalue l2is real.) Because l16¼l2, we gethu;vi¼0. Thus, the theorem is proved. 13.5 Orthogonal and Unitary Operators LetUbe a linear operator on a ﬁnite-dimensional inner product space V. Suppose U*¼U/C01or equivalently UU*¼U*U¼I Recall that Uis said to be orthogonal or unitary according as the underlying ﬁeld is real or complex. The next theorem (proved in Problem 13.10) gives alternative characterizations of these operators. THEOREM 13.6: The following conditions on an operator Uare equivalent: (i) U*¼U/C01; that is, UU*¼U*U¼I.[Uis unitary (orthogonal).] (ii) Upreserves inner products; that is, for every v;w2V, hUðvÞ,UðwÞi¼h v;wi. (iii) Upreserves lengths; that is, for every v2V,kUðvÞk¼k vk. EXAMPLE 13.2 (a) Let T:R3!R3be the linear operator that rotates each vector vabout the z-axis by a ﬁxed angle yas shown in Fig. 10-1 (Section 10.3). That is, Tis deﬁned by Tðx;y;zÞ¼ð xcosy/C0ysiny;xsinyþycosy;zÞ We note that lengths (distances from the origin) are preserved under T. Thus, Tis an orthogonal operator. (b) Let Vbel2-space (Hilbert space), deﬁned in Section 7.3. Let T:V!Vbe the linear operator deﬁned by Tða1;a2;a3;...Þ¼ð 0;a1;a2;a3;...Þ Clearly, Tpreserves inner products and lengths. However, Tis not surjective, because, for example, ð1;0;0;...Þ does not belong to the image of T; hence, Tis not invertible. Thus, we see that Theorem 13.6 is not valid for spaces of inﬁnite dimension. An isomorphism from one inner product space into another is a bijective mapping that preserves the three basic operations of an inner product space: vector addition, scalar multiplication, and inner380 CHAPTER 13 Linear Operators on Inner Product Spaces products. Thus, the above mappings (orthogonal and unitary) may also be characterized as the isomorphisms of Vinto itself. Note that such a mapping Ualso preserves distances, because kUðvÞ/C0UðwÞk¼k Uðv/C0wÞk¼k v/C0wk Hence, Uis called an isometry . 13.6 Orthogonal and Unitary Matrices LetUbe a linear operator on an inner product space V. By Theorem 13.1, we obtain the following results. THEOREM 13.7A: A complex matrix Arepresents a unitary operator U(relative to an orthonormal basis) if and only if A*¼A/C01. THEOREM 13.7B: A real matrix Arepresents an orthogonal operator U(relative to an orthonormal basis) if and only if AT¼A/C01. The above theorems motivate the following deﬁnitions (which appeared in Sections 2.10 and 2.11). DEFINITION: A complex matrix Afor which A*¼A/C01is called a unitary matrix . DEFINITION: A real matrix Afor which AT¼A/C01is called an orthogonal matrix . We repeat Theorem 2.6, which characterizes the above matrices. THEOREM 13.8: The following conditions on a matrix Aare equivalent: (i) Ais unitary (orthogonal). (ii) The rows of Aform an orthonormal set. (iii) The columns of Aform an orthonormal set. 13.7 Change of Orthonormal Basis Orthonormal bases play a special role in the theory of inner product spaces V. Thus, we are naturally interested in the properties of the change-of-basis matrix from one such basis to another. The followingtheorem (proved in Problem 13.12) holds. THEOREM 13.9: Letfu1;...;ungbe an orthonormal basis of an inner product space V. Then the change-of-basis matrix from fuiginto another orthonormal basis is unitary (orthogonal). Conversely, if P¼½aij/C138is a unitary (orthogonal) matrix, then the following is an orthonormal basis: fu0 i¼a1iu1þa2iu2þ/C1/C1/C1þ aniun:i¼1;...;ng Recall that matrices AandBrepresenting the same linear operator Tare similar; that is, B¼P/C01AP, where Pis the (nonsingular) change-of-basis matrix. On the other hand, if Vis an inner product space, we are usually interested in the case when Pis unitary (or orthogonal) as suggested by Theorem 13.9. (Recall thatPis unitary if the conjugate tranpose P*¼P/C01, and Pis orthogonal if the transpose PT¼P/C01.) This leads to the following deﬁnition. DEFINITION: Complex matrices AandBareunitarily equivalent if there exists a unitary matrix P for which B¼P*AP. Analogously, real matrices AandBareorthogonally equivalent if there exists an orthogonal matrix Pfor which B¼PTAP. Note that orthogonally equivalent matrices are necessarily congruent.CHAPTER 13 Linear Operators on Inner Product Spaces 381 13.8 Positive Deﬁnite and Positive Operators LetPbe a linear operator on an inner product space V. Then (i)Pis said to be positive deﬁnite ifP¼S*Sfor some nonsingular operators S: (ii)Pis said to be positive (ornonnegative orsemideﬁnite )i fP¼S*Sfor some operator S: The following theorems give alternative characterizations of these operators. THEOREM 13.10A: The following conditions on an operator Pare equivalent: (i) P¼T2for some nonsingular self-adjoint operator T. (ii) Pis positive deﬁnite. (iii) Pis self-adjoint and hPðuÞ;ui>0 for every u6¼0i n V. The corresponding theorem for positive operators (proved in Problem 13.21) follows. THEOREM 13.10B: The following conditions on an operator Pare equivalent: (i) P¼T2for some self-adjoint operator T. (ii) Pis positive; that is, P¼S/C3S: (iii) Pis self-adjoint and hPðuÞ;ui/C210 for every u2V. 13.9 Diagonalization and Canonical Forms in Inner Product Spaces LetTbe a linear operator on a ﬁnite-dimensional inner product space Vover K.R e p r e s e n t i n g Tby a diagonal matrix depends upon the eigenvectors and eigenvalues of T, and hence, upon the roots of the characteristic polynomial DðtÞofT.N o wDðtÞalways factors into linear polynomials over the complex ﬁeld Cbut may not have any linear polynomials over the real ﬁeld R. Thus, the situation for real inner product spaces (sometimes called Euclidean spaces) is inherently different than thesituation for complex inner product spaces (sometimes called unitary spaces). Thus, we treat themseparately. Real Inner Product Spaces, Symmetric and Orthogonal Operators The following theorem (proved in Problem 13.14) holds. THEOREM 13.11: LetTbe a symmetric (self-adjoint) operator on a real ﬁnite-dimensional product space V. Then there exists an orthonormal basis of Vconsisting of eigenvectors of T; that is, Tcan be represented by a diagonal matrix relative to an orthonormal basis. We give the corresponding statement for matrices. THEOREM 13.11: (Alternative Form) Let Abe a real symmetric matrix. Then there exists an orthogonal matrix Psuch that B¼P/C01AP¼PTAPis diagonal. We can choose the columns of the above matrix Pto be normalized orthogonal eigenvectors of A; then the diagonal entries of Bare the corresponding eigenvalues. On the other hand, an orthogonal operator Tneed not be symmetric, and so it may not be represented by a diagonal matrix relative to an orthonormal matrix. However, such a matrix Tdoes have a simple canonical representation, as described in the following theorem (proved in Problem 13.16).382 CHAPTER 13 Linear Operators on Inner Product Spaces THEOREM 13.12: LetTbe an orthogonal operator on a real inner product space V. Then there exists an orthonormal basis of Vin which Tis represented by a block diagonal matrix M of the form M¼diag Is;/C0It;cosy1/C0siny1 siny1 cosy1/C20/C21 ;...;cosyr/C0sinyr sinyr cosyr/C20/C21 /C18/C19 The reader may recognize that each of the 2 /C22 diagonal blocks represents a rotation in the corresponding two-dimensional subspace, and each diagonal entry /C01 represents a reﬂection in the corresponding one-dimensional subspace. Complex Inner Product Spaces, Normal and Triangular Operators A linear operator Tis said to be normal if it commutes with its adjoint—that is, if TT*¼T*T. We note that normal operators include both self-adjoint and unitary operators. Analogously, a complex matrix Ais said to be normal if it commutes with its conjugate transpose— that is, if AA*¼A*A. EXAMPLE 13.3 LetA¼11 i3þ2i/C20/C21 . Then A*¼1/C0i 13/C02i/C20/C21 . Also AA*¼23/C03i 3þ3i 14/C20/C21 ¼A*A. Thus, Ais normal. The following theorem (proved in Problem 13.19) holds. THEOREM 13.13: LetTbe a normal operator on a complex ﬁnite-dimensional inner product space V. Then there exists an orthonormal basis of Vconsisting of eigenvectors of T; that is,Tcan be represented by a diagonal matrix relative to an orthonormal basis. We give the corresponding statement for matrices. THEOREM 13.13: (Alternative Form) Let Abe a normal matrix. Then there exists a unitary matrix Psuch that B¼P/C01AP¼P*APis diagonal. The following theorem (proved in Problem 13.20) shows that even nonnormal operators on unitary spaces have a relatively simple form. THEOREM 13.14: LetTbe an arbitrary operator on a complex ﬁnite-dimensional inner product space V. Then Tcan be represented by a triangular matrix relative to an orthonormal basis of V. THEOREM 13.14: (Alternative Form) Let Abe an arbitrary complex matrix. Then there exists a unitary matrix Psuch that B¼P/C01AP¼P*APis triangular. 13.10 Spectral Theorem The Spectral Theorem is a reformulation of the diagonalization Theorems 13.11 and 13.13. THEOREM 13.15: (Spectral Theorem) Let Tbe a normal (symmetric) operator on a complex (real) ﬁnite-dimensional inner product space V. Then there exists linear operators E1;...;EronVand scalars l1;...;lrsuch that (i) T¼l1E1þl2E2þ/C1/C1/C1þ lrEr, (iii) E2 1¼E1;E2 2¼E2;...;E2 r¼Er, (ii) E1þE2þ/C1/C1/C1þ Er¼I, (iv) EiEj¼0 for i6¼j.CHAPTER 13 Linear Operators on Inner Product Spaces 383 The above linear operators E1;...;Erareprojections in the sense that E2 i¼Ei. Moreover, they are said to be orthogonal projections because they have the additional property that EiEj¼0 for i6¼j. The following example shows the relationship between a diagonal matrix representation and the corresponding orthogonal projections. EXAMPLE 13.4 Consider the following diagonal matrices A;E1;E2;E3: A¼2 3 3 52 6643 775;E1¼1 0 0 02 6643 775;E2¼0 1 1 02 6643 775;E3¼0 0 0 12 6643 775 The reader can verify that (i) A¼2E1þ3E2þ5E3, (ii) E1þE2þE3¼I, (iii) E2 i¼Ei, (iv) EiEj¼0 for i6¼j. SOLVED PROBLEMS Adjoints 13.1. Find the adjoint of F:R3!R3deﬁned by Fðx;y;zÞ¼ð 3xþ4y/C05z;2x/C06yþ7z;5x/C09yþzÞ First ﬁnd the matrix Athat represents Fin the usual basis of R3—that is, the matrix Awhose rows are the coefﬁcients of x;y;z—and then form the transpose ATofA. This yields A¼34/C05 2/C067 5/C0912 43 5 and then AT¼325 4/C06/C09 /C05712 43 5 The adjoint F* is represented by the transpose of A; hence, F*ðx;y;zÞ¼ð 3xþ2yþ5z;4x/C06y/C09z;/C05xþ7yþzÞ 13.2. Find the adjoint of G:C3!C3deﬁned by Gðx;y;zÞ¼½ 2xþð1/C0iÞy;ð3þ2iÞx/C04iz;2ixþð4/C03iÞy/C03z/C138 First ﬁnd the matrix Bthat represents Gin the usual basis of C3, and then form the conjugate transpose B*o f B. This yields B¼21/C0i 0 3þ2i 0/C04i 2i 4/C03i/C032 43 5 and then B*¼23/C02i/C02i 1þi 04þ3i 04 i/C032 43 5 Then G*ðx;y;zÞ¼½ 2xþð3/C02iÞy/C02iz;ð1þiÞxþð4þ3iÞz;4iy/C03z/C138: 13.3. Prove Theorem 13.3: Let fbe a linear functional on an n-dimensional inner product space V. Then there exists a unique vector u2Vsuch that fðvÞ¼h v;uifor every v2V. Letfw1;...;wngbe an orthonormal basis of V. Set u¼fðw1Þw1þfðw2Þw2þ/C1/C1/C1þ fðwnÞwn Let ^ube the linear functional on Vdeﬁned by ^uðvÞ¼h v;uifor every v2V. Then, for i¼1;...;n, ^uðwiÞ¼h wi;ui¼h wi;fðw1Þw1þ/C1/C1/C1þ fðwnÞwni¼fðwiÞ384 CHAPTER 13 Linear Operators on Inner Product Spaces Because ^uandfagree on each basis vector, ^u¼f. Now suppose u0is another vector in Vfor which fðvÞ¼h v;u0ifor every v2V. Thenhv;ui¼h v;u0i orhv;u/C0u0i¼0. In particular, this is true for v¼u/C0u0, and sohu/C0u0;u/C0u0i¼0. This yields u/C0u0¼0 and u¼u0. Thus, such a vector uis unique, as claimed. 13.4. Prove Theorem 13.1: Let Tbe a linear operator on an n-dimensional inner product space V. Then (a) There exists a unique linear operator T*o n Vsuch that hTðuÞ;vi¼h u;T*ðvÞifor all u;v2V: (b) Let Abe the matrix that represents Trelative to an orthonormal basis S¼fuig. Then the conjugate transpose A*o f Arepresents T* in the basis S. (a) We ﬁrst deﬁne the mapping T*. Let vbe an arbitrary but ﬁxed element of V. The map u7!hTðuÞ;vi is a linear functional on V. Hence, by Theorem 13.3, there exists a unique element v02Vsuch thathTðuÞ;vi¼h u;v0ifor every u2V. We deﬁne T*:V!Vby T*ðvÞ¼v0. Then hTðuÞ;vi¼h u;T*ðvÞifor every u;v2V. We next show that T* is linear. For any u;vi2V, and any a;b2K, hu;T*ðav1þbv2Þi¼h TðuÞ;av1þbv2i¼ /C22ahTðuÞ;v1iþ /C22bhTðuÞ;v2i ¼/C22ahu;T*ðv1Þiþ /C22bhu;T*ðv2Þi¼h u;aT*ðv1ÞþbT*ðv2Þi But this is true for every u2V; hence, T*ðav1þbv2Þ¼aT*ðv1ÞþbT*ðv2Þ. Thus, T* is linear. (b) The matrices A¼½aij/C138andB¼½bij/C138that represent TandT*, respectively, relative to the orthonormal basis Sare given by aij¼hTðujÞ;uiiandbij¼hT*ðujÞ;uii(Problem 13.67). Hence, bij¼hT*ðujÞ;uii¼hui;T*ðujÞi¼hTðuiÞ;uji¼aji Thus, B¼A*, as claimed. 13.5. Prove Theorem 13.2: (i)ðT1þT2Þ*¼T1*þT2*, (iii)ðT1T2Þ*¼T2*T1*, (ii)ðkTÞ*¼/C22kT*, (iv) ðT*Þ*¼T. (i) For any u;v2V, hðT1þT2ÞðuÞ;vi¼h T1ðuÞþT2ðuÞ;vi¼h T1ðuÞ;viþh T2ðuÞ;vi ¼hu;T1*ðvÞiþh u;T2*ðvÞi¼h u;T1*ðvÞþT2*ðvÞi ¼hu;ðT1*þT2*ÞðvÞi The uniqueness of the adjoint implies ðT1þT2Þ*¼T1*þT2*. (ii) For any u;v2V, hðkTÞðuÞ;vi¼h kTðuÞ;vi¼khTðuÞ;vi¼khu;T*ðvÞi¼h u;/C22kT*ðvÞi¼h u;ð/C22kT*ÞðvÞi The uniqueness of the adjoint implies ðkTÞ*¼/C22kT*. (iii) For any u;v2V, hðT1T2ÞðuÞ;vi¼h T1ðT2ðuÞÞ;vi¼h T2ðuÞ;T1*ðvÞi ¼hu;T2*ðT1*ðvÞÞi¼h u;ðT2*T1*ÞðvÞi The uniqueness of the adjoint implies ðT1T2Þ*¼T2*T1*. (iv) For any u;v2V, hT*ðuÞ;vi¼hv;T*ðuÞi¼hTðvÞ;ui¼h u;TðvÞi The uniqueness of the adjoint implies ðT*Þ*¼T.CHAPTER 13 Linear Operators on Inner Product Spaces 385 13.6. Show thatðaÞI*¼I, andðbÞ0*¼0. (a) For every u;v2V,hIðuÞ;vi¼h u;vi¼h u;IðvÞi; hence, I*¼I. (b) For every u;v2V,h0ðuÞ;vi¼h 0;vi¼0¼hu;0i¼h u;0ðvÞi; hence, 0*¼0. 13.7. Suppose Tis invertible. Show that ðT/C01Þ*¼ðT*Þ/C01. I¼I*¼ðTT/C01Þ*¼ðT/C01Þ*T*;hence ;ðT/C01Þ*¼ðT*Þ/C01: 13.8. LetTbe a linear operator on V, and let Wbe a T-invariant subspace of V. Show that W?is invariant under T*. Letu2W?.I f w2W, then TðwÞ2Wand sohw;T*ðuÞi¼h TðwÞ;ui¼0. Thus, T*ðuÞ2W? because it is orthogonal to every w2W. Hence, W?is invariant under T*. 13.9. LetTbe a linear operator on V. Show that each of the following conditions implies T¼0: (i)hTðuÞ;vi¼0 for every u;v2V. (ii) Vis a complex space, and hTðuÞ;ui¼0 for every u2V. (iii) Tis self-adjoint and hTðuÞ;ui¼0 for every u2V. Give an example of an operator Ton a real space Vfor whichhTðuÞ;ui¼0 for every u2VbutT6¼0. [Thus, (ii) need not hold for a real space V.] (i) Set v¼TðuÞ. ThenhTðuÞ;TðuÞi¼ 0, and hence, TðuÞ¼0, for every u2V. Accordingly, T¼0. (ii) By hypothesis, hTðvþwÞ;vþwi¼0 for any v;w2V. Expanding and setting hTðvÞ;vi¼0 and hTðwÞ;wi¼0, we ﬁnd hTðvÞ;wiþh TðwÞ;vi¼0 ð1Þ Note wis arbitrary in (1). Substituting iwforw, and usinghTðvÞ;iwi¼ /C22ihTðvÞ;wi¼/C0 ihTðvÞ;wiand hTðiwÞ;vi¼h iTðwÞ;vi¼ihTðwÞ;vi, we ﬁnd /C0ihTðvÞ;wiþihTðwÞ;vi¼0 Dividing through by iand adding to (1), we obtain hTðwÞ;vi¼0 for any v;w;2V. By (i), T¼0. (iii) By (ii), the result holds for the complex case; hence we need only consider the real case. Expanding hTðvþwÞ;vþwi¼0, we again obtain (1). Because Tis self-adjoint and as it is a real space, we havehTðwÞ;vi¼h w;TðvÞi¼h TðvÞ;wi. Substituting this into (1), we obtain hTðvÞ;wi¼0 for any v;w2V. By (i), T¼0. For an example, consider the linear operator TonR2deﬁned by Tðx;yÞ¼ð y;/C0xÞ. Then hTðuÞ;ui¼0 for every u2V, but T6¼0. Orthogonal and Unitary Operators and Matrices 13.10. Prove Theorem 13.6: The following conditions on an operator Uare equivalent: (i) U*¼U/C01; that is, Uis unitary. (ii)hUðvÞ;UðwÞi¼h u;wi. (iii)kUðvÞk¼k vk. Suppose (i) holds. Then, for every v;w;2V, hUðvÞ;UðwÞi¼h v;U*UðwÞi¼h v;IðwÞi¼h v;wi Thus, (i) implies (ii). Now if (ii) holds, then kUðvÞk¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hUðvÞ;UðvÞip ¼ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ hv;vip ¼kvk Hence, (ii) implies (iii). It remains to show that (iii) implies (i). Suppose (iii) holds. Then for every v2V, hU*UðvÞi¼h UðvÞ;UðvÞi¼h v;vi¼h IðvÞ;vi Hence,hðU*U/C0IÞðvÞ;vi¼0 for every v2V. But U*U/C0Iis self-adjoint (Prove!); then, by Problem 13.9, we have U*U/C0I¼0 and so U*U¼I. Thus, U*¼U/C01, as claimed.386 CHAPTER 13 Linear Operators on Inner Product Spaces 13.11. LetUbe a unitary (orthogonal) operator on V, and let Wbe a subspace invariant under U. Show thatW?is also invariant under U. Because Uis nonsingular, UðWÞ¼W; that is, for any w2W, there exists w02Wsuch that Uðw0Þ¼w. Now let v2W?. Then, for any w2W, hUðvÞ;wi¼h UðvÞ;Uðw0Þi¼h v;w0i¼0 Thus, UðvÞbelongs to W?. Therefore, W?is invariant under U. 13.12. Prove Theorem 13.9: The change-of-basis matrix from an orthonormal basis fu1;...;unginto another orthonormal basis is unitary (orthogonal). Conversely, if P¼½aij/C138is a unitary (ortho- gonal) matrix, then the vectors ui0¼P jajiujform an orthonormal basis. Supposefvigis another orthonormal basis and suppose vi¼bi1u1þbi2u2þ/C1/C1/C1þ binun;i¼1;...;n ð1Þ Becausefvigis orthonormal, dij¼hvi;vji¼bi1bj1þbi2bj2þ/C1/C1/C1þ binbjn ð2Þ LetB¼½bij/C138be the matrix of coefﬁcients in (1). (Then BTis the change-of-basis matrix from fuigto fvig.) Then BB*¼½cij/C138, where cij¼bi1bj1þbi2bj2þ/C1/C1/C1þ binbjn. By (2), cij¼dij, and therefore BB*¼I. Accordingly, B, and hence, BT, is unitary. It remains to prove that fu0 igis orthonormal. By Problem 13.67, hu0 i;u0 ji¼a1ia1jþa2ia2jþ/C1/C1/C1þ anianj¼hCi;Cji where Cidenotes the ith column of the unitary (orthogonal) matrix P¼½aij/C138:Because Pis unitary (orthogonal), its columns are orthonormal; hence, hu0 i;u0 ji¼h Ci;Cji¼dij. Thus,fu0 igis an orthonormal basis. Symmetric Operators and Canonical Forms in Euclidean Spaces 13.13. LetTbe a symmetric operator. Show that (a) The characteristic polynomial DðtÞofTis a product of linear polynomials (over R); (b) Thas a nonzero eigenvector. (a) Let Abe a matrix representing Trelative to an orthonormal basis of V; then A¼AT. LetDðtÞbe the characteristic polynomial of A. Viewing Aas a complex self-adjoint operator, Ahas only real eigenvalues by Theorem 13.4. Thus, DðtÞ¼ð t/C0l1Þðt/C0l2Þ/C1/C1/C1ð t/C0lnÞ where the liare all real. In other words, DðtÞis a product of linear polynomials over R. (b) By (a), Thas at least one (real) eigenvalue. Hence, Thas a nonzero eigenvector. 13.14. Prove Theorem 13.11: Let Tbe a symmetric operator on a real n-dimensional inner product space V. Then there exists an orthonormal basis of Vconsisting of eigenvectors of T. (Hence, T can be represented by a diagonal matrix relative to an orthonormal basis.) The proof is by induction on the dimension of V. If dim V¼1, the theorem trivially holds. Now suppose dim V¼n>1. By Problem 13.13, there exists a nonzero eigenvector v1ofT. Let Wbe the space spanned by v1, and let u1be a unit vector in W, e.g., let u1¼v1=kv1k. Because v1is an eigenvector of T, the subspace WofVis invariant under T. By Problem 13.8, W?is invariant under T*¼T. Thus, the restriction ^TofTtoW?is a symmetric operator. By Theorem 7.4, V¼W/C8W?. Hence, dim W?¼n/C01, because dim W¼1. By induction, there exists an orthonormal basisfu2;...;ungofW?consisting of eigenvectors of ^Tand hence of T. Buthu1;uii¼0 for i¼2;...;n because ui2W?. Accordinglyfu1;u2;...;ungis an orthonormal set and consists of eigenvectors of T. Thus, the theorem is proved.CHAPTER 13 Linear Operators on Inner Product Spaces 387 13.15. Letqðx;yÞ¼3x2/C06xyþ11y2. Find an orthonormal change of coordinates (linear substitution) that diagonalizes the quadratic form q. Find the symmetric matrix Arepresenting qand its characteristic polynomial DðtÞ. We have A¼3/C03 /C031 1/C20/C21 and DðtÞ¼t2/C0trðAÞtþjAj¼t2/C014tþ24¼ðt/C02Þðt/C012Þ The eigenvalues are l¼2 and l¼12. Hence, a diagonal form of qis qðs;tÞ¼2s2þ12t2 (where we use sandtas new variables). The corresponding orthogonal change of coordinates is obtained by ﬁnding an orthogonal set of eigenvectors of A. Subtract l¼2 down the diagonal of Ato obtain the matrix M¼1/C03 /C039/C20/C21 corresponding tox/C03y¼0 /C03xþ9y¼0or x/C03y¼0 A nonzero solution is u1¼ð3;1Þ. Next subtract l¼12 down the diagonal of Ato obtain the matrix M¼/C09/C03 /C03/C01/C20/C21 corresponding to/C09x/C03y¼0 /C03x/C0y¼0or/C03x/C0y¼0 A nonzero solution is u2¼ð/C0 1;3Þ. Normalize u1andu2to obtain the orthonormal basis ^u1¼ð3=ﬃﬃﬃﬃﬃ 10p ;1=ﬃﬃﬃﬃﬃ 10p Þ; ^u2¼ð/C0 1=ﬃﬃﬃﬃﬃ 10p ;3=ﬃﬃﬃﬃﬃ 10p Þ Now let Pbe the matrix whose columns are ^u1and ^u2. Then P¼3=ﬃﬃﬃﬃﬃ 10p /C01=ﬃﬃﬃﬃﬃ 10p 1=ﬃﬃﬃﬃﬃ 10p 3=ﬃﬃﬃﬃﬃ 10p"# and D¼P/C01AP¼PTAP¼20 01 2/C20/C21 Thus, the required orthogonal change of coordinates is x y/C20/C21 ¼Ps t/C20/C21 or x¼3s/C0tﬃﬃﬃﬃﬃ 10p ; y¼sþ3tﬃﬃﬃﬃﬃ 10p One can also express sandtin terms of xandyby using P/C01¼PT; that is, s¼3xþyﬃﬃﬃﬃﬃ 10p ; t¼/C0xþ3yﬃﬃﬃﬃﬃ 10p 13.16. Prove Theorem 13.12: Let Tbe an orthogonal operator on a real inner product space V. Then there exists an orthonormal basis of Vin which Tis represented by a block diagonal matrix Mof the form M¼diag 1 ;...;1;/C01;...;/C01;cosy1/C0siny1 siny1 cosy1/C20/C21 ;...;cosyr/C0sinyr sinyr cosyr/C20/C21 /C18/C19 LetS¼TþT/C01¼TþT*. Then S*¼ðTþT*Þ*¼T*þT¼S. Thus, Sis a symmetric operator onV. By Theorem 13.11, there exists an orthonormal basis of Vconsisting of eigenvectors of S.I f l1;...;lmdenote the distinct eigenvalues of S, then Vcan be decomposed into the direct sum V¼V1/C8V2/C8/C1/C1/C1/C8 Vmwhere the Viconsists of the eigenvectors of Sbelonging to li. We claim that each Viis invariant under T. For suppose v2V; then SðvÞ¼livand SðTðvÞÞ¼ð TþT/C01ÞTðvÞ¼TðTþT/C01ÞðvÞ¼TSðvÞ¼TðlivÞ¼liTðvÞ That is, TðvÞ2Vi. Hence, Viis invariant under T. Because the Viare orthogonal to each other, we can restrict our investigation to the way that Tacts on each individual Vi. On a given Vi;we haveðTþT/C01Þv¼SðvÞ¼liv. Multiplying by T, we get ðT2/C0liTþIÞðvÞ¼0 ð1Þ388 CHAPTER 13 Linear Operators on Inner Product Spaces We consider the cases li¼/C62 and li6¼/C62 separately. If li¼/C62, thenðT/C6IÞ2ðvÞ¼0, which leads to ðT/C6IÞðvÞ¼0o r TðvÞ¼/C6 v. Thus, Trestricted to this Viis either Ior/C0I. Ifli6¼/C62, then Thas no eigenvectors in Vi, because, by Theorem 13.4, the only eigenvalues of Tare 1o r/C01. Accordingly, for v6¼0, the vectors vandTðvÞare linearly independent. Let Wbe the subspace spanned by vandTðvÞ. Then Wis invariant under T, because using (1) we get TðTðvÞÞ¼ T2ðvÞ¼liTðvÞ/C0v2W By Theorem 7.4, Vi¼W/C8W?. Furthermore, by Problem 13.8, W?is also invariant under T. Thus, we can decompose Viinto the direct sum of two-dimensional subspaces Wjwhere the Wjare orthogonal to each other and each Wjis invariant under T. Thus, we can restrict our investigation to the way in which T acts on each individual Wj. Because T2/C0liTþI¼0, the characteristic polynomial DðtÞofTacting on Wjis DðtÞ¼t2/C0litþ1. Thus, the determinant of Tis 1, the constant term in DðtÞ. By Theorem 2.7, the matrix Arepresenting Tacting on Wjrelative to any orthogonal basis of Wjmust be of the form cosy/C0siny siny cosy/C20/C21 The union of the bases of the Wjgives an orthonormal basis of Vi, and the union of the bases of the Vigives an orthonormal basis of Vin which the matrix representing Tis of the desired form. Normal Operators and Canonical Forms in Unitary Spaces 13.17. Determine which of the following matrices is normal: (a) A¼1i 01/C20/C21 , (b) B¼1 i 12þi/C20/C21 (a) AA*¼1i 01/C20/C21 10 /C0i1/C20/C21 ¼2i /C0i1/C20/C21 , A*A¼10 /C0i1/C20/C21 1i 01/C20/C21 ¼1i /C0i2/C20/C21 Because AA*6¼A*A, the matrix Ais not normal. (b) BB*1 i 12þi/C20/C21 11 /C0i2/C0i/C20/C21 ¼22þ2i 2/C02i 6/C20/C21 ¼11 /C0i2/C0i/C20/C21 1 i 12þi/C20/C21 ¼B*B Because BB*¼B*B, the matrix Bis normal. 13.18. LetTbe a normal operator. Prove the following: (a)TðvÞ¼0 if and only if T*ðvÞ¼0. (b) T/C0lIis normal. (c) If TðvÞ¼lv, then T*ðvÞ¼ /C22lv; hence, any eigenvector of Tis also an eigenvector of T*. (d) If TðvÞ¼l1vandTðwÞ¼l2wwhere l16¼l2, thenhv;wi¼0; that is, eigenvectors of T belonging to distinct eigenvalues are orthogonal. (a) We show that hTðvÞ;TðvÞi¼h T*ðvÞ;T*ðvÞi: hTðvÞ;TðvÞi¼h v;T*TðvÞi¼h v;TT*ðvÞi¼h T*ðvÞ;T*ðvÞi Hence, by½I3/C138in the deﬁnition of the inner product in Section 7.2, TðvÞ¼0 if and only if T*ðvÞ¼0. (b) We show that T/C0lIcommutes with its adjoint: ðT/C0lIÞðT/C0lIÞ*¼ðT/C0lIÞðT*/C0/C22lIÞ¼TT*/C0lT*/C0/C22lTþl/C22lI ¼T*T/C0/C22lT/C0lT*þ/C22llI¼ðT*/C0/C22lIÞðT/C0lIÞ ¼ðT/C0lIÞ*ðT/C0lIÞ Thus, T/C0lIis normal.CHAPTER 13 Linear Operators on Inner Product Spaces 389 (c) If TðvÞ¼lv, thenðT/C0lIÞðvÞ¼0. Now T/C0lIis normal by (b); therefore, by (a), ðT/C0lIÞ*ðvÞ¼0. That is,ðT*/C0lIÞðvÞ¼0; hence, T*ðvÞ¼ /C22lv. (d) We show that l1hv;wi¼l2hv;wi: l1hv;wi¼h l1v;wi¼h TðvÞ;wi¼h v;T*ðwÞi¼h v;/C22l2wi¼l2hv;wi Butl16¼l2; hence,hv;wi¼0. 13.19. Prove Theorem 13.13: Let Tbe a normal operator on a complex ﬁnite-dimensional inner product space V. Then there exists an orthonormal basis of Vconsisting of eigenvectors of T. (Thus, T can be represented by a diagonal matrix relative to an orthonormal basis.) The proof is by induction on the dimension of V. If dim V¼1, then the theorem trivially holds. Now suppose dim V¼n>1. Because Vis a complex vector space, Thas at least one eigenvalue and hence a nonzero eigenvector v. Let Wbe the subspace of Vspanned by v, and let u1be a unit vector in W. Because vis an eigenvector of T, the subspace Wis invariant under T. However, vis also an eigenvector of T* by Problem 13.18; hence, Wis also invariant under T*. By Problem 13.8, W?is invariant under T**¼T. The remainder of the proof is identical with the latter part of the proof of Theorem 13.11 (Problem 13.14). 13.20. Prove Theorem 13.14: Let Tbe any operator on a complex ﬁnite-dimensional inner product space V. Then Tcan be represented by a triangular matrix relative to an orthonormal basis of V. The proof is by induction on the dimension of V. If dim V¼1, then the theorem trivially holds. Now suppose dim V¼n>1. Because Vis a complex vector space, Thas at least one eigenvalue and hence at least one nonzero eigenvector v. Let Wbe the subspace of Vspanned by v, and let u1be a unit vector in W. Then u1is an eigenvector of Tand, say, Tðu1Þ¼a11u1. By Theorem 7.4, V¼W/C8W?. Let Edenote the orthogonal projection VintoW?. Clearly W?is invariant under the operator ET. By induction, there exists an orthonormal basis fu2;...;ungofW?such that, for i¼2;...;n, ETðuiÞ¼ai2u2þi3u3þ/C1/C1/C1þ aiiui (Note thatfu1;u2;...;ungis an orthonormal basis of V.) But Eis the orthogonal projection of Vonto W?; hence, we must have TðuiÞ¼ai1u1þai2u2þ/C1/C1/C1þ aiiui fori¼2;...;n. This with Tðu1Þ¼a11u1gives us the desired result. Miscellaneous Problems 13.21. Prove Theorem 13.10B: The following are equivalent: (i) P¼T2for some self-adjoint operator T. (ii) P¼S*Sfor some operator S; that is, Pis positive. (iii) Pis self-adjoint and hPðuÞ;ui/C210 for every u2V. Suppose (i) holds; that is, P¼T2where T¼T*. Then P¼TT¼T*T, and so (i) implies (ii). Now suppose (ii) holds. Then P*¼ðS*SÞ*¼S*S**¼S*S¼P, and so Pis self-adjoint. Furthermore, hPðuÞ;ui¼h S*SðuÞ;ui¼h SðuÞ;SðuÞi/C21 0 Thus, (ii) implies (iii), and so it remains to prove that (iii) implies (i). Now suppose (iii) holds. Because Pis self-adjoint, there exists an orthonormal basis fu1;...;ungofV consisting of eigenvectors of P; say, PðuiÞ¼liui. By Theorem 13.4, the liare real. Using (iii), we show that the liare nonnegative. We have, for each i, 0/C20hPðuiÞ;uii¼h liui;uii¼lihui;uii Thus,hui;uii/C210 forces li/C210;as claimed. Accordingly,ﬃﬃﬃﬃ lip is a real number. Let Tbe the linear operator deﬁned by TðuiÞ¼ﬃﬃﬃﬃ lip uifori¼1;...;n390 CHAPTER 13 Linear Operators on Inner Product Spaces Because Tis represented by a real diagonal matrix relative to the orthonormal basis fuig,Tis self-adjoint. Moreover, for each i, T2ðuiÞ¼Tðﬃﬃﬃﬃ lip uiÞ¼ﬃﬃﬃﬃ lip TðiiÞ¼ﬃﬃﬃﬃ lip ﬃﬃﬃﬃ lip ui¼liui¼PðuiÞ Because T2andPagree on a basis of V;P¼T2. Thus, the theorem is proved. Remark: The above operator Tis the unique positive operator such that P¼T2; it is called the positive square root ofP. 13.22. Show that any operator Tis the sum of a self-adjoint operator and a skew-adjoint operator. SetS¼1 2ðTþT*ÞandU¼1 2ðT/C0T*Þ:Then T¼SþU;where S*¼½1 2ðTþT*Þ/C138*¼1 2ðT*þT**Þ¼1 2ðT*þTÞ¼S U*¼½1 2ðT/C0T*Þ/C138*¼1 2ðT*/C0TÞ¼/C01 2ðT/C0T*Þ¼/C0 Uand that is, Sis self-adjoint and Uis skew-adjoint. 13.23. Prove: Let Tbe an arbitrary linear operator on a ﬁnite-dimensional inner product space V. Then Tis a product of a unitary (orthogonal) operator Uand a unique positive operator P; that is, T¼UP. Furthermore, if Tis invertible, then Uis also uniquely determined. By Theorem 13.10, T*Tis a positive operator; hence, there exists a (unique) positive operator Psuch thatP2¼T*T(Problem 13.43). Observe that kPðvÞk2¼hPðvÞ;PðvÞi¼h P2ðvÞ;vi¼h T*TðvÞ;vi¼h TðvÞ;TðvÞi¼k TðvÞk2ð1Þ We now consider separately the cases when Tis invertible and noninvertible. IfTis invertible, then we set ^U¼PT/C01. We show that ^Uis unitary: ^U*¼ðPT/C01Þ*¼T/C01/C3P*¼ðT*Þ/C01P and ^U/C3^U¼ðT*Þ/C01PPT/C01¼ðT*Þ/C01T*TT/C01¼I Thus, ^Uis unitary. We next set U¼^U/C01. Then Uis also unitary, and T¼UPas required. To prove uniqueness, we assume T¼U0P0, where U0is unitary and P0is positive. Then T*T¼P0*U0*U0P0¼P0IP0¼P2 0 But the positive square root of T*Tis unique (Problem 13.43); hence, P0¼P. (Note that the invertibility ofTis not used to prove the uniqueness of P.) Now if Tis invertible, then Pis also invertible by (1). Multiplying U0P¼UPon the right by P/C01yields U0¼U. Thus, Uis also unique when Tis invertible. Now suppose Tis not invertible. Let Wbe the image of P;t h a ti s , W¼ImP.W ed e ﬁ n e U1:W!Vby U1ðwÞ¼TðvÞ; where PðvÞ¼w ð2Þ We must show that U1is well deﬁned; that is, that PðvÞ¼Pðv0Þimplies TðvÞ¼Tðv0Þ. This follows from the fact that Pðv/C0v0Þ¼0 is equivalent tokPðv/C0v0Þk¼ 0, which forceskTðv/C0v0Þk¼ 0 by (1). Thus, U1is well deﬁned. We next deﬁne U2:W!V. Note that, by (1), PandThave the same kernels. Hence, the images of PandThave the same dimension; that is, dim ðImPÞ¼dimW¼dimðImTÞ. Consequently, W?andðImTÞ?also have the same dimension. We let U2be any isomorphism between W?andðImTÞ?. We next set U¼U1/C8U2. [Here Uis deﬁned as follows: If v2Vand v¼wþw0, where w2W, w02W?, then UðvÞ¼U1ðwÞþU2ðw0Þ.] Now Uis linear (Problem 13.69), and, if v2VandPðvÞ¼w, then, by (2), TðvÞ¼U1ðwÞ¼UðwÞ¼UPðvÞ Thus, T¼UP, as required. It remains to show that Uis unitary. Now every vector x2Vcan be written in the form x¼PðvÞþw0, where w02W?.T h e n UðxÞ¼UPðvÞþU2ðw0Þ¼TðvÞþU2ðw0Þ,w h e r ehTðvÞ;U2ðw0Þi¼ 0 by deﬁnitionCHAPTER 13 Linear Operators on Inner Product Spaces 391 ofU2.A l s o ,hTðvÞ;TðvÞi¼h PðvÞ;PðvÞiby (1). Thus, hUðxÞ;UðxÞi¼h TðvÞþU2ðw0Þ;TðvÞþU2ðw0Þi¼h TðvÞ;TðvÞiþh U2ðw0Þ;U2ðw0Þi ¼hPðvÞ;PðvÞiþh w0;w0i¼h PðvÞþw0;PðvÞþw0Þ¼h x;xi [We also used the fact that hPðvÞ;w0i¼0:/C138Thus, Uis unitary, and the theorem is proved. 13.24. LetVbe the vector space of polynomials over Rwith inner product deﬁned by hf;gi¼ð1 0fðtÞgðtÞdt Give an example of a linear functional fonVfor which Theorem 13.3 does not hold—that is, for which there is no polynomial hðtÞsuch that fðfÞ¼h f;hifor every f2V. Letf:V!Rbe deﬁned by fðfÞ¼fð0Þ; that is, fevaluates fðtÞat 0, and hence maps fðtÞinto its constant term. Suppose a polynomial hðtÞexists for which fðfÞ¼fð0Þ¼ð1 0fðtÞhðtÞdt ð1Þ for every polynomial fðtÞ. Observe that fmaps the polynomial tfðtÞinto 0; hence, by (1), ð1 0tfðtÞhðtÞdt¼0 ð2Þ for every polynomial fðtÞ. In particular (2) must hold for fðtÞ¼thðtÞ; that is, ð1 0t2h2ðtÞdt¼0 This integral forces hðtÞto be the zero polynomial; hence, fðfÞ¼h f;hi¼h f;0i¼0 for every polynomial fðtÞ. This contradicts the fact that fis not the zero functional; hence, the polynomial hðtÞ does not exist. SUPPLEMENTARY PROBLEMS Adjoint Operators 13.25. Find the adjoint of: (a) A¼5/C02i3þ7i 4/C06i8þ3i/C20/C21 ; (b) B¼35 i i/C02i/C20/C21 ; (c) C¼11 23/C20/C21 13.26. LetT:R3!R3be deﬁned by Tðx;y;zÞ¼ð xþ2y;3x/C04z;yÞ:Find T*ðx;y;zÞ: 13.27. LetT:C3!C3be deﬁned by Tðx;y;zÞ¼½ ixþð2þ3iÞy;3xþð3/C0iÞz;ð2/C05iÞyþiz/C138: Find T*ðx;y;zÞ: 13.28. For each linear function fonV;ﬁndu2Vsuch that fðvÞ¼h v;uifor every v2V: (a) f:R3!Rdeﬁned by fðx;y;zÞ¼xþ2y/C03z: (b) f:C3!Cdeﬁned by fðx;y;zÞ¼ixþð2þ3iÞyþð1/C02iÞz: 13.29. Suppose Vhas ﬁnite dimension. Prove that the image of T* is the orthogonal complement of the kernel of T; that is, Im T*¼ðKerTÞ?:Hence, rankðTÞ¼rankðT*Þ: 13.30. Show that T*T¼0 implies T¼0:392 CHAPTER 13 Linear Operators on Inner Product Spaces 13.31. LetVbe the vector space of polynomials over Rwith inner product deﬁned by hf;gi¼Ð1 0fðtÞgðtÞdt:Let Dbe the derivative operator on V; that is, DðfÞ¼df=dt:Show that there is no operator D*o n Vsuch that hDðfÞ;gi¼h f;D*ðgÞifor every f;g2V:That is, Dhas no adjoint. Unitary and Orthogonal Operators and Matrices 13.32. Find a unitary (orthogonal) matrix whose ﬁrst row is (a)ð2=ﬃﬃﬃﬃﬃ 13p ;3=ﬃﬃﬃﬃﬃ 13p Þ, (b) a multiple of ð1;1/C0iÞ, (c) a multiple of ð1;/C0i;1/C0iÞ: 13.33. Prove that the products and inverses of orthogonal matrices are orthogonal. (Thus, the orthogonal matrices form a group under multiplication, called the orthogonal group .) 13.34. Prove that the products and inverses of unitary matrices are unitary. (Thus, the unitary matrices form a group under multiplication, called the unitary group .) 13.35. Show that if an orthogonal (unitary) matrix is triangular, then it is diagonal. 13.36. Recall that the complex matrices AandBare unitarily equivalent if there exists a unitary matrix Psuch that B¼P*AP. Show that this relation is an equivalence relation. 13.37. Recall that the real matrices AandBare orthogonally equivalent if there exists an orthogonal matrix Psuch thatB¼PTAP. Show that this relation is an equivalence relation. 13.38. LetWbe a subspace of V. For any v2V, let v¼wþw0, where w2W,w02W?. (Such a sum is unique because V¼W/C8W?.) Let T:V!Vbe deﬁned by TðvÞ¼w/C0w0. Show that Tis self-adjoint unitary operator on V. 13.39. LetVbe an inner product space, and suppose U:V!V(not assumed linear) is surjective (onto) and preserves inner products; that is, hUðvÞ;UðwÞi¼h u;wifor every v;w2V. Prove that Uis linear and hence unitary. Positive and Positive Deﬁnite Operators 13.40. Show that the sum of two positive (positive deﬁnite) operators is positive (positive deﬁnite). 13.41. LetTbe a linear operator on Vand let f:V/C2V!Kbe deﬁned by fðu;vÞ¼h TðuÞ;vi. Show that fis an inner product on Vif and only if Tis positive deﬁnite. 13.42. Suppose Eis an orthogonal projection onto some subspace WofV. Prove that kIþEis positive (positive deﬁnite) if k/C210ðk>0Þ. 13.43. Consider the operator Tdeﬁned by TðuiÞ¼ﬃﬃﬃﬃ lip ui;i¼1;...;n, in the proof of Theorem 13.10A. Show thatTis positive and that it is the only positive operator for which T2¼P. 13.44. Suppose Pis both positive and unitary. Prove that P¼I. 13.45. Determine which of the following matrices are positive (positive deﬁnite): ðiÞ11 11/C20/C21 ;ðiiÞ0i /C0i0/C20/C21 ;ðiiiÞ01 /C010/C20/C21 ;ðivÞ11 01/C20/C21 ;ðvÞ21 12/C20/C21 ;ðviÞ12 21/C20/C21 13.46. Prove that a 2/C22 complex matrix A¼ab cd/C20/C21 is positive if and only if (i) A¼A*, and (ii) a;dand jAj¼ad/C0bcare nonnegative real numbers.CHAPTER 13 Linear Operators on Inner Product Spaces 393 13.47. Prove that a diagonal matrix Ais positive (positive deﬁnite) if and only if every diagonal entry is a nonnegative (positive) real number. Self-adjoint and Symmetric Matrices 13.48. For any operator T, show that TþT* is self-adjoint and T/C0T* is skew-adjoint. 13.49. Suppose Tis self-adjoint. Show that T2ðvÞ¼0 implies TðvÞ¼0. Using this to prove that TnðvÞ¼0 also implies that TðvÞ¼0 for n>0. 13.50. LetVbe a complex inner product space. Suppose hTðvÞ;viis real for every v2V. Show that Tis self- adjoint. 13.51. Suppose T1andT2are self-adjoint. Show that T1T2is self-adjoint if and only if T1andT1commute; that is, T1T2¼T2T1. 13.52. For each of the following symmetric matrices A, ﬁnd an orthogonal matrix Pand a diagonal matrix Dsuch thatPTAPis diagonal: (a) A¼12 2/C02/C20/C21 ;(b) A¼54 4/C01/C20/C21 , (c) A¼73 3/C01/C20/C21 13.53. Find an orthogonal change of coordinates X¼PX0that diagonalizes each of the following quadratic forms and ﬁnd the corresponding diagonal quadratic form qðx0Þ: (a) qðx;yÞ¼2x2/C06xyþ10y2, (b) qðx;yÞ¼x2þ8xy/C05y2 (c) qðx;y;zÞ¼2x2/C04xyþ5y2þ2xz/C04yzþ2z2 Normal Operators and Matrices 13.54. LetA¼2i i2/C20/C21 . Verify that Ais normal. Find a unitary matrix Psuch that P*APis diagonal. Find P*AP. 13.55. Show that a triangular matrix is normal if and only if it is diagonal. 13.56. Prove that if Tis normal on V, thenkTðvÞk¼k T*ðvÞkfor every v2V. Prove that the converse holds in complex inner product spaces. 13.57. Show that self-adjoint, skew-adjoint, and unitary (orthogonal) operators are normal. 13.58. Suppose Tis normal. Prove that (a) Tis self-adjoint if and only if its eigenvalues are real. (b) Tis unitary if and only if its eigenvalues have absolute value 1. (c) Tis positive if and only if its eigenvalues are nonnegative real numbers. 13.59. Show that if Tis normal, then TandT* have the same kernel and the same image. 13.60. Suppose T1andT2are normal and commute. Show that T1þT2andT1T2are also normal. 13.61. Suppose T1is normal and commutes with T2. Show that T1also commutes with T2*. 13.62. Prove the following: Let T1andT2be normal operators on a complex ﬁnite-dimensional vector space V. Then there exists an orthonormal basis of Vconsisting of eigenvectors of both T1andT2. (That is, T1and T2can be simultaneously diagonalized.) Isomorphism Problems for Inner Product Spaces 13.63. LetS¼fu1;...;ungbe an orthonormal basis of an inner product space Vover K. Show that the mapping v7!½v/C138sis an (inner product space) isomorphism between VandKn. (Here½v/C138Sdenotes the coordinate vector of vin the basis S.)394 CHAPTER 13 Linear Operators on Inner Product Spaces 13.64. Show that inner product spaces VandWover Kare isomorphic if and only if VandWhave the same dimension. 13.65. Supposefu1;...;ungandfu0 1;...;u0 ngare orthonormal bases of VandW, respectively. Let T:V!Wbe the linear map deﬁned by TðuiÞ¼u0 ifor each i. Show that Tis an isomorphism. 13.66. LetVbe an inner product space. Recall that each u2Vdetermines a linear functional ^uin the dual space V* by the deﬁnition ^uðvÞ¼h v;uifor every v2V. (See the text immediately preceding Theorem 13.3.) Show that the map u7!^uis linear and nonsingular, and hence an isomorphism from Vonto V*. Miscellaneous Problems 13.67. Supposefu1;...;ungis an orthonormal basis of V:Prove (a)ha1u1þa2u2þ/C1/C1/C1þ anun;b1u1þb2u2þ/C1/C1/C1þ bnuni¼a1/C22b1þa2/C22b2þ.../C22an/C22bn (b) Let A¼½aij/C138be the matrix representing T:V!Vin the basisfuig:Then aij¼hTðuiÞ;uji: 13.68. Show that there exists an orthonormal basis fu1;...;ungofVconsisting of eigenvectors of Tif and only if there exist orthogonal projections E1;...;Erand scalars l1;...;lrsuch that (i) T¼l1E1þ/C1/C1/C1þ lrEr, (ii) E1þ/C1/C1/C1þ Er¼I, (iii) EiEj¼0 for i6¼j 13.69. Suppose V¼U/C8Wand suppose T1:U!VandT2:W!Vare linear. Show that T¼T1/C8T2is also linear. Here Tis deﬁned as follows: If v2Vand v¼uþwwhere u2U,w2W, then TðvÞ¼T1ðuÞþT2ðwÞ ANSWERS TO SUPPLEMENTARY PROBLEMS Notation:½R1;R2; ...; Rn/C138denotes a matrix with rows R1;R2;...;Rn. 13.25. (a)½5þ2i;4þ6i;3/C07i;8/C03i/C138, (b)½3;/C0i;/C05i;2i/C138, (c)½1;2;1;3/C138 13.26. T*ðx;y;zÞ¼ð xþ3y;2xþz;/C04yÞ 13.27. T*ðx;y;zÞ¼½/C0 ixþ3y;ð2/C03iÞxþð2þ5iÞz;ð3þiÞy/C0iz/C138 13.28. (a) u¼ð1;2;/C03Þ, (b) u¼ð/C0 i;2/C03i;1þ2iÞ 13.32. (a)ð1=ﬃﬃﬃﬃﬃ 13p Þ½2;3;3;/C02/C138, (b)ð1=ﬃﬃﬃ 3p Þ½1;1/C0i;1þi;/C01/C138, (c)1 2½1;/C0i;1/C0i;ﬃﬃﬃ 2p i;/C0ﬃﬃﬃ 2p ;0;1;/C0i;/C01þi/C138 13.45. Only (i) and (v) are positive. Only (v) is positive deﬁnite. 13.52. (a and b) P¼ð1=ﬃﬃﬃ 5p Þ½2;/C01;1;2/C138, (c) P¼ð1=ﬃﬃﬃﬃﬃ 10p Þ½3;/C01;1;3/C138 (a) D¼½2;0;0;/C03/C138; (b) D¼½7;0;0;/C03/C138; (c) D¼½8;0;0;/C02/C138 13.53. (a) x¼ð3x0/C0y0Þ=ﬃﬃﬃﬃﬃ 10p ;y¼ðx0þ3y0Þ=ﬃﬃﬃﬃﬃ 10p ; (b) x¼ð2x0/C0y0Þ=ﬃﬃﬃ 5p ;y¼ðx0þ2y0Þ=ﬃﬃﬃ 5p ; (c) x¼x0=ﬃﬃﬃ 3p þy0=ﬃﬃﬃ 2p þz0=ﬃﬃﬃ 6p ;y¼x0=ﬃﬃﬃ 3p /C02z0=ﬃﬃﬃ 6p ;z¼x0=ﬃﬃﬃ 3p /C0y0=ﬃﬃﬃ 2p þz0=ﬃﬃﬃ 6p ; (a) qðx0Þ¼diagð1;11Þ; (b) qðx0Þ¼diagð3;/C07Þ;(c) qðx0Þ¼diagð1;17Þ 13.54. (a) P¼ð1=ﬃﬃﬃ 2p Þ½1;/C01;1;1/C138;P*AP¼diagð2þi;2/C0iÞCHAPTER 13 Linear Operators on Inner Product Spaces 395 Multilinear Products A.1 Introduction The material in this appendix is much more abstract than that which has previously appeared. Accordingly, many of the proofs will be omitted. Also, we motivate the material with the following observation. LetSbe a basis of a vector space V. Theorem 5.2 may be restated as follows. THEOREM 5.2: Letg:S!Vbe the inclusion map of the basis SintoV. Then, for any vector space Uand any mapping f:S!U;there exists a unique linear mapping f/C3:V!Usuch thatf¼f/C3/C1g: Another way to state the fact that f¼f/C3/C1gis that the diagram in Fig. A-1(a) commutes. A.2 Bilinear Mapping and Tensor Products LetU,V,Wbe vector spaces over a ﬁeld K. Consider a map f:V/C2W!U Then fis said to be bilinear if, for each v2V;the map fv:W!Udeﬁned by fvwðÞ¼ fv;wðÞ is linear; and, for each w2W;the map fw:V!Udeﬁned by fwvðÞ¼ fv;wðÞ is linear. That is, fis linear in each of its two variables. Note that fis similar to a bilinear form except that the values of the map fare in a vector space Urather than the ﬁeld K. DEFINITION A.1: LetVandWbe vector spaces over the same ﬁeld K. The tensor product ofVand Wis a vector space Tover Ktogether with a bilinear map g:V/C2W!T; denoted by gv;wðÞ ¼ v/C10w;with the following property: (*) For any vector space UoverKand any bilinear map f:V/C2W!Uthere exists a unique linear map f/C3:T!Usuch that f/C3/C1g¼f: The tensor product ( T, g) [or simply Twhen gis understood] of VandWis denoted by V/C10W;and the element v/C10wis called the tensor ofvandw. Another way to state condition (*) is that the diagram in Fig. A-1(b) commutes. The fact that such a unique linear map f*exists is called the ‘‘Universal Mapping Principle’’ (UMP). As illustrated in Fig. A-1(b), condition (*) also says that any bilinear map f:V/C2W!U‘‘factors through’’ the tensor product T¼V/C10W:The uniqueness in (*) implies that the image of gspans T;t h a ti s ,s p a n v/C10wfgðÞ ¼ T: APPENDIX A Figure A-1 396 THEOREM A.1: (Uniqueness of Tensor Products) Let ( T,g) and T0;g0ðÞ be tensor products of V andW. Then there exists a unique isomorphism h:T!T0such that hg¼g0: Proof . Because Tis a tensor product, and g0:V/C10W!T0is bilinear, there exists a unique linear map h:T!T0such that hg¼g0:Similarly, because T0is a tensor product, and g:V/C10W!T0is bilinear, there exists a unique linear map h0:T0!Tsuch that h0g0¼g:Using hg¼g0, we get h0hg¼g:Also, because Tis a tensor product, and g:V/C10W!Tis bilinear, there exists a unique linear map h/C3:T!T such that h/C3g¼g:But 1Tg¼g:Thus, h0h¼h/C3¼1T. Similarly, hh0¼1T0:Therefore, his an isomorphism from TtoT0: THEOREM A.2: (Existence of Tensor Product) The tensor product T¼V/C10Wof vector spaces V andWover Kexists. Let v1;...;vmfg be a basis of Vand let w1;...;wnfg be a basis of W. Then the mnvectors vi/C10wii¼1;...;m;j¼1;...;n ðÞ form a basis of T. Thus, dim T¼mn¼dimVðÞ dimWðÞ : Outline of Proof . Suppose v1;...;vm/C8/C9 is a basis of V, and suppose w1;...;wnfg is a basis of W. Consider the mnsymbols tijji¼i;...;m;j¼1;...;n/C8/C9 . Let Tbe the vector space generated by the tij. That is, Tconsists of all linear combinations of the tijwith coefﬁcients in K. [See Problem 4.137.] Letv2Vandw2W. Say v¼a1v1þa2v2þ/C1/C1/C1þ amvmand w¼b1w1þb2w2þ/C1/C1/C1þ bmwm Letg:V/C2W!Tbe deﬁned by gv;wðÞ ¼X iX jaibjtij Then gis bilinear. [Proof left to reader.] Now let f:V/C2W!Ube bilinear. Because the tijform a basis of T, Theorem 5.2 (stated above) tells us that there exists a unique linear map f/C3:T!Usuch that f/C3tij/C0/C1 ¼fvi;wj/C0/C1 . Then, for v¼P iaiviand w¼P jbjwj, we have fðv;wÞ¼fX iaivi;X jbjwj ! ¼X iX jaibjfvi;wj/C0/C1 ¼X iX jaibjtij¼f/C3gv;wðÞðÞ : Therefore, f¼f/C3gwhere f* is the required map in Deﬁnition A.1. Thus, Tis a tensor product. Letfv0 1;...;v0 mgbe any basis of Vandfw0 1;...;w0 mgbe any basis of W. Letv2Vandw2Wand say v¼a0 1v01þ/C1/C1/C1þ a0 mv0mand w¼b0 1w01þ/C1/C1/C1þ b0 mw0m Then v/C10w¼gv;wðÞ ¼X iX ja0 ib0igv0 i;w0 iðÞ ¼X iX ja0 ib0jv0i/C10w0 j/C0/C1 Thus, the elements v0 i/C10w0 jspan T. There are mnsuch elements. They cannot be linearly dependent because tij/C8/C9 is a basis of T, and hence, dim T¼mn. Thus, the v0 i/C10w0 jform a basis of T. Next we give two concrete examples of tensor products. EXAMPLE A.1 LetVbe the vector space of polynomials Pr/C01xðÞand let Wbe the vector space of polynomials Ps/C01yðÞ. Thus, the following from bases of VandW, respectively, 1;x;x2;...;xr/C01and 1 ;y;y2;...;ys/C01 In particular, dim V¼rand dim W¼s:LetTbe the vector space of polynomials in variables xandy with basis xiyj/C8/C9 where i¼0;1;...;r/C01;j¼0;1;...;s/C01Appendix A Multilinear Products 397 Then Tis the tensor product V/C10Wunder the mapping xi/C10yj¼xiyi For example, suppose v¼2/C05xþ3x3andw¼7yþ4y2. Then v/C10w¼14yþ8y2/C035xy/C020xy2þ21x3yþ12x3y2 Note, dim T¼rs¼dimVðÞ dimWðÞ : EXAMPLE A.2 LetVbe the vector space of m/C2nmatrices over a ﬁeld Kand let Wbe the vector space of p/C2qmatrices over K. Suppose A¼½a11/C138belongs to V, and Bbelongs to W. Let Tbe the vector space of mp/C2nq matrices over K. Then Tis the tensor product of VandWwhere A/C10Bis the block matrix A/C10B¼aijB/C2/C3 ¼a11Ba12B/C1/C1/C1 a1nB a21Ba22B/C1/C1/C1 a2nB am1Bam2B/C1/C1/C1 amnB2 6643 775 For example, suppose A¼12 34/C20/C21 andB¼123 456/C20/C21 :Then A/C10B¼123246 4568 1 0 1 236948 1 2 12 15 18 16 20 242 66643 7775 Isomorphisms of Tensor Products First we note that tensoring is associative in a cannonical way. Namely, THEOREM A.3: LetU,V,Wbe vector spaces over a ﬁeld K. Then there exists a unique isomorphism U/C10VðÞ /C10 W!U/C10V/C10WðÞ such that, for every u2U;v2V;w2W; u/C10vðÞ /C10 w7!u/C10v/C10wðÞ Accordingly, we may omit parenthesis when tensoring any number of factors. Speciﬁcally, given vectors spaces V1;V2;...;Vmover a ﬁeld K, we may unambiguously form their tensor product V1/C10V2/C10.../C10Vm and, for vectors vjinVj, we may unambiguously form the tensor product v1/C10v2/C10.../C10vm Moreover, given a vector space Vover K, we may unambiguously deﬁne the following tensor product: /C10rV¼V/C10V/C10.../C10VrfactorsðÞ Also, there is a canonical isomorphism /C10rVðÞ /C10 /C10sVðÞ ! /C10rþsV Furthermore, viewing Kas a vector space over itself, we have the canonical isomorphism K/C10V!V where we deﬁne a/C10v¼av:/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1/C1398 Appendix A Multilinear Products A.3 Alternating Multilinear Maps Letf:Vr!Uwhere VandUare vector spaces over K. [Recall Vr¼V/C2V/C2.../C2V,rfactors.] (1) The mapping fis said to be multilinear or r-linear if fv1;...;vrðÞ is linear as a function of each vj when the other vi’s are held ﬁxed. That is, fð...;vjþv0 j;...Þ¼fð...;vj;...Þþfð...;v0 j;...Þ fð...;kvj;...Þ¼kfð...;vj;...Þ where only the jth position changes. (2) The mapping fis said to be alternating if fv1;...;vrðÞ ¼ 0 whenever vi¼vjwith i6¼j One can easily show (Prove!) that if fis an alternating multilinear mapping on Vr, then f...;vi;...;vj;.../C0/C1 ¼/C0f...;vj;...;vi;.../C0/C1 That is, if two of the vectors are interchanged, then the associated value changes sign. EXAMPLE A.3 (Determinants) The determinant function D:M!Kon the space Mofn/C2nmatrices may be viewed as an n-variable function DAðÞ¼ DR1;R2;...;RnðÞ deﬁned on the rows R1;R2;...;RnofA. Recall (Chapter 8) that, in this context, Dis both n-linear and alternating. We now need some additional notation. Let K¼k1;k2;...;kr½/C138 denote an r-list ( r-tuple) of elements from In¼1;2;...;nðÞ . We will then use the following notation where the vk’s denote vectors and the aik’s denote scalars: vK¼ðvk1;vk2;...;vkrÞandaK¼a1k1a2k2...arkr Note vKis a list of rvectors, and aKis a product of rscalars. Now suppose the elements in K¼k1;k2;...;kr½/C138 are distinct. Then Kis a permutation sKof an r-list J¼i1;i2;...;ir½/C138 instandard form , that is, where i1<i2<...<ir. The number of such standard-form r-lists Jfrom Inis the binomial coefﬁcient: n r/C18/C19 ¼n! r!n/C0rðÞ ! [Recall sign sKðÞ ¼/C0 1ðÞmKwhere mKis the number of interchanges that transforms KintoJ.] Now suppose A¼aij/C2/C3 is an r/C2nmatrix. For a given ordered r-listJ, we deﬁne DJAðÞ¼a1i1a1i2... a1ir a2i1a2i2... a2ir ari1ari2... arir/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12/C12 That is, D J(A) is the determinant of the r/C2rsubmatrix of Awhose column subscripts belong to J. Our main theorem below uses the following ‘‘shufﬂing’’ lemma. LEMMA A.4 LetVandUbe vector spaces over K, and let f:Vr!Ube an alternating r-linear mapping. Let v1;v2;...;vnbe vectors in Vand let A¼aij/C2/C3 be an r/C2nmatrix over K where r/C20n. For i¼1;2;...;r, let ui¼ai1viþai2v2þ/C1/C1/C1þ ainvn......................Appendix A Multilinear Products 399 Then fu1;...;urðÞ ¼X fDJAðÞfðvi1;vi2;...;virÞ where the sum is over all standard-form r-lists J¼i1;i2;...;irfg . The proof is technical but straightforward. The linearity of fgives us the sum fu1;...;urðÞ ¼X KaKfvKðÞ where the sum is over all r-lists Kfrom 1 ;...;nfg . The alternating property of ftells us that fvKðÞ ¼ 0 when Kdoes not contain distinct integers. The proof now mainly uses the fact that as we interchange the vj’s to transform fvKðÞ ¼ fðvk1;vk2;...;vkrÞto fvj/C0/C1 ¼fðvi1;vi2;...;virÞ so that i1</C1/C1/C1<ir, the associated sign of aK, will change in the same way as the sign of the corresponding permutation sKchanges when it is transformed to the identity permutation using transpositions. We illustrate the lemma below for r¼2 and n¼3. EXAMPLE A.4 Suppose f:V2!Uis an alternating multilinear function. Let v1;v2;v32Vand let u;w2V. Suppose u¼a1v1þa2v2þa3v3andw¼b1v1þb2v2þb3v3 Consider fu;wðÞ ¼ fa1v1þa2v2þa3v3;b1v1þb2v2þb3v3 ðÞ Using multilinearity, we get nine terms: fu;wðÞ ¼ a1b1fv1;vrðÞ þ a1b2fv1;v2ðÞ þ a1b3fv1;v3ðÞ þa2b1fv2;v1ðÞ þ a2b2fv2;v2ðÞ þ a2b3fv2;v3ðÞ þa3b1fv3;v1ðÞ þ a3b2fv3;v2ðÞ þ a3b3fv3;v3ðÞ (Note that J¼1;2½/C138 ;J0¼1;3½/C138 andJ00¼2;3½/C138 are the three standard-form 2-lists of I¼1;2;3½/C138 .) The alternating property of ftells us that each fvi;viðÞ ¼ 0;hence, three of the above nine terms are equal to 0. The alternating property also tells us that fvi;vf/C0/C1 ¼/C0fvf;vr/C0/C1 . Thus, three of the terms can be transformed so their subscripts form a standard-form 2-list by a single interchange. Finally we obtain fu;wðÞ ¼ a1b2/C0a2b1ðÞ fv1;v2ðÞ þ a1b3/C0a3b1ðÞ fv1;v3ðÞ þ a2b3/C0a3b2ðÞ fv2;v3ðÞ ¼a1a2 b1b2/C12/C12/C12/C12/C12/C12/C12/C12fv1;v2ðÞ þa1a3 b1b3/C12/C12/C12/C12/C12/C12/C12/C12fv1;v3ðÞ þa2a3 b2b3/C12/C12/C12/C12/C12/C12/C12/C12fv2;v3ðÞ which is the content of Lemma A.4. A.4 Exterior Products The following deﬁnition applies. DEFINITION A.2: LetVbe an n-dimensionmal vector space over a ﬁeld K,a n dl e t rbe an integer such that 1/C20r/C20n.T h e r-fold exterior product (or simply exterior product when ris understood) is a vector space EoverKtogether with an alternating r-linear mapping g:Vr!E, denoted by gv1;...;vrðÞ ¼ v1^...^vr, with the following property: (*) For any vector space Uover Kand any alternating r-linear map f:Vr!U there exists a unique linear map f/C3:E!Usuch that f/C3/C1g¼f.400 Appendix A Multilinear Products Ther-fold tensor product ( E,g) (or simply Ewhen gis understood) of Vis denoted by^rV, and the element v1^/C1/C1/C1^ vris called the exterior product orwedge product of the vi’s. Another way to state condition (*) is that the diagram in Fig. A-1(c) commutes. Again, the fact that such a unique linear map f* exists is called the ‘‘Universal Mapping Principle (UMP)’’. As illustrated in Fig. A-1(c), condition (*) also says that any alternating r-linear map f:Vr!U‘‘factors through’’ the exterior product E¼^rV. Again, the uniqueness in (*) implies that the image of gspans E; that is, span v1^/C1/C1/C1^ vrðÞ ¼ E. THEOREM A.5: (Uniqueness of Exterior Products) Let ( E,g) and E0;g0ðÞ ber-fold exterior products ofV. Then there exists a unique isomorphism h:E!E0such that hg¼g0. The proof is the same as the proof of Theorem A.1, which uses the UMP. THEOREM A.6: (Existence of Exterior Products) Let Vbe an n-dimensional vector space over K. Then the exterior product E¼^rVexists. If r>n, then E¼0fg.I fr/C20n, then dimE¼n r/C18/C19 . Moreover, if v1;...;vn½/C138 is a basis of V, then the vectors vi1^vi2^/C1/C1/C1^ vir; where 1/C20i1<i2</C1/C1/C1<ir/C20n, form a basis of E. We give a concrete example of an exterior product. EXAMPLE A.5 (Cross Product) Consider V¼R3with the usual basis ( i,j,k). Let E¼^2V. Note dim V¼3:Thus, dim E¼3 with basis i^j;i^k;j^k:We identify Ewith R3under the correspondence i¼j^k;j¼k^i¼/C0i^k;k¼i^j Letuandwbe arbitrary vectors in V¼R3, say u¼a1;a2;a3ðÞ ¼ a1iþa2jþa3kandw¼b1;b2;b3ðÞ ¼ b1iþb2jþb3k Then, as in Example A.3, u^w¼a1b2/C0a2b1ðÞ ð i^jÞþ a1b3/C0a3b1ðÞ ð i^kÞþ a2b3/C0a3b2ðÞ ð j^kÞ Using the above identiﬁcation, we get u^w¼a2b3/C0a3b2ðÞ i/C0a1b3/C0a3b1ðÞ jþa1b2/C0a2b1ðÞ k ¼a2a3 b2b3/C12/C12/C12/C12/C12/C12/C12/C12i/C0a 1a3 b1b3/C12/C12/C12/C12/C12/C12/C12/C12jþa 1a2 b1b2/C12/C12/C12/C12/C12/C12/C12/C12k The reader may recognize that the above exterior product is precisely the well-known cross product inR 3. Our last theorem tells us that we are actually able to ‘‘multiply’’ exterior products, which allows us to form an ‘‘exterior algebra’’ that is illustrated below. THEOREM A.7: LetVbe a vector space over K. Let randsbe positive integers. Then there is a unique bilinear mapping ^rV/C2^sV!^rþsV such that, for any vectors ui;wjinV, u1^/C1/C1/C1^ urðÞ /C2 w1^/C1/C1/C1^ wsðÞ7!u1^/C1/C1/C1^ ur^w1^/C1/C1/C1^ wsAppendix A Multilinear Products 401 EXAMPLE A.6 We form an exterior algebra Aover a ﬁeld Kusing noncommuting variables x,y,z. Because it is an exterior algebra, our variables satisfy: x^x¼0;y^y¼0;z^z¼0;and y^x¼/C0x^y;z^x¼/C0x^z;z^y¼/C0y^z Every element of Ais a linear combination of the eight elements 1;x;y;z;x^y;x^z;y^z;x^y^z We multiply two ‘‘polynomials’’ in Ausing the usual distributive law, but now we also use the above conditions. For example, 3þ4y/C05x^yþ6x^z ½/C138 ^ 5x/C02y½/C138 ¼ 15x/C06y/C020x^yþ12x^y^z Observe we use the fact that 4y½/C138 ^ 5x½/C138 ¼ 20y^x¼/C020x^yand 6 x^z½/C138 ^ /C0 2y½/C138 ¼ /C0 12x^z^y¼12x^y^z402 Appendix A Multilinear Products Algebraic Structures B.1 Introduction We deﬁne here algebraic structures that occur in almost all branches of mathematics. In particular, we will deﬁne a ﬁeld that appears in the deﬁnition of a vector space. We begin with the deﬁnition of a group , which is a relatively simple algebraic structure with only one operation and is used as a building block formany other algebraic systems. B.2 Groups LetGbe a nonempty set with a binary operation; that is, to each pair of elements a;b2Gthere is assigned an element ab2G. Then Gis called a group if the following axioms hold: G1½/C138 For any a;b;c2G, we have abðÞc¼ab cðÞ (theassociative law ). G2½/C138 There exists an element e2G, called the identity element, such that ae¼ea¼afor every a2G. G3½/C138 For each a2Gthere exists an element a/C012G, called the inverse ofa, such that aa/C01¼a/C01a¼e. A group Gis said to be abelian (or:commutative ) if the commutative law holds—that is, if ab¼bafor every a;b2G. When the binary operation is denoted by juxtaposition as above, the group Gis said to be written multiplicatively . Sometimes, when Gis abelian, the binary operation is denoted by + and Gis said to be written additively . In such a case, the identity element is denoted by 0 and is called the zero element; the inverse is denoted by /C0aand it is called the negative ofa. IfAandBare subsets of a group G, then we write AB¼abja2A;b2Bfg orAþB¼aþbja2A;b2B fg We also write afor { a}. A subset Hof a group Gis called a subgroup ofGifHforms a group under the operation of G.I fHis a subgroup of Ganda2G, then the set Hais called a right coset ofHand the set aHis called a left coset ofH. DEFINITION: A subgroup HofGis called a normal subgroup if a/C01Ha/C18Hfor every a2G. Equivalently, His normal if aH¼Hafor every a2G—that is, if the right and left cosets of Hcoincide. Note that every subgroup of an abelian group is normal. THEOREM B.1: LetHbe a normal subgroup of G. Then the cosets of HinGform a group under coset multiplication. This group is called the quotient group and is denoted by G/H. APPENDIX B 403 EXAMPLE B.1 The set Zof integers forms an abelian group under addition. (We remark that the even integers form a subgroup of Zbut the odd integers do not.) Let Hdenote the set of multiples of 5; that is, H¼f ...;/C010;/C05;0;5;10;...g. Then His a subgroup (necessarily normal) of Z. The cosets of HinZfollow: /C220¼0þH¼H¼ ...;/C010;/C05;0;5;10;... fg /C221¼1þH¼f ...;/C09;/C04;1;6;11;...g /C222¼2þH¼ ...;/C08;/C03;2;7;12;... fg /C223¼3þH¼ ...;/C07;/C02;3;8;13;... fg /C224¼4þH¼ ...;/C06;/C01;4;9;14;... fg For any other integer n2Z,/C22n¼nþHcoincides with one of the above cosets. Thus, by the above theorem, Z=H¼ /C220;/C221;/C222;/C223;/C224fg forms a group under coset addition; its addition table follows: þ /C220 /C221 /C222 /C223 /C224 /C220 /C220 /C221 /C222 /C223 /C224 /C221 /C221 /C222 /C223 /C224 /C220 /C222 /C222 /C223 /C224 /C220 /C221 /C223 /C223 /C224 /C220 /C221 /C222 /C224 /C224 /C220 /C221 /C222 /C223 This quotient group Z/His referred to as the integers modulo 5 and is frequently denoted by Z5. Analogeusly, for any positive integer n, there exists the quotient group Zncalled the integers modulo n. EXAMPLE B.2 The permutations of nsymbols (see page 267) form a group under composition of mappings; it is called the symmetric group of degree nand is denoted by Sn. We investigate S3here; its elements are E¼123 123/C18/C19 s2¼123 321/C18/C19 f1¼123 231/C18/C19 s1¼123 132/C18/C19 s3¼123 213/C18/C19 f2¼123 312/C18/C19 Here123 ijk/C18/C19 is the permutation that maps 1 7!i;27!j;37!k. The multiplication table of S3is Es1s2s3f1f2 EEs1s2s3f1f2 s1s1Ef1f2s2s3 s2s2f2Ef1f3s1 s3s3f1f2Es1s2 f1f1s3s1s2f2E f2f2s2s3s1Ef1 (The element in the ath row and bth column is ab.) The set H¼E;s1fg is a subgroup of S3; its right and left cosets are Right Cosets Left Cosets H¼E;s1fg H¼E;s1fg Hf1¼f1;s2fg f2H¼f1;s3fg Hf2¼f2;s3fg f2H¼f2;s2fg Observe that the right cosets and the left cosets are distinct; hence, His not a normal subgroup of S3. A mapping ffrom a group Ginto a group G0is called a homomorphism iffa bðÞ ¼ faðÞfbðÞ. For every a;b2G. (Iffis also bijective, i.e., one-to-one and onto, then fis called an isomorphism andGandG0are404 Appendix B Algebraic Structures said to be isomorphic .) Iff:G!G0is a homomorphism, then the kernel of fis the set of elements of G that map into the identity element e02G0: kernel of f¼a2GjfaðÞ¼ e0fg (As usual, f(G) is called the image of the mapping f:G!G0.) The following theorem applies. THEOREM B.2: Letf:G!Gbe a homomorphism with kernel K. Then Kis a normal subgroup of G, and the quotient group G/Kis isomorphic to the image of f. EXAMPLE B.3 LetGbe the group of real numbers under addition, and let G0be the group of positive real numbers under multiplication. The mapping f:G!G0deﬁned by faðÞ¼ 2ais a homomorphism because faþbðÞ ¼ 2aþb¼2a2b¼faðÞfbðÞ In particular, fis bijective, hence, GandG0are isomorphic. EXAMPLE B.4 LetGbe the group of nonzero complex numbers under multiplication, and let G0be the group of nonzero real numbers under multiplication. The mapping f:G!G0deﬁned by fzðÞ¼jzjis a homomorphism because fz1z2ðÞ ¼ j z1z2j¼jz1jjz2j¼fz1ðÞfz2ðÞ The kernel Koffconsists of those complex numbers zon the unit circle—that is, for which jzj¼1. Thus, G=Kis isomorphic to the image of f—that is, to the group of positive real numbers under multiplication. B.3 Rings, Integral Domains, and Fields LetRbe a nonempty set with two binary operations, an operation of addition (denoted by +) and an operation of multiplication (denoted by juxtaposition). Then Ris called a ringif the following axioms are satisﬁed: R1½/C138For any a;b;c2R, we have aþbðÞ þ c¼aþbþcðÞ . R2½/C138There exists an element 0 2R;called the zero element, such that aþ0¼0þa¼afor every a2R: R3½/C138For each a2Rthere exists an element /C0a2R, called the negative ofa, such that aþ/C0 aðÞ ¼ /C0 aðÞ þ a¼0. R4½/C138For any a;b2R;we have aþb¼bþa: R5½/C138For any a;b;c2R;we have abðÞc¼ab cðÞ: R6½/C138For any a;b;c2R;we have (i)abþcðÞ ¼ abþac;and (ii) bþcðÞ a¼baþca: Observe that the axioms R1½/C138through R4½/C138may be summarized by saying that Ris an abelian group under addition. Subtraction is deﬁned in Rbya/C0b/C17aþ/C0 bðÞ . It can be shown (see Problem B.25) that a/C10¼0/C1a¼0 for every a2R: Ris called a commutative ring ifab¼bafor every a;b2R:We also say that Ris aring with a unit element if there exists a nonzero element 1 2Rsuch that a/C11¼1/C1a¼afor every a2R: A nonempty subset SofRis called a subring ofRifSforms a ring under the operations of R. We note thatSis a subring of Rif and only if a;b2Simplies a/C0b2Sandab2S. A nonempty subset IofRis called a left ideal inRif (i) a/C0b2Iwhenever a;b2I;and (ii) ra2I whenever r2R;a2I:Note that a left ideal IinRis also a subring of R. Similarly, we can deﬁne a right ideal and a two-sided ideal . Clearly all ideals in commutative rings are two sided. The term ideal shall mean two-sided ideal uniess otherwise speciﬁed.Appendix B Algebraic Structures 405 THEOREM B.3: LetIbe a (two-sided) ideal in a ring R. Then the cosets aþIja2Rfg form a ring under coset addition and coset multiplication. This ring is denoted by R=Iand is called the quotient ring . Now let Rbe a commutative ring with a unit element. For any a2R, the set aðÞ¼ rajr2Rfg is an ideal; it is called the principal ideal generated by a. If every ideal in Ris a principal ideal, then Ris called aprincipal ideal ring . DEFINITION: A commutative ring Rwith a unit element is called an integral domain ifRhas no zero divisors —that is, if ab¼0 implies a¼0o r b¼0. DEFINITION: A commutative ring Rwith a unit element is called a ﬁeld if every nonzero a2Rhas a multiplicative inverse ; that is, there exists an element a/C012Rsuch that aa/C01¼a/C01a¼1: A ﬁeld is necessarily an integral domain; for if ab¼0 and a6¼0;then b¼1/C1b¼a/C01ab¼a/C01/C10¼0 We remark that a ﬁeld may also be viewed as a commutative ring in which the nonzero elements form a group under multiplication. EXAMPLE B.5 The set Zof integers with the usual operations of addition and multiplication is the classical example of an integral domain with a unit element. Every ideal IinZis a principal ideal; that is, I¼nðÞfor some integer n. The quotient ring Zn¼Z=nðÞis called the ring of integers module n .I fnis prime, then Znis a ﬁeld. On the other hand, if nis not prime then Znhas zero divisors. For example, in the ring Z6;/C222/C223¼/C220 and /C2226¼/C220 and /C2236¼/C220: EXAMPLE B.6 The rational numbers Qand the real numbers Reach form a ﬁeld with respect to the usual operations of addition and multiplication. EXAMPLE B.7 LetCdenote the set of ordered pairs of real numbers with addition and multiplication deﬁned by a;bðÞ þ c;dðÞ ¼ aþc;bþdðÞ a;bðÞ /C1 c;dðÞ ¼ ac/C0bd;adþbcðÞ Then Csatisﬁes all the required properties of a ﬁeld. In fact, Cis just the ﬁeld of complex numbers (see page 4). EXAMPLE B.8 The set Mof all 2 62 matrices with real entries forms a noncommutative ring with zero divisors under the operations of matrix addition and matrix multiplication. EXAMPLE B.9 LetRbe any ring. Then the set Rx½/C138of all polynomials over Rforms a ring with respect to the usual operations of addition and multiplication of polynomials. Moreover, if Ris an integral domain then Rx½/C138is also an integral domain. Now let Dbe an integral domain. We say that b divides a inDifa¼bcfor some c2D. An element u2Dis called a unitifudivides 1—that is, if uhas a multiplicative inverse. An element b2Dis called anassociate ofa2Difb¼uafor some unit u2D. A nonunit p2Dis said to be irreducible ifp¼ab implies aorbis a unit. An integral domain Dis called a unique factorization domain if every nonunit a2Dcan be written uniquely (up to associates and order) as a product of irreducible elements. EXAMPLE B.10 The ring Zof integers is the classical example of a unique factorization domain. The units of Z are 1 and/C01. The only associates of n2Zarenand/C0n. The irreducible elements of Zare the prime numbers. EXAMPLE B.11 The set D¼aþbﬃﬃﬃﬃﬃ 13p ja;bintegers/C8/C9 is an integral domain. The units of Dare/C61; 18/C65ﬃﬃﬃﬃﬃ 13p and/C018/C65ﬃﬃﬃﬃﬃ 13p . The elements 2 ;3/C0ﬃﬃﬃﬃﬃ 13p and/C03/C0ﬃﬃﬃﬃﬃ 13p are irreducible in D. Observe that 4¼2/C12¼3/C0ﬃﬃﬃﬃﬃ 13p/C0/C1 /C03/C0ﬃﬃﬃﬃﬃ 13p/C0/C1 :Thus, Dis not a unique factorization domain. (See Problem B.40.)406 Appendix B Algebraic Structures B.4 Modules LetMbe an additive abelian group and let Rbe a ring with a unit element. Then Mis said to be a (left) R- module if there exists a mapping R/C2M!Mthat satisﬁes the following axioms: M1½/C138 rm1þm2ðÞ ¼ rm1þrm2 M2½/C138 rþsðÞ m¼rmþsm M3½/C138 rsðÞm¼rs mðÞ M4½/C138 1/C1m¼m for any r;s2Rand any mi2M. We emphasize that an R-module is a generalization of a vector space where we allow the scalars to come from a ring rather than a ﬁeld. EXAMPLE B.12 LetGbe any additive abelian group. We make Ginto a module over the ring Zof integers by deﬁning ng¼gþgþ/C1/C1/C1þ g;zﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ}|ﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄﬄ{ntimes 0g¼0;/C0nðÞ g¼/C0ng where nis any positive integer. EXAMPLE B.13 LetRbe a ring and let Ibe an ideal in R. Then Imay be viewed as a module over R. EXAMPLE B.14 LetVbe a vector space over a ﬁeld Kand let T:V!Vbe a linear mapping. We make Vinto a module over the ring Kx½/C138of polynomials over Kby deﬁning fxðÞv¼fTðÞ vðÞ:The reader should check that a scalar multiplication has been deﬁned. LetMbe a module over R. An additive subgroup NofMis called a submodule ofMifu2Nand k2Rimply ku2N:(Note that Nis then a module over R.) LetMandM0beR-modules. A mapping T:M!M0is called a homomorphism (or:R-homomorphism orR-linear )i f (i)TuþvðÞ¼TuðÞþTvðÞ and (ii) Tk uðÞ¼kT uðÞ for every u;v2Mand every k2R. PROBLEMS Groups B.1. Determine whether each of the following systems forms a group G: (i)G¼set of integers ;operation subtraction; (ii)G¼f1;/C01g, operation multiplication; (iii) G¼set of nonzero rational numbers, operation division; (iv) G¼set of nonsingular n/C2nmatrices, operation matrix multiplication; (v)G¼faþbi:a;b2Zg, operation addition. B.2. Show that in a group G: (i) the identity element of Gis unique; (ii) each a2Ghas a unique inverse a/C012G; (iii) a/C01ðÞ/C01¼a;and abðÞ/C01¼b/C01a/C01; (iv) ab¼acimplies b¼c, and ba¼caimplies b¼c.Appendix B Algebraic Structures 407 B.3. In a group G, the powers of a2Gare deﬁned by a0¼e;an¼aan/C01;a/C0n¼anðÞ/C01;where n2N Show that the following formulas hold for any integers r;s;t2Z:(i)aras¼arþs;(ii) arðÞs¼ars; (iii) arþsðÞt¼arsþst. B.4. Show that if Gis an abelian group, then abðÞn¼anbnfor any a;b2Gand any integer n2Z: B.5. Suppose Gis a group such that abðÞ2¼a2b2for every a;b2G. Show that Gis abelian. B.6. Suppose His a subset of a group G. Show that His a subgroup of Gif and only if (i) His nonempty, and (ii) a;b2Himplies ab/C012H: B.7. Prove that the intersection of any number of subgroups of Gis also a subgroup of G. B.8. Show that the set of all powers of a2Gis a subgroup of G; it is called the cyclic group generated bya. B.9. A group Gis said to be cyclic ifGis generated by some a2G; that is, G¼an:n2ZðÞ . Show that every subgroup of a cyclic group is cyclic. B.10. Suppose Gis a cyclic subgroup. Show that Gis isomorphic to the set Zof integers under addition or to the set Zn(of the integers module n) under addition. B.11. LetHbe a subgroup of G. Show that the right (left) cosets of Hpartition Ginto mutually disjoint subsets. B.12. The order of a group G, denoted byjGj;is the number of elements of G. Prove Lagrange’s theorem: If His a subgroup of a ﬁnite group G, thenjHjdividesjGj. B.13. SupposejGj¼pwhere pis prime. Show that Gis cyclic. B.14. Suppose HandNare subgroups of Gwith Nnormal. Show that (i) HNis a subgroup of Gand (ii)H\Nis a normal subgroup of G. B.15. LetHbe a subgroup of Gwith only two right (left) cosets. Show that His a normal subgroup of G. B.16. Prove Theorem B.1: Let Hbe a normal subgroup of G. Then the cosets of HinGform a group G=Hunder coset multiplication. B.17. Suppose Gis an abelian group. Show that any factor group G=His also abelian. B.18. Letf:G!G0be a group homomorphism. Show that (i)feðÞ¼e0where eande0are the identity elements of GandG0, respectively; (ii)fa/C01ðÞ ¼ faðÞ/C01for any a2G. B.19. Prove Theorem B.2: Let f:G!G0be a group homomorphism with kernel K. Then Kis a normal subgroup of G, and the quotient group G=Kis isomorphic to the image of f. B.20. LetGbe the multiplicative group of complex numbers zsuch thatjzj¼1;and let Rbe the additive group of real numbers. Prove that Gis isomorphic to R=Z:408 Appendix B Algebraic Structures B.21. For a ﬁxed g2G, let ^g:G!Gbe deﬁned by ^gaðÞ¼ g/C01ag:Show that Gis an isomorphism of Gonto G. B.22. LetGbe the multiplicative group of n/C2nnonsingular matrices over R. Show that the mapping A7!jAjis a homomorphism of Ginto the multiplicative group of nonzero real numbers. B.23. LetGbe an abelian group. For a ﬁxed n2Z;show that the map a7!anis a homomorphism of G intoG. B.24. Suppose HandNare subgroups of Gwith Nnormal. Prove that H\Nis normal in Hand H=H\NðÞ is isomorphic to HN=N. Rings B.25. Show that in a ring R: (i) a/C10¼0/C1a¼0;(ii)a/C0bðÞ ¼ /C0 aðÞ b¼/C0ab, (iii)/C0aðÞ /C0 bðÞ ¼ ab: B.26. Show that in a ring Rwith a unit element: (i) /C01ðÞ a¼/C0a;(ii)/C01ðÞ /C0 1ðÞ ¼ 1. B.27. LetRbe a ring. Suppose a2¼afor every a2R:Prove that Ris a commutative ring. (Such a ring is called a Boolean ring .) B.28. LetRbe a ring with a unit element. We make Rinto another ring ^Rby deﬁning a/C8b¼aþbþ1 anda/C1b¼abþaþb. (i) Verify that ^Ris a ring. (ii) Determine the 0-element and 1-element of ^R. B.29. LetGbe any (additive) abelian group. Deﬁne a multiplication in Gbya/C1b¼0. Show that this makes Ginto a ring. B.30. Prove Theorem B.3: Let Ibe a (two-sided) ideal in a ring R. Then the cosets aþIja2RðÞ form a ring under coset addition and coset multiplication. B.31. LetI1andI2be ideals in R. Prove that I1þI2andI1\I2are also ideals in R. B.32. LetRandR0be rings. A mapping f:R!R0is called a homomorphism (or:ring homomorphism )i f (i) faþbðÞ ¼ faðÞþ fbðÞ and (ii) fa bðÞ ¼ faðÞfbðÞ, for every a;b2R. Prove that if f:R!R0is a homomorphism, then the set K¼r2RjfrðÞ¼ 0fg is an ideal in R. (The set Kis called the kernel off.) Integral Domains and Fields B.33. Prove that in an integral domain D,i fab¼ac;a6¼0;then b¼c. B.34. Prove that F¼aþbﬃﬃﬃ 2p ja;brational/C8/C9 is a ﬁeld. B.35. Prove that D¼aþbﬃﬃﬃ 2p ja;bintegers/C8/C9 is an integral domain but not a ﬁeld. B.36. Prove that a ﬁnite integral domain Dis a ﬁeld. B.37. Show that the only ideals in a ﬁeld Kare 0fgandK. B.38. A complex number aþbiwhere a,bare integers is called a Gaussian integer . Show that the set G of Gaussian integers is an integral domain. Also show that the units in Gare/C61 and/C6i.Appendix B Algebraic Structures 409 B.39. LetDbe an integral domain and let Ibe an ideal in D. Prove that the factor ring D=Iis an integral domain if and only if Iis a prime ideal. (An ideal Iisprime ifab2Iimplies a2Iorb2I:) B.40. Consider the integral domain D¼aþbﬃﬃﬃﬃﬃ 13p ja;bintegers/C8/C9 (see Example B.11). If a¼aþbﬃﬃﬃﬃﬃ 13p , we deﬁne NaðÞ¼ a2/C013b2. Prove: (i) NabðÞ ¼ NaðÞNbðÞ;(ii)ais a unit if and only if NaðÞ¼/C6 1; (iii) the units of Dare/C61;18/C65ﬃﬃﬃﬃﬃ 13p and/C018/C65ﬃﬃﬃﬃﬃ 13p ; (iv) the numbers 2 ;3/C0ﬃﬃﬃﬃﬃ 13p and/C03/C0ﬃﬃﬃﬃﬃ 13p are irreducible. Modules B.41. LetMbe an R-module and let AandBbe submodules of M. Show that AþBandA\Bare also submodules of M. B.42. LetMbe an R-module with submodule N. Show that the cosets uþN:u2Mfg form an R-module under coset addition and scalar multiplication deﬁned by ruþNðÞ¼ruþN. (This module is denoted by M=Nand is called the quotient module .) B.43. LetMandM0beR-modules and let f:M!M0be an R-homomorphism. Show that the set K¼u2M:fuðÞ¼ 0fg is a submodule of f. (The set Kis called the kernel off.) B.44. LetMbe an R-module and let EMðÞ denote the set of all R-homomorphism of Minto itself. Deﬁne the appropriate operations of addition and multiplication in EMðÞ so that EMðÞ becomes a ring.410 Appendix B Algebraic Structures Polynomials over a Field C.1 Introduction We will investigate polynomials over a ﬁeld Kand show that they have many properties that are analogous to properties of the integers. These results play an important role in obtaining canonical forms for a linear operator Ton a vector space Vover K. C.2 Ring of Polynomials LetKbe a ﬁeld. Formally, a polynomial of fover Kis an inﬁnite sequence of elements from Kin which all except a ﬁnite number of them are 0: f¼ ...;0;an;...;a1;a0 ðÞ (We write the sequence so that it extends to the left instead of to the right.) The entry akis called the kth coefﬁcient of f.I fnis the largest integer for which an6¼0, then we say that the degree offisn, written degf¼n We also call antheleading coefﬁcient off, and if an¼1 we call famonic polynomial . On the other hand, if every coefﬁcient of fis 0 then fis called the zero polynomial , written f¼0. The degree of the zero polynomial is not deﬁned. Now if gis another polynomial over K, say g¼ ...;0;bm;...;b1;b0 ðÞ then the sum fþgis the polynomial obtained by adding corresponding coefﬁcients. That is, if m/C20n,t h e n fþg¼ ...;0;an;...;amþbm;...;a1þb1;a0þb0 ðÞ Furthermore, the product fg is the polynomial fg¼ ...;0;anbm;...;a1b0þa0b1;a0b0 ðÞ that is, the kth coefﬁcient ckoffgis ck¼Xk t¼0a1bk/C01¼a0bkþa1bk/C01þ/C1/C1/C1þ akb0 The following theorem applies. THEOREM C.1: The set Pof polynomials over a ﬁeld Kunder the above operations of addition and multiplication forms a commutative ring with a unit element and with no zerodivisors—an integral domain. If fand gare nonzero polynomials in P, then deg fgðÞ ¼ degfðÞ deggðÞ . APPENDIX C 411 Notation We identify the scalar a02Kwith the polynomial a0¼ ...;0;a0ðÞ We also choose a symbol, say t, to denote the polynomial t¼ ...;0;1;0ðÞ We call the symbol tanindeterminant . Multiplying twith itself, we obtain t2¼ ...;0;1;0;0ðÞ ;t3¼ ...;0;1;0;0;0ðÞ ;... Thus, the above polynomial fcan be written uniquely in the usual form f¼antnþ/C1/C1/C1þ astþa0 When the symbol tis selected as the indeterminant, the ring of polynomials over Kis denoted by Kt½/C138 and a polynomial fis frequently denoted by ftðÞ. We also view the ﬁeld Kas a subset of Kt½/C138under the above identiﬁcation. This is possible because the operations of addition and multiplication of elements of Kare preserved under this identiﬁcation: ð...;0;a0Þþð ...;0;b0Þ¼ð ...;0;a0þb0Þ ð...;0;a0Þ/C1ð ...;0;b0Þ¼ð ...;0;a0b0Þ We remark that the nonzero elements of Kare the units of the ring Kt½/C138. We also remark that every nonzero polynomial is an associate of a unique monic polynomial. Hence, if dandd0are monic polynomials for which ddivides d0andd0divides d, then d¼d0. (A polynomial g divides a polynomial fif there is a polynomial hsuch that f¼hg:) C.3 Divisibility The following theorem formalizes the process known as ‘‘long division.’’ THEOREM C.2 (Division Algorithm): Let fandgbe polynomials over a ﬁeld Kwith g6¼0. Then there exist polynomials qandrsuch that f¼qgþr where either r¼0 or deg r<degg. Proof :I ff¼0 or if deg f<degg, then we have the required representation f¼0gþf Now suppose deg f/C21degg, say f¼antnþ/C1/C1/C1þ a1tþa0and g¼bmtmþ/C1/C1/C1þ b1tþb0 where an;bm6¼0 and n/C21m. We form the polynomial f1¼f/C0an bmtn/C0mg ð1Þ Then deg f1<degf. By induction, there exist polynomials q1andrsuch that f1¼q1gþr412 Appendix C Polynomials over a Field where either r¼0 or deg r<degg. Substituting this into (1) and solving for f, f¼q1þan bmtn/C0m/C18/C19 gþr which is the desired representation. THEOREM C.3: The ring Kt½/C138of polynomials over a ﬁeld Kis a principal ideal ring. If Iis an ideal in Kt½/C138, then there exists a unique monic polynomial dthat generates I, such that d divides every polynomial f2I. Proof . Let dbe a polynomial of lowest degree in I. Because we can multiply dby a nonzero scalar and still remain in I, we can assume without loss in generality that dis a monic polynomial. Now suppose f2I. By Theorem C.2 there exist polynomials qandrsuch that f¼qdþrwhere either r¼0 or deg r<degd Now f;d2Iimplies qd2I;and hence, r¼f/C0qd2I. But dis a polynomial of lowest degree in I. Accordingly, r¼0 and f¼qd;that is, ddivides f. It remains to show that dis unique. If d0is another monic polynomial that generates I, then ddivides d0andd0divides d. This implies that d¼d0, because d andd0are monic. Thus, the theorem is proved. THEOREM C.4: Letfandgbe nonzero polynomials in Kt½/C138. Then there exists a unique monic polynomial dsuch that (i)ddivides fandg; and (ii) d0divides fandg, then d0divides d. DEFINITION: The above polynomial dis called the greatest common divisor offandg.I fd¼1, then fandgare said to be relatively prime . Proof of Theorem C.4 . The set I¼mfþngjm;n2Kt½/C138 fg is an ideal. Let dbe the monic polynomial that generates I. Note f;g2I; hence, ddivides fandg. Now suppose d0divides fandg. Let Jbe the ideal generated by d0. Then f;g2J, and hence, I/C26J. Accordingly, d2Jand so d0divides das claimed. It remains to show that dis unique. If d1is another (monic) greatest common divisor of fandg, then d divides d1andd1divides d. This implies that d¼d1because dandd1are monic. Thus, the theorem is proved. COROLLARY C.5: Letdbe the greatest common divisor of the polynomials fandg. Then there exist polynomials mandnsuch that d¼mfþng. In particular, if fandgare relatively prime, then there exist polynomials mandnsuch that mfþng¼1. The corollary follows directly from the fact that dgenerates the ideal I¼mfþngjm;n2Kt½/C138 fg C.4 Factorization A polynomial p2Kt½/C138of positive degree is said to be irreducible if p¼fgimplies forgis a scalar. LEMMA C.6: Suppose p2Kt½/C138is irreducible. If pdivides the product fgof polynomials f;g2Kt½/C138, then pdivides forpdivides g. More generally, if pdivides the product of n polynomials f1f2...fn, then pdivides one of them. Proof . Suppose pdivides fgbut not f. Because pis irreducible, the polynomials fandpmust then be relatively prime. Thus, there exist polynomials m;n2Kt½/C138such that mfþnp¼1. Multiplying thisAppendix C Polynomials over a Field 413 equation by g, we obtain mfgþnpg¼g. But pdivides fgand so mfg, and pdivides npg; hence, pdivides the sum g¼mfgþnpg. Now suppose pdivides f1f2/C1/C1/C1fn:Ifpdivides f1, then we are through. If not, then by the above result p divides the product f2/C1/C1/C1fn:By induction on n,pdivides one of the polynomials f2;...fn:Thus, the lemma is proved. THEOREM C.7: (Unique Factorization Theorem) Let fbe a nonzero polynomial in Kt½/C138:Then fcan be written uniquely (except for order) as a product f¼kp1p2/C1/C1/C1pn where k2Kand the piare monic irreducible polynomials in Kt½/C138: Proof : We prove the existence of such a product ﬁrst. If fis irreducible or if f2K, then such a product clearly exists. On the other hand, suppose f¼ghwhere fandgare nonscalars. Then gandhhave degrees less than that of f. By induction, we can assume g¼k1g1g2/C1/C1/C1grand h¼k2h1h2/C1/C1/C1hs where k1;k22Kand the giandhjare monic irreducible polynomials. Accordingly, f¼k1k2ðÞ g1g2/C1/C1/C1grk1h2/C1/C1/C1hs is our desired representation. We next prove uniqueness (except for order) of such a product for f. Suppose f¼kp1p2/C1/C1/C1pn¼k0q1q2/C1/C1/C1qm where k;k02Kand the p1;...;pn;q1;...;qmare monic irreducible polynomials. Now p1divides k0q1/C1/C1/C1qm:Because p1is irreducible, it must divide one of the qiby the above lemma. Say p1divides q1. Because p1andq1are both irreducible and monic, p1¼q1. Accordingly, kp2/C1/C1/C1pn¼k0q2/C1/C1/C1qm By induction, we have that n¼mandp2¼q2;...;pn¼qmfor some rearrangement of the qi. We also have that k¼k0. Thus, the theorem is proved. If the ﬁeld Kis the complex ﬁeld C, then we have the following result that is known as the fundamental theorem of algebra; its proof lies beyond the scope of this text. THEOREM C.8: (Fundamental Theorem of Algebra) LetftðÞbe a nonzero polynomial over the complex ﬁeld C. Then ftðÞcan be written uniquely (except for order) as a product ftðÞ¼ kt/C0r2ðÞ t/C0r2ðÞ /C1 /C1 /C1 t/C0rnðÞ where k;ri2C—as a product of linear polynomials. In the case of the real ﬁeld Rwe have the following result. THEOREM C.9: LetftðÞbe a nonzero polynomial over the real ﬁeld R.Then ftðÞcan be written uniquely (except for order) as a product ftðÞ¼ kp1tðÞp2tðÞ/C1/C1/C1 pmtðÞ where k2Rand the pitðÞare monic irreducible polynomials of degree one or two.414 Appendix C Polynomials over a Field Odds and Ends D.1 Introduction This appendix discusses various topics, such as equivalence relations, determinants and block matrices, and the generalized MP (Moore–Penrose) inverse. D.2 Relations and Equivalence Relations Abinary relation or simply relation R from a set Ato a set Bassigns to each ordered pair a;bðÞ 2 A/C2B exactly one of the following statements: (i) ‘‘ ais related to b,’’ written aRb , (ii) ‘‘ ais not related to b’’ written aRb . A relation from a set Ato the same set Ais called a relation on A . Observe that any relation Rfrom AtoBuniquely deﬁnes a subset ^RofA/C2Bas follows: ^R¼ a;bðÞ j aRbfg Conversely, any subset ^RofA/C2Bdeﬁnes a relation from AtoBas follows: aRb if and only if a;bðÞ 2 R In view of the above correspondence between relations from AtoBand subsets of A/C2B, we redeﬁne a relation from AtoBas follows: DEFINITION D.1: A relation Rfrom AtoBis a subset of A/C2B. Equivalence Relations Consider a nonempty set S. A relation RonSis called an equivalence relation ifRis reﬂexive, symmetric, and transitive; that is, if Rsatisﬁed the following three axioms: [E1](Reﬂexivity) Every a2Ais related to itself. That is, for every a2A,aRa . [E2](Symmetry) If ais related to b, then bis related to a. That is, if aRb , then bRa . [E3](Transitivity) If ais related to bandbis related to c, then ais related to c. That is, ifaRb andbRc , then aRc : The general idea behind an equivalence relation is that it is a classiﬁcation of objects that are in some way ‘‘alike.’’ Clearly, the relation of equality is an equivalence relation. For this reason, one frequently uses ~ or/C17to denote an equivalence relation. EXAMPLE D.1 (a) In Euclidean geometry, similarity of triangles is an equivalence relation. Speciﬁcally, suppose a;b;gare triangles. Then (i) ais similar to itself. (ii). If ais similar to b, then bis similar to a. (iii) If ais similar to bandb is similar to g, then ais similar to g. APPENDIX D 415 (b) The relation/C18of set inclusion is not an equivalence relation. It is reﬂexive and transitive, but it is not symmetric because A/C18Bdoes not imply B/C18A. Equivalence Relations and Partitions LetSbe a nonempty set. Recall ﬁrst that a partition PofSis a subdivision of Sinto nonempty, nonoverlapping subsets; that is, a collection P¼fAjgof nonempty subsets of Ssuch that (i) Each a2S belong to one of the Aj, (ii) The setsfAjgare mutually disjoint. The subsets in a partition Pare called cells. Thus, each a2Sbelongs to exactly one of the cells. Also, any element b2Ajis called a representative of the cell Aj, and a subset BofSis called a system of representatives ifBcontains exactly one element in each of the cells in fAjg. Now suppose Ris an equivalence relation on the nonempty set S. For each a2S, the equivalence class ofa, denoted by [ a], is the set of elements of Sto which ais related: a½/C138¼ xjaR xfg : The collection of equivalence classes, denoted by S=R, is called the quotient ofSbyR: S=R¼a½/C138ja2Sfg The fundamental property of an equivalence relation and its quotient set is contained in the following theorem: THEOREM D.1: LetRbe an equivalence relation on a nonempty set S. Then the quotient set S=Ris a partition of S. EXAMPLE D.2 Let/C17be the relation on the set Zof integers deﬁned by x/C17ymod 5ðÞ which reads ‘‘ xis congruent to ymodulus 5’’ and which means that the difference x/C0yis divisible by 5. Then/C17is an equivalence relation on Z. Then there are exactly ﬁve equivalence classes in the quotient set Z=/C17as follows: A0¼ ...;/C010;/C05;0;5;10;... fg A1¼ ...;/C09;/C04;1;6;11;... fg A2¼ ...;/C08;/C03;2;7;12;... fg A3¼ ...;/C07;/C02;3;8;13;... fg A4¼ ...;/C06;/C01;4;9;14;... fg Note that any integer x, which can be expressed uniquely in the form x¼5qþrwhere 0/C20r<5, is a member of the equivalence class Arwhere ris the remainder. As expected, the equivalence classes are disjoint and their union is Z: Z¼A0[A1[A2[A3[A4 This quotient set Z=/C17, called the integers modulo 5 , is denoted Z=5Zor simply Z5: Usually one chooses 0 ;1;2;3;4fg or/C02;/C01;0;1;2fg as a system of representatives of the equiva- lence classes. Analagously, for any positive integer m, there exists the congruence relation /C17deﬁned by x/C17ymod mðÞ and the quotient set Z=/C17is called the integers modulo m .416 Appendix D Odds and Ends D.3 Determinants and Block Matrices Recall ﬁrst: THEOREM 8.12: Suppose Mis an upper (lower) triangular block matrix with diagonal blocks Aj;A2;...;An:Then det MðÞ ¼ detAj/C0/C1 detA2ðÞ ...detAnðÞ: Accordingly, if M¼AB 0D/C20/C21 where Aisr/C2randDiss/C2s. Then det MðÞ ¼ detAðÞdetDðÞ: THEOREM D.2: Consider the block matrix M¼AB CD/C20/C21 where Ais nonsingular, Aisr/C2randD iss/C2s:Then det MðÞ ¼ detAðÞdetD/C0CA/C01BðÞ Proof: Follows from the fact that M¼I 0 CA/C01I/C20/C21 AB 0D/C0CA/C01B/C20/C21 and the above result. D.4 Full Rank Factorization A matrix Bis said to have full row rank r ifBhasrrows that are linearly independent, and a matrix Cis said to have full column rank r ifChasrcolumns that are linearly independent. DEFINITION D.2: LetAbe am/C2nmatrix of rank r. Then Ais said to have the full rank factorization A¼BC where Bhas full-column rank randChas full-row rank r. THEOREM D.3: Every matrix Awith rank r>0 has a full rank factorization. There are many full rank factorizations of a matrix A. Fig. D-1 gives an algorithm to ﬁnd one such factorization. EXAMPLE D.3 LetA¼11/C012 22/C013 /C01/C012/C032 43 5where M¼110 1 001/C01 000 02 43 5is the row cannonical form of A. We set B¼1/C01 2/C01 /C0122 43 5and C¼110 1 001/C01/C20/C21 Then A¼BCis a full rank factorization of A.Algorithm D-1: The input is a matrix Aof rank r>0. The output is a full rank factorization of A. Step 1. Find the row cannonical form MofA. Step 2. LetBbe the matrix whose columns are the columns of Acorresponding to the columns of M with pivots. Step 3. LetCbe the matrix whose rows are the nonzero rows of M. Then A¼BCis a full rank factorization of A. Figure D-1Appendix D Odds and Ends 417 D.5 Generalized (Moore–Penrose) Inverse Here we assume that the ﬁeld of scalars is the complex ﬁeld Cwhere the matrix AHis the conjugate transpose of a matrix A. [If Ais a real matrix, then AH¼AT.] DEFINITION D.3: LetAbe an m/C2nmatrix over C. A matrix, denoted by Aþ, is called the pseudoinverse or Morre–Penrose inverse or MP-inverse of AifAsatisﬁes the following four equations: [MP1] AXA¼A;[MP3] AXðÞH¼AX; [MP2] XAX¼X;[MP4] XAðÞH¼XA; Clearly, Aþis an n/C2mmatrix. Also, Aþ¼A/C01ifAis nonsingular. LEMMA D.4: Aþis unique (when it exists). Proof. Suppose XandYsatisfy the four MP equations. Then AY¼AYðÞH¼AXAYðÞH¼AYðÞHAXðÞH¼AYAX¼AYAðÞ X¼AX The ﬁrst and fourth equations use [MP3] , and the second and last equations use [MP1] . Similarly, YA¼XA(which uses [MP4] and[MP1] ). Then, Y¼YAY¼YAðÞ Y¼XAðÞ Y¼XA YðÞ ¼ XA XðÞ ¼ X where the ﬁrst equation uses [MP2]. LEMMA D.5: Aþexists for any matrix A. Fig. D-2 gives an algorithm that ﬁnds an MP-inverse for any matrix A. Combining the above two lemmas we obtain: THEOREM D.6: Every matrix Aover Chas a unique Moore–Penrose matrix Aþ. There are special cases when Ahas full-row rank or full-column rank. THEOREM D.7: LetAbe a matrix over C. (a) If Ahas full column rank (columns are linearly independent), then Aþ¼AHAðÞ/C01AH: (b) If Ahas full row rank (rows are linearly independent), then Aþ¼AHAAHðÞ/C01: THEOREM D.8: LetAbe a matrix over C. Suppose A¼BCis a full rank factorization of A. Then Aþ¼CþBþ¼CHCCH/C0/C1/C01BHB/C0/C1/C01BH Moreover, AAþ¼BBþandAþA¼CþC:Algorithm D-2. Input is an m/C2nmatrix Aover Cor rank r. Output is Aþ. Step 1. Interchange rows and columns of Aso that PAQ¼A11A12 A21A22/C20/C21 where A11is a nonsingular r/C2rblock. [Here PandQare the products of elementary matrices corresponding to the interchanges of the rows and columns.] Step 2. SetB¼A11 A21/C20/C21 andC¼Ir;A/C01 11A12/C2/C3 where Iris the r/C2ridentity matrix. Step 3. SetAþ¼QCHCCHðÞ/C01BHBðÞ/C01B11hi P: Figure D-2418 Appendix D Odds and Ends EXAMPLE D.4 Consider the full rank factorization A¼BCin Example D.1; that is, A¼11/C012 22/C013 /C01/C012/C032 43 5¼1/C01 2/C01 /C0122 43 5110 1 001/C01/C20/C21 ¼BC Then CCH/C0/C1/C01¼1 521 13/C20/C21 ;CC CH/C0/C1/C01¼1 521 21 13 1/C022 6643 775;B HB/C0/C1/C01¼1 1165 56/C20/C21 ;BBHB/C0/C1/C01¼1 11174 /C0147/C20/C21 Accordingly, the following is the Moore–Penrose inverse of A: Aþ¼1 5511 8 1 5 11 8 1 5 /C021 9 2 5 3/C01/C0102 6643 775 D.6 Least-Square Solution Consider a system AX¼Bof linear equations. A least-square solution ofAX¼Bis the vector of smallest Euclidean norm that minimizes AX/C0Bkk2:That vector is X¼AþB [In case Ais invertible, so Aþ¼A/C01, then X¼A/C01B, which is the unique solution of the system.] EXAMPLE D.5 Consider the following system AX¼Bof linear equations: xþy/C0zþ2t¼1 2xþ2y/C0zþ3t¼3 /C0x/C0yþ2z/C03t¼2 Then, using Example D.4, A¼11/C012 22/C013 /C01/C012/C032 43 5;B¼1 3 22 43 5;Aþ¼1 5511 8 1 5 11 8 1 5 /C021 9 2 5 3/C01/C0102 6643 775 Accordingly, X¼AþB¼1=55ðÞ 85;85;105;/C020½/C138T¼17=11;17=11;21=11;/C04=11 ½/C138T is the vector of smallest Euclidean norm which minimizes AX/C0Bkk2:Appendix D Odds and Ends 419 LIST OF SYMBOLS A¼½aij/C138, matrix, 27 /C22A¼½/C22aij/C138, conjugate matrix, 38 jAj, determinant, 264, 268 A*, adjoint, 377 AH, conjugate transpose, 38 AT, transpose, 33 Aþ, Moore–Penrose inverse, 418 Aij, minor, 269 AðI;JÞ, minor, 273 AðVÞ, linear operators, 174 adjA, adjoint (classical), 271 A/C24B, row equivalence, 72 A’B, congruence, 360 C, complex numbers, 11 Cn, complex n-space, 13 C½a;b/C138, continuous functions, 228 CðfÞ, companion matrix, 304 colspðAÞ, column space, 120 dðu;vÞ, distance, 5, 241 diagða11;...;annÞ, diagonal matrix, 35 diagðA11;...;AnnÞ, block diagonal, 40 detðAÞ, determinant, 268 dimV, dimension, 124 fe1;...;eng, usual basis, 125 Ek, projections, 384 f:A!B, mapping, 164 FðXÞ, function space, 114 G/C14F, composition, 173 HomðV;UÞ, homomorphisms, 174 i,j,k,9 In, identity matrix, 33 Im F, image, 169 JðlÞ, Jordan block, 329 K, ﬁeld of scalars, 112 KerF, kernel, 169 mðtÞ, minimal polynomial, 303 Mm;n;m/C2nmatrices, 114n-space, 5, 13, 227, 240 P(t), polynomials, 114 PnðtÞ;polynomials, 114 projðu;vÞ, projection, 6, 234 projðu;VÞ, projection, 235 Q, rational numbers, 11 R, real numbers, 1 Rn, real n-space, 2 rowspðAÞ, row-space, 120 S?, orthogonal complement, 231 sgns, sign, parity, 267 spanðSÞ, linear span, 119 trðAÞ, trace, 33 ½T/C138S, matrix representation, 195 T*, adjoint, 377 T-invariant, 327 Tt, transpose, 351 kuk, norm, 5, 13, 227, 241 ½u/C138S, coordinate vector, 130 u/C1v, dot product, 4, 13 hu;vi, inner product, 226, 238 u/C2v, cross product, 10 u/C10v, tensor product, 396 u^v, exterior product, 401 u/C8v, direct sum, 129, 327 VﬃU, isomorphism, 132, 169 V/C10W, tensor product, 396 V*, dual space, 349 V**, second dual space, 350VrV, exterior product, 401 W0, annihilator, 351 /C22z, complex conjugate, 12 Zðv;TÞ,T-cyclic subspace, 330 dij, Kronecker delta, 37 DðtÞ, characteristic polynomial, 294 l, eigenvalue, 296P, summation symbol, 29 LIST OF SYMBOLS 420 A Absolute value (complex), 12 Abelian group, 403Adjoint, classical, 271 operator, 377, 384 Algebraic multiplicity, 298 Alternating mappings, 276, 360, 399Angle between vectors, 6, 230 Annihilator, 330, 351, 354 Associate, 406Associated homogeneous system, 83 Associative, 174, 403 Augmented matrix, 59 B Back-substitution, 63, 65, 67Basis, 82, 124, 139 change of, 199, 211 dual, 350, 352 orthogonal, 243orthonormal, 243 second dual, 367 standard, 125usual, 125 Basis-ﬁnding algorithm, 127 Bessel inequality, 264Bijective mapping, 166 Bilinear form, 359, 396 alternating, 276matrix representation of, 360polar form of, 363 real symmetric, 363 symmetric, 361 Bilinear mapping, 359, 396 Block matrix, 39, 50 determinants, 417Jordan, 344 square, 40 Bounded, 156 C Cancellation law, 113 Canonical forms, 205, 325 Jordan, 329, 336 rational, 331 row, 74triangular, 325 Casting-out algorithm, 128 Cauchy–Schwarz inequality, 5, 229, 240Cayley–Hamilton theorem, 294, 308 Cells, 39, 415 Change of basis, 199, 211Change-of-basis (transition) matrix, 199 Change-of-coordinate matrix, 221 Characteristic polynomial, 294, 305 value, 296 Classical adjoint, 271 Coefﬁcient, 57, 58, 411 Fourier, 233, 244matrix, 59 Cofactor, 269 Column, 27 matrix, 27 operations, 89 space, 120vector, 3 Colsp(A), column space, 126 Commutative law, 403 group, 113 Commuting (diagram), 396 Companion matrix, 304 Complement, orthogonal, 242Complementary minor, 273 Completing the square, 393 Complex: conjugate, 13 inner product, 239 matrix, 38, 49n-space, 13 numbers, 1, 11, 13 plane, 12 Complexity, 88Components, 2 Composition of mappings, 165 Congruent matrices, 360 diagonalization, 61 Conjugate: complex, 12linearity, 239 matrix, 38 symmetric, 239 Consistent system, 59 Constant term, 57, 58 Convex set, 193 Coordinates, 2, 130 vector, 130 Coset, 182, 332, 403 Cramer’s rule, 272 421 INDEX Cross product, 10 Curves, 8 Cyclic subspaces, 330, 342 group, 408 D dij, Kronecker delta function, 33 Decomposable, 327Decomposition: direct-sum, 129 primary, 238 Degenerate, 360 bilinear form, 360 linear equations, 59 Dependence, linear, 133Derivative, 168 Determinant, 63, 264, 267 computation of, 66, 270linear operator, 275order, 3, 266 Diagonal, 32 blocks, 40matrix, 35, 47 quadratic form, 302 Diagonal (of a matrix), 10Diagonalizable, 203, 292, 296 Diagonalization: algorithm, 299in inner product space, 382 Dimension of solution spaces, 82 Dimension of vector spaces, 82, 139 ﬁnite, 124inﬁnite, 124 subspaces, 126 Direct sum, 129, 327 decomposition, 327 Directed line segment, 7 Distance, 5, 241Divides, 412 Division algorithm, 412 Domain, 164, 406Dot product, 4 Dual: basis, 350, 352space, 349, 352 E Echelon: form, 65, 72 matrices, 70 Eigenline, 296Eigenspace, 299 Eigenvalue, 296, 298, 312 Eigenvector, 296, 298, 312Elementary divisors, 331Elementary matrix, 84 Elementary operations, 61 column, 86row, 72, 120Elimination, Gaussian, 67 Empty set,;, 112 Equal: functions, 164matrices, 27 vectors, 2 Equations ( SeeLinear equations) Equivalence: classes, 416 matrix, 87relation, 73, 415 row, 72 Equivalent systems, 61Euclidean n-space, 5, 228 Exterior product, 401 F Field of scalars, 11, 406 Finite dimension, 124 Form: bilinear, 359 linear, 349 quadratic, 363 Forward elimination 63, 67, 73 Fourier coefﬁcient, 81, 233 series, 233 Free variable, 65, 66 Full rank, 41 factorization, 417 Function, 154 space F(X), 114 Functional, linear, 349 Fundamental Theorem of Algebra, 414 G Gaussian elimination, 61, 67, 73Gaussian integer, 409 Gauss–Jordan algorithm, 74 General solution, 58Geometric multiplicity, 298Gram–Schmidt orthogonalization, 235 Graph, 164 Greatest common divisor, 413Group, 113, 403 H Hermitian: form, 364 matrix, 38, 49quadratic form, 364 Hilbert space, 229 Homogeneous system, 58, 81Homomorphism, 173, 404, 407 Hom( V,U), 173 Hyperplane, 7, 358 I i, imaginary, 12 Ideal, 405422 Index Identity: mapping, 166, 168 matrix, 33 ijknotation, 9 Image, 164, 169, 170 Imaginary part, 12 ImF, image, 169 Imz, imaginary part, 12 Inclusion mapping, 190 Inconsistent systems, 59Independence, linear, 133 Index, 30 Index of nilpotency, 328Inertia, Law of, 364 Inﬁnite dimension, 124 Inﬁnity-norm, 241Injective mapping, 166 Inner product, 4 complex, 239 Inner product spaces, 226 linear operators on, 377 Integral, 168 domain, 406 Invariance, 224 Invariant subspaces, 224, 326, 332 direct-sum, 327 Inverse image, 164 Inverse mapping, 164 Inverse matrix, 34, 46, 85 computing, 85 inversion, 267 Invertible: matrices, 34, 46 Irreducible, 406 Isometry, 381 Isomorphic vector spaces, 169, 404 J Jordan: block, 304canonical form, 329, 336 K KerF, kernel, 169 Kernel, 169, 170Kronecker delta function d ij,3 3 L l2-space, 229 Laplace expansion, 270 Law of inertia, 363 Leading: coefﬁcient, 60 nonzero element, 70 unknown, 60 Least square solution, 419Legendre polynomial, 237 Length, 5, 227 Limits (summation), 30Line, 8, 192Linear: combination, 3, 29, 60, 79, 115 dependence, 121form, 349functional, 349 independence, 121 span, 119 Linear equation, 57 Linear equations (system), 58 consistent, 59echelon form, 65 triangular form, 64 Linear mapping (function), 164, 167 image, 164, 169 kernel, 169 nullity, 171rank, 171 Linear operator: adjoint, 377 characteristic polynomial, 304determinant, 275 on inner product spaces, 377 invertible, 175matrix representation, 195 Linear transformation ( Seelinear mappings), 167 Located vectors, 7LUdecomposition, 87, 104 M M m,n, matrix vector space, 114 Mappings (maps), 164 bilinear, 359, 396composition of, 165 linear, 167 matrix, 168 Matrices: congruent, 360 equivalent, 87 similar, 203 Matrix, 27 augmented, 59change-of-basis, 199 coefﬁcient, 59 companion, 304diagonal, 35echelon, 65, 70 elementary, 84 equivalence, 87Hermitian, 38, 49 identity, 33 invertible, 34nonsingular, 34 normal, 38 orthogonal, 237positive deﬁnite, 238rank, 72, 87 space, M m,n, 114 square root, 296triangular, 36Index 423 Matrix mapping, 165 Matrix multiplication, 30 Matrix representation, 195, 238, 360 adjoint operator, 377, 384bilinear form, 359 change of basis, 199 linear mapping, 195 Metric space, 241 Minimal polynomial, 303, 305 Minkowski’s inequality, 5Minor, 269, 273 principle, 273 Module, 407Monic polynomial 303, 411 Moore–Penrose inverse, 418 Multilinearity, 276, 399Multiplicity, 298 Multiplier, 67, 73, 87 N n-linear, 276 n-space, 2 complex, 13 real, 2 Natural mapping, 351 New basis, 199 Nilpotent, 328, 336Nonnegative semideﬂnite, 226 Nonsingular, 112 linear maps, 172matrices, 34 Norm, 5, 227, 241 Normal, 7 matrix, 38operator, 380, 383 Normalized, 227 Normalizing, 5, 227, 233Normed vector space, 241 Nullity, 171 nullsp( A), 170 Null space, 170 O Old basis, 199 One-norm, 241 One-to-one: correspondence, 166 mapping, 166 Onto mapping (function), 166 Operators ( SeeLinear operators) Order, n: determinant, 264of a group, 408 Orthogonal, 4, 37, 80 basis, 231 complement, 231matrix, 237 operator, 380 projection, 384substitution, 302Orthogonalization, Gram–Schmidt, 235 Orthogonally equivalent, 381 Orthonormal, 233Outer product, 10 P Parameter, 64 form, 65 Particular solution, 58 Partition, 416Permutations, 8, 267Perpendicular, 4 Pivot, 67, 71 row reduction, 94variables, 65 Pivoting (row reduction), 94 Polar form, 363Polynomial, 411 characteristic, 294, 305 minimum, 303space, P n(t), 114 Positive deﬁnite, 226 matrices, 238operators, 336, 382 Positive operators, 226 square root, 391 Primary decomposition theorem, 328 Prime ideal, 410 Principle ideal ring, 406Principle minor, 273 Product: exterior, 401inner, 4tensor, 396 Projections, 167, 234, 344, 384 orthogonal, 384 Proper value, 296 vector, 296 Pythagorean theorem, 233 Q Q, rational numbers, 11 Quadratic form, 301, 315, 363Quotient group, 403 ring, 406spaces, 332, 416 R R, real numbers, 1, 12 R n, real n-space, 2 Range, 164, 169 Rank, 72, 87, 126, 171, 364 Rational: canonical form, 331numbers, Q,1 1 Real: numbers, R,1 part (complex number), 12424 Index Real symmetric bilinear form, 363 Reduce, 73 Relation, 415Representatives, 416Restriction mapping, 192 Right-handed system, 11 Right inverse, 189Ring, 405 quotient, 406 Root, 293Rotation, 169 Row, 27 canonical form, 72equivalence, 72 operations, 72 rank, 72reduce, 73 reduced echelon form, 73 space, 120 S S n, symmetric group, 267, 404 Scalar, 1, 12 matrix, 33 multiplication, 33product, 27 Scaling factor, 296 Schwarz inequality, 5, 229, 240 (SeeCauchy–Schwarz inequality) Second dual space, 350 Self-adjoint operator, 380Sign of permutation, 267Signature, 364 Similar, 203, 224 Similarity transformation, 203Singular, 172 Size (matrix), 27 Skew-adjoint operator, 380Skew-Hermitian, 38 Skew-symmetric, 360 matrix, 36, 48 Solution, (linear equations), 57 zero, 121 Spatial vectors, 9Span, 116Spanning sets, 116 Spectral theorem, 383 Square: matrix, 32, 44 system of linear equations, 58, 72 Square root of a matrix, 391Standard: basis, 125 form, 57, 399inner product, 228 Subdiagonal, 304 Subgroup, 403 Subset, 112Subspace, 117, 133 Sum of vector spaces, 129Summation symbol, 29 Superdiagonal, 304 Surjective map, 166Sylvester’s theorem, 364Symmetric: bilinear form, 361 matrices, 4, 36 Systems of linear equations, 58 T Tangent vector, T(t), 9 Target set, 164 Tensor product, 396Time complexity, 88 Top-down, 73 Trace, 33Transformation (linear), 167 Transition matrix, 199 Transpose: linear functional (dual space), 351 matrix, 32 Triangle inequality, 230Triangular form, 64Triangular matrix, 36, 47 block, 40 Triple product, 11Two-norm, 241 U Unique factorization domain, 406, 414Unit vector, 5, 227 matrix, 33 Unitary, 38, 49, 380Universal mapping principle (UMP), 396Usual: basis, 125 inner product, 228 V Vandermonde determinant, 290Variable, free, 65 Vector, 2 coordinates, 130located, 7 product, 10 spatial, 9 Vector space, 112, 226 basis, 124 dimension, 124 Volume, 274 W Wedge (exterior) product, 401 Z Z, integers, 406 Zero: mapping, 128, 168, 173 matrix, 27 polynomial, 411solution, 121 vector, 2Index 425