hefferon linear algebra

Linear Algebra Jim Hefferon 2 11 3 1 2 3 1 2 1x1
1 3 x1
1 2 x1
3 1 2 16 8 6 2 8 1 Notation R,R+,Rnreal numbers, reals greater than 0, n-tuples of reals Nnatural numbers: f0;1;2;:::g Ccomplex numbers f::::::gset of . . . such that . . . (a::b), [a::b]interval (open or closed) of reals between aandb h:::isequence; like a set but order matters V;W;U vector spaces ~ v;~ w vectors ~0,~0Vzero vector, zero vector of V B;D bases En=h~ e1; :::;~ enistandard basis for Rn ~;~basis vectors RepB(~ v)matrix representing the vector Pnset ofn-th degree polynomials Mnmset ofnmmatrices [S]span of the set S MN direct sum of subspaces V=W isomorphic spaces h;g homomorphisms, linear maps H;G matrices t;s transformations; maps from a space to itself T;S square matrices RepB;D(h)matrix representing the map h hi;j matrix entry from row i, columnj Znm;Z;Inn;I zero matrix, identity matrix jTjdeterminant of the matrix T R(h);N(h)rangespace and nullspace of the map h R1(h);N1(h)generalized rangespace and nullspace Lower case Greek alphabet name character name character name character alpha iota rho beta kappa sigma gamma lambda tau delta mu upsilon epsilon nu phi zeta xi chi eta omicrono psi theta pi omega! Cover. This is Cramer's Rule for the system x1+ 2x2= 6, 3x1+x2= 8. The size of the rst box is the determinant shown (the absolute value of the size is the area). The size of the second box is x1times that, and equals the size of the nal box. Hence, x1 is the nal determinant divided by the rst determinant. Preface This book helps students to master the material of a standard US undergraduate linear algebra course. The material is standard in that the topics covered are Gaussian reduction, vector spaces, linear maps, determinants, and eigenvalues and eigenvectors. An- other standard is book's audience: sophomores or juniors, usually with a back- ground of at least one semester of calculus. The help that it gives to students comes from taking a developmental approach | this book's presentation empha- sizes motivation and naturalness, driven home by a wide variety of examples and by extensive and careful exercises. The developmental approach is the feature that most recommends this book so I will say more. Courses in the beginning of most mathematics programs focus less on understanding theory and more on correctly applying formulas and algorithms. Later courses ask for mathematical maturity: the ability to follow dierent types of arguments, a familiarity with the themes that underlie many mathematical investigations such as elementary set and function facts, and a capacity for some independent reading and thinking. Linear algebra is an ideal spot to work on the transition. It comes early in a program so that progress made here pays o later, but also comes late enough that students are serious about mathematics, often majors and minors. The material is accessible, coherent, and elegant. There are a variety of argument styles, including proofs by contradiction, if and only if statements, and proofs by induction. And, examples are plentiful. Helping readers start the transition to being serious students of the subject of mathematics itself means taking the mathematics seriously, so all of the results in this book are proved. On the other hand, we cannot assume that students have already arrived and so in contrast with more abstract texts, we give many examples and they are often quite detailed. Some linear algebra books begin with extensive computations of linear sys- tems, matrix multiplications, and determinants. Then, when the concepts | vector spaces and linear maps | nally appear, and denitions and proofs start, often the abrupt change brings students to a stop. In this book, while we start with a computational topic, linear reduction, from the rst we do more than compute. We do linear systems quickly but completely, including the proofs needed to justify what we are computing. Then, with the linear systems work as motivation and at a point where the study of linear combinations seems nat- ural, the second chapter starts with the denition of a real vector space. In the iii schedule below, this occurs by the end of the third week. Another example of our emphasis on motivation and naturalness is that the third chapter on linear maps does not begin with the denition of homomor- phism, but with isomorphism. The denition of isomorphism is easily motivated by the observation that some spaces are \just like" others. After that, the next section takes the reasonable step of dening homomorphism by isolating the operation-preservation idea. This approach loses mathematical slickness, but it is a good trade because it gives to students a large gain in sensibility. One aim of our developmental approach is to present the material in such a way that students can see how the ideas arise, and perhaps can picture them- selves doing the same type of work. The clearest example of the developmental approach is the exercises. A stu- dent progresses most while doing the exercises, so the ones included here have been selected with great care. Each problem set ranges from simple checks to reasonably involved proofs. Since an instructor usually assigns about a dozen ex- ercises after each lecture, each section ends with about twice that many, thereby providing a selection. There are even a few problems that are challenging puz- zles taken from various journals, competitions, or problems collections. (These are marked with a ` ?' and as part of the fun, the original wording has been retained as much as possible.) In total, the exercises are aimed to both build an ability at, and help students experience the pleasure of, doing mathematics. Applications and computers. The point of view taken here, that students should think of linear algebra as about vector spaces and linear maps, is not taken to the complete exclusion of others. Applications and computing are important and vital aspects of the subject. Consequently, each of this book's chapters closes with a few application or computer-related topics. Some are: net- work ows, the speed and accuracy of computer linear reductions, Leontief In- put/Output analysis, dimensional analysis, Markov chains, voting paradoxes, analytic projective geometry, and dierence equations. These topics are brief enough to be done in a day's class or to be given as independent projects. Most simply give a reader a taste of the subject, discuss how linear algebra comes in, point to some further reading, and give a few exercises. In short, these topics invite readers to see for themselves that linear algebra is a tool that a professional must have. The license. This book is freely available. You can download and read it without restriction. Class instructors can print copies for students and charge for those. See http://joshua.smcvt.edu/linearalgebra for more license in- formation. That page also contains the latest version of this book, and the latest version of the worked answers to every exercise. Also there, I provide the L ATEX source of the text and some instructors may wish to add their own material. If you like, you can send such additions to me and I may possibly incorporate them into future editions. I am very glad for bug reports. I save them and periodically issue updates; people who contribute in this way are acknowledged in the text's source les. iv For people reading this book on their own. This book's emphasis on motivation and development make it a good choice for self-study. But while a professional instructor can judge what pace and topics suit a class, if you are an independent student then you may nd some advice helpful. Here are two timetables for a semester. The rst focuses on core material. week Monday Wednesday Friday 1One.I.1 One.I.1, 2 One.I.2, 3 2One.I.3 One.II.1 One.II.2 3One.III.1, 2 One.III.2 Two.I.1 4Two.I.2 Two.II Two.III.1 5Two.III.1, 2 Two.III.2 exam 6Two.III.2, 3 Two.III.3 Three.I.1 7Three.I.2 Three.II.1 Three.II.2 8Three.II.2 Three.II.2 Three.III.1 9Three.III.1 Three.III.2 Three.IV.1, 2 10 Three.IV.2, 3, 4 Three.IV.4 exam 11 Three.IV.4, Three.V.1 Three.V.1, 2 Four.I.1, 2 12 Four.I.3 Four.II Four.II 13 Four.III.1 Five.I Five.II.1 14 Five.II.2 Five.II.3 review The second timetable is more ambitious. It supposes that you know One.II, the elements of vectors, usually covered in third semester calculus. week Monday Wednesday Friday 1One.I.1 One.I.2 One.I.3 2One.I.3 One.III.1, 2 One.III.2 3Two.I.1 Two.I.2 Two.II 4Two.III.1 Two.III.2 Two.III.3 5Two.III.4 Three.I.1 exam 6Three.I.2 Three.II.1 Three.II.2 7Three.III.1 Three.III.2 Three.IV.1, 2 8Three.IV.2 Three.IV.3 Three.IV.4 9Three.V.1 Three.V.2 Three.VI.1 10 Three.VI.2 Four.I.1 exam 11 Four.I.2 Four.I.3 Four.I.4 12 Four.II Four.II, Four.III.1 Four.III.2, 3 13 Five.II.1, 2 Five.II.3 Five.III.1 14 Five.III.2 Five.IV.1, 2 Five.IV.2 In the table of contents I have marked subsections as optional if some instructors will pass over them in favor of spending more time elsewhere. You might pick one or two topics that appeal to you from the end of each chapter. You'll get more from these if you have access to computer software that can do any big calculations. I recommend Sage, freely available from http://sagemath.org . v My main advice is: do many exercises. I have marked a good sample with X's in the margin. For all of them, you must justify your answer either with a computation or with a proof. Be aware that few inexperienced people can write correct proofs. Try to nd someone with training to work with you on this. Finally, if I may, a caution for all students, independent or not: I cannot overemphasize how much the statement that I sometimes hear, \I understand the material, but it's only that I have trouble with the problems" is mistaken. Being able to do things with the ideas is their entire point. The quotes below express this sentiment admirably. They state what I believe is the key to both the beauty and the power of mathematics and the sciences in general, and of linear algebra in particular; I took the liberty of formatting them as verse. I know of no better tactic than the illustration of exciting principles by well-chosen particulars. {Stephen Jay Gould If you really wish to learn then you must mount the machine and become acquainted with its tricks by actual trial. {Wilbur Wright Jim Hefferon Mathematics, Saint Michael's College Colchester, Vermont USA 05439 http://joshua.smcvt.edu 2011-Jan-01 Author's Note. Inventing a good exercise, one that enlightens as well as tests, is a creative act, and hard work. The inventor deserves recognition. But for some reason texts have traditionally not given attributions for questions. I have changed that here where I was sure of the source. I would be glad to hear from anyone who can help me to correctly attribute others of the questions. vi Contents Chapter One: Linear Systems 1 I Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Gauss' Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Describing the Solution Set . . . . . . . . . . . . . . . . . . . . 11 3 General = Particular + Homogeneous . . . . . . . . . . . . . . 20 II Linear Geometry of n-Space . . . . . . . . . . . . . . . . . . . . . 32 1 Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2 Length and Angle Measures. . . . . . . . . . . . . . . . . . . 39 III Reduced Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . 46 1 Gauss-Jordan Reduction . . . . . . . . . . . . . . . . . . . . . . 46 2 Row Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Topic: Computer Algebra Systems . . . . . . . . . . . . . . . . . . . 61 Topic: Input-Output Analysis . . . . . . . . . . . . . . . . . . . . . . 63 Topic: Accuracy of Computations . . . . . . . . . . . . . . . . . . . . 67 Topic: Analyzing Networks . . . . . . . . . . . . . . . . . . . . . . . . 71 Chapter Two: Vector Spaces 77 I Denition of Vector Space . . . . . . . . . . . . . . . . . . . . . . 78 1 Denition and Examples . . . . . . . . . . . . . . . . . . . . . . 78 2 Subspaces and Spanning Sets . . . . . . . . . . . . . . . . . . . 89 II Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . 99 1 Denition and Examples . . . . . . . . . . . . . . . . . . . . . . 99 III Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . 110 1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 3 Vector Spaces and Linear Systems . . . . . . . . . . . . . . . . 122 4 Combining Subspaces. . . . . . . . . . . . . . . . . . . . . . . 129 Topic: Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Topic: Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Topic: Voting Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . 144 Topic: Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . 150 vii Chapter Three: Maps Between Spaces 157 I Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 157 2 Dimension Characterizes Isomorphism . . . . . . . . . . . . . . 166 II Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 2 Rangespace and Nullspace . . . . . . . . . . . . . . . . . . . . . 181 III Computing Linear Maps . . . . . . . . . . . . . . . . . . . . . . . 193 1 Representing Linear Maps with Matrices . . . . . . . . . . . . . 193 2 Any Matrix Represents a Linear Map. . . . . . . . . . . . . . 203 IV Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . 210 1 Sums and Scalar Products . . . . . . . . . . . . . . . . . . . . . 210 2 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . 213 3 Mechanics of Matrix Multiplication . . . . . . . . . . . . . . . . 220 4 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 V Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 1 Changing Representations of Vectors . . . . . . . . . . . . . . . 236 2 Changing Map Representations . . . . . . . . . . . . . . . . . . 240 VI Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 1 Orthogonal Projection Into a Line. . . . . . . . . . . . . . . . 248 2 Gram-Schmidt Orthogonalization. . . . . . . . . . . . . . . . 252 3 Projection Into a Subspace. . . . . . . . . . . . . . . . . . . . 258 Topic: Line of Best Fit . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Topic: Geometry of Linear Maps . . . . . . . . . . . . . . . . . . . . 272 Topic: Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . 279 Topic: Orthonormal Matrices . . . . . . . . . . . . . . . . . . . . . . 285 Chapter Four: Determinants 291 I Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 1 Exploration. . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . 297 3 The Permutation Expansion . . . . . . . . . . . . . . . . . . . . 301 4 Determinants Exist. . . . . . . . . . . . . . . . . . . . . . . . 309 II Geometry of Determinants . . . . . . . . . . . . . . . . . . . . . . 317 1 Determinants as Size Functions . . . . . . . . . . . . . . . . . . 317 III Other Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 1 Laplace's Expansion. . . . . . . . . . . . . . . . . . . . . . . . 324 Topic: Cramer's Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Topic: Speed of Calculating Determinants . . . . . . . . . . . . . . . 332 Topic: Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . 335 Chapter Five: Similarity 347 I Complex Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 347 1 Factoring and Complex Numbers; A Review. . . . . . . . . . 348 2 Complex Representations . . . . . . . . . . . . . . . . . . . . . 349 II Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 viii 1 Denition and Examples . . . . . . . . . . . . . . . . . . . . . . 351 2 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . 353 3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . 357 III Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 1 Self-Composition. . . . . . . . . . . . . . . . . . . . . . . . . 365 2 Strings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 IV Jordan Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 1 Polynomials of Maps and Matrices. . . . . . . . . . . . . . . . 379 2 Jordan Canonical Form. . . . . . . . . . . . . . . . . . . . . . 386 Topic: Method of Powers . . . . . . . . . . . . . . . . . . . . . . . . . 399 Topic: Stable Populations . . . . . . . . . . . . . . . . . . . . . . . . 403 Topic: Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . . 405 Appendix A-1 Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1 Quantiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-3 Techniques of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . A-5 Sets, Functions, and Relations . . . . . . . . . . . . . . . . . . . . . A-7 Note: starred subsections are optional. ix Chapter One Linear Systems I Solving Linear Systems Systems of linear equations are common in science and mathematics. These two examples from high school science [Onan] give a sense of how they arise. The rst example is from Physics. Suppose that we are given three objects, one with a mass known to be 2 kg, and are asked to nd the unknown masses. Suppose further that experimentation with a meter stick produces these two balances. ch 2 1540 50 ch 225 50 25 We know that the moment of each object is its mass times its distance from the balance point. We also know that for balance we must have that the sum of moments on the left equals the sum of moments on the right. That gives a system of two equations. 40h+ 15c= 100 25c= 50 + 50h The second example of a linear system is from Chemistry. We can mix, under controlled conditions, toluene C 7H8and nitric acid HNO 3to produce trinitrotoluene C 7H5O6N3along with the byproduct water (conditions have to be controlled very well | trinitrotoluene is better known as TNT). In what proportion should we mix those components? The number of atoms of each element present before the reaction xC7H8+yHNO 3!zC7H5O6N3+wH2O must equal the number present afterward. Applying that to the elements C, H, 1 2 Chapter One. Linear Systems N, and O in turn gives this system. 7x= 7z 8x+ 1y= 5z+ 2w 1y= 3z 3y= 6z+ 1w Finishing each of these examples requires solving a system of equations. In each system, the equations involve only the rst power of the variables. This chapter shows how to solve any such system. I.1 Gauss' Method 1.1 Denition Alinear combination ofx1;x2;:::;xnhas the form a1x1+a2x2+a3x3+


+anxn where the numbers a1;:::;an2Rare the combination's coecients . Alinear equation has the form a1x1+a2x2+a3x3+


+anxn=dwhered2Ris the constant . Ann-tuple (s1;s2;:::;sn)2Rnis asolution of, or satises , that equation if substituting the numbers s1, . . . ,snfor the variables gives a true statement: a1s1+a2s2+:::+ansn=d. Asystem of linear equations a1;1x1+a1;2x2+


+a1;nxn=d1 a2;1x1+a2;2x2+


+a2;nxn=d2 ... am;1x1+am;2x2+


+am;nxn=dm has the solution ( s1;s2;:::;sn) if thatn-tuple is a solution of all of the equa- tions in the system. 1.2 Example The combination 3 x1+ 2x2ofx1andx2is linear. The combi- nation 3x2 1+ 2 sin(x2) is not linear, nor is 3 x2 1+ 2x2. 1.3 Example The ordered pair ( 1;5) is a solution of this system. 3x1+ 2x2= 7 x1+x2= 6 In contrast, (5 ;1) is not a solution. Finding the set of all solutions is solving the system. No guesswork or good fortune is needed to solve a linear system. There is an algorithm that always Section I. Solving Linear Systems 3 works. The next example introduces that algorithm, called Gauss' method (or Gaussian elimination orlinear elimination ). It transforms the system, step by step, into one with a form that is easily solved. We will rst illustrate how it goes and then we will see the formal statement. 1.4 Example To solve this system 3x3= 9 x1+ 5x22x3= 2 1 3x1+ 2x2 = 3 we repeatedly transform it until it is in a form that is easy to solve. Below there are three transformations. The rst is to rewrite the system by interchanging the rst and third row. swap row 1 with row 3!1 3x1+ 2x2 = 3 x1+ 5x22x3= 2 3x3= 9 The second transformation is to rescale the rst row by multiplying both sides of the equation by 3. multiply row 1 by 3!x1+ 6x2 = 9 x1+ 5x22x3= 2 3x3= 9 The third transformation is the only nontrivial one. We mentally multiply both sides of the rst row by 1, mentally add that to the second row, and write the result in as the new second row. add1 times row 1 to row 2!x1+ 6x2 = 9 x22x3=7 3x3= 9 The point of this sucession of steps is that system is now in a form where we can easily nd the value of each variable. The bottom equation shows that x3= 3. Substituting 3 for x3in the middle equation shows that x2= 1. Substituting those two into the top equation gives that x1= 3 and so the system has a unique solution: the solution set is f(3;1;3)g. Most of this subsection and the next one consists of examples of solving linear systems by Gauss' method. We will use it throughout this book. It is fast and easy. But before we get to those examples, we will rst show that this method is also safe in that it never loses solutions or picks up extraneous solutions. 4 Chapter One. Linear Systems 1.5 Theorem (Gauss' method) If a linear system is changed to another by one of these operations (1) an equation is swapped with another (2) an equation has both sides multiplied by a nonzero constant (3) an equation is replaced by the sum of itself and a multiple of another then the two systems have the same set of solutions. Each of those three operations has a restriction. Multiplying a row by 0 is not allowed because that can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding 1 times the row to itself has the eect of multiplying the row by 0. Finally, swapping a row with itself is disallowed to make some results in the fourth chapter easier to state and remember. Proof .We will cover the equation swap operation here and save the other two cases for Exercise 30. Consider this swap of row iwith rowj. a1;1x1+a1;2x2+


a1;nxn=d1 ... ai;1x1+ai;2x2+


ai;nxn=di ... aj;1x1+aj;2x2+


aj;nxn=dj ... am;1x1+am;2x2+


am;nxn=dm!a1;1x1+a1;2x2+


a1;nxn=d1 ... aj;1x1+aj;2x2+


aj;nxn=dj ... ai;1x1+ai;2x2+


ai;nxn=di ... am;1x1+am;2x2+


am;nxn=dm Then-tuple (s1;::: ;sn) satises the system before the swap if and only if substituting the values, the s's, for the variables, the x's, gives true statements: a1;1s1+a1;2s2+


+a1;nsn=d1and . . .ai;1s1+ai;2s2+


+ai;nsn=diand . . . aj;1s1+aj;2s2+


+aj;nsn=djand . . .am;1s1+am;2s2+


+am;nsn=dm. In a requirement consisting of statements joined with `and' we can rearrange the order of the statements, so that this requirement is met if and only if a1;1s1+ a1;2s2+


+a1;nsn=d1and . . .aj;1s1+aj;2s2+


+aj;nsn=djand . . . ai;1s1+ai;2s2+


+ai;nsn=diand . . .am;1s1+am;2s2+


+am;nsn=dm. This is exactly the requirement that ( s1;::: ;sn) solves the system after the row swap. QED 1.6 Denition The three operations from Theorem 1.5 are the elementary reduction operations , or row operations , or Gaussian operations . They are swapping ,multiplying by a scalar (orrescaling ), and row combination . When writing out the calculations, we will abbreviate `row i' by `i'. For instance, we will denote a row combination operation by ki+j, with the row that is changed written second. We will also, to save writing, often list addition steps together when they use the same i. Section I. Solving Linear Systems 5 1.7 Example Gauss' method is to systemmatically apply those row operations to solve a system. Here is a typical case. x+y = 0 2xy+ 3z= 3 x2yz= 3 To start we use the rst row to eliminate the 2 xin the second row and the x in the third. To get rid of the 2 x, we mentally multiply the entire rst row by 2, add that to the second row, and write the result in as the new second row. To get rid of the x, we multiply the rst row by 1, add that to the third row, and write the result in as the new third row. (Using one entry to clear out the rest of a column is called pivoting on that entry.) 21+2! 1+3x+y = 0 3y+ 3z= 3 3yz= 3 In this version of the system, the last two equations involve only two unknowns. To nish we transform the second system into a third system, where the last equation involves only one unknown. We use the second row to eliminate yfrom the third row. 2+3!x+y = 0 3y+ 3z= 3 4z= 0 Now the third row shows that z= 0. Substitute that back into the second row to gety=1 and then substitute back into the rst row to get x= 1. 1.8 Example For the Physics problem from the start of this chapter, Gauss' method gives this. 40h+ 15c= 100 50h+ 25c= 505=41+2!40h+ 15c= 100 (175=4)c= 175 Soc= 4, and back-substitution gives that h= 1. (The Chemistry problem is solved later.) 1.9 Example The reduction x+y+z= 9 2x+ 4y3z= 1 3x+ 6y5z= 021+2! 31+3x+y+z= 9 2y5z=17 3y8z=27 (3=2)2+3!x+y+z= 9 2y 5z=17 (1=2)z=(3=2) shows that z= 3,y=1, andx= 7. 6 Chapter One. Linear Systems As these examples illustrate, the point of Gauss' method is to use the ele- mentary reduction operations to set up back-substitution. 1.10 Denition In each row of a system, the rst variable with a nonzero coecient is the row's leading variable . A system is in echelon form if each leading variable is to the right of the leading variable in the row above it (except for the leading variable in the rst row). 1.11 Example The only operation needed in the example above is row combi- nation. Here is a linear system that requires the operation of swapping equations to get it in echelon form. After the rst combination xy = 0 2x2y+z+ 2w= 4 y +w= 0 2z+w= 521+2!xy = 0 z+ 2w= 4 y +w= 0 2z+w= 5 the second equation has no leading y. To get one, we look lower down in the system for a row that has a leading yand swap it in. 2$3!xy = 0 y +w= 0 z+ 2w= 4 2z+w= 5 (Had there been more than one row below the second with a leading ythen we could have swapped in any one.) The rest of Gauss' method goes as before. 23+4!xy = 0 y+w= 0 z+ 2w= 4 3w=3 Back-substitution gives w= 1,z= 2 ,y=1, andx=1. Strictly speaking, the operation of rescaling rows is not needed to solve linear systems. We have included it because we will use it later in this chapter as part of a variation on Gauss' method, the Gauss-Jordan method. All of the systems seen so far have the same number of equations as un- knowns. All of them have a solution, and for all of them there is only one solution. We nish this subsection by seeing for contrast some other things that can happen. 1.12 Example Linear systems need not have the same number of equations as unknowns. This system x+ 3y= 1 2x+y=3 2x+ 2y=2 Section I. Solving Linear Systems 7 has more equations than variables. Gauss' method helps us understand this system also, since this 21+2! 21+3x+ 3y= 1 5y=5 4y=4 shows that one of the equations is redundant. Echelon form (4=5)2+3!x+ 3y= 1 5y=5 0 = 0 gives thaty= 1 andx=2. The `0 = 0' re ects the redundancy. That example's system has more equations than variables. Gauss' method is also useful on systems with more variables than equations. Many examples are in the next subsection. Another way that linear systems can dier from the examples shown earlier is that some linear systems do not have a unique solution. This can happen in two ways. The rst is that a system can fail to have any solution at all. 1.13 Example Contrast the system in the last example with this one. x+ 3y= 1 2x+y=3 2x+ 2y= 021+2! 21+3x+ 3y= 1 5y=5 4y=2 Here the system is inconsistent: no pair of numbers satises all of the equations simultaneously. Echelon form makes this inconsistency obvious. (4=5)2+3!x+ 3y= 1 5y=5 0 = 2 The solution set is empty. 1.14 Example The prior system has more equations than unknowns, but that is not what causes the inconsistency | Example 1.12 has more equations than unknowns and yet is consistent. Nor is having more equations than unknowns necessary for inconsistency, as is illustrated by this inconsistent system with the same number of equations as unknowns. x+ 2y= 8 2x+ 4y= 821+2!x+ 2y= 8 0 =8 The other way that a linear system can fail to have a unique solution is to have many solutions. 8 Chapter One. Linear Systems 1.15 Example In this system x+y= 4 2x+ 2y= 8 any pair of numbers satisfying the rst equation automatically satises the sec- ond. The solution set f(x;y)x+y= 4gis innite; some of its members are (0;4), (1;5), and (2:5;1:5). The result of applying Gauss' method here contrasts with the prior example because we do not get a contradictory equa- tion. 21+2!x+y= 4 0 = 0 Don't be fooled by the `0 = 0' equation in that example. It is not the signal that a system has many solutions. 1.16 Example The absence of a `0 = 0' does not keep a system from having many dierent solutions. This system is in echelon form x+y+z= 0 y+z= 0 has no `0 = 0', and yet has innitely many solutions. (For instance, each of these is a solution: (0 ;1;1), (0;1=2;1=2), (0;0;0), and (0;;). There are innitely many solutions because any triple whose rst component is 0 and whose second component is the negative of the third is a solution.) Nor does the presence of a `0 = 0' mean that the system must have many solutions. Example 1.12 shows that. So does this system, which does not have many solutions | in fact it has none | despite that when it is brought to echelon form it has a `0 = 0' row. 2x2z= 6 y+z= 1 2x+yz= 7 3y+ 3z= 01+3!2x2z= 6 y+z= 1 y+z= 1 3y+ 3z= 0 2+3! 32+42x2z= 6 y+z= 1 0 = 0 0 =3 We will nish this subsection with a summary of what we've seen so far about Gauss' method. Gauss' method uses the three row operations to set a system up for back substitution. If any step shows a contradictory equation then we can stop with the conclusion that the system has no solutions. If we reach echelon form without a contradictory equation, and each variable is a leading variable in its row, then the system has a unique solution and we nd it by back substitution. Section I. Solving Linear Systems 9 Finally, if we reach echelon form without a contradictory equation, and there is not a unique solution (at least one variable is not a leading variable) then the system has many solutions. The next subsection deals with the third case | we will see how to describe the solution set of a system with many solutions. Note For all exercises in this book, you must justify your answer. For instance, if a question asks whether a system has a solution then you must justify a yes response by producing the solution and must justify a no response by showing that no solution exists. Exercises X1.17 Use Gauss' method to nd the unique solution for each system. (a)2x+ 3y= 13 xy=1(b)xz= 0 3x+y = 1 x+y+z= 4 X1.18 Use Gauss' method to solve each system or conclude `many solutions' or `no solutions'. (a)2x+ 2y= 5 x4y= 0(b)x+y= 1 x+y= 2(c)x3y+z= 1 x+y+ 2z= 14 (d)xy= 1 3x3y= 2(e) 4y+z= 20 2x2y+z= 0 x +z= 5 x+yz= 10(f)2x +z+w= 5 yw=1 3xzw= 0 4x+y+ 2z+w= 9 X1.19 There are methods for solving linear systems other than Gauss' method. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. That step is repeated until there is an equation with only one variable. From that, the rst number in the solution is derived, and then back-substitution can be done. This method takes longer than Gauss' method, since it involves more arithmetic operations, and is also more likely to lead to errors. To illustrate how it can lead to wrong conclusions, we will use the system x+ 3y= 1 2x+y=3 2x+ 2y= 0 from Example 1.13. (a)Solve the rst equation for xand substitute that expression into the second equation. Find the resulting y. (b)Again solve the rst equation for x, but this time substitute that expression into the third equation. Find this y. What extra step must a user of this method take to avoid erroneously concluding a system has a solution? X1.20 For which values of kare there no solutions, many solutions, or a unique solution to this system? xy= 1 3x3y=k X1.21 This system is not linear, in some sense, 2 sincos+ 3 tan = 3 4 sin+ 2 cos2 tan = 10 6 sin3 cos+ tan = 9 10 Chapter One. Linear Systems and yet we can nonetheless apply Gauss' method. Do so. Does the system have a solution? X1.22 What conditions must the constants, the b's, satisfy so that each of these systems has a solution? Hint. Apply Gauss' method and see what happens to the right side. [Anton] (a)x3y=b1 3x+y=b2 x+ 7y=b3 2x+ 4y=b4(b)x1+ 2x2+ 3x3=b1 2x1+ 5x2+ 3x3=b2 x1 + 8x3=b3 1.23 True or false: a system with more unknowns than equations has at least one solution. (As always, to say `true' you must prove it, while to say `false' you must produce a counterexample.) 1.24 Must any Chemistry problem like the one that starts this subsection | a bal- ance the reaction problem | have innitely many solutions? X1.25 Find the coecients a,b, andcso that the graph of f(x) =ax2+bx+cpasses through the points (1 ;2), (1;6), and (2;3). 1.26 Gauss' method works by combining the equations in a system to make new equations. (a)Can the equation 3 x2y= 5 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? x+y= 1 4xy= 6 (b)Can the equation 5 x3y= 2 be derived, by a sequence of Gaussian reduction steps, from the equations in this system? 2x+ 2y= 5 3x+y= 4 (c)Can the equation 6 x9y+ 5z=2 be derived, by a sequence of Gaussian reduction steps, from the equations in the system? 2x+yz= 4 6x3y+z= 5 1.27 Prove that, where a;b;:::;e are real numbers and a6= 0, if ax+by=c has the same solution set as ax+dy=e then they are the same equation. What if a= 0? X1.28 Show that if adbc6= 0 then ax+by=j cx+dy=k has a unique solution. X1.29 In the system ax+by=c dx+ey=f each of the equations describes a line in the xy-plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are innitely many solutions. 1.30 Finish the proof of Theorem 1.5. Section I. Solving Linear Systems 11 1.31 Is there a two-unknowns linear system whose solution set is all of R2? X1.32 Are any of the operations used in Gauss' method redundant? That is, can any of the operations be made from a combination of the others? 1.33 Prove that each operation of Gauss' method is reversible. That is, show that if two systems are related by a row operation S1!S2then there is a row operation to go backS2!S1. ?1.34 A box holding pennies, nickels and dimes contains thirteen coins with a total value of 83 cents. How many coins of each type are in the box? [Anton] ?1.35 Four positive integers are given. Select any three of the integers, nd their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is: (a)19 (b)21 (c)23 (d)29 (e)17 [Con. Prob. 1955] ?X1.36 Laugh at this: AHAHA + TEHE = TEHAW. It resulted from substituting a code letter for each digit of a simple example in addition, and it is required to identify the letters and prove the solution unique. [Am. Math. Mon., Jan. 1935] ?1.37 The Wohascum County Board of Commissioners, which has 20 members, re- cently had to elect a President. There were three candidates ( A,B, andC); on each ballot the three candidates were to be listed in order of preference, with no abstentions. It was found that 11 members, a majority, preferred AoverB(thus the other 9 preferred BoverA). Similarly, it was found that 12 members preferred CoverA. Given these results, it was suggested that Bshould withdraw, to enable a runo election between AandC. However, Bprotested, and it was then found that 14 members preferred BoverC! The Board has not yet recovered from the re- sulting confusion. Given that every possible order of A,B,Cappeared on at least one ballot, how many members voted for Bas their rst choice? [Wohascum no. 2] ?1.38 \This system of nlinear equations with nunknowns," said the Great Math- ematician, \has a curious property." \Good heavens!" said the Poor Nut, \What is it?" \Note," said the Great Mathematician, \that the constants are in arithmetic progression." \It's all so clear when you explain it!" said the Poor Nut. \Do you mean like 6x+ 9y= 12 and 15 x+ 18y= 21?" \Quite so," said the Great Mathematician, pulling out his bassoon. \Indeed, the system has a unique solution. Can you nd it?" \Good heavens!" cried the Poor Nut, \I am baed." Are you? [Am. Math. Mon., Jan. 1963] I.2 Describing the Solution Set A linear system with a unique solution has a solution set with one element. A linear system with no solution has a solution set that is empty. In these cases the solution set is easy to describe. Solution sets are a challenge to describe only when they contain many elements. 12 Chapter One. Linear Systems 2.1 Example This system has many solutions because in echelon form 2x +z= 3 xyz= 1 3xy = 4(1=2)1+2! (3=2)1+32x +z= 3 y(3=2)z=1=2 y(3=2)z=1=2 2+3!2x +z= 3 y(3=2)z=1=2 0 = 0 not all of the variables are leading variables. The Gauss' method theorem showed that a triple ( x;y;z ) satises the rst system if and only if it satises the third. Thus, the solution set f(x;y;z )2x+z= 3 andxyz= 1 and 3xy= 4g can also be described as f(x;y;z )2x+z= 3 andy3z=2 =1=2g. How- ever, this second description is not much of an improvement. It has two equa- tions instead of three, but it still involves some hard-to-understand interaction among the variables. To get a description that is free of any such interaction, we take the vari- able that does not lead any equation, z, and use it to describe the variables that do lead, xandy. The second equation gives y= (1=2)(3=2)zand the rst equation gives x= (3=2)(1=2)z. Thus, the solution set can be de- scribed asf(x;y;z ) = ((3=2)(1=2)z;(1=2)(3=2)z;z)z2Rg. For instance, (1=2;5=2;2) is a solution because taking z= 2 gives a rst component of 1 =2 and a second component of 5=2. The advantage of this description over the ones above is that the only variable appearing, z, is unrestricted | it can be any real number. 2.2 Denition The non-leading variables in an echelon-form linear system arefree variables . In the echelon form system derived in the above example, xandyare leading variables and zis free. 2.3 Example A linear system can end with more than one variable free. This row reduction x+y+zw= 1 yz+w=1 3x + 6z6w= 6 y+zw= 131+3!x+y+zw= 1 yz+w=1 3y+ 3z3w= 3 y+zw= 1 32+3! 2+4x+y+zw= 1 yz+w=1 0 = 0 0 = 0 ends withxandyleading, and with both zandwfree. To get the description that we prefer we will start at the bottom. We rst express yin terms of the free variables zandwwithy=1 +zw. Next, moving up to the Section I. Solving Linear Systems 13 top equation, substituting for yin the rst equation x+ (1 +zw) +z w= 1 and solving for xyieldsx= 22z+ 2w. Thus, the solution set is f22z+ 2w;1 +zw;z;w )z;w2Rg. We prefer this description because the only variables that appear, zandw, are unrestricted. This makes the job of deciding which four-tuples are system solutions into an easy one. For instance, taking z= 1 andw= 2 gives the solution (4;2;1;2). In contrast, (3 ;2;1;2) is not a solution, since the rst component of any solution must be 2 minus twice the third component plus twice the fourth. 2.4 Example After this reduction 2x2y = 0 z+ 3w= 2 3x3y = 0 xy+ 2z+ 6w= 4(3=2)1+3! (1=2)1+42x2y = 0 z+ 3w= 2 0 = 0 2z+ 6w= 4 22+4!2x2y = 0 z+ 3w= 2 0 = 0 0 = 0 xandzlead,yandware free. The solution set is f(y;y;23w;w)y;w2Rg. For instance, (1 ;1;2;0) satises the system | take y= 1 andw= 0. The four- tuple (1;0;5;4) is not a solution since its rst coordinate does not equal its second. We refer to a variable used to describe a family of solutions as a parameter and we say that the set above is parametrized withyandw. (The terms `parameter' and `free variable' do not mean the same thing. Above, yandw are free because in the echelon form system they do not lead any row. They are parameters because they are used in the solution set description. We could have instead parametrized with yandzby rewriting the second equation as w= 2=3(1=3)z. In that case, the free variables are still yandw, but the parameters are yandz. Notice that we could not have parametrized with xand y, so there is sometimes a restriction on the choice of parameters. The terms `parameter' and `free' are related because, as we shall show later in this chapter, the solution set of a system can always be parametrized with the free variables. Consequently, we shall parametrize all of our descriptions in this way.) 2.5 Example This is another system with innitely many solutions. x+ 2y = 1 2x +z = 2 3x+ 2y+zw= 421+2! 31+3x+ 2y = 1 4y+z = 0 4y+zw= 1 2+3!x+ 2y = 1 4y+z = 0 w= 1 14 Chapter One. Linear Systems The leading variables are x,y, andw. The variable zis free. (Notice here that, although there are innitely many solutions, the value of one of the variables is xed |w=1.) Writewin terms of zwithw=1 + 0z. Theny= (1=4)z. To express xin terms of z, substitute for yinto the rst equation to get x= 1(1=2)z. The solution set is f(1(1=2)z;(1=4)z;z;1)z2Rg. We nish this subsection by developing the notation for linear systems and their solution sets that we shall use in the rest of this book. 2.6 Denition Anmnmatrix is a rectangular array of numbers with mrows andncolumns . Each number in the matrix is an entry , Matrices are usually named by upper case roman letters, e.g. A. Each entry is denoted by the corresponding lower-case letter, e.g. ai;jis the number in row i and column jof the array. For instance, A=1 2:2 5 3 47 has two rows and three columns, and so is a 2 3 matrix. (Read that as \two- by-three"; the number of rows is always stated rst.) The entry in the second row and rst column is a2;1= 3. Note that the order of the subscripts matters: a1;26=a2;1sincea1;2= 2:2. (The parentheses around the array are a typo- graphic device so that when two matrices are side by side we can tell where one ends and the other starts.) Matrices occur throughout this book. We shall use Mnmto denote the collection of nmmatrices. 2.7 Example We can abbreviate this linear system x+ 2y = 4 yz= 0 x + 2z= 4 with this matrix. 0 @1 2 0 4 0 110 1 0 2 41 A The vertical bar just reminds a reader of the dierence between the coecients on the systems's left hand side and the constants on the right. When a bar is used to divide a matrix into parts, we call it an augmented matrix. In this notation, Gauss' method goes this way. 0 @1 2 0 4 0 110 1 0 2 41 A1+3!0 @1 2 0 4 0 110 02 2 01 A22+3!0 @1 2 0 4 0 110 0 0 0 01 A The second row stands for yz= 0 and the rst row stands for x+ 2y= 4 so the solution set is f(42z;z;z )z2Rg. One advantage of the new notation is that the clerical load of Gauss' method | the copying of variables, the writing of +'s and ='s, etc. | is lighter. Section I. Solving Linear Systems 15 We will also use the array notation to clarify the descriptions of solution sets. A description like f(22z+ 2w;1 +zw;z;w )z;w2Rgfrom Ex- ample 2.3 is hard to read. We will rewrite it to group all the constants together, all the coecients of ztogether, and all the coecients of wtogether. We will write them vertically, in one-column wide matrices. f0 BB@2 1 0 01 CCA+0 BB@2 1 1 01 CCA
z+0 BB@2 1 0 11 CCA
wz;w2Rg For instance, the top line says that x= 22z+ 2wand the second line says thaty=1 +zw. The next section gives a geometric interpretation that will help us picture the solution sets when they are written in this way. 2.8 Denition Avector (orcolumn vector ) is a matrix with a single column. A matrix with a single row is a row vector . The entries of a vector are its components . Vectors are an exception to the convention of representing matrices with capital roman letters. We use lower-case roman or greek letters overlined with an arrow:~ a,~b, . . . or~ ,~, . . . (boldface is also common: aor). For instance, this is a column vector with a third component of 7. ~ v=0 @1 3 71 A 2.9 Denition The linear equation a1x1+a2x2+


+anxn=dwith unknownsx1;::: ;xnissatised by ~ s=0 B@s1 ... sn1 CA ifa1s1+a2s2+


+ansn=d. A vector satises a linear system if it satises each equation in the system. The style of description of solution sets that we use involves adding the vectors, and also multiplying them by real numbers, such as the zandw. We need to dene these operations. 2.10 Denition The vector sum of~ uand~ vis this. ~ u+~ v=0 B@u1 ... un1 CA+0 B@v1 ... vn1 CA=0 B@u1+v1 ... un+vn1 CA 16 Chapter One. Linear Systems Note that the vectors have to have the same number of entries for the addi- tion to be dened. This entry-by-entry addition works for any pair of matrices, not just vectors, provided that they have the same number of rows and columns. 2.11 Denition The scalar multiplication of the real number rand the vector ~ vis this. r
~ v=r
0 B@v1 ... vn1 CA=0 B@rv1 ... rvn1 CA As with the addition operation, this entry-by-entry scalar multiplication operation extends beyond vectors to any matrix. Scalar multiplication can be written in either order: r
~ vor~ v
r, or without the `
' symbol:r~ v. (Do not refer to scalar multiplication as `scalar product' because that name is used for a dierent operation.) 2.12 Example 0 @2 3 11 A+0 @3 1 41 A=0 @2 + 3 31 1 + 41 A=0 @5 2 51 A 7
0 BB@1 4 1 31 CCA=0 BB@7 28 7 211 CCA Notice that the denitions of vector addition and scalar multiplication agree where they overlap, for instance, ~ v+~ v= 2~ v. With the notation dened, we can now solve systems in the way that we will use throughout this book. 2.13 Example This system 2x+yw = 4 y +w+u= 4 xz+ 2w = 0 reduces in this way. 0 @2 1 01 0 4 0 1 0 1 1 4 1 01 2 0 01 A(1=2)1+3!0 @2 1 01 0 4 0 1 0 1 1 4 01=21 5=2 021 A (1=2)2+3!0 @2 1 01 0 4 0 1 0 1 1 4 0 01 3 1=201 A The solution set is f(w+ (1=2)u;4wu;3w+ (1=2)u;w;u )w;u2Rg. We write that in vector form. f0 BBBB@x y z w u1 CCCCA=0 BBBB@0 4 0 0 01 CCCCA+0 BBBB@1 1 3 1 01 CCCCAw+0 BBBB@1=2 1 1=2 0 11 CCCCAuw;u2Rg Section I. Solving Linear Systems 17 Note again how well vector notation sets o the coecients of each parameter. For instance, the third row of the vector form shows plainly that if uis held xed thenzincreases three times as fast as w. That format also shows plainly that there are innitely many solutions. For example, we can x uas 0, letwrange over the real numbers, and consider the rst component x. We get innitely many rst components and hence innitely many solutions. Another thing shown plainly is that setting both wanduto zero gives that this vector 0 BBBB@x y z w u1 CCCCA=0 BBBB@0 4 0 0 01 CCCCA is a particular solution of the linear system. 2.14 Example In the same way, this system xy+z= 1 3x +z= 3 5x2y+ 3z= 5 reduces 0 @11 1 1 3 0 1 3 52 3 51 A31+2! 51+30 @11 1 1 0 320 0 3201 A2+3!0 @11 1 1 0 320 0 0 0 01 A to a one-parameter solution set. f0 @1 0 01 A+0 @1=3 2=3 11 Azz2Rg Before the exercises, we pause to point out some things that we have yet to do. The rst two subsections have been on the mechanics of Gauss' method. Except for one result, Theorem 1.5 | without which developing the method doesn't make sense since it says that the method gives the right answers | we have not stopped to consider any of the interesting questions that arise. For example, can we always describe solution sets as above, with a particular solution vector added to an unrestricted linear combination of some other vec- tors? The solution sets we described with unrestricted parameters were easily seen to have innitely many solutions so an answer to this question could tell us something about the size of solution sets. An answer to that question could also help us picture the solution sets, in R2, or in R3, etc. Many questions arise from the observation that Gauss' method can be done in more than one way (for instance, when swapping rows, we may have a choice 18 Chapter One. Linear Systems of which row to swap with). Theorem 1.5 says that we must get the same solution set no matter how we proceed, but if we do Gauss' method in two dierent ways must we get the same number of free variables both times, so that any two solution set descriptions have the same number of parameters? Must those be the same variables (e.g., is it impossible to solve a problem one way and get yandwfree or solve it another way and get yandzfree)? In the rest of this chapter we answer these questions. The answer to each is `yes'. The rst question is answered in the last subsection of this section. In the second section we give a geometric description of solution sets. In the nal section of this chapter we tackle the last set of questions. Consequently, by the end of the rst chapter we will not only have a solid grounding in the practice of Gauss' method, we will also have a solid grounding in the theory. We will be sure of what can and cannot happen in a reduction. Exercises X2.15 Find the indicated entry of the matrix, if it is dened. A=1 3 1 21 4 (a)a2;1(b)a1;2(c)a2;2(d)a3;1 X2.16 Give the size of each matrix. (a)1 0 4 2 1 5 (b)0 @1 1 1 1 311 A (c)5 10 10 5 X2.17 Do the indicated vector operation, if it is dened. (a)0 @2 1 11 A+0 @3 0 41 A (b)54 1 (c)0 @1 5 11 A0 @3 1 11 A (d)72 1 + 93 5 (e)1 2 +0 @1 2 31 A (f)60 @3 1 11 A40 @2 0 31 A+ 20 @1 1 51 A X2.18 Solve each system using matrix notation. Express the solution using vec- tors. (a)3x+ 6y= 18 x+ 2y= 6(b)x+y= 1 xy=1(c)x1 +x3= 4 x1x2+ 2x3= 5 4x1x2+ 5x3= 17 (d)2a+bc= 2 2a +c= 3 ab = 0(e)x+ 2yz = 3 2x+y +w= 4 xy+z+w= 1(f)x +z+w= 4 2x+yw= 2 3x+y+z = 7 X2.19 Solve each system using matrix notation. Give each solution set in vector notation. (a)2x+yz= 1 4xy = 3(b)xz = 1 y+ 2zw= 3 x+ 2y+ 3zw= 7(c)xy+z = 0 y +w= 0 3x2y+ 3z+w= 0 yw= 0 (d)a+ 2b+ 3c+de= 1 3ab+c+d+e= 3 Section I. Solving Linear Systems 19 X2.20 The vector is in the set. What value of the parameters produces that vec- tor? (a)5 5 ,f1 1 kk2Rg (b)0 @1 2 11 A,f0 @2 1 01 Ai+0 @3 0 11 Aji;j2Rg (c)0 @0 4 21 A,f0 @1 1 01 Am+0 @2 0 11 Anm;n2Rg 2.21 Decide if the vector is in the set. (a)3 1 ,f6 2 kk2Rg (b)5 4 ,f5 4 jj2Rg (c)0 @2 1 11 A,f0 @0 3 71 A+0 @1 1 31 Arr2Rg (d)0 @1 0 11 A,f0 @2 0 11 Aj+0 @3 1 11 Akj;k2Rg 2.22 Parametrize the solution set of this one-equation system. x1+x2+


+xn= 0 X2.23 (a) Apply Gauss' method to the left-hand side to solve x+ 2yw=a 2x +z =b x+y + 2w=c forx,y,z, andw, in terms of the constants a,b, andc. (b)Use your answer from the prior part to solve this. x+ 2yw= 3 2x +z = 1 x+y + 2w=2 X2.24 Why is the comma needed in the notation ` ai;j' for matrix entries? X2.25 Give the 44 matrix whose i;j-th entry is (a)i+j;(b)1 to thei+jpower. 2.26 For any matrix A, the transpose ofA, writtenAtrans, is the matrix whose columns are the rows of A. Find the transpose of each of these. (a)1 2 3 4 5 6 (b)23 1 1 (c)5 10 10 5 (d)0 @1 1 01 A X2.27 (a) Describe all functions f(x) =ax2+bx+csuch thatf(1) = 2 and f(1) = 6. (b)Describe all functions f(x) =ax2+bx+csuch thatf(1) = 2. 2.28 Show that any set of ve points from the plane R2lie on a common conic section, that is, they all satisfy some equation of the form ax2+by2+cxy+dx+ ey+f= 0 where some of a; ::: ;f are nonzero. 2.29 Make up a four equations/four unknowns system having (a)a one-parameter solution set; 20 Chapter One. Linear Systems (b)a two-parameter solution set; (c)a three-parameter solution set. ?2.30 (a) Solve the system of equations. ax+y=a2 x+ay= 1 For what values of adoes the system fail to have solutions, and for what values ofaare there innitely many solutions? (b)Answer the above question for the system. ax+y=a3 x+ay= 1 [USSR Olympiad no. 174] ?2.31 In air a gold-surfaced sphere weighs 7588 grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerine its respective weights are 6588, 6688, 6778, and 6328 grams. How much, if any, of the forenamed metals does it contain if the specic gravities of the designated substances are taken to be as follows? Aluminum 2 :7 Alcohol 0.81 Copper 8 :9 Benzene 0 :90 Gold 19 :3 Glycerine 1 :26 Lead 11 :3 Water 1 :00 Silver 10 :8 [Math. Mag., Sept. 1952] I.3 General = Particular + Homogeneous The prior subsection has many descriptions of solution sets. They all t a pattern. They have a vector that is a particular solution of the system added to an unrestricted combination of some other vectors. The solution set from Example 2.13 illustrates. f0 BBBB@0 4 0 0 01 CCCCA |{z} particular solution+w0 BBBB@1 1 3 1 01 CCCCA+u0 BBBB@1=2 1 1=2 0 11 CCCCA |{z} unrestricted combinationw;u2Rg The combination is unrestricted in that wanducan be any real numbers | there is no condition like \such that 2 wu= 0" that would restrict which pairs w;u can be used to form combinations. That example shows an innite solution set conforming to the pattern. We can think of the other two kinds of solution sets as tting the same pattern. A one-element solution set ts the pattern in that it has a particular solution, and Section I. Solving Linear Systems 21 the unrestricted combination part is a trivial sum. (That is, instead of being a combination of two vectors, as above, or a combination of one vector, it is a combination of no vectors. We will use the convention that the sum of an empty set of vectors is the vector of all zeros.) A zero-element solution set ts the pattern since there is no particular solution, and so there are no sums of that form. This subsection formally proves what the prior paragraph outlines: every solution set can be written as a vector that is a particular solution of the system added to an unrestricted combination of some other vectors. 3.1 Theorem Any linear system's solution set can be described as f~ p+c1~1+


+ck~kc1; ::: ;ck2Rg where~ pis any particular solution, and where the number of vectors ~1, . . . , ~kequals the number of free variables that the system has after a Gaussian reduction. The solution description has two parts, the particular solution ~ pand also the unrestricted linear combination of the ~'s. We shall prove the theorem in two corresponding parts, with two lemmas. We will focus rst on the unrestricted combination part. To do that, we consider systems that have the vector of zeroes as one of the particular solutions, so that~ p+c1~1+


+ck~kcan be shortened to c1~1+


+ck~k. 3.2 Denition A linear equation is homogeneous if it has a constant of zero, that is, if it can be put in the form a1x1+a2x2+


+anxn= 0. 3.3 Example With any linear system like 3x+ 4y= 3 2xy= 1 we associate a system of homogeneous equations by setting the right side to zeros. 3x+ 4y= 0 2xy= 0 Our interest in the homogeneous system associated with a linear system can be understood by comparing the reduction of the system 3x+ 4y= 3 2xy= 1(2=3)1+2!3x+ 4y= 3 (11=3)y=1 with the reduction of the associated homogeneous system. 3x+ 4y= 0 2xy= 0(2=3)1+2!3x+ 4y= 0 (11=3)y= 0 Obviously the two reductions go in the same way. We can study how linear sys- tems are reduced by instead studying how the associated homogeneous systems are reduced. 22 Chapter One. Linear Systems Studying the associated homogeneous system has a great advantage over studying the original system. Nonhomogeneous systems can be inconsistent. But a homogeneous system must be consistent since there is always at least one solution, the vector of zeros. 3.4 Denition A column or row vector of all zeros is a zero vector , denoted ~0. There are many dierent zero vectors, e.g., the one-tall zero vector, the two-tall zero vector, etc. Nonetheless, people often refer to \the" zero vector, expecting that the size of the one being discussed will be clear from the context. 3.5 Example Some homogeneous systems have the zero vector as their only solution. 3x+ 2y+z= 0 6x+ 4y = 0 y+z= 021+2!3x+ 2y+z= 0 2z= 0 y+z= 02$3!3x+ 2y+z= 0 y+z= 0 2z= 0 3.6 Example Some homogeneous systems have many solutions. One example is the Chemistry problem from the rst page of this book. 7x7z = 0 8x+y5z2w= 0 y3z = 0 3y6zw= 0(8=7)1+2!7x7z = 0 y+ 3z2w= 0 y3z = 0 3y6zw= 0 2+3! 32+47x 7z = 0 y+ 3z2w= 0 6z+ 2w= 0 15z+ 5w= 0 (5=2)3+4!7x7z = 0 y+ 3z2w= 0 6z+ 2w= 0 0 = 0 The solution set: f0 BB@1=3 1 1=3 11 CCAww2Rg has many vectors besides the zero vector (if we interpret was a number of molecules then solutions make sense only when wis a nonnegative multiple of 3). We now have the terminology to prove the two parts of Theorem 3.1. The rst lemma deals with unrestricted combinations. Section I. Solving Linear Systems 23 3.7 Lemma For any homogeneous linear system there exist vectors ~1, . . . , ~ksuch that the solution set of the system is fc1~1+


+ck~kc1;:::;ck2Rg wherekis the number of free variables in an echelon form version of the system. Before the proof, we will recall the back substitution calculations that were done in the prior subsection. Imagine that we have brought a system to this echelon form. x+ 2yz+ 2w= 0 3y+z = 0 w= 0 We next perform back-substitution to express each variable in terms of the free variable z. Working from the bottom up, we get rst that wis 0
z, next thatyis (1=3)
z, and then substituting those two into the top equation x+ 2((1=3)z)z+ 2(0) = 0 gives x= (1=3)
z. So, back substitution gives a parametrization of the solution set by starting at the bottom equation and using the free variables as the parameters to work row-by-row to the top. The proof below follows this pattern. Comment: That is, this proof just does a verication of the bookkeeping in back substitution to show that we haven't overlooked any obscure cases where this procedure fails, say, by leading to a division by zero. So this argument, while quite detailed, doesn't give us any new insights. Nevertheless, we have written it out for two reasons. The rst reason is that we need the result | the computational procedure that we employ must be veried to work as promised. The second reason is that the row-by-row nature of back substitution leads to a proof that uses the technique of mathematical induction.This is an important, and non-obvious, proof technique that we shall use a number of times in this book. Doing an induction argument here gives us a chance to see one in a setting where the proof material is easy to follow, and so the technique can be studied. Readers who are unfamiliar with induction arguments should be sure to master this one and the ones later in this chapter before going on to the second chapter. Proof .First use Gauss' method to reduce the homogeneous system to echelon form. We will show that each leading variable can be expressed in terms of free variables. That will nish the argument because then we can use those free variables as the parameters. That is, the ~'s are the vectors of coecients of the free variables (as in Example 3.6, where the solution is x= (1=3)w,y=w, z= (1=3)w, andw=w). We will proceed by mathematical induction, which has two steps. The base step of the argument will be to focus on the bottom-most non-`0 = 0' equation and write its leading variable in terms of the free variables. The inductive step of the argument will be to argue that if we can express the leading variables from More information on mathematical induction is in the appendix. 24 Chapter One. Linear Systems the bottom trows in terms of free variables, then we can express the leading variable of the next row up | the t+ 1-th row up from the bottom | in terms of free variables. With those two steps, the theorem will be proved because by the base step it is true for the bottom equation, and by the inductive step the fact that it is true for the bottom equation shows that it is true for the next one up, and then another application of the inductive step implies it is true for the third equation up, etc. For the base step, consider the bottom-most non-`0 = 0' equation (the case where all the equations are `0 = 0' is trivial). We call that the m-th row: am;`mx`m+am;`m+1x`m+1+


+am;nxn= 0 wheream;`m6= 0. (The notation here has ` `' stand for `leading', so am;`mmeans \the coecient from the row mof the variable leading row m".) Either there are variables in this equation other than the leading one x`mor else there are not. If there are other variables x`m+1, etc., then they must be free variables because this is the bottom non-`0 = 0' row. Move them to the right and divide byam;`m x`m= (am;`m+1=am;`m)x`m+1+


+ (am;n=am;`m)xn to express this leading variable in terms of free variables. If there are no free variables in this equation then x`m= 0 (see the \tricky point" noted following this proof). For the inductive step, we assume that for the m-th equation, and for the (m1)-th equation, . . . , and for the ( mt)-th equation, we can express the leading variable in terms of free variables (where 0 t<m ). To prove that the same is true for the next equation up, the ( m(t+ 1))-th equation, we take each variable that leads in a lower-down equation x`m;:::;x`mtand substitute its expression in terms of free variables. The result has the form am(t+1);`m(t+1)x`m(t+1)+ sums of multiples of free variables = 0 wheream(t+1);`m(t+1)6= 0. We move the free variables to the right-hand side and divide by am(t+1);`m(t+1), to end with x`m(t+1)expressed in terms of free variables. Because we have shown both the base step and the inductive step, by the principle of mathematical induction the proposition is true. QED We say that the set fc1~1+


+ck~kc1;:::;ck2Rgisgenerated by or spanned by the set of vectors f~1;:::;~kg. There is a tricky point to this. We rely on the convention that the sum of an empty set of vectors is the zero vector. In particular, we need this in the case where a homogeneous system has a unique solution. Then the homogeneous case ts the pattern of the other solution sets: in the proof above, the solution set is derived by taking the c's to be the free variables and if there is a unique solution then there are no free variables. Section I. Solving Linear Systems 25 The proof incidentally shows, as discussed after Example 2.4, that solution sets can always be parametrized using the free variables. The next lemma nishes the proof of Theorem 3.1 by considering the par- ticular solution part of the solution set's description. 3.8 Lemma For a linear system, where ~ pis any particular solution, the solution set equals this set. f~ p+~h~hsatises the associated homogeneous system g Proof .We will show mutual set inclusion, that any solution to the system is in the above set and that anything in the set is a solution to the system. For set inclusion the rst way, that if a vector solves the system then it is in the set described above, assume that ~ ssolves the system. Then ~ s~ psolves the associated homogeneous system since for each equation index i, ai;1(s1p1) +


+ai;n(snpn) = (ai;1s1+


+ai;nsn) (ai;1p1+


+ai;npn) =didi = 0 wherepjandsjare thej-th components of ~ pand~ s. We can write ~ s~ pas~h, where~hsolves the associated homogeneous system, to express ~ sin the required ~ p+~hform. For set inclusion the other way, take a vector of the form ~ p+~h, where~ p solves the system and ~hsolves the associated homogeneous system, and note that it solves the given system: for any equation index i, ai;1(p1+h1) +


+ai;n(pn+hn) = (ai;1p1+


+ai;npn) + (ai;1h1+


+ai;nhn) =di+ 0 =di wherehjis thej-th component of ~h. QED The two lemmas above together establish Theorem 3.1. We remember that theorem with the slogan \General = Particular + Homogeneous". 3.9 Example This system illustrates Theorem 3.1. x+ 2yz= 1 2x+ 4y = 2 y3z= 0 More information on equality of sets is in the appendix. 26 Chapter One. Linear Systems Gauss' method 21+2!x+ 2yz= 1 2z= 0 y3z= 02$3!x+ 2yz= 1 y3z= 0 2z= 0 shows that the general solution is a singleton set. f0 @1 0 01 Ag That single vector is, of course, a particular solution. The associated homoge- neous system reduces via the same row operations x+ 2yz= 0 2x+ 4y = 0 y3z= 021+2!2$3!x+ 2yz= 0 y3z= 0 2z= 0 to also give a singleton set. f0 @0 0 01 Ag As the theorem states, and as discussed at the start of this subsection, in this single-solution case the general solution results from taking the particular solu- tion and adding to it the unique solution of the associated homogeneous system. 3.10 Example Also discussed at the start of this subsection is that the case where the general solution set is empty ts the `General = Particular + Homogeneous' pattern. This system illustrates. Gauss' method x +z+w=1 2xy +w= 3 x+y+ 3z+ 2w= 121+2! 1+3x +z+w=1 y2zw= 5 y+ 2z+w= 2 shows that it has no solutions because the nal two equations are in con ict. The associated homogeneous system, of course, has a solution. x +z+w= 0 2xy +w= 0 x+y+ 3z+ 2w= 021+2! 1+32+3!x +z+w= 0 y2zw= 0 0 = 0 In fact, the solution set of the homogeneous system is innite. f0 BB@1 2 1 01 CCAz+0 BB@1 1 0 11 CCAwz;w2Rg However, because no particular solution of the original system exists, the general solution set is empty | there are no vectors of the form ~ p+~hbecause there are no~ p's. Section I. Solving Linear Systems 27 3.11 Corollary Solution sets of linear systems are either empty, have one element, or have innitely many elements. Proof .We've seen examples of all three happening so we need only prove that those are the only possibilities. First, notice a homogeneous system with at least one non- ~0 solution~ vhas innitely many solutions because the set of multiples s~ vis innite | if s6= 1 thens~ v~ v= (s1)~ vis easily seen to be non- ~0, and sos~ v6=~ v. Now, apply Lemma 3.8 to conclude that a solution set f~ p+~h~hsolves the associated homogeneous system g is either empty (if there is no particular solution ~ p), or has one element (if there is a~ pand the homogeneous system has the unique solution ~0), or is innite (if there is a~ pand the homogeneous system has a non- ~0 solution, and thus by the prior paragraph has innitely many solutions). QED This table summarizes the factors aecting the size of a general solution. number of solutions of the associated homogeneous system particular solution exists?one innitely many yesunique solutioninnitely many solutions nono solutionsno solutions The factor on the top of the table is the simpler one. When we perform Gauss' method on a linear system, ignoring the constants on the right side and so paying attention only to the coecients on the left-hand side, we either end with every variable leading some row or else we nd that some variable does not lead a row, that is, that some variable is free. (Of course, \ignoring the constants on the right" is formalized by considering the associated homogeneous system. We are simply putting aside for the moment the possibility of a contradictory equation.) A nice insight into the factor on the top of this table at work comes from con- sidering the case of a system having the same number of equations as variables. This system will have a solution, and the solution will be unique, if and only if it reduces to an echelon form system where every variable leads its row, which will happen if and only if the associated homogeneous system has a unique solution. Thus, the question of uniqueness of solution is especially interesting when the system has the same number of equations as variables. 3.12 Denition A square matrix is nonsingular if it is the matrix of coe- cients of a homogeneous system with a unique solution. It is singular otherwise, that is, if it is the matrix of coecients of a homogeneous system with innitely many solutions. 28 Chapter One. Linear Systems The word singular means \departing from general expectation" and here expresses that we could expect that systems with the same number of equations as unknowns will typically have a unique solution. (That `singular' applies to systems having more than one solution is ironic, but it is the standard term.) 3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9 each have an associated homogeneous system with a unique solution. Thus these matrices are nonsingular.  3 4 210 @3 2 1 64 0 0 1 11 A0 @1 21 2 4 0 0 131 A The Chemistry problem from Example 3.6 is a homogeneous system with more than one solution so its matrix is singular. 0 BB@7 07 0 8 152 0 13 0 0 3611 CCA 3.14 Example The rst of these matrices is nonsingular while the second is singular 1 2 3 4  1 2 3 6 because the rst of these homogeneous systems has a unique solution while the second has innitely many solutions. x+ 2y= 0 3x+ 4y= 0x+ 2y= 0 3x+ 6y= 0 We have made the distinction in the denition because a system (with the same number of equations as variables) behaves in one of two ways, depending on whether its matrix of coecients is nonsingular or singular. A system where the matrix of coecients is nonsingular has a unique solution for any constants on the right side: for instance, Gauss' method shows that this system x+ 2y=a 3x+ 4y=b has the unique solution x=b2aandy= (3ab)=2. On the other hand, a system where the matrix of coecients is singular never has a unique solution | it has either no solutions or else has innitely many, as with these. x+ 2y= 1 3x+ 6y= 2x+ 2y= 1 3x+ 6y= 3 Thus, `singular' can be thought of as connoting \troublesome", or at least \not ideal". Section I. Solving Linear Systems 29 The above table has two factors. We have already considered the factor along the top: we can tell which column a given linear system goes in solely by considering the system's left-hand side | the constants on the right-hand side play no role in this factor. The table's other factor, determining whether a particular solution exists, is tougher. Consider these two 3x+ 2y= 5 3x+ 2y= 53x+ 2y= 5 3x+ 2y= 4 with the same left sides but dierent right sides. Obviously, the rst has a solution while the second does not, so here the constants on the right side decide if the system has a solution. We could conjecture that the left side of a linear system determines the number of solutions while the right side determines if solutions exist, but that guess is not correct. Compare these two systems 3x+ 2y= 5 4x+ 2y= 43x+ 2y= 5 3x+ 2y= 4 with the same right sides but dierent left sides. The rst has a solution but the second does not. Thus the constants on the right side of a system don't decide alone whether a solution exists; rather, it depends on some interaction between the left and right sides. For some intuition about that interaction, consider this system with one of the coecients left as the parameter c. x+ 2y+ 3z= 1 x+y+z= 1 cx+ 3y+ 4z= 0 Ifc= 2 then this system has no solution because the left-hand side has the third row as a sum of the rst two, while the right-hand does not. If c6= 2 then this system has a unique solution (try it with c= 1). For a system to have a solution, if one row of the matrix of coecients on the left is a linear combination of other rows, then on the right the constant from that row must be the same combination of constants from the same rows. More intuition about the interaction comes from studying linear combina- tions. That will be our focus in the second chapter, after we nish the study of Gauss' method itself in the rest of this chapter. Exercises X3.15 Solve each system. Express the solution set using vectors. Identify the par- ticular solution and the solution set of the homogeneous system. (a)3x+ 6y= 18 x+ 2y= 6(b)x+y= 1 xy=1(c)x1 +x3= 4 x1x2+ 2x3= 5 4x1x2+ 5x3= 17 (d)2a+bc= 2 2a +c= 3 ab = 0(e)x+ 2yz = 3 2x+y +w= 4 xy+z+w= 1(f)x +z+w= 4 2x+yw= 2 3x+y+z = 7 3.16 Solve each system, giving the solution set in vector notation. Identify the particular solution and the solution of the homogeneous system. 30 Chapter One. Linear Systems (a)2x+yz= 1 4xy = 3(b)xz = 1 y+ 2zw= 3 x+ 2y+ 3zw= 7(c)xy+z = 0 y +w= 0 3x2y+ 3z+w= 0 yw= 0 (d)a+ 2b+ 3c+de= 1 3ab+c+d+e= 3 X3.17 For the system 2xyw= 3 y+z+ 2w= 2 x2yz =1 which of these can be used as the particular solution part of some general solu- tion? (a)0 BB@0 3 5 01 CCA(b)0 BB@2 1 1 01 CCA(c)0 BB@1 4 8 11 CCA X3.18 Lemma 3.8 says that any particular solution may be used for ~ p. Find, if possible, a general solution to this system xy +w= 4 2x+ 3yz = 0 y+z+w= 4 that uses the given vector as its particular solution. (a)0 BB@0 0 0 41 CCA(b)0 BB@5 1 7 101 CCA(c)0 BB@2 1 1 11 CCA 3.19 One of these is nonsingular while the other is singular. Which is which? (a)1 3 412 (b)1 3 4 12 X3.20 Singular or nonsingular? (a)1 2 1 3 (b)1 2 36 (c)1 2 1 1 3 1 (Careful!) (d)0 @1 2 1 1 1 3 3 4 71 A (e)0 @2 2 1 1 0 5 1 1 41 A X3.21 Is the given vector in the set generated by the given set? (a)2 3 ;f1 4 ;1 5 g (b)0 @1 0 11 A;f0 @2 1 01 A;0 @1 0 11 Ag (c)0 @1 3 01 A;f0 @1 0 41 A;0 @2 1 51 A;0 @3 3 01 A;0 @4 2 11 Ag (d)0 BB@1 0 1 11 CCA;f0 BB@2 1 0 11 CCA;0 BB@3 0 0 21 CCAg Section I. Solving Linear Systems 31 3.22 Prove that any linear system with a nonsingular matrix of coecients has a solution, and that the solution is unique. 3.23 To tell the whole truth, there is another tricky point to the proof of Lemma 3.7. What happens if there are no non-`0 = 0' equations? (There aren't any more tricky points after this one.) X3.24 Prove that if ~ sand~tsatisfy a homogeneous system then so do these vec- tors. (a)~ s+~t(b)3~ s(c)k~ s+m~tfork;m2R What's wrong with: \These three show that if a homogeneous system has one solution then it has many solutions | any multiple of a solution is another solution, and any sum of solutions is a solution also | so there are no homogeneous systems with exactly one solution."? 3.25 Prove that if a system with only rational coecients and constants has a solution then it has at least one all-rational solution. Must it have innitely many? 32 Chapter One. Linear Systems II Linear Geometry of n-Space For readers who have seen the elements of vectors before, in calculus or physics, this section is an optional review. However, later work will refer to this material so if it is not a review then it is not optional. In the rst section, we had to do a bit of work to show that there are only three types of solution sets | singleton, empty, and innite. But in the special case of systems with two equations and two unknowns this is easy to see with a picture. Draw each two-unknowns equation as a line in the plane and then the two lines could have a unique intersection, be parallel, or be the same line. Unique solution 3x+ 2y= 7 xy=1No solutions 3x+ 2y= 7 3x+ 2y= 4Innitely many solutions 3x+ 2y= 7 6x+ 4y= 14 These pictures don't prove the results from the prior section, which apply to any number of linear equations and any number of unknowns, but nonetheless they do help us to understand those results. This section develops the ideas that we need to express our results from the prior section, and from some future sections, geometrically. In particular, while the two-dimensional case is familiar, to extend to systems with more than two unknowns we shall need some higher- dimensional geometry. II.1 Vectors in Space \Higher-dimensional geometry" sounds exotic. It is exotic | interesting and eye-opening. But it isn't distant or unreachable. We begin by dening one-dimensional space to be the set R1. To see that denition is reasonable, draw a one-dimensional space and make the usual correspondence with R: pick a point to label 0 and another to label 1. 0 1 Now, with a scale and a direction, nding the point corresponding to, say +2 :17, is easy | start at 0 and head in the direction of 1 (i.e., the positive direction), but don't stop there, go 2 :17 times as far. Section II. Linear Geometry of n-Space 33 The basic idea here, combining magnitude with direction, is the key to ex- tending to higher dimensions. An object comprised of a magnitude and a direction is a vector (we will use the same word as in the previous section because we shall show below how to describe such an object with a column vector). We can draw a vector as having some length, and pointing somewhere. There is a subtlety here | these vectors are equal, even though they start in dierent places, because they have equal lengths and equal directions. Again: those vectors are not just alike, they are equal. How can things that are in dierent places be equal? Think of a vector as representing a displacement (`vector' is Latin for \carrier" or \traveler"). These squares undergo the same displacement, despite that those displacements start in dierent places. Sometimes, to emphasize this property vectors have of not being anchored, they are referred to as freevectors. Thus, these free vectors are equal as each is a displacement of one over and two up. More generally, vectors in the plane are the same if and only if they have the same change in rst components and the same change in second components: the vector extending from ( a1;a2) to (b1;b2) equals the vector from ( c1;c2) to (d1;d2) if and only if b1a1=d1c1andb2a2=d2c2. An expression like `the vector that, were it to start at ( a1;a2), would extend to (b1;b2)' is awkward. We instead describe such a vector as b1a1 b2a2 so that, for instance, the `one over and two up' arrows shown above picture this vector.  1 2 34 Chapter One. Linear Systems We often draw the arrow as starting at the origin, and we then say it is in the canonical position (ornatural position orstandard position ). When the vector b1a1 b2a2 is in its canonical position then it extends to the endpoint ( b1a1;b2a2). We typically just refer to \the point 1 2 " rather than \the endpoint of the canonical position of" that vector. Thus, we will call both of these sets R2. f(x1;x2)x1;x22Rg fx1 x2x1;x22Rg In the prior section we dened vectors and vector operations with an alge- braic motivation; r
v1 v2 =rv1 rv2 v1 v2 +w1 w2 =v1+w1 v2+w2 we can now interpret those operations geometrically. For instance, if ~ vrepre- sents a displacement then 3 ~ vrepresents a displacement in the same direction but three times as far, and 1~ vrepresents a displacement of the same distance as~ vbut in the opposite direction. ~ v ~ v3~ v And, where ~ vand~ wrepresent displacements, ~ v+~ wrepresents those displace- ments combined. ~ v~ w~ v+~ w The long arrow is the combined displacement in this sense: if, in one minute, a ship's motion gives it the displacement relative to the earth of ~ vand a passen- ger's motion gives a displacement relative to the ship's deck of ~ w, then~ v+~ wis the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule . Draw the parallelogram formed by the vectors ~ v1;~ v2and then the sum ~ v1+~ v2 extends along the diagonal to the far corner. Section II. Linear Geometry of n-Space 35 ~ v+~ w ~ v~ w The above drawings show how vectors and vector operations behave in R2. We can extend to R3, or to even higher-dimensional spaces where we have no pictures, with the obvious generalization: the free vector that, if it starts at (a1;:::;an), ends at ( b1;:::;bn), is represented by this column 0 B@b1a1 ... bnan1 CA (vectors are equal if they have the same representation), we aren't too careful to distinguish between a point and the vector whose canonical representation ends at that point, Rn=f0 B@v1 ... vn1 CAv1;:::;vn2Rg and addition and scalar multiplication are done component-wise. Having considered points, we now turn to the lines. In R2, the line through (1;2) and (3;1) is comprised of (the endpoints of) the vectors in this set. f1 2 +t
2 1t2Rg That description expresses this picture.  2 1 = 3 1  1 2 The vector associated with the parameter t 2 1 =3 1 1 2 has its whole body in the line | it is a direction vector for the line. Note that points on the line to the left of x= 1 are described using negative values of t. Note also that this description of lines generalizes the familiar y=b+mxform for lines in the plane. InR3, the line through (1 ;2;1) and (2;3;2) is the set of (endpoints of) vectors of this form 36 Chapter One. Linear Systems f0 @1 2 11 A+t
0 @1 1 11 At2Rg and lines in even higher-dimensional spaces work in the same way. InR3, a line uses one parameter so that there is freedom to move back and forth in one dimension, and a plane involves two parameters. For exam- ple, the plane through the points (1 ;0;5), (2;1;3), and (2;4;0:5) consists of (endpoints of) the vectors in f0 @1 0 51 A+t
0 @1 1 81 A+s
0 @3 4 4:51 At;s2Rg (the column vectors associated with the parameters 0 @1 1 81 A=0 @2 1 31 A0 @1 0 51 A0 @3 4 4:51 A=0 @2 4 0:51 A0 @1 0 51 A are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative t's or negative s's or both. In algebra and calculus we often use a description of planes involving a single equation as the condition that describes the relationship among the rst, second, and third coordinates of points in a plane. P=f0 @x y z1 A2x+y+z= 4g The translation from such a description to the vector description that we favor in this book is to think of the condition as a one-equation linear system and parametrize x= (1=2)(4yz). P=f0 @2 0 01 A+0 @0:5 1 01 Ay+0 @0:5 0 11 Azy;z2Rg Section II. Linear Geometry of n-Space 37 Generalizing from lines and planes, we dene a k-dimensional linear sur- face(ork- at) inRnto bef~ p+t1~ v1+t2~ v2+


+tk~ vkt1;:::;tk2Rgwhere ~ v1;:::;~ vk2Rn. For example, in R4, f0 BB@2  3 0:51 CCA+t0 BB@1 0 0 01 CCAt2Rg is a line, f0 BB@0 0 0 01 CCA+t0 BB@1 1 0 11 CCA+s0 BB@2 0 1 01 CCAt;s2Rg is a plane, and f0 BB@3 1 2 0:51 CCA+r0 BB@0 0 0 11 CCA+s0 BB@1 0 1 01 CCA+t0 BB@2 0 1 01 CCAr;s;t2Rg is a three-dimensional linear surface. Again, the intuition is that a line permits motion in one direction, a plane permits motion in combinations of two direc- tions, etc. (When kis one less than the dimension of the space, that is in Rn whenk=n1, then ak-dimensional linear surface is called a hyperplane .) The description of a linear surface can be misleading about the dimension | this L=f0 BB@1 0 1 21 CCA+t0 BB@1 1 0 11 CCA+s0 BB@2 2 0 21 CCAt;s2Rg is adegenerate plane because it is actually a line | the vectors are multiples of each other so we can merge the two into one. L=f0 BB@1 0 1 21 CCA+r0 BB@1 1 0 11 CCAr2Rg We shall see in the Linear Independence section of Chapter Two what relation- ships among vectors causes the linear surface they generate to be degenerate. We nish this subsection by restating our conclusions from the rst section in geometric terms. First, the solution set of a linear system with nunknowns is a linear surface in Rn. Specically, it is a k-dimensional linear surface, where kis the number of free variables in an echelon form version of the system. Second, the solution set of a homogeneous linear system is a linear surface passing through the origin. Finally, we can view the general solution set of any linear system as being the solution set of its associated homogeneous system oset from the origin by a vector, namely by any particular solution. 38 Chapter One. Linear Systems Exercises X1.1Find the canonical name for each vector. (a)the vector from (2 ;1) to (4;2) inR2 (b)the vector from (3 ;3) to (2;5) inR2 (c)the vector from (1 ;0;6) to (5;0;3) inR3 (d)the vector from (6 ;8;8) to (6;8;8) inR3 X1.2Decide if the two vectors are equal. (a)the vector from (5 ;3) to (6;2) and the vector from (1 ;2) to (1;1) (b)the vector from (2 ;1;1) to (3;0;4) and the vector from (5 ;1;4) to (6;0;7) X1.3Does (1;0;2;1) lie on the line through ( 2;1;1;0) and (5;10;1;4)? X1.4 (a) Describe the plane through (1 ;1;5;1), (2;2;2;0), and (3;1;0;4). (b)Is the origin in that plane? 1.5Describe the plane that contains this point and line.0 @2 0 31 Af0 @1 0 41 A+0 @1 1 21 Att2Rg X1.6Intersect these planes. f0 @1 1 11 At+0 @0 1 31 Ast;s2Rg f0 @1 1 01 A+0 @0 3 01 Ak+0 @2 0 41 Amk;m2Rg X1.7Intersect each pair, if possible. (a)f0 @1 1 21 A+t0 @0 1 11 At2Rg,f0 @1 3 21 A+s0 @0 1 21 As2Rg (b)f0 @2 0 11 A+t0 @1 1 11 At2Rg,fs0 @0 1 21 A+w0 @0 4 11 As;w2Rg 1.8When a plane does not pass through the origin, performing operations on vec- tors whose bodies lie in it is more complicated than when the plane passes through the origin. Consider the picture in this subsection of the plane f0 @2 0 01 A+0 @0:5 1 01 Ay+0 @0:5 0 11 Azy;z2Rg and the three vectors it shows, with endpoints (2 ;0;0), (1:5;1;0), and (1:5;0;1). (a)Redraw the picture, including the vector in the plane that is twice as long as the one with endpoint (1 :5;1;0). The endpoint of your vector is not (3 ;2;0); what is it? (b)Redraw the picture, including the parallelogram in the plane that shows the sum of the vectors ending at (1 :5;0;1) and (1:5;1;0). The endpoint of the sum, on the diagonal, is not (3 ;1;1); what is it? 1.9Show that the line segments (a1;a2)(b1;b2) and (c1;c2)(d1;d2) have the same lengths and slopes if b1a1=d1c1andb2a2=d2c2. Is that only if? 1.10 How should R0be dened? ?X1.11 A person traveling eastward at a rate of 3 miles per hour nds that the wind appears to blow directly from the north. On doubling his speed it appears to come from the north east. What was the wind's velocity? [Math. Mag., Jan. 1957] Section II. Linear Geometry of n-Space 39 1.12 Euclid describes a plane as \a surface which lies evenly with the straight lines on itself". Commentators (e.g., Heron) have interpreted this to mean \(A plane surface is) such that, if a straight line pass through two points on it, the line coincides wholly with it at every spot, all ways". (Translations from [Heath], pp. 171-172.) Do planes, as described in this section, have that property? Does this description adequately dene planes? II.2 Length and Angle Measures We've translated the rst section's results about solution sets into geometric terms for insight into how those sets look. But we must watch out not to be misled by our own terms; labeling subsets of Rkof the formsf~ p+t~ vt2Rg andf~ p+t~ v+s~ wt;s2Rgas \lines" and \planes" doesn't make them act like the lines and planes of our prior experience. Rather, we must ensure that the names suit the sets. While we can't prove that the sets satisfy our intuition | we can't prove anything about intuition | in this subsection we'll observe that a result familiar from R2andR3, when generalized to arbitrary Rk, supports the idea that a line is straight and a plane is at. Specically, we'll see how to do Euclidean geometry in a \plane" by giving a denition of the angle between twoRnvectors in the plane that they generate. 2.1 Denition The length of a vector ~ v2Rnis this. k~ vk=q v2 1+


+v2n 2.2 Remark This is a natural generalization of the Pythagorean Theorem. A classic discussion is in [Polya]. We can use that denition to derive a formula for the angle between two vectors. For a model of what to do, consider two vectors in R3. ~ v ~ u Put them in canonical position and, in the plane that they determine, consider the triangle formed by ~ u,~ v, and~ u~ v. 40 Chapter One. Linear Systems Apply the Law of Cosines, k~ u~ vk2=k~ uk2+k~ vk22k~ ukk~ vkcos, where is the angle between the vectors. Expand both sides (u1v1)2+ (u2v2)2+ (u3v3)2 = (u2 1+u2 2+u2 3) + (v2 1+v2 2+v2 3)2k~ ukk~ vkcos and simplify. = arccos(u1v1+u2v2+u3v3 k~ ukk~ vk) In higher dimensions no picture suces but we can make the same argument analytically. First, the form of the numerator is clear | it comes from the middle terms of the squares ( u1v1)2, (u2v2)2, etc. 2.3 Denition The dot product (orinner product , orscalar product ) of two n-component real vectors is the linear combination of their components. ~ u~ v=u1v1+u2v2+


+unvn Note that the dot product of two vectors is a real number, not a vector, and that the dot product of a vector from Rnwith a vector from Rmis dened only when nequalsm. Note also this relationship between dot product and length: dotting a vector with itself gives its length squared ~ u~ u=u1u1+


+unun=k~ uk2. 2.4 Remark The wording in that denition allows one or both of the two to be a row vector instead of a column vector. Some books require that the rst vector be a row vector and that the second vector be a column vector. We shall not be that strict. Still reasoning with letters, but guided by the pictures, we use the next theorem to argue that the triangle formed by ~ u,~ v, and~ u~ vinRnlies in the planar subset of Rngenerated by ~ uand~ v. 2.5 Theorem (Triangle Inequality) For any~ u;~ v2Rn, k~ u+~ vkk~ uk+k~ vk with equality if and only if one of the vectors is a nonnegative scalar multiple of the other one. This inequality is the source of the familiar saying, \The shortest distance between two points is in a straight line." ~ u~ v~ u+~ v startnish Section II. Linear Geometry of n-Space 41 Proof .(We'll use some algebraic properties of dot product that we have not yet checked, for instance that ~ u(~ a+~b) =~ u~ a+~ u~band that~ u~ v=~ v~ u. See Exercise 17.) The desired inequality holds if and only if its square holds. k~ u+~ vk2(k~ uk+k~ vk)2 (~ u+~ v)(~ u+~ v)k~ uk2+ 2k~ ukk~ vk+k~ vk2 ~ u~ u+~ u~ v+~ v~ u+~ v~ v~ u~ u+ 2k~ ukk~ vk+~ v~ v 2~ u~ v2k~ ukk~ vk That, in turn, holds if and only if the relationship obtained by multiplying both sides by the nonnegative numbers k~ ukandk~ vk 2 (k~ vk~ u)(k~ uk~ v)2k~ uk2k~ vk2 and rewriting 0k~ uk2k~ vk22 (k~ vk~ u)(k~ uk~ v) +k~ uk2k~ vk2 is true. But factoring 0(k~ uk~ vk~ vk~ u)(k~ uk~ vk~ vk~ u) shows that this certainly is true since it only says that the square of the length of the vectork~ uk~ vk~ vk~ uis not negative. As for equality, it holds when, and only when, k~ uk~ vk~ vk~ uis~0. The check thatk~ uk~ v=k~ vk~ uif and only if one vector is a nonnegative real scalar multiple of the other is easy. QED This result supports the intuition that even in higher-dimensional spaces, lines are straight and planes are at. For any two points in a linear surface, the line segment connecting them is contained in that surface (this is easily checked from the denition). But if the surface has a bend then that would allow for a shortcut (shown here grayed, while the segment from PtoQthat is contained in the surface is solid). P Q Because the Triangle Inequality says that in any Rn, the shortest cut between two endpoints is simply the line segment connecting them, linear surfaces have no such bends. Back to the denition of angle measure. The heart of the Triangle Inequal- ity's proof is the ` ~ u~ vk~ ukk~ vk' line. At rst glance, a reader might wonder if some pairs of vectors satisfy the inequality in this way: while ~ u~ vis a large number, with absolute value bigger than the right-hand side, it is a negative large number. The next result says that no such pair of vectors exists. 42 Chapter One. Linear Systems 2.6 Corollary (Cauchy-Schwartz Inequality) For any~ u;~ v2Rn, j~ u~ vjk~ ukk~ vk with equality if and only if one vector is a scalar multiple of the other. Proof .The Triangle Inequality's proof shows that ~ u~ vk~ ukk~ vkso if~ u~ vis positive or zero then we are done. If ~ u~ vis negative then this holds. j~ u~ vj=(~ u~ v) = (~ u)~ vk~ ukk~ vk=k~ ukk~ vk The equality condition is Exercise 18. QED The Cauchy-Schwartz inequality assures us that the next denition makes sense because the fraction has absolute value less than or equal to one. 2.7 Denition The angle between two nonzero vectors ~ u;~ v2Rnis = arccos(~ u~ v k~ ukk~ vk) (the angle between the zero vector and any other vector is dened to be a right angle). Thus vectors from Rnare orthogonal, that is, perpendicular, if and only if their dot product is zero. 2.8 Example These vectors are orthogonal. 1 1 1 1 = 0 The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. 2.9 Example TheR3angle formula given at the start of this subsection is a special case of the denition. Between these two 0 @0 3 21 A 0 @1 1 01 A Section II. Linear Geometry of n-Space 43 the angle is arccos((1)(0) + (1)(3) + (0)(2)p 12+ 12+ 02p 02+ 32+ 22) = arccos(3p 2p 13) approximately 0 :94 radians. Notice that these vectors are not orthogonal. Al- though the yz-plane may appear to be perpendicular to the xy-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. Exercises X2.10 Find the length of each vector. (a)3 1 (b)1 2 (c)0 @4 1 11 A (d)0 @0 0 01 A (e)0 BB@1 1 1 01 CCA X2.11 Find the angle between each two, if it is dened. (a)1 2 ;1 4 (b)0 @1 2 01 A;0 @0 4 11 A (c)1 2 ;0 @1 4 11 A X2.12 During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): 1 :2 miles north, 6 :1 miles 38 degrees east of south, 4 :0 miles at 89 degrees east of north, and 6 :5 miles at 31 degrees east of north. Find the distance between starting and ending positions. [Ohanian] 2.13 Findkso that these two vectors are perpendicular.k 1 4 3 2.14 Describe the set of vectors in R3orthogonal to this one.0 @1 3 11 A X2.15 (a) Find the angle between the diagonal of the unit square in R2and one of the axes. (b)Find the angle between the diagonal of the unit cube in R3and one of the axes. (c)Find the angle between the diagonal of the unit cube in Rnand one of the axes. (d)What is the limit, as ngoes to1, of the angle between the diagonal of the unit cube in Rnand one of the axes? 2.16 Is any vector perpendicular to itself? X2.17 Describe the algebraic properties of dot product. (a)Is it right-distributive over addition: ( ~ u+~ v)~ w=~ u~ w+~ v~ w? (b)Is it left-distributive (over addition)? (c)Does it commute? (d)Associate? (e)How does it interact with scalar multiplication? As always, any assertion must be backed by either a proof or an example. 44 Chapter One. Linear Systems 2.18 Verify the equality condition in Corollary 2.6, the Cauchy-Schwartz Inequal- ity. (a)Show that if ~ uis a negative scalar multiple of ~ vthen~ u~ vand~ v~ uare less than or equal to zero. (b)Show thatj~ u~ vj=k~ ukk~ vkif and only if one vector is a scalar multiple of the other. 2.19 Suppose that ~ u~ v=~ u~ wand~ u6=~0. Must~ v=~ w? X2.20 Does any vector have length zero except a zero vector? (If \yes", produce an example. If \no", prove it.) X2.21 Find the midpoint of the line segment connecting ( x1;y1) with (x2;y2) inR2. Generalize to Rn. 2.22 Show that if ~ v6=~0 then~ v=k~ vkhas length one. What if ~ v=~0? 2.23 Show that if r0 thenr~ visrtimes as long as ~ v. What ifr<0? X2.24 A vector~ v2Rnof length one is a unit vector. Show that the dot product of two unit vectors has absolute value less than or equal to one. Can `less than' happen? Can `equal to'? 2.25 Prove thatk~ u+~ vk2+k~ u~ vk2= 2k~ uk2+ 2k~ vk2: 2.26 Show that if ~ x~ y= 0 for every ~ ythen~ x=~0. 2.27 Isk~ u1+


+~ unkk~ u1k+


+k~ unk? If it is true then it would generalize the Triangle Inequality. 2.28 What is the ratio between the sides in the Cauchy-Schwartz inequality? 2.29 Why is the zero vector dened to be perpendicular to every vector? 2.30 Describe the angle between two vectors in R1. 2.31 Give a simple necessary and sucient condition to determine whether the angle between two vectors is acute, right, or obtuse. X2.32 Generalize to Rnthe converse of the Pythagorean Theorem, that if ~ uand~ v are perpendicular then k~ u+~ vk2=k~ uk2+k~ vk2. 2.33 Show thatk~ uk=k~ vkif and only if ~ u+~ vand~ u~ vare perpendicular. Give an example in R2. 2.34 Show that if a vector is perpendicular to each of two others then it is perpen- dicular to each vector in the plane they generate. ( Remark. They could generate a degenerate plane | a line or a point | but the statement remains true.) 2.35 Prove that, where ~ u;~ v2Rnare nonzero vectors, the vector ~ u k~ uk+~ v k~ vk bisects the angle between them. Illustrate in R2. 2.36 Verify that the denition of angle is dimensionally correct: (1) if k >0 then the cosine of the angle between k~ uand~ vequals the cosine of the angle between ~ uand~ v, and (2) if k < 0 then the cosine of the angle between k~ uand~ vis the negative of the cosine of the angle between ~ uand~ v. X2.37 Show that the inner product operation is linear : for~ u;~ v;~ w2Rnandk;m2R, ~ u(k~ v+m~ w) =k(~ u~ v) +m(~ u~ w). X2.38 The geometric mean of two positive reals x;yispxy. It is analogous to the arithmetic mean (x+y)=2. Use the Cauchy-Schwartz inequality to show that the geometric mean of any x;y2Ris less than or equal to the arithmetic mean. Section II. Linear Geometry of n-Space 45 ?2.39 A ship is sailing with speed and direction ~ v1; the wind blows apparently (judging by the vane on the mast) in the direction of a vector ~ a; on changing the direction and speed of the ship from ~ v1to~ v2the apparent wind is in the direction of a vector~b. Find the vector velocity of the wind. [Am. Math. Mon., Feb. 1933] 2.40 Verify the Cauchy-Schwartz inequality by rst proving Lagrange's identity: 0 @X 1jnajbj1 A2 =0 @X 1jna2 j1 A0 @X 1jnb2 j1 AX 1k<jn(akbjajbk)2 and then noting that the nal term is positive. (Recall the meaningX 1jnajbj=a1b1+a2b2+


+anbn and X 1jnaj2=a12+a22+


+an2 of the  notation.) This result is an improvement over Cauchy-Schwartz because it gives a formula for the dierence between the two sides. Interpret that dierence inR2. 46 Chapter One. Linear Systems III Reduced Echelon Form After developing the mechanics of Gauss' method, we observed that it can be done in more than one way. One example is that we sometimes have to swap rows and there can be more than one row to choose from. Another example is that from this matrix 2 2 4 3 Gauss' method could derive any of these echelon form matrices. 2 2 01 1 1 01 2 0 01 The rst results from 21+2. The second comes from following (1 =2)1with 41+2. The third comes from 21+2followed by 2 2+1(after the rst row combination the matrix is already in echelon form so the second one is extra work but it is nonetheless a legal row operation). The fact that the echelon form outcome of Gauss' method is not unique leaves us with some questions. Will any two echelon form versions of a system have the same number of free variables? Will they in fact have exactly the same variables free? In this section we will answer both questions \yes". We will do more than answer the questions. We will give a way to decide if one linear system can be derived from another by row operations. The answers to the two questions will follow from this larger result. III.1 Gauss-Jordan Reduction Gaussian elimination coupled with back-substitution solves linear systems, but it's not the only method possible. Here is an extension of Gauss' method that has some advantages. 1.1 Example To solve x+y2z=2 y+ 3z= 7 xz=1 we can start by going to echelon form as usual. 1+3!0 @1 122 0 1 3 7 01 1 11 A2+3!0 @1 122 0 1 3 7 0 0 4 81 A We can keep going to a second stage by making the leading entries into ones (1=4)3!0 @1 122 0 1 3 7 0 0 1 21 A Section III. Reduced Echelon Form 47 and then to a third stage that uses the leading entries to eliminate all of the other entries in each column by combining upwards. 33+2! 23+10 @1 1 0 2 0 1 0 1 0 0 1 21 A2+1!0 @1 0 0 1 0 1 0 1 0 0 1 21 A The answer is x= 1,y= 1, andz= 2. Note that the row combination operations in the rst stage proceed from column one to column three while the combination operations in the third stage proceed from column three to column one. 1.2 Example We often combine the operations of the middle stage into a single step, even though they are operations on dierent rows. 2 1 7 426 21+2!2 1 7 048 (1=2)1! (1=4)21 1=27=2 0 1 2 (1=2)2+1! 1 0 5=2 0 1 2 The answer is x= 5=2 andy= 2. This extension of Gauss' method is Gauss-Jordan reduction . It goes past echelon form to a more rened, more specialized, matrix form. 1.3 Denition A matrix is in reduced echelon form if, in addition to being in echelon form, each leading entry is a one and is the only nonzero entry in its column. The disadvantage of using Gauss-Jordan reduction to solve a system is that the additional row operations mean additional arithmetic. The advantage is that the solution set can just be read o. In any echelon form, plain or reduced, we can read o when a system has an empty solution set because there is a contradictory equation, we can read o when a system has a one-element solution set because there is no contradiction and every variable is the leading variable in some row, and we can read o when a system has an innite solution set because there is no contradiction and at least one variable is free. In reduced echelon form we can read o not just what kind of solution set the system has, but also its description. Whether or not the echelon form is reduced, we have no trouble describing the solution set when it is empty, of course. The two examples above show that when the system has a single solution then the solution can be read o from the right-hand column. In the case when the solution set is innite, its parametrization can also be read o 48 Chapter One. Linear Systems of the reduced echelon form. Consider, for example, this system that is shown brought to echelon form and then to reduced echelon form. 0 @2 6 1 2 5 0 3 1 4 1 0 3 1 2 51 A2+3!0 @2 6 1 2 5 0 3 1 4 1 0 0 0241 A (1=2)1! (1=3)2 (1=2)3(4=3)3+2! 3+132+1!0 @1 01=2 09=2 0 1 1=3 0 3 0 0 0 1 21 A Starting with the middle matrix, the echelon form version, back substitution produces2x4= 4 so that x4=2, then another back substitution gives 3x2+x3+ 4(2) = 1 implying that x2= 3(1=3)x3, and then the nal back substitution gives 2 x1+ 6(3(1=3)x3) +x3+ 2(2) = 5 implying that x1=(9=2) + (1=2)x3. Thus the solution set is this. S=f0 BB@x1 x2 x3 x41 CCA=0 BB@9=2 3 0 21 CCA+0 BB@1=2 1=3 1 01 CCAx3x32Rg Now, considering the nal matrix, the reduced echelon form version, note that adjusting the parametrization by moving the x3terms to the other side does indeed give the description of this innite solution set. Part of the reason that this works is straightforward. While a set can have many parametrizations that describe it, e.g., both of these also describe the above setS(taketto bex3=6 andsto bex31) f0 BB@9=2 3 0 21 CCA+0 BB@3 2 6 01 CCAtt2Rg f0 BB@4 8=3 1 21 CCA+0 BB@1=2 1=3 1 01 CCAss2Rg nonetheless we have in this book stuck to a convention of parametrizing using the unmodied free variables (that is, x3=x3instead ofx3= 6t). We can easily see that a reduced echelon form version of a system is equivalent to a parametrization in terms of unmodied free variables. For instance, x1= 42x3 x2= 3x3()0 @1 0 2 4 0 1 1 3 0 0 0 01 A (to move from left to right we also need to know how many equations are in the system). So, the convention of parametrizing with the free variables by solving each equation for its leading variable and then eliminating that leading variable from every other equation is exactly equivalent to the reduced echelon form conditions that each leading entry must be a one and must be the only nonzero entry in its column. Section III. Reduced Echelon Form 49 Not as straightforward is the other part of the reason that the reduced echelon form version allows us to read o the parametrization that we would have gotten had we stopped at echelon form and then done back substitution. The prior paragraph shows that reduced echelon form corresponds to some parametrization, but why the same parametrization? A solution set can be parametrized in many ways, and Gauss' method or the Gauss-Jordan method can be done in many ways, so a rst guess might be that we could derive many dierent reduced echelon form versions of the same starting system and many dierent parametrizations. But we never do. Experience shows that starting with the same system and proceeding with row operations in many dierent ways always yields the same reduced echelon form and the same parametrization (using the unmodied free variables). In the rest of this section we will show that the reduced echelon form version of a matrix is unique. It follows that the parametrization of a linear system in terms of its unmodied free variables is unique because two dierent ones would give two dierent reduced echelon forms. We shall use this result, and the ones that lead up to it, in the rest of the book but perhaps a restatement in a way that makes it seem more immediately useful may be encouraging. Imagine that we solve a linear system, parametrize, and check in the back of the book for the answer. But the parametrization there appears dierent. Have we made a mistake, or could these be dierent-looking descriptions of the same set, as with the three descriptions above of S? The prior paragraph notes that we will show here that dierent-looking parametrizations (using the unmodied free variables) describe genuinely dierent sets. Here is an informal argument that the reduced echelon form version of a matrix is unique. Consider again the example that started this section of a matrix that reduces to three dierent echelon form matrices. The rst matrix of the three is the natural echelon form version. The second matrix is the same as the rst except that a row has been halved. The third matrix, too, is just a cosmetic variant of the rst. The denition of reduced echelon form outlaws this kind of fooling around. In reduced echelon form, halving a row is not possible because that would change the row's leading entry away from one, and neither is combining rows possible, because then a leading entry would no longer be alone in its column. This informal justication is not a proof; the argument shows that no two dierent reduced echelon form matrices are related by a single row operation step, but the argument does not ruled out the possibility that two dierent reduced echelon form matrices could be related by multiple steps. Before we go to the proof, we nish this subsection by rephrasing our work in a terminology that will be enlightening. Many dierent matrices yield the same reduced echelon form matrix. The three echelon form matrices from the start of this section, and the matrix they were derived from, all give this reduced echelon form matrix. 1 0 0 1 50 Chapter One. Linear Systems We think of these matrices as related to each other. The next result speaks to this relationship. 1.4 Lemma Elementary row operations are reversible. Proof .For any matrix A, the eect of swapping rows is reversed by swapping them back, multiplying a row by a nonzero kis undone by multiplying by 1 =k, and adding a multiple of row ito rowj(withi6=j) is undone by subtracting the same multiple of row ifrom rowj. Ai$j!j$i!A Aki!(1=k)i!A Aki+j!ki+j!A (Thei6=jconditions is needed. See Exercise 13.) QED This lemma suggests that `reduces to' is misleading | where A!B, we shouldn't think of Bas \after"Aor \simpler than" A. Instead we should think of them as interreducible or interrelated. Below is a picture of the idea. The matrices from the start of this section and their reduced echelon form version are shown in a cluster. They are all interreducible; these relationships are shown also.  1 0 0 1 2 2 4 3 2 0 01  1 1 01  2 2 01 We say that matrices that reduce to each other are `equivalent with respect to the relationship of row reducibility'. The next result veries this statement using the denition of an equivalence. 1.5 Lemma Between matrices, `reduces to' is an equivalence relation. Proof .We must check the conditions (i) re exivity, that any matrix reduces to itself, (ii) symmetry, that if Areduces toBthenBreduces toA, and (iii) tran- sitivity, that if Areduces toBandBreduces toCthenAreduces toC. Re exivity is easy; any matrix reduces to itself in zero row operations. That the relationship is symmetric is Lemma 1.4 | if Areduces to Bby some row operations then also Breduces toAby reversing those operations. For transitivity, suppose that Areduces to Band thatBreduces to C. Linking the reduction steps from A!


!Bwith those from B!


!C gives a reduction from AtoC. QED 1.6 Denition Two matrices that are interreducible by the elementary row operations are row equivalent . More information on equivalence relations is in the appendix. Section III. Reduced Echelon Form 51 The diagram below shows the collection of all matrices as a box. Inside that box, each matrix lies in some class. Matrices are in the same class if and only if they are interreducible. The classes are disjoint | no matrix is in two distinct classes. The collection of matrices has been partitioned into row equivalence classes . . . .A B One of the classes in this partition is the cluster of matrices shown above, expanded to include all of the nonsingular 2 2 matrices. The next subsection proves that the reduced echelon form of a matrix is unique; that every matrix reduces to one and only one reduced echelon form matrix. Rephrased in terms of the row-equivalence relationship, we shall prove that every matrix is row equivalent to one and only one reduced echelon form matrix. In terms of the partition what we shall prove is: every equivalence class contains one and only one reduced echelon form matrix. So each reduced echelon form matrix serves as a representative of its class. After that proof we shall, as mentioned in the introduction to this section, have a way to decide if one matrix can be derived from another by row reduction. We just apply the Gauss-Jordan procedure to both and see whether or not they come to the same reduced echelon form. Exercises X1.7Use Gauss-Jordan reduction to solve each system. (a)x+y= 2 xy= 0(b)xz= 4 2x+ 2y = 1(c)3x2y= 1 6x+y= 1=2 (d)2xy =1 x+ 3yz= 5 y+ 2z= 5 X1.8Find the reduced echelon form of each matrix. (a)2 1 1 3 (b)0 @1 3 1 2 0 4 1331 A (c)0 @1 0 3 1 2 1 4 2 1 5 3 4 8 1 21 A (d)0 @0 1 3 2 0 0 5 6 1 5 1 51 A X1.9Find each solution set by using Gauss-Jordan reduction, then reading o the parametrization. (a)2x+yz= 1 4xy = 3(b)xz = 1 y+ 2zw= 3 x+ 2y+ 3zw= 7(c)xy+z = 0 y +w= 0 3x2y+ 3z+w= 0 yw= 0 (d)a+ 2b+ 3c+de= 1 3ab+c+d+e= 3 More information on partitions and class representatives is in the appendix. 52 Chapter One. Linear Systems 1.10 Give two distinct echelon form versions of this matrix.0 @2 1 1 3 6 4 1 2 1 5 1 51 A X1.11 List the reduced echelon forms possible for each size. (a)22(b)23(c)32(d)33 X1.12 What results from applying Gauss-Jordan reduction to a nonsingular matrix? 1.13 The proof of Lemma 1.4 contains a reference to the i6=jcondition on the row combination operation. (a)The denition of row operations has an i6=jcondition on the swap operation i$j. Show that in Ai$j!i$j!Athis condition is not needed. (b)Write down a 22 matrix with nonzero entries, and show that the 1
1+1 operation is not reversed by 1
1+1. (c)Expand the proof of that lemma to make explicit exactly where the i6=j condition on combining is used. III.2 Row Equivalence We will close this section and this chapter by proving that every matrix is row equivalent to one and only one reduced echelon form matrix. The ideas that appear here will reappear, and be further developed, in the next chapter. The underlying theme here is that one way to understand a mathematical situation is by being able to classify the cases that can happen. We have met this theme several times already. We have classied solution sets of linear systems into the no-elements, one-element, and innitely-many elements cases. We have also classied linear systems with the same number of equations as unknowns into the nonsingular and singular cases. We adopted these classications because they give us a way to understand the situations that we were investigating. Here, where we are investigating row equivalence, we know that the set of all matrices breaks into the row equivalence classes. When we nish the proof here, we will have a way to understand each of those classes | its matrices can be thought of as derived by row operations from the unique reduced echelon form matrix in that class. To understand how row operations act to transform one matrix into another, we consider the eect that they have on the parts of a matrix. The crucial observation is that row operations combine the rows linearly. 2.1 Lemma (Linear Combination Lemma) A linear combination of linear combinations is a linear combination. Proof .Given the linear combinations c1;1x1+


+c1;nxnthroughcm;1x1+


+cm;nxn, consider a combination of those d1(c1;1x1+


+c1;nxn) +


+dm(cm;1x1+


+cm;nxn) Section III. Reduced Echelon Form 53 where thed's are scalars along with the c's. Distributing those d's and regroup- ing gives = (d1c1;1+


+dmcm;1)x1+


+ (d1c1;n+


+dmcm;n)xn which is a linear combination of the x's. QED In this subsection we will use the convention that, where a matrix is named with an upper case roman letter, the matching lower-case greek letter names the rows. A=0 BBBB@


1





2


...


m


1 CCCCAB=0 BBBB@


1





2


...


m


1 CCCCA 2.2 Corollary Where one matrix reduces to another, each row of the second is a linear combination of the rows of the rst. The proof below uses induction on the number of row operations used to reduce one matrix to the other. Before we proceed, here is an outline of the ar- gument (readers unfamiliar with induction may want to compare this argument with the one used in the `General = Particular + Homogeneous' proof). First, for the base step of the argument, we will verify that the proposition is true when reduction can be done in zero row operations. Second, for the inductive step, we will argue that if being able to reduce the rst matrix to the second in some number t0 of operations implies that each row of the second is a linear combination of the rows of the rst, then being able to reduce the rst to the second in t+ 1 operations implies the same thing. Together, this base step and induction step prove the result because by the inductive step the fact that it is true in the zero operations case (that's shown in the base step) implies that it is true in the one operation case, and then the inductive step applied again gives that it is therefore true in the two operations case, etc. Proof .We proceed by induction on the minimum number of row operations that take a rst matrix Ato a second one B. In the base step, that zero reduction operations suce, the two matrices are equal and each row of Bis obviously a combination of A's rows:~i= 0
~ 1+


+ 1
~ i+


+ 0
~ m. For the inductive step, assume the inductive hypothesis: with t0, if a matrix can be derived from Aintor fewer operations then its rows are linear combinations of the A's rows. Consider a Bthat takest+1 operations. Because there are more than zero operations, there must be a next-to-last matrix Gso thatA!


! G!B. ThisGis onlytoperations away from Aand so the More information on mathematical induction is in the appendix. 54 Chapter One. Linear Systems inductive hypothesis applies to it, that is, each row of Gis a linear combination of the rows of A. If the last operation, the one from GtoB, is a row swap then the rows ofBare just the rows of Greordered and thus each row of Bis also a linear combination of the rows of A. The other two possibilities for this last operation, that it multiplies a row by a scalar and that it adds a multiple of one row to another, both result in the rows of Bbeing linear combinations of the rows of G. But therefore, by the Linear Combination Lemma, each row of Bis a linear combination of the rows of A. With that, we have both the base step and the inductive step, and so the proposition follows. QED 2.3 Example In the reduction 0 2 1 1 1$2!1 1 0 2 (1=2)2!1 1 0 1 2+1!1 0 0 1 call the matrices A,D,G, andB. The methods of the proof show that there are three sets of linear relationships. 1= 0
1+ 1
2 2= 1
1+ 0
2 1= 0
1+ 1
2 2= (1=2)1+ 0
21= (1=2)1+ 1
2 2= (1=2)1+ 0
2 The prior result gives us the insight that Gauss' method works by taking linear combinations of the rows. But to what end; why do we go to echelon form as a particularly simple, or basic, version of a linear system? The answer, of course, is that echelon form is suitable for back substitution, because we have isolated the variables. For instance, in this matrix R=0 BB@2 3 7 8 0 0 0 0 1 5 1 1 0 0 0 3 3 0 0 0 0 0 2 11 CCA x1has been removed from x5's equation. That is, Gauss' method has made x5's row independent of x1's row. Independence of a collection of row vectors, or of any kind of vectors, will be precisely dened and explored in the next chapter. But a rst take on it is that we can show that, say, the third row above is not comprised of the other rows, that36=c11+c22+c44. For, suppose that there are scalars c1,c2, andc4such that this relationship holds. 0 0 0 3 3 0
=c12 3 7 8 0 0
+c20 0 1 5 1 1
+c40 0 0 0 2 1
The rst row's leading entry is in the rst column and narrowing our considera- tion of the above relationship to consideration only of the entries from the rst Section III. Reduced Echelon Form 55 column 0 = 2 c1+0c2+0c4gives thatc1= 0. The second row's leading entry is in the third column and the equation of entries in that column 0 = 7 c1+1c2+0c4, along with the knowledge that c1= 0, gives that c2= 0. Now, to nish, the third row's leading entry is in the fourth column and the equation of entries in that column 3 = 8 c1+ 5c2+ 0c4, along with c1= 0 andc2= 0, gives an impossibility. The following result shows that this eect always holds. It shows that what Gauss' linear elimination method eliminates is linear relationships among the rows. 2.4 Lemma In an echelon form matrix, no nonzero row is a linear combination of the other rows. Proof .LetRbe in echelon form. Suppose, to obtain a contradiction, that some nonzero row is a linear combination of the others. i=c11+:::+ci1i1+ci+1i+1+:::+cmm We will rst use induction to show that the coecients c1, . . . ,ci1associated with rows above iare all zero. The contradiction will come from consideration ofiand the rows below it. The base step of the induction argument is to show that the rst coecient c1is zero. Let the rst row's leading entry be in column number `1and consider the equation of entries in that column. i;`1=c11;`1+:::+ci1i1;`1+ci+1i+1;`1+:::+cmm;`1 The matrix is in echelon form so the entries 2;`1, . . . ,m;`1, includingi;`1, are all zero. 0 =c11;`1+


+ci1
0 +ci+1
0 +


+cm
0 Because the entry 1;`1is nonzero as it leads its row, the coecient c1must be zero. The inductive step is to show that for each row index kbetween 1 and i2, if the coecient c1and the coecients c2, . . . ,ckare all zero then ck+1is also zero. That argument, and the contradiction that nishes this proof, is saved for Exercise 20. QED We can now prove that each matrix is row equivalent to one and only one reduced echelon form matrix. We will nd it convenient to break the rst half of the argument o as a preliminary lemma. For one thing, it holds for any echelon form whatever, not just reduced echelon form. 2.5 Lemma If two echelon form matrices are row equivalent then the leading entries in their rst rows lie in the same column. The same is true of all the nonzero rows | the leading entries in their second rows lie in the same column, etc. 56 Chapter One. Linear Systems For the proof we rephrase the result in more technical terms. Dene the form of anmnmatrix to be the sequence h`1;`2;::: ;`miwhere`iis the column number of the leading entry in row iand`i=1if there is no leading entry in that row. The lemma says that if two echelon form matrices are row equivalent then their forms are equal sequences. Proof .LetBandDbe echelon form matrices that are row equivalent. Because they are row equivalent they must be the same size, say mn. Let the column number of the leading entry in row iofBbe`iand let the column number of the leading entry in row jofDbekj. We will show that `1=k1, that`2=k2, etc., by induction. This induction argument relies on the fact that the matrices are row equiv- alent, because the Linear Combination Lemma and its corollary therefore give that each row of Bis a linear combination of the rows of Dand vice versa: i=si;11+si;22+


+si;mmandj=tj;11+tj;22+


+tj;mm where thes's andt's are scalars. The base step of the induction is to verify the lemma for the rst rows of the matrices, that is, to verify that `1=k1. If either row is a zero row then every entry in the matrix is a zero since it is in echelon form, and therefore both matrices consist solely of zero entries (by Corollary 2.2), and so both `1andk1 are1. For the case where neither 1nor1is a zero row, consider the i= 1 instance of the linear relationship above. 1=s1;11+s1;22+


+s1;mm0


b1;`1



=s1;10


d1;k1



+s1;20


0



... +s1;m0


0



First, note that `1<k1is impossible: in the columns of Dto the left of column k1the entries are all zeroes (as d1;k1leads the rst row) and so if `1<k1then the equation of entries from column `1would beb1;`1=s1;1
0 +


+s1;m
0, butb1;`1isn't zero since it leads its row and so this is an impossibility. Next, a symmetric argument shows that k1<`1also is impossible. Thus the `1=k1 base case holds. The inductive step is to show that if `1=k1, and`2=k2, . . . , and`r=kr, then also`r+1=kr+1(forrin the interval 1 ::m1). This argument is saved for Exercise 21. QED That lemma answers two of the questions that we have posed: (i) any two echelon form versions of a matrix have the same free variables, and consequently, and (ii) any two echelon form versions have the same number of free variables. There is no linear system and no combination of row operations such that, say, we could solve the system one way and get yandzfree but solve it another Section III. Reduced Echelon Form 57 way and get yandwfree, or solve it one way and get two free variables while solving it another way yields three. We nish now by specializing to the case of reduced echelon form matrices. 2.6 Theorem Each matrix is row equivalent to a unique reduced echelon form matrix. Proof .Clearly any matrix is row equivalent to at least one reduced echelon form matrix, via Gauss-Jordan reduction. For the other half, that any matrix is equivalent to at most one reduced echelon form matrix, we will show that if a matrix Gauss-Jordan reduces to each of two others then those two are equal. Suppose that a matrix is row equivalent to two reduced echelon form ma- tricesBandD, which are therefore row equivalent to each other. The Linear Combination Lemma and its corollary allow us to write the rows of one, say B, as a linear combination of the rows of the other i=ci;11+


+ci;mm. The preliminary result, Lemma 2.5, says that in the two matrices, the same collection of rows are nonzero. Thus, if 1throughrare the nonzero rows of Bthen the nonzero rows of Dare1throughr. Zero rows don't contribute to the sum so we can rewrite the relationship to include just the nonzero rows. i=ci;11+


+ci;rr () The preliminary result also says that for each row jbetween 1 and r, the leading entries of the j-th row ofBandDappear in the same column, denoted `j. Rewriting the above relationship to focus on the entries in the `j-th column


bi;`j



=ci;1


d1;`j



+ci;2


d2;`j



... +ci;r


dr;`j



gives this set of equations for i= 1 up toi=r. b1;`j=c1;1d1;`j+


+c1;jdj;`j+


+c1;rdr;`j... bj;`j=cj;1d1;`j+


+cj;jdj;`j+


+cj;rdr;`j... br;`j=cr;1d1;`j+


+cr;jdj;`j+


+cr;rdr;`j SinceDis in reduced echelon form, all of the d's in column `jare zero except for dj;`j, which is 1. Thus each equation above simplies to bi;`j=ci;jdj;`j=ci;j
1. ButBis also in reduced echelon form and so all of the b's in column `jare zero except forbj;`j, which is 1. Therefore, each ci;jis zero, except that c1;1= 1, andc2;2= 1, . . . , and cr;r= 1. We have shown that the only nonzero coecient in the linear combination labelled () iscj;j, which is 1. Therefore j=j. Because this holds for all nonzero rows, B=D. QED 58 Chapter One. Linear Systems We end with a recap. In Gauss' method we start with a matrix and then derive a sequence of other matrices. We dened two matrices to be related if one can be derived from the other. That relation is an equivalence relation, called row equivalence, and so partitions the set of all matrices into row equivalence classes. . . .1 3 2 7
1 3 0 1
(There are innitely many matrices in the pictured class, but we've only got room to show two.) We have proved there is one and only one reduced echelon form matrix in each row equivalence class. So the reduced echelon form is a canonical formfor row equivalence: the reduced echelon form matrices are representatives of the classes. . . .?? ??1 0 0 1
We can answer questions about the classes by translating them into questions about the representatives. 2.7 Example We can decide if matrices are interreducible by seeing if Gauss- Jordan reduction produces the same reduced echelon form result. Thus, these are not row equivalent 13 2 6 13 2 5 because their reduced echelon forms are not equal. 13 0 0 1 0 0 1 2.8 Example Any nonsingular 3 3 matrix Gauss-Jordan reduces to this. 0 @1 0 0 0 1 0 0 0 11 A More information on canonical representatives is in the appendix. Section III. Reduced Echelon Form 59 2.9 Example We can describe the classes by listing all possible reduced echelon form matrices. Any 2 2 matrix lies in one of these: the class of matrices row equivalent to this,0 0 0 0 the innitely many classes of matrices row equivalent to one of this type 1a 0 0 wherea2R(includinga= 0), the class of matrices row equivalent to this,  0 1 0 0 and the class of matrices row equivalent to this  1 0 0 1 (this is the class of nonsingular 2 2 matrices). Exercises X2.10 Decide if the matrices are row equivalent. (a)1 2 4 8 ;0 1 1 2 (b)0 @1 0 2 31 1 51 51 A;0 @1 0 2 0 2 10 2 0 41 A (c)0 @2 11 1 1 0 4 311 A;1 0 2 0 2 10 (d)1 1 1 1 2 2 ;0 31 2 2 5 (e)1 1 1 0 0 3 ;0 1 2 11 1 2.11 Describe the matrices in each of the classes represented in Example 2.9. 2.12 Describe all matrices in the row equivalence class of these. (a)1 0 0 0 (b)1 2 2 4 (c)1 1 1 3 2.13 How many row equivalence classes are there? 2.14 Can row equivalence classes contain dierent-sized matrices? 2.15 How big are the row equivalence classes? (a)Show that for any matrix of all zeros, the class is nite. (b)Do any other classes contain only nitely many members? X2.16 Give two reduced echelon form matrices that have their leading entries in the same columns, but that are not row equivalent. X2.17 Show that any two nnnonsingular matrices are row equivalent. Are any two singular matrices row equivalent? X2.18 Describe all of the row equivalence classes containing these. 60 Chapter One. Linear Systems (a)22 matrices (b)23 matrices (c)32 matrices (d)33 matrices 2.19 (a) Show that a vector ~0is a linear combination of members of the set f~1;:::;~ngif and only if there is a linear relationship ~0 =c0~0+


+cn~n wherec0is not zero. ( Hint. Watch out for the ~0=~0 case.) (b)Use that to simplify the proof of Lemma 2.4. X2.20 Finish the proof of Lemma 2.4. (a)First illustrate the inductive step by showing that c2= 0. (b)Do the full inductive step: where 1 n < i1, assume that ck= 0 for 1<k<n and deduce that cn+1= 0 also. (c)Find the contradiction. 2.21 Finish the induction argument in Lemma 2.5. (a)State the inductive hypothesis, Also state what must be shown to follow from that hypothesis. (b)Check that the inductive hypothesis implies that in the relationship r+1= sr+1;11+sr+2;22+


+sr+1;mmthe coecients sr+1;1; ::: ;sr+1;rare each zero. (c)Finish the inductive step by arguing, as in the base case, that `r+1< kr+1 andkr+1<`r+1are impossible. 2.22 Why, in the proof of Theorem 2.6, do we bother to restrict to the nonzero rows? Why not just stick to the relationship that we began with, i=ci;11+


+ci;mm, withminstead ofr, and argue using it that the only nonzero coecient is ci;i, which is 1? X2.23 Three truck drivers went into a roadside cafe. One truck driver purchased four sandwiches, a cup of coee, and ten doughnuts for $8 :45. Another driver purchased three sandwiches, a cup of coee, and seven doughnuts for $6 :30. What did the third truck driver pay for a sandwich, a cup of coee, and a doughnut? [Trono] 2.24 The fact that Gaussian reduction disallows multiplication of a row by zero is needed for the proof of uniqueness of reduced echelon form, or else every matrix would be row equivalent to a matrix of all zeros. Where is it used? X2.25 The Linear Combination Lemma says which equations can be gotten from Gaussian reduction from a given linear system. (1) Produce an equation not implied by this system. 3x+ 4y= 8 2x+y= 3 (2) Can any equation be derived from an inconsistent system? 2.26 Extend the denition of row equivalence to linear systems. Under your de- nition, do equivalent systems have the same solution set? [Homan & Kunze] X2.27 In this matrix 0 @1 2 3 3 0 3 1 4 51 A the rst and second columns add to the third. (a)Show that remains true under any row operation. (b)Make a conjecture. (c)Prove that it holds. Topic: Computer Algebra Systems 61 Topic: Computer Algebra Systems The linear systems in this chapter are small enough that their solution by hand is easy. But large systems are easiest, and safest, to do on a computer. There are special purpose programs such as LINPACK for this job. Another popular tool is a general purpose computer algebra system, including both commercial packages such as Maple, Mathematica, or MATLAB, or free packages such as Sage. For example, in the Topic on Networks, we need to solve this. i0i1i2 = 0 i1i3i5 = 0 i2i4+i5 = 0 i3+i4i6= 0 5i1 + 10i3 = 10 2i2 + 4i4 = 10 5i12i2 + 50i5 = 0 It can be done by hand, but it would take a while and be error-prone. Using a computer is better. We illustrate by solving that system under Maple (for another system, a user's manual would obviously detail the exact syntax needed). The array of coecients can be entered in this way > A:=array( [[1,-1,-1,0,0,0,0], [0,1,0,-1,0,-1,0], [0,0,1,0,-1,1,0], [0,0,0,1,1,0,-1], [0,5,0,10,0,0,0], [0,0,2,0,4,0,0], [0,5,-1,0,0,10,0]] ); (putting the rows on separate lines is not necessary, but is done for clarity). The vector of constants is entered similarly. > u:=array( [0,0,0,0,10,10,0] ); Then the system is solved, like magic. > linsolve(A,u); 7 2 5 2 5 7 [ -, -, -, -, -, 0, - ] 3 3 3 3 3 3 Systems with innitely many solutions are solved in the same way | the com- puter simply returns a parametrization. Exercises Answers for this Topic use Maple as the computer algebra system. In particular, all of these were tested on Maple Vrunning under MS-DOS NT version 4:0. (On all of them, the preliminary command to load the linear algebra package along with Maple's responses to the Enter key, have been omitted.) Other systems have similar commands. 62 Chapter One. Linear Systems 1Use the computer to solve the two problems that opened this chapter. (a)This is the Statics problem. 40h+ 15c= 100 25c= 50 + 50h (b)This is the Chemistry problem. 7h= 7j 8h+ 1i= 5j+ 2k 1i= 3j 3i= 6j+ 1k 2Use the computer to solve these systems from the rst subsection, or conclude `many solutions' or `no solutions'. (a)2x+ 2y= 5 x4y= 0(b)x+y= 1 x+y= 2(c)x3y+z= 1 x+y+ 2z= 14 (d)xy= 1 3x3y= 2(e) 4y+z= 20 2x2y+z= 0 x +z= 5 x+yz= 10(f)2x +z+w= 5 yw=1 3xzw= 0 4x+y+ 2z+w= 9 3Use the computer to solve these systems from the second subsection. (a)3x+ 6y= 18 x+ 2y= 6(b)x+y= 1 xy=1(c)x1 +x3= 4 x1x2+ 2x3= 5 4x1x2+ 5x3= 17 (d)2a+bc= 2 2a +c= 3 ab = 0(e)x+ 2yz = 3 2x+y +w= 4 xy+z+w= 1(f)x +z+w= 4 2x+yw= 2 3x+y+z = 7 4What does the computer give for the solution of the general 2 2 system? ax+cy=p bx+dy=q Topic: Input-Output Analysis 63 Topic: Input-Output Analysis An economy is an immensely complicated network of interdependences. Changes in one part can ripple out to aect other parts. Economists have struggled to be able to describe, and to make predictions about, such a complicated object. Mathematical models using systems of linear equations have emerged as a key tool. One is Input-Output Analysis, pioneered by W. Leontief, who won the 1973 Nobel Prize in Economics. Consider an economy with many parts, two of which are the steel industry and the auto industry. As they work to meet the demand for their product from other parts of the economy, that is, from users external to the steel and auto sectors, these two interact tightly. For instance, should the external demand for autos go up, that would lead to an increase in the auto industry's usage of steel. Or, should the external demand for steel fall, then it would lead to a fall in steel's purchase of autos. The type of Input-Output model we will consider takes in the external demands and then predicts how the two interact to meet those demands. We start with a listing of production and consumption statistics. (These numbers, giving dollar values in millions, are excerpted from [Leontief 1965], describing the 1958 U.S. economy. Today's statistics would be quite dierent, both because of in ation and because of technical changes in the industries.) used by steelused by autoused by others total value of steel5 395 2 664 25 448 value of auto48 9 030 30 346 For instance, the dollar value of steel used by the auto industry in this year is 2;664 million. Note that industries may consume some of their own output. We can ll in the blanks for the external demand. This year's value of the steel used by others this year is 17 ;389 and this year's value of the auto used by others is 21 ;268. With that, we have a complete description of the external demands and of how auto and steel interact, this year, to meet them. Now, imagine that the external demand for steel has recently been going up by 200 per year and so we estimate that next year it will be 17 ;589. Imagine also that for similar reasons we estimate that next year's external demand for autos will be down 25 to 21 ;243. We wish to predict next year's total outputs. That prediction isn't as simple as adding 200 to this year's steel total and subtracting 25 from this year's auto total. For one thing, a rise in steel will cause that industry to have an increased demand for autos, which will mitigate, to some extent, the loss in external demand for autos. On the other hand, the drop in external demand for autos will cause the auto industry to use less steel, and so lessen somewhat the upswing in steel's business. In short, these two industries form a system, and we need to predict the totals at which the system as a whole will settle. 64 Chapter One. Linear Systems For that prediction, let sbe next years total production of steel and let abe next year's total output of autos. We form these equations. next year's production of steel = next year's use of steel by steel + next year's use of steel by auto + next year's use of steel by others next year's production of autos = next year's use of autos by steel + next year's use of autos by auto + next year's use of autos by others On the left side of those equations go the unknowns sanda. At the ends of the right sides go our external demand estimates for next year 17 ;589 and 21;243. For the remaining four terms, we look to the table of this year's information about how the industries interact. For instance, for next year's use of steel by steel, we note that this year the steel industry used 5395 units of steel input to produce 25 ;448 units of steel output. So next year, when the steel industry will produce sunits out, we expect that doing so will take s
(5395)=(25 448) units of steel input | this is simply the assumption that input is proportional to output. (We are assuming that the ratio of input to output remains constant over time; in practice, models may try to take account of trends of change in the ratios.) Next year's use of steel by the auto industry is similar. This year, the auto industry uses 2664 units of steel input to produce 30346 units of auto output. So next year, when the auto industry's total output is a, we expect it to consume a
(2664)=(30346) units of steel. Filling in the other equation in the same way, we get this system of linear equation. 5 395 25 448
s+2 664 30 346
a+ 17 589 = s 48 25 448
s+9 030 30 346
a+ 21 243 = a Gauss' method on this system. (20 053=25 448)s(2 664=30 346)a= 17 589 (48=25 448)s+ (21 316=30 346)a= 21 243 givess= 25 698 and a= 30 311. Looking back, recall that above we described why the prediction of next year's totals isn't as simple as adding 200 to last year's steel total and subtract- ing 25 from last year's auto total. In fact, comparing these totals for next year to the ones given at the start for the current year shows that, despite the drop in external demand, the total production of the auto industry is predicted to rise. The increase in internal demand for autos caused by steel's sharp rise in business more than makes up for the loss in external demand for autos. One of the advantages of having a mathematical model is that we can ask \What if . . . ?" questions. For instance, we can ask \What if the estimates for Topic: Input-Output Analysis 65 next year's external demands are somewhat o?" To try to understand how much the model's predictions change in reaction to changes in our estimates, we can try revising our estimate of next year's external steel demand from 17 ;589 down to 17 ;489, while keeping the assumption of next year's external demand for autos xed at 21 ;243. The resulting system (20 053=25 448)s(2 664=30 346)a= 17 489 (48=25 448)s+ (21 316=30 346)a= 21 243 when solved gives s= 25 571 and a= 30 311. This kind of exploration of the model is sensitivity analysis . We are seeing how sensitive the predictions of our model are to the accuracy of the assumptions. Obviously, we can consider larger models that detail the interactions among more sectors of an economy. These models are typically solved on a computer, using the techniques of matrix algebra that we will develop in Chapter Three. Some examples are given in the exercises. Obviously also, a single model does not suit every case; expert judgment is needed to see if the assumptions un- derlying the model are reasonable for a particular case. With those caveats, however, this model has proven in practice to be a useful and accurate tool for economic analysis. For further reading, try [Leontief 1951] and [Leontief 1965]. Exercises Hint: these systems are easiest to solve on a computer. 1With the steel-auto system given above, estimate next year's total productions in these cases. (a)Next year's external demands are: up 200 from this year for steel, and un- changed for autos. (b)Next year's external demands are: up 100 for steel, and up 200 for autos. (c)Next year's external demands are: up 200 for steel, and up 200 for autos. 2In the steel-auto system, the ratio for the use of steel by the auto industry is 2 664=30 346, about 0 :0878. Imagine that a new process for making autos reduces this ratio to :0500. (a)How will the predictions for next year's total productions change compared to the rst example discussed above (i.e., taking next year's external demands to be 17;589 for steel and 21 ;243 for autos)? (b)Predict next year's totals if, in addition, the external demand for autos rises to be 21;500 because the new cars are cheaper. 3This table gives the numbers for the auto-steel system from a dierent year, 1947 (see [Leontief 1951]). The units here are billions of 1947 dollars. used by steelused by autoused by others total value of steel6:90 1:28 18 :69 value of autos0 4:40 14 :27 (a)Solve for total output if next year's external demands are: steel's demand up 10% and auto's demand up 15%. (b)How do the ratios compare to those given above in the discussion for the 1958 economy? 66 Chapter One. Linear Systems (c)Solve the 1947 equations with the 1958 external demands (note the dierence in units; a 1947 dollar buys about what $1 :30 in 1958 dollars buys). How far o are the predictions for total output? 4Predict next year's total productions of each of the three sectors of the hypothet- ical economy shown below used by farmused by railused by shippingused by others total value of farm25 50 100 800 value of rail25 50 50 300 value of shipping15 10 0 500 if next year's external demands are as stated. (a)625 for farm, 200 for rail, 475 for shipping (b)650 for farm, 150 for rail, 450 for shipping 5This table gives the interrelationships among three segments of an economy (see [Clark & Coupe]). used by foodused by wholesaleused by retailused by otherstotal value of food 0 2 318 4 679 11 869 value of wholesale 393 1 089 22 459 122 242 value of retail 3 53 75 116 041 We will do an Input-Output analysis on this system. (a)Fill in the numbers for this year's external demands. (b)Set up the linear system, leaving next year's external demands blank. (c)Solve the system where next year's external demands are calculated by tak- ing this year's external demands and in ating them 10%. Do all three sectors increase their total business by 10%? Do they all even increase at the same rate? (d)Solve the system where next year's external demands are calculated by taking this year's external demands and reducing them 7%. (The study from which these numbers are taken concluded that because of the closing of a local military facility, overall personal income in the area would fall 7%, so this might be a rst guess at what would actually happen.) Topic: Accuracy of Computations 67 Topic: Accuracy of Computations Gauss' method lends itself nicely to computerization. The code below illustrates. It operates on an nnmatrix a, doing row combinations using the rst row, then the second row, etc. for(row=1;row<=n-1;row++){ for(row_below=row+1;row_below<=n;row_below++){ multiplier=a[row_below,row]/a[row,row]; for(col=row; col<=n; col++){ a[row_below,col]-=multiplier*a[row,col]; } } } (This code is in the C language. Here is a brief translation. The loop construct for(row=1;row<=n-1;row++){


}setsrowto 1 and then iterates while rowis less than or equal to n1, each time through incrementing the variable row by one with the ` ++' operation. The other non-obvious construct is that the `-=' in the innermost loop amounts to the a[row below,col] =multiplier a[row,col] +a[row below,col] operation.) While this code provides a quick take on how Gauss' method can be mecha- nized, it is not ready to use. It is naive in many ways. The most glaring way is that it assumes that a nonzero number is always found in the row;rowposition. To make it practical, one way in which this code needs to be reworked is to cover the case where nding a zero in that location leads to a row swap, or to the conclusion that the matrix is singular. Adding some if


statements to cover those cases is not hard, but we will instead consider some more subtle ways in which the code is naive. There are pitfalls arising from the computer's reliance on nite-precision oating point arithmetic. For example, we have seen above that we must handle as a separate case a system that is singular. But systems that are nearly singular also require care. Consider this one. x+ 2y= 3 1:000 000 01x+ 2y= 3:000 000 01 By eye we get the solution x= 1 andy= 1. But a computer has more trouble. A computer that represents real numbers to eight signicant places (as is common, usually called single precision ) will represent the second equation internally as 1:000 000 0x+ 2y= 3:000 000 0, losing the digits in the ninth place. Instead of reporting the correct solution, this computer will report something that is not even close | this computer thinks that the system is singular because the two equations are represented internally as equal. For some intuition about how the computer could come up with something that far o, we can graph the system. 68 Chapter One. Linear Systems (1;1) At the scale of this graph, the two lines cannot be resolved apart. This system is nearly singular in the sense that the two lines are nearly the same line. Near- singularity gives this system the property that a small change in the system can cause a large change in its solution; for instance, changing the 3 :000 000 01 to 3:000 000 03 changes the intersection point from (1 ;1) to (3;0). This system changes radically depending on a ninth digit, which explains why the eight- place computer has trouble. A problem that is very sensitive to inaccuracy or uncertainties in the input values is ill-conditioned . The above example gives one way in which a system can be dicult to solve on a computer. It has the advantage that the picture of nearly-equal lines gives a memorable insight into one way that numerical diculties can arise. Unfortunately this insight isn't very useful when we wish to solve some large system. We cannot, typically, hope to understand the geometry of an arbitrary large system. In addition, there are ways that a computer's results may be unreliable other than that the angle between some of the linear surfaces is quite small. For an example, consider the system below, from [Hamming]. 0:001x+y= 1 xy= 0() The second equation gives x=y, sox=y= 1=1:001 and thus both variables have values that are just less than 1. A computer using two digits represents the system internally in this way (we will do this example in two-digit oating point arithmetic, but a similar one with eight digits is easy to invent). (1:0102)x+ (1:0100)y= 1:0100 (1:0100)x(1:0100)y= 0:0100 The computer's row reduction step 10001+2produces a second equation 1001y=999, which the computer rounds to two places as ( 1:0103)y= 1:0103. Then the computer decides from the second equation that y= 1 and from the rst equation that x= 0. Thisyvalue is fairly good, but the x is quite bad. Thus, another cause of unreliable output is a mixture of oating point arithmetic and a reliance on using leading entries that are small. An experienced programmer may respond that we should go to double pre- cision where sixteen signicant digits are retained. This will indeed solve many problems. However, there are some diculties with it as a general approach. For one thing, double precision takes longer than single precision (on a '486 Topic: Accuracy of Computations 69 chip, multiplication takes eleven ticks in single precision but fourteen in dou- ble precision [Programmer's Ref.]) and has twice the memory requirements. So attempting to do all calculations in double precision is just not practical. And besides, the above systems can obviously be tweaked to give the same trouble in the seventeenth digit, so double precision won't x all problems. What we need is a strategy to minimize the numerical trouble arising from solving systems on a computer, and some guidance as to how far the reported solutions can be trusted. Mathematicians have made a careful study of how to get the most reliable results. A basic improvement on the naive code above is to not simply take the entry in the row;rowposition to determine the factor to use for the row combination, but rather to look at all of the entries in the rowcolumn below therowrow, and take the one that is most likely to give reliable results (e.g., take one that is not too small). This strategy is called partial pivoting . For example, to solve the troublesome system ( ) above, we start by looking at both equations for a best entry to use, and taking the 1 in the second equation as more likely to give good results. Then, the combination step of :0012+ 1gives a rst equation of 1 :001y= 1, which the computer will represent as (1:0100)y= 1:0100, leading to the conclusion that y= 1 and, after back- substitution, x= 1, both of which are close to right. The code from above can be adapted to this purpose. for(row=1;row<=n-1;row++){ /* find the largest entry in this column (in row max) */ max=row; for(row_below=row+1;row_below<=n;row_below++){ if (abs(a[row_below,row]) > abs(a[max,row])); max = row_below; } /* swap rows to move that best entry up */ for(col=row;col<=n;col++){ temp=a[row,col]; a[row,col]=a[max,col]; a[max,col]=temp; } /* proceed as before */ for(row_below=row+1;row_below<=n;row_below++){ multiplier=a[row_below,row]/a[row,row]; for(col=row;col<=n;col++){ a[row_below,col]-=multiplier*a[row,col]; } } } A full analysis of the best way to implement Gauss' method is outside the scope of the book (see [Wilkinson 1965]), but the method recommended by most experts is a variation on the code above that rst nds the best entry among the candidates, and then scales it to a number that is less likely to give trouble. This is scaled partial pivoting . 70 Chapter One. Linear Systems In addition to returning a result that is likely to be reliable, most well-done code will return a number, called the conditioning number that describes the factor by which uncertainties in the input numbers could be magnied to become inaccuracies in the results returned (see [Rice]). The lesson of this discussion is that just because Gauss' method always works in theory, and just because computer code correctly implements that method, doesn't mean that the answer is reliable. In practice, always use a package where experts have worked hard to counter what can go wrong. Exercises 1Using two decimal places, add 253 and 2 =3. 2This intersect-the-lines problem contrasts with the example discussed above. (1;1)x+ 2y= 3 3x2y= 1 Illustrate that in this system some small change in the numbers will produce only a small change in the solution by changing the constant in the bottom equation to 1:008 and solving. Compare it to the solution of the unchanged system. 3Solve this system by hand ([Rice]). 0:000 3x+ 1:556y= 1:569 0:345 4x2:346y= 1:018 (a)Solve it accurately, by hand. (b)Solve it by rounding at each step to four signicant digits. 4Rounding inside the computer often has an eect on the result. Assume that your machine has eight signicant digits. (a)Show that the machine will compute (2 =3) + ((2=3)(1=3)) as unequal to ((2=3) + (2=3))(1=3). Thus, computer arithmetic is not associative. (b)Compare the computer's version of (1 =3)x+y= 0 and (2=3)x+ 2y= 0. Is twice the rst equation the same as the second? 5Ill-conditioning is not only dependent on the matrix of coecients. This example [Hamming] shows that it can arise from an interaction between the left and right sides of the system. Let "be a small real. 3x+ 2y+z= 6 2x+ 2"y+ 2"z= 2 + 4" x+ 2"y"z= 1 +" (a)Solve the system by hand. Notice that the "'s divide out only because there is an exact cancelation of the integer parts on the right side as well as on the left. (b)Solve the system by hand, rounding to two decimal places, and with "= 0:001. Topic: Analyzing Networks 71 Topic: Analyzing Networks The diagram below shows some of a car's electrical network. The battery is on the left, drawn as stacked line segments. The wires are drawn as lines, shown straight and with sharp right angles for neatness. Each light is a circle enclosing a loop. 12VDome Light Door Actuated Switch Brake LightsL RBrake Actuated SwitchLight Switch O Dimmer Hi Lo L R L R HeadlightsL R Rear LightsL R Parking Lights The designer of such a network needs to answer questions like: How much electricity ows when both the hi-beam headlights and the brake lights are on? Below, we will use linear systems to analyze simpler versions of electrical networks. For the analysis we need two facts about electricity and two facts about electrical networks. The rst fact about electricity is that a battery is like a pump: it provides a force impelling the electricity to ow through the circuits connecting the bat- tery's ends, if there are any such circuits. We say that the battery provides a potential to ow. Of course, this network accomplishes its function when, as the electricity ows through a circuit, it goes through a light. For instance, when the driver steps on the brake then the switch makes contact and a cir- cuit is formed on the left side of the diagram, and the electrical current owing through that circuit will make the brake lights go on, warning drivers behind. The second electrical fact is that in some kinds of network components the amount of ow is proportional to the force provided by the battery. That is, for each such component there is a number, it's resistance , such that the potential is equal to the ow times the resistance. The units of measurement are: potential is described in volts, the rate of ow is in amperes , and resistance to the ow is inohms . These units are dened so that volts = amperes
ohms. Components with this property, that the voltage-amperage response curve is a line through the origin, are called resistors . (Light bulbs such as the ones shown above are not this kind of component, because their ohmage changes as they heat up.) For example, if a resistor measures 2 ohms then wiring it to a 12 volt battery results in a ow of 6 amperes. Conversely, if we have ow of electrical current of 2 amperes through it then there must be a 4 volt potential 72 Chapter One. Linear Systems dierence between it's ends. This is the voltage drop across the resistor. One way to think of a electrical circuits like the one above is that the battery provides a voltage rise while the other components are voltage drops. The two facts that we need about networks are Kirchho's Laws. Current Law. For any point in a network, the ow in equals the ow out. Voltage Law. Around any circuit the total drop equals the total rise. In the above network there is only one voltage rise, at the battery, but some networks have more than one. For a start we can consider the network below. It has a battery that provides the potential to ow and three resistors (resistors are drawn as zig-zags). When components are wired one after another, as here, they are said to be in series . 20volt potential2ohm resistance 5ohm resistance 3ohm resistance By Kirchho's Voltage Law, because the voltage rise is 20 volts, the total voltage drop must also be 20 volts. Since the resistance from start to nish is 10 ohms (the resistance of the wires is negligible), we get that the current is (20 =10) = 2 amperes. Now, by Kirchho's Current Law, there are 2 amperes through each resistor. (And therefore the voltage drops are: 4 volts across the 2 oh m resistor, 10 volts across the 5 ohm resistor, and 6 volts across the 3 ohm resistor.) The prior network is so simple that we didn't use a linear system, but the next network is more complicated. In this one, the resistors are in parallel . This network is more like the car lighting diagram shown earlier. 20volt 12ohm 8ohm We begin by labeling the branches, shown below. Let the current through the left branch of the parallel portion be i1and that through the right branch be i2, and also let the current through the battery be i0. (We are following Kircho's Current Law; for instance, all points in the right branch have the same current, which we call i2. Note that we don't need to know the actual direction of ow | if current ows in the direction opposite to our arrow then we will simply get a negative number in the solution.) Topic: Analyzing Networks 73 "i0 i1# # i2 The Current Law, applied to the point in the upper right where the ow i0 meetsi1andi2, gives that i0=i1+i2. Applied to the lower right it gives i1+i2=i0. In the circuit that loops out of the top of the battery, down the left branch of the parallel portion, and back into the bottom of the battery, the voltage rise is 20 while the voltage drop is i1
12, so the Voltage Law gives that 12i1= 20. Similarly, the circuit from the battery to the right branch and back to the battery gives that 8 i2= 20. And, in the circuit that simply loops around in the left and right branches of the parallel portion (arbitrarily taken clockwise), there is a voltage rise of 0 and a voltage drop of 8 i212i1so the Voltage Law gives that 8 i212i1= 0. i0i1i2= 0 i0+i1+i2= 0 12i1 = 20 8i2= 20 12i1+ 8i2= 0 The solution is i0= 25=6,i1= 5=3, andi2= 5=2, all in amperes. (Incidentally, this illustrates that redundant equations do indeed arise in practice.) Kirchho's laws can be used to establish the electrical properties of networks of great complexity. The next diagram shows ve resistors, wired in a series- parallel way. 10volt5ohm 2ohm 50ohm 10ohm 4ohm This network is a Wheatstone bridge (see Exercise 4). To analyze it, we can place the arrows in this way. "i0i1. & i2 i5! i3& . i4 74 Chapter One. Linear Systems Kircho's Current Law, applied to the top node, the left node, the right node, and the bottom node gives these. i0=i1+i2 i1=i3+i5 i2+i5=i4 i3+i4=i0 Kirchho's Voltage Law, applied to the inside loop (the i0toi1toi3toi0loop), the outside loop, and the upper loop not involving the battery, gives these. 5i1+ 10i3= 10 2i2+ 4i4= 10 5i1+ 50i52i2= 0 Those suce to determine the solution i0= 7=3,i1= 2=3,i2= 5=3,i3= 2=3, i4= 5=3, andi5= 0. Networks of other kinds, not just electrical ones, can also be analyzed in this way. For instance, networks of streets are given in the exercises. Exercises Many of the systems for these problems are mostly easily solved on a computer. 1Calculate the amperages in each part of each network. (a)This is a simple network. 9volt3ohm 2ohm 2ohm (b)Compare this one with the parallel case discussed above. 9volt3ohm 2ohm 2ohm 2ohm (c)This is a reasonably complicated network. 9volt3ohm 3ohm 2ohm 2ohm3ohm 4ohm 2ohm Topic: Analyzing Networks 75 2In the rst network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10 ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed, 20volt 12ohm 8ohm the electric current through the battery is 25 =6 amperes. Thus, the parallel portion isequivalent to a single resistor of 20 =(25=6) = 4:8 ohms. (a)What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms? (b)What is the equivalent resistance if the two are each 8 ohms? (c)Find the formula for the equivalent resistance if the two resistors in parallel arer1ohms andr2ohms. 3For the car dashboard example that opens this Topic, solve for these amperages (assume that all resistances are 2 ohms). (a)If the driver is stepping on the brakes, so the brake lights are on, and no other circuit is closed. (b)If the hi-beam headlights and the brake lights are on. 4Show that, in this Wheatstone Bridge, r1 r3 rg r2 r4 r2=r1equalsr4=r3if and only if the current owing through rgis zero. (The way that this device is used in practice is that an unknown resistance at r4is compared to the other three r1,r2, andr3. Atrgis placed a meter that shows the current. The three resistances r1,r2, andr3are varied | typically they each have a calibrated knob | until the current in the middle reads 0, and then the above equation gives the value of r4.) There are networks other than electrical ones, and we can ask how well Kircho's laws apply to them. The remaining questions consider an extension to networks of streets. 5Consider this trac circle. Main StreetNorth Avenue Pier Boulevard 76 Chapter One. Linear Systems This is the trac volume, in units of cars per ve minutes. North Pier Main into 100 150 25 out of 75 150 50 We can set up equations to model how the trac ows. (a)Adapt Kircho's Current Law to this circumstance. Is it a reasonable mod- elling assumption? (b)Label the three between-road arcs in the circle with a variable. Using the (adapted) Current Law, for each of the three in-out intersections state an equa- tion describing the trac ow at that node. (c)Solve that system. (d)Interpret your solution. (e)Restate the Voltage Law for this circumstance. How reasonable is it? 6This is a network of streets. Shelburne St Willow Winooski AvewesteastJay Ln The hourly ow of cars into this network's entrances, and out of its exits can be observed. east Winooski west Winooski Willow Jay Shelburne into 80 50 65 { 40 out of 30 5 70 55 75 (Note that to reach Jay a car must enter the network via some other road rst, which is why there is no `into Jay' entry in the table. Note also that over a long period of time, the total in must approximately equal the total out, which is why both rows add to 235 cars.) Once inside the network, the trac may ow in dier- ent ways, perhaps lling Willow and leaving Jay mostly empty, or perhaps owing in some other way. Kirchho's Laws give the limits on that freedom. (a)Determine the restrictions on the ow inside this network of streets by setting up a variable for each block, establishing the equations, and solving them. Notice that some streets are one-way only. ( Hint: this will not yield a unique solution, since trac can ow through this network in various ways; you should get at least one free variable.) (b)Suppose that some construction is proposed for Winooski Avenue East be- tween Willow and Jay, so trac on that block will be reduced. What is the least amount of trac ow that can be allowed on that block without disrupting the hourly ow into and out of the network? Chapter Two Vector Spaces The rst chapter began by introducing Gauss' method and nished with a fair understanding, keyed on the Linear Combination Lemma, of how it nds the solution set of a linear system. Gauss' method systematically takes linear com- binations of the rows. With that insight, we now move to a general study of linear combinations. We need a setting for this study. At times in the rst chapter, we've com- bined vectors from R2, at other times vectors from R3, and at other times vectors from even higher-dimensional spaces. Thus, our rst impulse might be to work inRn, leavingnunspecied. This would have the advantage that any of the results would hold for R2and for R3and for many other spaces, simultaneously. But, if having the results apply to many spaces at once is advantageous then sticking only to Rn's is overly restrictive. We'd like the results to also apply to combinations of row vectors, as in the nal section of the rst chapter. We've even seen some spaces that are not just a collection of all of the same-sized column vectors or row vectors. For instance, we've seen a solution set of a homogeneous system that is a plane, inside of R3. This solution set is a closed system in the sense that a linear combination of these solutions is also a solution. But it is not just a collection of all of the three-tall column vectors; only some of them are in this solution set. We want the results about linear combinations to apply anywhere that linear combinations are sensible. We shall call any such set a vector space . Our results, instead of being phrased as \Whenever we have a collection in which we can sensibly take linear combinations . . . ", will be stated as \In any vector space . . . ". Such a statement describes at once what happens in many spaces. The step up in abstraction from studying a single space at a time to studying a class of spaces can be hard to make. To understand its advantages, consider this analogy. Imagine that the government made laws one person at a time: \Leslie Jones can't jay walk." That would be a bad idea; statements have the virtue of economy when they apply to many cases at once. Or, suppose that they ruled, \Kim Ke must stop when passing the scene of an accident." Contrast that with, \Any doctor must stop when passing the scene of an accident." More general statements, in some ways, are clearer. 77 78 Chapter Two. Vector Spaces I Denition of Vector Space We shall study structures with two operations, an addition and a scalar multi- plication, that are subject to some simple conditions. We will re ect more on the conditions later, but on rst reading notice how reasonable they are. For in- stance, surely any operation that can be called an addition (e.g., column vector addition, row vector addition, or real number addition) will satisfy conditions (1) through (5) below. I.1 Denition and Examples 1.1 Denition Avector space (overR) consists of a set Valong with two operations `+' and `
' subject to these conditions. Where~ v;~ w2V, (1) their vector sum ~ v+~ wis an element of V. If~ u;~ v;~ w2V then (2)~ v+~ w=~ w+~ vand (3) (~ v+~ w) +~ u=~ v+ (~ w+~ u). (4) There is a zero vector~02Vsuch that~ v+~0 =~ vfor all~ v2V. (5) Each~ v2Vhas an additive inverse~ w2Vsuch that~ w+~ v=~0. Ifr;sarescalars , members of R, and~ v;~ w2Vthen (6) each scalar multiple r
~ vis inV. Ifr;s2Rand~ v;~ w2Vthen (7) (r+s)
~ v=r
~ v+s
~ v, and (8)r
(~ v+~ w) =r
~ v+r
~ w, and (9) (rs)
~ v=r
(s
~ v), and (10) 1
~ v=~ v. 1.2 Remark Because it involves two kinds of addition and two kinds of mul- tiplication, that denition may seem confused. For instance, in condition (7) `(r+s)
~ v=r
~ v+s
~ v', the rst `+' is the real number addition operator while the `+' to the right of the equals sign represents vector addition in the structure V. These expressions aren't ambiguous because, e.g., randsare real numbers so `r+s' can only mean real number addition. The best way to go through the examples below is to check all ten conditions in the denition. That check is written out at length in the rst example. Use it as a model for the others. Especially important are the rst condition ` ~ v+~ w is inV' and the sixth condition ` r
~ vis inV'. These are the closure conditions. They specify that the addition and scalar multiplication operations are always sensible | they are dened for every pair of vectors, and every scalar and vector, and the result of the operation is a member of the set (see Example 1.4). 1.3 Example The set R2is a vector space if the operations `+' and `
' have their usual meaning. x1 x2 +y1 y2 =x1+y1 x2+y2 r
x1 x2 =rx1 rx2 We shall check all of the conditions. Section I. Denition of Vector Space 79 There are ve conditions in item (1). For (1), closure of addition, note that for anyv1;v2;w1;w22Rthe result of the sum v1 v2 +w1 w2 =v1+w1 v2+w2 is a column array with two real entries, and so is in R2. For (2), that addition of vectors commutes, take all entries to be real numbers and compute v1 v2 +w1 w2 =v1+w1 v2+w2 =w1+v1 w2+v2 =w1 w2 +v1 v2 (the second equality follows from the fact that the components of the vectors are real numbers, and the addition of real numbers is commutative). Condition (3), associativity of vector addition, is similar. (v1 v2 +w1 w2 ) +u1 u2 =(v1+w1) +u1 (v2+w2) +u2 = v1+ (w1+u1) v2+ (w2+u2) =v1 v2 + (w1 w2 +u1 u2 ) For the fourth condition we must produce a zero element | the vector of zeroes is it. v1 v2 +0 0 =v1 v2 For (5), to produce an additive inverse, note that for any v1;v22Rwe have v1 v2 +v1 v2 =0 0 so the rst vector is the desired additive inverse of the second. The checks for the ve conditions having to do with scalar multiplication are just as routine. For (6), closure under scalar multiplication, where r;v1;v22R, r
v1 v2 =rv1 rv2 is a column array with two real entries, and so is in R2. Next, this checks (7). (r+s)
v1 v2 =(r+s)v1 (r+s)v2 =rv1+sv1 rv2+sv2 =r
v1 v2 +s
v1 v2 For (8), that scalar multiplication distributes from the left over vector addition, we have this. r
(v1 v2 +w1 w2 ) =r(v1+w1) r(v2+w2) =rv1+rw1 rv2+rw2 =r
v1 v2 +r
w1 w2 80 Chapter Two. Vector Spaces The ninth (rs)
v1 v2 =(rs)v1 (rs)v2 =r(sv1) r(sv2) =r
(s
v1 v2 ) and tenth conditions are also straightforward. 1
v1 v2 =1v1 1v2 =v1 v2 In a similar way, each Rnis a vector space with the usual operations of vector addition and scalar multiplication. (In R1, we usually do not write the members as column vectors, i.e., we usually do not write `( )'. Instead we just write `'.) 1.4 Example This subset of R3that is a plane through the origin P=f0 @x y z1 Ax+y+z= 0g is a vector space if `+' and `
' are interpreted in this way. 0 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @x1+x2 y1+y2 z1+z21 Ar
0 @x y z1 A=0 @rx ry rz1 A The addition and scalar multiplication operations here are just the ones of R3, reused on its subset P. We say that Pinherits these operations from R3. This example of an addition in P 0 @1 1 21 A+0 @1 0 11 A=0 @0 1 11 A illustrates that Pis closed under addition. We've added two vectors from P| that is, with the property that the sum of their three entries is zero | and the result is a vector also in P. Of course, this example of closure is not a proof of closure. To prove that Pis closed under addition, take two elements of P 0 @x1 y1 z11 A0 @x2 y2 z21 A (membership in Pmeans that x1+y1+z1= 0 andx2+y2+z2= 0), and observe that their sum 0 @x1+x2 y1+y2 z1+z21 A Section I. Denition of Vector Space 81 is also inPsince its entries add ( x1+x2) + (y1+y2) + (z1+z2) = (x1+y1+ z1) + (x2+y2+z2) to 0. To show that Pis closed under scalar multiplication, start with a vector from P0 @x y z1 A (so thatx+y+z= 0) and then for r2Robserve that the scalar multiple r
0 @x y z1 A=0 @rx ry rz1 A satises that rx+ry+rz=r(x+y+z) = 0. Thus the two closure conditions are satised. Verication of the other conditions in the denition of a vector space are just as straightforward. 1.5 Example Example 1.3 shows that the set of all two-tall vectors with real entries is a vector space. Example 1.4 gives a subset of an Rnthat is also a vector space. In contrast with those two, consider the set of two-tall columns with entries that are integers (under the obvious operations). This is a subset of a vector space, but it is not itself a vector space. The reason is that this set is not closed under scalar multiplication, that is, it does not satisfy condition (6). Here is a column with integer entries, and a scalar, such that the outcome of the operation 0:5
4 3 =2 1:5 is not a member of the set, since its entries are not all integers. 1.6 Example The singleton set f0 BB@0 0 0 01 CCAg is a vector space under the operations 0 BB@0 0 0 01 CCA+0 BB@0 0 0 01 CCA=0 BB@0 0 0 01 CCAr
0 BB@0 0 0 01 CCA=0 BB@0 0 0 01 CCA that it inherits from R4. A vector space must have at least one element, its zero vector. Thus a one-element vector space is the smallest one possible. 1.7 Denition A one-element vector space is a trivial space. 82 Chapter Two. Vector Spaces Warning! The examples so far involve sets of column vectors with the usual operations. But vector spaces need not be collections of column vectors, or even of row vectors. Below are some other types of vector spaces. The term `vector space' does not mean `collection of columns of reals'. It means something more like `collection in which any linear combination is sensible'. 1.8 Example ConsiderP3=fa0+a1x+a2x2+a3x3a0;:::;a 32Rg, the set of polynomials of degree three or less (in this book, we'll take constant polynomials, including the zero polynomial, to be of degree zero). It is a vector space under the operations (a0+a1x+a2x2+a3x3) + (b0+b1x+b2x2+b3x3) = (a0+b0) + (a1+b1)x+ (a2+b2)x2+ (a3+b3)x3 and r
(a0+a1x+a2x2+a3x3) = (ra0) + (ra1)x+ (ra2)x2+ (ra3)x3 (the verication is easy). This vector space is worthy of attention because these are the polynomial operations familiar from high school algebra. For instance, 3
(12x+ 3x24x3)2
(23x+x2(1=2)x3) =1 + 7x211x3. Although this space is not a subset of any Rn, there is a sense in which we can think ofP3as \the same" as R4. If we identify these two spaces's elements in this way a0+a1x+a2x2+a3x3corresponds to0 BB@a0 a1 a2 a31 CCA then the operations also correspond. Here is an example of corresponding ad- ditions. 12x+ 0x2+ 1x3 + 2 + 3x+ 7x24x3 3 + 1x+ 7x23x3corresponds to0 BB@1 2 0 11 CCA+0 BB@2 3 7 41 CCA=0 BB@3 1 7 31 CCA Things we are thinking of as \the same" add to \the same" sum. Chapter Three makes precise this idea of vector space correspondence. For now we shall just leave it as an intuition. 1.9 Example The setM22of 22 matrices with real number entries is a vector space under the natural entry-by-entry operations. a b c d +w x y z =a+w b +x c+y d +z r
a b c d =ra rb rc rd As in the prior example, we can think of this space as \the same" as R4. Section I. Denition of Vector Space 83 1.10 Example The setfff:N!Rgof all real-valued functions of one natural number variable is a vector space under the operations (f1+f2) (n) =f1(n) +f2(n) (r
f) (n) =rf(n) so that if, for example, f1(n) =n2+ 2 sin(n) andf2(n) =sin(n) + 0:5 then (f1+ 2f2) (n) =n2+ 1. We can view this space as a generalization of Example 1.3 | instead of 2-tall vectors, these functions are like innitely-tall vectors. nf(n) =n2+ 1 0 1 1 2 2 5 3 10 ......corresponds to0 BBBBB@1 2 5 10 ...1 CCCCCA Addition and scalar multiplication are component-wise, as in Example 1.3. (We can formalize \innitely-tall" by saying that it means an innite sequence, or that it means a function from NtoR.) 1.11 Example The set of polynomials with real coecients fa0+a1x+


+anxnn2Nanda0;:::;an2Rg makes a vector space when given the natural `+' (a0+a1x+


+anxn) + (b0+b1x+


+bnxn) = (a0+b0) + (a1+b1)x+


+ (an+bn)xn and `
'. r
(a0+a1x+:::anxn) = (ra0) + (ra1)x+:::(ran)xn This space diers from the space P3of Example 1.8. This space contains not just degree three polynomials, but degree thirty polynomials and degree three hun- dred polynomials, too. Each individual polynomial of course is of a nite degree, but the set has no single bound on the degree of all of its members. This example, like the prior one, can be thought of in terms of innite-tuples. For instance, we can think of 1 + 3 x+ 5x2as corresponding to (1 ;3;5;0;0;:::). However, this space diers from the one in Example 1.10. Here, each member of the set has a nite degree, that is, under the correspondence there is no element from this space matching (1 ;2;5;10; :::). Vectors in this space correspond to innite-tuples that end in zeroes. 1.12 Example The setfff:R!Rgof all real-valued functions of one real variable is a vector space under these. (f1+f2) (x) =f1(x) +f2(x) (r
f) (x) =rf(x) The dierence between this and Example 1.10 is the domain of the functions. 84 Chapter Two. Vector Spaces 1.13 Example The setF=facos+bsina;b2Rgof real-valued functions of the real variable is a vector space under the operations (a1cos+b1sin) + (a2cos+b2sin) = (a1+a2) cos+ (b1+b2) sin and r
(acos+bsin) = (ra) cos+ (rb) sin inherited from the space in the prior example. (We can think of Fas \the same" asR2in thatacos+bsincorresponds to the vector with components aand b.) 1.14 Example The set ff:R!Rd2f dx2+f= 0g is a vector space under the, by now natural, interpretation. (f+g) (x) =f(x) +g(x) (r
f) (x) =rf(x) In particular, notice that closure is a consequence d2(f+g) dx2+ (f+g) = (d2f dx2+f) + (d2g dx2+g) and d2(rf) dx2+ (rf) =r(d2f dx2+f) of basic Calculus. This turns out to equal the space from the prior example | functions satisfying this dierential equation have the form acos+bsin| but this description suggests an extension to solutions sets of other dierential equations. 1.15 Example The set of solutions of a homogeneous linear system in n variables is a vector space under the operations inherited from Rn. For ex- ample, for closure under addition consider a typical equation in that system c1x1+


+cnxn= 0 and suppose that both these vectors ~ v=0 B@v1 ... vn1 CA~ w=0 B@w1 ... wn1 CA satisfy the equation. Then their sum ~ v+~ walso satises that equation: c1(v1+ w1) +


+cn(vn+wn) = (c1v1+


+cnvn) + (c1w1+


+cnwn) = 0. The checks of the other vector space conditions are just as routine. As we've done in those equations, we often omit the multiplication symbol `
'. We can distinguish the multiplication in ` c1v1' from that in ` r~ v' since if both multiplicands are real numbers then real-real multiplication must be meant, while if one is a vector then scalar-vector multiplication must be meant. The prior example has brought us full circle since it is one of our motivating examples. Section I. Denition of Vector Space 85 1.16 Remark Now, with some feel for the kinds of structures that satisfy the denition of a vector space, we can re ect on that denition. For example, why specify in the denition the condition that 1
~ v=~ vbut not a condition that 0
~ v=~0? One answer is that this is just a denition | it gives the rules of the game from here on, and if you don't like it, put the book down and walk away. Another answer is perhaps more satisfying. People in this area have worked hard to develop the right balance of power and generality. This denition has been shaped so that it contains the conditions needed to prove all of the interest- ing and important properties of spaces of linear combinations. As we proceed, we shall derive all of the properties natural to collections of linear combinations from the conditions given in the denition. The next result is an example. We do not need to include these properties in the denition of vector space because they follow from the properties already listed there. 1.17 Lemma In any vector space V, for any~ v2Vandr2R, we have (1) 0
~ v=~0, and (2) (1
~ v) +~ v=~0, and (3)r
~0 =~0. Proof .For (1), note that ~ v= (1 + 0)
~ v=~ v+ (0
~ v). Add to both sides the additive inverse of ~ v, the vector ~ wsuch that~ w+~ v=~0. ~ w+~ v=~ w+~ v+ 0
~ v ~0 =~0 + 0
~ v ~0 = 0
~ v The second item is easy: ( 1
~ v) +~ v= (1 + 1)
~ v= 0
~ v=~0 shows that we can write `~ v' for the additive inverse of ~ vwithout worrying about possible confusion with (1)
~ v. For (3), this r
~0 =r
(0
~0) = (r
0)
~0 =~0 will do. QED We nish with a recap. Our study in Chapter One of Gaussian reduction led us to consider collec- tions of linear combinations. So in this chapter we have dened a vector space to be a structure in which we can form such combinations, expressions of the formc1
~ v1+


+cn
~ vn(subject to simple conditions on the addition and scalar multiplication operations). In a phrase: vector spaces are the right context in which to study linearity. Finally, a comment. From the fact that it forms a whole chapter, and espe- cially because that chapter is the rst one, a reader could come to think that the study of linear systems is our purpose. The truth is, we will not so much use vector spaces in the study of linear systems as we will instead have linear systems start us on the study of vector spaces. The wide variety of examples from this subsection shows that the study of vector spaces is interesting and im- portant in its own right, aside from how it helps us understand linear systems. Linear systems won't go away. But from now on our primary objects of study will be vector spaces. 86 Chapter Two. Vector Spaces Exercises 1.18 Name the zero vector for each of these vector spaces. (a)The space of degree three polynomials under the natural operations (b)The space of 24 matrices (c)The spaceff: [0::1]!Rfis continuousg (d)The space of real-valued functions of one natural number variable X1.19 Find the additive inverse, in the vector space, of the vector. (a)InP3, the vector32x+x2. (b)In the space 22,11 0 3 : (c)Infaex+bexa;b2Rg, the space of functions of the real variable xunder the natural operations, the vector 3 ex2ex. X1.20 Show that each of these is a vector space. (a)The set of linear polynomials P1=fa0+a1xa0;a12Rgunder the usual polynomial addition and scalar multiplication operations. (b)The set of 22 matrices with real entries under the usual matrix operations. (c)The set of three-component row vectors with their usual operations. (d)The set L=f0 BB@x y z w1 CCA2R4x+yz+w= 0g under the operations inherited from R4. X1.21 Show that each of these is not a vector space. ( Hint. Start by listing two members of each set.) (a)Under the operations inherited from R3, this set f0 @x y z1 A2R3x+y+z= 1g (b)Under the operations inherited from R3, this set f0 @x y z1 A2R3x2+y2+z2= 1g (c)Under the usual matrix operations, fa1 b ca;b;c2Rg (d)Under the usual polynomial operations, fa0+a1x+a2x2a0;a1;a22R+g where R+is the set of reals greater than zero (e)Under the inherited operations, fx y 2R2x+ 3y= 4 and 2xy= 3 and 6x+ 4y= 10g 1.22 Dene addition and scalar multiplication operations to make the complex numbers a vector space over R. X1.23 Is the set of rational numbers a vector space over Runder the usual addition and scalar multiplication operations? Section I. Denition of Vector Space 87 1.24 Show that the set of linear combinations of the variables x;y;z is a vector space under the natural addition and scalar multiplication operations. 1.25 Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations. x1 y1 +x2 y2 =x1x2 y1y2 r
x y =rx ry 1.26 Prove or disprove that R3is a vector space under these operations. (a)0 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @0 0 01 Aandr0 @x y z1 A=0 @rx ry rz1 A (b)0 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @0 0 01 Aandr0 @x y z1 A=0 @0 0 01 A X1.27 For each, decide if it is a vector space; the intended operations are the natural ones. (a)The diagonal 22 matrices fa0 0ba;b2Rg (b)This set of 22 matrices fx x +y x+y yx;y2Rg (c)This set f0 BB@x y z w1 CCA2R4x+y+w= 1g (d)The set of functions ff:R!Rdf=dx + 2f= 0g (e)The set of functions ff:R!Rdf=dx + 2f= 1g X1.28 Prove or disprove that this is a vector space: the real-valued functions fof one real variable such that f(7) = 0. X1.29 Show that the set R+of positive reals is a vector space when ` x+y' is inter- preted to mean the product of xandy(so that 2 + 3 is 6), and ` r
x' is interpreted as ther-th power of x. 1.30 Isf(x;y)x;y2Rga vector space under these operations? (a)(x1;y1) + (x2;y2) = (x1+x2;y1+y2) andr
(x;y) = (rx;y) (b)(x1;y1) + (x2;y2) = (x1+x2;y1+y2) andr
(x;y) = (rx;0) 1.31 Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial. 1.32 At this point \the same" is only an intuition, but nonetheless for each vector space identify the kfor which the space is \the same" as Rk. (a)The 23 matrices under the usual operations (b)Thenmmatrices (under their usual operations) (c)This set of 22 matrices fa0 b ca;b;c2Rg 88 Chapter Two. Vector Spaces (d)This set of 22 matrices fa0 b ca+b+c= 0g X1.33 Using~+ to represent vector addition and ~
for scalar multiplication, restate the denition of vector space. X1.34 Prove these. (a)Any vector is the additive inverse of the additive inverse of itself. (b)Vector addition left-cancels: if ~ v;~ s;~t2Vthen~ v+~ s=~ v+~timplies that ~ s=~t. 1.35 The denition of vector spaces does not explicitly say that ~0+~ v=~ v(it instead says that~ v+~0 =~ v). Show that it must nonetheless hold in any vector space. X1.36 Prove or disprove that this is a vector space: the set of all matrices, under the usual operations. 1.37 In a vector space every element has an additive inverse. Can some elements have two or more? 1.38 (a) Prove that every point, line, or plane thru the origin in R3is a vector space under the inherited operations. (b)What if it doesn't contain the origin? X1.39 Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or innitely many elements. Assume that ~ v2Vis not~0. (a)Prove that r
~ v=~0 if and only if r= 0. (b)Prove that r1
~ v=r2
~ vif and only if r1=r2. (c)Prove that any nontrivial vector space is innite. (d)Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion. 1.40 Is this a vector space under the natural operations: the real-valued functions of one real variable that are dierentiable? 1.41 Avector space over the complex numbers Chas the same denition as a vector space over the reals except that scalars are drawn from Cinstead of from R. Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply: ( a0+a1i) + (b0+b1i) = (a0+b0) + (a1+b1)iand (a0+a1i)(b0+b1i) = (a0b0a1b1) + (a0b1+a1b0)i.) (a)The set of degree two polynomials with complex coecients (b)This set f0a b0a;b2Canda+b= 0 + 0ig 1.42 Name a property shared by all of the Rn's but not listed as a requirement for a vector space. X1.43 (a) Prove that a sum of four vectors ~ v1;:::;~ v 42Vcan be associated in any way without changing the result. ((~ v1+~ v2) +~ v3) +~ v4= (~ v1+ (~ v2+~ v3)) +~ v4 = (~ v1+~ v2) + (~ v3+~ v4) =~ v1+ ((~ v2+~ v3) +~ v4) =~ v1+ (~ v2+ (~ v3+~ v4)) This allows us to simply write ` ~ v1+~ v2+~ v3+~ v4' without ambiguity. Section I. Denition of Vector Space 89 (b)Prove that any two ways of associating a sum of any number of vectors give the same sum. ( Hint. Use induction on the number of vectors.) 1.44 Example 1.5 gives a subset of R2that is not a vector space, under the obvious operations, because while it is closed under addition, it is not closed under scalar multiplication. Consider the set of vectors in the plane whose components have the same sign or are 0. Show that this set is closed under scalar multiplication but not addition. 1.45 For any vector space, a subset that is itself a vector space under the inherited operations (e.g., a plane through the origin inside of R3) is a subspace . (a)Show thatfa0+a1x+a2x2a0+a1+a2= 0gis a subspace of the vector space of degree two polynomials. (b)Show that this is a subspace of the 2 2 matrices. fa b c0a+b= 0g (c)Show that a nonempty subset Sof a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever c1;c22R and~ s1;~ s22Sthen the combination c1~ v1+c2~ v2is inS. I.2 Subspaces and Spanning Sets One of the examples that led us to introduce the idea of a vector space was the solution set of a homogeneous system. For instance, we've seen in Example 1.4 such a space that is a planar subset of R3. There, the vector space R3contains inside it another vector space, the plane. 2.1 Denition For any vector space, a subspace is a subset that is itself a vector space, under the inherited operations. 2.2 Example The plane from the prior subsection, P=f0 @x y z1 Ax+y+z= 0g is a subspace of R3. As specied in the denition, the operations are the ones that are inherited from the larger space, that is, vectors add in Pas they add inR30 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @x1+x2 y1+y2 z1+z21 A and scalar multiplication is also the same as it is in R3. To show that Pis a subspace, we need only note that it is a subset and then verify that it is a space. Checking that Psatises the conditions in the denition of a vector space is routine. For instance, for closure under addition, just note that if the summands satisfy that x1+y1+z1= 0 andx2+y2+z2= 0 then the sum satises that (x1+x2) + (y1+y2) + (z1+z2) = (x1+y1+z1) + (x2+y2+z2) = 0. 90 Chapter Two. Vector Spaces 2.3 Example Thex-axis in R2is a subspace where the addition and scalar multiplication operations are the inherited ones. x1 0 +x2 0 =x1+x2 0 r
x 0 =rx 0 As above, to verify that this is a subspace, we simply note that it is a subset and then check that it satises the conditions in denition of a vector space. For instance, the two closure conditions are satised: (1) adding two vectors with a second component of zero results in a vector with a second component of zero, and (2) multiplying a scalar times a vector with a second component of zero results in a vector with a second component of zero. 2.4 Example Another subspace of R2is f0 0 g its trivial subspace. Any vector space has a trivial subspace f~0g. At the opposite extreme, any vector space has itself for a subspace. These two are the improper subspaces. Other subspaces are proper . 2.5 Example The condition in the denition requiring that the addition and scalar multiplication operations must be the ones inherited from the larger space is important. Consider the subset f1gof the vector space R1. Under the opera- tions 1+1 = 1 and r
1 = 1 that set is a vector space, specically, a trivial space. But it is not a subspace of R1because those aren't the inherited operations, since of course R1has 1 + 1 = 2. 2.6 Example All kinds of vector spaces, not just Rn's, have subspaces. The vector space of cubic polynomials fa+bx+cx2+dx3a;b;c;d2Rghas a sub- space comprised of all linear polynomials fm+nxm;n2Rg. 2.7 Example Another example of a subspace not taken from an Rnis one from the examples following the denition of a vector space. The space of all real-valued functions of one real variable f:R!Rhas a subspace of functions satisfying the restriction ( d2f=dx2) +f= 0. 2.8 Example Being vector spaces themselves, subspaces must satisfy the clo- sure conditions. The set R+is not a subspace of the vector space R1because with the inherited operations it is not closed under scalar multiplication: if ~ v= 1 then1
~ v62R+. The next result says that Example 2.8 is prototypical. The only way that a subset can fail to be a subspace (if it is nonempty and the inherited operations are used) is if it isn't closed. Section I. Denition of Vector Space 91 2.9 Lemma For a nonempty subset Sof a vector space, under the inherited operations, the following are equivalent statements. (1)Sis a subspace of that vector space (2)Sis closed under linear combinations of pairs of vectors: for any vectors ~ s1;~ s22Sand scalars r1;r2the vectorr1~ s1+r2~ s2is inS (3)Sis closed under linear combinations of any number of vectors: for any vectors~ s1;:::;~ sn2Sand scalars r1;:::;rnthe vectorr1~ s1+


+rn~ snis inS. Brie y, the way that a subset gets to be a subspace is by being closed under linear combinations. Proof .`The following are equivalent' means that each pair of statements are equivalent. (1)() (2) (2)() (3) (3)() (1) We will show this equivalence by establishing that (1) = )(3) =)(2) =) (1). This strategy is suggested by noticing that (1) = )(3) and (3) =)(2) are easy and so we need only argue the single implication (2) = )(1). For that argument, assume that Sis a nonempty subset of a vector space V and thatSis closed under combinations of pairs of vectors. We will show that Sis a vector space by checking the conditions. The rst item in the vector space denition has ve conditions. First, for closure under addition, if ~ s1;~ s22Sthen~ s1+~ s22S, as~ s1+~ s2= 1
~ s1+ 1
~ s2. Second, for any ~ s1;~ s22S, because addition is inherited from V, the sum~ s1+~ s2 inSequals the sum ~ s1+~ s2inV, and that equals the sum ~ s2+~ s1inV(because Vis a vector space, its addition is commutative), and that in turn equals the sum~ s2+~ s1inS. The argument for the third condition is similar to that for the second. For the fourth, consider the zero vector of Vand note that closure of S under linear combinations of pairs of vectors gives that (where ~ sis any member of the nonempty set S) 0
~ s+ 0
~ s=~0 is inS; showing that ~0 acts under the inherited operations as the additive identity of Sis easy. The fth condition is satised because for any ~ s2S, closure under linear combinations shows that the vector 0
~0 + (1)
~ sis inS; showing that it is the additive inverse of ~ s under the inherited operations is routine. The checks for item (2) are similar and are saved for Exercise 32. QED We usually show that a subset is a subspace with (2) = )(1). 2.10 Remark At the start of this chapter we introduced vector spaces as collections in which linear combinations are \sensible". The above result speaks to this. The vector space denition has ten conditions but eight of them | the con- ditions not about closure | simply ensure that referring to the operations as an `addition' and a `scalar multiplication' is sensible. The proof above checks that these eight are inherited from the surrounding vector space provided that the More information on equivalence of statements is in the appendix. 92 Chapter Two. Vector Spaces nonempty set Ssatises Theorem 2.9's statement (2) (e.g., commutativity of addition in Sfollows right from commutativity of addition in V). So, in this context, this meaning of \sensible" is automatically satised. In assuring us that this rst meaning of the word is met, the result draws our attention to the second meaning of \sensible". It has to do with the two remaining conditions, the closure conditions. Above, the two separate closure conditions inherent in statement (1) are combined in statement (2) into the single condition of closure under all linear combinations of two vectors, which is then extended in statement (3) to closure under combinations of any number of vectors. The latter two statements say that we can always make sense of an expression like r1~ s1+r2~ s2, without restrictions on the r's | such expressions are \sensible" in that the vector described is dened and is in the set S. This second meaning suggests that a good way to think of a vector space is as a collection of unrestricted linear combinations. The next two examples take some spaces and describe them in this way. That is, in these examples we parametrize, just as we did in Chapter One to describe the solution set of a homogeneous linear system. 2.11 Example This subset of R3 S=f0 @x y z1 Ax2y+z= 0g is a subspace under the usual addition and scalar multiplication operations of column vectors (the check that it is nonempty and closed under linear combi- nations of two vectors is just like the one in Example 2.2). To parametrize, we can takex2y+z= 0 to be a one-equation linear system and expressing the leading variable in terms of the free variables x= 2yz. S=f0 @2yz y z1 Ay;z2Rg=fy0 @2 1 01 A+z0 @1 0 11 Ay;z2Rg Now the subspace is described as the collection of unrestricted linear combi- nations of those two vectors. Of course, in either description, this is a plane through the origin. 2.12 Example This is a subspace of the 2 2 matrices L=fa0 b ca+b+c= 0g (checking that it is nonempty and closed under linear combinations is easy). To parametrize, express the condition as a=bc. L=fbc0 b cb;c2Rg=fb1 0 1 0 +c1 0 0 1b;c2Rg As above, we've described the subspace as a collection of unrestricted linear combinations (by coincidence, also of two elements). Section I. Denition of Vector Space 93 Parametrization is an easy technique, but it is important. We shall use it often. 2.13 Denition The span (orlinear closure ) of a nonempty subset Sof a vector space is the set of all linear combinations of vectors from S. [S] =fc1~ s1+


+cn~ snc1;:::;cn2Rand~ s1;:::;~ sn2Sg The span of the empty subset of a vector space is the trivial subspace. No notation for the span is completely standard. The square brackets used here are common, but so are `span( S)' and `sp(S)'. 2.14 Remark In Chapter One, after we showed that the solution set of a ho- mogeneous linear system can be written as fc1~1+


+ck~kc1;:::;ck2Rg, we described that as the set `generated' by the ~'s. We now have the technical term; we call that the `span' of the set f~1;:::;~kg. Recall also the discussion of the \tricky point" in that proof. The span of the empty set is dened to be the set f~0gbecause we follow the convention that a linear combination of no vectors sums to ~0. Besides, dening the empty set's span to be the trivial subspace is a convienence in that it keeps results like the next one from having annoying exceptional cases. 2.15 Lemma In a vector space, the span of any subset is a subspace. Proof .Call the subset S. IfSis empty then by denition its span is the trivial subspace. If Sis not empty then by Lemma 2.9 we need only check that the span [S] is closed under linear combinations. For a pair of vectors from that span,~ v=c1~ s1+


+cn~ snand~ w=cn+1~ sn+1+


+cm~ sm, a linear combination p
(c1~ s1+


+cn~ sn) +r
(cn+1~ sn+1+


+cm~ sm) =pc1~ s1+


+pcn~ sn+rcn+1~ sn+1+


+rcm~ sm (p,rscalars) is a linear combination of elements of Sand so is in [ S] (possibly some of the ~ si's from~ vequal some of the ~ sj's from~ w, but it does not matter). QED The converse of the lemma holds: any subspace is the span of some set, because a subspace is obviously the span of the set of its members. Thus a subset of a vector space is a subspace if and only if it is a span. This ts the intuition that a good way to think of a vector space is as a collection in which linear combinations are sensible. Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset Sof a vector space is the smallest subspace containing all the members of S. 2.16 Example In any vector space V, for any vector ~ v, the setfr
~ vr2Rg is a subspace of V. For instance, for any vector ~ v2R3, the line through the origin containing that vector, fk~ vk2Rgis a subspace of R3. This is true even when~ vis the zero vector, in which case the subspace is the degenerate line, the trivial subspace. 94 Chapter Two. Vector Spaces 2.17 Example The span of this set is all of R2. f1 1 ;1 1 g To check this we must show that any member of R2is a linear combination of these two vectors. So we ask: for which vectors (with real components xandy) are there scalars c1andc2such that this holds? c11 1 +c21 1 =x y Gauss' method c1+c2=x c1c2=y1+2!c1+c2=x 2c2=x+y with back substitution gives c2= (xy)=2 andc1= (x+y)=2. These two equations show that for any xandythat we start with, there are appropriate coecients c1andc2making the above vector equation true. For instance, for x= 1 andy= 2 the coecients c2=1=2 andc1= 3=2 will do. That is, any vector in R2can be written as a linear combination of the two given vectors. Since spans are subspaces, and we know that a good way to understand a subspace is to parametrize its description, we can try to understand a set's span in that way. 2.18 Example Consider, inP2, the span of the set f3xx2;2xg. By the denition of span, it is the set of unrestricted linear combinations of the two fc1(3xx2) +c2(2x)c1;c22Rg. Clearly polynomials in this span must have a constant term of zero. Is that necessary condition also sucient? We are asking: for which members a2x2+a1x+a0ofP2are therec1andc2 such thata2x2+a1x+a0=c1(3xx2) +c2(2x)? Since polynomials are equal if and only if their coecients are equal, we are looking for conditions on a2, a1, anda0satisfying these. c1 =a2 3c1+ 2c2=a1 0 =a0 Gauss' method gives that c1=a2,c2= (3=2)a2+ (1=2)a1, and 0 =a0. Thus the only condition on polynomials in the span is the condition that we knew of | as long as a0= 0, we can give appropriate coecients c1andc2to describe the polynomial a0+a1x+a2x2as in the span. For instance, for the polynomial 04x+ 3x2, the coecients c1=3 andc2= 5=2 will do. So the span of the given set isfa1x+a2x2a1;a22Rg. This shows, incidentally, that the set fx;x2galso spans this subspace. A space can have more than one spanning set. Two other sets spanning this sub- space arefx;x2;x+ 2x2gandfx;x+x2;x+ 2x2;:::g. (Naturally, we usually prefer to work with spanning sets that have only a few members.) Section I. Denition of Vector Space 95 2.19 Example These are the subspaces of R3that we now know of, the trivial subspace, the lines through the origin, the planes through the origin, and the whole space (of course, the picture shows only a few of the innitely many subspaces). In the next section we will prove that R3has no other type of subspaces, so in fact this picture shows them all. fx0 @1 0 01 A+y0 @0 1 01 A+z0 @0 0 11 Ag  fx0 @1 0 01 A+y0 @0 1 01 Ag fx0 @1 0 01 A+z0 @0 0 11 Ag fx0 @1 1 01 A+z0 @0 0 11 Ag . . .  fx0 @1 0 01 AgAA fy0 @0 1 01 AgHHHH fy0 @2 1 01 Ag fy0 @1 1 11 Ag . . . XXXXXXXXXXXXPPPPPPPPHHHHH@@ f0 @0 0 01 Ag The subsets are described as spans of sets, using a minimal number of members, and are shown connected to their supersets. Note that these subspaces fall naturally into levels | planes on one level, lines on another, etc. | according to how many vectors are in a minimal-sized spanning set. So far in this chapter we have seen that to study the properties of linear combinations, the right setting is a collection that is closed under these com- binations. In the rst subsection we introduced such collections, vector spaces, and we saw a great variety of examples. In this subsection we saw still more spaces, ones that happen to be subspaces of others. In all of the variety we've seen a commonality. Example 2.19 above brings it out: vector spaces and sub- spaces are best understood as a span, and especially as a span of a small number of vectors. The next section studies spanning sets that are minimal. Exercises X2.20 Which of these subsets of the vector space of 2 2 matrices are subspaces under the inherited operations? For each one that is a subspace, parametrize its description. For each that is not, give a condition that fails. (a)fa0 0ba;b2Rg (b)fa0 0ba+b= 0g (c)fa0 0ba+b= 5g (d)fa c 0ba+b= 0;c2Rg X2.21 Is this a subspace of P2:fa0+a1x+a2x2a0+ 2a1+a2= 4g? If it is then parametrize its description. 96 Chapter Two. Vector Spaces X2.22 Decide if the vector lies in the span of the set, inside of the space. (a)0 @2 0 11 A,f0 @1 0 01 A;0 @0 0 11 Ag, inR3 (b)xx3,fx2;2x+x2;x+x3g, inP3 (c)0 1 4 2 ,f1 0 1 1 ;2 0 2 3 g, inM22 2.23 Which of these are members of the span [ fcos2x;sin2xg] in the vector space of real-valued functions of one real variable? (a)f(x) = 1 (b)f(x) = 3 +x2(c)f(x) = sinx(d)f(x) = cos(2x) X2.24 Which of these sets spans R3? That is, which of these sets has the property that any three-tall vector can be expressed as a suitable linear combination of the set's elements? (a)f0 @1 0 01 A;0 @0 2 01 A;0 @0 0 31 Ag(b)f0 @2 0 11 A;0 @1 1 01 A;0 @0 0 11 Ag(c)f0 @1 1 01 A;0 @3 0 01 Ag (d)f0 @1 0 11 A;0 @3 1 01 A;0 @1 0 01 A;0 @2 1 51 Ag(e)f0 @2 1 11 A;0 @3 0 11 A;0 @5 1 21 A;0 @6 0 21 Ag X2.25 Parametrize each subspace's description. Then express each subspace as a span. (a)The subsetf a b c
ac= 0gof the three-wide row vectors (b)This subset ofM22 fa b c da+d= 0g (c)This subset ofM22 fa b c d2acd= 0 anda+ 3b= 0g (d)The subsetfa+bx+cx3a2b+c= 0gofP3 (e)The subset ofP2of quadratic polynomials psuch thatp(7) = 0 X2.26 Find a set to span the given subspace of the given space. ( Hint. Parametrize each.) (a)thexz-plane in R3 (b)f0 @x y z1 A3x+ 2y+z= 0ginR3 (c)f0 BB@x y z w1 CCA2x+y+w= 0 andy+ 2z= 0ginR4 (d)fa0+a1x+a2x2+a3x3a0+a1= 0 anda2a3= 0ginP3 (e)The setP4in the spaceP4 (f)M22inM22 2.27 IsR2a subspace of R3? X2.28 Decide if each is a subspace of the vector space of real-valued functions of one real variable. (a)The even functionsff:R!Rf(x) =f(x) for allxg. For example, two members of this set are f1(x) =x2andf2(x) = cos(x). Section I. Denition of Vector Space 97 (b)The oddfunctionsff:R!Rf(x) =f(x) for allxg. Two members are f3(x) =x3andf4(x) = sin(x). 2.29 Example 2.16 says that for any vector ~ vthat is an element of a vector space V, the setfr
~ vr2Rgis a subspace of V. (This is of course, simply the span of the singleton setf~ vg.) Must any such subspace be a proper subspace, or can it be improper? 2.30 An example following the denition of a vector space shows that the solution set of a homogeneous linear system is a vector space. In the terminology of this subsection, it is a subspace of Rnwhere the system has nvariables. What about a non-homogeneous linear system; do its solutions form a subspace (under the inherited operations)? 2.31 Example 2.19 shows that R3has innitely many subspaces. Does every non- trivial space have innitely many subspaces? 2.32 Finish the proof of Lemma 2.9. 2.33 Show that each vector space has only one trivial subspace. X2.34 Show that for any subset Sof a vector space, the span of the span equals the span [[S]] = [S]. (Hint. Members of [ S] are linear combinations of members of S. Members of [[ S]] are linear combinations of linear combinations of members of S.) 2.35 All of the subspaces that we've seen use zero in their description in some way. For example, the subspace in Example 2.3 consists of all the vectors from R2 with a second component of zero. In contrast, the collection of vectors from R2 with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition were that the three components add to one then it would not be a subspace (again, it would fail to be closed). This exercise shows that a reliance on zero is not strictly necessary. Consider the set f0 @x y z1 Ax+y+z= 1g under these operations.0 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @x1+x21 y1+y2 z1+z21 Ar0 @x y z1 A=0 @rxr+ 1 ry rz1 A (a)Show that it is not a subspace of R3. (Hint. See Example 2.5). (b)Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply. (c)Show that any subspace of R3must pass through the origin, and so any subspace of R3must involve zero in its description. Does the converse hold? Does any subset of R3that contains the origin become a subspace when given the inherited operations? 2.36 We can give a justication for the convention that the sum of zero-many vectors equals the zero vector. Consider this sum of three vectors ~ v1+~ v2+ ~ v3. (a)What is the dierence between this sum of three vectors and the sum of the rst two of these three? (b)What is the dierence between the prior sum and the sum of just the rst one vector? 98 Chapter Two. Vector Spaces (c)What should be the dierence between the prior sum of one vector and the sum of no vectors? (d)So what should be the denition of the sum of no vectors? 2.37 Is a space determined by its subspaces? That is, if two vector spaces have the same subspaces, must the two be equal? 2.38 (a) Give a set that is closed under scalar multiplication but not addition. (b)Give a set closed under addition but not scalar multiplication. (c)Give a set closed under neither. 2.39 Show that the span of a set of vectors does not depend on the order in which the vectors are listed in that set. 2.40 Which trivial subspace is the span of the empty set? Is it f0 @0 0 01 AgR3;orf0 + 0xgP 1; or some other subspace? 2.41 Show that if a vector is in the span of a set then adding that vector to the set won't make the span any bigger. Is that also `only if'? X2.42 Subspaces are subsets and so we naturally consider how `is a subspace of' interacts with the usual set operations. (a)IfA;B are subspaces of a vector space, must their interesction A\Bbe a subspace? Always? Sometimes? Never? (b)Must the union A[Bbe a subspace? (c)IfAis a subspace, must its complement be a subspace? (Hint. Try some test subspaces from Example 2.19.) X2.43 Does the span of a set depend on the enclosing space? That is, if Wis a subspace of VandSis a subset of W(and so also a subset of V), might the span ofSinWdier from the span of SinV? 2.44 Is the relation `is a subspace of' transitive? That is, if Vis a subspace of W andWis a subspace of X, mustVbe a subspace of X? X2.45 Because `span of' is an operation on sets we naturally consider how it interacts with the usual set operations. (a)IfSTare subsets of a vector space, is [ S][T]? Always? Sometimes? Never? (b)IfS;T are subsets of a vector space, is [ S[T] = [S][[T]? (c)IfS;T are subsets of a vector space, is [ S\T] = [S]\[T]? (d)Is the span of the complement equal to the complement of the span? 2.46 Reprove Lemma 2.15 without doing the empty set separately. 2.47 Find a structure that is closed under linear combinations, and yet is not a vector space. ( Remark. This is a bit of a trick question.) Section II. Linear Independence 99 II Linear Independence The prior section shows that a vector space can be understood as an unrestricted linear combination of some of its elements | that is, as a span. For example, the space of linear polynomials fa+bxa;b2Rgis spanned by the set f1;xg. The prior section also showed that a space can have many sets that span it. The space of linear polynomials is also spanned by f1;2xgandf1;x;2xg. At the end of that section we described some spanning sets as `minimal', but we never precisely dened that word. We could take `minimal' to mean one of two things. We could mean that a spanning set is minimal if it contains the smallest number of members of any set with the same span. With this meaning f1;x;2xgis not minimal because it has one member more than the other two. Or we could mean that a spanning set is minimal when it has no elements that can be removed without changing the span. Under this meaning f1;x;2xgis not minimal because removing the 2 xand gettingf1;xgleaves the span unchanged. The rst sense of minimality appears to be a global requirement, in that to check if a spanning set is minimal we seemingly must look at all the spanning sets of a subspace and nd one with the least number of elements. The second sense of minimality is local in that we need to look only at the set under discussion and consider the span with and without various elements. For instance, using the second sense, we could compare the span of f1;x;2xgwith the span off1;xg and note that the 2 xis a \repeat" in that its removal doesn't shrink the span. In this section we will use the second sense of `minimal spanning set' because of this technical convenience. However, the most important result of this book is that the two senses coincide; we will prove that in the section after this one. II.1 Denition and Examples We rst characterize when a vector can be removed from a set without changing the span of that set. For that, note that if a vector ~ vis not a member of a set S then the union S[f~ vgand the set Sdier only in that the former contains ~ v. 1.1 Lemma WhereSis a subset of a vector space V, [S] = [S[f~ vg] if and only if ~ v2[S] for any~ v2V. Proof .The left to right implication is easy. If [ S] = [S[f~ vg] then, since ~ v2[S[f~ vg], the equality of the two sets gives that ~ v2[S]. For the right to left implication assume that ~ v2[S] to show that [ S] = [S[ f~ vg] by mutual inclusion. The inclusion [ S][S[f~ vg] is obvious. For the other inclusion [S][S[f~ vg], write an element of [ S[f~ vg] asd0~ v+d1~ s1+


+dm~ sm and substitute ~ v's expansion as a linear combination of members of the same set d0(c0~t0+


+ck~tk) +d1~ s1+


+dm~ sm. This is a linear combination of linear 100 Chapter Two. Vector Spaces combinations and so distributing d0results in a linear combination of vectors fromS. Hence each member of [ S[f~ vg] is also a member of [ S]. QED 1.2 Example InR3, where ~ v1=0 @1 0 01 A~ v2=0 @0 1 01 A~ v3=0 @2 1 01 A the spans [f~ v1;~ v2g] and [f~ v1;~ v2;~ v3g] are equal since ~ v3is in the span [f~ v1;~ v2g]. The lemma says that if we have a spanning set then we can remove a ~ vto get a new set Swith the same span if and only if ~ vis a linear combination of vectors from S. Thus, under the second sense described above, a spanning set is minimal if and only if it contains no vectors that are linear combinations of the others in that set. We have a term for this important property. 1.3 Denition A subset of a vector space is linearly independent if none of its elements is a linear combination of the others. Otherwise it is linearly dependent . Here is an important observation: although this way of writing one vector as a combination of the others ~ s0=c1~ s1+c2~ s2+


+cn~ sn visually sets ~ s0o from the other vectors, algebraically there is nothing special in that equation about ~ s0. For any~ siwith a coecient cithat is nonzero, we can rewrite the relationship to set o ~ si. ~ si= (1=ci)~ s0+ (c1=ci)~ s1+


+ (cn=ci)~ sn When we don't want to single out any vector by writing it alone on one side of the equation we will instead say that ~ s0;~ s1;:::;~ snare in a linear relationship and write the relationship with all of the vectors on the same side. The next result rephrases the linear independence denition in this style. It gives what is usually the easiest way to compute whether a nite set is dependent or independent. 1.4 Lemma A subsetSof a vector space is linearly independent if and only if for any distinct ~ s1;:::;~ sn2Sthe only linear relationship among those vectors c1~ s1+


+cn~ sn=~0c1;:::;cn2R is the trivial one: c1= 0;:::; cn= 0. Proof .This is a direct consequence of the observation above. If the setSis linearly independent then no vector ~ sican be written as a linear combination of the other vectors from Sso there is no linear relationship where some of the ~ s's have nonzero coecients. If Sis not linearly independent then some~ siis a linear combination ~ si=c1~ s1+


+ci1~ si1+ci+1~ si+1+


+cn~ snof other vectors from S, and subtracting ~ sifrom both sides of that equation gives a linear relationship involving a nonzero coecient, namely the 1 in front of ~ si. QED Section II. Linear Independence 101 1.5 Example In the vector space of two-wide row vectors, the two-element set f40 15
;50 25
gis linearly independent. To check this, set c1
40 15
+c2
50 25
=0 0
and solving the resulting system 40c150c2= 0 15c1+ 25c2= 0(15=40)1+2!40c1 50c2= 0 (175=4)c2= 0 shows that both c1andc2are zero. So the only linear relationship between the two given row vectors is the trivial relationship. In the same vector space, f40 15
;20 7:5
gis linearly dependent since we can satisfy c140 15
+c2
20 7:5
=0 0
withc1= 1 andc2=2. 1.6 Remark Recall the Statics example that began this book. We rst set the unknown-mass objects at 40 cm and 15 cm and got a balance, and then we set the objects at50 cm and 25 cm and got a balance. With those two pieces of information we could compute values of the unknown masses. Had we instead rst set the unknown-mass objects at 40 cm and 15 cm, and then at 20 cm and 7:5 cm, we would not have been able to compute the values of the unknown masses (try it). Intuitively, the problem is that the20 7:5
information is a \repeat" of the40 15
information | that is,20 7:5
is in the span of the setf 40 15
g| and so we would be trying to solve a two-unknowns problem with what is essentially one piece of information. 1.7 Example The setf1 +x;1xgis linearly independent in P2, the space of quadratic polynomials with real coecients, because 0 + 0x+ 0x2=c1(1 +x) +c2(1x) = (c1+c2) + (c1c2)x+ 0x2 gives c1+c2= 0 c1c2= 01+2!c1+c2= 0 2c2= 0 since polynomials are equal only if their coecients are equal. Thus, the only linear relationship between these two members of P2is the trivial one. 1.8 Example InR3, where ~ v1=0 @3 4 51 A~ v2=0 @2 9 21 A~ v3=0 @4 18 41 A the setS=f~ v1;~ v2;~ v3gis linearly dependent because this is a relationship 0
~ v1+ 2
~ v21
~ v3=~0 where not all of the scalars are zero (the fact that some of the scalars are zero doesn't matter). 102 Chapter Two. Vector Spaces 1.9 Remark That example illustrates why, although Denition 1.3 is a clearer statement of what independence is, Lemma 1.4 is more useful for computations. Working straight from the denition, someone trying to compute whether Sis linearly independent would start by setting ~ v1=c2~ v2+c3~ v3and concluding that there are no such c2andc3. But knowing that the rst vector is not dependent on the other two is not enough. This person would have to go on to try~ v2=c1~ v1+c3~ v3to nd the dependence c1= 0,c3= 1=2. Lemma 1.4 gets the same conclusion with only one computation. 1.10 Example The empty subset of a vector space is linearly independent. There is no nontrivial linear relationship among its members as it has no mem- bers. 1.11 Example In any vector space, any subset containing the zero vector is linearly dependent. For example, in the space P2of quadratic polynomials, consider the subset f1 +x;x+x2;0g. One way to see that this subset is linearly dependent is to use Lemma 1.4: we have 0
~ v1+0
~ v2+1
~0 =~0, and this is a nontrivial relationship as not all of the coecients are zero. Another way to see that this subset is linearly dependent is to go straight to Denition 1.3: we can express the third member of the subset as a linear combination of the rst two, namely, c1~ v1+c2~ v2=~0 is satised by takingc1= 0 andc2= 0 (in contrast to the lemma, the denition allows all of the coecients to be zero). (There is subtler way to see that this subset is dependent. The zero vector is equal to the trivial sum, the sum of the empty set. So a set containing the zero vector has an element that can be written as a combination of a set of other vectors from the set, specically, the zero vector can be written as a combination of the empty set.) The above examples, especially Example 1.5, underline the discussion that begins this section. The next result says that given a nite set, we can produce a linearly independent subset by discarding what Remark 1.6 calls \repeats". 1.12 Theorem In a vector space, any nite subset has a linearly independent subset with the same span. Proof .If the setS=f~ s1;:::;~ sngis linearly independent then Sitself satises the statement, so assume that it is linearly dependent. By the denition of dependence, there is a vector ~ sithat is a linear combina- tion of the others. Call that vector ~ v1. Discard it | dene the set S1=Sf~ v1g. By Lemma 1.1, the span does not shrink [ S1] = [S]. Now, ifS1is linearly independent then we are nished. Otherwise iterate the prior paragraph: take a vector ~ v2that is a linear combination of other members ofS1and discard it to derive S2=S1f~ v2gsuch that [S2] = [S1]. Repeat this until a linearly independent set Sjappears; one must appear eventually because Sis nite and the empty set is linearly independent. (Formally, this argument uses induction on n, the number of elements in the starting set. Exercise 37 asks for the details.) QED Section II. Linear Independence 103 1.13 Example This set spans R3(the check of this is easy). S=f0 @1 0 01 A;0 @0 2 01 A;0 @1 2 01 A;0 @0 1 11 A;0 @3 3 01 Ag Looking for a linear relationship c10 @1 0 01 A+c20 @0 2 01 A+c30 @1 2 01 A+c40 @0 1 11 A+c50 @3 3 01 A=0 @0 0 01 A () gives a system c1 +c3+ + 3c5= 0 2c2+ 2c3c4+ 3c5= 0 c4+ = 0 with leading variables c1,c2, andc4and free variables c3andc5. We can paramatrize the solution set in this way. f0 BBBB@c1 c2 c3 c4 c51 CCCCA=c30 BBBB@1 1 1 0 01 CCCCA+c50 BBBB@3 3=2 0 0 11 CCCCAc3;c52Rg SoSis linearly dependent. To nd something to discard, consider the vectors associated with the free variablesc3andc5. Settingc3= 0 andc5= 1 shows that that c1=3, c2=3=2,c3= 0,c4= 0, andc5= 1 is a linear dependence in equation ( ) above, that is, c5's vector is a linear combination of the rst two. Lemma 1.1 says that discarding this fth vector S1=f0 @1 0 01 A;0 @0 2 01 A;0 @1 2 01 A;0 @0 1 11 Ag leaves the span unchanged [ S1] = [S]. Similarly, setting c3= 1 andc5= 0 gives a linear dependence in equation ( ) above. Since c5= 0 this is a relationship among the rst four vectors, the members of S1. Thus we can discard c3's vector from S1to get S2=f0 @1 0 01 A;0 @0 2 01 A;0 @0 1 11 Ag with the same span as S1, and therefore the same span as S, but with one dierence. We can easily check that S2is linearly independent and so discarding any of its elements will shrink the span. 104 Chapter Two. Vector Spaces That example makes clear the general method: given a nite set of vectors, we rst write the system to nd a linear dependence. Then discarding any vectors associated with the free variables of that system will leave the span unchanged. Theorem 1.12 describes producing a linearly independent set by shrinking, that is, by taking subsets. We nish this subsection by considering how linear independence and dependence, which are properties of sets, interact with the subset relation between sets. 1.14 Lemma Any subset of a linearly independent set is also linearly inde- pendent. Any superset of a linearly dependent set is also linearly dependent. Proof .This is clear. QED Restated, independence is preserved by subset and dependence is preserved by superset. Those are two of the four possible cases of interaction that we can consider. The third case, whether linear dependence is preserved by the subset operation, is covered by Example 1.13, which gives a linearly dependent set Swith a subset S1that is linearly dependent and another subset S2that is linearly independent. That leaves one case, whether linear independence is preserved by superset. The next example shows what can happen. 1.15 Example In each of these three paragraphs the subset Sis linearly independent. For the set S=f0 @1 0 01 Ag the span [S] is thexaxis. Here are two supersets of S, one linearly dependent and the other linearly independent. dependent:f0 @1 0 01 A;0 @3 0 01 Ag independent:f0 @1 0 01 A;0 @0 1 01 Ag Checking the dependence or independence of these sets is easy. For S=f0 @1 0 01 A;0 @0 1 01 Ag the span [S] is thexyplane. These are two supersets. dependent:f0 @1 0 01 A;0 @0 1 01 A;0 @3 2 01 Ag independent:f0 @1 0 01 A;0 @0 1 01 A;0 @0 0 11 Ag Section II. Linear Independence 105 If S=f0 @1 0 01 A;0 @0 1 01 A;0 @0 0 11 Ag then [S] =R3. A linearly dependent superset is dependent:f0 @1 0 01 A;0 @0 1 01 A;0 @0 0 11 A;0 @2 1 31 Ag but there are no linearly independent supersets of S. The reason is that for any vector that we would add to make a superset, the linear dependence equation 0 @x y z1 A=c10 @1 0 01 A+c20 @0 1 01 A+c30 @0 0 11 A has a solution c1=x,c2=y, andc3=z. So, in general, a linearly independent set may have a superset that is depen- dent. And, in general, a linearly independent set may have a superset that is independent. We can characterize when the superset is one and when it is the other. 1.16 Lemma WhereSis a linearly independent subset of a vector space V, S[f~ vgis linearly dependent if and only if ~ v2[S] for any~ v2Vwith~ v62S. Proof .One implication is clear: if ~ v2[S] then~ v=c1~ s1+c2~ s2+


+cn~ sn where each ~ si2Sandci2R, and so~0 =c1~ s1+c2~ s2+


+cn~ sn+ (1)~ vis a nontrivial linear relationship among elements of S[f~ vg. The other implication requires the assumption that Sis linearly independent. WithS[f~ vglinearly dependent, there is a nontrivial linear relationship c0~ v+ c1~ s1+c2~ s2+


+cn~ sn=~0 and independence of Sthen implies that c06= 0, or else that would be a nontrivial relationship among members of S. Now rewriting this equation as ~ v=(c1=c0)~ s1


(cn=c0)~ snshows that ~ v2[S]. QED (Compare this result with Lemma 1.1. Both say, roughly, that ~ vis a \repeat" if it is in the span of S. However, note the additional hypothesis here of linear independence.) 1.17 Corollary A subsetS=f~ s1;:::;~ sngof a vector space is linearly depen- dent if and only if some ~ siis a linear combination of the vectors ~ s1, . . . ,~ si1 listed before it. Proof .ConsiderS0=fg,S1=f~ s1g,S2=f~ s1;~ s2g, etc. Some index i1 is the rst one with Si1[f~ siglinearly dependent, and there ~ si2[Si1].QED 106 Chapter Two. Vector Spaces Lemma 1.16 can be restated in terms of independence instead of dependence: ifSis linearly independent and ~ v62Sthen the set S[f~ vgis also linearly independent if and only if ~ v62[S]:Applying Lemma 1.1, we conclude that if S is linearly independent and ~ v62SthenS[f~ vgis also linearly independent if and only if [ S[f~ vg]6= [S]. Brie y, when passing from Sto a superset S1, to preserve linear independence we must expand the span [ S1]]. Example 1.15 shows that some linearly independent sets are maximal | have as many elements as possible | in that they have no supersets that are linearly independent. By the prior paragraph, a linearly independent sets is maximal if and only if it spans the entire space, because then no vector exists that is not already in the span. This table summarizes the interaction between the properties of indepen- dence and dependence and the relations of subset and superset. S1S S 1S Sindependent S1must be independent S1may be either Sdependent S1may be either S1must be dependent In developing this table we've uncovered an intimate relationship between linear independence and span. Complementing the fact that a spanning set is minimal if and only if it is linearly independent, a linearly independent set is maximal if and only if it spans the space. In summary, we have introduced the denition of linear independence to formalize the idea of the minimality of a spanning set. We have developed some properties of this idea. The most important is Lemma 1.16, which tells us that a linearly independent set is maximal when it spans the space. Exercises X1.18 Decide whether each subset of R3is linearly dependent or linearly indepen- dent. (a)f0 @1 3 51 A;0 @2 2 41 A;0 @4 4 141 Ag (b)f0 @1 7 71 A;0 @2 7 71 A;0 @3 7 71 Ag (c)f0 @0 0 11 A;0 @1 0 41 Ag (d)f0 @9 9 01 A;0 @2 0 11 A;0 @3 5 41 A;0 @12 12 11 Ag X1.19 Which of these subsets of P3are linearly dependent and which are indepen- dent? (a)f3x+ 9x2;56x+ 3x2;1 + 1x5x2g (b)fx2;1 + 4x2g (c)f2 +x+ 7x2;3x+ 2x2;43x2g Section II. Linear Independence 107 (d)f8 + 3x+ 3x2;x+ 2x2;2 + 2x+ 2x2;82x+ 5x2g X1.20 Prove that each set ff;ggis linearly independent in the vector space of all functions from R+toR. (a)f(x) =xandg(x) = 1=x (b)f(x) = cos(x) andg(x) = sin(x) (c)f(x) =exandg(x) = ln(x) X1.21 Which of these subsets of the space of real-valued functions of one real vari- able is linearly dependent and which is linearly independent? (Note that we have abbreviated some constant functions; e.g., in the rst item, the `2' stands for the constant function f(x) = 2.) (a)f2;4 sin2(x);cos2(x)g(b)f1;sin(x);sin(2x)g(c)fx;cos(x)g (d)f(1 +x)2;x2+ 2x;3g(e)fcos(2x);sin2(x);cos2(x)g(f)f0;x;x2g 1.22 Does the equation sin2(x)=cos2(x) = tan2(x) show that this set of functions fsin2(x);cos2(x);tan2(x)gis a linearly dependent subset of the set of all real-valued functions with domain the interval ( =2::=2) of real numbers between =2 and =2)? 1.23 Why does Lemma 1.4 say \distinct"? X1.24 Show that the nonzero rows of an echelon form matrix form a linearly inde- pendent set. X1.25 (a) Show that if the set f~ u;~ v;~ wgis linearly independent set then so is the setf~ u;~ u+~ v;~ u+~ v+~ wg. (b)What is the relationship between the linear independence or dependence of the setf~ u;~ v;~ wgand the independence or dependence of f~ u~ v;~ v~ w;~ w~ ug? 1.26 Example 1.10 shows that the empty set is linearly independent. (a)When is a one-element set linearly independent? (b)How about a set with two elements? 1.27 In any vector space V, the empty set is linearly independent. What about all ofV? 1.28 Show that iff~ x;~ y;~ zgis linearly independent then so are all of its proper subsets:f~ x;~ yg,f~ x;~ zg,f~ y;~ zg,f~ xg,f~ yg,f~ zg, andfg. Is that `only if' also? 1.29 (a) Show that this S=f0 @1 1 01 A;0 @1 2 01 Ag is a linearly independent subset of R3. (b)Show that0 @3 2 01 A is in the span of Sby ndingc1andc2giving a linear relationship. c10 @1 1 01 A+c20 @1 2 01 A=0 @3 2 01 A Show that the pair c1;c2is unique. (c)Assume that Sis a subset of a vector space and that ~ vis in [S], so that~ vis a linear combination of vectors from S. Prove that if Sis linearly independent then a linear combination of vectors from Sadding to~ vis unique (that is, unique up to reordering and adding or taking away terms of the form 0
~ s). ThusS 108 Chapter Two. Vector Spaces as a spanning set is minimal in this strong sense: each vector in [ S] is \hit" a minimum number of times | only once. (d)Prove that it can happen when Sis not linearly independent that distinct linear combinations sum to the same vector. 1.30 Prove that a polynomial gives rise to the zero function if and only if it is the zero polynomial. ( Comment. This question is not a Linear Algebra matter, but we often use the result. A polynomial gives rise to a function in the obvious way:x7!cnxn+


+c1x+c0.) 1.31 Return to Section 1.2 and redene point, line, plane, and other linear surfaces to avoid degenerate cases. 1.32 (a) Show that any set of four vectors in R2is linearly dependent. (b)Is this true for any set of ve? Any set of three? (c)What is the most number of elements that a linearly independent subset of R2can have? X1.33 Is there a set of four vectors in R3, any three of which form a linearly inde- pendent set? 1.34 Must every linearly dependent set have a subset that is dependent and a subset that is independent? 1.35 InR4, what is the biggest linearly independent set you can nd? The smallest? The biggest linearly dependent set? The smallest? (`Biggest' and `smallest' mean that there are no supersets or subsets with the same property.) X1.36 Linear independence and linear dependence are properties of sets. We can thus naturally ask how those properties act with respect to the familiar elementary set relations and operations. In this body of this subsection we have covered the subset and superset relations. We can also consider the operations of intersection, complementation, and union. (a)How does linear independence relate to intersection: can an intersection of linearly independent sets be independent? Must it be? (b)How does linear independence relate to complementation? (c)Show that the union of two linearly independent sets need not be linearly independent. (d)Characterize when the union of two linearly independent sets is linearly in- dependent, in terms of the intersection of the span of each. X1.37 For Theorem 1.12, (a)ll in the induction for the proof; (b)give an alternate proof that starts with the empty set and builds a sequence of linearly independent subsets of the given nite set until one appears with the same span as the given set. 1.38 With a little calculation we can get formulas to determine whether or not a set of vectors is linearly independent. (a)Show that this subset of R2 fa c ;b d g is linearly independent if and only if adbc6= 0. (b)Show that this subset of R3 f0 @a d g1 A;0 @b e h1 A;0 @c f i1 Ag is linearly independent i aei+bfg+cdhhfaidbgec6= 0. Section II. Linear Independence 109 (c)When is this subset of R3 f0 @a d g1 A;0 @b e h1 Ag linearly independent? (d)This is an opinion question: for a set of four vectors from R4, must there be a formula involving the sixteen entries that determines independence of the set? (You needn't produce such a formula, just decide if one exists.) X1.39 (a) Prove that a set of two perpendicular nonzero vectors from Rnis linearly independent when n>1. (b)What ifn= 1?n= 0? (c)Generalize to more than two vectors. 1.40 Consider the set of functions from the open interval ( 1::1) toR. (a)Show that this set is a vector space under the usual operations. (b)Recall the formula for the sum of an innite geometric series: 1+ x+x2+


= 1=(1x) for allx2(1::1). Why does this not express a dependence inside of the setfg(x) = 1=(1x);f0(x) = 1;f1(x) =x;f2(x) =x2;:::g(in the vector space that we are considering)? ( Hint. Review the denition of linear combination.) (c)Show that the set in the prior item is linearly independent. This shows that some vector spaces exist with linearly independent subsets that are innite. 1.41 Show that, where Sis a subspace of V, if a subset TofSis linearly indepen- dent inSthenTis also linearly independent in V. Is that `only if'? 110 Chapter Two. Vector Spaces III Basis and Dimension The prior section ends with the statement that a spanning set is minimal when it is linearly independent and a linearly independent set is maximal when it spans the space. So the notions of minimal spanning set and maximal independent set coincide. In this section we will name this idea and study its properties. III.1 Basis 1.1 Denition Abasis for a vector space is a sequence of vectors that form a set that is linearly independent and that spans the space. We denote a basis with angle brackets h~1;~2;:::ito signify that this collec- tion is a sequence| the order of the elements is signicant. (The requirement that a basis be ordered will be needed, for instance, in Denition 1.13.) 1.2 Example This is a basis for R2. h2 4 ;1 1 i It is linearly independent c1 2 4 +c2 1 1 =0 0 =)2c1+ 1c2= 0 4c1+ 1c2= 0=)c1=c2= 0 and it spans R2. 2c1+ 1c2=x 4c1+ 1c2=y=)c2= 2xyandc1= (yx)=2 1.3 Example This basis for R2 h1 1 ;2 4 i diers from the prior one because the vectors are in a dierent order. The verication that it is a basis is just as in the prior example. 1.4 Example The space R2has many bases. Another one is this. h1 0 ;0 1 i The verication is easy. More information on sequences is in the appendix. Section III. Basis and Dimension 111 1.5 Denition For any Rn, En=h0 BBB@1 0 ... 01 CCCA;0 BBB@0 1 ... 01 CCCA;:::;0 BBB@0 0 ... 11 CCCAi is the standard (ornatural ) basis. We denote these vectors by ~ e1;:::;~ en. (Calculus books refer to R2's standard basis vectors ~ {and~ |instead of~ e1and ~ e2, and they refer to R3's standard basis vectors ~ {,~ |, and~kinstead of~ e1,~ e2, and~ e3.) Note that the symbol ` ~ e1' means something dierent in a discussion of R3than it means in a discussion of R2. 1.6 Example Consider the space fa
cos+b
sina;b2Rgof functions of the real variable . This is a natural basis. h1
cos+ 0
sin;0
cos+ 1
sini=hcos;sini Another, more generic, basis is hcossin;2 cos+ 3 sini. Vercation that these two are bases is Exercise 22. 1.7 Example A natural basis for the vector space of cubic polynomials P3is h1;x;x2;x3i. Two other bases for this space are hx3;3x2;6x;6iandh1;1+x;1+ x+x2;1 +x+x2+x3i. Checking that these are linearly independent and span the space is easy. 1.8 Example The trivial space f~0ghas only one basis, the empty one hi. 1.9 Example The space of nite-degree polynomials has a basis with innitely many elementsh1;x;x2;:::i. 1.10 Example We have seen bases before. In the rst chapter we described the solution set of homogeneous systems such as this one x+yw= 0 z+w= 0 by parametrizing. f0 BB@1 1 0 01 CCAy+0 BB@1 0 1 11 CCAwy;w2Rg That is, we described the vector space of solutions as the span of a two-element set. We can easily check that this two-vector set is also linearly independent. Thus the solution set is a subspace of R4with a two-element basis. 112 Chapter Two. Vector Spaces 1.11 Example Parameterization helps nd bases for other vector spaces, not just for solution sets of homogeneous systems. To nd a basis for this subspace ofM22 f a b c0a+b2c= 0g we rewrite the condition as a=b+ 2c. f b+ 2c b c 0b;c2Rg=fb 1 1 0 0 +c 2 0 1 0b;c2Rg Thus, this is a natural candidate for a basis. h1 1 0 0 ;2 0 1 0 i The above work shows that it spans the space. To show that it is linearly independent is routine. Consider again Example 1.2. It involves two verications. In the rst, to check that the set is linearly independent we looked at linear combinations of the set's members that total to the zero vector c1~1+c2~2=0 0
. The resulting calculation shows that such a combination is unique, that c1must be 0 andc2must be 0. The second verication, that the set spans the space, looks at linear combi- nations that total to any member of the space c1~1+c2~2=x y
. In Example 1.2 we noted only that the resulting calculation shows that such a combination ex- ists, that for each x;ythere is ac1;c2. However, in fact the calculation also shows that the combination is unique: c1must be (yx)=2 andc2must be 2xy. That is, the rst calculation is a special case of the second. The next result says that this holds in general for a spanning set: the combination totaling to the zero vector is unique if and only if the combination totaling to any vector is unique. 1.12 Theorem In any vector space, a subset is a basis if and only if each vector in the space can be expressed as a linear combination of elements of the subset in a unique way. We consider combinations to be the same if they dier only in the order of summands or in the addition or deletion of terms of the form `0
~'. Proof .By denition, a sequence is a basis if and only if its vectors form both a spanning set and a linearly independent set. A subset is a spanning set if and only if each vector in the space is a linear combination of elements of that subset in at least one way. Thus, to nish we need only show that a subset is linearly independent if and only if every vector in the space is a linear combination of elements from the subset in at most one way. Consider two expressions of a vector as a linear Section III. Basis and Dimension 113 combination of the members of the basis. We can rearrange the two sums, and if necessary add some 0 ~iterms, so that the two sums combine the same ~'s in the same order: ~ v=c1~1+c2~2+


+cn~nand~ v=d1~1+d2~2+


+dn~n. Now c1~1+c2~2+


+cn~n=d1~1+d2~2+


+dn~n holds if and only if (c1d1)~1+


+ (cndn)~n=~0 holds, and so asserting that each coecient in the lower equation is zero is the same thing as asserting that ci=difor eachi. QED 1.13 Denition In a vector space with basis Btherepresentation of ~ vwith respect toBis the column vector of the coecients used to express ~ vas a linear combination of the basis vectors: RepB(~ v) =0 BBB@c1 c2 ... cn1 CCCA whereB=h~1;:::;~niand~ v=c1~1+c2~2+


+cn~n. Thec's are the coordinates of ~ vwith respect to B. We will later do representations in contexts that involve more than one basis. To help with the bookkeeping, we shall often attach a subscript Bto the column vector. 1.14 Example InP3, with respect to the basis B=h1;2x;2x2;2x3i, the representation of x+x2is RepB(x+x2) =0 BB@0 1=2 1=2 01 CCA B (note that the coordinates are scalars, not vectors). With respect to a dierent basisD=h1 +x;1x;x+x2;x+x3i, the representation RepD(x+x2) =0 BB@0 0 1 01 CCA D is dierent. 114 Chapter Two. Vector Spaces 1.15 Remark This use of column notation and the term `coordinates' has both a down side and an up side. The down side is that representations look like vectors from Rn, which can be confusing when the vector space we are working with is Rn, especially since we sometimes omit the subscript base. We must then infer the intent from the context. For example, the phrase `in R2, where~ v=3 2
' refers to the plane vector that, when in canonical position, ends at (3 ;2). To nd the coordinates of that vector with respect to the basis B=h1 1 ;0 2 i we solve c11 1 +c20 2 =3 2 to get that c1= 3 andc2= 1=2. Then we have this. RepB(~ v) =3 1=2 Here, although we've ommited the subscript Bfrom the column, the fact that the right side is a representation is clear from the context. The up side of the notation and the term `coordinates' is that they generalize the use that we are familiar with: in Rnand with respect to the standard basisEn, the vector starting at the origin and ending at ( v1;:::;vn) has this representation. RepEn(0 B@v1 ... vn1 CA) =0 B@v1 ... vn1 CA En Our main use of representations will come in the third chapter. The de- nition appears here because the fact that every vector is a linear combination of basis vectors in a unique way is a crucial property of bases, and also to help make two points. First, we x an order for the elements of a basis so that coordinates can be stated in that order. Second, for calculation of coordinates, among other things, we shall restrict our attention to spaces with bases having only nitely many elements. We will see that in the next subsection. Exercises X1.16 Decide if each is a basis for R3. (a)h0 @1 2 31 A;0 @3 2 11 A;0 @0 0 11 Ai(b)h0 @1 2 31 A;0 @3 2 11 Ai(c)h0 @0 2 11 A;0 @1 1 11 A;0 @2 5 01 Ai (d)h0 @0 2 11 A;0 @1 1 11 A;0 @1 3 01 Ai Section III. Basis and Dimension 115 X1.17 Represent the vector with respect to the basis. (a)1 2 ,B=h1 1 ;1 1 iR2 (b)x2+x3,D=h1;1 +x;1 +x+x2;1 +x+x2+x3iP 3 (c)0 BB@0 1 0 11 CCA,E4R4 1.18 Find a basis forP2, the space of all quadratic polynomials. Must any such basis contain a polynomial of each degree: degree zero, degree one, and degree two? 1.19 Find a basis for the solution set of this system. x14x2+ 3x3x4= 0 2x18x2+ 6x32x4= 0 X1.20 Find a basis forM22, the space of 22 matrices. X1.21 Find a basis for each. (a)The subspacefa2x2+a1x+a0a22a1=a0gofP2 (b)The space of three-wide row vectors whose rst and second components add to zero (c)This subspace of the 2 2 matrices fa b 0cc2b= 0g 1.22 Check Example 1.6. X1.23 Find the span of each set and then nd a basis for that span. (a)f1 +x;1 + 2xginP2(b)f22x;3 + 4x2ginP2 X1.24 Find a basis for each of these subspaces of the space P3of cubic polynomi- als. (a)The subspace of cubic polynomials p(x) such that p(7) = 0 (b)The subspace of polynomials p(x) such that p(7) = 0 and p(5) = 0 (c)The subspace of polynomials p(x) such that p(7) = 0,p(5) = 0, and p(3) = 0 (d)The space of polynomials p(x) such that p(7) = 0,p(5) = 0,p(3) = 0, andp(1) = 0 1.25 We've seen that it is possible for a basis to remain a basis when it is reordered. Must it remain a basis? 1.26 Can a basis contain a zero vector? X1.27 Leth~1;~2;~3ibe a basis for a vector space. (a)Show thathc1~1;c2~2;c3~3iis a basis when c1;c2;c36= 0. What happens when at least one ciis 0? (b)Prove thath~ 1;~ 2;~ 3iis a basis where ~ i=~1+~i. 1.28 Find one vector ~ vthat will make each into a basis for the space. (a)h1 1 ;~ viinR2(b)h0 @1 1 01 A;0 @0 1 01 A;~ viinR3(c)hx;1 +x2;~ viinP2 X1.29 Whereh~1;:::;~niis a basis, show that in this equation c1~1+


+ck~k=ck+1~k+1+


+cn~n each of the ci's is zero. Generalize. 1.30 A basis contains some of the vectors from a vector space; can it contain them all? 116 Chapter Two. Vector Spaces 1.31 Theorem 1.12 shows that, with respect to a basis, every linear combination is unique. If a subset is not a basis, can linear combinations be not unique? If so, must they be? X1.32 A square matrix is symmetric if for all indices iandj, entryi;jequals entry j;i. (a)Find a basis for the vector space of symmetric 2 2 matrices. (b)Find a basis for the space of symmetric 3 3 matrices. (c)Find a basis for the space of symmetric nnmatrices. X1.33 We can show that every basis for R3contains the same number of vec- tors. (a)Show that no linearly independent subset of R3contains more than three vectors. (b)Show that no spanning subset of R3contains fewer than three vectors. Hint: recall how to calculate the span of a set and show that this method cannot yield all ofR3when it is applied to fewer than three vectors. 1.34 One of the exercises in the Subspaces subsection shows that the set f0 @x y z1 Ax+y+z= 1g is a vector space under these operations.0 @x1 y1 z11 A+0 @x2 y2 z21 A=0 @x1+x21 y1+y2 z1+z21 Ar0 @x y z1 A=0 @rxr+ 1 ry rz1 A Find a basis. III.2 Dimension In the prior subsection we dened the basis of a vector space, and we saw that a space can have many dierent bases. For example, following the denition of a basis, we saw three dierent bases for R2. So we cannot talk about \the" basis for a vector space. True, some vector spaces have bases that strike us as more natural than others, for instance, R2's basisE2orR3's basisE3orP2's basis h1;x;x2i. But, for example in the space fa2x2+a1x+a02a2a0=a1g, no particular basis leaps out at us as the most natural one. We cannot, in general, associate with a space any single basis that best describes that space. We can, however, nd something about the bases that is uniquely associated with the space. This subsection shows that any two bases for a space have the same number of elements. So, with each space we can associate a number, the number of vectors in any of its bases. This brings us back to when we considered the two things that could be meant by the term `minimal spanning set'. At that point we dened `minimal' as linearly independent, but we noted that another reasonable interpretation of the term is that a spanning set is `minimal' when it has the fewest number of elements of any set with the same span. At the end of this subsection, after we Section III. Basis and Dimension 117 have shown that all bases have the same number of elements, then we will have shown that the two senses of `minimal' are equivalent. Before we start, we rst limit our attention to spaces where at least one basis has only nitely many members. 2.1 Denition A vector space is nite-dimensional if it has a basis with only nitely many vectors. (One reason for sticking to nite-dimensional spaces is so that the representation of a vector with respect to a basis is a nitely-tall vector, and so can be easily written.) From now on we study only nite-dimensional vector spaces. We shall take the term `vector space' to mean `nite-dimensional vector space'. Other spaces are interesting and important, but they lie outside of our scope. To prove the main theorem we shall use a technical result, the Exchange Lemma. We rst illustrate its conclusion with an example. 2.2 Example Here is a basis for R3and a vector given as a linear combination of members of that basis. B=h0 @1 0 01 A;0 @1 1 01 A;0 @0 0 21 Ai0 @1 2 01 A= (1)
0 @1 0 01 A+ 20 @1 1 01 A+ 0
0 @0 0 21 A In that combination two of the basis vectors have non-zero coecients. We can pick either one, here we pick the rst. Replacing it with the vector we've expressed as the combination ^B=h0 @1 2 01 A;0 @1 1 01 A;0 @0 0 21 Ai gives a new basis for the space. 2.3 Lemma (Exchange Lemma) Assume that B=h~1;:::;~niis a basis for a vector space, and that for the vector ~ vthe relationship ~ v=c1~1+c2~2+


+cn~nhasci6= 0. Then exchanging ~ifor~ vyields another basis for the space. Proof .Call the outcome of the exchange ^B=h~1;:::;~i1;~ v;~i+1;:::;~ni. We rst show that ^Bis linearly independent. Any relationship d1~1+


+ di~ v+


+dn~n=~0 among the members of ^B, after substitution for ~ v, d1~1+


+di
(c1~1+


+ci~i+


+cn~n) +


+dn~n=~0 () gives a linear relationship among the members of B. The basis Bis linearly independent, so the coecient diciof~iis zero. Because ciis assumed to be nonzero,di= 0. Using this in equation ( ) above gives that all of the other d's are also zero. Therefore ^Bis linearly independent. 118 Chapter Two. Vector Spaces We nish by showing that ^Bhas the same span as B. Half of this argument, that [ ^B][B], is easy; any member d1~1+


+di~ v+


+dn~nof [^B] can be written d1~1+


+di
(c1~1+


+cn~n) +


+dn~n, which is a linear combination of linear combinations of members of B, and hence is in [ B]. For the [B][^B] half of the argument, recall that when ~ v=c1~1+


+cn~nwith ci6= 0, then the equation can be rearranged to ~i= (c1=ci)~1+


+(1=ci)~ v+


+ (cn=ci)~n. Now, consider any member d1~1+


+di~i+


+dn~n of [B], substitute for ~iits expression as a linear combination of the members of^B, and recognize (as in the rst half of this argument) that the result is a linear combination of linear combinations, of members of ^B, and hence is in [^B]. QED 2.4 Theorem In any nite-dimensional vector space, all of the bases have the same number of elements. Proof .Fix a vector space with at least one nite basis. Choose, from among all of this space's bases, one B=h~1;:::;~niof minimal size. We will show that any other basis D=h~1;~2;:::ialso has the same number of members, n. BecauseBhas minimal size, Dhas no fewer than nvectors. We will argue that it cannot have more than nvectors. The basisBspans the space and ~1is in the space, so ~1is a nontrivial linear combination of elements of B. By the Exchange Lemma, ~1can be swapped for a vector from B, resulting in a basis B1, where one element is ~and all of the n1 other elements are ~'s. The prior paragraph forms the basis step for an induction argument. The inductive step starts with a basis Bk(for 1k<n ) containing kmembers of D andnkmembers of B. We know that Dhas at least nmembers so there is a ~k+1. Represent it as a linear combination of elements of Bk. The key point: in that representation, at least one of the nonzero scalars must be associated with a~ior else that representation would be a nontrivial linear relationship among elements of the linearly independent set D. Exchange ~k+1for~ito get a new basisBk+1with one~more and one ~fewer than the previous basis Bk. Repeat the inductive step until no ~'s remain, so that Bncontains~1;:::;~n. Now,Dcannot have more than these nvectors because any ~n+1that remains would be in the span of Bn(since it is a basis) and hence would be a linear com- bination of the other ~'s, contradicting that Dis linearly independent. QED 2.5 Denition The dimension of a vector space is the number of vectors in any of its bases. 2.6 Example Any basis for Rnhasnvectors since the standard basis Enhas nvectors. Thus, this denition generalizes the most familiar use of term, that Rnisn-dimensional. 2.7 Example The spacePnof polynomials of degree at most nhas dimension n+1. We can show this by exhibiting any basis | h1;x;:::;xnicomes to mind | and counting its members. Section III. Basis and Dimension 119 2.8 Example A trivial space is zero-dimensional since its basis is empty. Again, although we sometimes say `nite-dimensional' as a reminder, in the rest of this book all vector spaces are assumed to be nite-dimensional. An instance of this is that in the next result the word `space' should be taken to mean `nite-dimensional vector space'. 2.9 Corollary No linearly independent set can have a size greater than the dimension of the enclosing space. Proof .Inspection of the above proof shows that it never uses that Dspans the space, only that Dis linearly independent. QED 2.10 Example Recall the subspace diagram from the prior section showing the subspaces of R3. Each subspace shown is described with a minimal spanning set, for which we now have the term `basis'. The whole space has a basis with three members, the plane subspaces have bases with two members, the line subspaces have bases with one member, and the trivial subspace has a basis with zero members. When we saw that diagram we could not show that these are the only subspaces that this space has. We can show it now. The prior corollary proves that the only subspaces of R3are either three-, two-, one-, or zero-dimensional. Therefore, the diagram indicates all of the subspaces. There are no subspaces somehow, say, between lines and planes. 2.11 Corollary Any linearly independent set can be expanded to make a basis. Proof .If a linearly independent set is not already a basis then it must not span the space. Adding to it a vector that is not in the span preserves linear independence. Keep adding, until the resulting set does span the space, which the prior corollary shows will happen after only a nite number of steps. QED 2.12 Corollary Any spanning set can be shrunk to a basis. Proof .Call the spanning set S. IfSis empty then it is already a basis (the space must be a trivial space). If S=f~0gthen it can be shrunk to the empty basis, thereby making it linearly independent, without changing its span. Otherwise, Scontains a vector ~ s1with~ s16=~0 and we can form a basis B1=h~ s1i. If [B1] = [S] then we are done. If not then there is a ~ s22[S] such that ~ s262[B1]. LetB2=h~ s1;~ s2i; if [B2] = [S] then we are done. We can repeat this process until the spans are equal, which must happen in at most nitely many steps. QED 2.13 Corollary In ann-dimensional space, a set of nvectors is linearly inde- pendent if and only if it spans the space. Proof .First we will show that a subset with nvectors is linearly independent if and only if it is a basis. `If' is trivially true | bases are linearly independent. `Only if' holds because a linearly independent set can be expanded to a basis, 120 Chapter Two. Vector Spaces but a basis has nelements, so this expansion is actually the set that we began with. To nish, we will show that any subset with nvectors spans the space if and only if it is a basis. Again, `if' is trivial. `Only if' holds because any spanning set can be shrunk to a basis, but a basis has nelements and so this shrunken set is just the one we started with. QED The main result of this subsection, that all of the bases in a nite-dimensional vector space have the same number of elements, is the single most important result in this book because, as Example 2.10 shows, it describes what vector spaces and subspaces there can be. We will see more in the next chapter. 2.14 Remark The case of innite-dimensional vector spaces is somewhat con- troversial. The statement `any innite-dimensional vector space has a basis' is known to be equivalent to a statement called the Axiom of Choice (see [Blass 1984]). Mathematicians dier philosophically on whether to accept or reject this statement as an axiom on which to base mathematics (although, the great majority seem to accept it). Consequently the question about innite- dimensional vector spaces is still somewhat up in the air. (A discussion of the Axiom of Choice can be found in the Frequently Asked Questions list for the Usenet group sci.math . Another accessible reference is [Rucker].) Exercises Assume that all spaces are nite-dimensional unless otherwise stated. X2.15 Find a basis for, and the dimension of, P2. 2.16 Find a basis for, and the dimension of, the solution set of this system. x14x2+ 3x3x4= 0 2x18x2+ 6x32x4= 0 X2.17 Find a basis for, and the dimension of, M22, the vector space of 2 2 matrices. 2.18 Find the dimension of the vector space of matricesa b c d subject to each condition. (a)a;b;c;d2R (b)ab+ 2c= 0 andd2R (c)a+b+c= 0,a+bc= 0, andd2R X2.19 Find the dimension of each. (a)The space of cubic polynomials p(x) such that p(7) = 0 (b)The space of cubic polynomials p(x) such that p(7) = 0 and p(5) = 0 (c)The space of cubic polynomials p(x) such that p(7) = 0,p(5) = 0, and p(3) = 0 (d)The space of cubic polynomials p(x) such that p(7) = 0,p(5) = 0,p(3) = 0, andp(1) = 0 2.20 What is the dimension of the span of the set fcos2;sin2;cos 2;sin 2g? This span is a subspace of the space of all real-valued functions of one real variable. 2.21 Find the dimension of C47, the vector space of 47-tuples of complex numbers. Section III. Basis and Dimension 121 2.22 What is the dimension of the vector space M35of 35 matrices? X2.23 Show that this is a basis for R4. h0 BB@1 0 0 01 CCA;0 BB@1 1 0 01 CCA;0 BB@1 1 1 01 CCA;0 BB@1 1 1 11 CCAi (The results of this subsection can be used to simplify this job.) 2.24 Refer to Example 2.10. (a)Sketch a similar subspace diagram for P2. (b)Sketch one forM22. X2.25 WhereSis a set, the functions f:S!Rform a vector space under the natural operations: the sum f+gis the function given by f+g(s) =f(s) +g(s) and the scalar product is given by r
f(s) =r
f(s). What is the dimension of the space resulting for each domain? (a)S=f1g(b)S=f1;2g(c)S=f1;:::;ng 2.26 (See Exercise 25.) Prove that this is an innite-dimensional space: the set of all functions f:R!Runder the natural operations. 2.27 (See Exercise 25.) What is the dimension of the vector space of functions f:S!R, under the natural operations, where the domain Sis the empty set? 2.28 Show that any set of four vectors in R2is linearly dependent. 2.29 Show thath~ 1;~ 2;~ 3iR3is a basis if and only if there is no plane through the origin containing all three vectors. 2.30 (a) Prove that any subspace of a nite dimensional space has a basis. (b)Prove that any subspace of a nite dimensional space is nite dimensional. 2.31 Where is the niteness of Bused in Theorem 2.4? X2.32 Prove that if UandWare both three-dimensional subspaces of R5thenU\W is non-trivial. Generalize. 2.33 A basis for a space consists of elements of that space. So we are naturally led to how the property `is a basis' interacts with operations and\and[. (Of course, a basis is actually a sequence in that it is ordered, but there is a natural extension of these operations.) (a)Consider rst how bases might be related by . Assume that U;W are subspaces of some vector space and that UW. Can there exist bases BUfor UandBWforWsuch thatBUBW? Must such bases exist? For any basis BUforU, must there be a basis BWforWsuch thatBUBW? For any basis BWforW, must there be a basis BUforUsuch thatBUBW? For any bases BU;BWforUandW, mustBUbe a subset of BW? (b)Is the\of bases a basis? For what space? (c)Is the[of bases a basis? For what space? (d)What about the complement operation? (Hint. Test any conjectures against some subspaces of R3.) X2.34 Consider how `dimension' interacts with `subset'. Assume UandWare both subspaces of some vector space, and that UW. (a)Prove that dim( U)dim(W). (b)Prove that equality of dimension holds if and only if U=W. (c)Show that the prior item does not hold if they are innite-dimensional. ?2.35 For any vector ~ vinRnand any permutation of the numbers 1, 2, . . . , n (that is,is a rearrangement of those numbers into a new order), dene (~ v) 122 Chapter Two. Vector Spaces to be the vector whose components are v(1),v(2), . . . , andv(n)(where(1) is the rst number in the rearrangement, etc.). Now x ~ vand letVbe the span of f(~ v)permutes 1, . . . , ng. What are the possibilities for the dimension of V? [Wohascum no. 47] III.3 Vector Spaces and Linear Systems We will now reconsider linear systems and Gauss' method, aided by the tools and terms of this chapter. We will make three points. For the rst point, recall the rst chapter's Linear Combination Lemma and its corollary: if two matrices are related by row operations A!


! Bthen each row of Bis a linear combination of the rows of A. That is, Gauss' method works by taking linear combinations of rows. Therefore, the right setting in which to study row operations in general, and Gauss' method in particular, is the following vector space. 3.1 Denition The row space of a matrix is the span of the set of its rows. The row rank is the dimension of the row space, the number of linearly independent rows. 3.2 Example If A=2 3 4 6 then Rowspace( A) is this subspace of the space of two-component row vectors. fc1
2 3
+c2
4 6
c1;c22Rg The linear dependence of the second on the rst is obvious and so we can simplify this description to fc
2 3
c2Rg. 3.3 Lemma If the matrices AandBare related by a row operation Ai$j!BorAki!BorAki+j!B (fori6=jandk6= 0) then their row spaces are equal. Hence, row-equivalent matrices have the same row space, and hence also, the same row rank. Proof .By the Linear Combination Lemma's corollary, each row of Bis in the row space of A. Further, Rowspace( B)Rowspace(A) because a member of the set Rowspace( B) is a linear combination of the rows of B, which means it is a combination of a combination of the rows of A, and hence, by the Linear Combination Lemma, is also a member of Rowspace( A). For the other containment, recall that row operations are reversible: A!B if and only if B!A. With that, Rowspace( A)Rowspace(B) also follows from the prior paragraph, and so the two sets are equal. QED Section III. Basis and Dimension 123 Thus, row operations leave the row space unchanged. But of course, Gauss' method performs the row operations systematically, with a specic goal in mind, echelon form. 3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly independent set. Proof .A result in the rst chapter, Lemma III.2.4, states that in an echelon form matrix, no nonzero row is a linear combination of the other rows. This is a restatement of that result into new terminology. QED Thus, in the language of this chapter, Gaussian reduction works by elim- inating linear dependences among rows, leaving the span unchanged, until no nontrivial linear relationships remain (among the nonzero rows). That is, Gauss' method produces a basis for the row space. 3.5 Example From any matrix, we can produce a basis for the row space by performing Gauss' method and taking the nonzero rows of the resulting echelon form matrix. For instance, 0 @1 3 1 1 4 1 2 0 51 A1+2! 21+362+3!0 @1 3 1 0 1 0 0 0 31 A produces the basis h1 3 1
;0 1 0
;0 0 3
ifor the row space. This is a basis for the row space of both the starting and ending matrices, since the two row spaces are equal. Using this technique, we can also nd bases for spans not directly involving row vectors. 3.6 Denition The column space of a matrix is the span of the set of its columns. The column rank is the dimension of the column space, the number of linearly independent columns. Our interest in column spaces stems from our study of linear systems. An example is that this system c1+ 3c2+ 7c3=d1 2c1+ 3c2+ 8c3=d2 c2+ 2c3=d3 4c1 + 4c3=d4 has a solution if and only if the vector of d's is a linear combination of the other column vectors, c10 BB@1 2 0 41 CCA+c20 BB@3 3 1 01 CCA+c30 BB@7 8 2 41 CCA=0 BB@d1 d2 d3 d41 CCA meaning that the vector of d's is in the column space of the matrix of coecients. 124 Chapter Two. Vector Spaces 3.7 Example Given this matrix, 0 BB@1 3 7 2 3 8 0 1 2 4 0 41 CCA to get a basis for the column space, temporarily turn the columns into rows and reduce. 0 @1 2 0 4 3 3 1 0 7 8 2 41 A31+2! 71+322+3!0 @1 2 0 4 03 112 0 0 0 01 A Now turn the rows back to columns. h0 BB@1 2 0 41 CCA;0 BB@0 3 1 121 CCAi The result is a basis for the column space of the given matrix. 3.8 Denition The transpose of a matrix is the result of interchanging the rows and columns of that matrix. That is, column jof the matrix Ais rowj ofAtrans, and vice versa. So the instructions for the prior example are \transpose, reduce, and transpose back". We can even, at the price of tolerating the as-yet-vague idea of vector spaces being \the same", use Gauss' method to nd bases for spans in other types of vector spaces. 3.9 Example To get a basis for the span of fx2+x4;2x2+ 3x4;x23x4g in the spaceP4, think of these three polynomials as \the same" as the row vectors0 0 1 0 1
,0 0 2 0 3
, and0 01 03
, apply Gauss' method 0 @0 0 1 0 1 0 0 2 0 3 0 01 031 A21+2! 1+322+3!0 @0 0 1 0 1 0 0 0 0 1 0 0 0 0 01 A and translate back to get the basis hx2+x4;x4i. (As mentioned earlier, we will make the phrase \the same" precise at the start of the next chapter.) Thus, our rst point in this subsection is that the tools of this chapter give us a more conceptual understanding of Gaussian reduction. For the second point of this subsection, consider the eect on the column space of this row reduction. 1 2 2 4 21+2!1 2 0 0 Section III. Basis and Dimension 125 The column space of the left-hand matrix contains vectors with a second compo- nent that is nonzero. But the column space of the right-hand matrix is dierent because it contains only vectors whose second component is zero. It is this knowledge that row operations can change the column space that makes next result surprising. 3.10 Lemma Row operations do not change the column rank. Proof .Restated, if Areduces to Bthen the column rank of Bequals the column rank of A. We will be done if we can show that row operations do not aect linear relationships among columns because the column rank is just the size of the largest set of unrelated columns. That is, we will show that a relationship exists among columns (such as that the fth column is twice the second plus the fourth) if and only if that relationship exists after the row operation. But this is exactly the rst theorem of this book: in a relationship among columns, c1
0 BBB@a1;1 a2;1 ... am;11 CCCA+


+cn
0 BBB@a1;n a2;n ... am;n1 CCCA=0 BBB@0 0 ... 01 CCCA row operations leave unchanged the set of solutions ( c1;:::;cn). QED Another way, besides the prior result, to state that Gauss' method has some- thing to say about the column space as well as about the row space is to consider again Gauss-Jordan reduction. Recall that it ends with the reduced echelon form of a matrix, as here. 0 @1 3 1 6 2 6 3 16 1 3 1 61 A!


!0 @1 3 0 2 0 0 1 4 0 0 0 01 A Consider the row space and the column space of this result. Our rst point made above says that a basis for the row space is easy to get: simply collect together all of the rows with leading entries. However, because this is a reduced echelon form matrix, a basis for the column space is just as easy: take the columns containing the leading entries, that is, h~ e1;~ e2i. (Linear independence is obvious. The other columns are in the span of this set, since they all have a third component of zero.) Thus, for a reduced echelon form matrix, bases for the row and column spaces can be found in essentially the same way | by taking the parts of the matrix, the rows or columns, containing the leading entries. 3.11 Theorem The row rank and column rank of a matrix are equal. Proof .First bring the matrix to reduced echelon form. At that point, the row rank equals the number of leading entries since each equals the number of nonzero rows. Also at that point, the number of leading entries equals the 126 Chapter Two. Vector Spaces column rank because the set of columns containing leading entries consists of some of the ~ ei's from a standard basis, and that set is linearly independent and spans the set of columns. Hence, in the reduced echelon form matrix, the row rank equals the column rank, because each equals the number of leading entries. But Lemma 3.3 and Lemma 3.10 show that the row rank and column rank are not changed by using row operations to get to reduced echelon form. Thus the row rank and the column rank of the original matrix are also equal. QED 3.12 Denition The rank of a matrix is its row rank or column rank. So our second point in this subsection is that the column space and row space of a matrix have the same dimension. Our third and nal point is that the concepts that we've seen arising naturally in the study of vector spaces are exactly the ones that we have studied with linear systems. 3.13 Theorem For linear systems with nunknowns and with matrix of co- ecientsA, the statements (1) the rank of Aisr (2) the space of solutions of the associated homogeneous system has dimen- sionnr are equivalent. So if the system has at least one particular solution then for the set of solutions, the number of parameters equals nr, the number of variables minus the rank of the matrix of coecients. Proof .The rank of Aisrif and only if Gaussian reduction on Aends withr nonzero rows. That's true if and only if echelon form matrices row equivalent toAhaver-many leading variables. That in turn holds if and only if there are nrfree variables. QED 3.14 Remark [Munkres] Sometimes that result is mistakenly remembered to say that the general solution of an nunknown system of mequations uses nm parameters. The number of equations is not the relevant gure, rather, what matters is the number of independent equations (the number of equations in a maximal independent set). Where there are rindependent equations, the general solution involves nrparameters. 3.15 Corollary Where the matrix Aisnn, the statements (1) the rank of Aisn (2)Ais nonsingular (3) the rows of Aform a linearly independent set (4) the columns of Aform a linearly independent set (5) any linear system whose matrix of coecients is Ahas one and only one solution are equivalent. Section III. Basis and Dimension 127 Proof .Clearly (1)() (2)() (3)() (4). The last, (4) () (5), holds because a set of ncolumn vectors is linearly independent if and only if it is a basis for Rn, but the system c10 BBB@a1;1 a2;1 ... am;11 CCCA+


+cn0 BBB@a1;n a2;n ... am;n1 CCCA=0 BBB@d1 d2 ... dm1 CCCA has a unique solution for all choices of d1;:::;dn2Rif and only if the vectors ofa's form a basis. QED Exercises 3.16 Transpose each. (a)2 1 3 1 (b)2 1 1 3 (c)1 4 3 6 7 8 (d)0 @0 0 01 A (e) 12
X3.17 Decide if the vector is in the row space of the matrix. (a)2 1 3 1 , 1 0
(b)0 @0 1 3 1 0 1 1 2 71 A, 1 1 1
X3.18 Decide if the vector is in the column space. (a)1 1 1 1 ,1 3 (b)0 @1 3 1 2 0 4 1331 A,0 @1 0 01 A X3.19 Find a basis for the row space of this matrix. 0 BB@2 0 3 4 0 1 11 3 1 0 2 1 0411 CCA X3.20 Find the rank of each matrix. (a)0 @2 1 3 11 2 1 0 31 A (b)0 @11 2 33 6 2 241 A (c)0 @1 3 2 5 1 1 6 4 31 A (d)0 @0 0 0 0 0 0 0 0 01 A X3.21 Find a basis for the span of each set. (a)f 1 3
; 1 3
; 1 4
; 2 1
gM 12 (b)f0 @1 2 11 A;0 @3 1 11 A;0 @1 3 31 AgR3 (c)f1 +x;1x2;3 + 2xx2gP 3 (d)f1 0 1 3 11 ;1 0 3 2 1 4 ;1 05 119 gM 23 3.22 Which matrices have rank zero? Rank one? 128 Chapter Two. Vector Spaces X3.23 Givena;b;c2R, what choice of dwill cause this matrix to have the rank of one? a b c d 3.24 Find the column rank of this matrix.1 31 5 0 4 2 0 1 0 4 1 3.25 Show that a linear system with at least one solution has at most one solution if and only if the matrix of coecients has rank equal to the number of its columns. X3.26 If a matrix is 59, which set must be dependent, its set of rows or its set of columns? 3.27 Give an example to show that, despite that they have the same dimension, the row space and column space of a matrix need not be equal. Are they ever equal? 3.28 Show that the set f(1;1;2;3);(1;1;2;0);(3;1;6;6)gdoes not have the same span asf(1;0;1;0);(0;2;0;3)g. What, by the way, is the vector space? X3.29 Show that this set of column vectors8 < :0 @d1 d2 d31 Athere arex,y, andzsuch that3x+ 2y+ 4z=d1 xz=d2 2x+ 2y+ 5z=d39 = ; is a subspace of R3. Find a basis. 3.30 Show that the transpose operation is linear : (rA+sB)trans=rAtrans+sBtrans forr;s2RandA;B2Mmn, X3.31 In this subsection we have shown that Gaussian reduction nds a basis for the row space. (a)Show that this basis is not unique | dierent reductions may yield dierent bases. (b)Produce matrices with equal row spaces but unequal numbers of rows. (c)Prove that two matrices have equal row spaces if and only if after Gauss- Jordan reduction they have the same nonzero rows. 3.32 Why is there not a problem with Remark 3.14 in the case that ris bigger thann? 3.33 Show that the row rank of an mnmatrix is at most m. Is there a better bound? X3.34 Show that the rank of a matrix equals the rank of its transpose. 3.35 True or false: the column space of a matrix equals the row space of its trans- pose. X3.36 We have seen that a row operation may change the column space. Must it? 3.37 Prove that a linear system has a solution if and only if that system's matrix of coecients has the same rank as its augmented matrix. 3.38 Anmnmatrix has full row rank if its row rank is m, and it has full column rank if its column rank is n. (a)Show that a matrix can have both full row rank and full column rank only if it is square. (b)Prove that the linear system with matrix of coecients Ahas a solution for anyd1, . . . ,dn's on the right side if and only if Ahas full row rank. Section III. Basis and Dimension 129 (c)Prove that a homogeneous system has a unique solution if and only if its matrix of coecients Ahas full column rank. (d)Prove that the statement \if a system with matrix of coecients Ahas any solution then it has a unique solution" holds if and only if Ahas full column rank. 3.39 How would the conclusion of Lemma 3.3 change if Gauss' method is changed to allow multiplying a row by zero? X3.40 What is the relationship between rank( A) and rank(A)? Between rank( A) and rank(kA)? What, if any, is the relationship between rank( A), rank(B), and rank(A+B)? III.4 Combining Subspaces This subsection is optional. It is required only for the last sections of Chapter Three and Chapter Five and for occasional exercises, and can be passed over without loss of continuity. This chapter opened with the denition of a vector space, and the mid- dle consisted of a rst analysis of the idea. This subsection closes the chapter by nishing the analysis, in the sense that `analysis' means \method of de- termining the . . . essential features of something by separating it into parts" [Macmillan Dictionary]. A common way to understand things is to see how they can be built from component parts. For instance, we think of R3as put together, in some way, from thex-axis, they-axis, andz-axis. In this subsection we will make this precise; we will describe how to decompose a vector space into a combination of some of its subspaces. In developing this idea of subspace combination, we will keep the R3example in mind as a benchmark model. Subspaces are subsets and sets combine via union. But taking the combi- nation operation for subspaces to be the simple union operation isn't what we want. For one thing, the union of the x-axis, they-axis, andz-axis is not all of R3, so the benchmark model would be left out. Besides, union is all wrong for this reason: a union of subspaces need not be a subspace (it need not be closed; for instance, this R3vector 0 @1 0 01 A+0 @0 1 01 A+0 @0 0 11 A=0 @1 1 11 A is in none of the three axes and hence is not in the union). In addition to the members of the subspaces, we must at least also include all of the linear combinations. 4.1 Denition WhereW1;:::;Wkare subspaces of a vector space, their sum is the span of their union W1+W2+


+Wk= [W1[W2[:::Wk]. 130 Chapter Two. Vector Spaces (The notation, writing the `+' between sets in addition to using it between vectors, ts with the practice of using this symbol for any natural accumulation operation.) 4.2 Example TheR3model ts with this operation. Any vector ~ w2R3can be written as a linear combination c1~ v1+c2~ v2+c3~ v3where~ v1is a member of thex-axis, etc., in this way 0 @w1 w2 w31 A= 1
0 @w1 0 01 A+ 1
0 @0 w2 01 A+ 1
0 @0 0 w31 A and so R3=x-axis +y-axis +z-axis. 4.3 Example A sum of subspaces can be less than the entire space. Inside of P4, letLbe the subspace of linear polynomials fa+bxa;b2Rgand letCbe the subspace of purely-cubic polynomials fcx3c2Rg. ThenL+Cis not all ofP4. Instead, it is the subspace L+C=fa+bx+cx3a;b;c2Rg. 4.4 Example A space can be described as a combination of subspaces in more than one way. Besides the decomposition R3=x-axis +y-axis +z-axis, we can also write R3=xy-plane +yz-plane. To check this, note that any ~ w2R3can be written as a linear combination of a member of the xy-plane and a member of theyz-plane; here are two such combinations. 0 @w1 w2 w31 A= 1
0 @w1 w2 01 A+ 1
0 @0 0 w31 A0 @w1 w2 w31 A= 1
0 @w1 w2=2 01 A+ 1
0 @0 w2=2 w31 A The above denition gives one way in which a space can be thought of as a combination of some of its parts. However, the prior example shows that there is at least one interesting property of our benchmark model that is not captured by the denition of the sum of subspaces. In the familiar decomposition of R3, we often speak of a vector's ` xpart' or `ypart' or `zpart'. That is, in this model, each vector has a unique decomposition into parts that come from the parts making up the whole space. But in the decomposition used in Example 4.4, we cannot refer to the \ xypart" of a vector | these three sums 0 @1 2 31 A=0 @1 2 01 A+0 @0 0 31 A=0 @1 0 01 A+0 @0 2 31 A=0 @1 1 01 A+0 @0 1 31 A all describe the vector as comprised of something from the rst plane plus some- thing from the second plane, but the \ xypart" is dierent in each. That is, when we consider how R3is put together from the three axes \in some way", we might mean \in such a way that every vector has at least one decomposition", and that leads to the denition above. But if we take it to mean \in such a way that every vector has one and only one decomposition" Section III. Basis and Dimension 131 then we need another condition on combinations. To see what this condition is, recall that vectors are uniquely represented in terms of a basis. We can use this to break a space into a sum of subspaces such that any vector in the space breaks uniquely into a sum of members of those subspaces. 4.5 Example The benchmark is R3with its standard basis E3=h~ e1;~ e2;~ e3i. The subspace with the basis B1=h~ e1iis thex-axis. The subspace with the basisB2=h~ e2iis they-axis. The subspace with the basis B3=h~ e3iis the z-axis. The fact that any member of R3is expressible as a sum of vectors from these subspaces0 @x y z1 A=0 @x 0 01 A+0 @0 y 01 A+0 @0 0 z1 A is a re ection of the fact that E3spans the space | this equation 0 @x y z1 A=c10 @1 0 01 A+c20 @0 1 01 A+c30 @0 0 11 A has a solution for any x;y;z2R. And, the fact that each such expression is unique re ects that fact that E3is linearly independent | any equation like the one above has a unique solution. 4.6 Example We don't have to take the basis vectors one at a time, the same idea works if we conglomerate them into larger sequences. Consider again the space R3and the vectors from the standard basis E3. The subspace with the basisB1=h~ e1;~ e3iis thexz-plane. The subspace with the basis B2=h~ e2iis they-axis. As in the prior example, the fact that any member of the space is a sum of members of the two subspaces in one and only one way 0 @x y z1 A=0 @x 0 z1 A+0 @0 y 01 A is a re ection of the fact that these vectors form a basis | this system 0 @x y z1 A= (c10 @1 0 01 A+c30 @0 0 11 A) +c20 @0 1 01 A has one and only one solution for any x;y;z2R. These examples illustrate a natural way to decompose a space into a sum of subspaces in such a way that each vector decomposes uniquely into a sum of vectors from the parts. The next result says that this way is the only way. 4.7 Denition The concatenation of the sequences B1=h~1;1;:::;~1;n1i, . . . ,Bk=h~k;1;:::;~k;nkiis their adjoinment. B1_B2_


_Bk=h~1;1;:::;~1;n1;~2;1;:::;~k;nki 132 Chapter Two. Vector Spaces 4.8 Lemma LetVbe a vector space that is the sum of some of its subspaces V=W1+


+Wk. LetB1, . . . ,Bkbe any bases for these subspaces. Then the following are equivalent. (1) For every ~ v2V, the expression ~ v=~ w1+


+~ wk(with~ wi2Wi) is unique. (2) The concatenation B1_


_Bkis a basis for V. (3) The nonzero members of f~ w1;:::;~ wkg(with~ wi2Wi) form a linearly independent set | among nonzero vectors from dierent Wi's, every linear relationship is trivial. Proof .We will show that (1) = )(2), that (2) =)(3), and nally that (3) =)(1). For these arguments, observe that we can pass from a combination of~ w's to a combination of ~'s d1~ w1+


+dk~ wk =d1(c1;1~1;1+


+c1;n1~1;n1) +


+dk(ck;1~k;1+


+ck;nk~k;nk) =d1c1;1
~1;1+


+dkck;nk
~k;nk () and vice versa. For (1) =)(2), assume that all decompositions are unique. We will show thatB1_


_Bkspans the space and is linearly independent. It spans the space because the assumption that V=W1+


+Wkmeans that every ~ v can be expressed as ~ v=~ w1+


+~ wk, which translates by equation ( ) to an expression of ~ vas a linear combination of the ~'s from the concatenation. For linear independence, consider this linear relationship. ~0 =c1;1~1;1+


+ck;nk~k;nk Regroup as in () (that is, take d1, . . . ,dkto be 1 and move from bottom to top) to get the decomposition ~0 =~ w1+


+~ wk. Because of the assumption that decompositions are unique, and because the zero vector obviously has the decomposition ~0 =~0+


+~0, we now have that each ~ wiis the zero vector. This means that ci;1~i;1+


+ci;ni~i;ni=~0. Thus, since each Biis a basis, we have the desired conclusion that all of the c's are zero. For (2) =)(3), assume that B1_


_Bkis a basis for the space. Consider a linear relationship among nonzero vectors from dierent Wi's, ~0 =


+di~ wi+


in order to show that it is trivial. (The relationship is written in this way because we are considering a combination of nonzero vectors from only some of theWi's; for instance, there might not be a ~ w1in this combination.) As in ( ), ~0 =


+di(ci;1~i;1+


+ci;ni~i;ni)+


=


+dici;1
~i;1+


+dici;ni
~i;ni+


and the linear independence of B1_


_Bkgives that each coecient dici;jis zero. Now, ~ wiis a nonzero vector, so at least one of the ci;j's is not zero, and thusdiis zero. This holds for each di, and therefore the linear relationship is trivial. Section III. Basis and Dimension 133 Finally, for (3) = )(1), assume that, among nonzero vectors from dierent Wi's, any linear relationship is trivial. Consider two decompositions of a vector ~ v=~ w1+


+~ wkand~ v=~ u1+


+~ ukin order to show that the two are the same. We have ~0 = (~ w1+


+~ wk)(~ u1+


+~ uk) = (~ w1~ u1) +


+ (~ wk~ uk) which violates the assumption unless each ~ wi~ uiis the zero vector. Hence, decompositions are unique. QED 4.9 Denition A collection of subspaces fW1;:::;Wkgisindependent if no nonzero vector from any Wiis a linear combination of vectors from the other subspacesW1;:::;Wi1;Wi+1;:::;Wk. 4.10 Denition A vector space Vis the direct sum (orinternal direct sum ) of its subspaces W1;:::;WkifV=W1+W2+


+Wkand the collection fW1;:::;Wkgis independent. We write V=W1W2:::Wk. 4.11 Example The benchmark model ts: R3=x-axisy-axisz-axis. 4.12 Example The space of 22 matrices is this direct sum. fa0 0da;d2Rgf0b 0 0b2Rgf0 0 c0c2Rg It is the direct sum of subspaces in many other ways as well; direct sum decom- positions are not unique. 4.13 Corollary The dimension of a direct sum is the sum of the dimensions of its summands. Proof .In Lemma 4.8, the number of basis vectors in the concatenation equals the sum of the number of vectors in the subbases that make up the concatena- tion. QED The special case of two subspaces is worth mentioning separately. 4.14 Denition When a vector space is the direct sum of two of its subspaces, then they are said to be complements . 4.15 Lemma A vector space Vis the direct sum of two of its subspaces W1 andW2if and only if it is the sum of the two V=W1+W2and their intersection is trivialW1\W2=f~0g. Proof .Suppose rst that V=W1W2. By denition, Vis the sum of the two. To show that the two have a trivial intersection, let ~ vbe a vector from W1\W2and consider the equation ~ v=~ v. On the left side of that equation is a member of W1, and on the right side is a linear combination of members 134 Chapter Two. Vector Spaces (actually, of only one member) of W2. But the independence of the spaces then implies that ~ v=~0, as desired. For the other direction, suppose that Vis the sum of two spaces with a trivial intersection. To show that Vis a direct sum of the two, we need only show that the spaces are independent | no nonzero member of the rst is expressible as a linear combination of members of the second, and vice versa. This is true because any relationship ~ w1=c1~ w2;1+


+dk~ w2;k(with~ w12W1and ~ w2;j2W2for allj) shows that the vector on the left is also in W2, since the right side is a combination of members of W2. The intersection of these two spaces is trivial, so ~ w1=~0. The same argument works for any ~ w2. QED 4.16 Example In the space R2, thex-axis and the y-axis are complements, that is,R2=x-axisy-axis. A space can have more than one pair of complementary subspaces; another pair here are the subspaces consisting of the lines y=xand y= 2x. 4.17 Example In the space F=facos+bsina;b2Rg, the subspaces W1=facosa2RgandW2=fbsinb2Rgare complements. In addition to the fact that a space like Fcan have more than one pair of complementary subspaces, inside of the space a single subspace like W1can have more than one complement | another complement of W1isW3=fbsin+bcosb2Rg. 4.18 Example InR3, thexy-plane and the yz-planes are not complements, which is the point of the discussion following Example 4.4. One complement of thexy-plane is the z-axis. A complement of the yz-plane is the line through (1;1;1). 4.19 Example Following Lemma 4.15, here is a natural question: is the simple sumV=W1+


+Wkalso a direct sum if and only if the intersection of the subspaces is trivial? The answer is that if there are more than two subspaces then having a trivial intersection is not enough to guarantee unique decompo- sition (i.e., is not enough to ensure that the spaces are independent). In R3, let W1be thex-axis, letW2be they-axis, and let W3be this. W3=f0 @q q r1 Aq;r2Rg The check that R3=W1+W2+W3is easy. The intersection W1\W2\W3is trivial, but decompositions aren't unique. 0 @x y z1 A=0 @0 0 01 A+0 @0 yx 01 A+0 @x x z1 A=0 @xy 0 01 A+0 @0 0 01 A+0 @y y z1 A (This example also shows that this requirement is also not enough: that all pairwise intersections of the subspaces be trivial. See Exercise 30.) In this subsection we have seen two ways to regard a space as built up from component parts. Both are useful; in particular, in this book the direct sum denition is needed to do the Jordan Form construction in the fth chapter. Section III. Basis and Dimension 135 Exercises X4.20 Decide if R2is the direct sum of each W1andW2. (a)W1=fx 0x2Rg,W2=fx xx2Rg (b)W1=fs ss2Rg,W2=fs 1:1ss2Rg (c)W1=R2,W2=f~0g (d)W1=W2=ft tt2Rg (e)W1=f1 0 +x 0x2Rg,W2=f1 0 +0 yy2Rg X4.21 Show that R3is the direct sum of the xy-plane with each of these. (a)thez-axis (b)the line f0 @z z z1 Az2Rg 4.22 IsP2the direct sum of fa+bx2a;b2Rgandfcxc2Rg? X4.23 InPn, the even polynomials are the members of this set E=fp2Pnp(x) =p(x) for allxg and the oddpolynomials are the members of this set. O=fp2Pnp(x) =p(x) for allxg Show that these are complementary subspaces. 4.24 Which of these subspaces of R3 W1: thex-axis,W2: they-axis,W3: thez-axis, W4: the plane x+y+z= 0,W5: theyz-plane can be combined to (a)sum to R3?(b)direct sum to R3? X4.25 Show thatPn=fa0a02Rg:::fanxnan2Rg. 4.26 What isW1+W2ifW1W2? 4.27 Does Example 4.5 generalize? That is, is this true or false: if a vector space V has a basish~1;:::;~nithen it is the direct sum of the spans of the one-dimensional subspacesV= [f~1g]:::[f~ng]? 4.28 CanR4be decomposed as a direct sum in two dierent ways? Can R1? 4.29 This exercise makes the notation of writing `+' between sets more natural. Prove that, where W1;:::;Wkare subspaces of a vector space, W1+


+Wk=f~ w1+~ w2+


+~ wk~ w12W1;:::;~ wk2Wkg; and so the sum of subspaces is the subspace of all sums. 4.30 (Refer to Example 4.19. This exercise shows that the requirement that pari- wise intersections be trivial is genuinely stronger than the requirement only that the intersection of all of the subspaces be trivial.) Give a vector space and three subspacesW1,W2, andW3such that the space is the sum of the subspaces, the intersection of all three subspaces W1\W2\W3is trivial, but the pairwise inter- sectionsW1\W2,W1\W3, andW2\W3are nontrivial. 136 Chapter Two. Vector Spaces X4.31 Prove that if V=W1:::WkthenWi\Wjis trivial whenever i6=j. This shows that the rst half of the proof of Lemma 4.15 extends to the case of more than two subspaces. (Example 4.19 shows that this implication does not reverse; the other half does not extend.) 4.32 Recall that no linearly independent set contains the zero vector. Can an independent set of subspaces contain the trivial subspace? X4.33 Does every subspace have a complement? X4.34 LetW1;W2be subspaces of a vector space. (a)Assume that the set S1spansW1, and that the set S2spansW2. CanS1[S2 spanW1+W2? Must it? (b)Assume that S1is a linearly independent subset of W1and thatS2is a linearly independent subset of W2. CanS1[S2be a linearly independent subset ofW1+W2? Must it? 4.35 When a vector space is decomposed as a direct sum, the dimensions of the subspaces add to the dimension of the space. The situation with a space that is given as the sum of its subspaces is not as simple. This exercise considers the two-subspace special case. (a)For these subspaces of M22ndW1\W2, dim(W1\W2),W1+W2, and dim(W1+W2). W1=f0 0 c dc;d2RgW2=f0b c0b;c2Rg (b)Suppose that UandWare subspaces of a vector space. Suppose that the sequenceh~1;:::;~kiis a basis for U\W. Finally, suppose that the prior sequence has been expanded to give a sequence h~ 1;:::;~ j;~1;:::;~kithat is a basis forU, and a sequenceh~1;:::;~k;~ !1;:::;~ !pithat is a basis for W. Prove that this sequence h~ 1;:::;~ j;~1;:::;~k;~ !1;:::;~ !pi is a basis for for the sum U+W. (c)Conclude that dim( U+W) = dim(U) + dim(W)dim(U\W). (d)LetW1andW2be eight-dimensional subspaces of a ten-dimensional space. List all values possible for dim( W1\W2). 4.36 LetV=W1:::Wkand for each index isuppose that Siis a linearly independent subset of Wi. Prove that the union of the Si's is linearly independent. 4.37 A matrix is symmetric if for each pair of indices iandj, thei;jentry equals thej;ientry. A matrix is antisymmetric if eachi;jentry is the negative of the j;i entry. (a)Give a symmetric 2 2 matrix and an antisymmetric 2 2 matrix. ( Remark. For the second one, be careful about the entries on the diagional.) (b)What is the relationship between a square symmetric matrix and its trans- pose? Between a square antisymmetric matrix and its transpose? (c)Show thatMnnis the direct sum of the space of symmetric matrices and the space of antisymmetric matrices. 4.38 LetW1;W2;W3be subspaces of a vector space. Prove that ( W1\W2)+(W1\ W3)W1\(W2+W3). Does the inclusion reverse? 4.39 The example of the x-axis and the y-axis in R2shows that W1W2=Vdoes not imply that W1[W2=V. CanW1W2=VandW1[W2=Vhappen? X4.40 Consider Corollary 4.13. Does it work both ways | that is, supposing that V=W1+


+Wk, isV=W1:::Wkif and only if dim( V) = dim(W1) +


+ dim(Wk)? Section III. Basis and Dimension 137 4.41 We know that if V=W1W2then there is a basis for Vthat splits into a basis forW1and a basis for W2. Can we make the stronger statement that every basis forVsplits into a basis for W1and a basis for W2? 4.42 We can ask about the algebra of the `+' operation. (a)Is it commutative; is W1+W2=W2+W1? (b)Is it associative; is ( W1+W2) +W3=W1+ (W2+W3)? (c)LetWbe a subspace of some vector space. Show that W+W=W. (d)Must there be an identity element, a subspace Isuch thatI+W=W+I= Wfor all subspaces W? (e)Does left-cancelation hold: if W1+W2=W1+W3thenW2=W3? Right cancelation? 4.43 Consider the algebraic properties of the direct sum operation. (a)Does direct sum commute: does V=W1W2imply that V=W2W1? (b)Prove that direct sum is associative: ( W1W2)W3=W1(W2W3). (c)Show that R3is the direct sum of the three axes (the relevance here is that by the previous item, we needn't specify which two of the threee axes are combined rst). (d)Does the direct sum operation left-cancel: does W1W2=W1W3imply W2=W3? Does it right-cancel? (e)There is an identity element with respect to this operation. Find it. (f)Do some, or all, subspaces have inverses with respect to this operation: is there a subspace Wof some vector space such that there is a subspace Uwith the property that UWequals the identity element from the prior item? 138 Chapter Two. Vector Spaces Topic: Fields Linear combinations involving only fractions or only integers are much easier for computations than combinations involving real numbers, because computing with irrational numbers is awkward. Could other number systems, like the rationals or the integers, work in the place of Rin the denition of a vector space? Yes and no. If we take \work" to mean that the results of this chapter remain true then an analysis of which properties of the reals we have used in this chapter gives the following list of conditions an algebraic system needs in order to \work" in the place of R. Denition. Aeld is a setFwith two operations `+' and `
' such that (1) for any a;b2F the result of a+bis inFand a+b=b+a ifc2F thena+ (b+c) = (a+b) +c (2) for any a;b2F the result of a
bis inFand a
b=b
a ifc2F thena
(b
c) = (a
b)
c (3) ifa;b;c2F thena
(b+c) =a
b+a
c (4) there is an element 0 2F such that ifa2F thena+ 0 =a for eacha2F there is an element a2F such that (a) +a= 0 (5) there is an element 1 2F such that ifa2F thena
1 =a for each element a6= 0 ofFthere is an element a12F such that a1
a= 1. The number system consisting of the set of real numbers along with the usual addition and multiplication operation is a eld, naturally. Another eld is the set of rational numbers with its usual addition and multiplication operations. An example of an algebraic structure that is not a eld is the integer number system|it fails the nal condition. Some examples are surprising. The set f0;1gunder these operations: +0 1 00 1 11 0
0 1 00 0 10 1 is a eld (see Exercise 4). Topic: Fields 139 We could develop Linear Algebra as the theory of vector spaces with scalars from an arbitrary eld, instead of sticking to taking the scalars only from R. In that case, almost all of the statements in this book would carry over by replacing `R' with `F', and thus by taking coecients, vector entries, and matrix entries to be elements of F(\almost" because statements involving distances or angles are exceptions). Here are some examples; each applies to a vector space Vover a eldF. For any~ v2Vanda2F, (i) 0
~ v=~0, and (ii)1
~ v+~ v=~0, and (iii)a
~0 =~0. The span (the set of linear combinations) of a subset of Vis a subspace ofV. Any subset of a linearly independent set is also linearly independent. In a nite-dimensional vector space, any two bases have the same number of elements. (Even statements that don't explicitly mention Fuse eld properties in their proof.) We won't develop vector spaces in this more general setting because the additional abstraction can be a distraction. The ideas we want to bring out already appear when we stick to the reals. The only exception is in Chapter Five. In that chapter we must factor polynomials, so we will switch to considering vector spaces over the eld of complex numbers. We will discuss this more, including a brief review of complex arithmetic, when we get there. Exercises 1Show that the real numbers form a eld. 2Prove that these are elds. (a)The rational numbers Q(b)The complex numbers C 3Give an example that shows that the integer number system is not a eld. 4Consider the set f0;1gsubject to the operations given above. Show that it is a eld. 5Give suitable operations to make the set f0;1;2ga eld. 140 Chapter Two. Vector Spaces Topic: Crystals Everyone has noticed that table salt comes in little cubes. Remarkably, the explanation for the cubical external shape is the simplest one: the internal shape, the way the atoms lie, is also cubical. The internal structure is pictured below. Salt is sodium cloride, and the small spheres shown are sodium while the big ones are cloride. To simplify the view, it only shows the sodiums and clorides on the front, top, and right. The specks of salt that we see when we spread a little out on the table consist of many repetitions of this fundamental unit. That is, these cubes of atoms stack up to make the larger cubical structure that we see. A solid, such as table salt, with a regular internal structure is a crystal . We can restrict our attention to the front face. There, we have the square repeated many times. The distance between the corners of the square cell is about 3 :34Angstroms (anAngstrom is 1010meters). Obviously that unit is unwieldly. Instead, the thing to do is to take as a unit the length of each side of the square. That is, we naturally adopt this basis. h3:34 0 ;0 3:34 i Then we can describe, say, the corner in the upper right of the picture above as 3~1+ 2~2. Topic: Crystals 141 Another crystal from everyday experience is pencil lead. It is graphite, formed from carbon atoms arranged in this shape. This is a single plane of graphite. A piece of graphite consists of many of these planes layered in a stack. (The chemical bonds between the planes are much weaker than the bonds inside the planes, which explains why pencils write | the graphite can be sheared so that the planes slide o and are left on the paper.) We can get a convienent unit of length by decomposing the hexagonal ring into three regions that are rotations of this unit cell . Then a natural basis consists of the vectors that form the sides of that unit cell. The distance along the bottom and slant is 1 :42Angstroms, so this h 1:42 0 ; 1:23 :71 i is a good basis. The selection of convienent bases extends to three dimensions. Another familiar crystal formed from carbon is diamond. Like table salt, it is built from cubes, but the structure inside each cube is more complicated than salt's. In addition to carbons at each corner, there are carbons in the middle of each face. 142 Chapter Two. Vector Spaces (To show the added face carbons clearly, the corner carbons have been reduced to dots.) There are also four more carbons inside the cube, two that are a quarter of the way up from the bottom and two that are a quarter of the way down from the top. (As before, carbons shown earlier have been reduced here to dots.) The dis- tance along any edge of the cube is 2 :18Angstroms. Thus, a natural basis for describing the locations of the carbons, and the bonds between them, is this. h0 @2:18 0 01 A;0 @0 2:18 01 A;0 @0 0 2:181 Ai Even the few examples given here show that the structures of crystals is com- plicated enough that some organized system to give the locations of the atoms, and how they are chemically bound, is needed. One tool for that organization is a convienent basis. This application of bases is simple, but it shows a context where the idea arises naturally. The work in this chapter just takes this simple idea and develops it. Exercises 1How many fundamental regions are there in one face of a speck of salt? (With a ruler, we can estimate that face is a square that is 0 :1 cm on a side.) 2In the graphite picture, imagine that we are interested in a point 5 :67Angstroms over and 3:14Angstroms up from the origin. (a)Express that point in terms of the basis given for graphite. (b)How many hexagonal shapes away is this point from the origin? (c)Express that point in terms of a second basis, where the rst basis vector is the same, but the second is perpendicular to the rst (going up the plane) and of the same length. 3Give the locations of the atoms in the diamond cube both in terms of the basis, and in Angstroms. 4This illustrates how the dimensions of a unit cell could be computed from the shape in which a substance crystalizes ([Ebbing], p. 462). (a)Recall that there are 6 :0221023atoms in a mole (this is Avagadro's number). From that, and the fact that platinum has a mass of 195 :08 grams per mole, calculate the mass of each atom. (b)Platinum crystalizes in a face-centered cubic lattice with atoms at each lattice point, that is, it looks like the middle picture given above for the diamond crystal. Find the number of platinums per unit cell (hint: sum the fractions of platinums that are inside of a single cell). (c)From that, nd the mass of a unit cell. (d)Platinum crystal has a density of 21 :45 grams per cubic centimeter. From this, and the mass of a unit cell, calculate the volume of a unit cell. Topic: Crystals 143 (e)Find the length of each edge. (f)Describe a natural three-dimensional basis. 144 Chapter Two. Vector Spaces Topic: Voting Paradoxes Imagine that a Political Science class studying the American presidential pro- cess holds a mock election. Members of the class are asked to rank, from most preferred to least preferred, the nominees from the Democratic Party, the Re- publican Party, and the Third Party, and this is the result ( >means `is preferred to'). preference ordernumber with that preference Democrat>Republican >Third 5 Democrat>Third>Republican 4 Republican >Democrat>Third 2 Republican >Third>Democrat 8 Third>Democrat>Republican 8 Third>Republican >Democrat 2 total 29 What is the preference of the group as a whole? Overall, the group prefers the Democrat to the Republican by ve votes; seventeen voters ranked the Democrat above the Republican versus twelve the other way. And, overall, the group prefers the Republican to the Third's nomi- nee, fteen to fourteen. But, strangely enough, the group also prefers the Third to the Democrat, eighteen to eleven. Democrat Third Republican7 voters 1 voter5 voters This is an example of a voting paradox , specically, a majority cycle . Voting paradoxes are studied in part because of their implications for practi- cal politics. For instance, the instructor can manipulate the class into choosing the Democrat as the overall winner by rst asking the class to choose between the Republican and the Third, and then asking the class to choose between the winner of that contest, the Republican, and the Democrat. By similar manipu- lations, any of the other two candidates can be made to come out as the winner. (In this Topic we will stick to three-candidate elections, but similar results apply to larger elections.) Voting paradoxes are also studied simply because they are mathematically interesting. One interesting aspect is that the group's overall majority cycle occurs despite that each single voters's preference list is rational |in a straight- line order. That is, the majority cycle seems to arise in the aggregate, without being present in the elements of that aggregate, the preference lists. Recently, Topic: Voting Paradoxes 145 however, linear algebra has been used [Zwicker] to argue that a tendency toward cyclic preference is actually present in each voter's list, and that it surfaces when there is more adding of the tendency than cancelling. For this argument, abbreviating the choices as D,R, andT, we can describe how a voter with preference order D>R>T contributes to the above cycle. D T R1 voter 1 voter1 voter (The negative sign is here because the arrow describes Tas preferred to D, but this voter likes them the other way.) The descriptions for the other preference lists are in the table on page 147. Now, to conduct the election we linearly combine these descriptions; for instance, the Political Science mock election 5
D T R1 11 + 4
D T R1 11 +


+ 2
D T R1 11 yields the circular group preference shown earlier. Of course, taking linear combinations is linear algebra. The above cycle no- tation is suggestive but inconvienent, so we temporarily switch to using column vectors by starting at the Dand taking the numbers from the cycle in coun- terclockwise order. Thus, the mock election and a single D > R > T vote are represented in this way.0 @7 1 51 Aand0 @1 1 11 A We will decompose vote vectors into two parts, one cyclic and the other acyclic. For the rst part, we say that a vector is purely cyclic if it is in this subspace ofR3. C=f0 @k k k1 Ak2Rg=fk
0 @1 1 11 Ak2Rg For the second part, consider the subspace (see Exercise 6) of vectors that are perpendicular to all of the vectors in C. C?=f0 @c1 c2 c31 A0 @c1 c2 c31 A0 @k k k1 A= 0 for allk2Rg =f0 @c1 c2 c31 Ac1+c2+c3= 0g=fc20 @1 1 01 A+c30 @1 0 11 Ac2;c32Rg 146 Chapter Two. Vector Spaces (Read that aloud as \ Cperp".) So we are led to this basis for R3. h0 @1 1 11 A;0 @1 1 01 A;0 @1 0 11 Ai We can represent votes with respect to this basis, and thereby decompose them into a cyclic part and an acyclic part. (Note for readers who have covered the optional section in this chapter: that is, the space is the direct sum of CandC?.) For example, consider the D>R>T voter discussed above. The represen- tation in terms of the basis is easily found, c1c2c3=1 c1+c2 = 1 c1 +c3= 11+2! 1+3(1=2)2+3!c1c2c3=1 2c2+c3= 2 (3=2)c3= 1 so thatc1= 1=3,c2= 2=3, andc3= 2=3. Then 0 @1 1 11 A=1 3
0 @1 1 11 A+2 3
0 @1 1 01 A+2 3
0 @1 0 11 A=0 @1=3 1=3 1=31 A+0 @4=3 2=3 2=31 A gives the desired decomposition into a cyclic part and and an acyclic part. D T R1 11 =D T R1=3 1=31=3 +D T R4=3 2=32=3 Thus, this D > R > T voter's rational preference list can indeed be seen to have a cyclic part. TheT >R>D voter is opposite to the one just considered in that the ` >' symbols are reversed. This voter's decomposition D T R1 11 =D T R1=3 1=31=3 +D T R4=3 2=32=3 shows that these opposite preferences have decompositions that are opposite. We say that the rst voter has positive spin since the cycle part is with the direction we have chosen for the arrows, while the second voter's spin is negative. The fact that that these opposite voters cancel each other is re ected in the fact that their vote vectors add to zero. This suggests an alternate way to tally an election. We could rst cancel as many opposite preference lists as possible, and then determine the outcome by adding the remaining lists. The rows of the table below contain the three pairs of opposite preference lists. The columns group those pairs by spin. For instance, the rst row contains the two voters just considered. Topic: Voting Paradoxes 147 positive spin negative spin Democrat>Republican >Third D T R1 11 =D T R1=3 1=31=3 +D T R4=3 2=32=3Third>Republican >Democrat D T R1 11 =D T R1=3 1=31=3 +D T R4=3 2=32=3 Republican >Third>Democrat D T R1 11 =D T R1=3 1=31=3 +D T R2=3 2=34=3Democrat>Third>Republican D T R1 11 =D T R1=3 1=31=3 +D T R2=3 2=34=3 Third>Democrat>Republican D T R1 11 =D T R1=3 1=31=3 +D T R2=3 4=32=3Republican >Democrat>Third D T R1 11 =D T R1=3 1=31=3 +D T R2=3 4=32=3 If we conduct the election as just described then after the cancellation of as many opposite pairs of voters as possible, there will be left three sets of preference lists, one set from the rst row, one set from the second row, and one set from the third row. We will nish by proving that a voting paradox can happen only if the spins of these three sets are in the same direction. That is, for a voting paradox to occur, the three remaining sets must all come from the left of the table or all come from the right (see Exercise 3). This shows that there is some connection between the majority cycle and the decomposition that we are using|a voting paradox can happen only when the tendencies toward cyclic preference reinforce each other. For the proof, assume that opposite preference orders have been cancelled, and we are left with one set of preference lists from each of the three rows. Consider the sum of these three (here, the numbers a,b, andccould be positive, negative, or zero). D T Ra aa +D T Rb bb +D T Rc cc =D T Ra+b+c a+bcab+c A voting paradox occurs when the three numbers on the right, ab+cand a+bcanda+b+c, are all nonnegative or all nonpositive. On the left, at least two of the three numbers, aandbandc, are both nonnegative or both nonpositive. We can assume that they are aandb. That makes four cases: the cycle is nonnegative and aandbare nonnegative, the cycle is nonpositive and aandbare nonpositive, etc. We will do only the rst case, since the second is similar and the other two are also easy. So assume that the cycle is nonnegative and that aandbare nonnegative. The conditions 0 ab+cand 0a+b+cadd to give that 0 2c, which implies that cis also nonnegative, as desired. That ends the proof. This result says only that having all three spin in the same direction is a necessary condition for a majority cycle. It is not sucient; see Exercise 4. 148 Chapter Two. Vector Spaces Voting theory and associated topics are the subject of current research. There are many intriguing results, most notably the one produced by K. Arrow [Arrow], who won the Nobel Prize in part for this work, showing that no voting system is entirely fair (for a reasonable denition of \fair"). For more infor- mation, some good introductory articles are [Gardner, 1970], [Gardner, 1974], [Gardner, 1980], and [Neimi & Riker]. A quite readable recent book is [Taylor]. The long list of cases from recent American political history given in [Poundstone] show that manipulation of these paradoxes is routine in practice (and the author proposes a solution). This Topic is largely drawn from [Zwicker]. (Author's Note: I would like to thank Professor Zwicker for his kind and illuminating discussions.) Exercises 1Here is a reasonable way in which a voter could have a cyclic preference. Suppose that this voter ranks each candidate on each of three criteria. (a)Draw up a table with the rows labelled `Democrat', `Republican', and `Third', and the columns labelled `character', `experience', and `policies'. Inside each column, rank some candidate as most preferred, rank another as in the middle, and rank the remaining one as least preferred. (b)In this ranking, is the Democrat preferred to the Republican in (at least) two out of three criteria, or vice versa? Is the Republican preferred to the Third? (c)Does the table that was just constructed have a cyclic preference order? If not, make one that does. So it is possible for a voter to have a cyclic preference among candidates. The paradox described above, however, is that even if each voter has a straight-line preference list, a cyclic preference can still arise for the entire group. 2Compute the values in the table of decompositions. 3Do the cancellations of opposite preference orders for the Political Science class's mock election. Are all the remaining preferences from the left three rows of the table or from the right? 4The necessary condition that is proved above|a voting paradox can happen only if all three preference lists remaining after cancellation have the same spin|is not also sucient. (a)Continuing the positive cycle case considered in the proof, use the two in- equalities 0ab+cand 0a+b+cto show thatjabjc. (b)Also show that ca+b, and hence that jabjca+b. (c)Give an example of a vote where there is a majority cycle, and addition of one more voter with the same spin causes the cycle to go away. (d)Can the opposite happen; can addition of one voter with a \wrong" spin cause a cycle to appear? (e)Give a condition that is both necessary and sucient to get a majority cycle. 5A one-voter election cannot have a majority cycle because of the requirement that we've imposed that the voter's list must be rational. (a)Show that a two-voter election may have a majority cycle. (We consider the group preference a majority cycle if all three group totals are nonnegative or if all three are nonpositive|that is, we allow some zero's in the group preference.) (b)Show that for any number of voters greater than one, there is an election involving that many voters that results in a majority cycle. Topic: Voting Paradoxes 149 6LetUbe a subspace of R3. Prove that the set U?=f~ v~ v~ u= 0 for all~ u2Ug of vectors that are perpendicular to each vector in Uis also subspace of R3. Does this hold if Uis not a subspace? 150 Chapter Two. Vector Spaces Topic: Dimensional Analysis \You can't add apples and oranges," the old saying goes. It re ects our expe- rience that in applications the quantities have units and keeping track of those units is worthwhile. Everyone has done calculations such as this one that use the units as a check. 60sec min
60min hr
24hr day
365day year= 31 536 000sec year However, the idea of including the units can be taken beyond bookkeeping. It can be used to draw conclusions about what relationships are possible among the physical quantities. To start, consider the physics equation: distance = 16
(time)2. If the distance is in feet and the time is in seconds then this is a true statement about falling bodies. However it is not correct in other unit systems; for instance, it is not correct in the meter-second system. We can x that by making the 16 a dimensional constant . dist = 16ft sec2
(time)2 For instance, the above equation holds in the yard-second system. distance in yards = 16(1=3) yd sec2
(time in sec)2=16 3yd sec2
(time in sec)2 So our rst point is that by \including the units" we mean that we are restricting our attention to equations that use dimensional constants. By using dimensional constants, we can be vague about units and say only that all quantities are measured in combinations of some units of length L, massM, and timeT. We shall refer to these three as dimensions (these are the only three dimensions that we shall need in this Topic). For instance, velocity could be measured in feet =second or fathoms =hour, but in all events it involves some unit of length divided by some unit of time so the dimensional formula of velocity is L=T. Similarly, the dimensional formula of density is M=L3. We shall prefer using negative exponents over the fraction bars and we shall include the dimensions with a zero exponent, that is, we shall write the dimensional formula of velocity as L1M0T1and that of density as L3M1T0. In this context, \You can't add apples to oranges" becomes the advice to check that all of an equation's terms have the same dimensional formula. An ex- ample is this version of the falling body equation: dgt2= 0. The dimensional formula of the dterm isL1M0T0. For the other term, the dimensional for- mula ofgisL1M0T2(gis the dimensional constant given above as 16 ft =sec2) and the dimensional formula of tisL0M0T1, so that of the entire gt2term is L1M0T2(L0M0T1)2=L1M0T0. Thus the two terms have the same dimen- sional formula. An equation with this property is dimensionally homogeneous . Quantities with dimensional formula L0M0T0aredimensionless . For ex- ample, we measure an angle by taking the ratio of the subtended arc to the radius Topic: Dimensional Analysis 151 rarc which is the ratio of a length to a length L1M0T0=L1M0T0and thus angles have the dimensional formula L0M0T0. The classic example of using the units for more than bookkeeping, using them to draw conclusions, considers the formula for the period of a pendulum. p= {some expression involving the length of the string, etc.{ The period is in units of time L0M0T1. So the quantities on the other side of the equation must have dimensional formulas that combine in such a way that theirL's andM's cancel and only a single Tremains. The table on page 152 has the quantities that an experienced investigator would consider possibly relevant. The only dimensional formulas involving Lare for the length of the string and the acceleration due to gravity. For the L's of these two to cancel, when they appear in the equation they must be in ratio, e.g., as ( `=g)2, or as cos(`=g), or as (`=g)1. Therefore the period is a function of `=g. This is a remarkable result: with a pencil and paper analysis, before we ever took out the pendulum and made measurements, we have determined something about the relationship among the quantities. To do dimensional analysis systematically, we need to know two things (ar- guments for these are in [Bridgman], Chapter II and IV). The rst is that each equation relating physical quantities that we shall see involves a sum of terms, where each term has the form mp1 1mp2 2


mpk k for numbers m1, . . . ,mkthat measure the quantities. For the second, observe that an easy way to construct a dimensionally ho- mogeneous expression is by taking a product of dimensionless quantities or by adding such dimensionless terms. Buckingham's Theorem states that any complete relationship among quantities with dimensional formulas can be alge- braically manipulated into a form where there is some function fsuch that f(1;:::; n) = 0 for a complete set f1;:::; ngof dimensionless products. (The rst example below describes what makes a set of dimensionless products `complete'.) We usually want to express one of the quantities, m1for instance, in terms of the others, and for that we will assume that the above equality can be rewritten m1=mp2 2


mpk k
^f(2;:::; n) where  1=m1mp2 2


mpk kis dimensionless and the products  2, . . . , ndon't involvem1(as withf, here ^fis just some function, this time of n1 arguments). Thus, to do dimensional analysis we should nd which dimensionless products are possible. For example, consider again the formula for a pendulum's period. 152 Chapter Two. Vector Spaces quantitydimensional formula periodpL0M0T1 length of string `L1M0T0 mass of bob mL0M1T0 acceleration due to gravity gL1M0T2 arc of swing L0M0T0 By the rst fact cited above, we expect the formula to have (possibly sums of terms of) the form pp1`p2mp3gp4p5. To use the second fact, to nd which combinations of the powers p1, . . . ,p5yield dimensionless products, consider this equation. (L0M0T1)p1(L1M0T0)p2(L0M1T0)p3(L1M0T2)p4(L0M0T0)p5=L0M0T0 It gives three conditions on the powers. p2+p4= 0 p3 = 0 p12p4= 0 Note thatp3is 0 and so the mass of the bob does not aect the period. Gaussian reduction and parametrization of that system gives this f0 BBBB@p1 p2 p3 p4 p51 CCCCA=0 BBBB@1 1=2 0 1=2 01 CCCCAp1+0 BBBB@0 0 0 0 11 CCCCAp5p1;p52Rg (we've taken p1as one of the parameters in order to express the period in terms of the other quantities). Here is the linear algebra. The set of dimensionless products contains all termspp1`p2mp3ap4p5subject to the conditions above. This set forms a vector space under the `+' operation of multiplying two such products and the `
' operation of raising such a product to the power of the scalar (see Exercise 5). The term `complete set of dimensionless products' in Buckingham's Theorem means a basis for this vector space. We can get a basis by rst taking p1= 1,p5= 0 and then p1= 0,p5= 1. The associated dimensionless products are  1=p`1=2g1=2and  2=. Because the setf1;2gis complete, Buckingham's Theorem says that p=`1=2g1=2
^f() =p `=g
^f() where ^fis a function that we cannot determine from this analysis (a rst year physics text will show by other means that for small angles it is approximately the constant function ^f() = 2). Topic: Dimensional Analysis 153 Thus, analysis of the relationships that are possible between the quantities with the given dimensional formulas has produced a fair amount of informa- tion: a pendulum's period does not depend on the mass of the bob, and it rises with the square root of the length of the string. For the next example we try to determine the period of revolution of two bodies in space orbiting each other under mutual gravitational attraction. An experienced investigator could expect that these are the relevant quantities. quantitydimensional formula periodpL0M0T1 mean separation rL1M0T0 rst massm1L0M1T0 second mass m2L0M1T0 grav. constant GL3M1T2 To get the complete set of dimensionless products we consider the equation (L0M0T1)p1(L1M0T0)p2(L0M1T0)p3(L0M1T0)p4(L3M1T2)p5=L0M0T0 which results in a system p2 + 3p5= 0 p3+p4p5= 0 p12p5= 0 with this solution. f0 BBBB@1 3=2 1=2 0 1=21 CCCCAp1+0 BBBB@0 0 1 1 01 CCCCAp4p1;p42Rg As earlier, the linear algebra here is that the set of dimensionless prod- ucts of these quantities forms a vector space, and we want to produce a basis for that space, a `complete' set of dimensionless products. One such set, got- ten from setting p1= 1 andp4= 0, and also setting p1= 0 andp4= 1 isf1=pr3=2m1=2 1G1=2;2=m1 1m2g. With that, Buckingham's Theorem says that any complete relationship among these quantities is stateable this form. p=r3=2m1=2 1G1=2
^f(m1 1m2) =r3=2 pGm1
^f(m2=m1) Remark. An important application of the prior formula is when m1is the mass of the sun and m2is the mass of a planet. Because m1is very much greater thanm2, the argument to ^fis approximately 0, and we can wonder whether this part of the formula remains approximately constant as m2varies. One way to see that it does is this. The sun is so much larger than the planet that the 154 Chapter Two. Vector Spaces mutual rotation is approximately about the sun's center. If we vary the planet's massm2by a factor of x(e.g., Venus's mass is x= 0:815 times Earth's mass), then the force of attraction is multiplied by x, andxtimes the force acting on xtimes the mass gives, since F=ma, the same acceleration, about the same center (approximately). Hence, the orbit will be the same and so its period will be the same, and thus the right side of the above equation also remains unchanged (approximately). Therefore, ^f(m2=m1) is approximately constant asm2varies. This is Kepler's Third Law: the square of the period of a planet is proportional to the cube of the mean radius of its orbit about the sun. The nal example was one of the rst explicit applications of dimensional analysis. Lord Raleigh considered the speed of a wave in deep water and sug- gested these as the relevant quantities. quantitydimensional formula velocity of the wave vL1M0T1 density of the water dL3M1T0 acceleration due to gravity gL1M0T2 wavelength L1M0T0 The equation (L1M0T1)p1(L3M1T0)p2(L1M0T2)p3(L1M0T0)p4=L0M0T0 gives this system p13p2+p3+p4= 0 p2 = 0 p12p3 = 0 with this solution space f0 BB@1 0 1=2 1=21 CCAp1p12Rg (as in the pendulum example, one of the quantities dturns out not to be involved in the relationship). There is one dimensionless product,  1=vg1=21=2, and sovispgtimes a constant ( ^fis constant since it is a function of no arguments). As the three examples above show, dimensional analysis can bring us far toward expressing the relationship among the quantities. For further reading, the classic reference is [Bridgman]|this brief book is delightful. Another source is [Giordano, Wells, Wilde]. A description of dimensional analysis's place in modeling is in [Giordano, Jaye, Weir]. Exercises 1Consider a projectile, launched with initial velocity v0, at an angle . An in- vestigation of this motion might start with the guess that these are the relevant Topic: Dimensional Analysis 155 quantities. [de Mestre] quantitydimensional formula horizontal position xL1M0T0 vertical position yL1M0T0 initial speed v0L1M0T1 angle of launch L0M0T0 acceleration due to gravity gL1M0T2 timetL0M0T1 (a)Show thatfgt=v 0;gx=v2 0;gy=v2 0;gis a complete set of dimensionless prod- ucts. ( Hint. This can be done by nding the appropriate free variables in the linear system that arises, but there is a shortcut that uses the properties of a basis.) (b)These two equations of motion for projectiles are familiar: x=v0cos()tand y=v0sin()t(g=2)t2. Manipulate each to rewrite it as a relationship among the dimensionless products of the prior item. 2[Einstein] conjectured that the infrared characteristic frequencies of a solid may be determined by the same forces between atoms as determine the solid's ordanary elastic behavior. The relevant quantities are quantitydimensional formula characteristic frequency L0M0T1 compressibility kL1M1T2 number of atoms per cubic cm NL3M0T0 mass of an atom mL0M1T0 Show that there is one dimensionless product. Conclude that, in any complete relationship among quantities with these dimensional formulas, kis a constant times2N1=3m1. This conclusion played an important role in the early study of quantum phenomena. 3The torque produced by an engine has dimensional formula L2M1T2. We may rst guess that it depends on the engine's rotation rate (with dimensional formula L0M0T1), and the volume of air displaced (with dimensional formula L3M0T0). [Giordano, Wells, Wilde] (a)Try to nd a complete set of dimensionless products. What goes wrong? (b)Adjust the guess by adding the density of the air (with dimensional formula L3M1T0). Now nd a complete set of dimensionless products. 4Dominoes falling make a wave. We may conjecture that the wave speed vdepends on the the spacing dbetween the dominoes, the height hof each domino, and the acceleration due to gravity g. [Tilley] (a)Find the dimensional formula for each of the four quantities. (b)Show thatf1=h=d;2=dg=v2gis a complete set of dimensionless prod- ucts. (c)Show that if h=dis xed then the propagation speed is proportional to the square root of d. 5Prove that the dimensionless products form a vector space under the ~+ operation of multiplying two such products and the ~
operation of raising such the product to the power of the scalar. (The vector arrows are a precaution against confusion.) That is, prove that, for any particular homogeneous system, this set of products 156 Chapter Two. Vector Spaces of powers of m1, . . . ,mk fmp1 1:::mpk kp1, . . . ,pksatisfy the system g is a vector space under: mp1 1:::mpk k~+mq1 1:::mqk k=mp1+q1 1:::mpk+qk k and r~
(mp1 1:::mpk k) =mrp1 1:::mrpk k (assume that all variables represent real numbers). 6The advice about apples and oranges is not right. Consider the familiar equations for a circle C= 2randA=r2. (a)Check that CandAhave dierent dimensional formulas. (b)Produce an equation that is not dimensionally homogeneous (i.e., it adds apples and oranges) but is nonetheless true of any circle. (c)The prior item asks for an equation that is complete but not dimensionally homogeneous. Produce an equation that is dimensionally homogeneous but not complete. (Just because the old saying isn't strictly right, doesn't keep it from being a useful strategy. Dimensional homogeneity is often used as a check on the plausibility of equations used in models. For an argument that any complete equation can easily be made dimensionally homogeneous, see [Bridgman], Chapter I, especially page 15.) Chapter Three Maps Between Spaces I Isomorphisms In the examples following the denition of a vector space we developed the intuition that some spaces are \the same" as others. For instance, the space of two-tall column vectors and the space of two-wide row vectors are not equal because their elements | column vectors and row vectors | are not equal, but we have the idea that these spaces dier only in how their elements appear. We will now make this idea precise. This section illustrates a common aspect of a mathematical investigation. With the help of some examples, we've gotten an idea. We will next give a formal denition, and then we will produce some results backing our contention that the denition captures the idea. We've seen this happen already, for instance, in the rst section of the Vector Space chapter. There, the study of linear systems led us to consider collections closed under linear combinations. We dened such a collection as a vector space, and we followed it with some supporting results. Of course, that denition wasn't an end point, instead it led to new insights such as the idea of a basis. Here too, after producing a denition, and supporting it, we will get two surprises (pleasant ones). First, we will nd that the denition applies to some unforeseen, and interesting, cases. Second, the study of the denition will lead to new ideas. In this way, our investigation will build a momentum. I.1 Definition and Examples We start with two examples that suggest the right denition. 1.1 Example Consider the example mentioned above, the space of two-wide row vectors and the space of two-tall column vectors. They are \the same" in that if we associate the vectors that have the same components, e.g., 1 2
! 1 2 157 158 Chapter Three. Maps Between Spaces then this correspondence preserves the operations, for instance this addition 1 2
+3 4
=4 6
! 1 2 + 3 4 = 4 6 and this scalar multiplication. 5
1 2
=5 10
! 5
1 2 =5 10 More generally stated, under the correspondence a0a1
!a0 a1 both operations are preserved: a0a1
+b0b1
=a0+b0a1+b1
!a0 a1 +b0 b1 =a0+b0 a1+b1 and r
a0a1
=ra0ra1
!r
 a0 a1 = ra0 ra1 (all of the variables are real numbers). 1.2 Example Another two spaces we can think of as \the same" are P2, the space of quadratic polynomials, and R3. A natural correspondence is this. a0+a1x+a2x2 !0 @a0 a1 a21 A (e.g., 1 + 2x+ 3x2 !0 @1 2 31 A) The structure is preserved: corresponding elements add in a corresponding way a0+a1x+a2x2 +b0+b1x+b2x2 (a0+b0) + (a1+b1)x+ (a2+b2)x2 !0 @a0 a1 a21 A+0 @b0 b1 b21 A=0 @a0+b0 a1+b1 a2+b21 A and scalar multiplication corresponds also. r
(a0+a1x+a2x2) = (ra0) + (ra1)x+ (ra2)x2 !r
0 @a0 a1 a21 A=0 @ra0 ra1 ra21 A Section I. Isomorphisms 159 1.3 Denition Anisomorphism between two vector spaces VandWis a mapf:V!Wthat (1) is a correspondence: fis one-to-one and onto; (2)preserves structure: if~ v1;~ v22Vthen f(~ v1+~ v2) =f(~ v1) +f(~ v2) and if~ v2Vandr2Rthen f(r~ v) =rf(~ v) (we writeV=W, read \Vis isomorphic to W", when such a map exists). (\Morphism" means map, so \isomorphism" means a map expressing sameness.) 1.4 Example The vector space G=fc1cos+c2sinc1;c22Rgof func- tions ofis isomorphic to the vector space R2under this map. c1cos+c2sinf7!c1 c2 We will check this by going through the conditions in the denition. We will rst verify condition (1), that the map is a correspondence between the sets underlying the spaces. To establish that fis one-to-one, we must prove that f(~ a) =f(~b) only when ~ a=~b. If f(a1cos+a2sin) =f(b1cos+b2sin) then, by the denition of f, a1 a2 = b1 b2 from which we can conclude that a1=b1anda2=b2because column vectors are equal only when they have equal components. We've proved that f(~ a) =f(~b) implies that ~ a=~b, which shows that fis one-to-one. To check that fis onto we must check that any member of the codomain R2 is the image of some member of the domain G. But that's clear | any x y 2R2 is the image under fofxcos+ysin2G. Next we will verify condition (2), that fpreserves structure. More information on one-to-one and onto maps is in the appendix. 160 Chapter Three. Maps Between Spaces This computation shows that fpreserves addition. f (a1cos+a2sin) + (b1cos+b2sin)
=f (a1+b1) cos+ (a2+b2) sin
=a1+b1 a2+b2 = a1 a2 + b1 b2 =f(a1cos+a2sin) +f(b1cos+b2sin) A similar computation shows that fpreserves scalar multiplication. f r
(a1cos+a2sin)
=f(ra1cos+ra2sin) =ra1 ra2 =r
a1 a2 =r
f(a1cos+a2sin) With that, conditions (1) and (2) are veried, so we know that fis an isomorphism and we can say that the spaces are isomorphic G=R2. 1.5 Example LetVbe the spacefc1x+c2y+c3zc1;c2;c32Rgof linear combinations of three variables x,y, andz, under the natural addition and scalar multiplication operations. Then Vis isomorphic to P2, the space of quadratic polynomials. To show this we will produce an isomorphism map. There is more than one possibility; for instance, here are four. c1x+c2y+c3zf17!c1+c2x+c3x2 f27!c2+c3x+c1x2 f37! c1c2xc3x2 f47!c1+ (c1+c2)x+ (c1+c3)x2 The rst map is the more natural correspondence in that it just carries the coecients over. However, below we shall verify that the second one is an iso- morphism, to underline that there are isomorphisms other than just the obvious one (showing that f1is an isomorphism is Exercise 12). To show that f2is one-to-one, we will prove that if f2(c1x+c2y+c3z) = f2(d1x+d2y+d3z) thenc1x+c2y+c3z=d1x+d2y+d3z. The assumption thatf2(c1x+c2y+c3z) =f2(d1x+d2y+d3z) gives, by the denition of f2, that c2+c3x+c1x2=d2+d3x+d1x2. Equal polynomials have equal coecients, so c2=d2,c3=d3, andc1=d1. Thusf2(c1x+c2y+c3z) =f2(d1x+d2y+d3z) implies that c1x+c2y+c3z=d1x+d2y+d3zand therefore f2is one-to-one. Section I. Isomorphisms 161 The mapf2is onto because any member a+bx+cx2of the codomain is the image of some member of the domain, namely it is the image of cx+ay+bz. For instance, 2 + 3 x4x2isf2(4x+ 2y+ 3z). The computations for structure preservation are like those in the prior ex- ample. This map preserves addition f2 (c1x+c2y+c3z) + (d1x+d2y+d3z)
=f2 (c1+d1)x+ (c2+d2)y+ (c3+d3)z
= (c2+d2) + (c3+d3)x+ (c1+d1)x2 = (c2+c3x+c1x2) + (d2+d3x+d1x2) =f2(c1x+c2y+c3z) +f2(d1x+d2y+d3z) and scalar multiplication. f2 r
(c1x+c2y+c3z)
=f2(rc1x+rc2y+rc3z) =rc2+rc3x+rc1x2 =r
(c2+c3x+c1x2) =r
f2(c1x+c2y+c3z) Thusf2is an isomorphism and we write V=P2. We are sometimes interested in an isomorphism of a space with itself, called anautomorphism . An identity map is an automorphism. The next two examples show that there are others. 1.6 Example Adilation mapds:R2!R2that multiplies all vectors by a nonzero scalar sis an automorphism of R2. ~ u ~ vd1:5(~ u) d1:5(~ v)d1:5! Arotation orturning map t:R2!R2that rotates all vectors through an angle is an automorphism. ~ ut=6(~ u)t=6! A third type of automorphism of R2is a mapf`:R2!R2that ips orre ects all vectors over a line `through the origin. 162 Chapter Three. Maps Between Spaces ~ uf`(~ u) f`! See Exercise 29. 1.7 Example Consider the space P5of polynomials of degree 5 or less and the mapfthat sends a polynomial p(x) top(x1). For instance, under this map x27!(x1)2=x22x+1 andx3+2x7!(x1)3+2(x1) =x33x2+5x3. This map is an automorphism of this space; the check is Exercise 21. This isomorphism of P5with itself does more than just tell us that the space is \the same" as itself. It gives us some insight into the space's structure. For instance, below is shown a family of parabolas, graphs of members of P5. Each has a vertex at y=1, and the left-most one has zeroes at 2:25 and1:75, the next one has zeroes at 1:25 and0:75, etc. p0p1 Geometrically, the substitution of x1 forxin any function's argument shifts its graph to the right by one. Thus, f(p0) =p1andf's action is to shift all of the parabolas to the right by one. Notice that the picture before fis applied is the same as the picture after fis applied, because while each parabola moves to the right, another one comes in from the left to take its place. This also holds true for cubics, etc. So the automorphism fgives us the insight that P5has a certain horizontal-homogeneity; this space looks the same near x= 1 as near x= 0. As described in the preamble to this section, we will next produce some results supporting the contention that the denition of isomorphism above cap- tures our intuition of vector spaces being the same. Of course the denition itself is persuasive: a vector space consists of two components, a set and some structure, and the denition simply requires that the sets correspond and that the structures correspond also. Also persuasive are the examples above. In particular, Example 1.1, which gives an isomorphism between the space of two-wide row vectors and the space of two-tall column vectors, dramatizes our intuition that isomorphic spaces are the same in all relevant respects. Sometimes people say, where V=W, that \Wis justV painted green" | any dierences are merely cosmetic. Further support for the denition, in case it is needed, is provided by the following results that, taken together, suggest that all the things of interest in a Section I. Isomorphisms 163 vector space correspond under an isomorphism. Since we studied vector spaces to study linear combinations, \of interest" means \pertaining to linear combina- tions". Not of interest is the way that the vectors are presented typographically (or their color!). As an example, although the denition of isomorphism doesn't explicitly say that the zero vectors must correspond, it is a consequence of that denition. 1.8 Lemma An isomorphism maps a zero vector to a zero vector. Proof .Wheref:V!Wis an isomorphism, x any ~ v2V. Thenf(~0V) = f(0
~ v) = 0
f(~ v) =~0W. QED The denition of isomorphism requires that sums of two vectors correspond and that so do scalar multiples. We can extend that to say that all linear combinations correspond. 1.9 Lemma For any map f:V!Wbetween vector spaces these statements are equivalent. (1)fpreserves structure f(~ v1+~ v2) =f(~ v1) +f(~ v2) andf(c~ v) =cf(~ v) (2)fpreserves linear combinations of two vectors f(c1~ v1+c2~ v2) =c1f(~ v1) +c2f(~ v2) (3)fpreserves linear combinations of any nite number of vectors f(c1~ v1+


+cn~ vn) =c1f(~ v1) +


+cnf(~ vn) Proof .Since the implications (3) = )(2) and (2) =)(1) are clear, we need only show that (1) = )(3). Assume statement (1). We will prove statement (3) by induction on the number of summands n. The one-summand base case, that f(c~ v1) =cf(~ v1), is covered by the as- sumption of statement (1). For the inductive step assume that statement (3) holds whenever there are k or fewer summands, that is, whenever n= 1, orn= 2, . . . , or n=k. Consider thek+ 1-summand case. The rst half of (1) gives f(c1~ v1+


+ck~ vk+ck+1~ vk+1) =f(c1~ v1+


+ck~ vk) +f(ck+1~ vk+1) by breaking the sum along the nal `+'. Then the inductive hypothesis lets us break up the k-term sum. =f(c1~ v1) +


+f(ck~ vk) +f(ck+1~ vk+1) Finally, the second half of statement (1) gives =c1f(~ v1) +


+ckf(~ vk) +ck+1f(~ vk+1) when applied k+ 1 times. QED 164 Chapter Three. Maps Between Spaces In addition to adding to the intuition that the denition of isomorphism does indeed preserve the things of interest in a vector space, that lemma's second item is an especially handy way of checking that a map preserves structure. We close with a summary. The material in this section augments the chapter on Vector Spaces. There, after giving the denition of a vector space, we infor- mally looked at what dierent things can happen. Here, we dened the relation `=' between vector spaces and we have argued that it is the right way to split the collection of vector spaces into cases because it preserves the features of interest in a vector space | in particular, it preserves linear combinations. That is, we have now said precisely what we mean by `the same', and by `dierent', and so we have precisely classied the vector spaces. Exercises X1.10 Verify, using Example 1.4 as a model, that the two correspondences given before the denition are isomorphisms. (a)Example 1.1 (b)Example 1.2 X1.11 For the map f:P1!R2given by a+bxf7!ab b Find the image of each of these elements of the domain. (a)32x(b)2 + 2x(c)x Show that this map is an isomorphism. 1.12 Show that the natural map f1from Example 1.5 is an isomorphism. X1.13 Decide whether each map is an isomorphism (if it is an isomorphism then prove it and if it isn't then state a condition that it fails to satisfy). (a)f:M22!Rgiven bya b c d 7!adbc (b)f:M22!R4given by a b c d 7!0 BB@a+b+c+d a+b+c a+b a1 CCA (c)f:M22!P 3given bya b c d 7!c+ (d+c)x+ (b+a)x2+ax3 (d)f:M22!P 3given bya b c d 7!c+ (d+c)x+ (b+a+ 1)x2+ax3 1.14 Show that the map f:R1!R1given byf(x) =x3is one-to-one and onto. Is it an isomorphism? X1.15 Refer to Example 1.1. Produce two more isomorphisms (of course, you must also verify that they satisfy the conditions in the denition of isomorphism). 1.16 Refer to Example 1.2. Produce two more isomorphisms (and verify that they satisfy the conditions). Section I. Isomorphisms 165 X1.17 Show that, although R2is not itself a subspace of R3, it is isomorphic to the xy-plane subspace of R3. 1.18 Find two isomorphisms between R16andM44. X1.19 For whatkisMmnisomorphic to Rk? 1.20 For whatkisPkisomorphic to Rn? 1.21 Prove that the map in Example 1.7, from P5toP5given byp(x)7!p(x1), is a vector space isomorphism. 1.22 Why, in Lemma 1.8, must there be a ~ v2V? That is, why must Vbe nonempty? 1.23 Are any two trivial spaces isomorphic? 1.24 In the proof of Lemma 1.9, what about the zero-summands case (that is, if n is zero)? 1.25 Show that any isomorphism f:P0!R1has the form a7!kafor some nonzero real number k. X1.26 These prove that isomorphism is an equivalence relation. (a)Show that the identity map id: V!Vis an isomorphism. Thus, any vector space is isomorphic to itself. (b)Show that if f:V!Wis an isomorphism then so is its inverse f1:W!V. Thus, ifVis isomorphic to Wthen alsoWis isomorphic to V. (c)Show that a composition of isomorphisms is an isomorphism: if f:V!Wis an isomorphism and g:W!Uis an isomorphism then so also is gf:V!U. Thus, ifVis isomorphic to WandWis isomorphic to U, then alsoVis isomor- phic toU. 1.27 Suppose that f:V!Wpreserves structure. Show that fis one-to-one if and only if the unique member of Vmapped by fto~0Wis~0V. 1.28 Suppose that f:V!Wis an isomorphism. Prove that the set f~ v1;:::;~ vkg Vis linearly dependent if and only if the set of images ff(~ v1);:::;f (~ vk)gWis linearly dependent. X1.29 Show that each type of map from Example 1.6 is an automorphism. (a)Dilationdsby a nonzero scalar s. (b)Rotationtthrough an angle . (c)Re ectionf`over a line through the origin. Hint. For the second and third items, polar coordinates are useful. 1.30 Produce an automorphism of P2other than the identity map, and other than a shift map p(x)7!p(xk). 1.31 (a) Show that a function f:R1!R1is an automorphism if and only if it has the form x7!kxfor somek6= 0. (b)Letfbe an automorphism of R1such thatf(3) = 7. Find f(2). (c)Show that a function f:R2!R2is an automorphism if and only if it has the form x y 7!ax+by cx+dy for somea;b;c;d2Rwithadbc6= 0. Hint. Exercises in prior subsections have shown that b d is not a multiple ofa c if and only if adbc6= 0. 166 Chapter Three. Maps Between Spaces (d)Letfbe an automorphism of R2with f(1 3 ) =2 1 andf(1 4 ) =0 1 : Find f(0 1 ): 1.32 Refer to Lemma 1.8 and Lemma 1.9. Find two more things preserved by isomorphism. 1.33 We show that isomorphisms can be tailored to t in that, sometimes, given vectors in the domain and in the range we can produce an isomorphism associating those vectors. (a)LetB=h~1;~2;~3ibe a basis forP2so that any ~ p2P 2has a unique representation as ~ p=c1~1+c2~2+c3~3, which we denote in this way. RepB(~ p) =0 @c1 c2 c31 A Show that the RepB(
) operation is a function from P2toR3(this entails showing that with every domain vector ~ v2P 2there is an associated image vector in R3, and further, that with every domain vector ~ v2P 2there is at most one associated image vector). (b)Show that this RepB(
) function is one-to-one and onto. (c)Show that it preserves structure. (d)Produce an isomorphism from P2toR3that ts these specications. x+x27!0 @1 0 01 Aand 1x7!0 @0 1 01 A 1.34 Prove that a space is n-dimensional if and only if it is isomorphic to Rn. Hint. Fix a basis Bfor the space and consider the map sending a vector over to its representation with respect to B. 1.35 (Requires the subsection on Combining Subspaces, which is optional.) LetU andWbe vector spaces. Dene a new vector space, consisting of the set UW= f(~ u;~ w)~ u2Uand~ w2Wgalong with these operations. (~ u1;~ w1) + (~ u2;~ w2) = (~ u1+~ u2;~ w1+~ w2) andr
(~ u;~ w) = (r~ u;r~ w ) This is a vector space, the external direct sum ofUandW. (a)Check that it is a vector space. (b)Find a basis for, and the dimension of, the external direct sum P2R2. (c)What is the relationship among dim( U), dim(W), and dim( UW)? (d)Suppose that UandWare subspaces of a vector space Vsuch thatV= UW(in this case we say that Vis the internal direct sum ofUandW). Show that the map f:UW!Vgiven by (~ u;~ w)f7!~ u+~ w is an isomorphism. Thus if the internal direct sum is dened then the internal and external direct sums are isomorphic. Section I. Isomorphisms 167 I.2 Dimension Characterizes Isomorphism In the prior subsection, after stating the denition of an isomorphism, we gave some results supporting the intuition that such a map describes spaces as \the same". Here we will formalize this intuition. While two spaces that are isomorphic are not equal, we think of them as almost equal | as equivalent. In this subsection we shall show that the relationship `is isomorphic to' is an equivalence relation. 2.1 Theorem Isomorphism is an equivalence relation between vector spaces. Proof .We must prove that this relation has the three properties of being sym- metric, re exive, and transitive. For each of the three we will use item (2) of Lemma 1.9 and show that the map preserves structure by showing that it preserves linear combinations of two members of the domain. To check re exivity, that any space is isomorphic to itself, consider the iden- tity map. It is clearly one-to-one and onto. The calculation showing that it preserves linear combinations is easy. id(c1
~ v1+c2
~ v2) =c1~ v1+c2~ v2=c1
id(~ v1) +c2
id(~ v2) To check symmetry, that if Vis isomorphic to Wvia some map f:V!W then there is an isomorphism going the other way, consider the inverse map f1:W!V. As stated in the appendix, such an inverse function exists and it is also a correspondence. Thus we have reduced the symmetry issue to checking that, because fpreserves linear combinations, so also does f1. Assume that ~ w1=f(~ v1) and~ w2=f(~ v2), i.e., that f1(~ w1) =~ v1andf1(~ w2) =~ v2. f1(c1
~ w1+c2
~ w2) =f1 c1
f(~ v1) +c2
f(~ v2)
=f1(f c1~ v1+c2~ v2)
=c1~ v1+c2~ v2 =c1
f1(~ w1) +c2
f1(~ w2) Finally, we must check transitivity, that if Vis isomorphic to Wvia some mapfand ifWis isomorphic to Uvia some map gthen alsoVis isomorphic toU. Consider the composition gf:V!U. The appendix notes that the composition of two correspondences is a correspondence, so we need only check that the composition preserves linear combinations. gf c1
~ v1+c2
~ v2
=g f(c1
~ v1+c2
~ v2)
=g c1
f(~ v1) +c2
f(~ v2)
=c1
g f(~ v1)) +c2
g(f(~ v2)
=c1
(gf) (~ v1) +c2
(gf) (~ v2) Thusgf:V!Uis an isomorphism. QED More information on equivalence relations and equivalence classes is in the appendix. 168 Chapter Three. Maps Between Spaces As a consequence of that result, we know that the universe of vector spaces is partitioned into classes: every space is in one and only one isomorphism class. All nite dimensional vector spaces:. . .V WV=W 2.2 Theorem Vector spaces are isomorphic if and only if they have the same dimension. This follows from the next two lemmas. 2.3 Lemma If spaces are isomorphic then they have the same dimension. Proof .We shall show that an isomorphism of two spaces gives a correspondence between their bases. That is, where f:V!Wis an isomorphism and a basis for the domain VisB=h~1;:::;~ni, then the image set D=hf(~1);:::;f (~n)i is a basis for the codomain W. (The other half of the correspondence | that for any basis of Wthe inverse image is a basis for V| follows on recalling that iffis an isomorphism then f1is also an isomorphism, and applying the prior sentence to f1.) To see that DspansW, x any~ w2W, note that fis onto and so there is a~ v2Vwith~ w=f(~ v), and expand ~ vas a combination of basis vectors. ~ w=f(~ v) =f(v1~1+


+vn~n) =v1
f(~1) +


+vn
f(~n) For linear independence of D, if ~0W=c1f(~1) +


+cnf(~n) =f(c1~1+


+cn~n) then, since fis one-to-one and so the only vector sent to ~0Wis~0V, we have that~0V=c1~1+


+cn~n, implying that all of the c's are zero. QED 2.4 Lemma If spaces have the same dimension then they are isomorphic. Proof .To show that any two spaces of dimension nare isomorphic, we can simply show that any one is isomorphic to Rn. Then we will have shown that they are isomorphic to each other, by the transitivity of isomorphism (which was established in Theorem 2.1). LetVben-dimensional. Fix a basis B=h~1;:::;~nifor the domain V. Consider the representation of the members of that domain with respect to the basis as a function from VtoRn ~ v=v1~1+


+vn~nRepB7!0 B@v1 ... vn1 CA Section I. Isomorphisms 169 (it is well-denedsince every ~ vhas one and only one such representation | see Remark 2.5 below). This function is one-to-one because if RepB(u1~1+


+un~n) = RepB(v1~1+


+vn~n) then 0 B@u1 ... un1 CA=0 B@v1 ... vn1 CA and sou1=v1, . . . ,un=vn, and therefore the original arguments u1~1+


+ un~nandv1~1+


+vn~nare equal. This function is onto; any n-tall vector ~ w=0 B@w1 ... wn1 CA is the image of some ~ v2V, namely~ w= RepB(w1~1+


+wn~n). Finally, this function preserves structure. RepB(r
~ u+s
~ v) = RepB( (ru1+sv1)~1+


+ (run+svn)~n) =0 B@ru1+sv1 ... run+svn1 CA =r
0 B@u1 ... un1 CA+s
0 B@v1 ... vn1 CA =r
RepB(~ u) +s
RepB(~ v) Thus the RepBfunction is an isomorphism and thus any n-dimensional space is isomorphic to the n-dimensional space Rn. Consequently, any two spaces with the same dimension are isomorphic. QED 2.5 Remark The parenthetical comment in that proof about the role played by the `one and only one representation' result requires some explanation. We need to show that (for a xed B) each vector in the domain is associated by RepBwith one and only one vector in the codomain. A contrasting example, where an association doesn't have this property, is illuminating. Consider this subset of P2, which is not a basis. A=f1 + 0x+ 0x2;0 + 1x+ 0x2;0 + 0x+ 1x2;1 + 1x+ 2x2g More information on well-denedness is in the appendix. 170 Chapter Three. Maps Between Spaces Call those four polynomials ~ 1, . . . ,~ 4. If, mimicing above proof, we try to write the members of P2as~ p=c1~ 1+c2~ 2+c3~ 3+c4~ 4, and associate ~ pwith the four-tall vector with components c1, . . . ,c4then there is a problem. For, consider~ p(x) = 1 +x+x2. The setAspans the spaceP2, so there is at least one four-tall vector associated with ~ p. ButAis not linearly independent and so vectors do not have unique decompositions. In this case, both ~ p(x) = 1~ 1+ 1~ 2+ 1~ 3+ 0~ 4and~ p(x) = 0~ 1+ 0~ 21~ 3+ 1~ 4 and so there is more than one four-tall vector associated with ~ p. 0 BB@1 1 1 01 CCAand0 BB@0 0 1 11 CCA That is, with input ~ pthis association does not have a well-dened (i.e., single) output value. Any map whose denition appears possibly ambiguous must be checked to see that it is well-dened. For RepBin the above proof that check is Exercise 18. That ends the proof of Theorem 2.2. We say that the isomorphism classes arecharacterized by dimension because we can describe each class simply by giving the number that is the dimension of all of the spaces in that class. This subsection's results give us a collection of representatives of the isomor- phism classes. 2.6 Corollary A nite-dimensional vector space is isomorphic to one and only one of the Rn. The proofs above pack many ideas into a small space. Through the rest of this chapter we'll consider these ideas again, and ll them out. For a taste of this, we will expand here on the proof of Lemma 2.4. 2.7 Example The spaceM22of 22 matrices is isomorphic to R4. With this basis for the domain B=h1 0 0 0 ;0 1 0 0 ;0 0 1 0 ;0 0 0 1 i the isomorphism given in the lemma, the representation map f1= RepB, simply carries the entries over. a b c d f17!0 BB@a b c d1 CCA One way to think of the map f1is: x the basis Bfor the domain and the basis E4for the codomain, and associate ~1with~ e1, and~2with~ e2, etc. Then extend Section I. Isomorphisms 171 this association to all of the members of two spaces. a b c d =a~1+b~2+c~3+d~4f17!a~ e1+b~ e2+c~ e3+d~ e4=0 BB@a b c d1 CCA We say that the map has been extended linearly from the bases to the spaces. We can do the same thing with dierent bases, for instance, taking this basis for the domain. A=h2 0 0 0 ;0 2 0 0 ;0 0 2 0 ;0 0 0 2 i Associating corresponding members of AandE4and extending linearly a b c d = (a=2)~ 1+ (b=2)~ 2+ (c=2)~ 3+ (d=2)~ 4 f27! (a=2)~ e1+ (b=2)~ e2+ (c=2)~ e3+ (d=2)~ e4=0 BB@a=2 b=2 c=2 d=21 CCA gives rise to an isomorphism that is dierent than f1. The prior map arose by changing the basis for the domain. We can also change the basis for the codomain. Starting with BandD=h0 BB@1 0 0 01 CCA;0 BB@0 1 0 01 CCA;0 BB@0 0 0 11 CCA;0 BB@0 0 1 01 CCAi associating ~1with~1, etc., and then linearly extending that correspondence to all of the two spaces a b c d =a~1+b~2+c~3+d~4f37!a~1+b~2+c~3+d~4=0 BB@a b d c1 CCA gives still another isomorphism. So there is a connection between the maps between spaces and bases for those spaces. Later sections will explore that connection. We will close this section with a summary. Recall that in the rst chapter we dened two matrices as row equivalent if they can be derived from each other by elementary row operations (this was the meaning of same-ness that was of interest there). We showed that is an 172 Chapter Three. Maps Between Spaces equivalence relation and so the collection of matrices is partitioned into classes, where all the matrices that are row equivalent fall together into a single class. Then, for insight into which matrices are in each class, we gave representatives for the classes, the reduced echelon form matrices. In this section, except that the appropriate notion of same-ness here is vector space isomorphism, we have followed much the same outline. First we dened isomorphism, saw some examples, and established some properties. Then we showed that it is an equivalence relation, and now we have a set of class repre- sentatives, the real vector spaces R1,R2, etc. All nite dimensional vector spaces:. . .?R2?R0?R3 ?R1One representative per class As before, the list of representatives helps us to understand the partition. It is simply a classication of spaces by dimension. In the second chapter, with the denition of vector spaces, we seemed to have opened up our studies to many examples of new structures besides the familiar Rn's. We now know that isn't the case. Any nite-dimensional vector space is actually \the same" as a real space. We are thus considering exactly the structures that we need to consider. The rest of the chapter lls out the work in this section. In particular, in the next section we will consider maps that preserve structure, but are not necessarily correspondences. Exercises X2.8Decide if the spaces are isomorphic. (a)R2,R4(b)P5,R5(c)M23,R6(d)P5,M23(e)M2k,Ck X2.9Consider the isomorphism RepB(
):P1!R2whereB=h1;1 +xi. Find the image of each of these elements of the domain. (a)32x;(b)2 + 2x;(c)x X2.10 Show that if m6=nthenRm6=Rn. X2.11 IsMmn=Mnm? X2.12 Are any two planes through the origin in R3isomorphic? 2.13 Find a set of equivalence class representatives other than the set of Rn's. 2.14 True or false: between any n-dimensional space and Rnthere is exactly one isomorphism. 2.15 Can a vector space be isomorphic to one of its (proper) subspaces? X2.16 This subsection shows that for any isomorphism, the inverse map is also an iso- morphism. This subsection also shows that for a xed basis Bof ann-dimensional vector space V, the map RepB:V!Rnis an isomorphism. Find the inverse of this map. X2.17 Prove these facts about matrices. (a)The row space of a matrix is isomorphic to the column space of its transpose. (b)The row space of a matrix is isomorphic to its column space. Section I. Isomorphisms 173 2.18 Show that the function from Theorem 2.2 is well-dened. 2.19 Is the proof of Theorem 2.2 valid when n= 0? 2.20 For each, decide if it is a set of isomorphism class representatives. (a)fCkk2Ng(b)fPkk2f 1;0;1;:::gg (c)fMmnm;n2Ng 2.21 Letfbe a correspondence between vector spaces VandW(that is, a map that is one-to-one and onto). Show that the spaces VandWare isomorphic via f if and only if there are bases BVandDWsuch that corresponding vectors have the same coordinates: RepB(~ v) = RepD(f(~ v)). 2.22 Consider the isomorphism RepB:P3!R4. (a)Vectors in a real space are orthogonal if and only if their dot product is zero. Give a denition of orthogonality for polynomials. (b)The derivative of a member of P3is inP3. Give a denition of the derivative of a vector in R4. X2.23 Does every correspondence between bases, when extended to the spaces, give an isomorphism? 2.24 (Requires the subsection on Combining Subspaces, which is optional.) Suppose thatV=V1V2and thatVis isomorphic to the space Uunder the map f. Show thatU=f(V1)f(U2). 2.25 Show that this is not a well-dened function from the rational numbers to the integers: with each fraction, associate the value of its numerator. 174 Chapter Three. Maps Between Spaces II Homomorphisms The denition of isomorphism has two conditions. In this section we will con- sider the second one, that the map must preserve the algebraic structure of the space. We will focus on this condition by studying maps that are required only to preserve structure; that is, maps that are not required to be correspondences. Experience shows that this kind of map is tremendously useful in the study of vector spaces. For one thing, as we shall see in the second subsection below, while isomorphisms describe how spaces are the same, these maps describe how spaces can be thought of as alike. II.1 Denition 1.1 Denition A function between vector spaces h:V!Wthat preserves the operations of addition if~ v1;~ v22Vthenh(~ v1+~ v2) =h(~ v1) +h(~ v2) and scalar multiplication if~ v2Vandr2Rthenh(r
~ v) =r
h(~ v) is ahomomorphism orlinear map . 1.2 Example The projection map :R3!R2 0 @x y z1 A7!x y is a homomorphism. It preserves addition (0 @x1 y1 z11 A+0 @x2 y2 z21 A) =(0 @x1+x2 y1+y2 z1+z21 A) =x1+x2 y1+y2 =(0 @x1 y1 z11 A) +(0 @x2 y2 z21 A) and scalar multiplication. (r
0 @x1 y1 z11 A) =(0 @rx1 ry1 rz11 A) =rx1 ry1 =r
(0 @x1 y1 z11 A) This map is not an isomorphism since it is not one-to-one. For instance, both ~0 and~ e3inR3are mapped to the zero vector in R2. Section II. Homomorphisms 175 1.3 Example Of course, the domain and codomain might be other than spaces of column vectors. Both of these are homomorphisms; the verications are straightforward. (1)f1:P2!P 3given by a0+a1x+a2x27!a0x+ (a1=2)x2+ (a2=3)x3 (2)f2:M22!Rgiven bya b c d 7!a+d 1.4 Example Between any two spaces there is a zero homomorphism , mapping every vector in the domain to the zero vector in the codomain. 1.5 Example These two suggest why we use the term `linear map'. (1) The map g:R3!Rgiven by 0 @x y z1 Ag7!3x+ 2y4:5z is linear (i.e., is a homomorphism). In contrast, the map ^ g:R3!Rgiven by0 @x y z1 A^g7!3x+ 2y4:5z+ 1 is not; for instance, ^g(0 @0 0 01 A+0 @1 0 01 A) = 4 while ^ g(0 @0 0 01 A) + ^g(0 @1 0 01 A) = 5 (to show that a map is not linear we need only produce one example of a linear combination that is not preserved). (2) The rst of these two maps t1;t2:R3!R2is linear while the second is not.0 @x y z1 At17!5x2y x+y and0 @x y z1 At27!5x2y xy Finding an example that the second fails to preserve structure is easy. What distinguishes the homomorphisms is that the coordinate functions are linear combinations of the arguments. See also Exercise 23. 176 Chapter Three. Maps Between Spaces Obviously, any isomorphism is a homomorphism | an isomorphism is a ho- momorphism that is also a correspondence. So, one way to think of the `ho- momorphism' idea is that it is a generalization of `isomorphism', motivated by the observation that many of the properties of isomorphisms have only to do with the map's structure preservation property and not to do with it being a correspondence. As examples, these two results from the prior section do not use one-to-one-ness or onto-ness in their proof, and therefore apply to any homomorphism. 1.6 Lemma A homomorphism sends a zero vector to a zero vector. 1.7 Lemma Each of these is a necessary and sucient condition for f:V!W to be a homomorphism. (1)f(c1
~ v1+c2
~ v2) =c1
f(~ v1) +c2
f(~ v2) for anyc1;c22Rand~ v1;~ v22V (2)f(c1
~ v1+


+cn
~ vn) =c1
f(~ v1) +


+cn
f(~ vn) for anyc1;:::;cn2R and~ v1;:::;~ vn2V Part (1) is often used to check that a function is linear. 1.8 Example The mapf:R2!R4given by x y f7!0 BB@x=2 0 x+y 3y1 CCA satises (1) of the prior result 0 BB@r1(x1=2) +r2(x2=2) 0 r1(x1+y1) +r2(x2+y2) r1(3y1) +r2(3y2)1 CCA=r10 BB@x1=2 0 x1+y1 3y11 CCA+r20 BB@x2=2 0 x2+y2 3y21 CCA and so it is a homomorphism. However, some of the results that we have seen for isomorphisms fail to hold for homomorphisms in general. Consider the theorem that an isomorphism be- tween spaces gives a correspondence between their bases. Homomorphisms do not give any such correspondence; Example 1.2 shows that there is no such cor- respondence, and another example is the zero map between any two nontrivial spaces. Instead, for homomorphisms a weaker but still very useful result holds. 1.9 Theorem A homomorphism is determined by its action on a basis. That is, ifh~1;:::;~niis a basis of a vector space Vand~ w1;:::;~ wnare (perhaps not distinct) elements of a vector space Wthen there exists a homomorphism fromVtoWsending~1to~ w1, . . . , and~nto~ wn, and that homomorphism is unique. Section II. Homomorphisms 177 Proof .We will dene the map by associating ~1with~ w1, etc., and then ex- tending linearly to all of the domain. That is, where ~ v=c1~1+


+cn~n, the maph:V!Wis given by h(~ v) =c1~ w1+


+cn~ wn. This is well-dened because, with respect to the basis, the representation of each domain vector ~ v is unique. This map is a homomorphism since it preserves linear combinations; where ~ v1=c1~1+


+cn~nand~ v2=d1~1+


+dn~n, we have this. h(r1~ v1+r2~ v2) =h((r1c1+r2d1)~1+


+ (r1cn+r2dn)~n) = (r1c1+r2d1)~ w1+


+ (r1cn+r2dn)~ wn =r1h(~ v1) +r2h(~ v2) And, this map is unique since if ^h:V!Wis another homomorphism such that ^h(~i) =~ wifor eachithenhand^hagree on all of the vectors in the domain. ^h(~ v) =^h(c1~1+


+cn~n) =c1^h(~1) +


+cn^h(~n) =c1~ w1+


+cn~ wn =h(~ v) Thus,hand^hare the same map. QED 1.10 Example This result says that we can construct a homomorphism by xing a basis for the domain and specifying where the map sends those basis vectors. For instance, if we specify a map h:R2!R2that acts on the standard basisE2in this way h(1 0 ) =1 1 andh(0 1 ) =4 4 then the action of hon any other member of the domain is also specied. For instance, the value of hon this argument h(3 2 ) =h(3
1 0 2
0 1 ) = 3
h(1 0 )2
h(0 1 ) =5 5 is a direct consequence of the value of hon the basis vectors. Later in this chapter we shall develop a scheme, using matrices, that is convienent for computations like this one. Just as the isomorphisms of a space with itself are useful and interesting, so too are the homomorphisms of a space with itself. 1.11 Denition A linear map from a space into itself t:V!Vis a linear transformation . 178 Chapter Three. Maps Between Spaces 1.12 Remark In this book we use `linear transformation' only in the case where the codomain equals the domain, but it is widely used in other texts as a general synonym for `homomorphism'. 1.13 Example The map on R2that projects all vectors down to the x-axis x y 7!x 0 is a linear transformation. 1.14 Example The derivative map d=dx :Pn!Pn a0+a1x+


+anxnd=dx7!a1+ 2a2x+ 3a3x2+


+nanxn1 is a linear transformation, as this result from calculus notes: d(c1f+c2g)=dx= c1(df=dx ) +c2(dg=dx ). 1.15 Example The matrix transpose map a b c d 7!a c b d is a linear transformation of M22. Note that this transformation is one-to-one and onto, and so in fact it is an automorphism. We nish this subsection about maps by recalling that we can linearly com- bine maps. For instance, for these maps from R2to itself  x y f7! 2x 3x2y and x y g7! 0 5x the linear combination 5 f2gis also a map from R2to itself. x y 5f2g7!10x 5x10y 1.16 Lemma For vector spaces VandW, the set of linear functions from V toWis itself a vector space, a subspace of the space of all functions from Vto W. It is denotedL(V;W ). Proof .This set is non-empty because it contains the zero homomorphism. So to show that it is a subspace we need only check that it is closed under linear combinations. Let f;g:V!Wbe linear. Then their sum is linear (f+g)(c1~ v1+c2~ v2) =f(c1~ v1+c2~ v2) +g(c1~ v1+c2~ v2) =c1f(~ v1) +c2f(~ v2) +c1g(~ v1) +c2g(~ v2) =c1 f+g
(~ v1) +c2 f+g
(~ v2) and any scalar multiple is also linear. (r
f)(c1~ v1+c2~ v2) =r(c1f(~ v1) +c2f(~ v2)) =c1(r
f)(~ v1) +c2(r
f)(~ v2) HenceL(V;W ) is a subspace. QED Section II. Homomorphisms 179 We started this section by isolating the structure preservation property of isomorphisms. That is, we dened homomorphisms as a generalization of iso- morphisms. Some of the properties that we studied for isomorphisms carried over unchanged, while others were adapted to this more general setting. It would be a mistake, though, to view this new notion of homomorphism as derived from, or somehow secondary to, that of isomorphism. In the rest of this chapter we shall work mostly with homomorphisms, partly because any state- ment made about homomorphisms is automatically true about isomorphisms, but more because, while the isomorphism concept is perhaps more natural, ex- perience shows that the homomorphism concept is actually more fruitful and more central to further progress. Exercises X1.17 Decide if each h:R3!R2is linear. (a)h(0 @x y z1 A) =x x+y+z (b)h(0 @x y z1 A) =0 0 (c)h(0 @x y z1 A) =1 1 (d)h(0 @x y z1 A) =2x+y 3y4z X1.18 Decide if each map h:M22!Ris linear. (a)h(a b c d ) =a+d (b)h(a b c d ) =adbc (c)h(a b c d ) = 2a+ 3b+cd (d)h(a b c d ) =a2+b2 X1.19 Show that these two maps are homomorphisms. (a)d=dx :P3!P 2given bya0+a1x+a2x2+a3x3maps toa1+ 2a2x+ 3a3x2 (b)R :P2!P 3given byb0+b1x+b2x2maps tob0x+ (b1=2)x2+ (b2=3)x3 Are these maps inverse to each other? 1.20 Is (perpendicular) projection from R3to thexz-plane a homomorphism? Pro- jection to the yz-plane? To the x-axis? The y-axis? The z-axis? Projection to the origin? 1.21 Show that, while the maps from Example 1.3 preserve linear operations, they are not isomorphisms. 1.22 Is an identity map a linear transformation? X1.23 Stating that a function is `linear' is dierent than stating that its graph is a line. (a)The function f1:R!Rgiven byf1(x) = 2x1 has a graph that is a line. Show that it is not a linear function. (b)The function f2:R2!Rgiven by x y 7!x+ 2y does not have a graph that is a line. Show that it is a linear function. 180 Chapter Three. Maps Between Spaces X1.24 Part of the denition of a linear function is that it respects addition. Does a linear function respect subtraction? 1.25 Assume that his a linear transformation of Vand thath~1;:::;~niis a basis ofV. Prove each statement. (a)Ifh(~i) =~0 for each basis vector then his the zero map. (b)Ifh(~i) =~ifor each basis vector then his the identity map. (c)If there is a scalar rsuch thath(~i) =r
~ifor each basis vector then h(~ v) =r
~ vfor all vectors in V. X1.26 Consider the vector space R+where vector addition and scalar multiplication are not the ones inherited from Rbut rather are these: a+bis the product of aandb, andr
ais ther-th power of a. (This was shown to be a vector space in an earlier exercise.) Verify that the natural logarithm map ln: R+!Ris a homomorphism between these two spaces. Is it an isomorphism? X1.27 Consider this transformation of R2.x y 7!x=2 y=3 Find the image under this map of this ellipse. fx y(x2=4) + (y2=9) = 1g X1.28 Imagine a rope wound around the earth's equator so that it ts snugly (sup- pose that the earth is a sphere). How much extra rope must be added to raise the circle to a constant six feet o the ground? X1.29 Verify that this map h:R3!R0 @x y z1 A7!0 @x y z1 A0 @3 1 11 A= 3xyz is linear. Generalize. 1.30 Show that every homomorphism from R1toR1acts via multiplication by a scalar. Conclude that every nontrivial linear transformation of R1is an isomor- phism. Is that true for transformations of R2?Rn? 1.31 (a) Show that for any scalars a1;1;:::;am;nthis maph:Rn!Rmis a ho- momorphism.0 B@x1 ... xn1 CA7!0 B@a1;1x1+


+a1;nxn ... am;1x1+


+am;nxn1 CA (b)Show that for each i, thei-th derivative operator di=dxiis a linear trans- formation ofPn. Conclude that for any scalars ck;:::;c 0this map is a linear transformation of that space. f7!dk dxkf+ck1dk1 dxk1f+


+c1d dxf+c0f 1.32 Lemma 1.16 shows that a sum of linear functions is linear and that a scalar multiple of a linear function is linear. Show also that a composition of linear functions is linear. X1.33 Wheref:V!Wis linear, suppose that f(~ v1) =~ w1, . . . ,f(~ vn) =~ wnfor some vectors ~ w1, . . . ,~ wnfromW. (a)If the set of ~ w's is independent, must the set of ~ v's also be independent? Section II. Homomorphisms 181 (b)If the set of ~ v's is independent, must the set of ~ w's also be independent? (c)If the set of ~ w's spansW, must the set of ~ v's spanV? (d)If the set of ~ v's spansV, must the set of ~ w's spanW? 1.34 Generalize Example 1.15 by proving that the matrix transpose map is linear. What is the domain and codomain? 1.35 (a) Where~ u;~ v2Rn, the line segment connecting them is dened to be the set`=ft
~ u+ (1t)
~ vt2[0::1]g. Show that the image, under a homo- morphismh, of the segment between ~ uand~ vis the segment between h(~ u) and h(~ v). (b)A subset of Rnisconvex if, for any two points in that set, the line segment joining them lies entirely in that set. (The inside of a sphere is convex while the skin of a sphere is not.) Prove that linear maps from RntoRmpreserve the property of set convexity. X1.36 Leth:Rn!Rmbe a homomorphism. (a)Show that the image under hof a line in Rnis a (possibly degenerate) line inRm. (b)What happens to a k-dimensional linear surface? 1.37 Prove that the restriction of a homomorphism to a subspace of its domain is another homomorphism. 1.38 Assume that h:V!Wis linear. (a)Show that the rangespace of this mapfh(~ v)~ v2Vgis a subspace of the codomainW. (b)Show that the nullspace of this mapf~ v2Vh(~ v) =~0Wgis a subspace of the domain V. (c)Show that if Uis a subspace of the domain Vthen its imagefh(~ u)~ u2Ug is a subspace of the codomain W. This generalizes the rst item. (d)Generalize the second item. 1.39 Consider the set of isomorphisms from a vector space to itself. Is this a subspace of the space L(V;V) of homomorphisms from the space to itself? 1.40 Does Theorem 1.9 need that h~1;:::;~niis a basis? That is, can we still get a well-dened and unique homomorphism if we drop either the condition that the set of~'s be linearly independent, or the condition that it span the domain? 1.41 LetVbe a vector space and assume that the maps f1;f2:V!R1are lin- ear. (a)Dene a map F:V!R2whose component functions are the given linear ones. ~ v7!f1(~ v) f2(~ v) Show thatFis linear. (b)Does the converse hold | is any linear map from VtoR2made up of two linear component maps to R1? (c)Generalize. II.2 Rangespace and Nullspace Isomorphisms and homomorphisms both preserve structure. The dierence is 182 Chapter Three. Maps Between Spaces that homomorphisms needn't be onto and needn't be one-to-one. This means that homomorphisms are a more general kind of map, subject to fewer restric- tions than isomorphisms. We will examine what can happen with a homomor- phism that is prevented by the extra restrictions satised by an isomorphism. We rst consider the eect of dropping the onto requirement, of not requir- ing as part of the denition that a homomorphism be onto its codomain. For instance, the injection map :R2!R3 x y 7!0 @x y 01 A is not an isomorphism because it is not onto. Of course, being a function, a homomorphism is onto some set, namely its range; the map is onto the xy- plane subset of R3. 2.1 Lemma Under a homomorphism, the image of any subspace of the domain is a subspace of the codomain. In particular, the image of the entire space, the range of the homomorphism, is a subspace of the codomain. Proof .Leth:V!Wbe linear and let Sbe a subspace of the domain V. The image h(S) is a subset of the codomain W. It is nonempty because Sis nonempty and thus to show that h(S) is a subspace of Wwe need only show that it is closed under linear combinations of two vectors. If h(~ s1) andh(~ s2) are members of h(S) thenc1
h(~ s1)+c2
h(~ s2) =h(c1
~ s1)+h(c2
~ s2) =h(c1
~ s1+c2
~ s2) is also a member of h(S) because it is the image of c1
~ s1+c2
~ s2fromS.QED 2.2 Denition The rangespace of a homomorphism h:V!Wis R(h) =fh(~ v)~ v2Vg sometimes denoted h(V). The dimension of the rangespace is the map's rank. (We shall soon see the connection between the rank of a map and the rank of a matrix.) 2.3 Example Recall that the derivative map d=dx :P3!P 3given bya0+ a1x+a2x2+a3x37!a1+ 2a2x+ 3a3x2is linear. The rangespace R(d=dx ) is the set of quadratic polynomials fr+sx+tx2r;s;t2Rg. Thus, the rank of this map is three. 2.4 Example With this homomorphism h:M22!P 3 a b c d 7!(a+b+ 2d) + 0x+cx2+cx3 an image vector in the range can have any constant term, must have an x coecient of zero, and must have the same coecient of x2as ofx3. That is, the rangespace is R(h) =fr+ 0x+sx2+sx3r;s2Rgand so the rank is two. Section II. Homomorphisms 183 The prior result shows that, in passing from the denition of isomorphism to the more general denition of homomorphism, omitting the `onto' requirement doesn't make an essential dierence. Any homomorphism is onto its rangespace. However, omitting the `one-to-one' condition does make a dierence. A homomorphism may have many elements of the domain that map to one element of the codomain. Below is a \bean" sketch of a many-to-one map between sets.It shows three elements of the codomain that are each the image of many members of the domain. Recall that for any function h:V!W, the set of elements of Vthat are mapped to~ w2Wis the inverse image h1(~ w) =f~ v2Vh(~ v) =~ wg. Above, the three sets of many elements on the left are inverse images. 2.5 Example Consider the projection :R3!R2 0 @x y z1 A7!x y which is a homomorphism that is many-to-one. In this instance, an inverse image set is a vertical line of vectors in the domain. R3R2 ~ w 2.6 Example This homomorphism h:R2!R1 x y h7!x+y is also many-to-one; for a xed w2R1, the inverse image h1(w) R2R1 w More information on many-to-one maps is in the appendix. 184 Chapter Three. Maps Between Spaces is the set of plane vectors whose components add to w. The above examples have only to do with the fact that we are considering functions, specically, many-to-one functions. They show the inverse images as sets of vectors that are related to the image vector ~ w. But these are more than just arbitrary functions, they are homomorphisms; what do the two preservation conditions say about the relationships? In generalizing from isomorphisms to homomorphisms by dropping the one- to-one condition, we lose the property that we've stated intuitively as: the domain is \the same as" the range. That is, we lose that the domain corresponds perfectly to the range in a one-vector-by-one-vector way. What we shall keep, as the examples below illustrate, is that a homomorphism describes a way in which the domain is \like", or \analgous to", the range. 2.7 Example We think of R3as being like R2, except that vectors have an extra component. That is, we think of the vector with components x,y, andz as like the vector with components xandy. In dening the projection map , we make precise which members of the domain we are thinking of as related to which members of the codomain. Understanding in what way the preservation conditions in the denition of homomorphism show that the domain elements are like the codomain elements is easiest if we draw R2as thexy-plane inside of R3. (Of course, R2is a set of two-tall vectors while the xy-plane is a set of three-tall vectors with a third component of zero, but there is an obvious correspondence.) Then, (~ v) is the \shadow" of ~ vin the plane and the preservation of addition property says that 0 @x1 y1 z11 Aabovex1 y1 plus0 @x2 y2 z21 Aabovex2 y2 equals0 @x1+y1 y1+y2 z1+z21 Aabovex1+x2 y1+y2 Brie y, the shadow of a sum (~ v1+~ v2) equals the sum of the shadows (~ v1) + (~ v2). (Preservation of scalar multiplication has a similar interpretation.) Redrawing by separating the two spaces, moving the codomain R2to the right, gives an uglier picture but one that is more faithful to the \bean" sketch. ~ w1~ w2 ~ w1+~ w2 Section II. Homomorphisms 185 Again in this drawing, the vectors that map to ~ w1lie in the domain in a vertical line (only one such vector is shown, in gray). Call any such member of this inverse image a \ ~ w1vector". Similarly, there is a vertical line of \ ~ w2vectors" and a vertical line of \ ~ w1+~ w2vectors". Now, has the property that if (~ v1) =~ w1and(~ v2) =~ w2then(~ v1+~ v2) =(~ v1) +(~ v2) =~ w1+~ w2. This says that the vector classes add, in the sense that any ~ w1vector plus any ~ w2vector equals a ~ w1+~ w2vector, (A similar statement holds about the classes under scalar multiplication.) Thus, although the two spaces R3andR2are not isomorphic, describes a way in which they are alike: vectors in R3add as do the associated vectors in R2| vectors add as their shadows add. 2.8 Example A homomorphism can be used to express an analogy between spaces that is more subtle than the prior one. For the map x y h7!x+y from Example 2.6 x two numbers w1;w2in the range R. A~ v1that maps to w1has components that add to w1, that is, the inverse image h1(w1) is the set of vectors with endpoint on the diagonal line x+y=w1. Call these the \ w1 vectors". Similarly, we have the \ w2vectors" and the \ w1+w2vectors". Then the addition preservation property says that ~ v1~ v2~ v1+~ v2 a \w1vector" plus a \ w2vector" equals a \ w1+w2vector". Restated, if a w1vector is added to a w2vector then the result is mapped by hto aw1+w2vector. Brie y, the image of a sum is the sum of the images. Even more brie y, h(~ v1+~ v2) =h(~ v1) +h(~ v2). (The preservation of scalar multiplication condition has a similar restatement.) 2.9 Example The inverse images can be structures other than lines. For the linear map h:R3!R20 @x y z1 A7!x x the inverse image sets are planes x= 0,x= 1, etc., perpendicular to the x-axis. 186 Chapter Three. Maps Between Spaces We won't describe how every homomorphism that we will use is an analogy because the formal sense that we make of \alike in that . . . " is `a homomorphism exists such that . . . '. Nonetheless, the idea that a homomorphism between two spaces expresses how the domain's vectors fall into classes that act like the range's vectors is a good way to view homomorphisms. Another reason that we won't treat all of the homomorphisms that we see as above is that many vector spaces are hard to draw (e.g., a space of polynomials). However, there is nothing bad about gaining insights from those spaces that we are able to draw, especially when those insights extend to all vector spaces. We derive two such insights from the three examples 2.7, 2.8, and 2.9. First, in all three examples, the inverse images are lines or planes, that is, linear surfaces. In particular, the inverse image of the range's zero vector is a line or plane through the origin | a subspace of the domain. 2.10 Lemma For any homomorphism, the inverse image of a subspace of the range is a subspace of the domain. In particular, the inverse image of the trivial subspace of the range is a subspace of the domain. Proof .Leth:V!Wbe a homomorphism and let Sbe a subspace of the rangespace h. Consider h1(S) =f~ v2Vh(~ v)2Sg, the inverse image of the setS. It is nonempty because it contains ~0V, sinceh(~0V) =~0W, which is an elementS, asSis a subspace. To show that h1(S) is closed under linear combinations, let ~ v1and~ v2be elements, so that h(~ v1) andh(~ v2) are elements ofS, and thenc1~ v1+c2~ v2is also in the inverse image because h(c1~ v1+c2~ v2) = c1h(~ v1) +c2h(~ v2) is a member of the subspace S. QED 2.11 Denition The nullspace orkernel of a linear map h:V!Wis the inverse image of 0 W N(h) =h1(~0W) =f~ v2Vh(~ v) =~0Wg: The dimension of the nullspace is the map's nullity . 0V 0W 2.12 Example The map from Example 2.3 has this nullspace N(d=dx ) = fa0+ 0x+ 0x2+ 0x3a02Rg. 2.13 Example The map from Example 2.4 has this nullspace. N(h) =fa b 0(a+b)=2a;b2Rg Section II. Homomorphisms 187 Now for the second insight from the above pictures. In Example 2.7, each of the vertical lines is squashed down to a single point | , in passing from the domain to the range, takes all of these one-dimensional vertical lines and \zeroes them out", leaving the range one dimension smaller than the domain. Similarly, in Example 2.8, the two-dimensional domain is mapped to a one-dimensional range by breaking the domain into lines (here, they are diagonal lines), and compressing each of those lines to a single member of the range. Finally, in Example 2.9, the domain breaks into planes which get \zeroed out", and so the map starts with a three-dimensional domain but ends with a one-dimensional range | this map \subtracts" two from the dimension. (Notice that, in this third example, the codomain is two-dimensional but the range of the map is only one-dimensional, and it is the dimension of the range that is of interest.) 2.14 Theorem A linear map's rank plus its nullity equals the dimension of its domain. Proof .Leth:V!Wbe linear and let BN=h~1;:::;~kibe a basis for the nullspace. Extend that to a basis BV=h~1;:::;~k;~k+1;:::;~nifor the en- tire domain. We shall show that BR=hh(~k+1);:::;h (~n)iis a basis for the rangespace. Then counting the size of these bases gives the result. To see that BRis linearly independent, consider the equation ck+1h(~k+1)+


+cnh(~n) =~0W. This gives that h(ck+1~k+1+


+cn~n) =~0Wand so ck+1~k+1+


+cn~nis in the nullspace of h. AsBNis a basis for this nullspace, there are scalars c1;:::;ck2Rsatisfying this relationship. c1~1+


+ck~k=ck+1~k+1+


+cn~n ButBVis a basis for Vso each scalar equals zero. Therefore BRis linearly independent. To show that BRspans the rangespace, consider h(~ v)2R(h) and write ~ v as a linear combination ~ v=c1~1+


+cn~nof members of BV. This gives h(~ v) =h(c1~1+


+cn~n) =c1h(~1)+


+ckh(~k)+ck+1h(~k+1)+


+cnh(~n) and since~1, . . . ,~kare in the nullspace, we have that h(~ v) =~0 +


+~0 + ck+1h(~k+1) +


+cnh(~n). Thus,h(~ v) is a linear combination of members of BR, and soBRspans the space. QED 2.15 Example Whereh:R3!R4is 0 @x y z1 Ah7!0 BB@x 0 y 01 CCA the rangespace and nullspace are R(h) =f0 BB@a 0 b 01 CCAa;b2RgandN(h) =f0 @0 0 z1 Az2Rg 188 Chapter Three. Maps Between Spaces and so the rank of his two while the nullity is one. 2.16 Example Ift:R!Ris the linear transformation x7! 4x;then the range is R(t) =R1, and so the rank of tis one and the nullity is zero. 2.17 Corollary The rank of a linear map is less than or equal to the dimension of the domain. Equality holds if and only if the nullity of the map is zero. We know that an isomorphism exists between two spaces if and only if their dimensions are equal. Here we see that for a homomorphism to exist, the dimension of the range must be less than or equal to the dimension of the domain. For instance, there is no homomorphism from R2ontoR3. There are many homomorphisms from R2intoR3, but none is onto all of three-space. The rangespace of a linear map can be of dimension strictly less than the dimension of the domain (Example 2.3's derivative transformation on P3has a domain of dimension four but a range of dimension three). Thus, under a homomorphism, linearly independent sets in the domain may map to linearly dependent sets in the range (for instance, the derivative sends f1;x;x2;x3gto f0;1;2x;3x2g). That is, under a homomorphism, independence may be lost. In contrast, dependence stays. 2.18 Lemma Under a linear map, the image of a linearly dependent set is linearly dependent. Proof .Suppose that c1~ v1+


+cn~ vn=~0V, with some cinonzero. Then, becauseh(c1~ v1+


+cn~ vn) =c1h(~ v1)+


+cnh(~ vn) and because h(~0V) =~0W, we have that c1h(~ v1) +


+cnh(~ vn) =~0Wwith some nonzero ci. QED When is independence not lost? One obvious sucient condition is when the homomorphism is an isomorphism. This condition is also necessary; see Exercise 35. We will nish this subsection comparing homomorphisms with isomorphisms by observing that a one-to-one homomorphism is an isomorphism from its domain onto its range. 2.19 Denition A linear map that is one-to-one is nonsingular . (In the next section we will see the connection between this use of `nonsingular' for maps and its familiar use for matrices.) 2.20 Example This nonsingular homomorphism :R2!R3 x y 7!0 @x y 01 A gives the obvious correspondence between R2and thexy-plane inside of R3. The prior observation allows us to adapt some results about isomorphisms to this setting. Section II. Homomorphisms 189 2.21 Theorem In ann-dimensional vector space V, these: (1)his nonsingular, that is, one-to-one (2)hhas a linear inverse (3)N(h) =f~0g, that is, nullity( h) = 0 (4) rank(h) =n (5) ifh~1;:::;~niis a basis for Vthenhh(~1);:::;h (~n)iis a basis for R(h) are equivalent statements about a linear map h:V!W. Proof .We will rst show that (1) () (2). We will then show that (1) = ) (3) =)(4) =)(5) =)(2). For (1) =)(2), suppose that the linear map his one-to-one, and so has an inverse. The domain of that inverse is the range of hand so a linear combina- tion of two members of that domain has the form c1h(~ v1) +c2h(~ v2). On that combination, the inverse h1gives this. h1(c1h(~ v1) +c2h(~ v2)) =h1(h(c1~ v1+c2~ v2)) =h1h(c1~ v1+c2~ v2) =c1~ v1+c2~ v2 =c1h1h(~ v1) +c2h1h(~ v2) =c1
h1(h(~ v1)) +c2
h1(h(~ v2)) Thus the inverse of a one-to-one linear map is automatically linear. But this also gives the (2) =)(1) implication, because the inverse itself must be one-to-one. Of the remaining implications, (1) = )(3) holds because any homomor- phism maps ~0Vto~0W, but a one-to-one map sends at most one member of V to~0W. Next, (3) =)(4) is true since rank plus nullity equals the dimension of the domain. For (4) =)(5), to show thathh(~1);:::;h (~n)iis a basis for the rangespace we need only show that it is a spanning set, because by assumption the range has dimension n. Consider h(~ v)2R(h). Expressing ~ vas a linear combination of basis elements produces h(~ v) =h(c1~1+c2~2+


+cn~n), which gives that h(~ v) =c1h(~1) +


+cnh(~n), as desired. Finally, for the (5) = )(2) implication, assume that h~1;:::;~niis a basis forVso thathh(~1);:::;h (~n)iis a basis for R(h). Then every ~ w2R(h) a the unique representation ~ w=c1h(~1) +


+cnh(~n). Dene a map from R(h) to Vby ~ w7!c1~1+c2~2+


+cn~n (uniqueness of the representation makes this well-dened). Checking that it is linear and that it is the inverse of hare easy. QED We've now seen that a linear map shows how the structure of the domain is like that of the range. Such a map can be thought to organize the domain space into inverse images of points in the range. In the special case that the map is 190 Chapter Three. Maps Between Spaces one-to-one, each inverse image is a single point and the map is an isomorphism between the domain and the range. Exercises X2.22 Leth:P3!P 4be given by p(x)7!x
p(x). Which of these are in the nullspace? Which are in the rangespace? (a)x3(b)0(c)7(d)12x0:5x3(e)1 + 3x2x3 X2.23 Find the nullspace, nullity, rangespace, and rank of each map. (a)h:R2!P 3given bya b 7!a+ax+ax2 (b)h:M22!Rgiven bya b c d 7!a+d (c)h:M22!P 2given bya b c d 7!a+b+c+dx2 (d)the zero map Z:R3!R4 X2.24 Find the nullity of each map. (a)h:R5!R8of rank ve (b)h:P3!P 3of rank one (c)h:R6!R3, an onto map (d)h:M33!M 33, onto X2.25 What is the nullspace of the dierentiation transformation d=dx :Pn!Pn? What is the nullspace of the second derivative, as a transformation of Pn? The k-th derivative? 2.26 Example 2.7 restates the rst condition in the denition of homomorphism as `the shadow of a sum is the sum of the shadows'. Restate the second condition in the same style. 2.27 For the homomorphism h:P3!P 3given byh(a0+a1x+a2x2+a3x3) = a0+ (a0+a1)x+ (a2+a3)x3nd these. (a)N(h)(b)h1(2x3)(c)h1(1 +x2) X2.28 For the map f:R2!Rgiven by f(x y ) = 2x+y sketch these inverse image sets: f1(3),f1(0), andf1(1). X2.29 Each of these transformations of P3is nonsingular. Find the inverse function of each. (a)a0+a1x+a2x2+a3x37!a0+a1x+ 2a2x2+ 3a3x3 (b)a0+a1x+a2x2+a3x37!a0+a2x+a1x2+a3x3 (c)a0+a1x+a2x2+a3x37!a1+a2x+a3x2+a0x3 (d)a0+a1x+a2x2+a3x37!a0+(a0+a1)x+(a0+a1+a2)x2+(a0+a1+a2+a3)x3 2.30 Describe the nullspace and rangespace of a transformation given by ~ v7!2~ v. 2.31 List all pairs (rank( h);nullity(h)) that are possible for linear maps from R5 toR3. 2.32 Does the dierentiation map d=dx :Pn!Pnhave an inverse? X2.33 Find the nullity of the map h:Pn!Rgiven by a0+a1x+


+anxn7!Zx=1 x=0a0+a1x+


+anxndx: Section II. Homomorphisms 191 2.34 (a) Prove that a homomorphism is onto if and only if its rank equals the dimension of its codomain. (b)Conclude that a homomorphism between vector spaces with the same di- mension is one-to-one if and only if it is onto. 2.35 Show that a linear map is nonsingular if and only if it preserves linear inde- pendence. 2.36 Corollary 2.17 says that for there to be an onto homomorphism from a vector spaceVto a vector space W, it is necessary that the dimension of Wbe less than or equal to the dimension of V. Prove that this condition is also sucient; use Theorem 1.9 to show that if the dimension of Wis less than or equal to the dimension of V, then there is a homomorphism from VtoWthat is onto. X2.37 Recall that the nullspace is a subset of the domain and the rangespace is a subset of the codomain. Are they necessarily distinct? Is there a homomorphism that has a nontrivial intersection of its nullspace and its rangespace? 2.38 Prove that the image of a span equals the span of the images. That is, where h:V!Wis linear, prove that if Sis a subset of Vthenh([S]) equals [h(S)]. This generalizes Lemma 2.1 since it shows that if Uis any subspace of Vthen its image fh(~ u)~ u2Ugis a subspace of W, because the span of the set UisU. X2.39 (a) Prove that for any linear map h:V!Wand any~ w2W, the set h1(~ w) has the form f~ v+~ n~ n2N(h)g for~ v2Vwithh(~ v) =~ w(ifhis not onto then this set may be empty). Such a set is a coset ofN(h) and is denoted ~ v+N(h). (b)Consider the map t:R2!R2given byx y t7!ax+by cx+dy for some scalars a,b,c, andd. Prove that tis linear. (c)Conclude from the prior two items that for any linear system of the form ax+by=e cx+dy=f the solution set can be written (the vectors are members of R2) f~ p+~h~hsatises the associated homogeneous system g where~ pis a particular solution of that linear system (if there is no particular solution then the above set is empty). (d)Show that this map h:Rn!Rmis linear0 B@x1 ... xn1 CA7!0 B@a1;1x1+


+a1;nxn ... am;1x1+


+am;nxn1 CA for any scalars a1;1, . . . ,am;n. Extend the conclusion made in the prior item. (e)Show that the k-th derivative map is a linear transformation of Pnfor each k. Prove that this map is a linear transformation of that space f7!dk dxkf+ck1dk1 dxk1f+


+c1d dxf+c0f for any scalars ck, . . . ,c0. Draw a conclusion as above. 2.40 Prove that for any transformation t:V!Vthat is rank one, the map given by composing the operator with itself tt:V!Vsatisestt=r
tfor some real number r. 192 Chapter Three. Maps Between Spaces 2.41 Leth:V!Rbe a homomorphism, but not the zero homomorphism. Prove that ifh~1;:::;~niis a basis for the nullspace and if ~ v2Vis not in the nullspace thenh~ v;~1;:::;~niis a basis for the entire domain V. 2.42 Show that for any space Vof dimension n, the dual space L(V;R) =fh:V!Rhis linearg is isomorphic to Rn. It is often denoted V. Conclude that V=V. 2.43 Show that any linear map is the sum of maps of rank one. 2.44 Is `is homomorphic to' an equivalence relation? ( Hint: the diculty is to decide on an appropriate meaning for the quoted phrase.) 2.45 Show that the rangespaces and nullspaces of powers of linear maps t:V!V form descending VR(t)R(t2)::: and ascending f~0gN(t)N(t2)::: chains. Also show that if kis such that R(tk) =R(tk+1) then all following rangespaces are equal: R(tk) =R(tk+1) =R(tk+2):::. Similarly, if N(tk) = N(tk+1) thenN(tk) =N(tk+1) =N(tk+2) =:::. Section III. Computing Linear Maps 193 III Computing Linear Maps The prior section shows that a linear map is determined by its action on a basis. In fact, the equation h(~ v) =h(c1
~1+


+cn
~n) =c1
h(~1) +


+cn
h(~n) shows that, if we know the value of the map on the vectors in a basis, then we can compute the value of the map on any vector ~ vat all. We just need to nd thec's to express ~ vwith respect to the basis. This section gives the scheme that computes, from the representation of a vector in the domain RepB(~ v), the representation of that vector's image in the codomain RepD(h(~ v)), using the representations of h(~1), . . . ,h(~n). III.1 Representing Linear Maps with Matrices 1.1 Example Consider a map hwith domain R2and codomain R3, xing B=h2 0 ;1 4 iandD=h0 @1 0 01 A;0 @0 2 01 A;0 @1 0 11 Ai as the bases for these spaces, that is determined by this action on the vectors in the domain's basis. 2 0 h7!0 @1 1 11 A1 4 h7!0 @1 2 01 A To compute the action of this map on any vector at all from the domain, we rst express h(~1) andh(~2) with respect to the codomain's basis: 0 @1 1 11 A= 00 @1 0 01 A1 20 @0 2 01 A+ 10 @1 0 11 Aso RepD(h(~1)) =0 @0 1=2 11 A D and 0 @1 2 01 A= 10 @1 0 01 A10 @0 2 01 A+ 00 @1 0 11 Aso RepD(h(~2)) =0 @1 1 01 A D (these are easy to check). Then, as described in the preamble, for any member 194 Chapter Three. Maps Between Spaces ~ vof the domain, we can express the image h(~ v) in terms of the h(~)'s. h(~ v) =h(c1
2 0 +c2
1 4 ) =c1
h(2 0 ) +c2
h(1 4 ) =c1
(00 @1 0 01 A1 20 @0 2 01 A+ 10 @1 0 11 A) +c2
(10 @1 0 01 A10 @0 2 01 A+ 00 @1 0 11 A) = (0c1+ 1c2)
0 @1 0 01 A+ (1 2c11c2)
0 @0 2 01 A+ (1c1+ 0c2)
0 @1 0 11 A Thus, with RepB(~ v) =c1 c2 then RepD(h(~ v) ) =0 @0c1+ 1c2 (1=2)c11c2 1c1+ 0c21 A. For instance, with RepB( 4 8 ) = 1 2 Bthen RepD(h( 4 8 ) ) =0 @2 5=2 11 A. We will express computations like the one above with a matrix notation. 0 @0 1 1=21 1 01 A B;D c1 c2 B=0 @0c1+ 1c2 (1=2)c11c2 1c1+ 0c21 A D In the middle is the argument ~ vto the map, represented with respect to the domain's basis Bby a column vector with components c1andc2. On the right is the value h(~ v) of the map on that argument, represented with respect to the codomain's basis Dby a column vector with components 0 c1+ 1c2, etc. The matrix on the left is the new thing. It consists of the coecients from the vector on the right, 0 and 1 from the rst row, 1=2 and1 from the second row, and 1 and 0 from the third row. This notation simply breaks the parts from the right, the coecients and the c's, out separately on the left, into a vector that represents the map's argument and a matrix that we will take to represent the map itself. Section III. Computing Linear Maps 195 1.2 Denition Suppose that VandWare vector spaces of dimensions nand mwith bases BandD, and thath:V!Wis a linear map. If RepD(h(~1)) =0 BBB@h1;1 h2;1 ... hm;11 CCCA D:::RepD(h(~n)) =0 BBB@h1;n h2;n ... hm;n1 CCCA D then RepB;D(h) =0 BBB@h1;1h1;2::: h 1;n h2;1h2;2::: h 2;n ... hm;1hm;2::: hm;n1 CCCA B;D is the matrix representation of hwith respect to B;D . Brie y, the vectors representing the h(~)'s are adjoined to make the matrix representing the map. RepB;D(h) =0 BB@...... RepD(h(~1) )


RepD(h(~n) ) ......1 CCA Observe that the number of columns nof the matrix is the dimension of the domain of the map and the number of rows mis the dimension of the codomain. 1.3 Example Ifh:R3!P 1is given by 0 @a1 a2 a31 Ah7!(2a1+a2) + (a3)x then where B=h0 @0 0 11 A;0 @0 2 01 A;0 @2 0 01 AiandD=h1 +x;1 +xi the action of honBis given by 0 @0 0 11 Ah7!x0 @0 2 01 Ah7!20 @2 0 01 Ah7!4 and a simple calculation gives RepD(x) = 1=2 1=2 DRepD(2) = 1 1 DRepD(4) = 2 2 D 196 Chapter Three. Maps Between Spaces showing that this is the matrix representing hwith respect to the bases. RepB;D(h) =1=2 1 2 1=212 B;D We will use lower case letters for a map, upper case for the matrix, and lower case again for the entries of the matrix. Thus for the map h, the matrix representing it is H, with entries hi;j. 1.4 Theorem Assume that VandWare vector spaces of dimensions nand mwith basesBandD, and thath:V!Wis a linear map. If his represented by RepB;D(h) =0 BBB@h1;1h1;2::: h 1;n h2;1h2;2::: h 2;n ... hm;1hm;2::: hm;n1 CCCA B;D and~ v2Vis represented by RepB(~ v) =0 BBB@c1 c2 ... cn1 CCCA B then the representation of the image of ~ vis this. RepD(h(~ v) ) =0 BBB@h1;1c1+h1;2c2+


+h1;ncn h2;1c1+h2;2c2+


+h2;ncn ... hm;1c1+hm;2c2+


+hm;ncn1 CCCA D Proof .This formalizes Example 1.1; see Exercise 28. QED 1.5 Denition The matrix-vector product of amnmatrix and a n1 vector is this. 0 BBB@a1;1a1;2::: a 1;n a2;1a2;2::: a 2;n ... am;1am;2::: am;n1 CCCA0 B@c1 ... cn1 CA=0 BBB@a1;1c1+a1;2c2+


+a1;ncn a2;1c1+a2;2c2+


+a2;ncn ... am;1c1+am;2c2+


+am;ncn1 CCCA The point of Denition 1.2 is to generalize Example 1.1. That is, the point of the denition is Theorem 1.4: the product of the matrix RepB;D(h) and the vector RepB(~ v) is the vector RepD(h(~ v)). Brie y, application of a linear map is represented by the matrix-vector product of the map's representative and the vector's representative. Section III. Computing Linear Maps 197 1.6 Example With the matrix from Example 1.3 we can calculate where that map sends this vector. ~ v=0 @4 1 01 A This vector is represented, with respect to the domain basis B, by RepB(~ v) =0 @0 1=2 21 A B and so this is the representation of the value h(~ v) with respect to the codomain basisD. RepD(h(~ v)) =1=2 1 2 1=212 B;D0 @0 1=2 21 A B =(1=2)
0 + 1
(1=2) + 2
2 (1=2)
01
(1=2)2
2 D=9=2 9=2 D To ndh(~ v) itself, not its representation, take (9 =2)(1+x)(9=2)(1+x) = 9. 1.7 Example Let:R3!R2be projection onto the xy-plane. To give a matrix representing this map, we rst x bases. B=h0 @1 0 01 A;0 @1 1 01 A;0 @1 0 11 AiD=h2 1 ;1 1 i For each vector in the domain's basis, we nd its image under the map. 0 @1 0 01 A7! 1 00 @1 1 01 A7! 1 10 @1 0 11 A7! 1 0 Then we nd the representation of each image with respect to the codomain's basis RepD( 1 0 ) = 1 1 RepD( 1 1 ) = 0 1 RepD(1 0 ) =1 1 (these are easily checked). Finally, adjoining these representations gives the matrix representing with respect to B;D . RepB;D() =1 01 1 1 1 B;D 198 Chapter Three. Maps Between Spaces We can illustrate Theorem 1.4 by computing the matrix-vector product repre- senting the following statement about the projection map. (0 @2 2 11 A) =2 2 Representing this vector from the domain with respect to the domain's basis RepB(0 @2 2 11 A) =0 @1 2 11 A B gives this matrix-vector product. RepD((0 @2 1 11 A) ) =1 01 1 1 1 B;D0 @1 2 11 A B=0 2 D Expanding this representation into a linear combination of vectors from D 0
 2 1 + 2
 1 1 = 2 2 checks that the map's action is indeed re ected in the operation of the matrix. (We will sometimes compress these three displayed equations into one 0 @2 2 11 A=0 @1 2 11 A Bh7! H0 2 D=2 2 in the course of a calculation.) We now have two ways to compute the eect of projection, the straight- forward formula that drops each three-tall vector's third component to make a two-tall vector, and the above formula that uses representations and matrix- vector multiplication. Compared to the rst way, the second way might seem complicated. However, it has advantages. The next example shows that giving a formula for some maps is simplied by this new scheme. 1.8 Example To represent a rotation mapt:R2!R2that turns all vectors in the plane counterclockwise through an angle  ~ ut=6(~ u)t=6! Section III. Computing Linear Maps 199 we start by xing bases. Using E2both as a domain basis and as a codomain basis is natural, Now, we nd the image under the map of each vector in the domain's basis. 1 0 t7!cos sin 0 1 t7!sin cos Then we represent these images with respect to the codomain's basis. Because this basis isE2, vectors are represented by themselves. Finally, adjoining the representations gives the matrix representing the map. RepE2;E2(t) = cossin sincos The advantage of this scheme is that just by knowing how to represent the image of the two basis vectors, we get a formula that tells us the image of any vector at all; here a vector rotated by ==6. 3 2 t=67!p 3=21=2 1=2p 3=2 3 2  3:598 0:232 (Again, we are using the fact that, with respect to E2, vectors represent them- selves.) We have already seen the addition and scalar multiplication operations of matrices and the dot product operation of vectors. Matrix-vector multiplication is a new operation in the arithmetic of vectors and matrices. Nothing in De- nition 1.5 requires us to view it in terms of representations. We can get some insight into this operation by turning away from what is being represented, and instead focusing on how the entries combine. 1.9 Example In the denition the width of the matrix equals the height of the vector. Hence, the rst product below is dened while the second is not.  1 0 0 4 3 10 @1 0 21 A=1 6 1 0 0 4 3 11 0 One reason that this product is not dened is purely formal: the denition requires that the sizes match, and these sizes don't match. Behind the formality, though, is a reason why we will leave it undened | the matrix represents a map with a three-dimensional domain while the vector represents a member of a two- dimensional space. A good way to view a matrix-vector product is as the dot products of the rows of the matrix with the column vector. 0 BB@... ai;1ai;2::: ai;n ...1 CCA0 BBB@c1 c2 ... cn1 CCCA=0 BB@... ai;1c1+ai;2c2+:::+ai;ncn ...1 CCA 200 Chapter Three. Maps Between Spaces Looked at in this row-by-row way, this new operation generalizes dot product. Matrix-vector product can also be viewed column-by-column. 0 BBB@h1;1h1;2::: h 1;n h2;1h2;2::: h 2;n ... hm;1hm;2::: hm;n1 CCCA0 BBB@c1 c2 ... cn1 CCCA=0 BBB@h1;1c1+h1;2c2+


+h1;ncn h2;1c1+h2;2c2+


+h2;ncn ... hm;1c1+hm;2c2+


+hm;ncn1 CCCA =c10 BBB@h1;1 h2;1 ... hm;11 CCCA+


+cn0 BBB@h1;n h2;n ... hm;n1 CCCA 1.10 Example  1 01 2 0 30 @2 1 11 A= 2 1 2 1 0 0 + 1 1 3 = 1 7 The result has the columns of the matrix weighted by the entries of the vector. This way of looking at it brings us back to the objective stated at the start of this section, to compute h(c1~1+


+cn~n) asc1h(~1) +


+cnh(~n). We began this section by noting that the equality of these two enables us to compute the action of hon any argument knowing only h(~1), . . . ,h(~n). We have developed this into a scheme to compute the action of the map by taking the matrix-vector product of the matrix representing the map and the vector representing the argument. In this way, any linear map is represented with respect to some bases by a matrix. In the next subsection, we will show the converse, that any matrix represents a linear map. Exercises X1.11 Multiply the matrix0 @1 3 1 01 2 1 1 01 A by each vector (or state \not dened"). (a)0 @2 1 01 A (b)2 2 (c)0 @0 0 01 A 1.12 Perform, if possible, each matrix-vector multiplication. (a)2 1 31=24 2 (b)1 1 0 2 1 00 @1 3 11 A (c)1 1 2 10 @1 3 11 A X1.13 Solve this matrix equation.0 @2 1 1 0 1 3 11 21 A0 @x y z1 A=0 @8 4 41 A Section III. Computing Linear Maps 201 X1.14 For a homomorphism from P2toP3that sends 17!1 +x; x7!1 + 2x;andx27!xx3 where does 13x+ 2x2go? X1.15 Assume that h:R2!R3is determined by this action. 1 0 7!0 @2 2 01 A0 1 7!0 @0 1 11 A Using the standard bases, nd (a)the matrix representing this map; (b)a general formula for h(~ v). X1.16 Letd=dx :P3!P 3be the derivative transformation. (a)Representd=dx with respect to B;B whereB=h1;x;x2;x3i. (b)Representd=dx with respect to B;D whereD=h1;2x;3x2;4x3i. X1.17 Represent each linear map with respect to each pair of bases. (a)d=dx :Pn!Pnwith respect to B;B whereB=h1;x;:::;xni, given by a0+a1x+a2x2+


+anxn7!a1+ 2a2x+


+nanxn1 (b)R :Pn!Pn+1with respect to Bn;Bn+1whereBi=h1;x;:::;xii, given by a0+a1x+a2x2+


+anxn7!a0x+a1 2x2+


+an n+ 1xn+1 (c)R1 0:Pn!Rwith respect to B;E1whereB=h1;x;:::;xniandE1=h1i, given by a0+a1x+a2x2+


+anxn7!a0+a1 2+


+an n+ 1 (d)eval 3:Pn!Rwith respect to B;E1whereB=h1;x;:::;xniandE1=h1i, given by a0+a1x+a2x2+


+anxn7!a0+a1
3 +a2
32+


+an
3n (e)slide1:Pn!Pnwith respect to B;B whereB=h1;x;:::;xni, given by a0+a1x+a2x2+


+anxn7!a0+a1
(x+ 1) +


+an
(x+ 1)n 1.18 Represent the identity map on any nontrivial space with respect to B;B, whereBis any basis. 1.19 Represent, with respect to the natural basis, the transpose transformation on the spaceM22of 22 matrices. 1.20 Assume that B=h~1;~2;~3;~4iis a basis for a vector space. Represent with respect toB;B the transformation that is determined by each. (a)~17!~2,~27!~3,~37!~4,~47!~0 (b)~17!~2,~27!~0,~37!~4,~47!~0 (c)~17!~2,~27!~3,~37!~0,~47!~0 1.21 Example 1.8 shows how to represent the rotation transformation of the plane with respect to the standard basis. Express these other transformations also with respect to the standard basis. (a)thedilation mapds, which multiplies all vectors by the same scalar s (b)there ection mapf`, which re ects all all vectors across a line `through the origin X1.22 Consider a linear transformation of R2determined by these two.1 1 7!2 0 1 0 7!1 0 (a)Represent this transformation with respect to the standard bases. 202 Chapter Three. Maps Between Spaces (b)Where does the transformation send this vector? 0 5 (c)Represent this transformation with respect to these bases. B=h1 1 ;1 1 iD=h2 2 ;1 1 i (d)UsingBfrom the prior item, represent the transformation with respect to B;B. 1.23 Suppose that h:V!Wis nonsingular so that by Theorem 2.21, for any basisB=h~1;:::;~niVthe imageh(B) =hh(~1);:::;h (~n)iis a basis for W. (a)Represent the map hwith respect to B;h(B). (b)For a member ~ vof the domain, where the representation of ~ vhas components c1, . . . ,cn, represent the image vector h(~ v) with respect to the image basis h(B). 1.24 Give a formula for the product of a matrix and ~ ei, the column vector that is all zeroes except for a single one in the i-th position. X1.25 For each vector space of functions of one real variable, represent the derivative transformation with respect to B;B. (a)facosx+bsinxa;b2Rg,B=hcosx;sinxi (b)faex+be2xa;b2Rg,B=hex;e2xi (c)fa+bx+cex+dxexa;b;c;d2Rg,B=h1;x;ex;xexi 1.26 Find the range of the linear transformation of R2represented with respect to the standard bases by each matrix. (a)1 0 0 0 (b)0 0 3 2 (c)a matrix of the forma b 2a2b X1.27 Can one matrix represent two dierent linear maps? That is, can RepB;D(h) = Rep ^B;^D(^h)? 1.28 Prove Theorem 1.4. X1.29 Example 1.8 shows how to represent rotation of all vectors in the plane through an angleabout the origin, with respect to the standard bases. (a)Rotation of all vectors in three-space through an angle about thex-axis is a transformation of R3. Represent it with respect to the standard bases. Arrange the rotation so that to someone whose feet are at the origin and whose head is at (1;0;0), the movement appears clockwise. (b)Repeat the prior item, only rotate about the y-axis instead. (Put the person's head at~ e2.) (c)Repeat, about the z-axis. (d)Extend the prior item to R4. (Hint: `rotate about the z-axis' can be restated as `rotate parallel to the xy-plane'.) 1.30 (Schur's Triangularization Lemma) (a)LetUbe a subspace of Vand x bases BUBV. What is the relationship between the representation of a vector from Uwith respect to BUand the representation of that vector (viewed as a member of V) with respect to BV? (b)What about maps? (c)Fix a basis B=h~1;:::;~niforVand observe that the spans [f~0g] =f~0g[f~1g][f~1;~2g]


 [B] =V Section III. Computing Linear Maps 203 form a strictly increasing chain of subspaces. Show that for any linear map h:V!Wthere is a chain W0=f~0gW1


Wm=Wof subspaces of Wsuch that h([f~1;:::;~ig])Wi for eachi. (d)Conclude that for every linear map h:V!Wthere are bases B;D so the matrix representing hwith respect to B;D is upper-triangular (that is, each entryhi;jwithi>j is zero). (e)Is an upper-triangular representation unique? III.2 Any Matrix Represents a Linear Map The prior subsection shows that the action of a linear map his described by a matrixH, with respect to appropriate bases, in this way. ~ v=0 B@v1 ... vn1 CA Bh7! H0 B@h1;1v1+


+h1;nvn ... hm;1v1+


+hm;nvn1 CA D=h(~ v) In this subsection, we will show the converse, that each matrix represents a linear map. Recall that, in the denition of the matrix representation of a linear map, the number of columns of the matrix is the dimension of the map's domain and the number of rows of the matrix is the dimension of the map's codomain. Thus, for instance, a 23 matrix cannot represent a map from R5toR4. The next result says that, beyond this restriction on the dimensions, there are no other limitations: the 23 matrix represents a map from any three-dimensional space to any two-dimensional space. 2.1 Theorem Any matrix represents a homomorphism between vector spaces of appropriate dimensions, with respect to any pair of bases. Proof .For the matrix H=0 BBB@h1;1h1;2::: h 1;n h2;1h2;2::: h 2;n ... hm;1hm;2::: hm;n1 CCCA x anyn-dimensional domain space Vand anym-dimensional codomain space W. Also x bases B=h~1;:::;~niandD=h~1;:::;~mifor those spaces. Dene a function h:V!Wby: where~ vin the domain is represented as RepB(~ v) =0 B@v1 ... vn1 CA B 204 Chapter Three. Maps Between Spaces then its image h(~ v) is the member the codomain represented by RepD(h(~ v) ) =0 B@h1;1v1+


+h1;nvn ... hm;1v1+


+hm;nvn1 CA D that is,h(~ v) =h(v1~1+


+vn~n) is dened to be ( h1;1v1+


+h1;nvn)
~1+


+ (hm;1v1+


+hm;nvn)
~m. (This is well-dened by the uniqueness of the representation RepB(~ v).) Observe that hhas simply been dened to make it the map that is repre- sented with respect to B;D by the matrix H. So to nish, we need only check thathis linear. If ~ v;~ u2Vare such that RepB(~ v) =0 B@v1 ... vn1 CAand RepB(~ u) =0 B@u1 ... un1 CA andc;d2Rthen the calculation h(c~ v+d~ u) = h1;1(cv1+du1) +


+h1;n(cvn+dun)

~1+


+ hm;1(cv1+du1) +


+hm;n(cvn+dun)

~m =c
h(~ v) +d
h(~ u) provides this verication. QED 2.2 Example Which map the matrix represents depends on which bases are used. If H=1 0 0 0 ; B 1=D1=h1 0 ;0 1 i;andB2=D2=h0 1 ;1 0 i; thenh1:R2!R2represented by Hwith respect to B1;D1maps c1 c2 =c1 c2 B17!c1 0 D1=c1 0 whileh2:R2!R2represented by Hwith respect to B2;D2is this map. c1 c2 =c2 c1 B27!c2 0 D2=0 c2 These two are dierent. The rst is projection onto the xaxis, while the second is projection onto the yaxis. So not only is any linear map described by a matrix but any matrix describes a linear map. This means that we can, when convenient, handle linear maps entirely as matrices, simply doing the computations, without have to worry that Section III. Computing Linear Maps 205 a matrix of interest does not represent a linear map on some pair of spaces of interest. (In practice, when we are working with a matrix but no spaces or bases have been specied, we will often take the domain and codomain to be Rn andRmand use the standard bases. In this case, because the representation is transparent | the representation with respect to the standard basis of ~ vis~ v| the column space of the matrix equals the range of the map. Consequently, the column space of His often denoted by R(H).) With the theorem, we have characterized linear maps as those maps that act in this matrix way. Each linear map is described by a matrix and each matrix describes a linear map. We nish this section by illustrating how a matrix can be used to tell things about its maps. 2.3 Theorem The rank of a matrix equals the rank of any map that it represents. Proof .Suppose that the matrix Hismn. Fix domain and codomain spaces VandWof dimension nandm, with bases B=h~1;:::;~niandD. ThenH represents some linear map hbetween those spaces with respect to these bases whose rangespace fh(~ v)~ v2Vg=fh(c1~1+


+cn~n)c1;:::;cn2Rg =fc1h(~1) +


+cnh(~n)c1;:::;cn2Rg is the span [fh(~1);:::;h (~n)g]. The rank of his the dimension of this range- space. The rank of the matrix is its column rank (or its row rank; the two are equal). This is the dimension of the column space of the matrix, which is the span of the set of column vectors [ fRepD(h(~1));:::; RepD(h(~n))g]. To see that the two spans have the same dimension, recall that a represen- tation with respect to a basis gives an isomorphism RepD:W!Rm. Under this isomorphism, there is a linear relationship among members of the range- space if and only if the same relationship holds in the column space, e.g, ~0 = c1h(~1)+


+cnh(~n) if and only if ~0 =c1RepD(h(~1))+


+cnRepD(h(~n)). Hence, a subset of the rangespace is linearly independent if and only if the cor- responding subset of the column space is linearly independent. This means that the size of the largest linearly independent subset of the rangespace equals the size of the largest linearly independent subset of the column space, and so the two spaces have the same dimension. QED 2.4 Example Any map represented by 0 BB@1 2 2 1 2 1 0 0 3 0 0 21 CCA must, by denition, be from a three-dimensional domain to a four-dimensional codomain. In addition, because the rank of this matrix is two (we can spot this 206 Chapter Three. Maps Between Spaces by eye or get it with Gauss' method), any map represented by this matrix has a two-dimensional rangespace. 2.5 Corollary Lethbe a linear map represented by a matrix H. Thenh is onto if and only if the rank of Hequals the number of its rows, and his one-to-one if and only if the rank of Hequals the number of its columns. Proof .For the rst half, the dimension of the rangespace of his the rank of h, which equals the rank of Hby the theorem. Since the dimension of the codomain ofhis the number of rows in H, if the rank of Hequals the number of rows, then the dimension of the rangespace equals the dimension of the codomain. But a subspace with the same dimension as its superspace must equal that superspace (a basis for the rangespace is a linearly independent subset of the codomain, whose size is equal to the dimension of the codomain, and so this set is a basis for the codomain). For the second half, a linear map is one-to-one if and only if it is an isomor- phism between its domain and its range, that is, if and only if its domain has the same dimension as its range. But the number of columns in his the dimension ofh's domain, and by the theorem the rank of Hequals the dimension of h's range. QED The above results end any confusion caused by our use of the word `rank' to mean apparently dierent things when applied to matrices and when applied to maps. We can also justify the dual use of `nonsingular'. We've dened a matrix to be nonsingular if it is square and is the matrix of coecients of a linear system with a unique solution, and we've dened a linear map to be nonsingular if it is one-to-one. 2.6 Corollary A square matrix represents nonsingular maps if and only if it is a nonsingular matrix. Thus, a matrix represents an isomorphism if and only if it is square and nonsingular. Proof .Immediate from the prior result. QED 2.7 Example Any map from R2toP1represented with respect to any pair of bases by1 2 0 3 is nonsingular because this matrix has rank two. 2.8 Example Any mapg:V!Wrepresented by 1 2 3 6 is not nonsingular because this matrix is not nonsingular. Section III. Computing Linear Maps 207 We've now seen that the relationship between maps and matrices goes both ways: for a particular pair of bases, any linear map is represented by a matrix and any matrix describes a linear map. That is, by xing spaces and bases we get a correspondence between maps and matrices. In the rest of this chapter we will explore this correspondence. For instance, we've dened for linear maps the operations of addition and scalar multiplication and we shall see what the corresponding matrix operations are. We shall also see the matrix operation that represent the map operation of composition. And, we shall see how to nd the matrix that represents a map's inverse. Exercises X2.9Decide if the vector is in the column space of the matrix. (a)2 1 2 5 ,1 3 (b)48 24 ,0 1 (c)0 @11 1 1 11 11 11 A,0 @2 0 01 A X2.10 Decide if each vector lies in the range of the map from R3toR2represented with respect to the standard bases by the matrix. (a)1 1 3 0 1 4 ,1 3 (b)2 0 3 4 0 6 ,1 1 X2.11 Consider this matrix, representing a transformation of R2, and these bases for that space. 1 2
1 1 1 1 B=h0 1 ;1 0 iD=h1 1 ;1 1 i (a)To what vector in the codomain is the rst member of Bmapped? (b)The second member? (c)Where is a general vector from the domain (a vector with components xand y) mapped? That is, what transformation of R2is represented with respect to B;D by this matrix? 2.12 What transformation of F=facos+bsina;b2Rgis represented with respect toB=hcossin;siniandD=hcos+ sin;cosiby this matrix?0 0 1 0 X2.13 Decide if 1 + 2 xis in the range of the map from R3toP2represented with respect toE3andh1;1 +x2;xiby this matrix. 0 @1 3 0 0 1 0 1 0 11 A 2.14 Example 2.8 gives a matrix that is nonsingular, and is therefore associated with maps that are nonsingular. (a)Find the set of column vectors representing the members of the nullspace of any map represented by this matrix. (b)Find the nullity of any such map. (c)Find the set of column vectors representing the members of the rangespace of any map represented by this matrix. (d)Find the rank of any such map. (e)Check that rank plus nullity equals the dimension of the domain. 208 Chapter Three. Maps Between Spaces X2.15 Because the rank of a matrix equals the rank of any map it represents, if one matrix represents two dierent maps H= RepB;D(h) = Rep ^B;^D(^h) (where h;^h:V!W) then the dimension of the rangespace of hequals the dimension of the rangespace of ^h. Must these equal-dimensioned rangespaces actually be the same? X2.16 LetVbe ann-dimensional space with bases BandD. Consider a map that sends, for~ v2V, the column vector representing ~ vwith respect to Bto the column vector representing ~ vwith respect to D. Show that map is a linear transformation ofRn. 2.17 Example 2.2 shows that changing the pair of bases can change the map that a matrix represents, even though the domain and codomain remain the same. Could the map ever not change? Is there a matrix H, vector spaces VandW, and associated pairs of bases B1;D1andB2;D2(withB16=B2orD16=D2or both) such that the map represented by Hwith respect to B1;D1equals the map represented by Hwith respect to B2;D2? X2.18 A square matrix is a diagonal matrix if it is all zeroes except possibly for the entries on its upper-left to lower-right diagonal | its 1 ;1 entry, its 2 ;2 entry, etc. Show that a linear map is an isomorphism if there are bases such that, with respect to those bases, the map is represented by a diagonal matrix with no zeroes on the diagonal. 2.19 Describe geometrically the action on R2of the map represented with respect to the standard bases E2;E2by this matrix.3 0 0 2 Do the same for these.1 0 0 0 0 1 1 0 1 3 0 1 2.20 The fact that for any linear map the rank plus the nullity equals the dimension of the domain shows that a necessary condition for the existence of a homomor- phism between two spaces, onto the second space, is that there be no gain in dimension. That is, where h:V!Wis onto, the dimension of Wmust be less than or equal to the dimension of V. (a)Show that this (strong) converse holds: no gain in dimension implies that there is a homomorphism and, further, any matrix with the correct size and correct rank represents such a map. (b)Are there bases for R3such that this matrix H=0 @1 0 0 2 0 0 0 1 01 A represents a map from R3toR3whose range is the xyplane subspace of R3? 2.21 LetVbe ann-dimensional space and suppose that ~ x2Rn. Fix a basis BforVand consider the map h~ x:V!Rgiven~ v7!~ xRepB(~ v) by the dot product. (a)Show that this map is linear. (b)Show that for any linear map g:V!Rthere is an~ x2Rnsuch thatg=h~ x. (c)In the prior item we xed the basis and varied the ~ xto get all possible linear maps. Can we get all possible linear maps by xing an ~ xand varying the basis? Section III. Computing Linear Maps 209 2.22 LetV;W;X be vector spaces with bases B;C;D . (a)Suppose that h:V!Wis represented with respect to B;C by the matrix H. Give the matrix representing the scalar multiple rh(wherer2R) with respect toB;C by expressing it in terms of H. (b)Suppose that h;g:V!Ware represented with respect to B;C byHand G. Give the matrix representing h+gwith respect to B;C by expressing it in terms ofHandG. (c)Suppose that h:V!Wis represented with respect to B;C byHand g:W!Xis represented with respect to C;D byG. Give the matrix repre- sentingghwith respect to B;D by expressing it in terms of HandG. 210 Chapter Three. Maps Between Spaces IV Matrix Operations The prior section shows how matrices represent linear maps. A good strategy, on seeing a new idea, is to explore how it interacts with some already-established ideas. In the rst subsection we will ask how the representation of the sum of two mapsf+gis related to the representations of the two maps, and how the representation of a scalar product r
hof a map is related to the representation of that map. In later subsections we will see how to represent map composition and map inverse. IV.1 Sums and Scalar Products Recall that for two maps fandgwith the same domain and codomain, the map sumf+ghas this denition. ~ vf+g7!f(~ v) +g(~ v) The easiest way to see how the representations of the maps combine to represent the map sum is with an example. 1.1 Example Suppose that f;g:R2!R3are represented with respect to the basesBandDby these matrices. F= RepB;D(f) =0 @1 3 2 0 1 01 A B;DG= RepB;D(g) =0 @0 0 12 2 41 A B;D Then, for any ~ v2Vrepresented with respect to B, computation of the repre- sentation of f(~ v) +g(~ v) 0 @1 3 2 0 1 01 Av1 v2 +0 @0 0 12 2 41 Av1 v2 =0 @1v1+ 3v2 2v1+ 0v2 1v1+ 0v21 A+0 @0v1+ 0v2 1v12v2 2v1+ 4v21 A gives this representation of f+g(~ v). 0 @(1 + 0)v1+ (3 + 0)v2 (21)v1+ (02)v2 (1 + 2)v1+ (0 + 4)v21 A=0 @1v1+ 3v2 1v12v2 3v1+ 4v21 A Thus, the action of f+gis described by this matrix-vector product. 0 @1 3 12 3 41 A B;Dv1 v2 B=0 @1v1+ 3v2 1v12v2 3v1+ 4v21 A D This matrix is the entry-by-entry sum of original matrices, e.g., the 1 ;1 entry of RepB;D(f+g) is the sum of the 1 ;1 entry ofFand the 1;1 entry ofG. Section IV. Matrix Operations 211 Representing a scalar multiple of a map works the same way. 1.2 Example Iftis a transformation represented by RepB;D(t) =1 0 1 1 B;Dso that~ v=v1 v2 B7!v1 v1+v2 D=t(~ v) then the scalar multiple map 5 tacts in this way. ~ v=v1 v2 B7!5v1 5v1+ 5v2 D= 5
t(~ v) Therefore, this is the matrix representing 5 t. RepB;D(5t) = 5 0 5 5 B;D 1.3 Denition The sum of two same-sized matrices is their entry-by-entry sum. The scalar multiple of a matrix is the result of entry-by-entry scalar multiplication. 1.4 Remark These extend the vector addition and scalar multiplication oper- ations that we dened in the rst chapter. 1.5 Theorem Leth;g:V!Wbe linear maps represented with respect to basesB;D by the matrices HandG, and letrbe a scalar. Then the map h+g:V!Wis represented with respect to B;D byH+G, and the map r
h:V!Wis represented with respect to B;D byrH. Proof .Exercise 9; generalize the examples above. QED A special case of scalar multiplication is multiplication by zero. For any map 0
his the zero homomorphism and for any matrix 0
His the matrix with all entries zero. 1.6 Denition Azero matrix has all entries 0. We write Znm, or simply Z (another, very common, notation is to use 0 nmor just 0). 1.7 Example The zero map from any three-dimensional space to any two- dimensional space is represented by the 2 3 zero matrix Z=0 0 0 0 0 0 no matter which domain and codomain bases are used. 212 Chapter Three. Maps Between Spaces Exercises X1.8Perform the indicated operations, if dened. (a)51 2 6 1 1 +2 1 4 3 0 5 (b)6
211 1 2 3 (c)2 1 0 3 +2 1 0 3 (d)41 2 31 + 51 4 2 1 (e)32 1 3 0 + 21 1 4 3 0 5 1.9Prove Theorem 1.5. (a)Prove that matrix addition represents addition of linear maps. (b)Prove that matrix scalar multiplication represents scalar multiplication of linear maps. X1.10 Prove each, where the operations are dened, where G,H, andJare matrices, whereZis the zero matrix, and where randsare scalars. (a)Matrix addition is commutative G+H=H+G. (b)Matrix addition is associative G+ (H+J) = (G+H) +J. (c)The zero matrix is an additive identity G+Z=G. (d)0
G=Z (e)(r+s)G=rG+sG (f)Matrices have an additive inverse G+ (1)
G=Z. (g)r(G+H) =rG+rH (h)(rs)G=r(sG) 1.11 Fix domain and codomain spaces. In general, one matrix can represent many dierent maps with respect to dierent bases. However, prove that a zero matrix represents only a zero map. Are there other such matrices? X1.12 LetVandWbe vector spaces of dimensions nandm. Show that the space L(V;W ) of linear maps from VtoWis isomorphic toMmn. X1.13 Show that it follows from the prior questions that for any six transformations t1;:::;t 6:R2!R2there are scalars c1;:::;c 62Rsuch thatc1t1+


+c6t6is the zero map. ( Hint: this is a bit of a misleading question.) 1.14 The trace of a square matrix is the sum of the entries on the main diagonal (the 1;1 entry plus the 2 ;2 entry, etc.; we will see the signicance of the trace in Chapter Five). Show that trace( H+G) = trace(H) + trace(G). Is there a similar result for scalar multiplication? 1.15 Recall that the transpose of a matrix Mis another matrix, whose i;jentry is thej;ientry ofM. Veriy these identities. (a)(G+H)trans=Gtrans+Htrans (b)(r
H)trans=r
Htrans X1.16 A square matrix is symmetric if eachi;jentry equals the j;ientry, that is, if the matrix equals its transpose. (a)Prove that for any H, the matrix H+Htransis symmetric. Does every symmetric matrix have this form? (b)Prove that the set of nnsymmetric matrices is a subspace of Mnn. X1.17 (a) How does matrix rank interact with scalar multiplication | can a scalar product of a rank nmatrix have rank less than n? Greater? Section IV. Matrix Operations 213 (b)How does matrix rank interact with matrix addition | can a sum of rank n matrices have rank less than n? Greater? IV.2 Matrix Multiplication After representing addition and scalar multiplication of linear maps in the prior subsection, the natural next map operation to consider is composition. 2.1 Lemma A composition of linear maps is linear. Proof . (This argument has appeared earlier, as part of the proof that isomor- phism is an equivalence relation between spaces.) Leth:V!Wandg:W!U be linear. The calculation gh c1
~ v1+c2
~ v2
=g h(c1
~ v1+c2
~ v2)
=g c1
h(~ v1) +c2
h(~ v2)
=c1
g h(~ v1)) +c2
g(h(~ v2)
=c1
(gh)(~ v1) +c2
(gh)(~ v2) shows that gh:V!Upreserves linear combinations. QED To see how the representation of the composite arises out of the representa- tions of the two compositors, consider an example. 2.2 Example Leth:R4!R2andg:R2!R3, x basesBR4,CR2, DR3, and let these be the representations. H= RepB;C(h) = 4 6 8 2 5 7 9 3 B;CG= RepC;D(g) =0 @1 1 0 1 1 01 A C;D To represent the composition gh:R4!R3we x a~ v, represent hof~ v, and then represent gof that. The representation of h(~ v) is the product of h's matrix and~ v's vector. RepC(h(~ v) ) =4 6 8 2 5 7 9 3 B;C0 BB@v1 v2 v3 v41 CCA B=4v1+ 6v2+ 8v3+ 2v4 5v1+ 7v2+ 9v3+ 3v4 C The representation of g(h(~ v) ) is the product of g's matrix and h(~ v)'s vector. RepD(g(h(~ v)) )=0 @1 1 0 1 1 01 A C;D4v1+ 6v2+ 8v3+ 2v4 5v1+ 7v2+ 9v3+ 3v4 C =0 @1
(4v1+ 6v2+ 8v3+ 2v4) + 1
(5v1+ 7v2+ 9v3+ 3v4) 0
(4v1+ 6v2+ 8v3+ 2v4) + 1
(5v1+ 7v2+ 9v3+ 3v4) 1
(4v1+ 6v2+ 8v3+ 2v4) + 0
(5v1+ 7v2+ 9v3+ 3v4)1 A D 214 Chapter Three. Maps Between Spaces Distributing and regrouping on the v's gives =0 @(1
4 + 1
5)v1+ (1
6 + 1
7)v2+ (1
8 + 1
9)v3+ (1
2 + 1
3)v4 (0
4 + 1
5)v1+ (0
6 + 1
7)v2+ (0
8 + 1
9)v3+ (0
2 + 1
3)v4 (1
4 + 0
5)v1+ (1
6 + 0
7)v2+ (1
8 + 0
9)v3+ (1
2 + 0
3)v41 A D which we recognizing as the result of this matrix-vector product. =0 @1
4 + 1
5 1
6 + 1
7 1
8 + 1
9 1
2 + 1
3 0
4 + 1
5 0
6 + 1
7 0
8 + 1
9 0
2 + 1
3 1
4 + 0
5 1
6 + 0
7 1
8 + 0
9 1
2 + 0
31 A B;D0 BB@v1 v2 v3 v41 CCA D Thus, the matrix representing ghhas the rows of Gcombined with the columns ofH. 2.3 Denition The matrix-multiplicative product of themrmatrixGand thernmatrixHis themnmatrixP, where pi;j=gi;1h1;j+gi;2h2;j+


+gi;rhr;j that is, the i;j-th entry of the product is the dot product of the i-th row and thej-th column. GH=0 BB@... gi;1gi;2::: gi;r ...1 CCA0 BBB@h1;j ::: h 2;j::: ... hr;j1 CCCA=0 BB@... ::: pi;j::: ...1 CCA 2.4 Example The matrices from Example 2.2 combine in this way. 0 @1
4 + 1
5 1
6 + 1
7 1
8 + 1
9 1
2 + 1
3 0
4 + 1
5 0
6 + 1
7 0
8 + 1
9 0
2 + 1
3 1
4 + 0
5 1
6 + 0
7 1
8 + 0
9 1
2 + 0
31 A=0 @9 13 17 5 5 7 9 3 4 6 8 21 A 2.5 Example 0 @2 0 4 6 8 21 A1 3 5 7 =0 @2
1 + 0
5 2
3 + 0
7 4
1 + 6
5 4
3 + 6
7 8
1 + 2
5 8
3 + 2
71 A=0 @2 6 34 54 18 381 A 2.6 Theorem A composition of linear maps is represented by the matrix product of the representatives. Proof . (This argument parallels Example 2.2.) Leth:V!Wandg:W!X be represented by HandGwith respect to bases BV,CW, andDX, of sizesn,r, andm. For any~ v2V, thek-th component of RepC(h(~ v) ) is hk;1v1+


+hk;nvn Section IV. Matrix Operations 215 and so the i-th component of RepD(gh(~ v) ) is this. gi;1
(h1;1v1+


+h1;nvn) +gi;2
(h2;1v1+


+h2;nvn) +


+gi;r
(hr;1v1+


+hr;nvn) Distribute and regroup on the v's. = (gi;1h1;1+gi;2h2;1+


+gi;rhr;1)
v1 +


+ (gi;1h1;n+gi;2h2;n+


+gi;rhr;n)
vn Finish by recognizing that the coecient of each vj gi;1h1;j+gi;2h2;j+


+gi;rhr;j matches the denition of the i;jentry of the product GH. QED The theorem is an example of a result that supports a denition. We can picture what the denition and theorem together say with this arrow diagram (`wrt' abbreviates `with respect to'). Vwrt BWwrt C Xwrt Dh Hg G gh GH Above the arrows, the maps show that the two ways of going from VtoX, straight over via the composition or else by way of W, have the same eect ~ vgh7!g(h(~ v))~ vh7!h(~ v)g7!g(h(~ v)) (this is just the denition of composition). Below the arrows, the matrices indi- cate that the product does the same thing | multiplying GHinto the column vector RepB(~ v) has the same eect as multiplying the column rst by Hand then multiplying the result by G. RepB;D(gh) =GH= RepC;D(g) RepB;C(h) The denition of the matrix-matrix product operation does not restrict us to view it as a representation of a linear map composition. We can get insight into this operation by studying it as a mechanical procedure. The striking thing is the way that rows and columns combine. One aspect of that combination is that the sizes of the matrices involved is signicant. Brie y, mrtimesrnequalsmn. 2.7 Example This product is not dened 1 2 0 0 10 1:10 0 0 2 because the number of columns on the left does not equal the number of rows on the right. 216 Chapter Three. Maps Between Spaces In terms of the underlying maps, the fact that the sizes must match up re ects the fact that matrix multiplication is dened only when a corresponding function composition dimensionnspaceh! dimensionrspaceg! dimensionmspace is possible. 2.8 Remark The order in which these things are written can be confusing. In the `mrtimesrnequalsmn' equation, the number written rst mis the dimension of g's codomain and is thus the number that appears last in the map dimension description above. The explanation is that while fis done rst and thengis applied, that composition is written gf, from the notation ` g(f(~ v))'. (Some people try to lessen confusion by reading ` gf' aloud as \ gfollowing f".) That order then carries over to matrices: gfis represented by GF. Another aspect of the way that rows and columns combine in the matrix product operation is that in the denition of the i;jentry pi;j=gi;1h1;j+gi;2h2;j+


+gi;rhr;j the boxed subscripts on the g's are column indicators while those on the h's indicate rows. That is, summation takes place over the columns of Gbut over the rows of H; left is treated dierently than right, so GHmay be unequal to HG. Matrix multiplication is not commutative. 2.9 Example Matrix multiplication hardly ever commutes. Test that by mul- tiplying randomly chosen matrices both ways.  1 2 3 4 5 6 7 8 = 19 22 43 50  5 6 7 8 1 2 3 4 = 23 34 31 46 2.10 Example Commutativity can fail more dramatically:  5 6 7 8 1 2 0 3 4 0 = 23 34 0 31 46 0 while 1 2 0 3 4 05 6 7 8 isn't even dened. 2.11 Remark The fact that matrix multiplication is not commutative may be puzzling at rst sight, perhaps just because most algebraic operations in elementary mathematics are commutative. But on further re ection, it isn't so surprising. After all, matrix multiplication represents function composition, which is not commutative | if f(x) = 2xandg(x) =x+1 thengf(x) = 2x+1 whilefg(x) = 2(x+ 1) = 2x+ 2. True, this gis not linear and we might have hoped that linear functions commute, but this perspective shows that the failure of commutativity for matrix multiplication ts into a larger context. Section IV. Matrix Operations 217 Except for the lack of commutativity, matrix multiplication is algebraically well-behaved. Below are some nice properties and more are in Exercise 23 and Exercise 24. 2.12 Theorem IfF,G, andHare matrices, and the matrix products are dened, then the product is associative ( FG)H=F(GH) and distributes over matrix addition F(G+H) =FG+FHand (G+H)F=GF+HF. Proof .Associativity holds because matrix multiplication represents function composition, which is associative: the maps ( fg)handf(gh) are equal as both send ~ vtof(g(h(~ v))). Distributivity is similar. For instance, the rst one goes f(g+h) (~ v) = f (g+h)(~ v)
=f g(~ v) +h(~ v)
=f(g(~ v)) +f(h(~ v)) =fg(~ v) +fh(~ v) (the third equality uses the linearity of f). QED 2.13 Remark We could alternatively prove that result by slogging through the indices. For example, associativity goes: the i;j-th entry of ( FG)His (fi;1g1;1+fi;2g2;1+


+fi;rgr;1)h1;j + (fi;1g1;2+fi;2g2;2+


+fi;rgr;2)h2;j ... + (fi;1g1;s+fi;2g2;s+


+fi;rgr;s)hs;j (whereF,G, andHaremr,rs, andsnmatrices), distribute fi;1g1;1h1;j+fi;2g2;1h1;j+


+fi;rgr;1h1;j +fi;1g1;2h2;j+fi;2g2;2h2;j+


+fi;rgr;2h2;j ... +fi;1g1;shs;j+fi;2g2;shs;j+


+fi;rgr;shs;j and regroup around the f's fi;1(g1;1h1;j+g1;2h2;j+


+g1;shs;j) +fi;2(g2;1h1;j+g2;2h2;j+


+g2;shs;j) ... +fi;r(gr;1h1;j+gr;2h2;j+


+gr;shs;j) to get thei;jentry ofF(GH). Contrast these two ways of verifying associativity, the one in the proof and the one just above. The argument just above is hard to understand in the sense that, while the calculations are easy to check, the arithmetic seems unconnected to any idea (it also essentially repeats the proof of Theorem 2.6 and so is ine- cient). The argument in the proof is shorter, clearer, and says why this property \really" holds. This illustrates the comments made in the preamble to the chap- ter on vector spaces | at least some of the time an argument from higher-level constructs is clearer. 218 Chapter Three. Maps Between Spaces We have now seen how the representation of the composition of two linear maps is derived from the representations of the two maps. We have called the combination the product of the two matrices. This operation is extremely important. Before we go on to study how to represent the inverse of a linear map, we will explore it some more in the next subsection. Exercises X2.14 Compute, or state \not dened". (a)3 1 4 20 5 0 0:5 (b)1 11 4 0 30 @211 3 1 1 3 1 11 A (c)27 7 40 @1 0 5 1 1 1 3 8 41 A (d)5 2 3 11 2 35 X2.15 Where A=11 2 0 B=5 2 4 4 C=2 3 4 1 compute or state `not dened'. (a)AB (b)(AB)C (c)BC (d)A(BC) 2.16 Which products are dened? (a)32 times 23(b)23 times 32(c)22 times 33 (d)33 times 22 X2.17 Give the size of the product or state \not dened". (a)a 23 matrix times a 3 1 matrix (b)a 112 matrix times a 12 1 matrix (c)a 23 matrix times a 2 1 matrix (d)a 22 matrix times a 2 2 matrix X2.18 Find the system of equations resulting from starting with h1;1x1+h1;2x2+h1;3x3=d1 h2;1x1+h2;2x2+h2;3x3=d2 and making this change of variable (i.e., substitution). x1=g1;1y1+g1;2y2 x2=g2;1y1+g2;2y2 x3=g3;1y1+g3;2y2 2.19 As Denition 2.3 points out, the matrix product operation generalizes the dot product. Is the dot product of a 1 nrow vector and a n1 column vector the same as their matrix-multiplicative product? X2.20 Represent the derivative map on Pnwith respect to B;B whereBis the natural basish1;x;:::;xni. Show that the product of this matrix with itself is dened; what the map does it represent? 2.21 Show that composition of linear transformations on R1is commutative. Is this true for any one-dimensional space? 2.22 Why is matrix multiplication not dened as entry-wise multiplication? That would be easier, and commutative too. X2.23 (a) Prove that HpHq=Hp+qand (Hp)q=Hpqfor positive integers p;q. (b)Prove that ( rH)p=rp
Hpfor any positive integer pand scalarr2R. X2.24 (a) How does matrix multiplication interact with scalar multiplication: is r(GH) = (rG)H? IsG(rH) =r(GH)? Section IV. Matrix Operations 219 (b)How does matrix multiplication interact with linear combinations: is F(rG+ sH) =r(FG) +s(FH)? Is (rF+sG)H=rFH +sGH ? 2.25 We can ask how the matrix product operation interacts with the transpose operation. (a)Show that ( GH)trans=HtransGtrans. (b)A square matrix is symmetric if eachi;jentry equals the j;ientry, that is, if the matrix equals its own transpose. Show that the matrices HHtransand HtransHare symmetric. X2.26 Rotation of vectors in R3about an axis is a linear map. Show that linear maps do not commute by showing geometrically that rotations do not commute. 2.27 In the proof of Theorem 2.12 some maps are used. What are the domains and codomains? 2.28 How does matrix rank interact with matrix multiplication? (a)Can the product of rank nmatrices have rank less than n? Greater? (b)Show that the rank of the product of two matrices is less than or equal to the minimum of the rank of each factor. 2.29 Is `commutes with' an equivalence relation among nnmatrices? X2.30 (This will be used in the Matrix Inverses exercises.) Here is another property of matrix multiplication that might be puzzling at rst sight. (a)Prove that the composition of the projections x;y:R3!R3onto thex andyaxes is the zero map despite that neither one is itself the zero map. (b)Prove that the composition of the derivatives d2=dx2; d3=dx3:P4!P 4is the zero map despite that neither is the zero map. (c)Give a matrix equation representing the rst fact. (d)Give a matrix equation representing the second. When two things multiply to give zero despite that neither is zero, each is said to be a zero divisor . 2.31 Show that, for square matrices, ( S+T)(ST) need not equal S2T2. X2.32 Represent the identity transformation id: V!Vwith respect to B;B for any basisB. This is the identity matrix I. Show that this matrix plays the role in matrix multiplication that the number 1 plays in real number multiplication: HI= IH=H(for all matrices Hfor which the product is dened). 2.33 In real number algebra, quadratic equations have at most two solutions. That is not so with matrix algebra. Show that the 2 2 matrix equation T2=Ihas more than two solutions, where Iis the identity matrix (this matrix has ones in its 1;1 and 2;2 entries and zeroes elsewhere; see Exercise 32). 2.34 (a) Prove that for any 2 2 matrixTthere are scalars c0;:::;c 4that are not all 0 such that the combination c4T4+c3T3+c2T2+c1T+c0Iis the zero matrix (where Iis the 22 identity matrix, with 1's in its 1 ;1 and 2;2 entries and zeroes elsewhere; see Exercise 32). (b)Letp(x) be a polynomial p(x) =cnxn+


+c1x+c0. IfTis a square matrix we dene p(T) to be the matrix cnTn+


+c1T+I(whereIis the appropriately-sized identity matrix). Prove that for any square matrix there is a polynomial such that p(T) is the zero matrix. (c)The minimal polynomial m(x) of a square matrix is the polynomial of least degree, and with leading coecient 1, such that m(T) is the zero matrix. Find the minimal polynomial of this matrix.p 3=21=2 1=2p 3=2 220 Chapter Three. Maps Between Spaces (This is the representation with respect to E2;E2, the standard basis, of a rotation through=6 radians counterclockwise.) 2.35 The innite-dimensional space Pof all nite-degree polynomials gives a mem- orable example of the non-commutativity of linear maps. Let d=dx :P!P be the usual derivative and let s:P!P be the shift map. a0+a1x+


+anxns7! 0 +a0x+a1x2+


+anxn+1 Show that the two maps don't commute d=dxs6=sd=dx ; in fact, not only is (d=dxs)(sd=dx ) not the zero map, it is the identity map. 2.36 Recall the notation for the sum of the sequence of numbers a1;a2;:::;an. nX i=1ai=a1+a2+


+an In this notation, the i;jentry of the product of GandHis this. pi;j=rX k=1gi;khk;j Using this notation, (a)reprove that matrix multiplication is associative; (b)reprove Theorem 2.6. IV.3 Mechanics of Matrix Multiplication In this subsection we consider matrix multiplication as a mechanical process, putting aside for the moment any implications about the underlying maps. As described earlier, the striking thing about matrix multiplication is the way rows and columns combine. The i;jentry of the matrix product is the dot product of rowiof the left matrix with column jof the right one. For instance, here a second row and a third column combine to make a 2 ;3 entry. 0 B@1 1 0 1 1 01 CA 4 56 78 92 3! =0 @9 13 17 5 5 7 93 4 6 8 21 A We can view this as the left matrix acting by multiplying its rows, one at a time, into the columns of the right matrix. Of course, another perspective is that the right matrix uses its columns to act on the left matrix's rows. Below, we will examine actions from the left and from the right for some simple matrices. The rst case, the action of a zero matrix, is very easy. 3.1 Example Multiplying by an appropriately-sized zero matrix from the left or from the right 0 0 0 01 3 2 1 11 =0 0 0 0 0 0 2 3 1 40 0 0 0 =0 0 0 0 results in a zero matrix. Section IV. Matrix Operations 221 After zero matrices, the matrices whose actions are easiest to understand are the ones with a single nonzero entry. 3.2 Denition A matrix with all zeroes except for a one in the i;jentry is ani;junit matrix. 3.3 Example This is the 1 ;2 unit matrix with three rows and two columns, multiplying from the left. 0 @0 1 0 0 0 01 A5 6 7 8 =0 @7 8 0 0 0 01 A Acting from the left, an i;junit matrix copies row jof the multiplicand into rowiof the result. From the right an i;junit matrix copies column iof the multiplicand into column jof the result. 0 @1 2 3 4 5 6 7 8 91 A0 @0 1 0 0 0 01 A=0 @0 1 0 4 0 71 A 3.4 Example Rescaling these matrices simply rescales the result. This is the action from the left of the matrix that is twice the one in the prior example. 0 @0 2 0 0 0 01 A5 6 7 8 =0 @14 16 0 0 0 01 A And this is the action of the matrix that is minus three times the one from the prior example.0 @1 2 3 4 5 6 7 8 91 A0 @03 0 0 0 01 A=0 @03 012 0211 A Next in complication are matrices with two nonzero entries. There are two cases. If a left-multiplier has entries in dierent rows then their actions don't interact. 3.5 Example 0 @1 0 0 0 0 2 0 0 01 A0 @1 2 3 4 5 6 7 8 91 A= (0 @1 0 0 0 0 0 0 0 01 A+0 @0 0 0 0 0 2 0 0 01 A)0 @1 2 3 4 5 6 7 8 91 A =0 @1 2 3 0 0 0 0 0 01 A+0 @0 0 0 14 16 18 0 0 01 A =0 @1 2 3 14 16 18 0 0 01 A 222 Chapter Three. Maps Between Spaces But if the left-multiplier's nonzero entries are in the same row then that row of the result is a combination. 3.6 Example 0 @1 0 2 0 0 0 0 0 01 A0 @1 2 3 4 5 6 7 8 91 A= (0 @1 0 0 0 0 0 0 0 01 A+0 @0 0 2 0 0 0 0 0 01 A)0 @1 2 3 4 5 6 7 8 91 A =0 @1 2 3 0 0 0 0 0 01 A+0 @14 16 18 0 0 0 0 0 01 A =0 @15 18 21 0 0 0 0 0 01 A Right-multiplication acts in the same way, with columns. These observations about matrices that are mostly zeroes extend to arbitrary matrices. 3.7 Lemma In a product of two matrices GandH, the columns of GHare formed by taking Gtimes the columns of H G
0 BB@...... ~h1


~hn ......1 CCA=0 BB@...... G
~h1


G
~hn ......1 CCA and the rows of GHare formed by taking the rows of GtimesH 0 BB@


~ g1


...


~ gr


1 CCA
H=0 BB@


~ g1
H


...


~ gr
H


1 CCA (ignoring the extra parentheses). Proof .We will show the 2 2 case and leave the general case as an exercise. GH=g1;1g1;2 g2;1g2;2h1;1h1;2 h2;1h2;2 =g1;1h1;1+g1;2h2;1g1;1h1;2+g1;2h2;2 g2;1h1;1+g2;2h2;1g2;1h1;2+g2;2h2;2 The right side of the rst equation in the result  Gh1;1 h2;1 G h1;2 h2;2 = g1;1h1;1+g1;2h2;1 g2;1h1;1+g2;2h2;1 g1;1h1;2+g1;2h2;2 g2;1h1;2+g2;2h2;2 is indeed the same as the right side of GH, except for the extra parentheses (the ones marking the columns as column vectors). The other equation is similarly easy to recognize. QED Section IV. Matrix Operations 223 An application of those observations is that there is a matrix that just copies out the rows and columns. 3.8 Denition The main diagonal (orprinciple diagonal ordiagonal ) of a square matrix goes from the upper left to the lower right. 3.9 Denition Anidentity matrix is square and has with all entries zero except for ones in the main diagonal. Inn=0 BBB@1 0::: 0 0 1::: 0 ... 0 0::: 11 CCCA 3.10 Example Here is the 22 identity matrix leaving its multiplicand un- chaged when it acts from the right. 0 BB@12 02 11 4 31 CCA1 0 0 1 =0 BB@12 02 11 4 31 CCA 3.11 Example Here the 33 identity leaves its multiplicand unchanged both from the left 0 @1 0 0 0 1 0 0 0 11 A0 @2 3 6 1 3 8 7 1 01 A=0 @2 3 6 1 3 8 7 1 01 A and from the right. 0 @2 3 6 1 3 8 7 1 01 A0 @1 0 0 0 1 0 0 0 11 A=0 @2 3 6 1 3 8 7 1 01 A In short, an identity matrix is the identity element of the set of nnmatrices with respect to the operation of matrix multiplication. We next see two ways to generalize the identity matrix. The rst is that if the ones are relaxed to arbitrary reals, the resulting matrix will rescale whole rows or columns. 3.12 Denition Adiagonal matrix is square and has zeros o the main diagonal.0 BBB@a1;10::: 0 0a2;2::: 0 ... 0 0::: an;n1 CCCA 224 Chapter Three. Maps Between Spaces 3.13 Example From the left, the action of multiplication by a diagonal matrix is to rescales the rows. 2 0 012 1 41 1 3 4 4 =4 2 82 1344 From the right such a matrix rescales the columns. 1 2 1 2 2 20 @3 0 0 0 2 0 0 021 A=3 42 6 44 The second generalization of identity matrices is that we can put a single one in each row and column in ways other than putting them down the diagonal. 3.14 Denition Apermutation matrix is square and is all zeros except for a single one in each row and column. 3.15 Example From the left these matrices permute rows. 0 @0 0 1 1 0 0 0 1 01 A0 @1 2 3 4 5 6 7 8 91 A=0 @7 8 9 1 2 3 4 5 61 A From the right they permute columns. 0 @1 2 3 4 5 6 7 8 91 A0 @0 0 1 1 0 0 0 1 01 A=0 @2 3 1 5 6 4 8 9 71 A We nish this subsection by applying these observations to get matrices that perform Gauss' method and Gauss-Jordan reduction. 3.16 Example We have seen how to produce a matrix that will rescale rows. Multiplying by this diagonal matrix rescales the second row of the other by a factor of three. 0 @1 0 0 0 3 0 0 0 11 A0 @0 2 1 1 0 1=3 11 1 0 2 01 A=0 @0 2 1 1 0 1 33 1 0 2 01 A We have seen how to produce a matrix that will swap rows. Multiplying by this permutation matrix swaps the rst and third rows. 0 @0 0 1 0 1 0 1 0 01 A0 @0 2 1 1 0 1 33 1 0 2 01 A=0 @1 0 2 0 0 1 33 0 2 1 11 A Section IV. Matrix Operations 225 To see how to perform a row combination, we observe something about those two examples. The matrix that rescales the second row by a factor of three arises in this way from the identity. 0 @1 0 0 0 1 0 0 0 11 A32!0 @1 0 0 0 3 0 0 0 11 A Similarly, the matrix that swaps rst and third rows arises in this way. 0 @1 0 0 0 1 0 0 0 11 A1$3!0 @0 0 1 0 1 0 1 0 01 A 3.17 Example The 33 matrix that arises as 0 @1 0 0 0 1 0 0 0 11 A22+3!0 @1 0 0 0 1 0 02 11 A will, when it acts from the left, perform the combination operation 22+3. 0 @1 0 0 0 1 0 02 11 A0 @1 0 2 0 0 1 33 0 2 1 11 A=0 @1 0 2 0 0 1 33 0 05 71 A 3.18 Denition The elementary reduction matrices are obtained from iden- tity matrices with one Gaussian operation. We denote them: (1)Iki!Mi(k) fork6= 0; (2)Ii$j!Pi;jfori6=j; (3)Iki+j!Ci;j(k) fori6=j. 3.19 Lemma Gaussian reduction can be done through matrix multiplication. (1) IfHki!GthenMi(k)H=G. (2) IfHi$j!GthenPi;jH=G. (3) IfHki+j!GthenCi;j(k)H=G. Proof .Clear. QED 226 Chapter Three. Maps Between Spaces 3.20 Example This is the rst system, from the rst chapter, on which we performed Gauss' method. 3x3= 9 x1+ 5x22x3= 2 (1=3)x1+ 2x2 = 3 It can be reduced with matrix multiplication. Swap the rst and third rows, 0 @0 0 1 0 1 0 1 0 01 A0 @0 0 3 9 1 522 1=3 2 0 31 A=0 @1=3 2 0 3 1 522 0 0 3 91 A triple the rst row, 0 @3 0 0 0 1 0 0 0 11 A0 @1=3 2 0 3 1 522 0 0 3 91 A=0 @1 6 0 9 1 522 0 0 3 91 A and then add1 times the rst row to the second. 0 @1 0 0 1 1 0 0 0 11 A0 @1 6 0 9 1 522 0 0 3 91 A=0 @1 6 0 9 0127 0 0 3 91 A Now back substitution will give the solution. 3.21 Example Gauss-Jordan reduction works the same way. For the matrix ending the prior example, rst adjust the leading entries 0 @1 0 0 01 0 0 0 1=31 A0 @1 6 0 9 0127 0 0 3 91 A=0 @1 6 0 9 0 1 2 7 0 0 1 31 A and to nish, clear the third column and then the second column. 0 @16 0 0 1 0 0 0 11 A0 @1 0 0 0 12 0 0 11 A0 @1 6 0 9 0 1 2 7 0 0 1 31 A=0 @1 0 0 3 0 1 0 1 0 0 1 31 A We have observed the following result, which we shall use in the next sub- section. 3.22 Corollary For any matrix Hthere are elementary reduction matrices R1, . . . ,Rrsuch thatRr
Rr1


R1
His in reduced echelon form. Until now we have taken the point of view that our primary objects of study are vector spaces and the maps between them, and have adopted matrices only for computational convenience. This subsection show that this point of view isn't the whole story. Matrix theory is a fascinating and fruitful area. In the rest of this book we shall continue to focus on maps as the primary objects, but we will be pragmatic | if the matrix point of view gives some clearer idea then we shall use it. Section IV. Matrix Operations 227 Exercises X3.23 Predict the result of each multiplication by an elementary reduction matrix, and then check by multiplying it out. (a)3 0 0 01 2 3 4 (b)4 0 0 21 2 3 4 (c)1 0 2 11 2 3 4 (d)1 2 3 411 0 1 (e)1 2 3 40 1 1 0 X3.24 The need to take linear combinations of rows and columns in tables of numbers arises often in practice. For instance, this is a map of part of Vermont and New York. In part because of Lake Champlain, there are no roads directly connect- ing some pairs of towns. For in- stance, there is no way to go from Winooski to Grand Isle without go- ing through Colchester. (Of course, many other roads and towns have been left o to simplify the graph. From top to bottom of this map is about forty miles.) BurlingtonColchesterGrand IsleSwanton Winooski (a)The incidence matrix of a map is the square matrix whose i;jentry is the number of roads from city ito cityj. Produce the incidence matrix of this map (take the cities in alphabetical order). (b)A matrix is symmetric if it equals its transpose. Show that an incidence matrix is symmetric. (These are all two-way streets. Vermont doesn't have many one-way streets.) (c)What is the signicance of the square of the incidence matrix? The cube? X3.25 This table gives the number of hours of each type done by each worker, and the associated pay rates. Use matrices to compute the wages due. regular overtime Alan 40 12 Betty 35 6 Catherine 40 18 Donald 28 0wage regular $25:00 overtime $45:00 (Remark. This illustrates, as did the prior problem, that in practice we often want to compute linear combinations of rows and columns in a context where we really aren't interested in any associated linear maps.) 3.26 Find the product of this matrix with its transpose.cossin sincos 228 Chapter Three. Maps Between Spaces X3.27 Prove that the diagonal matrices form a subspace of Mnn. What is its dimension? 3.28 Does the identity matrix represent the identity map if the bases are unequal? 3.29 Show that every multiple of the identity commutes with every square matrix. Are there other matrices that commute with all square matrices? 3.30 Prove or disprove: nonsingular matrices commute. X3.31 Show that the product of a permutation matrix and its transpose is an identity matrix. 3.32 Show that if the rst and second rows of Gare equal then so are the rst and second rows of GH. Generalize. 3.33 Describe the product of two diagonal matrices. 3.34 Write 1 0 3 3 as the product of two elementary reduction matrices. X3.35 Show that if Ghas a row of zeros then GH(if dened) has a row of zeros. Does that work for columns? 3.36 Show that the set of unit matrices forms a basis for Mnm. 3.37 Find the formula for the n-th power of this matrix.1 1 1 0 X3.38 The trace of a square matrix is the sum of the entries on its diagonal (its signicance appears in Chapter Five). Show that trace( GH) = trace(HG). X3.39 A square matrix is upper triangular if its only nonzero entries lie above, or on, the diagonal. Show that the product of two upper triangular matrices is upper triangular. Does this hold for lower triangular also? 3.40 A square matrix is a Markov matrix if each entry is between zero and one and the sum along each row is one. Prove that a product of Markov matrices is Markov. X3.41 Give an example of two matrices of the same rank with squares of diering rank. 3.42 Combine the two generalizations of the identity matrix, the one allowing en- tires to be other than ones, and the one allowing the single one in each row and column to be o the diagonal. What is the action of this type of matrix? 3.43 On a computer multiplications are more costly than additions, so people are interested in reducing the number of multiplications used to compute a matrix product. (a)How many real number multiplications are needed in formula we gave for the product of a mrmatrix and a rnmatrix? (b)Matrix multiplication is associative, so all associations yield the same result. The cost in number of multiplications, however, varies. Find the association requiring the fewest real number multiplications to compute the matrix product of a 510 matrix, a 1020 matrix, a 205 matrix, and a 5 1 matrix. (c)(Very hard.) Find a way to multiply two 2 2 matrices using only seven multiplications instead of the eight suggested by the naive approach. ?3.44 IfAandBare square matrices of the same size such that ABAB = 0, does it follow that BABA = 0? [Putnam, 1990, A-5] Section IV. Matrix Operations 229 3.45 Demonstrate these four assertions to get an alternate proof that column rank equals row rank. [Am. Math. Mon., Dec. 1966] (a)~ y
~ y=~0 i~ y=~0. (b)A~ x=~0 iAtransA~ x=~0. (c)dim(R(A)) = dim( R(AtransA)). (d)col rank(A) = col rank( Atrans) = row rank( A). 3.46 Prove (where Ais annnmatrix and so denes a transformation of any n-dimensional space Vwith respect to B;B whereBis a basis) that dim( R(A)\ N(A)) = dim( R(A))dim(R(A2)). Conclude (a)N(A)R(A) i dim( N(A)) = dim( R(A))dim(R(A2)); (b)R(A)N(A) iA2= 0; (c)R(A) =N(A) iA2= 0 and dim( N(A)) = dim( R(A)) ; (d)dim(R(A)\N(A)) = 0 i dim( R(A)) = dim( R(A2)) ; (e)(Requires the Direct Sum subsection, which is optional.) V=R(A)N(A) i dim( R(A)) = dim( R(A2)). [Ackerson] IV.4 Inverses We now consider how to represent the inverse of a linear map. We start by recalling some facts about function inverses.Some functions have no inverse, or have an inverse on the left side or right side only. 4.1 Example Where:R3!R2is the projection map 0 @x y z1 A7!x y and:R2!R3is the embedding x y 7!0 @x y 01 A the composition is the identity map on R2. x y 7!0 @x y 01 A7!x y We sayis aleft inverse map ofor, what is the same thing, that is aright inverse map of. However, composition in the other order doesn't give the identity map | here is a vector that is not sent to itself under . 0 @0 0 11 A7!0 0 7!0 @0 0 01 A More information on function inverses is in the appendix. 230 Chapter Three. Maps Between Spaces In fact, the projection has no left inverse at all. For, if fwere to be a left inverse ofthen we would have 0 @x y z1 A7!x y f7!0 @x y z1 A for all of the innitely many z's. But no function fcan send a single argument to more than one value. (An example of a function with no inverse on either side is the zero transfor- mation on R2.) Some functions have a two-sided inverse map , another function that is the inverse of the rst, both from the left and from the right. For in- stance, the map given by ~ v7!2
~ vhas the two-sided inverse ~ v7!(1=2)
~ v. In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. The appendix also shows that if a function fhas a two-sided inverse then it is unique, and so it is called `the' inverse, and is denoted f1. So our purpose in this subsection is, where a linear map hhas an inverse, to nd the rela- tionship between RepB;D(h) and RepD;B(h1) (recall that we have shown, in Theorem 2.21 of Section II of this chapter, that if a linear map has an inverse then the inverse is a linear map also). 4.2 Denition A matrixGis aleft inverse matrix of the matrix HifGHis the identity matrix. It is a right inverse matrix ifHGis the identity. A matrix Hwith a two-sided inverse is an invertible matrix . That two-sided inverse is called the inverse matrix and is denoted H1. Because of the correspondence between linear maps and matrices, statements about map inverses translate into statements about matrix inverses. 4.3 Lemma If a matrix has both a left inverse and a right inverse then the two are equal. 4.4 Theorem A matrix is invertible if and only if it is nonsingular. Proof . (For both results.) Given a matrix H, x spaces of appropriate dimen- sion for the domain and codomain. Fix bases for these spaces. With respect to these bases, Hrepresents a map h. The statements are true about the map and therefore they are true about the matrix. QED 4.5 Lemma A product of invertible matrices is invertible | if GandHare invertible and if GHis dened then GHis invertible and ( GH)1=H1G1. Proof . (This is just like the prior proof except that it requires two maps.) Fix appropriate spaces and bases and consider the represented maps handg. Note thath1g1is a two-sided map inverse of ghsince (h1g1)(gh) =h1(id)h= h1h= id and (gh)(h1g1) =g(id)g1=gg1= id. This equality is re ected in the matrices representing the maps, as required. QED Section IV. Matrix Operations 231 Here is the arrow diagram giving the relationship between map inverses and matrix inverses. It is a special case of the diagram for function composition and matrix multiplication. Vwrt BWwrt C Vwrt Bh Hh1 H1 id I Beyond its place in our general program of seeing how to represent map operations, another reason for our interest in inverses comes from solving linear systems. A linear system is equivalent to a matrix equation, as here. x1+x2= 3 2x1x2= 2()1 1 21x1 x2 =3 2 () By xing spaces and bases (e.g., R2;R2andE2;E2), we take the matrix Hto represent some map h. Then solving the system is the same as asking: what domain vector ~ xis mapped by hto the result ~d? If we could invert hthen we could solve the system by multiplying RepD;B(h1)
RepD(~d) to get RepB(~ x). 4.6 Example We can nd a left inverse for the matrix just given m n p q1 1 21 =1 0 0 1 by using Gauss' method to solve the resulting linear system. m+ 2n = 1 mn = 0 p+ 2q= 0 pq= 1 Answer:m= 1=3,n= 1=3,p= 2=3, andq=1=3. This matrix is actually the two-sided inverse of H, as can easily be checked. With it we can solve the system () above by applying the inverse. x y =1=3 1=3 2=31=33 2 =5=3 4=3 4.7 Remark Why solve systems this way, when Gauss' method takes less arithmetic (this assertion can be made precise by counting the number of arith- metic operations, as computer algorithm designers do)? Beyond its conceptual appeal of tting into our program of discovering how to represent the various map operations, solving linear systems by using the matrix inverse has at least two advantages. 232 Chapter Three. Maps Between Spaces First, once the work of nding an inverse has been done, solving a system with the same coecients but dierent constants is easy and fast: if we change the entries on the right of the system ( ) then we get a related problem 1 1 21x y =5 1 with a related solution method. x y =1=3 1=3 2=31=35 1 =2 3 In applications, solving many systems having the same matrix of coecients is common. Another advantage of inverses is that we can explore a system's sensitivity to changes in the constants. For example, tweaking the 3 on the right of the system () to1 1 21x1 x2 =3:01 2 can be solved with the inverse. 1=3 1=3 2=31=33:01 2 =(1=3)(3:01) + (1=3)(2) (2=3)(3:01)(1=3)(2) to show that x1changes by 1 =3 of the tweak while x2moves by 2 =3 of that tweak. This sort of analysis is used, for example, to decide how accurately data must be specied in a linear model to ensure that the solution has a desired accuracy. We nish by describing the computational procedure usually used to nd the inverse matrix. 4.8 Lemma A matrix is invertible if and only if it can be written as the product of elementary reduction matrices. The inverse can be computed by applying to the identity matrix the same row steps, in the same order, as are used to Gauss-Jordan reduce the invertible matrix. Proof .A matrixHis invertible if and only if it is nonsingular and thus Gauss- Jordan reduces to the identity. By Corollary 3.22 this reduction can be done with elementary matrices Rr
Rr1:::R 1
H=I. This equation gives the two halves of the result. First, elementary matrices are invertible and their inverses are also elemen- tary. Applying R1 rto the left of both sides of that equation, then R1 r1, etc., givesHas the product of elementary matrices H=R1 1


R1 r
I(theIis here to cover the trivial r= 0 case). Second, matrix inverses are unique and so comparison of the above equation withH1H=Ishows that H1=Rr
Rr1:::R 1
I. Therefore, applying R1 to the identity, followed by R2, etc., yields the inverse of H. QED Section IV. Matrix Operations 233 4.9 Example To nd the inverse of 1 1 21 we do Gauss-Jordan reduction, meanwhile performing the same operations on the identity. For clerical convenience we write the matrix and the identity side- by-side, and do the reduction steps together. 1 1 1 0 210 1 21+2!1 1 1 0 032 1 1=32!1 1 1 0 0 1 2=31=3 2+1!1 0 1=3 1=3 0 1 2=31=3 This calculation has found the inverse.  1 1 211 = 1=3 1=3 2=31=3 4.10 Example This one happens to start with a row swap. 0 @0 311 0 0 1 0 1 0 1 0 11 0 0 0 11 A1$2!0 @1 0 1 0 1 0 0 311 0 0 11 0 0 0 11 A 1+3!0 @1 0 1 0 1 0 0 311 0 0 01101 11 A ... !0 @1 0 0 1=4 1=4 3=4 0 1 0 1=4 1=41=4 0 0 11=4 3=43=41 A 4.11 Example A non-invertible matrix is detected by the fact that the left half won't reduce to the identity. 1 1 1 0 2 2 0 1 21+2!1 1 1 0 0 02 1 This procedure will nd the inverse of a general nnmatrix. The 22 case is handy. 4.12 Corollary The inverse for a 2 2 matrix exists and equals a b c d1 =1 adbcdb c a if and only if adbc6= 0. 234 Chapter Three. Maps Between Spaces Proof .This computation is Exercise 22. QED We have seen here, as in the Mechanics of Matrix Multiplication subsection, that we can exploit the correspondence between linear maps and matrices. So we can fruitfully study both maps and matrices, translating back and forth to whichever helps us the most. Over the entire four subsections of this section we have developed an algebra system for matrices. We can compare it with the familiar algebra system for the real numbers. Here we are working not with numbers but with matrices. We have matrix addition and subtraction operations, and they work in much the same way as the real number operations, except that they only combine same-sized matrices. We also have a matrix multiplication operation and an operation inverse to multiplication. These are somewhat like the familiar real number operations (associativity, and distributivity over addition, for example), but there are dierences (failure of commutativity, for example). And, we have scalar multiplication, which is in some ways another extension of real number multiplication. This matrix system provides an example that algebra systems other than the elementary one can be interesting and useful. Exercises 4.13 Supply the intermediate steps in Example 4.10. X4.14 Use Corollary 4.12 to decide if each matrix has an inverse. (a)2 1 1 1 (b)0 4 13 (c)23 4 6 X4.15 For each invertible matrix in the prior problem, use Corollary 4.12 to nd its inverse. X4.16 Find the inverse, if it exists, by using the Gauss-Jordan method. Check the answers for the 2 2 matrices with Corollary 4.12. (a)3 1 0 2 (b)2 1=2 3 1 (c)24 1 2 (d)0 @1 1 3 0 2 4 1 1 01 A (e)0 @0 1 5 02 4 2 321 A (f)0 @2 2 3 123 4231 A X4.17 What matrix has this one for its inverse?1 3 2 5 4.18 How does the inverse operation interact with scalar multiplication and addi- tion of matrices? (a)What is the inverse of rH? (b)Is (H+G)1=H1+G1? X4.19 Is (Tk)1= (T1)k? 4.20 IsH1invertible? 4.21 For each real number lett:R2!R2be represented with respect to the standard bases by this matrix.cossin sincos Show thatt1+2=t1
t2. Show also that t1=t. Section IV. Matrix Operations 235 4.22 Do the calculations for the proof of Corollary 4.12. 4.23 Show that this matrix H=1 0 1 0 1 0 has innitely many right inverses. Show also that it has no left inverse. 4.24 In Example 4.1, how many left inverses has ? 4.25 If a matrix has innitely many right-inverses, can it have innitely many left-inverses? Must it have? X4.26 Assume that His invertible and that HGis the zero matrix. Show that Gis a zero matrix. 4.27 Prove that if His invertible then the inverse commutes with a matrix GH1= H1Gif and only if Hitself commutes with that matrix GH=HG. X4.28 Show that if Tis square and if T4is the zero matrix then ( IT)1= I+T+T2+T3. Generalize. X4.29 LetDbe diagonal. Describe D2,D3, . . . , etc. Describe D1,D2, . . . , etc. DeneD0appropriately. 4.30 Prove that any matrix row-equivalent to an invertible matrix is also invertible. 4.31 The rst question below appeared as Exercise 28. (a)Show that the rank of the product of two matrices is less than or equal to the minimum of the rank of each. (b)Show that if TandSare square then TS=Iif and only if ST=I. 4.32 Show that the inverse of a permutation matrix is its transpose. 4.33 The rst two parts of this question appeared as Exercise 25. (a)Show that ( GH)trans=HtransGtrans. (b)A square matrix is symmetric if eachi;jentry equals the j;ientry (that is, if the matrix equals its transpose). Show that the matrices HHtransandHtransH are symmetric. (c)Show that the inverse of the transpose is the transpose of the inverse. (d)Show that the inverse of a symmetric matrix is symmetric. X4.34 The items starting this question appeared as Exercise 30. (a)Prove that the composition of the projections x;y:R3!R3is the zero map despite that neither is the zero map. (b)Prove that the composition of the derivatives d2=dx2; d3=dx3:P4!P 4is the zero map despite that neither map is the zero map. (c)Give matrix equations representing each of the prior two items. When two things multiply to give zero despite that neither is zero, each is said to be a zero divisor . Prove that no zero divisor is invertible. 4.35 In real number algebra, there are exactly two numbers, 1 and 1, that are their own multiplicative inverse. Does H2=Ihave exactly two solutions for 2 2 matrices? 4.36 Is the relation `is a two-sided inverse of' transitive? Re exive? Symmetric? 4.37 Prove: if the sum of the elements of a square matrix is k, then the sum of the elements in each row of the inverse matrix is 1 =k. [Am. Math. Mon., Nov. 1951] 236 Chapter Three. Maps Between Spaces V Change of Basis Representations, whether of vectors or of maps, vary with the bases. For in- stance, with respect to the two bases E2and B=h1 1 ;1 1 i forR2, the vector ~ e1has two dierent representations. RepE2(~ e1) =1 0 RepB(~ e1) =1=2 1=2 Similarly, with respect to E2;E2andE2;B, the identity map has two dierent representations. RepE2;E2(id) =1 0 0 1 RepE2;B(id) =1=2 1=2 1=21=2 With our point of view that the objects of our studies are vectors and maps, in xing bases we are adopting a scheme of tags or names for these objects, that are convienent for computation. We will now see how to translate among these names | we will see exactly how representations vary as the bases vary. V.1 Changing Representations of Vectors In converting RepB(~ v) to RepD(~ v) the underlying vector ~ vdoesn't change. Thus, this translation is accomplished by the identity map on the space, de- scribed so that the domain space vectors are represented with respect to Band the codomain space vectors are represented with respect to D. Vw.r.t.B id??y Vw.r.t.D (The diagram is vertical to t with the ones in the next subsection.) 1.1 Denition The change of basis matrix for basesB;DVis the repre- sentation of the identity map id: V!Vwith respect to those bases. RepB;D(id) =0 BB@...... RepD(~1)


RepD(~n) ......1 CCA Section V. Change of Basis 237 1.2 Lemma Left-multiplication by the change of basis matrix for B;D converts a representation with respect to Bto one with respect to D. Conversly, if left- multiplication by a matrix changes bases M
RepB(~ v) = RepD(~ v) thenMis a change of basis matrix. Proof .For the rst sentence, for each ~ v, as matrix-vector multiplication repre- sents a map application, RepB;D(id)
RepB(~ v) = RepD( id(~ v) ) = RepD(~ v). For the second sentence, with respect to B;D the matrix Mrepresents some linear map, whose action is ~ v7!~ v, and is therefore the identity map. QED 1.3 Example With these bases for R2, B=h 2 1 ; 1 0 iD=h 1 1 ; 1 1 i because RepD( id(2 1 )) =1=2 3=2 DRepD( id(1 0 )) =1=2 1=2 D the change of basis matrix is this. RepB;D(id) = 1=21=2 3=2 1=2 We can see this matrix at work by nding the two representations of ~ e2 RepB( 0 1 ) = 1 2 RepD( 0 1 ) = 1=2 1=2 and checking that the conversion goes as expected. 1=21=2 3=2 1=21 2 =1=2 1=2 We nish this subsection by recognizing that the change of basis matrices are familiar. 1.4 Lemma A matrix changes bases if and only if it is nonsingular. Proof .For one direction, if left-multiplication by a matrix changes bases then the matrix represents an invertible function, simply because the function is inverted by changing the bases back. Such a matrix is itself invertible, and so nonsingular. To nish, we will show that any nonsingular matrix Mperforms a change of basis operation from any given starting basis Bto some ending basis. Because the matrix is nonsingular, it will Gauss-Jordan reduce to the identity, so there are elementatry reduction matrices such that Rr


R1
M=I. Elementary matrices are invertible and their inverses are also elementary, so multiplying from the left rst by Rr1, then byRr11, etc., gives Mas a product of 238 Chapter Three. Maps Between Spaces elementary matrices M=R11


Rr1. Thus, we will be done if we show that elementary matrices change a given basis to another basis, for then Rr1 changesBto some other basis Br, andRr11changesBrto someBr1, . . . , and the net eect is that MchangesBtoB1. We will prove this about elementary matrices by covering the three types as separate cases. Applying a row-multiplication matrix Mi(k)0 BBBBBB@c1 ... ci ... cn1 CCCCCCA=0 BBBBBB@c1 ... kci ... cn1 CCCCCCA changes a representation with respect to h~1;:::;~i;:::;~nito one with respect toh~1;:::; (1=k)~i;:::;~niin this way. ~ v=c1
~1+


+ci
~i+


+cn
~n 7!c1
~1+


+kci
(1=k)~i+


+cn
~n=~ v Similarly, left-multiplication by a row-swap matrix Pi;jchanges a representation with respect to the basis h~1;:::;~i;:::;~j;:::;~niinto one with respect to the basish~1;:::;~j;:::;~i;:::;~niin this way. ~ v=c1
~1+


+ci
~i+


+cj~j+


+cn
~n 7!c1
~1+


+cj
~j+


+ci
~i+


+cn
~n=~ v And, a representation with respect to h~1;:::;~i;:::;~j;:::;~nichanges via left-multiplication by a row-combination matrix Ci;j(k) into a representation with respect toh~1;:::;~ik~j;:::;~j;:::;~ni ~ v=c1
~1+


+ci
~i+cj~j+


+cn
~n 7!c1
~1+


+ci
(~ik~j) +


+ (kci+cj)
~j+


+cn
~n=~ v (the denition of reduction matrices species that i6=kandk6= 0 and so this last one is a basis). QED 1.5 Corollary A matrix is nonsingular if and only if it represents the identity map with respect to some pair of bases. In the next subsection we will see how to translate among representations of maps, that is, how to change RepB;D(h) to Rep ^B;^D(h). The above corollary is a special case of this, where the domain and range are the same space, and where the map is the identity map. Section V. Change of Basis 239 Exercises X1.6InR2, where D=h2 1 ;2 4 i nd the change of basis matrices from DtoE2and fromE2toD. Multiply the two. X1.7Find the change of basis matrix for B;DR2. (a)B=E2,D=h~ e2;~ e1i(b)B=E2,D=h1 2 ;1 4 i (c)B=h1 2 ;1 4 i,D=E2(d)B=h1 1 ;2 2 i,D=h0 4 ;1 3 i 1.8For the bases in Exercise 7, nd the change of basis matrix in the other direction, fromDtoB. X1.9Find the change of basis matrix for each B;DP 2. (a)B=h1;x;x2i;D=hx2;1;xi(b)B=h1;x;x2i;D=h1;1+x;1+x+x2i (c)B=h2;2x;x2i;D=h1 +x2;1x2;x+x2i X1.10 Decide if each changes bases on R2. To what basis is E2changed? (a)5 0 0 4 (b)2 1 3 1 (c)1 4 28 (d)11 1 1 1.11 Find bases such that this matrix represents the identity map with respect to those bases. 0 @3 1 4 21 1 0 0 41 A 1.12 Conside the vector space of real-valued functions with basis hsin(x);cos(x)i. Show thath2 sin(x)+cos(x);3 cos(x)iis also a basis for this space. Find the change of basis matrix in each direction. 1.13 Where does this matrixcos(2) sin(2) sin(2)cos(2) send the standard basis for R2? Any other bases? Hint. Consider the inverse. X1.14 What is the change of basis matrix with respect to B;B? 1.15 Prove that a matrix changes bases if and only if it is invertible. 1.16 Finish the proof of Lemma 1.4. X1.17 LetHbe annnonsingular matrix. What basis of RndoesHchange to the standard basis? X1.18 (a) InP3with basisB=h1 +x;1x;x2+x3;x2x3iwe have this repre- senatation. RepB(1x+ 3x2x3) =0 BB@0 1 1 21 CCA B Find a basis Dgiving this dierent representation for the same polynomial. RepD(1x+ 3x2x3) =0 BB@1 0 2 01 CCA D 240 Chapter Three. Maps Between Spaces (b)State and prove that any nonzero vector representation can be changed to any other. Hint. The proof of Lemma 1.4 is constructive | it not only says the bases change, it shows how they change. 1.19 LetV;W be vector spaces, and let B;^Bbe bases for VandD;^Dbe bases for W. Whereh:V!Wis linear, nd a formula relating RepB;D(h) to Rep ^B;^D(h). X1.20 Show that the columns of an nnchange of basis matrix form a basis for Rn. Do all bases appear in that way: can the vectors from any Rnbasis make the columns of a change of basis matrix? X1.21 Find a matrix having this eect.1 3 7!4 1 That is, nd a Mthat left-multiplies the starting vector to yield the ending vector. Is there a matrix having these two eects? (a)1 3 7!1 1 2 1 7!1 1 (b)1 3 7!1 1 2 6 7!1 1 Give a necessary and sucient condition for there to be a matrix such that ~ v17!~ w1 and~ v27!~ w2. V.2 Changing Map Representations The rst subsection shows how to convert the representation of a vector with respect to one basis to the representation of that same vector with respect to another basis. Here we will see how to convert the representation of a map with respect to one pair of bases to the representation of that map with respect to a dierent pair, how to change RepB;D(h) to Rep ^B;^D(h). That is, we want the relationship between the matrices in this arrow diagram. Vw.r.t.Bh! HWw.r.t.D id??y id??y Vw.r.t. ^Bh! ^HWw.r.t. ^D To move from the lower-left of this diagram to the lower-right we can either go straight over, or else up to VBthen over to WDand then down. So we can calculate ^H= Rep ^B;^D(h) either by simply using ^Band ^D, or else by rst changing bases with Rep ^B;B(id) then multiplying by H= RepB;D(h) and then changing bases with RepD;^D(id). This equation summarizes. ^H= RepD;^D(id)
H
Rep ^B;B(id) ( ) (To compare this equation with the sentence before it, remember that the equa- tion is read from right to left because function composition is read right to left and matrix multiplication represent the composition.) Section V. Change of Basis 241 2.1 Example The matrix T=cos(=6)sin(=6) sin(=6) cos(=6) =p 3=21=2 1=2p 3=2 represents, with respect to E2;E2, the transformation t:R2!R2that rotates vectors=6 radians counterclockwise.  1 3  (3 +p 3)=2 (1 + 3p 3)=2 t=6! We can translate that representation with respect to E2;E2to one with respect to ^B=h1 10 2 i ^D=h1 02 3 i by using the arrow diagram and formula ( ) above. R2 w.r.t.E2t! TR2 w.r.t.E2 id??y id??y R2 w.r.t. ^Bt! ^TR2 w.r.t. ^D^T= RepE2;^D(id)
T
Rep ^B;E2(id) Note that RepE2;^D(id) can be calculated as the matrix inverse of Rep ^D;E2(id). Rep ^B;^D(t) =1 2 0 31p 3=21=2 1=2p 3=21 0 1 2 =(5p 3)=6 (3 + 2p 3)=3 (1 +p 3)=6p 3=3 Although the new matrix is messier-appearing, the map that it represents is the same. For instance, to replicate the eect of tin the picture, start with ^B, Rep ^B( 1 3 ) = 1 1 ^B apply ^T, (5p 3)=6 (3 + 2p 3)=3 (1 +p 3)=6p 3=3 ^B;^D1 1 ^B=(11 + 3p 3)=6 (1 + 3p 3)=6 ^D and check it against ^D 11 + 3p 3 6
1 0 +1 + 3p 3 6
2 3 =(3 +p 3)=2 (1 + 3p 3)=2 to see that it is the same result as above. 242 Chapter Three. Maps Between Spaces 2.2 Example One reason to change bases is that the matrix may be simpler. OnR3the map0 @x y z1 At7!0 @y+z x+z x+y1 A that is represented with respect to the standard basis in this way RepE3;E3(t) =0 @0 1 1 1 0 1 1 1 01 A can also be represented with respect to another basis ifB=h0 @1 1 01 A;0 @1 1 21 A;0 @1 1 11 Ai then RepB;B(t) =0 @1 0 0 01 0 0 0 21 A in a way that is simpler, in that the action of a diagonal matrix is easy to understand. Naturally, we usually prefer basis changes that make the representation eas- ier to understand. When the representation with respect to equal starting and ending bases is a diagonal matrix we say the map or matrix has been diagonal- ized. In Chaper Five we shall see which maps and matrices are diagonalizable, and where one is not, we shall see how to get a representation that is nearly diagonal. We nish this subsection by considering the easier case where representa- tions are with respect to possibly dierent starting and ending bases. Recall that the prior subsection shows that a matrix changes bases if and only if it is nonsingular. That gives us another version of the above arrow diagram and equation (). 2.3 Denition Same-sized matrices Hand ^Harematrix equivalent if there are nonsingular matrices PandQsuch that ^H=PHQ . 2.4 Corollary Matrix equivalent matrices represent the same map, with re- spect to appropriate pairs of bases. Exercise 19 checks that matrix equivalence is an equivalence relation. Thus it partitions the set of matrices into matrix equivalence classes. All matrices: . . .H ^HHmatrix equivalent to^H Section V. Change of Basis 243 We can get some insight into the classes by comparing matrix equivalence with row equivalence (recall that matrices are row equivalent when they can be re- duced to each other by row operations). In ^H=PHQ , the matrices Pand Qare nonsingular and thus each can be written as a product of elementary reduction matrices (Lemma 4.8). Left-multiplication by the reduction matrices making up Phas the eect of performing row operations. Right-multiplication by the reduction matrices making up Qperforms column operations. Therefore, matrix equivalence is a generalization of row equivalence | two matrices are row equivalent if one can be converted to the other by a sequence of row reduction steps, while two matrices are matrix equivalent if one can be converted to the other by a sequence of row reduction steps followed by a sequence of column reduction steps. Thus, if matrices are row equivalent then they are also matrix equivalent (since we can take Qto be the identity matrix and so perform no column operations). The converse, however, does not hold: two matrices can be matrix equivalent but not row equivalent. 2.5 Example These two 1 0 0 0 1 1 0 0 are matrix equivalent because the second can be reduced to the rst by the column operation of taking 1 times the rst column and adding to the second. They are not row equivalent because they have dierent reduced echelon forms (in fact, both are already in reduced form). We will close this section by nding a set of representatives for the matrix equivalence classes. 2.6 Theorem Anymnmatrix of rank kis matrix equivalent to the mn matrix that is all zeros except that the rst kdiagonal entries are ones. 0 BBBBBBBBBB@1 0::: 0 0::: 0 0 1::: 0 0::: 0 ... 0 0::: 1 0::: 0 0 0::: 0 0::: 0 ... 0 0::: 0 0::: 01 CCCCCCCCCCA Sometimes this is described as a block partial-identity form. IZ ZZ More information on class representatives is in the appendix. 244 Chapter Three. Maps Between Spaces Proof .As discussed above, Gauss-Jordan reduce the given matrix and combine all the reduction matrices used there to make P. Then use the leading entries to do column reduction and nish by swapping columns to put the leading ones on the diagonal. Combine the reduction matrices used for those column operations intoQ. QED 2.7 Example We illustrate the proof by nding the PandQfor this matrix. 0 @1 2 11 0 0 11 2 4 221 A First Gauss-Jordan row-reduce. 0 @11 0 0 1 0 0 0 11 A0 @1 0 0 0 1 0 2 0 11 A0 @1 2 11 0 0 11 2 4 221 A=0 @1 2 0 0 0 0 11 0 0 0 01 A Then column-reduce, which involves right-multiplication. 0 @1 2 0 0 0 0 11 0 0 0 01 A0 BB@12 0 0 0 1 0 0 0 0 1 0 0 0 0 11 CCA0 BB@1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 11 CCA=0 @1 0 0 0 0 0 1 0 0 0 0 01 A Finish by swapping columns. 0 @1 0 0 0 0 0 1 0 0 0 0 01 A0 BB@1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 11 CCA=0 @1 0 0 0 0 1 0 0 0 0 0 01 A Finally, combine the left-multipliers together as Pand the right-multipliers together as Qto get thePHQ equation. 0 @11 0 0 1 0 2 0 11 A0 @1 2 11 0 0 11 2 4 221 A0 BB@1 02 0 0 0 1 0 0 1 0 1 0 0 0 11 CCA=0 @1 0 0 0 0 1 0 0 0 0 0 01 A 2.8 Corollary Two same-sized matrices are matrix equivalent if and only if they have the same rank. That is, the matrix equivalence classes are character- ized by rank. Proof .Two same-sized matrices with the same rank are equivalent to the same block partial-identity matrix. QED 2.9 Example The 22 matrices have only three possible ranks: zero, one, or two. Thus there are three matrix-equivalence classes. Section V. Change of Basis 245 All 22 matrices:?0 0 0 0 ?1 0 0 0 ?1 0 0 1Three equivalence classes Each class consists of all of the 2 2 matrices with the same rank. There is only one rank zero matrix, so that class has only one member, but the other two classes each have innitely many members. In this subsection we have seen how to change the representation of a map with respect to a rst pair of bases to one with respect to a second pair. That led to a denition describing when matrices are equivalent in this way. Finally we noted that, with the proper choice of (possibly dierent) starting and ending bases, any map can be represented in block partial-identity form. One of the nice things about this representation is that, in some sense, we can completely understand the map when it is expressed in this way: if the bases areB=h~1;:::;~niandD=h~1;:::;~mithen the map sends c1~1+


+ck~k+ck+1~k+1+


+cn~n7!c1~1+


+ck~k+~0 +


+~0 wherekis the map's rank. Thus, we can understand any linear map as a kind of projection.0 BBBBBBBB@c1 ... ck ck+1 ... cn1 CCCCCCCCA B7!0 BBBBBBBB@c1 ... ck 0 ... 01 CCCCCCCCA D Of course, \understanding" a map expressed in this way requires that we un- derstand the relationship between BandD. However, despite that diculty, this is a good classication of linear maps. Exercises X2.10 Decide if these matrices are matrix equivalent. (a)1 3 0 2 3 0 ,2 2 1 0 51 (b)0 3 1 1 ,4 0 0 5 (c)1 3 2 6 ,1 3 26 X2.11 Find the canonical representative of the matrix-equivalence class of each ma- trix. 246 Chapter Three. Maps Between Spaces (a)2 1 0 4 2 0 (b)0 @0 1 0 2 1 1 0 4 3 3 311 A 2.12 Suppose that, with respect to B=E2D=h1 1 ;1 1 i the transformation t:R2!R2is represented by this matrix.1 2 3 4 Use change of basis matrices to represent twith respect to each pair. (a) ^B=h0 1 ;1 1 i,^D=h1 0 ;2 1 i (b) ^B=h1 2 ;1 0 i,^D=h1 2 ;2 1 i X2.13 What sizes are PandQin the equation ^H=PHQ ? X2.14 Use Theorem 2.6 to show that a square matrix is nonsingular if and only if it is equivalent to an identity matrix. X2.15 Show that, where Ais a nonsingular square matrix, if PandQare nonsingular square matrices such that PAQ =IthenQP=A1. X2.16 Why does Theorem 2.6 not show that every matrix is diagonalizable (see Example 2.2)? 2.17 Must matrix equivalent matrices have matrix equivalent transposes? 2.18 What happens in Theorem 2.6 if k= 0? X2.19 Show that matrix-equivalence is an equivalence relation. X2.20 Show that a zero matrix is alone in its matrix equivalence class. Are there other matrices like that? 2.21 What are the matrix equivalence classes of matrices of transformations on R1? R3? 2.22 How many matrix equivalence classes are there? 2.23 Are matrix equivalence classes closed under scalar multiplication? Addition? 2.24 Lett:Rn!Rnrepresented by Twith respect toEn;En. (a)Find RepB;B(t) in this specic case. T=1 1 31 B=h1 2 ;1 1 i (b)Describe RepB;B(t) in the general case where B=h~1;:::;~ni. 2.25 (a) LetVhave basesB1andB2and suppose that Whas the basis D. Where h:V!W, nd the formula that computes RepB2;D(h) from RepB1;D(h). (b)Repeat the prior question with one basis for Vand two bases for W. 2.26 (a) If two matrices are matrix-equivalent and invertible, must their inverses be matrix-equivalent? (b)If two matrices have matrix-equivalent inverses, must the two be matrix- equivalent? (c)If two matrices are square and matrix-equivalent, must their squares be matrix-equivalent? (d)If two matrices are square and have matrix-equivalent squares, must they be matrix-equivalent? Section V. Change of Basis 247 X2.27 Square matrices are similar if they represent the same transformation, but each with respect to the same ending as starting basis. That is, RepB1;B1(t) is similar to RepB2;B2(t). (a)Give a denition of matrix similarity like that of Denition 2.3. (b)Prove that similar matrices are matrix equivalent. (c)Show that similarity is an equivalence relation. (d)Show that if Tis similar to ^TthenT2is similar to ^T2, the cubes are similar, etc. Contrast with the prior exercise. (e)Prove that there are matrix equivalent matrices that are not similar. 248 Chapter Three. Maps Between Spaces VI Projection This section is optional; only the last two sections of Chapter Five require this material. We have described the projection fromR3into itsxyplane subspace as a `shadow map'. This shows why, but it also shows that some shadows fall upward. 0 @1 2 21 A 0 @1 2 11 A So perhaps a better description is: the projection of ~ vis the~ pin the plane with the property that someone standing on ~ pand looking straight up or down sees ~ v. In this section we will generalize this to other projections, both orthogonal (i.e., `straight up and down') and nonorthogonal. VI.1 Orthogonal Projection Into a Line We rst consider orthogonal projection into a line. To orthogonally project a vector~ vinto a line `, darken a point on the line if someone on that line and looking straight up or down (from that person's point of view) sees ~ v. The picture shows someone who has walked out on the line until the tip of ~ vis straight overhead. That is, where the line is described as the span of some nonzero vector `=fc
~ sc2Rg, the person has walked out to nd the coecientc~ pwith the property that ~ vc~ p
~ sis orthogonal to c~ p
~ s. c~ p~ s~ v~ vc~ p~ s We can solve for this coecient by noting that because ~ vc~ p~ sis orthogonal to a scalar multiple of ~ sit must be orthogonal to ~ sitself, and then the consequent fact that the dot product ( ~ vc~ p~ s)~ sis zero gives that c~ p=~ v~ s=~ s~ s. Section VI. Projection 249 1.1 Denition The orthogonal projection of ~ vinto the line spanned by a nonzero~ sis this vector. proj[~ s](~ v) =~ v~ s ~ s~ s
~ s Exercise 19 checks that the outcome of the calculation depends only on the line and not on which vector ~ shappens to be used to describe that line. 1.2 Remark The wording of that denition says `spanned by ~ s' instead the more formal `the span of the set f~ sg'. This casual rst phrase is common. 1.3 Example To orthogonally project the vector2 3
into the line y= 2x, we rst pick a direction vector for the line. For instance, ~ s=1 2 will do. Then the calculation is routine. 0 @2 31 A0 @1 21 A 0 @1 21 A0 @1 21 A
1 2 =8 5
1 2 =8=5 16=5 1.4 Example InR3, the orthogonal projection of a general vector 0 @x y z1 A into they-axis is0 @x y z1 A0 @0 1 01 A 0 @0 1 01 A0 @0 1 01 A
0 @0 1 01 A=0 @0 y 01 A which matches our intuitive expectation. The picture above with the stick gure walking out on the line until ~ v's tip is overhead is one way to think of the orthogonal projection of a vector into a line. We nish this subsection with two other ways. 1.5 Example A railroad car left on an east-west track without its brake is pushed by a wind blowing toward the northeast at fteen miles per hour; what speed will the car reach? 250 Chapter Three. Maps Between Spaces For the wind we use a vector of length 15 that points toward the northeast. ~ v= 15p 1=2 15p 1=2 The car can only be aected by the part of the wind blowing in the east-west direction | the part of ~ vin the direction of the x-axis is this (the picture has the same perspective as the railroad car picture above). eastnorth ~ p= 15p 1=2 0 So the car will reach a velocity of 15p 1=2 miles per hour toward the east. Thus, another way to think of the picture that precedes the denition is that it shows~ vas decomposed into two parts, the part with the line (here, the part with the tracks, ~ p), and the part that is orthogonal to the line (shown here lying on the north-south axis). These two are \not interacting" or \independent", in the sense that the east-west car is not at all aected by the north-south part of the wind (see Exercise 11). So the orthogonal projection of ~ vinto the line spanned by ~ scan be thought of as the part of ~ vthat lies in the direction of ~ s. Finally, another useful way to think of the orthogonal projection is to have the person stand not on the line, but on the vector that is to be projected to the line. This person has a rope over the line and pulls it tight, naturally making the rope orthogonal to the line. That is, we can think of the projection ~ pas being the vector in the line that is closest to~ v(see Exercise 17). 1.6 Example A submarine is tracking a ship moving along the line y= 3x+2. Torpedo range is one-half mile. Can the sub stay where it is, at the origin on the chart below, or must it move to reach a place where the ship will pass within range? eastnorth Section VI. Projection 251 The formula for projection into a line does not immediately apply because the line doesn't pass through the origin, and so isn't the span of any ~ s. To adjust for this, we start by shifting the entire map down two units. Now the line is y= 3x, which is a subspace, and we can project to get the point ~ pof closest approach, the point on the line through the origin closest to ~ v=0 2 the sub's shifted position. ~ p=0 2 1 3 1 3 1 3
1 3 =3=5 9=5 The distance between ~ vand~ pis approximately 0 :63 miles and so the sub must move to get in range. This subsection has developed a natural projection map: orthogonal projec- tion into a line. As suggested by the examples, it is often called for in appli- cations. The next subsection shows how the denition of orthogonal projection into a line gives us a way to calculate especially convienent bases for vector spaces, again something that is common in applications. The nal subsection completely generalizes projection, orthogonal or not, into any subspace at all. Exercises X1.7Project the rst vector orthogonally into the line spanned by the second vec- tor. (a)2 1 ,3 2 (b)2 1 ,3 0 (c)0 @1 1 41 A,0 @1 2 11 A (d)0 @1 1 41 A,0 @3 3 121 A X1.8Project the vector orthogonally into the line. (a)0 @2 1 41 A;fc0 @3 1 31 Ac2Rg(b)1 1 , the liney= 3x 1.9Although the development of Denition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In R4project this vector into this line. ~ v=0 BB@1 2 1 31 CCA`=fc
0 BB@1 1 1 11 CCAc2Rg X1.10 Denition 1.1 uses two vectors ~ sand~ v. Consider the transformation of R2 resulting from xing ~ s=3 1 and projecting ~ vinto the line that is the span of ~ s. Apply it to these vec- tors. 252 Chapter Three. Maps Between Spaces (a)1 2 (b)0 4 Show that in general the projection tranformation is this. x1 x2 7!(x1+ 3x2)=10 (3x1+ 9x2)=10 Express the action of this transformation with a matrix. 1.11 Example 1.5 suggests that projection breaks ~ vinto two parts, proj[~ s](~ v) and ~ vproj[~ s](~ v), that are \not interacting". Recall that the two are orthogonal. Show that any two nonzero orthogonal vectors make up a linearly independent set. 1.12 (a) What is the orthogonal projection of ~ vinto a line if ~ vis a member of that line? (b)Show that if ~ vis not a member of the line then the set f~ v;~ vproj[~ s](~ v)gis linearly independent. 1.13 Denition 1.1 requires that ~ sbe nonzero. Why? What is the right denition of the orthogonal projection of a vector into the (degenerate) line spanned by the zero vector? 1.14 Are all vectors the projection of some other vector into some line? X1.15 Show that the projection of ~ vinto the line spanned by ~ shas length equal to the absolute value of the number ~ v~ sdivided by the length of the vector ~ s. 1.16 Find the formula for the distance from a point to a line. 1.17 Find the scalar csuch that ( cs1;cs2) is a minimum distance from the point (v1;v2) by using calculus (i.e., consider the distance function, set the rst derivative equal to zero, and solve). Generalize to Rn. X1.18 Prove that the orthogonal projection of a vector into a line is shorter than the vector. X1.19 Show that the denition of orthogonal projection into a line does not depend on the spanning vector: if ~ sis a nonzero multiple of ~ qthen (~ v~ s=~ s~ s)
~ sequals (~ v~ q=~ q~ q)
~ q. X1.20 Consider the function mapping to plane to itself that takes a vector to its projection into the line y=x. These two each show that the map is linear, the rst one in a way that is bound to the coordinates (that is, it xes a basis and then computes) and the second in a way that is more conceptual. (a)Produce a matrix that describes the function's action. (b)Show also that this map can be obtained by rst rotating everything in the plane=4 radians clockwise, then projecting into the x-axis, and then rotating =4 radians counterclockwise. 1.21 For~ a;~b2Rnlet~ v1be the projection of ~ ainto the line spanned by ~b, let~ v2be the projection of ~ v1into the line spanned by ~ a, let~ v3be the projection of ~ v2into the line spanned by ~b, etc., back and forth between the spans of ~ aand~b. That is, ~ vi+1is the projection of ~ viinto the span of ~ aifi+ 1 is even, and into the span of ~b ifi+ 1 is odd. Must that sequence of vectors eventually settle down | must there be a suciently large isuch that~ vi+2equals~ viand~ vi+3equals~ vi+1? If so, what is the earliest such i? Section VI. Projection 253 VI.2 Gram-Schmidt Orthogonalization This subsection is optional. It requires material from the prior, also optional, subsection. The work done here will only be needed in the nal two sections of Chapter Five. The prior subsection suggests that projecting into the line spanned by ~ s decomposes a vector ~ vinto two parts proj[~ s](~ p)~ v~ vproj[~ s](~ p) ~ v= proj[~ s](~ v) + ~ vproj[~ s](~ v) that are orthogonal and so are \not interacting". We will now develop that suggestion. 2.1 Denition Vectors~ v1;:::;~ vk2Rnaremutually orthogonal when any two are orthogonal: if i6=jthen the dot product ~ vi~ vjis zero. 2.2 Theorem If the vectors in a set f~ v1;:::;~ vkgRnare mutually orthog- onal and nonzero then that set is linearly independent. Proof .Consider a linear relationship c1~ v1+c2~ v2+


+ck~ vk=~0. Ifi2[1::k] then taking the dot product of ~ viwith both sides of the equation ~ vi(c1~ v1+c2~ v2+


+ck~ vk) =~ vi~0 ci
(~ vi~ vi) = 0 shows, since ~ viis nonzero, that ciis zero. QED 2.3 Corollary If the vectors in a size ksubset of a kdimensional space are mutually orthogonal and nonzero then that set is a basis for the space. Proof .Any linearly independent size ksubset of a kdimensional space is a basis. QED Of course, the converse of Corollary 2.3 does not hold | not every basis of every subspace of Rnis made of mutually orthogonal vectors. However, we can get the partial converse that for every subspace of Rnthere is at least one basis consisting of mutually orthogonal vectors. 2.4 Example The members ~1and~2of this basis for R2are not orthogonal. B=h4 2 ;1 3 i ~1~2 254 Chapter Three. Maps Between Spaces However, we can derive from Ba new basis for the same space that does have mutually orthogonal members. For the rst member of the new basis we simply use~1. ~ 1=4 2 For the second member of the new basis, we take away from ~2its part in the direction of ~ 1, ~ 2=1 3 proj[~ 1](1 3 ) =1 3 2 1 =1 2 ~ 2 which leaves the part, ~ 2pictured above, of ~2that is orthogonal to ~ 1(it is orthogonal by the denition of the projection into the span of ~ 1). Note that, by the corollary,f~ 1;~ 2gis a basis for R2. 2.5 Denition Anorthogonal basis for a vector space is a basis of mutually orthogonal vectors. The next result gives a way to produce an orthogonal basis from any given starting basis. We rst see an example. 2.6 Example To turn this basis for R3 h0 @1 1 11 A;0 @0 2 01 A;0 @1 0 31 Ai into an orthogonal basis, we take the rst vector as it is given. ~ 1=0 @1 1 11 A We get~ 2by starting with the given second vector ~2and subtracting away the part of it in the direction of ~ 1. ~ 2=0 @0 2 01 Aproj[~ 1](0 @0 2 01 A) =0 @0 2 01 A0 @2=3 2=3 2=31 A=0 @2=3 4=3 2=31 A Finally, we get ~ 3by taking the third given vector and subtracting the part of it in the direction of ~ 1, and also the part of it in the direction of ~ 2. ~ 3=0 @1 0 31 Aproj[~ 1](0 @1 0 31 A)proj[~ 2](0 @1 0 31 A) =0 @1 0 11 A Section VI. Projection 255 Again the corollary gives that h0 @1 1 11 A;0 @2=3 4=3 2=31 A;0 @1 0 11 Ai is a basis for the space. The next result veries that the process used in those examples works with any basis for any subspace of an Rn(we are restricted to Rnonly because we have not given a denition of orthogonality for other vector spaces). 2.7 Theorem (Gram-Schmidt orthogonalization) Ifh~1;:::~kiis a basis for a subspace of Rnthen, where ~ 1=~1 ~ 2=~2proj[~ 1](~2) ~ 3=~3proj[~ 1](~3)proj[~ 2](~3) ... ~ k=~kproj[~ 1](~k)


proj[~ k1](~k) the~ 's form an orthogonal basis for the same subspace. Proof .We will use induction to check that each ~ iis nonzero, is in the span of h~1;:::~iiand is orthogonal to all preceding vectors: ~ 1~ i=


=~ i1~ i= 0. With those, and with Corollary 2.3, we will have that h~ 1;:::~ kiis a basis for the same space as h~1;:::~ki. We shall cover the cases up to i= 3, which give the sense of the argument. Completing the details is Exercise 23. Thei= 1 case is trivial | setting ~ 1equal to~1makes it a nonzero vector since~1is a member of a basis, it is obviously in the desired span, and the `orthogonal to all preceding vectors' condition is vacuously met. For thei= 2 case, expand the denition of ~ 2. ~ 2=~2proj[~ 1](~2) =~2~2~ 1 ~ 1~ 1
~ 1=~2~2~ 1 ~ 1~ 1
~1 This expansion shows that ~ 2is nonzero or else this would be a non-trivial linear dependence among the ~'s (it is nontrivial because the coecient of ~2is 1) and also shows that ~ 2is in the desired span. Finally, ~ 2is orthogonal to the only preceding vector ~ 1~ 2=~ 1(~2proj[~ 1](~2)) = 0 because this projection is orthogonal. 256 Chapter Three. Maps Between Spaces Thei= 3 case is the same as the i= 2 case except for one detail. As in the i= 2 case, expanding the denition ~ 3=~3~3~ 1 ~ 1~ 1
~ 1~3~ 2 ~ 2~ 2
~ 2 =~3~3~ 1 ~ 1~ 1
~1~3~ 2 ~ 2~ 2
~2~2~ 1 ~ 1~ 1
~1
shows that ~ 3is nonzero and is in the span. A calculation shows that ~ 3is orthogonal to the preceding vector ~ 1. ~ 1~ 3=~ 1~3proj[~ 1](~3)proj[~ 2](~3)
=~ 1~3proj[~ 1](~3)
~ 1proj[~ 2](~3) = 0 (Here's the dierence from the i= 2 case | the second line has two kinds of terms. The rst term is zero because this projection is orthogonal, as in the i= 2 case. The second term is zero because ~ 1is orthogonal to ~ 2and so is orthogonal to any vector in the line spanned by ~ 2.) The check that ~ 3is also orthogonal to the other preceding vector ~ 2is similar. QED Beyond having the vectors in the basis be orthogonal, we can do more; we can arrange for each vector to have length one by dividing each by its own length (we can normalize the lengths). 2.8 Example Normalizing the length of each vector in the orthogonal basis of Example 2.6 produces this orthonormal basis . h0 @1=p 3 1=p 3 1=p 31 A;0 @1=p 6 2=p 6 1=p 61 A;0 @1=p 2 0 1=p 21 Ai Besides its intuitive appeal, and its analogy with the standard basis EnforRn, an orthonormal basis also simplies some computations. See Exercise 17, for example. Exercises 2.9Perform the Gram-Schmidt process on each of these bases for R2. (a)h1 1 ;2 1 i(b)h0 1 ;1 3 i(c)h0 1 ;1 0 i Then turn those orthogonal bases into orthonormal bases. X2.10 Perform the Gram-Schmidt process on each of these bases for R3. Section VI. Projection 257 (a)h0 @2 2 21 A;0 @1 0 11 A;0 @0 3 11 Ai(b)h0 @1 1 01 A;0 @0 1 01 A;0 @2 3 11 Ai Then turn those orthogonal bases into orthonormal bases. X2.11 Find an orthonormal basis for this subspace of R3: the plane xy+z= 0. 2.12 Find an orthonormal basis for this subspace of R4. f0 BB@x y z w1 CCAxyz+w= 0 andx+z= 0g 2.13 Show that any linearly independent subset of Rncan be orthogonalized with- out changing its span. X2.14 What happens if we apply the Gram-Schmidt process to a basis that is already orthogonal? 2.15 Leth~ 1;:::;~ kibe a set of mutually orthogonal vectors in Rn. (a)Prove that for any ~ vin the space, the vector ~ v(proj[~ 1](~ v)+


+proj[~ vk](~ v)) is orthogonal to each of ~ 1, . . . ,~ k. (b)Illustrate the prior item in R3by using~ e1as~ 1, using~ e2as~ 2, and taking ~ vto have components 1, 2, and 3. (c)Show that proj[~ 1](~ v) +


+ proj[~ vk](~ v) is the vector in the span of the set of~ 's that is closest to ~ v.Hint. To the illustration done for the prior part, add a vector d1~ 1+d2~ 2and apply the Pythagorean Theorem to the resulting triangle. 2.16 Find a vector in R3that is orthogonal to both of these.0 @1 5 11 A0 @2 2 01 A X2.17 One advantage of orthogonal bases is that they simplify nding the represen- tation of a vector with respect to that basis. (a)For this vector and this non-orthogonal basis for R2 ~ v=2 3 B=h1 1 ;1 0 i rst represent the vector with respect to the basis. Then project the vector into the span of each basis vector [ ~1] and [~2]. (b)With this orthogonal basis for R2 K=h1 1 ;1 1 i represent the same vector ~ vwith respect to the basis. Then project the vector into the span of each basis vector. Note that the coecients in the representation and the projection are the same. (c)LetK=h~ 1;:::;~ kibe an orthogonal basis for some subspace of Rn. Prove that for any ~ vin the subspace, the i-th component of the representation RepK(~ v) is the scalar coecient ( ~ v~ i)=(~ i~ i) from proj[~ i](~ v). (d)Prove that ~ v= proj[~ 1](~ v) +


+ proj[~ k](~ v). 2.18 Bessel's Inequality . Consider these orthonormal sets B1=f~ e1gB2=f~ e1;~ e2gB3=f~ e1;~ e2;~ e3gB4=f~ e1;~ e2;~ e3;~ e4g along with the vector ~ v2R4whose components are 4, 3, 2, and 1. (a)Find the coecient c1for the projection of ~ vinto the span of the vector in B1. Check thatk~ vk2jc1j2. 258 Chapter Three. Maps Between Spaces (b)Find the coecients c1andc2for the projection of ~ vinto the spans of the two vectors in B2. Check thatk~ vk2jc1j2+jc2j2. (c)Findc1,c2, andc3associated with the vectors in B3, andc1,c2,c3, andc4 for the vectors in B4. Check thatk~ vk2jc1j2+


+jc3j2and thatk~ vk2 jc1j2+


+jc4j2. Show that this holds in general: where f~ 1;:::;~ kgis an orthonormal set and ciis coecient of the projection of a vector ~ vfrom the space then k~ vk2jc1j2+


+ jckj2.Hint. One way is to look at the inequality 0 k~ v(c1~ 1+


+ck~ k)k2 and expand the c's. 2.19 Prove or disprove: every vector in Rnis in some orthogonal basis. 2.20 Show that the columns of an nnmatrix form an orthonormal set if and only if the inverse of the matrix is its transpose. Produce such a matrix. 2.21 Does the proof of Theorem 2.2 fail to consider the possibility that the set of vectors is empty (i.e., that k= 0)? 2.22 Theorem 2.7 describes a change of basis from any basis B=h~1;:::;~kito one that is orthogonal K=h~ 1;:::;~ ki. Consider the change of basis matrix RepB;K(id). (a)Prove that the matrix RepK;B(id) changing bases in the direction opposite to that of the theorem has an upper triangular shape | all of its entries below the main diagonal are zeros. (b)Prove that the inverse of an upper triangular matrix is also upper triangular (if the matrix is invertible, that is). This shows that the matrix RepB;K(id) changing bases in the direction described in the theorem is upper triangular. 2.23 Complete the induction argument in the proof of Theorem 2.7. VI.3 Projection Into a Subspace This subsection, like the others in this section, is optional. It also requires material from the optional earlier subsection on Combining Subspaces. The prior subsections project a vector into a line by decomposing it into two parts: the part in the line proj[~ s](~ v) and the rest ~ vproj[~ s](~ v). To generalize projection to arbitrary subspaces, we follow this idea. 3.1 Denition For any direct sum V=MNand any~ v2V, the projection of~ vintoMalongNis projM;N(~ v) =~ m where~ v=~ m+~ nwith~ m2M;~ n2N. This denition doesn't involve a sense of `orthogonal' so we can apply it to spaces other than subspaces of an Rn. (Denitions of orthogonality for other spaces are perfectly possible, but we haven't seen any in this book.) 3.2 Example The spaceM22of 22 matrices is the direct sum of these two. M=f a b 0 0a;b2RgN=f 0 0 c dc;d2Rg Section VI. Projection 259 To project A=3 1 0 4 intoMalongN, we rst x bases for the two subspaces. BM=h1 0 0 0 ;0 1 0 0 iBN=h0 0 1 0 ;0 0 0 1 i The concatenation of these B=BM_BN=h 1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1 i is a basis for the entire space, because the space is the direct sum, so we can use it to represent A. 3 1 0 4 = 3
1 0 0 0 + 1
0 1 0 0 + 0
0 0 1 0 + 4
0 0 0 1 Now the projection of AintoMalongNis found by keeping the Mpart of this sum and dropping the Npart. projM;N(3 1 0 4 ) = 3
1 0 0 0 + 1
0 1 0 0 =3 1 0 0 3.3 Example Both subscripts on projM;N(~ v) are signicant. The rst sub- scriptMmatters because the result of the projection is an ~ m2M, and changing this subspace would change the possible results. For an example showing that the second subscript matters, x this plane subspace of R3and its basis M=f0 @x y z1 Ay2z= 0gBM=h0 @1 0 01 A;0 @0 2 11 Ai and compare the projections along two dierent subspaces. N=fk0 @0 0 11 Ak2Rg ^N=fk0 @0 1 21 Ak2Rg (Verication that R3=MNandR3=M^Nis routine.) We will check that these projections are dierent by checking that they have dierent eects on this vector. ~ v=0 @2 2 51 A For the rst one we nd a basis for N BN=h0 @0 0 11 Ai 260 Chapter Three. Maps Between Spaces and represent ~ vwith respect to the concatenation BM_BN. 0 @2 2 51 A= 2
0 @1 0 01 A+ 1
0 @0 2 11 A+ 4
0 @0 0 11 A The projection of ~ vintoMalongNis found by keeping the Mpart and dropping theNpart. projM;N(~ v) = 2
0 @1 0 01 A+ 1
0 @0 2 11 A=0 @2 2 11 A For the other subspace ^N, this basis is natural. B^N=h0 @0 1 21 Ai Representing ~ vwith respect to the concatenation 0 @2 2 51 A= 2
0 @1 0 01 A+ (9=5)
0 @0 2 11 A(8=5)
0 @0 1 21 A and then keeping only the Mpart gives this. projM;^N(~ v) = 2
0 @1 0 01 A+ (9=5)
0 @0 2 11 A=0 @2 18=5 9=51 A Therefore projection along dierent subspaces may yield dierent results. These pictures compare the two maps. Both show that the projection is indeed `into' the plane and `along' the line. MN M^N Notice that the projection along Nis not orthogonal | there are members of the planeMthat are not orthogonal to the dotted line. But the projection along ^Nis orthogonal. A natural question is: what is the relationship between the projection op- eration dened above, and the operation of orthogonal projection into a line? The second picture above suggests the answer | orthogonal projection into a line is a special case of the projection dened above; it is just projection along a subspace perpendicular to the line. Section VI. Projection 261 N M In addition to pointing out that projection along a subspace is a generalization, this scheme shows how to dene orthogonal projection into any subspace of Rn, of any dimension. 3.4 Denition The orthogonal complement of a subspace MofRnis M?=f~ v2Rn~ vis perpendicular to all vectors in Mg (read \Mperp"). The orthogonal projection projM(~ v) of a vector is its pro- jection into MalongM?. 3.5 Example InR3, to nd the orthogonal complement of the plane P=f0 @x y z1 A3x+ 2yz= 0g we start with a basis for P. B=h0 @1 0 31 A;0 @0 1 21 Ai Any~ vperpendicular to every vector in Bis perpendicular to every vector in the span ofB(the proof of this assertion is Exercise 19). Therefore, the subspace P?consists of the vectors that satisfy these two conditions. 0 @1 0 31 A0 @v1 v2 v31 A= 00 @0 1 21 A0 @v1 v2 v31 A= 0 We can express those conditions more compactly as a linear system. P?=f0 @v1 v2 v31 A1 0 3 0 1 20 @v1 v2 v31 A=0 0 g We are thus left with nding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. P?=f0 @v1 v2 v31 Av1+ 3v3= 0 v2+ 2v3= 0g=fk0 @3 2 11 Ak2Rg Instead of the term orthogonal complement, in some contexts this is called the linenormal to the plane. 262 Chapter Three. Maps Between Spaces 3.6 Example WhereMis thexy-plane subspace of R3, what isM?? A common rst reaction is that M?is theyz-plane, but that's not right. Some vectors from the yz-plane are not perpendicular to every vector in the xy-plane. 0 @1 1 01 A6?0 @0 3 21 A = arccos(1
0 + 1
3 + 0
2p 2
p 13)0:94 rad InsteadM?is thez-axis, since proceeding as in the prior example and taking the natural basis for the xy-plane gives this. M?=f0 @x y z1 A1 0 0 0 1 00 @x y z1 A=0 0 g=f0 @x y z1 Ax= 0 andy= 0g The two examples that we've seen since Denition 3.4 illustrate the rst sentence in that denition. The next result justies the second sentence. 3.7 Lemma LetMbe a subspace of Rn. The orthogonal complement of Mis also a subspace. The space is the direct sum of the two Rn=MM?. And, for any~ v2Rn, the vector ~ vprojM(~ v) is perpendicular to every vector in M. Proof .First, the orthogonal complement M?is a subspace of Rnbecause, as noted in the prior two examples, it is a nullspace. Next, we can start with any basis BM=h~ 1;:::;~ kiforMand expand it to a basis for the entire space. Apply the Gram-Schmidt process to get an orthog- onal basisK=h~ 1;:::;~ niforRn. ThisKis the concatenation of two bases h~ 1;:::;~ ki(with the same number of members as BM) andh~ k+1;:::;~ ni. The rst is a basis for M, so if we show that the second is a basis for M?then we will have that the entire space is the direct sum of the two subspaces. Exercise 17 from the prior subsection proves this about any orthogonal ba- sis: each vector ~ vin the space is the sum of its orthogonal projections onto the lines spanned by the basis vectors. ~ v= proj[~ 1](~ v) +


+ proj[~ n](~ v) ( ) To check this, represent the vector ~ v=r1~ 1+


+rn~ n, apply~ ito both sides ~ v~ i= (r1~ 1+


+rn~ n)~ i=r1
0 +


+ri
(~ i~ i) +


+rn
0, and solve to get ri= (~ v~ i)=(~ i~ i), as desired. Since obviously any member of the span of h~ k+1;:::;~ niis orthogonal to any vector in M, to show that this is a basis for M?we need only show the other containment | that any ~ w2M?is in the span of this basis. The prior paragraph does this. On projections into basis vectors from M, any~ w2M? gives proj[~ 1](~ w) =~0;:::; proj[~ k](~ w) =~0 and therefore ( ) gives that ~ wis a linear combination of ~ k+1;:::;~ n. Thus this is a basis for M?andRnis the direct sum of the two. Section VI. Projection 263 The nal sentence is proved in much the same way. Write ~ v= proj[~ 1](~ v) +


+ proj[~ n](~ v). Then projM(~ v) is gotten by keeping only the Mpart and dropping the M?part projM(~ v) = proj[~ k+1](~ v) +


+ proj[~ k](~ v). Therefore ~ vprojM(~ v) consists of a linear combination of elements of M?and so is perpendicular to every vector in M. QED We can nd the orthogonal projection into a subspace by following the steps of the proof, but the next result gives a formula. 3.8 Theorem Let~ vbe a vector in Rnand letMbe a subspace of Rn with basish~1;:::;~ki. IfAis the matrix whose columns are the ~'s then projM(~ v) =c1~1+


+ck~kwhere the coecients ciare the entries of the vector (AtransA)1Atrans
~ v. That is, projM(~ v) =A(AtransA)1Atrans
~ v. Proof .The vector projM(~ v) is a member of Mand so it is a linear combination of basis vectors c1
~1+


+ck
~k. SinceA's columns are the ~'s, that can be expressed as: there is a ~ c2Rksuch that projM(~ v) =A~ c(this is expressed compactly with matrix multiplication as in Example 3.5 and 3.6). Because ~ vprojM(~ v) is perpendicular to each member of the basis, we have this (again, expressed compactly). ~0 =Atrans ~ vA~ c
=Atrans~ vAtransA~ c Solving for ~ c(showing that AtransAis invertible is an exercise) ~ c= AtransA
1Atrans
~ v gives the formula for the projection matrix as projM(~ v) =A
~ c. QED 3.9 Example To orthogonally project this vector into this subspace ~ v=0 @1 1 11 AP=f0 @x y z1 Ax+z= 0g rst make a matrix whose columns are a basis for the subspace A=0 @0 1 1 0 011 A and then compute. A AtransA
1Atrans=0 @0 1 1 0 011 A 1 0 0 1=2 0 1 0 1 01 =0 @1=2 01=2 0 1 0 1=2 0 1=21 A 264 Chapter Three. Maps Between Spaces With the matrix, calculating the orthogonal projection of any vector into Pis easy. projP(~ v) =0 @1=2 01=2 0 1 0 1=2 0 1=21 A0 @1 1 11 A=0 @0 1 01 A Note, as a check, that this result is indeed in P. Exercises X3.10 Project the vectors into MalongN. (a)3 2 ; M =fx yx+y= 0g; N =fx yx2y= 0g (b)1 2 ; M =fx yxy= 0g; N =fx y2x+y= 0g (c)0 @3 0 11 A; M =f0 @x y z1 Ax+y= 0g; N =fc
0 @1 0 11 Ac2Rg X3.11 FindM?. (a)M=fx yx+y= 0g(b)M=fx y2x+ 3y= 0g (c)M=fx yxy= 0g(d)M=f~0g(e)M=fx yx= 0g (f)M=f0 @x y z1 Ax+ 3y+z= 0g(g)M=f0 @x y z1 Ax= 0 andy+z= 0g 3.12 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the planePspecied by 3 x+ 2yz= 0 in R3. (a)Find a basis for P. (b)FindP?and a basis for P?. (c)Represent this vector with respect to the concatenation of the two bases from the prior item. ~ v=0 @1 1 21 A (d)Find the orthogonal projection of ~ vintoPby keeping only the Ppart from the prior item. (e)Check that against the result from applying Theorem 3.8. X3.13 We have three ways to nd the orthogonal projection of a vector into a line, the Denition 1.1 way from the rst subsection of this section, the Example 3.2 and 3.3 way of representing the vector with respect to a basis for the space and then keeping the Mpart, and the way of Theorem 3.8. For these cases, do all three ways. (a)~ v=1 3 ; M =fx yx+y= 0g (b)~ v=0 @0 1 21 A; M =f0 @x y z1 Ax+z= 0 andy= 0g Section VI. Projection 265 3.14 Check that the operation of Denition 3.1 is well-dened. That is, in Exam- ple 3.2 and 3.3, doesn't the answer depend on the choice of bases? 3.15 What is the orthogonal projection into the trivial subspace? 3.16 What is the projection of ~ vintoMalongNif~ v2M? 3.17 Show that if MRnis a subspace with orthonormal basis h~ 1;:::;~ nithen the orthogonal projection of ~ vintoMis this. (~ v~ 1)
~ 1+


+ (~ v~ n)
~ n X3.18 Prove that the map p:V!Vis the projection into MalongNif and only if the map idpis the projection into NalongM. (Recall the denition of the dierence of two maps: (id p) (~ v) = id(~ v)p(~ v) =~ vp(~ v).) X3.19 Show that if a vector is perpendicular to every vector in a set then it is perpendicular to every vector in the span of that set. 3.20 True or false: the intersection of a subspace and its orthogonal complement is trivial. 3.21 Show that the dimensions of orthogonal complements add to the dimension of the entire space. X3.22 Suppose that ~ v1;~ v22Rnare such that for all complements M;NRn, the projections of ~ v1and~ v2intoMalongNare equal. Must ~ v1equal~ v2? (If so, what if we relax the condition to: all orthogonal projections of the two are equal?) X3.23 LetM;N be subspaces of Rn. The perp operator acts on subspaces; we can ask how it interacts with other such operations. (a)Show that two perps cancel: ( M?)?=M. (b)Prove that MNimplies that N?M?. (c)Show that ( M+N)?=M?\N?. X3.24 The material in this subsection allows us to express a geometric relationship that we have not yet seen between the rangespace and the nullspace of a linear map. (a)Representf:R3!Rgiven by0 @v1 v2 v31 A7!1v1+ 2v2+ 3v3 with respect to the standard bases and show that0 @1 2 31 A is a member of the perp of the nullspace. Prove that N(f)?is equal to the span of this vector. (b)Generalize that to apply to any f:Rn!R. (c)Representf:R3!R2 0 @v1 v2 v31 A7!1v1+ 2v2+ 3v3 4v1+ 5v2+ 6v3 with respect to the standard bases and show that0 @1 2 31 A;0 @4 5 61 A are both members of the perp of the nullspace. Prove that N(f)?is the span of these two. ( Hint. See the third item of Exercise 23.) 266 Chapter Three. Maps Between Spaces (d)Generalize that to apply to any f:Rn!Rm. This, and related results, is called the Fundamental Theorem of Linear Algebra in [Strang 93]. 3.25 Dene a projection to be a linear transformation t:V!Vwith the property that repeating the projection does nothing more than does the projection alone: ( t t) (~ v) =t(~ v) for all~ v2V. (a)Show that orthogonal projection into a line has that property. (b)Show that projection along a subspace has that property. (c)Show that for any such tthere is a basis B=h~1;:::;~niforVsuch that t(~i) =(~ii= 1;2;:::; r ~0i=r+ 1;r+ 2;:::; n whereris the rank of t. (d)Conclude that every projection is a projection along a subspace. (e)Also conclude that every projection has a representation RepB;B(t) =IZ ZZ in block partial-identity form. 3.26 A square matrix is symmetric if eachi;jentry equals the j;ientry (i.e., if the matrix equals its transpose). Show that the projection matrix A(AtransA)1Atrans is symmetric. [Strang 80] Hint. Find properties of transposes by looking in the index under `transpose'. Topic: Line of Best Fit 267 Topic: Line of Best Fit This Topic requires the formulas from the subsections on Orthogonal Projection Into a Line, and Projection Into a Subspace. Scientists are often presented with a system that has no solution and they must nd an answer anyway. More precisely stated, they must nd a best answer. For instance, this is the result of ipping a penny, including some interme- diate numbers. number of ips 30 60 90 number of heads 16 34 51 In an experiment we can expect that samples will vary | here, sometimes the experimental ratio of heads to ips overestimates this penny's long-term ratio and sometimes it underestimates. So we expect that the system derived from the experiment has no solution. 30m= 16 60m= 34 90m= 51 That is, the vector of experimental data is not in the subspace of solutions. 0 @16 34 511 A62fm0 @30 60 901 Am2Rg However, we want to nd the mthat most nearly works. An orthogonal projec- tion of the data vector into the line subspace gives our best guess. 0 @16 34 511 A0 @30 60 901 A 0 @30 60 901 A0 @30 60 901 A
0 @30 60 901 A=7110 12600
0 @30 60 901 A The estimate ( m= 7110=126000:56) is a bit high but not much, so probably the penny is fair enough. The line with the slope m0:56 is the line of best t for this data. ips30 60 90heads 3060 268 Chapter Three. Maps Between Spaces Minimizing the distance between the given vector and the vector used as the right-hand side minimizes the total of these vertical lengths, and consequently we say that the line has been obtained through tting by least-squares (the vertical scale here has been exaggerated ten times to make the lengths visible). We arranged the equation above so that the line must pass through (0 ;0) because we take it to be the line whose slope is this coin's true proportion of heads to ips. We can also handle cases where the line need not pass through the origin. For example, the dierent denominations of U.S. money have dierent aver- age times in circulation (the $2 bill is left o as a special case). How long should we expect a $25 bill to last? denomination 1 5 10 20 50 100 average life (years) 1:5 2 3 5 9 20 The plot (see below) looks roughly linear. It isn't a perfect line, i.e., the linear system with equations b+ 1m= 1:5, . . . ,b+ 100m= 20 has no solution, but we can again use orthogonal projection to nd a best approximation. Consider the matrix of coecients of that linear system and also its vector of constants, the experimentally-determined values. A=0 BBBBBB@1 1 1 5 1 10 1 20 1 50 1 1001 CCCCCCA~ v=0 BBBBBB@1:5 2 3 5 9 201 CCCCCCA The ending result in the subsection on Projection into a Subspace says that coecients bandmso that the linear combination of the columns of Ais as close as possible to the vector ~ vare the entries of ( AtransA)1Atrans
~ v. Some calculation gives an intercept of b= 1:05 and a slope of m= 0:18. denom10 30 50 70 90avg life 515 Pluggingx= 25 into the equation of the line shows that such a bill should last between ve and six years. Topic: Line of Best Fit 269 We close by considering the progression of world record times for the men's mile race.[Oakley & Baker] In the early 1900's many people wondered when this record would fall below the four minute mark. Here are the times that were in force on January rst of each decade through the rst half of that century. (Restricting ourselves to the times at the start of each decade reduces the data entry burden and gives much the same result. There are a number of dierent sequences of times from competing standards bodies but these are from [WikipediaMensMile].) year 1870 1880 1890 1900 1910 1920 1930 1940 1950 secs 268:8 264:5 258:4 255:6 255:6 252:6 250:4 246:4 241:4 We can use this data to predict the date for 240 seconds, and we can then compare to the actual date. A few minutes in Sage gives the slope and intercept. sage: data=[[1870,268.8], [1880,264.5], [1890,258.4], [1900,255.6], ....: [1910,255.6], [1920,252.6], [1930,250.4], [1940,246.4], ....: [1950,241.4]] sage: var('slope,intercept') (slope, intercept) sage: model(x) = slope*x+intercept sage: find_fit(data,model) [intercept == 837.0872267857003, slope == -0.30483333572258886] Plotting the data along with the line of best t sage: points(data)+plot(model(intercept=find_fit(data,model)[0].rhs(), ....: slope=find_fit(data,model)[1].rhs()),(x,1860,1960),color='red') gives this graph. Note that the progression is surprisingly linear. Our prediction is 1958 :73; the actual date of Roger Bannister's record was 1954-May-06. Exercises The calculations here are best done on a computer. Some of the problems require more data that is available in your library, on the Internet, or in the Answers to the Exercises. 270 Chapter Three. Maps Between Spaces 1Use least-squares to judge if the coin in this experiment is fair. ips 8 16 24 32 40 heads 4 9 13 17 20 2For the men's mile record, rather than give each of the many records and its exact date, we've \smoothed" the data somewhat by taking a periodic sample. Do the longer calculation and compare the conclusions. 3Find the line of best t for the men's 1500 meter run. How does the slope compare with that for the men's mile? (The distances are close; a mile is about 1609 meters.) 4Find the line of best t for the records for women's mile. 5Do the lines of best t for the men's and women's miles cross? 6When the space shuttle Challenger exploded in 1986, one of the criticisms made of NASA's decision to launch was in the way the analysis of number of O-ring failures versus temperature was made (of course, O-ring failure caused the explosion). Four O-ring failures will cause the rocket to explode. NASA had data from 24 previous ights. tempF 53 75 57 58 63 70 70 66 67 67 67 failures 3 2 1 1 1 1 1 0 0 0 0 68 69 70 70 72 73 75 76 76 78 79 80 81 0 0 0 0 0 0 0 0 0 0 0 0 0 The temperature that day was forecast to be 31F. (a)NASA based the decision to launch partially on a chart showing only the ights that had at least one O-ring failure. Find the line that best ts these seven ights. On the basis of this data, predict the number of O-ring failures when the temperature is 31, and when the number of failures will exceed four. (b)Find the line that best ts all 24 ights. On the basis of this extra data, predict the number of O-ring failures when the temperature is 31, and when the number of failures will exceed four. Which do you think is the more accurate method of predicting? (An excellent discussion appears in [Dalal, et. al.].) 7This table lists the average distance from the sun to each of the rst seven planets, using earth's average as a unit. Mercury Venus Earth Mars Jupiter Saturn Uranus 0:39 0:72 1:00 1:52 5:20 9:54 19:2 (a)Plot the number of the planet (Mercury is 1, etc.) versus the distance. Note that it does not look like a line, and so nding the line of best t is not fruitful. (b)It does, however look like an exponential curve. Therefore, plot the number of the planet versus the logarithm of the distance. Does this look like a line? (c)The asteroid belt between Mars and Jupiter is thought to be what is left of a planet that broke apart. Renumber so that Jupiter is 6, Saturn is 7, and Uranus is 8, and plot against the log again. Does this look better? (d)Use least squares on that data to predict the location of Neptune. (e)Repeat to predict where Pluto is. (f)Is the formula accurate for Neptune and Pluto? This method was used to help discover Neptune (although the second item is mis- leading about the history; actually, the discovery of Neptune in position 9 prompted people to look for the \missing planet" in position 5). See [Gardner, 1970] Topic: Line of Best Fit 271 8William Bennett has proposed an Index of Leading Cultural Indicators for the US ([Bennett], in 1993). Among the statistics cited are the average daily hours spent watching TV, and the average combined SAT scores. 1960 1965 1970 1975 1980 1985 1990 1992 TV 5:06 5:29 5:56 6:07 6:36 7:07 6:55 7:04 SAT 975 969 948 910 890 906 900 899 Suppose that a cause and eect relationship is proposed between the time spent watching TV and the decline in SAT scores (in this article, Mr. Bennett does not argue that there is a direct connection). (a)Find the line of best t relating the independent variable of average daily TV hours to the dependent variable of SAT scores. (b)Find the most recent estimate of the average daily TV hours (Bennett's cites Neilsen Media Research as the source of these estimates). Estimate the associ- ated SAT score. How close is your estimate to the actual average? (Warning: a change has been made recently in the SAT, so you should investigate whether some adjustment needs to be made to the reported average to make a valid comparison.) 272 Chapter Three. Maps Between Spaces Topic: Geometry of Linear Maps The pictures below contrast f1(x) =exandf2(x) =x2, which are nonlinear, withh1(x) = 2xandh2(x) =x, which are linear. Each of the four pictures shows the domain R1on the left mapped to the codomain R1on the right. Arrows trace out where each map sends x= 0,x= 1,x= 2,x=1, and x=2. Note how the nonlinear maps distort the domain in transforming it into the range. For instance, f1(1) is further from f1(2) than it is from f1(0) | the map is spreading the domain out unevenly so that an interval near x= 2 is spread apart more than is an interval near x= 0 when they are carried over to the range. -505 -505 -505 -505 The linear maps are nicer, more regular, in that for each map all of the domain is spread by the same factor. -505 -505 -505 -505 The only linear maps from R1toR1are multiplications by a scalar. In higher dimensions more can happen. For instance, this linear transformation of R2, rotates vectors counterclockwise, and is not just a scalar multiplication. Topic: Geometry of Linear Maps 273 x y 7!xcosysin xsin+ycos 7! The transformation of R3which projects vectors into the xz-plane is also not just a rescaling. 0 @x y z1 A7!0 @x 0 z1 A 7! Nonetheless, even in higher dimensions the situation isn't too complicated. Below, we use the standard bases to represent each linear map h:Rn!Rm by a matrix H. Recall that any Hcan be factored H=PBQ , wherePandQare nonsingular and Bis a partial-identity matrix. Further, recall that nonsingular matrices factor into elementary matrices PBQ =TnTn1


TjBTj1


T1, which are matrices that are obtained from the identity Iwith one Gaussian step Iki!Mi(k)Ii$j!Pi;jIki+j!Ci;j(k) (i6=j,k6= 0). So if we understand the eect of a linear map described by a partial-identity matrix, and the eect of linear mapss described by the elementary matrices, then we will in some sense understand the eect of any linear map. (The pictures below stick to transformations of R2for ease of drawing, but the statements hold for maps from any Rnto any Rm.) The geometric eect of the linear transformation represented by a partial- identity matrix is projection. 0 @x y z1 A0 @1 0 0 0 1 0 0 0 01 A E3;E3 !0 @x y 01 A For theMi(k) matrices, the geometric action of a transformation represented by such a matrix (with respect to the standard basis) is to stretch vectors by a factor ofkalong thei-th axis. This map stretches by a factor of 3 along the x-axis. x y 7!3x y 7! 274 Chapter Three. Maps Between Spaces Note that if 0k<1 or ifk<0 then thei-th component goes the other way; here, toward the left. x y 7!2x y 7! Either of these is a dilation . The action of a transformation represented by a Pi;jpermutation matrix is to interchange the i-th andj-th axes; this is a particular kind of re ection. x y 7!y x 7! In higher dimensions, permutations involving many axes can be decomposed into a combination of swaps of pairs of axes | see Exercise 5. The remaining case is that of matrices of the form Ci;j(k). Recall that, for instance, that C1;2(2) performs 2 1+2. x y 1 0 2 1 E2;E2 !x 2x+y In the picture below, the vector ~ uwith the rst component of 1 is aected less than the vector ~ vwith the rst component of 2 | h(~ u) is only 2 higher than ~ u whileh(~ v) is 4 higher than ~ v. x y 7!x 2x+y 7!~ u ~ vh(~ u)h(~ v) Any vector with a rst component of 1 would be aected as is ~ u; it would be slid up by 2. And any vector with a rst component of 2 would be slid up 4, as was ~ v. That is, the transformation represented by Ci;j(k) aects vectors depending on theiri-th component. Another way to see this same point is to consider the action of this map on the unit square. In the next picture, vectors with a rst component of 0, like the origin, are not pushed vertically at all but vectors with a positive rst component are slid up. Here, all vectors with a rst component of 1 | the entire right side of the square | is aected to the same extent. More generally, vectors on the same vertical line are slid up the same amount, namely, they are slid up by twice their rst component. The resulting shape, a rhombus, has the same base and height as the square (and thus the same area) but the right angles are gone. Topic: Geometry of Linear Maps 275 x y 7!x 2x+y 7! For contrast the next picture shows the eect of the map represented by C2;1(1). In this case, vectors are aected according to their second component. The vectorx y
is slid horozontally by twice y. x y 7!x+ 2y y 7! Because of this action, this kind of map is called a skew. With that, we have covered the geometric eect of the four types of com- ponents in the expansion H=TnTn1


TjBTj1


T1, the partial-identity projectionBand the elementary Ti's. Since we understand its components, we in some sense understand the action of any H. As an illustration of this assertion, recall that under a linear map, the image of a subspace is a subspace and thus the linear transformation hrepresented by Hmaps lines through the origin to lines through the origin. (The dimension of the image space cannot be greater than the dimension of the domain space, so a line can't map onto, say, a plane.) We will extend that to show that any line, not just those through the origin, is mapped by hto a line. The proof is simply that the partial- identity projection Band the elementary Ti's each turn a line input into a line output (verifying the four cases is Exercise 6), and therefore their composition also preserves lines. Thus, by understanding its components we can understand arbitrary square matrices H, in the sense that we can prove things about them. An understanding of the geometric eect of linear transformations on Rnis very important in mathematics. Here is a familiar application from calculus. On the left is a picture of the action of the nonlinear function y(x) =x2+x. As at that start of this Topic, overall the geometric eect of this map is irregular in that at dierent domain points it has dierent eects (e.g., as the domain pointxgoes from 2 to2, the associated range point f(x) at rst decreases, then pauses instantaneously, and then increases). 05 05 276 Chapter Three. Maps Between Spaces But in calculus we don't focus on the map overall, we focus instead on the local eect of the map. At x= 1 the derivative is y0(1) = 3, so that near x= 1 we have
y3

x. That is, in a neighborhood of x= 1, in carrying the domain to the codomain this map causes it to grow by a factor of 3 | it is, locally, approximately, a dilation. The picture below shows a small interval in the domain ( x
x::x +
x) carried over to an interval in the codomain (y
y::y +
y) that is three times as wide:
y3

x. x= 1y= 2 (When the above picture is drawn in the traditional cartesian way then the prior sentence about the rate of growth of y(x) is usually stated: the derivative y0(1) = 3 gives the slope of the line tangent to the graph at the point (1 ;2).) In higher dimensions, the idea is the same but the approximation is not just theR1-to-R1scalar multiplication case. Instead, for a function y:Rn!Rm and a point ~ x2Rn, the derivative is dened to be the linear map h:Rn!Rm best approximating how ychanges near y(~ x). So the geometry studied above applies. We will close this Topic by remarking how this point of view makes clear an often-misunderstood, but very important, result about derivatives: the deriva- tive of the composition of two functions is computed by using the Chain Rule for combining their derivatives. Recall that (with suitable conditions on the two functions) d(gf) dx(x) =dg dx(f(x))
df dx(x) so that, for instance, the derivative of sin( x2+3x) is cos(x2+3x)
(2x+3). How does this combination arise? From this picture of the action of the composition. xf(x)g(f(x)) Topic: Geometry of Linear Maps 277 The rst map fdilates the neighborhood of xby a factor of df dx(x) and the second map gdilates some more, this time dilating a neighborhood of f(x) by a factor of dg dx(f(x) ) and as a result, the composition dilates by the product of these two. In higher dimensions the map expressing how a function changes near a point is a linear map, and is expressed as a matrix. (So we understand the basic geometry of higher-dimensional derivatives; they are compositions of dila- tions, interchanges of axes, shears, and a projection). And, the Chain Rule just multiplies the matrices. Thus, the geometry of linear maps h:Rn!Rmis appealing both for its simplicity and for its usefulness. Exercises 1Leth:R2!R2be the transformation that rotates vectors clockwise by =4 ra- dians. (a)Find the matrix Hrepresenting hwith respect to the standard bases. Use Gauss' method to reduce Hto the identity. (b)Translate the row reduction to to a matrix equation TjTj1


T1H=I(the prior item shows both that His similar to I, and that no column operations are needed to derive IfromH). (c)Solve this matrix equation for H. (d)Sketch the geometric eect matrix, that is, sketch how His expressed as a combination of dilations, ips, skews, and projections (the identity is a trivial projection). 2What combination of dilations, ips, skews, and projections produces a rotation counterclockwise by 2 =3 radians? 3What combination of dilations, ips, skews, and projections produces the map h:R3!R3represented with respect to the standard bases by this matrix?0 @1 2 1 3 6 0 1 2 21 A 4Show that any linear transformation of R1is the map that multiplies by a scalar x7!kx. 5Show that for any permutation (that is, reordering) pof the numbers 1, . . . , n, the map0 BBB@x1 x2 ... xn1 CCCA7!0 BBB@xp(1) xp(2) ... xp(n)1 CCCA can be accomplished with a composition of maps, each of which only swaps a single pair of coordinates. Hint: it can be done by induction on n. (Remark: in the fourth 278 Chapter Three. Maps Between Spaces chapter we will show this and we will also show that the parity of the number of swaps used is determined by p. That is, although a particular permutation could be accomplished in two dierent ways with two dierent numbers of swaps, either both ways use an even number of swaps, or both use an odd number.) 6Show that linear maps preserve the linear structures of a space. (a)Show that for any linear map from RntoRm, the image of any line is a line. The image may be a degenerate line, that is, a single point. (b)Show that the image of any linear surface is a linear surface. This generalizes the result that under a linear map the image of a subspace is a subspace. (c)Linear maps preserve other linear ideas. Show that linear maps preserve \betweeness": if the point Bis betweenAandCthen the image of Bis between the image of Aand the image of C. 7Use a picture like the one that appears in the discussion of the Chain Rule to answer: if a function f:R!Rhas an inverse, what's the relationship between how the function | locally, approximately | dilates space, and how its inverse dilates space (assuming, of course, that it has an inverse)? Topic: Markov Chains 279 Topic: Markov Chains Here is a simple game: a player bets on coin tosses, a dollar each time, and the game ends either when the player has no money left or is up to ve dollars. If the player starts with three dollars, what is the chance that the game takes at least ve ips? Twenty-ve ips? At any point, this player has either $0, or $1, . . . , or $5. We say that the player is in the states0,s1, . . . , ors5. A game consists of moving from state to state. For instance, a player now in state s3has on the next ip a :5 chance of moving to state s2and a:5 chance of moving to s4. The boundary states are a bit dierent; once in state s0or states5, the player never leaves. Letpi(n) be the probability that the player is in state siaftern ips. Then, for instance, we have that the probability of being in state s0after ipn+ 1 is p0(n+ 1) =p0(n) + 0:5
p1(n). This matrix equation sumarizes. 0 BBBBBB@1:5 0 0 0 0 0 0:5 0 0 0 0:5 0:5 0 0 0 0:5 0:5 0 0 0 0 :5 0 0 0 0 0 0 :5 11 CCCCCCA0 BBBBBB@p0(n) p1(n) p2(n) p3(n) p4(n) p5(n)1 CCCCCCA=0 BBBBBB@p0(n+ 1) p1(n+ 1) p2(n+ 1) p3(n+ 1) p4(n+ 1) p5(n+ 1)1 CCCCCCA With the initial condition that the player starts with three dollars, calculation gives this. n= 0n= 1n= 2n= 3n= 4


n= 240 BBBBBB@0 0 0 1 0 01 CCCCCCA0 BBBBBB@0 0 :5 0 :5 01 CCCCCCA0 BBBBBB@0 :25 0 :5 0 :251 CCCCCCA0 BBBBBB@:125 0 :375 0 :25 :251 CCCCCCA0 BBBBBB@:125 :1875 0 :3125 0 :3751 CCCCCCA


0 BBBBBB@:39600 :00276 0 :00447 0 :596761 CCCCCCA As this computational exploration suggests, the game is not likely to go on for long, with the player quickly ending in either state s0or states5. For instance, after the fourth ip there is a probability of 0 :50 that the game is already over. (Because a player who enters either of the boundary states never leaves, they are said to be absorbtive .) This game is an example of a Markov chain , named for A.A. Markov, who worked in the rst half of the 1900's. Each vector of p's is a probability vector and the matrix is a transition matrix . The notable feature of a Markov chain model is that it is historyless in that with a xed transition matrix, the next state depends only on the current state, not on any prior states. Thus a player, say, who arrives at s2by starting in state s3, then going to state s2, then to s1, and then to s2has at this point exactly the same chance of moving next to states3as does a player whose history was to start in s3, then go to s4, and to s3, and then to s2. 280 Chapter Three. Maps Between Spaces Here is a Markov chain from sociology. A study ([Macdonald & Ridge], p. 202) divided occupations in the United Kingdom into upper level (executives and professionals), middle level (supervisors and skilled manual workers), and lower level (unskilled). To determine the mobility across these levels in a gen- eration, about two thousand men were asked, \At which level are you, and at which level was your father when you were fourteen years old?" This equation summarizes the results. 0 @:60:29:16 :26:37:27 :14:34:571 A0 @pU(n) pM(n) pL(n)1 A=0 @pU(n+ 1) pM(n+ 1) pL(n+ 1)1 A For instance, a child of a lower class worker has a :27 probability of growing up to be middle class. Notice that the Markov model assumption about history seems reasonable | we expect that while a parent's occupation has a direct in uence on the occupation of the child, the grandparent's occupation has no such direct in uence. With the initial distribution of the respondents's fathers given below, this table lists the distributions for the next ve generations. n= 0n= 1n= 2n= 3n= 4n= 50 @:12 :32 :561 A0 @:23 :34 :421 A0 @:29 :34 :371 A0 @:31 :34 :351 A0 @:32 :33 :341 A0 @:33 :33 :341 A One more example, from a very important subject, indeed. The World Series of American baseball is played between the team winning the American League and the team winning the National League (we follow [Brunner] but see also [Woodside]). The series is won by the rst team to win four games. That means that a series is in one of twenty-four states: 0-0 (no games won yet by either team), 1-0 (one game won for the American League team and no games for the National League team), etc. If we assume that there is a probability pthat the American League team wins each game then we have the following transition matrix. 0 BBBBBBBBB@0 0 0 0 ::: p 0 0 0 ::: 1p 0 0 0 ::: 0p 0 0::: 0 1p p 0::: 0 0 1p0::: ............1 CCCCCCCCCA0 BBBBBBBBB@p0-0(n) p1-0(n) p0-1(n) p2-0(n) p1-1(n) p0-2(n) ...1 CCCCCCCCCA=0 BBBBBBBBB@p0-0(n+ 1) p1-0(n+ 1) p0-1(n+ 1) p2-0(n+ 1) p1-1(n+ 1) p0-2(n+ 1) ...1 CCCCCCCCCA An especially interesting special case is p= 0:50; this table lists the resulting components of the n= 0 through n= 7 vectors. (The code to generate this table in the computer algebra system Octave follows the exercises.) Topic: Markov Chains 281 n= 0n= 1n= 2n= 3n= 4n= 5n= 6n= 7 00 10 01 20 11 02 30 21 12 03 40 31 22 13 04 41 32 23 14 42 33 24 43 341 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0:5 0:5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0:25 0:5 0:25 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0:125 0:375 0:375 0:125 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0:0625 0:25 0:375 0:25 0:0625 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0:0625 0 0 0 0:0625 0:125 0:3125 0:3125 0:125 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0:0625 0 0 0 0:0625 0:125 0 0 0:125 0:15625 0:3125 0:15625 0 00 0 0 0 0 0 0 0 0 0 0:0625 0 0 0 0:0625 0:125 0 0 0:125 0:15625 0 0:15625 0:15625 0:15625 Note that evenly-matched teams are likely to have a long series | there is a probability of 0 :625 that the series goes at least six games. One reason for the inclusion of this Topic is that Markov chains are one of the most widely-used applications of matrix operations. Another reason is that it provides an example of the use of matrices where we do not consider the signicance of the maps represented by the matrices. For more on Markov chains, there are many sources such as [Kemeny & Snell] and [Iosifescu]. Exercises Use a computer for these problems. You can, for instance, adapt the Octave script given below. 1These questions refer to the coin- ipping game. (a)Check the computations in the table at the end of the rst paragraph. (b)Consider the second row of the vector table. Note that this row has alter- nating 0's. Must p1(j) be 0 when jis odd? Prove that it must be, or produce a counterexample. (c)Perform a computational experiment to estimate the chance that the player ends at ve dollars, starting with one dollar, two dollars, and four dollars. 2We consider throws of a die, and say the system is in state siif the largest number yet appearing on the die was i. (a)Give the transition matrix. (b)Start the system in state s1, and run it for ve throws. What is the vector at the end? 282 Chapter Three. Maps Between Spaces [Feller], p. 424 3There has been much interest in whether industries in the United States are moving from the Northeast and North Central regions to the South and West, motivated by the warmer climate, by lower wages, and by less unionization. Here is the transition matrix for large rms in Electric and Electronic Equipment ([Kelton], p. 43) NE NC S W Z NE NC S W Z0:787 0 0 0 0:0210 0:966 0:063 0 0:0090 0:034 0:937 0:074 0:0050:111 0 0 0:612 0:0100:102 0 0 0:314 0:954 For example, a rm in the Northeast region will be in the West region next year with probability 0 :111. (The Zentry is a \birth-death" state. For instance, with probability 0 :102 a large Electric and Electronic Equipment rm from the North- east will move out of this system next year: go out of business, move abroad, or move to another category of rm. There is a 0 :021 probability that a rm in the National Census of Manufacturers will move into Electronics, or be created, or move in from abroad, into the Northeast. Finally, with probability 0 :954 a rm out of the categories will stay out, according to this research.) (a)Does the Markov model assumption of lack of history seem justied? (b)Assume that the initial distribution is even, except that the value at Zis 0:9. Compute the vectors for n= 1 through n= 4. (c)Suppose that the initial distribution is this. NE NC S W Z 0:0000 0:6522 0:3478 0:0000 0:0000 Calculate the distributions for n= 1 through n= 4. (d)Find the distribution for n= 50 andn= 51. Has the system settled down to an equilibrium? 4This model has been suggested for some kinds of learning ([Wickens], p. 41). The learner starts in an undecided state sU. Eventually the learner has to decide to do either response A(that is, end in state sA) or response B(ending insB). However, the learner doesn't jump right from being undecided to being sure Ais the correct thing to do (or B). Instead, the learner spends some time in a \tentative- A" state, or a \tentative- B" state, trying the response out (denoted here tAandtB). Imagine that once the learner has decided, it is nal, so once sAorsBis entered it is never left. For the other state changes, imagine a transition is made with probability pin either direction. (a)Construct the transition matrix. (b)Takep= 0:25 and take the initial vector to be 1 at sU. Run this for ve steps. What is the chance of ending up at sA? (c)Do the same for p= 0:20. (d)Graphpversus the chance of ending at sA. Is there a threshold value for p, above which the learner is almost sure not to take longer than ve steps? 5A certain town is in a certain country (this is a hypothetical problem). Each year ten percent of the town dwellers move to other parts of the country. Each year one percent of the people from elsewhere move to the town. Assume that there are two states sT, living in town, and sC, living elsewhere. (a)Construct the transistion matrix. Topic: Markov Chains 283 (b)Starting with an initial distribution sT= 0:3 andsC= 0:7, get the results for the rst ten years. (c)Do the same for sT= 0:2. (d)Are the two outcomes alike or dierent? 6For the World Series application, use a computer to generate the seven vectors forp= 0:55 andp= 0:6. (a)What is the chance of the National League team winning it all, even though they have only a probability of 0 :45 or 0:40 of winning any one game? (b)Graph the probability pagainst the chance that the American League team wins it all. Is there a threshold value | a pabove which the better team is essentially ensured of winning? (Some sample code is included below.) 7AMarkov matrix has each entry positive and each column sums to 1. (a)Check that the three transistion matrices shown in this Topic meet these two conditions. Must any transition matrix do so? (b)Observe that if A~ v0=~ v1andA~ v1=~ v2thenA2is a transition matrix from ~ v0to~ v2. Show that a power of a Markov matrix is also a Markov matrix. (c)Generalize the prior item by proving that the product of two appropriately- sized Markov matrices is a Markov matrix. Computer Code This script markov.m for the computer algebra system Octave was used to generate the table of World Series outcomes. (The sharp character #marks the rest of a line as a comment.) # Octave script file to compute chance of World Series outcomes. function w = markov(p,v) q = 1-p; A=[0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-0 p,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-0 q,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-1_ 0,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-0 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-1 0,0,q,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-2__ 0,0,0,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-0 0,0,0,q,p,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-1 0,0,0,0,q,p, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-2_ 0,0,0,0,0,q, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-3 0,0,0,0,0,0, p,0,0,0,1,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 4-0 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-1__ 0,0,0,0,0,0, 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-2 0,0,0,0,0,0, 0,0,q,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-3 0,0,0,0,0,0, 0,0,0,q,0,0, 0,0,1,0,0,0, 0,0,0,0,0,0; # 0-4_ 0,0,0,0,0,0, 0,0,0,0,0,p, 0,0,0,1,0,0, 0,0,0,0,0,0; # 4-1 0,0,0,0,0,0, 0,0,0,0,0,q, p,0,0,0,0,0, 0,0,0,0,0,0; # 3-2 0,0,0,0,0,0, 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0; # 2-3__ 0,0,0,0,0,0, 0,0,0,0,0,0, 0,q,0,0,0,0, 1,0,0,0,0,0; # 1-4 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,p,0, 0,1,0,0,0,0; # 4-2 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,q,p, 0,0,0,0,0,0; # 3-3_ 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,q, 0,0,0,1,0,0; # 2-4 284 Chapter Three. Maps Between Spaces 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,p,0,1,0; # 4-3 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,q,0,0,1]; # 3-4 w = A * v; endfunction Then the Octave session was this. > v0=[1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0] > p=.5 > v1=markov(p,v0) > v2=markov(p,v1) ... Translating to another computer algebra system should be easy | all have com- mands similar to these. Topic: Orthonormal Matrices 285 Topic: Orthonormal Matrices InThe Elements , Euclid considers two gures to be the same if they have the same size and shape. That is, the triangles below are not equal because they are not the same set of points. But they are congruent | essentially indistin- guishable for Euclid's purposes | because we can imagine picking the plane up, sliding it over and rotating it a bit, although not warping or stretching it, and then putting it back down, to superimpose the rst gure on the second. (Euclid never explicitly states this principle but he uses it often [Casey].) P1P2 P3Q1Q2 Q3 In modern terminology, \picking the plane up . . . " means considering a map from the plane to itself. Euclid has limited consideration to only certain trans- formations of the plane, ones that may possibly slide or turn the plane but not bend or stretch it. Accordingly, we dene a map f:R2!R2to be distance- preserving or a rigid motion or an isometry , if for all points P1;P22R2, the distance from f(P1) tof(P2) equals the distance from P1toP2. We also dene a plane gure to be a set of points in the plane and we say that two gures arecongruent if there is a distance-preserving map from the plane to itself that carries one gure onto the other. Many statements from Euclidean geometry follow easily from these deni- tions. Some are: (i) collinearity is invariant under any distance-preserving map (that is, if P1,P2, andP3are collinear then so are f(P1),f(P2), andf(P3)), (ii) betweeness is invariant under any distance-preserving map (if P2is between P1andP3then so isf(P2) betweenf(P1) andf(P3)), (iii) the property of being a triangle is invariant under any distance-preserving map (if a gure is a triangle then the image of that gure is also a triangle), (iv) and the property of being a circle is invariant under any distance-preserving map. In 1872, F. Klein suggested that Euclidean geometry can be characterized as the study of prop- erties that are invariant under these maps. (This forms part of Klein's Erlanger Program, which proposes the organizing principle that each kind of geometry | Euclidean, projective, etc. | can be described as the study of the properties that are invariant under some group of transformations. The word `group' here means more than just `collection', but that lies outside of our scope.) We can use linear algebra to characterize the distance-preserving maps of the plane. First, there are distance-preserving transformations of the plane that are not linear. The obvious example is this translation . x y 7!x y +1 0 =x+ 1 y 286 Chapter Three. Maps Between Spaces However, this example turns out to be the only example, in the sense that if fis distance-preserving and sends ~0 to~ v0then the map ~ v7!f(~ v)~ v0is linear. That will follow immediately from this statement: a map tthat is distance-preserving and sends~0 to itself is linear. To prove this equivalent statement, let t(~ e1) =a b t(~ e2) =c d for somea;b;c;d2R. Then to show that tis linear, we can show that it can be represented by a matrix, that is, that tacts in this way for all x;y2R. ~ v=x y t7!ax+cy bx+dy () Recall that if we x three non-collinear points then any point in the plane can be described by giving its distance from those three. So any point ~ vin the domain is determined by its distance from the three xed points ~0,~ e1, and~ e2. Similarly, any point t(~ v) in the codomain is determined by its distance from the three xed points t(~0),t(~ e1), andt(~ e2) (these three are not collinear because, as mentioned above, collinearity is invariant and ~0,~ e1, and~ e2are not collinear). In fact, because tis distance-preserving, we can say more: for the point ~ vin the plane that is determined by being the distance d0from~0, the distance d1from ~ e1, and the distance d2from~ e2, its image t(~ v) must be the unique point in the codomain that is determined by being d0fromt(~0),d1fromt(~ e1), andd2from t(~ e2). Because of the uniqueness, checking that the action in ( ) works in the d0,d1, andd2cases dist( x y ;~0) = dist(t( x y );t(~0)) = dist( ax+cy bx+dy ;~0) (tis assumed to send ~0 to itself) dist( x y ;~ e1) = dist(t( x y );t(~ e1)) = dist( ax+cy bx+dy ; a b ) and dist(x y ;~ e2) = dist(t(x y );t(~ e2)) = dist(ax+cy bx+dy ;c d ) suces to show that ( ) describes t. Those checks are routine. Thus, any distance-preserving f:R2!R2can be written f(~ v) =t(~ v) +~ v0 for some constant vector ~ v0and linear map tthat is distance-preserving. Not every linear map is distance-preserving, for example, ~ v7!2~ vdoes not preserve distances. But there is a neat characterization: a linear transformation tof the plane is distance-preserving if and only if both kt(~ e1)k=kt(~ e2)k= 1 and t(~ e1) is orthogonal to t(~ e2). The `only if' half of that statement is easy | because tis distance-preserving it must preserve the lengths of vectors, and because t is distance-preserving the Pythagorean theorem shows that it must preserve Topic: Orthonormal Matrices 287 orthogonality. For the `if' half, it suces to check that the map preserves lengths of vectors, because then for all ~ pand~ qthe distance between the two is preservedkt(~ p~ q)k=kt(~ p)t(~ q)k=k~ p~ qk. For that check, let ~ v=x y t(~ e1) =a b t(~ e2) =c d and, with the `if' assumptions that a2+b2=c2+d2= 1 andac+bd= 0 we have this. kt(~ v)k2= (ax+cy)2+ (bx+dy)2 =a2x2+ 2acxy +c2y2+b2x2+ 2bdxy +d2y2 =x2(a2+b2) +y2(c2+d2) + 2xy(ac+bd) =x2+y2 =k~ vk2 One thing that is neat about this characterization is that we can easily recognize matrices that represent such a map with respect to the standard bases. Those matrices have that when the columns are written as vectors then they are of length one and are mutually orthogonal. Such a matrix is called an orthonormal matrix ororthogonal matrix (the second term is commonly used to mean not just that the columns are orthogonal, but also that they have length one). We can use this insight to delimit the geometric actions possible in distance- preserving maps. Because kt(~ v)k=k~ vk, any~ vis mapped by tto lie somewhere on the circle about the origin that has radius equal to the length of ~ v. In particular, ~ e1and~ e2are mapped to the unit circle. What's more, once we x the unit vector ~ e1as mapped to the vector with components aandbthen there are only two places where ~ e2can be mapped if that image is to be perpendicular to the rst vector: one where ~ e2maintains its position a quarter circle clockwise from~ e1  a b b a RepE2;E2(t) = ab b a and one where is is mapped a quarter circle counterclockwise.  a b  b aRepE2;E2(t) = a b ba 288 Chapter Three. Maps Between Spaces We can geometrically describe these two cases. Let be the angle between thex-axis and the image of ~ e1, measured counterclockwise. The rst matrix above represents, with respect to the standard bases, a rotation of the plane by radians.  a b b a x y t7!xcosysin xsin+ycos The second matrix above represents a re ection of the plane through the line bisecting the angle between ~ e1andt(~ e1).  a b  b ax y t7!xcos+ysin xsinycos (This picture shows ~ e1re ected up into the rst quadrant and ~ e2re ected down into the fourth quadrant.) Note again: the angle between ~ e1and~ e2runs counterclockwise, and in the rst map above the angle from t(~ e1) tot(~ e2) is also counterclockwise, so the orientation of the angle is preserved. But in the second map the orientation is reversed. A distance-preserving map is direct if it preserves orientations and opposite if it reverses orientation. So, we have characterized the Euclidean study of congruence: it considers, for plane gures, the properties that are invariant under combinations of (i) a rotation followed by a translation, or (ii) a re ection followed by a translation (a re ection followed by a non-trivial translation is a glide re ection ). Another idea, besides congruence of gures, encountered in elementary ge- ometry is that gures are similar if they are congruent after a change of scale. These two triangles are similar since the second is the same shape as the rst, but 3=2-ths the size. P1P2 P3Q1Q2 Q3 From the above work, we have that gures are similar if there is an orthonormal matrixTsuch that the points ~ qon one are derived from the points ~ pby~ q= (kT)~ v+~ p0for some nonzero real number kand constant vector ~ p0. Topic: Orthonormal Matrices 289 Although many of these ideas were rst explored by Euclid, mathematics is timeless and they are very much in use today. One application of the maps studied above is in computer graphics. We can, for example, animate this top view of a cube by putting together lm frames of it rotating; that's a rigid motion. Frame 1 Frame 2 Frame 3 We could also make the cube appear to be moving away from us by producing lm frames of it shrinking, which gives us gures that are similar. Frame 1: Frame 2: Frame 3: Computer graphics incorporates techniques from linear algebra in many other ways (see Exercise 4). So the analysis above of distance-preserving maps is useful as well as inter- esting. A beautiful book that explores some of this area is [Weyl]. More on groups, of transformations and otherwise, can be found in any book on Modern Algebra, for instance [Birkho & MacLane]. More on Klein and the Erlanger Program is in [Yaglom]. Exercises 1Decide if each of these is an orthonormal matrix. (a)1=p 21=p 2 1=p 21=p 2 (b)1=p 31=p 3 1=p 31=p 3 (c)1=p 3p 2=p 3 p 2=p 31=p 3 2Write down the formula for each of these distance-preserving maps. (a)the map that rotates =6 radians, and then translates by ~ e2 (b)the map that re ects about the line y= 2x (c)the map that re ects about y=2xand translates over 1 and up 1 3 (a) The proof that a map that is distance-preserving and sends the zero vector to itself incidentally shows that such a map is one-to-one and onto (the point in the domain determined by d0,d1, andd2corresponds to the point in the codomain determined by those three). Therefore any distance-preserving map has an inverse. Show that the inverse is also distance-preserving. (b)Prove that congruence is an equivalence relation between plane gures. 4In practice the matrix for the distance-preserving linear transformation and the translation are often combined into one. Check that these two computations yield the same rst two components. a c b dx y +e f0 @a c e b d f 0 0 11 A0 @x y 11 A 290 Chapter Three. Maps Between Spaces (These are homogeneous coordinates ; see the Topic on Projective Geometry). 5 (a) Verify that the properties described in the second paragraph of this Topic as invariant under distance-preserving maps are indeed so. (b)Give two more properties that are of interest in Euclidean geometry from your experience in studying that subject that are also invariant under distance- preserving maps. (c)Give a property that is not of interest in Euclidean geometry and is not invariant under distance-preserving maps. Chapter Four Determinants In the rst chapter of this book we considered linear systems and we picked out the special case of systems with the same number of equations as unknowns, those of the form T~ x=~bwhereTis a square matrix. We noted a distinction between two classes of T's. While such systems may have a unique solution or no solutions or innitely many solutions, if a particular Tis associated with a unique solution in any system, such as the homogeneous system ~b=~0, then Tis associated with a unique solution for every ~b. We call such a matrix of coecients `nonsingular'. The other kind of T, where every linear system for which it is the matrix of coecients has either no solution or innitely many solutions, we call `singular'. Through the second and third chapters the value of this distinction has been a theme. For instance, we now know that nonsingularity of an nnmatrixT is equivalent to each of these: a systemT~ x=~bhas a solution, and that solution is unique; Gauss-Jordan reduction of Tyields an identity matrix; the rows of Tform a linearly independent set; the columns of Tform a basis for Rn; any map that Trepresents is an isomorphism; an inverse matrix T1exists. So when we look at a particular square matrix, the question of whether it is nonsingular is one of the rst things that we ask. This chapter develops a formula to determine this. (Since we will restrict the discussion to square matrices, in this chapter we will usually simply say `matrix' in place of `square matrix'.) More precisely, we will develop innitely many formulas, one for 1 1 ma- trices, one for 22 matrices, etc. Of course, these formulas are related | that is, we will develop a family of formulas, a scheme that describes the formula for each size. 291 292 Chapter Four. Determinants I Definition For 11 matrices, determining nonsingularity is trivial. a
is nonsingular i a6= 0 The 22 formula came out in the course of developing the inverse. a b c d is nonsingular i adbc6= 0 The 33 formula can be produced similarly (see Exercise 9). 0 @a b c d e f g h i1 Ais nonsingular i aei+bfg+cdhhfaidbgec6= 0 With these cases in mind, we posit a family of formulas, a,adbc, etc. For each nthe formula gives rise to a determinant function det nn:Mnn!Rsuch that annnmatrixTis nonsingular if and only if det nn(T)6= 0. (We usually omit the subscript because if Tisnnthen `det(T)' could only mean `det nn(T)'.) I.1 Exploration This subsection is optional. It brie y describes how an investigator might come to a good general denition, which is given in the next subsection. The three cases above don't show an evident pattern to use for the general nnformula. We may spot that the 1 1 termahas one letter, that the 2 2 termsadandbchave two letters, and that the 3 3 termsaei, etc., have three letters. We may also observe that in those terms there is a letter from each row and column of the matrix, e.g., the letters in the cdhterm 0 @c d h1 A come one from each row and one from each column. But these observations perhaps seem more puzzling than enlightening. For instance, we might wonder why some of the terms are added while others are subtracted. A good problem solving strategy is to see what properties a solution must have and then search for something with those properties. So we shall start by asking what properties we require of the formulas. At this point, our primary way to decide whether a matrix is singular is to do Gaussian reduction and then check whether the diagonal of resulting echelon form matrix has any zeroes (that is, to check whether the product down the diagonal is zero). So, we may expect that the proof that a formula Section I. Definition 293 determines singularity will involve applying Gauss' method to the matrix, to show that in the end the product down the diagonal is zero if and only if the determinant formula gives zero. This suggests our initial plan: we will look for a family of functions with the property of being unaected by row operations and with the property that a determinant of an echelon form matrix is the product of its diagonal entries. Under this plan, a proof that the functions determine singularity would go, \Where T!


! ^Tis the Gaussian reduction, the determinant of Tequals the determinant of ^T(because the determinant is unchanged by row operations), which is the product down the diagonal, which is zero if and only if the matrix is singular". In the rest of this subsection we will test this plan on the 2 2 and 33 determinants that we know. We will end up modifying the \unaected by row operations" part, but not by much. The rst step in checking the plan is to test whether the 2 2 and 33 formulas are unaected by the row operation of combining: if Tki+j! ^T then is det( ^T) = det(T)? This check of the 2 2 determinant after the k1+2 operation det(a b ka+c kb +d ) =a(kb+d)(ka+c)b=adbc shows that it is indeed unchanged, and the other 2 2 combination k2+1 gives the same result. The 3 3 combination k3+2leaves the determinant unchanged det(0 @a b c kg+d kh +e ki +f g h i1 A) =a(kh+e)i+b(ki+f)g+c(kg+d)h h(ki+f)ai(kg+d)bg(kh+e)c =aei+bfg+cdhhfaidbgec as do the other 3 3 row combination operations. So there seems to be promise in the plan. Of course, perhaps the 4 4 deter- minant formula is aected by row combinations. We are exploring a possibility here and we do not yet have all the facts. Nonetheless, so far, so good. The next step is to compare det( ^T) with det(T) for the operation Ti$j! ^T of swapping two rows. The 2 2 row swap 1$2 det(c d a b ) =cbad does not yield adbc. This1$3swap inside of a 3 3 matrix det(0 @g h i d e f a b c1 A) =gec+hfa+idbbfgcdhaei 294 Chapter Four. Determinants also does not give the same determinant as before the swap | again there is a sign change. Trying a dierent 3 3 swap1$2 det(0 @d e f a b c g h i1 A) =dbi+ecg+fahhcdiaegbf also gives a change of sign. Thus, row swaps appear to change the sign of a determinant. This mod- ies our plan, but does not wreck it. We intend to decide nonsingularity by considering only whether the determinant is zero, not by considering its sign. Therefore, instead of expecting determinants to be entirely unaected by row operations, will look for them to change sign on a swap. To nish, we compare det( ^T) to det(T) for the operation Tki! ^T of multiplying a row by a scalar k6= 0. One of the 2 2 cases is det( a b kc kd ) =a(kd)(kc)b=k
(adbc) and the other case has the same result. Here is one 3 3 case det(0 @a b c d e f kg kh ki1 A) =ae(ki) +bf(kg) +cd(kh) (kh)fa(ki)db(kg)ec =k
(aei+bfg+cdhhfaidbgec) and the other two are similar. These lead us to suspect that multiplying a row bykmultiplies the determinant by k. This ts with our modied plan because we are asking only that the zeroness of the determinant be unchanged and we are not focusing on the determinant's sign or magnitude. In summary, to develop the scheme for the formulas to compute determi- nants, we look for determinant functions that remain unchanged under the operation of row combination, that change sign on a row swap, and that rescale on the rescaling of a row. In the next two subsections we will nd that for each nsuch a function exists and is unique. For the next subsection, note that, as above, scalars come out of each row without aecting other rows. For instance, in this equality det(0 @3 3 9 2 1 1 5 1051 A) = 3
det(0 @1 1 3 2 1 1 5 1051 A) the 3 isn't factored out of all three rows, only out of the top row. The determi- nant acts on each row of independently of the other rows. When we want to use this property of determinants, we shall write the determinant as a function of the rows: `det( ~ 1;~ 2;:::~ n)', instead of as `det( T)' or `det(t1;1;:::;tn;n)'. The denition of the determinant that starts the next subsection is written in this way. Section I. Definition 295 Exercises X1.1Evaluate the determinant of each. (a)3 1 1 1 (b)0 @2 0 1 3 1 1 1 0 11 A (c)0 @4 0 1 0 0 1 1 311 A 1.2Evaluate the determinant of each. (a)2 0 1 3 (b)0 @2 1 1 0 52 13 41 A (c)0 @2 3 4 5 6 7 8 9 11 A X1.3Verify that the determinant of an upper-triangular 3 3 matrix is the product down the diagonal. det(0 @a b c 0e f 0 0i1 A) =aei Do lower-triangular matrices work the same way? X1.4Use the determinant to decide if each is singular or nonsingular. (a)2 1 3 1 (b)0 1 11 (c)4 2 2 1 1.5Singular or nonsingular? Use the determinant to decide. (a)0 @2 1 1 3 2 2 0 1 41 A (b)0 @1 0 1 2 1 1 4 1 31 A (c)0 @2 1 0 32 0 1 0 01 A X1.6Each pair of matrices dier by one row operation. Use this operation to compare det(A) with det(B). (a)A=1 2 2 3 B=1 2 01 (b)A=0 @3 1 0 0 0 1 0 1 21 AB=0 @3 1 0 0 1 2 0 0 11 A (c)A=0 @11 3 2 26 1 0 41 AB=0 @11 3 1 13 1 0 41 A 1.7Show this. det(0 @1 1 1 a b c a2b2c21 A) = (ba)(ca)(cb) X1.8Which real numbers xmake this matrix singular?12x 4 8 8x 1.9Do the Gaussian reduction to check the formula for 3 3 matrices stated in the preamble to this section.0 @a b c d e f g h i1 Ais nonsingular i aei+bfg+cdhhfaidbgec6= 0 1.10 Show that the equation of a line in R2thru (x1;y1) and (x2;y2) is expressed by this determinant. det(0 @x y 1 x1y11 x2y211 A) = 0x16=x2 296 Chapter Four. Determinants X1.11 Many people know this mnemonic for the determinant of a 3 3 matrix: rst repeat the rst two columns and then sum the products on the forward diagonals and subtract the products on the backward diagonals. That is, rst write0 @h1;1h1;2h1;3h1;1h1;2 h2;1h2;2h2;3h2;1h2;2 h3;1h3;2h3;3h3;1h3;21 A and then calculate this. h1;1h2;2h3;3+h1;2h2;3h3;1+h1;3h2;1h3;2 h3;1h2;2h1;3h3;2h2;3h1;1h3;3h2;1h1;2 (a)Check that this agrees with the formula given in the preamble to this section. (b)Does it extend to other-sized determinants? 1.12 The cross product of the vectors ~ x=0 @x1 x2 x31 A~ y=0 @y1 y2 y31 A is the vector computed as this determinant. ~ x~ y= det(0 @~ e1~ e2~ e3 x1x2x3 y1y2y31 A) Note that the rst row is composed of vectors, the vectors from the standard basis forR3. Show that the cross product of two vectors is perpendicular to each vector. 1.13 Prove that each statement holds for 2 2 matrices. (a)The determinant of a product is the product of the determinants det( ST) = det(S)
det(T). (b)IfTis invertible then the determinant of the inverse is the inverse of the determinant det( T1) = ( det(T) )1. MatricesTandT0aresimilar if there is a nonsingular matrix Psuch thatT0= PTP1. (This denition is in Chapter Five.) Show that similar 2 2 matrices have the same determinant. X1.14 Prove that the area of this region in the plane  x1 y1 x2 y2 is equal to the value of this determinant. det(x1x2 y1y2 ) Compare with this. det(x2x1 y2y1 ) 1.15 Prove that for 22 matrices, the determinant of a matrix equals the determi- nant of its transpose. Does that also hold for 3 3 matrices? X1.16 Is the determinant function linear | is det( x
T+y
S) =x
det(T)+y
det(S)? 1.17 Show that if Ais 33 then det(c
A) =c3
det(A) for any scalar c. Section I. Definition 297 1.18 Which real numbers makecossin sincos singular? Explain geometrically. ?1.19 If a third order determinant has elements 1, 2, . . . , 9, what is the maximum value it may have? [Am. Math. Mon., Apr. 1955] I.2 Properties of Determinants As described above, we want a formula to determine whether an nnmatrix is nonsingular. We will not begin by stating such a formula. Instead, we will begin by considering the function that such a formula calculates. We will dene the function by its properties, then prove that the function with these proper- ties exist and is unique and also describe formulas that compute this function. (Because we will show that the function exists and is unique, from the start we will say `det( T)' instead of `if there is a determinant function then det( T)' and `the determinant' instead of `any determinant'.) 2.1 Denition Anndeterminant is a function det: Mnn!Rsuch that (1) det(~ 1;:::;k
~ i+~ j;:::;~ n) = det(~ 1;:::;~ j;:::;~ n) fori6=j (2) det(~ 1;:::;~ j;:::;~ i;:::;~ n) =det(~ 1;:::;~ i;:::;~ j;:::;~ n) fori6=j (3) det(~ 1;:::;k~ i;:::;~ n) =k
det(~ 1;:::;~ i;:::;~ n) fork6= 0 (4) det(I) = 1 where Iis an identity matrix (the~ 's are the rows of the matrix). We often write jTjfor det(T). 2.2 Remark Property (2) is redundant since Ti+j!j+i!i+j!i! ^T swaps rows iandj. It is listed only for convenience. The rst result shows that a function satisfying these conditions gives a criteria for nonsingularity. (Its last sentence is that, in the context of the rst three conditions, (4) is equivalent to the condition that the determinant of an echelon form matrix is the product down the diagonal.) 2.3 Lemma A matrix with two identical rows has a determinant of zero. A matrix with a zero row has a determinant of zero. A matrix is nonsingular if and only if its determinant is nonzero. The determinant of an echelon form matrix is the product down its diagonal. 298 Chapter Four. Determinants Proof .To verify the rst sentence, swap the two equal rows. The sign of the determinant changes, but the matrix is unchanged and so its determinant is unchanged. Thus the determinant is zero. The second sentence is clearly true if the matrix is 1 1. If it has at least two rows then apply property (1) of the denition with the zero row as row j and withk= 1. det(:::;~ i;:::;~0;:::) = det(:::;~ i;:::;~ i+~0;:::) The rst sentence of this lemma gives that the determinant is zero. For the third sentence, where T!


! ^Tis the Gauss-Jordan reduction, by the denition the determinant of Tis zero if and only if the determinant of ^Tis zero (although they could dier in sign or magnitude). A nonsingular T Gauss-Jordan reduces to an identity matrix and so has a nonzero determinant. A singular Treduces to a ^Twith a zero row; by the second sentence of this lemma its determinant is zero. Finally, for the fourth sentence, if an echelon form matrix is singular then it has a zero on its diagonal, that is, the product down its diagonal is zero. The third sentence says that if a matrix is singular then its determinant is zero. So if the echelon form matrix is singular then its determinant equals the product down its diagonal. If an echelon form matrix is nonsingular then none of its diagonal entries is zero so we can use property (3) of the denition to factor them out (again, the vertical barsj


j indicate the determinant operation). t1;1t1;2t1;n 0t2;2t2;n ... 0 tn;n=t1;1
t2;2


tn;n
1t1;2=t1;1t1;n=t1;1 0 1 t2;n=t2;2 ... 0 1 Next, the Jordan half of Gauss-Jordan elimination, using property (1) of the denition, leaves the identity matrix. =t1;1
t2;2


tn;n
1 0 0 0 1 0 ... 0 1=t1;1
t2;2


tn;n
1 Therefore, if an echelon form matrix is nonsingular then its determinant is the product down its diagonal. QED That result gives us a way to compute the value of a determinant function on a matrix: do Gaussian reduction, keeping track of any changes of sign caused by row swaps and any scalars that are factored out, and then nish by multiplying down the diagonal of the echelon form result. This takes the same amount of time as Gauss' method and so is fast enugh to be practical on the matrices that we see in this book. Section I. Definition 299 2.4 Example Doing 22 determinants 2 4 1 3=2 4 0 5= 10 with Gauss' method won't give a big savings because the 2 2 determinant formula is so easy. However, a 3 3 determinant is usually easier to calculate with Gauss' method than with the formula given earlier. 2 2 6 4 4 3 03 5=2 2 6 0 09 03 5=2 2 6 03 5 0 09=54 2.5 Example Determinants of matrices any bigger than 3 3 are almost always most quickly done with this Gauss' method procedure. 1 0 1 3 0 1 1 4 0 0 0 5 0 1 0 1=1 0 1 3 0 1 1 4 0 0 0 5 0 013=1 0 1 3 0 1 1 4 0 013 0 0 0 5=(5) = 5 The prior example illustrates an important point. Although we have not yet found a 44 determinant formula, if one exists then we know what value it gives to the matrix | if there is a function with properties (1)-(4) then on the above matrix the function must return 5. 2.6 Lemma For eachn, if there is an nndeterminant function then it is unique. Proof .For anynnmatrix we can perform Gauss' method on the matrix, keeping track of how the sign alternates on row swaps, and then multiply down the diagonal of the echelon form result. By the denition and the lemma, all nn determinant functions must return this value on this matrix. Thus all nnde- terminant functions are equal, that is, there is only one input argument/output value relationship satisfying the four conditions. QED The `if there is an nndeterminant function' emphasizes that, although we can use Gauss' method to compute the only value that a determinant function could possibly return, we haven't yet shown that such a determinant function exists for all n. In the rest of the section we will produce determinant functions. Exercises For these, assume that an nndeterminant function exists for all n. X2.7Use Gauss' method to nd each determinant. (a)3 1 2 3 1 0 0 1 4(b)1 0 0 1 2 1 1 0 1 0 1 0 1 1 1 0 2.8Use Gauss' method to nd each. 300 Chapter Four. Determinants (a)21 11(b)1 1 0 3 0 2 5 2 2 2.9For which values of kdoes this system have a unique solution? x+zw= 2 y2z = 3 x+kz = 4 zw= 2 X2.10 Express each of these in terms of jHj. (a)h3;1h3;2h3;3 h2;1h2;2h2;3 h1;1h1;2h1;3 (b)h1;1h1;2h1;3 2h2;12h2;22h2;3 3h3;13h3;23h3;3 (c)h1;1+h3;1h1;2+h3;2h1;3+h3;3 h2;1h2;2h2;3 5h3;1 5h3;2 5h3;3 X2.11 Find the determinant of a diagonal matrix. 2.12 Describe the solution set of a homogeneous linear system if the determinant of the matrix of coecients is nonzero. X2.13 Show that this determinant is zero.y+z x +z x +y x y z 1 1 1 2.14 (a) Find the 11, 22, and 33 matrices with i;jentry given by (1)i+j. (b)Find the determinant of the square matrix with i;jentry (1)i+j. 2.15 (a) Find the 11, 22, and 33 matrices with i;jentry given by i+j. (b)Find the determinant of the square matrix with i;jentryi+j. X2.16 Show that determinant functions are not linear by giving a case where jA+ Bj6=jAj+jBj. 2.17 The second condition in the denition, that row swaps change the sign of a determinant, is somewhat annoying. It means we have to keep track of the number of swaps, to compute how the sign alternates. Can we get rid of it? Can we replace it with the condition that row swaps leave the determinant unchanged? (If so then we would need new 1 1, 22, and 33 formulas, but that would be a minor matter.) 2.18 Prove that the determinant of any triangular matrix, upper or lower, is the product down its diagonal. 2.19 Refer to the denition of elementary matrices in the Mechanics of Matrix Multiplication subsection. (a)What is the determinant of each kind of elementary matrix? (b)Prove that if Eis any elementary matrix then jESj=jEjjSjfor any appro- priately sized S. (c)(This question doesn't involve determinants.) Prove that if Tis singular then a productTSis also singular. (d)Show thatjTSj=jTjjSj. (e)Show that if Tis nonsingular then jT1j=jTj1. Section I. Definition 301 2.20 Prove that the determinant of a product is the product of the determinants jTSj=jTjjSjin this way. Fix the nnmatrixSand consider the function d:Mnn!Rgiven byT7!jTSj=jSj. (a)Check that dsatises property (1) in the denition of a determinant function. (b)Check property (2). (c)Check property (3). (d)Check property (4). (e)Conclude the determinant of a product is the product of the determinants. 2.21 Asubmatrix of a given matrix Ais one that can be obtained by deleting some of the rows and columns of A. Thus, the rst matrix here is a submatrix of the second. 3 1 2 50 @3 4 1 0 92 21 51 A Prove that for any square matrix, the rank of the matrix is rif and only if ris the largest integer such that there is an rrsubmatrix with a nonzero determinant. X2.22 Prove that a matrix with rational entries has a rational determinant. ?2.23 Find the element of likeness in (a) simplifying a fraction, (b) powdering the nose, (c) building new steps on the church, (d) keeping emeritus professors on campus, (e) putting B,C,Din the determinant 1a a2a3 a31a a2 B a31a C D a31: [Am. Math. Mon., Feb. 1953] I.3 The Permutation Expansion The prior subsection denes a function to be a determinant if it satises four conditions and shows that there is at most one nndeterminant function for eachn. What is left is to show that for each nsuch a function exists. How could such a function not exist? After all, we have done computations that start with a square matrix, follow the conditions, and end with a number. The diculty is that, as far as we know, the computation might not give a well-dened result. To illustrate this possibility, suppose that we were to change the second condition in the denition of determinant to be that the value of a determinant does not change on a row swap. By Remark 2.2 we know that this con icts with the rst and third conditions. Here is an instance of the con ict: here are two Gauss' method reductions of the same matrix, the rst without any row swap  1 2 3 4 31+2! 1 2 02 302 Chapter Four. Determinants and the second with a swap. 1 2 3 4 1$2!3 4 1 2 (1=3)1+2!3 4 0 2=3 Following Denition 2.1 gives that both calculations yield the determinant 2 since in the second one we keep track of the fact that the row swap changes the sign of the result of multiplying down the diagonal. But if we follow the supposition and change the second condition then the two calculations yield dierent values,2 and 2. That is, under the supposition the outcome would not be well-dened | no function exists that satises the changed second condition along with the other three. Of course, observing that Denition 2.1 does the right thing in this one instance is not enough; what we will do in the rest of this section is to show that there is never a con ict. The natural way to try this would be to dene the determinant function with: \The value of the function is the result of doing Gauss' method, keeping track of row swaps, and nishing by multiplying down the diagonal". (Since Gauss' method allows for some variation, such as a choice of which row to use when swapping, we would have to x an explicit algorithm.) Then we would be done if we veried that this way of computing the determinant satises the four properties. For instance, if Tand ^Tare related by a row swap then we would need to show that this algorithm returns determinants that are negatives of each other. However, how to verify this is not evident. So the development below will not proceed in this way. Instead, in this subsection we will dene a dierent way to compute the value of a determinant, a formula, and we will use this way to prove that the conditions are satised. The formula that we shall use is based on an insight gotten from property (3) of the denition of determinants. This property shows that determinants are not linear. 3.1 Example For this matrix det(2 A)6= 2
det(A). A=2 1 1 3 Instead, the scalar comes out of each of the two rows. 4 2 2 6= 2
2 1 2 6= 4
2 1 1 3 Since scalars come out a row at a time, we might guess that determinants are linear a row at a time. 3.2 Denition LetVbe a vector space. A map f:Vn!Rismultilinear if (1)f(~ 1;:::;~ v +~ w;:::;~ n) =f(~ 1;:::;~ v;:::;~  n) +f(~ 1;:::;~ w;:::;~  n) (2)f(~ 1;:::;k~ v;:::;~  n) =k
f(~ 1;:::;~ v;:::;~  n) for~ v;~ w2Vandk2R. Section I. Definition 303 3.3 Lemma Determinants are multilinear. Proof .The denition of determinants gives property (2) (Lemma 2.3 following that denition covers the k= 0 case) so we need only check property (1). det(~ 1;:::;~ v +~ w;:::;~ n) = det(~ 1;:::;~ v;:::;~  n) + det(~ 1;:::;~ w;:::;~  n) If the setf~ 1;:::;~ i1;~ i+1;:::;~ ngis linearly dependent then all three matrices are singular and so all three determinants are zero and the equality is trivial. Therefore assume that the set is linearly independent. This set of n-wide row vectors has n1 members, so we can make a basis by adding one more vector h~ 1;:::;~ i1;~;~ i+1;:::;~ ni. Express~ vand~ wwith respect to this basis ~ v=v1~ 1+


+vi1~ i1+vi~+vi+1~ i+1+


+vn~ n ~ w=w1~ 1+


+wi1~ i1+wi~+wi+1~ i+1+


+wn~ n giving this. ~ v+~ w= (v1+w1)~ 1+


+ (vi+wi)~+


+ (vn+wn)~ n By the denition of determinant, the value of det( ~ 1;:::;~ v +~ w;:::;~ n) is un- changed by the operation of adding (v1+w1)~ 1to~ v+~ w. ~ v+~ w(v1+w1)~ 1= (v2+w2)~ 2+


+ (vi+wi)~+


+ (vn+wn)~ n Then, to the result, we can add (v2+w2)~ 2, etc. Thus det(~ 1;:::;~ v +~ w;:::;~ n) = det(~ 1;:::; (vi+wi)
~;:::;~ n) = (vi+wi)
det(~ 1;:::;~;:::;~ n) =vi
det(~ 1;:::;~;:::;~ n) +wi
det(~ 1;:::;~;:::;~ n) (using (2) for the second equality). To nish, bring viandwiback inside in front of~and use row combination again, this time to reconstruct the expressions of ~ vand~ win terms of the basis, e.g., start with the operations of adding v1~ 1to vi~andw1~ 1towi~ 1, etc. QED Multilinearity allows us to expand a determinant into a sum of determinants, each of which involves a simple matrix. 3.4 Example We can use multilinearity to split this determinant into two, rst breaking up the rst row 2 1 4 3=2 0 4 3+0 1 4 3 and then separating each of those two, breaking along the second rows. =2 0 4 0+2 0 0 3+0 1 4 0+0 1 0 3 304 Chapter Four. Determinants We are left with four determinants, such that in each row of each matrix there is a single entry from the original matrix. 3.5 Example In the same way, a 3 3 determinant separates into a sum of many simpler determinants. We start by splitting along the rst row, producing three determinants (the zero in the 1 ;3 position is underlined to set it o visually from the zeroes that appear in the splitting). 2 11 4 3 0 2 1 5=2 0 0 4 3 0 2 1 5+0 1 0 4 3 0 2 1 5+0 01 4 3 0 2 1 5 Each of these three will itself split in three along the second row. Each of the resulting nine splits in three along the third row, resulting in twenty seven determinants =2 0 0 4 0 0 2 0 0+2 0 0 4 0 0 0 1 0+2 0 0 4 0 0 0 0 5+2 0 0 0 3 0 2 0 0+


+0 01 0 0 0 0 0 5 such that each row contains a single entry from the starting matrix. So annndeterminant expands into a sum of nndeterminants where each row of each summands contains a single entry from the starting matrix. How- ever, many of these summand determinants are zero. 3.6 Example In each of these three matrices from the above expansion, two of the rows have their entry from the starting matrix in the same column, e.g., in the rst matrix, the 2 and the 4 both come from the rst column. 2 0 0 4 0 0 0 1 00 01 0 3 0 0 0 50 1 0 0 0 0 0 0 5 Any such matrix is singular, because in each, one row is a multiple of the other (or is a zero row). Thus, any such determinant is zero, by Lemma 2.3. Therefore, the above expansion of the 3 3 determinant into the sum of the twenty seven determinants simplies to the sum of these six. 2 11 4 3 0 2 1 5=2 0 0 0 3 0 0 0 5+2 0 0 0 0 0 0 1 0 +0 1 0 4 0 0 0 0 5+0 1 0 0 0 0 2 0 0 +0 01 4 0 0 0 1 0+0 01 0 3 0 2 0 0 Section I. Definition 305 We can bring out the scalars. = (2)(3)(5)1 0 0 0 1 0 0 0 1+ (2)(0 )(1)1 0 0 0 0 1 0 1 0 + (1)(4)(5)0 1 0 1 0 0 0 0 1+ (1)(0 )(2)0 1 0 0 0 1 1 0 0 + (1)(4)(1)0 0 1 1 0 0 0 1 0+ (1)(3)(2)0 0 1 0 1 0 1 0 0 To nish, we evaluate those six determinants by row-swapping them to the identity matrix, keeping track of the resulting sign changes. = 30
(+1) + 0
(1) + 20
(1) + 0
(+1) 4
(+1)6
(1) = 12 That example illustrates the key idea. We've applied multilinearity to a 3 3 determinant to get 33separate determinants, each with one distinguished entry per row. We can drop most of these new determinants because the matrices are singular, with one row a multiple of another. We are left with the one- entry-per-row determinants also having only one entry per column (one entry from the original determinant, that is). And, since we can factor scalars out, we can further reduce to only considering determinants of one-entry-per-row-and- column matrices where the entries are ones. These are permutation matrices. Thus, the determinant can be computed in this three-step way (Step 1) for each permutation matrix, multiply together the entries from the original matrix where that permutation matrix has ones, (Step 2) multiply that by the determinant of the permutation matrix and (Step 3) do that for all permutation matrices and sum the results together. To state this as a formula, we introduce a notation for permutation matrices. Letjbe the row vector that is all zeroes except for a one in its j-th entry, so that the four-wide 2is0 1 0 0
. We can construct permutation matrices by permuting | that is, scrambling | the numbers 1, 2, . . . , n, and using them as indices on the 's. For instance, to get a 4 4 permutation matrix matrix, we can scramble the numbers from 1 to 4 into this sequence h3;2;1;4iand take the corresponding row vector 's. 0 BB@3 2 1 41 CCA=0 BB@0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 11 CCA 3.7 Denition Ann-permutation is a sequence consisting of an arrangement of the numbers 1, 2, . . . , n. 306 Chapter Four. Determinants 3.8 Example The 2-permutations are 1=h1;2iand2=h2;1i. These are the associated permutation matrices. P1=1 2 =1 0 0 1 P2=2 1 =0 1 1 0 We sometimes write permutations as functions, e.g., 2(1) = 2, and 2(2) = 1. Then the rows of P2are2(1)=2and2(2)=1. The 3-permutations are 1=h1;2;3i,2=h1;3;2i,3=h2;1;3i,4= h2;3;1i,5=h3;1;2i, and6=h3;2;1i. Here are two of the associated permu- tation matrices. P2=0 @1 3 21 A=0 @1 0 0 0 0 1 0 1 01 AP5=0 @3 1 21 A=0 @0 0 1 1 0 0 0 1 01 A For instance, the rows of P5are5(1)=3,5(2)=1, and5(3)=2. 3.9 Denition The permutation expansion for determinants is t1;1t1;2::: t 1;n t2;1t2;2::: t 2;n ... tn;1tn;2::: tn;n=t1;1(1)t2;1(2)


tn;1(n)jP1j +t1;2(1)t2;2(2)


tn;2(n)jP2j... +t1;k(1)t2;k(2)


tn;k(n)jPkj where1;:::;kare all of the n-permutations. This formula is often written in summation notation jTj=X permutations t1;(1)t2;(2)


tn;(n)jPj read aloud as \the sum, over all permutations , of terms having the form t1;(1)t2;(2)


tn;(n)jPj". This phrase is just a restating of the three-step process (Step 1) for each permutation matrix, compute t1;(1)t2;(2)


tn;(n) (Step 2) multiply that by jPjand (Step 3) sum all such terms together. 3.10 Example The familiar formula for the determinant of a 2 2 matrix can be derived in this way. t1;1t1;2 t2;1t2;2=t1;1t2;2
jP1j+t1;2t2;1
jP2j =t1;1t2;2
1 0 0 1+t1;2t2;1
0 1 1 0 =t1;1t2;2t1;2t2;1 Section I. Definition 307 (the second permutation matrix takes one row swap to pass to the identity). Similarly, the formula for the determinant of a 3 3 matrix is this. t1;1t1;2t1;3 t2;1t2;2t2;3 t3;1t3;2t3;3=t1;1t2;2t3;3jP1j+t1;1t2;3t3;2jP2j+t1;2t2;1t3;3jP3j +t1;2t2;3t3;1jP4j+t1;3t2;1t3;2jP5j+t1;3t2;2t3;1jP6j =t1;1t2;2t3;3t1;1t2;3t3;2t1;2t2;1t3;3 +t1;2t2;3t3;1+t1;3t2;1t3;2t1;3t2;2t3;1 Computing a determinant by permutation expansion usually takes longer than Gauss' method. However, here we are not trying to do the computation eciently, we are instead trying to give a determinant formula that we can prove to be well-dened. While the permutation expansion is impractical for computations, we will nd it useful in the proofs below. 3.11 Theorem For eachnthere is anndeterminant function. The proof is deferred to the following subsection. Also there is the proof of the next result (they share some features). 3.12 Theorem The determinant of a matrix equals the determinant of its transpose. The consequence of this theorem is that, while we have so far stated results in terms of rows (e.g., determinants are multilinear in their rows, row swaps change the sign, etc.), all of the results also hold in terms of columns. The nal result gives examples. 3.13 Corollary A matrix with two equal columns is singular. Column swaps change the sign of a determinant. Determinants are multilinear in their columns. Proof .For the rst statement, transposing the matrix results in a matrix with the same determinant, and with two equal rows, and hence a determinant of zero. The other two are proved in the same way. QED We nish with a summary (although the nal subsection contains the un- nished business of proving the two theorems). Determinant functions exist, are unique, and we know how to compute them. As for what determinants are about, perhaps these lines [Kemp] help make it memorable. Determinant none, Solution: lots or none. Determinant some, Solution: just one. 308 Chapter Four. Determinants Exercises These summarize the notation used in this book for the 2- and 3- permutations. i 1 2 1(i)1 2 2(i)2 1i 1 2 3 1(i)1 2 3 2(i)1 3 2 3(i)2 1 3 4(i)2 3 1 5(i)3 1 2 6(i)3 2 1 X3.14 Compute the determinant by using the permutation expansion. (a)1 2 3 4 5 6 7 8 9(b)2 2 1 31 0 2 0 5 X3.15 Compute these both with Gauss' method and with the permutation expansion formula. (a)2 1 3 1(b)0 1 4 0 2 3 1 5 1 X3.16 Use the permutation expansion formula to derive the formula for 3 3 deter- minants. 3.17 List all of the 4-permutations. 3.18 A permutation, regarded as a function from the set f1;::;ngto itself, is one- to-one and onto. Therefore, each permutation has an inverse. (a)Find the inverse of each 2-permutation. (b)Find the inverse of each 3-permutation. 3.19 Prove that fis multilinear if and only if for all ~ v;~ w2Vandk1;k22R, this holds. f(~ 1;:::;k 1~ v1+k2~ v2;:::;~ n) =k1f(~ 1;:::;~ v 1;:::;~ n) +k2f(~ 1;:::;~ v 2;:::;~ n) 3.20 How would determinants change if we changed property (4) of the denition to read thatjIj= 2? 3.21 Verify the second and third statements in Corollary 3.13. X3.22 Show that if an nnmatrix has a nonzero determinant then any column vector ~ v2Rncan be expressed as a linear combination of the columns of the matrix. 3.23 True or false: a matrix whose entries are only zeros or ones has a determinant equal to zero, one, or negative one. [Strang 80] 3.24 (a) Show that there are 120 terms in the permutation expansion formula of a 55 matrix. (b)How many are sure to be zero if the 1 ;2 entry is zero? 3.25 How many n-permutations are there? 3.26 A matrixAisskew-symmetric ifAtrans=A, as in this matrix. A=0 3 3 0 Show thatnnskew-symmetric matrices with nonzero determinants exist only for evenn. X3.27 What is the smallest number of zeros, and the placement of those zeros, needed to ensure that a 4 4 matrix has a determinant of zero? Section I. Definition 309 X3.28 If we have ndata points ( x1;y1);(x2;y2);::: ; (xn;yn) and want to nd a polynomial p(x) =an1xn1+an2xn2+


+a1x+a0passing through those points then we can plug in the points to get an nequation/nunknown linear system. The matrix of coecients for that system is called the Vandermonde matrix . Prove that the determinant of the transpose of that matrix of coecients 1 1 ::: 1 x1x2::: xn x12x22::: xn2 ... x1n1x2n1::: xnn1 equals the product, over all indices i;j2f1;:::;ngwithi < j , of terms of the formxjxi. (This shows that the determinant is zero, and the linear system has no solution, if and only if the xi's in the data are not distinct.) 3.29 A matrix can be divided into blocks , as here, 0 @1 2 0 3 4 0 0 021 A which shows four blocks, the square 2 2 and 11 ones in the upper left and lower right, and the zero blocks in the upper right and lower left. Show that if a matrix can be partitioned as T=JZ2 Z1K whereJandKare square, and Z1andZ2are all zeroes, then jTj=jJj
jKj. X3.30 Prove that for any nnmatrixTthere are at most ndistinct reals rsuch that the matrix TrIhas determinant zero (we shall use this result in Chapter Five). ?3.31 The nine positive digits can be arranged into 3 3 arrays in 9! ways. Find the sum of the determinants of these arrays. [Math. Mag., Jan. 1963, Q307] 3.32 Show that x2x3x4 x+ 1x1x3 x4x7x10= 0: [Math. Mag., Jan. 1963, Q237] ?3.33 LetSbe the sum of the integer elements of a magic square of order three and letDbe the value of the square considered as a determinant. Show that D=S is an integer. [Am. Math. Mon., Jan. 1949] ?3.34 Show that the determinant of the n2elements in the upper left corner of the Pascal triangle 1 1 1 1 : : 1 2 3: : 1 3: : 1: : : : has the value unity. [Am. Math. Mon., Jun. 1931] 310 Chapter Four. Determinants I.4 Determinants Exist This subsection is optional. It consists of proofs of two results from the prior subsection. These proofs involve the properties of permutations, which will not be used later, except in the optional Jordan Canonical Form subsection. The prior subsection attacks the problem of showing that for any size there is a determinant function on the set of square matrices of that size by using multilinearity to develop the permutation expansion. t1;1t1;2::: t 1;n t2;1t2;2::: t 2;n ... tn;1tn;2::: tn;n=t1;1(1)t2;1(2)


tn;1(n)jP1j +t1;2(1)t2;2(2)


tn;2(n)jP2j ... +t1;k(1)t2;k(2)


tn;k(n)jPkj =X permutations t1;(1)t2;(2)


tn;(n)jPj This reduces the problem to showing that there is a determinant function on the set of permutation matrices of that size. Of course, a permutation matrix can be row-swapped to the identity matrix and to calculate its determinant we can keep track of the number of row swaps. However, the problem is still not solved. We still have not shown that the result is well-dened. For instance, the determinant of P=0 BB@0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 11 CCA could be computed with one swap P1$2!0 BB@1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 11 CCA or with three. P3$1!0 BB@0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 11 CCA2$3!0 BB@0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 11 CCA1$3!0 BB@1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 11 CCA Both reductions have an odd number of swaps so we gure that jPj=1 but how do we know that there isn't some way to do it with an even number of swaps? Corollary 4.6 below proves that there is no permutation matrix that can be row-swapped to an identity matrix in two ways, one with an even number of swaps and the other with an odd number of swaps. Section I. Definition 311 4.1 Denition Two rows of a permutation matrix 0 BBBBBBB@... k ... j ...1 CCCCCCCA such thatk>j are in an inversion of their natural order. 4.2 Example This permutation matrix 0 @3 2 11 A=0 @0 0 1 0 1 0 1 0 01 A has three inversions: 3precedes1,3precedes2, and2precedes1. 4.3 Lemma A row-swap in a permutation matrix changes the number of in- versions from even to odd, or from odd to even. Proof .Consider a swap of rows jandk, wherek > j . If the two rows are adjacent P=0 BBBB@... (j) (k) ...1 CCCCAk$j!0 BBBB@... (k) (j) ...1 CCCCA then the swap changes the total number of inversions by one | either removing or producing one inversion, depending on whether (j)> (k) or not, since inversions involving rows not in this pair are not aected. Consequently, the total number of inversions changes from odd to even or from even to odd. If the rows are not adjacent then they can be swapped via a sequence of adjacent swaps, rst bringing row kup 0 BBBBBBBBBBB@... (j) (j+1) (j+2) ... (k) ...1 CCCCCCCCCCCAk$k1!k1$k2!:::j+1$j!0 BBBBBBBBBBB@... (k) (j) (j+1) ... (k1) ...1 CCCCCCCCCCCA 312 Chapter Four. Determinants and then bringing row jdown. j+1$j+2!j+2$j+3!:::k1$k!0 BBBBBBBBBBB@... (k) (j+1) (j+2) ... (j) ...1 CCCCCCCCCCCA Each of these adjacent swaps changes the number of inversions from odd to even or from even to odd. There are an odd number ( kj) + (kj1) of them. The total change in the number of inversions is from even to odd or from odd to even. QED 4.4 Denition The signum of a permutation sgn( ) is +1 if the number of inversions in Pis even, and is1 if the number of inversions is odd. 4.5 Example With the subscripts from Example 3.8 for the 3-permutations, sgn(1) = 1 while sgn( 2) =1. 4.6 Corollary If a permutation matrix has an odd number of inversions then swapping it to the identity takes an odd number of swaps. If it has an even number of inversions then swapping to the identity takes an even number of swaps. Proof .The identity matrix has zero inversions. To change an odd number to zero requires an odd number of swaps, and to change an even number to zero requires an even number of swaps. QED We still have not shown that the permutation expansion is well-dened be- cause we have not considered row operations on permutation matrices other than row swaps. We will nesse this problem: we will dene a function d:Mnn!R by altering the permutation expansion formula, replacing jPjwith sgn() d(T) =X permutations t1;(1)t2;(2):::tn;(n)sgn() (this gives the same value as the permutation expansion because the prior result shows that det( P) = sgn()). This formula's advantage is that the number of inversions is clearly well-dened | just count them. Therefore, we will show that a determinant function exists for all sizes by showing that dis it, that is, thatdsatises the four conditions. 4.7 Lemma The function dis a determinant. Hence determinants exist for everyn. Section I. Definition 313 Proof .We'll must check that it has the four properties from the denition. Property (4) is easy; in d(I) =X perms1;(1)2;(2)


n;(n)sgn() all of the summands are zero except for the product down the diagonal, which is one. For property (3) consider d(^T) whereTki!^T. X perms^t1;(1)


^ti;(i)


^tn;(n)sgn() =X t1;(1)


kti;(i)


tn;(n)sgn() Factor thekout of each term to get the desired equality. =k
X t1;(1)


ti;(i)


tn;(n)sgn() =k
d(T) For (2), let Ti$j! ^T. d(^T) =X perms^t1;(1)


^ti;(i)


^tj;(j)


^tn;(n)sgn() To convert to unhatted t's, for each consider the permutation that equals  except that the i-th andj-th numbers are interchanged, (i) =(j) and(j) = (i). Replacing the in^t1;(1)


^ti;(i)


^tj;(j)


^tn;(n)with thisgives t1;(1)


tj;(j)


ti;(i)


tn;(n). Now sgn( ) =sgn() (by Lemma 4.3) and so we get =X t1;(1)


tj;(j)


ti;(i)


tn;(n)
sgn()
=X t1;(1)


tj;(j)


ti;(i)


tn;(n)
sgn() where the sum is over all permutations derived from another permutation  by a swap of the i-th andj-th numbers. But any permutation can be derived from some other permutation by such a swap, in one and only one way, so this summation is in fact a sum over all permutations, taken once and only once. Thusd(^T) =d(T). To do property (1) let Tki+j! ^Tand consider d(^T) =X perms^t1;(1)


^ti;(i)


^tj;(j)


^tn;(n)sgn() =X t1;(1)


ti;(i)


(kti;(j)+tj;(j))


tn;(n)sgn() 314 Chapter Four. Determinants (notice: that's kti;(j), notktj;(j)). Distribute, commute, and factor. =X  t1;(1)


ti;(i)


kti;(j)


tn;(n)sgn() +t1;(1)


ti;(i)


tj;(j)


tn;(n)sgn() =X t1;(1)


ti;(i)


kti;(j)


tn;(n)sgn() +X t1;(1)


ti;(i)


tj;(j)


tn;(n)sgn() =k
X t1;(1)


ti;(i)


ti;(j)


tn;(n)sgn() +d(T) We nish by showing that the terms t1;(1)


ti;(i)


ti;(j):::tn;(n)sgn() add to zero. This sum represents d(S) whereSis a matrix equal to Texcept that rowjofSis a copy of row iofT(because the factor is ti;(j), nottj;(j)). Thus,Shas two equal rows, rows iandj. Since we have already shown that d changes sign on row swaps, as in Lemma 2.3 we conclude that d(S) = 0. QED We have now shown that determinant functions exist for each size. We already know that for each size there is at most one determinant. Therefore, the permutation expansion computes the one and only determinant value of a square matrix. We end this subsection by proving the other result remaining from the prior subsection, that the determinant of a matrix equals the determinant of its trans- pose. 4.8 Example Writing out the permutation expansion of the general 3 3 matrix and of its transpose, and comparing corresponding terms a b c d e f g h i=


+cdh
0 0 1 1 0 0 0 1 0+


(terms with the same letters) a d g b e h c f i=


+dhc
0 1 0 0 0 1 1 0 0+


shows that the corresponding permutation matrices are transposes. That is, there is a relationship between these corresponding permutations. Exercise 16 shows that they are inverses. 4.9 Theorem The determinant of a matrix equals the determinant of its transpose. Section I. Definition 315 Proof .Call the matrix Tand denote the entries of Ttranswiths's so that ti;j=sj;i. Substitution gives this jTj=X permst1;(1):::tn;(n)sgn() =X s(1);1:::s(n);nsgn() and we can nish the argument by manipulating the expression on the right to be recognizable as the determinant of the transpose. We have written all permutation expansions (as in the middle expression above) with the row indices ascending. To rewrite the expression on the right in this way, note that because is a permutation, the row indices in the term on the right (1), . . . ,(n) are just the numbers 1, . . . , n, rearranged. We can thus commute to have these ascend, giving s1;1(1)


sn;1(n)(if the column index is jand the row index is(j) then, where the row index is i, the column index is 1(i)). Substituting on the right gives =X 1s1;1(1)


sn;1(n)sgn(1) (Exercise 15 shows that sgn( 1) = sgn()). Since every permutation is the inverse of another, a sum over all 1is a sum over all permutations  =X permss1;(1):::sn;(n)sgn() =Ttrans as required. QED Exercises These summarize the notation used in this book for the 2- and 3- permutations. i 1 2 1(i)1 2 2(i)2 1i 1 2 3 1(i)1 2 3 2(i)1 3 2 3(i)2 1 3 4(i)2 3 1 5(i)3 1 2 6(i)3 2 1 4.10 Give the permutation expansion of a general 2 2 matrix and its transpose. X4.11 This problem appears also in the prior subsection. (a)Find the inverse of each 2-permutation. (b)Find the inverse of each 3-permutation. X4.12 (a) Find the signum of each 2-permutation. (b)Find the signum of each 3-permutation. 4.13 Find the only nonzero term in the permutation expansion of this matrix.0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 Compute that determinant by nding the signum of the associated permutation. 316 Chapter Four. Determinants 4.14 What is the signum of the n-permutation =hn;n1;:::; 2;1i? [Strang 80] 4.15 Prove these. (a)Every permutation has an inverse. (b)sgn(1) = sgn() (c)Every permutation is the inverse of another. 4.16 Prove that the matrix of the permutation inverse is the transpose of the matrix of the permutation P1=Ptrans, for any permutation . X4.17 Show that a permutation matrix with minversions can be row swapped to the identity in msteps. Contrast this with Corollary 4.6. X4.18 For any permutation letg() be the integer dened in this way. g() =Y i<j[(j)(i)] (This is the product, over all indices iandjwithi < j , of terms of the given form.) (a)Compute the value of gon all 2-permutations. (b)Compute the value of gon all 3-permutations. (c)Prove that g() is not 0. (d)Prove this. sgn() =g() jg()j Many authors give this formula as the denition of the signum function. Section II. Geometry of Determinants 317 II Geometry of Determinants The prior section develops the determinant algebraically, by considering what formulas satisfy certain properties. This section complements that with a geo- metric approach. One advantage of this approach is that, while we have so far only considered whether or not a determinant is zero, here we shall give a mean- ing to the value of that determinant. (The prior section handles determinants as functions of the rows, but in this section columns are more convenient. The nal result of the prior section says that we can make the switch.) II.1 Determinants as Size Functions This parallelogram picture  x1 y1 x2 y2 is familiar from the construction of the sum of the two vectors. One way to compute the area that it encloses is to draw this rectangle and subtract the area of each subregion. y1y2 x2x1AB CD EFarea of parallelogram = area of rectangle area ofAarea ofB


area ofF = (x1+x2)(y1+y2)x2y1x1y1=2 x2y2=2x2y2=2x1y1=2x2y1 =x1y2x2y1 The fact that the area equals the value of the determinant x1x2 y1y2=x1y2x2y1 is no coincidence. The properties in the denition of determinants make rea- sonable postulates for a function that measures the size of the region enclosed by the vectors in the matrix. For instance, this shows the eect of multiplying one of the box-dening vectors by a scalar (the scalar used is k= 1:4). ~ v~ w k~ v~ w 318 Chapter Four. Determinants The region formed by k~ vand~ wis bigger, by a factor of k, than the shaded region enclosed by ~ vand~ w. That is, size( k~ v;~ w ) =k
size(~ v;~ w) and in general we expect of the size measure that size( :::;k~ v;::: ) =k
size(:::;~ v;::: ). Of course, this postulate is already familiar as one of the properties in the dention of determinants. Another property of determinants is that they are unaected by combining rows. Here are before-combining and after-combining boxes (the scalar used is k= 0:35). ~ v~ w ~ vk~ v+~ w Although the region on the right, the box formed by vandk~ v+~ w, is more slanted than the shaded region, the two have the same base and the same height and hence the same area. This illustrates that size( ~ v;k~ v +~ w) = size(~ v;~ w). Generalized, size( :::;~ v;:::;~ w;::: ) = size(:::;~ v;:::;k~ v +~ w;::: ), which is a restatement of the determinant postulate. Of course, this picture ~ e1~ e2 shows that size( ~ e1;~ e2) = 1, and we naturally extend that to any number of dimensions size( ~ e1;:::;~ en) = 1, which is a restatement of the property that the determinant of the identity matrix is one. With that, because property (2) of determinants is redundant (as remarked right after the denition), we have that all of the properties of determinants are reasonable to expect of a function that gives the size of boxes. We can now cite the work done in the prior section to show that the determinant exists and is unique to be assured that these postulates are consistent and sucient (that is, we do not need any more postulates). So we've got an intuitive justication to interpret det( ~ v1;:::;~ vn) as the size of the box formed by the vectors. ( Comment. An even more basic approach, which also leads to the denition below, is in [Weston].) 1.1 Example The volume of this parallelepiped, which can be found by the usual formula from high school geometry, is 12. 0 @2 0 21 A0 @0 3 11 A0 @1 0 11 A2 01 0 3 0 2 1 1= 12 Section II. Geometry of Determinants 319 1.2 Remark Although property (2) of the denition of determinants is redun- dant, it raises an important point. Consider these two. ~ u~ v ~ u~ v 4 1 2 3= 101 4 3 2=10 The only dierence between them is in the order in which the vectors are taken. If we take~ urst and then go to ~ v, follow the counterclockwise arc shown, then the sign is positive. Following a clockwise arc gives a negative sign. The sign returned by the size function re ects the orientation orsense of the box. We see the same thing if we picture the eect of scalar multiplication by a negative scalar. Although it is both interesting and important, we don't need the idea of ori- entation for the development below and so we will pass it by. (See Exercise 27.) 1.3 Denition InRnthebox(orparallelepiped ) formed byh~ v1;:::;~ vniin- cludes all of the set ft1~ v1+


+tn~ vnt1;:::;tn2[0::1]g. The volume of a box is the absolute value of the determinant of the matrix with those vectors as columns. 1.4 Example Volume, because it is an absolute value, does not depend on the order in which the vectors are given. The volume of the parallelepiped in Exercise 1.1, can also be computed as the absolute value of this determinant. 0 2 0 3 0 3 1 2 1=12 The denition of volume gives a geometric interpretation to something in the space, boxes made from vectors. The next result relates the geometry to the functions that operate on spaces. 1.5 Theorem A transformation t:Rn!Rnchanges the size of all boxes by the same factor, namely the size of the image of a box jt(S)jisjTjtimes the size of the boxjSj, whereTis the matrix representing twith respect to the standard basis. That is, for all nnmatrices, the determinant of a product is the product of the determinants jTSj=jTj
jSj. The two sentences state the same idea, rst in map terms and then in matrix terms. Although we tend to prefer a map point of view, the second sentence, the matrix version, is more convienent for the proof and is also the way that we shall use this result later. (Alternate proofs are given as Exercise 23 and Exercise 28.) 320 Chapter Four. Determinants Proof .The two statements are equivalent because jt(S)j=jTSj, as both give the size of the box that is the image of the unit box Enunder the composition ts(wheresis the map represented by Swith respect to the standard basis). First consider the case that jTj= 0. A matrix has a zero determinant if and only if it is not invertible. Observe that if TSis invertible, so that there is an Msuch that (TS)M=I, then the associative property of matrix multiplication T(SM) =Ishows that Tis also invertible (with inverse SM). Therefore, if T is not invertible then neither is TS| ifjTj= 0 thenjTSj= 0, and the result holds in this case. Now consider the case that jTj6= 0, thatTis nonsingular. Recall that any nonsingular matrix can be factored into a product of elementary matrices, so thatTS=E1E2


ErS. In the rest of this argument, we will verify that if E is an elementary matrix then jESj=jEj
jSj. The result will follow because thenjTSj=jE1


ErSj=jE1j


jErj
jSj=jE1


Erj
jSj=jTj
jSj. If the elementary matrix EisMi(k) thenMi(k)SequalsSexcept that row i has been multiplied by k. The third property of determinant functions then gives thatjMi(k)Sj=k
jSj. ButjMi(k)j=k, again by the third property becauseMi(k) is derived from the identity by multiplication of row ibyk, and sojESj=jEj
jSjholds forE=Mi(k). TheE=Pi;j=1 andE=Ci;j(k) checks are similar. QED 1.6 Example Application of the map trepresented with respect to the stan- dard bases by1 1 2 0 will double sizes of boxes, e.g., from this ~ w~ v2 1 1 2= 3 to this t(~ w)t(~ v) 3 3 42= 6 1.7 Corollary If a matrix is invertible then the determinant of its inverse is the inverse of its determinant jT1j= 1=jTj. Proof .1 =jIj=jTT1j=jTj
jT1j QED Recall that determinants are not additive homomorphisms, det( A+B) need not equal det( A) + det(B). The above theorem says, in contrast, that determi- nants are multiplicative homomorphisms: det( AB) does equal det( A)
det(B). Section II. Geometry of Determinants 321 Exercises 1.8Find the volume of the region formed. (a)h1 3 ;1 4 i (b)h0 @2 1 01 A;0 @3 2 41 A;0 @8 3 81 Ai (c)h0 BB@1 2 0 11 CCA;0 BB@2 2 2 21 CCA;0 BB@1 3 0 51 CCA;0 BB@0 1 0 71 CCAi X1.9Is 0 @4 1 21 A inside of the box formed by these three?0 @3 3 11 A0 @2 6 11 A0 @1 0 51 A X1.10 Find the volume of this region. X1.11 Suppose thatjAj= 3. By what factor do these change volumes? (a)A(b)A2(c)A2 X1.12 By what factor does each transformation change the size of boxes? (a)x y 7!2x 3y (b)x y 7!3xy 2x+y (c)0 @x y z1 A7!0 @xy x+y+z y2z1 A 1.13 What is the area of the image of the rectangle [2 ::4][2::5] under the action of this matrix? 2 3 41 1.14 Ift:R3!R3changes volumes by a factor of 7 and s:R3!R3changes vol- umes by a factor of 3 =2 then by what factor will their composition changes volumes? 1.15 In what way does the denition of a box dier from the dention of a span? X1.16 Why doesn't this picture contradict Theorem 1.5? 2 1 0 1
! area is 2 determinant is 2 area is 5 X1.17 DoesjTSj=jSTj?jT(SP)j=j(TS)Pj? 1.18 (a) Suppose thatjAj= 3 and thatjBj= 2. FindjA2
Btrans
B2
Atransj. (b)Assume thatjAj= 0. Prove thatj6A3+ 5A2+ 2Aj= 0. X1.19 LetTbe the matrix representing (with respect to the standard bases) the map that rotates plane vectors counterclockwise thru radians. By what factor doesTchange sizes? X1.20 Must a transformation t:R2!R2that preserves areas also preserve lengths? 322 Chapter Four. Determinants X1.21 What is the volume of a parallelepiped in R3bounded by a linearly dependent set? X1.22 Find the area of the triangle in R3with endpoints (1 ;2;1), (3;1;4), and (2;2;2). (Area, not volume. The triangle denes a plane | what is the area of the triangle in that plane?) X1.23 An alternate proof of Theorem 1.5 uses the denition of determinant func- tions. (a)Note that the vectors forming Smake a linearly dependent set if and only if jSj= 0, and check that the result holds in this case. (b)For thejSj6= 0 case, to show that jTSj=jSj=jTjfor all transformations, consider the function d:Mnn!Rgiven byT7!jTSj=jSj. Show that dhas the rst property of a determinant. (c)Show thatdhas the remaining three properties of a determinant function. (d)Conclude thatjTSj=jTj
jSj. 1.24 Give a non-identity matrix with the property that Atrans=A1. Show that ifAtrans=A1thenjAj=1. Does the converse hold? 1.25 The algebraic property of determinants that factoring a scalar out of a single row will multiply the determinant by that scalar shows that where His 33, the determinant of cHisc3times the determinant of H. Explain this geometrically, that is, using Theorem 1.5. (The observation that increasing the linear size of a three-dimensional object by a factor of cwill increase its volume by a factor of c3 (while only increasing its surface area by an amount proportional to a factor of c2) is the Square-cube law [Wikipedia].) X1.26 MatricesHandGare said to be similar if there is a nonsingular matrix P such thatH=P1GP(we will study this relation in Chapter Five). Show that similar matrices have the same determinant. 1.27 We usually represent vectors in R2with respect to the standard basis so vectors in the rst quadrant have both coordinates positive. ~ v RepE2(~ v) =+3 +2 Moving counterclockwise around the origin, we cycle thru four regions:


! + + ! + !  ! +  !


: Using this basis B=h0 1 ;1 0 i ~2~1 gives the same counterclockwise cycle. We say these two bases have the same orientation . (a)Why do they give the same cycle? (b)What other congurations of unit vectors on the axes give the same cycle? (c)Find the determinants of the matrices formed from those (ordered) bases. (d)What other counterclockwise cycles are possible, and what are the associated determinants? (e)What happens in R1? (f)What happens in R3? A fascinating general-audience discussion of orientations is in [Gardner]. Section II. Geometry of Determinants 323 1.28 This question uses material from the optional Determinant Functions Exist subsection. Prove Theorem 1.5 by using the permutation expansion formula for the determinant. X1.29 (a) Show that this gives the equation of a line in R2thru (x2;y2) and (x3;y3).x x 2x3 y y 2y3 1 1 1= 0 (b)[Petersen] Prove that the area of a triangle with vertices ( x1;y1), (x2;y2), and (x3;y3) is 1 2x1x2x3 y1y2y3 1 1 1: (c)[Math. Mag., Jan. 1973] Prove that the area of a triangle with vertices at (x1;y1), (x2;y2), and (x3;y3) whose coordinates are integers has an area of N orN=2 for some positive integer N. 324 Chapter Four. Determinants III Other Formulas (This section is optional. Later sections do not depend on this material.) Determinants are a fount of interesting and amusing formulas. Here is one that is often seen in calculus classes and used to compute determinants by hand. III.1 Laplace's Expansion 1.1 Example In this permutation expansion t1;1t1;2t1;3 t2;1t2;2t2;3 t3;1t3;2t3;3=t1;1t2;2t3;31 0 0 0 1 0 0 0 1+t1;1t2;3t3;21 0 0 0 0 1 0 1 0 +t1;2t2;1t3;30 1 0 1 0 0 0 0 1+t1;2t2;3t3;10 1 0 0 0 1 1 0 0 +t1;3t2;1t3;20 0 1 1 0 0 0 1 0+t1;3t2;2t3;10 0 1 0 1 0 1 0 0 we can, for instance, factor out the entries from the rst row =t1;1
2 4t2;2t3;31 0 0 0 1 0 0 0 1+t2;3t3;21 0 0 0 0 1 0 1 03 5 +t1;2
2 4t2;1t3;30 1 0 1 0 0 0 0 1+t2;3t3;10 1 0 0 0 1 1 0 03 5 +t1;3
2 4t2;1t3;20 0 1 1 0 0 0 1 0+t2;2t3;10 0 1 0 1 0 1 0 03 5 and swap rows in the permutation matrices to get this. =t1;1
2 4t2;2t3;31 0 0 0 1 0 0 0 1+t2;3t3;21 0 0 0 0 1 0 1 03 5 t1;2
2 4t2;1t3;31 0 0 0 1 0 0 0 1+t2;3t3;11 0 0 0 0 1 0 1 03 5 +t1;3
2 4t2;1t3;21 0 0 0 1 0 0 0 1+t2;2t3;11 0 0 0 0 1 0 1 03 5 Section III. Other Formulas 325 The point of the swapping (one swap to each of the permutation matrices on the second line and two swaps to each on the third line) is that the three lines simplify to three terms. =t1;1
t2;2t2;3 t3;2t3;3t1;2
t2;1t2;3 t3;1t3;3+t1;3
t2;1t2;2 t3;1t3;2 The formula given in Theorem 1.5, which generalizes this example, is a recur- rence | the determinant is expressed as a combination of determinants. This formula isn't circular because, as here, the determinant is expressed in terms of determinants of matrices of smaller size. 1.2 Denition For anynnmatrixT, the (n1)(n1) matrix formed by deleting row iand column jofTis thei;jminor ofT. Thei;jcofactor Ti;jofTis (1)i+jtimes the determinant of the i;jminor ofT. 1.3 Example The 1;2 cofactor of the matrix from Example 1.1 is the negative of the second 22 determinant. T1;2=1
t2;1t2;3 t3;1t3;3 1.4 Example Where T=0 @1 2 3 4 5 6 7 8 91 A these are the 1 ;2 and 2;2 cofactors. T1;2= (1)1+2
4 6 7 9= 6T2;2= (1)2+2
1 3 7 9=12 1.5 Theorem (Laplace Expansion of Determinants) WhereTis annn matrix, the determinant can be found by expanding by cofactors on row ior columnj. jTj=ti;1
Ti;1+ti;2
Ti;2+


+ti;n
Ti;n =t1;j
T1;j+t2;j
T2;j+


+tn;j
Tn;j Proof .Exercise 27. QED 1.6 Example We can compute the determinant jTj=1 2 3 4 5 6 7 8 9 by expanding along the rst row, as in Example 1.1. jTj= 1
(+1)5 6 8 9+ 2
(1)4 6 7 9+ 3
(+1)4 5 7 8=3 + 129 = 0 326 Chapter Four. Determinants Alternatively, we can expand down the second column. jTj= 2
(1)4 6 7 9+ 5
(+1)1 3 7 9+ 8
(1)1 3 4 6= 1260 + 48 = 0 1.7 Example A row or column with many zeroes suggests a Laplace expansion. 1 5 0 2 1 1 31 0= 0
(+1)2 1 31+ 1
(1)1 5 31+ 0
(+1)1 5 2 1= 16 We nish by applying this result to derive a new formula for the inverse of a matrix. With Theorem 1.5, the determinant of an nnmatrixTcan be calculated by taking linear combinations of entries from a row and their associated cofactors. ti;1
Ti;1+ti;2
Ti;2+


+ti;n
Ti;n=jTj () Recall that a matrix with two identical rows has a zero determinant. Thus, for any matrix T, weighing the cofactors by entries from the \wrong" row | row k withk6=i| gives zero ti;1
Tk;1+ti;2
Tk;2+


+ti;n
Tk;n= 0 ( ) because it represents the expansion along the row kof a matrix with row iequal to rowk. This equation summarizes ( ) and (). 0 BBB@t1;1t1;2::: t 1;n t2;1t2;2::: t 2;n ... tn;1tn;2::: tn;n1 CCCA0 BBB@T1;1T2;1::: Tn;1 T1;2T2;2::: Tn;2 ... T1;nT2;n::: Tn;n1 CCCA=0 BBB@jTj0::: 0 0jTj::: 0 ... 0 0:::jTj1 CCCA Note that the order of the subscripts in the matrix of cofactors is opposite to the order of subscripts in the other matrix; e.g., along the rst row of the matrix of cofactors the subscripts are 1 ;1 then 2;1, etc. 1.8 Denition The matrix adjoint to the square matrix Tis adj(T) =0 BBB@T1;1T2;1::: Tn;1 T1;2T2;2::: Tn;2 ... T1;nT2;n::: Tn;n1 CCCA whereTj;iis thej;icofactor. 1.9 Theorem WhereTis a square matrix, T
adj(T) = adj(T)
T=jTj
I. Proof .Equations () and (). QED Section III. Other Formulas 327 1.10 Example If T=0 @1 0 4 2 11 1 0 11 A then the adjoint adj( T) is 0 @T1;1T2;1T3;1 T1;2T2;2T3;2 T1;3T2;3T3;31 A=0 BBBBBBB@11 0 10 4 0 10 4 11 21 1 11 4 1 11 4 21 2 1 1 01 0 1 01 0 2 11 CCCCCCCA=0 @1 04 33 9 1 0 11 A and taking the product with Tgives the diagonal matrix jTj
I. 0 @1 0 4 2 11 1 0 11 A0 @1 04 33 9 1 0 11 A=0 @3 0 0 03 0 0 031 A 1.11 Corollary IfjTj6= 0 thenT1= (1=jTj)
adj(T). 1.12 Example The inverse of the matrix from Example 1.10 is (1 =3)
adj(T). T1=0 @1=3 0=34=3 3=33=3 9=3 1=3 0=3 1=31 A=0 @1=3 0 4=3 1 13 1=3 01=31 A The formulas from this section are often used for by-hand calculation and are sometimes useful with special types of matrices. However, they are not the best choice for computation with arbitrary matrices because they require more arithmetic than, for instance, the Gauss-Jordan method. Exercises X1.13 Find the cofactor. T=0 @1 0 2 1 1 3 0 211 A (a)T2;3(b)T3;2(c)T1;3 X1.14 Find the determinant by expanding3 0 1 1 2 2 1 3 0 (a)on the rst row (b)on the second row (c)on the third column. 1.15 Find the adjoint of the matrix in Example 1.6. X1.16 Find the matrix adjoint to each. 328 Chapter Four. Determinants (a)0 @2 1 4 1 0 2 1 0 11 A (b)31 2 4 (c)1 1 5 0 (d)0 @1 4 3 1 0 3 1 8 91 A X1.17 Find the inverse of each matrix in the prior question with Theorem 1.9. 1.18 Find the matrix adjoint to this one.0 BB@2 1 0 0 1 2 1 0 0 1 2 1 0 0 1 21 CCA X1.19 Expand across the rst row to derive the formula for the determinant of a 2 2 matrix. X1.20 Expand across the rst row to derive the formula for the determinant of a 3 3 matrix. X1.21 (a) Give a formula for the adjoint of a 2 2 matrix. (b)Use it to derive the formula for the inverse. X1.22 Can we compute a determinant by expanding down the diagonal? 1.23 Give a formula for the adjoint of a diagonal matrix. X1.24 Prove that the transpose of the adjoint is the adjoint of the transpose. 1.25 Prove or disprove: adj(adj( T)) =T. 1.26 A square matrix is upper triangular if eachi;jentry is zero in the part above the diagonal, that is, when i>j . (a)Must the adjoint of an upper triangular matrix be upper triangular? Lower triangular? (b)Prove that the inverse of a upper triangular matrix is upper triangular, if an inverse exists. 1.27 This question requires material from the optional Determinants Exist subsec- tion. Prove Theorem 1.5 by using the permutation expansion. 1.28 Prove that the determinant of a matrix equals the determinant of its transpose using Laplace's expansion and induction on the size of the matrix. ?1.29 Show that Fn=11 11 11::: 1 1 0 1 0 1 ::: 0 1 1 0 1 0 ::: 0 0 1 1 0 1 ::: : : : : : : ::: whereFnis then-th term of 1 ;1;2;3;5;:::;x;y;x +y;::: , the Fibonacci sequence, and the determinant is of order n1. [Am. Math. Mon., Jun. 1949] Topic: Cramer's Rule 329 Topic: Cramer's Rule We have introduced determinant functions algebraically by looking for a formula to decide whether a matrix is nonsingular. After that introduction we saw a geometric interpretation, that the determinant function gives the size of the box with sides formed by the columns of the matrix. This Topic makes a connection between the two views. First, a linear system x1+ 2x2= 6 3x1+x2= 8 is equivalent to a linear relationship among vectors. x1
 1 3 +x2
2 1 =6 8 The picture below shows a parallelogram with sides formed from1 3
and2 1
nested inside a parallelogram with sides formed from x11 3
andx22 1
. 2 1x2
2 11 3x1
1 36 8 So even without determinants we can state the algebraic issue that opened this book, nding the solution of a linear system, in geometric terms: by what factors x1andx2must we dilate the vectors to expand the small parallegram to ll the larger one? However, by employing the geometric signicance of determinants we can get something that is not just a restatement, but also gives us a new insight and sometimes allows us to compute answers quickly. Compare the sizes of these shaded boxes. 2 11 3 2 1x1
1 3 2 16 8 The second is formed from x11 3
and2 1
, and one of the properties of the size function | the determinant | is that its size is therefore x1times the size of the 330 Chapter Four. Determinants rst box. Since the third box is formed from x11 3
+x22 1
=6 8
and2 1
, and the determinant is unchanged by adding x2times the second column to the rst column, the size of the third box equals that of the second. We have this. 6 2 8 1=x1
1 2 x1
3 1=x1
1 2 3 1 Solving gives the value of one of the variables. x1=6 2 8 1 1 2 3 1=10 5= 2 The theorem that generalizes this example, Cramer's Rule , is: ifjAj6= 0 then the system A~ x=~bhas the unique solution xi=jBij=jAjwhere the matrix Biis formed from Aby replacing column iwith the vector ~b. Exercise 3 asks for a proof. For instance, to solve this system for x2 0 @1 0 4 2 11 1 0 11 A0 @x1 x2 x31 A=0 @2 1 11 A we do this computation. x2=1 2 4 2 11 11 1 1 0 4 2 11 1 0 1=18 3 Cramer's Rule allows us to solve many two equations/two unknowns systems by eye. It is also sometimes used for three equations/three unknowns systems. But computing large determinants takes a long time, so solving large systems by Cramer's Rule is not practical. Exercises 1Use Cramer's Rule to solve each for each of the variables. (a)xy= 4 x+ 2y=7(b)2x+y=2 x2y=2 2Use Cramer's Rule to solve this system for z. 2x+y+z= 1 3x +z= 4 xyz= 2 3Prove Cramer's Rule. Topic: Cramer's Rule 331 4Suppose that a linear system has as many equations as unknowns, that all of its coecients and constants are integers, and that its matrix of coecients has determinant 1. Prove that the entries in the solution are all integers. ( Remark. This is often used to invent linear systems for exercises. If an instructor makes the linear system with this property then the solution is not some disagreeable fraction.) 5Use Cramer's Rule to give a formula for the solution of a two equations/two unknowns linear system. 6Can Cramer's Rule tell the dierence between a system with no solutions and one with innitely many? 7The rst picture in this Topic (the one that doesn't use determinants) shows a unique solution case. Produce a similar picture for the case of inntely many solutions, and the case of no solutions. 332 Chapter Four. Determinants Topic: Speed of Calculating Determinants The permutation expansion formula for computing determinants is useful for proving theorems, but the method of using row operations is a much better for nding the determinants of a large matrix. We can make this statement precise by considering, as computer algorithm designers do, the number of arithmetic operations that each method uses. The speed of an algorithm is measured by nding how the time taken by the computer grows as the size of its input data set grows. For instance, how much longer will the algorithm take if we increase the size of the input data by a factor of ten, from a 1000 row matrix to a 10 ;000 row matrix or from 10;000 to 100;000? Does the time taken grow by a factor of ten, or by a factor of a hundred, or by a factor of a thousand? That is, is the time taken by the algorithm proportional to the size of the data set, or to the square of that size, or to the cube of that size, etc.? Recall the permutation expansion formula for determinants. t1;1t1;2::: t 1;n t2;1t2;2::: t 2;n ... tn;1tn;2::: tn;n=X permutations t1;(1)t2;(2)


tn;(n)jPj =t1;1(1)
t2;1(2)


tn;1(n)jP1j +t1;2(1)
t2;2(2)


tn;2(n)jP2j ... +t1;k(1)
t2;k(2)


tn;k(n)jPkj There aren! =n
(n1)
(n2)


2
1 dierentn-permutations. For numbers nof any size at all, this is a large value; for instance, even if nis only 10 then the expansion has 10! = 3 ;628;800 terms, all of which are obtained by multiplying nentries together. This is a very large number of multiplications (for instance, [Knuth] suggests 10! steps as a rough boundary for the limit of practical calculation). The factorial function grows faster than the square function. It grows faster than the cube function, the fourth power function, or any polynomial function. (One way to see that the factorial function grows faster than the square is to note that multiplying the rst two factors in n! givesn
(n1), which for large nis approximately n2, and then multiplying in more factors will make it even larger. The same argument works for the cube function, etc.) So a computer that is programmed to use the permutation expansion formula, and thus to perform a number of operations that is greater than or equal to the factorial of the number of rows, would take very long times as its input data set grows. In contrast, the time taken by the row reduction method does not grow so fast. This fragment of row-reduction code is in the computer language FOR- TRAN, which is widely used for numeric code. The matrix is stored in the NN array A. For each ROW between 1 and Nparts of the program not shown here Topic: Speed of Calculating Determinants 333 have already found the leading entry A(ROW;COL ). Now the program does a row combination. PIVINV
ROW +i (This code fragment is for illustration only and is incomplete. Still, analysis of a nished version that includes all of the tests and subcases is messier but gives essentially the same conclusion.) PIVINV=1.0/A(ROW,COL) DO 10 I=ROW+1, N DO 20 J=I, N A(I,J)=A(I,J)-PIVINV*A(ROW,J) 20 CONTINUE 10 CONTINUE The outermost loop (not shown) runs through N1 rows. For each row, the nestedIandJloops shown perform arithmetic on the entries in Athat are below and to the right of the leading entry. Assume that this entry is found in the expected place, that is, that COL =ROW . Then there are ( NROW )2 entries below and to the right of it. On average, ROW will beN=2. Thus, we estimate that the arithmetic will be performed about ( N=2)2times, that is, will run in a time proportional to the square of the number of equations. Taking into account the outer loop that is not shown, we get the estimate that the running time of the algorithm is proportional to the cube of the number of equations. Finding the fastest algorithm to compute the determinant is a topic of cur- rent research. Algorithms are known that run in time between the second and third power. Speed estimates like these help us to understand how quickly or slowly an algorithm will run. Algorithms that run in time proportional to the size of the data set are fast, algorithms that run in time proportional to the square of the size of the data set are less fast, but typically quite usable, and algorithms that run in time proportional to the cube of the size of the data set are still reasonable in speed for not-too-big input data. However, algorithms that run in time (greater than or equal to) the factorial of the size of the data set are not practical for input of any appreciable size. There are other methods besides the two discussed here that are also used for computation of determinants. Those lie outside of our scope. Nonetheless, this contrast of the two methods for computing determinants makes the point that although in principle they give the same answer, in practice the idea is to select the one that is fast. Exercises Most of these problems presume access to a computer. 1Computer systems generate random numbers (of course, these are only pseudo- random, in that they are generated by an algorithm, but they pass a number of reasonable statistical tests for randomness). (a)Fill a 55 array with random numbers (say, in the range [0 ::1)). See if it is singular. Repeat that experiment a few times. Are singular matrices frequent or rare (in this sense)? 334 Chapter Four. Determinants (b)Time your computer algebra system at nding the determinant of ten 5 5 arrays of random numbers. Find the average time per array. Repeat the prior item for 1515 arrays, 2525 arrays, 3535 arrays, etc. You may nd that you need to get above a certain size to get a timing that you can use. (Notice that, when an array is singular, it can sometimes be found to be so quite quickly, for instance if the rst row equals the second. In the light of your answer to the rst part, do you expect that singular systems play a large role in your average?) (c)Graph the input size versus the average time. 2Compute the determinant of each of these by hand using the two methods dis- cussed above. (a)2 1 53(b)3 1 1 1 0 5 1 22(c)2 1 0 0 1 3 2 0 012 1 0 02 1 Count the number of multiplications and divisions used in each case, for each of the methods. (On a computer, multiplications and divisions take much longer than additions and subtractions, so algorithm designers worry about them more.) 3What 1010 array can you invent that takes your computer system the longest to reduce? The shortest? 4Write the rest of the FORTRAN program to do a straightforward implementation of calculating determinants via Gauss' method. (Don't test for a zero leading entry.) Compare the speed of your code to that used in your computer algebra system. 5The FORTRAN language specication requires that arrays be stored \by col- umn", that is, the entire rst column is stored contiguously, then the second col- umn, etc. Does the code fragment given take advantage of this, or can it be rewritten to make it faster, by taking advantage of the fact that computer fetches are faster from contiguous locations? Topic: Projective Geometry 335 Topic: Projective Geometry There are geometries other than the familiar Euclidean one. One such geometry arose in art, where it was observed that what a viewer sees is not necessarily what is there. This is Leonardo da Vinci's The Last Supper . What is there in the room, for instance where the ceiling meets the left and right walls, are lines that are parallel. However, what a viewer sees is lines that, if extended, would intersect. The intersection point is called the vanishing point . This aspect of perspective is also familiar as the image of a long stretch of railroad tracks that appear to converge at the horizon. To depict the room, da Vinci has adopted a model of how we see, of how we project the three dimensional scene to a two dimensional image. This model is only a rst approximation | it does not take into account that our retina is curved and our lens bends the light, that we have binocular vision, or that our brain's processing greatly aects what we see | but nonetheless it is interesting, both artistically and mathematically. The projection is not orthogonal, it is a central projection from a single point, to the plane of the canvas. A B C (It is not an orthogonal projection since the line from the viewer to Cis not orthogonal to the image plane.) As the picture suggests, the operation of central projection preserves some geometric properties | lines project to lines. How- ever, it fails to preserve some others | equal length segments can project to segments of unequal length; the length of ABis greater than the length of 336 Chapter Four. Determinants BCbecause the segment projected to ABis closer to the viewer and closer things look bigger. The study of the eects of central projections is projective geometry. We will see how linear algebra can be used in this study. There are three cases of central projection. The rst is the projection done by a movie projector. projectorP sourceS imageI We can think that each source point is \pushed" from the domain plane out- ward to the image point in the codomain plane. This case of projection has a somewhat dierent character than the second case, that of the artist \pulling" the source back to the canvas. painterP imageI sourceS In the rst case Sis in the middle while in the second case Iis in the middle. One more conguration is possible, with Pin the middle. An example of this is when we use a pinhole to shine the image of a solar eclipse onto a piece of paper. sourceS pinholeP imageI We shall take each of the three to be a central projection by PofStoI. Topic: Projective Geometry 337 Consider again the eect of railroad tracks that appear to converge to a point. We model this with parallel lines in a domain plane Sand a projection via aPto a codomain plane I. (The gray lines are parallel to SandI.) S IP All three projection cases appear here. The rst picture below shows Pacting like a movie projector by pushing points from part of Sout to image points on the lower half of I. The middle picture shows Pacting like the artist by pulling points from another part of Sback to image points in the middle of I. In the third picture, Pacts like the pinhole, projecting points from Sto the upper part ofI. This picture is the trickiest | the points that are projected near to the vanishing point are the ones that are far out on the bottom left of S. Points inSthat are near to the vertical gray line are sent high up on I. S IPS IPS IP There are two awkward things about this situation. The rst is that neither of the two points in the domain nearest to the vertical gray line (see below) has an image because a projection from those two is along the gray line that is parallel to the codomain plane (we sometimes say that these two are projected \to innity"). The second awkward thing is that the vanishing point in Iisn't the image of any point from Sbecause a projection to this point would be along the gray line that is parallel to the domain plane (we sometimes say that the vanishing point is the image of a projection \from innity"). S IP 338 Chapter Four. Determinants For a better model, put the projector Pat the origin. Imagine that Pis covered by a glass hemispheric dome. As Plooks outward, anything in the line of vision is projected to the same spot on the dome. This includes things on the line between Pand the dome, as in the case of projection by the movie projector. It includes things on the line further from Pthan the dome, as in the case of projection by the painter. It also includes things on the line that lie behindP, as in the case of projection by a pinhole. `=fk
0 @1 2 31 Ak2Rg From this perspective P, all of the spots on the line are seen as the same point. Accordingly, for any nonzero vector ~ v2R3, we dene the associated pointv in the projective plane to be the setfk~ vk2Randk6= 0gof nonzero vectors lying on the same line through the origin as ~ v. To describe a projective point we can give any representative member of the line, so that the projective point shown above can be represented in any of these three ways. 0 @1 2 31 A0 @1=3 2=3 11 A0 @2 4 61 A Each of these is a homogeneous coordinate vector forv. This picture, and the above denition that arises from it, claries the de- scription of central projection but there is something awkward about the dome model: what if the viewer looks down? If we draw P's line of sight so that the part coming toward us, out of the page, goes down below the dome then we can trace the line of sight backward, up past Pand toward the part of the hemisphere that is behind the page. So in the dome model, looking down gives a projective point that is behind the viewer. Therefore, if the viewer in the picture above drops the line of sight toward the bottom of the dome then the projective point drops also and as the line of sight continues down past the equator, the projective point suddenly shifts from the front of the dome to the back of the dome. This discontinuity in the drawing means that we often have to treat equatorial points as a separate case. That is, while the railroad track discussion of central projection has three cases, the dome model has two. We can do better than this. Consider a sphere centered at the origin. Any line through the origin intersects the sphere in two spots, which are said to be antipodal . Because we associate each line through the origin with a point in the projective plane, we can draw such a point as a pair of antipodal spots on the sphere. Below, the two antipodal spots are shown connected by a dashed line Topic: Projective Geometry 339 to emphasize that they are not two dierent points, the pair of spots together make one projective point. While drawing a point as a pair of antipodal spots is not as natural as the one- spot-per-point dome mode, on the other hand the awkwardness of the dome model is gone, in that if as a line of view slides from north to south, no sudden changes happen on the picture. This model of central projection is uniform | the three cases are reduced to one. So far we have described points in projective geometry. What about lines? What a viewer at the origin sees as a line is shown below as a great circle, the intersection of the model sphere with a plane through the origin. (One of the projective points on this line is shown to bring out a subtlety. Because two antipodal spots together make up a single projective point, the great circle's behind-the-paper part is the same set of projective points as its in-front-of-the-paper part.) Just as we did with each projective point, we will also describe a projective line with a triple of reals. For instance, the members of this plane through the origin in R3 f0 @x y z1 Ax+yz= 0g project to a line that we can described with the triple1 11
(we use row vectors to typographically distinguish lines from points). In general, for any nonzero three-wide row vector ~Lwe dene the associated line in the projective plane , to be the set L=fk~Lk2Randk6= 0gof nonzero multiples of ~L. The reason that this description of a line as a triple is convienent is that in the projective plane, a point vand a line Lareincident | the point lies on the line, the line passes throught the point | if and only if a dot product of their representatives v1L1+v2L2+v3L3is zero (Exercise 4 shows that this is independent of the choice of representatives ~ vand~L). For instance, the projective point described above by the column vector with components 1, 2, and 3 lies in the projective line described by1 11
, simply because any 340 Chapter Four. Determinants vector in R3whose components are in ratio 1 : 2 : 3 lies in the plane through the origin whose equation is of the form 1 k
x+ 1k
y1k
z= 0 for any nonzero k. That is, the incidence formula is inherited from the three-space lines and planes of whichvandLare projections. Thus, we can do analytic projective geometry. For instance, the projective lineL=1 11
has the equation 1 v1+ 1v21v3= 0, because points incident on the line are characterized by having the property that their repre- sentatives satisfy this equation. One dierence from familiar Euclidean anlaytic geometry is that in projective geometry we talk about the equation of a point. For a xed point like v=0 @1 2 31 A the property that characterizes lines through this point (that is, lines incident on this point) is that the components of any representatives satisfy 1 L1+ 2L2+ 3L3= 0 and so this is the equation of v. This symmetry of the statements about lines and points brings up the Duality Principle of projective geometry: in any true statement, interchanging `point' with `line' results in another true statement. For example, just as two distinct points determine one and only one line, in the projective plane, two distinct lines determine one and only one point. Here is a picture showing two lines that cross in antipodal spots and thus cross at one projective point. () Contrast this with Euclidean geometry, where two distinct lines may have a unique intersection or may be parallel. In this way, projective geometry is simpler, more uniform, than Euclidean geometry. That simplicity is relevant because there is a relationship between the two spaces: the projective plane can be viewed as an extension of the Euclidean plane. Take the sphere model of the projective plane to be the unit sphere in R3and take Euclidean space to be the plane z= 1. This gives us a way of viewing some points in projective space as corresponding to points in Euclidean space, because all of the points on the plane are projections of antipodal spots from the sphere. () Topic: Projective Geometry 341 Note though that projective points on the equator don't project up to the plane. Instead, these project `out to innity'. We can thus think of projective space as consisting of the Euclidean plane with some extra points adjoined | the Eu- clidean plane is embedded in the projective plane. These extra points, the equatorial points, are the ideal points orpoints at innity and the equator is theideal line orline at innity (note that it is not a Euclidean line, it is a projective line). The advantage of the extension to the projective plane is that some of the awkwardness of Euclidean geometry disappears. For instance, the projective lines shown above in ( ) cross at antipodal spots, a single projective point, on the sphere's equator. If we put those lines into ( ) then they correspond to Euclidean lines that are parallel. That is, in moving from the Euclidean plane to the projective plane, we move from having two cases, that lines either intersect or are parallel, to having only one case, that lines intersect (possibly at a point at innity). The projective case is nicer in many ways than the Euclidean case but has the problem that we don't have the same experience or intuitions with it. That's one advantage of doing analytic geometry, where the equations can lead us to the right conclusions. Analytic projective geometry uses linear algebra. For instance, for three points of the projective plane t,u, andv, setting up the equations for those points by xing vectors representing each, shows that the three are collinear | incident in a single line | if and only if the resulting three- equation system has innitely many row vector solutions representing that line. That, in turn, holds if and only if this determinant is zero. t1u1v1 t2u2v2 t3u3v3 Thus, three points in the projective plane are collinear if and only if any three representative column vectors are linearly dependent. Similarly (and illustrating the Duality Principle), three lines in the projective plane are incident on a single point if and only if any three row vectors representing them are linearly dependent. The following result is more evidence of the `niceness' of the geometry of the projective plane, compared to the Euclidean case. These two triangles are said to be in perspective fromPbecause their corresponding vertices are collinear. O T1 U1V1 T2 U2V2 Consider the pairs of corresponding sides: the sides T1U1andT2U2, the sides T1V1andT2V2, and the sides U1V1andU2V2. Desargue's Theorem is that 342 Chapter Four. Determinants when the three pairs of corresponding sides are extended to lines, they intersect (shown here as the point TU, the point TV, and the point UV), and further, those three intersection points are collinear. TUTVUV We will prove this theorem, using projective geometry. (These are drawn as Euclidean gures because it is the more familiar image. To consider them as projective gures, we can imagine that, although the line segments shown are parts of great circles and so are curved, the model has such a large radius compared to the size of the gures that the sides appear in this sketch to be straight.) For this proof, we need a preliminary lemma [Coxeter]: if W,X,Y,Zare four points in the projective plane (no three of which are collinear) then there are homogeneous coordinate vectors ~ w,~ x,~ y, and~ zfor the projective points, and a basis BforR3, satisfying this. RepB(~ w) =0 @1 0 01 ARepB(~ x) =0 @0 1 01 ARepB(~ y) =0 @0 0 11 ARepB(~ z) =0 @1 1 11 A The proof is straightforward. Because W; X; Y are not on the same projective line, any homogeneous coordinate vectors ~ w0;~ x0;~ y0do not line on the same plane through the origin in R3and so form a spanning set for R3. Thus any homogeneous coordinate vector for Zcan be written as a combination ~ z0= a
~ w0+b
~ x0+c
~ y0. Then, we can take ~ w=a
~ w0,~ x=b
~ x0,~ y=c
~ y0, and ~ z=~ z0, where the basis is B=h~ w;~ x;~ yi. Now, to prove of Desargue's Theorem, use the lemma to x homogeneous coordinate vectors and a basis. RepB(~t1) =0 @1 0 01 ARepB(~ u1) =0 @0 1 01 ARepB(~ v1) =0 @0 0 11 ARepB(~ o) =0 @1 1 11 A Because the projective point T2is incident on the projective line OT1, any homogeneous coordinate vector for T2lies in the plane through the origin in R3 that is spanned by homogeneous coordinate vectors of OandT1: RepB(~t2) =a0 @1 1 11 A+b0 @1 0 01 A Topic: Projective Geometry 343 for some scalars aandb. That is, the homogenous coordinate vectors of members T2of the lineOT1are of the form on the left below, and the forms for U2and V2are similar. RepB(~t2) =0 @t2 1 11 A RepB(~ u2) =0 @1 u2 11 A RepB(~ v2) =0 @1 1 v21 A The projective line T1U1is the image of a plane through the origin in R3. A quick way to get its equation is to note that any vector in it is linearly dependent on the vectors for T1andU1and so this determinant is zero. 1 0x 0 1y 0 0z= 0 =)z= 0 The equation of the plane in R3whose image is the projective line T2U2is this. t21x 1u2y 1 1z= 0 =) (1u2)
x+ (1t2)
y+ (t2u21)
z= 0 Finding the intersection of the two is routine. T1U1\T2U2=0 @t21 1u2 01 A (This is, of course, the homogeneous coordinate vector of a projective point.) The other two intersections are similar. T1V1\T2V2=0 @1t2 0 v211 AU1V1\U2V2=0 @0 u21 1v21 A The proof is nished by noting that these projective points are on one projective line because the sum of the three homogeneous coordinate vectors is zero. Every projective theorem has a translation to a Euclidean version, although the Euclidean result is often messier to state and prove. Desargue's theorem illustrates this. In the translation to Euclidean space, the case where Olies on the ideal line must be treated separately for then the lines T1T2,U1U2, andV1V2 are parallel. The parenthetical remark following the statement of Desargue's Theorem suggests thinking of the Euclidean pictures as gures from projective geometry for a model of very large radius. That is, just as a small area of the earth appears at to people living there, the projective plane is also `locally Euclidean'. Although its local properties are the familiar Euclidean ones, there is a global property of the projective plane that is quite dierent. The picture below shows 344 Chapter Four. Determinants a projective point. At that point is drawn an xy-axis. There is something interesting about the way this axis appears at the antipodal ends of the sphere. In the northern hemisphere, where the axis are drawn in black, a right hand put down with ngers on the x-axis will have the thumb point along the y-axis. But the antipodal axis has just the opposite: a right hand placed with its ngers on thex-axis will have the thumb point in the wrong way, instead, it is a left hand that works. Brie y, the projective plane is not orientable: in this geometry, left and right handedness are not xed properties of gures. The sequence of pictures below dramatizes this non-orientability. They sketch a trip around this space in the direction of the ypart of the xy-axis. (Warning: the trip shown is not halfway around, it is a full circuit. True, if we made this into a movie then we could watch the northern hemisphere spots in the drawing above gradually rotate about halfway around the sphere to the last picture below. And we could watch the southern hemisphere spots in the picture above slide through the south pole and up through the equator to the last picture. But: the spots at either end of the dashed line are the same projective point. We don't need to continue on much further; we are pretty much back to the projective point where we started by the last picture.) =) =) At the end of the circuit, the xpart of the xy-axes sticks out in the other direction. Thus, in the projective plane we cannot describe a gure as right- or left-handed (another way to make this point is that we cannot describe a spiral as clockwise or counterclockwise). This exhibition of the existence of a non-orientable space raises the question of whether our universe is orientable: is is possible for an astronaut to leave right-handed and return left-handed? An excellent nontechnical reference is [Gardner]. An classic science ction story about orientation reversal is [Clarke]. So projective geometry is mathematically interesting, in addition to the nat- ural way in which it arises in art. It is more than just a technical device to shorten some proofs. For an overview, see [Courant & Robbins]. The approach we've taken here, the analytic approach, leads to quick theorems and | most importantly for us | illustrates the power of linear algebra (see [Hanes], [Ryan], and [Eggar]). But another approach, the synthetic approach of deriving the Topic: Projective Geometry 345 results from an axiom system, is both extraordinarily beautiful and is also the historical route of development. Two ne sources for this approach are [Coxeter] or [Seidenberg]. An interesting and easy application is [Davies] Exercises 1What is the equation of this point?0 @1 0 01 A 2 (a) Find the line incident on these points in the projective plane.0 @1 2 31 A;0 @4 5 61 A (b)Find the point incident on both of these projective lines. 1 2 3
; 4 5 6
3Find the formula for the line incident on two projective points. Find the formula for the point incident on two projective lines. 4Prove that the denition of incidence is independent of the choice of the rep- resentatives of pandL. That is, if p1,p2,p3, andq1,q2,q3are two triples of homogeneous coordinates for p, andL1,L2,L3, andM1,M2,M3are two triples of homogeneous coordinates for L, prove that p1L1+p2L2+p3L3= 0 if and only ifq1M1+q2M2+q3M3= 0. 5Give a drawing to show that central projection does not preserve circles, that a circle may project to an ellipse. Can a (non-circular) ellipse project to a circle? 6Give the formula for the correspondence between the non-equatorial part of the antipodal modal of the projective plane, and the plane z= 1. 7(Pappus's Theorem) Assume that T0,U0, andV0are collinear and that T1,U1, andV1are collinear. Consider these three points: (i) the intersection V2of the lines T0U1andT1U0, (ii) the intersection U2of the lines T0V1andT1V0, and (iii) the intersection T2ofU0V1andU1V0. (a)Draw a (Euclidean) picture. (b)Apply the lemma used in Desargue's Theorem to get simple homogeneous coordinate vectors for the T's andV0. (c)Find the resulting homogeneous coordinate vectors for U's (these must each involve a parameter as, e.g., U0could be anywhere on the T0V0line). (d)Find the resulting homogeneous coordinate vectors for V1. (Hint: it involves two parameters.) (e)Find the resulting homogeneous coordinate vectors for V2. (It also involves two parameters.) (f)Show that the product of the three parameters is 1. (g)Verify that V2is on theT2U2line. Chapter Five Similarity While studying matrix equivalence, we have shown that for any homomorphism there are bases BandDsuch that the representation matrix has a block partial- identity form. RepB;D(h) =Identity Zero Zero Zero This representation describes the map as sending c1~1+


+cn~ntoc1~1+


+ck~k+~0 +


+~0, wherenis the dimension of the domain and kis the dimension of the range. So, under this representation the action of the map is easy to understand because most of the matrix entries are zero. This chapter considers the special case where the domain and the codomain are equal, that is, where the homomorphism is a transformation. In this case we naturally ask to nd a single basis Bso that RepB;B(t) is as simple as possible (we will take `simple' to mean that it has many zeroes). A matrix having the above block partial-identity form is not always possible here. But we will develop a form that comes close, a representation that is nearly diagonal. I Complex Vector Spaces This chapter requires that we factor polynomials. Of course, many polynomials do not factor over the real numbers; for instance, x2+ 1 does not factor into the product of two linear polynomials with real coecients. For that reason, we shall from now on take our scalars from the complex numbers. That is, we are shifting from studying vector spaces over the real numbers to vector spaces over the complex numbers | in this chapter vector and matrix entries are complex. Any real number is a complex number and a glance through this chapter shows that most of the examples use only real numbers. Nonetheless, the critical theorems require that the scalars be complex numbers, so the rst section below is a quick review of complex numbers. 347 348 Chapter Five. Similarity In this book we are moving to the more general context of taking scalars to be complex only for the pragmatic reason that we must do so in order to develop the representation. We will not go into using other sets of scalars in more detail because it could distract from our goal. However, the idea of taking scalars from a structure other than the real numbers is an interesting one. Delightful presentations taking this approach are in [Halmos] and [Homan & Kunze]. I.1 Factoring and Complex Numbers; A Review This subsection is a review only and we take the main results as known. For proofs, see [Birkho & MacLane] or [Ebbinghaus]. Just as integers have a division operation | e.g., `4 goes 5 times into 21 with remainder 1' | so do polynomials. 1.1 Theorem (Division Theorem for Polynomials) Letc(x) be a polyno- mial. Ifm(x) is a non-zero polynomial then there are quotient and remainder polynomials q(x) andr(x) such that c(x) =m(x)
q(x) +r(x) where the degree of r(x) is strictly less than the degree of m(x). In this book constant polynomials, including the zero polynomial, are said to have degree 0. (This is not the standard denition, but it is convienent here.) The point of the integer division statement `4 goes 5 times into 21 with remainder 1' is that the remainder is less than 4 | while 4 goes 5 times, it does not go 6 times. In the same way, the point of the polynomial division statement is its nal clause. 1.2 Example Ifc(x) = 2x33x2+ 4xandm(x) =x2+ 1 thenq(x) = 2x3 andr(x) = 2x+ 3. Note that r(x) has a lower degree than m(x). 1.3 Corollary The remainder when c(x) is divided by xis the constant polynomial r(x) =c(). Proof .The remainder must be a constant polynomial because it is of degree less than the divisor x, To determine the constant, take m(x) from the theorem to bexand substitute forxto getc() = ()
q() +r(x). QED If a divisor m(x) goes into a dividend c(x) evenly, meaning that r(x) is the zero polynomial, then m(x) is a factor ofc(x). Any root of the factor (any 2Rsuch thatm() = 0) is a root of c(x) sincec() =m()
q() = 0. The prior corollary immediately yields the following converse. 1.4 Corollary Ifis a root of the polynomial c(x) thenxdividesc(x) evenly, that is, xis a factor of c(x). Section I. Complex Vector Spaces 349 Finding the roots and factors of a high-degree polynomial can be hard. But for second-degree polynomials we have the quadratic formula: the roots of ax2+ bx+care 1=b+p b24ac 2a2=bp b24ac 2a (if the discriminant b24acis negative then the polynomial has no real number roots). A polynomial that cannot be factored into two lower-degree polynomials with real number coecients is irreducible over the reals . 1.5 Theorem Any constant or linear polynomial is irreducible over the reals. A quadratic polynomial is irreducible over the reals if and only if its discrimi- nant is negative. No cubic or higher-degree polynomial is irreducible over the reals. 1.6 Corollary Any polynomial with real coecients can be factored into linear and irreducible quadratic polynomials. That factorization is unique; any two factorizations have the same powers of the same factors. Note the analogy with the prime factorization of integers. In both cases, the uniqueness clause is very useful. 1.7 Example Because of uniqueness we know, without multiplying them out, that (x+ 3)2(x2+ 1)3does not equal ( x+ 3)4(x2+x+ 1)2. 1.8 Example By uniqueness, if c(x) =m(x)
q(x) then where c(x) = (x 3)2(x+ 2)3andm(x) = (x3)(x+ 2)2, we know that q(x) = (x3)(x+ 2). Whilex2+ 1 has no real roots and so doesn't factor over the real numbers, if we imagine a root | traditionally denoted iso thati2+ 1 = 0 | then x2+ 1 factors into a product of linears ( xi)(x+i). So we adjoin this root ito the reals and close the new system with respect to addition, multiplication, etc. (i.e., we also add 3 + i, and 2i, and 3 + 2i, etc., putting in all linear combinations of 1 and i). We then get a new structure, the complex numbers , denoted C. InCwe can factor (obviously, at least some) quadratics that would be irre- ducible if we were to stick to the real numbers. Surprisingly, in Cwe can not only factor x2+ 1 and its close relatives, we can factor any quadratic. ax2+bx+c=a
xb+p b24ac 2a

xbp b24ac 2a
1.9 Example The second degree polynomial x2+x+1 factors over the complex numbers into the product of two rst degree polynomials. x1 +p3 2
x1p3 2
= x(1 2+p 3 2i)
x(1 2p 3 2i)
1.10 Corollary (Fundamental Theorem of Algebra) Polynomials with complex coecients factor into linear polynomials with complex coecients. The factorization is unique. 350 Chapter Five. Similarity I.2 Complex Representations Recall the denitions of the complex number addition (a+bi) + (c+di) = (a+c) + (b+d)i and multiplication. (a+bi)(c+di) =ac+adi+bci+bd(1) = (acbd) + (ad+bc)i 2.1 Example For instance, (12i) + (5+4i) = 6+2iand (23i)(40:5i) = 6:513i. Handling scalar operations with those rules, all of the operations that we've covered for real vector spaces carry over unchanged. 2.2 Example Matrix multiplication is the same, although the scalar arithmetic involves more bookkeeping. 1 + 1i20i i2 + 3i1 + 0i10i 3ii =(1 + 1i)
(1 + 0i) + (20i)
(3i) (1 + 1i)
(10i) + (20i)
(i) (i)
(1 + 0i) + (2 + 3i)
(3i) (i)
(10i) + (2 + 3i)
(i) = 1 + 7i11i 95i3 + 3i Everything else from prior chapters that we can, we shall also carry over unchanged. For instance, we shall call this h0 BBB@1 + 0i 0 + 0i ... 0 + 0i1 CCCA;:::;0 BBB@0 + 0i 0 + 0i ... 1 + 0i1 CCCAi thestandard basis forCnas a vector space over Cand again denote it En. Section II. Similarity 351 II Similarity II.1 Denition and Examples We've dened Hand ^Hto be matrix-equivalent if there are nonsingular ma- tricesPandQsuch that ^H=PHQ . That denition is motivated by this diagram Vw.r.t.Bh! HWw.r.t.D id??y id??y Vw.r.t. ^Bh! ^HWw.r.t. ^D showing that Hand ^Hboth represent hbut with respect to dierent pairs of bases. We now specialize that setup to the case where the codomain equals the domain, and where the codomain's basis equals the domain's basis. Vw.r.t.Bt!Vw.r.t.B id??y id??y Vw.r.t.Dt!Vw.r.t.D To move from the lower left to the lower right we can either go straight over, or up, over, and then down. In matrix terms, RepD;D(t) = RepB;D(id) RepB;B(t) RepB;D(id)
1 (recall that a representation of composition like this one reads right to left). 1.1 Denition The matrices TandSaresimilar if there is a nonsingular P such thatT=PSP1. Since nonsingular matrices are square, the similar matrices TandSmust be square and of the same size. 1.2 Example With these two, P=2 1 1 1 S=23 11 calculation gives that Sis similar to this matrix. T=01 1 1 352 Chapter Five. Similarity 1.3 Example The only matrix similar to the zero matrix is itself: PZP1= PZ=Z. The only matrix similar to the identity matrix is itself: PIP1= PP1=I. Since matrix similarity is a special case of matrix equivalence, if two ma- trices are similar then they are equivalent. What about the converse: must matrix equivalent square matrices be similar? The answer is no. The prior example shows that the similarity classes are dierent from the matrix equiv- alence classes, because the matrix equivalence class of the identity consists of all nonsingular matrices of that size. Thus, for instance, these two are matrix equivalent but not similar. T=1 0 0 1 S=1 2 0 3 So some matrix equivalence classes split into two or more similarity classes | similarity gives a ner partition than does equivalence. This picture shows some matrix equivalence classes subdivided into similarity classes. . . .A B To understand the similarity relation we shall study the similarity classes. We approach this question in the same way that we've studied both the row equivalence and matrix equivalence relations, by nding a canonical form for representativesof the similarity classes, called Jordan form. With this canon- ical form, we can decide if two matrices are similar by checking whether they reduce to the same representative. We've also seen with both row equivalence and matrix equivalence that a canonical form gives us insight into the ways in which members of the same class are alike (e.g., two identically-sized matrices are matrix equivalent if and only if they have the same rank). Exercises 1.4For S=1 3 26 T=0 0 11=25 P=4 2 3 2 check thatT=PSP1. X1.5Example 1.3 shows that the only matrix similar to a zero matrix is itself and that the only matrix similar to the identity is itself. (a)Show that the 11 matrix (2), also, is similar only to itself. (b)Is a matrix of the form cIfor some scalar csimilar only to itself? (c)Is a diagonal matrix similar only to itself? 1.6Show that these matrices are not similar.0 @1 0 4 1 1 3 2 1 71 A0 @1 0 1 0 1 1 3 1 21 A More information on representatives is in the appendix. Section II. Similarity 353 1.7Consider the transformation t:P2!P 2described by x27!x+ 1,x7!x21, and 17!3. (a)FindT= RepB;B(t) whereB=hx2;x;1i. (b)FindS= RepD;D(t) whereD=h1;1 +x;1 +x+x2i. (c)Find the matrix Psuch thatT=PSP1. X1.8Exhibit an nontrivial similarity relationship in this way: let t:C2!C2act by1 2 7!3 0 1 1 7!1 2 and pick two bases, and represent twith respect to then T= RepB;B(t) and S= RepD;D(t). Then compute the PandP1to change bases from BtoDand back again. 1.9Explain Example 1.3 in terms of maps. X1.10 Are there two matrices AandBthat are similar while A2andB2are not similar? [Halmos] X1.11 Prove that if two matrices are similar and one is invertible then so is the other. X1.12 Show that similarity is an equivalence relation. 1.13 Consider a matrix representing, with respect to some B;B, re ection across thex-axis in R2. Consider also a matrix representing, with respect to some D;D , re ection across the y-axis. Must they be similar? 1.14 Prove that similarity preserves determinants and rank. Does the converse hold? 1.15 Is there a matrix equivalence class with only one matrix similarity class inside? One with innitely many similarity classes? 1.16 Can two dierent diagonal matrices be in the same similarity class? X1.17 Prove that if two matrices are similar then their k-th powers are similar when k>0. What if k0? X1.18 Letp(x) be the polynomial cnxn+


+c1x+c0. Show that if Tis similar to Sthenp(T) =cnTn+


+c1T+c0Iis similar to p(S) =cnSn+


+c1S+c0I. 1.19 List all of the matrix equivalence classes of 1 1 matrices. Also list the sim- ilarity classes, and describe which similarity classes are contained inside of each matrix equivalence class. 1.20 Does similarity preserve sums? 1.21 Show that if TIandNare similar matrices then TandN+Iare also similar. II.2 Diagonalizability The prior subsection denes the relation of similarity and shows that, although similar matrices are necessarily matrix equivalent, the converse does not hold. Some matrix-equivalence classes break into two or more similarity classes (the nonsingular nnmatrices, for instance). This means that the canonical form for matrix equivalence, a block partial-identity, cannot be used as a canonical form for matrix similarity because the partial-identities cannot be in more than one similarity class, so there are similarity classes without one. This picture illustrates. As earlier in this book, class representatives are shown with stars. 354 Chapter Five. Similarity . . .? ??????? ? We are developing a canonical form for representatives of the similarity classes. We naturally try to build on our previous work, meaning rst that the partial identity matrices should represent the similarity classes into which they fall, and beyond that, that the representatives should be as simple as possible. The simplest extension of the partial-identity form is a diagonal form. 2.1 Denition A transformation is diagonalizable if it has a diagonal repre- sentation with respect to the same basis for the codomain as for the domain. Adiagonalizable matrix is one that is similar to a diagonal matrix: Tis diag- onalizable if there is a nonsingular Psuch thatPTP1is diagonal. 2.2 Example The matrix42 1 1 is diagonalizable. 2 0 0 3 =1 2 1142 1 11 2 111 2.3 Example Not every matrix is diagonalizable. The square of N=0 0 1 0 is the zero matrix. Thus, for any map nthatNrepresents (with respect to the same basis for the domain as for the codomain), the composition nnis the zero map. This implies that no such map ncan be diagonally represented (with respect to any B;B) because no power of a nonzero diagonal matrix is zero. That is, there is no diagonal matrix in N's similarity class. That example shows that a diagonal form will not do for a canonical form | we cannot nd a diagonal matrix in each matrix similarity class. However, the canonical form that we are developing has the property that if a matrix can be diagonalized then the diagonal matrix is the canonical representative of the similarity class. The next result characterizes which maps can be diagonalized. 2.4 Corollary A transformation tis diagonalizable if and only if there is a basisB=h~1;:::;~niand scalars 1;:::;nsuch thatt(~i) =i~ifor eachi. Proof .This follows from the denition by considering a diagonal representation matrix. RepB;B(t) =0 BB@...... RepB(t(~1))


RepB(t(~n)) ......1 CCA=0 B@1 0 ......... 0 n1 CA Section II. Similarity 355 This representation is equivalent to the existence of a basis satisfying the stated conditions simply by the denition of matrix representation. QED 2.5 Example To diagonalize T=3 2 0 1 we take it as the representation of a transformation with respect to the standard basisT= RepE2;E2(t) and we look for a basis B=h~1;~2isuch that RepB;B(t) =10 02 that is, such that t(~1) =1~1andt(~2) =2~2. 3 2 0 1 ~1=1
~13 2 0 1 ~2=2
~2 We are looking for scalars xsuch that this equation 3 2 0 1b1 b2 =x
b1 b2 has solutions b1andb2, which are not both zero. Rewrite that as a linear system. (3x)
b1+ 2
b2= 0 (1x)
b2= 0() In the bottom equation the two numbers multiply to give zero only if at least one of them is zero so there are two possibilities, b2= 0 andx= 1. In the b2= 0 possibility, the rst equation gives that either b1= 0 orx= 3. Since the case of bothb1= 0 andb2= 0 is disallowed, we are left looking at the possibility of x= 3. With it, the rst equation in ( ) is 0
b1+ 2
b2= 0 and so associated with 3 are vectors with a second component of zero and a rst component that is free. 3 2 0 1b1 0 = 3
b1 0 That is, one solution to ( ) is1= 3, and we have a rst basis vector. ~1=1 0 In thex= 1 possibility, the rst equation in ( ) is 2
b1+ 2
b2= 0, and so associated with 1 are vectors whose second component is the negative of their rst component.3 2 0 1b1 b1 = 1
b1 b1 356 Chapter Five. Similarity Thus, another solution is 2= 1 and a second basis vector is this. ~2=1 1 To nish, drawing the similarity diagram R2 w.r.t.E2t! TR2 w.r.t.E2 id??y id??y R2 w.r.t.Bt! DR2 w.r.t.B and noting that the matrix RepB;E2(id) is easy leads to this diagonalization. 3 0 0 1 =1 1 0113 2 0 11 1 01 In the next subsection, we will expand on that example by considering more closely the property of Corollary 2.4. This includes seeing another way, the way that we will routinely use, to nd the 's. Exercises X2.6Repeat Example 2.5 for the matrix from Example 2.2. 2.7Diagonalize these upper triangular matrices. (a)2 1 0 2 (b)5 4 0 1 X2.8What form do the powers of a diagonal matrix have? 2.9Give two same-sized diagonal matrices that are not similar. Must any two dierent diagonal matrices come from dierent similarity classes? 2.10 Give a nonsingular diagonal matrix. Can a diagonal matrix ever be singular? X2.11 Show that the inverse of a diagonal matrix is the diagonal of the the inverses, if no element on that diagonal is zero. What happens when a diagonal entry is zero? 2.12 The equation ending Example 2.51 1 0113 2 0 11 1 01 =3 0 0 1 is a bit jarring because for Pwe must take the rst matrix, which is shown as an inverse, and for P1we take the inverse of the rst matrix, so that the two 1 powers cancel and this matrix is shown without a superscript 1. (a)Check that this nicer-appearing equation holds.3 0 0 1 =1 1 013 2 0 11 1 011 (b)Is the previous item a coincidence? Or can we always switch the Pand the P1? 2.13 Show that the Pused to diagonalize in Example 2.5 is not unique. 2.14 Find a formula for the powers of this matrix Hint: see Exercise 8.3 1 4 2 X2.15 Diagonalize these. Section II. Similarity 357 (a)1 1 0 0 (b)0 1 1 0 2.16 We can ask how diagonalization interacts with the matrix operations. Assume thatt;s:V!Vare each diagonalizable. Is ctdiagonalizable for all scalars c? What about t+s?ts? X2.17 Show that matrices of this form are not diagonalizable. 1c 0 1 c6= 0 2.18 Show that each of these is diagonalizable. (a)1 2 2 1 (b)x y y z x;y;z scalars II.3 Eigenvalues and Eigenvectors In this subsection we will focus on the property of Corollary 2.4. 3.1 Denition A transformation t:V!Vhas a scalar eigenvalueif there is a nonzero eigenvector ~2Vsuch thatt(~) =
~. (\Eigen" is German for \characteristic of" or \peculiar to"; some authors call these characteristic values and vectors. No authors call them \peculiar".) 3.2 Example The projection map 0 @x y z1 A7!0 @x y 01 Ax;y;z2C has an eigenvalue of 1 associated with any eigenvector of the form 0 @x y 01 A wherexandyare non-0 scalars. On the other hand, 2 is not an eigenvalue of since no non- ~0 vector is doubled. That example shows why the `non- ~0' appears in the denition. Disallowing ~0 as an eigenvector eliminates trivial eigenvalues. (Note, however, that a matrix can have an eigenvalue = 0.) 3.3 Example The only transformation on the trivial space f~0gis~07!~0. This map has no eigenvalues because there are no non- ~0 vectors~ vmapped to a scalar multiple
~ vof themselves. 358 Chapter Five. Similarity 3.4 Example Consider the homomorphism t:P1!P 1given byc0+c1x7! (c0+c1) + (c0+c1)x. The range of tis one-dimensional. Thus an application of tto a vector in the range will simply rescale that vector: c+cx7!(2c) + (2c)x. That is,thas an eigenvalue of 2 associated with eigenvectors of the form c+cx wherec6= 0. This map also has an eigenvalue of 0 associated with eigenvectors of the form ccxwherec6= 0. 3.5 Denition A square matrix Thas a scalar eigenvalueassociated with the non-~0eigenvector ~ifT~=
~. 3.6 Remark Although this extension from maps to matrices is obvious, there is a point that must be made. Eigenvalues of a map are also the eigenvalues of matrices representing that map, and so similar matrices have the same eigen- values. But the eigenvectors are dierent | similar matrices need not have the same eigenvectors. For instance, consider again the transformation t:P1!P 1given byc0+ c1x7!(c0+c1)+(c0+c1)x. It has an eigenvalue of 2 associated with eigenvectors of the form c+cxwherec6= 0. If we represent twith respect to B=h1 + 1x;11xi T= RepB;B(t) =2 0 0 0 then 2 is an eigenvalue of T, associated with these eigenvectors. f c0 c1 2 0 0 0 c0 c1 = 2c0 2c1 g=f c0 0c02C; c06= 0g On the other hand, representing twith respect to D=h2 + 1x;1 + 0xigives S= RepD;D(t) =3 1 31 and the eigenvectors of Sassociated with the eigenvalue 2 are these. fc0 c13 1 31c0 c1 =2c0 2c1 g=f0 c1c12C; c16= 0g Thus similar matrices can have dierent eigenvectors. Here is an informal description of what's happening. The underlying trans- formation doubles the eigenvectors ~ v7!2
~ v. But when the matrix representing the transformation is T= RepB;B(t) then it \assumes" that column vectors are representations with respect to B. In contrast, S= RepD;D(t) \assumes" that column vectors are representations with respect to D. So the vectors that get doubled by each matrix look dierent. The next example illustrates the basic tool for nding eigenvectors and eigen- values. Section II. Similarity 359 3.7 Example What are the eigenvalues and eigenvectors of this matrix? T=0 @1 2 1 2 02 1 2 31 A To nd the scalars xsuch thatT~=x~for non-~0 eigenvectors ~, bring every- thing to the left-hand side 0 @1 2 1 2 02 1 2 31 A0 @z1 z2 z31 Ax0 @z1 z2 z31 A=~0 and factor ( TxI)~=~0. (Note that it says TxI; the expression Txdoesn't make sense because Tis a matrix while xis a scalar.) This homogeneous linear system0 @1x 2 1 2 0x2 1 2 3x1 A0 @z1 z2 z31 A=0 @0 0 01 A has a non-~0 solution if and only if the matrix is singular. We can determine when that happens. 0 =jTxIj =1x 2 1 2 0x2 1 2 3x =x34x2+ 4x =x(x2)2 The eigenvalues are 1= 0 and2= 2. To nd the associated eigenvectors, plug in each eigenvalue. Plugging in 1= 0 gives 0 @10 2 1 2 002 1 2 301 A0 @z1 z2 z31 A=0 @0 0 01 A =)0 @z1 z2 z31 A=0 @a a a1 A for a scalar parameter a6= 0 (ais non-0 because eigenvectors must be non- ~0). In the same way, plugging in 2= 2 gives 0 @12 2 1 2 022 1 2 321 A0 @z1 z2 z31 A=0 @0 0 01 A =)0 @z1 z2 z31 A=0 @b 0 b1 A withb6= 0. 360 Chapter Five. Similarity 3.8 Example If S=1 0 3 (hereis not a projection map, it is the number 3 :14:::) then  x 1 0 3x= (x)(x3) soShas eigenvalues of 1=and2= 3. To nd associated eigenvectors, rst plug in1forx:  1 0 3z1 z2 =0 0 =)z1 z2 =a 0 for a scalar a6= 0, and then plug in 2: 3 1 0 33z1 z2 =0 0 =)z1 z2 =b=(3) b whereb6= 0. 3.9 Denition The characteristic polynomial of a square matrix Tis the determinant of the matrix TxI, wherexis a variable. The characteristic equation isjTxIj= 0. The characteristic polynomial of a transformation t is the polynomial of any RepB;B(t). Exercise 30 checks that the characteristic polynomial of a transformation is well-dened, that is, any choice of basis yields the same polynomial. 3.10 Lemma A linear transformation on a nontrivial vector space has at least one eigenvalue. Proof .Any root of the characteristic polynomial is an eigenvalue. Over the complex numbers, any polynomial of degree one or greater has a root. (This is the reason that in this chapter we've gone to scalars that are complex.) QED Notice the familiar form of the sets of eigenvectors in the above examples. 3.11 Denition The eigenspace of a transformation tassociated with the eigenvalueisV=f~t(~) =~g. The eigenspace of a matrix is dened analogously. 3.12 Lemma An eigenspace is a subspace. Proof .An eigenspace must be nonempty | for one thing it contains the zero vector since a. linear transformation maps the zero vector to the zero vector. Section II. Similarity 361 Thus we need only check closure. Take vectors ~1;:::;~nfromV, to show that any linear combination is in V t(c1~1+c2~2+


+cn~n) =c1t(~1) +


+cnt(~n) =c1~1+


+cn~n =(c1~1+


+cn~n) (the second equality holds even if any ~iis~0 sincet(~0) =
~0 =~0). QED 3.13 Example In Example 3.8 the eigenspace associated with the eigenvalue and the eigenspace associated with the eigenvalue 3 are these. V=fa 0a2RgV3=fb=3 bb2Rg 3.14 Example In Example 3.7, these are the eigenspaces associated with the eigenvalues 0 and 2. V0=f0 @a a a1 Aa2Rg; V 2=f0 @b 0 b1 Ab2Rg: 3.15 Remark The characteristic equation is 0 = x(x2)2so in some sense 2 is an eigenvalue \twice". However there are not \twice" as many eigenvectors, in that the dimension of the eigenspace is one, not two. The next example shows a case where a number, 1, is a double root of the characteristic equation and the dimension of the associated eigenspace is two. 3.16 Example With respect to the standard bases, this matrix 0 @1 0 0 0 1 0 0 0 01 A represents projection. 0 @x y z1 A7!0 @x y 01 Ax;y;z2C Its eigenspace associated with the eigenvalue 0 and its eigenspace associated with the eigenvalue 1 are easy to nd. V0=f0 @0 0 c31 Ac32CgV1=f0 @c1 c2 01 Ac1;c22Cg 362 Chapter Five. Similarity By the lemma, if two eigenvectors ~ v1and~ v2are associated with the same eigenvalue then any linear combination of those two is also an eigenvector as- sociated with that same eigenvalue. But, if two eigenvectors ~ v1and~ v2are associated with dierent eigenvalues then the sum ~ v1+~ v2need not be related to the eigenvalue of either one. In fact, just the opposite. If the eigenvalues are dierent then the eigenvectors are not linearly related. 3.17 Theorem For any set of distinct eigenvalues of a map or matrix, a set of associated eigenvectors, one per eigenvalue, is linearly independent. Proof .We will use induction on the number of eigenvalues. If there is no eigen- value or only one eigenvalue then the set of associated eigenvectors is empty or is a singleton set with a non- ~0 member, and in either case is linearly independent. For induction, assume that the theorem is true for any set of kdistinct eigen- values, suppose that 1;:::;k+1are distinct eigenvalues, and let ~ v1;:::;~ vk+1 be associated eigenvectors. If c1~ v1+


+ck~ vk+ck+1~ vk+1=~0 then after multi- plying both sides of the displayed equation by k+1, applying the map or matrix to both sides of the displayed equation, and subtracting the rst result from the second, we have this. c1(k+11)~ v1+


+ck(k+1k)~ vk+ck+1(k+1k+1)~ vk+1=~0 The induction hypothesis now applies: c1(k+11) = 0;:::;ck(k+1k) = 0. Thus, as all the eigenvalues are distinct, c1; :::; ckare all 0. Finally, now ck+1 must be 0 because we are left with the equation ~ vk+16=~0. QED 3.18 Example The eigenvalues of 0 @22 2 0 1 1 4 8 31 A are distinct: 1= 1,2= 2, and3= 3. A set of associated eigenvectors like f0 @2 1 01 A;0 @9 4 41 A;0 @2 1 21 Ag is linearly independent. 3.19 Corollary Annnmatrix with ndistinct eigenvalues is diagonalizable. Proof .Form a basis of eigenvectors. Apply Corollary 2.4. QED Exercises 3.20 For each, nd the characteristic polynomial and the eigenvalues. Section II. Similarity 363 (a)109 42 (b)1 2 4 3 (c)0 3 7 0 (d)0 0 0 0 (e)1 0 0 1 X3.21 For each matrix, nd the characteristic equation, and the eigenvalues and associated eigenvectors. (a)3 0 81 (b)3 2 1 0 3.22 Find the characteristic equation, and the eigenvalues and associated eigenvec- tors for this matrix. Hint. The eigenvalues are complex.21 5 2 3.23 Find the characteristic polynomial, the eigenvalues, and the associated eigen- vectors of this matrix. 0 @1 1 1 0 0 1 0 0 11 A X3.24 For each matrix, nd the characteristic equation, and the eigenvalues and associated eigenvectors. (a)0 @32 0 2 3 0 0 0 51 A (b)0 @0 1 0 0 0 1 417 81 A X3.25 Lett:P2!P 2be a0+a1x+a2x27!(5a0+ 6a1+ 2a2)(a1+ 8a2)x+ (a02a2)x2: Find its eigenvalues and the associated eigenvectors. 3.26 Find the eigenvalues and eigenvectors of this map t:M2!M 2.a b c d 7!2c a +c b2c d X3.27 Find the eigenvalues and associated eigenvectors of the dierentiation operator d=dx :P3!P 3. 3.28 Prove that the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal. X3.29 Find the formula for the characteristic polynomial of a 2 2 matrix. 3.30 Prove that the characteristic polynomial of a transformation is well-dened. X3.31 (a) Can any non- ~0 vector in any nontrivial vector space be a eigenvector? That is, given a ~ v6=~0 from a nontrivial V, is there a transformation t:V!V and a scalar 2Rsuch thatt(~ v) =~ v? (b)Given a scalar , can any non- ~0 vector in any nontrivial vector space be an eigenvector associated with the eigenvalue ? X3.32 Suppose that t:V!VandT= RepB;B(t). Prove that the eigenvectors of T associated with are the non- ~0 vectors in the kernel of the map represented (with respect to the same bases) by TI. 3.33 Prove that if a;:::; d are all integers and a+b=c+dthena b c d has integral eigenvalues, namely a+bandac. 364 Chapter Five. Similarity X3.34 Prove that if Tis nonsingular and has eigenvalues 1;:::;nthenT1has eigenvalues 1 =1;:::; 1=n. Is the converse true? X3.35 Suppose that Tisnnandc;dare scalars. (a)Prove that if Thas the eigenvalue with an associated eigenvector ~ vthen~ v is an eigenvector of cT+dIassociated with eigenvalue c+d. (b)Prove that if Tis diagonalizable then so is cT+dI. X3.36 Show thatis an eigenvalue of Tif and only if the map represented by TI is not an isomorphism. 3.37 [Strang 80] (a)Show that if is an eigenvalue of Athenkis an eigenvalue of Ak. (b)What is wrong with this proof generalizing that? \If is an eigenvalue of A andis an eigenvalue for B, thenis an eigenvalue for AB, for, ifA~ x=~ x andB~ x=~ xthenAB~ x =A~ x =A~ x =~ x"? 3.38 Do matrix-equivalent matrices have the same eigenvalues? 3.39 Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue. 3.40 Diagonalize.0 @1 2 2 2 2 2 3661 A 3.41 Suppose that Pis a nonsingular nnmatrix. Show that the similarity trans- formation maptP:Mnn!MnnsendingT7!PTP1is an isomorphism. ?3.42 Show that if Ais annsquare matrix and each row (column) sums to cthen cis a characteristic root of A. [Math. Mag., Nov. 1967] Section III. Nilpotence 365 III Nilpotence The goal of this chapter is to show that every square matrix is similar to one that is a sum of two kinds of simple matrices. The prior section focused on the rst simple kind, diagonal matrices. We now consider the other kind. III.1 Self-Composition This subsection is optional, although it is necessary for later material in this section and in the next one. A linear transformations t:V!V, because it has the same domain and codomain, can be iterated.That is, compositions of twith itself such as t2=tt andt3=tttare dened. ~ v t(~ v) t2(~ v) Note that this power notation for the linear transformation functions dovetails with the notation that we've used earlier for their square matrix representations because if RepB;B(t) =Tthen RepB;B(tj) =Tj. 1.1 Example For the derivative map d=dx :P3!P 3given by a+bx+cx2+dx3d=dx7!b+ 2cx+ 3dx2 the second power is the second derivative a+bx+cx2+dx3d2=dx2 7! 2c+ 6dx the third power is the third derivative a+bx+cx2+dx3d3=dx3 7! 6d and any higher power is the zero map. 1.2 Example This transformation of the space of 2 2 matrices a b c d t7!b a d0 More information on function interation is in the appendix. 366 Chapter Five. Similarity has this second powera b c d t2 7!a b 0 0 and this third power.a b c d t3 7!b a 0 0 After that, t4=t2andt5=t3, etc. These examples suggest that on iteration more and more zeros appear until there is a settling down. The next result makes this precise. 1.3 Lemma For any transformation t:V!V, the rangespaces of the powers form a descending chain VR(t)R(t2)


and the nullspaces form an ascending chain. f~0gN(t)N(t2)


Further, there is a ksuch that for powers less than kthe subsets are proper (if j <k thenR(tj)R(tj+1) andN(tj)N(tj+1)), while for powers greater thankthe sets are equal (if jkthenR(tj) =R(tj+1) andN(tj) =N(tj+1)). Proof .We will do the rangespace half and leave the rest for Exercise 13. Recall, however, that for any map the dimension of its rangespace plus the dimension of its nullspace equals the dimension of its domain. So if the rangespaces shrink then the nullspaces must grow. That the rangespaces form chains is clear because if ~ w2R(tj+1), so that ~ w=tj+1(~ v), then~ w=tj(t(~ v) ) and so~ w2R(tj). To verify the \further" property, rst observe that if any pair of rangespaces in the chain are equal R(tk) =R(tk+1) then all subsequent ones are also equal R(tk+1) =R(tk+2), etc. This is because if t:R(tk+1)!R(tk+2) is the same map, with the same domain, as t:R(tk)!R(tk+1) and it therefore has the same range: R(tk+1) = R(tk+2) (and induction shows that it holds for all higher powers). So if the chain of rangespaces ever stops being strictly decreasing then it is stable from that point onward. But the chain must stop decreasing. Each rangespace is a subspace of the one before it. For it to be a proper subspace it must be of strictly lower dimension (see Exercise 11). These spaces are nite-dimensional and so the chain can fall for only nitely-many steps, that is, the power kis at most the dimension of V. QED 1.4 Example The derivative map a+bx+cx2+dx3d=dx7!b+ 2cx+ 3dx2of Example 1.1 has this chain of rangespaces P3P 2P 1P 0f~0g=f~0g=


Section III. Nilpotence 367 and this chain of nullspaces. f~0gP 0P 1P 2P 3=P3=


1.5 Example The transformation :C3!C3projecting onto the rst two coordinates 0 @c1 c2 c31 A7!0 @c1 c2 01 A hasC3R() =R(2) =


andf~0gN() =N(2) =


. 1.6 Example Lett:P2!P 2be the map c0+c1x+c2x27!2c0+c2x:As the lemma describes, on iteration the rangespace shrinks R(t0) =P2R(t) =fa+bxa;b2CgR(t2) =faa2Cg and then stabilizes R(t2) =R(t3) =


, while the nullspace grows N(t0) =f0gN(t) =fcxc2CgN(t2) =fcx+dc;d2Cg and then stabilizes N(t2) =N(t3) =


. This graph illustrates Lemma 1.3. The horizontal axis gives the power j of a transformation. The vertical axis gives the dimension of the rangespace oftjas the distance above zero | and thus also shows the dimension of the nullspace as the distance below the gray horizontal line, because the two add to the dimension nof the domain. 012jnn rank(tj) Powerjof the transformation As sketched, on iteration the rank falls and with it the nullity grows until the two reach a steady state. This state must be reached by the n-th iterate. The steady state's distance above zero is the dimension of the generalized rangespace and its distance below nis the dimension of the generalized nullspace. 1.7 Denition Lettbe a transformation on an n-dimensional space. The generalized rangespace (or the closure of the rangespace ) isR1(t) =R(tn) The generalized nullspace (or the closure of the nullspace ) isN1(t) =N(tn). 368 Chapter Five. Similarity Exercises 1.8Give the chains of rangespaces and nullspaces for the zero and identity trans- formations. 1.9For each map, give the chain of rangespaces and the chain of nullspaces, and the generalized rangespace and the generalized nullspace. (a)t0:P2!P 2,a+bx+cx27!b+cx2 (b)t1:R2!R2,a b 7!0 a (c)t2:P2!P 2,a+bx+cx27!b+cx+ax2 (d)t3:R3!R3,0 @a b c1 A7!0 @a a b1 A 1.10 Prove that function composition is associative ( tt)t=t(tt) and so we can writet3without specifying a grouping. 1.11 Check that a subspace must be of dimension less than or equal to the dimen- sion of its superspace. Check that if the subspace is proper (the subspace does not equal the superspace) then the dimension is strictly less. (This is used in the proof of Lemma 1.3.) 1.12 Prove that the generalized rangespace R1(t) is the entire space, and the generalized nullspace N1(t) is trivial, if the transformation tis nonsingular. Is this `only if' also? 1.13 Verify the nullspace half of Lemma 1.3. 1.14 Give an example of a transformation on a three dimensional space whose range has dimension two. What is its nullspace? Iterate your example until the rangespace and nullspace stabilize. 1.15 Show that the rangespace and nullspace of a linear transformation need not be disjoint. Are they ever disjoint? III.2 Strings This subsection is optional, and requires material from the optional Direct Sum subsection. The prior subsection shows that as jincreases, the dimensions of the R(tj)'s fall while the dimensions of the N(tj)'s rise, in such a way that this rank and nullity split the dimension of V. Can we say more; do the two split a basis | is V=R(tj)N(tj)? The answer is yes for the smallest power j= 0 sinceV=R(t0)N(t0) = Vf~0g. The answer is also yes at the other extreme. 2.1 Lemma Wheret:V!Vis a linear transformation, the space is the direct sumV=R1(t)N1(t). That is, both dim( V) = dim( R1(t)) + dim( N1(t)) andR1(t)\N1(t) =f~0g. Section III. Nilpotence 369 Proof .We will verify the second sentence, which is equivalent to the rst. The rst clause, that the dimension nof the domain of tnequals the rank of tnplus the nullity of tn, holds for any transformation and so we need only verify the second clause. Assume that ~ v2R1(t)\N1(t) =R(tn)\N(tn), to prove that ~ vis~0. Because~ vis in the nullspace, tn(~ v) =~0. On the other hand, because R(tn) = R(tn+1), the mapt:R1(t)!R1(t) is a dimension-preserving homomorphism and therefore is one-to-one. A composition of one-to-one maps is one-to-one, and sotn:R1(t)!R1(t) is one-to-one. But now | because only ~0 is sent by a one-to-one linear map to ~0 | the fact that tn(~ v) =~0 implies that ~ v=~0.QED 2.2 Note Technically we should distinguish the map t:V!Vfrom the map t:R1(t)!R1(t) because the domains or codomains might dier. The second one is said to be the restrictionofttoR(tk). We shall use later a point from that proof about the restriction map, namely that it is nonsingular. In contrast to the j= 0 andj=ncases, for intermediate powers the space Vmight not be the direct sum of R(tj) andN(tj). The next example shows that the two can have a nontrivial intersection. 2.3 Example Consider the transformation of C2dened by this action on the elements of the standard basis. 1 0 n7!0 1 0 1 n7!0 0 N= RepE2;E2(n) =0 0 1 0 The vector ~ e2= 0 1 is in both the rangespace and nullspace. Another way to depict this map's action is with a string . ~ e17!~ e27!~0 2.4 Example A map ^n:C4!C4whose action onE4is given by the string ~ e17!~ e27!~ e37!~ e47!~0 hasR(^n)\N(^n) equal to the span [ f~ e4g], hasR(^n2)\N(^n2) = [f~ e3;~ e4g], and hasR(^n3)\N(^n3) = [f~ e4g]. The matrix representation is all zeros except for some subdiagonal ones. ^N= RepE4;E4(^n) =0 BB@0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 01 CCA More information on map restrictions is in the appendix. 370 Chapter Five. Similarity 2.5 Example Transformations can act via more than one string. A transfor- mationtacting on a basis B=h~1;:::;~5iby ~17!~27!~37!~0 ~47!~57!~0 is represented by a matrix that is all zeros except for blocks of subdiagonal ones RepB;B(t) =0 BBBB@0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 01 CCCCA (the lines just visually organize the blocks). In those three examples all vectors are eventually transformed to zero. 2.6 Denition Anilpotent transformation is one with a power that is the zero map. A nilpotent matrix is one with a power that is the zero matrix. In either case, the least such power is the index of nilpotency . 2.7 Example In Example 2.3 the index of nilpotency is two. In Example 2.4 it is four. In Example 2.5 it is three. 2.8 Example The dierentiation map d=dx :P2!P 2is nilpotent of index three since the third derivative of any quadratic polynomial is zero. This map's action is described by the string x27!2x7!27!0 and taking the basis B=hx2;2x;2igives this representation. RepB;B(d=dx ) =0 @0 0 0 1 0 0 0 1 01 A Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones. 2.9 Example With the matrix ^Nfrom Example 2.4, and this four-vector basis D=h0 BB@1 0 1 01 CCA;0 BB@0 2 1 01 CCA;0 BB@1 1 1 01 CCA;0 BB@0 0 0 11 CCAi a change of basis operation produces this representation with respect to D;D . 0 BB@1 0 1 0 0 2 1 0 1 1 1 0 0 0 0 11 CCA0 BB@0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 01 CCA0 BB@1 0 1 0 0 2 1 0 1 1 1 0 0 0 0 11 CCA1 =0 BB@1 0 1 0 32 5 0 21 3 0 2 12 01 CCA Section III. Nilpotence 371 The new matrix is nilpotent; it's fourth power is the zero matrix since (P^NP1)4=P^NP1
P^NP1
P^NP1
P^NP1=P^N4P1 and ^N4is the zero matrix. The goal of this subsection is Theorem 2.13, which shows that the prior example is prototypical in that every nilpotent matrix is similar to one that is all zeros except for blocks of subdiagonal ones. 2.10 Denition Lettbe a nilpotent transformation on V. At-string gener- ated by~ v2Vis a sequenceh~ v;t(~ v);:::;tk1(~ v)i. This sequence has lengthk. At-string basis is a basis that is a concatenation of t-strings. 2.11 Example In Example 2.5, the t-stringsh~1;~2;~3iandh~4;~5i, of length three and two, can be concatenated to make a basis for the domain of t. 2.12 Lemma If a space has a t-string basis then the longest string in it has length equal to the index of nilpotency of t. Proof .Suppose not. Those strings cannot be longer; if the index is kthen tksends any vector | including those starting the string | to ~0. So suppose instead that there is a transformation tof indexkon some space, such that the space has a t-string basis where all of the strings are shorter than length k. Becausethas indexk, there is a vector ~ vsuch thattk1(~ v)6=~0. Represent ~ vas a linear combination of basis elements and apply tk1. We are supposing thattk1sends each basis element to ~0 but that it does not send ~ vto~0. That is impossible. QED We shall show that every nilpotent map has an associated string basis. Then our goal theorem, that every nilpotent matrix is similar to one that is all zeros except for blocks of subdiagonal ones, is immediate, as in Example 2.5. Looking for a counterexample, a nilpotent map without an associated string basis that is disjoint, will suggest the idea for the proof. Consider the map t:C5!C5with this action. ~ e1 ~ e27! 7!~ e37!~0 ~ e47!~ e57!~0RepE5;E5(t) =0 BBBB@0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 01 CCCCA Even after ommitting the zero vector, these three strings aren't disjoint, but that doesn't end hope of nding a t-string basis. It only means that E5will not do for the string basis. To nd a basis that will do, we rst nd the number and lengths of its strings. Since t's index of nilpotency is two, Lemma 2.12 says that at least one 372 Chapter Five. Similarity string in the basis has length two. Thus the map must act on a string basis in one of these two ways. ~17!~27!~0 ~37!~47!~0 ~57!~0~17!~27!~0 ~37!~0 ~47!~0 ~57!~0 Now, the key point. A transformation with the left-hand action has a nullspace of dimension three since that's how many basis vectors are sent to zero. A transformation with the right-hand action has a nullspace of dimension four. Using the matrix representation above, calculation of t's nullspace N(t) =f0 BBBB@x x z 0 r1 CCCCAx;z;r2Cg shows that it is three-dimensional, meaning that we want the left-hand action. To produce a string basis, rst pick ~2and~4fromR(t)\N(t) ~2=0 BBBB@0 0 1 0 01 CCCCA~4=0 BBBB@0 0 0 0 11 CCCCA (other choices are possible, just be sure that f~2;~4gis linearly independent). For~5pick a vector from N(t) that is not in the span of f~2;~4g. ~5=0 BBBB@1 1 0 0 01 CCCCA Finally, take ~1and~3such thatt(~1) =~2andt(~3) =~4. ~1=0 BBBB@0 1 0 0 01 CCCCA~3=0 BBBB@0 0 0 1 01 CCCCA Section III. Nilpotence 373 Now, with respect to B=h~1;:::;~5i, the matrix of tis as desired. RepB;B(t) =0 BBBB@0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 01 CCCCA 2.13 Theorem Any nilpotent transformation tis associated with a t-string basis. While the basis is not unique, the number and the length of the strings is determined by t. Proof .This illustrates the argument below, which describes three kinds of basis vectors (these basis vectors are shown as squares or circles, according to whether they are in the nullspace or not). k37!k17!





7! k17!17!~0 k37!k17!





7!k17!17!~0 ... k37!k17!


7!k17!17!~0 27!~0... 27!~0 Fix a vector space V; we will argue by induction on the index of nilpotency oft:V!V. If that index is 1 then tis the zero map and any basis is a string basis~17!~0, . . . ,~n7!~0. For the inductive step, assume that the theorem holds for any transformation with an index of nilpotency between 1 and k1 and consider the index kcase. First observe that the restriction to the rangespace t:R(t)!R(t) is also nilpotent, of index k1. Apply the inductive hypothesis to get a string basis forR(t), where the number and length of the strings is determined by t. B=h~1;t(~1);:::;th1(~1)i_h~2;:::;th2(~2)i_


_h~i;:::;thi(~i)i (In the illustration these are the basis vectors of kind 1, so there are istrings shown with this kind of basis vector.) Second, note that taking the nal nonzero vector in each string gives a basis C=hth1(~1);:::;thi(~i)iforR(t)\N(t). (These are illustrated with 1's in squares.) For, a member of R(t) is mapped to zero if and only if it is a linear combination of those basis vectors that are mapped to zero. Extend Cto a basis for all of N(t). ^C=C_h~1;:::;~pi (The~'s are the vectors of kind 2 so that ^Cis the set of squares.) While many choices are possible for the ~'s, their number pis determined by the map tas it is the dimension of N(t) minus the dimension of R(t)\N(t). 374 Chapter Five. Similarity Finally,B_^Cis a basis for R(t)+N(t) because any sum of something in the rangespace with something in the nullspace can be represented using elements ofBfor the rangespace part and elements of ^Cfor the part from the nullspace. Note that dim R(t) +N(t)
= dim(R(t)) + dim( N(t))dim(R(t)\N(t)) = rank(t) + nullity(t)i = dim(V)i and soB_^Ccan be extended to a basis for all of Vby the addition of imore vectors. Specically, remember that each of ~1;:::;~iis inR(t), and extend B_^Cwith vectors ~ v1;:::;~ visuch thatt(~ v1) =~1;:::;t (~ vi) =~i. (In the illustration, these are the 3's.) The check that linear independence is preserved by this extension is Exercise 29. QED 2.14 Corollary Every nilpotent matrix is similar to a matrix that is all zeros except for blocks of subdiagonal ones. That is, every nilpotent map is repre- sented with respect to some basis by such a matrix. This form is unique in the sense that if a nilpotent matrix is similar to two such matrices then those two simply have their blocks ordered dierently. Thus this is a canonical form for the similarity classes of nilpotent matrices provided that we order the blocks, say, from longest to shortest. 2.15 Example The matrix M=11 11 has an index of nilpotency of two, as this calculation shows. pMpN(Mp) 1M= 11 11 f x xx2Cg 2M2=0 0 0 0 C2 The calculation also describes how a map mrepresented by Mmust act on any string basis. With one map application the nullspace has dimension one and so one vector of the basis is sent to zero. On a second application, the nullspace has dimension two and so the other basis vector is sent to zero. Thus, the action of the map is ~17!~27!~0 and the canonical form of the matrix is this. 0 0 1 0 We can exhibit such a m-string basis and the change of basis matrices wit- nessing the matrix similarity. For the basis, take Mto represent mwith respect Section III. Nilpotence 375 to the standard bases, pick a ~22N(m) and also pick a ~1so thatm(~1) =~2. ~2=1 1 ~1=1 0 (If we take Mto be a representative with respect to some nonstandard bases then this picking step is just more messy.) Recall the similarity diagram. C2 w.r.t.E2m! MC2 w.r.t.E2 id??yP id??yP C2 w.r.t.Bm! C2 w.r.t.B The canonical form equals RepB;B(m) =PMP1, where P1= RepB;E2(id) =1 1 0 1 P= (P1)1=11 0 1 and the verication of the matrix calculation is routine.  11 0 1 11 11 1 1 0 1 =0 0 1 0 2.16 Example The matrix 0 BBBB@0 0 0 0 0 1 0 0 0 0 1 1 11 1 0 1 0 0 0 1 01 111 CCCCA is nilpotent. These calculations show the nullspaces growing. p NpN(Np) 10 BBBB@0 0 0 0 0 1 0 0 0 0 1 1 11 1 0 1 0 0 0 1 01 111 CCCCAf0 BBBB@0 0 uv u v1 CCCCAu;v2Cg 20 BBBB@0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 01 CCCCAf0 BBBB@0 y z u v1 CCCCAy;z;u;v2Cg 3 {zero matrix{ C5 That table shows that any string basis must satisfy: the nullspace after one map application has dimension two so two basis vectors are sent directly to zero, 376 Chapter Five. Similarity the nullspace after the second application has dimension four so two additional basis vectors are sent to zero by the second iteration, and the nullspace after three applications is of dimension ve so the nal basis vector is sent to zero in three hops. ~17!~27!~37!~0 ~47!~57!~0 To produce such a basis, rst pick two independent vectors from N(n) ~3=0 BBBB@0 0 1 1 01 CCCCA~5=0 BBBB@0 0 0 1 11 CCCCA then add~2;~42N(n2) such that n(~2) =~3andn(~4) =~5 ~2=0 BBBB@0 1 0 0 01 CCCCA~4=0 BBBB@0 1 0 1 01 CCCCA and nish by adding ~12N(n3) =C5) such that n(~1) =~2. ~1=0 BBBB@1 0 1 0 01 CCCCA Exercises X2.17 What is the index of nilpotency of the left-shift operator, here acting on the space of triples of reals? (x;y;z )7!(0;x;y) X2.18 For each string basis state the index of nilpotency and give the dimension of the rangespace and nullspace of each iteration of the nilpotent map. (a)~17!~27!~0 ~37!~47!~0 (b)~17!~27!~37!~0 ~47!~0 ~57!~0 ~67!~0 (c)~17!~27!~37!~0 Also give the canonical form of the matrix. 2.19 Decide which of these matrices are nilpotent. Section III. Nilpotence 377 (a)2 4 1 2 (b)3 1 1 3 (c)0 @3 2 1 3 2 1 3 2 11 A (d)0 @1 1 4 3 01 5 2 71 A (e)0 @452219 331614 6934291 A X2.20 Find the canonical form of this matrix.0 BBBB@0 1 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 CCCCA X2.21 Consider the matrix from Example 2.16. (a)Use the action of the map on the string basis to give the canonical form. (b)Find the change of basis matrices that bring the matrix to canonical form. (c)Use the answer in the prior item to check the answer in the rst item. X2.22 Each of these matrices is nilpotent. (a)1=21=2 1=21=2 (b)0 @0 0 0 01 1 01 11 A (c)0 @1 11 1 0 1 11 11 A Put each in canonical form. 2.23 Describe the eect of left or right multiplication by a matrix that is in the canonical form for nilpotent matrices. 2.24 Is nilpotence invariant under similarity? That is, must a matrix similar to a nilpotent matrix also be nilpotent? If so, with the same index? X2.25 Show that the only eigenvalue of a nilpotent matrix is zero. 2.26 Is there a nilpotent transformation of index three on a two-dimensional space? 2.27 In the proof of Theorem 2.13, why isn't the proof's base case that the index of nilpotency is zero? X2.28 Lett:V!Vbe a linear transformation and suppose ~ v2Vis such that tk(~ v) =~0 buttk1(~ v)6=~0. Consider the t-stringh~ v;t(~ v);:::;tk1(~ v)i. (a)Prove thattis a transformation on the span of the set of vectors in the string, that is, prove that trestricted to the span has a range that is a subset of the span. We say that the span is a t-invariant subspace. (b)Prove that the restriction is nilpotent. (c)Prove that the t-string is linearly independent and so is a basis for its span. (d)Represent the restriction map with respect to the t-string basis. 2.29 Finish the proof of Theorem 2.13. 2.30 Show that the terms `nilpotent transformation' and `nilpotent matrix', as given in Denition 2.6, t with each other: a map is nilpotent if and only if it is represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an only if there is a basis such that the map's representation with respect to that basis is a nilpotent matrix, or that any representation is a nilpotent matrix?) 2.31 LetTbe nilpotent of index four. How big can the rangespace of T3be? 2.32 Recall that similar matrices have the same eigenvalues. Show that the converse does not hold. 2.33 Prove a nilpotent matrix is similar to one that is all zeros except for blocks of super-diagonal ones. 378 Chapter Five. Similarity X2.34 Prove that if a transformation has the same rangespace as nullspace. then the dimension of its domain is even. 2.35 Prove that if two nilpotent matrices commute then their product and sum are also nilpotent. 2.36 Consider the transformation of Mnngiven bytS(T) =STTSwhereSis annnmatrix. Prove that if Sis nilpotent then so is tS. 2.37 Show that if Nis nilpotent then INis invertible. Is that `only if' also? Section IV. Jordan Form 379 IV Jordan Form This section uses material from three optional subsections: Direct Sum, Deter- minants Exist, and Other Formulas for the Determinant. The chapter on linear maps shows that every h:V!Wcan be represented by a partial-identity matrix with respect to some bases BVandDW. This chapter revisits this issue in the special case that the map is a linear transformation t:V!V. Of course, the general result still applies but with the codomain and domain equal we naturally ask about having the two bases also be equal. That is, we want a canonical form to represent transformations as RepB;B(t). After a brief review section, we began by noting that a block partial identity form matrix is not always obtainable in this B;B case. We therefore considered the natural generalization, diagonal matrices, and showed that if its eigenvalues are distinct then a map or matrix can be diagonalized. But we also gave an example of a matrix that cannot be diagonalized and in the section prior to this one we developed that example. We showed that a linear map is nilpotent | if we take higher and higher powers of the map or matrix then we eventually get the zero map or matrix | if and only if there is a basis on which it acts via disjoint strings. That led to a canonical form for nilpotent matrices. Now, this section concludes the chapter. We will show that the two cases we've studied are exhaustive in that for any linear transformation there is a basis such that the matrix representation RepB;B(t) is the sum of a diagonal matrix and a nilpotent matrix in its canonical form. IV.1 Polynomials of Maps and Matrices Recall that the set of square matrices is a vector space under entry-by-entry addition and scalar multiplication and that this space Mnnhas dimension n2. Thus, for any nnmatrixTthen2+1-member setfI;T;T2;:::;Tn2gis linearly dependent and so there are scalars c0;:::;cn2such thatcn2Tn2+


+c1T+c0I is the zero matrix. 1.1 Remark This observation is small but important. It says that every transformation exhibits a generalized nilpotency: the powers of a square matrix cannot climb forever without a \repeat". 1.2 Example Rotation of plane vectors =6 radians counterclockwise is rep- resented with respect to the standard basis by T=p 3=21=2 1=2p 3=2 and verifying that 0 T4+ 0T3+ 1T22T1Iequals the zero matrix is easy. 380 Chapter Five. Similarity 1.3 Denition For any polynomial f(x) =cnxn+


+c1x+c0, wheretis a linear transformation then f(t) is the transformation cntn+


+c1t+c0(id) on the same space and where Tis a square matrix then f(T) is the matrix cnTn+


+c1T+c0I. 1.4 Remark If, for instance, f(x) =x3, then most authors write in the identity matrix: f(T) =T3I. But most authors don't write in the identity map:f(t) =t3. In this book we shall also observe this convention. Of course, if T= RepB;B(t) thenf(T) = RepB;B(f(t)), which follows from the relationships Tj= RepB;B(tj), andcT= RepB;B(ct), andT1+T2= RepB;B(t1+t2). As Example 1.2 shows, there may be polynomials of degree smaller than n2 that zero the map or matrix. 1.5 Denition The minimal polynomial m(x) of a transformation tor a square matrix Tis the polynomial of least degree and with leading coecient 1 such that m(t) is the zero map or m(T) is the zero matrix. A minimal polynomial always exists by the observation opening this subsec- tion. A minimal polynomial is unique by the `with leading coecient 1' clause. This is because if there are two polynomials m(x) and ^m(x) that are both of the minimal degree to make the map or matrix zero (and thus are of equal degree), and both have leading 1's, then their dierence m(x)^m(x) has a smaller de- gree than either and still sends the map or matrix to zero. Thus m(x)^m(x) is the zero polynomial and the two are equal. (The leading coecient requirement also prevents a minimal polynomial from being the zero polynomial.) 1.6 Example We can see that m(x) =x22x1 is minimal for the matrix of Example 1.2 by computing the powers of Tup to the power n2= 4. T2=1=2p 3=2p 3=2 1=2 T3=01 1 0 T4=1=2p 3=2p 3=21=2 Next, putc4T4+c3T3+c2T2+c1T+c0Iequal to the zero matrix (1=2)c4 + (1=2)c2+ (p 3=2)c1+c0= 0 (p 3=2)c4c3(p 3=2)c2(1=2)c1 = 0 (p 3=2)c4+c3+ (p 3=2)c2+ (1=2)c1 = 0 (1=2)c4 + (1=2)c2+ (p 3=2)c1+c0= 0 and use Gauss' method. c4c2p 3c12c0= 0 c3+p 3c2+ 2c1+p 3c0= 0 Settingc4,c3, andc2to zero forces c1andc0to also come out as zero. To get a leading one, the most we can do is to set c4andc3to zero. Thus the minimal polynomial is quadratic. Section IV. Jordan Form 381 Using the method of that example to nd the minimal polynomial of a 3 3 matrix would mean doing Gaussian reduction on a system with nine equations in ten unknowns. We shall develop an alternative. To begin, note that we can break a polynomial of a map or a matrix into its components. (For this lemma, recall that we are using complex numbers in this chapter so all polynomials break completely into linear factors.) 1.7 Lemma Suppose that the polynomial f(x) =cnxn+


+c1x+c0factors ask(x1)q1


(x`)q`. Iftis a linear transformation then these two are equal maps. cntn+


+c1t+c0=k
(t1)q1


 (t`)q` Consequently, if Tis a square matrix then f(T) andk
(T1I)q1


(T`I)q` are equal matrices. Proof .This argument is by induction on the degree of the polynomial. The cases where the polynomial is of degree 0 and 1 are clear. The full induction argument is Exercise 1.7 but the degree two case gives its sense. A quadratic polynomial factors into two linear terms f(x) =k(x1)
(x 2) =k(x2+ (1+2)x+12) (the roots 1and2might be equal). We can check that substituting tforxin the factored and unfactored versions gives the same map. k
(t1)(t2)
(~ v) = k
(t1)
(t(~ v)2~ v) =k
t(t(~ v))t(2~ v)1t(~ v)12~ v
=k
tt(~ v)(1+2)t(~ v) +12~ v
=k
(t2(1+2)t+12) (~ v) The third equality holds because the scalar 2comes out of the second term, as tis linear. QED In particular, if a minimial polynomial m(x) for a transformation tfactors asm(x) = (x1)q1


(x`)q`thenm(t) = (t1)q1


 (t`)q`is the zero map. Since m(t) sends every vector to zero, at least one of the maps tisends some nonzero vectors to zero. So, too, in the matrix case | if mis minimal for Tthenm(T) = (T1I)q1


(T`I)q`is the zero matrix and at least one of the matrices TiIsends some nonzero vectors to zero. Rewording both cases: at least some of the iare eigenvalues. (See Exercise 29.) Recall how we have earlier found eigenvalues. We have looked for such that T~ v=~ vby considering the equation ~0 =T~ vx~ v= (TxI)~ vand computing the determinant of the matrix TxI. That determinant is a polynomial in x, the characteristic polynomial, whose roots are the eigenvalues. The major result of this subsection, the next result, is that there is a connection between this characteristic polynomial and the minimal polynomial. This results expands on the prior paragraph's insight that some roots of the minimal polynomial are eigenvalues by asserting that every root of the minimal polynomial is an 382 Chapter Five. Similarity eigenvalue and further that every eigenvalue is a root of the minimal polynomial (this is because it says `1 qi' and not just `0qi'). 1.8 Theorem (Cayley-Hamilton) If the characteristic polynomial of a transformation or square matrix factors into k
(x1)p1(x2)p2


(x`)p` then its minimal polynomial factors into (x1)q1(x2)q2


(x`)q` where 1qipifor eachibetween 1 and `. The proof takes up the next three lemmas. Although they are stated only in matrix terms, they apply equally well to maps. We give the matrix version only because it is convenient for the rst proof. The rst result is the key | some authors call it the Cayley-Hamilton Theo- rem and call Theorem 1.8 above a corollary. For the proof, observe that a matrix of polynomials can be thought of as a polynomial with matrix coecients. 2x2+ 3x1x2+ 2 3x2+ 4x+ 1 4x2+x+ 1 =2 1 3 4 x2+3 0 4 1 x+1 2 1 1 1.9 Lemma IfTis a square matrix with characteristic polynomial c(x) then c(T) is the zero matrix. Proof .LetCbeTxI, the matrix whose determinant is the characteristic polynomial c(x) =cnxn+


+c1x+c0. C=0 BBB@t1;1x t 1;2::: t2;1t2;2x ...... tn;nx1 CCCA Recall that the product of the adjoint of a matrix with the matrix itself is the determinant of that matrix times the identity. c(x)
I= adj(C)C= adj(C)(TxI) = adj(C)Tadj(C)
x () The entries of adj( C) are polynomials, each of degree at most n1 since the minors of a matrix drop a row and column. Rewrite it, as suggested above, as adj(C) =Cn1xn1+


+C1x+C0where each Ciis a matrix of scalars. The left and right ends of equation ( ) above give this. cnIxn+cn1Ixn1+


+c1Ix+c0I= (Cn1T)xn1+


+ (C1T)x+C0T Cn1xnCn2xn1


C0x Section IV. Jordan Form 383 Equate the coecients of xn, the coecients of xn1, etc. cnI=Cn1 cn1I=Cn2+Cn1T ... c1I=C0+C1T c0I=C0T Multiply (from the right) both sides of the rst equation by Tn, both sides of the second equation by Tn1, etc. Add. The result on the left is cnTn+ cn1Tn1+


+c0I, and the result on the right is the zero matrix. QED We sometimes refer to that lemma by saying that a matrix or map satises its characteristic polynomial. 1.10 Lemma Wheref(x) is a polynomial, if f(T) is the zero matrix then f(x) is divisible by the minimal polynomial of T. That is, any polynomial satised byTis divisable by T's minimal polynomial. Proof .Letm(x) be minimal for T. The Division Theorem for Polynomials givesf(x) =q(x)m(x) +r(x) where the degree of ris strictly less than the degree ofm. Plugging Tin shows that r(T) is the zero matrix, because T satises both fandm. That contradicts the minimality of munlessris the zero polynomial. QED Combining the prior two lemmas gives that the minimal polynomial divides the characteristic polynomial. Thus, any root of the minimal polynomial is also a root of the characteristic polynomial. That is, so far we have that if m(x) = (x1)q1:::(xi)qithenc(x) must has the form ( x1)p1:::(x i)pi(xi+1)pi+1:::(x`)p`where each qjis less than or equal to pj. The proof of the Cayley-Hamilton Theorem is nished by showing that in fact the characteristic polynomial has no extra roots i+1, etc. 1.11 Lemma Each linear factor of the characteristic polynomial of a square matrix is also a linear factor of the minimal polynomial. Proof .LetTbe a square matrix with minimal polynomial m(x) and assume thatxis a factor of the characteristic polynomial of T, that is, assume that is an eigenvalue of T. We must show that xis a factor of m, that is, that m() = 0. In general, where is associated with the eigenvector ~ v, for any polyno- mial function f(x), application of the matrix f(T) to~ vequals the result of multiplying ~ vby the scalar f(). (For instance, if Thas eigenvalue associ- ated with the eigenvector ~ vandf(x) =x2+ 2x+ 3 then (T2+ 2T+ 3) (~ v) = T2(~ v) + 2T(~ v) + 3~ v=2
~ v+ 2
~ v+ 3
~ v= (2+ 2+ 3)
~ v.) Now, asm(T) is the zero matrix, ~0 =m(T)(~ v) =m()
~ vand therefore m() = 0. QED 384 Chapter Five. Similarity 1.12 Example We can use the Cayley-Hamilton Theorem to help nd the minimal polynomial of this matrix. T=0 BB@2 0 0 1 1 2 0 2 0 0 21 0 0 0 11 CCA First, its characteristic polynomial c(x) = (x1)(x2)3can be found with the usual determinant. Now, the Cayley-Hamilton Theorem says that T's minimal polynomial is either ( x1)(x2) or (x1)(x2)2or (x1)(x2)3. We can decide among the choices just by computing: (T1I)(T2I) =0 BB@1 0 0 1 1 1 0 2 0 0 11 0 0 0 01 CCA0 BB@0 0 0 1 1 0 0 2 0 0 01 0 0 011 CCA=0 BB@0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 01 CCA and (T1I)(T2I)2=0 BB@0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 01 CCA0 BB@0 0 0 1 1 0 0 2 0 0 01 0 0 011 CCA=0 BB@0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 CCA and som(x) = (x1)(x2)2. Exercises X1.13 What are the possible minimal polynomials if a matrix has the given charac- teristic polynomial? (a)8
(x3)4(b)(1=3)
(x+ 1)3(x4) (c)1
(x2)2(x5)2 (d)5
(x+ 3)2(x1)(x2)2 What is the degree of each possibility? X1.14 Find the minimal polynomial of each matrix. (a)0 @3 0 0 1 3 0 0 0 41 A (b)0 @3 0 0 1 3 0 0 0 31 A (c)0 @3 0 0 1 3 0 0 1 31 A (d)0 @2 0 1 0 6 2 0 0 21 A (e)0 @2 2 1 0 6 2 0 0 21 A (f)0 BBBB@1 4 0 0 0 0 3 0 0 0 041 0 0 394 21 1 5 4 1 41 CCCCA 1.15 Find the minimal polynomial of this matrix. 0 @0 1 0 0 0 1 1 0 01 A X1.16 What is the minimal polynomial of the dierentiation operator d=dx onPn? Section IV. Jordan Form 385 X1.17 Find the minimal polynomial of matrices of this form 0 BBBBBBB@0 0::: 0 10 0 0 1 ... 0 0 0::: 11 CCCCCCCA where the scalar is xed (i.e., is not a variable). 1.18 What is the minimal polynomial of the transformation of Pnthat sendsp(x) top(x+ 1)? 1.19 What is the minimal polynomial of the map :C3!C3projecting onto the rst two coordinates? 1.20 Find a 33 matrix whose minimal polynomial is x2. 1.21 What is wrong with this claimed proof of Lemma 1.9: \if c(x) =jTxIjthen c(T) =jTTIj= 0"? [Cullen] 1.22 Verify Lemma 1.9 for 2 2 matrices by direct calculation. X1.23 Prove that the minimal polynomial of an nnmatrix has degree at most n(notn2as might be guessed from this subsection's opening). Verify that this maximum, n, can happen. X1.24 The only eigenvalue of a nilpotent map is zero. Show that the converse state- ment holds. 1.25 What is the minimal polynomial of a zero map or matrix? Of an identity map or matrix? X1.26 Interpret the minimal polynomial of Example 1.2 geometrically. 1.27 What is the minimal polynomial of a diagonal matrix? X1.28 Aprojection is any transformation tsuch thatt2=t. (For instance, the transformation of the plane R2projecting each vector onto its rst coordinate will, if done twice, result in the same value as if it is done just once.) What is the minimal polynomial of a projection? 1.29 The rst two items of this question are review. (a)Prove that the composition of one-to-one maps is one-to-one. (b)Prove that if a linear map is not one-to-one then at least one nonzero vector from the domain is sent to the zero vector in the codomain. (c)Verify the statement, excerpted here, that preceeds Theorem 1.8. . . . if a minimial polynomial m(x) for a transformation tfactors as m(x) = (x1)q1


(x`)q`thenm(t) = (t1)q1


 (t`)q` is the zero map. Since m(t) sends every vector to zero, at least one of the maps tisends some nonzero vectors to zero. . . . Rewording . . . : at least some of the iare eigenvalues. 1.30 True or false: for a transformation on an ndimensional space, if the minimal polynomial has degree nthen the map is diagonalizable. 1.31 Letf(x) be a polynomial. Prove that if AandBare similar matrices then f(A) is similar to f(B). (a)Now show that similar matrices have the same characteristic polynomial. (b)Show that similar matrices have the same minimal polynomial. 386 Chapter Five. Similarity (c)Decide if these are similar. 1 3 2 3 41 1 1 1.32 (a) Show that a matrix is invertible if and only if the constant term in its minimal polynomial is not 0. (b)Show that if a square matrix Tis not invertible then there is a nonzero matrixSsuch thatSTandTSboth equal the zero matrix. X1.33 (a) Finish the proof of Lemma 1.7. (b)Give an example to show that the result does not hold if tis not linear. 1.34 Any transformation or square matrix has a minimal polynomial. Does the converse hold? IV.2 Jordan Canonical Form This subsection moves from the canonical form for nilpotent matrices to the one for all matrices. We have shown that if a map is nilpotent then all of its eigenvalues are zero. We can now prove the converse. 2.1 Lemma A linear transformation whose only eigenvalue is zero is nilpotent. Proof .If a transformation ton ann-dimensional space has only the single eigenvalue of zero then its characteristic polynomial is xn. The Cayley-Hamilton Theorem says that a map satises its characteristic polynimial so tnis the zero map. Thus tis nilpotent. QED We have a canonical form for nilpotent matrices, that is, for each matrix whose single eigenvalue is zero: each such matrix is similar to one that is all zeroes except for blocks of subdiagonal ones. (To make this representation unique we can x some arrangement of the blocks, say, from longest to shortest.) We next extend this to all single-eigenvalue matrices. Observe that if t's only eigenvalue is thent's only eigenvalue is 0 becauset(~ v) =~ vif and only if ( t) (~ v) = 0
~ v. The natural way to extend the results for nilpotent matrices is to represent tin the canonical form N, and try to use that to get a simple representation Tfort. The next result says that this try works. 2.2 Lemma If the matrices TIandNare similar then TandN+Iare also similar, via the same change of basis matrices. Proof .WithN=P(TI)P1=PTP1P(I)P1we haveN= PTP1PP1(I) since the diagonal matrix Icommutes with anything, and soN=PTP1I. Therefore N+I=PTP1, as required. QED Section IV. Jordan Form 387 2.3 Example The characteristic polynomial of T=21 1 4 is (x3)2and soThas only the single eigenvalue 3. Thus for T3I=11 1 1 the only eigenvalue is 0, and T3Iis nilpotent. The null spaces are routine to nd; to ease this computation we take Tto represent the transformation t:C2!C2with respect to the standard basis (we shall maintain this convention for the rest of the chapter). N(t3) =fy yy2CgN((t3)2) =C2 The dimensions of these null spaces show that the action of an associated map t3 on a string basis is ~17!~27!~0. Thus, the canonical form for t3 with one choice for a string basis is RepB;B(t3) =N=0 0 1 0 B=h1 1 ;2 2 i and by Lemma 2.2, Tis similar to this matrix. Rept(B;B) =N+ 3I=3 0 1 3 We can produce the similarity computation. Recall from the Nilpotence section how to nd the change of basis matrices PandP1to expressNas P(T3I)P1. The similarity diagram C2 w.r.t.E2t3! T3IC2 w.r.t.E2 id??yP id??yP C2 w.r.t.Bt3! NC2 w.r.t.B describes that to move from the lower left to the upper left we multiply by P1= RepE2;B(id)
1= RepB;E2(id) =12 1 2 and to move from the upper right to the lower right we multiply by this matrix. P= 12 1 21 = 1=2 1=2 1=4 1=4 388 Chapter Five. Similarity So the similarity is expressed by 3 0 1 3 =1=2 1=2 1=4 1=421 1 412 1 2 which is easily checked. 2.4 Example This matrix has characteristic polynomial ( x4)4 T=0 BB@4 1 01 0 3 0 1 0 0 4 0 1 0 0 51 CCA and so has the single eigenvalue 4. The nullities of t4 are: the null space of t4 has dimension two, the null space of ( t4)2has dimension three, and the null space of ( t4)3has dimension four. Thus, t4 has the action on a string basis of~17!~27!~37!~0 and~47!~0. This gives the canonical form Nfor t4, which in turn gives the form for t. N+ 4I=0 BB@4 0 0 0 1 4 0 0 0 1 4 0 0 0 0 41 CCA An array that is all zeroes, except for some number down the diagonal and blocks of subdiagonal ones, is a Jordan block . We have shown that Jordan block matrices are canonical representatives of the similarity classes of single- eigenvalue matrices. 2.5 Example The 33 matrices whose only eigenvalue is 1 =2 separate into three similarity classes. The three classes have these canonical representatives. 0 @1=2 0 0 0 1=2 0 0 0 1 =21 A0 @1=2 0 0 1 1=2 0 0 0 1 =21 A0 @1=2 0 0 1 1=2 0 0 1 1 =21 A In particular, this matrix0 @1=2 0 0 0 1=2 0 0 1 1 =21 A belongs to the similarity class represented by the middle one, because we have adopted the convention of ordering the blocks of subdiagonal ones from the longest block to the shortest. We will now nish the program of this chapter by extending this work to cover maps and matrices with multiple eigenvalues. The best possibility for general maps and matrices would be if we could break them into a part involving Section IV. Jordan Form 389 their rst eigenvalue 1(which we represent using its Jordan block), a part with 2, etc. This ideal is in fact what happens. For any transformation t:V!V, we shall break the space Vinto the direct sum of a part on which t1is nilpotent, plus a part on which t2is nilpotent, etc. More precisely, we shall take three steps to get to this section's major theorem and the third step shows that V=N1(t1)


 N1(t`) where1;:::;`aret's eigenvalues. Suppose that t:V!Vis a linear transformation. Note that the restriction oftto a subspace Mneed not be a linear transformation on Mbecause there may be an~ m2Mwitht(~ m)62M. To ensure that the restriction of a transformation to a `part' of a space is a transformation on the partwe need the next condition. 2.6 Denition Lett:V!Vbe a transformation. A subspace Mistin- variant if whenever ~ m2Mthent(~ m)2M(shorter:t(M)M). Two examples are that the generalized null space N1(t) and the generalized range space R1(t) of any transformation tare invariant. For the generalized null space, if~ v2N1(t) thentn(~ v) =~0 wherenis the dimension of the underlying space and so t(~ v)2N1(t) becausetn(t(~ v) ) is zero also. For the generalized range space, if ~ v2R1(t) then~ v=tn(~ w) for some~ wand thent(~ v) =tn+1(~ w) = tn(t(~ w) ) shows that t(~ v) is also a member of R1(t). Thus the spaces N1(ti) andR1(ti) aretiinvariant. Observe also thattiis nilpotent on N1(ti) because, simply, if ~ vhas the property that some power of timaps it to zero | that is, if it is in the generalized null space | then some power of timaps it to zero. The generalized null spaceN1(ti) is a `part' of the space on which the action of tiis easy to understand. The next result is the rst of our three steps. It establishes that tjleaves ti's part unchanged. 2.7 Lemma A subspace is tinvariant if and only if it is tinvariant for any scalar. In particular, where iis an eigenvalue of a linear transformation t, then for any other eigenvalue j, the spaces N1(ti) andR1(ti) are tjinvariant. Proof .For the rst sentence we check the two implications of the `if and only if' separately. One of them is easy: if the subspace is tinvariant for any  then taking = 0 shows that it is tinvariant. For the other implication suppose that the subspace is tinvariant, so that if ~ m2Mthent(~ m)2M, and let be any scalar. The subspace Mis closed under linear combinations and so if t(~ m)2Mthent(~ m)~ m2M. Thus if~ m2Mthen (t) (~ m)2M, as required. The second sentence follows straight from the rst. Because the two spaces aretiinvariant, they are therefore tinvariant. From this, applying the rst sentence again, we conclude that they are also tjinvariant. QED More information on restrictions of functions is in the appendix. 390 Chapter Five. Similarity The second step of the three that we will take to prove this section's major result makes use of an additional property of N1(ti) andR1(ti), that they are complementary. Recall that if a space is the direct sum of two others V=NRthen any vector ~ vin the space breaks into two parts ~ v=~ n+~ r where~ n2Nand~ r2R, and recall also that if BNandBRare bases for N andRthen the concatenation BN_BRis linearly independent (and so the two parts of~ vdo not \overlap"). The next result says that for any subspaces N andRthat are complementary as well as tinvariant, the action of ton~ vbreaks into the \non-overlapping" actions of ton~ nand on~ r. 2.8 Lemma Lett:V!Vbe a transformation and let NandRbetinvariant complementary subspaces of V. Thentcan be represented by a matrix with blocks of square submatrices T1andT2 T1Z2 Z1T2gdim(N)-many rows gdim(R)-many rows whereZ1andZ2are blocks of zeroes. Proof .Since the two subspaces are complementary, the concatenation of a basis forNand a basis for Rmakes a basis B=h~ 1;:::;~ p;~ 1;:::;~ qiforV. We shall show that the matrix RepB;B(t) =0 BB@...... RepB(t(~ 1))


RepB(t(~ q)) ......1 CCA has the desired form. Any vector ~ v2Vis inNif and only if its nal qcomponents are zeroes when it is represented with respect to B. AsNistinvariant, each of the vectors RepB(t(~ 1)), . . . , RepB(t(~ p)) has that form. Hence the lower left of RepB;B(t) is all zeroes. The argument for the upper right is similar. QED To see that thas been decomposed into its action on the parts, observe that the restrictions of tto the subspaces NandRare represented, with respect to the obvious bases, by the matrices T1andT2. So, with subspaces that are invariant and complementary, we can split the problem of examining a linear transformation into two lower-dimensional subproblems. The next result illustrates this decomposition into blocks. 2.9 Lemma IfTis a matrices with square submatrices T1andT2 T=T1Z2 Z1T2 where theZ's are blocks of zeroes, then jTj=jT1j
jT2j. Section IV. Jordan Form 391 Proof .Suppose that Tisnn, thatT1ispp, and that T2isqq. In the permutation formula for the determinant jTj=X permutations t1;(1)t2;(2)


tn;(n)sgn() each term comes from a rearrangement of the column numbers 1 ;:::;n into a new order(1);:::; (n). The upper right block Z2is all zeroes, so if a has at least one of p+ 1;:::;n among its rst pcolumn numbers (1);:::; (p) then the term arising from is zero, e.g., if (1) =nthent1;(1)t2;(2):::tn;(n)= 0
t2;(2):::tn;(n)= 0. So the above formula reduces to a sum over all permutations with two halves: any signicant is the composition of a 1that rearranges only 1 ;:::;p and a2that rearranges only p+ 1;:::;p +q. Now, the distributive law (and the fact that the signum of a composition is the product of the signums) gives that this jT1j
jT2j=X perms1 of 1;:::;pt1;1(1)


tp;1(p)sgn(1)
X perms2 ofp+1;:::;p+qtp+1;2(p+1)


tp+q;2(p+q)sgn(2) equalsjTj=P signicantt1;(1)t2;(2)


tn;(n)sgn(). QED 2.10 Example 2 0 0 0 1 2 0 0 0 0 3 0 0 0 0 3=2 0 1 2
3 0 0 3= 36 From Lemma 2.9 we conclude that if two subspaces are complementary and tinvariant then tis nonsingular if and only if its restrictions to both subspaces are nonsingular. Now for the promised third, nal, step to the main result. 2.11 Lemma If a linear transformation t:V!Vhas the characteristic poly- nomial (x1)p1:::(x`)p`then (1)V=N1(t1)


 N1(t`) and (2) dim( N1(ti)) =pi. Proof .Because dim( V) is the degree p1+


+p`of the characteristic poly- nomial, to establish statement (1) we need only show that statement (2) holds and that N1(ti)\N1(tj) is trivial whenever i6=j. For the latter, by Lemma 2.7, both N1(ti) andN1(tj) aretinvariant. Notice that an intersection of tinvariant subspaces is tinvariant and so the restriction of ttoN1(ti)\N1(tj) is a linear transformation. But both tiandtjare nilpotent on this subspace and so if thas any eigenvalues 392 Chapter Five. Similarity on the intersection then its \only" eigenvalue is both iandj. That cannot be, so this restriction has no eigenvalues: N1(ti)\N1(tj) is trivial (Lemma 3.10 shows that the only transformation without any eigenvalues is on the trivial space). To prove statement (2), x the index i. Decompose VasN1(ti) R1(ti) and apply Lemma 2.8. T= T1Z2 Z1T2 gdim(N1(ti) )-many rows gdim(R1(ti) )-many rows By Lemma 2.9,jTxIj=jT1xIj
jT2xIj. By the uniqueness clause of the Fundamental Theorem of Arithmetic, the determinants of the blocks have the same factors as the characteristic polynomial jT1xIj= (x1)q1:::(x`)q` andjT2xIj= (x1)r1:::(x`)r`, and the sum of the powers of these factors is the power of the factor in the characteristic polynomial: q1+r1=p1, . . . ,q`+r`=p`. Statement (2) will be proved if we will show that qi=piand thatqj= 0 for allj6=i, because then the degree of the polynomial jT1xIj| which equals the dimension of the generalized null space | is as required. For that, rst, as the restriction of titoN1(ti) is nilpotent on that space, the only eigenvalue of ton it isi. Thus the characteristic equation of t onN1(ti) isjT1xIj= (xi)qi. And thus qj= 0 for allj6=i. Now consider the restriction of ttoR1(ti). By Note II.2.2, the map tiis nonsingular on R1(ti) and soiis not an eigenvalue of ton that subspace. Therefore, xiis not a factor of jT2xIj, and soqi=pi.QED Our major result just translates those steps into matrix terms. 2.12 Theorem Any square matrix is similar to one in Jordan form 0 BBBBB@J1 {zeroes{ J2 ... J`1 {zeroes{ J`1 CCCCCA where each Jis the Jordan block associated with the eigenvalue of the original matrix (that is, is all zeroes except for 's down the diagonal and some subdiagonal ones). Proof .Given annnmatrixT, consider the linear map t:Cn!Cnthat it represents with respect to the standard bases. Use the prior lemma to write Cn=N1(t1)


 N1(t`) where1;:::;`are the eigenvalues of t. Because each N1(ti) istinvariant, Lemma 2.8 and the prior lemma show thattis represented by a matrix that is all zeroes except for square blocks along the diagonal. To make those blocks into Jordan blocks, pick each Bito be a string basis for the action of tionN1(ti). QED Section IV. Jordan Form 393 Jordan form is a canonical form for similarity classes of square matrices, provided that we make it unique by arranging the Jordan blocks from least eigenvalue to greatest and then arranging the subdiagonal 1 blocks inside each Jordan block from longest to shortest. 2.13 Example This matrix has the characteristic polynomial ( x2)2(x6). T=0 @2 0 1 0 6 2 0 0 21 A We will handle the eigenvalues 2 and 6 separately. Computation of the powers, and the null spaces and nullities, of T2Iis routine. (Recall from Example 2.3 the convention of taking Tto represent a transformation, here t:C3!C3, with respect to the standard basis.) powerp (T2I)pN((t2)p) nullity 10 B@0 0 1 0 4 2 0 0 01 CAf0 B@x 0 01 CAx2Cg 1 20 B@0 0 0 0 16 8 0 0 01 CAf0 B@x z=2 z1 CAx;z2Cg 2 30 B@0 0 0 0 64 32 0 0 01 CA {same{ | So the generalized null space N1(t2) has dimension two. We've noted that the restriction of t2 is nilpotent on this subspace. From the way that the nullities grow we know that the action of t2 on a string basis ~17!~27!~0. Thus the restriction can be represented in the canonical form N2=0 0 1 0 = RepB;B(t2)B2=h0 @1 1 21 A;0 @2 0 01 Ai where many choices of basis are possible. Consequently, the action of the re- striction of ttoN1(t2) is represented by this matrix. J2=N2+ 2I= RepB2;B2(t) = 2 0 1 2 The second eigenvalue's computations are easier. Because the power of x6 in the characteristic polynomial is one, the restriction of t6 toN1(t6) must be nilpotent of index one. Its action on a string basis must be ~37!~0 and since it is the zero map, its canonical form N6is the 11 zero matrix. Consequently, 394 Chapter Five. Similarity the canonical form J6for the action of tonN1(t6) is the 11 matrix with the single entry 6. For the basis we can use any nonzero vector from the generalized null space. B6=h0 @0 1 01 Ai Taken together, these two give that the Jordan form of Tis RepB;B(t) =0 @2 0 0 1 2 0 0 0 61 A whereBis the concatenation of B2andB6. 2.14 Example Contrast the prior example with T=0 @2 2 1 0 6 2 0 0 21 A which has the same characteristic polynomial ( x2)2(x6). While the characteristic polynomial is the same, powerp (T2I)pN((t2)p) nullity 10 B@0 2 1 0 4 2 0 0 01 CAf0 B@x z=2 z1 CAx;z2Cg 2 20 B@0 8 4 0 16 8 0 0 01 CA {same{ | here the action of t2 is stable after only one application | the restriction of of t2 toN1(t2) is nilpotent of index only one. (So the contrast with the prior example is that while the characteristic polynomial tells us to look at the action of thet2 on its generalized null space, the characteristic polynomial does not describe completely its action and we must do some computations to nd, in this example, that the minimal polynomial is ( x2)(x6).) The restriction of t2 to the generalized null space acts on a string basis as ~17!~0 and~27!~0, and we get this Jordan block associated with the eigenvalue 2. J2= 2 0 0 2 For the other eigenvalue, the arguments for the second eigenvalue of the prior example apply again. The restriction of t6 toN1(t6) is nilpotent of index one (it can't be of index less than one, and since x6 is a factor of Section IV. Jordan Form 395 the characteristic polynomial to the power one it can't be of index more than one either). Thus t6's canonical form N6is the 11 zero matrix, and the associated Jordan block J6is the 11 matrix with entry 6. Therefore,Tis diagonalizable. RepB;B(t) =0 @2 0 0 0 2 0 0 0 61 AB=B2_B6=h0 @1 0 01 A;0 @0 1 21 A;0 @3 4 01 Ai (Checking that the third vector in Bis in the nullspace of t6 is routine.) 2.15 Example A bit of computing with T=0 BBBB@1 4 0 0 0 0 3 0 0 0 041 0 0 394 21 1 5 4 1 41 CCCCA shows that its characteristic polynomial is ( x3)3(x+ 1)2. This table powerp (T3I)pN((t3)p) nullity 10 BBBBBB@4 4 0 0 0 0 0 0 0 0 044 0 0 39411 1 5 4 1 11 CCCCCCAf0 BBBBBB@(u+v)=2 (u+v)=2 (u+v)=2 u v1 CCCCCCAu;v2Cg2 20 BBBBBB@1616 0 0 0 0 0 0 0 0 0 16 16 0 0 16 32 16 0 0 01616 0 01 CCCCCCAf0 BBBBBB@z z z u v1 CCCCCCAz;u;v2Cg 3 30 BBBBBB@64 64 0 0 0 0 0 0 0 0 06464 0 0 6412864 0 0 0 64 64 0 01 CCCCCCA{same{ | shows that the restriction of t3 toN1(t3) acts on a string basis via the two strings ~17!~27!~0 and~37!~0. A similar calculation for the other eigenvalue 396 Chapter Five. Similarity powerp (T+ 1I)pN((t+ 1)p) nullity 10 BBBBBB@0 4 0 0 0 0 4 0 0 0 04 0 0 0 394 31 1 5 4 1 51 CCCCCCAf0 BBBBBB@(u+v) 0 v u v1 CCCCCCAu;v2Cg 2 20 BBBBBB@0 16 0 0 0 0 16 0 0 0 016 0 0 0 84016 88 8 24 16 8 241 CCCCCCA{same{ | shows that the restriction of t+ 1 to its generalized null space acts on a string basis via the two separate strings ~47!~0 and~57!~0. ThereforeTis similar to this Jordan form matrix. 0 BBBB@1 0 0 0 0 01 0 0 0 0 0 3 0 0 0 0 1 3 0 0 0 0 0 31 CCCCA We close with the statement that the subjects considered earlier in this Chapter are indeed, in this sense, exhaustive. 2.16 Corollary Every square matrix is similar to the sum of a diagonal matrix and a nilpotent matrix. Exercises 2.17 Do the check for Example 2.3. 2.18 Each matrix is in Jordan form. State its characteristic polynomial and its minimal polynomial. (a)3 0 1 3 (b)1 0 01 (c)0 @2 0 0 1 2 0 0 01=21 A (d)0 @3 0 0 1 3 0 0 1 31 A (e)0 BB@3 0 0 0 1 3 0 0 0 0 3 0 0 0 1 31 CCA(f)0 BB@4 0 0 0 1 4 0 0 0 04 0 0 0 141 CCA(g)0 @5 0 0 0 2 0 0 0 31 A (h)0 BB@5 0 0 0 0 2 0 0 0 0 2 0 0 0 0 31 CCA(i)0 BB@5 0 0 0 0 2 0 0 0 1 2 0 0 0 0 31 CCA X2.19 Find the Jordan form from the given data. (a)The matrix Tis 55 with the single eigenvalue 3. The nullities of the powers are:T3Ihas nullity two, ( T3I)2has nullity three, ( T3I)3has nullity four, and (T3I)4has nullity ve. Section IV. Jordan Form 397 (b)The matrix Sis 55 with two eigenvalues. For the eigenvalue 2 the nullities are:S2Ihas nullity two, and ( S2I)2has nullity four. For the eigenvalue 1 the nullities are: S+ 1Ihas nullity one. 2.20 Find the change of basis matrices for each example. (a)Example 2.13 (b)Example 2.14 (c)Example 2.15 X2.21 Find the Jordan form and a Jordan basis for each matrix. (a)10 4 25 10 (b)54 97 (c)0 @4 0 0 2 1 3 5 0 41 A (d)0 @5 4 3 1 03 12 11 A (e)0 @9 7 3 974 4 4 41 A (f)0 @2 21 11 1 12 21 A (g)0 BB@7 1 2 2 1 411 2 1 51 1 1 2 81 CCA X2.22 Find all possible Jordan forms of a transformation with characteristic poly- nomial (x1)2(x+ 2)2. 2.23 Find all possible Jordan forms of a transformation with characteristic poly- nomial (x1)3(x+ 2). X2.24 Find all possible Jordan forms of a transformation with characteristic poly- nomial (x2)3(x+ 1) and minimal polynomial ( x2)2(x+ 1). 2.25 Find all possible Jordan forms of a transformation with characteristic poly- nomial (x2)4(x+ 1) and minimal polynomial ( x2)2(x+ 1). X2.26 Diagonalize these. (a)1 1 0 0 (b)0 1 1 0 X2.27 Find the Jordan matrix representing the dierentiation operator on P3. X2.28 Decide if these two are similar.11 43 1 0 11 2.29 Find the Jordan form of this matrix.01 1 0 Also give a Jordan basis. 2.30 How many similarity classes are there for 3 3 matrices whose only eigenvalues are3 and 4? 398 Chapter Five. Similarity X2.31 Prove that a matrix is diagonalizable if and only if its minimal polynomial has only linear factors. 2.32 Give an example of a linear transformation on a vector space that has no non-trivial invariant subspaces. 2.33 Show that a subspace is t1invariant if and only if it is t2invariant. 2.34 Prove or disprove: two nnmatrices are similar if and only if they have the same characteristic and minimal polynomials. 2.35 The trace of a square matrix is the sum of its diagonal entries. (a)Find the formula for the characteristic polynomial of a 2 2 matrix. (b)Show that trace is invariant under similarity, and so we can sensibly speak of the `trace of a map'. ( Hint: see the prior item.) (c)Is trace invariant under matrix equivalence? (d)Show that the trace of a map is the sum of its eigenvalues (counting multi- plicities). (e)Show that the trace of a nilpotent map is zero. Does the converse hold? 2.36 To use Denition 2.6 to check whether a subspace is tinvariant, we seemingly have to check all of the innitely many vectors in a (nontrivial) subspace to see if they satisfy the condition. Prove that a subspace is tinvariant if and only if its subbasis has the property that for all of its elements, t(~) is in the subspace. X2.37 Istinvariance preserved under intersection? Under union? Complementation? Sums of subspaces? 2.38 Give a way to order the Jordan blocks if some of the eigenvalues are complex numbers. That is, suggest a reasonable ordering for the complex numbers. 2.39 LetPj(R) be the vector space over the reals of degree jpolynomials. Show that ifjkthenPj(R) is an invariant subspace of Pk(R) under the dierentiation operator. InP7(R), does any ofP0(R), . . . ,P6(R) have an invariant complement? 2.40 InPn(R), the vector space (over the reals) of degree npolynomials, E=fp(x)2Pn(R)p(x) =p(x) for allxg and O=fp(x)2Pn(R)p(x) =p(x) for allxg are the even and the oddpolynomials; p(x) =x2is even while p(x) =x3is odd. Show that they are subspaces. Are they complementary? Are they invariant under the dierentiation transformation? 2.41 Lemma 2.8 says that if MandNare invariant complements then thas a representation in the given block form (with respect to the same ending as starting basis, of course). Does the implication reverse? 2.42 A matrixSis the square root of anotherTifS2=T. Show that any nonsin- gular matrix has a square root. Topic: Method of Powers 399 Topic: Method of Powers In practice, calculating eigenvalues and eigenvectors is a dicult problem. Find- ing, and solving, the characteristic polynomial of the large matrices often en- countered in applications is too slow and too hard. Other techniques, indirect ones that avoid the characteristic polynomial, are used. Here we shall see such a method that is suitable for large matrices that are `sparse' (the great majority of the entries are zero). Suppose that the nnmatrixThas thendistinct eigenvalues 1,2, . . . ,n. ThenRnhas a basis that is composed of the associated eigenvectors h~1;:::;~ni. For any~ v2Rn, writing~ v=c1~1+


+cn~nand iterating Ton~ vgives these. T~ v=c11~1+c22~2+


+cnn~n T2~ v=c12 1~1+c22 2~2+


+cn2 n~n T3~ v=c13 1~1+c23 2~2+


+cn3 n~n ... Tk~ v=c1k 1~1+c2k 2~2+


+cnk n~n If one of the eigenvalues has a larger absolute value than any of the other eigenvalues then its term will dominate the above expression. Put another way, assuming that the absolute value of 1is the largest and dividing through Tk~ v k 1=c1~1+c2k 2 k 1~2+


+cnk n k 1~n shows that as kgets larger the fractions go to zero. Thus, the entire expression goes toc1~1. That is (as long as c1is not zero), as kincreases, the vectors Tk~ vwill tend toward the direction of the eigenvectors associated with the dominant eigenvalue, and, consequently, the ratios of the lengths kTk~ vk=kTk1~ vkwill tend toward that dominant eigenvalue. For example (sample computer code for this follows the exercises), because the matrix T=3 0 81 is triangular, its eigenvalues are just the entries on the diagonal, 3 and 1. Arbitrarily taking ~ vto have the components 1 and 1 gives ~ vT~ v T2~ v


T9~ v T10~ v1 13 7 9 17


19 683 39 367 59 049 118 097 and the ratio between the lengths of the last two is 2 :999 9. Two implementation issues must be addressed. The rst issue is that, instead of nding the powers of Tand applying them to ~ v, we will compute ~ v1asT~ vand then compute ~ v2asT~ v1, etc. (i.e., we never separately calculate T2,T3, etc.). 400 Chapter Five. Similarity These matrix-vector products can be done quickly even if Tis large, provided that it is sparse. The second issue is that, to avoid generating numbers that are so large that they over ow our computer's capability, we can normalize the ~ vi's at each step. For instance, we can divide each ~ viby its length (other possibilities are to divide it by its largest component, or simply by its rst component). We thus implement this method by generating ~ w0=~ v0=k~ v0k ~ v1=T~ w0 ~ w1=~ v1=k~ v1k ~ v2=T~ w2 ... ~ wk1=~ vk1=k~ vk1k ~ vk=T~ wk until we are satised. Then the vector ~ vkis an approximation of an eigenvector, and the approximation of the dominant eigenvalue is the ratio k~ vkk=k~ wk1k. One way we could be `satised' is to iterate until our approximation of the eigenvalue settles down. We could decide, for instance, to stop the iteration process not after some xed number of steps, but instead when k~ vkkdiers fromk~ vk1kby less than one percent, or when they agree up to the second signicant digit. The rate of convergence is determined by the rate at which the powers of k2=1kgo to zero, where 2is the eigenvalue of second largest norm. If that ratio is much less than one then convergence is fast, but if it is only slightly less than one then convergence can be quite slow. Consequently, the method of powers is not the most commonly used way of nding eigenvalues (although it is the simplest one, which is why it is here as the illustration of the possibility of computing eigenvalues without solving the characteristic polynomial). Instead, there are a variety of methods that generally work by rst replacing the given matrixTwith another that is similar to it and so has the same eigenvalues, but is in some reduced form such as tridiagonal form : the only nonzero entries are on the diagonal, or just above or below it. Then special techniques can be used to nd the eigenvalues. Once the eigenvalues are known, the eigenvectors of T can be easily computed. These other methods are outside of our scope. A good reference is [Goult, et al. ] Exercises 1Use ten iterations to estimate the largest eigenvalue of these matrices, starting from the vector with components 1 and 2. Compare the answer with the one obtained by solving the characteristic equation. (a)1 5 0 4 (b)3 2 1 0 2Redo the prior exercise by iterating until k~ vkkk~ vk1khas absolute value less than 0:01 At each step, normalize by dividing each vector by its length. How many iterations are required? Are the answers signicantly dierent? Topic: Method of Powers 401 3Use ten iterations to estimate the largest eigenvalue of these matrices, starting from the vector with components 1, 2, and 3. Compare the answer with the one obtained by solving the characteristic equation. (a)0 @4 0 1 2 1 0 2 0 11 A (b)0 @1 2 2 2 2 2 3661 A 4Redo the prior exercise by iterating until k~ vkkk~ vk1khas absolute value less than 0:01. At each step, normalize by dividing each vector by its length. How many iterations does it take? Are the answers signicantly dierent? 5What happens if c1= 0? That is, what happens if the initial vector does not to have any component in the direction of the relevant eigenvector? 6How can the method of powers be adopted to nd the smallest eigenvalue? Computer Code This is the code for the computer algebra system Octave that was used to do the calculation above. (It has been lightly edited to remove blank lines, etc.) >T=[3, 0; 8, -1] T= 3 0 8 -1 >v0=[1; 2] v0= 1 1 >v1=T*v0 v1= 3 7 >v2=T*v1 v2= 9 17 >T9=T**9 T9= 19683 0 39368 -1 >T10=T**10 T10= 59049 0 118096 1 >v9=T9*v0 v9= 19683 39367 >v10=T10*v0 v10= 59049 118096 402 Chapter Five. Similarity >norm(v10)/norm(v9) ans=2.9999 Remark: we are ignoring the power of Octave here; there are built-in func- tions to automatically apply quite sophisticated methods to nd eigenvalues and eigenvectors. Instead, we are using just the system as a calculator. Topic: Stable Populations 403 Topic: Stable Populations Imagine a reserve park with animals from a species that we are trying to protect. The park doesn't have a fence and so animals cross the boundary, both from the inside out and in the other direction. Every year, 10% of the animals from inside of the park leave, and 1% of the animals from the outside nd their way in. We can ask if we can nd a stable level of population for this park: is there a population that, once established, will stay constant over time, with the number of animals leaving equal to the number of animals entering? To answer that question, we must rst establish the equations. Let the year npopulation in the park be pnand in the rest of the world be rn. pn+1=:90pn+:01rn rn+1=:10pn+:99rn We can set this system up as a matrix equation (see the Markov Chain topic). pn+1 rn+1 =:90:01 :10:99pn rn Now, \stable level" means that pn+1=pnandrn+1=rn, so that the matrix equation~ vn+1=T~ vnbecomes~ v=T~ v. We are therefore looking for eigenvectors forTthat are associated with the eigenvalue 1. The equation ( IT)~ v=~0 is :10:01 :10:01p r =0 0 which gives the eigenspace: vectors with the restriction that p=:1r. Coupled with additional information, that the total world population of this species is is p+r= 110 000, we nd that the stable state is p= 10;000 andr= 100;000. If we start with a park population of ten thousand animals, so that the rest of the world has one hundred thousand, then every year ten percent (a thousand animals) of those inside will leave the park, and every year one percent (a thousand) of those from the rest of the world will enter the park. It is stable, self-sustaining. Now imagine that we are trying to gradually build up the total world pop- ulation of this species. We can try, for instance, to have the world population grow at a rate of 1% per year. In this case, we can take a \stable" state for the park's population to be that it also grows at 1% per year. The equation ~ vn+1= 1:01
~ vn=T~ vnleads to ((1 :01
I)T)~ v=~0, which gives this system. :11:01 :10:02p r =0 0 The matrix is nonsingular, and so the only solution is p= 0 andr= 0. Thus, there is no (usable) initial population that we can establish at the park and expect that it will grow at the same rate as the rest of the world. 404 Chapter Five. Similarity Knowing that an annual world population growth rate of 1% forces an un- stable park population, we can ask which growth rates there are that would allow an initial population for the park that will be self-sustaining. We consider ~ v=T~ vand solve for . 0 =:9:01 :10:99= (:9)(:99)(:10)(:01) =21:89+:89 A shortcut to factoring that quadratic is our knowledge that = 1 is an eigen- value ofT, so the other eigenvalue is :89. Thus there are two ways to have a stable park population (a population that grows at the same rate as the popu- lation of the rest of the world, despite the leaky park boundaries): have a world population that is does not grow or shrink, and have a world population that shrinks by 11% every year. So this is one meaning of eigenvalues and eigenvectors | they give a sta- ble state for a system. If the eigenvalue is 1 then the system is static. If the eigenvalue isn't 1 then the system is either growing or shrinking, but in a dynamically-stable way. Exercises 1What initial population for the park discussed above should be set up in the case where world populations are allowed to decline by 11% every year? 2What will happen to the population of the park in the event of a growth in world population of 1% per year? Will it lag the world growth, or lead it? Assume that the inital park population is ten thousand, and the world population is one hunderd thousand, and calculate over a ten year span. 3The park discussed above is partially fenced so that now, every year, only 5% of the animals from inside of the park leave (still, about 1% of the animals from the outside nd their way in). Under what conditions can the park maintain a stable population now? 4Suppose that a species of bird only lives in Canada, the United States, or in Mexico. Every year, 4% of the Canadian birds travel to the US, and 1% of them travel to Mexico. Every year, 6% of the US birds travel to Canada, and 4% go to Mexico. From Mexico, every year 10% travel to the US, and 0% go to Canada. (a)Give the transition matrix. (b)Is there a way for the three countries to have constant populations? (c)Find all stable situations. Topic: Linear Recurrences 405 Topic: Linear Recurrences In 1202 Leonardo of Pisa, also known as Fibonacci, posed this problem. A certain man put a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive? This moves past an elementary exponential growth model for population in- crease to include the fact that there is an initial period where newborns are not fertile. However, it retains other simplyng assumptions, such as that there is no gestation period and no mortality. To get the total number of pairs we will have next month, we add this month's total to the number of pairs that will be newborn next month. The number of pairs that will be productive next month, that will then be in their \second month on," is the number that were alive last month. f(n+ 1) =f(n) +f(n1) where f(0) = 0,f(1) = 1 The is an example of a recurrence relation , becausefrecurs in its own dening equation. From it, we can easily answer Fibonacci's twelve-month question. month 0 1 2 3 4 5 6 7 8 9 10 11 12 pairs 1 1 2 3 5 8 13 21 34 55 89 144 233 The sequence of numbers dened by the above equation (of which the rst few are listed) is the Fibonacci sequence . The material of this chapter can be used to give a formula with which we can can calculate f(n+ 1) without having to rst ndf(n),f(n1), etc. For that, observe that the recurrence is a linear relationship and so we can give a suitable matrix formulation of it.  1 1 1 0 f(n) f(n1) = f(n+ 1) f(n) where f(1) f(0) = 1 1 Then, where we write Tfor the matrix and ~ vnfor the vector with components f(n+1) andf(n), we have that ~ vn=Tn~ v0. The advantage of this matrix formu- lation is that by diagonalizing Twe get a fast way to compute its powers: where T=PDP1we haveTn=PDnP1, and then-th power of the diagonal matrixDis the diagonal matrix whose entries that are the n-th powers of the entries ofD. The characteristic equation of Tis21. The quadratic formula gives its roots as (1 +p 5)=2 and (1p 5)=2. Diagonalizing gives this. 1 1 1 0 =1+p 5 21p 5 2 1 1 1+p 5 20 01p 5 2! 1p 51p 5 2p 5 1p 51+p 5 2p 5! 406 Chapter Five. Similarity Introducing the vectors and taking the n-th power, we have f(n+ 1) f(n) =1 1 1 0nf(1) f(0) =1+p 5 21p 5 2 1 10 @ 1+p 5 2n 0 0 1p 5 2n1 A 1p 51p 5 2p 5 1p 51+p 5 2p 5!1 0 The calculation is ugly but not hard. f(n+ 1) f(n) =1+p 5 21p 5 2 1 10 @ 1+p 5 2n 0 0 1p 5 2n1 A 1p 5 1p 5! =1+p 5 21p 5 2 1 10 @1p 5 1+p 5 2n 1p 5 1p 5 2n1 A =0 @1p 5 1+p 5 2n+1 1p 5 1p 5 2n+1 1p 5 1+p 5 2n 1p 5 1p 5 2n1 A We want the second component of that equation. f(n) =1p 5" 1 +p 5 2!n 1p 5 2!n# Notice that (1p 5)=20:618 has absolute value less than one and so its powers go to zero. Thus the expression is dominated by its rst term. Although we have extended the elementary model of population growth by adding a delay period before the onset of fertility, we nonetheless still get an asmyptotically exponential function. In general, a linear recurrence relation has the form f(n+ 1) =anf(n) +an1f(n1) +


+ankf(nk) (it is also called a dierence equation ). This recurrence relation is homogeneous because there is no constant term; i.e, it can be put into the form 0 = f(n+1)+ anf(n)+an1f(n1)+


+ankf(nk). This is said to be a relation of order k. The relation, along with the initial conditions f(0), . . . ,f(k) completely determine a sequence. For instance, the Fibonacci relation is of order 2 and it, along with the two initial conditions f(0) = 1 and f(1) = 1, determines the Fibonacci sequence simply because we can compute any f(n) by rst computing f(2),f(3), etc. In this Topic, we shall see how linear algebra can be used to solve linear recurrence relations. First, we dene the vector space in which we are working. Let Vbe the set of functions ffrom the natural numbers N=f0;1;2;:::gto the real numbers. Topic: Linear Recurrences 407 (Below we shall have functions with domain f1;2;:::g, that is, without 0, but it is not an important distinction.) Putting the initial conditions aside for a moment, for any recurrence, we can consider the subset SofVof solutions. For example, without initial conditions, in addition to the function fgiven above, the Fibonacci relation is also solved by the function gwhose rst few values are g(0) = 1,g(1) = 2,g(2) = 3,g(3) = 4, andg(4) = 7. The subset Sis a subspace of V. It is nonempty because the zero function is a solution. It is closed under addition since if f1andf2are solutions, then an+1(f1+f2)(n+ 1) +


+ank(f1+f2)(nk) = (an+1f1(n+ 1) +


+ankf1(nk)) + (an+1f2(n+ 1) +


+ankf2(nk)) = 0: And, it is closed under scalar multiplication since an+1(rf1)(n+ 1) +


+ank(rf1)(nk) =r(an+1f1(n+ 1) +


+ankf1(nk)) =r
0 = 0: We can give the dimension of S. Consider this map from the set of functions S to the set of vectors Rk. f7!0 BBB@f(0) f(1) ... f(k)1 CCCA Exercise 3 shows that this map is linear. Because, as noted above, any solution of the recurrence is uniquely determined by the kinitial conditions, this map is one-to-one and onto. Thus it is an isomorphism, and thus Shas dimension k, the order of the recurrence. So (again, without any initial conditions), we can describe the set of solu- tions of any linear homogeneous recurrence relation of degree kby taking linear combinations of only klinearly independent functions. It remains to produce those functions. For that, we express the recurrence f(n+ 1) =anf(n) +


+ankf(nk) with a matrix equation. 0 BBBBBBB@anan1an2::: ank+1ank 1 0 0 ::: 0 0 0 1 0 0 0 1 ............ 0 0 0 ::: 1 01 CCCCCCCA0 BBB@f(n) f(n1) ... f(nk)1 CCCA=0 BBB@f(n+ 1) f(n) ... f(nk+ 1)1 CCCA 408 Chapter Five. Similarity In trying to nd the characteristic function of the matrix, we can see the pattern in the 22 casean an1 1 =2anan1 and 33 case. 0 @an an1an2 1 0 0 11 A=3+an2+an1+an2 Exercise 4 shows that the characteristic equation is this. an an1an2::: ank+1ank 1 0::: 0 0 0 1 0 0 1 ............ 0 0 0 ::: 1 =(k+ank1+an1k2+


+ank+1+ank) We call that the polynomial `associated' with the recurrence relation. (We will be nding the roots of this polynomial and so we can drop the as irrelevant.) Ifk+ank1+an1k2+


+ank+1+ankhas no repeated roots then the matrix is diagonalizable and we can, in theory, get a formula for f(n) as in the Fibonacci case. But, because we know that the subspace of solutions has dimension k, we do not need to do the diagonalization calculation, provided that we can exhibit klinearly independent functions satisfying the relation. Wherer1,r2, . . . ,rkare the distinct roots, consider the functions fr1(n) =rn 1 throughfrk(n) =rn kof powers of those roots. Exercise 5 shows that each is a solution of the recurrence and that the kof them form a linearly independent set. So, given the homogeneous linear recurrence f(n+ 1) =anf(n) +


+ ankf(nk) (that is, 0 =f(n+1)+anf(n)+


+ankf(nk)) we consider the associated equation 0 = k+ank1+


+ank+1+ank. We nd its rootsr1, . . . ,rk, and if those roots are distinct then any solution of the relation has the form f(n) =c1rn 1+c2rn 2+


+ckrn kforc1;:::;cn2R. (The case of repeated roots is also easily done, but we won't cover it here | see any text on Discrete Mathematics.) Now, given some initial conditions, so that we are interested in a particular solution, we can solve for c1, . . . ,cn. For instance, the polynomial associated with the Fibonacci relation is 2++ 1, whose roots are (1 p 5)=2 and so any solution of the Fibonacci equation has the form f(n) =c1((1 +p 5)=2)n+ c2((1p 5)=2)n. Including the initial conditions for the cases n= 0 andn= 1 gives c1+ c2= 1 (1 +p 5=2)c1+ (1p 5=2)c2= 1 Topic: Linear Recurrences 409 which yields c1= 1=p 5 andc2=1=p 5, as was calculated above. We close by considering the nonhomogeneous case, where the relation has the formf(n+1) =anf(n)+an1f(n1)+


+ankf(nk)+bfor some nonzero b. As in the rst chapter of this book, only a small adjustment is needed to make the transition from the homogeneous case. This classic example illustrates. In 1883, Edouard Lucas posed the following problem. In the great temple at Benares, beneath the dome which marks the center of the world, rests a brass plate in which are xed three diamond needles, each a cubit high and as thick as the body of a bee. On one of these needles, at the creation, God placed sixty four disks of pure gold, the largest disk resting on the brass plate, and the others getting smaller and smaller up to the top one. This is the Tower of Bramah. Day and night unceasingly the priests transfer the disks from one diamond needle to another according to the xed and immutable laws of Bramah, which require that the priest on duty must not move more than one disk at a time and that he must place this disk on a needle so that there is no smaller disk below it. When the sixty-four disks shall have been thus transferred from the needle on which at the creation God placed them to one of the other needles, tower, temple, and Brahmins alike will crumble into dusk, and with a thunderclap the world will vanish. (Translation of [De Parville] from [Ball & Coxeter].) How many disk moves will it take? Instead of tackling the sixty four disk problem right away, we will consider the problem for smaller numbers of disks, starting with three. To begin, all three disks are on the same needle. After moving the small disk to the far needle, the mid-sized disk to the middle needle, and then moving the small disk to the middle needle we have this. 410 Chapter Five. Similarity Now we can move the big disk over. Then, to nish, we repeat the process of moving the smaller disks, this time so that they end up on the third needle, on top of the big disk. So the thing to see is that to move the very largest disk, the bottom disk, at a minimum we must: rst move the smaller disks to the middle needle, then move the big one, and then move all the smaller ones from the middle needle to the ending needle. Those three steps give us this recurence. T(n+ 1) =T(n) + 1 +T(n) = 2T(n) + 1 where T(1) = 1 We can easily get the rst few values of T. n1 2 3 4 5 6 7 8 9 10 T(n)1 3 7 15 31 63 127 255 511 1023 We recognize those as being simply one less than a power of two. To derive this equation instead of just guessing at it, we write the original relation as1 =T(n+ 1) + 2T(n), consider the homogeneous relation 0 = T(n) + 2T(n1), get its associated polynomial + 2, which obviously has the single, unique, root of r1= 2, and conclude that functions satisfying the homogeneous relation take the form T(n) =c12n. That's the homogeneous solution. Now we need a particular solution. Because the nonhomogeneous relation 1 =T(n+ 1) + 2T(n) is so simple, in a few minutes (or by remembering the table) we can spot the particular solutionT(n) =1 (there are other particular solutions, but this one is easily spotted). So we have that | without yet considering the initial condition | any solution of T(n+ 1) = 2T(n) + 1 is the sum of the homogeneous solution and this particular solution: T(n) =c12n1. The initial condition T(1) = 1 now gives that c1= 1, and we've gotten the formula that generates the table: the n-disk Tower of Hanoi problem requires a minimum of 2n1 moves. Finding a particular solution in more complicated cases is, naturally, more complicated. A delightful and rewarding, but challenging, source on recur- rence relations is [Graham, Knuth, Patashnik]., For more on the Tower of Hanoi, [Ball & Coxeter] or [Gardner 1957] are good starting points. So is [Hofstadter]. Some computer code for trying some recurrence relations follows the exercises. Exercises 1Solve each homogeneous linear recurrence relations. (a)f(n+ 1) = 5f(n)6f(n1) (b)f(n+ 1) = 4f(n1) (c)f(n+ 1) = 5f(n)2f(n1)8f(n2) 2Give a formula for the relations of the prior exercise, with these initial condi- tions. (a)f(0) = 1,f(1) = 1 (b)f(0) = 0,f(1) = 1 (c)f(0) = 1,f(1) = 1,f(2) = 3. Topic: Linear Recurrences 411 3Check that the isomorphism given betwween SandRkis a linear map. It is argued above that this map is one-to-one. What is its inverse? 4Show that the characteristic equation of the matrix is as stated, that is, is the polynomial associated with the relation. (Hint: expanding down the nal column, and using induction will work.) 5Given a homogeneous linear recurrence relation f(n+ 1) =anf(n) +


+ ankf(nk), letr1, . . . ,rkbe the roots of the associated polynomial. (a)Prove that each function fri(n) =rn ksatises the recurrence (without initial conditions). (b)Prove that no riis 0. (c)Prove that the set ffr1;:::;frkgis linearly independent. 6(This refers to the value T(64) = 18;446;744;073;709;551;615 given in the com- puter code below.) Transferring one disk per second, how many years would it take the priests at the Tower of Hanoi to nish the job? Computer Code This code allows the generation of the rst few values of a function dened by a recurrence and initial conditions. It is in the Scheme dialect of LISP (specically, it was written for A. Jaer's free scheme interpreter SCM, although it should run in any Scheme implementation). First, the Tower of Hanoi code is a straightforward implementation of the recurrence. (define (tower-of-hanoi-moves n) (if (= n 1) 1 (+ (* (tower-of-hanoi-moves (- n 1)) 2) 1) ) ) (Note for readers unused to recursive code: to compute T(64), the computer is told to compute 2 T(63)1, which requires, of course, computing T(63). The computer puts the `times 2' and the `plus 1' aside for a moment to do that. It computesT(63) by using this same piece of code (that's what `recursive' means), and to do that is told to compute 2 T(62)1. This keeps up (the next step is to try to do T(62) while the other arithmetic is held in waiting), until, after 63 steps, the computer tries to compute T(1). It then returns T(1) = 1, which now means that the computation of T(2) can proceed, etc., up until the original computation of T(64) nishes.) The next routine calculates a table of the rst few values. (Some language notes: '()is the empty list, that is, the empty sequence, and cons pushes something onto the start of a list. Note that, in the last line, the procedure proc is called on argument n.) (define (first-few-outputs proc n) (first-few-outputs-helper proc n '()) ) ; (define (first-few-outputs-aux proc n lst) (if (< n 1) 412 Chapter Five. Similarity lst (first-few-outputs-aux proc (- n 1) (cons (proc n) lst)) ) ) The session at the SCM prompt went like this. >(first-few-outputs tower-of-hanoi-moves 64) Evaluation took 120 mSec (1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 65535 131071 262143 524287 1048575 2097151 4194303 8388607 16777215 33554431 67108863 134217727 268435455 536870911 1073741823 2147483647 4294967295 8589934591 17179869183 34359738367 68719476735 137438953471 274877906943 549755813887 1099511627775 2199023255551 4398046511103 8796093022207 17592186044415 35184372088831 70368744177663 140737488355327 281474976710655 562949953421311 1125899906842623 2251799813685247 4503599627370495 9007199254740991 18014398509481983 36028797018963967 72057594037927935 144115188075855871 288230376151711743 576460752303423487 1152921504606846975 2305843009213693951 4611686018427387903 9223372036854775807 18446744073709551615) This is a list of T(1) through T(64). (The 120 mSec came on a 50 mHz '486 running in an XTerm of XWindow under Linux. The session was edited to put line breaks between numbers.) Appendix Mathematics is made of arguments (reasoned discourse that is, not crockery- throwing). This section is a reference to the most used techniques. A reader having trouble with, say, proof by contradiction, can turn here for an outline of that method. But this section gives only a sketch. For more, these are classics: Methods of Logic by Quine, Induction and Analogy in Mathematics by P olya, and Naive Set Theory by Halmos. IV.3 Propositions The point at issue in an argument is the proposition . Mathematicians usually write the point in full before the proof and label it either Theorem for major points, Corollary for points that follow immediately from a prior one, or Lemma for results chie y used to prove other results. The statements expressing propositions can be complex, with many subparts. The truth or falsity of the entire proposition depends both on the truth value of the parts, and on the words used to assemble the statement from its parts. Not. For example, where Pis a proposition, `it is not the case that P' is true provided that Pis false. Thus, ` nis not prime' is true only when nis the product of smaller integers. We can picture the `not' operation with a Venn diagram . P Where the box encloses all natural numbers, and inside the circle are the primes, the shaded area holds numbers satisfying `not P'. To prove that a `not P' statement holds, show that Pis false. A-1 A-2 And. Consider the statement form ` PandQ'. For the statement to be true both halves must hold: `7 is prime and so is 3' is true, while `7 is prime and 3 is not' is false. Here is the Venn diagram for ` PandQ'. PQ To prove `PandQ', prove that each half holds. Or. A `PorQ' is true when either half holds: `7 is prime or 4 is prime' is true, while `7 is not prime or 4 is prime' is false. We take `or' inclusively so that if both halves are true `7 is prime or 4 is not' then the statement as a whole is true. (In everyday speech, sometimes `or' is meant in an exclusive way | \Eat your vegetables or no dessert" does not intend both halves to hold | but we will not use `or' in that way.) The Venn diagram for `or' includes all of both circles. PQ To prove `PorQ', show that in all cases at least one half holds (perhaps sometimes one half and sometimes the other, but always at least one). If-then. An `ifPthenQ' statement (sometimes written ` Pmaterially implies Q' or just `PimpliesQ' or `P=)Q') is true unless Pis true while Qis false. Thus `if 7 is prime then 4 is not' is true while `if 7 is prime then 4 is also prime' is false. (Contrary to its use in casual speech, in mathematics `if PthenQ' does not connote that PprecedesQor causesQ.) More subtly, in mathematics `if PthenQ' is true when Pis false: `if 4 is prime then 7 is prime' and `if 4 is prime then 7 is not' are both true statements, sometimes said to be vacuously true . We adopt this convention because we want statements like `if a number is a perfect square then it is not prime' to be true, for instance when the number is 5 or when the number is 6. The diagram PQ A-3 shows that Qholds whenever Pdoes (another phrasing is ` Pis sucient to give Q'). Notice again that if Pdoes not hold, Qmay or may not be in force. There are two main ways to establish an implication. The rst way is direct: assume that Pis true and, using that assumption, prove Q. For instance, to show `if a number is divisible by 5 then twice that number is divisible by 10', assume that the number is 5 nand deduce that 2(5 n) = 10n. The second way is indirect: prove the contrapositive statement: `if Qis false then Pis false' (rephrased, ` Qcan only be false when Pis also false'). As an example, to show `if a number is prime then it is not a perfect square', argue that if it were a squarep=n2then it could be factored p=n
nwheren<p and so wouldn't be prime (of course p= 0 orp= 1 don't give n<p but they are nonprime by denition). Note two things about this statement form. First, an `if PthenQ' result can sometimes be improved by weakening P or strengthening Q. Thus, `if a number is divisible by p2then its square is also divisible by p2' could be upgraded either by relaxing its hypothesis: `if a number is divisible by pthen its square is divisible by p2', or by tightening its conclusion: `if a number is divisible by p2then its square is divisible by p4'. Second, after showing `if PthenQ', a good next step is to look into whether there are cases where Qholds butPdoes not. The idea is to better under- stand the relationship between PandQ, with an eye toward strengthening the proposition. Equivalence. An if-then statement cannot be improved when not only does PimplyQ, but alsoQimpliesP. Some ways to say this are: ` Pif and only if Q', `PiQ', `PandQare logically equivalent', ` Pis necessary and sucient to giveQ', `P()Q'. For example, `a number is divisible by a prime if and only if that number squared is divisible by the prime squared'. The picture here shows that PandQhold in exactly the same cases. PQ Although in simple arguments a chain like \ Pif and only if R, which holds if and only if S. . . " may be practical, typically we show equivalence by showing the `ifPthenQ' and `ifQthenP' halves separately. IV.4 Quantiers Compare these two statements about natural numbers: `there is an xsuch thatxis divisible by x2' is true, while `for all numbers x, thatxis divisible by x2' is false. We call the `there is' and `for all' prexes quantiers . A-4 For all. The `for all' prex is the universal quantier , symbolized8. Venn diagrams aren't very helpful with quantiers, but in a sense the box we draw to border the diagram shows the universal quantier since it dilineates the universe of possible members. To prove that a statement holds in all cases, we must show that it holds in each case. Thus, to prove `every number divisible by phas its square divisible byp2', take a single number of the form pnand square it ( pn)2=p2n2. This is a \typical element" or \generic element" proof. This kind of argument requires that we are careful to not assume properties for that element other than those in the hypothesis | for instance, this type of wrong argument is a common mistake: \if nis divisible by a prime, say 2, so thatn= 2kthenn2= (2k)2= 4k2and the square of the number is divisible by the square of the prime". That is an argument about the case p= 2, but it isn't a proof for general p. There exists. We will also use the existential quantier , symbolized9and read `there exists'. As noted above, Venn diagrams are not much help with quantiers, but a picture of `there is a number such that P' would show both that there can be more than one and that not all numbers need satisfy P. P An existence proposition can be proved by producing something satisfying the property: once, to settle the question of primality of 225+1, Euler produced its divisor 641. But there are proofs showing that something exists without say- ing how to nd it; Euclid's argument given in the next subsection shows there are innitely many primes without naming them. In general, while demon- strating existence is better than nothing, giving an example is better, and an exhaustive list of all instances is great. Still, mathematicians take what they can get. Finally, along with \Are there any?" we often ask \How many?" That is why the issue of uniqueness often arises in conjunction with questions of existence. Many times the two arguments are simpler if separated, so note that just as proving something exists does not show it is unique, neither does proving something is unique show that it exists. (Obviously `the natural number with A-5 more factors than any other' would be unique, but in fact no such number exists.) IV.5 Techniques of Proof Induction. Many proofs are iterative, \Here's why the statement is true for for the case of the number 1, it then follows for 2, and from there to 3, and so on . . . ". These are called proofs by induction . Such a proof has two steps. In thebase step the proposition is established for some rst number, often 0 or 1. Then in the inductive step we assume that the proposition holds for numbers up to some kand deduce that it then holds for the next number k+ 1. Here is an example. We will prove that 1 + 2 + 3 +


+n=n(n+ 1)=2. For the base step we must show that the formula holds when n= 1. That's easy, the sum of the rst 1 number does indeed equal 1(1 + 1) =2. For the inductive step, assume that the formula holds for the numbers 1;2;:::;k . That is, assume all of these instances of the formula. 1 = 1(1 + 1) =2 and 1 + 2 = 2(2 + 1) =2 and 1 + 2 + 3 = 3(3 + 1) =2 ... and 1 +


+k=k(k+ 1)=2 From this assumption we will deduce that the formula therefore also holds in thek+ 1 next case. The deduction is straightforward algebra. 1 + 2 +


+k+ (k+ 1) =k(k+ 1) 2+ (k+ 1) =(k+ 1)(k+ 2) 2 We've shown in the base case that the above proposition holds for 1. We've shown in the inductive step that if it holds for the case of 1 then it also holds for 2; therefore it does hold for 2. We've also shown in the inductive step that if the statement holds for the cases of 1 and 2 then it also holds for the next case 3, etc. Thus it holds for any natural number greater than or equal to 1. Here is another example. We will prove that every integer greater than 1 is a product of primes. The base step is easy: 2 is the product of a single prime. For the inductive step assume that each of 2 ;3;:::;k is a product of primes, aiming to show k+ 1 is also a product of primes. There are two A-6 possibilities: (i) if k+ 1 is not divisible by a number smaller than itself then it is a prime and so is the product of primes, and (ii) if k+ 1 is divisible then its factors can be written as a product of primes (by the inductive hypothesis) and so k+1 can be rewritten as a product of primes. That ends the proof. (Remark. The Prime Factorization Theorem of Number Theory says that not only does a factorization exist, but that it is unique. We've shown the easy half.) There are two things to note about the `next number' in an induction argu- ment. For one thing, while induction works on the integers, it's no good on the reals. There is no `next' real. The other thing is that we sometimes use induction to go down, say, from 10 to 9 to 8, etc., down to 0. So `next number' could mean `next lowest number'. Of course, at the end we have not shown the fact for all natural numbers, only for those less than or equal to 10. Contradiction. Another technique of proof is to show something is true by showing it can't be false. The classic example is Euclid's, that there are innitely many primes. Suppose there are only nitely many primes p1;:::;pk. Consider p1
p2:::pk+1. None of the primes on this supposedly exhaustive list divides that number evenly, each leaves a remainder of 1. But every number is a product of primes so this can't be. Thus there cannot be only nitely many primes. Every proof by contradiction has the same form: assume that the proposition is false and derive some contradiction to known facts. Another example is this proof thatp 2 is not a rational number. Suppose thatp 2 =m=n. 2n2=m2 Factor out any 2's: n= 2kn
^nandm= 2km
^mand rewrite. 2
(2kn
^n)2= (2km
^m)2 The Prime Factorization Theorem says that there must be the same num- ber of factors of 2 on both sides, but there are an odd number 1 + 2 knon the left and an even number 2 kmon the right. That's a contradiction, so a rational with a square of 2 cannot be. Both of these examples aimed to prove something doesn't exist. A negative proposition often suggests a proof by contradiction. A-7 IV.6 Sets, Functions, and Relations Sets. Mathematicians work with collections, called sets. A set can be given as a listing between curly braces as in f1;4;9;16g, or, if that's unwieldy, by using set-builder notation as in fxx53x3+ 2 = 0g(read \the set of all x such that . . . "). We name sets with capital roman letters as with the primes P=f2;3;5;7;11;:::g, except for a few special sets such as the real numbers R, and the complex numbers C. To denote that something is an element (or member ) of a set we use ` 2', so that 72f3;5;7gwhile 862f3;5;7g. What distinguishes a set from any other type of collection is the Principle of Extensionality, that two sets with the same elements are equal. Because of this principle, in a set repeats collapse f7;7g=f7gand order doesn't matter f2;g=f;2g. We use `' for the proper subset relationship: Ais a subset of B, so that any element ofAis an element of B, butA6=B. An example isf2;gf 2;;7g. We use `' if eitherABor two sets are equal. These symbols may be ipped, for instancef2;;5gf 2;5g. Because of Extensionality, to prove that two sets are equal A=B, just show that they have the same members. Usually we show mutual inclusion, that both ABandAB. Set operations. Venn diagrams are handy here. For instance, x2Pcan be pictured P x and `PQ' looks like this. PQ Note that this is a repeat of the diagram for `if . . . then . . . ' propositions. That's because `PQ' means `ifx2Pthenx2Q'. In general, for every propositional logic operator there is an associated set operator. For instance, the complement ofPisPcomp=fxnot(x2P)g A-8 P theunion isP[Q=fx(x2P) or (x2Q)g PQ and the intersection isP\Q=fx(x2P) and (x2Q)g: PQ When two sets share no members their intersection is the empty setfg, symbolized ?. Any set has the empty set for a subset, by the `vacuously true' property of the denition of implication. Sequences. We shall also use collections where order does matter and where repeats do not collapse. These are sequences , denoted with angle brackets: h2;3;7i6=h2;7;3i. A sequence of length 2 is sometimes called an ordered pair and written with parentheses: ( ;3). We also sometimes say `ordered triple', `ordered 4-tuple', etc. The set of ordered n-tuples of elements of a set Ais denotedAn. Thus the set of pairs of reals is R2. Functions. We rst see functions in elementary Algebra, where they are pre- sented as formulas (e.g., f(x) = 16x2100), but progressing to more advanced Mathematics reveals more general functions | trigonometric ones, exponential and logarithmic ones, and even constructs like absolute value that involve piec- ing together parts | and we see that functions aren't formulas, instead the key idea is that a function associates with its input xa single output f(x). Consequently, a function ormap is dened to be a set of ordered pairs (x;f(x) ) such that xsuces to determine f(x), that is: if x1=x2thenf(x1) = f(x2) (this requirement is referred to by saying a function is well-dened ). Each input xis one of the function's arguments and each output f(x) is a value . The set of all arguments is f'sdomain and the set of output values is itsrange . Usually we don't need know what is and is not in the range and we instead work with a superset of the range, the codomain . The notation for a functionfwith domain Xand codomain Yisf:X!Y. More on this is in the section on isomorphisms A-9 We sometimes instead use the notation xf7!16x2100, read `xmaps under fto 16x2100', or `16 x2100 is the image ofx'. Some maps, like x7!sin(1=x), can be thought of as combinations of simple maps, here, g(y) = sin(y) applied to the image of f(x) = 1=x. The composition ofg:Y!Zwithf:X!Y, is the map sending x2Xtog(f(x) )2Z. It is denotedgf:X!Z. This denition only makes sense if the range of fis a subset of the domain of g. Observe that the identity map id:Y!Ydened by id( y) =yhas the property that for any f:X!Y, the composition id fis equal to f. So an identity map plays the same role with respect to function composition that the number 0 plays in real number addition, or that the number 1 plays in multiplication. In line with that analogy, dene a left inverse of a mapf:X!Yto be a functiong: range(f)!Xsuch thatgfis the identity map on X. Of course, aright inverse offis ah:Y!Xsuch thatfhis the identity. A map that is both a left and right inverse of fis called simply an inverse . An inverse, if one exists, is unique because if both g1andg2are inverses of f theng1(x) =g1(fg2)(x) = (g1f)g2(x) =g2(x) (the middle equality comes from the associativity of function composition), so we often call it \the" inverse, written f1. For instance, the inverse of the function f:R!Rgiven byf(x) = 2x3 is the function f1:R!Rgiven byf1(x) = (x+ 3)=2. The superscript ` f1' notation for function inverse can be confusing | it doesn't mean 1 =f(x). It is used because it ts into a larger scheme. Func- tions that have the same codomain as domain can be iterated, so that where f:X!X, we can consider the composition of fwith itself: ff, andfff, etc. Naturally enough, we write ffasf2andfffasf3, etc. Note that the familiar exponent rules for real numbers obviously hold: fifj=fi+j and (fi)j=fi
j. The relationship with the prior paragraph is that, where fis invertible, writing f1for the inverse and f2for the inverse of f2, etc., gives that these familiar exponent rules continue to hold, once f0is dened to be the identity map. If the codomain Yequals the range of fthen we say that the function is onto. A function has a right inverse if and only if it is onto (this is not hard to check). If no two arguments share an image, if x16=x2implies that f(x1)6=f(x2), then the function is one-to-one . A function has a left inverse if and only if it is one-to-one (this is also not hard to check). By the prior paragraph, a map has an inverse if and only if it is both onto and one-to-one; such a function is a correspondence . It associates one and only one element of the domain with each element of the range (for example, nite A-10 sets must have the same number of elements to be matched up in this way). Because a composition of one-to-one maps is one-to-one, and a composition of onto maps is onto, a composition of correspondences is a correspondence. We sometimes want to shrink the domain of a function. For instance, we may take the function f:R!Rgiven byf(x) =x2and, in order to have an inverse, limit input arguments to nonnegative reals ^f:R+!R. Technically, ^fis a dierent function than f; we call it the restriction offto the smaller domain. A nal point on functions: neither xnorf(x) need be a number. As an example, we can think of f(x;y) =x+yas a function that takes the ordered pair (x;y) as its argument. Relations. Some familiar operations are obviously functions: addition maps (5;3) to 8. But what of ` <' or `='? We here take the approach of rephrasing `3<5' to `(3;5) is in the relation <'. That is, dene a binary relation on a set Ato be a set of ordered pairs of elements of A. For example, the <relation is the setf(a;b)a<bg; some elements of that set are (3 ;5), (3;7), and (1;100). Another binary relation on the natural numbers is equality; this relation is formally written as the set f:::;(1;1);(0;0);(1;1);:::g. Still another example is `closer than 10', the set f(x;y)jxyj<10g. Some members of that relation are (1 ;10), (10;1), and (42 ;44). Neither (11 ;1) nor (1;11) is a member. Those examples illustrate the generality of the denition. All kinds of re- lationships (e.g., `both numbers even' or `rst number is the second with the digits reversed') are covered under the denition. Equivalence Relations. We shall need to say, formally, that two objects are alike in some way. While these alike things aren't identical, they are related (e.g., two integers that `give the same remainder when divided by 2'). A binary relation f(a;b);:::gis an equivalence relation when it satises (1)re exivity : any object is related to itself; (2)symmetry : ifais related to bthenbis related to a; (3)transitivity : ifais related to bandbis related to cthenais related to c. (To see that these conditions formalize being the same, read them again, replac- ing `is related to' with `is like'.) Some examples (on the integers): `=' is an equivalence relation, ` <' does not satisfy symmetry, `same sign' is a equivalence, while `nearer than 10' fails transitivity. Partitions. In `same sign'f(1;3);(5;7);(1;1);:::gthere are two kinds of pairs, the rst with both numbers positive and the second with both negative. So integers fall into exactly one of two classes, positive or negative. Apartition of a setSis a collection of subsets fS1;S2;:::gsuch that every element ofSis in one and only one Si:S1[S2[:::=S, and ifiis not equal tojthenSi\Sj=?. PictureSbeing decomposed into distinct parts. A-11 . . .S0S1S2 S3 Thus, the rst paragraph says `same sign' partitions the integers into the pos- itives and the negatives. Similarly, the equivalence relation `=' partitions the integers into one-element sets. Another example is the fractions. Of course, 2 =3 and 4=6 are equivalent fractions. That is, for the set S=fn=dn;d2Zandd6= 0g, we dene two elementsn1=d1andn2=d2to be equivalent if n1d2=n2d1. We can check that this is an equivalence relation, that is, that it satises the above three conditions. With that, Sis divided up into parts. . . .:0=1 :0=3:1=1 :2=2:2=4:2=4 :4=3 :8=6 Before we show that equivalence relations always give rise to partitions, we rst illustrate the argument. Consider the relationship between two in- tegers of `same parity', the set f(1;3);(2;4);(0;0);:::g(i.e., `give the same remainder when divided by 2'). We want to say that the natural numbers split into two pieces, the evens and the odds, and inside a piece each mem- ber has the same parity as each other. So for each xwe dene the set of numbers associated with it: Sx=fy(x;y)2`same parity'g. Some exam- ples areS1=f:::;3;1;1;3;:::g, andS4=f:::;2;0;2;4;:::g, andS1= f:::;3;1;1;3;:::g. These are the parts, e.g., S1is the odds. Theorem. An equivalence relation induces a partition on the underlying set. Proof .Call the set Sand the relation R. In line with the illustration in the paragraph above, for each x2SdeneSx=fy(x;y)2Rg. Observe that, as xis a member if Sx, the union of all these sets is S. So we will be done if we show that distinct parts are disjoint: if Sx6=Sythen Sx\Sy=?. We will verify this through the contrapositive, that is, we wlll assume that Sx\Sy6=?in order to deduce that Sx=Sy. Letpbe an element of the intersection. Then by denition of SxandSy, the two (x;p) and (y;p) are members of R, and by symmetry of this relation ( p;x) and (p;y) are also members of R. To show that Sx=Sywe will show each is a subset of the other. Assume that q2Sxso that (q;x)2R. Use transitivity along with ( x;p)2R to conclude that ( q;p) is also an element of R. But (p;y)2Rso another use of transitivity gives that ( q;y)2R. Thusq2Sy. Therefore q2Sximplies q2Sy, and soSxSy. The same argument in the other direction gives the other inclusion, and so the two sets are equal, completing the contrapositive argument. QED A-12 We call each part of a partition an equivalence class (or informally, `part'). We somtimes pick a single element of each equivalence class to be the class representative . . . .??? ? Usually when we pick representatives we have some natural scheme in mind. In that case we call them the canonical representatives. An example is the simplest form of a fraction. We've dened 3 =5 and 9=15 to be equivalent fractions. In everyday work we often use the `simplest form' or `reduced form' fraction as the class representatives. . . .?0=1?1=1?1=2 ?4=3 Bibliography [Abbot] Edwin Abbott, Flatland , Dover. [Ackerson] R. H. Ackerson, A Note on Vector Spaces , American Mathematical Monthly, volume 62 number 10 (Dec. 1955), p. 721. [Agnew] Jeanne Agnew, Explorations in Number Theory , Brooks/Cole, 1972. [Am. Math. Mon., Jun. 1931] C. A. Rupp (proposer), H. T. R. Aude (solver), problem 3468, American Mathematical Monthly, volume 37 number 6 (June-July 1931), p. 355. [Am. Math. Mon., Feb. 1933] V. F. Ivano (proposer), T. C. Esty (solver), problem 3529, American Mathematical Monthly, volume 39 number 2 (Feb. 1933), p. 118. [Am. Math. Mon., Jan. 1935] W. R. Ransom (proposer), Hansraj Gupta (solver), El- ementary problem 105, American Mathematical Monthly, volume 42 number 1 (Jan. 1935), p. 47. [Am. Math. Mon., Jan. 1949] C. W. Trigg (proposer), R. J. Walker (solver), Elemen- tary problem 813, American Mathematical Monthly, volume 56 number 1 (Jan. 1949), p. 33. [Am. Math. Mon., Jun. 1949] Don Walter (proposer), Alex Tytun (solver), Elemen- tary problem 834, American Mathematical Monthly, volume 56 number 6 (June-July 1949), p. 409. [Am. Math. Mon., Nov. 1951] Albert Wilansky, The Row-Sums of the Inverse Matrix , American Mathematical Monthly, volume 58 number 9 (Nov. 1951), p. 614. [Am. Math. Mon., Feb. 1953] Norman Anning (proposer), C. W. Trigg (solver), Ele- mentary problem 1016, American Mathematical Monthly, volume 60 number 2 (Feb. 1953), p. 115. [Am. Math. Mon., Apr. 1955] Vern Haggett (proposer), F. W. Saunders (solver), El- ementary problem 1135, American Mathematical Monthly, volume 62 number 4 (Apr. 1955), p. 257. [Am. Math. Mon., Jan. 1963] Underwood Dudley, Arnold Lebow (proposers), David Rothman (solver), Elemantary problem 1151, American Mathematical Monthly, volume 70 number 1 (Jan. 1963), p. 93. [Am. Math. Mon., Dec. 1966] Hans Liebeck, A Proof of the Equality of Column Rank and Row Rank of a Matrix American Mathematical Monthly, volume 73 num- ber 10 (Dec. 1966), p. 1114. [Anton] Howard Anton, Elementary Linear Algebra , John Wiley & Sons, 1987. [Arrow] Kenneth J. Arrow, Social Choice and Individual Values , Wiley, 1963. [Ball & Coxeter] W.W. Rouse Ball, Mathematical Recreations and Essays , revised by H.S.M. Coxeter, MacMillan, 1962. [Bennett] William J. Bennett, Quantifying America's Decline , inWall Street Journal , 15 March, 1993. [Birkho & MacLane] Garrett Birkho, Saunders MacLane, Survey of Modern Alge- bra, third edition, Macmillan, 1965. [Blass 1984] A. Blass, Existence of Bases Implies the Axiom of Choice , pp. 31 { 33, Ax- iomatic Set Theory, J. E. Baumgartner, ed., American Mathematical Society, Providence RI, 1984. [Bridgman] P.W. Bridgman, Dimensional Analysis , Yale University Press, 1931. [Casey] John Casey, The Elements of Euclid, Books I to VI and XI , ninth edition, Hodges, Figgis, and Co., Dublin, 1890. [Clark & Coupe] David H. Clark, John D. Coupe, The Bangor Area Economy Its Present and Future , report to the city of Bangor ME, Mar. 1967. [Clarke] Arthur C. Clarke, Technical Error , Fantasy, December 1946, reprinted in Great SF Stories 8 (1946), DAW Books, 1982. [Con. Prob. 1955] The Contest Problem Book , 1955 number 38. [Coxeter] H.S.M. Coxeter, Projective Geometry , second edition, Springer-Verlag, 1974. [Courant & Robbins] Richard Courant, Herbert Robbins, What is Mathematics? , Ox- ford University Press, 1978. [Cullen] Charles G. Cullen, Matrices and Linear Transformations , second edition, Dover, 1990. [Dalal, et. al.] Siddhartha R. Dalal, Edward B. Fowlkes, & Bruce Hoadley, Lesson Learned from Challenger: A Statistical Perspective , Stats: the Magazine for Students of Statistics, Fall 1989, p. 3. [Davies] Thomas D. Davies, New Evidence Places Peary at the Pole , National Geo- graphic Magazine, vol. 177 no. 1 (Jan. 1990), p. 44. [de Mestre] Neville de Mestre, The Mathematics of Projectiles in Sport , Cambridge University Press, 1990. [De Parville] De Parville, La Nature , Paris, 1884, part I, p. 285 { 286 (citation from [Ball & Coxeter]). [Duncan] W.J. Duncan, Method of Dimensions , in Encyclopaedic Dictionary of Physics Macmillan, 1962. [Ebbing] Darrell D. Ebbing, General Chemistry , fourth edition, Houghton Miin, 1993. [Ebbinghaus] H. D. Ebbinghaus, Numbers , Springer-Verlag, 1990. [Einstein] A. Einstein, Annals of Physics, v. 35, 1911, p. 686. [Eggar] M.H. Eggar, Pinhole Cameras, Perspecitve, and Projective Geometry , Amer- ican Mathematicsl Monthly, August-September 1998, p. 618 { 630. [Feller] William Feller, An Introduction to Probability Theory and Its Applications (vol. 1, 3rd ed.), John Wiley, 1968. [Finkbeiner] Daniel T. Finkbeiner III, Introduction to Matrices and Linear Transfor- mations , third edition, W. H. Freeman and Company, 1978. [Fraleigh & Beauregard] Fraleigh & Beauregard, Linear Algebra . [Gardner] Martin Gardner, The New Ambidextrous Universe , third revised edition, W. H. Freeman and Company, 1990. [Gardner 1957] Martin Gardner, Mathematical Games: About the remarkable similar- ity between the Icosian Game and the Tower of Hanoi , Scientic American, May 1957, p. 150 { 154. [Gardner, 1970] Martin Gardner, Mathematical Games, Some mathematical curiosi- ties embedded in the solar system , Scientic American, April 1970, p. 108 { 112. [Gardner, 1980] Martin Gardner, Mathematical Games, From counting votes to mak- ing votes count: the mathematics of elections , Scientic American, October 1980. [Gardner, 1974] Martin Gardner, Mathematical Games, On the paradoxical situations that arise from nontransitive relations , Scientic American, October 1974. [Giordano, Wells, Wilde] Frank R. Giordano, Michael E. Wells, Carroll O. Wilde, Di- mensional Analysis , UMAP Unit 526, in UMAP Modules, 1987 , COMAP, 1987. [Giordano, Jaye, Weir] Frank R. Giordano, Michael J. Jaye, Maurice D. Weir, The Use of Dimensional Analysis in Mathematical Modeling , UMAP Unit 632, in UMAP Modules, 1986 , COMAP, 1986. [Glazer] A.M. Glazer, The Structure of Crystals , Adam Hilger, 1987. [Goult, et al. ] R.J. Goult, R.F. Hoskins, J.A. Milner, M.J. Pratt, Computational Methods in Linear Algebra , Wiley, 1975. [Graham, Knuth, Patashnik] Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics , Addison-Wesley, 1988. [Halmos] Paul P. Halmos, Finite Dimensional Vector Spaces , second edition, Van Nos- trand, 1958. [Hamming] Richard W. Hamming, Introduction to Applied Numerical Analysis , Hemi- sphere Publishing, 1971. [Hanes] Kit Hanes, Analytic Projective Geometry and its Applications , UMAP Unit 710, UMAP Modules, 1990, p. 111. [Heath] T. Heath, Euclid's Elements , volume 1, Dover, 1956. [Homan & Kunze] Kenneth Homan, Ray Kunze, Linear Algebra , second edition, Prentice-Hall, 1971. [Hofstadter] Douglas R. Hofstadter, Metamagical Themas: Questing for the Essence of Mind and Pattern , Basic Books, 1985. [Iosifescu] Marius Iofescu, Finite Markov Processes and Their Applications , John Wi- ley, 1980. [Kelton] Christina M.L. Kelton, Trends on the Relocation of U.S. Manufacturing , UMI Research Press, 1983. [Kemeny & Snell] John G. Kemeny, J. Laurie Snell, Finite Markov Chains , D. Van Nostrand, 1960. [Kemp] Franklin Kemp Linear Equations , American Mathematical Monthly, volume 89 number 8 (Oct. 1982), p. 608. [Knuth] Donald E. Knuth, The Art of Computer Programming , Addison Wesley, 1988. [Lang] Serge Lang, Linear Algebra , Addison-Wesley, 1966. [Leontief 1951] Wassily W. Leontief, Input-Output Economics , Scientic American, volume 185 number 4 (Oct. 1951), p. 15. [Leontief 1965] Wassily W. Leontief, The Structure of the U.S. Economy , Scientic American, volume 212 number 4 (Apr. 1965), p. 25. [Lieberman] David Lieberman, The Big Bucks Ad Battles Over TV's Most Expensive Minutes , TV Guide, Jan. 26 1991, p. 11. [Macdonald & Ridge] Kenneth Macdonald, John Ridge, Social Mobility , in British Social Trends Since 1900 , A.H. Halsey, Macmillian, 1988. [Macmillan Dictionary] William D. Halsey, Macmillan, 1979. [Math. Mag., Sept. 1952] Dewey Duncan (proposer), W. H. Quelch (solver), Mathe- matics Magazine, volume 26 number 1 (Sept-Oct. 1952), p. 48. [Math. Mag., Jan. 1957] M. S. Klamkin (proposer), Trickie T-27, Mathematics Mag- azine, volume 30 number 3 (Jan-Feb. 1957), p. 173. [Math. Mag., Jan. 1963, Q237] D. L. Silverman (proposer), C. W. Trigg (solver), Quickie 237, Mathematics Magazine, volume 36 number 1 (Jan. 1963). [Math. Mag., Jan. 1963, Q307] C. W. Trigg (proposer). Quickie 307, Mathematics Magazine, volume 36 number 1 (Jan. 1963), p. 77. [Math. Mag., Nov. 1967] Clarence C. Morrison (proposer), Quickie, Mathematics Magazine, volume 40 number 4 (Nov. 1967), p. 232. [Math. Mag., Jan. 1973] Marvin Bittinger (proposer), Quickie 578, Mathematics Magazine, volume 46 number 5 (Jan. 1973), p. 286, 296. [Munkres] James R. Munkres, Elementary Linear Algebra , Addison-Wesley, 1964. [Neimi & Riker] Richard G. Neimi, William H. Riker, The Choice of Voting Systems , Scientic American, June 1976, p. 21 { 27. [Nering] Evar D. Nering, Linear Algebra and Matrix Theory , second edition, John Wiley, 1970. [Niven & Zuckerman] I. Niven, H. Zuckerman, An Introduction to the Theory of Num- bers, third edition, John Wiley, 1972. [Oakley & Baker] Cletus O. Oakley, Justine C. Baker, Least Squares and the 3 : 40 Mile, Mathematics Teacher, Apr. 1977. [Ohanian] Hans O'Hanian, Physics , volume one, W. W. Norton, 1985. [Onan] Onan, Linear Algebra . [Petersen] G. M. Petersen, Area of a Triangle , American Mathematical Monthly, vol- ume 62 number 4 (Apr. 1955), p. 249. [Polya] G. Polya, Patterns of Plausible Inference , [Poundstone] W. Poundstone, Gaming the Vote , Hill and Wang, 2008. ISBN-13: 978- 0-8090-4893-9 [Programmer's Ref.] Microsoft Programmers Reference Microsoft Press. [Putnam, 1988, A-6] William Lowell Putnam Mathematical Competition, Problem A- 6, 1988, solution: American Mathematical Monthly, volume 96 number 8 (Oct. 1989), p. 688. [Putnam, 1990, A-5] William Lowell Putnam Mathematical Competition, Problem A- 5, 1990. [Rice] John R. Rice, Numerical Mathods, Software, and Analysis , second edition, Academic Press, 1993. [Quine] W. V. Quine, Methods of Logic , fourth edition, Harvard University Press, 1982. [Rucker] Rudy Rucker, Innity and the Mind , Birkhauser, 1982. [Ryan] Patrick J. Ryan, Euclidean and Non-Euclidean Geometry: an Analytic Ap- proach , Cambridge University Press, 1986. [Seidenberg] A. Seidenberg, Lectures in Projective Geometry , Van Nostrand, 1962. [Smith] Larry Smith, Linear Algebra , second edition, Springer-Verlag, 1984. [Strang 93] Gilbert Strang The Fundamental Theorem of Linear Algebra , American Mathematical Monthly, Nov. 1993, p. 848 { 855. [Strang 80] Gilbert Strang, Linear Algebra and its Applications , second edition, Har- court Brace Jovanovich, 1980. [Taylor] Alan D. Taylor, Mathematics and Politics: Strategy, Voting, Power, and Proof , Springer-Verlag, 1995. [Tilley] Burt Tilley, private communication, 1996. [Trono] Tony Trono, compiler, University of Vermont Mathematics Department High School Prize Examinations 1958-1991 , mimeographed printing, 1991. [USSR Olympiad no. 174] The USSR Mathematics Olympiad , number 174. [Weston] J. D. Weston, Volume in Vector Spaces , American Mathematical Monthly, volume 66 number 7 (Aug./Sept. 1959), p. 575 { 577. [Weyl] Hermann Weyl, Symmetry , Princeton University Press, 1952. [Wickens] Thomas D. Wickens, Models for Behavior , W.H. Freeman, 1982. [Wilkinson 1965] Wilkinson, 1965 [Wikipedia] The Square-cube law ,http://en.wikipedia.org/wiki/Square-cube_ law, 2011-Jan-17. [WikipediaMensMile] Mile run world record progression ,http://en.wikipedia.org/ wiki/Mile_run_world_record_progression , 2011-Apr-09. [Wohascum no. 2] The Wohascum County Problem Book problem number 2. [Wohascum no. 47] The Wohascum County Problem Book problem number 47. [Yaglom] I.M. Yaglom, Felix Klein and Sophus Lie: Evolution of the Idea of Symmetry in the Nineteenth Century , translated by Sergei Sossinsky, Birkh auser, 1988. [Zwicker] William S. Zwicker, The Voters' Paradox, Spin, and the Borda Count , Math- ematical Social Sciences, vol. 22 (1991), p. 187 { 227. Index accuracy of Gauss' method, 67{70 rounding error, 68 adding rows, 4 addition vector, 78 additive inverse, 78 adjoint matrix, 326 angle, 42 antipodal, 338 antisymmetric matrix, 136 argument, A-8 arrow diagram, 215, 231, 236, 240, 351 augmented matrix, 14 automorphism, 161 dilation, 161 re ection, 161 rotation, 161 back-substitution, 5 base step of induction, A-5 basis, 110{122 change of, 236 denition, 110 orthogonal, 254 orthogonalization, 255 orthonormal, 256 standard, 111, 350 standard over the complex numbers, 350 string, 371 best t line, 267 block matrix, 309 box, 319 orientation, 319 sense, 319 volume, 319 C language, 67canonical form for matrix equivalence, 243 for nilpotent matrices, 374 for row equivalence, 58 for similarity, 392 canonical representative, A-12 Cauchy-Schwartz Inequality, 42 Cayley-Hamilton theorem, 382 central projection, 335 change of basis, 236{247 characteristic vectors, values, 357 characteristic equation, 360 characteristic polynomial, 360 characterized, 170 characterizes, 244 Chemistry problem, 1, 10 chemistry problem, 22 circuits parallel, 72 series, 72 series-parallel, 73 class equivalence, A-12 closure, 93 of nullspace, 367 of rangespace, 367 codomain, A-8 cofactor, 325 column, 14 rank, 123 vector, 15 column rank full, 128 column space, 123 combining rows, 4 complement, A-7 complementary subspaces, 133 orthogonal, 261 complex numbers vector space over, 88 component, 15 composition, A-9 self, 365 computer algebra systems, 61{62 concatenation, 131 conditioning number, 70 congruent gures, 285 congruent plane gures, 285 contradiction, A-6 contrapositive, A-3 convex set, 181 coordinates homogeneous, 338 with respect to a basis, 113 corollary, A-1 correspondence, 159, A-9 coset, 191 Cramer's rule, 329{331 cross product, 296 crystals, 140{143 diamond, 141 graphite, 141 salt, 140 unit cell, 141 da Vinci, Leonardo, 335 determinant, 292, 297{316 cofactor, 325 Cramer's rule, 330 denition, 297 exists, 307, 312 Laplace expansion, 325 minor, 325 permutation expansion, 306, 310, 332 diagonal matrix, 208, 223 diagonalizable, 354{357 dierence equation, 406 homogeneous, 406 dilation, 161, 274 representing, 201 dimension, 118 physical, 150 direct map, 288 direct sum, 129 ), 137 denition, 133 external, 166 internal, 166 of two subspaces, 133direction vector, 35 distance-preserving, 285 division theorem, 348 domain, A-8 dot product, 40 double precision, 68 dual space, 192 echelon form, 6 free variable, 12 leading variable, 6 reduced, 47 eigenspace, 360 eigenvalue, eigenvector of a matrix, 358 of a transformation, 357 element, A-7 elementary matrix, 225, 273 elementary reduction operations, 4 rescaling, 4 row combination, 4 swapping, 4 elementary row operations, 4 elimination, 3 empty set, A-8 entry, 14 equivalence class, A-12 canonical representative, A-12 relation, A-10 representative, A-12 equivalence relation, A-10, A-11 isomorphism, 167 matrix equivalence, 242 matrix similarity, 351 row equivalence, 50 equivalent statements, A-3 Erlanger Program, 285 Euclid, 285 even functions, 96, 135 even polynomials, 398 external direct sum, 166 Fibonacci sequence, 405 eld, 138{139 denition, 138 nite-dimensional vector space, 117 at, 37 form, 56 free variable, 12 full column rank, 128 full row rank, 128 function, A-8 inverse image, 183 argument, A-8 codomain, A-8 composition, 214, A-9 correspondence, A-9 domain, A-8 even, 96 identity, A-9 inverse, 230, A-9 left inverse, 229 multilinear, 302 odd, 97 one-to-one, A-9 onto, A-9 range, A-8 restriction, A-10 right inverse, 229 structure preserving, 159, 163 seehomomorphism, 174 two-sided inverse, 230 value, A-8 well-dened, A-8 zero, 175 Fundamental Theorem of Linear Algebra, 266 Gauss' method, 3 accuracy, 67{70 back-substitution, 5 elementary operations, 4 Gauss-Jordan, 47 Gauss-Jordan, 47 Gaussian elimination, 3 generalized nullspace, 367 generalized rangespace, 367 Geometry of Linear Maps, 272{278 Gram-Schmidt process, 253{258 historyless Markov chain, 279 homogeneous coordinate vector, 338 homogeneous coordinates, 290 homogeneous equation, 21 homomorphism, 174 composition, 214 matrix representing, 193{203nonsingular, 188, 206 nullity, 186 nullspace, 186 rangespace, 182 rank, 205 zero, 175 hyperplane, 37 ideal line, 341 ideal point, 341 identity function, A-9 matrix, 223 if-then statement, A-2 ill-conditioned, 68 image under a function, A-9 improper subspace, 90 incidence matrix, 227 index of nilpotency, 370 induction, 23, A-5 inductive step of induction, A-5 inherited operations, 80 inner product, 40 Input-Output Analysis, 63{66 internal direct sum, 133, 166 intersection, A-8 invariant subspace, 377 invariant subspace denition, 389 inverse, 230, A-9 additive, 78 exists, 230 left, 230, A-9 matrix, 327 right, 230, A-9 two-sided, A-9 inverse function, 230 inverse image, 183 inversion, 311 isometry, 285 isomorphism, 157{173 characterized by dimension, 170 denition, 159 of a space with itself, 161 Jordan block, 388 Jordan form, 379{398 represents similarity classes, 392 kernel, 186 Kirchho's Laws, 72 Klein, F., 285 Laplace expansion, 324{328 computes determinant, 325 leading variable, 6 least squares, 267{271 lemma, A-1 length, 39 Leontief, W., 63 line best t, 267 in projective plane, 339 line at innity, 341 line of best t, 267{271 linear transpose operation, 128 linear combination, 2 Linear Combination Lemma, 52 linear elimination, 3 linear equation, 2 coecients, 2 constant, 2 homogeneous, 21 inconsistent systems, 267 satised by a vector, 15 solution of, 2 Gauss' method, 4 Gauss-Jordan, 47 solutions of Cramer's rule, 330 system of, 2 linear map dilation, 274 re ection, 288 rotation, 272, 288 seehomomorphism, 174 skew, 275 trace, 398 linear recurrence, 406 linear recurrences, 405{412 linear relationship, 100 linear surface, 37 linear transformation seetransformation, 177 linearly dependent, 100linearly independent, 100 LINPACK, 61 map, A-8 distance-preserving, 285 extended linearly, 171 self composition, 365 Maple, 61 Markov chain, 279 historyless, 279 Markov chains, 279{284 Markov matrix, 283 material implication, A-2 Mathematica, 61 mathematical induction, 23, A-5 MATLAB, 61 matrix, 14 adjoint, 326 antisymmetric, 136 augmented, 14 block, 243, 309 change of basis, 236 characteristic polynomial, 360 cofactor, 325 column, 14 column space, 123 conditioning number, 70 determinant, 292, 297 diagonal, 208, 223 diagonalizable, 354 diagonalized, 242 elementary reduction, 225, 273 entry, 14 equivalent, 242 identity, 219, 223 incidence, 227 inverse, 327 main diagonal, 223 Markov, 228, 283 matrix-vector product, 196 minimal polynomial, 219, 380 minor, 325 multiplication, 214 nilpotent, 370 nonsingular, 27, 206 orthogonal, 287 orthonormal, 285{290 permutation, 224 rank, 205 representation, 195 row, 14 row equivalence, 50 row rank, 122 row space, 122 scalar multiple, 211 scalar multiplication, 16 similar, 322 similarity, 351 singular, 27 skew-symmetric, 308 submatrix, 301 sum, 16, 211 symmetric, 116, 136, 212, 219, 227, 266 trace, 212, 228, 398 transition, 279 transpose, 19, 124, 212 triangular, 202, 228, 328 unit, 221 Vandermonde, 309 zero, 211 matrix equivalence, 240{247 canonical form, 243 denition, 242 matrix:form, 56 mean arithmetic, 44 geometric, 44 member, A-7 method of powers, 399{402 minimal polynomial, 219, 380 minor, 325 morphism, 159 multilinear, 302 multiplication matrix-matrix, 214 matrix-vector, 196 mutual inclusion, A-7 natural representative, A-12 networks, 71{76 Kirchho's Laws, 72 nilpotent, 368{378 canonical form for, 374 denition, 370 matrix, 370 transformation, 370 nilpotentcy index, 370 nonsingular, 206, 230homomorphism, 188 matrix, 27 normal, 261 normalize, 256 nullity, 186 nullspace, 186 closure of, 367 generalized, 367 odd functions, 97, 135 one-to-one function, A-9 onto function, A-9 opposite map, 288 order of a recurrence, 406 ordered pair, A-8 orientation, 319, 322 orthogonal, 42 basis, 254 complement, 261 mutually, 253 projection, 261 orthogonal matrix, 287 orthogonalization, 255 orthonormal basis, 256 orthonormal matrix, 285{290 pair ordered, A-8 parallelepiped, 319 parallelogram rule, 34 parameter, 13 partial pivoting, 69 partition, A-10{A-12 matrix equivalence classes, 242, 244 row equivalence classes, 51 partitions into isomorphism classes, 168 permutation, 305 inversions, 311 matrix, 224 signum, 312 permutation expansion, 306, 310, 332 perp, 261 perpendicular, 42 perspective triangles, 341 Physics problem, 1 pivoting, 5 full, 69 partial scaled, 69 plane gure, 285 congruence, 285 point at innity, 341 in projective plane, 338 polynomial even, 398 minimal, 380 of map, matrix, 379 polynomials division theorem, 348 populations, stable, 403{404 potential, 71 powers, method of, 399{402 preserves structure, 174 probability vector, 279 projection, 174, 183, 248, 266, 385 along a subspace, 258 central, 335 vanishing point, 335 into a line, 249 into a subspace, 258 orthogonal, 249, 261 Projective Geometry, 335{345 projective geometry Duality Principle, 340 projective plane ideal line, 341 ideal point, 341 lines, 339 proof techniques induction, 23 proper subset, A-7 proper subspace, 90 proposition, A-1 propositions equivalent, A-3 quantier, A-3 existential, A-4 universal, A-4 quantiers, A-3 range, A-8 rangespace, 182 closure of, 367 generalized, 367rank, 126, 205 column, 123 of a homomorphism, 182, 187 recurrence, 325, 406 homogeneous, 406 initial conditions, 406 reduced echelon form, 47 re ection, 288 glide, 288 re ection (or ip) about a line, 161 re exivity, A-10 relation, A-10 equivalence, A-10 re exive, A-10 symmetric, A-10 transitive, A-10 relationship linear, 100 representation of a matrix, 195 of a vector, 113 representative, A-12 canonical, A-12 for row equivalence classes, 58 of matrix equivalence classes, 243 of similarity classes, 393 rescaling rows, 4 resistance, 71 resistance:equivalent, 75 resistor, 71 restriction, A-10 rigid motion, 285 rotation, 272, 288 rotation (or turning), 161 represented, 198 row, 14 rank, 122 vector, 15 row equivalence, 50 row rank full, 128 row space, 122 Sage, 61 scalar, 78 scalar multiple matrix, 211 vector, 16, 34, 78 scalar multiplication matrix, 16 scalar product, 40 scaled partial pivoting, 69 Schwartz Inequality, 42 self composition of maps, 365 sense, 319 sequence, A-8 concatenation, 131 set, A-7 complement, A-7 element, A-7 empty, A-8 intersection, A-8 member, A-7 union, A-8 sets, A-7 dependent, independent, 100 empty, 102 mutual inclusion, A-7 proper subset, A-7 span of, 93 subset, A-7 sgn seesignum, 312 signum, 312 similar, 296, 322 canonical form, 392 similar matrices, 351 similar triangles, 288 similarity, 351{364 similarity transformation, 364 single precision, 67 singular matrix, 27 size, 317, 319 skew, 275 skew-symmetric, 308 span, 93 of a singleton, 97 spin, 146 square root, 398 stable populations, 403{404 standard basis, 111 state, 279 absorbtive, 279 Statics problem, 5 string, 371 basis, 371 of basis vectors, 369 structurepreservation, 174 submatrix, 301 subspace, 89{98 closed, 91 complementary, 133 denition, 89 direct sum, 133 improper, 90 independence, 133 invariant, 389 proper, 90 sum, 129 sum matrix, 16 of matrices, 211 of subspaces, 129 vector, 15, 34, 78 summation notation for permutation expansion, 306 swapping rows, 4 symmetric matrix, 116, 136, 212, 219 symmetry, A-10 system of linear equations, 2 elimination, 3 Gauss' method, 3 Gaussian elimination, 3 linear elimination, 3 solving, 2 theorem, A-1 trace, 212, 228, 398 transformation characteristic polynomial, 360 composed with itself, 365 diagonalizable, 354 eigenspace, 360 eigenvalue, eigenvector, 357 Jordan form for, 392 minimal polynomial, 380 nilpotent, 370 canonical representative, 374 projection, 385 size change, 319 transition matrix, 279 transitivity, A-10 translation, 285 transpose, 19, 124 determinant, 307, 314 interaction with sum and scalar mul- tiplication, 212 Triangle Inequality, 40 triangles similar, 288 triangular matrix, 228 Triangularization, 202 trivial space, 81, 111 turning map, 161 union, A-8 unit matrix, 221 vacuously true, A-2 value, A-8 Vandermonde matrix, 309 vanishing point, 335 vector, 15, 33 angle, 42 canonical position, 34 column, 15 component, 15 cross product, 296 direction, 35 dot product, 40 free, 33 homogeneous coordinate, 338 length, 39 natural position, 34 orthogonal, 42 probability, 279 representation of, 113, 236 row, 15 satises an equation, 15 scalar multiple, 16, 34, 78 standard position, 34 sum, 15, 34, 78 unit, 44 zero, 22, 78 vector space, 78{98 basis, 110 closure, 78 complex scalars, 88 denition, 78 dimension, 118 dual, 192 nite dimensional, 117 homomorphism, 174 isomorphism, 159 map, 174 over complex numbers, 347 subspace, 89trivial, 81, 111 Venn diagram, A-1 voltage drop, 72 volume, 319 voting paradox, 144 majority cycle, 144 rational preference order, 144 voting paradoxes, 144{149 spin, 146 well-dened, A-8 Wheatstone bridge, 73 zero divisor, 219 zero divison, 235 zero divisor, 219 zero homomorphism, 175 zero matrix, 211 zero vector, 22, 78