chapter8

Chapter 8 Jordan Normal Form 8.1 Minimal Polynomials Recall pA(x)=d e t ( xI−A) is called the characteristic polynomial of the matrix A. Theorem 8.1.1. LetA∈Mn. Then there exists a unique monic polyno- mial qA(x)of minimum degree for which qA(A)=0 .I fp(x)is any polyno- mial such that p(A)=0 ,t h e n qA(x)divides p(x). Proof. Since there is a polynomial pA(x)f o rw h i c h pA(A) = 0, there is one of minimal degree, which we can assume is monic. by the Euclidean algorithm pA(x)=qA(x)h(x)+r(x) where deg r(x)<degqA(x). We know pA(A)=qA(A)h(A)+r(A). Hence r(A) = 0, and by the minimality assumption r(x)≡0. Thus qA divided pA(x) and also any polynomial for which p(A) = 0. to establish that qAis unique, suppose q(x) is another monic polynomial of the same degree for which q(A)=0 . T h e n r(x)=q(x)−qA(x) is a polynomial of degree less than qA(x)f o rw h i c h r(a)=q(A)−qA(A)=0 . This cannot be unless r(x)≡0=0 q=qA. Definition 8.1.1. The polynomial qA(x) in the theorem above is called the minimal polynomial. 221 222 CHAPTER 8. JORDAN NORMAL FORM Corollary 8.1.1. IfA, B∈Mnare similar, then they have the same min- imal polynomial. Proof. B=S−1AS qA(B)=qA(S−1AS)=S−1qA(A)S=qA(A)=0. If there is a minimal polynomial for Bof smaller degree, say qB(x), then qB(A) = 0 by the same argument. This contradicts the minimality of qA(x). Now that we have a minimum polynomial for any matrix, can we find a matrix with a given polynomial as its minimum polynomial? Can the degreethe polynomial and the size of the matrix match? The answers to both questions are a ffirmative and presented below in one theorem. Theorem 8.1.2. For any n thdegree polynomial p(x)=xn+an−1xn−1+an−2xn−2+···+a1x+a0 there is a matrix A∈Mn(C)for which it is the minimal polynomial. Proof. Consider the matrix given by A= 00 ... ... −a 0 10 −a1 01 0...−a2 ... 0... 0... 01−an−1 . Observe that Ie 1=e1=A0e1 Ae1=e2=Ae1 Ae2=e3=A2e1 ... Aen−1=en=An−1e1 8.1. MINIMAL POLYNOMIALS 223 and Aen=−an−1en−an−2en−1−···−a1e2−a0e1 Since Aen=Ane1, it follows that p(A)e1=Ane1+an−1An−1e1+an−2A−2e1+···+a1Ae1+a0Ie1=0 Also p(A)ek=p(A)Ak−1e1=Ak−1p(A)e1=Ak−1(0) = 0 k=2,... ,n . Hence p(A)ej=0 f o r j=1...n.T h u s p(A) = 0. We know also that p(x) is monic. Suppose now that q(x)=xm+bm−1xm−1···+b1x+b0 where m<n andq(A)=0 . T h e n q(A)e1=Ame1+bm−1Am−1e1+···+b1Ae1+b0e1 =em+1+bm−1em+···+b1e2+b0e1=0. But the vectors em+1...e 1are linear independent from which we conclude thatq(A) = 0 is impossible. Thus p(x)i sm i n i m a l . Definition 8.1.2. For a given monic polynomial p(x), the matrix Acon- structed above is called the companion matrix to p. The transpose of the companion matrix can also be used to generate a linear differential system which has the same characteristic polynomial as a given nthorder differential equation. Consider the linear di fferential equation y(n)+an−1y(n−1)+···+a1y0+a0=0. This nthorder ODE can be converted to a first order system as follows: u1=y u2=u0 1 =y0 u3=u0 2 =y00 ...... un=u0 n−1=y(n−1) 224 CHAPTER 8. JORDAN NORMAL FORM Then we have  u 1 u2 ... un 0 = 01 01 0 01 ...... 0... 1 a 0−a1 ... ... −an−1  u 1 u2 ... un  8.2 Invariant subspaces There seems to be no truly simple way to the Jordan normal form. The approach taken here is intended to reveal a number of features of a matrix,interesting in their own right. In particular, we will construct “generalized eigenspaces” that envelop the entire connection of a matrix with its eigen- values. We have in various ways considered subspaces VofC nthat are invariant under the matrix A∈Mn(C). Recall this means that AV⊂V. For example, eigenvectors can be used to create invariant subspaces. Nullspaces, the eigenspace of the zero eigenvalue, are invariant as well. Triangu-lar matrices furnish an easily recognizable sequence of invariant subspaces. Assuming T∈M n(C) is upper triangular, it is easy to see that the sub- spaces generated by the coordinate vectors {e1,...,e m}form=1,...,n are invariant under T. We now consider a speci fic type of invariant subspace that will lead the so-called Jordan normal form of a ma trix, the closest matrix similar to A that resembles a diagonal matrix. Definition 8.2.1 (Generalized Eigenspace). LetA∈Mn(C)w i t hs p e c - trumσ(A)={λ1,...,λk}.D e fine the generalized eigenspace pertaining to λiby Vλi={x∈Cn|(A−λiI)nx=0} Observe that all the eigenvectors pertaining to λiare contained in Vλi. If the span of the eigenvectors pertaining to λiis not equal to Vλithen there must be a positive power pand a vector xsuch that ( A−λiI)px= 0 but that y=(A−λiI)p−1x6=0 . T h u s yis an eigenvector pertaining to λi.F o rt h i s reason we will call Vλithe space of generalized eigenvectors pertaining to λi.O u r first result, that Vλiis invariant under A,i ss i m p l et op r o v e ,n o t i n g that only closure under vector addition and scalar multiplication need be established. 8.2. INVARIANT SUBSPACES 225 Theorem 8.2.1. LetA∈Mn(C)with spectrum σ(A)= {λ1,...,λk}. Then for each i=1,...,k ,Vλiis an invariant subspace of A. One might question as to whether Vλicould be enlarged by allowing higher powers than nin the de finition. The negative answer is most simply expressed by evoking the Hamilton-Cayley theorem. We write the charac- teristic polynomial pA(λ)=Q(λ−λi)mA(λi),w h e r e mA(λi) is the algebraic multiplicity of λi.S i n c e pA(A)=Q(A−λiI)mA(λi)= 0, it is an easy mat- ter to see that we exhaust all of Cnwith the spaces Vλi.T h i s i s t o s a y t h a t allowing higher powers in the de finition will not increase the subspaces Vλi. Indeed, as we shall see, the power of ( λ−λi) can be decreased to the geomet- ric multiplicity mg(λi)t h ep o w e rf o r λi. For now the general power nwill suffice. One very important result, and an essential fir s ts t e pi nd e r i v i n g the Jordan form, is to establish that any square matrix Ais similar to a block diagonal matrix, with each block carrying a single eigenvalue. Theorem 8.2.2. LetA∈Mn(C)with spectrum σ(A)= {λ1,...,λk}and with invariant subspaces Vλi,i=1,2,...,k .T h e n ( i ) T h e s p a c e s Vλi,j= 1,...,k are mutually linearly independent. (ii)Lk i=1Vλi=Cn(alternatively Cn=S(Vλ1,...,V λk)) (iii) dimVλi=mA(λi).( i v ) Ais similar to a block diagonal matrix with kblocks A1,...,A k.M o r e o v e r , σ(Ai)= {λi}and dimAi=mA(λi). Proof. (i) It should be clear that the subspaces Vλiare linearly independent of each other. For if there is a vector xin both VλiandVλjthen there is a vector for some integer q,it must be true that ( A−λjI)q−1x6= 0 but (A−λjI)qx=0.This means that y=(A−λjI)q−1xis an eigenvector pertaining to λj.S i n c e ( A−λiI)nx= 0 we must also have that (A−λjI)q−1(A−λiI)nx=(A−λiI)n(A−λjI)q−1x =(A−λiI)ny=0 =nX k=0µn k¶ (−λi)n−kAky =nX k=0µn k¶ (−λi)n−kλjky =(λj−λi)ny=0 This is impossible unless λj=λi. (ii) The key part of the proof is to block diagonalize Awith respect to these invariant subspaces. To that end, let S 226 CHAPTER 8. JORDAN NORMAL FORM be the matrix with columns generated from bases of the individual Vλitaken in the order of the indices. Supposing there are more linearly independentvectors in C nother than those already selected, fill out the matrix Swith vectors linearly independent to the subspaces Vλi,i=1,...,k .N o w d e fine ˜A=S−1AS. We conclude by the invariance of the subspaces and their mutual linear independence that ˜Ahas the following block structure. ˜A=S−1AS= A 10 ··· 0∗ 0 A2 0∗ ......... Ak∗ 0 ··· 0 B  It follows that p A(λ)=p˜A(λ)=³Y pAi(λ)´ pB(λ) Any root rofpB(λ) must be an eigenvalue of A,sayλj,and there must be an eigenvector xpertaining to λj. Moreover, due to the block structure we can assume that x=[ 0,..., 0,x]T,where there are kblocked zeros of the sizes of theAirespectively. Then it is easy to see that ASx =λjSx, and this implies thatSx∈Vλj. Thus there is another vector in Vλj, which contradicts its definition. Therefore Bis null, or what is the same thing, ⊕k i=1Vλi=Cn. (iii) Let di=d i m Vλi.F r o m ( i i ) k n o wPdi=n. Suppose that λi∈σ(Aj). Then there is another eigenvector xpertaining to λiand for which Ajx= λix.Moreover, this vector has the form x=[ 0,..., 0,x ,0,...0]T, analogous to the argument above. By construction Sx /∈Vλi, but ASx =λiSx,and this contradicts the de finition of Vλi. W et h u sh a v et h a t pAi(λ)=(λ−λi)di. Since pA(λ)=QpAi(λ)=Q(λ−λi)mA(λi), it follows that di=mA(λi) (iv) Putting (ii), and (iii) together gives the block diagonal structure as required. On account of the mutual linear independence of the invariant subspacesV λiand the fact that they exhaust Cnthe following corollary is immediate. Corollary 8.2.1. LetA∈Mn(C)with spectrum σ(A)=λ1,...,λkand with generalized eigenspaces Vλi,i=1,2,...,k .T h e n e a c h x∈Cnhas a unique representation x=Pk i=1xiwhere xi∈Vλi. Another interesting result which reveals how the matrix works as a linear transformation is to decompose the it into components with respect to the 8.2. INVARIANT SUBSPACES 227 generalized eigenspaces. In particular, viewing the block diagonal form ˜A=S−1AS= A10 ··· 0 0 A2 0 ...... Ak  the space Cncan be split into a direct sum of subspaces E1,...,E kbased on coordinate blocks. This is accomplished in such that any vector y∈Cncan be written uniquely as y=Pk i=1yiwhere the yi∈Ei. (Keep in mind that eachyi∈Cn; its coordinates are zero outside the coordinate block pertaining Ei.) Then ˜Ay=˜APk i=1yi=Pk i=1˜Ayi=Pk i=1Aiyi.This provides a computational tool – when this block diagonal form is known. Note that the blocks correspond directly to the invariant subspaces by SEi=Vλi.W e can use these invariant subspaces to get at the minimal polynomial. Foreach i=1,...,k define m i=m i n j{(A−λiI)jx=0 |x∈Vλi} Theorem 8.2.3. LetA∈Mn(C)with spectrum σ(A)=λ1,...,λkand with invariant subspaces Vλi,i=1,2,...,k . Then the minimal polynomial ofAis given by q(λ)=kY i=1(λ−λi)mi Proof. Certainly we see that for any vector x∈Vλj q(A)x=ÃkY i=1(A−λiI)mi! x=0 Hence, the minimal polynomial qA(λ)d i v i d e s q(A). To see that indeed they are in fact equal, suppose that the minimal polynomial has the form qA(λ)=kY i=1(λ−λi)ˆmi where ˆ mi≤mi,fori=1,...,k a n di np a r t i c u l a r ˆ mj<m j.B y c o n s t r u c t i o n there must exist a vector x∈Vλjsuch that ( A−λjI)mjx= 0 but y= 228 CHAPTER 8. JORDAN NORMAL FORM (A−λjI)mj−1x6=0.Then if q(A)x=ÃkY i=1(A−λiI)ˆmi! x = kY i=1 i6=j(A−λiI)ˆmi y =0 This cannot be because the contrary implies that there is another vector in one of the invariant subspaces Vλk. Just one more step is needed before the Jordan normal form can be derived. For a given Vλiwe can interpret the spaces in a heirarchical viewpoint. We know that Vλicontains all the eigenvectors pertaining to λi.C a l l t h e s e eigenvectors the first order generalized eigenvectors . If the span of these is not equal to Vλi, then there must be a vector x∈Vλifor which y= (A−λiI)2x= 0 but ( A−λiI)x6=0 . T h a ti st os a y yis an eigenvector of Apertaining to λi. Call such vectors second order generalized eigenvectors . In general we call an x∈Vλia generalized eigenvector of order jify= (A−λiI)jx= 0 but ( A−λiI)j−1x6= 0. In light of our previous discussion Vλicontains generalized eigenvectors of order up to but not greater than mλi. Theorem 8.2.4. LetA∈Mn(C)with spectrum σ(A)= {λ1,...,λk}and with invariant subspaces Vλi,i=1,2,...,k . (i) Let x∈Vλibe a generalized eigenvector of order p. Then the vectors x,(A−λiI)x,(A−λiI)2x ,..., (A−λiI)p−1x (1) are linearly independent. (ii) The subspace of Cngenerated by the vectors in (1) is an invariant subspace of A. Proof. (i) To prove linear independence of a set of vectors we suppose linear dependence. That is there is a smallest integer kand constants bjsuch that kX j=0xj=kX j=0bj(A−λiI)jx=0 8.2. INVARIANT SUBSPACES 229 where bk6=0.Solving we obtain bk(A−λiI)kx=−Pk−1 j=0bj(A−λiI)jx. Now apply ( A−λiI)p−kto both sides and obtain a new linearly dependent set as the following calculation shows. 0= bk(A−λiI)p−k+kx=−k−1X j=0bj(A−λiI)j+p−kx =−p−1X j=p−kbj+p−k(A−λiI)jx T h ek e yp o i n tt on o t eh e r ei st h a tt h el o w e rl i m i to ft h es u mi si n c r e a s e d . This new linearly dependent set, which we denote with the notationPp−1 j=p−kcj(A−λiI)jx can be split in the same way as before, where we assume with no loss in gen-erality that c p−16=0 . T h e n cp−1(A−λiI)p−1x=−p−2X j=p−kcj(A−λiI)jx Apply ( A−λiI) to both sides to get 0= cp−1(A−λiI)px=−p−2X j=p−kcj(A−λiI)j+1x =−p−1X j=p−k+1cj−1(A−λiI)jx Thus we have obtained another linearly independent set with the lower limit of powers increased by one. Continue this process until the linear depen-dence of ( A−λ iI)p−1xand ( A−λiI)p−2xis achieved. Thus we have c(A−λiI)p−1x=d(A−λiI)p−2x (A−λiI)y=d cy where y=(A−λiI)p−2x.T h i s i m p l i e s t h a t λi+d cis a new eigenvalue with eigenvector y∈Vλi, and of course this is a contradiction. (ii) The invariance under Ais more straightforward. First note that while x1=x, 230 CHAPTER 8. JORDAN NORMAL FORM x2=(A−λI)x=Ax−λxso that Ax=x2−λx1.Consider any vector y defined by y=Pp−1 j=0bj(A−λiI)jxIt follows that Ay =Ap−1X j=0bj(A−λiI)jx =p−1X j=0bj(A−λiI)jAx =p−1X j=0bj(A−λiI)j(x2−λx1) =p−1X j=0bj(A−λiI)j[(A−λiI)x1−λx1] =p−1X j=0cj(A−λiI)jx where cj=bj−1−λforj>0a n d c0=−λ, which proves the result. 8.3 The Jordan Normal Form We need a lemma that points in the direction we are headed, that being the use of invariant subspaces as a basis for the (Jordan) block diagonalization of any matrix. These results were discussed in detail in the Section 8.2. Therestatement here illustrates the “invariant subspace” nature of the result,irrespective of generalized eigenspac es. Its proof is elementary and is left to the reader. Lemma 8.3.1. LetA∈M n(C)with invariant subspace V⊂Cn. (i) Suppose v1...v kis a basis for VandSis an invertible matrix with thefirstkcolumns given by v1...v k.T h e n 1...k S−1AS=·∗∗ 0∗¸ . 8.3. THE JORDAN NORMAL FORM 231 (ii) Suppose that V1,V2⊂Cnare two invariant subspaces of AandCn= V1⊕V2. Let the (invertible) matrix Sconsist respectively of bases from V1 andV2as its columns. Then S−1AS=·∗0 0∗¸ . Definition 8.3.1. Letλ∈C.A Jordan block Jk(λ)i sa k×kupper triangular matrix of the form Jk(λ)= λ1 0 λ1 0...1 λ . AJordan matrix is any matrix of the form J= J n1(λ1)0 ... 0 Jnk(λk) . where the matrices Jn1are Jordan blocks. If J∈Mn(C), then n1+n2···+ nk=n. Theorem 8.3.1 (Jordan normal form). LetA∈Mn(C). Then there is a nonsingular matrix S∈Mnsuch that A=S J n1(λ1) 0 ... 0 Jnk(λk) S −1=SJS−1 where Jni(λi)is a Jordan block, where n1+n2+···+nk=n.Jis unique up to permutations of the blocks. The eigenvalues λ1,... ,λkare not necessarily distinct. If Ais real with real eigenvalues, then Scan be taken as real. Proof. This result is proved in four steps. (1) Block diagonalize (by similarity) into invariant subspaces pertaining to σ(A). This is accomplished as follows. First block diagonalize the ma- trix according to the generalized eigenspaces Vλi={x∈Cn|(A−λiI)nx= 232 CHAPTER 8. JORDAN NORMAL FORM 0}as discussed in the previous section. Beginning with the highest or- der eigenvector in x∈Vλi, construct the invariant subspace as in (1) of Section 8.2. Repeat this process until all generalized eigenvectorshave been included in an invariant subspace. This includes of coursefirst order eigenvectors that are not a ssociated with higher order eigen- vectors. These invariant subspaces have dimension one. Each of these invariant subspaces is linearly independent from the others. Continuethis process for all the generalized eigenspaces. This exhausts C n. Each of the blocks contains exactly one eigenvector. The dimensionsof these invariant subspaces can range from one to m λi,t h e r eb e i n ga t least one subspace of dimension mλi. (2) Triangularize each block by Schur’s theorem, so that each block has the form K(λ)= λ∗ ... 0λ  You will note that K(λ)=λI+N where Nis nilpotent, or K(λ)i s1 ×1. (3) “Jordanize” each triangular block. Assume that K1(λ)i sm×m,w h e r e m> 1. By construction K1(λ) pertains to an invariant subspace for which there is a unique vector xfor which Nm−1x6=0 a n d Nmx=0. Thus Nm−1xis an eigenvector of K1(λ), the unique eigenvector. De fine yi=Ni−1xi =1,2,... ,m . Expand the set {yi}m i=1as a basis of Cm.D efine S1=" ymym−1···y1 ...... ···...# . Then NS 1=" 0ymym−1... y 2 ......... ···...# . 8.3. THE JORDAN NORMAL FORM 233 So S−1 1NS 1= 01 0 01 0... ...1 00 . We conclude that S −1 1K1(λ)S1= λ10 λ1 λ1 ...... ...1 0 λ  (4) Assemble all of the blocks to form the Jordan form. For example, the block K 1(λ) and the corresponding similarity transformation S1 studied above can be treated in the assembly process as follows: De fine then×nmatrix ˆS1= I00 0S10 00 I  where S1i st h eb l o c kc o n s t r u c t e da b o v ea n dp l a c e di nt h e n×nmatrix in the position that K1(λ) was extracted from the block triangular form of A. Repeat this for each of the blocks pertaining to minimally invariant subspaces. This gives a sequence of block diagonal matrices ˆS1,ˆS2..., ˆSk.D efineT=ˆS1ˆS2...ˆSk. It has the form T= ˆS 1 0 ˆS2 ... 0 ˆSk  Together with the original matrix Pthat transformed the matrix to the minimal invariant subspace blocked form and the unitary matrix 234 CHAPTER 8. JORDAN NORMAL FORM Vused to triangularize A, it follows that A=PVT J n1(λ1) 0 ... 0 Jnk(λk) (PVT ) −1=SJS−1 with S=PVT . Example 8.3.1. Let J= 2 1 02 2 310 031 003 −1  In this example, there are four blocks, with two of the blocks pertaining to the single eigenvalue 2. For the first block there is the single eigenvector e 1, but the invariant subspace is S(e1,e2).For the second block, the eigenvector, e3,generates the one dimensional invariant subspace. The block pertaining to the eigenvector 3 has the single eigenvector e4while the minimal invariant subspace is S(e4,e5,e6). Finally, the one dimensional subspace pertaining to the eigenvector −1 is spanned by e7.The minimal invariant polynomial isq(λ)=(λ−2)2(λ−3)3(λ+1 ) . 8.4 Convergent matrices Using the Jordan normal form, the study of convergent matrices becomes relatively straightforward and simpler. Theorem 8.4.1. IfA∈Mnandρ(A)<1.T h e n lim k→∞A=0. 8.5. EXERCISES 235 Proof. We assume Ais a Jordan matrix. Each Jordan block Jk(λ)c a nb e written as Jk(λ)=λIk+Nk where Nk= 01 0 01 ...... ...1 00 is nilpotent. Now A= J n1(λ1) ... Jmk(λk) . We compute, for m>n k (Jnk(λk))M=(λI+N)m =λmI+mkX j=0λm−jNjµm j¶ because Nj=0f o r j>n k.W eh a v e λm−jµm j¶ →0a sm→∞ since |λ|<1. The results follows. 8.5 Exercises 1. Prove Theorem 8.2.1. 2. Find a 3 ×3 matrix that has the same eigenvalues are the squares of the roots of the equation λ3−3λ2+4λ−5=0 . 3. Suppose that Ais a square matrix with σ(A)= {3},ma(3) = 6, and mg(3) = 3. Up to permutations of the blocks show all possible Jordan normal forms for A. 236 CHAPTER 8. JORDAN NORMAL FORM 4. Let A∈Mn(C)a n dl e t x1∈Cn.Definexi+1=Axifori=1,...,n−1. Show that V=S({x1,...,x n}) is an invariant subspace of A. Show thatVcontains an eigenvector of A. 5. Referring to the previous problem, let A∈Mn(R)b eap e r m u t a t i o n matrix. (i) Find starting vectors so that dim V=n.( i i ) F i n d s t a r t - ing vectors so that dim V= 1. (iii) Show that if λ= 1 is a simple eigenvalue of Athen dim V=1o rd i m V=n. Chapter 9 Hermitian and Symmetric Matrices Example 9.0.1. Letf:D→R,D⊂Rn.T h e Hessian is defined by H(x)=hij(x)≡∂f ∂xi∂xj∈Mn. Since for functions f∈C2it is known that ∂2f ∂xi∂xj=∂2f ∂xj∂xi it follows that H(x)i ss y m m e t r i c . Definition 9.0.1. A function f:R→Risconvex if f(λx+( 1−λ)y)≤λf(x)+( 1−λ)f(y) forx, y∈D(domain) and 0 ≤λ≤1. Proposition 9.0.1. Iff∈C2(D)andf00(x)≥0onDthenf(x)is convex. Proof. Because f00≥0, this implies that f0(x) is increasing. Therefore if x<x m<ywe must have f(xm)≤f(x)+f0(xm)(xm−x) and f(xm)≤f(y)+f0(xm)(xm−y) 237