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# Intro Lin Algebra MATH 513

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This 7 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 513 at University of Michigan taught by Igor Dolgachev in Fall. Since its upload, it has received 9 views. For similar materials see /class/231484/math-513-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15

MATH 513 JORDAN FORM Let A17 Ak be square matrices of size 711 Wm respectively with entries in a eld F We de ne the matrix A1 63 63 Ak of size n m nk as the block matrix 000 000 d I d It is called the direct sum of the matrices A17 7Ak A matrix of the form 1 0 0 0 1 0 0 1 0 0 is called a Jordan block If k is its size7 it is denoted by Jk A direct sum J Jk1 JkTT of Jordan blocks is called a Jordan matrix Theorem Let T V a V be a linear operator in a nitedimensional vector space over a eld F Assume that the characteristic polynomial of T is a product of linear polynimials Then there exists a basis 5 in V such that Tlg is a Jordan matrix Corollary Let A E Assume that its characteristic polynomial is a product of linear polynomials Then there exists a Jordan matrix J and an invertible matrix C such that A CJC I Notice that the Jordan matrix J which is called a Jordan form of A is not de ned uniquely For example7 we can permute its Jordan blocks Otherwise the matrix J is de ned uniquely see Problem 7 On the other hand7 there are many choices for C We have seen this already in the diagonalization process What is good about it We have7 as in the case when A is diagonalizalole7 AN CJNC l So7 if we can compute JN7 we can compute AN It follows from the matrix multiplication that A1AkN AVA V Thus it is enough to learn how to raise a Jordan block in Nth power First consider the case when 0 We have 010 02 0010 0 0010 00 01 0 2 II 39 I Mo 39IIquot 0 001 0 01 0 000 0 00 0 000 We see that the ones go to the right until they disappear Continuing in this way we see that Jk0k 0 Now we have JkN Mn Jk0N N N N ANIn 1gtAN1M0 gtANZJk0Z N71 This is proved in the same way as one proves the Newton formula a bN i an ibi 720 1 Wm MOW lt1 We look at the product of N factors a b a b To get a monomial an ibi we choose 1 brackets from which we will take b The number of choices is 11V Notice that in formula 1 the powers Jk0i are equal to zero as soon as i 2 k So we get AN JDN l JDW AN ktl 0 AN me WM crew Alv ANA 0 0 0 N where by de nition x 0 if N lt m Before we go to the proof of the Theorem let us explain how to nd J and C Notice that rankJk0 k 71 rankJk02 k 7 2 rankJkOk 0 Let us introduce the notion of the comnk of a matrix A E MatnF by setting corankA n 7 rankA Then we see that corankJk0i i and is equal to the number of the rst zero columns Now for any Jordan matrix J Jk11 63 Jk AT we have T corankJ 7 ALLY Z coranksz0q ieIA where A is the set of indices 1 for which A Let dm the number of Jordan blocks Jm in J cq corankJ 7 ADJ corankA 7 ADJ 2 The last equality follows from the fact that M7Ah UCA7AMV0XJ7MQCAVCU7AMVCA So7 A 7 Abby and J 7 ALLY are matrices of the same operator and hence have the same rank We have go dmoy mZI MM7MMZNMM m22 Solving this system for dm we nd d1 201k 7 02 jZHWn71 a 7 dj Cj7l 2613 Cj1 dn cn 7 on1 This gives the answer for J Remark For matrices of small size we can list all possible Jordan forms7 and then choose the right one by applying formula Example Let 1 1 1 A 1 1 2 0 0 0 The characteristic polynomial is PA 7M3 272 So the eigenvalues are 0 and 2 Since 0 is a root of multiplicity 27 it must appear twice at the diagonal Possible Jordan forms are 0 0 0 010 J1 0 0 0 7 J2 0 0 0 0 0 2 0 0 2 Since corankA corankA 7 012 corankJ2 7 012 1 we have to choose J J2 Now let us give the answer for C We have A C39JCquot17 hence the columns CZ of C form a basis 3 in F such that the matrix TAlg J By definition7 A011017 1402 102 1 017 7 ACk110k10k1717 3 ACk11 2Ck117 AC2 2Ck12 Ck117 7 ACkg A201 010717 and so on We solve the rst equation for Cl This is a homogeneous system of linear equations A 7 A1109 0 We take any non zero solution 01 of this system for which A 7 Alma 01 is solvable Then we solve this equation to nd 02 Continuing in this way we nd 01 Ck1 Then we go to the next block with eigenvalue A2 lf 1 A2 when we solve A 7 A1109 0 to nd Oh we choose a solution which is not proportional to Cl If there are more Jordan blocks with the same eigenvalue A1 each time we solve A 7 A1109 0 we nd a solution which is not a linear combination of the previous solutions In fact7 we nd a basis of NA 7 All and take its elements for the columns 0101 Ck51 if 1 A5 7 i7 1 gt s It follows from the proof of the theorem about the existence of a Jordan form that one can always nd C de ned by the above equations Example cont Let us nd the matrix C such that A CJC l First we solve A9 O The space of solutions is one dimensional The vector 717 17 0 forms a basis in the space of solutions This will be our column Cl The second column 02 is found by solving A02 Cl A solution is 37 O7 72 It is not a unique solution We can add to it any solution of A9 0 to obtain another solution Finally we nd the third column 03 by solving the equation A7 23Cg O A solution is 03 110 So 71 73 1 C 1 0 1 0 2 0 We leave to the reader to verify that A CJC l Now let us go to the proof of the main theorem For any linear operator T V 7 V and a non negative integer i we denote by T the compo sition of T with itself i times By de nition7 T0 idV A linear subspace L C V is called inuan39ant with respect to T if Tv C L for any 1 C L Let LlCLgCCLkCCV be a sequence of linear subspaces of V such that each is a subset of the next one Let L U Lk k be the union of these subspaces it could be in nitely many of them I claim that L is a linear subspace of V lndeed7 take 1210 C L Let U C Lkw C Lm Without loss of generality we may assume that k g m Since Lk C Lm7 we get 1210 C Lm Since Lm is a linear subspace7 we get u 1 cu C Lm C L for any 0 C F Thus L is a linear subspace We apply this to the following situation Let U be a linear operator7 and Lk NUk We have NUk C NUkH since Nkv 0 implies NUkH NNkv NO 0 Thus we have a sequence of linear subspaces NU CNU2 c CNUk c cv It follows from above that UNUk u e V Um 0 for some k gt 0 k 4 is a linear subspace of V It is also invariant with respect to U ln fact7 if v E NUk for some k then Uk 1Uv Ukv O7 hence Uv E NUk 1 if k 07 we have u 0 so Uv 0 belongs to any subspace Recall that the eigensubspace of T corresponding to an eigenvalue A is the kernel of the operator T 7 AidV De ne the 39 J 39 of T LULLU ponding to an A by VT7 A v E V T 7 AidViv 0 for some i gt 0 Take U T 7 AV in above7 we obtain the proof of the following Lemma 1 VT7 A is a linear susbspace of V It is invariant with respect to T Let us restrict the operator T to the invariant subspace VT7 A that is consider the same rule for T only applied to vectors from VT7 We shall exhibit a basis in VT7 A such that the matrix of T with respect to this basis is the direct sum of Jordan block matrices with A at the diagonal Notice that the operator U T7AidV when restricted to VT7 A satis es Um 0 for some m gt 0 ln fact7 every vector v E VT7 A satis es Uiv 0 for some i gt 0 Choose a basis v1 Wok in VT7 A and let m be chosen such that Umvi 0 for all i 17 Wk This can be done since Ujv 0 imlplies U902 Us jUjv 0 for s 2 j Now writing any 1 E VT7 A as a linear combination of the basis7 and using that U is a linear operator7 we obtain that Umv 0 for all u 6 WT A Let us consider any nite dimensional vector space W and a linear operator U W 7 W satisfying Um 0 for some m 2 0 a linear operator with such property is called a nilpotent operator Observe that RUi1 c RUi lndeed Ui1v UiUv7 so if a vector w is equal to the value of UPr1 at some vector u then it is also equal to the value of Ui at Uv E VT7 A So we have a chain of linear subspaces 0 RUm C RUm 1 C C RU C W 4 Observe that I I URUZ MUM To see this7 use that Ui1v UUiv7 so each vector in RUi1 is equal to the value of U at some vector in RUi Lemma 2 Let U be a nilpotent linear operator on a vector space W 7 Let m be the minimal positive integer such that U 0 the level of the nilpotency of U Then all inclusions in 4 are strict Proof We use induction on m If m 17 the assertion is obvious since U OidV7 hence RU 0 7 W Assume the assertion is true for linear operators with level of nilpotency lt m Take W1 RU and let U W1 7 W1 be the restriction of U to W1 Then RUk RU k 1 and we have the sequence 0 RUm c RUm 1 c c RU w1 c W 5 equal to the sequence 0 RU m 1 C RU m 2 C C RU C W1 C W Since U is not invertible because Um O7 we have dim RU dim W 7 dim NU lt dim W So7 RU 7 W By induction7 starting from W17 all the inclusions are strict Thus all the inclusions are strict Let us go back to our situation when U T 7 idV restricted to VT7 A Let M dim RUm 1 Since RUm 07 U sends all vectors from RUm 1 to Let 125171291 be a basis of this space Since U RUm 2 a RUm 1 is surjective7 we can nd 11 712 in RUm 2 with UU 2gtU 1gt i1n1 I claim that 1 2 14 lu L11gtu H443 are linearly independent ln fact7 if 111151 amvgl l b11152 bmvgi 07 we apply U to obtain that 0 a1Uu 1gt am UM b1Uv 2 mev L21 5ng bnlug This gives b1 bn1 07 and hence a1 an1 0 Notice that 125171291 belong to NU7 so we can nd a basis of NU RUm 2 of the form 1251 v 117v 2117 quot712 Together with the vectors v52 712 we get a basis of RUm 2 ln fact7 by the formula for the dimension of the range space of a linear transformation7 the dimensions of the subspaces spanv 27 7129 and NU RUm 2 add up to the dimension of RUm 2 Also their intersection is the zero subspace ln fact7 if Ll1212 E NU7 applying U we get aiUv 2 aivgl 7 hence a1 an1 0 because the vectors v1 7 7127 are linearly independent Next we nd vf 1255 E RUm 3 which are mapped to 12 4252 respectively Then we nd a basis of NU RUm 3 which includes the previous basis 12 712212712221 712222 ofNU RUm 2 The union of this basis and the set 142 7123 vf 712232 is a basis of RUm 3 Proceeding in this way7 we nd a basis in V Ugh Ugly 42gt Us we 6 m m m Um satisfying the following property ii 711 m dim RUm i7 in particular7 m nm dim VT7 A iii T 7 Aidvu 1 1111 if 1nj Let us nd the matrix of VT7 A of T with respect to this basis We rst reorder the vectors by taking the rst m vectors from the rst column in 6 starting from the top7 then go to the second column and so on Since T 7 AidVv1 07 T 7 AidVvZ 1217171 27 7m7 we obtain TU1 A1117 T UZ Am 1 1114 i 2 7m This shows that the rst m columns of the matrix of T look like A 1 0 0 0 A 1 0 0 0 0 A 1 0 0 0 A Continuing in this way we easily convince ourselves that the matrix of T in our basis is equal to the direct sum of 711 Jordan blocks of size m7 712 7 711 Jordan blocks of size m 7 17 and7 nally7 nm 7 nm1 Jordan blocks of size 1 All of them of course have A at the diagonal Remark The proof shows for which eigenvectors Cl with eigenvalue A you can solve the equation A02 A02 1 Cl The vector 01 must come from the range space RA 7 AI of the matrix A 7 AInm 17 where m is the nilpotency level ofA 7 AI To nish the proof we use Lemma 3 Let A be an eigenvalue of T Then v 7 W A ea W where W is invariant with respect to T and T 7 AidV is invertible when restricted to W Proof We know that VT7 A NT7AidVm for some m gt 0 De ne W RT7AidVm Then the dimensions of the spaces VT7 A and W add up to dim V It remains to show that VT7 A O W If v is in the intersection we have u T 7 Aidymw7 for some w 6 V7 and hence 0 T 7 Aidvmv T 7 AidV2mw This implies that w E VT7 A But then T 7 AidVmw 0 and thus 1 0 Now we can nish the proof Take a Jordan basis in VT7 A and extend it to some basis of V The matrix of T is the direct sum of a Jordan matrix and a matrix of T restricted to W It is easy to see that the determinant of a block matrix is equal to the product of determinants of the blocks This shows that the characteristic polynomial of an operator restricted to W divides the characteristic polynomial of T this is true for any invariant subspace7 see Theorem 521 from the book By assumption it factors into the product of linear polynomials By induction on dimension of the vector space7 we may assume that the theorem is true for W Since the restriction of T 7 AidV to W is inveritible7 its eigenvalues are different from A Thus T restricted to W has a basis such that the matrix of T is the sum of Jordan blocks with no A at the diagonal Taking this basis and adding to this the basis for VT7 A which we have just constructed we see that the matrix of T is the sum of Jordan blocks The theorem is proven 7

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