Class Note for MATH 790 at KU
Popular in Course
Popular in Department
This 20 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 25 views.
Reviews for Class Note for MATH 790 at KU
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 02/06/15
The Rational and Jordan Forms Linear Algebra Notes Satya Mandal November 57 2005 1 Cyclic Subspaces In a given context7 a cyclic thing77 is an one generated 77thing For example7 a cyclic groups is a one generated group Likewise7 a module M over a ring R is said to be a cyclic module if M is one generated or M Rm for some in E M We do not use the expression 77cyclic vector spaces77 because one generated vector spaces are zero or one dimensional vector spaces 11 De nition and Facts Suppose V is a vector space over a eld l7 With nite dimV 72 Fix a linear operator T E LV7 V 1 Write R FlTl fT 1 00 6 Fle E LV7V Then R lFT is a commutative ring We did considered this ring in last chapter in the proof of Coley Hamilton Theorem 2 NOW V acquires Rimodule structure With scalar multiplication as fol lows Define fTv fTv E V V fT E MT7 v E V 3 For an element 1 E V de ne ZvT lFTv fTv fT E R Note that ZvT is the cyclic Resubmodule generated by v I like the notation FlTlv the textbook uses the notation ZvT We say7 ZvT is the Ticyclic subspace generated by v 4 If V Zv7 T lFTv7 we say that that V is a Ticyclic space and v is called the Ticyclic generator of V Here I di er a little from the textbook 5 Obviosly7 ZvT is also a vector subspace over lF ln fact7 ZvT SpanawvTwine gt Since dimZv7 T S dim V n7 is nite7 a nite subset of U7 Tv7 T2vT3v7 will from a basis of Zv7 T 6 Also for v 6 V7 de ne annv fX E EX fTv 0 So7 mm 1 is an ideal of the polynomial ring EX and is called Tiannihilator of v 7 Note that7 if v a 07 then annv a EX So7 there polynomial pv such that annv Flepv As we know7 pv is the non constant monic polynomial polynomial in annv This polynomial pv is called the minimal momic polynomial MMP of annv 8 In the next theorem 127 we will give a baisis of Zv7T 12 Theorem Suppose V is a vector space over a eld l7 with nite dimV n Fix a linear operator T E LV7 V Let i E V be a non zero element and pv is the mimimal momic polynomial MMP of annv 1 lfk degreepv then v7 Tv7 T2v7 7Tkquotlv is abasis of Zv7 2 degreepv dimZv7T 3 Note that ZvT is invariant under T Also7 if U TKZWT is the restriction of T to Zv7 T then the MMP of T is pv 2 4 Write FAX Co 01X 0194in1 Xk where Ci 6 F Also write 60 U 1 T U 2 T2vek1 Tk 1v T 07 T 17T 2 7T k71 0 0 0 0 700 l 0 0 0 701 6076176277 k71 0 1 0 0 02 0 0 0 1 761971 This gives the matrix of U TKZWT with respect to the basis 60 61 62 ek1 of ZvT Proof First we prove Recall ZvT lFTv Let m E ZvT Then m fTv for some polynomial fX Using devison algorithm we have qXp1X i TX7 where qr E EX and either 7 0 or degreer lt k degredpv So m fTv qTpvTv TT U TT U Therefore ZvT SpanvTv Tk 1v Now suppose 101 a1Tv ak1Tk 1v 0 for some 1239 E F Then fX a0 a1X ak1Xk 1 E annv By minimality of 1 we have fX 0 and hence a 0 for all 239 0k 71 Therefore vTv Tk 1v are linearly independent and hence a basis of ZvT This establishes Now 2 follows from Now we will prove Clearly TZvT TlFTv Q lFTv ZvT Therefore ZvT is invariant under T To prove that p is the MMP of U we prove that FXpv annv annU In fact for any polynomial y we have 9X E annv gt yTv 0 gt yUv 0 gt 9U 0 gt g E annU This completes the proof of The proof of the 4 is obvious 13 De nition Given a polynomial pX CO c1X cIHX 1 Xk e M with Ci 6 F The matrix 0 0 0 700 1 0 0 0 701 1 0 0 702 0 0 0 1 70164 is called the companion matrix of p 14 Theorem Suppose W is a vector space over a eld F with nite dimW 72 Fix a linear operator T E LW7 Then W is Ticyclic if and only if there is a basis E of W such that the matrix of T is given by the companion matrix of MMP p of T Proof i This part follows from the 4 of theorem 12 To prove the converse let E 607617 76711 and the matrix of T is given by the companion matrix of the MMP pX CD 01X 011inl X Therefore7 we have T 07T 17T 2 7T n71 0 0 0 0 700 1 0 0 0 701 6076176277en1 0 1 0 0 702 0 0 0 1 707171 Forz391n 71 we have 6239 T6 1 Tieo Therefore V Span 07 7611 lFX60 Hence V is Ticyclic This completes the proof The following is the matrix version of the above theorem 14 15 Theorem Suppose p E EX is a monic polynomial and A is the companion matrix of p Then both the characteristic polynomial and the MMP of A is p Proof Write pX CD 01X cn1X 1 X Then 0 0 0 0 700 1 0 0 701 A 0 1 0 0 702 0 0 0 1 ian Therefore the characteristic polynomial of A is q detXIn 7 A Expand the determinant along the rst row and use induction to see q detXIn 7 A p Now consider the linear operator T l a F given by TX AX By theorem 147 we have F is Ticyclic By 3 of theorem 127 MMP of T is p So7 MMP of A is also p This completes the proof 2 Cyclic Decomposition and Rational Forms Given a nite dimensional vector space V and an operator T E LV7 V7 the main goal of this section is to decompose V Zv1T 69 69 Zvk7T as direct sum of Ticyclic subspaces 21 Remark Suppose T E LV7 V is an operator on a nite dimensional vector space V Also suppose V has a direct sum decomposition V W 69 W where both VVW are invariant under T In this case7 we say W is Tiinvariant complement of W Now7 let 1 w 117 with w E W7w E W and fX be a polynomial In this case7 fTv E W fTv fTw fTw fT1Ul So7 we have the the following de nition 22 De nition Suppose T E LV7 V is an operator on a nite dimen sional vector space V A subspace W of V is said to be Tiadmissible if 1 W is invariant under T 2 For a polymonial fX and v 6 V7 fTv E W gt fTv fTw for some 10 E W 23 Remark We have7 1 Obviously7 0 and V are Tiadmissible 2 Above remark 217 asserts that if an invariant subspace W has an in variant complement then W is Tiadmissible The following is the main theorem 24 Cyclic Decomposition Theorem Let V be a nite dimensional vector space with dimV n and T E LV V be an operator Sup pose W0 is a proper Tiadmissible subspace of V Then there exists non zero 101 w E V such that 1 V W0 Zw1T G9 ZwT 2 Let pk be the MMP of annw see Part 7 of 11 Then pk l p19L for k 2 7 Further the integer 7 and p1 p are uniquely determined by 1 and Proof We will complete the proof in several steps StepI Conductors and all Suppose u E V and W is an invariant suspace V Assume 1 W 1 Write IvW fX E 1FXfTv E 2 Since W is invariant under T we have 1 W is an ideal of EX Also since 1 W we have 1 IvW and IvW is a proper ideal The textbook uses the notation 81 I use the notation IvW because it reminds us that it is an ideal 3 This ideal 1 W is called the Ticonductor of u in W 4 Let 1 W 1FXp We say that p is the MMP of the conductor 5 Exercise We have dimW 1FTv dimW degreep So degreep S dimV n StepII If V W0 there is nothing to prove So we assume V a W0 Let d1 madegreep 1 W0 1FXpu W0 Note that the set on the right hand side is bounded by n So 11 is well de ned and there is a U1 6 V such that u1W0 1FXp1 and degreep1 11 Write W1 W0 1 Z UlT Then dimW0 lt dimW1 S n and W1 is Tiinvariant Therefore this process can be repeated and we can nd v112 uk such that 1 We write Wk W194 Z39Uk7T7 2 vk Wkih 3 pk is the MMP ofthe conductor oka in W194 That means 1197 W194 FlePk 4 Also dk degreepk madegreep 17 W194 FXpv W194 5 dimW0 lt dimW1 lt lt dimWT n and so WT V 6 VW0Zv1TZv2TZvT7T StepIII Here we prove the following for later use Let v17vT be as above Fix k with 1 S k S 7 Let 1 E V and U7Wk1 FXf Suppose 1971 fv v0 Zgivi with v0 6 ngi E EX 2 1 Then f l g and v0 f0 for some 0 6 W0 To prove this7 use division algorithm and let gi fhi n where n 0 or degree i lt degreef We will prove n 0 for all 239 Let 1971 u v 7 2 him 2 1 Note 1 7 u E W194L and hence 717 W194 vvwkil Flef Also fufvfu7v 1971 1971 1971 v0 Zth T U2 thivi 00 272 21 21 21 Assume not all n 0 Then i fuv0vai Eqnil 2 1 where j lt k and T7 a 0 Let WIV771 FleP Since WWI71 Q u7Wk1 we have p f9 for some polynomial 9 Apply 9 to Eqn l and get jil Pu 9f 900 297215 973 21 Therefore 9 E 1 Wm FlePj By maximality of pi we have degpj 2 1691 Therefore deggrj Z degpj 2 1691 degfg Hence 16903 2 1690 with is a contradiction So7 n 0 for alli17r So7 1971 fvvOZfTivi 21 Now7 by admissibility of W07 we have v0 fmo for some 0 6 W0 So7 Stap lll is complete StepIV We will pick wl W0 such that 1 V W0Zw17TZv2T Z UT7T and W1 W0Zwl7T7 2 The ideals FXp1 117W0 annw1 3 W0 Zw1T 4 W1 WO Zw1T W0 Zw1T To see this rst note that by choice plvl 6 W0 and by admissibility we have plvl pal for some 1 6 W0 Take 101 v1 7 1 Proof of 1 follows from the fact that 101 7 v1 6 W0 By choice pl 6 annw1 Therefore v1W0 Flem Q annw1 Now suppose f E annw1 Then fvl funL fv1 7 101 fv1 7 101 6 W0 Hence annw1 Q v1 W0 Therefore 2 is established To prove 3 rst note that plwl 0 Now let m 6 W0 0 Zw1T Therefore m fwl for some polynomial f Using division algorithm we have f qpl So m funL qplwl 0 So 3 is established The last part 4 follows from StepV I wish W1 was admissible so that we could repeat the process But that does not seem correct In any case for 1 S k S r we will use Step lll and use induct to pick LU2 wk 1 V W0Zw1TZw2T ZwkTZvk1T ZvT and Wk W0 Zw1T Zw2T ZwkT 2 The ideals FXpk vkWk1 annwk 3 Wkil ZwkT 4 Wk W0 69 Zw1T e9 Zw2T 69 69 ZwkT To prove this we assume the statements holds for the previous step and prove it fome the kthstep So we are assuming that we have already picked 101 wk1 with all the above properties and will pick wk We use Step lll with f pk and vk W194 We have pkvk E W194L and 1971 pkvk v0 291 with v0 6 W0g E EX i1 So 92 pkhi and v0 pkg0 for some yo 6 W0 Write 1971 wk 1 10 E hm 21 10 Therefore wk 7 Uk 6 W194 Therefore 1 and 2 follow immediately Also 4 will follow from Proof of 3 follows as in the previous step because pkwk 0 StepVI In this step we prove that pk l p194 We will use Step lll We have Iwk W194 FXpk By our choice 1ka 191971101971 39 39 39 192102 191101 0 Therefore we have 1ka 0 191971101971 192102 191101 Hence by Step lll whe have pk l p194 StepVII Uniqueness Now we prove the uniqueness part Suppose 103 w E V such that 1 VW0ZwiTEBEBZwT 2 Let qk be the MMP of annw and qk lqk1 for k 2 3 We will prove that r 3 and pi qi First write I f E lFX fTV Q W0 Clearly I is an ideal We claim that I annw1 Note that piwi 0 and p l p1 Therefore p110 0 for 239 1 7 Therefore P1TV P1TW0 G9 Z1U17T39 G9 Zw57T P1TW0 Q W0 Therefore pl 6 I and hence annw1 FXp1 Q I Conversely suppose f E I Then fTw1 6 W0 0 Zw1T Therefore f E annw1 Hence I Q annw1 lFXp1 So we have I annw1 lFXp1 Similarly I annwi FXq1 11 Since both p1 and q1 are monic we have p1 ql Write W1 W0 69 ZwiT Now assume that 2 S r S 3 We proceed to prove p2 12 We start with the following observations 1 annw1 annwi Proof Suppose f E annw1 Then funL 0 Write 103 100 yTw1 Then fw L fwo 6 W0 0 ZwiT So f LUl 0 and annw1 Q annwi Similarly we prove the other inclusion and hence the equality 2 annpTw1 annpTwi for any polynomial p Proof The proof is similar to the above proof but we will give a proof Suppose f E annpTw1 Then fpwl 0 Write 103 w0gTw1i Then fpw L fpwo E W00ZwiT So fpw L 0 and annpw1 Q annpwi Similarly we prove the other inclusion and hence the equality 3 dimZfTw1T dimZfTwiT for any polynomial f This part follows from the above and 2 of theorem 12 We have V W0 69 Zw1T 69 69 ZwTT Apply p2T and we have 1920 192W0 P2Zw17T G9 39 39 39 P2Z1UT7T Since p2Zw T 0 for 239 Z 1 we have 1920 192W0 G9 p2Z1U17 T 192W0 G9 ZP21U17 T Similar arguments also shows that 1920 192W0 p2Zu1T 619 p2Zw T 619 69 p2Zw 7T So 1920 szwo QB prszivT 99 Z P2w 7T 99 39 39 39 99 ZP2WL7T By 37 dimfpszo QB prsziv dimfpszo 99 ZP2w17T dimp2V Therefore Zp2wT 0 for 239 2 3 So pg 6 annwg FXq2 and hence 12 l p2 Similarly p2 l 12 and hence p2 12 The proof is completed by continuing this process 12 25 Corollary admissible complement Let V be a nite dimensional vector space with dimV n and T E LV7 V be an operator Suppose W is a Tiadmissible subspace of V Then W has a complementary subspace W which is also invariant under T Proof By the above therem 247 we have V W Zw17T a ZwTT The proof is complete by taking W Zw17T 69 69 ZwTT 26 Corollary Cyclicity Let V be a nite dimensional vector space with dimV n and T E LV7 V be an operator Let p be the MMP of T and P be the characteristic polynomial of T 1 There is a vector 10 E V such that annw FXp 2 V is Ticyclic if and only if P p Proof If V 0 then the theorem is obviously true So7 assmue V a By theorem 247 with W0 07 we have VZw17TZwT7T where wk 6 V and annwk FXpk and pk l p194 From the proof of uniqueness part of theorem 247 it follows that FXp1 annw1 f E EX fT 0 annT This means that p1 p is the MMP of T This establishes To prove 27 suppose V Zw7T By theorem 147 the matrix of T7 with respect to a basis E7 is given by the companion matrix of the MMP p of T In this case7 the characteristic polynomial P p the MMP of T Conversely7 suppose p P By 17 there is a a w E V such that annw FXp It follows that dimZw7 T degreep degreeP dimV So7 V ZwT and the proof is complete 27 Generalized CaleyHarnilton Theorem Let V be a nite dimen sional vector space with dimV n and T E LV7 V be an operator Let p be the MMP of T and P be the characteristic polynomial of T 1 Thenp l P 13 2 The prime factors of p and P are same 3 If p q qln is the prime factorization of p then P qfl1qldl where nullity q T degreeq Proof If V 0 then the theorem is obviously true So assmue V a By theorem 24 with W0 0 we have VZw1TZwTT where wk 6 V and annwk FXpk and pk l p194 From the proof of uniqueness part of theorem 24 it follows that FXp1 annw1 f E EX fT 0 annT This means that p1 p is the MMP of T Let T TWWT Again by cyclicity of Zw T both the MMP of Ti and characteristic polynomial of T is pi Hence the characteristic polynomial P of T is given by P p1p2 pT So p p1 l P and 1 is established Since p p1 l P prime factors of p are also the factors of P Also since pk l p1 it follows that the prime factors of P are also the factors of p So 2 is established By an application of Primary Decomposition Theorem see end of Chapter 6 we have if W is the null space of q then VW139VV1 and the MMP of the restriction Ti T WZ is If B is the characteristic polynomial of Ti then by 2 B qfi Since MMP divides B we have 1 Z n Also dimVlQ degreeB degreeq d and dimVlQ Nullity q T So proof of 3 is complete 28 Corollary Let V be a nite dimensional vector space with dimV n and T E LV V be an operator Suppose T a 0 and is nilpotent that means TN 0 for some integer N Z 1 Then characteristic polynomial of T is X 14 Proof Let T 0 and Tm l a 0 Then annT FXXm This needs a small proof7 Which you can do So MMP of T is p Xm By Generalized Caley Hamilton Theorem 277 the characteristic polynomial of T is q XN7 for some N 2 m Since degreeq dimV we have N n So7 the proof is complete 29 Exercise 1 Find a matrix A With characteristic polynomial X2 2 Find a matrix A With characteristic polynomial 1 7 X2 3 Read Exercise 1 3 from page 239 21 Rational Form We Wish to consider the matrix version of the cyclic decomposition theorem 24 First7 we give a de nition 210 De nition Suppose m7 pT E EX are non constant monic poly nomials Let Ai be the companion matric of pi Also assume that P 1lp f0 117r71 Write A1 0 0 0 0 A2 0 0 A 0 0 A3 0 0 0 0 AT In this case7 we say that the matrix A is in rational form 211 Theorem Rational Form Suppose l is a eld as always in this section Let A an n X 72 matrix Then A is similar to a matrix B Which is in rational form More over7 for a given matrix A7 there is only one such matrix B that means B is unique Proof Let T F a F be the operator given by TX AX By cyclic decomposition theoren 247 There are elements 101 wT E F such that 15 1 F decomposes as F Zw1T G9 Zw2T 69 69 ZwTT and 2 if pi is the MMP of annw then p 1lp for 117r71 Now suppose degredpi k and Ai is the companion matrix of pi Let EZ 7Tki7 and E E1UET Then 1 E2 is a basis of Zw 7T and E is a basis of of lF 2 Then Ai matrix of the restriction Timmy With respect to the basis E2 3 So7 the matrix of T With respect to E is A1 0 0 0 0 A2 0 0 B 0 0 A3 0 0 0 0 AT Clearly7 B is in Jordan form and A is similar to B So7 it only remains to establish the uniqueness part Suppose A is similar to another matric C Which is in rational form So7 A PCP l for some matric P E GLnlF Let E0 61776n be the standard basis and 617 76n 617 7enP Then T 177T 1 177 n0 This equation gives a cyclic decomposition of lF Now the uniqueness follows from the uniqueness part of the cyclic decomposition theorem 24 This completes the proof of this theorem 16 3 The Jordan Form 31 Facts Let l be a eld and V be a nite dimensional vector space With dimV 71 Suppose N E LV7 V be a linear operator By cuclic decomposition theorem 247 there are elments 1017 wT E V such that 1 10 P17 V decomposes as VZw17N aZw27N aZwT7N lf pi is the MMP of annw then pk l p19L for k 277 7pT are unique Now assume that N is nilpotent Since N is nilpotent7 let Nk 0 and N19quot1 a 0 So7 MMP of N is X Note pl 6 annN lFXXk and Xk E annw1 lFXp1 Therefore P1 Xk 8071 X with k1 k 2 kg 2 2 kT 2 1 and k1k2mm n The rational form of N is determined by k1 k 2 kg 2 2 CT 2 1 The companion matrix of X1 i is the ki gtlt ki matrix 0 0 0 0 0 1 0 0 0 0 Ai 0 1 0 0 0 0 0 0 1 0 Let E W Nw 7 7 Nki 1wi and E E1 u u E Then E2 is a basis of Zw 7 N and E is a basis of V With respect to the basis E the matrix of N is A1 0 0 0 0 0 A2 0 0 0 0 0 A3 0 0 0 0 0 0 AT 17 11 In fact nullityN 7 ln deed Nk1 1w Nk2 1w2 Nkr 1w forms a basis of Null space of N Proof Let N be the null space of N and W SpanNk1 1wiNk2 1w2 N1 1w Clearly Nki lhui E N and hence W E N To see the converse let 1 E N From the decomposition v f1Nw1 f2Nw2 39 39 39 fTNwT for some f E EX We can assume degredfi S k 7 1 or f 0 So 0 Nltvgt Nf1N1U1 Nf2Nwz WWW Therefore from decomposition Nf Nw 0 So Xfi E annw 1FXin Therefore f giin l for some 9239 E EX From degree consideration f Ciin l Where Ci E 1 Therefore 1 ClNkrlwl CgNk2 1w2 CTNkT lhuT is in the spanSo N E W Before we proceed we de ne elementary Jordan matrices 32 De nition Let 1 be a eld and C E 1 De ne the n X 72 matrix C 0 0 0 0 1 C 0 0 0 JJCJCn 0 1 C 0 0 0 0 0 1 C This matrix is called an elementary Jordan matrix With eigen value C 18 Following is the jordan decomposition theorem 33 Theorem Jordan Decomposition Let 1 be a eld and V be vec tor space over 1 Suppose T E LV7V be a linear operator on V Assume tha characteristic polynomial q of T factorizes completely as q X 7 cld1 X 7 02d2 X 7 ck where 017 70k are the distinct eigen values 1 In this case7 the MMP p of T is given by p X 7 cl X 7 02 X 7 ck where 1 S n 3 di 2 Let VVZ be the null space of T 7 Ci 3 By primary decomposition theorem V W1 W2 6 Wk and MMP of the restriction Ti T WZ is X 7 ck 4 Write Ni 7 Ci 6 LVQ7 Then Ni is nilpotent The matrix of Ni was computed in 10 of Facts 31 So there is a basis E2 of VVi and the matrix of Ni is given by A1 0 0 0 0 0 A2 0 0 0 0 0 A3 0 0 0 0 0 0 Ali where A has the form as in 8 of Facts 31 5 Now since Ti Ci Ni the matrix of Ti with respect to E2 is given by J1 0 0 0 0 0 J2 0 0 0 Bi 0 0 J3 0 0 0 0 0 0 J12 where Jj Ci A are elementary Jordan matrices with eigen values Ci 19 6 Now E E1 U U Ek is a basis of V With respect to E the matrix of T is given by B1 0 0 0 0 0 B2 0 0 0 B 0 0 B3 0 0 0 0 0 0 Bk Where Bi are as in 7 Here are a few de nitions a The matrix Bi is called the Jordan block matrix of T7 corre sponding to the eigen value Ci b The matrix B above 6 is the called the Jordan matrix of T c Any matrix of the form in 6 is said to be in Jordan form 34 Theorem Main Jordan Decomposition Let l be a eld and V be vector space over F Suppose T E LV7 V be a linear operator on V Assume tha characteristic polynomial q of T factorizes completely as q X Cld1X 02W 39 39 39 X 7 Ckdk Where 017 70k are the distinct eigen values Then T has Jordan matrix representation With respect to some basis E Also note that there is one Jordan block Bi corresponding to each eigen value Ci and Bi is a di gtlt di matrix The textbook talks about uniqueness that feel is unnecessary and we will skip 35 Exercise Suppose A is a commutative ring and a E A is a nilpotent element Then 1 a is an unit Likewise7 c is an unit for any unit 0 E A 20
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'