M. Sc. MATHEMATICS MAL-521 (ADVANCE ABSTRACT ALGEBRA)

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1 . Sc. ATHEATICS AL-52 (ADVANCE ABSTRACT ALGEBRA) Lesso No &Lesso Name Wrer Veer Lear Trasformaos Dr. Pakaj Kumar Dr. Nawee Hooda 2 Caocal Trasformaos Dr. Pakaj Kumar Dr. Nawee Hooda 3 odules I Dr. Pakaj Kumar Dr. Nawee Hooda 4 odules II Dr. Pakaj Kumar Dr. Nawee Hooda DIRECTORATE OF DISTANCE EDUCATIONS GURU JABHESHWAR UNIVERSITY OF SCIENCE & TECHNOLOGY HISAR 25

2 AL-52:. Sc. ahemacs (Algebra) Lesso No. Lesso: Lear Trasformaos Wre by Dr. Pakaj Kumar Veed by Dr. Nawee Hooda STRUCTURE. OBJECTIVE. INTRODUCTION.2 LINEAR TRANSFORATIONS.3 ALGEBRA OF LINEAR TRANSFORATIONS.4 CHARACTERISTIC ROOTS.5 CHARACTERISTIC VECTORS.6 ATRIX OF TRANSFORATION.7 SIILAR TRANSFORATIONS.8 CANONICAL FOR(TRIANGULAR FOR).9 KEY WORDS. SUARY. SELF ASSESENT QUESTIONS.2 SUGGESTED READINGS. OBJECTIVE Objecve of hs Chaper s o sudy Lear Trasformao o he fe dmesoal vecor space V over he feld F.. INTRODUCTION Le U ad V be wo gve fe dmesoal vecor spaces over he same feld F. Our eres s o fd a relao (geerally called as lear rasformao) bewee he elemes of U ad V whch sasfes cera codos ad, how hs relao from U o V becomes a vecor space over he feld F. The se of all rasformao o U o self s of much eres. O fe dmesoal vecor space V over F, for gve bass of V, here always exs a marx ad for gve bass ad gve marx of order here always exs a lear rasformao. I hs Chaper, Seco.2, we sudy abou lear rasformaos. I Seco.3, Algebra of lear rasformaos s suded. I ex wo

3 secos characersc roos ad characersc vecors of lear rasformaos are suded. I Seco.6, marx of rasformao s suded. I Seco.7 caocal rasformaos are suded ad las seco we come o kow abou caocal form (Tragular form)..2 LINEAR TRANSFORATIONS.2. Defo. Vecor Space. Le F be a feld. A o empy se V wh wo bary operaos, addo (+)ad scalar mulplcaos(. ), s called a vecor space over F f V s a abela group uder + ad for followg codos are also sasfed: () α. (v+w) = αv+ αw for all α F ad v, w V, (2) ( α + β).v = αv+β v, (3) (αβ).v = (4).v = v α.( β v) v V, α. v V. The For all α, β F ad v, w belogg o V. Here v ad w are called vecors ad α, β are called scalar..2.2 Defo. Homomorphsm. Le V ad W are wo vecor space over he same feld F he he mappg T from V o W s called homomorphsm f () (v +v 2 )T= v T+v 2 T () (αv )T= α(v T) for all v, v 2 belogg o V ad α belogg o F. Above wo codos are equvale o (αv +βv 2 )T=α(v T)+ β(v 2 T). If T s oe-oe ad oo mappg from V o W, he T s called a somorphsm ad he wo spaces are somorphc. Se of all homomorphsm from V o W s deoed by Hom(V, W) or Hom R (V, W).2.3 Defo. Le S ad T Hom(V, W), he S+T ad λs s defed as: () v(s+t)= vs+vt ad () v(λs)= λ(vs) for all v V ad λ F.2.4 Problem. S+T ad λs are elemes of Hom(V, W).e. S+T ad λs are homomorphsms from V o W.

4 Proof. For () we have o show ha (αu+βv)(s+t)= α(u(s+t))+ β(v(s+t)) By Defo.2.3, (αu+βv)(s+t)=(αu+βv)s+(αu+βv)t. Sce S ad T are lear rasformaos, herefore, (αu+βv)(s+t)=α(us)+β(vs)+α(ut)+β(vt) =α((us)+α(ut))+β((vs)+(vt)) Aga by defo.2.3, we ge ha (αu+βv)(s+t)=α(u(s+t))+β(v(s+t)). I proves he resul. () Smlarly we ca show ha (αu+βv)(λs)=α(u(λs))+β(v(λs)).e. λs s also lear rasformao..2.5 Theorem. Prove ha Hom(V, W) becomes a vecor space uder he wo operao operaos v(s+t)= vs + vt ad v(λs)= λ(vs) for all v V, λ F ad S, T Hom(V, W). Proof. As s clear ha boh operaos are bary operaos o Hom(V, W). We wll show ha uder +, Hom(V,W) becomes a abela group. As Hom(V,W) such ha v= v V( s call zero rasformao), herefore, v(s+)= vs+v = vs = +vs= v+vs= v(+s) v V.e. dey eleme exss Hom(V, W). Furher for S Hom(V, W), here exs -S Hom(V, W) such ha v(s+(-s))= vs+v(-s)= vs-vs== v v V.e. S+(-S)=. Hece verse of every eleme exs Hom(V, W). I s easy o see ha T +(T 2 +T 3 )= (T +T 2 )+T 3 ad T +T 2 = T 2 +T T, T 2, T 3 Hom(V, W). Hece Hom(V, W) s a abela group uder +. Furher s easy o see ha for all S, T Hom(V, W) ad α, β F, we have α(s+t)= αs+αt, (α+β)s= αs+βs, (αβ)s= α(βs) ad.s=s. I proves ha Hom(V, W) s a vecor space over F..2.6 Theorem. If V ad W are vecor spaces over F of dmesos m ad respecvely, he Hom(V, W) s of dmeso m over F. Proof. Sce V ad W are vecor spaces over F of dmesos m ad respecvely, le v, v 2,, v m be bass of V over F ad w, w 2,, w be bass

5 of W over F. Sce v = δv + δ2v2 + + δmvm where deermed for v V. Le us defe T j from V o W by v T j = δ w j.e. v w j T kj = δ F are uquely f = k. I s easy o see ha T j f k Hom(V,W). Now we wll show ha m elemes T j m ad j form he bass for Hom(V, W). Take β T + β2t2 + + β T + + βt + β2t2 + + βt + + β T + β T + + β T = m m m2 m2 m (Sce a lear rasformao o V ca be deermed compleely f mage of every bass eleme of s deermed) v ( β T + β2t2 + + β T + + βt + β2t2 + + βt + m + β T + β T + + β T )=v = m m m2 m2 w j f = k β w + β2w2 + + βw = ( v T kj = ) f k m Bu w, w 2,, w are learly depede over F, herefore, β = β2 = = β =. Ragg m, we ge each β j =. Hece T j are learly depede over F. Now we clam ha every eleme of Hom(V,W) s lear combao of T j over F. Le S Hom(V,W) such ha Take S m vs = αw + α2w2 + + α w, vs = αw + α2w2 + + αw vms = αmw + αm2w αmw. αt + α2t2 + + α T + + αt + α2t2 + + αt = + α mtm + αm2tm2 + + αmtm.the v S = v( α T + α2t2 + + α T + + αt + α2t2 + + αt + α mtm + αm2tm2 + + αmtm ) = w + α2w2 + + αw α = v S. Smlarly we ca see ha v S = v S for every, m. Therefore, vs = vs v V. Hece S =S. I shows ha every eleme of Hom(V,W) s a lear combao of T j over F. I proves he resul.

6 .2.7 Corollary. If dmeso of V over F s, he dmeso of Hom(V,V) over F = 2 ad dmeso of Hom(V,F) s over F..2.8 Noe. Hom(V, F) s called dual space ad s elemes are called lear fucoal o V o F. Le v, v 2,, v be bass of V over F he defed by vˆ (v j) = vˆ, vˆ 2,, vˆ f = j are lear fucoals o V whch acs as f j bass elemes for V. If v s o zero eleme of V he choose v =v, v 2,, v as he bass for V. The here exs vˆ (v) = vˆ (v) =. I oher words we have show ha for gve o zero vecor v V we have a lear rasformao f(say) such ha f(v)..3 ALGEBRA OF LINEAR TRANSFORATIONS.3. Defo. Algebra. A assocave rg A whch s a vecor space over F such ha α(ab)= (αa)b= a(αb) for all a, b A ad α F s called a algebra over F..3.2 Noe. I s easy o see ha se of all Hom(V, V) becomes a algebra uder he mulplcao of S ad T Hom(V, V) defed as: v(st)= (vs)t for all v V. we wll deoe Hom(V, V)=A(V). If dmeso of V over F.e. dm F V=, he dm F A(V)= 2 over F..3.3 Theorem. Le A be a algebra wh u eleme ad dm F A=, he every eleme of A sasfes some polyomal of degree a mos. I parcular f dm F V=, he every eleme of A(V) sasfes some polyomal of degree a mos 2. Proof. Le e be he u eleme of A. As dm F A=, herefore, for a A, he + elemes e, a, a 2,,a are all A ad are learly depede over F,.e. here exs β, β,, β F, o all zero, such ha β e+β a+ + β a =. Bu he a sasfes a polyomal β +β x+ + β x over F. I proves he resul. Sce he dm F A(V)= 2, herefore, every eleme of A(V) sasfes some polyomal of degree a mos 2.

7 .3.4 Defo. A eleme T A(V) s called rgh verble f here exs S A(V) such ha TS=I. Smlarly ST=I (Here I s dey mappg) mples ha T s lef verble. A eleme T s called verble or regular f boh rgh as well as lef verble. If T s o regular he s called sgular rasformao. I may be ha a eleme of A(V) s rgh verble bu o lef. For example, Le F be he feld of real umbers ad V be he space of all polyomal x over F. Defe T o V by f (x)s = x f (x)dx df (x) f (x)t = ad S by dx. Boh S ad T are lear rasformaos. Sce f (x)(st) f (x).e. ST I ad f (x)(ts) = f (x).e. TS =I. Here T s rgh verble whle s o lef verble..3.5 Noe. Sce T A(V) sasfes some polyomal over F, he polyomal of mmum degree sasfed by T s called he mmal polyomal of T over F.3.6 Theorem. If V s fe dmesoal over F, he T A(V) s verble f ad oly f he cosa erm of he mmal polyomal for T s o zero. Proof. Le p(x)= β +β x+ + β x, β, be he mmal polyomal for T over F. Frs suppose ha β, he = p(t)= β +β T+ + β T mples ha -β I=T(β T+ +β T - ) or β β β β I = T( ) ( β β T = T ) T. β β β β T T β β β Therefore, S ( β β T = T ) s he verse of T. β β β Coversely suppose ha T s verble, ye β =. The β T+ + β T = (β T+ + β T - )T=. As T s verble, o operag T - o boh sdes of above equaos we ge (β T+ + β T - )=.e. T sasfes a polyomal of degree less he he degree of mmal polyomal of T, coradcg o our assumpo ha β =. Hece β resul.. I proves he

8 .3.7 Corollary. If V s fe dmesoal over F ad f T A(V) s sgular, he here exs o zero eleme S of A(V) such ha ST=TS=. Proof. Le p(x)= β +β x+ + β x, β be he mmal polyomal for T over F. Sce T s sgular, herefore, cosa erm of p(x) s zero. Hece (β T+ + β T - )T=T(β T+ + β T - )=. Choose S=(β T+ + β T - ), he S (f S=, he T sasfes he polyomal of degree less ha he degree of mmal polyomal of ) fulfll he requreme of he resul..3.8 Corollary. If V s fe dmesoal over F ad f T belogg o A(V) s rgh verble, he s lef verble also. I oher words f T s rgh verble he s verble. Proof. Le U A(V) be he rgh verse of T.e. TU=I. If possble suppose T s sgular, he here exs o-zero rasformao S such ha ST=TS=. As S(TU)= (ST)U SI=U S=, a coradco ha S s o zero. Ths coradco proves ha T s verble..3.9 Theorem. For a fe dmesoal vecor space over F, T A(V) s sgular f ad oly f here exs a v V such ha vt=. Proof. By Corollary.3.7, T s sgular f ad oly f here exs o zero eleme S A(V) such ha ST=TS=. As S s o zero, herefore, here exs a eleme u V such ha us. ore over =u=u(st)=(us)t. Choose v=us, he v ad vt=. I prove he resul..4 CHARACTERISTIC ROOTS I res of he resuls, V s always fe dmesoal vecor space over F..4. Defo. For T A(V), λ F s called Characersc roo of T f λi-t s sgular where I s dey rasformao A(V). If T s sgular, he clearly s characersc roo of T..4.2 Theorem. The eleme λ F s called characersc roo of T f ad oly here exs a eleme v V such ha vt=λv.

9 Proof. Sce λ s characersc roo of T, herefore, by defo he mappg λi-t s sgular. Bu he by Theorem.3.9, λi-t s sgular f ad oly f v(λi-t)= for some v V. As v(λi-t)= vλ-vt= vt= λv. Hece λ F s characersc roo of T f ad oly here exs a eleme v V such ha vt=λv..4.3 Theorem. If λ F s a characersc roo of T, he for ay polyomal q(x) over F[x], q(λ) s a characersc roo of q[t]. Proof. By Theorem.4.2, f λ F s characersc roo of T he here exs a eleme v V such ha vt=λv. Bu he vt 2 =(vt)t=(λv)t=λλv= λ 2 v..e. vt 2 =λ 2 v. Coug hs way we ge, vt k =λ k v. Le q(x)=β +β x+ + β x, he q(t)= β +β T+ + β T. Now by above dscusso, vq(t)=v(β +β T+ + β T )= β v+β (vt)+ + β (vt )= β v+β λ 2 v + + β λ v = (β +β λ β λ )v=q(λ)v. Hece q(λ) s characersc roo of q(t)..4.4 Theorem. If λ s characersc roo of T, he λ s a roo of mmal polyomal of T. I parcular, T has a fe umber of characersc roos F. Proof. As we kow ha f λ s a characersc roo of T, he for ay polyomal q(x) over F, here exs a o zero vecor v such ha vq(t)=q(λ)v. If we ake q(x) as mmal polyomal of T he q(t)=. Bu he vq(t)=q(λ)v q(λ)v=. As v s o zero, herefore, q(λ)=.e. λ s roo of mmal polyomal of T..5 CHARACTERISTIC VECTORS.5. Defo. The o zero vecor v V s called characersc vecor belogg o characersc roo λ F f vt=λv..5.2 Theorem. If v, v 2,,v are dffere characersc vecors belogg o dsc characersc roos λ, λ 2,, λ respecvely, he v, v 2,,v k are learly depede over F.

10 Proof. Le f possble v, v 2,,v are learly depede over F, he here exs a relao β v + + β v =, where β,+ + β are all F ad o all of hem are zero. I all such relao, here s oe relao havg as few o zero coeffce as possble. By suably reumberg he vecors, le us assume ha hs shores relao be β v + + β k v k =, where β,, β k. () Applyg T o boh sdes ad usg v T=λ v () we ge λ β v + + λ k β k v k = () ulplyg () by λ ad subracg from (), we oba (λ 2 -λ )β 2 v (λ k -λ )β k v k = Now (λ -λ ) for > ad β 2, herefore, (λ -λ )β. Bu he we oba a shorer relao ha ha () bewee v, v 2,,v. Ths coradco proves he heorem..5.3 Corollary. If dm F V=, he T A(V) ca have a mos dsc characersc roos F. Proof. Le f possble T has more ha dsc characersc roos F, he here wll be more ha dsc characersc vecors belogg o hese dsc characersc roos. By Theorem.5.2, hese vecors wll be learly depede over F. Sce dm F V=, hese + eleme wll be learly depede, a coradco. Ths coradco proves T ca have a mos dsc characersc roos F..5.4 Corollary. If dm F V= ad T A(V) has dsc characersc roos F. The here s a bass of V over F whch cosss of characersc vecors of T. Proof. As T has dsc characersc roos F, herefore, characersc vecors belogg o hese characersc roos wll be learly depede over F. As we kow ha f dm F V= he every se of learly depede vecors acs as bass of V(prove ). Hece se of characersc vecors wll ac as bass of V over F. I proves he resul. Example. If T A(V) ad f q(x) F[x] s such ha q(t)=, s rue ha every roo of q(x) F s a characersc roo of T? Eher prove ha hs s rue or gve a example o show ha s false.

11 Soluo. I s o rue always. For ake V, a vecor space over F wh dm F V=2 wh v ad v 2 as bass eleme. I s clear ha for v V, we have uque α, β F such ha v=αv +βv 2. Defe a rasformao T A(V) by v T=v 2 ad v 2 T=. le λ be characersc roo of T F, he λi-t s sgular. I mea here exs a vecor v( ) V such ha vt=λv (αv +βv 2 )T=λαv +λβv 2 α(v T)+β(v 2 T)=λαv +λβv 2 αv 2 +β.=λαv +λβv 2. As v s ozero vecor, herefore, a leas oe of α or β s ozero. Bu he αv 2 +β.=λαv +λβv 2 mples ha λ=. Hece zero s he oly characersc roo of T F. If We ake a polyomal q(x)=x 2 (x-), he q(t)=t 2 (T-I). Now v q(t)= ((v T)T)(T-I) =(v 2 T)(T-I)=(T-I)=, v 2 q(t)= ((v 2 T)T)(T-I) =(T)(T-I)=, herefore, vq(t)= v V. Hece q(t)=. As every roo of q(x) les F ye every roo of T s o a characersc roo of T. Example. If T A(V) ad f p(x) F[x] s he mmal polyomal for T over F, suppose ha p(x) has all s roos F. Prove ha every roo of p(x) s a characersc roo of T. Soluo. Le p(x)= x + β x - + +β be he mmal polyomal for T ad λ be s roo. The p(x)= (x-λ)(x - + γ x γ ). Sce p(t)=, herefore, (T-λ)(T - + γ T γ )=. If (T-λ) s regular he (T - +γ T γ )=, coradcg he fac ha he mmal polyomal of T s of degree over F. Hece (T-λ) s o regular.e. (T-λ) s sgular ad hece here exs a o zero vecor v V such ha v(t-λ)=.e. vt=λv. Cosequely λ s characersc roo of T..6 ATRIX OF TRANSFORATIONS.6. Noao. The marx of T uder gve bass of V s deoed by m(t). We kow ha for deermg a rasformao T A(V) s suffce o fd ou he mage of every bass eleme of V. Le v, v 2,,v be he bass of V over F ad le v T = αv + α2v2 + + α v vt = αv + α2v2 + + αv

12 vt = αv + α2v2 + + αv The marx of T uder hs bass s α m(t)= α α α α α α. α α Example. Le F be he feld ad V be he se of all polyomals x of degree - or less. I s clear ha V s a vecor space over F. The dmeso of hs vecor space s. Le {, x, x 2,, x - } be s bass. For β +β x+ + β - x - V, Defe (β +β x+ + β - x - )D=β +2β 2 x β - x -2. The D s a lear rasformao o V. Now we calculae he marx of D uder he bass v (=), v 2 (=x), v 3 (=x 2 ),.., v (=x - ) as: v D=D==.v +.v v v 2 D=xD==.v +.v v v 3 D=x 2 D=2x=.v + 2.v + +. v D= x - D=x - = 2 v.v + v D= x - D=-x -2 = The marx of D s m(d)=. +.v2 + v +. v.v +.v2 + + ( )v +. v Smlarly we ake aoher bass v (=x - ), v 2 (=x -2 ),, v (=), he marx of D uder hs bass s

13 m (D)= 2 3. If we ake he bass v (=), v 2 (=+x), v 3 (=+x 2 ),.., v (=+x - ) he he marx of D uder hs bass s obaed as: v D=D==.v +.v v v 2 D=(+x)D==.v +.v v v 3 D=(+x 2 )D=2x=-2+2(+x)= 2.v + 2.v + +. v D=x - D=-x -2 =-(-)+-(+x -2 )= The marx of D s 2 m 3 (D)= 3. ( 2) ( ) 2. 2 v ( ).v + + ( )v +. v Theorem. If V s dmesoal over F ad f T A(V) has a marx m (T) he bass v, v 2,,v ad he marx he bass he bass w, w 2,,w of V over F. The here s a eleme C F such ha m 2 (T)= Cm (T)C -. I fac C s marx of rasformao S A(V) where S s defed by v S=w ;. Proof. Le m (T)=(α j ), herefore, for, v T=α v +α 2 v 2 + +α v = j= Smlarly, f m 2 (T)=(β j ), herefore, for, w T=β w +β 2 w 2 + +β w = β α j= jv j () jw j (2)

14 Sce v S=w, he mappg oe oe ad oo. Usg v S=w (2) we ge v ST=β (v S)+β 2 (v 2 S)+ +β (v S) =(β.v +β 2 v 2 + +β v )S As S s verble, herefore, o applyg S - o boh sdes of above equao we ge v (STS - )=(β.v +β 2 v 2 + +β v ). The by defo of marx we ge m (STS - )=(β j )= m 2 (T). As he mappg T m(t) s a somorphsm from A(V) o F, herefore, m (STS - )= m (S)m (T)m (S - )= m (S)m (T)m (S) - = m 2 (T). Choose C= m (S), he he resul follows. Example. Le V be he vecor space of all polyomal of degree 3 or less over he feld of reals. Le T A(V) s defed as: (β +β x+β 2 x 2 +β 3 x 3 )T =β +2β 2 x+3β 3 x 2. The D s a lear rasformao o V. The marx of T he bass v (=), v 2 (=x), v 3 (=x 2 ), v 4 (=x 3 ) as: v T=T==.v +.v2 + v3 +. v4 v 2 T=xT ==.v +.v2 + v3 +. v4 v 3 T=x 2 T=2x=.v + 2.v2 + v3 +. v4 v 4 T= x 3 T=3x 2 =.v +.v2 + 3v3 +. v4 The marx of s m (D)= 2 3 Smlarly marx of T he bass w (=), w 2 (=+x), w 3 (=+x 2 ), w 4 (=+x 3 ), s m 2 (D)= 2 3 If We se v S=w, he 2 3. v S= w = =.v +.v2 + v3 +. v4 v 2 S=w 2 = +x=.v +.v2 + v3 +. v4 v 3 S=w 3 =+x 2 =.v +.v2 + v3 +. v4 v 4 T= w 4 =+x 3 =.v +.v2 + v3 +. v4

15 Bu he C=m(S)= ad C - = ad Cm (D)C - = 3 2 = =m 2 (D) as requred..6.3 Noe. I above example we see ha for gve bass of V here always exs a square marx of order equal o he dm F V. Coverse par s also rue..e. for gve bass ad gve marx here always exs a lear rasformao. Le V be he vecor space of all -uples over he feld F, he F he se of all marx s a algebra over F. I fac f v =(,,,), v 2 =(,,,),, v =(,,,), he (α j ) F acs as: v (α j )= frs row of (α j ),, v (α j )= h row of (α j ). We deoe s a square marx of order such ha s each super dagoal ery s oe ad he res of he eres are zero. For example 3 = 3 3 ad 4 = SIILAR TRANSFORATIONS..7. Defo (Smlar rasformaos). Trasformaos S ad T belogg o A(V) are sad o smlar f here exs R A(V) such ha RSR - =T..7.2 Defo. A subspace W of vecor space V s vara uder T A(V) f WT W. I oher words wt W w W..7.3 Theorem. If subspace W of vecor space s vara uder T, he T duces a lear rasformao T o W V, defed by W vt W)T v ( + = +. Furher f T sasfes he polyomal q(x) over F, he so does T.

16 Proof. Sce he elemes of W V are he coses of W V, herefore, T defed by defed as herefore, V ( v + W)T = vt + W s a mappg o. The mappg s well W v W = v W v W. Sce W s vara uder T, v 2 v + W = v2 + W ( v v2)t W whch furher mples ha T + W = v T W.e. ( v + W)T = (v2 + W) T. Furher v 2 + ( α(v + W) + β(v2 + W))T = (( αv + βv2) + W))T = ( αv + βv2)t + W. Sce T s lear rasformao, herefore, α v + βv )T + W = α(v T) + (v T) ( 2 β 2 + W = α(v T) + β(v2t) + W = α(vt + W) + β(v2t + W) = α( v + W) T V + β( v 2 + W)T.e. T s a lear rasformao o. W Now we wll show ha for gve polyomal q(x) over F, ( 2 2 V q (T) = q(t). For gve eleme v+w of, v + W)T = vt + W W V = ( vt)t + W = (vt + W)T = (v + W)TT = (v + W)T v+w..e. W T T =. Smlarly we ca see ha T T =. If q(x) + αx + + αx = α, he q(t) + αt + + αt = α ad (v + W)q(T) = (v + W)( α + α T + + T ) = v( α + α T + + α T ) W α + v = α + W + α (vt + W) + + α (vt + W) = α (v + W) + α (v + W) T + α (v + W) T +. Usg T T = we ge ( v + W)q(T) = α (v + W) + α ( v + W)T + + α (v + W) T = (v + W)( α + αt + + αt ) = ( v + W)q(T).e. q (T) = q(t). Sce by gve codo q(t)=, herefore, = q (T) = q(t). Hece T sasfes he same polyomal as sasfed by T..7.4 Corollary. If subspace W of vecor space s vara uder T, he T duces a lear rasformao T o W V, defed by ( v + W)T = vt + W ad

17 mmal polyomal p (x)(say) of T dvdes he mmal polyomal p(x) of T. Proof. Sce p(x) s mmal polyomal of T, herefore, p(t)=. Bu he by Theorem.7.3, p( T)=. Furher, p (x) s mmal polyomal of T, herefore, p (x) dvdes p(x)..8 CANONICAL FOR(TRIANGULAR FOR).8. Defo. Le T be a lear rasformao o V over F. The marx of T he bass v,v2,, v s called ragular f vt = α v, v2t = α2 v + α22 v2. vt = α v + α2 v2 + α v.... vt = α v + α2 v2 + α v.8.2 Theorem. If T A(V) has all s characersc roos F, he here exs a bass of V whch he marx of T s ragular. Proof. We wll prove he resul by duco o dm F V=. Le =. By Corollary.5.3, T has exacly oe dsc roo λ(say) F. Le v( ) be correspodg characersc roo V. The vt= λv. Sce =. ake {v} as a bass of V. Now he marx of T hs bass s [λ]. Hece he resul s rue for =. Choose > ad suppose ha he resul holds for all rasformaos havg all s roos F ad are defed o vecor space V* havg dmeso less he. Sce T has all s characersc roos F; le λ be he roo characersc roos F ad v be he correspodg characersc vecor. Hece v T=λ v. Choose W={αv α F}. The W s oe dmesoal subspace of V. Sce (αv )T=α(v T)= αλ v W, herefore, W s vara uder T. Le V Vˆ =. The Vˆ s a subspace of V such ha dm F Vˆ = dm F V- W

18 dm F W=-. By Corollary.7.4, all he roos of mmal polyomal of duced rasformao T beg he roos of mmal polyomal of T, les F. Hece he lear rasformao T s aco o Vˆ sasfes hypohess of he heorem. Furher dm F Vˆ <, here fore by duco hypohess, here s a bass ( = v W), ( = v W),, v ( = v W) of Vˆ over F such ha v2 2 + v v v T α22v2 =, 3 T α32v2 + α33 v3 =,... v T = α 2 v 2 + α 3 v α... v T = α.e marx of s ragular Take a se B={ v 2 v,v2,, v 2 + α 3 v 3 + v + + α v }. We wll show ha B s he requred bass whch fulflls he requreme of he heorem. As he mappg V Vˆ defed by v v ( = v + W) v V s a oo homomorphsm uder whch v 2, v 3,, v are he mages of v 2, v 3,, v respecvely. Sce v 2, v 3,, v are learly depede over F, he here pre-mage vecors.e. v 2, v 3,, v are also learly depede over F. ore over v ca o be leal combao of vecors v 2, v 3,, v because f s so he v 2, v 3,, v wll be learly depede over F. Hece he vecors v, v 2,, v are learly depede vecors over F. Choose hs se as he bass of V. Sce v T=λ v = =α v for α =λ. Sce v2 T = α22v2 or ( v2 + W) T = α22v2 + W or v2 T + W = α22v2 + W. Bu he v2t α22v2 W ad hece v2t α22v2 = α2v Smlarly v 3 T α32v2 + α33 v3. Equvalely, v2t = α2v + α22v2. = v3t = α3v + α32v2 + α33v3. Coug hs way we ge ha v T = α 2 v 2 + α 3 v α v

19 v T = α v + α v + + α v for all,. 2 2 Hece B={v, v 2,, v } s he requred bass whch he marx of T s ragular..8.3 Theorem. If he marx A F (=se of all order square marces over F) has all s characersc roos F, he here s a marx C F such ha CAC - s a ragular marx. Proof. Le A=[a j ] F. Furher le F = {(α, α 2,,α ) α F} be a vecor space over F ad e, e 2,, e be a bass of bass of V over F. Defe T:V V by e T = ae + a2e2 + + ae + + ae. The T s a lear rasformao o V ad he marx of T hs bass s m (T)= [a j ]=A. Sce he mappg A(V) F defed by T m (T) s a algebra somorphsm, herefore all he characersc roos of A are F. Equvalely all he characersc roo of T are F. Therefore, by Theorem.8.2, here exs a bass of V whch he marx of T s ragular. Le be m 2 (T). By Theorem.6.3, here exs a verble marx C F such ha m 2 (T)= Cm (T)C - = CAC -. Hece CAC - s ragular..8.4 Theorem. If V s dmesoal vecor space over F ad le he marx A F has dsc characersc roos F, he here s a marx C F such ha CAC - s a dagoal marx. Proof. Sce all he characersc roos of marx A are dsc, he lear rasformao T correspodg o hs marx uder a gve bass, also has dsc characersc roos say λ, λ 2,, λ F. Le v, v 2,, v be he correspodg characersc vecors V. Bu he vt = λv () We kow ha vecors correspodg o dsc characersc roo are learly depede over F. Sce hese are learly depede vecors over F ad dmeso of V over F s, herefore, se B={ v, v 2,, v } ca be ake as bass se of V over F. Now he marx of T hs bass s

20 λ λ2. Now By above Theorem, here λ λ exs C F such ha CAC - = λ2 λ s dagoal marx..8.5 Theorem. If V s dmesoal vecor space over F ad T A(V) has all s characersc roos F, he T sasfes a polyomal of degree over F. Proof. By Theorem.8.3, we ca fd ou a bass of V whch marx of T s ragular.e. we have a bass v, v 2,, v of V over F such ha v T = λv v2t = α2v + λ2v2 v T = αv + α2 v2 + + α( ) v + λv v T = αv + α2 v2 + + α ( ) v + λ v Equvalely, v (T λ) = v2(t λ2) = α2v.. v (T λ) = αv + α2 v2 + + α ( ) v Take he rasformao v (T λ ) = αv + α2 v2 + + α ( ) v. S= T λ )(T λ )(T ). ( 2 λ The v S= v (T λ )(T λ )(T λ ) = (T λ )(T λ ) 2 2 = v 2 S= v (T λ )(T λ )(T λ ) = v (T λ )(T λ )(T ) λ = α v (T λ )(T λ ). 2 =

21 Smlarly we ca see ha v S= for. Equvalely, vs= v V. Hece S= T λ )(T λ )(T ) =.e. S s zero rasformao o V. ( 2 λ Cosequely T sasfes he polyomal x λ )(x λ )(x ) of degree ( 2 λ over F..9 KEY WORDS Trasformaos, smlar rasformaos, characersc roos, caocal forms.. SUARY I hs chaper, we sudy abou lear rasformaos, Algebra of lear rasformaos, characersc roos ad characersc vecors of lear rasformaos, marx of rasformao ad caocal form (Tragular form).. SELF ASSESENT QUESTIONS () If V s a fe dmesoal vecor space over he feld of real umbers wh bass v ad v 2. Fd he characersc roos ad correspodg characersc vecors for T defed by () v T = v + v 2, v 2 T = v - v 2 () v T = 5v + 6v 2, v 2 T = -7v 2 () v T = v + 2v 2, v 2 T = 3v + 6v 2 (2) If V s wo-dmesoal vecor space over F, prove ha every eleme A(V) sasfes a polyomal of degree 2 over F.2 SUGGESTED READINGS: () Topcs Algebra; I.N HERSTEIN, Joh wley ad sos, New York. (2) oder Algebra; SURJEET SINGH ad QAZI ZAEERUDDIN, Vkas Publcaos. (3) Basc Absrac Algebra; P.B. BHATTARAYA, S.K.JAIN, S.R. NAGPAUL, Cambrdge Uversy Press, Secod Edo.

22 AL-52:. Sc. ahemacs (Advace Absrac Algebra) Lesso No. 2 Wre by Dr. Pakaj Kumar Lesso: Caocal forms Veed by Dr. Nawee Hooda STRUCTURE 2. OBJECTIVE 2. INTRODUCTION 2.2 NILPOTENT TRANSFORATION 2.3 CANONICAL FOR(JORDAN FOR) 2.4 CANONICAL FOR( RATIONAL FOR) 2.5 KEY WORDS 2.6 SUARY 2.7 SELF ASSESENT QUESTIONS 2.8 SUGGESTED READINGS 2. OBJECTIVE Objecve of hs Chaper s o sudy Nlpoe Trasformaos ad caocal forms of some rasformaos o he fe dmesoal vecor space V over he feld F. 2. INTRODUCTION Le T A(V), V s fe dmesoal vecor space over F. I frs chaper, we see ha every T sasfes some mmal polyomal over F. If T s lpoe rasformao o V, he all he characersc roo of T les F. Therefore, here exss a bass of V uder whch marx of T has ce form. Some me all he roo of mmal polyomal of T does o les F. I ha case we sudy, raoal caocal form of T. I hs Chaper, Seco 2.2, we sudy abou Nlpoe rasformaos. I ex Seco, Jorda forms of a rasformao are suded. A he ed of hs chaper, we sudy, raoal caocal forms. 2.2 NILPOTENT TRANSFORATION 2.2. Defo. Nlpoe rasformao. A rasformao T A(V) s called

23 lpoe f T = for some posve eger. Furher f T r = ad T k for k<r, he T s lpoe rasformao wh dex of lpoece r Theorem. Prove ha all he characersc roos of a lpoe rasformao T A(V) les F. Proof. Sce T s lpoe, le r be he dex of lpoece of T. The T r =. Le λ be he characersc roo of T, he here exs v( ) V such ha vt=λv. As vt 2 =(vt)t= (λv)t=λ(vt)= λλv =λ 2 v. Therefore, coug hs way we ge vt 3 =λ 3 v,, vt r =λ r v. Sce T r =, hece vt r =v = ad hece λ r v=. Bu v, herefore, λ r = ad hece λ=, whch all les F Theorem. If T A(V) s lpoe ad β, he β +β T+ + β m T m ; β F s verble. Proof. If S s lpoe he S r = for some eger r. Le β, he ( β I + S)( β S β 2 S + β ( ) r S r β r ) S = I β 2 S S + β β 2 S + β ( ) = I. Hece ( β + S) s verble. r Now f T k =, he for he rasformao S=β T+ + β m T m, S β r r ( ) r S β r r + ( ) vs k =v(β T+ + β m T m ) k =vt k (β + + β m T m- ) k v V. Sce T k =, herefore, vt k = ad hece vs k = v V.e. S k =. Equvalely, S k s a lpoe rasformao. Bu he by above dscusso β +S=β +β T+ + β m T m s verble f β. I proves he resul. r r S β r Theorem. If V= V V 2 V k where each subspace V of V s of dmeso ad s vara uder T A(V). The a bass of V ca be foud so ha he

24 A marx of T hs bass s of he form A2 A3 K K K K where each A k A s a marx ad s he marx of lear rasformao T duced by T o V. () () () 2 Proof. Sce each V s of dmeso, le { v,v,,v }, (2) (2) (2) 2 2 { v,v,,v },, { v () () (),v2,,v },,{ v (k) (k) (k),v2,,v k } are he bass of V, V 2,, V,, V k respecvely, over F. We wll show ha () () () 2 { v,v,,v, v (2) (2) (2),v2,,v 2,, () () () v,v2,,v (k) (k) (k) 2 k,, v,v,,v } s he bass of V. Frs we wll show ha hese vecors are learly depede over F. Le α () v () () () () () + α 2 v α + α (2) v (2) (2) (2) (2) (2) + α 2 v α + + v v α () v () () () () () + α 2 v α + + α (k) v (k) (k) (k) (k) (k) + α 2 v α =. v v k k Bu V s drec sum of V s herefore, zero has uque represeao.e. = Hece α () v () () () () () + α 2 v α v = for k. Bu for k, () () (),v2,,v v are learly depede over F. Hece () () () α = α2 = = α = ad hece () () () v,v2,,v, (2) (2) (2) v,v2,,v 2,, () () () v,v2,,v (k) (k) 2 (k),, v,v,,v are learly depede over F. ore k over for v V, here exs v V such ha v=v + v 2 + +v + +v k. Bu for k, v () () () () () () () = α v + α2 v2 + + α v ; for, α j F. Hece v= α () () () () (k) (k) (k) (k) v + + α v + + α v + + α v k k. I oher words we ca say ha every eleme of V s lear combao of () () () v,v2,,v,

25 (2) (2) (2) v,v2,,v 2,, v () () (),v2,,v (k) (k) (k) 2 k,, v,v,,v over F. Hece () () { v,, v, v (2) (2) (2),v2,,v 2,, () () () v,v2,,v,, v (k) (k) (k),v2,,v k } s a bass for V over F. Defe T o V by seg v T =v T v V. The T s a lear rasformao o V. Sce V are lealy depede, herefore, For obag m(t) we proceed as: v ( ) T = α () () () () () () v + α 2 v + α v () () () () () () = α v + α v + α + (2) (2) (k) (k).v + +.v +.v + +.v. 2 v v ( 2 ) T = α () () () () () () 2 v + α 22 v + α 2 v 2 k () () () () () () = α v + α v + α + (2) (2) (k) (k).v + +.v +.v + +.v v 2 k.. v ( ) () () () () () () T = α v + α v 2 + α v () () () () () () (2) (2) (k) (k) = α v + α v + α v +.v + +.v +.v + +.v. 2 2 k Sce s easy o see ha m(t )= () [ j ] α =A. Therefore, role of T o V produces a par of m(t) gve by [A ], here s a zero marx of order. Smlarly par of m(t) obaed by he roll of T o V 2 s [ A 2 ], here frs s a zero marx of order, A 2 = (2) [ αj ] 2 2 marx of order Coug hs way we ge ha ad he las zero s a zero A A2 A3 K K K K as requred. A k Theorem. If T A(V) s lpoe wh dex of lpoece, he here always exss subspaces V ad W vara uder T so ha V =V W.

26 Proof. For provg he heorem, frs we prove some lemmas: Lemma. If T A(V) s lpoe wh dex of lpoece, he here always exss subspace V of V of dmeso whch s vara uder T. Proof. Sce dex of lpoece of T s, herefore, T = ad T k for k -. Le v( ) V. Cosder he elemes v, vt, vt 2, vt of V. (s ) Take α v + α vt + + α vt + + α vt = 2 s, α F ad le αs be he frs o zero eleme above equao. Hece (s ) αsvt + + α vt ( s ) s =. Bu he vt ( αs + + α T ) =. As s α s ad T s lpoe, herefore, ( α + + α T ) s verble ad s hece vt (s ) = v V.e. (s ) T = for some eger less ha, a coradco. Hece each α =. I meas elemes v, vt, vt2,, vt are learly depede over F. Le V be he space geeraed by he elemes v, vt, vt 2,, vt. The he dmeso of V over F s. Le u V, he 2 u= β v + + β vt + β vt ad = β v + + β vt ut= βv + + β vt + β vt.e. ut s also a lear combao of v, vt, vt 2,, vt over F. Hece ut V..e. V s vara uder T. Lemma(2). If V s subspace of V spaed by v, vt, vt 2,, vt, T A(V) k = s lpoe wh dex of lpece ad u V s such ha ut ; < k k, he u= ut for some u V. (k ) Proof. For u V, u= α v + + αk vt + αk+ vt + α vt ; α F. k ad= ut =( (k ) k v k vt k vt vt α + + α + α + + α ) T = α = k k k 2 k vt + + αkvt + αk+ vt + α vt k α vt + + αkvt. Sce vt k + + vt are learly depede over F, herefore, α = = αk =. Bu he

27 k k+ k+ α k u= α vt + + α vt = ( α v + + vt ) T. Pu k αk+ v + + α vt = u. The u=u T k. I proves he lemma. Proof of Theorem. Sce T s lpoe wh dex of lpoece, he by Lemma 3, here always exs a subspace V of V geeraed by v, vt, vt 2,, vt. Le W be he subspace of V of maxmal dmeso such ha () V W=() ad () W s vara uder T. We wll show ha V=V +W. Le f possble V V +W. he here exs z V k such ha z V +W. Sce T =, herefore, zt =. Bu he here exs a eger < k such ha k zt V + W ad zt V + W for <<k. Le zt k = u + w. Sce = zt = z(t k k k k T k T ) (zt )T = (u + w) = = k k ut + wt, herefore, k k ut = wt. Bu he ut k V ad k = k W. Hece ut. By Lemma 3, u= ut for some u V. Hece zt k k k = ut + w or (z u)t W. Take z =z-u, he zt W. Furher, k for <k, zt W because f zt W, he zt ut W. Equvalely, zt V + W, a coradco o our earler assumpo ha <k, zt V + W. k z T Le W be he subspace geeraed by W, z, z T, z T 2,,. Sce z does o belogs o W, herefore, W s properly coaed W ad hece dm F W > dm F W. Sce W s vara uder T, herefore, W s also vara uder T. Now by duco hypohess, V W (). Le k z + α2zt + + αkzt w + α be a o zero eleme belogg o V W. Here all zero w + α α s are o zero because he V W (). Le α. The s z T s + + α k z T k = w + z T s ( α s k s + + α k T α s be he frs o k s ) V. Sce α s, herefore, R = ( αs + + αkt ) s verble ad hece

28 wr + z T s V R V. Equvalely, T V W, a coradco. s z + Ths coradco proves ha V=V +W. Hece V=V W Theorem. If T A(V) s lpoe wh dex of lpoece, he here exs subspace V, V 2,, V r, of dmesos, 2,, r respecvely, each V s vara uder T such ha V= V V 2 V r, 2 r ad dm V = r. ore over we ca fd a bass of V over F whch marx of T s of he form 2 3 K K K K. r Proof. Frs we prove a lemma. If T A(V) s lpoe wh dex of lpoece, V s a subspace of V spaed by v, vt, vt 2,, vt where v V. The v 2 = vt,, v Proof. Sce wll be he marx of T o V uder he bass v = v, = vt. v T =.v +.v v v 2 T=(vT)T=vT 2 = v 3 =.v +.v 2 +.v v 2 T = (vt )T = vt = v =.v +.v2 +. v v + ad T = (vt )T = vt = =.v +.v2 +. v v +, herefore, he marx of T uder he bass v, vt, vt 2,, vt K K K K =. s

29 Proof of ma heorem. Sce by Theorem 2.2.5, If T A(V) s lpoe wh dex of lpoece, he here always exss subspaces V ad W, vara uder T so ha V =V W. Now le T 2 be he rasformao 2 duced by T o W. The T = o W. Bu he here exs a eger 2 such ha 2 ad 2 s dex of lpoee of T 2. Bu he we ca wre W= V 2 W where V 2 s subspace of V spaed by u, ut 2, ut,, ut where u V ad W s vara subspace of V. Coug hs way we ge ha V= V V 2 V k Where each V s dmesoal vara subspace of V o whch he marx of T (.e. marx of T obaed by usg bass of V ) s where 2 k ad k ==dm V. Sce V= V V 2 V k, herefore, by Theorem 2.2.4, he marx of T.e. A m(t)= A2 A3 K K K K where each A =. I proves he heorem. A k Defo. Le T A(V) s lpoe rasformao wh dex of lpoece. The here exs subspace V, V 2,,V k of dmesos, 2,, k respecvely, each V s vara uder T such ha V= V V 2 V k, 2 k ad dm V = k. These egers, 2,, k are called varas of T Defo. Cyclc subspace. A subspace of dmeso m s called cyclc wh respec o T A(V) f () T m =, T m- () here exs x such ha x, xt,, xt m- forms bass of Theorem. If s cyclc subspace wh respec o T he he dmeso of T k s m-k for all k m.

30 Proof. Sce s cyclc wh respec o T, herefore, here exs x such ha x, xt,, xt m- s a bass of. Bu he z, z= a x+ a 2 xt+ + a m xt m- ; a F Equvalely, zt k = a xt k + a 2 xt k+ + + a m-k xt m- +..+a m xt m+k = a xt k + a 2 xt k+ + + a m-k xt m-. Hece every eleme z of T k s lear combao of m-k elemes xt k, xt k+,,xt m-. Beg a subse of learly depede se hese are learly depede also. Hece he dmeso of T k s m-k for all k Theorem. Prove ha varas of a lpoe rasformao are uque. Proof. Le f possble here are wo ses of vara, 2,, r ad m, m 2,, m r of T. The V= V V 2 V r ad V= W W 2 W s, where each V ad W s are cyclc subspaces of V of dmeso ad m respecvely, We wll show ha r=s ad =m. Suppose ha k be he frs eger such ha k m k..e. =m, 2 =m 2,, k- =m k-. Whou loss of geeraly suppose ha k >m k. Cosder m VT k. The VT mk mk 2 mk r mk = V T V T V T ad dm( VT mk m m ) = dm(v T k ) + dm(v T k ) + + dm(v T k ). As by Theorem m 2.2., dm( V T k ) = mk, herefore, dm( VT m k 2 ) ( m ) + + ( m ) () m k m k Smlarly dm( VT k ) = dm(w T k ) + dm(w T k ) + + dm(w T k ). As m j m k for j k, herefore, Wj T k = {} subspace ad he m m k dm( W T k j ) =. Hece dm( VT ) (m mk ) + + (m k mk ). Sce =m, 2 =m 2,, k- =m k-, herefore, dm( VT m k ) = m ) + + ( m ), coradcg (). Hece =m. ( k k k Furher r = dm V=m +m 2 + +m s ad =m for all mples ha r=s. I proves he heorem. k 2 m m r m s m Theorem. Prove ha rasformaos S ad T A(V) are smlar ff hey have same varas.

31 Proof. Frs suppose ha S ad T are smlar.e. here exs a regular mappg R such ha RTR - =S. Le, 2,, r be he varas of S ad m, m 2,, m s are ha of T. The V= V V 2 V r ad V= W W 2 W s, where each V ad W s are cyclc ad vara subspaces of V of dmeso ad m respecvely, We wll show ha r=s ad =m. As V S V, herefore, V (RTR - ) V (V R)(TR - ) V. Pu V R= U. Sce R s regular, herefore, dm U =dmv =. Furher U T= V RT= V SR. As V S V, herefore, U T U. Equvalely we have show ha U s vara uder T. ore over V=VR= V R V 2 R V r R=U U 2 U r. Now we wll show ha each U s cyclc wh respec o T. Sce each V s cyclc wh respec o S ad s of dmeso, herefore, for v V, v, vs,, vs s bass of V over F. As R s regular rasformao o V, herefore, vr, vsr,, vs R s also a bass of V. Furher S=RTR - SR=RT S 2 R=S(SR)=S(RT)=(SR)T=RTT= RT 2. Smlarly we have S R=RT. Hece {vr, vsr,, vs R} = {vr, vrt,, vrt }. Now vr les U whose dmeso s ad vr, vrt,, vrt are elemes learly depede U, he se {vr, vrt,, vrt } becomes a bass of U. Hece U s cyclc wh respec o T. Hece vara of T are, 2,, r. As by Theorem 2.2., he varas of lpoes rasformaos are uque, herefore, =m ad r=s. Coversely, suppose ha wo lpoe rasformaos R ad S have same varas. We wll show ha hey are smlar. As hey have same varas, herefore, here exs wo bass say X={x, x 2,, x } ad Y={y, y 2,, y }of V such ha he marx of S uder X s equal o marx of T uder Y s same. Le be A=[a j ]. Defe a regular mappg R:V V by x R=y. = j= As x (RTR - )= x R(TR - )= y TR - = (y T)R - = ( j (y jr ) = a jx j j= a j= a j y ) R = x S. Hece RTR - =S.e. S ad T are smlar. j =

32 2.3 CANONICAL FOR(JORDAN FOR) 2.3. Defo. Le W be a subspace of V vara uder T A(V), he he mappg T defed by wt =wt s called he rasformao duced by T o W Noe.() Sce W s vara uder T ad wt=wt, herefore, wt 2 =(wt)t= 2 (wt)t =(wt )T =wt w W. Hece T 2 2 =T. Coug hs way we ge T k k =T. Hece o W, q(t)=q(t ) for all q(x) F[x]. () Furher s easy o see ha f p(x) s mmal polyomal of T ad r(t)=, he p(x) always dvdes r(x) Lemma. Le V ad V 2 be wo vara subspaces of fe dmesoal vecor space V over F such ha V=V V 2. Furher le T ad T 2 be he lear rasformaos duced by T o V ad V 2 respecvely. If p(x) ad q(x) are mmal polyomals of T ad T 2 respecvely, he he mmal polyomal for T over F s he leas commo mulple of p(x) ad q(x). Proof. Le h(x)= lcm(p(x), q(x)) ad r(x) be he mmal polyomal of T. The r(t)=. By Noe 3.2(), r(t )= ad r(t 2 )=. By Noe 3.2(), p(x) r(x) ad q(x) r(x). Hece h(x) r(x). Now we wll show ha r(x) h(x). By he assumpos made he saeme of lemma we have p(t )= ad q(t 2 )=. Sce h(x) = lcm(p(x), q(x)), herefore, h(x)= p(x) (x) ad h(x)= p(x) 2 (x), where (x) ad 2 (x) belogs o F[x]. As V=V V 2, herefore, for v V we have uque v V ad v 2 V 2 such ha v = v + v 2. Now vh(t) = v h(t) + v 2 h(t) = v h(t ) + v 2 h(t 2 ) = v p(t ) (T ) +v 2 p(t 2 ) 2 (T 2 )=+=. Sce he resul holds for all v V, herefore, h(t)= o V. Bu he by Noe 2.3.2(), r(x) h(x). Now h(x) r(x) ad r(x) h(x) mples ha h(x)=r(x). I proves he lemma Corollary. Le V, V 2,, V k are vara subspaces of fe dmesoal vecor space V over F such ha V=V V 2 V k.. Furher le T, T 2,, T k be he lear rasformaos duced by T o V, V 2,, V k respecvely. If p (x), p 2 (x),, p k (x) are her respecve mmal polyomals. The he

33 mmal polyomal for T over F s he leas commo mulple of p (x), p 2 (x),, p k (x). Proof. I s proof s rval. 2 k ) Theorem. If p(x)= p (x) p (x) 2 p (x k ; p (x) are rreducble facors of p(x) over F, s he mmal polyomal of T, he for k, he se = V = {v V vp (T) } s o empy subspace of V vara uder T. Proof. We wll show ha V s a subspace of V. Le v ad v 2 are wo elemes of V. The by defo, v p (T) = ad v p (T) 2 =. Now usg 2 2 = leary propery of T we ge ( v v )p (T) = v p (T) v p (T). Hece v - v 2 V. Sce mmal polyomal of T over F s p(x), herefore, h = + k (T) p(t) p (T) p+ (T) pk (T). Hece here exs u V such ha uh (T). Bu uh (T)p (T) =, herefore, uh (T) V. Hece V. = ore over for v V, vt(p (T) ) = vp (T) (T) = T. Hece vtv for all v V. Hece V s vara uder T. I proves he lemma. 2 k ) Theorem. If p(x)= p (x) p (x) 2 p (x k ; p (x) are rreducble facors of p(x) over F, s he mmal polyomal of T, he for k, = V = {v V vp (T) } (), V= V V 2 V k. ad he mmal polyomal for T s p (x). Proof. If k=.e. umber of rreducble facors p(x) s oe he V=V ad ) he mmal polyomal of T s p (x.e. he resul holds rvally. Therefore, suppose k >. By Theorem 2.3.5, each V s o zero subspace of V vara uder T. Defe 2 3 h k (x) = p2(x) p3(x) pk (x), 3 k 2 (x) p(x) p3(x) pk (x) h =,.

34 j h (x) p (x). = k = j j j j The polyomals h (x), h 2 (x),, h k (x) are relavely prme. Hece we ca fd polyomals a (x), a 2 (x),,a k (x) F[x] such ha Now for v V, Sce a (x) h (x)+ a 2 (x) h 2 (x)+ + a k (x) h k (x)=. Equvalely, we ge a (T) h (T)+ a 2 (T) h 2 (T)+ + a k (T) h k (T)=I(dey rasformao). v=vi=v( a (T) h (T)+ a 2 (T) h 2 (T)+ + a k (T) h k (T)) = va (T) h (T)+ va 2 (T) h 2 (T)+ + va k (T) h k (T). ) va (T)h (T)p (T =, herefore, (T)h(T) V va. Le va (T)h (T) = v. The v=v +v 2 + +v k. Thus V=V +V 2 + +V k. Now we wll show ha f u +u 2 + +u k =, u V he each u =. As u +u 2 + +u k = u h (T)+u 2 h (T)+ +u k h (T)=h (T)=. Sce (T) p2(t) p3(t) pk (T) h =, herefore, u j h (T) = for all j=2,3,,k. 2 3 k Bu he u h (T)+u 2 h (T)+ +u k h (T)= u h (T)=. Furher u p (T) =. Sce gcd(h (x), p (x))=, herefore, we ca fd polyomals r(x) ad g(x) such ha h (x)r(x) p (x) + g(x) =. Equvalely, (T)r(T) + p(t) g(t) =I. Hece u =u I= u (h (T)r(T) p (T) + g(t)) h = u h (T)r(T) u p (T) + g(t) =. Smlarly we ca show ha f u +u 2 + +u k = he each u =. I proves ha V= V V 2 V k. Now we wll prove ha p s he mmal polyomal of T o V. (x) Sce V p (T) = (), herefore, p (T) = o V. Hece he mmal polyomal of T dvdes p (x). Bu he he mmal polyomal of T s p (x) r ; r for each =, 2,,k. By Corollary 2.3.4, he mmal polyomal r ) r 2 ) r k ) of T o V s leas commo mulple of p (x, p (x 2,, p (x k whch s r p (x r (x 2 p ) p p (x k. Bu he mmal polyomal s fac 2 ) r k ) (x) p2(x) pk (x) 2 k ha he mmal polyomal of T o V s, herefore, r for each =, 2,,k. Hece we ge p (x). I proves he resul.

35 2.3.7 Corollary. If all he dsc characersc roos λ, λ 2,,λ k of T les F, he V ca be wre as V= V V 2 V k where = V = {v V v(t λ ) } ad where T has oly oe characersc roo λ I o V. Proof. As we kow ha f all he dsc characersc roos of T les F, he every characersc roo of T s a roo of s mmal polyomal ad vce versa. Sce he dsc characersc roos λ, λ 2,,λ k of T les F. Le he mulplcy of hese roos are, 2,, k. The he mmal polyomal of 2 k ) T over F s ( x λ ) (x λ ) 2 (x λ k. If we defe = V = {v V v(t λ ) }, he by Theorem 3.6, he corollary follows. λ K λ K Defo. The marx λ K K λ of order s called Jorda λ block of order belogg o λ. For example, s he Jorda block of λ order 2 belogg o λ Theorem. If all he dsc characersc roos λ, λ 2,,λ k of T A(V) les F, he a bass of V ca be foud whch he marx of T s of he form J J2 B where each J = Jk Br are basc Jorda block belogg o λ. B 2 ad where B, B 2,, Br Proof. Sce all he characersc roos of T les F, he mmal polyomal of T over F wll be of he form ) (x λ2) (x k ) 2 k ( x λ λ. If we defe = V = {v V v(t λ ) }, he for each, V () s a subspace of V whch s vara uder T ad V= V V 2 V k such ha ( x λ ) wll be he

36 mmal polyomal of T. As we kow ha f V s drec sum of s subspaces vara uder T, he we ca fd a bass of V whch he marx of T s of J he form J2, where each J s he marx of T (he Jk rasformao duced by T o V ) uder he bass of V. Sce he mmal polyomal of T o V s ( x λ ), herefore, T λ I) ( s lpoe rasformao o V wh dex of lpoece. Bu he we ca oba a bass X of V whch he marx of T λ I) s of he form. ( 2 r where 2 r ; r = =dm V. Sce T =λ I+ T -λ I, herefore, he marx of T he bass X of V s J = marx of λ I uder he bass X + marx of T -λ I uder he bass λ X. Hece J = λ λ + 2 r B = B 2, B j are basc Jorda blocks. I proves he resul. Br 2.4 CANONICAL FOR(RATIONAL FOR) 2.4. Defo. A abela group s called module over a rg R or R-module f rm for all r R ad m ad () (r + s)m=rm + rs () r(m + m 2 ) = rm + rm 2 () (rs)m = r(sm) for all r, s R ad m, m, m 2.

37 2.4.2 Defo. Le V be a vecor space over he feld F ad T A(V). For f(x) F[x], defe, f(x)v=vf(t), f(x) F[x] ad v V. Uder hs mulplcao V becomes a F[x]-module Defo. A R-module s called cyclc module f ={rm r R ad some m Resul. If s fely geeraed module over a prcpal deal doma R. The ca be wre as drec sum of fe umber of cyclc R-modules..e. here exs x, x 2,, x such ha =Rx Rx 2 Rx Defo. Le f(x)= a + a x + +a m- x m- + x m be a polyomal over he feld F. The he compao marx of f(x) s a O a am m m I s a square marx [b j ] of order m such ha b, + = for m-, b m, j = a j- for j m ad for he res of eres b j =. The above marx s called compao marx of f(x). I s deoed by C(f(x)). For example compao marx of +2x -5x 2 +4x 3 + x 4 s Noe. Every F[x]-module becomes a vecor space over F.Uder he mulplcao f(x)v = vf(t), T A(V) ad v V, V becomes a vecor space over F Theorem. Le V be a vecor space over F ad T A(V). If f(x) = a + a x + +a m- x m- + x m s mmal polyomal of T over F ad V s cyclc F[x]- module, he here exs a bass of V uder whch he marx of T s compao marx of f(x).

38 Proof. Clearly V becomes F[x]-module uder he mulplcao defed by f(x)v= vf(t) for all v V, T A(V). As V s cyclc F[x]-module, herefore, here exs v V such ha V = F[x]v ={ f(x)v f(x) F[x]}= { v f(t) f(x) F[x]}. Now we wll show ha f v s(t)=, he s(t) s zero rasformao o V. Sce v = f(x)v, he vs(t) = (f(x)v )s(t)= (v f(t))s(t) = (v s(t))f(t)= f(t)=..e. every eleme of v s ake o by s(t). Hece s(t) s zero rasformao o V. I oher words T also sasfes s(t). Bu he f(x) dvdes s(x). Hece we have show ha for a polyomal s(x) F[x], f v s(t) =, he f(x) s(x). Now cosder he se A={v, v T,, v T m- } of elemes of V. We wll show ha s requred bass of V. Take r v + r (v T) + + r m- ( v T m- ) =, r F. Furher suppose ha a leas oe of r s o zero. The r v + r (v T) + + r m- ( v T m- ) = v (r + r T + + r m- T m- ) =. The by above dscusso f(x) (r + r T + + r m- T m- ), a coradco. Hece f r v + r (v T) + + r m- ( v T m- ) = he each r =. e he se A s learly depede over F. Take v V. The v = (x)v for some (x) F[x]. As we ca wre (x)= f(x)q(x) + r(x), r(x) = r + r x + + r m- x m-, herefore, (T)= f(t)q(t) + r(t) where r(t)= r + r T + + r m- T m-. Hece v = (x)v = v (T) = v (f(t)q(t) +r(t)) = v f(t)q(t) + v r(t) = v r(t) = v (r + r T + + r m- T m- )= r v + r (v T) + + r m- ( v T m- ). Hece every eleme of V s lear combao of eleme of he se A over F. Therefore, A s a bass of V over F. Le v =v, v 2 =v T, v 3 =v T 2,, v m- =v T m-2, v m =v T m-. The v T= v 2 =.v +.v 2 +.v v m- +v m, v 2 T= v 3 =.v +.v 2 +.v v m- +v m,, v m- T= v m =.v +.v 2 +.v v m- +v m. Sce f(t)= v f(t) = v (a + a T + +a m- T m- + T m ) = a v + a v T + +a m- v T m- + v T m = v T m = -a v - a v T - - a m- v T m-. As v m T= v T m- T=v T m = -a v - a v T - - a m- v T m-

39 = -a v - a v a m- v m. Hece he marx uder he bass v =v, v 2 =v T, v 3 =v T 2,, v m- =v T m-2, v m =v T m- s a O a L am m m = C(f(x)). I proves he resul Theorem. Le V be a fe dmesoal vecor space over F ad T A(V). Suppose q(x) s he mmal polyomal for T over F, where q(x) s rreducble moc polyomal over F. The here exs a bass of V such ha he marx of T uder hs bass s of he form C(q(x) ) C(q(x) 2 ) L L L L C(q(x) k where = 2 k. ) Proof. Sce we kow ha f s a fely geeraed module over a prcpal deal doma R, he ca be wre as drec sum of fe umber of cyclc R-submodules. We kow ha V s a vecor space over F[x] wh he scalar mulplcao defed by f(x)v=vf(t). As V s a fe dmesoal vecor space over F, herefore, s fely dmesoal vecor space over F[x] also. Thus, s fely geeraed module over F[x] (because each vecor space s a module also). Bu he we ca oba cyclc submodules of V say F[x]v, F[x]v 2,, F[x]v k such ha V = F[x]v F[x]v 2 F[x]v k, v V. Sce (F(x)v ) T =(v F[T]) T = =v (F[T] T) = =(v g(t))= g(x)v F[x]v. Hece each F[x]v s vara uder T. Bu he we ca fd a bass of V whch he marx of T s A A2 L L L L Ak where A s he marx of T uder he bass of V. Now we clam ha A C(q(x) ). Le p (x) be he mmal polyomal of T (.e of T o V ). Sce w q(t) = for all w F[x]v, herefore, p (x) dvdes q(x). Thus = p = q(x).. Re dexg V, we ca fd 2 k. Sce V = F[x]v F[x]v 2

40 F[x]v k, herefore, he mmal polyomal of T o V s lcm( q (x), q(x) 2,,q(x) k )= q (x). The q (x) = q (x). Hece =. By Theorem 2.4.7, he marx of T o V s compao marx of moc mmal polyomal of T o V. Hece A = C(q(x) ). I proves he resul Theorem. Le V be a fe dmesoal vecor space over F ad T A(V). Suppose 2 k (x) q2(x) qk (x) q s he mmal polyomal for T over F, where q (x) are rreducble moc polyomal over F. The here exs a bass of V such ha he marx of T uder hs bass s of he form A A2 L L L L Ak C(q(x) where A = ) C(q(x) 2 ) L L L L C(q(x) r ) where = r 2 r for each, k, j = j= ad r = =. Proof. Le V ={ v V v q (T) =}. The each V s o zero vara (uder T) subspace of V ad V = V V 2 V k. Also he mmal polyomal of T o V s q (x). For such a V, we ca fd a bass of V uder whch he marx of T s of he form A A2 L L L L Ak. I hs marx, each A s a square marx ad s he marx of T V. As T has q as s mmal (x) polyomal, herefore, by Theorem, 2.4.8, A = C(q(x) ) C(q(x) 2 ) L L L L C(q(x) r. Res par of he resul s easy o ) prove. k k ) r kr 2.4. Defo. The polyomals q (x),,q (x),., q (x) k,,q (x k called elemeary dvsors of T. are

41 2.4. Theorem. Prove ha elemeary dvsors of T are uque. Proof. Le l q x q x q x q x) l l ( ) = ( ) ( ) 2 k 2 k ( be he mmal polyomal of T where each q (x) s rreducble ad l. Le V = { v V vq ) l ( =}. The T V s a o zero vara subspace of V, V=V V 2 V k ad he mmal polyomal of T o V.e. of T, s q l (x). ore over we ca fd a bass of V such ha he marx of T s R, where R s he marx of T o V. Rk Sce V becomes a F[x] module uder he operao f(x)v=vf(t), herefore, each V s also a F[x]-module. Hece here exs v, v 2,, such ha V = F[x]v + +F[x] v = V + V r vr V V r where each V j s a subspace of V ad hece of V. ore over V j s cyclc F[x] module also. Le l j j q ( x) be he mmal polyomals of T o V j. The q ( x) becomes elemeary dvsors of T, k ad j r. Thus o prove ha elemeary dvsors of T are uque, s suffce o prove ha for all, k, he l polyomals ( ) l q x, q ( ) 2 x,, q ( x) are uque. Equvalely, we have o prove he resul for T A(V), wh q(x) l, q(x) s rreducble as he mmal polyomal have uque elemeary dvsor. l r Suppose V = V V 2 V r ad V = W W 2 W s where each V ad W s a cyclc F[x]-module. The mmal polyomal of T o V s have uque elemeary dvsors r l q x) l* s. Also l d = = dm V ad = ( where l=l l 2 l r ad l=l* l* 2 s = * d = dm V, d s he degree of q(x). We wll sow ha l = l* ad r=s. Suppose s frs eger such ha l =l*, l 2 =l* 2,, l - = l* - ad l l*. Sce each V ad W are vara uder T, herefore, dmeso l * l * l * l Vq( T ) = V q( T ) V q( T ). Bu he he r Vq( T ) = dmv q( T ) j= j l* j= j r dmv q( T ) l * l * l. Sce l l*, whou loss of geeraly, suppose ha l > l*. As V q( T ) = d(l j -l* ), j l*