Bi tp i s tuyn tnh

1. Bi tp v khng gian vectorBi 1.1 Gi s A l mt ma trn vung cp n, v C(A) = {B | BA= AB} l tphp tt c cc ma trn vung phc cp n giao hon c vi A. Chng minh rng:C(A) l khng gian vector con ca khng gian vector Mnn v dimC(A) n.Bi 1.2 Cho S l khng gian con ca khng gian cc ma trn vung phc cpn Mnn sinh bi tp tt c cc ma trn c dng ABBA. Chng minh rng:dimS= n21.Bi 1.3 Cho A,B l cc khng gian vector con ca khng gian vector hu hnchiu V sao cho A+B = V. Gi n = dimV,a = dimA,b = dimB. Ly S l tptt c cc t ng cu f ca V m f (A) A, f (B) B. Chng minh rng S lkhng gian con ca khng gian tt c cc t ng cu ca V v hy biu th schiu ca S qua a,b,n.

Bi 1.4 Cho T l t ng cu ca khng gian vector V. Gi s x V m Tmx=0,Tm1x 6= 0 vi m l s nguyn no . Chng minh rng: x,Tx,T 2x, . . . ,Tm1xc lp tuyn tnh.

Bi 1.5 Cho E l mt khng gian Euclide n chiu. Chng ta ni hai c s (ai)v (bi) cng hng nu ma trn chuyn t c s (ai) sang c s (bi) c nhthc dng. Gi s (ai) v (bi) l hai c s trc chun cng hng. Chngminh rng (ai+2bi) cng l mt c s ca E cng hng vi (ai).

Bi 1.6 Cho l nh x tuyn tnh t V vo W , trong V v W l cc khnggian vector hu hn chiu. Gi L,Z l khng gian vector con ca V v W .Chng minh rng:

a) dim(L)+dim(kerL) = dimLb) dimLdimker dim(L) dimLc) dimZ dim1Z dimZ+dimker

Bi 1.7 Cho cc ng cu ca cc IK-khng gian vector hu hn chiu :V W, :W Z. Chng minh rng:

a) dimker(.) = dimker+dim(Imker)b) dimker(.) dimker+dimkerc) rank(.) = rankdim(ker Im)d) rank(.) rank+ rankdimW

Bi 1.8 Gi s P,Q,R l cc ma trn vung cp n. Chng minh rng:rank(PQ)+ rank(QR) rankQ+ rank(PQR).

Bi 1.9 Cho V v W l cc khng gian vector hu hn chiu. T : V W lnh x tuyn tnh, X l khng gian vector con ca khng gian vector W Chng

1

minh: dim(T1X) dimV dimW +dimX . Hn na nu T ton nh th ta cng thc.

Bi 1.10 Cho A v B l cc ma trn vung cp n. Chng minh rng khng giannghim ca hai phng trnh AX = 0 v BX = 0 bng nhau khi v ch khi tnti ma trn C kh nghch sao cho A=CB.

Bi 1.11 Cho A l ma trn vung phc cp n sao cho trAk = 0 vi k = 1, . . . ,n.Chng minh rng A l ma trn lu linh.

Hint Gi s A c dng cho ho Jordan vi cc khi Jordan tng ng vi ccgi tr ring 1, . . . ,m phn bit. Khi Ak l ma trn c cc phn t trnng cho chnh l cc gi tr ring ki . T gi thuyt tr(A

k) = 0,1 km tac h phng trnh:

m

i=1

ki = 0,k = 1, ...,n.

T h ny ta suy ra i = 0,1 i m. Vy A s l ma trn lu linh.Bi 1.12 Cho A l ma trn phc cp m sao cho dy (An)n=1 hi t n ma trnB. Chng minh rng B ng dng vi ma trn ng cho m cc phn t trnng cho chnh bng 0 hoc 1.

Hint: Do A2n = An.An suy ra B2 = B. Vy ta c iu cn chng minh.

Bi 1.13 Cho W l khng gian vector n-chiu, U v V l cc khng gian conca W sao cho U V = {0}. Gi s u1,u2, . . . ,uk U v v1,v2, . . . ,uk V vik > dimU + dimV . Chng minh rng tn ti cc s 1,2, . . . ,k khng ngthi bng 0 sao cho

k

i=1

iui =k

i=1

ivi = 0.

Khng nh trn cn ng khng nu k dimU+dimV.Hint Ch rng ta c n cu UV W nn s chiu ca UV khng qun.

2. Dng chnh tcBi 2.1 Cho ma trn:

A=

2 1 01 2 10 1 2

2

Chng minh rng: mi ma trn B sao cho AB= BA c dng:

B= aI+bA+ cA2,

vi a,b,c l cc s thc no .

Bi 2.2 Cho A l ma trn cp n c n gi tr ring phn bit. Chng minh rng:mi ma trn B giao hon c vi ma trn A c dng: B= f (A), vi f l mta thc h s thc, bc khng qu n1.Bi 2.3 Cho

A=(1 21 1

).

Hy biu th A1 nh l mt a thc ca A vi h s thc.

Bi 2.4 Vi x R, t

Ax =

x 1 1 11 x 1 11 1 x 11 1 1 x

.a) Chng minh rng detAx = (x1)3(x+3).b) Chng minh rng nu x 6= 1,3, th A1x = (x1)1(x+3)1Ax2.

Bi 2.5 Tnh A10 vi

A=

3 1 12 4 21 1 1

.Bi 2.6 Chng minh hoc a ra phn v d: Vi mi ma trn vung phc Acp 2, tn ti ma trn vung phc B cp 2 sao cho A= B2.

Bi 2.7 Cho

A=

0 0 0 10 0 0 00 0 0 00 0 0 0

.Vi s nguyn n no th s tn ti ma trn vung phc X cp 4 sao cho Xn = A.

Bi 2.8 Khng nh sau ng hay khng:Tn ti ma trn vung thc A cp n sao cho

A2+2A+5I = 0,

nu v ch nu n l s chn.

3

Bi 2.9 Phng trnh no c nghim l mt ma trn vung thc (khng nhtthit phi ch ra nghim):

X3 =

0 0 01 0 02 3 0

2X5+X =

3 5 05 1 90 9 0

X6+2X4+10X =

(0 11 0

)

X4 =

3 4 00 3 00 0 3

.Bi 2.10 Cho x l s thc dng. Hi c tn ti hay khng mt ma trn vungthc cp 2 sao cho

A2004 =(1 0

0 1 x).

3. Vector ring v gi tr ring

Bi 3.1 Cho M l ma trn vung thc cp 3, M3 = I v M 6= I.a) Tm cc gi tr ring ca M.b) Cho mt ma trn c tnh cht nh th.

Bi 3.2 Cho F l mt trng, n v m l cc s nguyn v A l mt ma trnvung cp n vi cc phn t trong F sao cho Am = 0. Chng minh rng: An = 0.

Bi 3.3 Cho V l khng gian vector hu hn chiu trn trng s hu t Q, Ml mt t ng cu ca V, M(x) 6= x, x V \0. Gi s Mp = IdV , vi p l mts nguyn t. Chng minh rng s chiu ca V chia ht cho p1.Bi 3.4 Chng minh rng ma trn 1 1,00001 11,00001 1 1.00001

1 1,00001 1

.c mt gi tr ring dng v mt gi tr ring m.

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Bi 3.5 Cho a,b,c l cc phn t bt k ca trng F, hy tnh a thc ti tiuca ma trn 0 0 a1 0 b

0 1 c

.Bi 3.6 Gi s A,B l cc t ng cu ca khng gian vector hu hn chiu Vtrn trng F. ng hay sai cc khng nh sau:

a) Mi vector ring ca AB l mt vector ring ca BA.b) Mi gi ring ca AB l mt gi ring ca BA.

Bi 3.7 Cho

A=(a bc d

)l mt ma trn thc vi a,b,c,d > 0. Chng minh rng A c mt vector ring(

xy

) R2,

vi x,y> 0.

Bi 3.8 Cho A l ma trn vung phc cp n v P(t) l mt a thc bc m.Chng minh rng nu 1,2, . . . ,n l cc gi tr ring ca ma trn A th:

1) |P(A)|= P(1).P(2) . . .P(n).2) P(1),P(2), . . . ,P(n) l cc gi tr ring ca P(A).

Bi 3.9 Cho A v B l cc ma trn i xng thc tho mn AB = BA. Chngminh rng A v B c chung1 vector ring trong Rn.

Bi 3.10 Gi S l tp khng rng gm cc ma trn phc cp n giao hon cvi nhau tng i mt. Chng minh rng cc phn t ca S c chung mt vectorring

Bi 3.11 Gi A v B l cc ma trn phc cp n sao cho AB= BA2. Gi s rngA khng c cc gi tr ring c moun bng 1, chng minh rng A v B cchung mt vect ring.

Bi 3.12 Cho l t ng cu tuyn tnh cho ho c ca Rn. Chng minhrng khng gian con W ca Rn l bt bin i vi khi v ch khi trong Wchn c mt c s gm cc vector ring ca .

Bi 3.13 Cho A v B l hai ma trn cho ho c v giao hon c vi nhau.Chng minh rng tn ti mt c s ca Rn gm ton cc vector ring ca A vB.

Bi 3.14 Cho A l ma trn phc cp n v a thc ti tiu c bc k.

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1) Chng minh rng nu khng l gi tr ring ca A th tn ti mt athc p bc k1 sao cho p(A) = (AE)1.

2) Gi 1,2, . . . ,k l cc s phc phn bit v khng l gi tr ring ca A.Chng minh rng: tn ti cc s phc c1,c2, . . . ,ck sao cho

k

i=1

ck(AkE)1 = E.

Hint Xt ng thc p(A)(AE) = p(A) p()E = p()E suy ra c a thcp. Vi mi i tn ti cc pi tng ng. Xt h pt theo cc n ci ta thu ch Cramer do tn ti cc ci cn tm.

4. Hng v nh thc

Bi 4.1 Cho A l ma trn vung thc cp n v At l ma trn chuyn v ca n.Chng minh rng AtA v A cng hng.

Bi 4.2 Gi s P v Q l cc ma trn vung cp n tha mn cc iu kin sau:P2 = P,Q2 = Q v I (P+Q) kh nghch. Chng minh rng P v Q c hngbng nhau.

Bi 4.3 Cho

T =

a1 b1 0 0 . . . 0 0b1 a2 b2 0 . . . 0 00 b2 a3 b3 . . . 0 0...

......

... . . ....

...0 0 0 0 . . . an1 bn10 0 0 0 . . . bn1 an

.

Gi s bi 6= 0, vi mi i. Chng minh rng:a) rankT n1,b) T c n gi tr ring phn bit.

Bi 4.4 Cho (ai j) l ma trn vung cp n vi cc ai j l cc s nguyn.a) Chng minh rng nu s nguyn k l mt gi tr ring ca A th nh thc

ca A chia ht cho k.b) Gi s m l mt s nguyn v mi dng ca A c tng bng m

n

j=1

ai j = m, i= 1,2, . . . ,n.

6

Chng minh rng nh thc ca A chia ht cho m.

Bi 4.5 Cho nh thc Vandermonde (phc)

A=

1 a0 a20 . . . a

n0

1 a1 a21 . . . an1

......

... . . ....

1 an a2n . . . ann

,vi ai l cc s phc.

a) Chng minh rng A kh nghch khi v ch khi cc ai i mt khc nhau.b) Nu cc ai i mt khc nhau v b1,b2, . . . ,bn l cc s phc ty .

Chng minh rng tn ti duy nht a thc f bc n vi h s phc sao chof (ai) = bi, i= 1,2, . . . ,n.Bi 4.6 Cho v d mt hm lin tc f :RR3 vi tnh cht l f (v1), f (v2), f (v3)lp thnh mt c s ca R3, trong v1,v2,v3 l cc s thc phn bit.

Bi 4.7 Cho f1, f2, . . . , fn l cc hm nhn cc gi tr thc lin tc trn [a,b].Chng minh rng { f1, f2, . . . , fn} ph thuc tuyn tnh khi v ch khi

det( b

a fi(x) f j(x)dx)= 0.

Bi 4.8 K hiu M22 l khng gian cc ma trn vung thc cp 2. Cho

A=(

1 21 3

), B=

(2 10 4

).

Xt php bin i tuyn tnh L :M22 M22 xc nh bi L(X) = AXB. Hytnh vt v nh thc ca L.

Bi 4.9 K hiu M33 l khng gian cc ma trn vung thc cp 3. Cho

A=

1 0 00 2 00 0 1

Xt php bin i tuyn tnh L :M33 M33 xc nh bi L(X) = 12(AX +XA). Hy tnh nh thc ca L.

Bi 4.10 K hiu M33 l khng gian cc ma trn vung thc cp 3. Gi sA M33,detA = 32 v a thc ti tiu ca A l ( 4)( 2). Xt nh xtuyn tnh: LA :M33 M33 xc inh bi LA(X) = AX . Hy tnh vt ca LA.

7

Bi 4.11 K hiu M77 l khng gian cc ma trn vung thc cp 7. Gi sAM77 l mt ma trn cho vi ng cho chnh gm 4 hng t +1 v 3 hngt -1. Xt nh x tuyn tnh LA :M77M77 xc nh bi LA(X) =AXXA.Hy tnh dimLA.

Bi 4.12 Cho F l mt trng, n v m l hai s nguyn, Mmn l khng giancc ma trn cp mn trn trng F . Gi s A v B l hai ma trn c nh caMmn. Xt nh x tuyn tnh L : Mmn Mmn xc nh bi L(X) = AXB.Chng minh rng nu m 6= n th L suy bin.Bi 4.13 Gi s A1,A2, . . . ,An+1 l cc ma trn cp n. Chng minh rng tmc n+ 1 s x1,x2, . . . ,xn+1 khng ng thi bng 0 sao cho ma trn x1A1+x2A2 + xn+1An+1 suy bin.Bi 4.14 Gi s A l ma trn cp n hng r. Tm s nghim c lp tuyn tnhca phng trnh AX = 0 vi X l ma trn cp n.

Bi 4.15 Cho A l ma trn vung cp n. Chng minh rng nu A2 = E th tnghng ca cc ma trn AE v A+E bng n (E l ma trn n v).Bi 4.16 Cho A l ma trn vung thc cp n. Chng minh rng: det(A2+E) 0.Khi no th ng thc xy ra.

Bi 4.17 Cho tam thc bc hai p(x) = x2+ax+b tho mn p(x) 0, x R vA l mt ma trn vung thc cp n. Chng minh rng: det p(A) 0.Bi 4.18 Cho f (x) l a thc h s thc c bc dng, h s dn u bng1 v f (x) 0, x R, A l mt ma trn vung thc cp n. Chng minh rngdet f (A) 0.Bi 4.19 Cho A l ma trn vung cp n. Chng minh rng: det(AAt +E) > 0,trong At l ma trn chuyn v ca ma trn A v E l ma trn n v cng cpvi A.

Bi 4.20 Cho A v B l cc ma trn thc cp n. Chng minh rng: det(AAt +BBt) 0.

Cc thi Olympic

Olympic ngh 2003

8

Bi 1: Cho

A=

2 1 0 0 . . . 0 0 01 2 1 0 . . . 0 0 00 1 2 1 . . . 0 0 00 0 1 2 . . . 0 0 0...

......

... . . ....

......

0 0 0 0 . . . 2 1 00 0 0 0 . . . 1 2 10 0 0 0 . . . 0 1 2

Chng minh rng mi gi tr ring ca A l mt s thc dng.

Bi 2: Cho A l ma trn vung thc cp n v At l ma trn chuyn v ca n.Chng minh AtA v A cng hng.

thi chn i tuyn Olympic ca Trng nm 2003

s 1:

Bi 1: nh thc ca mt ma trn vung thay i nh th no khi thay miphn t bng phn t i xng vi n qua ng cho th hai.

Bi 2: Gi s xi 6= 0,i= 1,2, . . . ,n. Hy tnh nh thc sau:

a1 a2 a3 . . . anx1 x2 0 . . . 00 x2 x3 . . . 0...

...... . . .

...0 0 0 . . . xn

.

Bi 3: Xc nh cc s nguyn dng m,n, p sao cho a thc x3m+x3n+1+x3p+2

chia ht cho a thc x2 x+1.Bi 4: Cho

A=( 3

212

12 12

).

Hy tnh A100 v A7.

Bi 5: Cho A l ma trn vung cp 2. Chng minh rng Ak = 0 khi v ch khiA2 = 0.

Bi 6: K hiu M33 lkhng gian cc ma trn vung thc cp 3. Cho

A=

1 0 00 2 00 0 1

,9

Xt php bin i tuyn tnh L :M33 M33 xc nh bi L(X) = 12(AX XA). Hy tnh nh thc ca L.

s 2:

Bi 1: Tnh nh thc cp n m phn t dng i ct j l |i j|.Bi 2: Gi s P v Q l cc ma trn vung cp n tho mn cc iu kin sau:P2 = P;Q2 = Q v I (P+Q) kh nghch. Chng minh rng P v Q c hngbng nhau.

Bi 3: K hiu M33 l khng gian cc ma trn vung thc cp 3. Gi sA M33, detA = 32 v a thc ti tiu ca A l ( 4)( 2). Xt nh xtuyn tnh LA :M33 M33 xc nh bi LA(X) = AX . Hy tnh vt ca matrn A.

Bi 4: K hiu M22 l khng gian cc ma trn vung thc cp 2. Cho

A=(

1 21 3

), B=

(2 10 4

).

Xt php bin i tuyn tnh L :M22 M22 xc nh bi L(X) = AXB. Hytnh vt v nh thc ca L.

Bi 5: Cho m1,m2, . . . ,mr l nhng s nguyn tng i mt phn bit, r 2.Chng minh rng a thc

f (x) = (xm1)(xm2) . . .(xmr)1khng c nghim nguyn.

Bi 6: Chng minh rng vi mi ma trn A cp mn ta lun lun c bt ngthc sau:

|AtA| m

k=1

n

i=1

a2ik.

bi tp i s i cngBi 1 Cho R l mt vnh c n v 1. Gi s rng A1,A2, . . . ,An l cc Ideal trica R sao cho R= A1

A2 An (xem nh mt nhm cng). Chng minh

rng tn ti cc phn t ui Ai sao cho vi mi ai Ai, aiui Ai v aiu j = 0nu i 6= j.Bi 2 Chng t rng nhm G ng cu vi nhm con (nhm cng) cc s hut nu v ch nu G m c v mi tp con hu hn ca G u cha trongmt nhm con xyclic v hn ca G.

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Li gii

1. Khng gian vector

Bi 1.1 Xt nh x tuyn tnh:

T :Mnn MnnB 7 ABBA.

Khi S = kerT l khng gian vector con ca khng gian cc ma trn Mnn. rng, nu C l ma trn kh nghch th

AB= BA

khi v ch khi C1ACC1BC = C1BCC1AC. Nu D1, . . . ,Dn l cc ma trnc lp tuyn tnh th C1D1C, . . . ,C1DnC cng c lp tuyn tnh. Do n gin ta gi s A c dng Jordan, vi khi Jordan th i cp k l:

Ai =

a 1 . . . 0

. . . . . .0 a 10 0 a

.Khi Ai giao hon vi

Bi =

b1 b2 . . . bk

. . . . . .0 b1 b20 0 b1

.Do A giao hon vi

B=

B1 . . .Br

.V trong B c n bin nn dimC(A) n.Bi 1.2 Ta cn ch ra S c n21 vector c lp tuyn tnh. l cc ma trn:Mi j =MikMk jMk jMik i 6= j (c n2n phn t)

M11M j j = Mi jM j1M j1Mi j j 6= 1 (c n 1 phn t), trong ma trnMi j l ma trn c phn t 1 v tr i j, cc v tr khc u bng 0. Do dimS n21, mt khc S 6=Mnn nn dimS< n2. Suy ra: dimS= n21.

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Bi 1.3 Ly f ,g S v r,s R. Khi ta c: v A,(r f + sg)(v) = f (rv)+g(sv) A v f ,g bt bin i vi A. Tng t ta cng c (r f + sg)(v) B. Vyr f + sg S, hay S l khng gian vector con ca khng gian vector cc t ngcu ca V. tnh s chiu ca S ta ch cn tnh s chiu ca khng gian ccma trn bt bin vi A v B. Gi A1,B1 l khng gian vector con ca V sao choA = ABA1,B = ABB1. Khi dim(AB) = r = a+ b n,dimA1 =a r,dimB1 = b r. Ly {u1, ...,uar} l c s ca A1, {v1, ...,vr} l c sca AB, {w1, ...,wbr} l c s ca B1, Mi t ng cu bt bin i viA,B th phi bt bin i vi AB. Do f (ui) c biu th tuyn tnh qua{u1, ...,uar,v1, ...,vr}, f (vi) ch c th biu din tuyn tnh qua {v1, ...,vr},f (wi) c biu din tuyn tnh qua {v1, ...,vr,w1, ...,wbr}. Suy ra ma trn caf c dng:

a r r b ra r M1 0 0

r M2 M3 M4b r 0 0 M5

trong s phn t khc 0 nhiu nht l (a r)2+ rn+(b r)2 = a2+ b2+n2 (a+b)n. Vy dimS= a2+b2+n2 (a+b)n.

Bi 1.4 Gi s rng c:

a0x+a1Tx+ +akT kx+ +am1Tm1x= 0.

Tc ng Tm1 vo hai v ta c: a0Tm1x= 0, suy ra a0 = 0. Bng quy np tac ak = 0,k = 0,m1 suy ra iu phi chng minh

Bi 1.5 Gi P l ma trn chuyn t (ai) sang (bi). Khi I+ 2P l ma trnchuyn t (ai) sang (ai+2bi). Ta c l gi tr ring ca I+2P khi v ch khi12(1) l gi tr ring ca P. Do (ai) v (bi) l cc c s trc chun nn P l

ma trn trc giao v cc gi tr ring ca P l 1, suy ra cc gi tr ring caI+2P l 3,1. Do 0 khng phi l gi tr ring ca I+2P nn I+2P khnghch v (ai+2bi) l c s. Hn na detP= (1)1 vi , l bi ca ccgi tr ring 1,1 ca P. Do det(I+2P) = (1)3. V det p> 0 nn ls chn. Vy det(I+2P)> 0, hay (ai) v (ai+2bi) cng hng vi nhau.

Bi 1.6 a) Xt nh x tuyn tnh hn ch ca ln L ta c:

|L : L L,

ker|L = kerL. Do : dim(L)+dim(kerL) = dimL.b) Suy ra t a) vi ch rng dim(kerL) dimker.

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c) t L = 1Z v ch rng: L Z. T cu b) ta c: dim1Z dim(1Z)+dimker dimZ+dimker.

Mt khc: ker L nn t a) ta c:

dim(L)+dimker= dimL (1).

Ta cng c: (L) = Z(V ) nn

dim(L) =dim(Z(V ))=dimZ+dim(V )dim(Z+(V ))dimZ+dim(V )dimW=dimZdimker. (2)

T (1) v (2) ta c iu phi chng minh.

Bi 1.7 a) t L= Im v p dng bi tp 1.6.a ta c:

dim(L)+dim(kerL) = dimL

haydimIm(.)+dim(kerL) = dimV dimker

dimker+dim(kerL) = dimV dimIm(.) = dimker(..b) Suy ra t cu a) vi ch rng: kerL kerc) Suy ra t lp lun chng minh ca cu a).d) Suy ra t cu c) vi ch rng: ker Im ker.

Bi 1.8 S dng bi tp 1.7 cu c) ta c:

rank(PQR) = rank(PQ)dim(ker(PQ) ImR)

rank(QR) = rankQdim(kerQ ImR)Suy ra:

rank(PQ)+ rank(QR) = rank(PQR)+ rankQ+dim(kerQ ImR)dim(ker(PQ) ImR) rank(PQR)+ rankQ

Bi 1.9 Xt nh x tuyn tnh: F :V/T1X W/X c cho bi: F(x) = T (x).Khi F l n nh. Tht vy, nu F(y) = 0 th T (y) X do y T1X hayy= 0. T suy ra:

dim(V/T1X) dim(W/X)

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haydimV dimT1X dimW dimX .

VydimT1X dimV dimW +dimX .

2. Dng chnh tc

Bi 2.2 Do A c n gi tr ring phn bit nn A cho ha c, tc l tn ti matrn C kh nghch sao cho C1AC = P l ma trn cho. Khi , ma trn B giaohon c vi A khi v ch khi ma trn Q=C1BC giao hon c vi P. Gis:

P=

1 0 00 2 0 0 0 n

trong i l cc gi tr thc khc nhau tng i mt. Bng cch th trc tipta c: Q giao hon c vi P khi v ch khi Q c dng:

Q=

1 0 00 2 0 0 0 n

trong i l cc gi tr thc no . By gi ta cn tm cc s thc 0,1, ...,n1sao cho

Q= 0I+1P+ +n1Pn1iu ny thc hin c nh vic gii h phng trnh tuyn tnh:

x0+1x1+ +n11 xn1 = 1x0+2x1+ +n12 xn1 = 2 x0+nx1+ +n1n xn1 = n

T ta suy ra:B= 0I+1A+ +n1An1

14

(pcm).

Bi 2.3 Ta c a thc c trng ca A l:

A() = 23

. Do : A23I = 0 hay A2 = 3I, suy ra A kh nghch v A1 = 13A.

Bi 2.4 a) Tnh ton trc tip ta c detAx = (x1)3(x+3).b) Nu x 6= 1,3 th Ax kh nghch v a thc c trng ca Ax l:

(t) = (x t1)3(x t+3).Suy ra a thc ti tiu ca Ax l: m(t) = (x t1)(x t+3), do : ((x1)IAx)((x+3)IAx) = 0, khai trin ta c c: (x1)(x+3)I2(x1)Ax+A2x =0. Nhn hai v vi A1x v bin i ta c

A1x =(x1)1(x+3)1Ax2.

Bi 2.6 (Gii vn tt) Chn A=(0 10 0

)th s khng c mt ma trn vung phc

B cp 2 no m A= B2.

Bi 2.8 Khng nh ng.Gi s A tn ti, suy ra A c a thc ti tiu chia ht t2+2t+5 l a thc

bt kh qui trn R Vy mA(t) = t2+2t+5. V a thc c trng v a thc titiu c cng nhn t bt kh qui nn

A(t) = mA(t)k

suy ra n= degA(t) phi l s chn.Ngc li, n chn, ta thy A0 =

(0 51 2

)l mt nghim ca phng trnh

t2+ 2t+ 5 = 0. Do ma trn khi gmn2khi A0 trn ng cho chnh l

ma trn tha mn yu cu ca bi.

hoa 3. Vector ring v gi tr ring

Bi 3.1a) Do M l nghim ca a thc x31 nn a thc ti tiu ca M phi l c

ca x31. Mt khc, M c t nht mt gi tr ring thc, nn a thc ti tiu

15

c nhn t (x-1). V M 6= I nn a thc ti tiu ca M khng th l x1. Do a thc ti tiu ca M l m(x) = x31. Vy M c 1 gi tr ring 1

b) Mt ma trn c tnh cht nh vy l:

M =

1 0 00 12 320

32

12

Bi 3.2 Do An = 0 nn a thc ti tiu p(x) ca A phi l c ca xm. Suy rap(x) = xk, vi k n. Vy An = 0.

Bi 3.3 Do Mp = I nn a thc ti tiu p(x) ca M phi l c ca

xp1= (x1)(xp1+ . . .+1)

Do M(x) 6= x vi mi x 6= 0 nn 1 khng l gi tr ring, suy ra p(x) l c ca(xp1+ . . .+ 1). Nhng (xp1+ . . .+ 1) l a thc kh qui trn trng Q nnp(x) = (xp1+ . . .+1).

Mt khc, a thc c trng M v a thc ti tiu c chung nhn t bt khqui. Do M(x) = (p(x))k, k 1. Vy dimV = rankM = degM = k(p1).(pcm)

Bi 3.5 a thc t trng l

(t) = t3 ct2bta.

Ta s chng t y l a thc ti tiu. Tht vy, chn x0 = (1,0,0), khi x0,Ax0 = (0,1,0),A2x0 = (0,0,1) l c lp tuyn tnh. Gi s A l nghim camt a thc bc 2, tc l k1A2+k2A+k3I = 0, suy ra k1A2x0+k2Ax0+k3x0 = 0v ta c k1 = k2 = k3 = 0, iu ny l v l. Vy a thc ti tiu phi c bc 3,hay (t) = t3 ct2bta.

Bi 3.6 a) Sai, chn hn A=(1 11 1

), B=

(1 10 1).

b) ng. Gi s 6= 0 l gi tr ring ng vi vector ring x ca AB. Khi BA(Bx) = B(ABx) = Bx nn s l gi tr ring ca BA (v B(x) 6= 0). Nu= 0 l mt gi tr ring ca AB th BA cng suy bin, do BA cng c gitr ring l 0.

Bi 3.7 a thc c trng ca A:

A(t) = t2 (a+d)t+adbc

16

c nghim

t1,2 =12(a+d) 1

2

=

12(a+d

(ad)2+4bc).

t = 12(a+d+

(ad)2+4bc) v v= (x,y) l vector ring ng vi x> 0.Biu din hng t u tin ca Av ta c:

ax+by=12(a+d+

)x

2by= (da+)x.

Do b> 0 v da+> 0 nn y> 0. pcm

Bi 3.8 1) Gi () = |AE| l a thc t trng ca ma trn A. Gi P(t) la thc bc m v 1,2, . . .m l cc nghim (thc hoc phc k c bi) caP(t). Ta c:

() = (1)n(1)(2)...(n)P(t) = c(t1)(t2)...(tm).

Do P(A) = c(A1E)(A2E)...(AmE),

|P(A)|= cn|A1E|.|A2E|...|AmE|= cnm

i=1

(i).

Mt khc:

(i) = (1)n(i1)(i2)...(in) =n

j=1

( ji)

V vy

|P(A)|=cnm

i=1

(i) = cnm

i=1

n

j=1

( ji)

=n

j=1

cm

i=1

( ji) =n

j=1

P( j).

2) t p(t) = P(t) v p dng kt qu trn ta c:

|p(A)|= p(1).p(2)...p(n)

hay|P(A)E|= (1)n(P(1))(P(2))...(P(n)).

17

Vy cc gi tr ring ca P(A) l P(1),P(2), . . . ,P(n).

4. Hng v nh thc

Bi 4.1 Trc ht ta chng minh: dim(kerAtA) = dimkerA. R rng: kerA kerAtA, ngc li gi s v kerAtA th AtAv= 0, suy ra AtAv,v= Av,Av= 0hay Av = 0, tc l v kerA. Do vy dim(kerAtA) = dimkerA, t ta crank(AtA) = rankA.

Bi 4.2 Ta c:rankP= rankP(IPQ) = rankPQrankQ= rank(IPQ)Q= rankPQ

Vy ta c iu phi chng minh.

Bi 4.3 a) Ma trn con c c bng cch b dng 1, ct n c hng bng (n1).b) Gi s l gi tr ring ca A tc l det(A I) = 0. Theo cu a)

rank(AI) = n1 nn dimker(AI) = 1, suy ra khng gian con ring ngvi gi tr ring l mt chiu. Do A l ma trn i xng nn A c n gi trring k c bi. Vy A c n gi tr ring khc nhau.

Bi 4.4 a) Ta c det(AI) = (1)nn+ ...+ ci(1)ii+ ...+ cn trong cn =detA (ai j nguyn nn ci nguyn). Nu k l gi tr ring nn

(1)nkn+ ...+ ci(1)iki+ ...+detA= 0suy ra k l c ca detA.

b) Ly x= (1, ...,1) ta c Ax=mx nn m l gi tr ring ca A. Theo cu a)ta c m l c ca detA.

Bi 4.5 a) Ta c: detA= i> j

(aia j), do A kh nghch khi v ch khi cc aikhc nhau tng i mt.

b) Gi s f = c0+ c1x+ + cnxn l mt a thc bc n h s phc sao chof (ai) = bi, ta c h phng trnh n l ci, i= 0,n

c0+ c1a1+ + cnan1 = b1c0+ c1a2+ + cnan2 = b2

c0+ c1an+ + cnann = bn

18

h phng trnh trn c nh thc Crame khc 0 nn c nghim duy nht. Vytn ti duy nht a thc f bc n vi h s phc sao cho f (ai) = bi.

Bi 4.6 Xt hm f (t) = (1, t, t2) th f l hm lin tc. Khi nu ti, i= 1,2,3khc nhau tng i mt th

det

1 t1 t211 t2 t221 t3 t231 t3 t21

6= 0.Bi 4.8 Xt cc nh x tuyn tnh

LA(X) = AX

LB(X) = XB.

Ma trn ca LA v LB ln lc l:

MA =

= 1 0 2 00 1 0 21 0 3 00 1 0 3

MB =2 0 0 01 4 0 00 0 2 00 0 1 4

.Suy ra detL= detLA.detLB = 26.52, Tr(L) = Tr(MA.MB) = 24

Bi 4.9 Ly X = (xi j), ta c:

L(X) =

x11 32x12 x1332x21 2x22

32x23

x31 32x32 x33

.D thy mi ma trn Mi j u l vector ring ca L. Suy ra detL= 2.(

32)4 =

818.

Bi 4.12 Trng hp m > n. Ta vit T = T1 T2, trong T2 :Mnm Mnnc xc nh bi: T2(X) = XB v T1 :Mnn Mmn c cho bi: T1(Y ) =AY . V dimMnm = nm> n2 = dimMnn nn T2 khng n nh, suy ra T cngkhng n nh hay T khng kh nghch.

Trng hp m< n xt tng t.

Bi 4.13 Gi v1,v2, . . . ,vn+1 l cc vector c to l ct u tin ca cc matrn A1,A2, . . . ,An+1 tng ng. Khi n+1 vector ny ph thuc tuyn tnh.Do tn ti n+1 s thc x1,x2, . . . ,xn+1 khng ng thi bng 0 sao cho

x1v1+ x2v2+ + vn+1xn+1 = 0.

19

Lc ma trn x1A1+x2A2+ +xn+1An+1 c ct u tin bng 0 nn ma trnx1A1+ x2A2+ + xn+1An+1 suy bin.Bi 4.14 Do A l ma trn cp n c hng r nn tn ti cc ma trn kh nghchP,Q sao cho A= PIn,rQ vi In,r l ma trn c dng:

In,r =(Ir 00 0

),

(tc l ma trn c r phn t u tin trn ng cho chnh bng 1 cc phnt cn li bng 0). Ta c nhn xt sau: k ma trn X1, . . . ,Xk c lp khi v chkhi cc ma trn QX1, . . . ,QXk c lp tuyn tnh (do Q l ma trn kh nghch).Phng trnh AX = 0 tng ng vi In,rQX = 0, nn t nhn xt trn tms nghim c lp tuyn tnh ca phng trnh AX = 0 ta ch cn i tm snghim c lp tuyn tnh ca phng trnh In,rY = 0. Ma trn Y tho phngtrnh In,rY = 0 phi c dng sau:

Y =( r n r

r 0 0n r Y1 Y2

)Suy ra s nghim c lp tuyn tnh ca phng trnh AX = 0 l n(n r).Bi 4.15 Xem A l t ng cu tuyn tnh ca Rn. iu cn chng minh rank(AE)+ rank(A+E) = n tng ng vi dim(ker(AE))+dim(ker(A+E)) = n.Tht vy, vi mi x Rn ta c

x=12(x+Ax)+

12(xAx)

trong 12(x+Ax) ker(AE) v 1

2(xAx) ker(A+E).

Mt khc ker(A+E)ker(AE) = {0} nnRn = ker(A+E)

ker(AE),

suy ra dim(ker(AE))+dim(ker(A+E)) = n (pcm).Bi 4.16 Ta vit

A2+E = (A+ iE)(A iE) = (A+ iE)(A+ iE).Suy ra

det(A2+E) = det(A+ iE)det((A+ iE))

= det(A+ iE)det(A+ iE) = |det(A+ iE)|2 0.

20

Vy det(A2+E) 0 ng thc xy ra khi v ch khi a thc c trng ca Anhn i lm nghim.Bi 4.17 T gi thit ta c p(x) c hai nghim phc lin hp v , do

p(x) = (x)(x),

p(A) = (AE)(AE) = (AE)(AE).Suy ra

det p(A) = |det(AE)|2 0.

Bi 4.18 Do f (x) 0 x R v h s dn u bng 1 nn f (x) l tch ca cctam thc bc hai c dng x2+ax+b khng m vi mi x. Theo bi 4.17 ta cpcm.

Bi 4.19 Ta c (AAt +E) l ma trn i xng nn n l ma trn ca mt dngton phng. Hn na, dng ton phng ny xc nh dng. Tht vy, vimi x Rn ta c

(AAt +E)x,x= AAtx,x+ x,x= Ax,Ax+ x,x> 0.

Do cc gi tr ring ca A u dng, v vy nh thc ca A bng tch ccgi tr ring ca A cng dng.

Bi 4.20 Gii tng t nh bi 4.19

Bi tp b sung

Bi 1 Cho A l ma trn vung cp n, gi B v C l cc ma trn to bi k ct uv n k ct cui tng ng ca ma trn A. MCR, det(A)2 det(BtB)det(AtA).

Bi 4: Cho E l khng gian vector hu hn chiu v A Aut(E). Chng t cciu kin sau l tng ng:

(i) A= I+N, trong N l t ng cu lu linh.(ii) Tn ti mt c s ca E sao cho ma trn ca t ng cu A i vi c

s c mi phn t nm trn ng cho chnh bng 1 cn mi phn t nmngoi ng cho chnh u bng 0.

(iii) Tt c cc nghim ca a thc c trng ca t ng cu A (trong trngng i s) u bng 1.

21

Bi 6: Cho E l khng gian vector hu hn chiu trn trng phc. A Aut(E).Chng t rng t ng cu A c th phn tch di dng tng:

A= S+N,

trong S cho ho c, N lu linh v SN = NS. Chng t rng S v N cth biu din di dng cc a thc theo A.

Hng dn: Gi s PA(t) =s

i=1

(t ti)mi , Ei l ht nhn ca(A tiI)mi . Th th E l tng trc tip ca cc Ei. Xc nh S trn E sao choSv= tivi, t N = AS. Xt a thc g(t) = tigi(t), trong gi(t) c chnsao cho thnh phn ca Av trong Ei bng gi(t)vi. Khi S= g(A).Bi 7 Cho A,B l cc ma trn vung cp n, tho mn iu kin: AB= BA= 0v ImA kerA = {0}, ImB kerB = {0}. Chng minh rng: rank(A+ B) =rank(A)+ rank(B).

Hng dn Ta c rank(A+B) rank(A)+ rank(B). Gi s e1,e2, . . . ,ek vu1,u2, . . . ,us l cc c s ca Im(A) v Im(B) tng ng. Ta chng minh hvector e1,e2, . . . ,ek,u1,u2, . . . ,us c lp tuyn tnh trong Im(A+B). Tht vy,gi s iei+ ju j = 0, ta suy ra iAei+ jAu j = 0. T git thuyt AB= 0ta c Im(B) ker(A), do ta suy ra iAei = 0, hay A(iei) = 0. T tac iei = 0. Vy i = 0. Tng t ta cng c j = 0. Tm li ta c h vectore1,e2, . . . ,ek,u1,u2, . . . ,us l c s ca Im(A+B).

Vy rank(A+B) = rank(A)+ rank(B).

Bi 8: Cho A1,A2, . . . ,Am l cc ma trn vung i xng cp n tho mn iukin AiA j = 0,i 6= j. Chng minh rng:

rank(A1)+ rank(A2)+ + rank(Am) n.

Bi 9 Cho f ,g l cc t ng cu tuyn tnh ca khng gian vector V n-chiutho mn iu kin f g = g f , f lu linh v rank( f g) = rank( f ). Chngminh cc khng nh sau:

a) Im( f )ker(g f ) = {0},b) Im( f )ker(g2 f ) = {0},c) T suy ra f = 0.

Bi 9 Cho f l mt ng cu tuyn tnh ca khng gian vector V n-chiu. Gis V = L

N, dim(N) = m,0 < m < n. Chng minh rng tn ti s nguyn

k,(k n2m) sao cho V = f k(L)N.22

Bi 10 Cho l mt t ng cu tuyn tnh ca khng gian vector hu hnchiu V .

a) Gi s a thc ti tiu ca c phn tch p(t) = h(t)g(t), trong h,gl cc a thc nguyn t cng nhau. Chng minh rng: V = L1

L2, vi

L1 = ker(h()),L2 = ker(g()).b) Gi s a thc ti tiu ca c phn tch p(t) = h1(t) . . .hk(t), trong

hi(t),1 i k l cc a thc i mt nguyn t cng nhau. Chng minh rng:

V =k

i=1Li,

vi Li = ker(hi()),1 i k.

Hng dn a) Do h(t) v g(t) l hai a thc nguyn t cng nhau nn tn ticc a thc u(t) v v(t) sao cho 1 = h(t)u(t)+ g(t)v(t). Khi mi vectorx u c phn tch duy nht thnh x = h()u()(x) + g()v()(x) trong h()u()(x) L2 v g()v()(x) L1.

Bi 11 Chng minh rng nu v l cc php bin i i xng, trong xc nh dng, th cc gi tr ring ca u thc v cho ho c.

Hint Do xc nh dng nn tn ti php bin i to cng a v vdng cho. T ta c kt lun.

Bi 12 (Problem in net)

I have the following PROBLEM IN LINEAR ALGEBRA, I do not know theanswer. Assume that d and n are natural numbers and define f : Rd R by

f (x) = (d

l=1

cos2(xl))1/n,

where x = (x1, ...,xd). Hence xl is the lth component of the vector x. Prove ordisprove the following CONJECTURE: For any given x1, ...,xn Rd the (n,n)-matrix A given by

ai j = f (xi x j)is positive semidefinite, i.e., the eigenvalues are nonnegative. (Comment: I knowthat this is true for n 2d . So the interesting case would be n< 2d .)

Bi 13 Cho A,B l hai ma trn c tnh cht A2 = A,B2 = B. Chng minh rngA ng dng vi B khi v ch khi rank(A) = rank(B).

23

Bi 14 Cho A v B l hai ma trn thc cp n tho mn iu kin tn ti ma trnphc V sao cho A=VBV1. Chng minh rng tn ti mt ma trn thc U saocho A=UBU1.

Bi 15 Cho A l ma trn vung cp n tho mn iu kin A2 = A. Hy tnh athc c trng ca A.

Bi 16 Cho A,B l 2 ma trn vung thc cp n, gi s det(A+B) v det(AB)khc khng. Chng minh rng ma trn

M =(A BB A

)kh nghch.

Bi 17 Cho A l ma trn thc cp nm. Chng minh rng tn ti ma trn thcB cp n sao cho AAt = B2004

Bi 18 Cho phng trnh AX = B, trong A l hai ma trn cho trc cp n, Xl n (X l ma trn cp n). Chng minh rng phng trnh trn c nghim khiv ch khi rank(A) = rank(A|B), trong (A|B) l ma trn cp n2n c cbng cch ghp ma trn B vo bn phi ma trn A.

Bi 19 Cho A l ma trn cp n tho A2 = A. Chng minh rng phng trnhAX XA = 0 c nghim, cn v l: tn ti ma trn X0 sao cho X = AX0+X0AX0.Bi 20 Cho f l a thc h s thc c bc n > 0 v p0, p1, p2, . . . , pn l cca thc h s thc v c bc dng. CMR, tn ti cc s thc a0,a1,a2, . . . ,an

khng ng thi bng khng sao cho a thc Q(x) =n

i=0

ai(pi(x))i chia ht cho

f .

Bi 21 Cho A v B l hai ma trn lu linh, AB= BA. CMRa) IA kh nghch v A+B l ma trn lu linh.b) det(I+A) = 1.c) I+A+B kh nghch.

Bi 22 Cho N l ma trn (phc) lu linh v r l mt s nguyn dng. Chngminh rng tn ti ma trn phc A sao cho Ar = I+N.

Bi 23 Cho A,B,C,D l cc ma trn cp n, AC=CA. t M =(A BC D

). Chng

minh rng det(M) = det(ADBC).

24

Hint Gi s A kh nghch, ta phn tch: M =(I 0X I

)(A B0 Y

), vi Y = D

CA1B. Nu A tu th thay A bi AI v p dng lp lun trn.Bi 24 Cho khng gian vector E v E =MN, gi p l php chiu ln M theophng N. Cho u l ton t tuyn tnh ca E. Chng minh rng:

a) M l khng gian con bt bin ca u nu v ch nu pup= up.b) M v N u bt bin qua u khi v ch khi pu= up.

Bi 25 Nu u l ton t tuyn tnh vi trn khng gian vector hu hn chiu vnu u giao hon vi mi php chiu c hng 1, th u= I.

Bi 26 Cho u l ton t tuyn tnh trn khng gian vector hu hn chiu. CMRa) Nu u cho ho c v tn ti n N sao cho um+1 = um, nu v ch nu

u l php chiu.b) Nu u cho ho c v um = I vi mt gi tr m N, th u2 = I.

Bi 27 Cho u l ton t trn khng gian vector phc n-chiu. Ma trn ca u ivi mt c s no c dng:

M =

0 0 . . 0 10 0 . . 2 0. . . . . .. . . . . .0 n1 . . 0 0n 0 . . 0 0

CMR, u cho ho c khi v ch khi vi mi k {1,2, . . . ,n}, nu k = 0, thn+1k = 0. Tm a thc ti tiu ca u2.

Bi 28 Cho u v v l cc ton t cho ho c ca khng gian vector hu hnchiu E. CMR, tn ti ng cu tuyn tnh f ca E sao cho f u= v f khi vchi khi u v v c tp cc gi tr ring trng nhau v cc khng gian ring ngvi tng gi tr ring ca u v v c cng s chiu.

Bi 29 Cho u v v l cc ton t cho ho c trn khng gian vector E n-chiu.CMR, cc khng nh sau l tng ng.

a) uv= vu.b) Tn ti mt c s ca E gm ton cc vector ring ca u v v.c) Tn ti mt ton t w cho ho c ca E v cc a thc f ,g R[x],h

R[x,y] sao cho u= f (w),v= g(w),w= h(u,v).T suy ra, mt ton t trn E giao hon c vi u v v khi v ch khi n

giao hon c vi w.

25

Bi 30 Cho u1,u2, . . . ,um l cc ton t cho ho c ca khng gian vector En-chiu. CMR, cc khng nh sau l tng ng:

a) uiu j = u jui vi mi i, j [1,m].b) Tn ti mt c s ca E gm ton cc vector ring ca ui.c) Tn ti ton t w cho ho c ca E v cc a thc f1, f2 , . . . , fm

R[X ],hR[X1,X2, . . . ,Xm] sao cho fi(w) = ui,1 im v h(u1,u2, . . . ,um) =w.

Bi 31 Chng minh tnh cht sau ca nh thc Gram

G(a1,a2, . . . ,ak,b1, . . . ,bk) G(a1, . . . ,ak)G(b1, . . . ,bl).

ng thc xy ra khi v ch khi

ai,b j= 0 (i= 1, . . . ,k; j = 1, . . . , l)

hoc mt trong hai h vector {a1, . . . ,ak};{b1, . . . ,bl} l ph thuc tuyn tnh.Hng dn Trc giao ha h vector {a1, ...,ak,b1, ...,kl} thnh h vector trcgiao {1, ...,k,1, ...,l} v {b1, ...,bl} thnh {1, ...,l}.

Gi Li = a1, ...,ak,b1, ...,bk1 v Ni l phn b trc giao ca Li trong V. Tac

V = Li

Ni.

Qu trnh trc giao ha ta c

bi = yi+i,

vi yi =i1j=1

j b1, ...,bi1 v yii.Mt khc, ta c phn tch

i = yi+ zi,

vi yi Li, xi Ni.Hn na, ta c bi = i+ xi, vi xi Li v i trc giao vi Li nn i Ni.

Vy ta c 2 biu din bi = xi+i v bi = (yi+ yi)+ zi. Suy ra i = zi v do i= zi i.

Ta li c

Gr(a1, ...,ak,b1, ...,bl) = 1,1 ...k,k1,1 ...l,l=Gr(a1, ...,ak).1,1 ...l,lGr(a1, ...,ak).1,1 ...l,l=Gr(a1, ...,ak).Gr(1, ...,l) = Gr(a1, ...,ak).Gr(b1, ...,bl)

26

Bi 32 Cho A l ma trn i xng thc cp n vi cc nh thc con chnh ukhng m, A1 l mt ma trn con cp k (k < n) gc trn tri ca ma trn Av A2 l ma trn con cp kn gc di phi ca ma trn A. CMR,

det(A) det(A1)det(A2).

Bi 33 Cho V l khng gian vector n chiu vW l mt khng gian con m chiuca V , (m< n). CMR, tn ti mt c s ca V khng cha mt vector no caW.

Hint Gi {v1, . . . ,vm} l c s ca W v {u1, . . . ,unm} l c s ca phn btuyn tnh caW trong V. Khi c s {v1+u1, . . . ,vm+u1,u1, . . . ,unm} chnhl c s cn tm.

27

11th Vietnamese Mathematics Olympiad for College Students 2003

A1. A is the 4 4 matrix a11 = a22 = a33 = a44 = a,a12 = a21 = a34 = a43 =b,a23 = a32 = 1, other entries 0, where a,b are real with a > |b| . Show thatthe eigenvalues of A are positive reals.

A2. B is the 33 matrix with b11 = a,b22 = d,b33 = q,b12 = b ,b13 = c

,b21 =

b ,b23 = p

,b31 = c

,b32 = p

, where a,b,c,d, p,q are reals and ,, are non-

zero reals. Show that B has real eigenvalues.

A3. Dk is the kk matrix with 0s down the main diagonal, 1s for all other entriesin the first row and first column, and x for all other entries. Find detD2+detD3+ +detDn.

A4. In denotes the nn unit matrix (so I11 = I22 = . . . = Inn = 1, other entries0). P and Q are nn matrices such that PQ = QP and Pr = Qs = 0 for somepositive integers r,s. Show that In+(P+Q) and In (P+Q) are inverses.

A5. A is a square matrix such that A2003 = 0. Show that rank(A) = rank(A+A2+ +An) for all n.

A6. A is the 44 matrix with a11 = 1+x1,a22 = 1+x2,a33 = 1+x3,a4 = 1+x4,and all other entries 1, where xi are the roots of x4 x+1. Find det(A).

A7. p(x) is a polynomial of order n> 1 with real coefficients and m real roots.Show that (x2+1)p(x)+ p(x) has at least m real roots.

28