Ebook bài tập Đại số tuyến tính

nghiem dac trung cim A dgu dvong (Wang ung dgu am).. Vi tray ngu dal IA ma tran ma cam cot la Y„ thi P IA ma trait trvc giao.. Chung mink rang ten 41 ma teen khong suy Bien C Ct. [r]

PHAN HUY PHIJ • NGUYEN DOAN TUAN

BAI TAP

DAI SO TUYEN TINH

Chin trach nhiem xual bcin

Gicim doe: NGUYEN VAN THOA

Tong bien Op: NGUYEN THIEF N GIAP

Bien tap: HUY CHU

DOAN 'MAN NGOC QUYEN

Trinh bay Ilia: NGOC ANH

BAI TAP DAI sq TUYEN TINH Ma s6: 01.249.0K.2002

Lai NOI DAU

Mon Dai s$ tuygn tinh dude dua vao giang day a hau hat cac trUnng dai hoc va cao dang nhtt 1a mot mon hoc cd se; can thigt d@ tigp thu nhUng mon hoc khan Nham cung cap them mot tai lieu tham khao phut vu cho sinh vien nganh Toan vi cac nganh Ki thuat, chting Col Bien soan cugn "BM tap Dai so tuygn tinh" Cugn each dude chia lam ba chudng bao g6m nhUng van d6 Cd ban cna Dal so tuygn tinh: Dinh thfic va ma

trail - Khong gian tuygn tinh, anh xa tuygn tinh, he phticing trinh tuygn tinh - Dang than phttdng

Trong mOi chudng chung toi trinh bay phan torn tat lY thuyat, cac vi du, cac hal tap W giai va cugi mOi chudng c6 phan hudng dan (HD) hoac dap s6 (DS) Cac vi du va bai tap &roc chon be a mac an to trung binh den kh6, c6 nhUng bai tap mang tinh 1± thuygt va nhUng bai tap ran luyen ki nang nham gain sinh vien higu sau them mon lice

Chung toi xin cam on Ban bien tap nha xugt ban Dai hoc Qugc gia Ha Nei da Lao digt, kien de cugn sach som dude mat ban doe

Mac du chting tea da sa dung 'Lai lieu nhigu narn cho sinh vien Toan Dal hoc Su pham Ha NOi va da co nhieu co gang bier, soon, nhUng chat than có khigm khuygt Cluing toi rat mong nhan dude nhUng y kin clang gap cna dee gia

rvikic LUC

Chubhg .1: DINH THOC - MA TRA:N

A - Tom tat ly thuyeet

§1 Phep th6

§ Dinh thitc

§ Ma tram 10

B - Vi dn 12

C - Bei tap 35

D HtiOng dein hoac clap so 43

Chudng KHONG GIAN VECTO - ANH XA TUYEN TINH

• PHUGNG TRINH TUYEN TINH 57

A - TOrn tat ly thuyeet 57

§1 Kh8ng gian vec to 57

§2 Anh xa tuyeen tinh 61

§ He phydng trinh tuy6n tinh 64

§4 Can true caa tai ding cku 67

B Vi dtt 71

C - Biti tap 96

§1 'thong gian vec to va anh xa tuyeen tinh 96

§2 He pinking trinh tuy6n tinh 104

§3 Cau tit cna melt tu thing calu 106

§1 Khong gian vec td va anh xn tuyin tinh 11(

§ He phudng trinh tuyeit tinh 12';

§3 Cau trite dm mot tg ang cau 12Z

Chtedng DANG TOAN PHUONG - KHONG GIAN VEC TO

OCLIT VA KHONG GIAN VEC TO UNITA 134

A Tom Vitt 1t thuyeet 134

§1 Dang song tuy6n tinh aol xUng va dang town phuong 139

§ Killing gian vec to gent 135

§3 Khong gian vec to Unita 142

B Vi du 14E

C - Bai DM 174

D Hitting dan hotic ditp so 179

Chuang

DINH THUG - MA TRAN

A - TOM TAT Lt THUYET §1 PHEP THE

Met song anh o tit tap 11, 2, met phep the bac n, ki hieu la

'1

n} len chinh no duet goi la

\GI a2 G3

15 del a, = a(1), 02 = a(2), , a„ = a(n)

Tap cac phep the bac n yeti phep nhan anh xa lap met nhom, goi la nh6m del xeing bac n, ki hieu S S6 cac Olen t3 cua nhom S„ bang n! = 1, n

Khi n > 1, cap s6 j} (khong thu tv) dude pi IA met nghich

the cem a n6u s6 - j) (a,- a) am Phep the a &foe goi la than

ndeM s6 nghich thg cim a chan, a &toe goi la phep the le n6u s6

nghich the ciaa a le

1 neM s la phep the chan

Ki hieu sgna =

-1 net} a la phep th6 le

va sgna goi IA deu am, phep the a Neu a vat la hai phOp the

cling bac, thi sgn(a = sgn(a) sgn( )

va a(i) = i vdi moi i x i„ i k Vong )(felt dttoc ki hieu IA ik) M9i phep th6 dau &tan tfch the tfch nhung yang xfch doe lap

Met vOng xfch dal dude goi IA met chuygn trf Vong xfch ••• , i k ) phan tfch chive tfch ,

§ DINH THUG

I Gia sit K IA met trueng (trong cuan sich to din yau xet K la &Ong s6thvc K hoac truang s6 phitc C) Ma tran kidu (m, n) vdi cox phan tit troll twang IC la met bang chit nhat gfim m hang, n cet cac phan tit K, i = 1,m, j = 1,n Tap cac ma tran kidu (m, n) chive kf hieu M(m, n, R) Ma trail vuong cap n IA ma tran co n dong, n cot Tap cac ma trail vu8ng cap n vdi cac phan tit thuoc truong K ki hiOu IA Mat(n, K)

2 Cho ma tr4n A vuong cap n, A = (ad, i, j = 1, 2, , n Dinh thitc ciia ma tran A, kf hieu det A la met flan tit dm K dude xac dinh nhu sau:

detA = zsgn(a)a mo)

E Sn

3 Tinh eh& ceta Binh that

a) Neu dgi cho hai dong (hoac hai cot) nao cim ma tram A, thi dinh auk cim no ddi da:u

b) N6u them met dong (hoac met cot) cim ma tran A met to hdp tuygn tinh cim nhUng thing (hoac nhung khac, thi dinh auk khong thay ddi

fan ail

de a21 +alci

‘a n„ a,,, +ani a ll al; .a 1,„ all = det a21 21 +det a21

—a1111/

an,

a2„

a2 n

" S 'Ill " S IM /

c) Ngu mot Bong (hay mot phan tfch tong, thi dinh thitc dU9c phan tfch tong hai dinh thfic, cv th6:

d) Cho A = (It o) E Mat(n, K), thi = b) a do = aij &toe

goi la ma tran chuy6n vi cim A Ta co detA = detA t

4 Cdch tinh dinh that

a) Cho ma tran A E Mat(n, K) Kf hi'911 Mi; la dinh that cua

ma trail alp (n-1) nhan dine bAng cach gach be clOng thU i, cot

thu j cut ma tram A vb Aij = (-1)H M u clucic g9i la pha'n phu dai s6cUa phgn to a ii cna ma trait A Ta có CAC tong thtic:

O ngu i k

det A ngu i = k

O ngu i x k det A ngu i = k Nhu fly detA = EamAki (k = 1, 2, n)

1=1 heat detA = Z

•

CUT thac tit throe goi la cang thdc khai trim dinh tilde theo (long hay theo cot

b) Dinh 1ST Laplace

Cho ma Iran A = (a, J) c Mat(n, K) Vo; rn6i bQ ;2.••, ix), s i, <1 < <

ik va Oh ik), =11 < j2 < ••• <jk n, s k< n, dat A la

11-11,

dinh thac caa ma tran vuong cap k nam d cae &nig i„ i k va cac

cot j, jk cim ma tran A; M h la dinh thdc cita ma tran vuong cap (n - k) c6 dude bang each gach the clang thu ik va cac cot thu j , , jk caa ma trail A Ta c6 kgt qua:

detA = ZED' Si k+31+ A ik

do j1. jk la k cot cgdinh Tgng &toe lgy theo tat ca cac bQ ik) cho < i2 < < i k Cong thdc troll dude goi la c8ng thdc khai trim dinh thae theo k cot ji, 'along tV, to có gong thdc khai trim theo k clang Khi k = 1, to &roc gong tilde da not muc a

§ MA TRANI

811 a19 aln

A= a,1 a99 aon

amt amt arnn,

2 Cac phep todn tren Mat(m, n,

Cho A = (a y ), B= (b,j ) thuOc Mat(m n, K) Ta có:

a) Ma tran C = (cg) a do cy = a ii +

&toe goi la tong cua hai ma tran A va B va ki hien la A +B Ma tran D= (d,,) a do di; = a ij -

dude goi la hiOu cila ma trail A va B va Id hi'eu la A - B

b) Vdi k E At, ma trail kA c8 cac phAn to la (ka ii ) duoc goi la tick cua ma tran A vdi ph&n td k cua trudng K

c) Neu A = (aii ) c Mat(m, n, K) va B = (bp() e Mat(n, p, K) thi

ma tran A B E Mat(m, p, K) ma cac phAn tit &tele xac dinh INN AB = (c, k), a do

eik = Zaijbjk

5=1 &toe goi litich caa hai ma tram B ye A

d) Tap Mat(n, K) ma tran yang cap n vdi phep toan cOng lap mot nhom giao hoan, vdi phep Wan rang ma tran va phep nhan ma trail lap mat vanh khong giao hodn, co don vi

3 Hang ctia ma tran; Ma trim nghich ddo

Gig A E Mat(m, n, K), ta dinh nghia hang ciat ma tran A

la cap cao nhgt cua dinh thric khgc khong rut W ma tran A KM A E Mat(n, K) va hang A = n (ta cling dung ki hi3u hang

A la rang A) thi ma tran A goi la khong suy bign, detA * va ton tai nhgt ma tran B thuOc M(n, K) A.B = B.A = I„; d I lit ma trail don vi Ma tran B &roc goi 11 ma tran nghich dgo cna ma tran A va ki hi3u la A'

Gig su A= (A u )la ma trail plw hpp cim ma tran A = (ad, Ab la Olga Ow dal see mitt phgn ht aii ; A t la ma tran chuya'n vi cua A Khi do:

At detA

B- VI DTI

Vi cla 1.1 Xac dinh clgu rim cac phep th6 saw a) a 11

2

b) 5=r1 I n n+1 n+2 211 2n+1

ll ai-2

Lai gidi

a) Phan tick) a tich cac chuydn tri: =

(1 4 5)

= (1 5) = (1, 5) (1, 3) (1, 2)

(chi) la pile)) nhtin cae chuydn tri dude thuc hign tii phai sang trai nhu hap cua the Anh xa)

Vay sgna = (-1) s = -1

Co the lam each khan: Cam nghich the cua a la (1, 5), (2, 5), (3, 5), (4, 5), (3, 4)

Vay a có nghich the nen sgna = -1

b) Ta hay tinh sS nghich the cua boa)) vi (1, 4, 3n-2, 2,

1 khong tham gia vao nghich the

4 tham gia yen nghich the voi the s6 thing sau no 7 tham gia van nghich the

3n - tham gia vao 2(n - 1) nghich the voi the se dung sau no khong tham gia vao nghich the nao vdi the se dung sau no tham gia yen nghich the voi the se dung sau n6

S tham gia vao nghich the vdi cae s6 dUng sau n6

Vay co tat ca + + 2(n-1) + + + (n 1) - 3(n -1)n

(n-1 )n

nghich the hoot vi da neu va d6 sgn S = (-1) Khi n = 4k hoac n = 4k + thi sgn =

con neu n = 4k + ho4 n = 4k + thi sgn = -1 Vi du 1.2

_

Cho phep th'e' f - en dgu la (-1)

12 fn

Hay gag dinh da"u dm: a) 1-1

b) g = (In

2

fn-) • n ) Lift( gidi:

a) Vi sgn f sgn = sgn (f

= sgn(Id) = non sgn (e 1)= sgn (f) = (-1) k

b) X4t phep the"a = nj n -1 thi g = f a

Do gay sgn g = sgnf sgn a n(n-1)

Vida 1.3

ChUng minh rang vier nhan mat phep th6 vdi ehuy6n trf j) v6 ben trai Wring throng v6i viac dai cha car s6 i, j a clang

drah cna phep the Cling nhu vay, nhan mat phop th6 Nth ehuynn trf (i, j) v6 been phai tunng during voi del eh?, I, j a dong tit 66a phep th6

Lo gidi Gia sif a la truong hop nhan

Gitisi2 a=

Theo d nh nghia

phop th6 cho j) la phep ben trai tile la f = (i, j) a

(3 n

1 (i, j) = (

9

_ (1 2a

ai

nj n

)

chuy6n tri Xet

Trunng hop nhan ben phai dude ant Wring fib

Vi dy 1.4 Cho f va g la hal phop th6cua n strtn nhien clAu tien a) Chung minh rang có the' cilia f va g bang khong qua (n-1) phop chuyan trf (nghia la ton tai k phop chuyan trf a l , cr2 , ak , k 5.11 - g = ak ak _, a, f)

La gicii

a) Xet phep the g o f', phan tich g o e' thanh tfch cac vang xich dOc lap T , Tp

g o e' = T T i Neu kf hiOu m i nt do, dai cart yang 'dell T i thi

± m2 + =

rang mOt vong xfch (a l , a2, am) la mOt plop the a cac s6 tv nhien Ui den n cho a(a) = a 1+1 (i = m-1) va a(a„,) = a l , a(l) = 1 nen yen moi i = 1, ,., m Vong xfch (a l , a2, u„) goi la ce do, dai m

Ta da hiet rad yang xfch do, dai m deu phan tich duo m - chuyen trf Vi vay g o e' phan tfch duo 'Lich caa i(m i -1)= n -p = k phep chuyen trf

i=1

Nhungpa n-1:115n-1

Nhu vay g o f -1 = ak (s, - chuyen tri) TV g = ak ak.,, 0 f, k n - va cac a, la cac phep chuyen trf

2 nj

ri

b) Cho g = la phep the ddng nhA

(1

n Oa n-1)

f ve g duo Wang it hdn n - phep chuyen trf

' WA met phep the" h =

1

ta not rang la

n

phan tit chinh quy ngu i De" rang neat nhgn vao ben trai cua h met chuygn Lri thi A:C 1)111in tit chinh quy tang cling lAm la met don vi That way, ngu ngudc kg, cheng hen i, j la hai phan tit kheng chinh quy cim h ma nob d6i dig h i voi hj ta Jai &tog hai phign tu chinh guy (cum phep th6 m6i) th6 thi: h, < i h, < j nhang hj i, h; j vti 1Y

Do f chi co met phAn tei chinh quy ye g co n phan tV chinh quy, vi vgy khong thg clua f vg g bring it hon n - phep chuygn tri

Vi du 1.5

Chung minh rang vdi mei so k (0 <k < C;) t6n tai met phep thg a e S„ co dung k nghich thg

1231 giai

Ditch 1: Ta hay cluing minh met It& gug manh

Nigu a = (a,, la met hogn vi cda 1, 2, n va ace 1: nghich thg, < k < , thi có thg ct6i che hai phfin tri nao de thu ridge hog') vi c-3 co k + nghich thg That Nay, int& hgt ta nhan thy rang negi > vdi moi i = 1, 2, , n-1 thi

co

a nghich the' Vi vay, so' nghich th6 caa a la k < , nen ten tai i o dg oc ii) < a i0+1 ;

Xet hoan vi p = 0„) 111 i = a, ngu i # i„, i„ + 1,

con p io = a ;0+1 , p i+, =a,„ thi HI rang 13 co nhigu hon a met nghich the" Nghia la s6 nghich th6 ciga p la k +

Ta tha'y A= B ca do

1

1

B= vi C=

11

t1

Nhan cot thu nhal ciia ma tran A vdi -k rdi cOng vac) cot

this k, ta dude:

1 -1 -2 -(n-1)

1 -1 - (n - 2) 0 - (n - 3)

1 0

det A =

Khai trio'n the() dung Ulu n, ta ea: -1 -2 -(n -1)

-1

= (-1)"+1 (-1)"-'=1

Cdch

ma detB =1, detC = nen detA = detB detC = Vi du 1.9 Hay tinh

cosa 0

1 2cosa

= 2cosa 0

2cosa

Lai gidi:

Khai trim dinh thac Lheo cot cub" to co D„ = 2cosa - 17,1_2

De thay D, = cosa

cosa

2cosa, = 2coi2 a - = cos2a Gia sa D1 = cosia \TM moi = 1, k

Ta có

Dk,I = 2cosa Dk - Dk_ i

= 2.cosa coska - cos(k -Da

= (cos(k+Da + cos(k-1)x) - cos(k-1)a = cos(k+1)a Nhn vay D„ = cosna

Vi 1.10 Hay Dull

1 eP + e -(1)

a the phan to tren &tang choo chinh bang va band eq) +e -9 ; the phan tit tren hai &tong xien Win nhat \TM (Mang the() chinh bang 1, the phAn ta khac bang

+ a-9 1 e" +e

A„ =

0

0

Khai trin theo cot tht nhEt, to c6: A n = (e P +e -P)A n _i:

e 21' - e -2(P Nlinn xet rang = +Cc =

A2 = ((i e ro e

636 - -39 e (P - e (1+1)6 - e -(lrv1),p

Girt sit AR - e(-0

, k - 1, 2, , n - -e (P

Ta c6 An = (e c e - P)A n _ i -A n _,

=(eP +e w)enip - e npe(

n-1)n4) - e e(:+1)o - e-(n+1)4'

ew e q) - —

e(n+1)p - e-(n+1),p

Nhu v 6}.: An =

Vi du 1.11 Tinh:

ll a a, a n

1 a l: +h i a, a n D = dot a a, +b., a n

a a.9 a n +b n -

Lai gidi:

LAY ciOng dAu nhan vOi -1 r6i Ong vao cac thing 1ai, to có D = b, 1) .b„

Vi du 1.12

Cho da thric P(x)=x(x+1) (x+n) Hay tinh Binh thdc:

P(x) P(x)

P(x +1) P(x +1)

P(x + n) P(x +n) d =

P(n-1) (x) P th-1) (x+1) P th-1) (x + n) P thl (x) Pthl (x +1) P (n) (x +n) gidi:

Ta b6 sung de' dude ma Han dip (n+2):

P(x) P(x +1) P(x +n)

P(x) P4x+1) P4x+n)

D =

P(n ) (x) Pthd(x +1) P0P(x + n) P(„+l) (x) 1301+1) (x +1) pg+0( x +n ) RO rang det D = d

Nhan dOng 1111 k cua ma trail D vdi (x +n

dc-ix( 1) k-1 r6i (k-1)!

Ong vao clang Hirt nhgt vgi tat ca k=2, n+2) Khi do, phAn tii dung dau có clang:

poc + + P(k) (x +0.(x +11.) k o k = n) k!

Pkx+1.) P(x+n)

PTUx +1) P(n)(x+n) PT+I kx+1) PT+I kx+n)

d = del D - (x+n)°+i

(n+1)! (-1) 1T1 (x +n) n+i

n; phan tit a cot cuoi bang nghia 13 clang thg nhaa c6 dang:

(n+1)!

(0, 0, , (+1)"+1(x+n)ni ) (n+1)! Do

Ta ki hieu dinh thge a v6 phai bai C va ma tr5n Wong ring

hai (6-1 Vi da thge P(x) = n(x + i) Ken P'" 0(x) (n+1) !, vi vay

i=0

the s6 hang a dOng cu6i &au bang (n+1) ! dgn gian ki higu va each viek ta dirt xk= x+ k, k = 0, 1, n d dOng thg hai tit &leg len ciaa ( ta co:

(Pw(x0), P("ax,), PT)(x.,)) ((n+1) hco + a l , , (n+1) tx„+

a do a l la hang s6 &do Khi nhan (long cue"' ding voi r6i Ong vao clang trail no, ta dtta ma trgn ( ) va dang

P(x ) P(x ] ) P (x i) )

P (11-1) (x ) Pth-e (x l ) PT-1) (x n ) (n+1)!x (n+1)!x (n+1)!x n

(n+1)! (n+1)! (n+1)!

a l

(n+1)!

Deng this ba tit dUdi len caa ma Dan (*) co dang (n +1)!

xo +a i x o +a,, „(n+1)! x n2 +a i x n +a.,

2

Cong vac) (long hai clang °Ma sau nhan voi cac s a,

va a1 ta nhan dude clang (n +1)! (n+1)!

(n+1)! (n+1)! (n +1)

xo ,

2 2

Bang each bidn ct6i nhu vay vdi cac dong lai, ta clan m Dan ( '61) ve clang sau ma khang thay d6i clinh thfic caa no

c = det = det

x on X n

n-1 x n1

X0

(n+1)! = (n+vn

k=1 IC!

X0

n(n+1)

(-1) ((n +1)!6'1 D n

Jk!

6 do D„ la dinh thiic Vandermonde cua clic so" x„, x„ n-1

D6' thay D„= n(x k - x i )= n (c_i) = 11(n-o!= 1

k>i>0 k>1 1=0 k=1

Vay

d = detD = (-1) [(n +1)! i n (x +n)n+1 Vi du 1.13

Gia sU A e Mat(n, K), A= (au) a do ali = vdi moil= 1, , aij bang hoac 2001 \Ted i t j Chung to rang ne"u n chan, thi det A #

Lbi gidi:

Nhan xet rang neh to them vim mit phan to a jj nao cha ma tran vuemg A mot s6 than, thi dinh thfic cha ma tran nhan dticic se sai khac vat dinh thiic cha ma tran A mot s6 than Vi the ne'u to hat di 2000 don vi a nhUng phan t,i bang 2001 cha A, thi tinh than le cha dinh thfic cila A khong thay d6i, nghia la:

detA = detB (mod 2) a B = (14 hi; = n6u i= j va = ngu it j Ta có:

0 1 det B =

nhan Bong ddu veli -1 fee cling vao cac dOng lei ta dttuc:

detB =

1 0 -1

Deng vao cot thu nhat ca car cot lai ta cO: detB = (n-1)

Vey n than thi detB la sidle, vey detA # Vi du 1.14

Tinh Binh theft:

' 1 1

0 C C I C i

2 n+1

D = det q C., C 241 Cn+2

n-1 n-1

c c C.!;1-11-.) C2-11

n n+1 2n-1 >

LtlL gill

1 0 0

C12 1 1

D = det Cy c?c, C1 C1„1

" •

011 C1,1,12 C112 cn-2 2n-2

lai lam nhtt tren, to co

1 0

C l2- 0

D = det C3 C13 1

c^-3 n-3 2n-4

sau n -1 budc nhit v'ay, to

1 0 0'

1 0

D = det

cn-1

CI3

Lin-2

1

cn-3 CI

0

1

= 1

VEty D =

1.71 du 1.15

Cho A=

cos9 - sin9 sin9 cow ,

fray tinh A n

Lai gidi:

cosy -saw cos9 -sine cos2p -sin29 Ta co A

Gia A k = cosky -sinkp vOi k = 1, n-1 sink cosk(p

Ta ttnh

cos(n - lap - sin(n cosy - situp A" = A n-1 A =

sin(n -1)9 cos(n -1)9 \ sing cosy , cos (n -1)9 cosy - sin(n -1)9 since -cos(n - lap since- sin(n -1 cosy sin(n - lap cosy + cos(n - lap situp cos(n - lap cosy - sin(n -1)9 sirup

cosmp - sinmp sinmp cosny

Vay

= cosny - sinny

vdi 11191 n E N

sinny cosny , Vi (11 t 1.16

0

Cho A = hay tinh A 200 "

Lbi gidi:

-1 0 -1 (1

Ta c6 A =

0 -1 0

ma 2000 chia het cho

\ray A 200° =(A) 500 =I, (I, la ma tran thin vi cep hai) Vi du 1.17

Hay chang minh: 'rich caa hai ma tran phan doi xang A va la mat ma tran phan doi 'ding va chi AB = -BA

Lai gidi:

Gia sid A = (ad, B = (b id d ait + ai, = 0, b y = di mm L j = 1, , n

Dat C = A B = (c,); = =1 D = B A = (c1,0 = Ib ija jk

id]

Ta ca.: c,k = Za ji b ik - Eb ji - d ki

Nha vay AB phan del xang <=> c, = V i, k

tt=> = vet mot i, k c> AB = -BA Vi chi 1.18

Cho A, B la hai ma tran vuong c9p n Hay minh let(A.B)= detA detB

gidi:

G1a sit A = (a,,) , B = (It o ) yea i, j = Ta lap ma tram

an a1.,a l n 0

a91 a22 a 2n 0

ant a

n11

-1 bth

0 b21 b2n

C=

Khai tri6n theo n clang dal) (thee dinh 12) Laplace), ta co

detC = detA detB ( )

Mat khac, bi6n &it ma Iran C bai phep bi6n ct6i sd ca) sau: Nhan cot thu nhal vat b 1, cot thil hat Null cot thin n vet btl; r6i Ong vac) Ot thu n + j (j = 1, 2, n), ta dude ma tran D Bang sau ma dinh thiic cua D va cua C bang nhau:

a l a a d ta d

a21

a m a m a nn d m d,, d nn

-1 0

0 -1

0

a = ria ik b ki , nghia la (dij) = A B

k=1

Khai tri6n theo n cot cuoi (theo dinh 157 Laplace) ta có detD = det(dij) = det(A B) (2) Tit (1) va (2) va detC = detD nen ta co:

det(A0 B) = detA detB

Virtu 1.19

Cho X la ma tran vueng cap n Chung minh rang X giao hean vOi moi ma tran vueng Ong ca) = X co clang XI,,, a I„ la ma trail dun vi cap n

Lari gidi:

Ni111u X = 1 thi re rang X giao ho an vdi moi ma Han \along cling cap

Ngnoc lai, gia sit X = (X„) giao haul vdi moi ma tran yang cdp n

VOi i„ # j 0, to chiing minh x inio = Mudn stay chon A = (ad a joio =1 one phan tit khac ddu bAng Thong Phan to ding i o cot j„ cim ma tran XA Wing x •' can phan tit a thing io cot j, cim AX la Tir di6u kien AX = XA suy =O Nhu y X co dang:

k r

„

0 k„

Cho ma tran A = (a) a do a ] , = vii moi i, j Khi phan tit

o dung i cat j cim ma Han XA la X , phan to ding i cot j dia ma tran AX la nen k, =

Vi vay:

X= I,, Vi du 1.20

Cho ma Iran cap n:

a b b a A=

b

a) Chung minh detA = (a-b)° (a + (n-1)b)

b) Trong trudng hOp detA s Hay Huh ma trdn nghich dao A+ oda ma tran A

Lai giai:

a) Gang the clang vao dOng this nhat ro'i nit a + (n-1)b a clang dau, ta (Woe

1 b a b detA = (a + (n-1)b) x

b b a

Lay dOng chin aria (Anil th.fie tit nhan voi -b r6i Ong vao the dong sau, ta eó:

1 0 a - b detA = (a + (n-1)b) x

0 a - b detA = (a + (n-1)b) (a-b)" -i

h) A kha nghieh <=> detA (=> a b va a + (n-Gb O Goi B = (b ii) la ma tran nghich (lac) cila A = (a„)

1

Ta Wet rang b id = detA A„ a A.; la phAn Phu dai so etia phan ma tran A

a b b a

Vol mdi i thi = la nh thiic cap (n-1)

b b a

A a+(n-2)1( Vay b„ -

det A (a 1- (n -1)1).(a - b)

Vol i # j = (-1)H , d kl„ la chub thue e6p n-1, co dupe ba' lig each x6a &Ong thu i va cat tha j eila ma tran A Do A dovi ming nen = Gia sU rang i < j, cot thfi i va dung thu j-1 cua M o gain town nhang phan t& b N6u d6i eh Bong j - len tren Ming dau (Mu nguyen cac dung khac), rei lai MS( nit i len 6;4 thu nhai (va van gib nguyen the cat khac), thi to (Moe:

m = = (-1)' 34 b.(a-b)" -2

Nha v(6y b

dot A (a +(n -1)b)(a - b) Do

a +(n -2)b -b

(a +(n -1)b)(a -b)

-11 a +(n -2)b

C BAI TAP

1.1 Chung to rang mai phep chuyan tri dm (n 2) lh mot phep the le

1.2 Hay phan Lich eau pile') the! sau tieh ene phep chuy6n tri

1 ' a)

8 ) (

b)

6

1.3 Tim s6 tat ca the phep the a e S„ oho a(i) x i vol

mm i = 1, 2, n Cluing t8 rang n chan, so" one phep th6 clang tren la m54 3616

1.4 Ki hi8u (n, k) la secac hoan vj rim 1, n ce dung k nghich the Chiing minh cong thile truy 146i sau:

(n+1, k) = (n, k) + (n k-1) + + (n, k-n) vdi guy vac (n, j) = nevu j < hoac j >

1.5 Ta goi 45 giam ena phdp the f Ia hi5u cna s6 the phan tic khting bat dog (nghia la s6 the pilau tit i ma f(i) i) va s6 clic Yong xich <IQ dai ion him phan tich cua f tich car yang xich 458 lap

a) Chung minh f cotang tinh chat chan la vOi giam cua no h) Chling minh rang s6 161 thidu one nhan tic phan tich cna f tich the chuydri tri bgng 45 giAm cua f

1.6 DM \h hai s6 x va n nguyen, n x 0, to ki hi5u r(x, n) IA s6 chi chia x cho n : r(x, n) < n Chiang minh rang nal n > va a la so? nguyen, nguytin to' d61 vdi n, thi tthing Ung ki—>r(ak, n) la mot phan tit cua S„,, k e 11, , n-1}

1.8 Xac dinh davu cua cac phop the' sau: a) I

n n +1 n + 2n 2n +1 3n

43 3n 3n-2

b) n+1 n+2

2n 2n 2n-1, 1.9 Chgng mink rang:

a) Dinh thiec cap ma cac Phan to Wang +1 hoac -1 le mot s6

b) Dinh attic ay Ion nhig la

c) Dinh thew c)41) ba ma cac pga'n tit la hoac dot gia tri IOn nheit biing

1.10 Tinh dinh thew cap 2n D = det(cM), a vai i= j

dO di; = b yen i + j= 2n +1 i=j vet i+j 42n+1

1<i, j S 2n

1.11 Cho A = (an) la ma trn cap n, a1 e R Chung minh rang detA khong thay dei, neit cac phein tei a i; ma i +j 15 dooc thay bai so' dal cga no

1.12 Chung to cac dinh thew sau day bang khong: cosa costa cos3a

cosa costa cos3a cos4a a)

a)

c+-1 b+-1

c b

2 a + —1 a

1 c+-

c

b + a+

b a

(abc -1)

b) = 1

-1 a) Dn = -1 -1 -1 -1 -1

0 -1 -1 -1 1

-b1 a1 - b2 a l -b n

a, -b1 a2-b2 a, -b a

h) ; n >

a n -b a n -b.,-b n n 1.13 Cho hai ma tran A B cho:

5 11 14

A B = I, B A -

J1 25 ) y

Hay tam x, y va A, B

1.14 Hay khai tridn dinh thiic va chilng minh:

b)

(a+b)2 b2 a2

b (b+c) 29 c - a = c = (a + c)

= 2(ab be + ca)

1.16 Hay tinh dinh thdc sau:

1+x'2 x 0

x 1+x x

a) D„= x 1+x

0 1+x

nghla Dn la dinh thiic cap n ma cac phan tit tran during cheo chinh bang + x2 , the plidn tit thuOc hai [Wang chat) grin during cheo chinh bang x, the phOn tit lai bang

2 0 0

1.17 Cho da thite P x) = (x - al)(x - (x - a„)

a d6 a, la cac so thvc dot mot phan biet Hay tinh climb thde sau: P(x) P(x) P(x)

x-a x-a, x-a n

1 1

a l a, a n a;

a

an-2 n-2 an -2

1.18 Xet hai ma tram ph& cap ha:

'a b c ' 1

A= c a b va J= j

b c a, v1 j-

2n

dO j = e = cos— + sm-2n =

3 2

Way chang minh det J # Hay tinh ma tr'nn AJ tit suv gin tri detA Hay nen hal Loan Wong to kin A ya J cite ma 1.rn cap n

1.19 a) Hay tinh dinh thing cap n san:

a+13 (A.13 0 a+11 a.{3

14„ = a+^

0 0 a+

1)) Chung to rang dinh tithe sal' khong plat thuhe vao y,, ) ••• Ynt

1 g1+ Yl

(x1 1111 1)(x1 11 379)

1

x2 +y1 (x9 + N 1)(x9 + y2)

(x n +y )

(x„ +y )(x +y2)

D„=

n-1

11 (X + Yi )

1.20 Cho ma 14.'0

Hay tinh A u1°

n-I

Fl(x.)+yi) n(xn +)'i)

1=1 i=1

( -2 \ A= -1

1.21 (ha si:t ma trgn A e Mat(m, n, va rang A =

Jhisng minh rang cac ma trgn B e M(m, 1, K) va C e Mat(1,

1g A= B C

1.22 Chung minh rang ma trap A = a b ) ( d thOa man pIntong trinh:

X2 - (a + d)X + (ad - bc) = 0, la ma Iran don vi cap hai

1.23 Cho ma trgn 9 v

4 A=

90

vOi x Hay tinh A- '

1.24 Cho ma tran vuong c5p

cosa sincx cosa sina

cns2a :;in2a 2cos2a 2sin2a

cos3o sin3a 3cos3a 3sin3a cos4a sin1a 4cos4a 4sin4a

Chung minh rang A khg nghich va chi a ♦ kg (Ice Z) 1.25 Cho ma trgn A = (a) e Mat (n, 1H) ma the phan tit

doge cho bai tong that:

(-1)H- CV yin

vOi i = j vai i>j 6 do Chi.,'

1.26 Gth sa X = ())) e Mat(n, R), 1-1

a do x ii 4- (-1)")! Chitng minh X' = I 01- 4,

Chit Se voi aeR k e N, ki hiqu

)a,

a(a - 1) (a - k +1)

- a ())) )0

)-

k! ;

1.27 Gia = yXx„ x.„„ x„) vdi (i = 1, 2, , n) la ham

[ cua cac bi8n doe lap x , x2, x„ Ma tran J = J(Y, X) = aYi

dx• 1 i

dliciC goi ]a ma tran Jacobi cua phdp bi6n din, Binh thud dm

no duck goi la Jacobien dm phep biers den

Bay gin /cot m6i quan hq gilla n ham vii va n bien xii dune cho Isdi ding thiic:

Y= A.X.B, a do Y= (y i) , X = (xi), A B e Mat(n, R) la hai ma tran the trn6c Chung minh rAng det(J(Y, X)) = (detA)" (detB)"

1.28 Cho X = (x„) e Mat(n, R) la ma tran tam gide dudi; va Y = X X

Chung minh rAng det(J(Y, X)) =

1.29 Cho Z lA tap cac sqinguyen; A, S hai ma tran vu8ng cap n, cac phAn t> la nhilng s6nguyen (ta vie) A, S e Mat(n, Z)) Hun nua detA = 1, det S x

1.30 GM sit ma tran A = (a u) c Mat(n, R) da cho rude tat ea the phan ti a d (i # j) Chang minh rang có thk dien Mo &rang char) chinh the s60 hoc de ma tran A kheng suy Bien

1.31 Tim tat ca cac ma tran A e Mat(n, K), A= (dj), na tan toi ma tran nghich dao A-' cling có the phiin t5 khOng am

1.32 Cho ma tran vuong A co the phan to la s6 nguyen rim diet' kien can va du de ma train nghich dao cung c6 the

-Men to la s6 nguyen

D - HDONG DAN HOAC DAP S6

Xet T e S„ (n 2) gie sii i < j va T = j, T (j) = T (k) = k vdi moi k s i, j Khi the nglach the eaa

ji, k} Nob < k < j

j/, j} Nob / i + 1, j -

With vey co tat ek la U - i) + (j - i - 1) = 2U - I) - nghich the Vi so nghich the le nen t la phop the le

1.2 a) Phop th6 da cho phan Deb hai yang xich chic lap (1 2) (4 7) = (1, 2) (1, 8) (4, 7) (4, 5) (4, 6)

b) (1, 6, 3) (2 4) -= (1, 3) (1, 6) (2, 4) (2, 5)

1.3 DS S6 Dat ca cac phep the a e a(i) # i vai moi

f n

(-

n +1 NMI yky A, (i =1, 2, , n-F1) i =1,n la n! E

k0 = kl

1.4 Xet tap A gam at ce eke Minn vj (Ma 1, 2, , 14, n+1 co thing k nghich the Ki hieu A i la b° phkn eaa A gdm cac imam NO

an, an+1) ma =

Xet Anki = {a = (a p a , , an+1)1 = n +1}

NMI 0y so cac nghich the cna a bAng se cac nghich the cue

nghia la bling 1+ Dieu cheng to so ea(

U Ch2

Jhfin to cria A„,, bang (n, k)

Xet tap Ai (i = 1, , n) cis su a e Ai, a = (a, a j , a„ ,) Theo dinh nghia a ; = n+1 Nhti vay (aj, khong la nghich th( vdi j < i va (x„ ai) la nghich the vdi j > i Do a, tham gia vac n+1-i nghich the Xet hoan vi a' = (a l — aa,_„ (>61, cCia S2 S5 nghich the maa a' bang sernghich the cna a trii NM; vay ta có met song anh tit A ; len tap cac hofin vi cua S„ co thing k-n-l+i nghich the., do so' phan t& cera A ; IA (n, k-n-1+i) Td

do, sephan to cUa A la:

(n + 1, k) = (n, k) + (n, k-1) + (n, k-2)+ + (n, k-n)

1.5 a) Xet phan tich f = a l o a, o a,„ tich cac veng xich dec lap di) dai > Gia sa dai a l, la d k; ta they f(i) # i va chi i thuoc met cac yang xich Vay giam coal IA d,+€1.2 M6i vOng xich dai d k phan tich dude dk -1 chuydn trf Do vay f phan tick deck thhnh d + + (.16-1n chuyen trf Do kha'ng dinh a) driec chting mink

Thep the f, thi nhan viii chuyen tri (a, (i), hai yang xich se Thep lai lam met (136 de a thing chfing minh) Tr/ neu g a phop the va T la met chuyen tri tin dp Mem cua g oT khong vire); qua giam cfia g ceng them Vi the nen f phAn tich luec h chuyen tri thi giam cUa f khong \wet qua h

1.6 H.D Do a va n nguyen t6 vfii nhau, nen a k khong chic het cho n vdi mm k = 1, n-1 de r(ak, n) la nhfing so phan hied

1.7 Vi moi phep the phan tich due() tich cat ghee chuyOn tri, nen chi can chfing minh bai town cho phep chuyen tri(1,j)e Sk vei 1, j # Ta ce (i j) = (1, i) (1, ) (1, i)

1.8 Xem vi du 1.1 - 1-1(n+D a) DS: (-1)

r(1121) h) DS: (-1)

1.10 Iasi trien cot dau, to ca D ykk =(a2 _ " " 12n-2 •

Do D2 = -13.2 nen D 2n = (a

1.11 A = (aid , detA = E sgna ai,(1) ana(N

E S

Trong moi tich a ko(k) a ngfrik , tong E(k +cr(k) )= n(n +1) k=1

la sir than; nen re met se cliSn the thfia so ak.,0.;) ma tong k+G(k)

1.12 a) COng dOng thu nhat vdi (long the./ ba, ta dude don

tY le vdi (long thu hai

b) Nhan dong thd nhat vdi (-1), rdi cUng vao die (long th hai va Ulf( ba, ta dupe hai dOng CY 15

1.13 Ma tran A, B e Mat(2, K) Ta có hai bat bian detAB

detBA va tr(AB) = tr(BA) Vi vay:

fx+y=30 {x = 20 {x =10

hoar

x.y= 200 {y=10 {y=20

20 14 \ a) x = 20 va y = 10 o BA=

14 10 20 14' 11

Ta co ABA = A A

14 10 , 11 25

Tit ta tinh dude A = 5k -14X , A E R

-11k va B = A - 11

11 25,

10 14 b) x = 10 va y = 20 BA=

:14 20,

10 14 11 -3 l

ABA = A X A= P

14 20 11 25, X 55 32

1.15 a) D i, =

b) A n = -A„ + 4,1 =

1.16 a) Khai tri4n D„ theo cot thu nhat, to cee D T, = (1+ x )D n_i -,4 211)„,, , D, = + x 2,

D2 = (1 + x') - x' = + x + x

Taco: Da -D„-1= x (D„_i -Dn _2 )= x (13 1-2 - D n _3 )

x201-2) (D9 _ D1) x21 Tct D n = Dn _1 + x 2”

Do vay D r, =1+ x + x + + x 2" b) Ap clung eau a) vdi x = DS (n+1)

1.17 Khai tri on A(x) then (long tha nhat, to co:

P(x) P(x)

A(x) = D(a , a n ) x - a l -

+ ( 1)"7 I3(x) ; a 46 D(a , x -a n

la dInh Dane Vandermoncle eaa cac s6 a , , an;

Cho x = a,, to co A(a l ) = (a, - a ) (a, - a„) n(a

j>i22

A(a,) = (-1)" n(a i -a ) j>141

Thong ta A(a 2) = = A(a„) = A(a ) Da tithe A(x) bac nhO hOn hoar bang (n-1), co the gia Da hang tai n diem phan biat, vi vay A(x) la hIing

Tit do: A(x) = (-1) 11 ' Fr x (a -a.)

1.18 J In ma tran Vandormonde vat (Me s6 biet nen dot J *

Ta

a+b+c a+ bj+ cj a + bj + ej A J = a+b+c c+aj+bj c+aj +bj • a+b+c b+cj+aj b+cj +ttj dotAJ = (a + b + c) (a + bj + cj ) (a + bj + cj)det 1.19 a) Khai trien theo cot the] ula La

D„ = (a + a0D«-s Ta co: D, = a + 13, 1/2 = az + an + 13 ,

k

Gia sit Dk = Ea i f3 k-i (k = 1, 2, n-1) =o

n-1

Ta co = (a +p) a i r-1-i - an Ia'0 2-2-I

i=o i=o

n-1 n-1 n-2

= ai+lpn-i- ork ira a al nn-i-1

1.0 i=0 i=0

n-1 n-1

= r3n cti v _i

-

i=1 ]=1

= E

=o

b) Cheng minh qui nap thee n

1 I,

n = o D = = x2 - x, kheng phi thuec y, + y i x, +y i

n = D„= (x, - x,)(x,, - xi)( 119 - x1)- Gia stir Dk =

11(Xj -x i ) vdi moi k = 2, n-1 Si

i <MX

Ta chting minh ding thim tren dung voi

Khai triL 1) 1, theo clang flu./ nhal, r6i rut cac thira soi chung die cot ye sit (King gia thiel qui nap, La cc):

D i, = ( - 1) " n( Yi) n(X j -X1) (1); k=1 /=k 151<“n

i,j #1:

Neu ee i x j ma x, = xp thID„ = 0; ve ming theft quy nap Van dung

Neu x ; x x vol moi i #1; thi to Ming thee (1), to co:

D T, =

1 i<gnI

(4)k-1 n (xi +

44

11(x, -x k ) Fl(x k -x j ) j>k

Net 14(y )

• (-Dkjjl 11(x2 +y )

lxk

k=1 n _ x, n (x k k=1

(x / +y )

ixk

(x j -x k )

jxk

1.20 agA=B+I L la ma tran don vi,

0 - 1' 0 \

B= -1 0 , t3do B = -1 ; W =

-2 0 \ -2

Ding khai trie'n nhi thric Niu-ton to co:

A100 = (0 , 13 ) 100

13 +100.B+ 100;99 u 1-

2 -200 100 Suy A 100 = -100 9901 -4950

-200 19800 -9899

1.21 Vi rang A = 1, nen tat ca car (long tS , le vb7 mot dong Wiy ma tran C E Mat(1, n, K) la ma trqn dong Giq sv clang thd icim ma trqn A bang b, C, lay B = (b) la ma Lean cqt Ta co A = BC

1.23

A -1 =

1.24 Tinh detA = Tit db suy ket qua 1.25 Bat A2 = (b (), b i; = ia ik a k;

k=1

+ VOi i <j bli = Za ik a ki =

= E(_ 1) k_icry-1)Hc i(

k=i k=i

•

0-0!

_ z( 1))+1; (J-1)•

k=i (i -1)!(j - i)! (k - i)!(j- k)!

(j-1)! (j - k)!(i -1)!(k - i)!

= (-1) 1+1 c

- 11 (-1)kIck• : fiat k - = / J-J

= (-1)1+3 cl-1• Z_.r -1/ cJ- (i =0, vi 1-0

(- 1,c = - =-0 1-0

+ Vert = j, ta co b ii = Ea ik ki =1 • k=i

NhLt vi,ty A7 = I„

1.26 Triloc h6t to có nhan xet sau: Cho a, b, C e N, c b Khi do:

b (2.-1,' E(-1) k

kzo - ‘a., - a,

Cong thtic tam co the thong minh guy nal theo a = 0, 1, 2, va hiu y la = voi k > a

flat X- =(3.7.5) th y ] , = [_,xlkxkj

k=1

•• =

• =

k=r n-k, - i ,

xot tren)

Gia sit = (20, to ed z a = Zy ik x kj kst

kblin-i‘ -

Ora

k=1 Tr -1, „n - j,

k „ iEnn+1+1+k n-i

ki=1 „k -1 n

Neui<j zu =0

Neu i= j =

Ngu i > j z,i = E Enh-Ft

aXt<l) t b day t = k -1 a = n - b = n - j)

Nha vSy zij = D_ o b+t t! b!

a<t<b a!(t-a)! t!(b-t)!

b

b ! x (_ i )1)-0

a 1)+1 t-a Cb 1) C b_ik

a<t<b (b - t)! (t -a)! t=a Ixa

= (-1) am

s=0 Nha vky = 8,j nen Xx = I„

1.27 Trade het to Chang mink rang ne!u Z = AX,

a (Id Z, A, X e Mat(n, R.) thi det(J(Z, X)) = (detAr

Tif a ik n6u / = j

0 nau / j • (1)

Ta coi ma tran J(Z, X) E Mat(n 1K) vdi can dOng va can cot

duct danh s6bai can cap c6 thit Lu (i, j) , < i , j n Thii ti cac thing clang nhu cac ctit la (1, 1), (2, 1) , (n, 1), (1, 2), (2, 2),

azi (n, 2), (1, 3) (n, n) Phgn t& a clang (i, j), cot (k, /) la Tn

ax k/ do, (1) ta en:

A A J(Z, X) =

0 va nhn vay det(J(Z, X)) = (detA)"

Turing to n6u Y = Z B thi ta clang ce):

' det J(Y, Z) = (detB) ° Xet Y = A X B = Z B Ta co: J(Y, X) = J(Y, Z) J(Z, X) Mut vay

det(J(Y, X)) = det(J(Y, Z)) det (J(Z, X)) = (detB)n (det(A)) n 1.28 Gin s>1 X = (xi) e Mat(n, K); x,, = vdi i < j Ta có Y = X Xt = (Y,)

Y;i= x i (xk Exik jk k=1

OY • Xet

ax

13 Do (*) to có

,j aX

2X vat i = j 11

X JJ

j

xis ngu i=j,r=j

Ta có ay 'J ngu r = = j, j (*)

4

2/c i„, neat r = i = j

0 the twang hop loi

Ma trOn (1(Y, X) e Mat(n 2, R), the (Ping nä cac cot dttoc darth s6 nhu bai 1.27 Phan tn a hang (i, j), cot (r, s) Pt 13

ors Ta nhan tha'y vdi (i, j) < (r, s) (nghia Itt i < r ngu j = s hotic

dy j < s) thi -

Mt n;

Nhu vay J(Y, X) la ma tran tam gitic &RR

Nhu \ray det(J(Y, X)) = aXij _ - 'J=1 jX j1

=1l fl - n(2xn)=-

JJ j1

J

1.29 Goi A = (AO la ma tram phu hOp cat) A

Ta co A41 = A t - A l e Mat(n, Z) det A

Ta co B = SI A S Brn = S4 S

VI ma tran S g6m cac so nguyen nen k = Ida S e N* (do gia thigt dotS = 0) Vol m6i se; nguyen dming p, )(et ma tran AP e Mat(n, Z), gia sit A P (ap ; goi R la set du chia a!; cho

k = Idet S Idat = 0'0 vi < r i; < k, nguyen nen so' dile ma tran dang Rp la huu hen VI Oy t&n Lai p, q c N*, p > q cho the ph&n tii taring ung cad AP va A P bang thee modk

Do AP = +(detS).0 , C e Mat(n, Z)

Tit do: = +(detS).C.(A ql that m = p - q Rhi do: I - A"'S=1„ +detS.S -1 C.A -4 S;

Vi S e Mat(n, Z), (detS) Sl' = S i la ma Iran phu hop cua S, nen S e Mat(n, Z) va vi vity: B" c Mat(n, Z)

1.30 Ta chfing minh quy nap then n n = : hien nhion,

a ll a l p a21 22

Neh cho track a , a t 0, to chon a, = a 22 = Con nein a 21 a u, = 0, La chon a„ = a 22 = Nha vay n = 1, thi bad Wan dung

Chi sit bai toan dung laid moi dinh tithc cap < n - Ta chiing minh no dung ved dinh at& cap n

Gia silt A = (ap) e Mat(n Xet A„ la phan phu dai so' caa ph dn tit an Theo gia thi6t quy nap, to da Hein dude a22, a„„ de? A,

DAt a ll = x, khai trien theo clang thu nhAt ta co:

detA = x A, + Za li A li

Ta co I:A uA li la hang set NAu hang so kluic khang, ta J=2

chon x = a„ = 0; nAu hang s6 bang killing ta chon x = a t , = se clime detA

1.31 Trade he'd ta chting t3 rang nen ma Iran A va A-' E Mat(n, co the phAn t& kheing Am Chi a m61 cot cila A co dung met plan

ThAt vAy, gia sa a cot thu j cua A co secdim/1g, a clang i, vA dOng i Chon clang k vdi k # j cila ma tran A' VI A-1A = I„ non tich cua dOng k cua A' vdi cot j CEla A bang Nhu va t), cac phzin t0 cfm cot thil i, va i nam tren clang k deli bang ki:Mg (k x j) Do hai cot i va i2 cua A-1 tY le vdi Diu clan deIn mau thuan

NMI vAy m61 cot cim ma tram A co dung met s6 clueing (con lai dgu bang 0) Dicing te, m6i deng cila ma tran A clang có dung met s6 throng Giao hoan cac demg (hoac cac get) ta duoc ma tran chg °

Nguoc lai, tat ca the ma tran 'Then dude tt ma tran cheo (ali > 0) bang each chuygn clang host cot deu kha nghich Ma trAn A-I nhAn dude to ma tran A bang each Iay nghich dao cac phan to khac kitting rdi chuyAn vi

Chuang

KHONG GIAN VECTO - ANH XA TUYE-N TINH

- HE PHUONG TRINH TUYEN TINH

A - TOM TAT Lt THUlt

§1 KHONG WAN VECTCS

1 Dinh nghin: Gin sit K la mot traong Mt tap hop V kirk rung cimg vdi hai phep town "+" : V x V —> V

(a , p) 1—* + va phep than " •" : K x V —> V

(k, a) > k a

dime goi la mot kheng gian vec to tren twang K n6u no thOa man cac tinh chgt sau mot k, /, c K va moi a, p, y, c V, to CO:

a) a +p=p+ a

b) (a+ p)+y= a+(p+ y)

c) CO phlin tit e V cho:

a + = + a = a vin moi a c V

d) Vdi a e V, ton tai (-a) e V cho: a + (-a) = e) k(a + is) = ka +103

g) (k + /)a = ka + la h) (k / )a = It (/ a)

Mot kh8ng gian vec CO tren throng 1K &foe goi K -

kKong gian \ecto

Khi 1K = R, kh8ng gian vac to &loc goi la kh8ng gian vec to

thee Khi K = C, kh8ng gian vac to dude goi la kh8ng gian vecto

phiie

2 - Sit crOc 15p tuygn tinh vit phu thqe tuye'n tinh

Vec to 13= k,a, + + k„,a„„ (a, c V, k e K)

goi Ht met to' hop tuyo'n tinh cua cac vec to = 1, Ta ding not p bieu thi tuy6n tinh qua cac vac to a t , , an„

MOt he vac to la,, •, caa V dude goi la he phu thmic tuyeIn tinh nen c6 cac s6 = 1, m) kh8ng deng thoi bang kheng cna ]K sad cho = Met Oat hkeu along &king:

1=1

he (st„ am) ka he pha thuec tuy6n tinh nen c6 met vec Lo nac de cim he bleu thi tuyen tinh qua cac vec to can lai oda he

Met he cac vec to kh8ng phu thueic tuy6n tinh doge goi he doe lap tuyen tinh Nhu \ray, he {a„ am} la he dOc lar tuyen tinh ned.1 moi t6 hop tuy6n tinh k i a i =0 to suy k, = = k m =

Cho he 14„ aid cac vec to doe lap tuygn tinh caa kh8ng gian vec to V ma m6i vec to caa no la t6 hop tuyeIn tinh caa ca( vec td cim he ([3„ thi k <

Cho he vec to {di } ; e kheng gian vac to V, I la tap chi

lap tuyen tinh t6i dai cim he da cho neu n6 la he dec lap tuyen tinh, va nen them bet kST vec td ak nao (k E I \J) thi ta dude met

he phu thuec tuyen tinh

Cho he hau han vec td {a, , , a m} khong gian vec td V, thi s6 phain tit cim moi he dee lap tuyen tinh t6i dal ciia he tren deal bang nhau, so/ duoc goi la hang cim he vec td {ao ,

3 - Cd sd va so clueu cua khong gian vec td

Met he {e„ , en} the vec td doe lap tuyen tinh am killing gian vec td V duos goi la met co sa oda V nen mm vec to cim V deli la t8 hop tuygn tinh cim vec to {c„ , e n} Khi V c6 ed sa g6m n vec td thi moi co sa cim V deu c6 dung n vac td S6 n goi la so" chigu cim V, Id hieu dimV Neu kleing tan tai mat cd sa Om him han vec to, thi V goi la khong gian vec td v8 han chieu

Cho co sa e = le i , , ej, yen vec tO bet kS7 a e V, ta co a , , x„) dupe goi la cac toa doo cim vec td a del vOi ed sa {e„ , e n}, ; la toa de thu i cim a doa vdi cd 56

Gia sit co ed sa khac c = {c,, , E„} ciaa V, ma si = ]=1 = 1, n) Nen vec td a có toa di) (x i) cd s6 va c6 toe de' (x1 1) ed sa thi ta c6:

= Ec oxl i , , n

4 - Klaang gian vec tei va klaing gian vec td thacing Tap khong r6ng W ciaa khong gian vec to V duoc goi IA khong gian vec to ciaa V nen W la khOng gian vec to, voi cac phep toan ciaa V han dig tren W

Tap khong rang W dia V la khong gian vec to ciaa V va chi W on climb dna vdi hai phep toan dm V, nghia la vdi a, 13 E W va k E K, thi +p E W va k a E W

Cho W, W EI la cac khong gian vec to cria khong gian

11

vec to V, nW, IA mOt, khong gian vec td Gem V, IA khong gian ldn nhat nam moi W i , i = 1, 2, , n,,

Cho tap hop X c V, khong gian vec la be nhat cna V chga X duck goi 1a bao tuygn tinh cila tap hop X, ki MO <X> hay Vect(X) Neu X = , bao tuygn tinh ciaa X thick ki hieu la <a t , , a„,>

Cho W„ , ho nhUng khong gian ciaa V Khi de bao tuygn tinh ciao, tap hop W,U UW E, dude goi la tang cim cac khong gian W, ,W„, ki hien Ia W, + + W„ hay ZW;

Ta thay rang a e EW; va chi a = Za i , a, e i=1

Neu moi e /113/4 , a vigt chicle mat each nhat a i=1

dang a = a, + + a„ , a, e thi t8ng W, ch.toe goi la tang I=t

true tigp cua n khong gian vec td (i = 1, 2, , n) va ki

hieu la W, e W2 ED ®W„ hay S W,

Gia sit V la klafing gian hitu han chikt,W, va W2 la hai khfing gian vec to dm V, do:

dim(W,+ W2) = dimW, + dimW2 - dim(W, fl W7)

Cho W la khong gian vac to cim khong gian vec to V Ket quan he Wong throng tran V: a-Paa-Pe W Lop Wong :lining cam vec to a dirge ki hiau la [a]

Tap thudng V/W vdi hai [top toan:

[a] + [in = [a + [I] va k[cx]= [kal

vdi moi a, 13 E V va k e K, lam mat klihng gian vac td,

durie got la kitting gian vec to thtfong (ciia V chia cho W):

dimV = n, dimW = in (0 m n), thi dim V/W = n - m Anh xa it : V -> V/W ma a(a) = [al (Inge goi la phdp chi6u

tdc

§2 ANH XA TUYEN TINH

1 Dinh nghia Cho V va W lh cac khong gian vec to tram twang K; Anh xa f: V -> W duo° goi la anh xa tuye"n tinh (hay itIng ca'u tuy6n tinh, hay toan tit tuyan tinh) n6u no bao

phop toan cua khong gian vec to, cu the' la: veil moi a p e V 1119i k e IK, to ce:

P) = f(a) f(P) f(k a) = k f(a)

Hai khong gian vac td V va W chive goi la clang cdu vol nAu ce met ding cau f tla V len W

2 - Cac phep than tren cac anh xa tuy6n tinh

Ta ki hieu tap cac anh xa tuy6n tinh tit khong gian vac to V dAn khong gian vac to W la Hom(V, W) hay Hom K(V,W) de chi rd K la tniang co sir

HomK(V,W) la met khong gian vec to tit truong K vbi hai phep town nhtt sau:

Vai f, g e Hom K(V,W); anh xa f + g e Hom K(V,W) the dinh

ben (f + g) (a) = f(a) + g(a) vdi moi a e V

Vol k e K, f e Hom K(V,W), thi kf: V —> W xac dinh boi (kf)(a) = kf(a) vdi moi a e V

Anh xa f + g dupe goi la tong Kin hai anh xa f va g Anh xa k f ddoc goi la tich eda anh xa f vdi vo hdong k - Di6u ki6n xac dinh anh xa tuy6n tinh

Anh xa tuydn tinh f: V —> W hohn toan chidc :Mc dinh Mei anh cOm met co so

NMI le„ , e„) la co se, cOta khong gian vac to V va a,, a„ la n vec to cent MI:Ong gian vec to W (V, W la cac khong gian vec to tren clang rant trnong K), thi Kin tai nhdt met anh xa f e flom K(V, W) de f(e i) = a, yea j = I, f la don cdu va chi he , a„} doe lap tuyetn tinh, f lA clang cdu va chi he M I , , a„) la co sa oda W Gia = EJ la cd

in

tan dm anh xa tuy6n tinh f dOi vOi co sa {e,) va {EJ Nhg vay 16u cho cd sa E cua W va co sa e cua V, thi dinh 19 tren chring to :Ang co mOt song anh girla tap Hom K(V, W) va tap 1),E#t(m, n, K)

FlOp cua hai anh xa tuy6n tinh la mot anh xa tuy6n rhh, nghia la nOu f: V —> W va g: W —> Z la the anh 'ea tuygn

thi g f: V > Z cling la mot anh xa tuygn tinh - Anh ca hat nhan cua anh xa tuygn tinh

Cho f: V —> W la thing thu tuy6n tinh gifla cac khong gian 7ec to, neh X la khong gian vec to cua V, thi f(X) = { f(a) I a E X) a khong gian cua W, va n6u Y c W, Y la khong gian eon

W thi f-'(Y) = e VI f(a) e Y} la mot kWh:1g gian cua V Ta goi Kerf = f-1 101 la hat nhan cda anh xa f va Imf = f(V) la inh cua anh xa f S6 dim Imf doge goi la hang cua anh xa f, kf lieu rang E

Gia sit f e Hom K(V, W),

la don eau va chi Kerf = {0} Nth dimV Fa hfiu han thi IimV = dim Kerf + dim Imf

Cho ma Wan A e Mat(m n, 1K), xem A nInt ma tran cua inh xa tuy6n Unh f: K n —> Km thc od sei chinh tdc Khi

tang cua ma trail A (da dude dinh nghia chudng I) hang

rang cua f va chinh la hang cua he vec td cot cua ma tran A - Ma tran aim t‘i thing cgu the cd ad khac Cho f E Hom K (V, W), co se: e = (e„ , e n) f cO ma Wan

Gie sii & = (6 o••ogs) le mot cd sa &bac, ma E, = /Cijej , tron€ i=1

cd s6 6, f co ma tren B = (b ii) ughia = Eb ii Ma trar i=1

C = (co) chide goi la ma tren chuyen co sa Ta ca: B=C' AC

Hai ma tran A ya B dude goi la deng deng neiu có ma trey khong suy bign C de B = C -' A C Nhu yey hai ma Win cue ding mot phep bign (lei tuygn tinh hai cd sd khic &dug clang

Ta goi vet ma tren vueng A la t6ng cac phen to trer • &tang Oleo chinh Hai ma tran tieing deng co vet bang

Vet cua mot to deng cdu tuygn tinh la vet cila ma trail cfla ni mot cd sd nal] Vet clad ma tran A dude ki hieu la trace A hay trA V6t cua td d6ng cliu f thick ki hieu la tracef hay trf

§ HE PHUtiNd TRINH TUYEN TINH - He pinking trinh

He Ea ki xi = b k (k = 1: 2, , i=1

dO aki , b k E K cho trUoc, k = 1, , m; it= 1, , n

xi la cac en, dude goi la h0 phudng trinh tuyen tinh (hay 14( phudng trinh dai s6 tuygn tinh) g6m m phudng trinh, n an so Khi K la truang s6 (nhu R hoac C), thi cac a to goi la the he 66,

hQ se" to

all a 12 a ln

a21 a22 a 2n

b

b

Ma trail Abs = goi la ma tran

b6 sung, no có ducic tii ma tran A bang each them cot cac he so' tti vac, cat thu (n + 1)

b

Neu ki hieu X = va B = la ma tran cot,

xni bini

thi he phtiOng trinh (1) co the vieit duoi clung: AX =B

2 - HQ Cramer

He n phudng trinh tuyein tfnh n an s6, ma ma Iran cac he s6khong suy bi6n goi la he Cramer

HO Cramer c6 nghiem nhdt each tam nghiem nhu sau: Cdch 1: Xet phasing trinh ma trdn AX = B, vi detA # nen tan tai ma trdn nghich dao Al', va ta co:

X = B

Cdch 2: Xet he vec td cat an}, ma a i = (aii ,a si , ,a mj )

va b(b„ , b 0,) la vac to cat td do, the thi he viat dude dU6i clang =b N6u ta goi D i la dinh thiic cda ma trail nhain dude bang each thay cat thit i cda ma trdn A bat cot cac he si6 do, thi x

I = D ado D la Binh thiic cim ma tran A

3 - He phtiong trinh tuygn tinh thuAn nhiIt HO phudng trinh tuy6n tinh thuan nhal ce clang:

AX = (2)

Xet ma tran A = (a o) nhu ma Wm cua anh xa tuyen tinh f: K" —> Km cAc co s6 chinh tac cua 1K'l va Km, thi tap hop nghiem cilia (2) chinh la Kerf Mei cci sa caa Kerf goi la met hee, nghiem co ban cilia (2) HO (2) ludo có nghiem x, = = x„ = 0, nghiem goi la nghiem tam thuong KM rang A = n, thi he chi co nghiem tam thuang Khi rang A = p < n, thi tap hop nghiem la kh8ng gian vec to n - p chieu (n la s6 an ciia phudng trinh)

4 - He phudng trinh tuye'n tinh tong gnat

a) Dinh b./ (Gauss hay Kronecker - Cape 11i)

He phudng trinh Za ii x j = h i (i= 1, , m) ( ) i=1

ce nghiem va chi rangA = rangA hs b) Phining phap khet nem Gauss

Cho he pinking trinh (1), n6u dung cac pilau Men dpi sau day thi La van nhan dude met he phudng trinh thong doing \TM he (1), nghla la he cO ding tap hop nghiem nhit he (1)

+ :Than hai ye cUa met phudng trinh nao cem hee, vat s6 k #0 + Geng vao met phudng trinh cua he sau da nhan met s6 bat ky vao hai ye cem phudng trinh khac -

Cap Oen bi6n ling during cl6i vOi he phuong trinh chinh la cac Olen bin d6i so cap thy(' hien Den cac clang dm ma trail 1)6 sung Abs cUa he

Dung phuong phap khil Gauss la Dille hien cac phen bign ckii Liking during de chia he phuong trinh (1) v6 he phuong trinh ma ma trail 06 dung:

P

0

I 131

h ip p+i m-p

0 b'm

p n-13

phan t>i a phan goch (*) co the khac Khi rid n6u 13,;+1 + + > thi he ye nghiem,

n6u = b'„,= thi he có nghiem phn thuOc n-p tham s6 §4 CAU TRUC CUA T11 DoNG CAU

1 Khong gian rieng - da thitc dac thing

Vec to a # thuec V dude goi la vec to rieng cim f e EndK(V) ling vat gia tri rieng X ngu f(a) = Ira, A e K Khi khong gian met chigu sinh bai vec to a Et met kh8ng gian vec bgt bign cim f

Val A E 1K, tap ker(f - AId) no khac {(5} la khong gian

con cim V, gam vec to khong va tat ca cac vec to rieng caa f ling vdi gia tri rieng X Kitting gian goi la khong gian rieng cim f

v6i gia tri rieng A, ki hieu

Gia sit ta cl6ng eau f e End(V) met co sa Mao cim V co ma tran A, thi det(A - XI n) IA met da fink bac n dovi v6i bign X, khong phu thuec vao vice, e chon co sa, va &lac goi la da thitc (lac trang caa ta (tang eau f (ta cling nOi IA da that da, e trang caa ma tran A), ki hieu M.(x) = det I A - X I n I Nha vay A la met gia tri rieng cim f va chi X11 nghiem &la da that dac trang eim f

GM sit , Ak la cac gia tri rieng cigi met phan biet caa I Px Pik la cac khong gian rieng Wang ling via ale gia tri rieng do, thi t6ng Pr + Px Pxk la tong true tigp

2 -Ong gian rieng suy Ong

+ Vdi moi khong gian rieng suy rang k 7r IA met gia tri rieng cim f va Y, c Rs

+ V6i X IA gia tri rieng cua f, dim2h x bAng bei cern nghiem

cua da thitc dac trung / f(X)

+ Mot g?",, la met khring gian cila V bait bi6n qua anh xa f + Gia sf1 , "f.k IA cac gia tri rieng phan biet tiing cap

va u, e \ {0} = 1, 2, , k) thi he cac vec td {u„ u k} doc lap tuy6n tfnh

3 - Tit citing clu luy linh

a) Ta not ring f e End K(V) IA tu ding caiu luy linh nau tan tai s6nguy'en k > de'f = f o of= 0, hdn nua, nau # va

k hin

0' = thi k goi la bac luy linh cua E

Tu dang udu f e End K(V) ma co cd sa te„ , e n} cho Re) = (i = 1, , n-1) va Re p) = 0, thi hay linh bOg n va ed sa

{e i, , en} dude goi 11 cd so xyclic d6i v6i f Trong cd SO xyclic ma

trail cua f co clang:

0 0'

1 0

1

0

b) f e EndK(V) U la khong gian the to eon cem V, U chide gui IA khong gian vec to xyclic chid vdi f nEu U IA f- bas biEn va co mot ca so xyclic dee vdi f/U: U -> U

c) NEu f e End K(V), dimV = n, thi V phAn tich dude

tang trip tiEp cua cac khong gian vec to xyclic doi vdi f Vdi mdi so nguyen s 1, sE cac khong gian vac to s chiEu xyclic doi vdi f moi each phan tich dEu bdng va bAng:

ran g(f 8') - 2rang(f s) rang(f

d) N'Eu ) IA khong gian rung suy rang cua f dng vdi A, thi

(f - A.Ick) la td d6ng eau Iuy linh cern V

4 - Ma tran clang chutha lac Jordan caa tat d6ng eau GiA sei V IA khong gian vec td hisiu han chigu tren truang K, f e Endx(V) ma da that ddc trung 94(X) cd dang:

i(X) = (>1 (X

, -X)`2 -X) sk

(cac &Ai mat phAn biet, i = 1, 2, , k), V la King true tiEp tha cac khong gian riAng suy rang V = x G3.? X2 e Xk va V co mot co sa (ei) = 1, n; (n = dimV) dA co ad ma trdn din f tao bdi the khung Jordan

nam doe (Wang cheo chfnh so khung jordan cAp s vai phan eh& X, bang:

rang(f - - 2rang(f - kildv)s + rang(f -

Ma tram clang tren cua f xac dinh nhAt sai khae each sap xap cac khung Jordan Ma tran dude goi la ma trail dang chuan tac Jordan cila (tang eau f

B- VI DV

Vi d4 2.1: Xet R", R!" la the khOng gian vec td tren R Cho

x„ x , , x„, la m vec td thuee R" ]E la met khong gian vac td eila ChUng minh rang tap IF the vec td ena K" dang

Zt i x i (t , , e E Fa khong gian vec to ena K', dimF = dimE - dim(E nN), trong de N la khbng gian cac nghiem tha phtlong trinh:

E ti , =0 (a t i , tn, la cac An) i=1

Lai gidi:

Xet anh xa tuyan tinh u c Hom(R m, R") a do u(t,, t„,) =

EtiXj Khi F = u(E), vi vay F la khong gian vec to cna

R' dimF = dimE - dim(Keru'), u' = uI E Kern' = (KerU) fl E = E (1 N

Vi du 2.2 Cho x i , , xi, la nhUng vec to khac khong khong gian tuy6n tinh V Gia sit c6 Oleg bi6n d6i f e End V cho f(x 1) = x , f(xk) = xk + vdi moi k = 2, , n

Chung L6 rang he {x 1, , xj la doe lap tuy6n tinh Lb gidi:

Ta chiing minh quy nap theo n

Via n = 1, thi {x 1} dee lap tuy6n tinh gia thiat x # Gia sit menh de dung vdi moi he fx„ xki; k < n - Ta chimg minh menh d6 dung vdi k = n

Xet t6 hqp tuyen tinh

ECiX i = (1)

i=1

Khi do:

) = CIX ] Zcif(xi) = c,x, + Ec i (x i +xi_1 ) =

i=1 i=2 1=2

= ZeiXi (2)

i=1 i=2 n-1

Tit (1) va (2) to suy ra: Ec i x,„ = = 0

i=2 i=1

Do gin thik quy nap hee, fx,, , doe lap tuyen tinh nen

C2 =e2 = = cn =

Vi du 2.3 Trong khong gian vec to V cho he ak} cac vec to doe lap tuygn tinh ma m8i vec to dm no la t6 hop tuyen

tinh cua cac vec to cim he 113„ , Hay ehung minh k < 1

Ta the' gia thigt ••• • la doe lap tuy6n tinh, neu khong to se lay he doe lap tuyen tinh t6i dal eim {p,, , pi} g6m m vec to va se elnYng minh k s m < 1

Theo gia thigt a„ , a k bieu thi tuyern tinh qua D i , p, nen ten tai cac vS hudng a.„ e K cho:

a ; ; =1 k) (1)

Ta gia sit k > Ta se cluing minh tuygn tinh Xet t6 hop euyen tinh

k

ak phty thuec

/Xiai

1=1

I j=1

<=> i=1

( k

.i=1

j=1

pi =o

=0

<=> EauXi =0 j=1

(2)

vdi j = 1 he pi leriic lap tuyen tinh

ton tai nghigm (x„ , x k) khac khong Tn d6 suy he la,, ,a k phu thuOc tuydn tinh Dieu trai vai gia thi6t Do k < /

Vi 2.4 Xet hai khong gian E I va E2 cua khong giar vac to E Gia e>i E/E IA khong gian thacing va h: E, —> E/E li han chd caa anh xa chiou chInh tde troll E,

1) Tim didu kien can va chi dd a) h 1a toan anh

b) h la don anh

2) Chung tO khong gian (E, + E,)/E (fang eau v6i kholu gian E i /E i nE2

L of Rich:

1) a) X& E —> E/E2 la anh xa chidu chinh cac, h = h toan anh a ME,/ =11(E)

E, + Kern = E + Ker n = E E, + E2 = E (vi kerb = E2)

b) Ta en Kerh = E, r1 Kern = E, n E2

Nhtt vey h don anh a Kerh = a E, E, = {O} 2) GiA se F = E, + E2;

Xet F F/E va k = 1-1/E, •

Do phAn 1) k la toan anh va kerk = E l C Kern = E l 11 E2

Vi du 2.5 GM sit V la khong gian vec td tren truing K Ty Bong can p: V -+ V ducic goi la met phep chien ngu p = p

1) Chung to rang ngu p, q la hai phep chigu, thi p + q la phep chigu va chi pq + qp =

2) Chiang t6 rang p.q IA phep chi gu va chi [p,qI = qp - pq la anh xa tuyern tinh chuye'n Imq van Kerp

3) Vol p,q Fa hai phep chigu cho p+q la phep chigu, hay cluing to Im (p+q) = Imp + Imq va Ker(p+q) = Kerp n Kerq

Lei gied:

1) p+q la phep chigu (p+q) = p+q <=> p + p.q + q.p + = p+q <=> p.q + q.p =

2) p.q la phep chigu <=> (p.q) = p.q.p.q = p.q = p q Er> p (qp — pq) q q (pq — pq) (Imq) = (=> (qp — pq) (Imq) c Kerp

3) Vi mai phep chigu p co rang p = trace p,

va vet cim tong hai anh xa tuygn tinh bang tong cac vet cim no, cho nen ngu p + q la Agri chigu thi:

trace (p + q) = trace p + trace q va rang (p + q) = rang p + rang q

Tit dim Im (p + q) = dim Imp + dim Imq, suy Im (p + q) = Imp ED Imq

Nguuc lai vdi x e Ker (p + q) thi p(x) + q(x) = Do Im (p.+ q) = Imp $ Imq nen MI p(x) + q(x) = suy p(x) = q(x) = 0, hay x E Kerp n Kerq

Vi du as Gis sit E va Fla hai khong gian vec td tren trulang K; Horn (E, F) IA tap cac anh xa tuyeal tinh tit F Mn ]F; Ft la khong gian vec to cna F

a) Chiang to rang £ = e Horn (E, F) / Imf c F 1) la khong gian cria Hom (E, F)

b) Gia s5 F= K la mot phAn tich cua F t"o"ng tryc tip

i=1

ciaa khong gian F Vbi mdi F, xet 2, = if E Horn (E, F) I Imf c Fd

Chung to rang Horn (E, F) = 2

WL gia'i:

a) Do g ding kin vai phep town tong anh xa va nhan anh xA vdi mat vet hriong thuac K, nen hieln nhien la khdng gian vec to crIm Horn (E, F)

b) Vdi h E Horn (E, F), vbi m6i x e E, to phan tich h(x) = y theo eec thinh ph'An y = h(x) = E yi , yi e F1

id

Va vdi moi x e E, thi h(x) = Eh; (x) i=1 Do 05 h= , e

Gia six Eh ; = nghia la vial moi x e E,

( ) (x) = h i (x) = 0, h,(x) e F, 1=1 1=1

to suy 11,(x) = voi moi i vi F = e K I=1 Nhu \ray Horn (E, F) = e g

=1

Vi du a 7 GO Q la Huang s6 h> u ti va WI la mot tkp hap khong rkng Ki hiku E la tap cac anh xa tif c32 vao Q E la Q - khong gian vac to viii hai phep than: f, g e E, a e cc9, k E Q the

(f + g) (a) = f(a) + g (a) K f) (a) = k f (a) Gia sif V la khong gian vac to ciaa E

1) Chang to rang: nku f e V va co n dim x„ x„E Ca cho

ingui=j —

k(x,) = = ej=1,n

Ongui#j

Goi W la khong gian cea V sinh bed f„), m81 g E V 11611 viet dude met each nhat dudi clang: g = h, +112,

a h, e W, h E V va h = 2(x)= yea mai i = 1,n

2) Chung CO rang hai tinh chat eau day la Wong clueing: a) dim V n

b) Ten tai n ham g„ thuec V va n diem x„ x„ ciaa W( cho g, (x,) = 8,; vet moi i, j = I,n

Lai gidi:

1) Xet to hop tuyen tfnh EXif, = 0, c Q 1-1

Khi vdi moi x E cam, LW; (x) = 0;

Chon x= x„ thi (xi) = 7.j = ( = 1,2, n)

suy he {f„ , f„) doe lap tuyen tinh

Val g e V, ham h, e W cAn tim phai thaa man g(x) = (h, + h 2) (xi) = h r (x,) vdi moi

Do If„ f„) la co sa cem W nen h, :11 (x, = E gfrA lei 1=1

Dat =g - h, eV

2) Theo phan 1) neu b) clime th6a man, thi he (g„ g„) doe lap tuyen tinh, \ay dim V n; nglila la menh de b) a) thing

Ta chung minh a) suy b) bang phydng phap quy nap theo n

Vdi n = 1; n6u dim V 1, tan tai fi x va vi vAy có x, e c14 de fi (x,) = A x Chon g, = thi (x,) =

Gia sii menh de dung vdi moi k = 1,2, n Ta chiing minh nd dung vdi n

Theo gia thi6t quy nap dim V n + dim V ? n nen tan tai n ham g, E V va n diem e A de gi (x,) = 8,, (i, j = 1,2, , n)

Goi W la khong gian sinh bid {g„ gn}, to cd dim W< dim V

Theo cau 1), vdi f e V \ W, to cd f = h, + h 2, vi h, e W nen h2 0; vi th6 c6 x,,, d'e' h1 0; (x,„., x x, vdi i = 1, 2, , n)

h

hi

-

) t bin+, vdi MOt i = 1,2, , n+1 Bay gia = g - =1, , n),

thi vdi j = 1, n to co gi (xj )= (x J )= s i; va gi (xn+1)

=g (x„.1) - x; = n6u 2, = gi (x„,)_

Nhu Nty co n4-1 ham {g-i } th6a man digu kien bai town

Vi du 2.8 Gia sit 98 la ho dem doe cac anh xa tuy6n tinh tit R" (16n R"I; vdi mai a e R" xet 0(a) = {f(a) / f e :43} GM sit g la Anh xa tuyein tinh to den âR" cho g(a) E :0304 vdi

Lo gidi:

Ccieh \TM m6i x e R, xet vec to dx (1, x, x2 , H9 eac vee to d x c6 tinh chat:

a) vdi x„ doi mat phan biet, thi ta xi dPe ld tuy6n tinh, vi Binh

# (Dinh thite Vandermonde)

b) V6i m6i d x e R', tan tai f E cho g(Ci;( ) = f( ) Ta phai chiing minh g c 93, nghia la c6 f e d f g@t„ i ) tren bb la co so cua R"

Gia sii node Lai, vdi m6i f E c6 khong qui (n - 1) sd the phan biet xi cl6 g(d xi ) = * xi )

(rich (Mtn}, cho doe der hied mot sa sri kien cilia khong gnin metric)

Theo gia thiet vdi moi a e der f e de" f(a) = g(a) nhu vay — g) rad= => a e Ker (rig)

NInt vay U (Ker (f — g) f E gq} = R

Hia sir g vSy vdi mai f E thi f — g x O vay dim Key (1— g) n — i Vi vay Ker (1 - g) lie met kliong gian thing, khbng dau tra mat cent R" Dieu gay nen man thuan RS la khong gian metric day vdi metric dieing Hwang vii dinh 19 Haire aid rang: met khong gian metric day kheng the bang hop dem dude mita nheing rap hop khong dim Hu mat

III du 2.9 Cho hai so nguyen dining r, n ma r < n: (it c Mat (Th R) rg." = ran k rang cl = r m& (CP =

Hay chdng minh vet ()f = a,, + a i„ + + a„,, = r (Vet can ma Man red thudng duo( ki Heti bdi trace cel)

Xet f e End (IR") co ma trhn cd ed tly &nen e = le,, e„) i(e,)= apie

Theo gia Hired to cd f2 = f

Ta co Imf n Kerf = {0), that Gay, gia sit p e Imf Kerf, thi co a e R" d f(a) = p va 13 E Kerf nen f(0)= suy ra:

f(p) = 12(a) = f(a) = p = 0, vay Imf n Kerf = {0) Nhu vay R" Imf O Kern dim Imf = r va f I Imf = id, Chon ed sa s Rua R" sac) cho {E,, £,.} c Imf, {E„,„ E„) e Kerf, thi trace f= trace al= r

Cho Nhaat lai rang n6u f la anh toa tuyeat tinh cie'n Jrco ma Ran oat = (ad cd sa e = (e,, nao ciaa Tthi so' a„ +, ,+ a„„ khong phu thuoc vita vice chop cd so oda °L va clack goi la vat ciaa anh xa tuyan tinh f va chiac kf hieu la trace f

Vi du 2d0 Gia sa °W la hai khbng gian vac td treat twang K, f c HomK CP; °Tf)

Xet anh xa f: Kerr Gil" [a]-a f [al = f (a),

Hay chttng to f la dun eau tuyen tinh va Imf = Imf , ta f la dang cau to /Kerf len Imf

ianh xa f lh sac dinh, khong phu thuR vao dai dien That way, vdi [al = thi a - a' e Kerf, W f(a) = f(a") hay II-al = f

thd thay f la Maya tinh va Imf =Imf.Ta chung to f la ddn cau That Ray RR [a] # [a'] thi a - a Kerf f(a - = f(a) - f(a) # do f(a) # f(a) suy f [a] # f Nhu way f don cau va do f la clang cau tit cliKerf len Imf

Chu" y: neat so chieu Irau hen, thi tfx vi du Hen to dim tillierf = dim Imf say dim "P= dim Kerf + dim Imf

du 2.11 Giei he phudng trtnh

+2x +3x +4x =30 -x i +2x., - 3x + 4x -10

x, - X3 +x =3

x i + +x3 + X =10

1 -1 -3 D =

0 -1 1 1 Day 1a he phudng Huh Cramer

gthi:

= -4

80 30

10 -3 -1 10 -3

1), = D,= = -8

3 -1 -1

10 1 I 10 1

1 30 30

- -1 10 -1 -3 10

D a = -12 D'= = -16

0 1 -1

1 10 1 1 10

Vi dii 2.12 Gihi va bran luhn theo thaw s6

A.X1 x, + x•

Xx, + x 2, + 4x

I) =

a) Veil X s vic a x -2 Day la he Cramer

it +1

va ta ce, - , x -

X +2 2, + +2

b) vdi A = 1, to c6 he Wring during vdi:

x i + x2 + x,, =1 hay x = I - x, -x., x i , x, lay trtyl,

c) Vol X = -2 He co dang:

-2x + x., + x = x i 2x, + x g - x i + x, - 2x

Vi du 2.13 Dung phitong phap khn hay giai he phuang trinh:

x t + 3x, + x + 2x )(5 = 3x + 10x + 5x + 7)( + 5)( =

2x + 8x, + 6x + 8x + 10x = (I) 2x + 9x„ + Sx + 8)41 + 10)( =

2x + 8x, + 6x + 9x + 12x =

Lm giai:

Nhan hai ye aim phtiong trinh (tau vdi Inning so/ thich hop, I.()) acing vac) eac phtaing trinh khae, ta clia; he pinning trinh ()rang throng vdi he (I):

x + :3x + x,, + 2x + X5 - X9 + 2X3 x + 2x5 =

2x, + 4x + 4x + 8x = -2 (II) 3x, + 6x3 + 4)( + 8x = -1

2x, + 4x + 5x + 10x = -3

Nhan pIntong trinh thit hai cna he (II) vdi gag s6 thich hop roi tong viva du; phuong trinh kink cita he, ta &toe he Wong dyeing -

x + 3x + x + 2x + x = (I) x, + 2x + x, + 2x = (2)

2x + 4x =-2 (3) (Iii) x + 2x5 = -1 (4)

(

Car phtfting trinh thit (3), (4) (5) he (III) la tudng during Vi 04 he (III) tudng during vol he

Giai

x i + 3x

y +

he (IV), to (little

xn = —1 — X2 = —

x i = +

2x 2x 2x 5x i + + + 2x x

- I

x

3x

+ 2x = + 2x =

a do x4 x -1

Inv 31

(IV)

Vi dv 2.14 Cho hai ma lien A, B thuOr Mat (n, K), A = (ad, B = 09

Khi A + B =(a i +1)0 (bloc goi la tang hai ma tren A va B Chang minh rang:

1) I rang A -rang B l < rang (A + B) a rang A + rang B 2) rang A + rang B -n a rang (A B) a (rang A, rang B) 3) Nan A' = E, tin rang (E + A) + rang (E - A) = n (3 E la ma tran don vi cap n)

1) Gia stir f, g la hai ph&n tit cilia End (â "), co ma 'Iran A, B Wong fing m(llt cc; se( s = (s 1) di( cho Khi f + g c6 ma tren A + B VI Im(f + g) c Imf Img, nen:

Tit de suy ra:

rang(A + B) < rang A + rangB Mat khdc:

rangA = rang(A+B-B) < rang(A+B) + rang(-B) suy ra: rangA < rang(A + B) + rangB

Tit do:

ningA - rangB < rang (A +

ta: rangB - rangA < rang(A + B) VI yay rangA - rangB 15 rang (A + B)

2) Ta co f: K" , g: K" la hai anh xa tuy6n tinh, Im(f o g) = Im(f img) c Imf nen rang(AB) < rangA

Mat khan: rang(AB) = dim(lm fog) < dimlm f = rangB Do yay rang(A o B) < min(rangA, rangB)

Bay gio ta churig minh

rangA rangB - n < rang(AB) Ta co dim lle = n = dimIm(f o g) + dimKer(f o g) Mat khde 26t anh xa Kor(f g)/Kerg Kerf (1 Img

a do tI/4xl = g(2), yin X E Ker(f g) De they (To la (Tang au tuyeal

tinh Vi

dimKer(f o g) = dimKerg + dim(Kerf fl Img) Tit dim Im (f g) = n - dim Ker(f o g)

Nhu vay:

rang(AB) = rangB - dim(Kerf (I Iing) je rangB - dimKerf rang(AB) -e rangB+rangA- n

3) Vi A = E (E la ma (ran den vi), nen: (A - E) (A + = Vt vay, then phan 2), to co:

rangA -E) + rang(A + E) - n <0 hay rang(A - E) t- rang(A + E) n M3t khac, then phfin 1)

rang(A+E) + rang(A-E) rang(A = rangA = n Do vay rang(A + + rang(A - E) = n

V( du 2.15

Cho ma tran A e Mat(n, R), cac phan to (ran duang cheo chinh bang 0, cac phan to khac hang hoac bang p, d (16 p la mot eel nguyen lan hdn Chung to rang A e n-1

142i

)(et ma tran II = (h ij ) e Mat(n, 1R), b ij = -1 vol moi i, j Khi do rangB =1

va A + B = (CO, vdi C,, = -1, thc phfin to khOng thuOc during cheo chinh bang hofic bang p -

Do p-1 > nen det(A +B) # hay rang(A+13)= n NhUng Chao vi du 14, to co:

rang(A+B)< rangA + rangB = rangA + Do vay rangA n-1

Vi du 2.16

Gui sit (p la 1-1101, L1.1 long cau run khang gian vac td phis n

chieh V Ching (Dinh rang tan tai (ig khong gian vec td - hat bin k chieu Vk:

0= V„ V, c c V„ = V

gidi:

9: V —> V la Di dOng cau clic( C - khon g gian vec (5 V, 11611

ce mot vec td Hong e t

Bat V, = Vect< e l >, thlkhong gian v@c td chi6u (p - brilt biOn

Gia sv (ili xily dung duo() car khong gian (4) - bat Heal V, V2, Vk ma V, = Vect <e l >,

V2 = VITA <ei , > Vk = Vect e h 8: dimVk =k

Tel get long cau Wk : y vk \wk

cam sinh bai clang cau p Do tir k IA ta clang can caa khong gian vac to V/Vk tren truang se" phtic C, nen Co vac td rieng rea +i

ling vdi gil tri rieng can chfing minh:

Ubl Vect le, ,e k , e k*, I

khong gian k+1 chieu dm V va la - bat Man kik

Xet re hop tuyan tinh e i = i=1

Neu ?L k+ , thi ek,i e Vk [ ek+1 = [6], trai vei vice k

chon{ a k ,, I la vac td rieng, vay k,l = tit =0 nen iat

= 12 0 he {a ; .6 k doe lap tuyan tinh Nha way he

lei , eki doe lap tuyan tinh = k +

Ta chUng minh Vkki lap - bat Bien Chi can chang minh:

{14 ) e Vik*I-

Ta cif) pk k k-k+lk = im(rp )1 )

`T‘ -k+1 kk = •`k-hlk k-kk = `- k+1 k

p (e +, ) - X e k+, e Vk C Vk+l, tit suy Bang each do, ta say dung clam cat khbng gian:

Vo = c V, c c Vn = V p - bat hien

Nhain xet: Ma tran tha ta clang cgu p co sa 1, e „}

)(ay clang a tran c6 clang tam giac tren Dung ngan ngil ma [ran, ta c6 the phat hie: moi ma tran phew vuong cap n dgu d6ng Bang vOl mat ma tran tam gile tren

Vi du 2.17 (Dinh It Hamilton - Cayley)

Cho V la mot K - khong gian vac to; cp e End(V); 9 c6 ma tran A cd so e = (ed i = 1, n; 1,, - la ma trail dun vi, det(A - xI„) IA ma da thdc bac n, doe gut la da thitc da, c trung cUa tii (ItIng cau cp D6- thay da thuc khong phu thuc van ed so e = (u) da chon Ta cling goi da tilde tron la da HI& dac trung dm ma trail A

Gia sit det(A - xI„) = a ox" + a l x"- '+ +a„_,x +a„ Khi to co: a 0A" + a,A" - ' + +a n_,A +a„ = hay 42'1 + a,""-1 + + an Id =

Lai gidi:

Xef ma luau B la ma trail chuyi1n vi ci'm ma tran phil hop am A - xI„ Cac phan ;I'M ma Iran B IA nhfing da thilc cua x v6i he tiI truong K, c6 hac khong qua n-1 Ta

B = B0 +13,x + „ + x" -2 + B n_ x"-1

trong Bo , Bi , B, la nhung ma trait vuong cap n, khong phu thuiic x Theo tinh chat dm ma tran phu 110p, to co:

(B0 + 11,x + +B R, x"1') (A - xI„) = det(A - x1,) =

= (aox" + + + an_i x + a„) I„ Ta co mot he cac clang thitc sau:

B0 A = a n I

B„-,A - 13,12, = a, 1„ -B„., = a,,T„

NhAn vg voi vacua the clang their trcn Bin Met vdi I n , A X- A" rOi enng Jai, to (kith

a n A n + + a, ^A+a n I,=0

Chu Dinh 19 Hamilton-Cayloy thadng dude ghat binu:

MO ma tran vuting dela la nghiem maa da tithe dar trnn, ua ne

Vi rig 2.18 GiA s t B la ma trap ley linh, la ma tran giro

halm vdi B Hay chtIng mink

det(A + B) detA

Leh gidi:

a) Xet twang hen detA =

cia stY n la bac lily linh rem B, nghia la Bn = Ta re: = zcnkAkBn-k = Ecnk A k w _k =

(A + B)" A IckAk-1 n k

k=0 k=1 k=1

Vi detA = nen det(A+B)n = Vi way det(A + B) = = detA b) Ttheng Mk) detA =

Theo gii thiat B lay linh, nghla Isl co se n de 13" = O tit

-1 B)" = vi vay A -1 B lay linh NM( vay tan tai ma tran kheng

ay Men C C - 911 -1 B).0 la ma tran dime tan hal nhning ma An inking Bang:

' o

1

0

am riot thco during cher) chinh tr.( ta co det(1 + A -1 B) = 1, a la ma Ban don vi (tang dip vdi B va A

Nhu vay det(A+ B) = det[A(I + A -1 13)I= detA (-Iot(I +2.1 11 B)= detA Vi dy 2.19 GiA six V la met khong gian vec td tren truang va f e End,(V) Chiang minh rang Kerf = Imf va chi = vh ten tai h e Enci,(V) h o f+ foh= Hy

Didu ki(es chi: Tu f = say Imf c Kerr

Bay gia ta chUng minh Kerf c Imf VOi IV la Itheng gian ni bat 14 cna V, to co:

(1) f + f II) (NV) c hof (W) + f h(W) Do ta co of + f oh= Id, nen

Digit ki•en din

Gia si7 Imf = Kerf = V, Goi V2 la phein bet tuygn tinh ego

V, V nghia la V = V, $ V2; p l : V V I la phep chi& chinh tac, nghia la ven x e V, x = x, + x , x, e V i thi p,(x) = x l

Xot anh xa ?: V2 -)

x2 1—› (x2) = nx2)

De thgy If la Bang ego Dat h = I -1 o P I , to en: f h(x)= f-' pi(x) = pi(x) = xi

Thong tki h f(x) = ho f(x, x2) = ho f(x2)

= -I Pi f(g2)= f((2)= -1 • ?(x) = x.2- Do vgy (10 h + h (x) = x l + x2 = x

Do foh+hoN Id o

Tu gia thing Imf = Kerf suy f = O

Vi du 2.2a GiA si V la khong gian vac to tit' truong 1K v? f e End K(V) Vol m61 da thfic P e IK[x], P = ±a i x i , dat P(f)

=0

'a; f' thu6c End k(V) De thyIKPc] EndK(V) (13(P) = P(f =o

la mat d6ng cau tuyen tinh

1) Gia sia V I la kh8ng gian oda V bar bin do=i vdi f va f, = f I V, la clYng au cam sinh Chung to rang da thfic to) tiaai P, ena f, la Pdc cua da thfic to) tiPu P r

2) Gia sfi rang V phfin tfch throe tong cac kheng gian baat bign del vdi f: V = la dying (Au cam sinh trim

=1

V, va P, la da thuc tot tiPu cPa f, Chung LO rang P, la boi chung nhO nhavt cua cic da thfic P

Liii gidi:

1) Vi f = f1 nen vdi moi Q E K[X], La co Q(f) = WO fit VI kreh Pf lA da thfic ten tiUu rim f, thi P r(f1 ) = Do vhy Pf la bOi

da thfic toi ti6u P, oda f,

Ta cling nhyn flay rang P, khang nhal thi6t la uoc thpc sp Lim P,, V, la klieg-1g gian thpc so cim V Vi du ne).1 '= %H", thi La c6 Pf = P, = x - vdi moi V, c V

2) V, la khong gian f bi6n, nen can 1), da thfic

:61 thi6u Pr chia hect cho vdi moi i = 1, 2, p Do vyy Pf chia Mich° bOi chung nhO nhiit P cua Pi

Ta chUng minh P(f) = That vyy, vdi moi i = p to c6 I)

= Q P, tit c16 vdi moi x c V, x = Ix ; , I'(f)(x) = LP(f)(x,) =

1=1 =1

P p

P(f; )(x i ) = EQ ; (fi )13; (fi )(x ; ) = P r(f) = vdi i r=-1 i=1

X = (a,), A =

2.1 Cho ma tran X E in ]K)

7 0,

)1 - H la

cot tha j ca» ma tran X, =1,2, 1)

C - BAT TAP

§1 KHONG GIAN VEC TO VA ANH XA TUYEN TINH

a

)

B,, B., , B„ la cite cot cUa ma tran X X'

Chang minh rang cite khong gian rim R" sinh boa Al, A„ va sinh bai 13 1, , B„ trimg nhau, tit de suy rangX = rang(X

2.2 Chang minh rang mei ma ran hang r dou via dude

duai clang tang cua r ma tran hang

2.3 Cho E 19 la hai K - khang gian vec to va u e Hom w (E, F) Cac vec tel x , , x, thuoc Kern; y l , y, la nhang vec to brit k9 thuec E Chang minh rang hai khang dinh sau keo theo khang dinh MEI

1) lx,, , la cd sa can Ecru

2) B(y,), u(ydl la cd sa cUa Imu

3) K r , y s} la ed so ciia E Tit suy nen E hau han chieu thi

2.4 a) GM sit E va Fla hai kheng gian vec to him han chieu tren truring IK, dimE = dimF = n u e Hom K(E, F) Chung minh rang u don cgu va chi u town

b) Neu vi du chUng to rang neeu E va F có so" chie"u vo han, menh da tren kfrong dung MM

2.5 Cho a la mat ph6p thgbac n va u E End(C) xac dinh bar u(Z,, Za) = (Z.K0 , Zem •• • ;II) )-

a) Hay tam ma tran A cim u co sa to nhien te r , &la C

It) Tim tat ca the ma trait B e Mat(n, giao hoan duple vdi A

2.6 X6t kh8ng gian vec td E him han chigu tren truang K a) GM s>i fe„ , e„) la mot co so cim E, a„ , a,, la nhiing vo hudng doi mat klMc nhau, u e End(E) the dinh bai

u(e i) = ° ; e; = 1, , n) Chung minh rang nth v e End(E) ye u.v = v d thi ton tai nhUng vo hudng p„ ,j3„ cho v(ei) = [3 ;e;

b) ChUng minh rang nth to clang cgu tuygn tinh u giao hoan vdi moi to clang eau tuyeIn tinh tha E, thi t6n tai ve hitting

E K u(x) = 7rx vdi moi x e E

2.7 Cho A =(aIii)e Mat(n, K); det(A) x 0; V la mat killing *4, vec to tren truang K va uj e End(V), j = 1, , n Chting minh

2.8 Gia sil A e Mat(n, R), detA # va moi (long efn A co dung mot s6 khac khong, bang ± Chung minh rang:

a) Al = A-1

b) Co seta nhien m de Arl = A'

2.9 Cho V la khong gian vac to Hen truing K va u End(V), x la vec to cria V them man IOW = va u° (x) # v6 mat se' nguyen throng q nao Chimg minh rang he {x = u°(x)

u(x), , 10-1 (x)} lap mat he [lac lap tuygn tinh

2.10 Gin sir) V la lit - khong gian vec to n chialt; f, g End(V), Id la anh xa clang nhdt cim V Chung minh rang nei Id - g o f la clang cdu dm V thi fog - Id cling lA (tang cdu ciaa V 2.11 a) Hai tu citing cdu u, v E EndK(V) duac goi la Worn during ngu c6 die dang cdu p, q cem V cho uep=q v Chung CO rang u va v taking during va chi chung NS cum hang

b) Tit a) suy rang nalu X E Mat(n, K), hang X = r, thi tar tai cac ma trnn khong suy bP6n P, Q e Mat(n, K) cho:

'Ir I,

X = Q P a la ma tran vuong cap n

0 0 ,

al gee tren b8n trai la ma trnn don vi h cap r

2.12 a) Cho V a khong gian vac to n chigu tren trUang Ira u, v e End(V) cho u o v =

Chung minh rang hang(u) + hang(v) < n

(^ri±ng 11111111 rang vo mai hi ci6ng eau u e End(V) deu

Ong cdu V e End(V) cho u v = va hang(u)+ hang(v) = n

2.11 Cho E, F, G, H la cac khong gian vec td hitu Mtn cht4u men trodng K, u c Hom(F 0), u ce hang r Hay tim hang dm cac anh xa tuyeho t(nh sau:

a) cp: Hom(E F) —> Horn(E, G) u v b) di; Hom(G, H) —> Bom(F, H)

F > 0)01.1

2.14 Cho E la khong gian vec to n chien, u, v e End(E)

cho rang (u - Id E ) = r, rang(V - Id E) = s Chiing minh rang range o v - id s) r + s

2.15 Cho E la khOng gian vec to tren trtiang K Nth u e End(E), rang u = ChUng minh rang t6n tai e 1K a u2 =

Min ntm, neAl X # thi Id E - u la clang mill

2.16 Gia s>i V la khong gian vac td tren trtfong so thuc R; dim V = n; f e End (V)

Chdng minh rang en sdN nguyen ding cho rang (fk) = rang (F`') vol moi k N

2.17 Cho ma tran A e Mat (n, K), A = (a id ma a ii = vol moi i = 1, 2, n Chimg minh rang t6n tai ma trail B, C e Mat (n, R) de A = BC — CB

2.18 Cho hai ma tram A, B E Mat (n, K)

a) Chiang minh rang ma tren A d6ng clang vol X In thi A=X I„ X e ]K, In la ma tran ddn vi

b) Neu A, B la hai ma train d6ng dang, thi A2 deing clang vet B2, At &Ong clang vdi 13' va ndu A, B kha nghich, thi A -1 &ling dang vdi B-'

2.19 Chung minh rang m8i ma trail vuong cep hai tren truring K &et' d6ng clang vdi ma tran chuydn vi eila no

2.20 a) Chung minh rang to d6ng ceu tuyen trnh u e End(V) thCla man didu kien

u2 (Id — u) = u (Id — u) = thi u lily clang, nghia la u = u b) Flay tim phep bign del tuygn tinh kluing lily clang u thda man (Id — u) =

c) Hay dm phep bign ddi tuygn tinh khong lily ding u th6a man u (Id — u)2 =

2.21 Cho V la kh8ng gian tree td tren trtiang K

a) Neu f, g la nhfing dang tuyen tinh tren klreng gian vac td V thOa man f(x) = nett g(x)-e 0, thi co v8 Inking a e K M f=a g

b) Cho f, f„ f„ la nhung dang tuyen tinh trail khong gian •vdc td V cho f(x) = neM fl(x) = 0, i = 1, 2, , n Chung minh rang ton tai nhung vo hudng a l , c(2 , , an cho f = a f, +a2f2 + + a„f n

2.22 rot khong gian vec td n chigu V veil cd sa le l , e„) vk

tuyen Utah tren V Vdi moi = 1, 2, n xec dinh clang Wyk tinh e V* bait

Qej) = Sij =

j i = j

Chung minh rang (£ , ,1„1 la cd sa cua V* va do.dim V* = dim V = n Cd sa fb, f„} ciudc goi la cd sa lien hop hay ed sa d6i ngau N:i ed sa ••.,

2.23 Chang minh rang nen fil, la nhfing dang tuy6n tinh tren khong gian vec to n chieu V, vdi m < n thi ton tai phiin ti x t 0 V cho fi (x) = vdi moi i = 1, 2, , m TU

do hay suy met k6t qua v6 nghiem cua he phudng trinh thuan nht

2.24 Cho V* Fa khong gian lien hdp cna khong gian vec td V; u e End(V) Xet u*: V* —> V* the dinh bat u* (y) = you

a) Chling minh rang u la phep Men dot tuyen tinh cua V*, u* chive goi la phep Bien d6i tuyen tinh lien hop vdi u

b) co sa{e , en} (1) cila V va cd sa{f i , •••, in} (2) cua V* lien hdp vdi ed sa (1) Chung minh rang ma tran (l u), cua u* cd sa (2) la ma trait chuye'n vi cua (ct ii)„ cua u Cd sa

(1), nghia la Po = a3; vdi moi i, j n

2.25 Cho V* la khong gian lien hop vdi khong gian vac td V, S la be phan dm V RI hie, u 8° la tap tal ca the phan to f E V* cho f(x) = vdi moi x e S

a) ChUng minh S° la kitting gian cua V*

2.26 GiA s& V* IA khong gian lien hop vdi V, T la met be phan ciia V Ki hieu T ° IA tap tat ca cat x e V cho f(x) = vdi moi f E T

a) Ch'ing minh I"' la khong gian ciaa V

h) Xet S c V va S' ) c V* duct dinh nghia a bai 2.25 Chung minh rang (St la khong gian oils V shill bal S Dao biet

S la khOng gian thi (S")" = S

c) Gia st V hfiu han chi6u Chung minh (T") ) la Xining gian cfm V* sinh boi T

D4c biet, neal T la khong gian cUa V* thi (T") ) = T d) Chting t6 rang Meal V ve han °hien thi Wee khong gian T cho (11°)° # T

2.27 Cho V* la khong gian lien hop ciia khong gian vet td V IV,I, €1 la ho nhfing khong gian oda V Chiang minh:

a)

n VID = ier Biel

(

.0

1, fly; = IVi° neat V Mitt han chi6u

IET / ki

c) Khang dinh b) van ching n6u I Mill han, V ce the vo han chigu

d) Khang dinh b) khong dting neal I ve han va V ve hat chi6u

2.28 Cho (1',), e i IA ho the khong gian ciaa khong gian V* lien hop vdi khong gian vec to V

Chung minh:

a) ETi I = nrE

\ iel iE=.1

n = n6u V heti hen chiau

ieL

c) NC, u N.2 va han, thi V* cia hai khong gian

can T, (T,11T,r, Ti r12

2.29 Cho E la te'ng true tipcim hai khong gian V va W; x e E viet dupe nhat dUdi clang x= y+ z; y c V va z e W Ta goi phep chigu cea E len V song song vdi W la clang eau tuyen tinh p: E —> E xAc dinh bai p(x) = y

Ching minh rang cl8ng cad tuyan tinh p: E E la phep chi6u ve chi p= = p

2.30 Cho E la khong gian vec td va p e End (E) Chiing minh rang p la phep chi6u va chi Id — p la phep chi6u Khi Imp = Ker (1d-p); Keep = Im (Id-p)

2.31 Cho V Ea khong gian vec td tren tnidng K va f E End(V)

Chung minh rang P = va chi Imf c Kerf

ChUng minh rang trUCing hap g = hi + f la melt tkt deng ceu dm V

a) Cluing minh rang co hai co sa {e„ e„) va {E D ., en} cna V via soi nguyen s (0 s n) cho u(e,) = 6, vdi i i< s va u(e) =

vii s < i <

b) Tit suy bin tai to clang caiu tuygn tinh v mart V cho you la phep chigu

2.33 a) Cho p la met phep chigu caa kheng gian huu han chigu V Chung minh rang V co cd sa ma Wm A

= (a 3) ena p co clang dac biet: a,4 = ngu i # j; ail = neu 1<i< s va a;; = ngu i> s

b) Tu (Icing Pau tuygn tinh u duos got la del hop ngu u = id ChUng minh rang clang thuc u = 2p - id thigt lap mat song finh tit tap tail ca cac phep chigu p V len tap Cal ca cac doi hop u 6 day V la khong gian vec to tren truing K voi dac s6 khac hai

c) Tit a) va b) co thd not gi lid ma tran cua phdp del hop teen mat khang gian huu han chidu

§2 HE PHUCNG TRINH TUYEN TiNH

2.34 Tim met he nghiem cd ban caa he phuong trinh sau:

x - 7x + 4x3 + 2x -0 2x -5x +3x +2x + X =0 5x -8x.2 +5x +4x +3x5 =0 4x1 -x2 + X + 2x + 3X =

2.35 Cho hee, phudng trinh thuan nhat:

2x 5x, + x + 2x4 = 5x - 9x 2x3 7x4 =

-3x1 7X9 X3 + 4x4 =

4x1 6X2 + X3 - kX4 =

Hay giai va bien Man then A he phudng trinh Wen 2.36 ret he phudng trinh:

1 a(b -c)x + b(c -4 + c(a -13) z = (b-c)x + (c -4 + (a -b)z =

GM sit a, b, c cloi met phan Wet Chung to rang he tren co ghiem hoac x = y = z hoac

ab -be + ea bc -ca +ab

-

ca -ab +bc 2.37 Gigi he, plutAng trinh:

X

-X1 + 1 4- X3 x a

- x2

▪

X3 + X

xl + X9 + X3 + x4 X1 + X2 + X3 + x4

2.38 Giai va bien Man he phudng trinh tuyen tinh sau:

2x + y z = a

X + my z

2.39 Tim digu kien din va du a 4 dim A,(x l , VI), Ai(xi, 312), A3(x3, 310, Ai(xi y1 ) khong ce digm nao thang hang ngm tren met &tang trOn

2.40 Tim da the bac vdi he so thoc f(x) cho £(-1) = 0, f(1) = f(2) = va f(3) = 16

2.41, Ch9 A = (a ig e Mat (n K) cho dot A = Goi la phan phu dal ce cUa phgn to a,,, gia thigt A 11 #

Hay tam mat he nghiem co ban cila he phtiong trinh thuan nhal sau:

LAuxi

1=1

§3 CAU TRUC CUA MOT TV DONG CAU

2.42 Ching minh rang m6i gig tri rieng ciga phep chiegi dgu bang hoar va m6i gia tri rieng oda phep dM hop dgu bang hogc bang —1 (xem bai 2.29 va 2.33)

2.43 Cho A, B la hai ma trail vuong gang

a) ChUng minh rang neu A, B deng Bang thi cluing ca clang met da fink dac trting

b) Chung minh rang n6u A, B ce cling da thug dac thi det A = det B

PTO,) = (-2,)" + c,(-2,)" + + + 'Prong ck Yang cua tat ca the dinh thew chin)) cap k cua ma Han A, )inh thew eon chinh la dinh thfic lap nen tit the (long va

re c.0t vdi chi s6giang nhau)

2.45 Gia sa ( la to deng Ha.' cua khong gian yen to V n

nen trru trtfong K

Gni la mot nF,Thi em cua da ',hue dac Hang cua 9, 2,„ co hieu r= rang (9-2, 0M)

Hay cluang minh I < n — r p

Y.46 Havain) nghiem dac trung cua ma tran AeMat (n, C) -1

1 -1

0

1

2.47 Tim da the dac bung cua ma tran vuong cap n

0

0

ava a

1 an

Tit suy m6; da thin; bac n yea he t>i cao nhat (-1)" den a da their dac Hung cua met ma trdn valuing cap n nao

2.48 GM V lk khong gian vec W n chieu troll trUang K,

cp E End (V), X° la mOt gia tri rieng ena gyp, 700 la khong gian rieng cna V, ling vdi gia tri rieng A ChUng minh rang so chigu cim g1 /.0 khOng vuyt qua sei bOi p cem nghiem 700 cua da thdc (lac trung dm T Hay chi VI du dim "91 A, < p

2.49 Cho o la mot phop thg bac n va u e End (0 1) xac dinh ban

u (Z„ Z„) = (74,„„ ), ,

Hay tim da thfic dac trung va the gia tri rieng ctia tu clemg tau u

2.50 Cho f e End (R"), co se, to nhien {o„ , e n} co ma tran A=(a d),tldoa1 = a vOi moi i va a d = b yen ix j; a # b

Hay tim mOt ccI so cem R" a ma tran cua f cd sa do co Bang

2.51 Cho A, B e Mat (n, C)

a) Chung minh rang the ma train AB va BA co cling tap hdp the gia tri rieng

b) HOi AB va BA co clang da thne dac trung hay killing? 2.52 Gia sU V la khOng gian vec td phito him han chieu, f e End (V) ma co se N, nguyen during dg fN = Id Chtng minh rang V co co sä gom nhfing vec td rieng cua f

dual dang A = A, + A2, a A, giao haw vdi S va A2 phan giao hoan vdi S, nghia la A,S = SA„ A 2S = -SA2

Chung minh rang 52 = X I„ la ma tran dun vi

2.54 Cho E la khOng gian vec td huu han chieu va f e End(E) Vai m8i i = 1, 2, dat V; = kern va W, = Im(P)

a) Chung minh V, c Vm , voi moi i = 1, va t6n tai s6 m V„, = v„,+, va V = V,„„ vdi moi j x

b) Chung minh E = V„, W„, vdi m tim chide can a) c) Vm va tim duo a) la nhUng khong gian bOt Bien doi vdi f Hon nua, f Iny linh tren va kha nghich tren W„,

2.55 Cho V la khong gian vec to tren truOng 1K (1K bang R hoac C), e End (V) va p lily linh Chting minh rang:

a) Moi gia tri rieng cua cp deli bang

b) Neu dim V = n thi da thlic dac trung cem la oon 2.56 Chiang minh rang neu W la tv d6ng ca'u cua khong gian vec td phac han han chieu ma moi gia tri rieng deli bang khong thi linh

2.57 Gia s>t V la Haling gian vec td tren &Jiang ]K va dim V = n f e End (V) la tv clang cali lily linh dm V ChUng minh rang PI = O

Chung minh rang ton tai met co so dm V de r tran cern f c6 dang:

hoac

0 0 \ 0

Neu f co ma Den a dang thu hai, thi f lay linh va c2 = O 2.59 Chung to rang neu hai deing cau L g gem kh6) gian V Dacia man he thew f.g—g f = Id (0 thi vai moi k e ta — &c.f = k.g" (2)

Tit d6 suy rang khong the t6n tai the dgng cit'u tiv man he theic (1)

2.60 Chung minh rang, d6i yeti ho bat kY CAC torn to tinh giao hoan vdi Ding doi mgt khong gian vac hem han chigu- tren trang so" Ode, tan tai yen to rieng chui cho tat cA the Loan to tuyen tinh cria ho

D HU6NG DAN HOAC DAP S6

§1 KHONG GIAN VEC TO VA ANH XA TUYEN TINI

2.1 GM sU V, W Ian hurt 1a cac kheng gian sinh bai A„ ka B , , Dr Veli mei = 1, 2, n thi B, = a,,A, + a 2A2 + u„,A„ Do W c V De chUng mini) W = V, ta chi can char minh dim W dim V

Ta c6 dim W = rang (X X') = n — d, d bang chigu cua khong gian H the nghiem ena phudng trinh:

'a '

0

to p

0 0

t„) (X Xt) = (0, 0, , 0) Neat Y = (t,, t) E H Y XX' =

X.)4' = 0o Y X X' = (Y.X) (Y.X)' =

Y.X = Nhtt vay Y e M la khOng gian nghiem cim

phuong trinh t„) X = to dim W=n—cln— dim M = rang X = dim V

Do W = V

2.2 Gia su A„ , la the vec to cot (-Ala ma tran X hang r C6 th6 gia thi6t A„ (lac lap tuy6n tinh Khi cac Ai ; r < n bie'u thi tuyett tinh qua A„ , A,

Al = a,,A, + cii2A2 + + a„ A,

N611 dung ki hieu (A„ A2, , A,) (16 chi ma tran co cac vec to cot la A l , A2, A„ thi

aciAl)

Mei ma tran v6phili c6 hang 'Pa co digu phai cheing minh 2.4 b) Xet R [x] la khong gian vec td the da thew sr& he s6 thitc Xet u: R [x] —> R[x]

f(x) ) x

116 rang u la chin caM, nhting khong than eau, vi Imu khong chiia nhiing da theic bac khOng

2.5 a) Theo each the dinh caa u thi ea so tu nhien (e l , , en} cern C", to ca u(e) = e,„„ Vay A = (ad la ma trait cua u co so {e„ thi

{1 i (j) auo =

0 voi # a (j) b) Gia sii B = e Mat (n, C)

Xem B IA ma tram cila phep bien den tuygn tinh cd to nhien cna e n : v(e i )= Eb na Khi AB = BA a uV = vu < uv(ej) = vu(a) j = 1, 2, in n Eb u(a)= v(e a0)), j = 1, n <

n n n

Ebijecro) = Ebi,wei = <=> b ;,= bc,(;) „„, i=1

1, , n, nhu vay, the ma tran giao hoan dime vo tran B = (130 sao cho b u = bum on) voi myi i, j = 1, 2,

2.6 a) De dang

11) Gia stl u e End s (E) giao hoan voi

End K (E) Theo a) u (e) = p ie, voi {e„ cd so

Gia sV i # j, g 3, j < n to hay chfing minh 13, = End K (E) cho v(e k) = Vk 1, 2, , n Tit vou vou(e) = uov(e,) Nhu vay j3, e3 = u (a) = rija a (D i P: -pj=A Tit dou(x)=Axvoimoixe E

2.7 Ta co the bieu din n clang (laic Ea nu jet n) &MI dang clang (Mk ma tran sant

an1 ant ann

( ail 312 aIn ‘ Ill

[a21 a22 v2

/ n

, v

Vn

voi mci i, j

i A la cic m n

d6ng eau v nao dm I

p i Clam v = uov say -13i) = =

i=

VI ma Han A = (a n ) khong suy men, nen co ma tran nghich dge 11 = A- ' = (bd

Do b(au) = (ba) u voi nun vo Imang a, b va voi moi phep bi6n den tuyen tinh u nen La co:

u l (ui

B =[B A] •

Aun j

Tie to co:

1

0

0

l

ul u2

Lon)

=

1 bin

621 b 2n ••-

bn b nn v2

v

ni

Hay u ; = ybi,v;

Theo gig thigt, eac vi giao &ban voi nhau, nen suy the u, thing giao honn voi

2.8 a) Vi A khOng suy bign nen mei cot ena no cling co dung met phAn to klthe khong

Gig set A = (a n), At = fa t) a t _ a ,, ij

Hat A.AL = B = (b 0) Ta c6

bii = L a ik -aki = L a ikaik=sij

k=1

b) Cacti 1: Goi G la tap cac ma tran clang troll

Ta thdy A e G, B E G thi A B e G, A-' =A° e G Do C

lam nhgm cim nhian GL (n, R) cite ma Dan vuong cdp n khong suy bign Vi s6 cac phial to °Pa nhOm G la lulu han, nen vdi A e G, t6n tai k e N k 21, dd = Ak - ' = =

At, nhtt vay A"' = At vdi m = k — e N

Cad' 2: Gia sit (e ) , , en) la co sa W nhien cim f e End(R") c6 ma trait A co sa {e l , enL Ta e6 f(e,) = ± no) d6 a la phep thg bac n Vdi m61 i = 1, 2, n xet a(i), a 2(i), tai k,deg ak'(i)=i, f(e,) = e ap o = e, PLi(6) = e,

Dat k = k, k n thi gc(e) = e, V ; = 1, n hay f" = Id Ak = In, ldy m = k — to c6 = AL-1 = At

2.10 GM sit f.g — Id khong la clang cdu claa V, t6n tai

x E V, x # a (fog — id) (x) = Hay fog (x) = x Dant g(x) = y e V,

to co f(y) = xz 0= yx

Xet (Id — gof) (y) = y — g (fly)) = y — g(x) = y — y = Digu chfing to rang c6 y x dO (Id — gof) y = 0, trai vdi gia thief Id — gof la (fang cdu

2.11 a) D6 thdy u, v Wong ding thi co gang hang Ngudc lai, n6u hang u = hang v = r, theo bai 2.3, V c6 cac cd sd {6, , en}, (e),, e'„1 t(ei) = e't = 1, r), u(u) = > r), va c6 cac cc' sa e„}, e'„} v(e) = e,' vdi i = 1, r, v(el) = 0, j > r Xgt p E End (V) p (e,) = e; voi moi i= 1, n; Xet q e End (V) cho q(s) = e', (i = 1, n) thi uop = qov

2.12 a) De dang Tiny c Keru va dim Imu + dim keru = dim V

b) Xet co so ., en/ va {c , c„} rim V cho u(e,) = e, (i = 1, 2, r) va u(e,) = (j > r) Xet v E End (V): v (c,) = (i 1, r); v(c,) = ei vai r + j < n Khi (id vou = va hang u + hang

2.13 a) rang rp= dim E x r b) rang ty = dim FI x r

2.14 WE( u, = u — Id E , v, = v —

Tit u.v —Id E = + Id E) (v, + IdE) — Id E =

= u, v + u, + v, = u, (v i-lId E) + v, nhung rang u, (v, + IdE) ic rang u, va rang fu l (v, + Id E) + rang u, (v, +14) + rang v

Do vdy: rang (uov — H E ) < rang u, + rang v, = r + s (xem vi du 2.14)

2.15 Vi rang u = dim (Imu) = Gia sii {x,4 la ed sd GM Imu Do u(x 0) e Imu nen co 2, e 1K: u(x o) = Ax„

Ta hay thing minh u = = Au Vol x e E u(x) = a x o u2(x) = a u (x0) = a Xxo = X a xo = Au(x) vdi moi x e E u2 = Au

Gia # 1, tim 71 a (Id — u) (Id - nu) = Id

Di5u tudng &tong vdi (fik x_11 14.y

2.17 HD: Chon B la ma Dan cheo vdi cac phan to tren during cheo chinh khac doi mgt: b,, Chon C = (au)

a

dsao Cho C

v itt va Cli y, tin A = BC - CB b i - bi

a b' 2.19 GM X =

d1

E Mat (2, K) Hay chUng to c6 P =

x y

e Mat (2, K), det P t 0 de XP = P.Xt z t

bz = by (a-d)y= cx - bt

-d)z = cx - bt cy= cz , xt zy

(1)

Xet cac trudng hop sau:

+) b = c = 0, chon y = z = 0, x t = +) a = d, chon y = z = x = t = +) b +cz> va a t d: chAng ban b s 0, chon x = 0, t = 1,y= b -z

a -d

Nhtt gay moi trUdng hop, hP (1) dgu c6 nghigm 2.20 a) Tit u (Id - u) = suy 112 =

Tit u (Id - u)2 = suy u - u - u2 + = W vgy uz = u

=

b) Vi du V = K3 , u (x i , x2, x3) = (x„ x 3, 0) Ta thSy u2 * u nhung u =u3

e) V = K3, v(x„ x2, x3) = ( , x2 - x3, xa) Ta có v * v nhting v (Id - v) =

Chu Y: Co the neu nhieu vi du khac nfia

2.21 a) Ngu g(x) = voi moi x c V thi f(x) = \raj moi x e V va f = a g voi a e K Neu có xo e V ma g(x0)*

Data-

f(x) g(x{,)

Ta chUng minh f(x) = a g (x) voi moi x c V Vi g(xo)* nen g (x) = g(x), do: g(x -;xo) = f (x - )( 0) = f(x) - 4f(x0) =

f(x) = 1(x0) = g(x) f(x ) = a.g(x) g(x )

b) HD: chtlng minh quy nap theo n n = dilng vi day la trUang help a)

Gia eu bai town dung voi k = 1, 2, n - Ta chUng minh dung voi n

Trubng help 1: Neu co i de fl Kerf c Ker f, thi fl Kerfj =

joi j#i

fl Kerf, c Kerf Nhu vay theo gia thief quy nap, to co

i=1

f= ifJ=Eaifi (ai= 0)

Truang kip 2: Vol mdi i, n Ker fi ¢ Ker j#i

Khi vdi mdi i (i = 1, 2, , n) co xi e V de? f(xi) # Ova fi(x,)

= ( # i) D'.1t a,

-f(x ) f

Ta chting minh dude f = + + a„ f„

2.23 Ta chUng minh fl Kerf, 101

Hay thong minh quy nap rang dim ( n Redd n - vat i=1

1 < t < n

TU suy ket qua: Alai he phudng trinh tuyen tinh thuan nhdt vdi se; pinmng trinh ft hon s5 dgu co nghiem khan khong

2.25 a) De' clang

b) HD: xet 1x , x„) la cd sa cua S, {x„ x„„ x,„„, xn} la co se) clia V Voi moi i = m+1, n, set E V * : ii(Xi) = — ki hieu Kronecker) Ta thdy f, E S°

Hay chUng minh {f„,,,, f„} is co so cfm TU dim S ° = n - m

2.26 a) Hidn nhien

Nhung f(x) ± nen x e (5 °)° Nhit vay vdi x e W x (S°)° tit (S°) ° c W va nbut vay (S°) ° la khong gian sinh bai S

c) Ta cd (TY la khOng gian cim V* va T c (T°)°

Gia sa W* la khong gian cim V* va W* = T

Ta °hang to (r) ° c W

Gia sit (fi , f„,} la co sa cim W Neu f e V* \ W* thi f hong la t8 hop tuyen tinh cim cac LI Do do, theo hal 21, cO x € V (x) = 0(i= 1, , m) nhung f(x)t O Vi {f} la co

V1?'" (x) = vdi moi i = 1, 2, m nen g(x) = vdi moi g e via do, vdi moi g c T (vi T c W*) Tit x e T°, Nhung f(x)# f e (T°)°

Chiang minh tren chiing to (T°)° c W*, nghia la (T°)') la khong gian cim V*, sinh bai T

d) HD Gia sit (x i) c i la co sa cim khong gian V, I — tap vo ban; vdi i e I, Kat f, e V*: fi(xj) = S i (Su — ki hieu Kronecker)

X6t T la khong gian cim V*, sinh bai (fJ ; c

Hay chUng minh (T°)° = V* nhung T x V (vi (In f E V* ma f(x) = Niel moi i f T)

2.27 a) va b) (16 clang

c) chi can thing minh netu V„ V2 la hai khong gian cim V, thi (V n vo° = vi° + v2"

Ta cO V, n V2 C VI, V2 VI °

v,0 c (NT, n v2)°

D6 chiing minh (V fl V2 )° c V 14) , lay f e (V1 n V2)°

f(x) = vol mot x E v, n V2 Ta cluing minh f e VI ° + V20 )(et WI la khong gian bit maa V, id V2 V, o V, =

WI ED (VI n V2); Wong ta vet W2 cho: V2 = W2 $ (V, n V2) via V = (V, n V2) e W Khi V la teng trip tip gun V, n V2, WI, W2, W

Chon g, h e V* the dinh bat:

lb nab x E Vt n V2 hoitc X E WI

.if(x)ngu x e W2 boas x e W

t0 ngu x e VI I)V., hoac x e W2 boas x e W f(x)ngu x e WI

WO rang g e V

1 0, he V2° va f = h + g Nhu vgy

(VI n V2)" t-_- W 1° + V2° Tit dO (V, n V2)(' = Vi ± 1.72°- d) Vi du cluing to khAng dinh b) khong dung ngu I v6 han va V co chigu ve han

Gia sa (e) i e I la co so v6 han phis to caa V, mOt phAln to

caa V co dang Ecc,e, a i hau hgt bang khOng, tit mot jel

so hilt! han

Vol mOi i e I, ky hiQu V, la khong gian

COD cila V gam cac

phan tit E cgs, vgi al = ieI

Taco n o n vir = v* ieI

Trong Vi° F {f E I f(e) = V,

g(x)

h(x)c

Do V ic' la khong gian sinh bai f, (NhAc lai f, (e) = 6,j — 'hi 2.26) ma EV? x V* (bai 2.26)

idl

2.28 c) Gia sit {6} i E I la ed so vo han dm V; f, e V* cho , (e) = §i

T, la khong gian sinh btli {f1}, T, V* (Kern bai 2.26) f e V* cho Re) = vOi moi i E I, xet T2 la khong gian inh bai f T2 (1 T1=101 (Ty n T1P = V; nhung T 10 =401 cV

T2° t V vi T 2° chi chita cac MAI) ti) El:t i e, ma Ea ; =0

lei lei

Slut very (T2 (1 T1) ° a V+ T20

2.30 Ta c6 p la phep chigu (r) p = p <=> Id — p — p + p' = Id — 2P + p = Id — p

(Id — p) = (Id — p).;=, Id — p la phep chigu Bay gib n6u y e Imp thi y= p(x) (Id — p) (y) =

y e Ker (Id — p)

Node lai, n6u y e Ker (Id — p) (Id — p) (y) = p(y) = y= y e Imp

Nhti very, neM p la phep chi6u thi Imp = Ker (Id — p)

D6 y rang p = Id - (Id — p), to co Kerp = Im (Id — p)

GiO sit {e l , , es} la cd sd cim Imu Khi en e, (i = 1, s) cho u (e,)= c i Ta có dim (Keru) = n — s LO {e sti , e„} la co sa dm V (xem bai tap 2.3) Chi can LIS sung de" co co sä ., E s ,

£ } Oil to dude hai cd sa can tim

b) Xet v e End (V) ma v(c) = e„ i = n thi you la phop chi6u ciia V len khong gian sinh cj song song voi khong gian sinh btli {e„„ e„}

§ HE PHUT$NG TRiNH TUYEN TiNH -7 0'

2.34 A = -5 , rang A = -8

4 -1 3,

Nhu vdy khong gian nghiem co so" chiOu FLO da cho tUdng dudng vOi

7x + 4x + 2x4 =

2x 5x + 3x + 2x + x =

1

xi

9 — X 9x5

5

X =

9

—

9

X5 + —

9

X4 — —

'

Xr

'

Ta có mot he nghiem cd bin, chAng han la: Et i =(4, 2, 3, 3, 3)

a2 = (-7 , -1, 0, 0, 9)

2.35 HD He c6 nghiem khong tam thang va chi ih thiic cua he Wang khong a X = 35 Khi d6 hang Mitt ma n cite he s6 Wang a khong gian nghiem thigh

2.36 HD Coi he da cho la he phudng trinh thudn ntat viii biers (It - c)x, (c - a)y, (a - b)z Ta dan Mit he thuc:

(I) (c-a)y -b)z bz -cy cz - ay ay -bx k

De he c6 nghiem khong tam thitang, to phai c6 k = ±1 2.37 = ,

4 a + b + c +

x2

=-4(a-b+c+d)

x4 = —1 + b + c-d)

2.38 A=

1 m

I -1

1 -m

, det A = 2m (2-m)

a) v6i m* va m # a He la he Cramer

=0 1 1 Yt y2 32 Y4

Digu kien có nghi5m la

2 a b c

=a+ b - c =

x=b-z Khi

{,y ,=a-2b+3z, z my c) m = =4 rang A =

Digu kien có nghiem: -5a + b + 3c =

Khi

2a -b x -z +

3 y- z+ 2b

3

22 x +y x

2

X3 + Y3 2

x4 +Y4 2.39 PS:

x3

x4

2.40, f(x) = ax + bx2 + cx + d

0-1) = -a+b-c+d = 1(1) = a+b+c+d = f(2) = 8a+4b+2c+d = 03) = 27a+ 914+3c+d = 16 Tito: a = b = -5, c = , d =

f(x) 2x3 - 5x2 +

Do A„ # 0, nen rang A = n-1 va khong gian nghiOni cna (1) co so chigu Hay rang (A11) =

TV hO EA ii x i =0 (i = 1, 2, n) Wong doing voi j=1

X2 X2 ± Al2X, = O

Ta có mot he nghiOm co ban, riling ban = (-Al2 , A 11 , 0, n2= An,

0) 0)

Et„_i =(-Alloo,o, A„)

§3 CA.0 TRUC CUA MOT TV DoNG CAU

2.42 Gia s& u la Mt chigu u2 = u Ngu x la vec to rioting ung voi gia tri rieng ? u(x) = Ax

u2(x) = u(x) = Xu(x) = X2x Nhung u2 = u nen: A 2x = Xx (1.2 - ),.)x = Do x s nen - X = X= hoar =

Tudng to del vdi phep doi hop v2 = Id 2.43 eau a), b) dO dang

c) Dao ciia a) khong dung Vi du:

Cho A = In (ma tran don vi), B = ma (On tam gihc (Judi ma the phan t& dgu bang Khi det (A-AIn) = det (B-AIn) = (1 - Xr Nhung A AM B khong &Ong dang, vi A chi ding dung vii chinh no (xem bai 2.18)

Dao cua b) cling khong dung Vi du: Chan Ala ma trar (cap n > 1) va B e Mat (n, It); B= (b id ma b 11 = 1, cent; = mm (I, j) # (1, 1)

Khi det A = det B = 0, nhung det (A - XIn) = (-k)° I (B - Mn) = (1 - A) (-X)11-1

2.44.Vi N e End V, cd sa nao co ma tram A nen thtic dac trung cUa N co

IA - XIn I = (-X)" + C1(-X) 111 + + C

De thgy C„ = detA (cho X = 0) Ck 11 he so cim cac se; hang

s6 mu (-X)^4, (Woe tao thAnh bai tich cua n-k phan to tr &fang cheo chinh dang: (a yi - X) (a ind,,,, -X) vdi cac phgn khai trie'n dinh thtic IA - I, khOng chVa T va ni cac dang, cot ba vdi cac dong, cOt c6 chi se; {i Do dinh thtic chinh cap k ciia A co cac phgn tv tren &tang ch chinh veil chi so" ba .•• VI 60 {II • Lk} la thy $, nen

s6 cua (4)"4 chinh bang te'ng tat ca cac dinh thuic chi dang tren

2.45 Ccich Gia su A la ma tran cila cp cd sa nao dm V, thi = rang (9 - XDId) = rang (A- koI) CIA sii X la vac cOt, X thuoc khOng gian rieng P, ling vdi gia trt rieng X chi ( A - A01)X = > dim P kn = n-r Goi R la kitting git rieng suy Ong ling vdi gia tri rieng Xo, thi dim E xci = p (130i cl ko), ma Pxo c R ko nen n - r < p Mat khac det I A-X.011=0, ni ran= S n-r Vay S n-r p

Cdch Dat B = A - kJ Khi d6 = I A - (p+X, D)I I = A - XI I , voi A = µ + X Nhd vay A = Xo la nghiem bet p cua IA - XI I = thi µ= la nghiem bei p cna IB-µII = Theo bid 2.44, I B -µII = (-0" + C1( -0"1 + + Ck (- 11) "a C, a

do Ck 11 tong cga cac Binh tilde chinh cap k dia B Do rang B = r nen cac dinh there chinh cap k > r dgu triet lieu Ck = voi moi k> r

Vey I B - I = (-P)"+ + Doclop=n-s?.n-r

2i

2.46 DS Ak = 21 cos k n +1

C (-1,)" 5, s < r

HD Xem bai 1.19, cam a) dat a + p = -A, a.p = -1 2.47 Goi p(x) le da tilde dac tning cga ma trap A

-A G '

-A

an-1 an-2

thing coot to cg:

(1

P(A)=(-1)""a2,2det :v +(-1)"*2 a,, det

P(x) = det Khai trim theo

o'

1,

f- k

0'

1)

(-1)2" (a - A) det

Nhu v0y:

P(X) = (-1)' (a._, + a„ X + + asXn- ') + (- 1)11 l n Ke't qua tiep theo suy tit ceng dr&

2.49 HD: Phan Hell a tich cac yang xfch doe 10 a = T2 T I ; gia {e1 , , en} la cd s6; tn nhien cua Ca, tt

u(e) =e a Sap x6p lai co sa fe;} theo thft ty thich hdp ta dude c sa mdi cua C" sap P nhom, m6i nh6m g6m m a vet t (mk bang de dai Tk), k = 1, p Chang han nhom thit nhat gen fe' l , e',„ 1} thl u (e';) = e't+1, u(e'nil) = e' I Tuong t>i cho ca

nho khac v0y Da flak cl0c trung cua u co clang

I A - kin I= D I D„ a do Dk la da thile b•c m k dang:

- x 0

1 -X 0

Dk = - X =(—OnIk+I-P(

- Ark

0 0 1-1

= E lrk ()Link Nhu vay

A-xin Hoy (Am! -1) VP -1)

2.50 HD Theo vi chi 1.20 (chttdng 1), ta có det (A - AI) = - b - X) 111 (a + (n-1) b - X)

Com gia tr} rieng cua f la: a - b (bel n - 1) va a + (n-1) b vó X = a - b, co (n-1) vet to rieng doc 10p tuyen tfnh thaa man

VOi it = a + (n-1) VS'.y c6 the than Cd va xn (1, 1, , 1) gem

2.51 HD Xet dang (XIn - AB A

0 AIn i

b, co vec sox, = cac vec

-aide

'In 0'

2B In,

to rieng: t i (1, -1, 0), to rieng cita ma tran:

In

[ B In

= t2 = ., x2 1

()Lin

O

= t„ = (1, 0, 0, ,

A XIn - BA,

-1)

a) Ta có det (XIn - AB) An = A" det (Mn - BA)

+) N a3i X # 0, Ta có det (AB - Mn) = det (BA - Mn) nen AB va BA c6 cling gin tri rieng khac khong

+) Vol X = 0, dot AB = det BA, nen no cling cO nghiem chung X =

b) Do det (AB - Mn) = det (BA - Mn) vdi moi A, nen hai da thnc dac trung eim AB va cna BA trung

2.52 Xet cci so (6,, , e n} ciia V, ma tran f duac tao heti m khung Giooc clang Ung vdi m gia tri rieng phan hi@ A l , Atm

(2`k

1 Xk

Ak = 1 ' , cap Gila Ak bang dim/Kt)

0 X i<

(Rtk -khong gian rieng suy rang Ung yea gil tri rieng VI la khung gian bit bie-n cua f, nen theo gie thiet to co Ak =I nk (I„ k la ma trMa thin vi can dim Rkk = Ilk Ta co Ak =

'0 0'

(Xki n r r_- EN ika nNk i , a cN x J nk =

1 i=0

Nein nk > 1, thi c6 cac phAn tit Oleo Wang vdi moi i, vi \ay AI co phAn tit cheo Wang vdi mm N, digu de trod gia thigt Vi \ray nk = moi k Nghia la ma Pearl cim f cd sd (or e„.} co dang then, hay {e l, , en} la co sa g6m nhang \Tee to rieng cim f

2.53 Ta c6 bd de: Vdi S e Mat (n, K), det S = thi co ma

tr6n M e Mat (n, K), M a MS = Trude het to sit dung be

de tren de chung minh bai than

Gia sit S e Mat (n, K) co tinh chat: moi ma trail A e Mat (n, K) den vigt dude mgt each nh6t clang A = A, + A2, SA, = A S, SA2 = - A2 S

Ta nhan thdy S phad khong suy bign, ngu ngdde lai S suy hien thi theo be de tit cO ma tran M # de MS = SM = 0; Khi = M + (-M) = + la hai phan tich ma tran 0, th6a man

digu kien bai town, trai vdi gra thiet va tinh nh6t

Vdi mr6 A e Mat (n, K), to co the tim dttdc A l , A2 W di6U (SA = SA1 + SA2

kien:

AS=A I S+ A 2S , SAk -SA

A l =j-k+S -1 AS)

A2 =

2 (A -5 AS) Vi A1 S = S.A nen to he thiic tren to co AS + S'A 52 = SA + AS to AS = S2A

{

VOi mm A e Mat (n, 111) S' in (xem vi du 1.19); A # S khOng any bien

(Meng minh bd de: xet P la da thiac tea tieu cith ma tr n S (chi xet S # 0) CAC, P = xk + + + a k 2x + a k , k21, va p(s) = sk aisk- , = ak_,S + ak In= O Voi S # O Ta nhgn

they k> 1, vi neu k = thi P(S) = S + a, I„ = S = a, I„; S suy bien nen S =

Ta lai thely ak = 0, vi neal a k # thi S kha nghich, trai gia thiea Nhu vay:

S (S 1' + S k22 + ± at.; - 10= 0

DM M = + ad1 1-1 + , I„ # 0, to co SM = MS = 2.54 a) DO (ang

b) Nhan they V n Wm = {0} That vay, gia se x e V„, n vv„,

[moo = va co y E E de fth(y) = x, tit el(y) = yeV2m= V„, l'"(y)= x =

Nhu vay dim (V, 1V„) = dim V„, + dim W n, = dim E 'E_ V„, W

2.56

Ccich 1: Da the da'c trting cua W co dang P,p(a.) = (-I)n A" Theo dinhb) Haminton — Cayley Ta có p" = luy linh

Ccich 2: Theo vi du 2.16, ta co the' du) mot co se ma tram A cna (;) cc) dang tam giac tren Theo gia thiel cac phial tU tren clang cheo chinh bang khong, d6 An = p" =0 cp

uy linh

2.57 Cach Gia q la se nguyen diking be nhal d5 f' = Vi f9-1 x neu co vac to x e V (I5 f^ - '(x) m 0, r(x) = Theo hal tap 2.9, ta ce {x, f(x), 0-1 (x)} dOc lap tuyen tinh TpY q n Vi = =

Cach 2: Dat W' = Ink(?) Ta co W' = W = to có m n d5 W" = vi ne'u kheng ta co n = dim V > dim W' > dim W' > > dim W6-1- dim W"+' < 0, ye 1.3"7 Mut Sy W" = = Wm+i suy W" = hay P"= 0 f" =

2.58 HD: rang f = I dim (Kerf) = n-1

Chen e, e Kerf va e,J la co sa cua Kerf xet-f(e l ) = ale, + a2e2 + + aue„, phan biet hai truong hop a, = va a, x

2.59 He tinic Igk — gk.f = k.g" dung vat k = 0,1

Ta chUng minh rang nen no dung voi k-1 va k, thi h& thiw dung vol k+I

Gia sit co f.gk — gk.f = ta suy gk+1 gk.Eg = k.gk va gigk — gickf.f = k.gk:

Tit f.gk± ,7, g'+' f + g (fgk - l- e-, f) g = 2k.gk

Do gia thiCi guy nap: gk-' — = (k-1) gk•2 nen: fel

g k+1 , fkti (k-i) gk = 2k.gk

hay (k+1).gk; nghia la he thtic (2) thing vdi k+1 Vi \Tay, he UMW (2) dung vdi k = 0,1, nen n6 dung vat moi k e N

Ta cluing td rang khAng ton tai cac ty dling lieu thaa man he thin: (1)

Thai vz;)), , )(et P(x) = xP + a p.,xP-1 + + a o la da thdc tot lieu cua g; nghia la da thtic kluic khbng, cd bac nh6 nhet cho P(g) = Khi viol he thug (2) vdi k = 0,1, P to nhan dude:

f P(g) — P(g) f P' (g) d 13' la dao ham dm P VI P(g)= nen P'(g) = 0, trai vdi gia thiet P la da thne t6i tint' clan g

Chi) y: CO the nhan xet rang khong the ton tai cac tq d8ng

Chuang III

DANG TOAN PHUONG - KHONG GIAN VEC TO OCLIT VA KHONG GIAN VEC TO UNITA

A TOM TAT lit THUYET

DANG SONG TUYEN TINH DOI XUNG VA DANG TOAN PHUONG

1 Dinh nghia

Gia sit V la khong gian vele to trial truang so th0c Wit dang song tuyeal tinh xac dinh tren V la mat tinh xa

0:VxV—>R (x, y) H) (x, y) cho vdi 662 kjc x, y, z e V va e 7f& ta ca:

0 (x + z, y) = O (x y) + (z, y)

0 (x, y + z) = (x, y) + (x z) (X x y) (x, y)

0 ( x, y) = (x y) (x, y) = (y, x)

Neu vdi moi x, y e V ta co 0(x, y) = 0(y, x) thi (bloc gyi la dth

2 Bieu thik toa cita clang song tuyen tinh

Gia sn dim V = n, e = (e) = 1, 2, n la cd sa cila V Anh xa song tuy6n tinh B hohn toan (Liao the dinh bai ma trail A = (a 0), a a i, = (0,, e i)

Khi voi x= , y= y, e„ , ta co (x, Y)= Lao xi Y1

i=1

hay yiet dual (bong ma tran 0(x, y) = .A.Y, a do X, Y la the

ma ran tr neat cac tya dO cua x, y co sa (e i)

Gia s& e = (E,), j = 1, 2, n la mOt cd sa khac ciaa V ma CiiCi Dang song tuygn tinh cd so e = (au có ma ,=1

trim A, va cd sa s = (8) có ma Ran A', to co A' = A C,

a do C = (c, J)

Rang song tuy6n tinh B la do'i xi:mg va chi co sa = (e) nao do, co ma tran del xiing

3 Dang -than pinning

N6u B: V x V —> R la dang song tuyeh tinh del xung, thi H: V R, H (x) = B (x, x) dude goi la clang loan phudng le6t hop voi 0, B dude goi la dang eqe vim H Trong mot cd sa da chon, ma tran cua B cling dude goi la ma tran cua dang toan phudng H ling voi nO N6u bi6t dang toan phudng H, thi clang cue dm H hoan Wan dude the dinh, Cu thO:

•

Gia sai co sd e = (a), (x, y)= Za ii x i yi , thi

H(4= Za ii x i yi a do (x„ xx ) va (y1 , yd IA toa dO cam x va

y Wong N6u 100 cd sä nao a = (e), dang than phudng H co dang Ii(4= D i x?, LW cd sa a = (th dime goi la cd

sd chinh tac d6i vdi H, va to not cd so e = (th, H co dang chinh the Ta cling not cd sa e = (ci) a tren IA cd sd true giao (180 voi dang cuc Ta có dinh 15/

Dinh 13i: Neu H IA mot dang than phuong bat kjI tren

khang gian vec to thuc n chigu V thi V luon ton tai mot ea sa a = ( a trong co stl do, H co dang chinh tac

Chu $': TR cling có dinh nghia dang toan phudng tren killing gian vec td V tren tthang K s ', gan viii dang song tuy6n Huh d6i /ding the chi-1h tren V

4 Rang va hach caa dang toan phudng

Cho clang Loan phudng H tren khang gian vec td thuc n chik Gia six mot cd sa nao do, dang toan piniong H co ma tran A vol rang A = r Trong mot cd sa khac vdi ma tran chuyk cd sa C, thi H có ma trait A' = A C, rang A = rang N = r S6 r kh8ng phu thutic vac) co sa dang xet va dupe goi IA hang ciia dang toan phttong H, cling dupe goi IA hang ctia dang cuc caa H

Khi hang H = n = dim V, dang than phudng H dude goi IA khang suy

N6u V = V, S V2 ma O (x, y) = vdi moi x E V, va y E V2 thi

a n6i V la t6ng trttc ti6p trtIc giao cern V, va V2 (d6i vei 0) va ki riOu la V =V, ® V2 T6ng quAt, to co khai niem tong trttc ti6p

r

rye giao: V = e V, e e Vk

Ta goi hoch cim H (hay hack) cim 0) la tap V = {x e (x, y) = vdi moi y e V4 Day la mot khong gian am V ma ang H = dim V - dim Vo va vdi moi phCin bu tuy6n tinh W ctia V, thi lion ch6 teen W la khong suy biers Va

3 = W

5 Dinh lY chi se quail tinh va dinh Hi Sylvester Gia su fi la mot don ft 'man phvong teen 14.-khOng gian vec to V H &toe goi la xac dinh n6u Mix) * v6i moi x s O

H dti0c goi la xac dioh doting nen I4(x) > \FM moi x x H dvoc goi IA xac dinh Am n6u 14(x) < veli moi x* Dinh 19

Cho V la R - khong gian vec to n chi6u H la dung town hiving tren V thi V la tong true 061) true giao (doi v6i H) clia a khong gian Vo, V,, Vs

V = V CS V ff) Vo ma HIV, la xac (huh during, HIV_ la xac inh am, H I V0 = Cach phan tich tren khong nhal nhvng o luon ]a hod) cem H, dim V = p, dim V = q khong (16i; p, q leo OM to goi la chi s6 dtiong van 'al-1h, chi s6 am (man tinh Oa H (hay ens clang eve cim no)

Dinh ly Sylvester

Gia sit V la R - khong gian vec to n chigu H la clang to pfuldng tren V, A la ma tran cua clang Loan phudng H tro mat cd sa nao Goi Ak la ma trail ;oiling cap k a goc tr ben trai cim ma trail A (Ak tao bai giao cua k clang <tau va k ( dau elm ma trail A) Khi do:

H Ea clang toan pinicing xac dinh dining va chi k det Ak > vdi rani k = 1, 2, n

H le dang toan phuong xac dinh am va chi det Ak %

vdi k than va det Ak < vdi k Le

Chu 9: Khi H có ma tran dti) 'clang A, n6u H )(lc dinh duct to cling noi ma trail A 'the dinh ducing

§ KH6NG WAN VEC TO OCLIT

1 Dinh nghia

Cho E la khong gian vec to tren truiing s6 tlnic R Ta mat tich ve hudng a tren E la mkt anh xa song tuy6n tinh, xilng va xac dinh dining lien E, ki hiou ban <,> hay ( , ), ngh la to co: < >; E x E R la anh xa song tuy6n tinh

thOa man (x, x) > vdi moi x E 1E va <x, x> = suy x =

Khong gian vec td E cling vdi mot tich va xac dii tren E dnpc goi la nEat khong gian vec Ed (Mit

2 Mc).t so tinh chat

Ta có halt clang theic sau, goi la bat clang these Cos Bunhiacopski: <, >2 < N2 MYM2 •

b) Hai vec td x, y &tee goi IA true giao vol n6u <x, y> = Khi to co:

)1 = D1M M2 •

3 Cd sa trip chua'n khong gian vec to dclit hitu han chi6u

Gia sit E IA khong gian vec to delft n chigu it vec td e l , e2, ,e„ i = j dupe goi la co so true chuitn cUa lE neu <e i e j > =

0 i # j Dinh 19: 'Prong bluing gian vec td Gclit n chieu holt kY loon ton tat mgt co sa true chuan

Chung minh: gia sti la„ a,i l3 mot co sa nao dm

khong gian vec td (kilt E Khi co the ray dung mot en sa true chuan { e,, e2, e„} nlut sau:

c o

c o = a = et, - <cti ,

ME2M

i

e = M

n-1 a

en ka

= , d a n =a n -I< a n ,e k >e k k=1

Thutit Loan chi a day &talc goi la qua trinh Hate chuA' hoa Gram - Schmidt co so {a„ a„} 1-56 thky khong gian sin bai {e l , , ek} trimg vat khong gian sinh bai vat (a l , ak} vi moi k = 1, , n

Nigu {et, ai=1,2 m) la co so true chuL x c V thi x

vdi = (x, N6u co y =E yi e i thi <x,y > = Ex i yi 1=1

Gia s& F la khfing gian vac to eim khong gian yea t Oclit E We to a E E goi lk true giao vdi F neru <a, p> = vt moi 13 E F Hai khong gian F, va F2 ena E goi la trip gia netu moi vec to cua F, trip giao vdi F1 WO khong gian F ei E, t*p Fi = E I a _L khong gian cam E v n6u dim E = n, dim F = m , thi dim F' = n - m Khi (F')' = va E = F F1

4 TV ding cau trip giao va tti ding eau dal xiing a) Dinh nghia 1: Anh xa tuyeM tinh f: E —> E' 1 the khong gian NT& to ()alit dine goi la inh xa tuy6n tinh trg giao n6u no bao ton tich vo bleing, nghia la vdi moi x, y e E, c6 <f(x), f(y)> = <x, y>

b) Tinh chat cna tg thing au trite giao

+) Ty &Ong eau f: E -, E la true giao va chi no bign 1Cit ed ad true chud'n ed so true chudn

+) f I 'a td &Ong eau true giao va chi ma tr8n A cim f rong ed sa true chudn la met ma trail true giao, nghia la =

+) f la td citing eau true giao thi moi gia tri rieng mid f ddu dng hoac -1

+) Ngu f la to citing eau true giao, va W la met kheng gian on f bad Morn, thi Wi cling la khong gian f &Kt bign

e) Dinh nghia

Tv King cau f: E ) E cna kh8ng gian yea td dclit E dude Ri la d61 xfing (hay to lien hop) ndu voi moi vac td x, y e E co 1(x), y> = <x, f(y)›

d) Tinh eh/it dm ttl ddng edit ded 'ding

Tti (long cdu f: E E cim kheing gian vac td debt la ddi va chi ma tran f met cd se) true chudn toy a met ma trdn itol xding

+) bigu f la tu (long cdu clod zing cim kitting gian vac td gait a chigu E va IF la met khbng gian f - bdt bign cna E, thi f I E ding la met to dOng cdu del 'ding, va tong la met khong gian vec td f - bdt bidn

+) Moi nghi8m (pink) cia da Odic ddc tiding ena ding +6,81 dal Jiang f dgu la dup

+) N6u f la to d6ng ciu del xeing cua khong gian vac to gclii luau han chiOu E thi E co nail co sa true chufin goer nhilng vec td rieng eim f

+) Cac Mating gian rieng ung via cae gia tri rieng phar, biet ena mot to d6ng c5u dal 'ding la true giao vdi

§3 KHONG GIAN VEC TO UNITA

Trong rule to xet cac kliling gian vec td tren twang se ph& C

I Dinh nghia

Cho U Ea mat khong gian vec td tren truang C Mlit dang Hecmit xac dinh during trim U la met anh xa <, >; U x U C thea man the tinh chit saw vdi moi x, y, z E U; C C to cd:

a) <Xx + rtz, y> = X <x, y> + p <z, y> b) <x, Jay + rtz> = X <x, y> + µ <z, z> a X rt la lien hdp phew clia X, u

c) <x, y> = < y,x >

d) <x, x> vdi moi x \TA <x, x> = suy x = Khdng gian vec td U tren trtiang C ming vdi mot dang Hecmit the dinh dating tren U dtidc goi la kh8ng gian vac to Unita

2 Tinh chift am khong gian vec td Unita a) Cid si U la khong gian vac to Unita n - chtiu

Cd sa {e,, e , e„} dude goi la ed sa trip chuan ngu i= j

c,,e, >=8-• =

1} i j

Ta có kgt qua: Vdi moi kheng gian vec td Unita dgu ton tai ed so true chufin

b) Ta goi chuan cua \Tee td x }cluing gian vec td Unita

114= j< x, x > Ta cd bat clang thilc Cosi - Bunhiacopski: H X, y H2.11312 moi x, y E U

d I <x, y> la modun dia s6phiic <x, y>

Chung minh: Vdi X la s6 phuc Lily 9, to eo <2}x - y, Xx-y> z Binh nghia dang flecmit, La ce:

<Xx - y, Xx - y> = X.?, <x, x> - 1<x, y> - <y, x> + <y, y>

= I/1 < x, x > -2„(x, y) < x,y > + < Y>

do: 12}I < x, x > y) -X < x, y > + < y, y >20 (1) Dat: <x, y> = I <x, y> I (cos + i sill 0),

de) arg <x, y> =

Ta lay A= t (cos q'- i sin co) vdi t c R + tuy9

Ta et; R I = t, k<x, y> = Iliac (1) co dang:

X.<x,y> = tl<x, y>I Bra clang

(2) xay vdi moi t N6u to 1Ny X = t (-cos p + isinp) raj t 2( thi I X I = t, X <x, y> = 7r < x,y > = -t 1<x, y> I ya bat dang thif (1) co clang:

t2 <x, x> + 2t Vx, y>I <y, y> (3) vdi moi t> Kat hop (2) va (3) to

t2 <x, x> + 2t 1<x, y>1 + <y, y> vdi moi t e Tit suy I <x, y> 12 < <x, x> <y, y> Ta co dieu phai chting minh

c) Voi moi x, y thuec khong gian tree to Unita U, to co: il x+ Yll = 11x11+11

3 Toan ter tuy6n firth tit lien hdp a) Khdi niem Loin to (tit &Mg eau) lien hop

Cho f la to deng cau cem kheng gian vac td Unita U Tt dOng eau g cem U dtioc goi la lien hop vdi 1, n6u yea moi x, y e U, to có <f(x), y> = <x, g(y)>

b) Dinh 5i: MOi mot tv deng cau cUa khong gian vec tc Unita co nhlt met tit deng cau lien hop

c) Tinh chat cua tit deng cau lien hop KY hieu la to tieing call lien hop dm Ta co

1°) Id* = Id

2°) (f + f* + g*

3) (kg)* n g*

d) Ti doting can f dna khong gian vec to Unita U die goi la td lien hop, nett f = f*

e) Dinh lj: Gin sii f la t-L7 d6ng cdu cua khong gian vec to

Unita U Khi to có f = f, + i fa, ado f, va fa la cac tg d6ng cdu WI lien hop, &too goi tticing nag la (than thnc va ph'Un a() cda td d6ng cdu I

Chiing minh:

Goi PIA td d6ng cdu lien hop end f

f f •

Wit f

2 =-i(f-f*)

Khi = , = fa va f = f, + if,

g) Dinh Gia sd A la ma tran ciaa tti d6ng cdu f e End(V)

trong mot cd sa true chud'n cda V, a do V la khong gian vec to Unita Khi f la tti (long cdu tti lien hop va chi A t =:(c

Chti 9: Ma tran phirc A co tinh chdt A t =A throe goi la ma trdn Hecmit hay ma trail tn lion hop

h) Dinh lj: Cdc gia riling cita tst d6ng cdu W lien hop

dell tilde

Chung minh: GM sd U la khong gian vec to Unita, f e End(U) la tq doing cdu td lien hop cua U, x e U la mat vec to riling cda f nng vdi gia tri rintralt

Ta c6 f(x) = Xx,

Tit do: - X) <x, x> = Do <x, x> > nen I = X hay X thee i) Dinh 0: Cae vec td rieng Ung vat ode gia tri rieng phan

biet caa met to deng au tai lien hdp la true giao vdi

°ding minh:

Gia sit f la met deng au W lien hdp caa 'cluing gian Unita U; x, y la hai vec to rieng Ung voi hai gia tri rieng phan biet X 1, X2

Ta co f(x) = f(y) = X2y

<f(x), y> = <X ix, y> = X i<x, y> = = <x, f(y)> = <x, X2y) = X2<x, y>

Tit (A1 - A2) <x, y> = suy ra<x, y> =

B VI DV

Vi du 3.1: Gia s& E va F la hai khong gian vac to tren

trtiong seithuc K

1) Chung minh rAng n6u 9: E x E Fla anh xa song tuy6n

tfnh thi viet dude met each nhait a dang = s + a, a do s: E x Fla anh xa song tuyen tinh doi xiing, a: E x E —> F la anh xa song tuygn tinh phan del xfing

2) Ki hieu (E, It) la khong gian cac dang song tuyin tinh tren E, S (tudng ling A) la khong gian eon am 22 (E, K) g6m

Li gidi

1) Xac dinh s, a: E x E -> F hal tong thitc

s(x, y) = -1(y(x, y(y, x))

a(x, = y) - 9(Y, 9)

kieIm tra s la song tuy6n tinh d6i xUng, a la song tuy6n tinh phan d6i xiing va y = s + a De' chUng minh bik din la nhA, gia sit cp = s' + a' s' del 'clang va a' phan del xung

Da't s - s' = a' - a = y VI = s - s' nen W dea xUng, y = a a nen y phan dayi ximg Vol moi (x, y) c E x E, to co

kv(x, y) = - 41(3', = - W(x, suy w(x, y)=0= = to s = s' va a = a'

2) Ta bigt rang 2'2 (E, R) clang cku vdi khong gian cac ma trAn vuong cap n tren K, (n = dim E) Do dim 9z(E, K) = n Vi S (ttiong ung A) dAng caIu vdi khong gian cox ma tram doi xiing (phan xiing) cap n Do

n(n +1) n(n -1) dim S - , dim A =

2

x e E) f &roc g9i la kheng suy bign ngu n6 khong suy bin trai va khong suy bign phai

Oiling minh rang:

1) Ngu f kh6ng suy bign trai va F hi u han chigu thi E cling Mtn han chigu va dim ]E < dim F

2) Neu f khong suy bign phi va E huu han chieu thi F cling huu han chigu va dimF < dimE

3) Ngu f kheng suy bin va met hai khong gian ]E, có s6 chigu Min han, thi khong gian cling co s6 chi6u hilu han ve dim E = dim F

Lo gidi:

1) Ta chang minh Wang phan chetng Gia sit f kheng suy bign trai, (yo yz, yo} la met cd set cua F va dim E > n Khi E et) he doe lap tuy6n anh gem n + vac to {x„ x o , xo, xozz } Dat a o = yeti < i < n+1, < j < n

He thuan nhal:

n+1

=

do s6 8n nhigu hon s6 plutdng trinh nen co nghiem khong tam

n+1

11111611g (C„ C„+1) Khi x„ = Ec i x i la vac to khac kheng cua E ma f(xo, = vol nail j = 1, , n Vi 1.Y1, = yij la cd SO cua F, nen ax„ y) = vdi Inca y e F Digu trai vdi gia thigt f kheng suy bign trai

2) Chiang minh Wong tv nhtt phan

3) Day la Re qua true ti6p ciao hai phan tren

Vi du 3.3 Gia sit E la kh8ng gian vee to tren trthang sen thvc g la (tang song tuy6n tinh tren 1E va g (y, x) = mot

g (x, y) = Chang minh rang g hoac del Ming, hoac phan d6i

xting Ldi

Gia sii g khOng phan dal xang, co x0 e E [le g(xo, x0) # Ta hay °hung minh g doi x3ng Vin m6i x E E g (x o, xo) #

nen co a e Yb de g (x, xo) = a g (xo, xo) Khi g (x - a xo, xo) = TIT gia thi6t suy g (xo, x - axo) = 0, do g(xo , x) = g(xo, axo) = a g(xo , xo) = g(x, x„)

Bay girt lay x, y e E N6u g(x, xo) # thi ce aeRa g(x, y) = a g (x, x o)

hay g (x, y - a xo) = = g (y - a x o, x) g(y, x) = a g(xo , x) = a g(x, x o) = g(x, y)

Tug:Mg -St neu g(xo , y) z thi to cang c6 g(x, y) = g(y, x) Cual cung, gia sit rang g(x, x o) = g(xo , y) = 0,

g(x, y) = g(x, y + xo) va g(y, x) = g(y + xo, x)

Ta c6 g(xo , y + x0) = g(xo, xo) # nen g(x, y) = a g (x o, y + xo) = a g(xo , xo) = g(x, Y xo)

Suy g (x - a xo, y + x o) =

g (y + x0, x) = a g (y + xo, x0) = a g(x„, x0) = g(x,y)

NMI va.37, moi &Jiang hop to clgu co g(x, y) = g(y, x) nghia la g dOl xQng

Vi du 3.4 Cho E la khong gian vac td thuc n chieu VA co Fa Bang song toyan tinh dal xung xac dinh throng teen E Gra sit x,, x2, xk la nhung vec td mia 1E Dal aid = xi), j s k Ta goi dinh Ulric det (a ) la dinh thuc Cram (Ma cac vac td x,, xk va ki hiOu la Gr xk)

Chang minh rang Gr (x i, xk) va Gr (x , xk) = va chi xl , xk phu tha0c tuyeln tinh

Lidi brick

Ta chgng to rang ngu xi , xk phu thuoc tuyan tinh thi Gr(x l , xk) = 0, ngu , xk clOc lap tuygn tinh thi Gr(x l , xk) >

Gia su x,, xk phu thul)c tuy6n tinh, th6 thi co vac td (2 < r k) bigu thi tuygn tinh qua x,, xj+ : x = a„ x i + +

cc„,

Khi aji = u(x r, = a, a lj + a2 aji + + nghia la thing thir r am ma trap (ad k tau thi tuythn tinh qua r-1 dOng dau Ta suy Gr (x l , xk) = det (ad k =

during nen theo Binh lY Sylvester, to co det (a, i)k > nghia la Gr{x l , >

Vi du 3.5: Cho A la ma tran yang del xiing dip n tren R (

xi

Xet khong gian R" cac vac to cot X = , x i e It; u

n

la phep bign del tuygn t;nh co ma tran co sa' to nhien la A Chung minh rang:

1) Neu X, Y la nhang vac to rieng cern u iing voi nhitng gia tri rieng khae nhau, thi X, Y true giao theo nghia V X =

2) Moi nghiem day trung cim A deli la s6 that

3) Nefu A xac climb duong (vac Binh am), thi mot nghigm da'e trung eim A dgu dutong (tttemg Ung: dgu am)

Lai gicii:

1) Gia sit X, Y la hat vec to rieng cua u Ung voi hai gia tri rieng X, n; n Ta co AX = XX, AY = HY Tii de

Yt.AX = VAX = X.Y.X; Xt.AY = Xt.p.Y = g.Xt.Y; NhUng Yt.A.X = (r.A.Y) t At = A, nen X Yt X = (µX t Y)` =N Y` X

to (X - YtX = YL X =

2) Gia sit X + in la nghtem dac trung cim A The thi ten tai vee td cot (phitc) X + iY X Y la nhUng vac to cot that

A (X + Y) = (7 + ip) (X + iY) day i la don vi ao

So sanh cac phan Uwe va pfin no a ca h v6, ta ducre

AX = XX - (1)

AY = XY + p X (2)

Tit (1) va (2) ta co:

VA X = Y`X - p Y'.Y X`AY=i.- p Nhung Yt A X= (XL A Y)L nen ta có:

AY`.X-pYt.Y=XYLX+gr.X

p (XL X + Y) = VI X, Y khong dang that bang khong, nen r X + Y.Y>Oviy*yn=0,docloA,+ip=X e R

3) Gia sit A la nghiem (lac trung cith A, theo tren X e Khi co vec td cot X = de' AX = XX, suy Xt.A.X = Vi X = nen Xt.X > TU do, ndu A the dinh (lacing thi X`AX > nen X > Ndu A the dinh am thi XtAX < nen X <

Vi du 3.6 Cho A la ma tran thong tithe ddi ximg cap u e End (R") có ma trgn A co sa tti ten Chung minh rang ndu X lit vec td khac khong tha R", thi ton tai vac to thong Y cua u thuOc khong gian sinh bai X, AX, A 2X, A" - ')C

LaPi

h6 nhat de AkX bleu thi tuygn tinh qua X, AX, Ak 'X Khi

0 tan tai the s6 thuc b k., , bo dg

AkX + b k, Akd + + b iAX +13 X = O Gia sit (x - p ) (x - i.tk) la &tan tich cim da tilde

xk + + +bo the nhan tit tuy5n tinh, vdi g , gk E C Khi

0 = (Ak + bk., Ak- ' + + boI) X = (A - (A - ukI) X hay (A - = vol Y = (A -1.1 2I) (A - tikI)X x0

Nhtt vay AY = suy g, la nghigm dac trung cim A 'heo vi du 3.5, to co p, e R 'Nang to go , , gk dgu thuc Do lh vec to rieng cUa u va Y thnOc khong gian sinh bat X, AX,

Ak- `X

Vi du 3.7: Gia su ,\ is ma trail vuong thuc dal xiing cal-) n, u End (Y") có ma tran A co so to nhign Chung mink ang tan tai ma trap trot giao P cho ma tran Pt A P = :1 ma tran cheo Cite phan to tren during cheo chinh cim hinh la cac nghigm ciao trung cim A (kg ca bOi) Tii suy ang A xac dinh throng (Wong Ung xac dinh am) va chi

nghiem dac trung cim A dgu dvong (Wang ung dgu am) Chu y: del chigu kect qua vi du 3.5)

Gia X,, s < n la mot hg trot giao gam nhung vec td

Ta cheing to rang có vec td rieng X„, trip giao vet cac i = 1, 2, s That tray, gia sat X x la mOt vec td true giao Xi (i = 1, s) Vei mai i= 1, s va r > 0, ta co:

X` A' X, = = (WIC) =

(X, 11 gia tri rieng eng vdi vec to rieng X 1) to (Alot X, = 0, s AIX true giao yea Xi Theo kat qua vi du 3.6, co vec rieng Xs,1 cern thueic khong gian sinh bei X, AX, A"- Nhung cac tree to AIX true giao \el cac X, (i = 1, s) vi tray

clang true giao yea Nhtt tray co X„ X„ trip g

gem toan vec to tieing ciaa u

Dat - X• thi he Y1 , Ya la he cac vec to rie

.X,

sao choY,` Y, = 8,, ki hiOu Kronecker) Vi tray ngu dal IA ma tran ma cam cot la Y„ thi P IA ma trait trvc giao

Mat khac, AX ; = 3,X ; nen AY, = X ; lc ti/ A Y; V, = X, va vie j thi Y,' A Y, = X; Yit Y; = O

Do PAP=A A la ma tren chdo yea cac Olen eh& a ., X Vi P = nen A thing dang vdi A, suy

I A - XIn I = - I = (X, - X) •-• -

Pu suy 3.„ chinh la tat ca Cite nghiem (lac tn./ cUa A, Ire ca bOi ,

Vi du 3.8: Cho A, B la hai ma tran that d6i xiing cap n, hen nem B the dinh during Chung mink rang ten 41 ma teen khong suy Bien C Ct A C=A va C' B C= In, de A la ma trait chef), va In la ma trail den vi cap n

Lai gicii:

Do B xac (huh &king nen t6n tai ma lit' khong suy bin T de 'FL B T = In Ma tren Tt A T doi thing nen theo kgt qua vi du 3.7, c6 ma trail tnic giao P de 13` (Tt A T) P = A la ma trail cheo Dat C = T.P, to chide Ct AC = A va Ct BC = In

Vi du 3.9: Vei mbi ma tran (pik) A, ki hieu A* la ma Oran lien help veil ma tran chuyen vi cim A, nghia la A* = A Ma trait vuong A throe goi la ma tran Unita n'etu A* A = In Cho A FA ma tract Unita, eheing minh:

a) Neu k la nghiem (Mc cern A thi I XI =

b) Ngu A la nghiem dac eUa A thi — cling la nghiem dac thing cim A

c) Ngu A ante vi co cap le thi A co ft nha't met nghiem dac trung bang ±

Lel gidi:

a) GM sei X la nghiem dac trting &la A, c6 ma tran eet X # AX = XX, tit de (AX)* = X*.A* = 7r X* Nhan vg vat v6 hai clang thtic tren, to co X*A* A X = X* X Vi X x 0, nen X* X x 0, vi A* A = In nen:

b) Gia sit X la mat nghiem dac cga A det (A-X I) = det A* det (A - XI) = det (A*A - ) = det (I-A =

det (-1 - A*) = det (-1 I - A*) =

det (A - —1 I) = nhu vay X — la nghiem dac thing dm ma tran A c) N6u A la ma trail thge co cap le, thi det (A - XI) la d thitc bac le voi he s6 thgc, vi vAy co it nhat, met nghiem thg Theo phan a), nghiem time phai bAng ±

Vi du 3.10 Cho E IA kitting gian vec to n chieu tren tradn so' thuc Tk va f IA met clang toan phudng iron E We to x e dude goi la clang huong nett f(x) = Chung minh rang n6u f di data, nghia la t6n tai x,, x, cho f(x,) > va f(x 2) < thi tron E t6n tai co sa g6m nhung vec to dang huOng

Ldi gidi:

Gia sii (e), < i < n la co so chudn tAc cua E, nghia la ed s ma f c6 dang chudn tac Hun nua, gia sit AO = vdi i 1, 2, , r; f(ei) = -1 vdi j = r + 1, r + s va f(e,) = voi t = r + + 1, n Do f da't ddu nensz 1, r21

Nhu vOy yen x = LU i e j E lE, taco:

too ai2 + arz _ ctsr_fri _ _ coo_s

Ta say dung cd se {v,} (i = 1, n) g6m nhung vec td clan Intang cim E nhu sau:

De thdy he {vi), < i < n gem nhiing vec to clang cua Ta chfing minh he NJ la cd se cita E Xet EX,v, = 0, to minh = vdi moi i

rrs

Ta co ZA.,(e, ei.+1 ) + e j ), + Ek te t =

j=r+1 t=r+s+1

(

r+s r

) + EX , '.1 j ei + EX , , e j=r+1 r 1=2 =1

yi X r+1 er+1

r+s

E A i e i Excet = j=r+1 t>r+s

Do he ted, <i < n doe lap tuydn tinh nen = vdi moi i =

Nhu vay he (v,) la cd s i eida E, gem nhung vec td clang tudng

Vi du 3.11 Ta goi non ding cna clang Loan phudng f

a tap K tdt ca cac vac td dang hUdng Chung minh rang non Vang hudng cita clang than phudng f tren khong gian n chieu E A khong gian cud E va chi f khong ddi clgu, nghia A f(x) vdi mm x e E hoac f(x) < vdi moi x e E

1,64 gidi:

Dieu kien Gia sit 1K la khong gian cim E Neu f

thi then vi did 3.10, E có met cd sa g6m nhfing vac td dang hang NMI vay K = E, nghia la moi vec td cim E den

dang hudng Digu mau thuan vat anti Witi clAu cita E Do vay khong dau vi digu kiln can chn}c cluing minh

Dieu kik gia si f kitting de) dAu, eking han f(x) voi moi x e E Xet co sä {n) chudn tic cart f, to ed

f(

Ici,e,)= + i=1

2

Nlw tray, non clang hudng la tap cac

vec to x = Ewe, me a, = = a, = O Do tap K cac vec td Bang hudng lk khong gian dim E sinh bai {a„„

Vi du 3.12 Giit sit f tia g la hai dang Wan pinking tren kitting gian tree W V, có bjeu auk toa da mOt cd {n} = 1, 2, n) nao In WS la: f = Za iix,xi , g = Ebo i x i

Ta goi hop cua hai dang toan Oath:Mg f, g la clang toan phudng (h g) = b ij x i x i

Chung minh rang:

a) Neu cac dang f, g khong am, thi dang (f, g) cling khong am b) Neu cac dang f, g xac dinh during, thi dang (f, g) cling xac dinh chidng

Ld gidi:

a) Hign nhien la (f, + f2, g) = (f„ g) + (f2, g)

Dua cac dang Wan phuang f, g ye dang clutAn tic:

EyF , g = , a do 37 , z3 la cite ham tuyen tinh dei vdi i=1

• Bien x Do la bieu tbiic toa do, cim NT* chuyen cd so Ta co

= X

i g? • E lf = EZ(37?' 4) •

g)

i=lj=1

Xet mot s6 hang (y2, z2 y= Ea k xk , z = Ebk xk

k=1 k=1

• y2 = Ea k a / xk xis z2 Ebk b i xk xi , va (2 z 2) = k,1=1

2

ai b k b / x k x / = Ea kb kx k I 20 Nhu vay khang dinh

i=1 k =1

dude chUng minh

b) Gia sit f > va g > Ta dua g v2 clang chua'n tAc

= Ey , yi = Eq 33 xi (1 = 1, 2, , n), det (n o) # KM 2

i =1 i=1

g)=E(f,

i=1

\

Nhung = Ecniq ik x 3x k , nen (f,y 32 )= Ea ik kq u x i lkk xk ),

do f = Ia ik xix k Do f xac dinh dtidng nen (f, n2) Neu X j,k=1

(xi) 0, thi co; a x; # 0, tea do có i qi; # 0 n8n ci n xy x 0, va o f xac dinh dudng nen (f, 37, 2) > Do \Tay (f, g) >

Vi du 3.13: Chung to rang kh6ng gian vec to Odclit chieu, c6 the tim dudc ho n vec to don vi u l , u2, cho di vai tat ca cap s6nguyen phan biet i, j, yen to - 11j cling la of

1 "

to don vi Ta {tat x= n +1 In, Hay tim goe giva vec to x -

1

va x - U

Ldi

Gth six {e1} (i = I, n) la mot co sa bye chuSn °Cm khan gian vec to dclit E Ta xay dung hen vec to {u„ u„) nhu sat he (ti t , u2) thuOc khong gian sinh bai {e„ e a} va {0, u,, u2} lar tam gine deu, nghia la ria l 11=11112 E1=1 va 14.u = Ta b sung tr, nhu sau:

u3 = + ).2u2 + X3e3, u3 u = 113 1.12 = —1 , u32 = I X? Xi

+

1 Tit co —2 = X, +=

2 ' 2 A2 A1 X2 - —3 IF

Va vi Al + x.22 + + Xi X2 = X3 = + —2

Bulk xay dung vec to UM n to (n-1) vec to ducte xi{ Wong ty Gia su to co (u1, u2,—,u.,-,) la he n-1 vec to nam klning gian sinh bai {e„ e„ { va th6a man (lieu kien bai than Ta tim vec to

U.,= X, + + .;

Sir dung dieu kiOn un u; =

— vdt j = 1, n-1 va u,12 = 1, to

tim duac = 12 = = An_ = —1 va In — n +1N hit vay to da

xdy dung dude hee {m, u„} cac vec to don vi kitting gian dent E" they man (lieu kiOn: goo gliva hai vec to bet kyt cim he bting 60°, tit I tt, -uli = (i j)

1

Bat x -n

+1 + u2 + + ti n ), goi 01a gee gala x-u, va x-u Ta ce = (u - 110 = [(x-u 1) - (x-u 2)] = (x-u )2 + (x-u2)2

-2(x-u 2) (x-u 2);

va (x-1.1 )2 = (x-u 2)• =

Tit cost/ =

1

tux + -

M +1r - 2M +1)

Vi du 3.14: Gia sii V la met kh8ng gian vec to Unita, B: VxV->C IA dang tuyen tinh rued tren V, nghia la B tuyen tinh dot vdi Men thil nhet, Ong tinh dei vdi Mon th>Y hai va B (x, ky) = (x, y) vdi moi x, y e V va e C Khi tan tai duy nhet met

phop del tuy6n tinh A cim V cho B(x, y) = <x, Ay> yin moi x, ye V

gidi:

Truisc ta xet bit sau:

BS de: Gia sit f la dang tuygn tinh tren khong gian vec to

Unita V, tan tai duy nhet met phdn tit h E V, cho vdi

moi x e V, ta co f(x) = < x, h>

Ghiing minh de: Net cd sa true chua'n te,, e 2, , ea} V ret phdn tit h e V, h = Eh k e k , a hk = fk) Gia sii

k=1

x= EXic ek e V, Ta co f(x)= Exk k = xk hk =

k =1 k=1 k=1

n n

< Ix kek , Ehkek > = < x,h >

Ta chfing minh phan tit h la nhat Gia sii co 11 , h2 li hai vac [Al cho f(x) = <x, h,> = <x, h 2> vOi moi x

c V Tit d

<x, h, - h2> = vdi moi x e V Dac biet, lay x = h, - h 2, ta ci

11h, - h211 = h2 = h2

Bay gia ta chung minh bli Loan: Gia sit y la phan 'eft c11 din} Mt Icy caa V Khi B(x, y) la clang tuyen tinh clSi vat than x Theo be de' co phan to h xac Binh nhat (phu thuOc y) sa( cho B (x, y) = <x, h>

Xet anh xa A: V —> V y —> A(y) = h

Ta chitng 6> A la anh xa tuyen tinh Ta co B (x, y, + y2) = B (x, y1) + B (x, y2) <x, A(Y(+Y2)> = <x, A(y,)> + <x, A(Y2)>

<x, A(371+Y2) - A(Y1) - A(y2)> = vdi moi x c V Do v'OY A(Y)+372) = A(37 1) + A(Y2)-

Tudng Lu B(x, 1y) = <x, A(Xy)> = y) = 7<x, A(y)> = <x, A(y)> <x, A(1y) - X A(y)> = vdi moi x e V

Tii suy A(1y) = X A(y)

Ta chiing minh tinh nhat maa A

GM sii A, va A2 la hai phep Man de4 tuyen tinh thea man B(x,y) =<x, A1y> = <x, A2y> vdi moi x, y e V Suy <x, (A, -A2) y> = vat moi x c V, ye V, to (A l -A2) y =0 vdi moi y E V, hay A l A2

Vi du 3.15: GM sit A NM B la hai ma trail thuc, xiing xac Binh during cap n

Chung minh rang det (A + B)?, det A + det B Lai gicii:

ton tai ma trail truc giao C cho C B C c6 clang cheo ma cac phan tii tren duang cheo chinh la cac gia tri cna B

Vi vay det (I + B) = det C - ' (I + B) C= det (I + C -1 B C) = (1 + (1 + a + X, X„ = det I + det B Ret truong hop t6ng gnat, vi ma tran A xac:dinh duong, nen c6 ma tran D cho A = D2, a do D the dinhducmg That vay, có ma trail D, de2 D,' A D, = A co clang cheo, D, la ma tit) true giao (vi du 3.7) A la ma tran cheo ma cac pith) tit tit)

cluong cheo chfnh la cac gia tri rieng cim A: p„ n„ > Dat P la ma tran cheo mA cac phan to trot) during cheo chfnh la P2 = A va A = D I A D l d = D,P1 D1 = D I PD1

D,PD -1 = D 2, vei D = D, P D, - ' = D,PD,t, D la ma tran d61 xang

Td de A + B = D2 + B D (I + D - ' B D- ') D

Vi vay det (A + B) = (det D) dot (I + B D-1 )

Do D - ' B del sung, the dinh dming nen set dung k6t qua

trot), to of):

Det (A + B) Y (det D) (1 + det D - ' B D -') = = det D1 + det B = det A + det B

NMI vay, bai town dude chting minh

Vi du 3.16 Ta ky hieu M la khang gian cac ma Iran yang cap hai tren truong C

a) Gia sit f: M C

X f(X) = det X

j al as

= a, X, + a2 X2 + a3 X3 + X4 a s •

a) Gial sit X =

4

b) Ta nhac lai rang f(X Y) = f(X) f(Y)

Gia su p la clang toan ph/dna tny y khac kh8ng tren 1M thOa man di6u ki6n p (X Y) = cp (X) p (Y) Hay chfing t6 rang

= va n6u X khong suy bieh, thi p (X) x Hay tinh gin tri cim p tren cac ma tran lily tinh va cac ma Han suy bi6n not Chung

c) Chung to rang p = f Lai

do {XI , X4} la cd so to nhien dm M

f(X) = a,a, - a2as Day la mot dang than phudng tren M Trong cd so tg nhien cem M, f có ma tran

0 0

0 0

2 - 0

2

0 0

1

det A = — 0, vay clang 'Loan phtidng f co hang 24

b) Vi p khac khong nen co X e M de? p(X) x Theo gia thiet to co p(X) = p(X I) = y(X) 9(I) Do 9(X) x nen p (I) = Neu det X x 0, thi co = I Vi \Tay

9(X Xd ) = 9(X) 9(X -1) = 9(I) = 1, cho nen p(X) + A =

Gia A E M, A x va Ala ma tran lily Kith Nhu vay A = 0, vay W (A') = MAW = tii (A) =

Gia s5 X e M, det X = th6 thi hang cem X bang hoac sang Neu hang X = thi X = va vi bay p(X) = N6u hang C = 1, thi eó ma trail khong suy Bien P, Q cle X = P A Q, a

'0 1`

= (xem bat tap 2.11) ,0

Tit de cp (X) = w(13) W(A) (p(Q) = 0, vi cp(A) = 0, A ley linh

c) Vol X Lily Y thuee M, ta ehiing minh 9(X) = f(X) That bay set ma trail X + AI e M, ta co

f(X + XI) = f (X) + k F(X, I) + )L2

9 (X + XI) = (X) + 27r4 (X, 1) + x2

6 F, (F la the dang cue wong fing cua f va cp Khi X + i, I suy bi6n nghla la k la gia tri Hong cim X, thi f(X + AI) = (p(X+Xl) = Til suy hai da thfic tren bang vat m9i X Dab biet, cho 7%= ta co tp(X) = f(X) Vi X tay ST nen f=T

Vi du 3.17 Khong tinh dinh thilc, hay giai thfch tai ma tran A sau day kha nghich:

Zd gidi:

Ta co A = A, - a A l = 1 12

3

3 -2

4 '

3 -2

Ma trail A I la ma trAn thvc dai xting cap 4, nen tat ca gin tri rieng dm A, deo thvc, (xem vi du 3.5), nghia IA da thfic POO = det (A, - ce, the nghiOm deu thuc, ter P (i)* 0, nghia lA det Ax

Vi du 3.18

a) cis sit {e h e2, e3} IA mot co sa cim R3 Cho clang Loan phttong Q tren Q (x) = x 12 + x22 + x32 - x1 x2 - x2x3; x = (x 1, x2, x2) la toa dO cua x co sa {e , e2 , e3}

Chang to (tang town phvong Q xac dinh mot can trtic delft fret) Re Hay tim moot cc( sa trvc chua'n cim R dot voi Q

b) Gia sit f e End (R2), co ma trap A co sa 02, e3}: '3 -1 '

A= -1

Hay kie7m tra rang A la den xiing doi yea c‘u trot delft phan a) (nghia la f la tv dang ca2u del xi:Mg)

Ldi gidi:

Taco H, = 1, H

1 det

2

H = — > Vi H la ma trail xac dinh during nen Q la dang town phudng xac dinh during Mut vay Q xac dinh mot Lich vo hudng tren Ra (do chinh la clang cip tudng ting vdi Q)

Bay gia to tim mot cd so tn.tc chuAn del vo Q Neu x = (x,, 12, x3) thi Q(x)=(x, x2 )2 + (1 x3 - X ) +

2

1

xl = '

1-2 22

1

x2 = x3

1 x3 = )r2

Hat

X =

x

+ x3

2

+ —x KM

e el

ez = e3 e3

2 —e

' + ne + '

thi eel sd fe'„ e' 2, thdc cua clang town phdong Q eó Bang Q x,, x )- x 12 + x'; +

Nhtt vay {e' , 8'3} la co so true ehuan

b) Goi n la dang eve cua Q, co sa e2, e2),

co ma tran H

ife" chdng minh f la tv d6ng ciiu del xdng, ta phai chtIng minh n(f(x), y) = n(x, f(y)) vdi moi x, y E R Goi X, Y la ma tran

cot vac toa de) cua x, y, ta phai el-Ong minh: (AX)'1-1.Y=Xt.H.A.Y hayrAt H.Y=Xt.H.A.Y

That Vey At H = ( -1 ' -1

3 i 2 2 -2 :`

= -2 -2 = H A -2

a) Chung to rang ngu H , la xac dinh (Wong, thi ten tai met cd so de co so ca H, ve H dgu co clang chinh tilt

b) Neu vi du chUng to rang kheng phai bao gio cling clang thai dua &toe hai dang toan phudng da cho vg clang chinh tat

c) Cho hai dang than phudng tren R2 cd so chinh tat có clang:

\

H i (X)= Xj.2 +54 +144 -F2X I X9+2XIX3+10X2X3 H2(X)= 4x 22 +104 + 6x,x +14x2 x3

Hay tim mot co sa trong ea H, va H dgu co dang chinh tac

L.& gidi

a) Gia sii dang cue eda dang toan phuong H Vi H, )(lc dinh dudng, nen W siuh mOt tich vo huang trail V: <x, y> = y) Ta bigt rang, khong gian vectd dclit V co co so trip chan clg clang toan phudng H2 co dang chinh

the H2(X)=- Trong có so true chua'n do, H (x) =O 1(x, x) = n

<x, x> = Exr Nhti \Tay, cd so {v , Vi co ca hai

i=1

clang toan phudng da cho dgu co clang chinh tilt

H2(x, y) = xy GM a co mot co so te l , dg' cd sã H, va

H2 dgu có ding chinh tac

GM ad: = e ll e l +C 21 e

62 rc +C22 62 Goi (x', y') lk toa dO co sä

IX c'Cii11+ Cl2

) Khi

(y= C21x'+ C22 r'

Ta nhan thdy H, c6 dang chinh the ca sa {6,, E 2} thi C„ C 12 = 0; H2 c6 clang chinh tac trong co so fe„ E 2) thi CI, C22 +

Cu Ti12

C.2 C = Tit suy ma tran C = suy hign

C21 C 22

Didu mall than vi C la ma trail chuygn cd so

c) Ta nhan thdy H la clang tan phudng xac Binh throng H, (x) = (x + x2 + x%)2 + (x2 + x2)2 + 9x32

1

Y3 = 3x

{YI = X I + x + x

Pat bign mai y2 = 2x + 2x3 hay

Trong ho toa do, do, I-I2 co bia thfic: H2 = 1722 + 213373

xl = Y1 Y2

1

x2 = Y3

(2)

(3)

X3 = 3y3 thi voi die toa dO (y„ y 2, , y3), H, co dang chinh tac:

2

'o 1' Ma tran maa H2 CO dang A =

,1 0

(4)

Bay gia ta (Ilia ma tran (A) vg dang the() bai ma tran tryc ;iao B&ng phnang pho.p quen thu0c, ta tim &No

1

0 -1 0'

n&

T= 0 116 A T = (5)

1

0

1

z yl =- r z i +

V2 a/2

Dung phep toa (.10 Y2 = (6)

1

22 Y3="' +

Trong ca hai dang town Oaring se c6 clang chinh tAc H, (z

H2 (Z

Tii (1) va (6) ta co

, z2 z3)

, z2, 23)

x =

x

xa —

9

= + +

— - +22

1

,z,

1/2 ' 3.h z1 + +22 + + z

1

— z3 lz + 2z3

(7) (8) (9) z2 1

3,5 3-,,& z2

HS {el, e2, C31 7a co sa K3, H, va H2 ce dang

1 1 1 el 3"2- e2

+

ea $ ,a i)

3-,N -2 + 3I_ -3 1

= +,7,

Chu if: Da dila dang thai hai clang town phydng H„ H2 d6 H, xac dinh diking v6 dang chinh talc, ta lam cac btafic eau:

Bieck 1: Dila dang toan phtidng H, v6 dang cheo, nghla IA tim mat co sa c , c„} true chuan dai vdi dang cue 0, cUa H„

aide 2: Vik biau iliac maaja cd sa fe l , c2 ,

Bade 3: Dua clang Loan plincing H2 vi dang cheo bang ma tran true giao, nghla la tim cd sa {a„ a„} true chuan doh vdi de' H2 = co dang cheo Khi co sq {a„ um} ea hai dang H i , H2 den có dang cheo

Vi du 3.20: Cho dang town phtiong f tren khOng gian thuc n chit E, có cac chi s6quan tinh throng va am Ian Itto la p va q Gia sa E l la khong gian cern E ma dim E, = n - thi chi see quan tinh cua dang than phydng f han ch6tren F 1a bao nhieu? Lo gidi:

Trude 116t ta nhan thay, neu F c E lh khong gian cila ]E ma f I F xac dinh diving (tticing tang xac dinh am) thi dim F < p, (tticing tang dim F < q) That vay, gia s f/F xac dinh &icing va dim F > p Xet co sa (e r , e„ crib e„+„fri, • , e„I ma

f(e,) = -1, p+1 < j p+q f(e/) = 1> p+q

Bat F, la khong gian sinh bai cle vac to {e„,, epop • , e n), thi f/ F, o Nhung dim F n F, = dim F + dim Ft - dim (F + F1 ) nen dim Fri F l >p+ (n-p) - n = 0, matt thuan vdi

gia thiat f/F > 0, f/F, o

Gia se dim E l = n - va p', q' la cac chi s6 dicing quart tinh va chi s6 am quail tinh maa f han the troll E l Ki hieu F2 la khOng gian sinh bai le l , , e p) tren d6 f xac Binh ducing

Ta co dim F2 (1 E, z p - 1, nhung dim (F n El) o Tit de p - p', nhu vay = p-1 hoac p'=q

Tuong to q' = q-1 hoac q' = q

Ta nhan thgy ea ban trudng hop

P‘= P IP P = I Pr= P -1 q -1 k r = '11 kir= q -1 den co the xay Ta xet cac vi du sau:

a) n'au p + q < n va E, la kh8ng gian vac to sinh bai le„ e„41, thi p' = p, q = q

b) neu q > 1, thi vdi E l sinh bai he fe l , ep, e p+ 2, to co le1 = P

q = q-1

d) n6u co p z 1, q 1, chnn E, sinh bai le, + e2 , e,, en+2, en} thi p' = p-1, q' = q-1

C - BAI TAP

3.1 Gia sii H la clang than phuong tren khong gian vec tc R3 Dang chuan titc cua dang toan phtiong H la dang chinh rac cac hG s6 deu bang ±1 hoac bang Hay tim dang chuan tac cua H cac trtiang hop sau:

a) H (x„ x2, x3) = -24 -Fx g +2x,x +4x1 x3 +2x x3 b) H (x , x2, x3) = x,x2 + x,x3 + x2x3

c) H (x„ x2 , x3) = 44+,, 3+4-4x1 x2 +4„„( -3x

3.2 Xac dinh X de dang toan phtiong sau xec Binh cling a) + 44 +4 + 2Xx, x +10x,,x + 6x.,x3

b) 24 +24 +4 +2Xxi x, +6x x3 +2x,x c) +24 +21/4,42 -2x,x +6x x -2x2x

3.3 Dua dang than phtiong sau v6 dang chinh tAc:

11

a) Q(x l , x„)= Dx i t + Ex i x j , 1<i<j<n i=1 i<j

Q(x , x„)= Exi x, <n

du 3.10 va 3.11) Chung minh rang khong gian t6i did &la E nam non clang hueing K có chigu la n - max (p,q) Ngu f khOng suy bign thi s6 ehigu la (p, q)

3.5 Cho E la khong gian vac to (Debt, fo„ e„} la co so trUc chuan cua e Vol {a,, a„} la he n vac to tny y ena E, goi A la dinh fink dm he tan ., an} dgi yea co sa trite chuan fe n •••, nghia la A = det (N) a do a i = Za ijei (j = 1, n) K1 hieu Grier, an} la dinh thile Gram cna {a„ aj Hay chUng minh Gr {ct„ ct„} = A2

3.6 Gia A1 , , An la the doan thong trgn during thong thvc h (khOng gian vee to (kilt mc5t ehigu R) Ki hieu a y la dal dont] A, 11 A; j s n Pat A = (a u) e Mat (n, R) Hay chUng minh det A a

3.7 Gia sit K,, K„ la nhring hinh troll mat phang KI hieu b„ la then Lich cue giao 1{, fl j Got ma trim B = e Mat (n, R.) Hay chiing minh det 0

3.8 Chiang minh rang moi ma tran Unita dgu eheo hoa duoc, va moi gia tri rieng ena no co modun bang

3.9 Cho ma trait Heemit A, nghia la A t = A, Chung minh rAng co ma trail Unita C d@ C' A la ma tran cheo

3.10 Chung minh rang ngu A = (a„) la ma tren deri xthig, xae dinh (Wong cap n thi (tr A) (tr A -') n2

phudng e6 ma trAn A - Xth the dinh &tong vdi moi Xo < a va xac dinh am vdi moi Xo > b

3.12 Gia sit A e Mat(n, R) la ma Win thud del xfing va u, la vec to rieng (cot) eim A vdi gia tri rieng X, DM B = A - a I UI Ut

a) Chung 6) rang c6 the chon dttOc U, de B, U, =

b) Xet met cd sa true chua'n Up U9, Un giSm cac vec tc rieng dm A, ting veil the gia tri rieng ChUng td rang

A=

3.13 Cho E la kh6ng gian vec td tren truang s6 thtic R, f lII Bang song tuy6n tinh del )(ling tren E Gia sit ton tai vec tc khac khong a e E cho f (a, = va to clang that f(x, = ( suy x ding phudng vdi a hoac f(x, x) <

Chung minh rang n6u x * va n6u co y de? f(x,y) = Nq' f(y, y) > thi f(x, x) <

3.14 Cho ma tran A E Mat (n, R), A phan dol xOng Chiim

minh rAng del A a

3.15 Cho ma trAn A e Mat (n, C ), A, la ma trAn lien hqi cim A Chung minh rAng det (AA + I) a

3.16 Cho U la khong gian vec td Unita, f e End (U) K lieu Pr la to &Ong cgu lien hop vdi f Hay chiing mink

c) (Xt)*= ti f* yak mai A e C d) (CI)* = f

e) (or = f*og*

3.17 Cho U la khang gian vec td Unita va f e End (U) Chung minh rang f la to citing ca.0 tg lien hop va ehi <1(x), x> thgc \Tdi moi x c U

3.18 Chang to rang gin tri rieng haft kg X aim (long tau to lien hop f caa khang gian vec td Unita U bang <f(x), x>, a x la vec to nal] oda U thOa man Pc11 =

3.19 Cho U la khang gian vac to Unita, f la to citing ca:u ty lien hop tren U {X„ la sac gift tri rieng cim f, {e„ la cc; sa trge chua'n go'm nhung vec td rieng cim f cho f(e 1) = ki e i ; i = 1, n Xet Pk; U U, Pk(x) = <X, ek> ek

Hay chatng minh: a) Pk =1)k

b) PioPk = nen k # j

c)f- XkPk

k=1

d) Gia sii Q(x) = Eckx, la da thac tag § vdi he 86 philc

k=0

Pat Q(f) = riC i r e End (U); f' = Id ]=4

Chiang minh rang Q(f) = Q (4)Pk

3.20 Hay cluing minh dinh lY Hamilton - Kelli cho &Jiang hop ma Whit A la Hecmit: Cho A la ma tran Hecmit va P(k) = det (A - XI) = (-X)" + C (4.)"- + + C k (-X)" + + C la da thuc dac thing cim ma tran A Hay chitng minh P(A) = (-A)"

C (-A)°4 + + C„ I n = O

3.21 Gia sit A = (ay) e Mat (n, R), a do n > 3, va t&t ca can ph&n tit an # Chung minh Ang oh ma tran Be Mat (n, > Nth mci i, j ma C = (a bd la ma tran suy bign

3.22 Gia sit E2 la khong gian vec to dent hai chigu Tu citing din dot xitng f e End (E2) co sa trite chuan fe l , e2} co

1

1

2 ,

ma tran cha f co clang cheo

3.23 Hay dua ma trail dot xiing sau vg clang cheo nha ma tran trhe giao

' 0 B=

1 0

3.24 Trong khong gian vec to R3, cho hai clang town phttong:

= 2x 12 -24 - 3x32 -10x x +2x x3

9 = 2x12 +34 +2x32+2xiX3

a) Chung to o la clang tnan phuong xac Binh during

b) Bang Olen biOn deit toa thfch hip, hay dua ca hai dang toan phudng tren vg dang chinh tat

3.25 Gra sit' E la khong gian vet to aclit vdi tich hitting < , > Tv ddng eau f E End (E) duoc goi la phan doei 'ding nefu

vdi moi e, y E E, to co <x, y (y)> = - <1(x), y>

a) Chung to rang f phan asi 'ding va chi <f(x), x> = vdi moi x e E

b) Trong Wit co so true chua'n bat Icy, ma trait cait to ddng eau phan d61 ming la ma trail phan 'ding Tit suy moi to ddng ca."u phan d8I xiing caa khang gian delft s6 chigu to de).1

khang kha nghich

c) Gia f la tti ddng c'au phan ddi xang ChUng to rang Imf

va Kerf la hai khOng gian ba trip giao vdi Chung to rang hang caa f la mot s6 than

D NCTONO DAN HOC DAP SO

9 32

3.1 a) H = (3, + x2 + 2302 - [1-3-x2 +731 x ) -(a33

y i = +x +2x

fat 2 =- 2 + I 3

Y 3x3

Taco H= b) H -37F2 - c) H = _2 32 Y3

3.2

a) Khong co nao th6a man b) Kh6ng nao th6a man c) A> 13

3.3

a) Q-602 -i612)2 + 673)2 + -+ n+1 67 :32

6 2n

1 That \ray, dat y, = x, + —(x2 +

2 x„), thi

2

-3-x,x ) ;2

Dat y2 = x2 + (x3 + + xn), tu

f

= +71 1374) +

-6 Ex; +—x i x ; <1 ;>3 i<j

Gia si c1641 blidc thtt p:

ExVE Ex i x j j , i=p+1 P ± i<j

Q = 671) +

adO(p.4i<j)

p+1(_ , )2 + p+2

2p VP! 2p +

Da, t = x,,, , + (xl,„2 + + x„) ta dude tong thiic tren, p +

dng voi p+ I_ Do sau (hsj) buoc ta co ket qua Ta ce the' viet gon quy tacd6i bie'n nhu sau:

1

1 1

2 2

0 1

3

(Yu, 0 0 b) Dua ve a) nhu sau

Q= x i x +x l Ex; +x2 Ex i +Exix i ; 3<i<j

i=3 i=3 i<j

= (x, + x s + + xs) (xs + x, + + xs) -

1 t

Q =-1 + x s + 2Z x,1

4 >3

Dang (n-2) bin x3, , xi, E .1

3.5+=j

dttqc xet phein a) Nhu vay ta co

Q=(Y1)2 -6,7)2 L

2k -1 —2) (Y1J2

1(23 (k

1 s V = —kx, + x ),- Lx,

2 =3

YI

y2

/ x1

1

Y2=2 + x9)

Y3 = x3 + kx +

' Y4 = X4 + —2/ +

x i

x n

y fi =

3.5 Gr (a l , an ) = det (<a1, ak>); < j, k n

a.j =Ea-e• a k = Ea ik ei Do {ea trite chuan nen

,=1 1=1

< a ja k >=Ea,a ;,, ; Dat A= (ao)

Ta có ma train (<ci, ak>) = At A Tit Gr (a„ an) = (det A)2 =

3.6 Gia s8 doan [a, hi chita tat ca cac = 1, n) Xet

e [a, hi la khong gian cac ham lien tuc (co the trii mat s6 frau han diem) tren loan [a, N WI* [a, 131 la kheing gian vec to vo han chiau Den truang sathac K

NMI T,i e C [(Lb], (tat = cf(t)g(t)dt la tich vo httdng tren C la,b1 (khong phu thuec vao dai dien f, g x;

{.1n6uteA; la ham dac trdng caa tile la X i(t) =

Oneitto6„

Thl > va det A = Gr (R I %^) >_ o 3.7 Thong to bat 3.6

3.10 Vi ma tran A d6i xfing, xac dinh &mug nen ton tai ma trail true giao C C-1 A C co clang cheo:

'a x,, la citc gia

X2

0 tri rieng caa ma tran A

1 xt

1 12

k„

0

= A-1 C Ta có: 131 =

1

NMI vay A-' có cac gia tr rieng —, , —1 ki

n

Va trace A = Ek ; trace A-' E -

i=1

n 2

Ta có (tr A) (tr 21/2- ') = E-

LI Xi)

3.11 I-ID Ta thky rang: rigu I la nghiem dac cda ma tran A thi - X (,) la nghiem dac ciia A - X0I (xem giai bei 2.45, each 2) Nhu vay A Ia nghiem dac tning cim A, X e [a, b], thi k - ?to IA nghiem dac cna A - - X0 e [a - k, b - X] Neu moi nghiem dac cda A deli thuec [a, b] va I, < a thi a - 10 > nen moi nghiem dac caa A - X0I, deo duong, vay A - Xid xic dinh duong vdi X < a (xem vi du 3.7) N6u b <10

thi b - < 0, nhu vay moi nghiem (lac thing cna A - 7.01 dgu am Do vay A - XeI xac dinh am Phan ngvdc Iai chting minh Wong tv

3.12 a) Vol B, = A -1 11dUI, La c6 B I U, = AU, -1,15 U1 U1 = X 1U, - X,U • IlUd1 = X(1 -111-11 1112)Up Do do, n6u 7L1 = thi = vdi mm vec to rieng U ; ngu X1 # thi B I U, = va chi U, la vec to rieng co chua'n dclit bang

b) Gi sii X la vec to true giao vdi U, (vdi tich ye Inking chinh tic) Khi .X = va B,X = AX - X1U1 1.1 X = AX Nhu vay, n6u U2 la tree to Hong don vi Gila A trip giao vdi U , thi U2 cling la vac to rieng cim B Xet clang eau 112 = A - a,U Ui -

12U2 M2 , ta co KerB2 chtla khong gian vec to sinh bai U2}; va thu hep cita B2 tren khong gian true giao vdi Vect(IJ I , U2)

trong vdi thu hep cila A tren (coi ma trait A la ma trail cua to clang eau khong gian vec to dclit R5 Sau n bade ta nhan

dude B,1=A-EX .111 , c6 tinh chat Ker(Ba) Vect(IJI,

i=1

Nhu vay B„ = hay A = E)< lU i U1 i=1

3.13 Gia sit co xx 0, yx ma f(x, y) = va f(y, y) >0 nhung (x, x) Ta thky x, y dec lap tuy6n tinh va vat z thuec khong ;tan vec to sinh bat x, y thi fez, z) = f(ax I by, ax + by) = a2 f(x, x) +

f(y, y) + 2ab f(x,y) ?0, (")

Theo gia thiet n6u f(x, a) =0 thi x cling phitong n6u f(y,a) = hi y cane phtiong voi a Nhung x, y khong tang phuong, nen (x, a) va f(y, khong deng that bang khong (**) Do co hai

6 thvc k, I cK cho k' +1' > va k f(x, + I f(y,a) = TU f(kx + ly, = Theo gia thiet kx + ly ding phucing 'di a, trudng hop f(kx + ly, kx + ly) < Coat nhan 'cot (*) Do to a= 'K ix + ltyx Theo gia thiet f(a, a) = k, 2f(x, x)-11,1 f(y, y) = O

)o f(y, y) > 0, f(x, x) '2 nen to co 1, = Ira f(x, x) = Nhtt t = k, x va ter [(a, y) = 0, f(a, x) = O Mau thuan 'got (*")

3.14 Xet A e M:11 (n, c Mat (n, C) Vi A phan di51 ximg

len ma Iran iA Hemnit - Min vt ao) Ta co det (A - kin) = <=> let (iA - ixIn) = Nhung mot nghiem da' c trung caa ma tran iecmit den thuc Tif suy mkti agh*n dac trung cila ma trail A

a thuAn ao hac bang khong Gia this cat nghiem khac khong

as da thud dee trung PA la: ja i , , tat , -iak, (cti E K, x 0)

Chi PA N=

PA W ) + a ?)

{0 vain >2k Do det A = PA(0) = k

II ot i nefun=2k t

Do det A a TU day, de thAy detA = ngu n le Di& cling có the suy bang cluing mirth trkic tigp

3.15 Bo dg: Cho V la khong gian vec to tren truang C, U li anh xa nem tuygn tinh: V —> V, nghia la u (ax + py) = u (x)+ 5 u (y) vol mgi x, y e V, a, p E C Khi u2 IA huh 3u tuygn tinh Gia sit X la mgt gia tri rieng thkic, am cua u , X la nghiem bOi than cim da thfic dac trung Put

Chung mink be, dg: De thy u e End (V) Gin) six 11 mg gia tri rieng time < cim u va a x la vec td rieng cua u2 vdi u2 (a) = Aa Khi u(a) va a la dec lap tuygn tinh V That vay n'elk u(a) = -4 a, e C thi u 2(a) = u(4a) = u(a) = 141

a=Xa

-

Do X = 141 a 0, trai vdi X< O

G9i W la khong gian vec td hai chigu cna V, sinh bai a u (a) De thAy moi vec td cim W den la vec to rieng

vol gia tri rieng X va u (W) c W Dal V = V/W, xet anh xa can sinh

u:V c —>

[x] —> u,[x] = [u(x)]

Ta c6 u, IA anh xa nUa tuygn tinh, tit u 12 la anh xa tuygr tinh va u1 2[x] = [u2(x)] VI vay, ki higu PIO va Pu, la cac dz thtc dac trung cliatt c End (V) va u1 c End (V1) ttiOng ling thi Pu2(t) = - A.) Pu 2(t) Niu y lai IA nghigm cf.a da thug 4( thing Pil l', lap lai qua trinh tren to có (t - )0 la ink Gila Pu l a

na (t - A) phan b.& Put la s6 than B6 de' dude cluing ainh

Chung mink bai Man:

Xet u: C° —> C11, u (x) = Al; u la Anh xa n&a tuyen tinh; t2(x) = u(u00) = u (Al) = K Ax voi moi x e Ta co Pu2 (t) = let (A A — tIn) Viii moi t e R, ta en det (A.A -t In) = let (A A - t In) = det k -t In) (xem bat 2.51) Nhu \Tay Pu (t)

a da dine vol he so' thve

Gin) Pu2(t) =(%14)a' • 0.2 - 0'2 - oak x dO sn ,sk muyen during, ., Irk e R;

Q e R Q khong co nghiem thne

Vi Q khong có nghiem thile nen deg Q = n -(s, + s + + sk) Alan He se cao nhat cua Q(t) la (-1)"" sk), nghia la bang Do Q(t) > vdi moi t e R, to Q(-1) >

Bay gid ta xet the nghiem (i = 1, 2, , k) N6u CO < 0, thl boi s, ena nghiem ?9 chart, (Xi + nsi z N6u 7u, z 0, thi

re rang (X,+1) 31 >0.Nhv vay Pu2(-1) a 0, nghia la det (A A+ In) a a

Chu $: Deu bang co th6 x637 ra, chamg han xet A =

3.17 Vol f e End (U) Neu f la t0 Tang caM tor lien h0p thi vdi moi x e U; ta co:

Nhu vay <x, f(x)> thole hay <f(x), x> thitc \TM moi x e U Node lai n6u moi x e U, to có <f(x), x> thoc, to cheini minh f to lien hOp Than Lich f tting can hai to deing cM to lien hop:

f = ft + i ft,, <f(x), x> = <f,(x), x> + i <ft,(x), x>

Nhung ft, 1, la nhang to thing chin to lien hop nen <Ii(x), xi

va <f,(x), x> thoc Vi vay <fa (x), x> = Im <f(x), x> N'eal <1(x), xi thitc thi <fa(x), x> = yen moi x, ta If„ = va f, = Do vay = f1 nghia la f la to lien hqp

3.18 Gia sa y la gia tri rieng c>ia f e End (U), ado f la tit long din to lien hop caa khong gian vec in Unita U Khi co

e U, z # d' f(z) = Az D4t x

z f(x) = ax, x = va <f(x), x> = <Xx,

=GI4=X

3.19 a) va b) hir3n nhi'en

c) 1(x) = ek > e k = E< x, f(e k )>e k

k=1 k=1

n

= Ex], < ek > ek = Xk Pk (X)

k=1 k=1

\ray f = Ek kPk

d) Theo trot) to co f = Xi, Pk, vgy f = 14 pk va

k=1 t=1

Is = En Ask Pk k=1

Nhu vgy Q(f)= iOi k )Pk

k=1

3.20 Gia sit f e End (C") la titan tit tit lien hdp kh8ng ;Mit vec to Unita C" vdi tfch vo hthing Hecmit tti nhien, f c6 ma rdn A c6 so chinh tag cria C Gia sit X„ la cac gid ri Hong (thtic) cua f, tfic 2., la nghiem mad da thitc dgc trung det A - = P(X) Goi (e t , co so trite chugn gam nhiing vec Hong ring voi cac gid tri rieng f(g) = ).1 e1 (i = 1, n)

Hat Pk: C" C", Pk(x) = <x, e k > ek Theo bat 3.19 Ta co P(f) = IP(21k)• Pk • Nhung P(1.k) =0 171c nen P(f)= 0, to P(A) =

k=1

3.22 Platong trinh flag trung ciia ma trdn c1 c6 dang: x

2 .F3

2

o -1

(2 - X)X ' 2x =0

2

1761 k = to co x1 = f x , churl vec to rieng

2

e E2 =

\TM X = -1 to co x =-Z xl , chon vec to rieng (

X2= e

1/3

N6u lKy cd sd true chan

,,75 ix, =

2 2

—e + -1e ; x = le l -‘1 e thi cd so do, r tran cim f co dqng eh&

JJJJ

3.23 HD Nghi'em dac cim ma tran B la: X = -1, =

(INN 2)

1

V2

Vdi 7r = 1, cac toa &la \roc td rieng thaa man: x, = x3 Ta

6 hai vec td rieng Dap

1

giao

V2

IA: X2 =

1 .`

va X3 = '0'

1

o

0

42

Pat C = 0 thi C la ma tram tnic giao vi 1

.%& -1 0`

B C =

0

3.25 c) HD Xet F = Imf c E, F1 ® F = E Kerf = Ta c6 f/Imf: F F lA clang au, c6 hang mtc dai, f - phan del rang nen theo b) dim Imf la s6 chan

TAI Lieu THAM KHAO

1 Nge TInie Lanh Dai s6 tuy'en tinh NXB Dai hoc va Trur hoc chuyen nghiep Ha net 1970

2 Doan Quynh Bien) Giao trinh Dal s6 tuyeln tinh hinh hoc giai Bch NXB Dai hoc Qu6c gia Ha NOi, 1998 Khu Que.c Anh, Nguyen Anh Kiet, Ta Man, Nguyen Doa

Turin BM tap Dai s6 tuy6n tinh va hinh hoc giii tic' NXB DHQG Ha mai, 1999

4 R Millman Introduction to matrix analysis MCGRAl book company, inc New-youk - Toronto - London, 1960 J Rivaud Exercices d'algebre lineaire Librairie Vuibei

1978

6 Proxcuriacop I V Tuyk tap the bli tap Dai s6 tuye tinh NXB "Khoo hoe", 1978 (Tielng Nga)