kg_vec_to_con

CHNG 4: KHNG GIAN VCT

TS. L Xun iTrng i hc Bch Khoa TP HCM

Khoa Khoa hc ng dng, b mn Ton ng dng

TP. HCM 2011.

TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 1 / 37

Ta ca vct, chuyn c s Ta ca vct

nh nghaCho K -kgv E , dim(E ) = n, n N. Gi sB = {e1, e2, . . . , en} l mt c s ca E . Nh vyx E ,x1, x2, . . . , xn K : x =

ni=1

xiei . Cc s

xi , (i = 1, 2, . . . , n) c xc nh duy nht vc gi l ta ca vct x trong c s B . K

hiu [x ]B =

x1x2...xn



nh lVi mi x E , B l mt c s ca E th

1 Ta [x ]B l duy nht.2 [x ]B = [x ]B , K .3 [x + y ]B = [x ]B + [y ]B , x , y E .




1 Ta [x ]B l duy nht.

2 [x ]B = [x ]B , K .3 [x + y ]B = [x ]B + [y ]B , x , y E .




1 Ta [x ]B l duy nht.2 [x ]B = [x ]B , K .

3 [x + y ]B = [x ]B + [y ]B , x , y E .




1 Ta [x ]B l duy nht.2 [x ]B = [x ]B , K .3 [x + y ]B = [x ]B + [y ]B , x , y E .


Ta ca vct, chuyn c s V d

V dTm ta ca vct x = (6, 5, 4) trong c s Bca R3: e1 = (1, 1, 0), e2 = (2, 1, 3), e3 = (1, 0, 2)

Tm x1, x2, x3

x = (6, 5, 4) = x1(1, 1, 0) + x2(2, 1, 3) + x3(1, 0, 2)

x1 + 2x2 + x3 = 6

x1 + x2 = 5

3x2 + 2x3 = 4

x1 = 3

x2 = 2

x3 = 1Vy [x ]B = (3, 2,1)T .



V dTm ta ca vct x = (6, 5, 4) trong c s Bca R3: e1 = (1, 1, 0), e2 = (2, 1, 3), e3 = (1, 0, 2)

Tm x1, x2, x3

x = (6, 5, 4) = x1(1, 1, 0) + x2(2, 1, 3) + x3(1, 0, 2)

x1 + 2x2 + x3 = 6

x1 + x2 = 5

3x2 + 2x3 = 4

x1 = 3

x2 = 2

x3 = 1Vy [x ]B = (3, 2,1)T .TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 4 / 37


V dTrong Rkgv P2(x) cho c sp1(x) = 1 + x , p2(x) = 1 x , p3(x) = x2 + x .Tm ta ca vct p(x) = x2 + 7x 2

p(x) = 1p1(x) + 2p2(x) + 3p3(x)

x2+7x2 = 1(1+x)+2(1x)+3(x2+x)

3 = 1

1 2 + 3 = 71 + 2 = 2

1 = 2

2 = 43 = 1

Vy [x ]B = (2,4, 1)T .



V dTrong Rkgv P2(x) cho c sp1(x) = 1 + x , p2(x) = 1 x , p3(x) = x2 + x .Tm ta ca vct p(x) = x2 + 7x 2p(x) = 1p1(x) + 2p2(x) + 3p3(x)

x2+7x2 = 1(1+x)+2(1x)+3(x2+x)

3 = 1

1 2 + 3 = 71 + 2 = 2

1 = 2

2 = 43 = 1

Vy [x ]B = (2,4, 1)T .TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 5 / 37

Ta ca vct, chuyn c s Chuyn c s

Cho K -kgv E , B = {e1, e2, . . . , en} vB = {e 1, e 2, . . . , e n} l 2 c s ca E . Gi s giaB v B c mi lin h

e i =n

k=1

skiek , i = 1, 2, . . . n.

e 1 = s11e1 + s21e2 + . . . + sn1en. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

e n = s1ne1 + s2ne2 + . . . + snnen


Ta ca vct, chuyn c s Chuyn c s

nh ngha

Ta gi ma trn S =

s11 s12 . . . s1ns21 s22 . . . s2n. . . . . . . . . . . .

sn1 sn2 . . . snn

cgi l ma trn chuyn t c s B sang B . K hiuS = Pass(B ,B ).


Ta ca vct, chuyn c s Mi lin h gia ta ca vct trong 2 c s khc nhau

Cho K -kgv E , B = {e1, e2, . . . , en} vB = {e 1, e 2, . . . , e n} l 2 c s ca E . Gi sx E ta cx =

nk=1

xkek hay [x ]B = (x1, x2, . . . , xn)T v

x =n

i=1

x i ei hay [x ]B = (x 1, x 2, . . . , x n)T

Ta tm mi lin h gia [x ]B v [x ]B


Ta ca vct, chuyn c s Mi lin h gia ta ca vct trong 2 c s khc nhau

x =n

i=1

x i ei

= x 1e1 + x

2e2 + . . . + x

nen

= x 1(s11e1+ s21e2+ . . .+ sn1en)+x2(s12e1+ s22e2+

. . .+ sn2en) + . . .+ xn(s1ne1 + s2ne2 + . . .+ snnen)

= (s11x1 + s12x

2 + . . .+ s1nx

n)e1 + (s21x

1 + s22x

2 +

. . .+ s2nxn)e2+ . . .+(sn1x

1+ sn2x

2+ . . .+ snnx

n)en

=n

k=1

xkek


Ta ca vct, chuyn c s Mi lin h gia ta ca vct trong 2 c s khc nhaux1 = s11x

1 + s12x

2 + . . . + s1nx

n

x2 = s21x1 + s22x

2 + . . . + s2nx

n

. . . . . . . . . . . . . . . . . . . . .

xn = sn1x1 + sn2x

2 + . . . + snnx

n

x1x2...xn

=

s11 s12 . . . s1ns21 s22 . . . s2n. . . . . . . . . . . .

sn1 sn2 . . . snn

x 1x 2...x n

[x ]B = S [x ]B , [x ]B = S1[x ]B .TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 10 / 37


V dTrong Rkgv P2(x) cho 2 c sB = {2x2 + x , x2 + 3, 1},B = {x2 + 1, x 2, x + 3} v vctp(x) = 8x2 4x + 6.

1 Tm ma trn chuyn c s S t c s B sangB .

2 Tm ta ca p(x) trong 2 c s B ,B .



Ta c e 1 = x2 + 1, e 2 = x 2, e 3 = x + 3 ve1 = 2x

2 + x , e2 = x2 + 3, e3 = 1. Ta s tm ta

ca e 1, e 2, e 3 theo c s B tc l

e 1 = s11e1 + s21e2 + s31e3e 2 = s12e1 + s22e2 + s32e3e 3 = s13e1 + s23e2 + s33e3



e 1 = s11e1 + s21e2 + s31e3 s11(2x2 + x) + s21(x2 + 3) + s31.1 = x2 + 1

2s11 + s21 = 1

s11 = 0

3s21 + s31 = 1 s11 = 0, s21 = 1, s31 = 2.



e 2 = s12e1 + s22e2 + s32e3 s12(2x2 + x) + s22(x2 + 3) + s32.1 = x 2

2s12 + s22 = 0

s12 = 1

3s22 + s32 = 2 s12 = 1, s22 = 2, s32 = 4.



e 3 = s13e1 + s23e2 + s33e3 s13(2x2 + x) + s23(x2 + 3) + s33.1 = x + 3

2s12 + s22 = 0

s12 = 1

3s22 + s32 = 3 s13 = 1, s23 = 2, s33 = 9.



Vy ma trn chuyn c s S t c s B sang B l 0 1 11 2 22 4 9



2. Tm ta ca p(x) trong 2 c s B ,B .Ta ca p(x) trong c s B l 1, 2, 3 thap(x) = 1e1 + 2e2 + 3e3 1(2x2+ x)+2(x2+3)+3.1 = 8x2 4x +6

21 + 2 = 8

1 = 432 + 3 = 6

1 = 4, 2 = 16, 3 = 42. [p(x)]B = (4, 16,42)T .

Ta ca p(x) trong c s B l[p(x)]B = S

1.[p(x)]B = (8,2,2)T



2. Tm ta ca p(x) trong 2 c s B ,B .Ta ca p(x) trong c s B l 1, 2, 3 thap(x) = 1e1 + 2e2 + 3e3 1(2x2+ x)+2(x2+3)+3.1 = 8x2 4x +6

21 + 2 = 8

1 = 432 + 3 = 6

1 = 4, 2 = 16, 3 = 42. [p(x)]B = (4, 16,42)T .Ta ca p(x) trong c s B l[p(x)]B = S

1.[p(x)]B = (8,2,2)TTS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 17 / 37

C s v s chiu ca khng gian vct con C s v s chiu ca khng gian vct con

H quCho E l mt K -kgv, dim(E ) = n, F l khnggian vct con ca E th dim(F ) 6 n.

Chng minh.

Do F E nn mi tp con c lp tuyn tnhca F u c s phn t 6 n.Gi B = {x1, x2, . . . , xk}(k 6 n) l 1 tp conc lp tuyn tnh ca F c s phn t lnnht. chng minh B l c s ca F ta chcn chng minh B l tp sinh ca F .



Vi mi x F , nu x khng l t hp tuyn tnhca nhng vct ca B th tp B {x} c lptuyn tnh.

Tht vy, gi s1x1 + 2x2 + . . . + kxk + k+1x = 0. Nuk+1 6= 0 th x l t hp tuyn tnh cax1, x2, . . . , xk (tri vi gi thit). Nu k+1 = 0 th1x1 + 2x2 + . . . + kxk = 0

1 = 2 = . . . = k = 0 (v x1, x2, . . . , xk clp tuyn tnh).



Vi mi x F , nu x khng l t hp tuyn tnhca nhng vct ca B th tp B {x} c lptuyn tnh. Tht vy, gi s1x1 + 2x2 + . . . + kxk + k+1x = 0. Nuk+1 6= 0 th x l t hp tuyn tnh cax1, x2, . . . , xk (tri vi gi thit).

Nu k+1 = 0 th1x1 + 2x2 + . . . + kxk = 0




Vi mi x F , nu x khng l t hp tuyn tnhca nhng vct ca B th tp B {x} c lptuyn tnh. Tht vy, gi s1x1 + 2x2 + . . . + kxk + k+1x = 0. Nuk+1 6= 0 th x l t hp tuyn tnh cax1, x2, . . . , xk (tri vi gi thit). Nu k+1 = 0 th1x1 + 2x2 + . . . + kxk = 0




Vy, B {x} F c lp tuyn tnh v s phnt ca n l k + 1 > k. (tri vi gi thit k lnnht).

Do , x F u l t hp tuyn tnhca nhng vct ca B B l tp sinh



Vy, B {x} F c lp tuyn tnh v s phnt ca n l k + 1 > k. (tri vi gi thit k lnnht). Do , x F u l t hp tuyn tnhca nhng vct ca B B l tp sinh


C s v s chiu ca khng gian vct con V d

V dTrong Rkgv P2(x) cho khng gian conF = {p(x) P2(x)\p(1) = 0, p(1) = 0}. Tmmt c s v s chiu ca khng gian con F .

p(x) = ax2 + bx + c F , ta cp(1) = a + b + c = 0 vp(1) = a b + c = 0. Gii h phng trnh{

a + b + c = 0

a b + c = 0 {

a = cb = 0



Vy p(x) = c(x2 + 1). Do {x2 + 1} l tpsinh ca F .x2 + 1 6= 0 nn lun c lp tuyn tnh.

Nh vy, x2 + 1 l 1 c s ca F v s chiudim(F ) = 1.



V dTm mt c s v s chiu ca khng gian con Wca R3 cho bi

W = {(x1, x2, x3)\x1 + x2 + x3 = 0}

tm c s ca W ta gii phng trnhx1 + x2 + x3 = 0 x1 = x2 x3. Nghim c sl (1, 1, 0) v (1, 0, 1). Ta s chng minh(1, 1, 0) v (1, 0, 1) l c s ca W .



V dTm mt c s v s chiu ca khng gian con Wca R3 cho bi

W = {(x1, x2, x3)\x1 + x2 + x3 = 0} tm c s ca W ta gii phng trnhx1 + x2 + x3 = 0 x1 = x2 x3. Nghim c sl (1, 1, 0) v (1, 0, 1). Ta s chng minh(1, 1, 0) v (1, 0, 1) l c s ca W .



Hai vct (1, 1, 0) v (1, 0, 1) c lptuyn tnh.Ta chng minh (1, 1, 0) v (1, 0, 1) sinh raW . Tht vy, x = (x1, x2, x3) W thx = x2(1, 1, 0) + x3(1, 0, 1).

Nh vy, (1, 1, 0) v (1, 0, 1) l 1 c s ca Wv s chiu dim(W ) = 2.


C s v s chiu ca khng gian vct con Khng gian nghim ca h phng trnh tuyn tnh thun nht

nh lCho h phng trnh tuyn tnh thun nht gmm phng trnh v n n AmnXn1 = 0m1. Khi cc nghim ca h phng trnh ny to thnhkhng gian vct con ca khng gian K n.

nh lKhng gian vct nghim ca h phng trnhtuyn tnh thun nht tng qut c s chiu bngn r trong r = rank(A) v n l s n.


C s v s chiu ca khng gian vct con Khng gian nghim ca h phng trnh tuyn tnh thun nht

nh lCho h phng trnh tuyn tnh thun nht gmm phng trnh v n n AmnXn1 = 0m1. Khi cc nghim ca h phng trnh ny to thnhkhng gian vct con ca khng gian K n.

nh lKhng gian vct nghim ca h phng trnhtuyn tnh thun nht tng qut c s chiu bngn r trong r = rank(A) v n l s n.TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 25 / 37


V dGii h tm nghim ca khng gian nghim

x1 + 2x2 x3 + x4 = 02x1 + 4x2 3x3 = 0x1 + 2x2 + x3 + 5x4 = 0

1 2 1 12 4 3 01 2 1 5

h2h22h1h3h3h1 1 2 1 10 0 1 2

0 0 2 4



V dGii h tm nghim ca khng gian nghim

x1 + 2x2 x3 + x4 = 02x1 + 4x2 3x3 = 0x1 + 2x2 + x3 + 5x4 = 0 1 2 1 12 4 3 01 2 1 5

h2h22h1h3h3h1 1 2 1 10 0 1 2

0 0 2 4



h3h3+2h2 1 2 1 10 0 1 2

0 0 0 0

x1, x3 l bin cs, x2, x4 l bin t do. t x2 = , x4 =

x1x2x3x4

=2 3

2

= 2100

+ 3021

Vy X1 = (2, 1, 0, 0)T v X2 = (3, 0,2, 1)T l c sca khng gian nghim. S chiu ca khng gian nghim

ca h ny l 2.


Hng ca mt h vct nh ngha

nh nghaCho tp M = {x1, x2, . . . , xp} E l mtK kgv . Tp N = {xi1, xi2, . . . , xir} c gi ltp con c lp tuyn tnh ti i ca M nu vch nu N c lp tuyn tnh v mi vct ca Mu l t hp tuyn tnh ca cc vct ca N .

nh nghaHng ca mt h vct ca mt K -kgv E l svct c lp tuyn tnh ti i ca n.

Nu M = {0} th coi hng ca M bng 0.


Hng ca mt h vct nh ngha

nh nghaCho tp M = {x1, x2, . . . , xp} E l mtK kgv . Tp N = {xi1, xi2, . . . , xir} c gi ltp con c lp tuyn tnh ti i ca M nu vch nu N c lp tuyn tnh v mi vct ca Mu l t hp tuyn tnh ca cc vct ca N .

nh nghaHng ca mt h vct ca mt K -kgv E l svct c lp tuyn tnh ti i ca n.

Nu M = {0} th coi hng ca M bng 0.TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 28 / 37

Hng ca mt h vct V d

V dTrong Rkgv P3(x) cho hH = {p1(x) = 5x , p2(x) = x + 3x2, p3(x) =4x 5x2, p4(x) = x2 + 6x}. Tm hng ca H .

p1(x), p2(x) c lp tuyn tnh. V t1p1(x) + 2p2(x) = 0

32x2 + (51 + 2)x = 0 1 = 2 = 0.p1(x), p2(x), p3(x), p4(x) u l t hp tuyntnh ca p1(x), p2(x)

Nn hng ca H bng 2


Hng ca mt h vct V d

V dTrong Rkgv P3(x) cho hH = {p1(x) = 5x , p2(x) = x + 3x2, p3(x) =4x 5x2, p4(x) = x2 + 6x}. Tm hng ca H .

p1(x), p2(x) c lp tuyn tnh. V t1p1(x) + 2p2(x) = 0

32x2 + (51 + 2)x = 0 1 = 2 = 0.p1(x), p2(x), p3(x), p4(x) u l t hp tuyntnh ca p1(x), p2(x)

Nn hng ca H bng 2TS. L Xun i (BK TPHCM) CHNG 4: KHNG GIAN VCT TP. HCM 2011. 29 / 37

Hng ca mt h vct C s v s chiu ca bao tuyn tnh

nh lGi s M = {x1, x2, . . . , xp} E l mt K -kgv chng r v W =< M > l khng gian vct consinh bi M . Khi dim(W ) = r .

Chng minh.

Gi s Mr = {xi1, xi2, . . . xir} l 1 tp con clp tuyn tnh ti i ca M .Chng minh Mr sinh ra W . Mi vct thuc Mu l t hp tuyn tnh ca cc vct ca Mr




Chng minh.

Gi s Mr = {xi1, xi2, . . . xir} l 1 tp con clp tuyn tnh ti i ca M .

Chng minh Mr sinh ra W . Mi vct thuc Mu l t hp tuyn tnh ca cc vct ca Mr




Chng minh.

Gi s Mr = {xi1, xi2, . . . xir} l 1 tp con clp tuyn tnh ti i ca M .Chng minh Mr sinh ra W .

Mi vct thuc Mu l t hp tuyn tnh ca cc vct ca Mr




Chng minh.

Gi s Mr = {xi1, xi2, . . . xir} l 1 tp con clp tuyn tnh ti i ca M .Chng minh Mr sinh ra W . Mi vct thuc Mu l t hp tuyn tnh ca cc vct ca Mr



mi vct ca W l t hp tuyn tnh ca ccvct ca M th cng l t hp tuyn tnh ca ccvct ca Mr . W =< M > W =< Mr > .

Mr c lp tuyn tnh.

Mr l c s ca W dim(W ) = r = rank(M).



mi vct ca W l t hp tuyn tnh ca ccvct ca M th cng l t hp tuyn tnh ca ccvct ca Mr . W =< M > W =< Mr > .

Mr c lp tuyn tnh. Mr l c s ca W dim(W ) = r = rank(M).



Tm c s v s chiu ca khng gian con M ca kgv Esinh bi m vct x1, x2, . . . , xm

1 Ly mt c s B = {e1, e2, . . . , en} bt k caE . Tm [x1]B , [x2]B , . . . , [xm]B

2 Xt khng gian ct ca ma trnA = ([x1]B , [x2]B , . . . , [xm]B)

3 Bin i A v dng bc thang t xc nhr(A) v c s ca M , s chiu ca M bngr(A).


Hng ca mt h vct V d

V dTrong Rkgv P2(x) cho p1(x) =x2 + 2x + 1, p2(x) = 2x

2 + x 1, p3(x) = 4x + 4.Tm c s v s chiu ca khng gian con sinh bi3 vct trn.

Xt c s chnh tc x2, x , 1 ca P2(x), vy ma

trn cc ct A l A =

1 2 02 1 41 1 4


Hng ca mt h vct V d

A

h2h22h1h3h3h1

1 2 00 3 40 3 4

h3h3h2 1 2 00 3 4

0 0 0

= B . Ma trn B c ct 1 v ct 2c lp tuyn tnh v l c s ca khng gian consinh bi 3 vct p1(x), p2(x), p3(x). Vyp1(x), p2(x) l c s v s chiu ca khng giancon sinh bi 3 vct trn l 2.


Hng ca mt h vct H cc vct ct v h cc vct hng

nh lCho ma trn A Mmn(K ). Khi nu gi rh vrc tng ng l hng ca cc vct hng v ccvct ct tng ng ca A th

rank(A) = rh = rc .


Hng ca mt h vct V d

V dTrong R4 tm hng ca h cc vct sau:x1 = (1, 2, 4, 0), x2 = (3, 2, 1, 2),

x3 = (2, 0,1, 4), x4 = (1,2,5, 4),x5 = (5, 2, 0, 6)

1 2 4 03 2 1 22 0 1 41 2 5 45 2 0 6

BSC hng

1 2 4 00 4 11 20 0 2 20 0 0 00 0 0 0

rA = 3 nn hng ca h cc vct cng bng 3.


Hng ca mt h vct V d

THANK YOU FOR ATTENTION


Ta ca vct, chuyn c sTa ca vctV dChuyn c sMi lin h gia ta ca vct trong 2 c s khc nhauV d

C s v s chiu ca khng gian vct conC s v s chiu ca khng gian vct conV dKhng gian nghim ca h phng trnh tuyn tnh thun nhtV d

Hng ca mt h vctnh nghaV dC s v s chiu ca bao tuyn tnhV dH cc vct ct v h cc vct hngV d

kg_vec_to_con

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