BigingmnhcisA1Chng3:[email protected]://www.math.hcmus.edu.vn/lvluyen/09ttihcKhoaHcTNhinTp.HChMinhLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010
1/86NidungChng3.KHNGGIANVECT1. Khnggianvect2. Thptuyntnh3.
Csvschiucakhnggianvect4. Khnggianvectcon5.
Khnggiannghimcahphngtrnhtuyntnh6. TavmatrnchuyncsLVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 2/861.
Khnggianvect1.Khnggianvectnhngha. ChoVl mt tp hp vi php ton +.Vc gi
lkhnggianvecttrn R nu miu, v, w Vv, R ta c 8tnh cht sau:(1) u+v =
v+u;(2) (u+v)+w = u+(v+w);(3) tn ti 0 V: u+0 = 0+u = u;(4) tn ti
u
V: u
+u = u+u
= 0;(5) ()u = (u);(6) ( + )u = u + u;(7) (u+v) = u+v;(8) 1.u =
u.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 3/861.
KhnggianvectKhi ta gi:mi phn tu Vl mtvect.mi s R l mtvhng.vect 0
lvectkhng.vect u
lvecticau.Vd. XtV= Rn= (x1, x2, . . . , xn) [xi R, i 1, n.Viu =
(a1, a2, . . . , an), v = (b1, b2, . . . , bn) Rnv R, ta nhngha php
cng + v nhn. v hng nh sau: u+v = (a1 + b1, a2 + b2, . . . , an +
bn); u = (a1, a2, . . . , an).Khi Rnl khng gian vect trn R. Trong
:Vect khng l 0 = (0, 0, . . . , 0);Vect i cau l u = (a1, a2, . . .
, an).LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 4/861.
KhnggianvectV d. Tp hpMmn(R) vi php cng ma trn v nhn ma trn vimt s
thc thng thng l mt khng gian vect trn R. Trong :Vect khng l ma trn
khngVect i caA l A.Vd. Tp hpR[x] = p(x) = anxn+ . . . + a1x + a0[ n
N, ai R, i 1, ngm cc a thc theox vi cc h s trong R l mt khng gian
vecttrn R vi php cng vect l php cng a thc thng thng v phpnhn v hng
vi vect l php nhn thng thng mt s vi athc.Vd. Tp hp Rn[x] gm cc a
thc bc nh hn hoc bngn theox vi cc h s trong R l mt khng gian vect
trn R.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 5/861.
KhnggianvectVd. ChoV= (x1, x2, x3) R3[ 2x1 + 3x2 + x3 = 0.Khi Vl
khng gian vect trn R.Vd. ChoW= (x1, x2, x3) R3[x1 + x2 2x3 = 1.Khi
Wkhng l khng gian vect, vu = (1, 2, 1) W, v = (2, 3, 2) W,nhngu + v
= (3, 5, 3)/ WMnh. ChoVl mt khng gian vect trn R. Khi vi miu Vv R,
ta ci) u = 0 ( = 0 hay u = 0);ii) (1)u = u.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 6/862.
Thptuyntnh2.Thptuyntnh1.1Thptuyntnh1.2clpvphthuctuyntnhLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 7/862.
Thptuyntnh2.1Thptuyntnhnhngha. Chou1, u2, . . . , um V .
Mtthptuyntnhcau1, u2, . . . , um l mt vect c dngu = 1u1 + 2u2 + . .
. + mumvii RKhi , ng thc trn c gi ldngbiudincau theo ccvectu1, u2,
. . . , um.Vd.Vectu = (4, 4, 2) l t hp tuyn tnh ca cc vectu1 = (1,
1, 2), u2 = (2, 3, 1), u3 = (0, 1, 2), vu = u1 + 2u2 u3.Vect 0 lun
lun l mt t hp tuyn tnh cau1, u2, ..., um v0 = 0u1 + 0u2 + . . . +
0um.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 8/862.
ThptuyntnhHi. Lm cch no bitu l t hp tuyn tnh cau1, u2, ..., um?Ta
cu l t hp tuyn tnh cau1, u2, ..., um khi phng trnhu = 1u1 + 2u2 + .
. . + mum()c nghim1, 2, . . . m R.Xt trng hp khng gian Rn. Gi su =
(b1, b2, . . . , bn)u1= (u11, u21 . . . , un1);u2= (u12, u22 . . .
, un2);. . . . . . . . . . . . . . . . . . . . . . . . . . . .um=
(u1m, u2m . . . , unm).Khi () ___u111 + u122 + . . . + u1mm=
b1;u211 + u222 + . . . + u2mm= b2;. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .un11 + un22 + . . . +
unmm= bn.()LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 9/862.
ThptuyntnhMa trn ha () ta c____u11u12. . . u1mb1u21u22. . . u2mb2.
. . . . . . . . . . . . . . . . . . . .un1un2. . . unmbn____Tc
l(u
1u
2. . . u
m [u
)Nh vy, kim trau l t hp tuyn tnh cau1, u2, ..., um trong Rnta lm
nh sau:Lp ma trn ha (u
1u
2. . . u
m [u
) (1)Nu (1) v nghim, kt lunu khng phi l t hp tuyn tnhcau1, u2,
..., um. Nu (1) c nghim1, 2, . . . m thu l t hp tuyn tnh vc dng biu
din theo lu1, u2, ..., um :u = 1u1 + 2u2 + . . . + mumLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 10/862. ThptuyntnhVd. Xt
xemu = (3, 1, 4) c l t hp tuyn tnh ca cc vectu1 = (1, 2, 1), u2 =
(1, 1, 1), u3 = (2, 1, 1) hay khng?Gii. (u
1u
2u
3[u
) =__1 1 2 32 1 1 11 1 1 4__d2:=d22d1d3:=d3d1__1 1 2 30 1 5 70 2
3 7__d1:=d1+d2d3:=d32d2__1 0 3 40 1 5 70 0 7
7__d3:=17d3d1:=d13d3d2:=d25d3__1 0 0 10 1 0 20 0 1 1__.H phng trnh
c nghim duy nht (1; 2; 3) = (1; 2; 1).Vyu l t hp tuyn tnh cau1, u2,
u3.Dng biu din cau lu = u1 + 2u2 + u3.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 11/862. ThptuyntnhVd. Xt xemu = (4, 3, 5) c
l t hp tuyn tnh ca cc vectu1 = (1, 2, 5), u2 = (1, 3, 7), u3 = (2,
3, 4) hay khng?Gii. (u
1u
2u
3[u
) =__1 1 2 42 3 3 35 7 4 5__d2:=d22d1d3:=d35d1__1 1 2 40 1 7 50
2 14 15__d1:=d1d2d3:=d32d2__1 0 9 90 1 7 50 0 0 5__H v nghim v 0x +
0y + 0z = 5. Vyu khng l t hp tuyntnh cau1, u2, u3.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 12/862. ThptuyntnhVd. Xt
xemu = (4, 3, 10) c l t hp tuyn tnh ca cc vectu1 = (1, 2, 5), u2 =
(1, 3, 7), u3 = (2, 3, 4) hay khng?Gii. (u
1u
2u
3[u
) =__1 1 2 42 3 3 35 7 4 10__d2:=d22d1d3:=d35d1__1 1 2 40 1 7 50
2 14 10__d1:=d1d2d3:=d32d2__1 0 9 90 1 7 50 0 0 0__Nghim ca h l (1;
2; 3) = (9 + 9t, 5 7t, t)Vyu l t hp tuyn tnh cau1, u2, u3.Dng biu
din cau lu = (9 + 9t)u1 + (5 7t)u2 + tu3.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 13/862. ThptuyntnhVd. Trong khng gian R4cho
cc vectu1 = (1, 1, 1, 1);u2 = (2, 3, 1, 0);u3 = (1, 1, 1, 1). Tm iu
kin vectu = (a, b, c, d) l mt t hp tuyn tnh cau1, u2, u3.Gii.(u
1u
2u
3[u
) =____1 2 1 a1 3 1 b1 1 1 c1 0 1 d________1 2 1 a0 1 0 b a0 3 2
c a0 2 2 d a________0 2 1 a0 1 0 a + b0 0 2 4a + 3b + c0 0 2 3a +
2b + d________0 2 1 a0 1 0 a + b0 0 2 4a + 3b + c0 0 0 a b c +
d____. u l mt t hp tuyn tnh ca u1, u2, u3 th h c nghim, tc la + d =
b + c.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 14/862.
Thptuyntnh2.2clpvphthuctuyntnhnhngha. Chou1, u2, . . . , um V . Xt
phng trnh1u1 + 2u2 + . . . + mum = 0. ()Nu () ch c nghim tm thng1 =
2 = . . . = m = 0 thta niu1, u2, . . . , um (hay u1, u2, . . . ,
um)clptuyntnh.Nu ngoi nghim tm thng, () cn c nghim khc th ta niu1,
u2, . . . , um (hay u1, u2, . . . , um)phthuctuyntnh.Ni cch khc,Nu
phng trnh () c nghim duy nht thu1, u2, . . . , um clp tuyn tnh.Nu
phng trnh () c v s nghim thu1, u2, . . . , um phthuc tuyn
tnh.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 15/862.
ThptuyntnhVd. Trong khng gian R3cho cc vectu1 = (1, 2, 3);u2 = (2,
5, 1);u3 = (1, 1, 9). Hiu1, u2, u3 c lp hay ph thuctuyn tnh?Gii. Xt
phng trnh1u1 + 2u2 + 3u3 = 0 1(1, 2, 3) + 2(2, 5, 1) + 3(1, 1, 9) =
(0, 0, 0)___1+ 22+ 3= 0;21+ 52+ 3= 0;31 2 93= 0.Ma trn ha h phng
trnh,A =__1 2 12 5 13 1 8__.Ta cr(A) = 3 nn h c nghim duy nht. Suy
rau1, u2, u3 c lptuyn tnh.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 16/862. ThptuyntnhV d. Trong khng gian R3cho cc vectu1 =
(1, 1, 1);u2 = (2, 1, 3);u3 = (1, 2, 0). Hiu1, u2, u3 c lp hay ph
thuc tuyn tnh?Gii. Xt phng trnh1u1 + 2u2 + 3u3 = 0 ( + 22 + 3, + 2
+ 23, + 32) = (0, 0, 0)___1+ 22+ 3= 01+ 2+ 23= 01+ 32= 0Ma trn ha h
phng trnh,A =__1 2 11 1 21 3 0__.Ta cr(A) = 2 nn h v s nghim. Suy
rau1, u2, u3 ph thuctuyn tnh.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 17/862. ThptuyntnhNhn xt. H vectu1, u2, . . . , umph
thuc tuyn tnh khi v ch khitn ti vectuil t hp tuyn tnh ca cc vect cn
li. Tht vy,Nuu1, u2, . . . , umph thuc tuyn tnh th c1,2,. . . ,m R
khng ng thi bng 0 sao chom
j=1juj = 0. Gi si ,= 0, khi ui = 1i
j=ijuj.Nu cuisao choui = j=ijujthm
j=1juj = 0, trong i = 1 ,= 0, iu ny chng tu1, u2, . . . , umph
thuc tuyntnh.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010
18/862. ThptuyntnhMnh. ChoVl khng gian vect trn R vS = u1, u2, . .
. , uml tp hp cc vect thucV . Khi NuSph thuc tuyn tnh th mi tp
chaSu ph thuctuyn tnh.NuSc lp tuyn tnh th mi tp con caSu c lptuyn
tnh.Hqu. Chou1, u2, . . . , umlm vect trong Rn. GiA l ma trn cc bng
cch xpu1, u2, . . . , umthnh cc dng. Khi u1, u2, . . . , umc lp
tuyn tnh khi v ch khiA c hng lr(A) = m.T H qu trn ta s xy dng thut
ton kim tra tnh c lptuyn tnh ca cc vect trong RnLVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 19/862.
ThptuyntnhThuttonkimtratnhclptuyntnhcaccvecttrong RnBc1: Lp ma trnA
bng cch xpu1, u2, . . . , um thnh cc dng.Bc2: Xc nh hngr(A)
caA.Nur(A) = m thu1, u2, . . . , um c lp tuyn tnh.Nur(A) < m
thu1, u2, . . . , um ph thuc tuyn tnh.Trng hpm = n, ta cA l ma trn
vung. Khi c th thay Bc2 bng Bc 2 sau y:Bc2: Tnh nh thc detA.Nu detA
,= 0 thu1, u2, . . . , um c lp tuyn tnh.Nu detA = 0 thu1, u2, . . .
, um ph thuc tuyn tnh.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 20/862. ThptuyntnhVd. Trong khng gian R5cho cc vectu1 =
(1, 2, 3, 5, 1);u2 = (1, 3, 13, 22, 1);u3 = (3, 5, 1, 2, 5). Hy xt
xemu1, u2, u3 clp tuyn tnh hay ph thuc tuyn tnh.Gii.Lp A
=__u1u2u3__=__1 2 3 5 11 3 13 22 13 5 1 2 5__d2:=d2d1d3:=d33d1__1 2
3 5 10 1 10 17 20 1 10 17 2__d3:=d3+d2__1 2 3 5 10 1 10 17 20 0 0 0
0__Ta cr(A) = 2 < 3. Suy rau1, u2, u3 ph thuc tuyn tnh.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 21/862. ThptuyntnhVd.
Trong khng gian R3cho cc vectu1 = (2m + 1, m, m + 1);u2 = (m2, m1,
m2);u3 = (2m1, m1, 2m1).Tm iu kin u1, u2, u3 c lp tuyn tnh.Gii. LpA
=__u1u2u3__=__2m + 1 m m + 1m2 m1 m22m1 m1 2m1__Ta c[A[ =2m + 1 m m
+ 1m2 m1 m22m1 m1 2m1c1:=c1c3======m m m + 10 m1 m20 m1
2m1ct1====mm1 m2m1 2m1= m(m1)(m + 1).Do u1, u2, u3 c lp tuyn tnh
khi v ch khi[A[ , = 0 m(m1)(m + 1) ,= 0 m ,= 0 vm ,= 1.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 22/863.
Csvschiucakhnggianvect3.Csvschiucakhnggianvect3.1Tpsinh3.2CsvschiuLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 23/863.
Csvschiucakhnggianvect3.1Tpsinhnhngha. ChoVl khng gian vect vS V.Sc
gi ltpsinhcaVnu mi vectu caVu l t hp tuyn tnh caS. Khi, ta niSsinh
raVhocVc sinh biS, k hiu V = S).Vd. Trong khng gian R3, choS = u1 =
(1, 1, 1); u2 = (1, 2, 1); u3 = (2, 3, 1).HiSc l tp sinh ca
R3khng?Gii. Viu = (x, y, z) R3, kim tra xemu c l t hp tuyn tnhcau1,
u2, u3 khng?Ta lp h phng trnh(u
1u
2u
3[u
) =__1 1 2 x1 2 3 y1 1 1 z____1 1 2 x0 1 1 x + y0 0 1 x +
z__.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 24/863.
CsvschiucakhnggianvectH c nghim. Suy rau l t hp tuyn tnh cau1, u2,
u3. VySltp sinh ca R3.Vd. Trong khng gian R3, choS = u1 = (1, 1,
1); u2 = (2, 3, 1); u3 = (3, 4, 0).HiSc l tp sinh ca R3khng?Gii.
Viu = (x, y, z) R3, ta lp h phng trnh(u
1u
2u
3[u
) =__1 2 3 x1 3 4 y1 1 0 z____1 2 3 x0 1 1 x + y0 0 0 4x 3y +
z__.Viu0 = (1, 1, 1) th h trn v nghim.Vyu0 khng l t hp tuyn tnh
cau1, u2, u3. Suy raSkhng ltp sinh ca R3.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 25/863. CsvschiucakhnggianvectVd. Trong
khng gian R2[x], choS = f1 = x2+ x + 1; f2 = 2x2+ 3x + 1; f3 = x2+
2x + 1.HiSc l tp sinh ca R2[x] khng?Gii. Vif= ax2+ bx + c R2[x],
kim tra xemfc l t hp tuyntnh caf1, f2, f3 khng?Xt phng trnh1f1 +
2f2 + 3f3 = f.___1+ 22+ 3= a;1+ 32+ 23= b;1+ 2+ 3= c.Ma trn ha,A
=__1 2 1 a1 3 2 b1 1 1 c____1 2 1 a0 1 1 a + b0 0 1 2a + b + c__H c
nghim. Vyfl t hp tuyn tnh caf1, f2, f3. Suy raSltp sinh ca
R2[x].LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 26/863.
Csvschiucakhnggianvect3.2Csvschiunhngha. ChoVl khng gian vect v B l
con caV. B cgi l mtcscaVnu B l mt tp sinh v B c lp tuyn tnh.Vd.
Trong khng gian R3, choB = u1 = (1, 1, 1); u2 = (1, 2, 1); u3 = (2,
3, 1).Kim tra B l c s ca R3.Gii. B l tp sinh ca R3. (theo v d
trn)Kim tra B c lp tuyn tnh.Lp ma trnA =__u1u2u3__=__1 1 11 2 12 3
1__.Ta cr(A) = 3 (hoc [A[ = 1). Suy ra B c lp tuyn tnh. Vy B lc s
ca R3.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 27/863.
CsvschiucakhnggianvectVd. Trong khng gian R3, choS = u1 = (1, 1,
2); u2 = (2, 3, 3); u3 = (5, 7, 4).HiSc l c s ca R3khng?Vd. Trong
khng gian R3, choS = u1 = (1, 1, 1); u2 = (2, 1, 0); u3 = (1, 1,
0); u4 = (1, 4, 1).HiSc l c s ca R3khng?Vd. Trong khng gian R2[x],
choS = f1 = x2+ x + 1; f2 = 2x2+ x + 1; f3 = x2+ 2x + 2HiSc l c s
caR2[x] khng?LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010
28/863. CsvschiucakhnggianvectSchiuB. Gi sVsinh bim vect,V= u1, u2,
. . . , um). Khi mitp hp con c lp tuyn tnh caVc khng qum phn t.Hqu.
Gi sVc mt c s Bgmn vect. Khi mi c skhc caVhu hn v c ngn
vect.nhngha. Cho V l khng gian vect,schiucaV , k hiu ldimV, l s
vect ca tp c s. Trong trng hp v hn chiu, ta kdimV= .Vd. Trong khng
gian R3, choB = u1 = (1, 1, 1); u2 = (1, 2, 1); u3 = (2, 3, 1).Khi
B l c s ca R3. Do dimR3= 3.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 29/863. CsvschiucakhnggianvectVd. Trong khng gian Rn, xt
B0 = e1, e2, . . . , en, trong e1= (1, 0, 0, . . . , 0),e2= (0, 1,
0, . . . , 0),. . . . . . . . . . . . . . . . . . . . . .en= (0, 0,
. . . , 0, 1).Viu = (x1, x2, . . . , xn) Rn. Ta cu = x1e1 + x2e2 +
+ xnen.Do B0 l tp sinh ca Rn. Mt khc B0 c lp tuyn tnh nn B0 lc s ca
Rn. B0 c gi lcschnhtcca Rn. Nh vydimRn= n.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 30/863. CsvschiucakhnggianvectVd. Khng gian
vectMmn(R) c c sB0 = Eij[ , i 1.m, j 1, n,trong Eijl ma trn loimn
ch c mt h s khc 0 duy nht lh s 1 dngi ctj. Do Mmn(R) hu hn chiu
vdimMmn(R) = mn.V d. Khng gian Rn[x] gm cc a thc theox bc khng qun
vih s trong R, l khng gian vect hu hn chiu trn R cdimRn[x] = n + 1
vi c s B0 = 1, x, . . . , xn.V d. Khng gian R[x] gm tt cc a thc
theox vi h s trong R,l khng gian vect v hn chiu trn R vi c s B0 =
1, x, x2, . . ..LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010
31/863. CsvschiucakhnggianvectHqu. ChoVl khng gian vect cdimV= n.
Khi i) Mi tp con caVcha nhiu hnn vect th ph thuc tuyntnh.ii) Mi tp
con caVcha t hn n vect khng sinh raV .B. ChoSl mt tp con c lp tuyn
tnh caVvu Vlmt vect sao chou khng l t hp tuyn tnh caS. Khi tp hpS1
= S u c lp tuyn tnh.nhl.ChoVl mt khng gian vect hu hn chiu vi dimV=
n. Khi i) Mi tp con c lp tuyn tnh gmn vect caVu l c scaV .ii) Mi tp
sinh caVgmn vect u l c s caV .LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 32/863.
CsvschiucakhnggianvectNhndincscakhnggianV cdimV= nV dimV= n nn mi c
s caVphi gmn vect. Hn na, nuS Vv s phn t caSbngn thSl c s caV Sc lp
tuyn tnh. Sl tp sinh caV.Vd. Kim tra tp hp no sau y l c s ca khng
gian vectca R3?a) B1 = u1 = (1, 2, 3), u2 = (2, 3, 4).b) B2 = u1 =
(2, 1, 3), u2 = (2, 1, 4), u3 = (2, 3, 1), u4 = (3, 4, 5).c) B3 =
u1 = (1, 2, 1), u2 = (1, 3, 2), u3 = (2, 1, 2)d) B4 = u1 = (2, 1,
0), u2 = (1, 2, 3), u3 = (5, 0, 3)LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 33/863. CsvschiucakhnggianvectGii.a) b)B1,
B2 khng phi l c s ca R3v s vect khng bng 3.c)B3 = u1 = (1, 2, 1),
u2 = (1, 3, 2), u3 = (2, 1, 2)LpA =__u1u2u3__=__1 2 11 3 22 1
2__.Ta c detA = 3. Suy raB3 c lp tuyn tnh. Mt khc s vectcaB3 bng 3
= dimR3nnB3 l c s ca R3d)B4 = u1 = (2, 1, 0), u2 = (1, 2, 3), u3 =
(5, 0, 3)LpA =__u1u2u3__=__2 1 01 2 35 0 3__.Ta c detA = 0. Suy
raB4 khng c lp tuyn tnh. V vyB4khng l c s ca R3LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 34/863. CsvschiucakhnggianvectVd.
Trong khng gian R2[x], choS = f1 = x2+ x + 1; f2 = 2x2+ 3x + 1; f3
= x2+ 2x + 1.HiSc l c s ca R2[x] khng?Gii. V dimR2[x] = 3 v s phn t
caSbng 3 nnSl c s caR2[x] khiSc lp tuyn tnh hocSl tp sinh.Cch 1.
Kim traSc lp tuyn tnh.Xt phng trnh1f1 + 2f2 + 3f3 = 0___1+ 22+ 3=
01+ 32+ 23= 01+ 2+ 3= 0Ma trn ha,A =__1 2 1 01 3 2 01 1 1 0____1 2
1 00 1 1 00 0 1 0__LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 35/863. CsvschiucakhnggianvectH c nghim duy nht1 = 2 = 3
= 0. VySc lp tuyntnh. Suy raSl c s ca R2[x].Cch 2. Kim traSl tp
sinh.Xemli v dVd. Trong khng gian R3, choS = u1 = (1, m2, 2), u2 =
(m1, 3, 3), u3 = (m, m + 2, 2).Tm iu kimm Sl c s ca R3.Gii. Do s
phn t caSbng 3 nnSl c s ca R3khiSc lptuyn tnh.LpA =__u1u2u3__=__1
m2 2m1 3 3m m + 2 2__. Ta c detA = mm2.Suy ra,Sc lp tuyn tnh khi
detA ,= 0. Nh vy, Sl c s caR3thm ,= 0 vm ,= 1.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 36/864.
Khnggianvectcon4.Khnggianvectcon4.1nhngha4.2Khnggiansinhbitphp4.3Khnggiandngcamatrn4.4KhnggiantngLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 37/864.
Khnggianvectcon4.1nhnghanhngha. ChoWl mt tp con khc caV . Ta niWl
mtkhnggianvectcon(gi tt,khnggiancon) caV , k hiuW V, nuWvi php ton
(+, .) c hn ch tVcng l mtkhng gian vect trn R.Vd. W= 0 vVl cc vect
con caV . Ta gi y l cckhnggiancontmthngcaV .nhl. ChoWl mt tp con
khc caV . Khi cc mnh sau tng ng:i) W V .ii) Vi miu, v W; R, ta cu +
v Wvu W.iii) Vi miu, v W; R, ta cu + v W.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 38/864. KhnggianvectconVd. ChoW=_(x1, x2,
x3) R3[ 2x1 + x2 x3 = 0 . HiWc lkhng gian con ca R3khng?Gii.Ta cW
R3.0 = (0, 0, 0) W(v 2.0 + 0 0 = 0). Suy raW ,= .Vi miu = (x1, x2,
x3) W, ngha l 2x1 + x2 x3 = 0,v = (y1, y2, y3) Wngha l 2y1 + y2 y3
= 0v R. Ta c u + v = (x1 + y1, x2 + y2, x3 + y3). Ta c2(x1 + y1) +
(x2 + y2) (x3 + y3) =(2x1 + x2 x3) + (2y1 + y2 y3) = 0 + 0 = 0.Suy
rau + v W. (1) u = (x1, x2, x3). Ta c2x1 + x2 x3 = (2x1 + x2 x3) =
0 = 0.Suy rau W. (2)T (1) v (2) suy raW R3.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 39/864. KhnggianvectconNhnxt. ChoVl
khng gian vect vW V. Khi :NuWl khng gian con caVth 0 W.Nu 0/
WthWkhng l khng gian con caV..Vd. ChoW=_(x1, x2, x3) R3[ 3x1 + 2x2
4x3 = 1 . HiWcl khng gian con ca R3khng?Gii. Ta c 0 = (0, 0, 0)/
W(v 3.0 + 2.0 4.0 = 0 ,= 1). Suy raWkhng l khng gian con ca R3.Vd.
ChoW=_(x1, x2, x3) R3[ x1 = 2x2x3 . HiWc l khnggian con ca
R3khng?Gii. Viu = (2, 1, 1) vv = (4, 2, 1). Ta cu, v W.u + v = (6,
3, 2)/ W(v 6 ,= 2.3.2). Suy raWkhng l khng giancon ca R3.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 40/864.
Khnggianvectconnhl. NuW1, W2l khng gian con caVthW1 W2cng lmt khng
gian con caV .Chngminh. W1 W2 V(vW1 V ,W2 V )0 W1 W2 (v 0 W1, 0
W2)Vi miu, v W1 W2; R.Vu, v W1 nnu + v W1 (vW1 V ).Vu, v W1 nnu + v
W2 (vW2 V ).Suy rau + v W1 W2.VyW1 W2 V.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 41/864. Khnggianvectconnhl. NuW1, W2l khng
gian con caV, ta nh nghaW1 + W2 = w1 + w2 [w1 W1, w2 W2 .Khi W1 +
W2cng l mt khng gian con caV .Chngminh. W1 + W2 V(vW1 V ,W2 V )0 =
0 + 0 W1 + W2 (v 0 W1, 0 W2)Vi miu = u1 + u2, v = v1 + v2 W1 + W2;
R.Vu1, v1 W1 nnu1 + v1 W1 (vW1 V ).Vu2, v2 W1 nnu2 + v2 W2 (vW2 V
).Ta cu+v = (u1+u2)+(v1+v2) = (u1+v1)+(u2+v2) W1+W2.Vyu + v W1 +
W2.VyW1 + W2 V.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010
42/864. Khnggianvectcon4.2Khnggianconsinhbitphpnhl. ChoVl khng gian
vect trn R vSl tp con khc rngcaV. Ta tWl tp hp tt c cc t tuyn tnh
caS. Khi :i) W V.ii) Wl khng gian nh nht trong tt c cc khng gian
con caVm chaS.Khng gianWc gi l khng gian con sinh biS, k hiu W =
S).C th, nuS = u1, u2, . . . , um thW= S) = 1u1 + 2u2 + + mum [i
RVd. Trong khng gian R2, ta xtS = u = (1, 2). Khi W= S) = a(1, 2)
[a R = (a, 2a) [a R.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 43/864. KhnggianvectconVd. Trong khng gian R3, ta xtS =
u1 = (1, 2, 1), u2 = (1, 2, 0).Khi S) = tu1 + su2 [t, s R = (t s,
2t + 2s, t) [t, s RNhnxt. V khng gian sinh biSl khng gian nh nht
chaSnn ta quy c ) = 0.Vd. Trong khng gian R3, choW= (a + 2b, a b, a
+ 2b) [a, b Ra) Chng minhWl khng gian con ca R3.b) Tm mt tp sinh
caW.Gii. a) Ta c 0 Wv 0 = (0, 0, 0) = (0 + 2.0, 0 0, 0 +
2.0)LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 44/864.
KhnggianvectconViu, v Wv R,u = (a1 + 2b1, a1 b1, a1 + 2b1) via1, b1
Rv = (a2 + 2b2, a2 b2, a2 + 2b2) via2, b2 R. Khi : u + v = ( (a1 +
a2) + 2(b1 + b2), (a1 + a2) (b1 + b2), (a1 + a2) + 2(b1 + b2) )
W(va1 + a2, b1 + b2 R). u = (a1 + 2b1, a1 b1, a1 + 2b1) W(va1, b1
R).Vyu + v, u W. Suy raW R3.b) Ta cW= (a + 2b, a b, a + 2b) [a, b
R= a(1, 1, 1) + b(2, 1, 2) [a, b RV mi vect thucWu l t hp tuyn tnh
cau1 = (1, 1, 1), u2 = (2, 1, 2) nnS = u1, u2 l tp sinh caW.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 45/864.
Khnggianvectconnhl. ChoVl khng gian vect vS1, S2l tp con caV . Khi,
nu mi vect caS1u l t hp tuyn tnh caS2v ngc lith S1) = S2)Chngminh.
V mi vect caS1 u l t hp tuyn tnh caS2nnS1 S2). Mt khc S1) l khng
gian nh nht chaS1 nnS1) S2). L lun tng t ta c S2) S1).Vd. Trong
khng gian R3choS1 = u1 = (1, 1, 4), u2 = (2, 1, 3),S2 = u3 = (1, 2,
1), u4 = (5, 1, 10).Chng minh S1) = S2).LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 46/864. Khnggianvectconnhl. [vcskhngtonvn]
ChoVl mt khng gian vecthu hn chiu vSl mt tp con c lp tuyn tnh caV .
Khi ,nuSkhng l c s caVth c th thm voSmt s vect cmt c s caV .Vd.
Trong khng gian R4, choS = u1 = (1, 0, 2, 1, u2 = (1, 0, 4, 4).Chng
tSc lp tuyn tnh v thm voSmt s vect Strthnh c s ca R4.Gii. LpA
=_u1u2_=_1 0 2 11 0 4 4__1 0 2 10 0 2 3_.Ta cr(A) = 2 bng s vect
caS. Suy raSc lp tuyn tnh.Da voA ta c th thm voShai vectu3 = (0, 1,
0, 0), u4 = (0, 0, 0, 1).R rngS = u1, u2, u3, u4 ltt. Suy raSl c s
ca R4.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 47/864.
Khnggianvectconnhl. ChoVl mt khng gian vect hu hn chiu sinh biS.Khi
tn ti mt c s BcaVsao cho B S. Ni cch khc, nuSkhng phi l mt c s
caVth ta c th loi b ra khiSmt svect c mt c s caV .Vd. Trong khng
gian R3, choWsinh biS = u1 = (1, 1, 1), u2 = (2, 1, 3), u3 = (1, 2,
0).Tm mt tp con caS l c s caW.Gii. Xt phng trnh1u1 + 2u2 + 3u3 = 0
( + 22 + 3, + 2 + 23, + 32) = (0, 0, 0)___1+ 22+ 3= 01+ 2+ 23= 01+
32= 0LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 48/864.
KhnggianvectconMa trn ha h phng trnh,A =__1 2 11 1 21 3 0____1 2 10
1 10 1 1____1 0 30 1 10 0 0__.Suy ra h c nghim l1 = 3t, 2 = t, 3 =
t. Vy3tu1 + tu2 + tu3 = 0.Chot = 1, ta c 3u1 + u2 + u3 = 0 nnu2 =
3u1 u3.Suy rau2 l t hp tuyn tnh cau1, u3. Do u1, u3 l tp sinhcaW,
hn na n c lp tuyn tnh nn n l c s caW.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 49/864.
Khnggianvectcon4.3Khnggiandngcamatrnnhngha. Cho ma trnA = (aij)
Mmn(R)A =____a11a12. . . a1na21a22. . . a2n. . . . . . . . . . . .
. . . . . . . .am1am2. . . amn____.tu1= (a11, a12, . . . , a1n);u2=
(a21, a22, . . . , a2n);. . . . . . . . . . . . . . . . . . . . . .
. . . . . .um= (am1, am2, . . . , amn)vWA = u1, u2, . . . , um).Ta
giu1, u2, . . . , um l ccvectdngcaA, vWA lkhnggiandngcaA.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 50/864. KhnggianvectconB.
NuA vBl hai ma trn tng ng dng thWA = WB,ngha l hai ma trn tng ng
dng c cng khng gian dng.nhl. Gi sA Mmn(R). Khi , dimWA = r(A) v tp
hpcc vect khc khng trong dng ma trn bc thang caA l c s caWA.Vd. Tm
s chiu v mt c s ca khng gian dng ca ma trnA =____1 2 1 12 5 1 45 11
2 89 20 3 14____.Gii. A =____1 2 1 12 5 1 45 11 2 89 20 3
14________1 2 1 10 1 3 20 0 0 10 0 0 0____.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 51/864. KhnggianvectconSuy ra dimWA
= r(A) = 3 v mt c s caWA lu1 = (1, 2, 1, 1); u2 = (0, 1, 3, 2); u3
= (0, 0, 0, 1).ThuttontmschiuvcscamtkhnggianconcaRnkhibitmttpsinhGi
sW= u1, u2, . . . , um) Rn(u1, u2, . . . , um khng nht thitc lp
tuyn tnh). tm s chiu v mt c s caWta tin hnhnh sau:Bc1. Lp ma trnA
bng cch xpu1, u2, . . . , um thnh cc dng.Bc2. Dng cc php BSCTD aA v
dng bc thangR.Bc3. S chiu caWbng s dng khc 0 caR (do bngr(A)) v cc
vect dng khc 0 caR to thnh mt c s caW.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 52/864. KhnggianvectconVd. ChoWsinh biS =
u1, u2, u3, u4 trong u1 = (1, 2, 1, 1);u2 = (3, 6, 5, 7);u3 = (4,
8, 6, 8);u4 = (8, 16, 12, 20). Tm mt c s cakhng gianW.Gii. LpA
=____u1u2u3u4____=____1 2 1 13 6 5 74 8 6 88 16 12 20________1 2 1
10 0 1 20 0 0 10 0 0 0____.Do Wc dimW= 3 v c mt c sv1 = (1, 2, 1,
1); v2 = (0, 0, 1, 2); v3 = (0, 0, 0, 1).Nhnxt. V dimW= 3, hn na, c
th kim chngu1, u2, u4 clp tuyn tnh nn ta cng c u1, u2, u3 l mt c s
caW.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 53/864.
KhnggianvectconVd. Tm mt c s cho khng gian con ca R4sinh bi cc
vectu1, u2, u3, trong u1 = (1, 2, 1, 3);u2 = (2, 4, 3, 0);u3 = (3,
6, 4, 4).Gii. LpA =__u1u2u3__=__1 2 1 32 4 3 03 6 4 4____1 2 1 30 0
1 60 0 0 1__.Do c dimW= 3 v c mt c sv1 = (1, 2, 1, 3); v2 = (0, 0,
1, 6); v3 = (0, 0, 0, 1).Nhnxt. Trong v d trn, vr(A) = 3 nnu1, u2,
u3 c lp tuyntnh, v do u1, u2, u3 cng l mt c s caW.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 54/864.
Khnggianvectcon4.4Khnggiantngnhl. ChoVl khng gian vect trn R vW1,
W2l khng giancon caV . Khi :i) W1 + W2l khng gian con caV.ii) NuW1
= S1) vW2 = S2) thW1 + W2 = S1 S2).Vd. Trong khng gian R4cho cc
vectu1 = (1, 2, 1, 1);u2 = (3, 6, 5, 7);u3 = (4, 8, 6, 8);u4 = (8,
16, 12, 16);u5 = (1, 3, 3, 3);u6 = (2, 5, 5, 6);u7 = (3, 8, 8,
9);u8 = (6, 16, 16, 18).tW1 = u1, u2, u3, u4) vW2 = u5, u6, u7,
u8). Tm mt c s vxc nh s chiu ca mi khng gianW1, W2 vW1 +
W2.Gii.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 55/864.
Khnggianvectcon Tm c s caW1LpA1 =____u1u2u3u4____=____1 2 1 13 6 5
74 8 6 88 16 12 16________1 2 1 10 0 1 20 0 0 00 0 0 0____.Do W1 c
s chiu l 2 v mt c s lv1 = (1, 2, 1, 1); v2 = (0, 0, 1, 2). Tm c s
caW2LpA2 =____u5u6u7u8____=____1 3 3 32 5 5 63 8 8 96 16 16
18________1 3 3 30 1 1 00 0 0 00 0 0 0____.Do W2 c s chiu l 2 v mt
c s lv3 = (1, 3, 3, 3); v4 = (0, 1, 1, 0)LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 56/864. Khnggianvectcon Tm c s caW1 + W2Ta
cW1 + W2 sinh bi cc vectv1 = (1, 2, 1, 1); v2 = (0, 0, 1, 2); v3 =
(1, 3, 3, 3); v4 = (0, 1, 1, 0).LpA =____v1v2v3v4____=____1 2 1 10
0 1 21 3 3 30 1 1 0________1 2 1 10 1 1 00 0 1 20 0 0 0____.Suy
raW1 + W2 c s chiu l 3 v mt c s lw1 = (1, 2, 1, 1); w2 = (0, 1, 1,
0); w3 = (0, 0, 1, 2).LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 57/865.
Khnggiannghimcahphngtrnhtuyntnh5.Khnggiannghimcahphngtrnhtuyntnh5.1Mu5.2Tmcscakhnggiannghim5.3KhnggiangiaoLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 58/865.
Khnggiannghimcahphngtrnhtuyntnh5.1MuVd. ChoWl tp tt c cc nghim (x1,
x2, x3, x4) ca h phngtrnh tuyn tnh thun nht:___x1+ 2x2 3x3+ 5x4=
0;x1+ 3x2 13x3+ 22x4= 0;3x1+ 5x2+ x3 2x4= 0;2x1+ 3x2+ 4x3 7x4= 0.Ma
trn ha h phng trnh, ta cA =____1 2 3 51 3 13 223 5 1 22 3 4
7________1 0 17 290 1 10 170 0 0 00 0 0 0____.Vy h c nghim l(x1,
x2, x3, x4) = (17t + 29s, 10t 17s, t, s) vit, s RLVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 59/865.
KhnggiannghimcahphngtrnhtuyntnhDo W = (17t + 29s, 10t 17s, t, s) [
t, s R= (17t, 10t, t, 0) + (29s, 17s, 0, s) [ t, s R= t(17, 10, 1,
0) + s(29, 17, 0, 1) [ t, s Rtu1 = (17, 10, 1, 0), u2 = (29, 17, 0,
1). Theo biu thc trn, viu Wthu l t hp tuyn tnh cau1, u2. Suy raW=
u1, u2).Hn na u1, u2 c lp tuyn tnh, nn u1, u2 l c s caW. Suyra
dimW= 2.Nhnxt. Vectu1 vu2 c c bng cch cho ln ltt = 1, s = 0vt = 0,
s = 1. Ta gi nghimu1, u2 c gi lnghimcbncah.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 60/865.
Khnggiannghimcahphngtrnhtuyntnhnhl. GiWl tp hp nghim (x1, x2, . . .
, xn) ca h phngtrnh tuyn tnh thun nht___a11x1+ a12x2+ . . . +
a1nxn= 0;a21x1+ a22x2+ . . . + a2nxn= 0;. . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.am1x1+ am2x2+ . . . + amnxn= 0.Khi , W l khng gian con ca Rnv s
chiu caWbng s n tdo ca h.Nh vyW= u Rn[Au
= 0 viA l ma trn cho trc vu = (x1, x2, . . . , xn)LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 61/865.
Khnggiannghimcahphngtrnhtuyntnh5.2TmcscakhnggiannghimThuttonBc1.
Gii h phng trnh, tm nghim tng qut.Bc2. Ln lt cho b n t do cc gi
tr(1, 0, . . . , 0), . . . , (0, 0, . . . , 1) ta c cc nghim c
bnu1, u2, . . . , um.Bc3. Khi khng gian nghim c c s l u1, u2, . . .
, um.Vd. Tm c s v s chiu ca khng gian nghim sau___x1+ 2x2 3x3+ 5x4=
0;x1+ 3x2 13x3+ 22x4= 0;3x1+ 5x2+ x3 2x4= 0;2x1+ 3x2+ 4x3 7x4=
0,Gii. Ma trn ha h phng trnh, ta cLVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 62/865. KhnggiannghimcahphngtrnhtuyntnhA
=____1 2 3 51 3 13 223 5 1 22 3 4
7____d2:=d2d1d3:=d33d1d4:=d42d1____1 2 3 50 1 10 170 1 10 170 1 10
17____d1:=d12d2d3:=d3+d2d4:=d4+d2____1 0 17 290 1 10 170 0 0 00 0 0
0____.Suy ra nghim ca h lu = (x1, x2, x3, x4) = (17t + 29s, 10t
17s, t, s) vit, s R.Cc nghim c bn ca h lu1 = (17, 10, 1, 0), u2 =
(29, 17, 0, 1).Do , nu W l khng gian nghim th B = u1, u2 c s ca W
vdimW= 2.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 63/865.
Khnggiannghimcahphngtrnhtuyntnh5.3KhnggiangiaoChoVl khng gian vect
vW1, W2 l khng gian con caV. Khi W1 W2 l khng gian con caV. Hn na
nuW1 = S1), W2 = S2)thu W1 W2 khi v ch khiu l t hp tuyn tnh caS1 vu
l thp tuyn tnh caS2.Vd. Trong khng gian R4cho cc vectu1 = (1, 2, 1,
1),u2 = (1, 2, 2, 3),u3 = (2, 4, 3, 4),u4 = (1, 3, 3, 3),u5 = (0,
1, 1, 0). tW1 = u1, u2, u3),W2 = u4, u5). Tm c s ca khng gianW1
W2.Gii. Giu = (x, y, z, t) W1 W2 Vu W1 nnu l t hp tuyn tnh cau1,
u2, u3._u
1u
2u
3u
_=____1 1 2 x2 2 4 y1 2 3 z1 3 4 t________1 1 2 x0 1 1 x z0 0 0
2x + y0 0 0 x 2z + t____Suy ra u W1 th 2x + y = 0 vx 2z + t = 0
(1)LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 64/865.
Khnggiannghimcahphngtrnhtuyntnh Vu W2 nnu l t hp tuyn tnh cau4,
u5._u
4u
5u
_=____1 0 x3 1 y3 1 z3 0 t________1 0 x0 1 3x + y0 0 y + z0 0 3x
+ t____Suy ra u W2 th y + z = 0 v 3x + t = 0 (2)T (1) v (2) ta
c___2x + y = 0x 2z + t = 0 y + z = 03x + t = 0Ma trn haA =____2 1 0
01 0 2 10 1 1 03 0 0 1____LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 65/865. KhnggiannghimcahphngtrnhtuyntnhA =____2 1 0 01 0
2 10 1 1 03 0 0 1____d1:=d1d4d2:=d2d1d4:=d43d1____1 1 0 10 1 2 20 1
1 00 3 0 2____d2:=d2d1:=d1d2d3:=d3d2d4:=d43d2____1 0 2 10 1 2 20 0
3 20 0 6 4____d3:=13d3d1:=d1+2d3d2:=d22d3d4:=d4+6d3____1 0 0 1/30 1
0 2/30 0 1 2/30 0 0 0____Suy ra nghim ca h lu = (x, y, z, t) =
(13a, 23a, 23a, a) via R.Nghim c bn ca h lu1 = (13, 23, 23, 1).Suy
raW1 W2 c c s l u1 = (13, 23, 23, 1).LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 66/865. Khnggiannghimcahphngtrnhtuyntnhnh
l. ChoW1, W2l hai khng gian con hu hn chiu caV . Khidim(W1 + W2) =
dimW1 + dimW2 dim(W1 W2).LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 67/866.
Tavmatrnchuyncs6.Tavmatrnchuyncs6.1Ta6.2MatrnchuyncsLVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 68/866.
Tavmatrnchuyncs6.1Tanhngha. ChoVl khng gian vect v B = u1, u2, . .
. , un lmt c s caV . Khi B c gi lcscspcaVnu tht cc vect trong B c c
nh. Ta thng dng k hiu(u1, u2, . . . , un) ch c s c sp theo th tu1,
u2, . . . , un.nhl. Cho B = (u1, u2, . . . , un) l c s caV . Khi mi
vectu Vu c biu din mt cch duy nht di dngu = 1u1 + 2u2 + +
nun.Chngminh.S tnti. V B l c s caVnn B l tp sinh. Viu Vthul t hp
tuyn tnh ca B. Suy ra, tn ti1, 2, . . . n R u = 1u1 + 2u2 + +
nun.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 69/866.
TavmatrnchuyncsS duynht.Gi su c mt dng biu din khcu = 1u1 + 2u2 + +
nun.Ngha l:u = 1u1 + + nun = 1u1 + 2u2 + + nun.Khi (1 1)u1 + (2
2)u2 + + (n n)un = 0.Do B l c s nn B c lp tuyn tnh, ta c1 1 = 2 2
== n n = 0hay1 = 1, 2 = 2, . . . , n = n.iu ny chng tu c mt dng biu
din duy nht.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010
70/866. TavmatrnchuyncsTaNh vy, nu B = (u1, u2, . . . , un) l c s
caVvu Vthu s cdng biu din duy nht l:u = 1u1 + 2u2 + + nun.Ta t[u]B
=_____12...n_____.Khi [u]Bc gi ltacau theo c s B.Vd. Trong khng
gian R3, ta c c s chnh tcB0 = e1 = (1, 0, 0), e2 = (0, 1, 0), e3 =
(0, 0, 1).Viu = (x1, x2, x3) ta c:u = x1e1 + x2e2 + x3e3.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 71/866.
TavmatrnchuyncsSuy ra [u]B0 =__x1x2x3__= u
.Nhnxt. i vi c s chnh tc B0 = (e1, e2, . . . , en) ca khnggian
Rnvu = (x1, x2, . . . , xn) Rnta c[u]B0 =_____x1x2...xn_____= u
.Vd. Khng gianR2[x] c c s chnh tc lB0 = x2, x, 1.Vif= ax2+ bx +
c, ta c [f]B0 =__abc__.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 72/866. TavmatrnchuyncsPhngphptm[u]BChoVl khng gian vect
c c s l B = (u1, u2, . . . , un) vu V . tm [u]Bta i gii phng trnhu
= 1u1 + 2u2 + + nun()vi n1, 2, . . . n R. Do B l c s nn phng trnh
() c nghimduy nht(1, 2, . . . , n) = (c1, c2, . . . , cn).Khi [u]B
=_____c1c2...cn_____.Lu. KhiV= Rn, gii phng trnh () ta lp h(u
1u
2. . . u
n [u
)LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 73/866.
TavmatrnchuyncsVd. Trong khng gian R3, cho cc vectu1 = (1, 2, 1),
u2 = (1, 3, 1), u3 = (2, 5, 3).a) Chng minh B = (u1, u2, u3) l mt c
s ca R3.b) Tm ta ca vectu = (a, b, c) R3theo c s B.Gii.a) LpA
=__u1u2u3__=__1 2 11 3 12 5 3__. Ta c |A|=1, suy rau1, u2, u3c lp
tuyn tnh. Vy B l c s ca R3.b) Viu = (a, b, c), tm [u]Bta lp h phng
trnh(u
1u
2u
3[u
) __1 1 2 a2 3 5 b1 1 3 c____1 0 0 4a b c0 1 0 a + b c0 0 1 a +
c__Vy [u]B =__4a b ca + b ca + c__.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 74/866. TavmatrnchuyncsVd. Trong khng
gianR2[x] chof1 = x2+ x + 1, f2 = 2x2+ 3x + 1, f3 = x2+ 2x + 1.a)
Chng minh B = (f1, f2, f3) l mt c s ca R2[x].b) Tm ta ca vectf= x2+
3x + 3 theo c s B.c) Cho [g]B =__234__, tm g?Gii.a) Kim tra B l c s
(t lm)b) Vif= x2+ 3x + 3, tm [f]Bta i gii phng trnhf= 1f1 + 2f2 +
3f3___1+ 22+ 3= 1;1+ 32+ 23= 3;1+ 2+ 3= 3.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 75/866. TavmatrnchuyncsMa trn ha,A =__1 2 1
11 3 2 31 1 1 3____1 0 0 10 1 0 20 0 1 4__Vy [f]B =__124__.c) Ta c
[g]B =__234__, suy rag = 2f1 + 3f2 4f3g = 2(x2+ x + 1) + 3(2x2+ 3x
+ 1) 4(x2+ 2x + 1)= 4x2+ 3x + 1.Mnh. Cho Bl c s caV. Khi , vi miu,
v V, R tac:[u + v]B = [u]B + [v]B.[u]B = [u]B.LVnLuyn (HKHTNHCM)
Chng3. Khnggianvect 22/05/2010 76/866.
Tavmatrnchuyncs6.2Matrnchuyncsnhngha. ChoVl mt khng gian vect vB1 =
(u1, u2, . . . , un), B2 = (v1, v2, . . . , vn).l hai c s caV . tP=
([v1]B1[v2]B1 . . . [vn]B1).Khi Pc gi lmatrnchuyncst c s B1 sang c
sB2 v c k hiu (B1 B2).Vd. Trong khng gian R3, choB = (u1 = (1, 2,
3), u2 = (2, 3, 1), u3 = (3, 1, 3))l c s ca R3. Gi B0 l c s chnh tc
ca R3. Khi (B0 B) = ([u1]B0[u2]B0[u3]B0) = (u
1u
2u
3 ) =__1 2 32 3 13 1 3__LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 77/866. TavmatrnchuyncsNhnxt. Nu B = (u1, u2, . . . ,
un) l mt c s ca Rnv B0l cs chnh tc ca Rnth(B0 B) = (u
1u
2. . . u
n)Phngphptm(B1 B2)Gi s B1 = (u1, u2, . . . un) v B2 = (v1, v2, .
. . vn) l hai c s caV.Ta thc hin nh sau:Chou l vect bt k caV , xc
nh [u]B1.Ln lt thay thu bngv1, v2, . . . vn ta xc nh c[v1]B1,
[v2]B1, . . . , [vn]B1.Khi (B1 B2) = ([v1]B1[v2]B1 . . .
[vn]B1)LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 78/866.
Tavmatrnchuyncsc bit, khiV= Rn, xc nh (B1 B2) ta c th lm nh
sau:Thnh lp ma trn m rng (u
1u
2. . . u
n [v
1v
2. . . v
n )Dng cc php bin i s cp trn dng a ma trn trn vdng (In[P).Khi
(B1 B2) = P.Vd. Trong khng gian R3, cho hai c sB1 = (u1 = (1, 1,
1), u2 = (1, 2, 1), u3 = (2, 3, 1))vB2 = (v1 = (1, 3, 2), v2 = (1,
2, 4), v3 = (3, 3, 2)).Tm ma trn chuyn c s t B1 sang B2.Gii. Chou =
(a, b, c) R3, xc nh [u]B1. Ta lp(u
1u
2u
3[u
) __1 1 2 a1 2 3 b1 1 1 c____1 0 0 a b + c0 1 0 2a + b + c0 0 1
a c__LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 79/866.
TavmatrnchuyncsNh vy [u]B1 =__a b + c2a + b + ca c__.Thay ln ltu
biv1, v2, v3 ta c[v1]B1 =__631__, [v2]B1 =__545__, [v3]B1
=__255__.Vy (B1 B2) =__6 5 23 4 51 5 5__.CchkhcLp ma trn m
rng(u
1u
2u
3[v
1v
2v
3 ) __1 1 2 1 1 31 2 3 3 2 31 1 1 2 4 2____1 0 0 6 5 20 1 0 3 4
50 0 1 1 5 5__. Suy ra (B1 B2) =__6 5 23 4 51 5 5__.LVnLuyn
(HKHTNHCM) Chng3. Khnggianvect 22/05/2010 80/866.
Tavmatrnchuyncsnhl. ChoVl mt khng gian vect hu hn chiu v B1, B2,
B3l ba c s caV . Khi i) (B1 B1) = In.ii) u V, [u]B1 = (B1
B2)[u]B2.iii) (B2 B1) = (B1 B2)1.iv) (B1 B3) = (B1 B2)(B2 B3).Hqu.
Cho B1 = (u1, u2, . . . , un); B2 = (v1, v2, . . . , vn) l hai c
sca khng gian Rn. Gi B0l c s chnh tc ca Rn. Ta ci) (B0 B1) = (u
1u
2. . . u
n).ii) (B1 B0) = (B0 B1)1.iii) u V, [u]B1 = (B0 B1)1[u]B0.iv)
(B1 B2) = (B0 B1)1(B0 B2).LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 81/866. TavmatrnchuyncsVd. ChoWl khng gian con ca R4sinh
bi cc vect:u1 = (1, 2, 2, 1), u2 = (0, 2, 0, 1), u3 = (2, 3, 4,
1).a) Chng minh B = (u1, u2, u3) l mt c s caW.b) Chou = (a, b, c,
d), tm iu kin u W. Khi tm [u]B.c) Chov1 = (1, 0, 2, 0); v2 = (0, 2,
0, 1); v3 = (0, 0, 0, 1). Chng minhB
= (v1, v2, v3) cng l mt c s caW. Tm ma trn chuyn c s tB sang
B
.Gii.a) Chng minh B = (u1, u2, u3) l mt c s caW.LpA
=__u1u2u3__=__1 2 2 10 2 0 12 3 4 1__. Ta cr(A) = 3, suy ra Bc lp
tuyn tnh. VW= B) nn B l c s caW.LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 82/866. Tavmatrnchuyncsb) Chou = (a, b, c,
d), tm iu kin u W. Khi tm [u]B.Ta cu Wkhiu l t hp tuyn tnh ca B.Lp
h phng trnhu
1u
2u
3[u
) ____1 0 2 a2 2 3 b2 0 4 c1 1 1 d________1 0 0 a + 2b 4d0 1 0 a
3b + 7d0 0 1 b 2d0 0 0 2a + c____Da vo h phng trnh, u Wth 2a + c =
0. Suy ra[u]B =__a + 2b 4da 3b + 7db 2d__LVnLuyn (HKHTNHCM) Chng3.
Khnggianvect 22/05/2010 83/866. Tavmatrnchuyncsc) Chov1 = (1, 0, 2,
0); v2 = (0, 2, 0, 1); v3 = (0, 0, 0, 1). Chng minhB
= (v1, v2, v3) cng l mt c s caW. Tm ma trn chuyn c s tB sang
B
.Ta thy cc vectv1, v2, v3 u tha iu kin 2a + c = 0 nn theocua),
cc vect ny thucW.Mt khc, d thy rng B
= (v1, v2, v3) c lp tuyn tnh nn B
cngl c s caW(do dimW= [B[ = 3 = [B
[ ). Dng kt qu cub) tac[v1]B =__110__, [v2]B =__010__, [v2]B
=__472__Suy ra (B B
) =__1 0 41 1 70 0 2__.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 84/866. TavmatrnchuyncsVd. Trong khng gian R3, choS =
(u1 = (1, 1, 3), u2 = (1, 2, 1), u3 = (1, 1, 2))T= (v1 = (1, 2, 2),
v2 = (1, 2, 1), v3 = (1, 1, 2))a) Chng tSvTl c s ca R3.b) Tm ma trn
chuyn c s tSsangT(k hiu (S T)).c) Chou R3tha [u]T=__232__. Tnh
[u]S.Chng tSvTl c s ca R3Ta c dimR3= 3 = s vec t caS. Do ,S l c s
ca R3khiS clp tuyn tnh. LpA =__u1u2u3__=__1 1 31 2 11 1 2__. Ta
cr(A) = 3,suy raSc lp tuyn tnh. VySl c s ca R3.Lm tng t
choT.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect 22/05/2010 85/866.
Tavmatrnchuyncsb) Tm ma trn chuyn c s tSsangT(k hiu (S T))Lp ma trn
m rng(u
1u
2u
3[v
1v
2v
3 ) __1 1 1 1 1 11 2 1 2 2 13 1 2 2 1 2____1 0 0 1 0 00 1 0 1 1
00 0 1 3 0 1__. Suy ra (S T) =__1 0 01 1 03 0 1__c) Chou R3tha
[u]T=__232__. Tnh [u]S.Ta c [u]S = (S T)[u]T=__1 0 01 1 03 0
1____232__=__254__.LVnLuyn (HKHTNHCM) Chng3. Khnggianvect
22/05/2010 86/86
