khong gian vecto.10.12

8/13/2019 Khong Gian Vecto.10.12

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Khng gian vect


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Vect

Vecto n chiu l b n s thc: (x1, x2,,xn)T

vi xi R. Rn =tp tt c cc vectn chiu =

{(x1,x2,,xn)T|xiR}. Trn R

nta nh ngha

cc php ton: Cng 2 vect: (x1,x2,,xn)

T+(y1,y2,,yn)T=

(x1+y1,x2+y2,,xn+yn)T.

Nhn s vi vect:k.(x1,x2,,xn)=(kx1,kx2,,xn)T.

Cng php ton, Rn l mt khng gian vect.


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Cho X=(1,3,7)T v Y=(-3,2,6)T. Tnh

3X+2Y

XY


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c lp tuyn tnh Hvect{v1,v2,,vn} c lp tuyn tnh nu

k1v1+k2v2+knvn= khi v chkhik1=k2==kn=0.

V d: Cho v1=(1,2,3)T, v2=(5,6,7)

T. CMR

{v1,v2} c lp tuyn tnh. Nhn xt: nu hvect{v1,v2,,vn} c lp

tuyn tnh hphng trnh

k1v1+k2v2+knvn= c nghim duy nht rank{v1,v2,,vn}=n.


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Ph thuc tuyn tnh

Hvect{v1, v

2, ,v

n} ph thuc tuyn tnh nu

chng khng c lp tuyn tnh.

Nhn xt: hvect{v1, v2, ,vn} ph thuctuyn tnh hphng trnh

k1v1+k2v2++knvn= c v s nghim rank{v1,v2,,vn}


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Hcon c lp tuyn tnh cc i Cho hvectA, v B l mt b phn (h con)

ca A. Ta ni h con B l c lp tuyn tnhcc i trong A nu:

B c lp tuyn tnh

Vi uA-B (nu c) th B{u} s l ph thuctuyn tnh.

Cho A={v1=(1,2)T, v2=(2,4)

T, v3=(0,1)T}. CM

B={v1,v3} clp tuyn tnh.


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Hng ca hvect Tha nhn rng mi hcon c lp tuyn

tnh cc i trong hvect{v1,v2,,vn} uc cng slng vect.

Hng ca mt hvectl svectca h

con c lp tuyn tnh cc i. V d: Cho A={v1=(1,2)

T, v2=(2,4)T, v3=(0,1)

T}.

Tm hng ca hvectny.

NX: hng ca hvect{v1,v2,,vn}=rank matrn [v1,v2,,vn].


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H

sinh

V l khng gian vectv S={v1,v2,,vn} l bphn ca V. Gi s mi vectvV u c thbiu din di dng t hp ca cc vecttrongS, tc l v c th vit c di dng v= 1 v1+

2v2++ nvn. Khi ta ni khng gian V sinhbi tp S.

Ta cn ni hvectS l h sinh ca khng gian

vectV.


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Khng gian con

Cho tp hp con V ca Rn

. V c gi lkhng gian con ca Rn nu:

Vi mi x,yV th u+vV

Vi k

R, x

V th kv

V. V d: Tp hp W={v=(x1 ,x2)

T |x1+x2=0} l

khng gian vectcon ca R2.


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Cs v s chiu

Cs ca khng gian V l hcon B c lptuyn tnh cc i trong V.

Mi khng gian vectc th c nhiu cs.

V d: CM hE={e1=(1,0,0)

T,e2=(01,0)T,e3=(0,0,1)

T} l csca khng gian vectR3.


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S chiu ca khng gian V l svectcacs B. K hiu dim V.

NX: s chiu ca Rn =dim Rn=n.

V d: Cho V={v=(x1,x2,x3,0)T

| xiR} l khnggian con ca R4. Tm dim V.


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T hp tuyn tnh Cho A={v1,v2,,vn} R

n v vRn. Ta ni v l

t hp tuyn tnh ca A nu v biu din cdi dng: v=k1v1+k2v2++knvn.

NX: nu v l t hp tuyn tnh ca

A={v1,v2,,vn} th A{v} l ph thuc tuyntnh.

NX: nu v l t hp tuyn tnh ca

A={v1,v2,,vn} th hphng trnhv=k1v1+k2v2++knvn c nghim.


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Bi tp Mt cng ty dng phng tin TV, i radio,

bo qung co sn phm. H thy rngtng ng c: 30000, 10000, v 20000ngi ch qung co sn phm vi chi ph

1 triu/ln mi phng tin. Lp phng n dng c3 phng tin c

4 triu ngi ch sn phm.

Lp phng n dng c3 phng tin sdng ht 120 triu.

Lp phng n tha mn chai iu trn.


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Cho V={v=(x1,x2,x3,2x3)T|xiR}.

CM V l khng gian con ca R4. Tm dim V.

Tm s chiu v cs ca khng gian V sinh

bi: {v1=(1,0,0,-1)T, v2=(2,1,1,0)T,v3=(1,1,1,1)

T,v4=(1,2,3,4)T, v5=(0,1,2,3)

T}

Bin lun theo m s chiu v cs ca

khng gian sinh bi {v1=(1,-1,1,0)T, v2=(-1,1,6,m)T, v3=(5,0,1,-1)

T, v4=(5,7,8,9)T}.


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NX: tm s chiu ca khng gian vectV

sinh bi hvect{v1,v2,,vn} (h sinh), thchin nhsau:

Tm rank ca ma trn [v1,v2,,vn]

Nu rank=k th dimV=k; v cs ca V gm kvecttrong sn vect.


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Ta ca mt vecti vi cs

Gi s S= {v1, v2, ,vn} l mt cs cakhng gian vectV; v l vectbt k v c sbiu din qua h S l v= 1v1+ 2v2++ nvn

Ta ni b s (1, 2,, n) l ta cavectv i vi cs S, v k hiu

[v] S=(1, 2,, n)T.


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V d Trong R3, cho cs

S={v1=(1,0,0)T,v2=(1,1,0)T,v3=(1,2,3)T}. Tmta ca v=(2,3,5)Ti vi cs S.

NX: tm ta ca vectv i vi cs

S={v1,v2,,vn} thc cht l i gii hphngtrnh. Hphng trnh ny c ma trn mrng l A=[v1,v2,,vn,v] v ma trn h s

A=[v1,v2,,vn]. Vit di dng ma trn l: A[v]S=v


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i cs

Gi s E={e1,e2,, en} v F={f1, f2, ,fn} l haics ca khng gian vectn chiu V.

Gi svectv biu din qua cs E lv=[e1,e2,,en][v]E. Hi khi biu din qua cs Fth ta ca v snhth no?

Biu din cc eiqua cs F. Gi s[e1,e2,,en]=[f1, f2, ,fn] {[e1]F,[e2]F,,[en]}.

Khi , ta ca v i vi cs F l[v]F={[e1]F,[e2]F,,[en]F} [v]E


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V d Trong R3, cho cs

S={v1=(1,0,0)T,v2=(1,1,0)T,v3=(1,2,3)T} vR={u1=(1,2,-1)

T,u2=(-1,2,0)T,u3=(3,0,0)

T}.

Tm ta ca v=(2,3,5)Ti vi cs S, v

biu din v di dng tch ma trn. Tm ta ca v i vi cs R.

khong gian vecto.10.12

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