lý thuyết ma trận chuong không gian vecto

8/11/2019 L Thuyt Ma Trn Chuong Khng Gian Vecto

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CHNG 3

KHNG GIAN VECT-----


2/28

Ni dung

1. Khng gian vect

2. Khng gian con ca khng gian vect

3. Ph thuc tuyn tnh, c lp tuyn tnh4. C s, s chiu v ta ca KGVT

5. H thc bin i ta ca vect khi c s thayi. Ma trn chuyn c s.

6. Khng gian nghim.

7. Khng gian dng ca ma trn.

Chng 3. Khng gian vect


3/28


1. Khng gian vect:

nh ngha 1: Cho V l mt tp khc rng, trong xc nh 2php ton:

i. Php ton cng (k hiu +)

V c gi l khng gian vect (KGVT) trn trng s thc R nu thamn cc tnh cht sau i vi php cng v nhn v hng:

,u v V u v V+

(Php hp thnh trong)

ii. Php nhn v hng:

(Php hp thnh ngoi)

, ,u V k ku V

R

Cc phn t ca V c gi l cc vect.


4/28


i. Tnh giao hon ca php cng

ii. Tnh kt hp ca php cng:

iii. Tn ti mt phn t khng, k hiu 0, tha mn:

iv. tn ti mt phn t i, k hiu l , tha mn:u V" u-

, ,u v V u v v u" + = +

( ) ( ), , ,u v w V u v w u v w" + + = + +

, 0u V u u" + =

( ) 0u u+ - =

v.

vi.

vii.

viii.

( ), , ,u v V k k u v ku kv" " + = +R( ), , ,u V k h h k u hu ku" " + = +R

( ) ( ), , ,u V k h h ku hk u" " =R

,1.u V u u" =


5/28


Tnh cht:

Php tr trong KGVT c nh ngha nh sau:

( )u v u v- = + -

i. Phn t 0 trong (iii) v phn t -u trong (iv) l duy nht.

ii.

iii.

, 0.u V u" = 0

,k V" R 0 .k =0 0

iv. Nu ku= 0th hoc k= 0 hoc u= 0

v. ( )1u u- = -


6/28

V d:

1. Khng gian vect Rn:


1 2 1 2; , , , ,..., , , ,...,n

n nk u v u u u u v v v v R R

1 1 2 2, ,..., n nu v u v u v u v 1 2, ,..., nku ku ku ku

0,0,...,00 phn t khng.

trong cc uiv vil cc s thc v c gi l cc thnh phnca vec t u v v.


7/28


8/28


4. Tp tt c cc ma trn cp mxn:

Php cng: cng ma trnPhp nhn v hng: nhn v hng vi mt ma trn

l mt KGVT trn trng s thc.

m n

m n

5. Trng s thc Rl KGVT trn chnh n.


9/28

1. Khng gian vect


3. Ph thuc tuyn tnh, c lp tuyn tnh

4. C s, s chiu v ta ca KGVT


6. Khng gian nghim.




10/28


2. Khng gian con ca KGVT:

nh ngha 2:

Khng gian con ca KGVT V trn trng s thc R (gi tt lkhng gian con) l mt tp hp W khc rng ca V tha 2 tchcht sau:

i. , ,u v W u v W " +

ii. , ,u W k ku W " " R

Nhn xt:

Hai tnh cht trn c th c thay bng tnh cht sau:

, , ,u v W k ku v W " " + R


11/28


nh l:

Phn giao ca mt s bt k cc khng gian con ca KGVT V lkhng gian con ca KGVT V.

nh l:

Tp hp nghim ca h phng trnh thun nht trn R:AX = 0

trong vl khng gian con ca KGVT Rn.

m nA 1nX

Chng minh: R1

;n

k X, Y

1nk + X Ycn cm cng l nghim ca h AX = 0

( ) { {k k+ = + =0 0

A X Y AX AY 0

vi X v Y l nghim ca AX =0

Suy ra iu phi chng minh.


12/28

1. Khng gian vect


3. Ph thuc tuyn tnh, c lp tuyn tnh



6. Khng gian nghim.




13/28

3. Ph thuc tuyn tnh, c lp tuyn tnh:

nh ngha 3:

V l KGVT trn R. Cho . Vect cdng


1 2, , ..., mv v v V u V

1 1 2 2 ... m mu v v va a a= + + +

trong , c gi l t hp tuyn tnh cacc vect , 1,i i ma =R

1 2, , ...,

mv v v

nh ngha 4:

H cc vect v1, v2, ,vmca KGVT V c gi l ph thuc

tuyn tnh, nu tn ti cc v hng (cc s thc), 1 2, ,..., ma a a khng ng thi bng khng, sao cho:

1 1 2 2 ... m mv v va a a+ + + = 0

H vect khng ph thuc tuyn tnh c gi l c lp tuyntnh.


14/28

nh l:

Cc vect ph thuc tuyn tnh khi v ch khic t nht mt vect l t hp tuyn tnh ca cc vect cn li.


1 2, , ..., mv v v V

Ch :

i. Cc vect c lp tuyn tnh nu v ch

nu1 2, , ..., mv v v V

1

1

,..., , 0 0, 1,...m

m i i i

i

v i ma a a a=

= = " =R

ii. Mi h hu hn cc vect, trong c vect 0 u ph thuctuyn tnh.

iii. , mt h vect gm 1 vect, k hiu c lp tuyntnh khi v ch khi .v V" { }v

v 0

Ch 3 h i


15/28

Phng php kim tra h cc vect LTT hay PTTT:


Bc 1:

Lp h phng trnh tuyn tnh thun nht:

=AX 0

1 2 mv v v =

A L

1 2

T

i i i niv v v v =

L

11 12 1

21 22 2

1

m

m

n nm

v v vv v v

v v

=

A

LL

M M O M

L L

trong A l ma trn c cc ct l cc vect v1, v2, ,vm.

Ch 3 Kh i


16/28


1 2 ma a a= X L

v vect Xc dng:

Bc 2:

Gii h phng trnh tuyn tnh thun nht trn ta c:

i. H c nghim tm thng suy ra h cc vect LTT

ii. H c v s nghim (c nghim khng tm thng) suy ra h

cc vect PTTT

Ch 3 Kh i t


17/28


1. Khng gian vect

2. Khng gian con ca khng gian vect3. Ph thuc tuyn tnh, c lp tuyn tnh



6. Khng gian nghim.


Ch 3 Kh i t


18/28


4. C s, s chiu v ta ca KGVT Rn:

nh ngha 5: (c s ca KGVT)

Tp gm m vect ca KGVT Rnlp thnh mt

h cc phn t sinh ca Rn, nu vi mi vect v bt k trong Rnl mt t hp tuyn tnh ca cc vect , tc l c thbiu din v di dng:

{ }1 2, , ..., mf f f=B

1 2, , ..., ma a a

1 1 2 ...

m m mv f f f a a a= + + +

trong l cc v hng.

1 2, , ...,

mf f f

C s ca KGVT Rnl mt h cc phn tsinh c lp tuyn tnh, tc l Btha mn hai tnh cht sau:

{ }1 2, , ..., nf f f=B

i) c biu din di dngn

v R

1 1 2 2 ...

n nv f f f a a a= + + +

ii) Phng trnh ch tha mn khi1 1 2 2 ... 0n nf f fl l l+ + + =

1 2 ... 0nl l l= = = =

(cng thc khai trin vect v thnh cc thnh phn)

Ch 3 Kh i t


19/28

Cc v hng c gi l cc ta ca vect v trong

c s


{ }1 2, ,..., .nf f f=B

1 2, , ...,

na a a

K hiu:1

2

n

v

a

a

a

=

B M

Ch 3 Kh i t


20/28


V d:

1) Trong KGVT R2: mi vect u c th biu din thng qua 2 vect

khng cng phng. V hai vect khng cng phng th LTT.Vy c s ca R2l mt h gm 2 vect khng cng phng.

2) Trong KGVT R3: mi vect u c th biu din thng qua 3 vectkhng ng phng (khng nm trn cng mt phng). V 3 vectkhng ng phng th LTT. Vy c s ca R3l mt h gm 3vect khng ng phng.

( ) ( )1, 2 ; 2, 0a b= =( )4, 4 2c c a b= = +

vy: 2

1c

=

B

Ch 3 Kh i t


21/28

Ch :


{ }0 1 2, ,..., .ne e e=B

i) Mi vect vtrong Rnc khai trin thnh cc thnh phn mt cchduy nht

ii) Vi mi c s khc nhau, mt vect c khai trin thnh cc thnhphn khc nhau (tr vect 0)

iii) C s chnh tc trong Rn: k hiu

1

2

3

1, 0, 0, , 0 ,

0,1, 0, , 0 ,

0, 0,1, , 0 ,

0, 0, 0, , 1 .n

e

e

e

e

= = =

=

K

K

K

M

K

( ) 1 2 31, 2, 3 2 3a a e e e= = + +0

1

2

3

c

=

B

V d:

Ch 3 Kh i t


22/28

nh ngha 6: (chiu ca ca KGVT)

Nu tn ti s nguyn dng n sao cho KGVT V c mt c s

gm n vect, s nguyn ny l duy nht v c gi l s chiuca KGVT.


K hiu: n = dimV

Nhn xt:

i) S chiu ca mt KGVT chnh l s vect ca mi c s ca V vcng l s ti i cc vect c lp tuyn tnh ca KGVT V

ii) KGVT c s chiu hu hn th gi l KGVT hu hn chiu. KGVTtrong c th tm c v s vect c lp tuyn tnh c gi lKGVT v hn chiu.

Chng 3 Khng gian vect


23/28

nh l:


Trong KGVT Rn, mt h bt k gm n vect c tuyn tnh th

to thnh mt c s

nh l:

H gm n vect trong KGVT Rnc lp tuyn tnh khi v chkhi nh thc ca ma trn to bi cc thnh phn ca vect khc khng.



24/28


1. Khng gian vect

2. Khng gian con ca khng gian vect3. Ph thuc tuyn tnh, c lp tuyn tnh



6. Khng gian nghim.




25/28

5. H thc bin i ta ca vect khi c s thay i. Ma trnchuyn c s.


l hai c s khc nhau ca KGVT Rn.{ }1 2, , ..., ne e e=B { }1 2, , ..., nf f f=B

Quy c Bl c s c v B l c s mi.

Ta ca cc vect trong c s mi c biu din trongc s c nh sau:

1 11 1 12 2 1

2 21 1 22 2 2

1 1 2 2

...

...

...

n n

n n

n n n nn n

f e e e

f e e e

f e e e

a a a

a a a

a a a

= + + +

= + + +

= + + +

M



26/28


Ma trn vung cp n:

11 21 1

12 22 2

1 2

...

...

...

n

n

B B

n n nn

P

a a a

a a a

a a a

=

M M M

c gi l ma trn chuyn c s t c s c Bsang c smi B (hoc ma trn chuyn).



27/28


nh l:

l ma trn chuyn t c s B={ei

} sang c s B={fi

}

v l ma trn chuyn c s t B sang c s B. Khi

kh nghch v

B BP

B BQ

B BP

1B B B BQ P- =



28/28


nh l:

l ma trn chuyn t c s B={ei} sang c s B={fi}

trong KGVT V. Khi i vi vect bt k v trong V:

B BP

B Bv P v = B B

1

B Bv P v

-

= B B

i)

ii)

lý thuyết ma trận chuong không gian vecto

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