toán đs
TRANSCRIPT
Chng II KHNG GIAN VECT
M
U
Trong chng I ta th y, nh nh th c ta gi i c h phng trnh Cramer. Song n u ch dng nh th c nghin c u vi c gi i h phng trnh tuy n tnh t ng qut (khi m n ho c khi m = n nhng nh th c c a h phng trnh b ng 0) th s c nhi u kh khn, ph c t p. Khng gian vect s gip ta v t qua nh ng kh khn y v cng gip ta trnh by l thuy t h phng trnh tuy n tnh m t cch sng s a. tr ng Ph thng trung h c ta dng vect nghin c u hnh h c. Vect cn c dng nghin c u nhi u ngnh ton h c khc v c nh ng mn khoa h c khc nh C h c, V t l, Ho h c, a l, v nhi u ngnh k thu t. N u xt t p h p V cc vect c chung i m g c O m ta h c tr ng Ph thng th ta th y t p V cng v i php c ng hai vect v php nhn m t vect v i m t s tho mn nh ng i u ki n sau: 1) ( + ) + = + ( + ); 2) + = + ; 3) c vect khng 0 tho mn i u ki n: + 0 = ; 4) m i c m t vect 5) r( + ) = r + r ; 6) (r + s) = r + s ; 7) (rs) = r(s ) ; 8) 1. = , trong r, s, 1 l nh ng s th c. i - tho mn i u ki n: + (- ) = 0 ;
Trong ton h c v nhi u khoa h c khc cn c nh ng t p h p m cc ph n t c a chng khng ph i l nh ng vect hnh h c nh ta v a ni, nhng cng c hai php ton tho mn 8 i u ki n nu trn. Chng c g i l nh ng khng gian vect. M c tiu c a chng ny l trnh by nh ngha khng gian vect, cc tnh ch t c a n v c u t o c a m t khng gian vect, chu n b cho vi c p d ng n vo l thuy t h phng trnh tuy n tnh v vi c nghin c u n su s c hn trong nh ng chng sau c th p d ng n nhi u hn vo nh ng b mn ton h c khc cng nh nh ng lnh v c khoa h c khc. V th ta c n: - N m v ng gian con: nh ngha v cc tnh ch t c a khng gian vect, khng
- Hi u r r ng m i khng gian vect c t o thnh t m t h t i thi u nh ng vect c a khng gian m ta g i l c s ; bi t cch tm c s v s chi u c a m t khng gian vect; - Bi t c m i lin h gi a to s khc nhau. c a cng m t vect trong hai c
Trong gio trnh ny ta ch xt cc khng gian vect trn cc tr ng s Tuy nhin nh ng i u trnh by sau y u ng trong m i tr ng tu .
1. 1.1.
NH NGHA V CC TNH CH T N GI N
nh ngha
nh ngha. Gi s V l m t t p h p m cc ph n t c k hi u b i , , ,..., K l m t tr ng s. Trn V c m t php ton g i l php c ng hai ph n t c a V (k hi u "+") v php ton th hai g i l php nhn m t ph n t c a V v i m t s thu c tr ng K (k hi u "."). T p h p V cng v i hai php ton ny c g i l m t khng gian vect trn tr ng K (hay m t K-khng gian vect) nn cc i u ki n sau c tho mn i v i m i , , , V v m i r, s, 1 K. 1) ( + ) + = +( + ); 2) + = + ; 3) c m t ph n t 0 V tho mn i u ki n: + 0 = ; 4) v i m i V c m t ph n t , k hi u b i - , cng thu c V tho mn i u ki n: + (- ) = 0 ; 5) r( + ) = r + r 6) (r + s) = r + s ; 7) (rs) = r (s ) ; 8) 1. = . V c g i l m t vect, 0 c g i l vect khng, - c g i l vect i c a .
B n c c th dng nh ngha c a khng gian vect ki m ch ng r ng cc t p h p cho trong cc v d d i y l nh ng khng gian vect. V d 1. T p h p V cc vect OA , OB , OC ?... chung g c O trong khng gian (m ta h c tr ng ph thng) cng v i php c ng hai vect v php nhn m t vect v i m t s th c l m t khng gian vect. N c g i l khng gian vect hnh h c.
V d 2. M i tr ng K l m t khng gian vect trn K c ng v php nhn trn K.
i v i php
V d 3. Tr ng s th c R l m t khng gian vect trn tr ng s h u t Q. V d 4. Tr ng s ph c C l m t khng gian vect trn tr ng s th c R v cng l m t khng gian vect trn tr ng Q. V d 5. Gi s K l m t tr ng s , t p h p K[x] cc a th c c a n x v i h s trong K, cng v i php c ng hai a th c v php nhn a th c v i m t s , l m t K-khng gian vect. V d 6. Kn = K x K x... x K l tch cc c a n phin b n K. Trn Kn xc nh php c ng hai ph n t v php nhn m t ph n t c a Kn v i m t s thu c K nh sau: V i = (a1, a2,..., an), = (b1, b2,..., bn) thu c Kn v s r K, (a1, a2,..., an) + (b1, b2,..., bn) = (a1, + b1, a2 + b2, an,..., bn), r(a1, a2, ..., an) = (ra1, ra2, ..., ran). Kn l m t K-khng gian vect. T y tr i, m i khi ni n khng gian Kn ta hi u r ng hai php ton trong c nh ngha nh trn. T nh ngha khng gian vect ta suy ra ngay m t s tnh ch t n gian cua n. 1.2. M t s tnh ch t n gi n Gi s V l m t K-khng gian vect. 1) V ch c m t vect khng 0 duy nh t. 2) V i m i V, vect i - duy nh t.
3) V i m i V, -(- ) = . 4) V i V v r K, = 0 khi v ch khi r = 0 ho c = 0 . 5) V i V v r K, ta c: (- = -( ) = ( ). Ch ng minh. 1) Gi s0 v 0' l nh ng vect khng c a V. Theo i u ki n 3)
trong nh ngha, v 0 l vect khng nn 0 + 0' = 0' . Tng t , v 0' l vect khng nn 0 + 0' = 0 . V y 0 = 0' . 2) Gi s V c nh ng ph n t i l - v ' . Theo i u ki n 4) trong nh ngha, + (- ) = 0 = + . Do , p d ng cc i u ki n 1) v 2), ta c : = + 0 = + [ + (- )] = ( ' + ) + (- ) = 0 + (- ) = - .
3) V -(- ) v 4)
u l vect
i c a - nn t 2) suy ra -(- ) = .
N u r = 0 th theo i u ki n 6), ta c: 0 = (0 + 0) = 0 + 0 . C ng -0 vo v u v v cu i ta c: 0 = 0 .
N u = 0 th theo i u ki n 5), ta c: r 0 = r( 0 + 0 ) = r 0 + r 0 . C ng -r 0 vo v u v v cu i ta c 0 = r 0 .
Gi s r = 0 . N u r 0 th theo i u ki n 7) v 8), ta c:1 1 1 = 1. = ( .r) = ( .r )= 0 = 0 r r r
5) V (r ) l vect i c a ra nn nh tnh ch t 2), ta ch c n ch ng minh (-r) v r(- ) u l vect i c a r . Ta c: (-r) + r = (-r + r) = 0 = 0 ; r(- ) + r = r(- + ) = r 0 = 0 . i u ch ng t r ng (-r) v r(- ) u l vect i c a r . V y
(-r) = -(r ) = r(- ). 1.3. Hi u c a hai vect nh ngha. + (- ) c g i l hi u c a v , k hi u b i v
c l tr .
T qu .
nh ngha ny v tnh ch t c a khng gian vect ta suy ra: H
1) ( - ) = p - c . 2) ( - ) = - . Ch ng minh. Xin dnh cho b n c.
2. KHNG GIAN CON 2.1. nh ngha
nh ngha. Gi s W l m t t p con c a khng gian vect V. N u W cng l m t khng gian vect i v i hai php ton cho trong V th W c g i l m t khng gian con c a V. Nh v y mu n ch ng minh t p con W l m t khng gian con c a khng gian vect V ta ph i ch ng t r ng cc php cho trong V cng l cc php ton trong W v ph i ki m tra r ng 8 i u ki n nu trong nh ngha khng gian vect u c tho mn. Song ta s th y r ng ch c n ki m tra m t s t i u ki n hn. 2.2. Tnh ch t c trng
nh l. Gi s V l m t khng gian vect trn tr ng K. W l m t t p con c a V. Cc m nh sau tng ng: (i) W l m t khng gian con c a V. (ii) W v v i m i , thu c W, m i r thu c tr ng K, ta c + W, W. (iii) W v v i m i , thu c W, m i r, s thu c tr ng K, ta c r + W. Ch ng minh. "(i) (ii)": N u W l m t khng gian con c a khng gian vect V th W ph i ch a m t vect 0 c a n. Do W . Cc i u ki n cn l i c a (ii) hi n nhin c tho mn. "(i) (iii)": Hi n nhin.
"(iii) (i)": Gi s cc i u ki n c a (iii) c tho mn. Khi , v i , thu c W v r = s = 1 K, + = 1 + 1 W; v i W, r K, ta c: r = r + 0 W ; ngha l cc php ton trong W cng l hai php ton trong V. Ta ph i ki m tra r ng 8 i u ki n trong nh ngha c a khng gian vect u c tho mn. Hi n nhin cc i u ki n 1), 2), 5), 6), 7), 8) c tho mn v hai php ton trong W chnh l hai php ton cho trong V. Ch cn c n ki m tra cc i u ki n 3) v 4). V W nn c m t W. Theo tnh ch t c a khng gian vect, 0 = 0 + 0 , m t khc, theo gi thi t 0 + 0 W. Do 0 W. Tng t , v i m i W ta u c - = (-1) + 0 W. V y W l m t khng gian vect trn tr ng K v do W l m t khng gian con c a V. B n c hy dng nh l 2.2 trong cc v d d i y: ch ng minh nh ng i u kh ng nh
V d 1. V i m i khng gian vect V, b n thn V v t p { 0 } l nh ng khng gian con c a V. Chng c g i l nh ng khng gian con t m th ng c a V. V d 2. T p Pn g m a th c 0 v cc a th c c b c b hn hay b ng n c a K[x], (xem v d 5, m c 1.1) l m t khng gian con c a khng gian vect K[x]. V d 3. Theo v d 6), m c 1.1, v i n = 4 v K = R l tr ng s th c, th R4 l m t R-khng gian vect. T p W = {(a1, a2, 0, 0}|ai R) l m t khng gian con c a khng gian R4. Th t v y, ta ch ng minh cho v d 3. R rng W v (0, 0, 0, 0) W. By gi v i = (a1, a2, 0, 0), = (b1, b2, 0, 0) thu c W v r R, ta c: + = (a1, a2, 0, 0) + (b1, b2, 0, 0) = (a1 + b1, a2 + b2, 0, 0) W,
r = r(a1, a2, 0, 0) = (ra1, ra2, 0, 0) W. W tho mn i u ki n (ii) trong gian con c a R4. nh l 2.2. V y W l m t khng
C nhi u cch t o thnh nh ng khng gian con c a m t khng gian
vect V. 2.3. T ng c a nh ng khng gian con M nh v nh ngha. Gi s W1, W2,... Wm l nh ng khng gian vect con c a K-khng gian vect V. Khi : T p h p W = { 1 + 2 + ... + n| i Wi, {1, 2,..., m }} l m t khng gian con c a V. Khng gian ny c g i l t ng c a m khng gian con Wi cho v c k hi u b i W1 + W2 +... + Wm hay
i =1
m
Wi .
Ch ng minh. V 0 Wi v i m i i {1, 2,..., m} nn 0 = 0 + 0 + ... + 0 W ; ngha l W . V i = 1 + 2 + .. +am W, = 1+ 2 +...+ m W v r K, ta c: + = 1 + 2 + 1+ 2 + ... + m = ( 1 + 1) + ( 2 + 2) + .. + ( m + m) V i, i Wi v Wi l khng gian con c a khng gian vect V nn i + i Wi, r i Wi, v i m i i {1, 2,..., m}. Do + W,
r W.
Theo
nh l 2.2, W l m t khng gian con c a V.
2.4. Giao c a nh ng khng gian con M nh v nh ngha. Gi s W1, W2,..., Wm l nh ng khng gian vect con c a K-khng gian vect V. T ph pU=
I W l mi i=1
m
t khng gian con c a V v c g i l giao c.
c a m khng gian con Wi. Ch ng minh. Xin dnh cho b n T m t h (m t s hay m t h ) vect c a khng gian V cng c th t o thnh m t khng gian con c a V.
2.5. Khng gian sinh b i m t h vect nh l. Gi s A = { 1, 2,..., m} l m t h vect c a K-khng gian vect V. Khi t p h p W = {r 1 + 2 2 + ...+ /ri K, v i m i i {1, 2,..., m}} l m t khng gian con c a V. W c g i l khng gian sinh b i h vect A, cn A sinh c a W. c g i l h
Ch ng minh. R rng W v 0 = 1 + 0 2 + ...+ 0 m W. Gi s , W v t K, ch ng h n:
T cc i u ki n trong
nh ngha c a khng gian vect, ta suy ra:
Theo
nh l 2.2, W l m t khng gian con c a V.
Ch . Khng gian sinh b i m t vect th ng c k hi u b i K . N u W l khng gian sinh b i h vect { 1, 2,..., m} th W =
Ki=1
n
1
.
Khng gian W trn y sinh b i m t h h u h n vect. Ng i ta g i n l khng gian h u h n sinh. C nh ng khng gian vect c h sinh v h n nhng khng c h sinh h u h n no. Trong gio trnh ny ta ch xt cc khng gian vect c h sinh h u h n V d 1. 1) Gi s V l khng gian vect hnh h c trong khng gian (xem v d li trong m c 1.2). OI l m t vect c nh. N u O I th t p U = {r OI | r R} ch ch a vect 0 , l m t khng
gian con t m th ng c a V. N u O I th t p U = {r OI | r R} g m cc vect g c O, n m trn ng th ng OI.
Gi s OJ l vect khng cng phng v i OI . Khi , t p W = {r1 OI + r2 OJ | r1 R, r2 R) l m t khng gian con c a V g m cc vect,,... n m trong m t ph ng (OIJ).
Gi s OK khng ng ph ng v i OI , OJ . Th th { OI , OJ , OK } l m t h sinh c a V. Th t v y, nh ta bi t m i vect OA trong khng gian u c d ng: OA = r1 OI + r2 OJ + r3 OK . (r1 OI = r1 OA1 , r2 OJ = OA 2 + r3 OK = OA 3
V d 2. Xt khng gian vect R4 v khng gian con W trong v d 3, m c 2.2. H hai vect 1 = (1, 0, 0, 0), 2 = (0, 1, 0, 0), c a R4 l m t h sinh c a W. ch ng minh i u ny ta ph i ch ng t r ng m i W c bi u di n d i d ng = r1 1 + r2 2 . Bi t r ng m i vect trong W c d ng = (a1, a2, 0, 0) W. Theo php c ng v php nhn v i m t s trong R4, ta c: = (a1, a2, 0, 0) = (a1, 0, 0, 0) + (0, a2, 0, 0)
= a1(1, 0, 0, 0) + a2(0, 1, 0, 0) = a1 1 + a2 2 . V y { 1 , 2 } l h sinh c a W. Ta hy th thm vect (2, 3, 0, 0) vo h vect { 1 , 2 } v xt khng gian con W' sinh b i h vect { 1 , 2 , }. M i = a1 1 + a2 2 + a W u c th vi t thnh:
l m t vect trong W. Nh v y, W W. Ng c l i, m i vect = b1 1 + b2 2 W = b1 1 + b2 2 + .
u c th vi t d i d ng u l h sinh
l m t vect thu c W.
V y W = W; ngha l hai h {, 2, } v {, 2, } c a khng gian vect W.
M t cu h i t ra l trong m t h sinh c a m t khng gian vect c th c m t s t i thi u vect sinh ra khng gian y hay khng? Tr l i c a cu h i ny lin quan n m t khi ni m g i l h vect c l p tuy n tnh.
3. S 3.1. Gi s
C L P TUY N TNH - S nh ngha
PH THU C TUY N TNH
A = { 1, 2,..., m-1, m}
(1)
l m t h vect c a K- khng gian vect V, (m > 0). nh ngha 1. N u = r1 1 + r2 2 + ... + rm-1 m-1 + r1 m th ta ni l m t t h p tuy n tnh c a h vect A hay bi u th tuy n tnh qua m vect cho. nh ngha 2. H vect A c g i l ph thu c tuy n tnh n u c m s r1, r2,..., rm-1, rm thu c tr ng K, khng ng th i b ng 0, sao cho r1 1 + r2 2 + ...+ rm-1 m-1 + rm m = 0 . nh ngha 3. H vect A c g i l c l p tuy n tnh n u n khng ph thu c tuy n tnh; ni cch khc, n u r1 1 + r2 2 + ...+ rm-1 m-1 + rm m = 0 . th r1 = r2 = ... = rm-1 = rm = 0. V d 1. Trong khng gian vect, m i vect khc 0 u l p thnh m t h vect c l p tuy n tnh. Th t v y, gi s l m t vect khc 0 trong K-khng gian vect V. T r = 0 v i r ~ K, nh tnh ch t 4), m c 1.2, suy ra r = 0; ngha l h vect { } c l p tuy n tnh. V d 2. M i h vect ch a 0 u l ph thu c tuy n tnh. Th t v y, gi s { 1, 2,..., m, 0 } l m t h vect b t k c a khng gian vect V. Ch n r1 = r2 = ...= rm = 0, rm+1 = 1, ta c: 0 1 + 0 2 + ... +0 m +1. 0 = 0 . i u ny ch ng t h cho ph thu c tuy n tnh. V d 3. Trong khng gian vect hnh h c V, (xem v d 1, m c 1.1), ba vect l p thnh m t h ph thu c tuy n tnh khi v ch khi chng ng ph ng; c l p tuy n tnh khi v ch khi chng khng ng ph ng. Th t v y, OI , OJ , OA ph thu c tuy n tnh khi v ch khi t n t i ba s th c r1, r2, r3 khng ng th i b ng 0 sao cho r1 OI + r2 OJ + r3 OA =
0 ; ch ng h n, r3 0. Khi OA = -
r1 r OI - 2 OK . r3 r3
i u ny ch ng t
ba vect
ng ph ng.
V d 4. Xt khng gian vect R4. H g m ba vect 1 = (1, 0, 0, 0), 2 = (0, 1, 0, 0), = (2, - 5, 0, 0) l ph thu c tuy n tnh, cn cc h vect { 1 , 2 }, { 1 , }, { 2 , } c l p tuy n tnh. Th t v y, = (2, -5, 0, 0) = (2, 0, 0, 0) + (0, -5, 0, 0) = 2(1, 0, 0, 0) (5, 1, 0, 0) = 2 1 - 5 2 hay 2 1 - 5 2 + (-1) = 0 ; ngha l h { 1 , 2 , } l ph thu c tuy n tnh v bi u th tuy n tnh qua 1 , 2 . By gi ta xt h vect { 1 , }. Gi s r1 1 + r2 = 0 , ngha l r1(1, 0, 0, 0) + r2(2, - 5, 0, 0) - (0, 0, 0, 0) hay (r1, 0, 0, 0) + (2r2, - 5r2, 0, 0) = (r1 + 2r2, - 5r2, 0, 0) = (0, 0, 0, 0). Suy ra:
H phng trnh hai n r1, r2 ny c nghi m duy nh t l r1 = 0, r2 = 0. V y h hai vect { 1 , } B n { 2 , }. T c hy t ki m tra s c l p tuy n tnh. c l p tuy n tnh c a hai h { 1 , 2 },
nh ngha suy ra cc tnh ch t sau.
3.2. Cc tnh ch t Theo nh ngha, hai khi ni m ph thu c tuy n tnh v c l p tuy n tnh c a h vect l hai khi ni m ph nh l n nhau. V th , khi ni m ny c m t tnh ch t g th l p t c suy ra m t tnh ch t tng ng c a khi ni m kia. Tnh ch t 1. 1) N u thm p vect vo m t h vect ph thu c tuy n tnh th c
m t h ph thu c tuy n tnh. 2) N u b t i p vect c a m t h vect m th c l p tuy n tnh. c l p tuy n tnh th c
Ch ng minh. 1) Gi s h vect A = { 1, 2,..., m-1, m} ph thu c tuy n tnh. Khi t n t i m s s1,..., sm khng ng th i b ng 0, ch ng h n si 0, sao cho: Th th:
Theo nh ngha, h vect { 1,..., i-1 i, i+1,..., m,..., m+1,..., m+p} ph thu c tuy n tnh. 2) Gi s t h vect c l p tuy n tnh A b t i p vect ta c h vect B. N u B ph thu c tuy n tnh th theo 1), thm p vect ni trn vo B l i c h A ph thu c tuy n tnh; tri v i gi thi t. V y B c l p tuy n tnh. Tnh ch t 2. 1) M t h g m m vect (m > 1) l ph thu c tuy n tnh khi v ch khi c m t vect c a h c bi u th qua cc vect cn l i. 2) M t h g m m vect (m > 1) l c l p tuy n tnh khi v ch khi khng c m t vect no c a h c bi u th qua cc vect cn l i. Ch ng minh. 1) "" Gi s h vect
C a K-khng gian vect V l ph thu c tuy n tnh. Theo nh ngha, t n t i m s ri K, i {1, 2,..., m) khng ng th i b ng 0, ch ng h n, ri 0, sao cho: r1 1 + ...+ ri-1 i-1 + ti+1 i+1 ... + rm m = 0 . Khi r1 1 = - r1 1 -...- ti-1 i-1 ri+1 i+1 - ... - rm 0 . V ri 0 nn t ng th c ny suy ra
ngha l i c bi u th tuy n tnh qua cc vect cn l i. Gi s trong h vect (1) c vect i; tho mn ng th c:
V c si = -1 0 nn tnh. Tnh ch t 3.
ng th c ny ch ng t h (1) ph thu c tuy n
2) Tr c ti p suy ra t 1). 1) M t h g m m vect (m > 0) l c l p tuy n tnh khi v ch khi m i t h p tuy n tnh c a h u ch c m t cch bi u th tuy n tnh duy nh t qua h . 2) M t h g m m vect (m > 0) c a khng gian vect V l ph thu c tuy n tnh khi v ch khi c m t vect c a V bi u th tuy n tnh c qua h theo hai cch khc nhau. Ch ng minh. 1) Gi s h vect { 1, 2,..., m} tnh v c l p tuy n
N u cn c cch bi u th tuy n tnh
th (b1 b1) 1 + (b2 - b'2 ) 2 + ... + (bm bm ) m = 0 . V h vect g cho c l p tuy n tnh nn theo b2- b'2 = ... = bm bm = 0. nh ngha, b1 b1 =
Suy ra: b1 = b1, b2 = b'2 ,..., bm = bm; ngha l cch bi u th tuy n tnh c a qua h vect cho l duy nh t. "": N u m i t h p tuy n tnh c a h vect { 1, 2,..., m} u ch c m t cch bi u th tuy n tnh duy nh t th 0 = 0 1 + 0 2 + ... +
0 m cng l cch bi u th tuy n tnh duy nh t c a 0 . Do , n u 0 = r1 + r2 2 + ...+ rm m th b t bu c r1 = r2 ... = rm = 0. V y h vect cho c l p tuy n tnh. 2) Suy ra t 1). Tnh ch t 4. 1) N u thm vo m t h c l p tuy n tnh m t vect khng bi u th tuy n tnh c qua h y th c m t h c l p tuy n tnh. 2) N u b t i m t h ph thu c tuy n tnh m t vect khng bi u th tuy n tnh c qua cc vect cn l i th c m t h ph thu c tuy n tnh. Ch ng ninh. 1) Gi s A = { 1, a2,..., m-1, m}l m t h vect c l p tuy n tnh c a K-khng gian vect V. V l m t vect khng bi u th tuy n tnh c qua h A. Ta ph i ch ng minh h vect B = { 1, a2,..., m-1, m , } c l p tuy n tnh. Gi s
N u r 0 th
tri v i gi thi t v . Do r = 0 v r= 1 +...+ rm m = 0 v h A c l p tuy n tnh. Suy ra r1 = ... = rm = 0. V y B l h vect c l p tuy n tnh. 2) Suy ra ngay t 1). Sau khi c khi ni m v h sinh c a m t khng gian vect v h vect c l p tuy n tnh ta nghin c u c u t o c a khng gian vect.
4. C S
C A KHNG GIAN VECT
Ta nh c l i r ng, trong gio trnh ny ta ch xt cc khng gian vect c h sinh h u h n (h u h n sinh) trn tr ng s . 4.1. nh ngha
M t h sinh c l p tuy n tnh c a m t khng gian vect khc { 0 } c g i l m t c s c a n. Khng gian vect { 0 } khng c c s ; hay c th ni, s vect trong c s c a khng gian { 0 } b ng 0. V d 1. Trong khng gian vect Pn g m a th c 0 v cc a th c thu c K[x] v i b c b hn hay b ng n, h vect {1, x, x2,..., xn) l m t c s . Th t v y, m i a th c f(x) Pn u c d ng f(x) = a0 + a1x + a2x2 +... + anxn, ai K, v i m i i {0, 1, 2,..., n). i u ch ng t {1, x, x2,..., xn) l m t h sinh c a Pn. M t khc, n u a0 + a1x + a2x2 +... + anxn = 0 th t nh ngha a th c suy ra a0 = a1 = a2 = ... = an = 0; ngha l {1, 2 n x, x ,..., x ) l h vect c l p tuy n tnh. V y n l m t c s c a Pn. V d 2. Trong khng gian vect R3, h ba vect 1 = (1, 0, 0), 2 = (0, 1, 0), 3 = (0, 0, 1) l m t c s ; ng i ta g i l c s chnh t c. B n c c th ch ng t i u . H ba vect 1 = (1, 1, 0), 2 = (0, 1, 1), 3 (1, 0, 1) cng l m t c s . kh ng nh i u ny ta s ch ng minh h vect { 1, 2, 3} l m t h sinh c a R3 v c l p tuy n tnh. Gi s = (a1, a2, a3) l m t vect b t k thu c R3. Ta tm ba s r1, r2, r3 R sao cho = r1 1 + r2 2 + r3 3 hay sao cho:
Gi i h phng trnh 3 n r1, r2, r3 ny ta c nghi m duy nh t
i u ny ch ng t { 1, 2, 3} l m t h sinh c a R3. M t khc, v ba s r1, r2, r3 c xc nh duy nh t nn m i u c cch bi u th tuy n tnh duy nh t qua h sinh ny. Theo tnh ch t 3, m c 3.2, h sinh ny c l p tuy n tnh. V y n l m t c s c a R3. M t cu h i t ra l m i khng gian vect u c c s hay khng? tr l i cu h i ny ta hy xt m i lin quan gi a h sinh v c s . 4.2. S t n t i c a c s Tr c h t ta xt b sau v m i lin quan gi a h sinh v c s B . N u khng gian vect c m t h sinh g m m vect th s vect c a m i h vect c l p tuy n tnh c a n khng v t qu m. Ch ng minh. Gi s K-khng gian vect V c m t h sinh A = { 1, 2,..., m}, 0 v i m i i {1, 2,..., m) v e = { 1, 2,..., n} l m t h vect c l p tuy n tnh c a V v i n > m. V A l m t h sinh nn
0 nn c m t a1j khc 0, ch ng h n a11 0. Do
Thay 1 trong h A b i 1 ta c h A1 = { 1, 2 ,..., m}. Gi s V = b1 1 + b2 2 + ... + bm m. Th th
Nh v y m i V
u bi u th tuy n tnh c qua h A1; do
A1 l m t h sinh c a V. Ni ring, 2 c d ng:
N u t t c cc h s c a cc i u b ng 0 th 2 = a21 1. Suy ra h e ph thu c tuy n tnh; tri v i gi thi t. V th c m t a2j 0, V i j 1. N u c n ta nh s l i cc i gi thi t r ng a22 0. Khi
Thay 2 trong A1 b i 2 ta c h A2 = { 1, 2,..., m }. L p lu n nh trn, A2 l m t h sinh c a V. C ti p t c nh th , ta l n l t thay m vect c a h A b i m vect u tin c a h e v c h sinh Am = { 1, 2,..., m} c a v. Theo gi thi t, n > m nn m+l Am. Nhng Am l h sinh c a V nn m+1 c bi u th tuy n tnh qua h vect ny; tri v i gi thi t c l p tuy n tnh c a h e. V y n m. H qu . S vect trong hai c s c a m t khng gian vect b ng nhau. Ch ng minh. Suy ra ngay t By gi ta tr l i cho cu h i nh l trn. t ra tr c m c 4.2. u c c s .
nh l 1. M i K - khng gian vect V { 0 }
Ch ng minh. Gi s 1 0 l m t vect thu c V. Theo v d 1, m c 3.1, h { 1} c l p tuy n tnh. N u m i vect c a V u bi u th tuy n tnh qua h ny th l m t c s c a V. N u tri l i, trong V c 2 khng bi u th tuy n c qua 1. Theo tnh ch t 4, m c 3.2, h vect { 1, 2} c l p tuy n tnh. N u h ny khng ph i l m t c s th
trong V c m t 3 khng bi u th tuy n tnh c qua h { 1, 2}. L i theo tnh ch t 4, m c 3.2, h vect { 1, 2, 3} c l p tuy n tnh. Ti p t c, b sung nh th ta c nh ng h vect c l p tuy n tnh c a V. V V c m t h sinh g m m vect no (c th ta khng bi t h sinh y) nn theo b , qu trnh ny ph i k t thc vect n no v i n m. Lc ta c h vect E = { 1, 2, 3 ,..., n} m m i vect c a v u bi u th tuy n tnh c qua h e. V y e = { 1, 2, 3 ,..., n} l m t c s c a V. H qu . Trong khng gian vect, m i h vect k u c th b sung thnh m t c s . c l p tuy n tnh b t
ngha c a nh l trn y l d cho khng bi t tr c h sinh c a khng gian vect ta v n c th d ng c m t c s c a n. Song khi bi t m t h sinh c a khng gian vect th nh l sau y cho th y c th ch n m t c s trong h sinh ny. l tr l i cho cu h i t ra tr c 3. nh l 2. T m t h sinh c a m t khng gian vect khc { 0 } c th ch n ra m t c s . Ch ng minh. Cch ch ng minh nh l ny gi ng nh cch ch ng minh nh l trn; ch khc ch l ng l ta ch n cc vect ; trong V th y ta ph i ch n chng trong h sinh cho.
5. S
CHI U C A KHNG GIAN VECT
H qu c a b , m c 4.2, cho th y s vect trong hai c s khc nhau c a m t khng gian vect th b ng nhau. i u cho php ta nh ngha: 5.1. nh ngha
S vect trong m t c s c a K-khng gian vect V c g i l s chi u c a V. K hi u: dimKV. N u khng c n ch r tr ng K c th , ta c vi t n gi n l dimV. V d 1. Khng gian Pn g m a th c 0 v cc a th c b c b hn hay b ng n c s chi u b ng n + 1; t c l dimKPn = n + 1. (Xem v d 1, m c 4.1). V d 2. dimRR3 - 3. V d 3. Khng gian V cc vect hnh h c trong khng gian c dimRV = 3. H qu . Trong khng gian vect n chi u m i h vect tnh g m n vect u l c s . c l p tuy n
Ch ng minh. Gi s dimKV = n v a = { 1, 2,..., n} l m t h vect c l p tuy n tnh c a V. Theo h qu c a nh l 1, m c 4.2, c th b sung vo a c m t c s c a V. V dimV = n, m i c s g m n vect cho nn khng c n b sung vect no vo a n a. V y a l m t h sinh c l p tuy n tnh, do l m t c s c a V. Ta hy tm hi u m i lin h gi a s chi u c a m t khng gian vect v i s chi u c a cc khng gian con c a n. 5.2. S chi u c a khng gian con nh l 1. Gi s W l m t khng gian con c a K-khng gian vect V. Th th: 1) dimKW dimKV. 2) dimKW = dimKV khi v ch khi W = V. Ch ng minh.
1) N u W = { 0 } th dimkw = 0 dimKV. By gi gi s dimKV = n, dimKW = m > 0. Khi W c m t c s , ch ng h n, () g m m vect. V () l h vect c l p tuy n tnh trong W v W V nn () cng l c l p tuy n tnh trong V. Theo b , m c 4.2, dimKW m n = dimKV. 2) Suy ra t h qu , m c 5.1. nh l 2. N u U, W l nh ng khng gian con c a K-khng gian vect V th: dim(U + W) = dimU + dimW - dim(U W). Ch ng minh. Gi s dim U = p, dimW = q, dim(U W) = r v { 1, ,.., r} (1) l m t c s c a U W. V c s ny l m t h vect c l p tuy n tnh trong U v trong W nn, theo h qu nh l 1, m c 4.2, c th b sung thnh c s :
c a U v thnh c s
c a W. Ta s ch ng minh r ng
l m t c s c a U + W. Mu n th , tr c h t, ta hy ch ng minh l m t h sinh c a U + W. V { 1,..., p-r, 1,..., r} U v { 1,..., q-r} W nn h (4) n m trong U + W. Gi s = + , v i
Th th ngha l h (4) l m t h sinh c a U + W. Hn n a, h (4) c l p tuy n tnh. Th t v y, gi s
V v tri l m t vect trong U cn v ph i l m t vect trong W. C s c a UW l h (1) nn c th vi t
V h (3)
c l p tuy n tnh nn t
ng th c ny suy ra (6) ng th c (5) ta c
t1 = ... = tr = z1 = ... = zq-r = 0; Thay cc gi tr ny c a zi vo
V h (2)
c l p tuy n tnh nn (7) c l p tuy n tnh. Do n l m t c s
x1 = ... = xp = y1 = ... = yr = 0 T (6) v (7) suy ra h (4) a U+W. V y
dim(u + W) = p - r + r + q - r : p + q - r = dimU + dimW - dim(U W).
6. T A
C A M T VECT
V c s l m t h sinh c l p tuy n tnh c a khng gian vect nn th i vect c a khng gian u c cch bi u th tuy n tnh duy nh t qua c s . 6.1. nh ngha
Gi s () = { 1, 2,..., n} l m t c s c a K-khng gian vect V, V c cch bi u di n duy nh t d i d ng
B n s s (a1, a2,..., an) c g i l cc t a (). Thay cho l i ni c cc t a a2,..., an).
c a
i v i c s
l (a1, a2,..., an) ta vi t: (a1,
V d . Trong v d 2, m c 4, 1 ta bi t h () = { 1, 2, 3), trong 1 = (1, 1, 0), 2 = (0, 1, 1), 3(1, 0, 1) l m t c s c a R3. Vect = 3 1 - 5 2 + 3 c t a i v i c s () l (3,- 5, 1). Cng nh i v i cc vect hnh h c bi t tr ng trung h c, c m t m i lin quan gi a to v cc php ton trn cc vect. nh l. N u k K, v c t a b2,..., bn) th: 1) To 2) To l n l t l (a1, a2,..., an) v (b1,
c a + l (a1 + b1, a2 + b2,..., an + bn); c a k l (ka1, ka2,..., kan). c. c a = 3 1 - 5 2 + 3 trong v d trn
Ch ng minh. Xin dnh cho b n By gi ta th tm to
y i v i c s chnh t c, t c l c s () = { 1, 2, n} trong 1 = (1, 0, 0), 2 = (0, 1, 0), 3 = (0, 0, 1). Ta c:
V y to
c a
i v i c s () l (4, - 2, - 4). i c s th to c a m t vect thay i.
i u ny ch ng t khi
Ta hy xem to c a cng m t vect trong hai c s khc nhau c quan h v i nhau nh th no. Tr c h t, m i lin quan gi a hai c s c di n t b i nh ngha sau. 6.2. Ma tr n chuy n nh ngha. Gi s () = { 1, 2,..., n} v { 1, 2,..., 3} l hai c s c a K-khng gian vect V,
(v m i vect i
u bi u th tuy n tnh qua c s ()).
Ta g i ma tr n vung c p n
l ma tr n chuy n t c s ( ) sang c s (). V d . Xt khng gian vect R3 v i hai c s () = { 1, 2, n} trong 1 (1, 0, 0), 2 = (0, 1, 0), 3 = (0, 0, 1), v { 1, 2, 3} trong 1 = (1, 1, 0), 2 = (0, 1, 1), 3 = (1, 0, 1), (Xem v d m c 6.1). a) Tm ma tr n chuy n t c s () sang c s ()
b) Tm ma tr n chuy n t c s () sang c s (). Gi i a) Ta c:
V y ma tr n chuy n t c s () sang c s () l
b) Ta ph i bi u di n cc vect i qua c s (). C th , ta vi t: (1, 0, 0) = 1 = b11 1 + b21 2 + b31 3
ng th c (1) cho ta (1, 0, 0) = b11(1, 1, 0) + b21(0, 1, 1) + b31(1, 0, 1) = (b11 + b31, b11 + b21, b21 + b31). Suy ra:
Gi i h ny ta tm c:
ng th c (2) cho ta m t h phng trnh; gi i n ta tm c:
Tng t , nh
ng th c (3) ta tm c:
V y ma tr n chuy n t c s () sang c s () l
By gi ta s tm ra cng th c lin h gi a cc t a vect trong hai c s khc nhau. 6.3. Lin h gi a cc t a nhau c a m t vect
c a cng m t
i v i hai c s khc
nh l. Gi s () = { 1, 2,..., n) v () = { 1, 2,..., n) l hai c s c a K-khng gian vect V, T = (tij) l ma tr n chuy n t c s () sang c s (), (x1, x2,..., xn) (y1, y2,...,yn) l n l t l t a c a vect i v i c s () v c s (). Th th:
Ch ng minh. Theo gi thi t ta c:
T s bi u th tuy n tnh duy nh t c a qua c s () suy ra:
T ng qut: xi = ti1y1 + t12y2 +... + t1nyn = n}.
tj=1
n
ij
y j , V i m i i {1, 2,...,
V d . Xt khng gian R3 v i hai c s () v () trong v d 6.2. Cho = (- 5, 0 , 1) R3. Hy tm t a c a vect R3 c s (). Gi i
m c iv i
T v d m c 6.2, ta bi t r ng ma tr n chuy n t c s () sang c s () l:
G it a
c a
i v i c s () l (x1, x2, x3). Theo gi thi t t a it a
c a i v i c s () l y1 = - 5, y2 = 0, y3 = 1. Theo cng th c ta c:
7. H NG C A H VECT- H NG C A MA TR N nghin c u cc chng sau ta c n bi t cch tm c s c a nh ng khng gian con sinh b i m t h vect c a m t khng gian vect. Tuy v phng di n l thuy t, nh l 4.6 cho th y t m t h sinh c a m t khng gian vect c th tm c m t c s c a n. Song kh c th dng n vo th c hnh. Trong m c ny ta s xt m t k thu t tm c s nh th . Tr c h t ta hy nh ngha m t khi ni m ti n di n t. l h ng c a h vect. Khi ni m ny cng c ng d ng trong nhi u v n khc. 7.1. H ng c a h vect nh ngha. S chi u c a khng gian vect sinh b i h vect a c g i l h ng c a h a K hi u h ng (a). H qu . H a g m m vect l (a) = m. Ch ng minh. Xin dnh cho b n c l p tuy n tnh khi v ch khi h ng c.
V d . Xt v d 2, m c 2.5. H vect a = { 1, 2} c l p tuy n tnh nn n l c s c a khng gian vect W sinh b i a. Theo nh ngha 7.1, h ng (a) = dimW = 2. H b = cng sinh ra khng gian vect W. Do h ng(b) = dimW = 2. C th gi i thch v sao m h ng = h ng hay khng? Hy xem m nh d i y. M nh . N u thm vo m t h vect m t t h p tuy n tnh c a h th h ng c a h m i b ng h ng c a h cho. Ch ng minh. Gi s a = { 1 , 2 ,... m }, h ng(a) = n. Thm vo a vect
G i W, W l n l t l nh ng khng gian sinh b i h a v h b. V a b nn W W'. Ng c l i, gi s W'. Khi l m t t h p tuy n tnh c a h b, ch ng h n,
Do W W. V y W = W. Suy ra h ng (B) = dim(W') - dim(W) h ng(A). 7.2. H ng c a ma tr n nh ngha. 1) Cho ma tr n
Coi cc thnh ph n trong m t dng c a ma tr n A nh cc t a c a m t vect trong khng gian vect V (n chi u) i v i m t c s no . Ta g i h vect a g m :
l h vect dng c a ma tr n A. H vect c t c a ma tr n A c ( nh ngha tng t . c a h vect a Ng c l i, ma tr n A c g i l ma tr n cc t a i v i c s cho).
2) Ta g i h ng c a h vect dng c a ma tr n A l h ng c a ma tr n A. K hi u h ng(A).
h ng (A) = 1 v h g m m t vect = (1, 0, - 2) 0
c l p tuy n tnh.
V dng th hai c a B l t h p tuy n tnh c a dng th nh t nt theo m nh m c 7.1, h ng(B) = h ng(A) = 1. i v i ma tr n C ta th y h vect dng c a n g m ba vect:
Chng
u bi u th tuy n tnh c qua 1 : 2 = 0 1, 3 = -
2 1. 5
Do khng gian sinh b i ba vect ny c c s l { 1}. V y h ng(C) = 1. Theo nh ngha h ng c a ma tr n th tm h ng c a ma tr n cng l tm h ng c a h vect dng tng ng v do bi t h ng c a ma tr n s suy ra c s v s chi u c a khng gian vect sinh b i h vect dng c a ma tr n y. V i ma tr n A = (aij) ki u (m, n), nu ch n r dng, r c t th cc thnh ph n n m giao c a r dng r c t y l p thnh m t nh th c c p r. Ta g i n l inh th c con c p r c a A. nh l. H ng c a ma tr n A b ng c p cao nh t c a cc con khc 0 c a A. nh th c
Ch ng minh. Gi s ma tr n A = (aij)(m.n) v c p cao nh t c a cc nh th c con khc 0 c a n b ng r. C th gi thi t r ng c m t nh th c con khc 0, c p r n m r dng u, ch ng h n:
(Th t v y, ta c th i ch cc dng t c i u v php i ch nh th khng lm thay i h ng c a h vect dng). Ta s ch ng minh r ng h g m r vect dng u
l c s c a khng gian vect sinh b i m vect dng c a ma tr n A. Tr c h t, h vect (2) c l p tuy n tnh. Th t v y, gi s
p d ng cc php ton trong khng gian vect Kn, ta c:
R rng (0, 0,..., 0) l m t ngh m c a h ny. V
nn y l m t h Cramer. Do (0, 0,..., 0) l nghi m duy nh t; ngha l b t bu c x1 = x2 =... = xi = 0. i u ny ch ng t h (2) c l p tuy n tnh. By gi ta ch ng minh r ng m i vect dng cn l i c a ma tr n A bi u th tuy n tnh c qua h (2); t c l ph i ch ng minh r ng v i m i
Mu n v y, ph i ch ng minh r ng: Gi s i c nh. i v i aij ta xt nh th c
y l m t nh th c con c p r + 1 c a ma tr n A nn theo gi thi t n b ng 0. Khai tri n n theo c t cu i ta c: trong As l ph n b i s c a thnh ph n ai trong m i s {1, 2,..., r}. V D 0 nn nh th c Dij, v i
Khi i c
nh, j thay
i, ma tr n
khng
i nn cc As khng
i v chng l nh ng
nh th c con c p r
c a ma tr n ny. V th nh ch n, ta c Do
A t ks = - s , v i m i s {1, 2,..., r}, v i i c D
ng th c (3) c ch ng minh.
V y h vect (2) l m t c s c a khng gian vect sinh b i m vect dng c a ma tr n A. Suy ra h ng(A) = r. Ch . Trong php ch ng minh nh l trn ta th y n u nh th c con c p cao nh t khc 0 c a ma tr n A n m r dng no th r vect dng y l c s c a khng gian vect sinh b i m vect dng c a ma tr n A. V d : Tm c s c a khng gian vect sinh b i h vect g m cc vect trong R3:
Gi i G i A l ma tr n m cc vect dng l cc vect cho:
Nh v y
1 5 l 2 1
nh th c con c p cao nh t khc 0 c a ma tr n A.
N n m dng th nh t v th ba c a ma tr n A. V y h vect { 1, 3} l c s c n tm. H qu 1. H m vect l c l p tuy n tnh khi v ch khi ma tr n cc t a c a chng c m t inh th c con khc 0, c p m. Ni ring, trong khng gian vect n chi u, h n vect l c l p tuy n tnh khi v ch khi nh th c c a ma tr n cc t a c a chng khc 0. Ch ng minh. Xin dnh cho b n c. H qu 2. H ng c a ma tr n b ng h ng c a h vect c t c a n. Ch ng minh. Gi s A = (aij) l ma tr n ki u (m, n). Xt ma tr n chuy n v t A. H vect dng c a tA l h vect c t c a A. Theo nh l trn, h ng(tA) b ng c p c a nh th c con c p cao nh t khc 0 c a tA. Nhng m i nh th c con c a tA l i l chuy n v c a m t nh th c con c a A v ng c l i. M t khc nh th c c a hai ma tr n chuy n v l n nhau b ng nhau. Do , |B| l m t nh th c con c p cao nh t khc 0 c a t A khi v ch khi |B| l m t nh th c con c p cao nh t khc 0 c a A. V y h ng (A) = h ng (tA) = h ng (h vect dng c a tA) = h ng (h vect c t c a A). 7.3. Cch tm h ng c a ma tr n Mu n tm h ng c a h vect ta tm h ng c a ma tr n cc t a h vect y. Mu n tm h ng c a ma tr n ta tm c a ma tr n y. c a
nh th c con c p cao nh t khc 0
tm nh th c con c p cao nh t khc 0 c a ma tr n A, khng c n xt t t c cc nh th c con c a n m ch c n xu t pht t m t nh th c con D1 0 c p s bi t r i xt cc nh th c con c p s + 1 ch a D1. N u t t c cc nh th c con ny u b ng 0 th D1 l m t nh th c con c p cao nh t khc 0; do h ng(A) = s. N u c m t nh th c D2 0, c p s + 1 th ti p t c xt cc nh th c c p s + 2 ch a D2. C nh th cho t i khi tm c m t nh th c D 0, c p r m m i nh th c c p r + 1 bao quanh n u b ng 0. Suy ra D l m t nh th c con c p cao nh t khc 0
c a ma tr n A. Th t v y, v D n m r dng no v m i nh th c con c p r + 1 bao quanh D u b ng 0 nn v i l p lu n nh ch ng minh c a nh l, h r vect ny l c s c a khng gian vect W sinh b i h vect ng c a ma tr n A. Do dim(W) = r. V i m i nh th c con D 0, c p t, v Di n m trong t dng no c a A, th h g m t vect dng ny c l p tuy n tnh trong W. V y t dim(W) = r. p d ng. Cho h vect a g m:
a) Tm h ng(a). b) Tm c s c a khng gian vect sinh b i a. Gi i a) tm h ng c a h vect, ta ph i tm h ng c a ma tr n
Theo
nh l, ta ph i tm
nh th c con c p cao nh t khc 0 c a A.
Xu t pht t m t
nh th c con khc 0 b t k, ch ng h n
Ta xt cc
nh th c c p 3 ch a D2.
Xt ti p cc th , l:
nh th c con c p 4 ch a D3. Ch cn hai
nh th c nh
V y h ng(A) = 3. b) V D3 n m c n tm. Cng c th c ba dng u nn h vect { 1, 2, 3} l m t c s
nh th c c p 3 khc ch a D2 v khc 0, ch ng h n,
V y h ng(A) = 3. V D n m ba dng: th nh t, th hai v th nm nn h vect { 1, 2, 3} cng l m t c s . 7.4. Tm h ng c a ma tr n b ng cc php bi n Ta cng c th dng m t s php bi n c a ma tr n. nh ngha. Cc php bi n i s c p trn cc ma tr n: 1) 0. 3) Nhn m i thnh ph n trong m t dng (c t) v i cng m t s r i c ng vo thnh ph n cng c t (dng) trong m t dng (c t) khc. i s c p tm h ng i trn ma tr n
i sau y c g i l cc php bi n
i ch hai dng (hai c t) cho nhau;
2) Nhn m i thnh ph n trong m t dng (c t) v i cng m t s khc
nh l. N u th c hi n cc php bi n i s c p trn m t ma tr n th h ng c a ma tr n thu c b ng h ng c a ma tr n cho. L m d ng ngn ng c th ni: Cc php bi n thay i h ng c a m t ma tr n. Ch ng minh. Xin dnh cho b n V d . Tm h ng c a ma tr n i s c p khng lm
c nh m t bi t p.
Gi i i ch dng th nh t v dng th t cho nhau:
C ng dng th nh t vo dng th hai; nhn dng th nh t v i - 4, r i c ng vo dng th ba; nhn dng th nh t v i - 3, r i c ng vo dng th t. ta c:
i ch dng th hai v dng th t cho nhau:
C ng dng th hai vo dng th ba:
C ng dng th ba vo dng th t ta c ma tr n:
Ma tr n cu i cng c
nh th c c p ba:
l nh th c con c p cao nh t khc 0 c a B. V y h ng(B) = 3. V cc php bi n i s c p khng lm thay i h ng c a ma tr n nn h ng(A) = h ng(B) = 3. N u dng php bi n i s c p tm c s c a khng gian sinh b i m t h vect th ta g p m t kh khn nh trong vi c xc nh nh ng vect no c a h l p nn c s , v qu trnh bi n i ta i ch cc dng, cc c t. B n c hy th tm cch kh c ph c kh khn y. 7.5. Tm c s , s chi u c a khng gian sinh b i m t h vect b ng my tnh i n t Mu n tm c s v s chi u c a khng gian W sinh b i m t h vect a ta ch c n tm nh th c con c p cao nh t khc 0 c a ma tr n A thi t l p b i h vect cho. dimW = h ng(A) v nh th c con c p cao nh t khc 0 n m nh ng dng no th nh ng vect dng y l p thnh m t c s . Nh cc php bi n i s c p trn ma tr n, my tnh i n t c th th c hi n cc php bi n i y bi n m t ma tr n cho thnh m t ma tr n cng h ng m ta c th nh n ra ngay nh th c c p cao nh t khc 0. T suy ra h ng c a ma tr n, cng l h ng c a h vect, v s chi u c n tm. V d 1. Tm s chi u c a khng gian W sinh b i h vect
Gi i t o ma tr n nh l nh:
Trn mn hnh xu t hi n: Out[1]: = {{11 -1, 2, 0, -1},{-2, 2, 0, 0, -2}, {2, -1, -1, 0, 1}, {-1, -1, 1, 2, 2},{1, - 1, -1, 0, 1} l p ma tr n thu g n nh ti p l nh: RowReduce[A]//MatrixForm Mn hnh xu t hi n: Out[2]:=MatrixForm
Trong ma tr n ny ta th y ngay
nh th c con c p cao nh t khc 0 l:
V y h ng c a ma tr n b ng 4. Nhng cc php bi n i m my th c hi n cho ta m t ma tr n cng h ng v i ma tr n A nn h ng = h ng(A) = 4 = dimW. tm c s c a khng gian ta c n ch r ng khi bi n i ma tr n, theo chng trnh Mathematica 4.0, my tnh c th i ch cc dng, do th t cc dng b thay i. V th nhn vo ma tr n thu c trn
mn hnh ta khng bi t c c s g m nh ng vect no trong cc vect cho. Tuy nhin my tnh khng thay i c t. V v y, ta hy l y cc vect cho l p thnh ma tr n c t. Song, my tnh l i khng t o ma tr n c t. trnh ph i l p m t ma tr n c t gi y nhp, ta v n l p ma tr n dng r i l y ma tr n chuy n v . V d 2. Tm c s c a khng gian W sinh b i h vect: a = { 1 = (-2, 4, 2, 5), 2= (3, 1, 0, 7), 3 = (-1, 9, 4, 17), (1, 0, 2, 1)}.
Gi i nh l nh t o ma tr n: A = {{-2, 4, 2, 5}, {3, 1, 0, 7},{-1, 9, 4, 17}, {1, 0, 2, 1}} Mn hnh xu t hi n: Out[1] ={{-2, 4, 2, 5} {3, 1, 0, 7},{- 1, 9, 4, 17}, {1, 0, 2, 1}}. l p ma tr n chuy n v , nh l nh: tA = Transpose[A] Mn hnh xu t hi n: Out[2]={{- 2, 3, - 1, 1},{4, 1, 9, 0),{2, 0, 4, 2},{5, 7, 17, 1}} tm ma tr n thu g n, nh l nh: RowReduce[tA]//1MatnxForm Mn hnh xu t hi n: Out[3]=MatrixForm
Ma tr n ny cho ta
nh th c con c p cao nh t khc 0 l:
V y h ng c a ma tr n ny b ng 3: h ng(tA) = h ng(A) = h ng(a). nh th c con c p cao nh t khc 0 n m cc c t th nh t, th hai, th t, do cc vect c t 1, 2, 4 l p thnh m t c s .
TM T T Chng II, trnh by khi ni m khng gian vect trn m t tr ng K. l m t t p h p V m m i ph n t g i l m t vect, trn c php ng hai vect v php nhn m t vect v i m t s thu c K tho mn 8 ti u ki n nu nh ngha 1.1. M t t p con W c a V c g i l m t khng gian con c a V n u b n thn W cng l m t khng gian i v i hai php ton cho trong V. Mu n ch ng minh W l khng gian con c a V ch c n ch ng minh r ng: 1) W , v i , thu c W v r K, ta c + W, r W; ho c 2) W , v i , thu c W v r, s K, ta c r + s W. T ng c a m khng gian con W1, W2,..., Wm c a khng gian vect V l i l m t khng gian con W = i | i Wi , i = 1, 2,...m . i =1 m
th
N u a = { 1, 2,..., m} l m t h vect c a K-khng gian vect V
l m t khng gian con c a V v c g i l khng gian con sinh b i h vect a. H a c g i l ra k1 = k2 =...= km = 0. M i khng gian vect u c m t h sinh c l p tuy n tnh, g i l c s c a n. N u () = { 1, 2,..., n} l m t c s c a K- khng gian vect V th m i V u c cch bi u di n duy nh t d i d ng = a1 1+ a2 2+...+ an n.
c l p tuy n tnh n u t
ng th c
k i i =1
m
i
= 0 suy
Cc s ai c g i l t a
c a
i v i c s ().
M t khng gian vect c th c nhi u c s khc nhau. T a c a m t vect trong c s ny khc v i t a c a n i v i c s kia. Bi t t a c a i v i c s () c th tm c t a c a n i v i c
s () n u bi t ma tr n T chuy n t c s () sang c s (). tr n c thi t l p nh sau:
l ma
Th th
Khi , n u (x1, x2,..., xn) v (y1, y2,..., yn) l n l t l t a v i c s () v c s () th
c a
i
Khi ni m h ng c a m t h vect cng nh h ng c a ma tr n r t c n thi t cho cc chng sau. Ta nh ngha h ng c a m t h vect l s chi u c a khng gian sinh b i h vect . V i A l m t ma tr n, ta c: h ng(A) = h ng(h vect dng) = h ng(h vect c t).
BI T P 1. 1. Dng NH NGHA V TNH CH T N GI N nh ngha c a khng gian vect ch ng t r ng:
a) T p s th c R cng v i php c ng hai s th c, php nhn m t s th c v i m t s h u t l m t Q-khng gian vect; b) T p s ph c C cng v i php c ng hai s ph c v php nhn m t s ph c v i m t s th c l m t R-khng gian vect; c) T p Q[x] cc a th c c a n x trn tr ng s h u t Q l m t Qkhng gian vect; d) V i Q l t p s h u t , R l t p s th c, V = Q x R, cng v i php c ng hai ph n t trong V v php nhn m t ph n t c a V v i m t s h u t xc nh nh sau: (a, b) + (c, d) = (a + c, b + d), r(a, b) = (ra, rb), a, c, r l nh ng s h u t b, d l nh ng s th c,l m t Q-khng gian vect. e) T p Q ( 2 ) = {a + b 2 | a, b Q}, v i php c ng v php nhn v i s h u t xc nh nh sau: (a + b 2 ) + (c + d2 ) = (a + c) + (b + d) 2,
r(a + b 2 ) = ra + r 2 , r Q l m t Q- khng gian vect. 2. Cc t p sau c ph i l R-khng gian vect khng? a) T p Q v i php c ng hai s h u t v php nhn m t s h u t v i m t s th c; b) T p s ph c C cng v i php c ng hai s ph c v php nhn m t s ph c v i m t s th c; c) T p s nguyn Z cng v i php c ng hai s nguyn v php nhn m t s nguyn v i m t s th c; d) T p V ni trong bi t p 1d) cng v i php c ng v php nhn v i m t s th c cng xc nh nh th , e) T p D cc a th c b c n v i h s th c.
3. Cho G l t p h p cc hm s xc nh trn t p s th c R c d ng f(x) = ax + b, trong a, b R, v i php c ng v php nhn v i m t s h u t xc nh nh sau: xc v i f, g G, v f(x) = ax + b, g(x) = cx + d th f + g l m t hm s nh b i (f + g)(x) = (a + c)x + (b + d), v i f G v r R, v f(x) = ax + b th rf l m t hm s xc (rf)(x) = rax + rb. Ch ng minh r ng G l m t R- khng gian vect. 4. Cho F l t p cc hm s c a bi n s x xc nh trn R (v l y cc gi tr trong R) v i php c ng v php nhn v i m t s th c xc nh nh sau: v i f, g F, f + g l m t hm s xc g(x); nh b i (f + g)(x) = f(x) + nh b i (rf)(x) = r.f(x). nh nh b i
v i f F v r R, rf l m t hm s xc Ch ng minh r ng F l m t R-khng gian vect.
5. Cho A v B l hai K-khng gian vect. Trn V = AB, xc php c ng v php nhn v i m t s thu c tr ng K nh sau:
( , ) + ( , ) = ( + , + ) , v1 1 2 2 1 2 1 2
i m i 1, 2 A, 1, 2 B;
k( , ) = (k , k ) , v i m i k K, m i A, m i B. Ch ng minh r ng V l m t K-khng gian vect. 6. Trn t p A = {a} xc th c nh sau: nh php c ng v php nhn v i m t s
a + a = a, ra = a, v i m i r R. Ch ng minh r ng A l m t R-khng gian vect... 2. KHNG GIAN CON 7. ch ng minh r ng: a) Q l m t khng gian con c a Q-khng gian vect R; b) R l m t khng gian con c a Q- khng gian vect C (C l t p s ph c);
c) Q 2 l m t khng gian con c a Q-khng gian vect R, (xem bi t p 2f, 1) ; d) R- khng gian vect G l khng gian con c a R-khng gian vect F, (v i G v F l nh ng khng gian trong cc bi t p 3 v 4, 1). e) T p Dn g m a th c 0 v cc a th c c a Q[x] c b c b hn hay b ng n, (n l s t nhin), l khng gian con c a Q-khng gian vect Q[x]. 8. Cc t p h p sau c ph i l nh ng khng gian con c a khng gian vect R3 khng? a) E = {(a1, a2, a3 | ai R, i = 1, 3}; b) F = {(a1, a2, - a1) | ai R, i = 1, 2}; c) B = {(a1, a2, a1a2) | ai R, i = 1, 2}; d) G = {(a1, a2, a1 + a2) | ai R, i - 1, 2}; e) C = {(a1, a2, a2) | ai R, i = 1, 2}; f) H = {(a1, a2, a3) | a1 + a2 + a3 = 0 ai R, i = 1, 2, 3}; 9. T p h p cc s nguyn c ph i l m t khng gian con c a khng gian vect R trn tr ng Q hay khng? 10. T p h p cc a th c b c ch n thu c R[x] c ph i l m t khng gian con c a R [x] khng? 11. Gi s W l m t khng gian con c a khng gian vect R4, (a1, a2, 4, a4) W. Ch ng minh r ng v i m i r R c m t (b1, b2, r, b4) W. 12. Gi s U v W l hai khng gian con c a K-khng gian vect V tho mn i u ki n V = U W. Ch ng minh r ng V = U ho c V = W. 13. Gi s W1, W2,..., Wm l nh ng khng gian vect con c a Kkhng gian vect V. Ch ng minh r ng U = con c a V. 14. Cho V l m t R-khng gian vect, V. Ch ng minh r ng t p R = {r | r R} l m t khng gian con c a V. R c g i l khng gian sinh b i vect . 15. Gi s U v W l hai khng gian con c a R-khng gian vect V.m
Ii =1
Wi l m t khng gian
Ch ng minh r ng U + W l giao c a t t c cc khng gian con c a V ch a U W. 3. S3
C L P TUY N TNH-S
PH THU C TUY N TNH
16. Cho ba vect 1 = (2, 0, 3), 2 = (0, 2, -1), 3 = (1, 2, 3) thu c R . Hy bi u th tuy n tnh vect = (5, - 2, 1) qua ba vect cho. 17. Cho ba vect 1 = x - 1, 2 = 1, 3 = x2 + 1 thu c R[x]. Hy bi u th tuy n tnh vect = x2 - x + 2 qua ba vect cho. 18. Xt xem cc h vect sau trong R3 h no a) 1 = (4, 0, 1), 2 = (2, 0, 1), 3 = (1, 2, 1); b) 1 = (1, 2, 3), 2 = (4, 5, 6), 3 = (5, 7, 9); c) 1 = (1, 2, 3), 2 = (4, 5, 6), 3 = (9, 8, 7); d) 1 = (1, 2, 3), 2 = (4, 5, 6), 3 = (2, -1, 0). 19. Xt xem cc h vect sau trong R[x] h no a) 1 = 1, 2 = x, 3 = x2; b) 1 = 1, 2 = x + 1, 3 = x2+ 1, 4 = 2x2 + x + 3. 20. Cho hai vect = (3, a + b, 5), = (a + 1, b - 2, 10) trong Q3. Tm a v b hai vect ny ph thu c tuy n tnh. 21. Cho ba vect 1, 2, 3 c a K-khng gian vect V, tuy n tnh. c l p c l p tuy n tnh: c l p tuy n tnh?
a) Ch ng minh r ng ba vect 1 = 1, 2 = 1 + 2 , 3 = 1 + 2 3 c l p tuy n tnh. b Ch ng minh r ng ba vect 1 = 1 + 2, 2 = 2 + 3, 3 = 3+ 1 c l p tuy n tnh. c) Ba vect = 1 + 2, 2 = 2 + 3, 4 = 3 - 1 c tuy n tnh khng? 22. Ch ng minh r ng: N u hai vect 1, 2 c a khng gian vect V l c l p tuy n tnh v cl p
R 1, R 2 l nh ng khng gian con c a V l n l t sinh b i 1, 2 th R1 R2 = 0 . 4. C S C A KHNG GIAN VECT
23. Cho { 1,..., i,..., n} l m t c s c a khng gian vect V. Ch ng minh r ng: a) N u thay i b i r i (v i r 0) th h vect thu c cng l m t c s c a V; b) N u ta c ng vo i m t t h p tuy n tnh c a cc vect cn l i th c m t c s m i c a V. 24. Ch ng minh r ng cc h vect sau l nh ng c s c a khng gian vect R3:
25. Cc h vect sau c ph i l c s c a khng gian vect R4 khng:
26. ch ng minh r ng n u ba vect 1, 2, 3 l p thnh m t c s c a K-khng gian vect V th ba vect 1 = 1 + 2, 2 = 2 + 3, 3 = 3+ 1 cng l p thnh m t c s c a V. 27. G i P3 l khng gian vect g m a th c 0 v cc a th c f(x) R[x] c b c f(x) 3. Ch ng minh r ng hai h vect:2 3 1 = 1, 2 = x, 3= x , 4= x ;
1 = 1, 2= x - 1, 3 = (x - 1) , 4 = (x - 1)
2
3
l hai c s c a P3. 28. Ch ng minh r ng n u { 1,..., 2,..., n} l m t c s c a Rkhng gian vect V th V =
i =1
n
R i v R i
j1
R j = 0 , v i m i i
{1, 2,..., n}, trong
j1
R j l t ng c a cc R j, (ch ng h n, R 3
(R 1 + R 2 + R 4 +...+ R n) = { 0 }. Ng i ta ni r ng V l t ng tr c ti p c a cc khng gian con R i, i {1, 2,..., n}. 5. S CHI U C A KHNG GIAN VECT
29. Tm s chi u c a khng gian vect V sinh b i cc h vect sau:
30. Gi s U, W l hai khng gian con c a khng gian vect V v V = U + W. Ch ng minh r ng dimV = dimU + dimW khi v ch khi U W = {0 } 31. Gi s U, W l hai khng gian con th c s c a khng gian vect V, U W. a) N u dimV = 3, dimU = dimW = 2 th dim(U W) b ng bao nhiu? b) n u dimV = 6, dimU = dimW = 4 th dim(U W) c th b ng bao nhiu? 32. Trong R4, khng gian con U sinh b i hai vect 1 = (-1, 1, 1, 1), 2 = (1, 2, 1, 0), khng gian con W sinh b i hai vect 1 = (2, -1, 0, 1), 2 = (0, - 5, 6, 0). Tm dim(U + W) v dim(U W). 33. Cho h vect { 1, 2, 3, 4} l m t c s c a khng gian vect V. U l khng gian con sinh b i { 1, 2, 3,}, W l khng gian con sinh b i { 2, 3, 4,}. a) ch ng minh r ng { 2, 3} l m t c s c a U W. b) Tm c s v s chi u c a U + W. 6. T A 34. Tm t a C A VECT i v i c s { 1, 2, 3, 4}:
c a cc vect sau
35. Tm t a
c a vect = (5, - 2, 4, 1)
i v i c s :
36. Bi t t a
c a cc vect
i v i m t c s () no nh sau:
(0, -5, 4, 1), (2, 7, 0, 9), (4, 0, 1, 2).
Tm t a
c a cc vect sau
i v i c s ():
37. G i P3 l khng gian vect g m a th c 0 v cc a th c f(x) R[x] c b c f(x) 3. a) Ch ng minh r ng: () :
l hai c s c a khng gian P3. b) Tm ma tr n chuy n t c s () sang c s (). c) Tm t a 5. S c a cc vect f(x) = 2x3 - x + 5 i v i c s ().
CHI U C A KHNG GIAN VECT
29. Tm s chi u c a khng gian vect V sinh b i cc h vect sau:
30. Gi s U, W l hai khng gian con c a khng gian vect V v V = U + W. Ch ng minh r ng dimV = dimU + dimW khi v ch khi U W = { 0 }. 31. Gi s U, W l hai khng gian con th c s c a khng gian vect( V, U W.
a) N u dimV = 3, dimU = dimW = 2 th dim(UW) b ng bao nhiu? b) N u dimV = 6, dimU = dimW = 4 th dim(UW) c th b ng bao nhiu? 32. Trong R4, khng gian con U sinh b i hai vect 1 = (1, 1, 1, 1) 2 = (1, 2, 1, 0), khng gian con W sinh b i hai vect 1 (2, -1, 0, 1) 2 = (0, - 5, 6, 0). Tm dim(U + W) v dim(U W). 33. Cho h vect { 1, 2, 3, 4} l m t c s c a khng gian vect V U l khng gian con sinh b i { 1, 2, 3}, W l khng gian con su b i { 2, 3, 4}. a) Ch ng minh r ng { 2, 3} l m t c s c a U W. b) Tm c s v s chi u c a U + W.
a) Dng h ng c a h vect c s c a R4. c) Tm t a
ch ng t r ng hai h vect ny l hai
b) Tm ma tr n chuy n t c s a sang c s b. c a = (2, 0, 4, 0) it a i v i c s b. c a i v i c s . i i s c p khng lm thay d) Dng cng th c tnh t a
42. a) Ch ng minh r ng cc php bi n h ng c a ma tr n.
b) B n hy t tm m t h vect trong khng gian R4 v dng cc php bi n i s c p ch ng t r ng h vect y c l p tuy n tnh. c) B n hy t ch n hai h , m i h g m 4 vect, r i dng cc php bi n i s c p ch ng t r ng m t h c h ng 3, m t h c h ng 2 v ch ra c s khng gian sinh b i m i h vect cho.