CH¯ NG 1: GI TR» RING – VECTOR RING – D NG CHU¨N T®C JORDAN

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1. CHNG 1: GI TR RING VECTOR RING DNG CHUN TC JORDAN _______________________________________________________ I. Gi tr ring v vector ring ca ma trn Cho ha ma trn: 1. Tm gi tr ring v vector ring ca mt ma trn: V d: Cho ma trn 7 2 4 1 A = - . a) Xc nh a thc c trng ca A . b) Xc nh cc gi tr ring i l ca A . c) Xc nh chiu v mt c s khng gian vect ring ( )A i E l . d) Xc nh mt c s S ca 2 gm cc vect ring ca A . Gii a) a thc c trng ( )AP t ca A l 2 2 ( ) tr( ) det 8 15.A t t A t A t tP = + = + b) Cc gi tr ring i ca A l cc nghim ca phng trnh c trng ( ) 0Af t = . Phng trnh c trng ( ) 0Af t = c cc nghim 3, 5. Vy 1 3 = v 2 5 = l cc gi tr ring ca ma trn A. c) Vi 1 3 = . Cc vc t ring ca ma trn A ng vi gi tr ring 1 3 = l cc nghim khng tm thng ca h phng trnh tuyn tnh thun nht 1 2 1 1 2 2 4 2 0 4 2 0 2 x x x a x x x a + = = = = Vy khng gian vc t ring (3)AE ca A ng vi gi tr ring 1 3 = l (3) {( , 2 ) | } { (1, 2) | } (1, 2)AE a a a a a= = = Vy dim (3) 1AE = v {(1, 2)} l mt c s ca (3)AE . * Vi 2 5 = . Cc vc t ring ca ma trn A ng vi gi tr ring 2 5 = l cc nghim khng tm thng ca h phng trnh tuyn tnh thun nht 1 2 1 1 2 2 2 2 0 4 4 0 x x x a x x x a + = = = = Vy khng gian vc t ring (5)AE ca A ng vi gi tr ring 2 5 = l (5) {( , ) | } { (1, 1) | } (1, 1)AE a a a a a= = = Vy dim (5) 1AE = v {(1, 1)} l mt c s ca (5)AE . 2. d) t {(1, 2),(1, 1)}S = gm cc vc t ring ca A c lp tuyn tnh trong 2 . Do S l mt c s ca 2 . Bi tp: 1) Cho ma trn 1 4 2 3 4 0 3 1 3 A - - = - - . a) Xc nh a thc c trng ca A . b) Xc nh cc gi tr ring i l ca A . c) Xc nh chiu v mt c s khng gian vect ring ( )A i E l . d) Xc nh mt c s S ca 3 gm cc vect ring ca A . Hng dn: Sinh vin lm tng t nh v d. 2) Cho ma trn 1 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 A = a) Xc nh a thc c trng ( )A f t ca A . b) Xc nh cc gi tr ring i l ca A . c) Xc nh chiu v mt c s khng gian vect ring ( )A i E l . d) Xc nh mt c s S ca 4 gm cc vect ring ca A . Hng dn: Sinh vin lm tng t v d a thc c trng ( )AP t ca A l 2 1 0 1 1 0 1 1 1 ( ) det( ) ( 1) ( 1)( 3). 1 1 1 0 1 1 0 1 A t t t A tI t t t t t P = = = + 2. Chng minh cc tnh cht i vi gi tr ring v vector ring: 1) Cho l gi tr ring ca M ( )nA K , K v k . Chng minh rng a) l gi tr ring ca ma trn A . b) k l gi tr ring ca ma trn k A . 3. c) + l gi tr ring ca ma trn A I+ . d) ( )f l gi tr ring ca ma trn a thc ( )f A . Hng dn: a) Do l gi tr ring ca M ( )nA K nn tn ti n v K sao cho Av v= . ( ) ( )( )A v Av v v = = = . Vy l gi tr ring ca ma trn A . b) Ta c ( ) ( )1 1 1 ( ) ...k k k k k A v A Av A v A v v = = = = = . Vy k l gi tr ring ca ma trn k A . c) Ta c ( ) ( )A I v Av Iv v v v + = + = + = + Vy + l gi tr ring ca ma trn A I+ . d) Gi s 1 ( ) [ ]. n i i i f t a t K t = = Khi , 1 1 ( ) , ( ) n n i i i i i i f a f A a A = = = = V ( ) ( ) 1 1 1 1 ( ) ( ) n n n n i i i i i i i i i i i i f A v a A v a A v a v a v f v = = = = = = = = = Vy ( )f l gi tr ring ca f (A). Sinh vin cho v d minh ha cho nhng kt qu trn. 2) Cho l gi tr ring ca M ( )nA K . Chng minh rng a) Nu A kh nghch th 1 l gi tr ring ca ma trn 1 A . b) Nu A kh nghch th 1 + l gi tr ring ca ma trn 1 A A + . Hng dn : a) V A kh nghch nn 0 . Ta c, ( )1 1 1 1 1 1 A v A v A Av v = = = Vy Nu A kh nghch th 1 l gi tr ring ca ma trn 1 A . b) V A kh nghch nn 1 1 A v v = . Khi , ta c 1 1 1 1 ( ) ( )A A v Av A v v v v + = + = + = + Nu A kh nghch th 1 + l gi tr ring ca ma trn 1 A A + . Sinh vin tm cc v d minh ha cho nhng kt qu trn. 3) Cho A l ma trn vung cp n trn K v 1 2, , , n l cc gi tr ring ca n. Chng minh rng 1 2det nA = L . Hng dn: Do 1 2, , , n l cc gi tr ring ca A nn 1 2, , , n l cc nghim ca a thc c trng ( )Af t . Do , 1 2( ) det( ) ( 1) ( )( )...( )n A nf t A I t t t = = . Ly t = 0, ta c: 1 2 1 2det (0) ( 1) (0 )(0 )...(0 ) ...n A n nA f = = = Sinh vin tm cc v d minh ha cho nhng kt qu trn. 4. 4) Cho A l ma trn vung cp n trn K v 1 2, , , n l cc gi tr ring ca n. Chng minh rng a) 1 2det( ) n nA = L . b) 1 2det k k k k nA = L . c) 1 2det( ) ( )( ) ( )nA I + = + + +L . d) 1 2det ( ) ( ) ( ) ( )nf A f f f = L . Hng dn: a) Do 1 2, , , n l cc gi tr ring A nn 1 2, , , n l cc gi tr ring ca ma trn A . Do 1 2 1 2det( ) ( )( ) ( ) .n n nA = = L L Sinh vin cho v d minh ha. b) Do 1 2, , , n l cc gi tr ring A nn 1 2, , ,k k k n l cc gi tr ring ca ma trn k A . Do 1 2det k k k k nA = L . c) Do 1 2, , , n l cc gi tr ring A nn 1 2, , n + + + l cc gi tr ring ca ma trn A I+ . Do 1 2det( ) ( )( ) ( )nA I+ = + + + L . d) Do 1 2, , , n l cc gi tr ring A nn 1 2( ), ( ), , ( )nf f f l cc gi tr ring ca ma trn ( )f A . Do 1 2det ( ) ( ) ( ) ( )nf A f f f= L . Sinh vin cho cc v d minh ha. 5) Cho A l ma trn vung cp n trn K v 1 2, , , n l cc gi tr ring ca n. Chng minh rng a) Nu A kh nghch th 1 1 1 1 1 2det nA = L . b) Nu A kh nghch th 1 1 1 1 1 1 2 2det( ) ( )( ) ( )n nA A + = + + +L . c) Nu K khng l gi tr ring ca A th ma trn A I kh nghch v 1 1 1 det( ) . n i i A I = = Hng dn: a) Do 1 2, , , n l cc gi tr ring A nn 1 1 1 1 2, , , n l cc gi tr ring ca ma trn 1 A . Do 1 1 1 1 1 2det nA = L . b) Do 1 2, , , n l cc gi tr ring A nn 1 1 1 1 1 2 2, , , n n + + + l cc gi tr ring ca ma trn 1 A A + . Do 1 1 1 1 1 1 2 2det( ) ( )( ) ( )n nA A + = + + + L 5. c) Do khng l gi tr ring ca A nn nh thc ca ma trn A I khc 0. Vy A I kh nghch. Theo gi thit 1 2, , , n l cc gi tr ring ca A nn 1 2, , , n l cc gi tr ring ca ma trn A I v do 1 1 1 1 2( ) ,( ) , ,( )n l cc gi tr ring ca 1 ( )A I . Vy 1 1 1 1 1 det( ) ( ) . n n i i i i A I = = = = Sinh vin cho v d minh ha. 3. Cho ha ma trn: Cch cho ha mt ma trn: Cho A l mt ma trn vung cp n. cho ha ma trn A ta lm nh sau: Tm cc gi tr ring v cc vector ring c lp tuyn tnh ca A, bng cch tm a thc c trng, gii phng trnh c trng tm cc gi tr ring sau ng vi tng gi tr ring tm cc vector ring. Khi xy ra mt trong hai kh nng sau: TH1: Nu tng s vector ring c lp tuyn tnh ca A b hn n th kt lun A khng cho ha c. TH2: Nu tng s vector ring c lp tuyn tnh ca A bng n th kt lun A cho ha c. Khi ma trn P cn tm l ma trn m cc ct ca n l cc vector ring c lp tuyn tnh ca A vit theo ct v khi 1 21 0 ... 0 0 ... 0 ... ... ... ... 0 0 ... n P AP = l ma trn cho trong cc i l cc gi tr ring ca A ng vi vector ring l vector ct th i ca ma trn P. 1. V d: Cho ha ma trn sau: 0 1 1 1 0 1 1 1 0 A = Hng dn: a thc c trng ca ma trn A l: 3 1 1 ( ) 1 1 3 2 1 1 AP = = + + 3 ( ) 0 3 2 0 1, 2AP = + + = = = Vy ma trn A c hai gi tr ring l 1, 2 = = . ng vi 1 = , gii h pt: 1 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 6. H c v s nghim ph thuc hai tham s ] 1 2 3 2 2 3 3 x t t x t x t = = = Khng gian con ring ng vi gi tr ring 1 = l 2 3 2 3 2 3( 1) {( , , ) | , }E t t t t t t = C s ca E(-1) gm hai vector 1 2( 1,1,0); ( 1,0,1) = = . ng vi gi tr ring 2 = , tm vector ring ta gii h pt: 2 1 1 0 1 1 2 0 1 1 2 0 1 1 2 0 1 2 1 0 1 2 1 0 0 3 3 0 0 3 3 0 1 1 2 0 2 1 1 0 0 3 3 0 0 0 0 0 H c v s nghim ph thuc 1 tham s 1 2 3 x t x t x t = = = Do , khng gian con ring ca A ng vi gi tr ring 2 = l {(2) ( , , ) | }E t t t t= C s ca (2)E gm 1 vector 3 (1,1,1) = . Nhn xt: Cc vector 1 2 3, , c lp tuyn tnh nn ma trn A cho ha c. Khi , tn ti ma trn kh nghch P sao cho 1 P AP D = vi D l ma trn cho. 1 1 1 1 0 1 0 1 1 P = v 1 0 0 0 1 0 0 0 2 D = 2. Bi tp: 1. Cho ma trn 1 2 3 0 2 3 0 0 3 A = . Hi ma trn A c cho ha c khng? Tm ma trn C lm cho ha A (nu c). Hng dn: SV. Lm tng t nh v d. 2. Cho A, B v P l cc ma trn sao cho 1 A PBP = . Chng minh rng 1k k A PB P = vi mi k . Hng dn: S dng tnh cht 1 1 1 . ...k A PBP PBP PBP = (k ln) v 1 .P P I = . Sinh vin cho v d minh ha. 3. Cho ma trn 4 3 2 1 A = a) Xc nh a thc c trng v cc gi tr ring ca A. b) Xc nh mt c s ca khng gian vector ring tng ng c) Chng t rng A cho ha c. Tm mt ma trn kh nghch P v ma trn ng cho D sao cho 1 A PDP = d) Tnh k A vi mi s nguyn dng k. 7. Hng dn: Cc cu a); b); c) lm tng t nh cc v d trong ti liu. Cu d) p dng tnh cht ca bi 2.(Tc l khi 1 A PDP = th 1k k A PD P = ). 4. Cho ma trn 2 2 1 1 3 1 1 2 2 A = a) Xc nh a thc c trng v cc gi tr ring ca A. b) Xc nh mt c s ca khng gian vector ring tng ng c) Chng t rng A cho ha c. Tm mt ma trn kh nghch P v ma trn ng cho D sao cho 1 A PDP = d) Tnh k A vi mi s nguyn dng k. Hng dn: Lm tng t nh bi 3. 5. Cho ma trn 3 12 2 7 A = , 1 2 3 2 , 1 1 u u = = . Chng minh rng 1 2,u u l cc vector ring ca A. Hy tm mt ma trn kh nghch P v ma trn ng cho D 1 A PDP = . Hng dn: chng minh 1 2,u u l cc vector ring ca A th cn tm cc gi tr 1 2; sao cho 1 1 2 2;Au u Au u = = . Khi , ma trn ng cho D c dng 1 2( , )diag . 6. Cho ma trn vung cp 4 A c cc gi tr ring l 5, 3, -2. Gi s khng gian vector ring ng vi gi tr ring 3 = c chiu l 2. Hi ma trn A c cho ha c khng? Hng dn: Da vo iu kin cho ha c ca ma trn. 7. Hy xc nh a thc c trng v mt c s khng gian vector ring ca cc ma trn sau. Trong s cc ma trn sau y ma trn no cho ha c, khi hy tm ma trn kh nghch P v ma trn ng cho D sao cho 1 A PDP = . 2 7 5 3 3 4 ) 7 2 4 3 4 8 a 1 0 1 6 2 0 0 3 1 ) 2 3 1 2 9 0 3 0 2 0 6 0 5 8 3 1 2 0 b 1 0 1 1 0 1 1 1 2 1 2 4 0 1 1 1 1 0 1 1 12 1 4 9 ) 1 1 1 0 1 1 0 1 6 5 2 4 1 1 0 1 1 1 1 0 3 4 5 10 c 8. Xc nh a thc c trng ca ma trn sau trn 1 1 1 a b A a c b c = v a b c d b a d c B c d a b d c b a = 9. Cho ha cc ma trn sau (nu c). 8. 3 1 1 5 5 1 0 5 2 3 4 1 4 2 2 2 4 2 2 2 4 1 4 2 3 4 0 3 1 3 7 4 16 2 5 8 2 2 5 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 5 3 0 9 0 3 1 2 0 0 2 0 0 0 0 2 10.Cho ma trn A trn trng s thc nh sau 9 1 5 7 8 3 2 4 0 0 3 6 0 0 1 8 A = a) Tnh det A b) Tnh 4det(