Baixe Respostas exercícios do Cap IV do livro do Boldrini e outras Exercícios em PDF para Geometria Analítica e Álgebra Linear, somente na Docsity! Seja , temos então, pela propriedade (iii) que:n + = + = u u v v u u 1 2 1 2 1 2 1 1 2 2 1 2 1 1 1 1 1 1 2 2 2 2 2 2 ( , , , ) ( , , , ) ( , , , ) ( , , , ) ( , , , ) 0 0 0 n n n n n n n n n n n n x x x y y y x x x x y x y x y x x x x y x y x x x y x y x x x y x y x x + = + + + = + = = − = + = = − = + = = − = 1 2( , , , ) (0,0, ,0) (vetor nulo)ny y y V = = = v 0 1 2 1 2 1 1 2 2 1 1 1 1 2 2 2 2 1 2 1 2 Seja , temos então, pela propriedade (iv) que: ( , , , ) ( , , , ) (0,0, ,0) ( , , , ) (0,0, ,0) 0 0 0 ( , , , ) ( , , , ) n n n n n n n n n n n x x x y y y x y x y x y x y y x x y y x x y y x y y y x x x + = + = + = + + + = + = = − + = = − + = = − = = − − − u u v v u 0 v v (vetor oposto)= −u 1 1 1 1 2 2 2 2 1 2 1 2 1 2 1 2 1 1 1 1 1 1 1 1 4 ) Sejam , , devemos mostrar que: i) ) , ( , , , ) ( , , , ) ( , , , ) ( , , , ) ( , , , ) Logo W é um subespaço vetorial de a W W ii W x x t t x x t t x x x x t t t t W x x t t x x t t W + = − = − + = + − − + + = − = − u v u v u u v u v u y x= − z t= 1 1 1 1 2 2 2 2 1 2 1 2 1 2 1 2 1 1 1 1 1 1 1 1 ) Sejam , , devemos mostrar que: i) ) , ( , ,0,2 ) ( , ,0,2 ) ( , ,0,2( ) ) ( , ,0,2 ) ( , ,0,2 ) Logo W é um subespaço vetorial b W W ii W x y x y x y x y x x y y x x y y W x y x y x y x y W + = + = + + = + + + + + = + = + u v u v u u v u v u 4 de 2t x y= + ) Sejam e 0 0 0 0 0 ) , , , , , , ) , , , , a V a b c d V a c m n a m c n i V a c d m n p c d n p c n d p a c a c ii V a c d c d c d = = = = + + + = + = + + = = A, B A B A (2,2)Logo é um subespaço vetorial de M .V 1 1 2 1 2 0 0 1 | 0 ( 0) | 0 01 | 0 | 0 a c x y b d a c L L a a b d c com a a L bL L b d + = → → − + 2 2 1 | 0 0 | 0 c a a L Lbc ad ad bc a − + − 0 2 e 2 0 1 e 2 ad bc p n LI ou ad bc p n LD − = = − = = = 2 2 1 | 0 0 1 | 0 c a a L L ad bc − OU 11 11 22 22 1 2 11 11 22 22 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 nn nn nn nn a b a b D D a b a b a b D D W a b + = + = + + + = + 11 11 22 22 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0nn nn a a a a D W a a = = ( , ) ( , )Como ( , ,.) é um subespaço vetorial de n n n nW M W M + 11 22 1 11 22 0 0 0 0 0 0 Continuação a) (gerador) [ ] 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 nn nn a a D a a a a = = + + + 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 , , ,0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 = ( ) ( )nDim W Dim D n= = Resolução: Pode-se notar que o conjunto em questão está contido no conjunto das matrizes 2x2 (vide página 104) Logo, se verificarmos que este conjunto é um subespaço das matrizes M2x2 ele será um Espaço Vetorial também.... (GERADOR) [ ] a a b C a b + = i) Sejam C1 e C2 matrizes em C, onde: 1 1 1 1 1 1 a a b a b + = C 2 2 2 2 2 2 a a b a b + = Ce 1 1 1 2 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 1 2 , , , , a a b a a b a b a b a a a b a b C a a b b a a b b + + + = + + + + + + + C C ii) Sejam e C1 C , temos: 1 1 1 1 1 1 1 1 1 1 1 1 1 ( ) , , . a a b a a b a b a b a b + + = = C C Podemos concluir que C é um subespaço de M(2,2) Resolução: Pode-se notar que o conjunto em questão está contido no conjunto das n-uplas (n, +, .) 2 ,1, 1,2 (1,1, 2,4) b(1,1, 1,2) c(1,4, 4,8) 3 a − = − + − + − O vetor pertencerá a S se, e somente se: 2 ,1, 1,2 3 = − v 2 3 4 1 2 4 1 4 2 8 2 a b c a b c c b c a b c + + = + + = − − − = − + + = 2 1 1 1 3 1 1 4 1 2 1 4 1 4 2 8 2 − − − − Matriz Ampliada 1 2 2 1 1 1 3 3 1 1 4 1 ~ 2 1 4 1 4 2 8 2 L L − − − − 2 1 2 3 1 3 4 1 4 1 1 4 1 33 3 3 2 22 1 4 1 44 2 8 2 L L L L L L L L L → − + → +− − − − → − + 2 3 1 1 4 1 0 0 9 1 ~ 0 1 4 1 0 2 8 2 L L − − − − − 1 4 1 1 4 1 0 1 4 1 0 0 9 1 20 2 8 2 L L − − +− − − 1 1 4 1 0 1 4 1 ~ 0 0 9 1 0 0 0 0 − − O posto de A = 3 e n = 3 (SPD) 9 1 1 9 c c − = − = 4 1 1 9 4 5 1 4 9 9 9 b c b + = − = − = = 4 1 5 1 9 9 1 4 1 4 0 9 9 9 a b c a b c + + = − = − − = − − = = 2 5 1 ,1, 1,2 0(1,1, 2,4) (1,1, 1,2) (1,4, 4,8) 3 9 9 − = − + − + − ( )0,0,1,1 (1,1, 2,4) b(1,1, 1,2) c(1,4, 4,8)a= − + − + − O vetor pertencerá a S se, e somente se: ( )0,0,1,1=v 0 4 0 2 4 1 4 2 8 1 a b c a b c c b c a b c + + = + + = − − − = + + = 1 1 1 0 1 1 4 0 2 1 4 1 4 2 8 1 − − − Matriz Ampliada Logo o vetor 0 2 3 1 não pertence ao conjunto W. 2 2 0 2 0 3 1 a a b a b + = − Observe que 0 3 0 2 0 0 0 0 0 1 3 4 1 1 0 1 0 0 5 0 0 0 1 0 0 0 a b c = + − + 0 0 0 0 0 0 2 0 0 3 0 4 0 0 5 0 0 0 0 c a a b b + + = + + = + + = − − = + + = + + = 2 3 5 4 3 5 4 (F) c a b a b = = = − = − = 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 x y z w + + + = 0 0 0 0 x y z w = = = = 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 + + + = Os vetores são LI e geram M(2,2) Logo, esse conjunto é uma base de M(2,2) 1 11 12 1 2 21 22 2 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 n n n n a a a a a a a a + + + + + + + + 0 0 0 0 0 0 0 0 1 nna + 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 , , , , , , , 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 , , 0 0 0 = 0 0 0 0 0 0 0 0 0 0 0 0 , , , 1 0 0 0 1 0 0 0 1 2( )Dim W n= (1,0,0) (1,1,1) ( 1,1,0) (1,0, 1)a b c= + − + − 2 1 2 3 1 3 1 1 1 1 1 0 1 1 0 0 0 1 0 1 0 a b c a b L L L a c L L L − + = − + = →− + − = − → − + 2 3 1 1 1 1 ~ 0 2 1 1 0 1 2 1 L L − − − → − − 3 2 3 1 1 1 1 ~ 0 1 2 1 0 2 1 1 2L L L − − − − − → − + 1 1 1 1 ~ 0 1 2 1 0 0 3 1 − − − 1 3 1 3 c c= = 2 1 1 2b c b c− = − = − + 1 2 1 1 2 3 3 2 1 3 3 b c b b − = − = − + − + = = − 1 1 1 1 1 3 3 a b c a b c a − + = = − + = − − + 2 3 1 3 3 a − + = = (1,0,0) (1,1,1) ( 1,1,0) (1,0, 1)a b c= + − + − 1 1 1 (1,0,0) (1,1,1) ( 1,1,0) (1,0, 1) 3 3 3 = − − + − 2 0 1 2( ( ) , ,.)n n nP x a a x a x a x= + + + + + 2{1, , , } ( ( )) 1n nx x x Dim P x n = = + 2 2 0 1 2 2 2 2 0 1 2 Note que: ( ) (1 0 0 ) (0 0 ) (0 0 ) P x a a x a x a x x a x x a x x = + + = + + + + + + + + 2{1, , }x x =