Algebra Lineal by Gabriela Jeronimo, Juan Sabia y Susana Tesauri

By Gabriela Jeronimo, Juan Sabia y Susana Tesauri

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Vs } es linealmente independiente, entonces es una base de V . Si no es linealmente independiente, alguno de los vectores del conjunto es combinaci´on lineal de los otros. Supongamos que vs ∈ < v1 , . . , vs−1 >. Consideramos ahora {v1 , . . , vs−1 }, que es un sistema de generadores de V , y procedemos inductivamente. ii) Sea B = {z1 , . . , zn } una base de V . Sea G0 = < w1 , . . , wr >. Consideramos G1 := {w1 , . . , wr , z1 } {w1 , . . , wr } si z1 ∈ / < G0 > si z1 ∈ < G0 >. Se procede inductivamente para 2 ≤ i ≤ n, es decir, Gi := Gi−1 ∪ {zi } si zi ∈ / < Gi−1 > Gi−1 si zi ∈ < Gi−1 >.

I. 2. {v1 , . . , vi , . . i. ⇐⇒ {v1 , . . , λvi , . . i. para λ ∈ K − {0}. 3. {v1 , . . , vi , . . , vj , . . i. i. para λ ∈ K. ⇐⇒ {v1 , . . , vi + λvj , . . , vj , . . , vn } ⊆ V es Demostraci´ on. 1. Se deduce del hecho que en un conjunto no interesa el orden de sus elementos. 2. Supongamos que {v1 , . . , vi , . . , vn } es linealmente independiente. Sean α1 , . . , αn ∈ K tales que α1 v1 + · · · + αi (λvi ) + · · · + αn vn = 0. λ = 0. Puesto que λ = 0, resulta que tambi´en αi = 0.

12 Para cada n ∈ N , se verifican las siguientes propiedades: 1. B ∈ GL(n, K). B)−1 = B −1 A−1 . En particular, el producto de matrices · es una operaci´ on en GL(n, K). 2. In ∈ GL(n, K). 3. Si A ∈ GL(n, K), entonces A−1 ∈ GL(n, K). Demostraci´ on. 1. Sean A, B ∈ GL(n, K). Entonces existen A−1 y B −1 . (B −1 . A−1 ) = In y (B −1 . B) = In . B)−1 = B −1 . A−1 . 2. In = In . 3. De la definici´on de inversa se deduce inmediatamente que si A ∈ GL(n, K), entonces (A−1 )−1 = A y por lo tanto A−1 ∈ GL(n, K).

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