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Download Algebra Lineal by Gabriela Jeronimo, Juan Sabia y Susana Tesauri PDF

By Gabriela Jeronimo, Juan Sabia y Susana Tesauri

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Sample text

Vi , . . , vj , . . , vn } implica que α1 = . . = αi = . . = αi λ + αj = . . = αn = 0, de donde αk = 0 para todo 1 ≤ k ≤ n. En consecuencia, el conjunto {v1 , . . , vi + λvj , . . , vj , . . , vn } es linealmente independiente. La otra implicaci´on se deduce de ´esta observando que el conjunto {v1 , . . , vn } se obtiene de {v1 , . . , vi + λvj , . . , vj , . . , vn } cambiando el i-´esimo vector vi + λvj por (vi + λvj ) + (−λ)vj = vi . Como consecuencia de la proposici´on anterior, para decidir si un subconjunto de vectores {v1 , .

Para cada n ∈ N consideraremos el conjunto de todas las matrices inversibles en K n×n : GL(n, K) = {A ∈ K n×n / A es inversible}. Nos interesa estudiar la estructura de este conjunto. 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 .

Como veremos m´as adelante, aplicando esta propiedad se trabajar´a en un K-espacio vectorial de dimensi´on n arbitrario como si fuese K n . 37 Sea V un K-espacio vectorial de dimensi´ on finita. Sea {v1 , . . , vn } una base de V . Entonces para cada x ∈ V existen u ´nicos α1 , . . , αn ∈ K tales que x = n i=1 αi vi . Demostraci´ on. La existencia se deduce de que, por ser una base de V , {v1 , . . , vn } es un sistema de generadores de V . Supongamos que n i=1 αi vi = n i=1 βi vi , entonces n (αi − βi )vi = 0.

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