February 10, 2018

Download PDF by Jorge Antezana y Demetrio Stojanoff: Analisis Matricial

By Jorge Antezana y Demetrio Stojanoff

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Sean λ ∈ σ (A) y x ∈ Cn tales que x = 0 y Ax = λx. Llamemos X a la matriz cuyas columnas son todas iguales al vector x. Luego, AX = λX, y por ende |λ| X = AX ≤ A X , de donde se deduce que |λ| ≤ A . Como el autovalor era cualquiera, ρ(A) ≤ A . 2, se tiene que σ (Am ) = σ (A)m , y entonces tambi´en ρ(Am ) = ρ(A)m . Por lo tanto, usando la parte ya probada, obtenemos que ρ(A) ≤ Am 1/m . 7. Dada una norma matricial · en Mn (C) y una matriz S ∈ Gl (n), la f´ormula A S := SAS −1 , A ∈ Mn (C), define otra norma matricial.

Vn } una BON de Cn adaptada a λ(A). 1, cualquier x ∈ Cn verifica la Eq. 3). Dado r ∈ In , notemos por Hr = Gen {v1 , . . , vr } y Kr = Gen {vr , . . , vn }. Notar que dim Hr = r y dim Kr = n − r + 1. Por la Eq. 3) vemos que, si x ∈ Kk , n λi (A) | x, vi | 2 =⇒ λk (A) = m´ın Ax, x ≤ Ax, x = x∈(Kk )1 i=k m´ax m´ın Ax, x . dim S=n−k+1 x∈S1 Por otro lado, si dim S = n − k + 1, entonces S ∩ Hk = {0}. Pero si y ∈ (S ∩ Hk )1 , la Eq. 3) asegura que Ay, y = λi (A) | y, vi | 2 y que y 2 i=1 k = | y, vi | 2 = 1 .

6 de bastante m´as adelante). 10). Sea H un C-espacio de Hilbert y sea A ∈ L(H). Entonces 1 Ax, y = 4 4 i k A (x + i k y), (x + i k y) para todo par x, y ∈ H . 2. 2 (sobre matrices triangulares). 3. 2 (sobre matrices de bloques). 4. 4 usando la definici´on del determinante con permutaciones de Sn . Usar que las permutaciones que no pasan por el bloque nulo de abajo, son todas las del tipo σ ∈ Sk × Sn−k . 5. 3 (sobre matrices tipo x y). 6. Sean A, ∈ Mm,n (C). 7. Sean A, ∈ Mm,n (C) |aij |2 . = j=1 para todo par i, j ∈ In .

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Analisis Matricial by Jorge Antezana y Demetrio Stojanoff

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