Analisis Matricial by Jorge Antezana y Demetrio Stojanoff PDF

By Jorge Antezana y Demetrio Stojanoff

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1 sabemos que 0 < λ1 (A[n−1] ) . 2 nos asegura que λ2 (A) ≥ λ1 (A[n−1] ) > 0. Luego n 0< n λk (A) y tamb´en 0 < det A[n] = det A = k=2 λk (A) . k=1 De ah´ı deducimos que λ1 (A) > 0. 1, podemos concluir r´apidamente que Ax, x > 0 para todo x = 0, o sea que A ∈ Gl (n)+ . La prueba de la equivalencia con el ´ıtem 3 es exactamente igual, pero usando para la inducci´on a A(1) ∈ Mn−1 (C) . 7 (dif´ıcil). Probar que, dada A ∈ H(n), entonces A ∈ Mn (C)+ ⇐⇒ det A[J] ≥ 0 para todo J ⊆ In . Se suguiere induccionar en n.

2). 6. Sea A ∈ Mn (C). Las siguientes condiciones son equivalentes: 1. A ∈ Mn (C)+ . 2. Existen y1 , . . , yr ∈ Cn tales que A = r r yi yi · yi∗ . yi = i=1 i=1 Demostraci´ on. La implicaci´on 2 → 1 es clara, porque cada matriz yi · yi∗ ∈ Mn (C)+ . Rec´ıprocamente, si A ∈ Mn (C)+ , sea B = {x1 , . . , xn } es una BON de Cn adaptada a µ(A). 12), para todo z ∈ Cn se tiene que n Az = A n z, xi xi = i=1 n z, xi Axi = i=1 n µi (A) z, xi xi = i=1 µi (A) xi xi z . i=1 Luego basta elegir yi = µi (A)1/2 xi para aquellos i ∈ In tales que µi (A) > 0.

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 . Luego i=1 Ay, y ≤ λk (A) =⇒ m´ın Ax, x ≤ λk (A) =⇒ λk (A) ≥ x∈S1 m´ax m´ın Ax, x . dim S=n−k+1 x∈S1 La otra f´ormula se demuestra en forma an´aloga: el m´ın se alcanza en M = Hk , y cualquier dim M=k otro tal M cumple que M ∩ Kk = {0}.

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Analisis Matricial by Jorge Antezana y Demetrio Stojanoff

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