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Analisis Matricial by Jorge Antezana y Demetrio Stojanoff

By Jorge Antezana y Demetrio Stojanoff

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Y para ver esto, alcanza con hacer la siguiente cuenta: v v∗ ◦ B = vi vj Bij i,j∈In = diag (v) B diag (v)∗ ∈ Mn (C)+ , donde la igualdad del medio se testea haciendo la cuenta, o mirando la Eq. 14). 3. Sean A, B ∈ Mn (C)+ , entonces 1. µn (A)µn (B) ≤ µn (A ◦ B). 2. A ◦ B sp = µ1 (A ◦ B) ≤ µ1 (A)µ1 (B) = A sp B sp . Demostraci´ on. 13), pero usando que A ≥ µn (A)I y B ≥ µn (B)I. 2). 4. Si A ∈ Mn (C)+ , entonces B = |Aij |2 Demostraci´ on. Se deduce de que AT = A = Aij i,j∈In i,j∈In ∈ Mn (C)+ . ∈ Mn (C)+ .

Es sabido que existe un u ´nico y ∈ Cm tal que ϕ(z) = ϕy (z) = z, y , para todo z ∈ Cm (basta poner yi = ϕ(ei ), para cada i ∈ Im ). Luego, por la Eq. 26), podemos concluir que A = x y. 2. 27). 3. Estudiaremos a continuaci´on las propiedades de las matrices x y. Enumeraremos los resultados, y las pruebas no escritas quedar´an como ejercicio para el lector. Tomemos dos vectores x ∈ Cn e y ∈ Cm . 10 Ejercicios 20 1. La norma espectral: x 2. El adjunto: (x y sp = x y , ya que ϕy y)∗ = (xy ∗ )∗ = y x∗ = y 3.

E. B − A ∈ Mn (C)+ . Entonces λj (A) ≤ λj (B) para todo j ∈ In . Demostraci´ on. Llamemos C = B − A. 5, tenemos que λj (A) + λ1 (C) ≤ λj (A + C) = λj (A + (B − A) ) = λj (B) . Por otra parte, como C ∈ Mn (C)+ , entonces λ1 (C) = m´ın Cx, x ≥ 0. x =1 Una consecuencia importante del Teorema de Weyl es el hecho de que, entre las autoadjuntas, matrices muy cercanas tienen autovalores muy cercanos. 8. Sean A, B ∈ H(n). Entonces: λ(A) − λ(B) ∞ := m´ax |λj (A) − λj (B)| ≤ ρ(A − B) = A − B j∈ In sp . 4 Entrelace de Cauchy 31 Demostraci´ on.

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