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By Jorge Antezana y Demetrio Stojanoff

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En otras palabras, se debe encontrar un ejemplo de una matriz A ∈ Mn (C)+ tal que B = |Aij | i,j∈In ∈ / Mn (C)+ . Observar que hay que buscar para n ≥ 3. 6. Si A ∈ Mn (C)+ y P (x) ∈ R[x] tiene coeficientes no negativos, entonces P◦ (A) := P (Aij ) i,j∈In ∈ Mn (C)+ . Demostraci´ on. Por una inducci´on directa, podemos ver que A[k] = A ◦ A ◦ · · · ◦ A ∈ Mn (C)+ (se multiplica k veces) para todo k ∈ N. Despu´es se usa lo que cumple P (x). 7. 7 eAij i,j∈In ∈ Mn (C)+ . El famoso truco 2 × 2 Cuando se trabaja con operadores y matrices, muchas veces una cuenta inmanejable termina saliendo “m´agicamente” y en un par de renglones, con el famoso truco de matrices de bloques de 2 × 2.

A1/2 = P (A) para culquier P ∈ C[x] tal que P (λ) = λ1/2 para todo λ ∈ σ(A). 3. Sea A ∈ Mn (C), 1. Llamaremos “m´odulo de A” a la matriz |A| = (A∗ A)1/2 ∈ Mn (C)+ . 2. Llamaremos valores singulares de A a los autovalores de |A| ordenados en forma decreciente, not´andolos s1 (A) ≥ · · · ≥ sn (A) ≥ 0. 2, si (A) = µi (|A|) = µi (A∗ A)1/2 , para todo i ∈ In . 3. Llamaremos s(A) = (s1 (A), . . , sn (A) ) = µ(|A|) y Σ(A) a la matriz diagonal   s1 (A) 0 0  ..  . Σ(A) = diag (s(A)) =  ...  0 0 sn (A) Observar que |A| ∼ = Σ(A).

Luego se verifican las siguientes propiedades: 1. Si B = {v1 , . . , vn } es una BON de Cn adaptada a µ(A), luego A∗ A = A2 , |A|, |A|1/2 y U son diagonales en la base B. Por lo tanto conmutan entre ellos (y con A). 2. En la base B, la matriz de U es diagonal con ±1’s en la diagonal. 4) dado que |A| vk = (µ2k (A) )1/2 vk = |µk (A)| vk para todo k ∈ In . Por lo tanto, U ∗ = U = U −1 y −I ≤U ≤I . 3. Podemos deducir que −|A| ≤ A ≤ |A|. En efecto, |A|1/2 U |A|1/2 = A, y −|A| = −|A|1/2 I|A|1/2 ≤ |A|1/2 U |A|1/2 ≤ |A|1/2 I|A|1/2 = |A|.

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