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Sea B ∈ Mn (C) tal que A = B ∗ B. 2. Entonces A = T ∗ Q∗ QT = T ∗ T . 1. 3. 2. Sea A ∈ Mn (C)+ . 3 se muestran las dos maneras usuales de describir a A1/2 : 1. Si A = U diag (λ(A) ) U ∗ , con U ∈ U(n), se tiene que A1/2 = U diag λ(A)1/2 U ∗ . 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.

23. Sean A, B ∈ Mn (C). Definamos las transformaciones lineales LA y RB : Mn (C) → Mn (C) dadas por LA (X) = AX y RB (X) = XB , X ∈ Mn (C) . 1. Probar que σ(LA ) = σ(A) y que σ(RB ) = σ(B). 2. Probar que σ(LA − RB ) = {λ − µ : λ ∈ σ(A) y µ ∈ σ(B)}. 3. Deducir que las siguientes condiciones son equivalentes: (a) Para todo Y ∈ Mn (C), existe un u ´nico X ∈ Mn (C) tal que AX − XB = Y . (b) σ(A) ∩ σ(B) = ∅. 24 (Proceso QR). Sea A ∈ Gl (n). Asumiremos que todos los autovalores de A tienen m´odulos distintos.

Hallar los autovalores de E2 y E3 . Generalizar para En . 19. Probar que cualquier familia de matrices que conmutan dos a dos tiene un autovector com´ un a todas, mediante los siguientes pasos: 1. Probar que si A, B ∈ Mn (C) conmutan, entonces tienen un autovector en com´ un. 2. Si F = {A1 , . . , Am } es una familia finita de matrices que conmutan dos a dos, usar inducci´on para probar que hay un autovector com´ un para todos. 3. Si la familia tiene cardinal no finito, encontrar alg´ un curro para que d´e.