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By Jorge Antezana y Demetrio Stojanoff

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6). 5. Sean A, B ∈ Mn (C) matrices normales. e. A ∼ =B . En otras palabras, si definimos la ´orbita unitaria U(A) := { U AU ∗ : U ∈ U(n)}, entonces U(A) = { B ∈ N (n) : λ(B) = λ(A) } . Demostraci´ on. La Eq. 1) asegura que si A ∼ = B, entonces λ(A) = λ(B). 2 dice que A ∼ =D∼ = B. 2 Matrices Hermitianas Por lo general no es f´acil calcular los autovalores de una matriz. Pero en muchos casos es suficiente saber que ellos est´an en un intervalo especificado. En el resto de este Cap´ıtulo estudiaremos algunas de las principales caracter´ısticas que distinguen a las matrices Hermitianas, en particular los principios variacionales que se utilizan para localizar su espectro, sin la necesidad de conocer los autovectores asociados en forma exacta.

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. 2: 1. Pongamos A1 = A = Q1 R1 .

As´ı seguimos definiendo y factorizando para todo m ∈ N. Probar que estas sucesiones cumplen lo siguiente. (a) A2 = Q∗1 AQ1 y A3 = Q∗2 A2 Q2 = Q∗2 Q∗1 A Q1 Q2 . m (b) Dado m ∈ N, sea Um = ∗ Qk ∈ U(n). Entonces Am+1 = Um A Um . k=1 (c) Se cumple que Qm −−−→ I. M´as a´ un, Um −−−→ U ∈ U(n). m→∞ m→∞ (d) Rm −−−→ T ∈ T S(n), y tambi´en Am = Qm Rm −−−→ T . m→∞ ∗ m→∞ (e) T = U AU , por lo que λ(T ) = λ(A). Este proceso es f´acil de aplicar, porque hacer QR es barato computacionalmente. 1, por lo que que permite calcular los autovalores de A, cosa que en general es bien complicada.

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