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Download Analisis Matricial by Jorge Antezana y Demetrio Stojanoff PDF

By Jorge Antezana y Demetrio Stojanoff

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El problema es que la matriz U ∈ U(n) que diagonaliza a A no es u ´nica. Sea otra C ∈ Mn (C)+ tal que C 2 = A. 1 asegura que existe V ∈ U(n) tal que V ∗ AV = D y V ∗ CV ∈ Mn (C)+ es diagonal . 2. Como (V ∗ CV )2 = V ∗ AV = D, es claro que V ∗ CV = D1/2 (entre diagonales la unicidad es trivial). Por otro lado, U DU ∗ = V DV ∗ = A =⇒ (V ∗ U )D = D(V ∗ U ) . Aqu´ı usaremos que D1/2 se puede escribir como P (D) para cierto polinomio P ∈ R[X]. En efecto, basta elegir un P tal que P (λi (A) ) = λi (A)1/2 , para todo i ∈ In .

3 (Aronszajn). Sea A = C X X∗ D k ∈ H(n). Probar que n−k µi+j−1 (A) ≤ µi (C) + µj (D) para todo par i ∈ Ik , j ∈ In−k . 4. Dado A ∈ H(n), mostrar que: 1. Si J ⊆ In cumple que |J| = r, entonces para cada k ∈ Ir , se tiene µk (A) ≥ µk A[J] ≥ µk+n−r (A) . En particular, si r ∈ In , entonces µk (A) ≥ µk Ar ≥ µk+1 (A) para todo k ∈ In−1 . 2. Sea P ∈ L(H)+ es un proyector autoadjunto (o sea ortogonal) sobre un subespacio S de dim S = r. Sea AS = P AP S ∈ L(S), la compresi´on de A a S. Luego µk (A) ≥ µk AS ) ≥ µk+n−r (A) , para cada k ∈ Ir .

7. 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 .

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