Download PDF by Jorge Antezana y Demetrio Stojanoff: Analisis Matricial

By Jorge Antezana y Demetrio Stojanoff

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Xn−1 ) ∈ Cn−1 a su parte significativa. Observar que n−1 Aij xj xi = Ax, x , dado que xn = 0. 3, que λk (An ) = m´ın m´ax An x0 , x0 ≥ dim M=k x∈M1 M⊆ Hn−1 m´ın m´ax Ax, x = λk (A). x∈M1 dim M=k n M⊆ C Tomemos ahora subespacios S ⊆ Hn−1 tales que dim S = n − k. Como n − k = n − (k + 1) + 1 y a la ves, n − k = (n − 1) − k + 1, obtenemos λk (An ) = m´ax m´ın An x0 , x0 ≤ dim S=n−k x∈S1 S⊆ Hn−1 m´ax m´ın Ax, x = λk+1 (A) , dim S=n−k x∈S1 lo que prueba el teorema. 3. En forma an´aloga se puede probar versiones m´as generales del Teorema anterior: Dado A ∈ H(n), 1.

Podemos definir (con buena definici´on) una isometr´ıa suryectiva U1 : R(|A|) → R(A) dada por U1 (|A|x) = Ax, para todo x ∈ Cn . De hecho, |A|x = |A|y ⇐⇒ x − y ∈ ker |A| = ker A ⇐⇒ Ax = Ay. Como dim R(A)⊥ = n − dim R(A) = dim ker(A) = dim ker(|A|) = dim R(|A|)⊥ , podemos extender la isometr´ıa U1 a una matriz unitaria U ∈ U(n), operando isom´etricamente desde R(|A|)⊥ sobre R(A)⊥ . Por la definici´on de U , se tiene que A = U |A|. 4. Notar que AA∗ = U |A|2 U ∗ = U A∗ AU ∗ . Sea P (x) ∈ C[x] tal que P (λ) = λ1/2 , para todo λ ∈ σ (AA∗ ) = σ (A∗ A) (acabamos de ver que AA∗ ∼ = A∗ A).

Demostraci´ on. 1. Sea x ∈ Cn tal que R(A) = Gen {x}. Luego existe una funcional lineal ϕ : Cm → C tal que Az = ϕ(z) · x , para todo z ∈ Cm . 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.

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Analisis Matricial by Jorge Antezana y Demetrio Stojanoff

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