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Bases Ortonormales en Espacios Reales y Complejos

Module by: Justin Romberg. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Orthonormal Bases in Real and Complex Spaces by Justin Romberg

Summary: Este módulo define los términos de la transpuesta, el producto interno y la transpuesta Hermitiana y su uso para encontrar una base ortonormal.

Notación

El operador de la Transpuesta AT A voltea la matriz a través de su diagonal. A=( a1,1a1,2 a2,1a2,2 ) A a 1 1 a 1 2 a 2 1 a 2 2 AT=( a1,1a2,1 a1,2a2,2 ) A a 1 1 a 2 1 a 1 2 a 2 2 La columna i i de A A es una fila i i de AT A

Recordando que el, producto interno x=( x 0 x 1 ⋮ x n - 1 ) x x 0 x 1 ⋮ x n - 1 y=( y 0 y 1 ⋮ y n - 1 ) y y 0 y 1 ⋮ y n - 1 xT⁢y=( x 0 x 1 … x n - 1 )⁢( y 0 y 1 ⋮ y n - 1 )=∑ i xi⁢yi=〈y,x〉 x y x 0 x 1 … x n - 1 y 0 y 1 ⋮ y n - 1 i x i y i y x en Rn n

Transpuesta Hermitiana AH A , transpuesta y conjugada AH=AT¯ A A 〈y,x〉=xH⁢y=∑ i xi⁢yi¯ y x x y i x i y i en Cn n

Sea b 0 b 1 … b n - 1 b 0 b 1 … b n - 1 una base ortonormal para Cn n ∀i,:〈 b i , b i 〉=1 i i 0 1 … n 1 b i b i 1 ⁢i≠j〈 b i , b j 〉= b j H⁢ b i =0 i j b i b j b j b i 0

Matriz de la base: B=( ⋮⋮⋮ b 0 b 1 … b n - 1 ⋮⋮⋮ ) B ⋮ ⋮ ⋮ b 0 b 1 … b n - 1 ⋮ ⋮ ⋮ Ahora, BH⁢B=( … b 0 H… … b 1 H… ⋮ … b n - 1 H… )⁢( ⋮⋮⋮ b 0 b 1 … b n - 1 ⋮⋮⋮ )=( b 0 H⁢ b 0 b 0 H⁢ b 1 … b 0 H⁢ b n - 1 b 1 H⁢ b 0 b 1 H⁢ b 1 … b 1 H⁢ b n - 1 ⋮ b n - 1 H⁢ b 0 b n - 1 H⁢ b 1 … b n - 1 H⁢ b n - 1 ) B B … b 0 … … b 1 … ⋮ … b n - 1 … ⋮ ⋮ ⋮ b 0 b 1 … b n - 1 ⋮ ⋮ ⋮ b 0 b 0 b 0 b 1 … b 0 b n - 1 b 1 b 0 b 1 b 1 … b 1 b n - 1 ⋮ b n - 1 b 0 b n - 1 b 1 … b n - 1 b n - 1

Para una base ortonormal con una matriz de la base B B BH=B-1 B B ( BT=B-1 B B in Rn n ) BH B es fácil calcular mientras que B-1 B es difícil de calcular.

Así que, para encontrar α 0 α 1 … α n - 1 α 0 α 1 … α n - 1 tal que x=∑ i α i ⁢ b i x i α i b i Calcular (α=B-1⁢x)⇒(α=BH⁢x) α B x α B x usando una base ortonormal nos libramos de la operación inversa.

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