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Bases Ortonormales en Espacios Reales y Complejos

Module by: Justin Romberg Translated by: Fara Meza, Erika JacksonBased on: Orthonormal Bases in Real and Complex Spaces por 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=a11a12a21a22 A a 1 1 a 1 2 a 2 1 a 2 2 AT=a11a21a12a22 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 xTy= x 0 x 1 x n - 1 y 0 y 1 y n - 1 =xiyi=<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 n n
Transpuesta Hermitiana AH A , transpuesta y conjugada AH=AT¯ A A <y,x>=xHy=xiyi¯ y x x y i x i y i en n n
Sea b0b1b n - 1 b 0 b 1 b n - 1 una base ortonormal para n n i,:i=01n-1<bi,bi>=1 i i 0 1 n 1 b i b i 1 ij <bi,bj>=bjHbi=0 i j b i b j b j b i 0
Matriz de la base: B=b0b1b n - 1 B b 0 b 1 b n - 1 Ahora, BHB=b0Hb1Hb n - 1 Hb0b1b n - 1 =b0Hb0b0Hb1b0Hb n - 1 b1Hb0b1Hb1b1Hb n - 1 b n - 1 Hb0b n - 1 Hb1b n - 1 Hb 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 n 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= α i b i x i α i b i Calcular α=B-1xα=BHx α B x α B x usando una base ortonormal nos libramos de la operación inversa.

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