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Diagonalización de Matrices

Module by: Michael Haag. E-mail the authorTranslated By: Fara Meza, Erika Jackson

Based on: Matrix Diagonalization by Michael Haag

Summary: (Blank Abstract)

De nuestro entendimiento de eigenvalores y eigenvectores hemos descubierto ciertas cosas sobre nuestro operador, la matriz AA. Sabemos que los eigenvectores de AA generan el espacio Cn n y sabemos como expresar cualquier vector xx en términos de v 1 v 2 … v n v 1 v 2 … v n , entonces tenemos el operador AA calculado. Si tenemos AA actuando en xx, después esto es igual a AA actuando en la combinación de los eigenvectores.

Todavía tenemos dos preguntas pendientes:

  1. ¿Cuándo los eigenvectores v 1 v 2 … v n v 1 v 2 … v n de AA generan el espacio Cn n (asumiendo que v 1 v 2 … v n v 1 v 2 … v n linealmente independientes)?
  2. ¿Cómo expresamos un vector dado xx en términos de v 1 v 2 … v n v 1 v 2 … v n ?

1 Respuesta a la Pregunta #1

Question #1:

¿Cuándo los eigenvectores v 1 v 2 … v n v 1 v 2 … v n de AA generan el espacio Cn n ?
Si AA tiene nn diferentes eigenvalores λ i ≠ λ j   ,   i≠j    i i j λ i λ j donde ii y jj son enteros, entonces AA tiene nn eigenvectores linealmente independientes. v 1 v 2 … v n v 1 v 2 … v n que generan el espacio Cn n .

nota:

La demostración de esta proposición no es muy difícil, pero no es interesante para incluirla aquí. Si desea investigar esta idea, léase Strang G., “Algebra Lineal y sus aplicaciones” para la demostración.
Además, nn diferentes eigenvalores significa que det(A−λ⁢I)= c n ⁢λn+ c n − 1 ⁢λn−1+…+ c 1 ⁢λ+ c 0 =0 A λ I c n λ n c n − 1 λ n 1 … c 1 λ c 0 0 tiene nn raíces diferentes.

Respuesta a la Pregunta #2

Question #2:

¿Cómo expresamos un vector dado xx en términos de v 1 v 2 … v n v 1 v 2 … v n ?
Queremos encontrar α 1 α 2 … α n ∈C α 1 α 2 … α n tal que
x= α 1 ⁢ v 1 + α 2 ⁢ v 2 +…+ α n ⁢ v n x α 1 v 1 α 2 v 2 … α n v n
(1)
Para poder encontrar el conjunto de variables, empezaremos poniendo los vectores v 1 v 2 … v n v 1 v 2 … v n como culumnas en una matriz V V de n×n. V=( ⋮⋮ ⋮ v 1 v 2 … v n ⋮⋮ ⋮ ) V ⋮ ⋮   ⋮ v 1 v 2 … v n ⋮ ⋮   ⋮ Ahora la ecuación 1 se convierte en x=( ⋮⋮ ⋮ v 1 v 2 … v n ⋮⋮ ⋮ )⁢ α 1 ⋮ α n x ⋮ ⋮   ⋮ v 1 v 2 … v n ⋮ ⋮   ⋮ α 1 ⋮ α n ó x=V⁢α x V α Lo que nos da una forma sencilla de resolver para la variable de nuestra pregunta αα: α=V-1⁢x α V -1 x Notese que VV es invertible ya que tiene nn columnas linealmnete independientes.

Comentarios Adicionales

Recordemos el conocimiento de funciones y sus bases y examinemos el papel de VV. x=V⁢α x V α x 1 ⋮ x n =V⁢ α 1 ⋮ α n x 1 ⋮ x n V α 1 ⋮ α n donde αα es solo xx expresada en una base diferente: x= x 1 ⁢10⋮0+ x 2 ⁢01⋮0+…+ x n ⁢00⋮1 x x 1 1 0 ⋮ 0 x 2 0 1 ⋮ 0 … x n 0 0 ⋮ 1 x= α 1 ⁢⋮ v 1 ⋮+ α 2 ⁢⋮ v 2 ⋮+…+ α n ⁢⋮ v n ⋮ x α 1 ⋮ v 1 ⋮ α 2 ⋮ v 2 ⋮ … α n ⋮ v n ⋮ VV transforma xx de la base canónica a la base v 1 v 2 … v n v 1 v 2 … v n

Diagonalización de Matrices y Salidas

También podemos usar los vectores v 1 v 2 … v n v 1 v 2 … v n para representar la salida b b, del sistema: b=A⁢x=A⁢( α 1 ⁢ v 1 + α 2 ⁢ v 2 +…+ α n ⁢ v n ) b A x A α 1 v 1 α 2 v 2 … α n v n A⁢x= α 1 ⁢ λ 1 ⁢ v 1 + α 2 ⁢ λ 2 ⁢ v 2 +…+ α n ⁢ λ n ⁢ v n =b A x α 1 λ 1 v 1 α 2 λ 2 v 2 … α n λ n v n b A⁢x=( ⋮⋮ ⋮ v 1 v 2 … v n ⋮⋮ ⋮ )⁢ λ 1 ⁢ α 1 ⋮ λ 1 ⁢ α n A x ⋮ ⋮   ⋮ v 1 v 2 … v n ⋮ ⋮   ⋮ λ 1 α 1 ⋮ λ 1 α n A⁢x=V⁢Λ⁢α A x V Λ α A⁢x=V⁢Λ⁢V-1⁢x A x V Λ V -1 x donde ΛΛ es la matriz con eigenvalores en la diagonal: Λ=( λ 1 0…0 0 λ 2 …0 ⋮⋮⋱⋮ 00… λ n ) Λ λ 1 0 … 0 0 λ 2 … 0 ⋮ ⋮ ⋱ ⋮ 0 0 … λ n Finalmente, podemos cancelar las x x y quedarnos con una ecuación final para AA: A=V⁢Λ⁢V-1 A V Λ V -1

1 Interpretación

Para nuestra interpretación, recordemos nuestra formulas: α=V-1⁢x α V -1 x b=∑i α i ⁢ λ i ⁢ v i b i α i λ i v i podemos interpretar el funcionamiento de xx con AA como: x 1 ⋮ x n → α 1 ⋮ α n → λ 1 ⁢ α 1 ⋮ λ 1 ⁢ α n → b 1 ⋮ b n x 1 ⋮ x n α 1 ⋮ α n λ 1 α 1 ⋮ λ 1 α n b 1 ⋮ b n Donde los tres pasos (las flechas) en la ilustración anterior representan las siguientes tres operaciones:

  1. Transformar x x usando V-1 V -1 , nos da αα
  2. Multiplicar por Λ Λ
  3. Transformada Inversa usando V V, lo que nos da bb
¡Este es el paradigma que usaremos para los sistemas LTI!

Figura 1: Ilustración simple del sistema LTI.
Figura 1 (eigv_sys.png)

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