15.12: Bases Ortonormales en Espacios Reales y Complejos

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Notación

Transpose operador\(A^T\) voltea la matriz a través de su diagonal.

\ [\ begin {array} {l}
A=\ left (\ begin {array} {ll}
a_ {1,1} & a_ {1,2}\
a_ {2,1} & a_ {2,2}
\ end {array}\ derecha)\\
A^ {T} =\ left (\ begin {array} {ll}
a_ {1,1} & a_ {2,1}\
a_ {1,2} & a_ _ {2,2}
\ end {array}\ right)
\ end {array}\ nonumber\]

Columna\(i\) de\(A\) es fila\(i\) de\(A^T\)

Recall, producto interno

\ [\ boldsymbol {x} =\ left (\ begin {array} {c}
x_ {0}\\
x_ {1}\\
\ vdots\\
x_ {n-1}
\ end {array}\ derecha)\ nonumber\]

\ [\ boldsymbol {y} =\ left (\ begin {array} {c}
y_ {0}\\
y_ {1}\\
\ vdots\\
y_ {n-1}
\ end {array}\ derecha)\ nonumber\]

\ [\ negridsymbol {x} ^ {T}\ negridsymbol {y} =\ left (\ begin {array} {cccc}
x_ {0} & x_ {1} &\ ldots & x_ {n-1}
\ end {array}\ derecha)\ left (\ begin {array} {c}
y_ {0}\
y_ {1}\\
\ vdots\
y_ _ {n-1}
\ end {array}\ derecha) =\ suma_ { i}\ negridsymbol {x} _ {i}\ negridsymbol {y} _ _ {i} =\ langle\ negridsymbol {y},\ negridsymbol {x}\ rangle\ nonumber\]

encendido\(\mathbb{R}^n\).

Transposición hermitiana\(A^H\), transposición y conjugación

\ [\ begin {array} {c}
A^ {\ mathrm {H}} =\ overline {A^ {T}}\\ langle
\ negridsymbol {y},\ negridsymbol {x}\ rangle=\ negridsymbol {x} ^ {\ mathrm {H}}\ negridsymbol {y} =\ sum_ {i}\ negridsymbol {x} _ i}\ bar {\ boldsymbol {y}} _ {i}
\ end {array}\ nonumber\]

encendido\(\mathbb{C}^n\).

Ahora, dejemos\(\left\{b_{0}, b_{1}, \dots, b_{n-1}\right\}\) ser una base ortonormal para\(\mathbb{C}^n\)

\ [\ begin {array} {c}
i=\ {0,1,\ ldots, n-1\}\ izquierda\ langle b_ {i}, b_ {i}\ derecha\ rangle=1\
\ izquierda (i\ neq j,\ izquierda\ langle b_ {i}, b_ {i}\ derecha\ rangle=b_ {j} ^ {H} b_ {i} =0\ derecha)
\ end {array}\ nonumber\]

Matriz de base:

\ [B=\ left (\ begin {array} {cccc}
\ vdots &\ vdots &\ vdots\\
b_ {0} & b_ {1} &\ dots & b_ {n-1}\\\ vdots &
\ vdots &\ vdots &\ vdots
\ end {array}\ right)\ nonumber\]

Ahora,

\ [B^ {\ mathrm {H}} B=\ left (\ begin {array} {ccc}
\ cdots & b_ {0} ^ {\ mathrm {H}} &\ cdots\
\ cdots & b_ {1} ^ {\ mathrm {H}} &\ cdots\\
&\ vdots\
\ cdots & b_ {n-1} ^ {\ mathrm {H}} &\ ldots
\ end {array}\ derecha)\ left (\ begin {array } {cccc}
\ vdots &\ vdots &\ vdots\\
b_ {0} & b_ {1} &\ cdots & b_ {n-1}\\
\ vdots &\ vdots & &\ vdots
\ end {array}\ derecha) =\ left (\ begin {array} {cccc}
b_ {0} ^ {\ mathrm {H}} b_ {0}} & b_ {0} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {0} ^ {\ mathrm {H}} b_ {n-1}\\
b_ {1} ^ {\ mathrm {H}} b_ {0} & b_ {1} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {1} ^ {\ mathrm {H}} b_ {n-1}\
\ vdots & &\\
b_ {n-1} ^ {\ mathrm {H}} b_ {0} & b_ {n-1} ^ {\ mathrm {H}} b_ {1} &\ ldots & b_ {n-1} ^ {\ mathrm {H}} b_ {n-1}
\ end {array}\ derecha)\ nonumber\]

Para base ortonormal con matriz de base\(B\)

\[B^H=B^{-1} \nonumber \]

(\(B^{T}=B^{-1} \text {in } \mathbb{R}^{n}\)in\(\mathbb{R}^n\))\(B^H\) es fácil de calcular mientras que\(B^{-1}\) es difícil de calcular.

Entonces, para encontrar\(\left\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{n-1}\right\}\) tal que

\[\boldsymbol{x}=\sum_{i} \alpha_{i} b_{i} \nonumber \]

Calcular

\[\left(\boldsymbol{\alpha}=B^{-1} \boldsymbol{x}\right) \Rightarrow\left(\boldsymbol{\alpha}=B^{H} \boldsymbol{x}\right) \nonumber \]

Utilizando una base ortonormal nos libramos de la operación inversa.