Vejamos a definição e alguns resultados sobre operadores auto-adjuntos , um tipo de operador normal. Os operadores auto-adjuntos são o objeto central do Teorema Espectral
Seja V V V T ∈ L ( V ) T\in \mathcal{L}(V) T ∈ L ( V ) auto-adjunto se ∀ v , w ∈ V \forall v,w\in V ∀ v , w ∈ V
⟨ T v , w ⟩ = ⟨ v , T w ⟩ . \langle Tv , w \rangle=\langle v , Tw \rangle. ⟨ T v , w ⟩ = ⟨ v , Tw ⟩ . Ou seja, T = T ∗ \boxed{T=T^{*}} T = T ∗
Como exemplo, considere o operador linear T ∈ L ( R 2 ) T\in \mathcal{L}(\mathbb{R}^{2}) T ∈ L ( R 2 ) T ( x , y ) = ( x , 2 y ) T(x,y)=(x,2y) T ( x , y ) = ( x , 2 y )
⟨ ( 1 , 0 ) , T ∗ ( x , y ) ⟩ = ⟨ T ( 1 , 0 ) , ( x , y ) ⟩ = ⟨ ( 1 , 0 ) , ( x , y ) ⟩ = x ⟨ ( 0 , 1 ) , T ∗ ( x , y ) ⟩ = ⟨ T ( 0 , 1 ) , ( x , y ) ⟩ = ⟨ ( 0 , 2 ) , ( x , y ) ⟩ = 2 y \begin{align}
&\langle (1,0) , T^{*}(x,y) \rangle=\langle T(1,0) , (x,y) \rangle=\langle (1,0) , (x,y) \rangle=x \\
& \langle (0,1) , T^{*}(x,y) \rangle=\langle T(0,1) , (x,y) \rangle=\langle (0,2) , (x,y) \rangle=2y
\end{align} ⟨( 1 , 0 ) , T ∗ ( x , y )⟩ = ⟨ T ( 1 , 0 ) , ( x , y )⟩ = ⟨( 1 , 0 ) , ( x , y )⟩ = x ⟨( 0 , 1 ) , T ∗ ( x , y )⟩ = ⟨ T ( 0 , 1 ) , ( x , y )⟩ = ⟨( 0 , 2 ) , ( x , y )⟩ = 2 y Ou seja, dado que ⟨ e i , ( x 1 , … , x n ) ⟩ \langle e_{i} , (x_{1},\dots,x_{n}) \rangle ⟨ e i , ( x 1 , … , x n )⟩ x i x_{i} x i ( x , y ) ∈ R 2 (x,y)\in \mathbb{R}^{2} ( x , y ) ∈ R 2 T ∗ ( x , y ) = ( x , 2 y ) = T ( x , y ) T^{*}(x,y)=(x,2y)=T(x,y) T ∗ ( x , y ) = ( x , 2 y ) = T ( x , y ) T = T ∗ T=T^{*} T = T ∗ T T T
Utilizando as propriedades da adjunta:
( A + B ) ∗ = A ∗ + B ∗ = A + B ( α A ) ∗ = α A ∗ = α A \begin{align}
&(A+B)^{*}=A^{*}+B^{*}=A+B \\
&(\alpha A)^{*}= \alpha A^{*}=\alpha A
\end{align} ( A + B ) ∗ = A ∗ + B ∗ = A + B ( α A ) ∗ = α A ∗ = α A
Dada a propriedade que ( T − 1 ) ∗ = ( T ∗ ) − 1 (T^{-1})^{*}=(T^{*})^{-1} ( T − 1 ) ∗ = ( T ∗ ) − 1 T ∗ T^{*} T ∗ T T T T = T ∗ T=T^{*} T = T ∗
( T − 1 ) ∗ = T − 1 . (T^{-1})^{*}=T^{-1}. ( T − 1 ) ∗ = T − 1 . Logo, T − 1 T^{-1} T − 1
Sejam v , w ∈ V v,w\in V v , w ∈ V λ \lambda λ μ \mu μ T v = λ v Tv=\lambda v T v = λ v T w = μ w Tw=\mu w Tw = μ w T T T
⟨ T v , w ⟩ = ⟨ v , T w ⟩ . \langle Tv , w \rangle=\langle v , Tw \rangle. ⟨ T v , w ⟩ = ⟨ v , Tw ⟩ . Logo,
⟨ λ v , w ⟩ = ⟨ v , μ w ⟩ λ ⟨ v , w ⟩ = μ ⟨ v , w ⟩ ( λ − μ ) ⟨ v , w ⟩ = 0 \begin{align}
\langle \lambda v , w \rangle &= \langle v , \mu w \rangle \\
\lambda \langle v , w \rangle &= \mu \langle v , w \rangle \\
(\lambda-\mu)\langle v , w \rangle &= 0
\end{align} ⟨ λ v , w ⟩ λ ⟨ v , w ⟩ ( λ − μ ) ⟨ v , w ⟩ = ⟨ v , μ w ⟩ = μ ⟨ v , w ⟩ = 0 Como λ ≠ μ \lambda \neq \mu λ = μ ⟨ v , w ⟩ = 0 \langle v , w \rangle= 0 ⟨ v , w ⟩ = 0 v v v w w w
Matrizes de operadores auto-adjuntos ¶ Como é de se esperar, operadores auto-adjuntos possuem uma equivalência matricial.
Uma matriz é dita simétrica quando é igual a sua transposta. Isto é, seja A ∈ M n ( R ) A \in \mathcal{M}_{n}(\mathbb{R}) A ∈ M n ( R ) A A A A T = A A^{T}=A A T = A
Seja T ∈ L ( V ) T\in \mathcal{L}(V) T ∈ L ( V ) T T T [ T ] β [T]_\beta [ T ] β β \beta β V V V
Seja β \beta β V V V
( ⟹ ) : (\implies): ( ⟹ ) : T T T
[ T ] β = [ T ∗ ] β = ( [ T ] β ) T . [T]_{\beta}=[T^{*}]_{\beta}=([T]_{\beta})^{T}. [ T ] β = [ T ∗ ] β = ([ T ] β ) T . Portanto, [ T ] β [T]_{\beta} [ T ] β
( ⟸ ) : (\impliedby): ( ⟸ ) : [ T ] β = ( [ T ] β ) T [T]_{\beta}=([T]_{\beta})^{T} [ T ] β = ([ T ] β ) T β = { v 1 , … , v n } \beta = \{ v_{1},\dots,v_{n} \} β = { v 1 , … , v n }
⟨ T v i , v j ⟩ = ⟨ T v j , v i ⟩ , \langle Tv_{i} , v_{j} \rangle=\langle Tv_{j} , v_{i} \rangle, ⟨ T v i , v j ⟩ = ⟨ T v j , v i ⟩ , para todo i , j i,j i , j [ T ] β [T]_\beta [ T ] β ( [ T ] β ) T ([T]_{\beta})^{T} ([ T ] β ) T
Sejam u , v ∈ V u,v\in V u , v ∈ V u = ∑ i x i v i u=\sum_{i}x_{i}v_{i} u = ∑ i x i v i v = ∑ j y j v j v=\sum_{j}y_{j}v_{j} v = ∑ j y j v j
⟨ u , T v ⟩ = ⟨ ∑ i x i v i , T ( ∑ j y j v j ) ⟩ = ∑ i x i ∑ j y j ⟨ v i , T v j ⟩ = ∑ i x i ∑ j y j ⟨ T v j , v i ⟩ = ∑ i x i ∑ j y j ⟨ T v i , v j ⟩ , \begin{align}
\langle u , Tv \rangle &= \left\langle \sum_{i}x_{i}v_{i} , T\left( \sum_{j}y_{j}v_{j} \right) \right\rangle \\
&= \sum_{i}x_{i}\sum_{j}y_{j}\langle v_{i} , Tv_{j} \rangle \\
&= \sum_{i}x_{i}\sum_{j}y_{j}\langle Tv_{j} , v_{i} \rangle \\
&= \sum_{i}x_{i}\sum_{j}y_{j}\langle Tv_{i} , v_{j} \rangle,
\end{align} ⟨ u , T v ⟩ = ⟨ i ∑ x i v i , T ( j ∑ y j v j ) ⟩ = i ∑ x i j ∑ y j ⟨ v i , T v j ⟩ = i ∑ x i j ∑ y j ⟨ T v j , v i ⟩ = i ∑ x i j ∑ y j ⟨ T v i , v j ⟩ , (na última igualdade usamos (8)
Portanto, temos que
⟨ u , T v ⟩ = ⟨ ∑ i x i T v i , ∑ j y j v j ⟩ = ⟨ T ( ∑ i x i v i ) , ∑ j y j v j ⟩ = ⟨ T u , v ⟩ , \langle u , Tv \rangle=\left\langle \sum_{i}x_{i}Tv_{i} , \sum_{j}y_{j}v_{j} \right\rangle=\left\langle T\left( \sum_{i}x_{i}v_{i} \right) , \sum_{j}y_{j}v_{j} \right\rangle=\langle Tu , v \rangle, ⟨ u , T v ⟩ = ⟨ i ∑ x i T v i , j ∑ y j v j ⟩ = ⟨ T ( i ∑ x i v i ) , j ∑ y j v j ⟩ = ⟨ T u , v ⟩ , concluindo que T = T ∗ T=T^{*} T = T ∗
Voltando ao exemplo dado após a Definição 1 T ( 1 , 0 ) = ( 1 , 0 ) T(1,0)=(1,0) T ( 1 , 0 ) = ( 1 , 0 ) T ( 0 , 1 ) = ( 0 , 2 ) T(0,1)=(0,2) T ( 0 , 1 ) = ( 0 , 2 ) c c c
[ T ] c = ( 1 0 0 2 ) [T]_{c}=\begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix} [ T ] c = ( 1 0 0 2 ) Além disso,
( [ T ] c ) T = ( 1 0 0 2 ) = [ T ] c ([T]_{c})^{T}=\begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix} = [T]_{c} ([ T ] c ) T = ( 1 0 0 2 ) = [ T ] c Portanto, [ T ] c [T]_{c} [ T ] c T T T