[proofplan] We prove invariance directly from the definition of orthogonal complement. Take a vector $x \in E_\lambda^\perp$ and test $Tx$ against an arbitrary vector $y \in E_\lambda$. Self-adjointness moves $T$ from the first inner-product slot to the second, and the eigenvector relation $Ty=\lambda y$ reduces the expression to a scalar multiple of $(x,y)_H$, which is zero. Since $Tx$ is orthogonal to every vector in $E_\lambda$, it lies in $E_\lambda^\perp$. [/proofplan] [step:Test $Tx$ against an arbitrary eigenvector] Let $x \in E_\lambda^\perp$ be arbitrary. To prove $Tx \in E_\lambda^\perp$, by the definition of orthogonal complement it is enough to show that \begin{align*} (Tx,y)_H=0 \end{align*} for every $y \in E_\lambda$. Fix $y \in E_\lambda$. Since $T$ is self-adjoint, we have \begin{align*} (Tx,y)_H=(x,Ty)_H. \end{align*} Since $y \in E_\lambda=\ker(T-\lambda I)$, we have $Ty=\lambda y$. Therefore, using conjugate-linearity of the second variable of the Hilbert-space [inner product](/page/Inner%20Product), \begin{align*} (Tx,y)_H=(x,\lambda y)_H=\overline{\lambda}(x,y)_H. \end{align*} Because $x \in E_\lambda^\perp$ and $y \in E_\lambda$, the definition of $E_\lambda^\perp$ gives \begin{align*} (x,y)_H=0. \end{align*} Hence \begin{align*} (Tx,y)_H=0. \end{align*} [guided] We want to prove that $E_\lambda^\perp$ is invariant under $T$, which means that applying $T$ to a vector already in $E_\lambda^\perp$ keeps it inside $E_\lambda^\perp$. So let $x \in E_\lambda^\perp$ be arbitrary. The definition of orthogonal complement says that a vector $v \in H$ belongs to $E_\lambda^\perp$ exactly when \begin{align*} (v,y)_H=0 \end{align*} for every $y \in E_\lambda$. Thus, to prove $Tx \in E_\lambda^\perp$, we must prove \begin{align*} (Tx,y)_H=0 \end{align*} for every $y \in E_\lambda$. Fix an arbitrary vector $y \in E_\lambda$. The point of self-adjointness is that it lets us move $T$ from the first input of the inner product to the second input: \begin{align*} (Tx,y)_H=(x,Ty)_H. \end{align*} Now the choice $y \in E_\lambda$ becomes useful. Since \begin{align*} E_\lambda=\ker(T-\lambda I), \end{align*} membership of $y$ in $E_\lambda$ means precisely that \begin{align*} (T-\lambda I)y=0, \end{align*} and hence \begin{align*} Ty=\lambda y. \end{align*} Substituting this into the inner product identity gives \begin{align*} (Tx,y)_H=(x,\lambda y)_H. \end{align*} With Androma's Hilbert-space convention, the inner product is linear in the first variable and conjugate-linear in the second variable, so \begin{align*} (x,\lambda y)_H=\overline{\lambda}(x,y)_H. \end{align*} Finally, because $x \in E_\lambda^\perp$ and $y \in E_\lambda$, orthogonality gives \begin{align*} (x,y)_H=0. \end{align*} Therefore \begin{align*} (Tx,y)_H=\overline{\lambda}(x,y)_H=0. \end{align*} Since this holds for the arbitrary vector $y \in E_\lambda$, the vector $Tx$ is orthogonal to all of $E_\lambda$. [/guided] [/step] [step:Conclude that the orthogonal complement is invariant] The preceding step shows that for the arbitrary vector $x \in E_\lambda^\perp$, \begin{align*} (Tx,y)_H=0 \end{align*} for every $y \in E_\lambda$. Hence $Tx \in E_\lambda^\perp$. Since $x \in E_\lambda^\perp$ was arbitrary, we have \begin{align*} T(E_\lambda^\perp)\subset E_\lambda^\perp. \end{align*} Thus $E_\lambda^\perp$ is invariant under $T$. [/step]