[proofplan]
The proof first uses normality to convert an eigenvector relation for $N$ into the corresponding eigenvector relation for $N^*$. Specifically, from $Nx=\lambda x$ we show $(N^*-\overline{\lambda}I)x=0$ by applying the normal-operator norm identity to $N-\lambda I$. We then use the adjoint identity and the convention that the Hilbert-space [inner product](/page/Inner%20Product) is linear in the first argument and conjugate-linear in the second to compare $(x,Ny)_H$ in two ways.
[/proofplan]
[step:Convert the eigenvector relation for $N$ into one for $N^*$]
Let $I := I_H$ denote the identity operator on $H$, and recall from the statement that $N^*\in\mathcal{L}(H)$ denotes the Hilbert-space adjoint of $N$. Define the shifted operator $T_\lambda \in \mathcal{L}(H)$ by
\begin{align*}T_\lambda = N-\lambda I.\end{align*}
Since $N$ is normal, $H$ is a complex [Hilbert space](/page/Hilbert%20Space), and $\lambda \in \mathbb{C}$, [citetheorem:8396] gives, for every $u \in H$,
\begin{align*}\|T_\lambda u\|_H=\|(N^*-\overline{\lambda}I)u\|_H.\end{align*}
Taking $u=x$ and using $Nx=\lambda x$, we get
\begin{align*}\|(N^*-\overline{\lambda}I)x\|_H=\|T_\lambda x\|_H=\|Nx-\lambda x\|_H=0.\end{align*}
Because a Hilbert-space norm vanishes only at the zero vector,
\begin{align*}
N^*x=\overline{\lambda}x.
\end{align*}
[guided]
We want to replace the assumption $Nx=\lambda x$ by an identity involving $N^*$, because the adjoint identity relates $Ny$ to $N^*x$ inside the inner product. Let $I := I_H$ denote the identity operator on $H$, and recall from the statement that $N^*\in\mathcal{L}(H)$ denotes the Hilbert-space adjoint of $N$. Define the shifted operator $T_\lambda \in \mathcal{L}(H)$ by
\begin{align*}T_\lambda=N-\lambda I.\end{align*}
Then $T_\lambda x=0$ is exactly the eigenvector equation $Nx=\lambda x$ rewritten as a kernel statement.
The result we need is [citetheorem:8396]. Its hypotheses are satisfied here: $H$ is a complex Hilbert space by the statement, $N \in \mathcal{L}(H)$ is normal by hypothesis, and $\lambda \in \mathbb{C}$. Therefore, for every $u \in H$,
\begin{align*}\|(N-\lambda I)u\|_H=\|(N^*-\overline{\lambda}I)u\|_H.\end{align*}
Applying this with $u=x$ gives
\begin{align*}\|(N^*-\overline{\lambda}I)x\|_H=\|(N-\lambda I)x\|_H.\end{align*}
Since $Nx=\lambda x$, the right-hand side is
\begin{align*}\|(N-\lambda I)x\|_H=\|Nx-\lambda x\|_H=0.\end{align*}
Hence
\begin{align*}\|(N^*-\overline{\lambda}I)x\|_H=0.\end{align*}
A norm is zero exactly on the zero vector, so
\begin{align*}(N^*-\overline{\lambda}I)x=0.\end{align*}
Equivalently,
\begin{align*}
N^*x=\overline{\lambda}x.
\end{align*}
This is the point where normality is used: for a general bounded operator, $Nx=\lambda x$ need not imply $N^*x=\overline{\lambda}x$.
[/guided]
[/step]
[step:Compare the same inner product using the adjoint relation]
By the adjoint identity from [citetheorem:8393], applied to $N \in \mathcal{L}(H)$ and the vectors $x,y \in H$,
\begin{align*}
(x,Ny)_H=(N^*x,y)_H.
\end{align*}
Using $Ny=\mu y$ and conjugate-linearity in the second argument,
\begin{align*}
(x,Ny)_H=(x,\mu y)_H=\overline{\mu}(x,y)_H.
\end{align*}
Using $N^*x=\overline{\lambda}x$ and linearity in the first argument,
\begin{align*}
(N^*x,y)_H=(\overline{\lambda}x,y)_H=\overline{\lambda}(x,y)_H.
\end{align*}
Therefore
\begin{align*}
\overline{\mu}(x,y)_H=\overline{\lambda}(x,y)_H.
\end{align*}
Subtracting the right-hand side from the left-hand side gives
\begin{align*}
(\overline{\mu}-\overline{\lambda})(x,y)_H=0.
\end{align*}
Since $\lambda \ne \mu$, we have $\overline{\mu}-\overline{\lambda}\ne 0$. Thus
\begin{align*}
(x,y)_H=0.
\end{align*}
This proves that the eigenspaces corresponding to the distinct eigenvalues $\lambda$ and $\mu$ are orthogonal.
[/step]
