TITLE: Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)
QUESTION [23 upvotes]: The problem statement is as follows:

Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem.  The gist of my proof is presented below.
If possible, I would like to find a simpler proof.  I was hoping that there might be some sort of manipulation along these lines, noting that
$$
\langle Av_1,A v_2\rangle 
= \langle v_1,A^*Av_2\rangle 
= \langle v_1,AA^*v_2\rangle 
= \langle A^* v_1,A^* v_2 \rangle
$$
Any ideas here would be appreciated.

My proof:
Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that
$$
A v_{\lambda,i} = \lambda v_{\lambda,i}
$$
Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$.  We may write
$
v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda}
.$
We then have
$$
0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}
$$
So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$.  Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$.  From there, we have
$$
\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}
$$
the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

REPLY [5 votes]: The answers to this question are very good but they are glossing over the fact that if $A \in \mathbb C^{n \times n}$ is normal then $Ax = \lambda x \implies A^*x = \bar \lambda x$. I think it's helpful to just spell out the whole thing explicitly.
Lemma: If $M \in \mathbb C^{n \times n}$ is normal, then $M$ and $M^*$ have the same null space.
Proof: Let $M \in \mathbb C^{n \times n}$ be a normal matrix. Then
\begin{align}
&M^*x = 0 \\
\iff & \| M^* x \|^2 = 0 \\
\iff & \langle M^*x, M^* x \rangle = 0 \\
\iff & \langle M M^* x, x \rangle = 0 \\
\iff & \langle M^* M x, x \rangle = 0 \\
\iff & \langle Mx, Mx \rangle = 0 \\
\iff & \| Mx \|^2 = 0 \\
\iff & Mx = 0.
\end{align}
Lemma: If $A \in \mathbb C^{n \times n}$ is normal then $Ax = \lambda x \implies A^* x = \bar \lambda x$.
Proof: Suppose that $A \in \mathbb C^{n \times n}$ is normal and $Ax = \lambda x$. So $x$ is a null vector of $M = A - \lambda I$.
Note that $M$ is normal, as you can check by expanding both $M^* M$ and $M M^*$.
By the above lemma, $x$ is also a null vector of $M^* = A^* - \bar \lambda I$. Thus $A^*x = \bar \lambda x$.
Theorem: If $x$ and $y$ are eigenvectors corresponding to distinct eigenvalues of a normal matrix $A \in \mathbb C^{n \times n}$, then $\langle x, y \rangle = 0$.
Proof:
Let $A \in \mathbb C^{n \times n}$ be a normal matrix, and suppose that $x$ and $y$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\gamma$, respectively.
Note that
$$\langle x, Ay \rangle = \langle x, \gamma y \rangle = \bar \gamma \langle x, y \rangle.$$
On the other hand,
$$
\langle x, Ay \rangle = \langle A^* x, y \rangle = \langle \bar \lambda x, y \rangle = \bar \lambda \langle x, y \rangle.
$$
So we find that $\bar \gamma \langle x, y \rangle = \bar \lambda \langle x, y \rangle$, which implies that $\langle x, y \rangle = 0$.