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 $b_{i,\lambda} = 0$.
Best Answer
Assume $\;\lambda\neq \mu\;$ and
$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$
From this we get:
$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\ \langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$
and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$
Question: Where did we use normality in the above?