Perpendicular eigenvectors

Let $A \in \mathbb{R}^{n \times n}$ be a matrix such that $A=A^t$. Prove that, if $u$ and $v$ are eigenvectors of $A$ corresponding to different eigenvalues , then $\langle u, v \rangle =0$.

Solution

Since the matrix is real and $A=A^t$ holds this means that $A$ is symmetric, hence diagonizable. Let $\ell_1, \; \ell_2$ be two different eigenvalues of $A$ and $u, \; v$ be their eigenvectors respectively. Then:

$$Au =\ell_1 u , \quad Av = \ell_2 v$$

Hence:

$$\begin{aligned}
\langle Au, v \rangle = \langle u , Av \rangle &\Leftrightarrow \langle \ell_1 u, v \rangle = \langle u, \ell_2 v \rangle \\
 &\Leftrightarrow \ell_1 \langle u, v \rangle = \ell_2 \langle u, v \rangle \\
 &\Leftrightarrow \langle u, v \rangle \left ( \ell_1 -\ell_2 \right )=0 \\
 &\overset{\ell_1 -\ell_2 \neq 0}{\Leftarrow \! =\! =\! =\! \Rightarrow } \; \langle u, v \rangle =0
\end{aligned}$$

completing the proof.