[proofplan]
We compute the real inner product $\langle Av, w \rangle$ in two ways: once using $Av = \lambda v$ and once using symmetry $A^\top = A$ to move $A$ onto $w$ and apply $Aw = \mu w$. Equating the two expressions and using $\lambda \neq \mu$ forces $\langle v, w \rangle = 0$.
[/proofplan]
[step:Compute $\langle Av, w \rangle$ using the eigenvalue equation for $v$]
Since $Av = \lambda v$, we have
\begin{align*}
\langle Av, w \rangle = \langle \lambda v, w \rangle = \lambda \langle v, w \rangle.
\end{align*}
Here $\lambda \in \mathbb{R}$ by the [Eigenvalues of Symmetric Matrices are Real](/theorems/3279), so the scalar factors out without conjugation.
[/step]
[step:Compute $\langle Av, w \rangle$ using symmetry and the eigenvalue equation for $w$]
Since $A$ is symmetric, $A^\top = A$, so for any $x, y \in \mathbb{R}^n$:
\begin{align*}
\langle Ax, y \rangle = (Ax)^\top y = x^\top A^\top y = x^\top A y = \langle x, Ay \rangle.
\end{align*}
Applying this with $x = v$, $y = w$:
\begin{align*}
\langle Av, w \rangle = \langle v, Aw \rangle = \langle v, \mu w \rangle = \mu \langle v, w \rangle.
\end{align*}
[guided]
We use the defining property of a symmetric matrix to transfer $A$ from the first argument to the second. The real inner product on $\mathbb{R}^n$ is $\langle x, y \rangle = x^\top y = \sum_{i=1}^n x_i y_i$. For $A = A^\top$:
\begin{align*}
\langle Ax, y \rangle = (Ax)^\top y = x^\top A^\top y = x^\top Ay = \langle x, Ay \rangle.
\end{align*}
This is precisely the statement that $A$ is self-adjoint with respect to the standard inner product on $\mathbb{R}^n$. Applying this with $x = v$ and $y = w$, and then using $Aw = \mu w$:
\begin{align*}
\langle Av, w \rangle = \langle v, Aw \rangle = \langle v, \mu w \rangle = \mu \langle v, w \rangle.
\end{align*}
Again $\mu \in \mathbb{R}$ by the [Eigenvalues of Symmetric Matrices are Real](/theorems/3279), so the scalar factors out directly.
[/guided]
[/step]
[step:Equate the two expressions and conclude $\langle v, w \rangle = 0$]
From the two computations:
\begin{align*}
\lambda \langle v, w \rangle = \langle Av, w \rangle = \mu \langle v, w \rangle.
\end{align*}
Rearranging:
\begin{align*}
(\lambda - \mu) \langle v, w \rangle = 0.
\end{align*}
Since $\lambda \neq \mu$, the factor $\lambda - \mu \neq 0$, so $\langle v, w \rangle = 0$.
[guided]
We have established two expressions for the same quantity:
\begin{align*}
\lambda \langle v, w \rangle = \langle Av, w \rangle = \mu \langle v, w \rangle.
\end{align*}
Subtracting:
\begin{align*}
(\lambda - \mu) \langle v, w \rangle = 0.
\end{align*}
Since $\lambda$ and $\mu$ are distinct eigenvalues, $\lambda - \mu \neq 0$. Dividing both sides by $\lambda - \mu$ gives $\langle v, w \rangle = 0$, i.e., $v$ and $w$ are orthogonal.
This result is the reason that symmetric (more generally, self-adjoint) operators are so well-behaved: not only are their eigenvectors linearly independent (which holds for any matrix by the [Linear Independence of Eigenvectors for Distinct Eigenvalues](/theorems/920)), but they are in fact orthogonal. This stronger conclusion is what ultimately drives the Spectral Theorem: the eigenspaces of a symmetric matrix are mutually orthogonal, allowing the construction of an orthonormal basis of eigenvectors.
[/guided]
[/step]
