Suppose that a real symmetric matrix $A$ has two distinct eigenvalues $\alpha$ and $\beta$.
Show that any eigenvector corresponding to $\alpha$ is orthogonal to any eigenvector corresponding to $\beta$.

(Nagoya University, Linear Algebra Final Exam Problem)
 

Hint.

Two vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if their inner (dot) product $\mathbf{u}\cdot \mathbf{v}:=\mathbf{u}^{\trans}\mathbf{v}=0$.

Here $\mathbf{u}^{\trans}$ is the transpose of $\mathbf{u}$.

A fact that we will use below is that for matrices $A$ and $B$, we have $(AB)^{\trans}=B^{\trans}A^{\trans}$.

Proof.

Let $\mathbf{u}, \mathbf{v}$ be eigenvectors corresponding to $\alpha, \beta$, respectively.
Namely we have
\[A\mathbf{u}=\alpha \mathbf{u} \text{ and } A\mathbf{v}=\beta \mathbf{v}. \tag{*}\]

To prove that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, we show that the inner product $\mathbf{u} \cdot \mathbf{v}=0$.
Keeping this in mind, we compute
\begin{align*}
&\alpha (\mathbf{u} \cdot \mathbf{v}) =(\alpha \mathbf{u}) \cdot \mathbf{v} \\
&\stackrel{(*)}{=} A\mathbf{u}\cdot \mathbf{v} =(A\mathbf{u})^{\trans} \mathbf{v}\\
&=\mathbf{u}^{\trans}A^{\trans}\mathbf{v} \text{ (This follows from the fact mentioned in the hint above)} \\
&=\mathbf{u}^{\trans}A\mathbf{v} \text{ (since $A$ is symmetric.)}\\
& \stackrel{(*)}{=} \mathbf{u}^{\trans}\beta \mathbf{v}=\beta (\mathbf{u}^{\trans} \mathbf{v})=\beta (\mathbf{u}\cdot \mathbf{v}).
\end{align*}

Therefore we obtain
\[\alpha (\mathbf{u} \cdot \mathbf{v})=\beta (\mathbf{u} \cdot \mathbf{v}),\] and thus
\[(\alpha-\beta)(\mathbf{u} \cdot \mathbf{v})=0.\]

Since $\alpha$ and $\beta$ are distinct, $\alpha-\beta \neq 0$.
Hence we must have
\[\mathbf{u} \cdot \mathbf{v}=0,\] and the eigenvectors $\mathbf{u}, \mathbf{v}$ are orthogonal.

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