> [!NOTE] Theorem > Let $A \in \operatorname{Mat}_{n n}$ be a symmetric matrix. If $\lambda, \mu$ are distinct eigenvalues of $A$ and $\underline{v}, \underline{w}$ are corresponding eigenvectors, then $\underline{v} \cdot \underline{w}=0$. Proof. Since $A \underline{v}=\lambda \underline{v}$ and $A \underline{w}=\mu \underline{w}$, we have $ \underline{v}^T A \underline{w}=\underline{v}^T(A \underline{w})=\underline{v}^T(\mu \underline{w})=\mu \underline{v}^T \underline{w} $ Similarly we have $\underline{w}^T A \underline{v}=\lambda \underline{w}^T \underline{v}$, which after transposing and using $A=A^T$ gives $ \underline{v}^T A \underline{w}=\left(\underline{w}^T A^T \underline{v}\right)^T=\left(\underline{w}^T A \underline{v}\right)^T=\left(\lambda \underline{w}^T \underline{v}\right)^T=\lambda \underline{v}^T \underline{w} $ Subtracting these two expressions gives $ \underline{0}=(\mu-\lambda) \underline{v}^T \underline{w} $ Since $\lambda \neq \mu$ and $\underline{v} \cdot \underline{w}=\underline{v}^T \underline{w}$, the result follows.