Let $A$ be a Hermitian matrix (i.e. equal to its conjugate transpose $A^H$) having eigenvectors $x, y$ with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_1 \neq \lambda_2$. Show $x$ and $y$ are orthogonal.
My proof is below. Please verify, critique, or improve.
Proof:
$$\begin{align} \lambda_1 x^Hy &= (Ax)^Hy \\ &= x^HAy \quad \text{(since }A=A^H\text{)} \\ &= x^H(\lambda_2y) \\ &= \lambda_2 x^Hy \quad \text{(by linearity)}. \end{align}$$
Since $\lambda_1 \neq \lambda_2, x^Hy = 0$.