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.


$$\begin{align} \lambda_1 x^Hy &= (Ax)^Hy \\ &= x^HAy \quad \text{(since }A=A^H\text{)} \\ &= x^H(\lambda_2y) \\ &= \lambda_2 x^Hy \quad \text{(by linearity)}. \end{align}$$

Since $\lambda_1 \neq \lambda_2, x^Hy = 0$.


1 Answer 1


What the proof is missing is to mention the fact that $\lambda_1,\lambda_2$ are real because $A$ is Hermitian. Otherwise you would have to start with $\overline\lambda_1$ and the argument doesn't work.

As a very minor point, if you are into writing the proof carefully, you should have one more step in $(Ax)^Hy=x^HA^Hy=x^HAy$.


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