Need verification - Given a Hermitian matrix and two eigenvectors corresponding to distinct eigenvalues, show x and y are orthogonal.

Question

Claim: Let $A \in \mathbb{C}^{mxm}$ be hermitian ($A = A^*)$. If $x$ and $y$ are eigenvectors corresponding to distinct eigenvalues, then x and y are orthogonal.

Proof: Let $x$ and $y$ correspond to the eigenvalues $\lambda_x$ and $\lambda_y$ respectively.

Firstly, as $$(Ax)^*y = (\lambda_xx)^*y,$$ $$(x^*A^*)y = (x^*\lambda_x^*)y,$$ $$(x^*\lambda_x^*)y = \lambda_x(x^*y).$$

Further, as $$(x^*A^*)y = x^*(A^*y),$$ $$x^*(A^*y) = x^*(\lambda_yy),$$ $$x^*(\lambda_yy) = \lambda_y(x^*y).$$

Thus, $$\lambda_x(x^*y) = \lambda_y(x^*y),$$ $$\lambda_x(x^*y)-\lambda_y(x^*y) = 0,$$ $$(\lambda_x - \lambda_y)(x^*y) = 0.$$

Finally, as $\lambda_x \neq \lambda_y$, $\lambda_x - \lambda_y \neq 0,$ so $x^*y = 0,$ showing $x$ and $y$ to be orthogonal.

Answer 1

Your proof looks good! Looks like there's nothing to add or correct here.

Answer 2

It's all right! Even is short!