# In a Hermitian Matrix, the Eigenvectors of Different Eigenvalues are Orthogonal

This is an elementary (yet important) fact in matrix analysis.

Statement

Let $M$ be an $n\times n$ complex Hermitian matrix which means $M=M^*$ where $*$ denotes the conjugate transpose operation. Let $\lambda_1, \neq \lambda_2$ be two different  eigenvalues of $M$. Let $x, y$ be the two eigenvectors of $M$ corresponding to the two eigenvalues $\lambda_1$ and $\lambda_2$, respectively.

Then the following is true:

$\boxed{\lambda_1 \neq \lambda_2 \iff \langle x, y \rangle = 0.}$

Here $\langle x,y \rangle := y^*x$ denotes the usual inner product of two vectors $x,y$.

Proof

It is given that

$Mx=\lambda_1x$,
$My=\lambda_2y$.

Since $M=M^*$, it follows that

$\langle x,My \rangle =y^*M^*x=y^*Mx=y^*\lambda_1x=\lambda_1y^*x=\lambda_1 \langle x,y \rangle$.

However, we have

$\langle x,My \rangle = \langle x, \lambda_2y \rangle =\lambda_2^* \langle x,y \rangle = \lambda_2 \langle x,y \rangle$

since $\lambda_1, \lambda_2$ must be real. Therefore,

$\lambda_1 \langle x,y \rangle =\lambda_2 \langle x,y \rangle$
$\implies (\lambda_1-\lambda_2) \langle x,y \rangle =0$.

Therefore, $\lambda_1 \neq \lambda_2 \implies \langle x,y \rangle =0$,

and

$\langle x,y \rangle \neq 0 \implies \lambda_1=\lambda_2.$

Thus the eigenvectors corresponding to different eigenvalues of a Hermitian matrix are orthogonal. Additionally, the eigenvalues corresponding to a pair of non-orthogonal eigenvectors are equal.

$\square$