# 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 $\langle a,b\rangle$ denotes the usual inner product of two vectors $x,y$, i.e.,

$\langle x,y \rangle := y^*x$.

We will show that the eigenvalues of the $n\times n$ Laplacian matrix $L$ of the complete graph $K_n$ are $\{0,1\}$ where  the eigenvalue $0$ has (algebraic) multiplicity $1$ and the eigenvalue $n$ has multiplicity $n-1$.