# Eigenvalues of a Hermitian Matrix are Real

In this post, we prove the following:

Statement: All eigenvalues of a Hermitian matrix are real.

Proof:

Since the matrix $A$ is Hermitian, by definition, $A^*=A\,.\ \ \ \ (1)$

Let $\lambda$ be an eignevalue of $A$, and $\mathbf{x}$ be the corresponding eigenvector. Let

$\mathbf{x}=\begin{bmatrix}a_1+ib_1\\a_2+ib_2\\\vdots\\a_n+ib_n\end{bmatrix}\,,$ and

$\mathbf{x}^*=\begin{bmatrix}a_1-ib_1\ \ a_2-ib_2\ \ \cdots\ \ a_n-ib_n\end{bmatrix}\,.$

Then,

$q=\mathbf{x}^*A\mathbf{x}=\mathbf{x}^*\lambda \mathbf{x}=\lambda \mathbf{x}^*\mathbf{x}=\lambda \sum_{i=i}^n{ (a_i^2+b_i^2)}\,.\ \ \ \ (2)$

The sum in the right hand side is real. It follows that $\lambda$ will be real if and only if $q$ is real.  Let us examine: Using $A = A^*$,

$q^*=\left(\mathbf{x}^*A\mathbf{x}\right)^*=\mathbf{x}^*A^*\mathbf{x}=\mathbf{x}^*A\mathbf{x}=q$,

thus $q$ must be real. Consequently, $\lambda$ must be real.

$\square$

In another post, we have shown that for a Hermitian matrix, the eigenvectors pertaining to different eigenvalues are orthogonal.