This is elementary, yet useful and interesting.

Statement: All eigenvalues of a Hermitian matrix are real.

Proof:

If 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,

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

Since $\mathbf{x}$ is an $n\times 1$ vector, $q$ is just a number (real or complex).

Now, it follows from (1) that $q^*=\left(\mathbf{x}^*A\mathbf{x}\right)^*=\mathbf{x}^*A^*\mathbf{x}=\mathbf{x}^*A\mathbf{x}=q$.

Any complex number which is equal to its conjugate must have the imaginary part equal to zero. Therefore, $q$ must be real, which implies $\displaystyle \lambda \sum_{i=i}^n{a_i^2+b_i^2}$ must also be real. Since the sum $\sum_{i=i}^n{a_i^2+b_i^2}$ is also real, it follows the $\lambda$ must be real.

$\square$