Hermitian Matrix has Real Eigenvalues

Theorem
Every hermitian matrix has real eigenvalues.

Corollary
Every real symmetric matrix has real eigenvalues.

Proof
Let $A$ be a Hermitian (or self-adjoint) matrix. Then, by definition, $A=A^*$, where * designates the conjugate transpose.

Let $\mathbf v$ be the eigenvector corresponding to the eigenvalue, $\lambda$, of the matrix $A$.

Now, by definition, $A\mathbf v=\lambda\mathbf v$.

Left-multiplying each equation by $\mathbf v^*$, we obtain:
 * $\mathbf v^*A\mathbf v=\mathbf v^*\lambda\mathbf v=\lambda\mathbf v^*\mathbf v$

Firstly, note that both $\mathbf v^*A\mathbf v$ and $\mathbf v^*\mathbf v$ are $1 \times 1$ matrices.

If we examine $\mathbf v^*A\mathbf v$, we find that it is also Hermitian. That is:
 * $(\mathbf v^*A\mathbf v)^*= \mathbf v^* A^*(\mathbf v^*)^*$

due to the cyclic property of the conjugate transpose.

Clearly, $\mathbf v^* A^*(\mathbf v^*)^* = \mathbf v^*A\mathbf v$ which means $\mathbf v^*A\mathbf v$ is Hermitian.

Also, $\mathbf v^*\mathbf v$ is Hermitian. That is:
 * $ (\mathbf v^*\mathbf v)^*=\mathbf v^*(\mathbf v^*)^*=\mathbf v^*\mathbf v $

So both $\mathbf v^*A\mathbf v$ and $\mathbf v^*\mathbf v$ are Hermitian $1 \times 1$ matrices. If we let a be the entry in $\mathbf v^*A\mathbf v$ and b the entry in $\mathbf v^*\mathbf v$:

By definition of Hermitian Matrices, $a=\bar{a}$ and $b=\bar{b}$ which can only be true if they are real entries.

From above, we have $\mathbf v^*A\mathbf v=\lambda\mathbf v^*\mathbf v$. This means that $\lambda$ must also be real.

Therefore, Hermitian matrices have real eigenvalues.