Hermitian Matrix has Real Eigenvalues

Theorem
Every hermitian 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.