Hermitian Matrix has Real Eigenvalues/Proof 1

Proof
Let $\mathbf A$ be a Hermitian matrix.

Then, by definition:
 * $\mathbf A = \mathbf A^*$

where $\mathbf A^*$ denotes the conjugate transpose of $\mathbf A$.

Let $\lambda$ be an eigenvalue of $\mathbf A$.

Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$.

By definition of eigenvector:
 * $\mathbf{A v} = \lambda \mathbf v$

Left-multiplying both sides by $\mathbf v^*$, we obtain:


 * $(1):\quad \mathbf v^* \mathbf{A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$

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

Now observe that, using Conjugate Transpose of Matrix Product: General Case:


 * $\left({\mathbf v^* \mathbf{A v}}\right)^* = \mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^*$

As $\mathbf A$ is Hermitian, and $\left({\mathbf v^*}\right)^* = \mathbf v$ by Conjugate Transpose is Involution, it follows that:


 * $\mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^* = \mathbf v^* \mathbf{A v}$

That is, $\mathbf v^* \mathbf{A v}$ is also Hermitian.

By Product with Conjugate Transpose Matrix is Hermitian, $\mathbf v^* \mathbf v$ is Hermitian.

So both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are Hermitian $1 \times 1$ matrices.

Now suppose that we have for some $a,b \in \C$:


 * $\mathbf v^* \mathbf{A v} = \begin{bmatrix}a\end{bmatrix}$
 * $\mathbf v^* \mathbf v = \begin{bmatrix}b\end{bmatrix}$

Note that $b \ne 0$ as an eigenvector is by definition non-zero.

By definition of a Hermitian matrix:
 * $a = \bar a$ and $b = \bar b$

where $\bar a$ denotes the complex conjugate of $a$.

By Complex Number equals Conjugate iff Wholly Real, it follows that $a, b \in \R$, that is, are real.

From equation $(1)$, it follows that $\begin{bmatrix}a\end{bmatrix} = \lambda \begin{bmatrix}b\end{bmatrix}$.

Thus, $a = \lambda b$, i.e. $\lambda = \dfrac a b$ (recall that $b \ne 0$).

Hence $\lambda$, being a quotient of real numbers, is real.