Hermitian Matrix has Real Eigenvalues/Proof 2

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

Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates the conjugate transpose.

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

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

Denote with $\left\langle{\cdot, \cdot}\right\rangle$ the inner product on $\C$.

We have that $v \ne 0$, and because of the positive definiteness, it must be that:
 * $\left\langle v, v\right\rangle \ne 0$

Thus:
 * $\left\langle v, v\right\rangle \ne 0$

So we can divide both sides by $\left\langle v, v\right\rangle$.

Thus $\lambda = \overline{\lambda}$.

By Complex Number equals Conjugate iff Wholly Real, $\lambda$ is a real number.

$\lambda$ was arbitrary, so it follows that every eigenvalue is a real number.

Hence the result.