# Hermitian Matrix has Real Eigenvalues/Proof 2

## Theorem

Every Hermitian matrix has eigenvalues which are all real numbers.

## Proof

Let $\mathbf A$ be a Hermitian matrix.

Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its Hermitian conjugate.

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

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

Denote with $\innerprod \cdot \cdot$ the inner product on $\C$.

 $\ds \lambda * \innerprod v v$ $=$ $\ds \innerprod {\lambda * v} v$ Properties of Complex Inner Product $\ds$ $=$ $\ds \innerprod {\mathbf A * v} v$ Definition of Eigenvector of Linear Operator: $\lambda * v = \mathbf A * v$ $\ds$ $=$ $\ds \innerprod v {\mathbf A^\dagger * v}$ Properties of Adjugate $\ds$ $=$ $\ds \innerprod v {\mathbf A * v}$ $\mathbf A$ is Hermitian, so $\mathbf A^\dagger = \mathbf A$ $\ds$ $=$ $\ds \innerprod v {\lambda * v}$ Definition of Eigenvector of Linear Operator: $\lambda*v = \mathbf A*v$ $\ds$ $=$ $\ds \overline \lambda * \innerprod v v$ Properties of Complex Inner Product

We have that $v \ne 0$.

Hence by non-negative definiteness of an inner product:

$\innerprod v v \ne 0$

and we can divide both sides by $\innerprod v v$.

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.

$\blacksquare$