Eigenvalues of Hermitian Operator are Real

Theorem
Let $\HH$ be a Hilbert space.

Let $A \in \map B \HH$ be a Hermitian operator.

Then all eigenvalues of $A$ are real.

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

Let $v \in H$ be an eigenvector for $\lambda$.

That is:
 * $A v = \lambda v$

Now compute:

Now $v$, being an eigenvector, is non-zero.

By property $(5)$ of an inner product, this implies $\innerprod v v \ne 0$.

Thence, dividing out $\innerprod v v$, obtain $\lambda = \bar \lambda$.

From Complex Number equals Conjugate iff Wholly Real, $\lambda \in \R$.