# 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:

 $\ds \lambda \innerprod v v$ $=$ $\ds \innerprod {\lambda v} v$ Property $(2)$ of an inner product $\ds$ $=$ $\ds \innerprod {A v} v$ $v$ is an eigenvector $\ds$ $=$ $\ds \innerprod v {A v}$ $A$ is Hermitian $\ds$ $=$ $\ds \overline {\innerprod {A v} v}$ Property $(1)$ of an inner product $\ds$ $=$ $\ds \overline {\innerprod {\lambda v} v}$ $v$ is an eigenvector $\ds$ $=$ $\ds \bar \lambda \innerprod v v$ Properties $(2)$, $(4)$ of an inner product

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, $\lambda \in \R$.

$\blacksquare$