Eigenvalues of Hermitian Operator are Real

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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, $\lambda \in \R$.

$\blacksquare$


Sources

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