Eigenvalues of Hermitian Operator are Real

From ProofWiki
Jump to navigation Jump to search


It has been suggested that this page or section be merged into Hermitian Operator has Real Eigenvalues.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Mergeto}} from the code.


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

Retrieved from "https://proofwiki.org/w/index.php?title=Eigenvalues_of_Hermitian_Operator_are_Real&oldid=616934"