Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.

Let $\struct {\map D T, T}$ be a densely-defined linear operator.

Let $\map {\sigma_p} T$ be the point spectrum of $T$.


Then $\lambda \in \map {\sigma_p} T$ if and only if $\lambda$ is an eigenvalue of $T$.


Proof

We have that $\lambda \in \map {\sigma_p} T$ if and only if:

$T - \lambda I$ is not injective.

That is, if and only if there exists $x \in \map D T \setminus \set 0$ such that:

$\paren {T - \lambda I} x = \map {\paren {T - \lambda I} } 0 = 0$

So $\lambda \in \map {\sigma_p} T$ if and only if there exists $x \in \map D T \setminus \set 0$ such that:

$T x = \lambda I x = \lambda x$

That is, $\lambda \in \map {\sigma_p} T$ if and only if $\lambda$ is an eigenvalue of $T$.

$\blacksquare$