Spectrum of Self-Adjoint Bounded Linear Operator is Real
Theorem
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded self-adjoint operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then:
- $\map \sigma T \subseteq \R$
Proof 1
This follows from:
- Spectrum of Self-Adjoint Densely-Defined Linear Operator is Real and Closed
- Spectrum of Bounded Linear Operator equal to Spectrum as Densely-Defined Linear Operator
$\blacksquare$
Proof 2
Let $\lambda := a + i b \in \C \setminus \R$.
Note that $b \ne 0$.
For all $\phi \in \HH$:
| \(\ds \norm {\paren {T - \lambda I} \phi}^2\) | \(=\) | \(\ds \norm {\paren {T - a I} \phi}^2 + b^2 \norm {\phi}^2\) | as $\Re \innerprod {\paren {T - a I} \phi} {- i b \phi} = 0$ | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(\ge\) | \(\ds b^2 \norm {\phi}^2\) |
In view of $(1)$, both $T - \lambda I$ and $T - \overline \lambda I$ are injective.
Moreover:
| \(\ds \paren {\Img {T - \lambda I} }^\perp\) | \(=\) | \(\ds \paren {\Img { \paren {T - \overline \lambda I}^\ast } }^\perp\) | as $T - \lambda I = \paren {T - \overline \lambda I}^\ast$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \ker \paren {T - \overline \lambda I}\) | Kernel of Linear Transformation is Orthocomplement of Image of Adjoint | |||||||||||
| \(\ds \) | \(=\) | \(\ds \set 0\) | as $T - \overline \lambda I$ is injective |
By Linear Subspace Dense iff Zero Orthocomplement, $\Img {T - \lambda I}$ is dense in $\HH$.
Let $\phi \in \HH$.
Then there exists an $\sequence {\psi _n} \subseteq \HH$ such that:
- $\paren {T - \lambda I} \psi_n \to \phi$
By $(1)$ there exists a $\psi \in \HH$ such that:
- $\psi_n \to \psi$
and:
- $\paren {T - \lambda I} \psi = \phi$
Thus $T - \lambda I$ is surjective.
In addition, by $(1)$ we have:
- $\norm \psi \le \size b^{-1} \norm \phi$
Therefore $T - \lambda I$ is invertible such that:
- $\norm {\paren {T - \lambda I}^{-1} } \le \size b^{-1}$
That is, $\lambda \not \in \map \sigma T$.
$\blacksquare$