Spectrum of Adjoint of Bounded Linear Operator
Theorem
Let $X$ be a Banach space over $\C$.
Let $T : X \to X$ be a bounded linear operator.
Let $T^\ast : X \to X$ be the adjoint of $T$.
Let $\map \sigma T$ and $\map \sigma {T^\ast}$ be the spectrum of $T$ and $T^\ast$ respectively.
Then:
- $\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$
where $\overline \lambda$ denotes the complex conjugate of $\lambda$.
Proof
We show that for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ if and only if $\overline \lambda \not \in \map \sigma {T^\ast}$.
Let $\lambda \in \C$ have $\lambda \not \in \map \sigma T$.
Then $T - \lambda I$ is invertible as a bounded linear operator.
Note that $I^\ast = I$ from Adjoint of Identity Transformation.
From Adjoining is Linear, we have that $T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$.
From Adjoining Commutes with Inverting, we have:
- $T^\ast - \overline \lambda I = \paren {T - \lambda I}^\ast$ is invertible
So we have $\overline \lambda \not \in \map \sigma {T^\ast}$.
So if $\lambda \not \in \map \sigma T$ then $\overline \lambda \not \in \map \sigma {T^\ast}$.
From Adjoint is Involutive, swapping $T$ for $T^\ast$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$.
From Complex Conjugation is Involution, swapping $\lambda$ for $\overline \lambda$ we have that $\lambda \not \in \map \sigma {T^\ast}$ implies $\overline \lambda \not \in \map \sigma T$.
So for $\lambda \in \C$, we have $\lambda \not \in \map \sigma T$ if and only if $\overline \lambda \not \in \map \sigma {T^\ast}$.
So we have:
- $\map \sigma {T^\ast} = \set {\overline \lambda : \lambda \in \map \sigma T}$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $14.1$: The Resolvent and Spectrum