Spectrum of Adjoint of Bounded Linear Operator

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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

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