I have been struggling to analyse the spectrum of the adjoint of a bounded linear operator on a hilbert space.
Throughout the internet I have found vague references that
$\sigma(T^*) = \sigma(T)$ but I fail to be able to prove that.
What I have succeeded in doing is to prove that $\rho(T^*) = \overline{\rho(T)}$ (where $\rho$ is the resolvent $\rho(T) = \mathbb{C} \backslash \sigma(T)$).
Thus imho $\sigma(T^*)$ should be equal to $\overline{\sigma(T)}$ and not the above…
[Math] Spectrum of adjoint bounded linear operator on hilbert space
hilbert-spaces
Best Answer
You're correct that $\sigma(T^*) = \overline{\sigma(T)}$, where the overline denotes complex conjugation (and not closure as a subset of $\mathbb{C}$ - recall that the resolvent set of a bounded operator is open, and hence the spectrum is closed). In order to conclude that $\sigma(T^*) = \sigma(T)$, you need to impose the stronger condition that the eigenvalues be real (by specifying that $T$ be self-adjoint, e.g.).