I am having a hard time getting my head around the Hilbert -Schmidt operators.
Now I have a question about them: why is the essential spectrum of Hilbert -Schmidt operators is empty? The definition of the notation is as follows:
If $H$ is a Hilbert space and $T\in\mathcal{L}(H)$ then $T$ is Hilbert-Schmidt if there is some orthonormal basis of $H$ such that:
$$\sum_n \lVert T(e_n)\rVert_{H}^2<\infty$$
And the essential spectrum consists of the continum spectrum and eigenvalues with infinite multiplicity.
I have trouble figuring out the reason why the Hilbert -Schmidt operators have no essential spectrums. Why do they have only discrete spectrums? Thanks very much!
Best Answer
A Hilbert-Schmidt operator is compact. This means that any nonzero eigenvalue necessarily has a finite-dimensional eigenspace; and this immediately implies that the essential spectrum of a compact operator is $\{0\}$. This is simply the fact that $\sigma_{\rm ess}(T)=\sigma(\pi(T))$, where $\pi$ is the quotient map onto the Calkin algebra.