I need to find the essential spectrum of an abitrary compact operator.
Let $T : X \to Y$ be a compact operator. The essential spectrum is the set of all $\lambda \in \mathbb{C}$ such that $T- \lambda I$ is not a Fredhlom operator. Hence, we need to find the set of $\lambda$'s such that $dim(Ker(T- \lambda I)) < \infty$ and $dim(Coker(T- \lambda I)) < \infty$, where $I$ is the identity map.
Until now I was able to show that $\lambda=0$ is in the essential spectrum for the simple reason that there are compact operators that are not Fredholm. Moreover, I proved that $dim(Ker(T- \lambda I)) < \infty$, but I am struggling in the last part. In Wikipedia I have read that in fact $dim(Ker(T- \lambda I)) = dim(Coker(T- \lambda I))$ for compact operators $T$ and non zero $\lambda$. I have even taken a look to the reference but I find it pretty hard to follow. Could you help me with simpler proof?
Thanks in advance!
Best Answer
For this you would usually use the characterization that $T$ is Fredholm if and only if $T$ is invertible modulo the compacts, i.e. iff there exists $S$ such that $TS-I$ and $ST-I$ are compact. In particular, $T$ is Fredholm if and only if $T+K$ is Fredholm for all compact $K$; indeed, if $TS-I$ is compact, then $$(T+K)S-I=(TS-I)+KS$$ is compact, and similarly $S(T+K)-I$ is compact.
When $T$ is already compact and $\lambda\ne0$, $$TS-\lambda I=\lambda(TS/\lambda-I)$$ is Fredholm (because $-I$ and we are adding a compact to it) so it cannot be compact.
So $\sigma_{\rm ess}(T)\subset\{0\}$, which forces $\sigma_{\rm ess}(T)=\{0\}$ since the essential spectrum is nonempty.