Normal plus compact is Fredholm with index 0

Question

Let $T, K \in B(H)$, $TT^{*} = T^{*}T$, $K$ compact. Why is $T+K$ Fredholm and $ind(T +K) = 0$? Is it even true? I have forgotten all I knew about Fredholm theory and I need that result for proving some stuff about exact sequence of $C^{*}$-algebras.

Answer 1

As Fred points out, this statement is not true in general. The issue here is that there exist normal operators which are not Fredholm (the normal non-Fredholm operator $0$ being the given example).

What is true is that if $T$ is normal and Fredholm and $K$ is compact, then $T+K$ is Fredholm and $\operatorname{index}(T+K)=0$. This of course follows from the stability of Fredholm operators (and their indices) under compact perturbation, and the fact that normal Fredholm operators have index $0$.

Answer 2

It is not true. If $ \dim H = \infty$ and $T=0$, then the compact operator $T+K=K$ is not Fredholm, in general.