Suppose $T$ is a compact operator from a Hilbert Space. Then for $a\neq 0$, we have that $\text{range}(T- aI)^{n-1}=\text{range}(T-aI)^{n}$ for some $n$ and $\ker(T-aI)^{m-1}=\ker(T-aI)^{m}$ for some $m$.
Firstly, if $a$ is a not an eigen value, then by Fredholm alternative we know that $T-aI$ is surjective.
So if $a$ is an eigenvalue, then we have that $\dim(\ker(T-aI))<\infty$. If I know look at the chain $\ker(T)\subseteq \ker(T^{2})\subseteq \ker(T^{3})...\subseteq$ I cannot use the finite dimensioinality in any helpful way because the chain is increasing and I only have an lower bound on the dimension.
I also know that if $\ker(T-aI)^{m}=\ker(T-aI)^{m-1}$ for some $m$, then we also will have that $\text{range}(T-aI)^{m-1}=\text{range}(T-aI)^{m}$ as this is a result from Linear Algebra. And also I can write $H=\ker(T-aI)^{m}\oplus \text{range}(T-aI)^{m}$.
But I am struggling to find a way in this problem. Any help is appreciated.
EDIT: From the link given in the comment, I can do it if I can show that $\text{range}(T-aI)$ is closed. I have answered my own question here