Why is $\ker(T-\lambda I)^n$ finite-dimensional?

Question

Let $X$ be a normed space and $T$ be a compact operator on $X$ and $\lambda \in \sigma(T)\setminus\{0\}$.

A closed unit ball in $\ker(T-\lambda I)$ always admit a convergent subsequence of a sequence, it is compact. Since a closed unit ball in an infinite-dimensional normed space is never compact, $\ker(T-\lambda I)$ is finite-dimensional.

However, this argument cannot be directly applied to $\ker(T-\lambda I)^n$ when $n>1$.

How do I prove that $\ker(T-\lambda I)^n$ is finite-dimensional?

Answer 1

If you you expand $(T-\lambda I)^n$ you get $R+(-\lambda)^nI$ with $R$ compact, and now you can apply the same argument as before.