Suppose that $X$ is a Banach space and $T$ is a compact operator in $\mathcal{B}(X)$. $\lambda\neq 0$ and let $S=\lambda I-T$.
Prove that:
(a) $\mathcal{N}(S^n)=\mathcal{N}(S^{n+1}) \Rightarrow \mathcal{N}(S^n)=\mathcal{N}(S^{n+k})$
(b) There must be some integer $n$ s.t. $\mathcal{N}(S^n)=\mathcal{N}(S^{n+1})$
(c) Let $n$ be the smallest integer that (a) holds, then $\dim\mathcal{N}(S^n)<\infty$, and $X=\mathcal{N}(S^n) \oplus \mathcal{R}(S^n)$, $S$ restricted on $\mathcal{R}(S^n)$ is a bijection.
I've proved (a)(b) and the 'bijection' part in (c) by showing that $\mathcal{N}(S^n) \cap \mathcal{R}(S^n)=\emptyset$, but I don't know how to prove the part $\dim\mathcal{N}(S^n)<\infty$ and $X=\mathcal{N}(S^n) + \mathcal{R}(S^n)$.
Also, I'm interested that if we have a smallest integer $n$ s.t. $\mathcal{N}(S^n)=\mathcal{N}(S^{n+1})$, then there will also be a smallest integer $m$ s.t. $\mathcal{R}(S^m)=\mathcal{R}(S^{m+1})$. So is it necessary that $m=n$? And if it's true, how can we prove that?
Thanks in advance.