Functional Analysis – Spectrum of a Nilpotent Operator

Question

Let $X$ be a Banach space and $A:X\to X$ be a bounded operator such that $A^n=0$ for some $n\in \mathbb{N}$. Is the spectrum of $A$ finite, countable ?

Answer 1

If $A$ is a bounded linear operator on a Banach space, then the spectral radius $r(A) = \sup\{|\lambda|\ |\lambda\in\sigma(A)\}$ satisfies $$ r(A) = \lim_{n\to\infty} \|A^n\|^{1/n}.$$ So the spectrum of a nilpotent operator $A$ is not only countable and finite, it contains only zero. (c.f. these notes, proposition 9.5, pp 220-1 --- the theorem is for Hilbert spaces, but the proof does not seem to rely on the use of the inner product. They refer to Reed & Simon for more details.)