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.)

Answer 2

Or you can proceed directly by algebraic methods. The spectrum of $A$ is $\{ \lambda \in \mathbb{C} : A- \lambda I$ is not invertible $\}$. If $A$ is nilpotent and $\lambda \neq 0$, then there is a positive integer $n$ such that $A^{n} - \lambda^{n}I = -\lambda^{n}I,$ which is an invertible operator. Since $A^{n} - \lambda^{n} I = (A-\lambda I) (A^{n-1} + \lambda A^{n-2} + \ldots \lambda^{n-2}A + \lambda^{n-1}I),$ we see that $A-\lambda I$ is invertible, and $\lambda$ is not in the spectrum of $A.$ On the other hand, if $A$ is nilpotent, then $A = A- 0I$ is certainly not invertible, and $0$ is in the spectrum of $A.$

Answer 3

The spectrum of $A$ is $\{0\}$ because there has to be something in the spectrum, and resolvent series converges everywhere except at $\lambda=0$: $$ (\lambda I-A)^{-1}=\sum_{n=0}^{\infty} \frac{1}{\lambda^{n+1}}A^{n} $$

Answer 4

For any polynomial $p$ we know $\sigma(p(A)) = p(\sigma(A))$.