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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 ?

user165633
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4 Answers4

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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.)

Neal
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    That formula for the spectral radius is certainly valid for any Banach algebra, so for the algebra of bounded linear operators from any Banach space to itself ( I think it is due to Gelfand). – Geoff Robinson Nov 15 '14 at 16:12
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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 $A.$

Geoff Robinson
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  • If $A$ is nilpotent, is it Fredholm? – Neal Nov 15 '14 at 17:48
  • @Neal: I don't think so: If $B$ is an infinite dimensional Banach Space, and $T: B \to B$ is a nilpotent linear operator, there an infinite-dimensional $T$-invariant subspace $V,$ and the kernel of the restriction $T:V \to V$ is infinite dimensional. To construct $V,$ choose an infinite linearly independent set of vectors ${v_{i} }$, and consider $V$, the span of ${ T^{j}(v_{i}) }$ where $j$ runs from $0$ to $k-1$ and $k$ is minimal subject to $T^{k} = 0.$ – Geoff Robinson Nov 15 '14 at 18:12
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The spectrum of $A$ is $\{0\}$ because there has to be something in the spectrum, and resolvent series converges everywhere except at $\lambda=0$: $$ (\lambda I-A)^{-1}=\sum_{n=0}^{\infty} \frac{1}{\lambda^{n+1}}A^{n} $$

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