# Thread: A property of the spectrum of a compact operator

1. ## A property of the spectrum of a compact operator

Hello!

I'm going through the proof of the fact, that given a compact operator $K$ on a Banach space $X$ its spectrum $\sigma(K)$ is at most countable and 0 as the only possible accumulation point (which may or may not belong to the spectrum).

The reading has been fairly easy so far, but at the moment I'm somehow stuck at one step, namely:
The overall objective is to show that all points $\lambda\in\sigma(K)\setminus\{0\}$ are isolated.
Let $N:=N_\lambda^m(K)=\text{Ker}(\lambda I-K)^m$ and $K|_N$ be the restriction of $K$ to $N$
There is a moment when it is evident that we need finiteness of $\sigma(K|_N)$.
So they argue that this follows from the fact, that $N$ is finite-dimensional.
I can't see at the moment how exactly it follows.
Does anybody have an idea?

P.S.:
I think I should mention that $m$ is chosen so that:
$X=R+N$ with $R\cap N=\{0\}$, where $R:=R_\lambda^m(K)=\text{Im}(\lambda I-K)^m$.

HAL

2. Two questions:

1. Are you working with compact linear operators?
2. By "finiteness of $\sigma(K|_{N})$", do you mean that the cardinality of the set is finite, or that the set is bounded in the complex plane?

3. 1. Yes, it's all about linear operators.
2. I mean the cardinality of the set.

4. Suppose for a second that we are dealing with $m=1.$ Then the null space in question is the space of eigenvectors corresponding to nonzero eigenvalues, correct? Suppose you had an infinite set of nonzero eigenvalues. Well, then it is a fact that eigenvectors corresponding to different eigenvalues are linearly independent. Therefore, in order to span the eigenspace, you'd have to have infinitely many eigenvectors, contradicting the finiteness of the dimension of the null space.

I think this idea will work for you, though you'll have to adapt it to take into account the mth power of your operator.