Can a Accumulation Point be an Eigenvalue?

Question

I have a discrete (separable) infinite dimensional Hilbert Space with a compact operator defined on it. So 0 is an accumulation point (some theorem says so). Can 0 also be an eigenvalue? And how would I prove it is/isnt?

Thanks

Answer 1

The consensus, expressed in comments, is that $0$ can indeed be an eigenvalue: for example, the zero operator has $0$ as the only eigenvalue. For a less trivial example, consider the multiplication operator on $\ell_2$ with entries $0,1/2,1/3,1/4,\dots$, that is, $$(x_n)\mapsto (0,x_2/2,x_3/3,x_4/4\dots)$$ However, I advise some caution in the interpretation of

So 0 is an accumulation point (some theorem says so).

You should understand accumulation point as the limit of the sequence of eigenvalues, not a limit point of the set of eigenvalues. For example, orthogonal projection onto a line has the spectrum $\{0,1\}$. Looking at it as a set, we would not say that $\{0\}$ is an accumulation point. But it is the limit of the sequence of eigenvalues, which is $1,0,0,0,0,\dots$.

And finally, you may want to include something about normality or self-adjointness of the operators in question.