Does anybody know if the following is true,

Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is finite it follows that $0$ is an eigenvalue.

I think it is wrong... but I am not able to construct a nice counter example.

Could someone help me?

Thanks in advance!

  • $\begingroup$ Is the following the definition: $z$ is NOT in the spectrum if $K-zI$ has bounded inverse? $\endgroup$ – user99914 Jun 2 '14 at 9:26
  • $\begingroup$ @John Yes exactly. $\endgroup$ – Thorben Jun 2 '14 at 9:34

As John noted below, the Volterra operator gives the desired counterexample.

The original assertion, on the other hand, is true whe $K$ is normal: the difference being that a basis of eigenvectors can be obtained.

So when $K$ is normal: All the nonzero elements of the spectrum of a compact operator are eigenvalues with finite-dimensional eigenspaces. Since there is only finitely many of them, the image of $K$ is finite-dimensional. By the first isomorphism theorem, the kernel of $K$ is nonzero (infinite-dimensional, actually). Then zero is an eigenvalue.

  • $\begingroup$ But the volterra-operator is compact with no eigenvalue (and spectrum = $\{0\}$)? $\endgroup$ – user99914 Jun 3 '14 at 6:02
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    $\begingroup$ You are right, of course. My mind tricks me into acting as if everything was normal. $\endgroup$ – Martin Argerami Jun 3 '14 at 12:42

This is actually a true statement if $K$ is self-adjoint and compact.

First, $0$ is in the spectrum because $K$ is a compact operator and thus is not invertible. For compact operators on a Hilbert Space, the eigenvalues are dense in the spectrum. So, if the spectrum is finite, then it is discrete and so it must be that the set of eigenvalues is equal to the spectrum. So $0$ is an eigenvalue.

  • $\begingroup$ Very smart answer! Thank you very much! $\endgroup$ – Thorben Jun 2 '14 at 10:16
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    $\begingroup$ Can you prove that the eigenvalues are dense in the spectrum? I think I need $K$ to be self-adjoint for this. Sigh.. $\endgroup$ – Ravi Jun 2 '14 at 10:24

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