## Countability of eigenvalues of a linear operator

Is it true that every closed operator on a separable Hilbert H space only has countably many eigenvalues?

Or put the other way around, if I want to ensure that a (not necessarily bounded) linear operator on a separable Hilbert space only has countably many eigenvalues, is closedness (or better said, closability) a sufficient condition?

(By the term eigenvalue, I do not only mean a point in the spectrum of course, but one that actually fulfills $Tx = \lambda x$.)

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So are there any other natural conditions that ensure that the number of eigenvalues is countable apart from normality of the operator? – Kofi Oct 15 2011 at 21:26
Compact operators have countably many eigenvalues. If you think about why normal operators do too, you'll see that it's essentially because eigenvectors corresponding to different eigenvalues are orthogonal. So any operator satisfying this condition will have countably many eigenvalues. For example subnormal operators--or more generally, hyponormal operators--fit the bill. – Faisal Oct 15 2011 at 23:14

Let $T:\ell^2\rightarrow\ell^2$ be the backwards shift operator, $T(a_n) = (a_2,a_3,\cdots)$. This is a contraction. For any $\lambda\in\mathbb C$, consider the sequence given by $a_n = \lambda^n$. Thus $(a_{n+1}) = (\lambda^2,\lambda^3,\cdots) = \lambda(\lambda,\lambda^2,\cdots)$ and so, if $(a_n)\in\ell^2$, then $(a_n)$ is an eigenvector of $T$, for eigenvalue $\lambda$. Of course, $\sum_n |\lambda^n|^2 = \sum_n |\lambda^2|^n <\infty$ if and only if $|\lambda|<1$.

So even a bounded operator can have a continuum of eigenvectors.

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Certainly not. In fact there are bounded operators with uncountably many eigenvalues. For example, the left shift $S^\ast$ defined on $\ell^2$ by $S^\ast(x_1,x_2,\ldots)=(x_2,x_3,\ldots)$ has point spectrum equal to the open unit disk.

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