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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? –  Matthias Ludewig Oct 15 '11 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 '11 at 23:14

3 Answers 3

up vote 10 down vote accepted

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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How this can be possible if we work in a separate Hilbert space ? Yes we may have a continuum of spectral values but certainly not a continuum of eigenvectors ! –  user36539 Sep 27 '13 at 18:16
Separable = countable dense subset... –  Matthew Daws Sep 27 '13 at 20:57
in a separable HS the basis is countable. you probably mean a continuum of "eigenvectors" which belong to a space wider than HS (Dirac) –  user36539 Sep 28 '13 at 10:20
The example shows that you are wrong, user36539. It is an operator with uncountably many ACTUAL eigenvectors to differnt eigenvalues in the SAME Hilbert Space! –  Matthias Ludewig Sep 29 '13 at 17:11

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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In fact, unbounded operators with compact resolvents have discrete spectrum and this occur when they are givend on compact manifolds.

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An unbounded operator "on" a compact manifold does not even make sense, although I can imagine what you mean. However, this is merely some example, and the question was way more general. –  Matthias Ludewig Sep 29 '13 at 17:13
Laplacian given on the sphere S, its domain is a dense subset of $L^2(S)$ but its resolvent is compact...Obviously the Laplacian acts on function belonging to $L^2(S)$ which take values on the sphere –  user36539 Sep 29 '13 at 20:32
@Kofi Computer scientists prefer high level language (C, Pascal) than Assembler contrary to the situation for (pure) mathematicians which preffer "assembler-like" language...If you preffer Laplacian given on sections of TS viewed as a Hilbert space of half-densities $L^2(S)$ –  user36539 Sep 29 '13 at 20:40

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