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1
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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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9
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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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7
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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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