Generally speaking, compact operators have countable eigenvalues and countable s-values (or singular values). What about the reverse? If I know that a (non-compact) operator has countable singular values, can I deduce that it also has countable eigenvalues?
2026-03-28 05:38:42.1774676322
For non compact operators, is countability of singular values equivalent to countability of eigenvalues?
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No. Consider a separable Hilbert, fix an orthonormal basis, and let $S$ be the unilateral shift. The operator $T=S^*$ has uncountably many eigenvalues, since every $\lambda\in\mathbb C$ with $|\lambda|<1$ is an eigenvalue. But $T^*T=SS^*$ is a projection, so in particular its eigenvalues are $0$ and $1$.
That is, $T$ has uncountably many eigenvalues and two singular values.