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Normal compact linear operators on Hilbert spaces have infinitely many (counting multiplicities) eigenvalues by the spectral theorem.

For non-normal operators this no longer has to be true. There exist even examples of compact operators without eigenvalues such as weighted shifts and the Volterra operator $Tf(t) = \int_0^t f(s) \ ds$ on $L^2(0,1),$ which when applied to a polynomial basis can also be interpreted as a shift. So a perhaps bold guess would be that we want to avoid some generalized infinite Jordan blocks.

I would therefore like to understand: Do there exist other criteria not assuming normality that imply the existence of infinitely many eigenvalues for compact operators on Hilbert spaces?

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  • $\begingroup$ It is still true (for non-normal $T$) that any $z\in\sigma(T)$, $z\not= 0$ is an eigenvalue, see here: en.wikipedia.org/wiki/… $\endgroup$ Commented Dec 3, 2021 at 22:09
  • $\begingroup$ So $T$ has infinitely many eigenvalues if and only if $\sigma(T)$ is infinite. $\endgroup$ Commented Dec 3, 2021 at 22:12
  • $\begingroup$ The zero operator has infinitely many eigenvalues in the sense of this query (quote: "counting multiplicities"). $\endgroup$ Commented Dec 4, 2021 at 8:05
  • $\begingroup$ @bathalf15320: Still the same thing essentially: Then $T$ has infinitely many eigenvalues if and only if the kernel is infinite-dimensional or $\sigma(T)$ is infinite. $\endgroup$ Commented Dec 4, 2021 at 13:30

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The only criterion I know is based on a Theorem in the second book of Dunford and Schwartz, see Theorem X1.6.29 and following. If the resolvent of an Hilbert-Schimdt operator satisfies some decay estimates on some rays dividing the complex plane, then the span of the generalized eigenfuntions is dense. The theorem generalizes to operators having trace-class powers and to closed operators with trace-class powers of the resolvent.

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