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I know that if an operator (on a Banach space with approximation property) is nuclear of order zero, then its eigenvalues are $p$-summable for any $p>0$. (I read it from Grothendieck’s book “Produits tensoriels topologiques et espaces nucléaires” II. P 16. Theorem 4, if I didn’t misunderstand the statement.) I wonder if the converse is true?

Or in general, is there any results on deducing the convergence rate of ”singular values” from the convergence rate of eigenvalues?

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  • $\begingroup$ You need some compactness assumptions. But even then, if there are no or finitely many eigenvalues, then they converge extremely quickly. $\endgroup$ Jul 6 '18 at 16:27
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    $\begingroup$ @AndrásBátkai Hey, thanks for your reply. I think when an operator is nuclear, it is compact automatically. And let’s suppose it is not of finite rank. $\endgroup$
    – Fan
    Jul 6 '18 at 16:31
  • $\begingroup$ I thought you want the converse and your conclusion should be that the operator is nuclear. Hence you cannot use it. And there are many compact operators with no eigenvalues which are not of finite rank... $\endgroup$ Jul 7 '18 at 17:47
  • $\begingroup$ @AndrásBátkai Sorry for not making it clear. What I meant to say was, assume an operator is nuclear, but we don’t know the asymptotic behaviour of the singular values. Can we deduce that from the asymptotic behaviour of its eigenvalues? I have checked the references given by lieder. And all of them are trying to do the converse. I guess given the same eigenvalues, singular values of operators can behave very differently as your example suggested. Thanks for that $\endgroup$
    – Fan
    Jul 8 '18 at 13:14
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There are several monographs available which treat this subject in some detail, e.g., by

A. Pietsch, Eigenvalues and $s$ numbers;

H. König, Eigenvalue distribution of compact operatores;

R. Retherford, Hilbert space compact operators and the trace theorem;

B. Simon, Trace ideals and their applications.

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