For a compact operator $A$ on a Hilbert space, it is said that $A$ is trace-class if for some (and hence any) orthonormal basis $\{e_n\}_{n \in \mathbb{N}}$, the series $$s_k = \sum_{n=1}^\infty\big\langle |A|e_n,e_n \big\rangle$$ is convergent.

If $A$ is diagonalizable, then it is very easy to see that, for large enough $t \in \mathbb{N}$, the operator $A^t$ is trace class.
Is this necessarily true for a non-diagonalizable compact operator?

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    $\begingroup$ Your definition of trace-class is wrong. You need $|A|$ there, as stated it is easy to construct both compact and non-compact, not trace-class, operators satisfying your definition. $\endgroup$ – Martin Argerami Feb 14 '19 at 18:20
  • $\begingroup$ @Martin: Thanks for pointing out the error. Do you mean it should satisfy $s_k == \sum_{n=1}^k \left<|A|(e_n),e_n \right> $ is a convergent sequence? $\endgroup$ – Pierre Dubois Feb 14 '19 at 22:43
  • $\begingroup$ Yes. $\ \ \ \ \ $ $\endgroup$ – Martin Argerami Feb 14 '19 at 23:06
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    $\begingroup$ I have fixed it $\endgroup$ – Pierre Dubois Feb 15 '19 at 11:22

Why do you think that $A^t$ is trace class for sufficiently large $t$ if $A$ is diagonalizable? What if the eigenvalues are $\lambda_n = \frac{1}{\log(n+1)}$?

  • $\begingroup$ Yes, of course! If the eigenvaluesgo slowly ebough to $0$, then no power raising can help it to converge! $\endgroup$ – Pierre Dubois Feb 14 '19 at 17:04

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