# Non-diagonalizable compact operators and the trace-class condition

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?

• 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. – Martin Argerami Feb 14 '19 at 18:20
• @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? – Pierre Dubois Feb 14 '19 at 22:43
• Yes. $\ \ \ \ \$ – Martin Argerami Feb 14 '19 at 23:06
• I have fixed it – 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 $$\lambda_n = \frac{1}{\log(n+1)}$$?
• Yes, of course! If the eigenvaluesgo slowly ebough to $0$, then no power raising can help it to converge! – Pierre Dubois Feb 14 '19 at 17:04