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