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Let $T$ be a compact operator on a Hilbert space. Let $(\lambda_n)$ be the sequence of its eigenvalues repeated with their algebraic multiplicity. The Theorem of Lidskii says that if $T$ is trace class, then the sum $\sum_n\lambda_n$ converges absolutely and equals the trace of $T$. Now my question is for the converse: suppose that $\sum_n|\lambda_n|<\infty$. Does it follow that $T$ is trace class?

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No, this is not true. Take for example the following operator on $l^2(\mathbb{N})$. $$T(e_{2n})=0,T(e_{2n+1})=\frac{1}{\sqrt{n}}e_{2n}$$One can check easily that $T^2=0$, so all $\lambda_n=0$ but $\mu_{2n}=0,\;\mu_{2n+1}=\frac{1}{n}$. Hence $T$ is compact but not trace class.

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  • $\begingroup$ Outch! Should have seen that. I was too much focussed on the non-zero-eigenvalues... $\endgroup$ – user1688 Feb 6 '17 at 11:06

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