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?



Converse for Lidskii's Theorem
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.


