Compact + absolute convergence of eigenvalues $\Rightarrow$ trace class?

Question

Let $H$ be a complex Hilbert space. Lidskii’s theorem says that if an operator $T \in B(H)$ is trace class, then $\operatorname{tr}(T) = \sum_i \lambda_i$, where the sum includes all nonzero eigenvalues repeated according to (algebraic) multiplicity. Since this quantity is well defined, the sum must converge absolutely (otherwise it could depend on the order of summation). Thus, trace class operators are compact and satisfy $\sum_i |\lambda_i| < \infty$. Does the converse hold? If $T$ is compact and satisfies $\sum_i |\lambda_i| < \infty$, is $T$ trace class?

Answer 1

Let $T: \ell^2\to \ell^2$ be defined by $$T(x_0,x_1,\ldots )=(0,a_0x_0,a_1x_1,\ldots )$$ where $a_n\to 0$ and $\sum |a_n|^2=\infty.$ Then $T$ is compact. Moreover $T$ does not admit any eigenvalues and $T$ is not a Hilbert-Schmidt operator. In particular $T$ is not a trace class operator.

Consider the orthonormal basis $f_{2n}=2^{-1/2}[e_{2n}+e_{2n+1}]$ and $f_{2n-1}=2^{-1/2}[e_{2n}-e_{2n+1}],$ where $\{e_n\}$ denotes the standard basis in $\ell^2.$ Then $$ \langle Tf_{2n}, f_{2n}\rangle =2^{-1}a_{2n}\\ \langle Tf_{2n-1}, f_{2n-1}\rangle =-2^{-1}a_{2n-1} $$ Therefore if $\sum a_{2n} $ is divergent while $\sum a_{2n-1}$ is convergent then the series $\sum \langle Tf_{n}, f_{n}\rangle $ is divergent. Another possibility: $a_{2n-1}={1\over 2}a_{2n}$ and $\sum a_{2n}$ is divergent.

The conclusion holds if we assume that $T$ is selfadjoint. Indeed if $T$ is compact then there is an orthonormal basis $\{v_n\}$ such that $Tv_n=\lambda_n.$ Then $|T|v_n=|\lambda_n|v_n.$ The positive operator $|T|$ is trace class iff $\sum |\lambda_n|<\infty.$ Thus if the series is convergent then $|T|$ is trace class and sinus $T$ by the polar decomposition $T=U|T|,$ where $U$ is a partial isometry.