This is a variation of the problem questioned some time ago.
For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} < \infty$$ converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.
Assume that $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$ converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?
The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$
The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.
For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to $$(UT U^{-1})(f(x)) = g(x) f(x)$$ where $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$
Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
The series is required to converge for any basis, and so it has to survive reordering, and thus absolute convergence.
Since the real and imaginary parts of $T$ will satisfy the hypothesis and linear combinations of trace-class are trace-class, we may assume that $T$ is selfadjoint. We may also assume that $T$ is compact; because if $T$ is not compact, there exists $\lambda\in\sigma(T)\setminus\{0\}$ and $\delta>0$ such that $\lambda-\delta>0$ and the spectral projection $E_T(\lambda-\delta,\lambda+\delta)$ is infinite, and so an orthonormal basis of its range, extended to an orthonormal basis of $H$, provides $\{f_j\}$ such that $\sum_n|\langle Te_n,e_n\rangle|=\infty$.
Knowing that $T$ is compact, by the Spectral Theorem, we know that $$\tag1T=\sum_j\lambda_jP_j,$$ where $\lambda_j\in\mathbb R\setminus\{0\}$ for all $j$, and the projections $P_j$ are rank-one and pairwise orthogonal.
If $T$ is not trace-class, then $$ \operatorname{Tr}(T)=\sum_j|\lambda_j|=\infty. $$ If we write $\lambda_j^+$ for the positive eigenvalues and $\lambda_j^-$ for the negative ones, at least one of $\sum_j\lambda_j^+$ and $\sum_j\lambda_j^-$ diverges. Then, with $\{e_j\}$ the orthonormal basis given by $(1)$ (i.e., $P_j=\langle\cdot,e_j\rangle\,e_j$), we have that $$ \sum_j\langle Te_j,e_j\rangle $$ cannot converge absolutely.