Self adjoint operators and trace class property 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?
 A: 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.
