TITLE: Matrix Representation of Trace Class Operators
QUESTION [7 upvotes]: Suppose we have a separable Hilbert space (thus with a countable basis) and that represent an operator in matrix form, i.e:
$A: H \rightarrow H $$$x \;\rightarrow \sum_{j \in \mathbb{N}}\left(\sum_{k \in \mathbb{N}} a(j,k)\cdot\langle x,e_k\rangle  \right)e_j$$
Given that:

The series of complex numbers $\sum_{k \in \mathbb{N}} a(j,k)\cdot\langle x,e_k\rangle $ converges.
That $\sum_{j \in \mathbb{N}}\left(\sum_{k \in \mathbb{N}} a(j,k)\cdot\langle x,e_k\rangle  \right)e_j$ converges.
That $\sum_{j \in \mathbb{N}}\sum_{k \in \mathbb{N}} \left |a(j,k) \right|^2 < \infty$ (the previous conditions + this one imply that A is a Hilbert-Schmidt operator).
And that $\sum_{k \in \mathbb{N}} \left |a(k,k) \right| < \infty$.

Prove that $A$ is a trace class operator.
My attempt at a solution
By these conditions we know that $\left | Tr (A) \right | \leq \displaystyle{\sum_{k \in \mathbb{N}}} \left | a(k,k) \right | < \infty$. But  I can't see the connection to $Tr(\left |A \right |)$. Any hints on how to proceed?

REPLY [8 votes]: I don't think that $A$ needs to be trace-class. 
Consider a sequence $b=\{b_j\}$ such $b\in \ell^2(\mathbb N)\setminus \ell^1(\mathbb N)$ (for example, $b_j=1/j$). Define the operator
$$
A=\bigoplus_j\begin{bmatrix}0&b_j\\ \overline {b_j}&0\end{bmatrix}.
$$
That is, in the canonical basis, 
$$
A=\begin{bmatrix}0&1\\ 1&0 \\& & 0 & 1/2\\ & & 1/2&0\\ &&&&\ddots
\end{bmatrix}
$$
where the empty entries are also $0$. 
Since every row contains a single nonzero element, the first condition is satisfied trivially. The second condition is also easy, because we end up with a sum over the canonical basis with coefficients in an $\ell^2$-sequence. 
Third condition holds, since $\sum_j\sum_k|a(j,k)|^2=2\sum_j|b_j|^2<\infty$. 
And the fourth condition holds because $a(k,k)=0$ for all $k$. 
But $A$ is not trace-class: because $|A|$ is the diagonal operator with diagonal $b_1,b_1,b_2,b_2\ldots$, and so its trace is infinite.