Matrix Representation of Trace Class Operators

Question

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?

Martin Argerami · Accepted Answer · 2017-07-16T13:11:14.587

9

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.

edited Jul 16 '17 at 13:11

answered Apr 18 '13 at 23:28

Martin Argerami

205,756

2

This seems to be a canonical counterexample, +1. – Julien Apr 19 '13 at 02:10
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I guess. I actually learnt it from an answer to a question of mine: http://math.stackexchange.com/questions/174821/selfadjoint-compact-operator-with-finite-trace – Martin Argerami Apr 19 '13 at 05:30
Nice counterexample. Thanks! – grizzlyjoker Apr 19 '13 at 07:35
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@MartinArgerami What does the notation $|A|$ mean here? – ManUtdBloke Jul 16 '17 at 11:43
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@eurocoder: $|A|=(A^*A)^{1/2}$. – Martin Argerami Jul 16 '17 at 11:46
@MartinArgerami How does this the trace of $|A|$ being infinite imply that the trace of $A$ itself is infinite? – ManUtdBloke Jul 16 '17 at 11:50
If the trace of $|A|$ is infinite, then $A $ is not trace class and its trace is not defined. – Martin Argerami Jul 16 '17 at 13:04

Matrix Representation of Trace Class Operators

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