Invariance of trace of an operator

Question

I am following Reed & Simon's text on functional analysis. They first introduce the trace of a positive operator in the following theorem:

Let $H$ be a separable Hilbert space, $\{\varphi_n\}_{n=1}^\infty$ an orthonormal basis. Then for any positive operator $A \in \mathscr{L}(H)$ we define $tr (A) = \sum_{n=1}^\infty (\varphi_n, A\varphi_n)$. The number $tr A$ is called the trace of $A$ and is independent of the orthonormal basis chosen.

They also give the following definition of a trace class operator:

An operator $A \in \mathscr{L}(H)$ is called trace class if and only if $tr |A| < \infty$. The family of all trace class operators is denoted by $\mathscr{T}_1$.

Afterwords, they define the trace on $\mathscr{T}_1$ instead of positive operators:

The map $tr: \mathscr{T_1} \rightarrow \mathbb{C}$ given by $tr (A) = \sum_{n=1}^\infty (\varphi_n, A\varphi_n)$ where $\{\varphi_n\}$ is any orthonormal basis is called the trace.

So far I understand all of the above. My confusion comes from the following remark in the book:

We remark that it is not true that $\sum_{n=1}^\infty |(\varphi_n, A\varphi_n)| < \infty$ for some orthonormal basis implies $A \in \mathscr{T}_1$. For $A$ to be in $\mathscr{T}_1$ the sum must be finite for all orthonormal bases.

I see how why the first part of this remark holds, since not every convergent series is absolutely convergent. However, I thought the trace is invariant under a change of basis so why is proving finiteness for all orthonormal bases necessary? This is not included as part of the definition of trace class that I've included above. Does the invariance of the trace operator only hold for positive operators? I am having some trouble seeing the difference between the trace of a positive operator and the trace of an operator in $\mathscr{T}_1$.

Answer 1

Consider the operator $$S\{x_n\}=\{ x_{n+1}\},\quad \{x_n\}\in \ell^2(\mathbb{Z})$$ Then $$\langle Se_n,e_n\rangle =\langle e_{n-1}, e_n\rangle =0$$ where $e_n$ denote the elements of the standard basis. Hence $\sum |\langle Se_n,e_n\rangle|<\infty $ but the operator $S$ is not in trace class. Indeed $S$ is unitary, thus $|S|=I.$

Remark If $\sum |\langle A\varphi_n,\varphi_n|<\infty$ for any orthonormal basis then $A$ is in trace class.

Answer 2

To elaborate on Ryszard's answer and address your concern about the invariance under basis changes, note that we don't expect the sum $\sum |(\phi_n, A\phi_n)|$ (i.e. the absolute sum) to be invariant under changes of basis, even if $A$ is trace class.

For example, in a two-dimensional space, the reflection operator $S$ that swaps $e_1$ with $e_2$ has trace 0 with respect to this basis: $$tr(S) = (e_1, e_2) + (e_2, e_1) = 0$$

Consider instead the basis $f_1 = \frac{e_1 - e_2}{\sqrt{2}}$ and $f_2 = \frac{e_1+e_2}{\sqrt{2}}$ (the basis you get by rotating the original basis vectors 45 degrees clockwise). For the trace with respect to this basis, we get $$tr(S) = (f_1, Sf_1) + (f_2, Sf_2) = -1 + 1 = 0$$ But note that $\sum_i |(f_i, Sf_i)| = 2$ in this case, though the absolute sum was zero for the $e$ basis. The trace hasn't changed, of course. This operator has trace zero, no matter what basis you pick. But we can see that the way in which you get a total of zero changes, depending on the basis. That's a point of concern in infinite dimensions.

If we now work in the setting of Ryszard's answer, where $S$ is now the shift operator, we can do a similar trick. If $n$ is odd, let $f_n = \frac{e_n - e_{n+1}}{\sqrt{2}}$ and $f_{n+1} = \frac{e_n + e_{n+1}}{\sqrt{2}}$ (similar to what we did before, but just viewing $\ell^2(\mathbb{Z})$ as a big sum of two-dimensional spaces). Then $\{f_n\}$ is again an orthonormal basis, and you can check that $(f_n, Sf_n) = (-1)^n\frac{1}{2}$.

So the "trace" of $S$ with respect to the basis $\{f_n\}$ would be $$tr(S) = \sum_{-\infty}^\infty (-1)^n \frac{1}{2}$$ This may converge to zero in some senses, but it's neither convergent nor absolutely convergent in the usual sense. So this basis witnesses the issue from the Reed & Simon remark you mention.

These conditional-convergence kinds of problems don't arise in the context of positive operators because all of the summands are non-negative and the series is absolutely convergent (or infinite). But you can see that we get convergence issues if we try defining the trace for non-positive operators in general - their trace isn't always well-defined.