It is well known that given two Hilbert-Schmidt operators $a$ and $b$ on a Hilbert space $H$, their product is trace class and $tr(ab)=tr(ba)$. A similar result holds for $a$ bounded and $b$ trace class.


The following attractive statement, however, is false:


**Non-theorem**:
*Let $a$ and $b$ be bounded operators on $H$. If $ab$ is trace class , then $ba$ is trace class and $tr(ab)=tr(ba)$.*


The counterexample is $a=\pmatrix{0&0&0\\0&0&1\\0&0&0}\otimes 1\_{\ell^2(\mathbb N)}$, $b=\pmatrix{0&1&0\\0&0&0\\0&0&0}\otimes 1\_{\ell^2(\mathbb N)}$.


I'm guessing that the following is also false, but I can't find a counterexample:


**Non-theorem?**:
*Let $a$ and $b$ be two bounded operators on $H$. If $ab$ and $ba$ are trace class, then $tr(ab)=tr(ba)$.*



tr(ab) = tr(ba)?
**EDIT:** Bill Johnson has pointed out a gap in my initial answer. It seems that bridging this gap is just as difficult as proving that $tr(AB)=tr(BA)$. Below I give two other proofs of this equality. The flawed proof is also reproduced at the end.


**Proof 1** (alluded to by Bill in his comment to Gjergji's answer): It's well-known that $AB$ and $BA$ have the same nonzero eigenvalues (with the same multiplicities). Lidskii's trace formula then implies that $tr(AB)=tr(BA)$. QED.


**Proof 2**: This proof doesn't use Lidskii's formula. It's taken from



> 
> Laurie, Nordgren, Radjavi, Rosenthal, *On triangularization of algebras of operators.* Reine Angew. Math. **327** (1981), 143–155.
> 
> 
> 


We'll need to rely on the fact that $tr(ST)=tr(TS)$ if one of $S$ and $T$ is trace class.


Let $A=UP$ be the polar decomposition of $A$. Then $PB=U^\ast AB$ is trace class. But then one knows that $tr(UPB)=tr(PBU)$. So it suffices to prove that $tr(PBU)=tr(BUP)$. Therefore, we might as well assume that $A$ is positive. In this case we can let $P\_n$ denote the spectral projection of $A$ onto $[1/n,\|A\|]$. Then $\lim\_{n\to\infty} tr(P\_nAB) = tr(AB)$. Now, for $T$ trace class $Q$ a projection we have $tr(QT)=tr(QTQ)=tr(TQ)$, whence $tr(P\_n AB) = tr(P\_n A P\_n B P\_n)$ since $P\_n$ and $A$ commute. But the contraction $P\_n B P\_n$ of $B$ is trace class, for $P\_n AB$ is trace class and the restriction of $P\_n A$ to the range of $P\_n$ is invertible. Thus $tr(P\_n A P\_n B P\_n) = tr(P\_n B P\_n A) = tr(BA P\_n)$ and consequently
$$ tr(AB) = \lim\_{n\to\infty} tr(P\_n AB) = \lim\_{n\to\infty} tr(BAP\_n) = tr(BA), $$
as desired.




---


Finally, here is the flawed proof.



> 
> Let $\{e\_i\}$ be an orthonormal basis for $H$. Then
>  $$ tr(AB) = \sum\_i \langle ABe\_i, e\_i \rangle = \sum\_i \langle Be\_i, A^\ast e\_i \rangle. $$
>  But $\langle Be\_i,A^\ast e\_i \rangle = \sum\_j \langle Be\_i, e\_j\rangle \overline{\langle A^\ast e\_i, e\_j\rangle}$, and therefore
>  $$\begin{align}
> tr(AB) &= \sum\_i \sum\_j \langle Be\_i, e\_j\rangle \overline{\langle A^\ast e\_i, e\_j\rangle} \\
> &= \sum\_i \sum\_j \overline{\langle B^\ast e\_j, e\_i\rangle} \langle A e\_j, e\_i \rangle \\
> &= \sum\_j \sum\_i \overline{\langle B^\ast e\_j, e\_i\rangle} \langle A e\_j, e\_i \rangle \qquad (???) \\
> &= \sum\_j \langle Ae\_j, B^\ast e\_j \rangle \\
> &= \sum\_j \langle BAe\_j, e\_j \rangle \\
> &= tr(BA).
> \end{align}$$
> 
> 
> 


