TITLE: tr(ab) = tr(ba)?
QUESTION [33 upvotes]: 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  $tr(ab)=tr(ba)$.

REPLY [20 votes]: This follows from proposition 7.3 in "Trace-class operators and commutators" by N.J. Kalton. The theorem actually proves something stronger, that for $AB-BA$, if you arrange the eigenvalues with algebraic multiplicity so that $|\lambda_n|\geq |\lambda_{n+1}|$, then $$\sum_{n=1}^{\infty}\frac{\lambda_1+\cdots+\lambda_{n}}{n}<\infty$$
which implies trace zero, but the converse is false.