# Traces and commutators

The following is a true (and well-known) theorem: $$\newcommand{\Tr}{\mathop{\rm Tr}}$$

Suppose $$A$$ and $$B$$ are bounded operators on a Hilbert space, and $$AB$$ and $$BA$$ are trace class.  Then $$\Tr(AB)=\Tr(BA)$$.

This is easy to prove if one of the operators $$A,B$$ is itself of trace class, or if they both are Hilbert-Schmidt (the obvious calculation works).  In the general case it is a bit harder.  The “usual” argument proceeds via Lidskii’s trace theorem – the trace of any trace-class operator is the sum of the eigenvalues – together with the purely algebraic fact that the nonzero eigenvalues of $$AB$$ and $$BA$$ are the same (including multiplicities).  Continue reading