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