Let $H$ Hilbert with orthonormal basis $\{e_k\}$, $B \colon H \to H$ linear and bounded, invertible. $Q \colon H \to H$ linear operator, not trace class, i.e.
$$tr Q =\sum_{k \in \mathbb{N}} \langle Qe_k, e_k \rangle = +\infty$$
Then does it follow that $BQB^*$ is not trace class?
Best Answer
You need to use that the trace class operators form a two-sided $*$-ideal in the bounded operators $H \to H$.
Concretely, assume to the contrary that $BQB^*$ is trace class. Then so is $B^{-1}BQB^*(B^*)^{-1} = Q$, which contradicts your assumption. Thus $BQB^*$ is never trace class.