# Is the following operator trace class?

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?

• no let $B=0$ or $Bx=\langle e_k, x\rangle e_k$ Sep 20, 2021 at 11:08
• Yes I mean invertible, anyway if B goes to zero in the directions of the eigenvalues of $Q$ should do the same right? Sep 20, 2021 at 11:10
• I'm not sure about the case when $B$ is inverteble. Sep 20, 2021 at 11:11
• If $B$ is unitary, I am sure the answer is affirmative (i.e. $BQB^\star$ is NOT trace class). Not sure about when $B$ is not unitary, though. In that case, $B^\star\ne B^{-1}$, so the conjugation $BQB^\star$ is not really a meaningful operation. Sep 20, 2021 at 11:13
• Btw, this is not the definition of a trace-class operator. This series may well converge for a single ONB and $Q$ still not be in the trace class. Sep 21, 2021 at 5:36

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.