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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?

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  • $\begingroup$ no let $B=0$ or $Bx=\langle e_k, x\rangle e_k$ $\endgroup$ Sep 20, 2021 at 11:08
  • $\begingroup$ Yes I mean invertible, anyway if B goes to zero in the directions of the eigenvalues of $Q$ should do the same right? $\endgroup$
    – carlos85
    Sep 20, 2021 at 11:10
  • $\begingroup$ I'm not sure about the case when $B$ is inverteble. $\endgroup$ Sep 20, 2021 at 11:11
  • $\begingroup$ 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. $\endgroup$ Sep 20, 2021 at 11:13
  • $\begingroup$ 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. $\endgroup$
    – MaoWao
    Sep 21, 2021 at 5:36

1 Answer 1

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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.

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  • $\begingroup$ I see it's correct so I give you the answer. Anyway can you point out a reference for the fact that trace class operators form a two sided *-ideal un the bounded operators? $\endgroup$
    – carlos85
    Sep 21, 2021 at 17:50
  • $\begingroup$ @carlos85 It is in Murphy's book "C*-algebras and operator theory", p65. $\endgroup$
    – J. De Ro
    Sep 21, 2021 at 21:14

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