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Suppose $T_1,T_2 \in B(H)$,where $H$ is an infinite dimensional Hilbert space. $S\in \mathcal{HS}(H)$,$\mathcal{HS}(H)$ is the set of Hilbert Schmidt operators on $H$.

Does there exist nonzero operators $T_3,T_4 \in B(H)$ such that $T_1T_3ST_4T_2=\alpha S$,where $\alpha$ is a nonzero complex number.

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Yes. $T_3$ or $T_4$ being the zero operator, we have $T_1T_3ST_4T_2=0=0S$.

If you want $\alpha\neq0$, then the answer is no. If either $T_1$ or $T_2$ is rank one, then $T_1T_3ST_4T_2$ will be rank one (or zero). But $\alpha S$ will have the same rank as $S$, which can be any natural number, or infinite.

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  • $\begingroup$ If $S$ is a finte rank projection,what about the conclusion? $\endgroup$ – math112358 Oct 18 '18 at 14:46
  • $\begingroup$ I want to take $T_3,T_4$ be the inverse element of $T1,T2$,so I asked the question of the criterion of invertibility of bounded operators $\endgroup$ – math112358 Oct 18 '18 at 14:49

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