# multiplication of Hilbert Schmidt operators and bounded operators

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.

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.
• If $S$ is a finte rank projection,what about the conclusion? – math112358 Oct 18 '18 at 14:46
• 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 – math112358 Oct 18 '18 at 14:49