# multiplication of Hilbert Schmidt operators

Suppose $H$ is an infinite dimensional Hilbert space,$B(H)$ is the set of bounded operators on $H$,$\mathcal{HS}(H)$ is the set of Hilbert-Schmidt operators on $H$. I have two questions:

1.If $T$ is nonzero in $B(H)$,does there exist an nonzero operator $S$ in $\mathcal{HS}(H)$ such that $TS\neq 0$.

2.If $S_1$ and $S_2$ are two nonzero operators in $\mathcal{HS}(H)$, I guess $S_1S_2\neq 0$ is not true.Can anyone show me a counterexample.

If $T\neq 0$, there exists a nonzero $x\in H$ such that $Tx\neq0$. Let $P$ be the projection onto $\operatorname{span}\{x\}$. Then $P$ is Hilbert-Schmidt and $TPx=Tx\neq0$, so $TP\neq0$.
Fix $x,y\in H$ nonzero with $\langle x,y\rangle=0$. Let $P$ be the projection onto $\operatorname{span}\{x\}$ and let $Q$ be the projection onto $\operatorname{span}\{y\}$. Then $P$ and $Q$ are Hilbert-Schmidt, and $PQ=0$.
• As for the second answer,why does there exist nonzero $x,y$ such that there innerproduct is zero? – math112358 Sep 19 '18 at 17:40
• Since $\dim H>1$, we can take any two linearly independent vectors and apply Gram-Schmidt to get orthogonal vectors. – Aweygan Sep 19 '18 at 17:42
• I have another question,if $T$ is any nonzero operator in $\mathcal{HS}(H)$,does there exist an nonzero operator $S$ in $B(H)$ such that $TS\neq 0$? – math112358 Sep 20 '18 at 2:17
• Of course. If $T\in B(H)$ is nonzero, then $T^*T$ is nonzero, regardless of whether or not $T$ is Hilbert-Schmidt. – Aweygan Sep 20 '18 at 2:26