Symmetry of the tensor product of the operator

Question

I am doing the following problem

The "if" parts are ok, but I do not know how to deal with the "only if" parts. Anyone here can help me?.

Answer 1

If $T_1\otimes T_2$ is symmetric, you have $$\tag1 \langle (T_1\otimes T_2) x\otimes x,y\otimes y\rangle =\langle x\otimes x,(T_1\otimes T_2) y\otimes y\rangle. $$ This gives the equality $$\tag2 \langle T_1x,x\rangle\langle T_2y,y\rangle=\langle x,T_1x\rangle\langle y,T_2y\rangle $$ for all $x,y$. Assume that $T_2\ne0$ (if $T_1=0$ or $T_2=0$, there is nothing to prove). Let $y_0$ such that $c=\langle T_2y_0,y_0\rangle$, and $|c|=1$. So, for this $y_0$, now $(2)$ reads (note that $\langle y_0,T_2y_0\rangle=\overline{\langle T_2y_0,y_0\rangle}=\bar c$) $$ \langle cT_1x,x\rangle=\langle x,cT_1x\rangle. $$ By polarization we get $\langle cT_1x,y\rangle=\langle x,cT_1y\rangle$ for all $x,y$, so $cT_1$ is symmetric.

If we know repeat the argument for $T_2$, we get $d$ with $|d|=1$, $d=\langle T_1x_0,x_0\rangle$, and $dT_2$ symmetric. If we know look at $(2)$ with $x=x_0$, $y=y_0$, we have $$ cd=\overline{cd}. $$ As $|cd|=1$, it follows that $cd=1$, so $d=c^{-1}$.