Existence of a special element in $B(H,K)$

Question

Let $H$ and $K$ be Hilbert spaces. Assume $B(H,K)$ denotes the space of bounded linear operators. Does there exist a $T_0 \in B(H,K)$ such that $T_0 T_0^* T=T=TT_0^*T_0$ for all $T \in B(H,K)$ Where $T_0^*$ denotes the adjoint of $T_0$

If $H$ and $K$ are finite dimensional then existence of $T_0$ is obvious but in general case I don’t see the existence. Any ideas?

Answer 1

Write $P=T_0^*T_0$, $Q=T_0T_0^*$. By taking $T$ to be rank-one operators, we easily see from $QT=T$ that $Q=I_K$. Then $T_0^*$ is an isometry: for all $w\in K$, $$ \|T_0^*w\|^2=\langle Qw,w\rangle=\langle w,w\rangle=\|w\|^2. $$ In particular $\dim K\leq \dim H$. From $TP=T$ for all $T\in B(H,K)$, taking adjoints we get $PT^*=T^*$, so $PS=S$ for all $S\in B(K,H)$. With the same argument as above, we get that $P=I_H$, and $\dim H\leq\dim K$.

So $T_0$ only exists when $\dim H=\dim K$, whether they are finite or infinite dimensional; being a co-isometry, it has to be a unitary.