If $P,Q$ orthogonal projections on $H$ and $T\colon H\to H$ extension of Hilbert isomorphism $S\colon P(H)\to Q(H)$, then $Q=TT^{*}$.

Question

Let $P,Q\colon H\to H$ orthogonal projections (i.e. $P^{2}=P^{*}=P$ and $Q^{2}=Q^{*}=Q$) on a Hilbert space $H$ and let $S\colon P(H)\to Q(H)$ be an isomorphism of Hilbert spaces. We can extend $S$ to a linear (bounded) operator $T\colon H\to H$ by defining $T=0$ on $p(H)^{\perp}=(I-P)(H)$. Then $$T(h)=T(P(h)+(I-P)(h))=T(P(h))=S(P(h)).$$ In particular $T$ maps $H$ into $Q(H)$. I want to prove that $Q=TT^{*}$. I tried to prove that $\langle TT^{*}x,y\rangle=\langle Qx,y\rangle$ for all $x,y\in H$. But don't know how to 'force' $Q$ into my equations when I start computing on the left hand side. Any suggestions or hints would be greatly appreciated!

Answer 1

Notice that we can write $T =\iota \circ S \circ \tilde{P}$ where $\iota: Q(H) \to H$ is the obvious inclusion map and $\tilde{P}: H \to P(H)$ is the map obtained by restricting the codomain of $P$. This means that $$TT^* = \iota \circ S \circ \tilde{P} \circ \tilde{P}^* \circ S^* \circ \iota^* = \iota \circ S \circ S^* \circ \iota^*$$ where I have used the easily checked fact that $\tilde{P}^*$ is the obvious inclusion $P(H) \to H$ and so $\tilde{P} \circ \tilde{P}^* = \operatorname{Id}:P(H) \to P(H)$.

Now $\iota^* : H \to Q(H)$ is easily seen to be the same thing as $\tilde{Q}$ (using the same convention as earlier). This means that $\iota \circ \iota^* = \iota \circ \tilde{Q} = Q$.

Finally, since $S$ is unitary (you assume it is isometric and invertible), $SS^* = \operatorname{Id}$ and so we have that $TT^* = Q$.