Proposition 5

Let $u \in \mathscr{L}(E; F)$, let $P$ be its initial subspace and $Q$ be the final subspace. Let $p$ (resp. $q$) denote the initial (resp. final) orthoprojector of $u$. Assume that $u$ is partially isometric.

(i) The mapping $u^* \in \mathscr{L}(F; E)$ is partially isometric, with initial subspace $Q$ and final subspace $P$. The isomorphism from $P$ onto $Q$ induced by $u$ is then the inverse of the isomorphism from $Q$ onto $P$ induced by $u^*$.

(ii) We have $u^*u = p$ and $uu^* = q$.

On account of prop. 4 (i), assertion (i) is a consequence of (ii).

We now prove (ii). Since $P$ contains the image of $u^*$, the mapping $u^*u$ maps $E$ into $P$. Let $x \in E$ and $y \in P$, then

If $x$ belongs to $P$, then $\langle u(x)|u(y) \rangle = \langle x|y \rangle$ by the definition of a partially isometric mapping; if $x$ belongs to the kernel $N$ of $u$, then $u(x) = 0$, hence $\langle u(x)|u(y) \rangle = 0$ and $\langle x|y \rangle = 0$ since $N$ and $P$ are orthogonal. Since $E = P \oplus N$, we have $\langle u^*u(x) - x|y \rangle = 0$ in all the cases, and so $u^*u$ is the orthoprojector $p$ from $E$ onto $P$. That $uu^* = q$ follows by interchanging $u$ and $u^*$ in the above.