Proposition 6

For every $u \in \mathscr{L}(E; F)$, the following conditions are equivalent :

(i) $u$ is partially isometric ; (ii) $u^*$ is partially isometric ; (iii) $u^*u$ is an orthoprojector ; (iv) $uu^*$ is an orthoprojector ; (v) *$uu^*u = u ;*$ (vi) *$u^*uu^* = u^* .*$

By prop. 5, (i) is equivalent to (ii).

(i) $\Rightarrow$ (v) : Suppose $u$ is partially isometric. Then $u^*u$ is the initial orthoprojector of $u$ by prop. 5. Hence for every $x \in E$, $u^*u(x) - x$ belongs to the kernel of $u$, that is, $uu^*u(x) = u(x)$.

(v) $\Rightarrow$ (iii) : Suppose that $uu^*u = u$ and let $p = u^*u$; then $p = p^*$ and $p^2 = p$. Let $M$ (resp. $N$) be the image (resp. the kernel) of $p$. For $x \in M$ and $y \in N$, we have $\langle x|y \rangle = \langle p(x)|y \rangle = \langle x|p^*(y) \rangle = \langle x|p(y) \rangle = 0$. Since $M$ and $N$ are orthogonal, $p$ is the orthoprojector from $E$ onto $M$.

(iii) $\Rightarrow$ (i) : Suppose $p = u^*u$ is an orthoprojector with image $M$ and kernel $N$.

For all $x \in E$, we have

Hence $u(x) = 0$ for $x \in N$ and $\|u(x)\| = \|x\|$ for $x \in M$, and so $u$ is partially isometric with kernel $N$ and initial subspace $M$.

We have proved the equivalence of (i), (iii) and (v). Replacing $u$ by $u^*$, we can deduce the equivalence of (ii), (iv) and (vi). This proves prop. 6.