Let $\,V\,$ be a partial isometry on a Hilbert space $\,\mathsf H$. This means by definition that $\,V^*V\,$ is idempotent which is equivalent to
$\,VV^*$ being idempotent. Thus both $\,V^*V\,$ and $\,VV^*$ are orthogonal projectors.
I'd like to show that $$Vx=x\iff V^*x=x\tag{$\star$}$$
where $x\in\mathsf H$.
Here is my reasoning:
Starting on the left we may conclude the RHS, if $\,V^*Vx=x\,$ is shown to hold. This means that $x$ lives in the support of $V$ which equals $(\operatorname{Ker}V)^\perp$. Consider the orthogonal decomposition
$$x\;=\;V^*Vx \;+\; (I-V^*V)x \\[1.5ex]
\implies\|x\|^2\,=\:\|V^*Vx\|^2+\:\|x-V^*Vx\|^2$$
and $\,\|V^*Vx\|=\|V^*x\|=\|x\|\,$ because $\,x=Vx\,$ is contained in the support of $\,V^*$ where it acts isometrically. It then follows that
$\|x-V^*Vx\|=0\,$.
Due to the symmetry of the involution it is sufficient to follow "one-way in ($\star$)".
Notice that the equivalence ($\star$) implies
$$\operatorname{Ker}(V-I)\;=\;\operatorname{Ker}(V^*-I)$$
and on this closed subspace $V$ and $V^*$ act like the identity operator $I$.
But I do not trust me enough which leaves 2 questions:
Do you consider the preceding reasoning to be correct?
If $(\star)$ holds true indeed: Is there another/straighter/more elegant way to derive it?
Best Answer
The reasoning indeed seems to be correct and here is a (maybe) more straightforward way to think about it.
Note that for any orthogonal projection operator $P$ on $H$, we have for any $v \in H$ $$\langle Pv, v \rangle \leq \langle v,v \rangle,$$ with equality happening if and only if $Pv = v$.
Now, in this case, by assumption we know $$\langle x, x \rangle = \langle Vx, Vx \rangle = \langle x, V^*Vx\rangle.$$ Thus be the above observation, $V^*Vx = x$, as needed.
This doesn't really contain any new ideas, but explains everything in the terms of the inner product on $H$.