Let $(\mathcal{H}, (\cdot, \cdot))$ be a Hilbert space over $\mathbb{L} = \mathbb{R}$ or $\mathbb{C}$.
Suppose that $U : \mathcal{H} \to\mathcal{H}$ is a bounded linear operator such that $UU^*$ and $U^*U$ are both projections. That is, we know $(UU^*)^2= UU^*$, $(U^*U)^2= U^*U.$
I would like to show that $U$ is a partial isometry. That is, I would like to show that $\|U(x)\| = \|x\|$ for all $x \in (\operatorname{Ker}U)^\perp$.
Here is my attempt: we want to show that $\|x\|^2 = (x,x) = \|Ux\|^2 = (Ux,Ux)$ for $x \in (\operatorname{Ker}U)^\perp$. We of course have $(Ux,Ux) = (x, U^*Ux)$. So it is enough to show that $U^*Ux = x$.
Now, since $U^*U$ is a projection, we know that $\operatorname{Ran}U^*U$ is closed, hence $x$ can be uniquely written as $x = y + z$ for $y \in \operatorname{Ran}U^*U$, $z \in (\operatorname{Ran}U^*U)^\perp$. We want to show that $z = 0$. This will imply $x = y$ and then $U^*Ux = U^*Uy = y =x$.
So I've gotten stuck on showing that $z =0$. It seems that it must depend on the fact that $x \in (\operatorname{ker}U)^\perp$, but I haven't figured out how to use this yet.
Hints or solutions are greatly appreciated!
Note that ${\rm Ker}\,U={\rm Ker}\, U^*U$. Thus $({\rm Ker}\,U)^\bot={\rm Im}\,(U^*U)$ because $U^*U$ is an orthogonal projection. So, for $x\in({\rm Ker}\,U)^\bot$ we have $U^*Ux=x$ that is $\langle x,U^*Ux\rangle=\langle x,x\rangle$ or equivalently $\Vert Ux\Vert=\Vert x\Vert$.