$ PQ ; text {orthogonal projection} ; Leftrightarrow PQ = QP $

Exercise:

To let $ H $ to be a Hilbert room and $ P, Q in mathcal {L} (H) $ are orthogonal projections, then show:
$$ PQ ; text {orthogonal projection} ; Leftrightarrow PQ = QP $$

When I was looking for a formal and rigorous proof, I came to the following elaboration:

$ ( Rightarrow) $ Since $ PQ $ is an orthogonal projection $ ker PQ offered PQ (H) $,

It is now $ PQ (u-PQu) = PQU-PQ ^ 2u = PQU-PQu = 0 Rightarrow u-PQU in ker PQ $,

For all $ u in H $ the operator $ (PQ) ^ * PQ $ is self-adjoint

It is beyond that $ langle u-PQu, u rangle = 0 $, But it is :
$$ langle u-PQu, PQu rangle = 0 Rightarrow langle u, PQu rangle = langle (PQ) ^ * PQu, u rangle = langle PQu, u rangle = langle u, (PQ ) ^ * u rangle $$
$$ Rightarrow PQ = (PQ) ^ * Equivalent QP $$

$ ( Leftarrow) $ Since $ PQ $ is self adjoint, it is:
$$ PQ = (PQ) ^ * Rightarrow PQ (PQ) ^ * = (PQ) ^ * PQ Rightarrow PQ ; text {normal} $$
But then :
$$ PQ ; text {normal} Rightarrow ker PQ = ker (PQ) ^ * = PQ (H) ^ bot Rightarrow ker PQ = PQ (H) ^ bot Rightarrow PQ ; text {orth. proj.} $$

If there are any errors or hints, I would be happy to be informed.

Note : I know that when $ P in mathcal {L} (H) $ If it's a projection, that's synonymous with it $ P $ is an orthogonal projection that is synonymous with it $ P $ is self-adjoint let now $ P: = PQ $ I'll do the trick for a very short and straightforward proof, but I think that's not so strict and complete. If not, please point out, as this would be an easy way out.