# \$ 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.

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