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Exercise :

Let $H$ be a Hilbert space and $P,Q \in \mathcal{L}(H)$ are orthogonal projections, then show that : $$PQ \; \text{orthogonal projection} \; \Leftrightarrow PQ = QP$$

Seeking a formal and rigorous proof, I came up with the following elaboration :

$(\Rightarrow)$ Since $PQ$ is an orthogonal projection, it is $\ker PQ \bot PQ(H)$.

Now, it is $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.

Moreover, it is $\langle u-PQu,u\rangle = 0$. But, it is : $$\langle u-PQu, PQu\rangle = 0 \Rightarrow \langle u,PQu\rangle = \langle (PQ)^*PQu,u\rangle = \langle PQu,u\rangle = \langle u, (PQ)^*u\rangle$$ $$\Rightarrow PQ = (PQ)^* \equiv QP$$

$(\Leftarrow)$ 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 any mistakes or hints, I would be pleased to be informed.

Note : I know that if $P \in \mathcal{L}(H)$ is a projection, then this is equivalent to saying that $P$ is an orthogonal projection, which is also equivalent to saying that $P$ is self adjoint. Now, letting $P := PQ$ will do the trick for a very short and straightforward proof, but I guess this is not as rigorous and complete. If not, please point out since that would be an easy way out.

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Recall that $A \in \mathcal{L}(H)$ is an orthogonal projection if and only if it is self-adjoint and $A^2=A$.

Assume $PQ$ is an orthogonal projection. Then in particular it is self-adjoint so $$PQ = (PQ)^* = Q^*P^* = QP$$

Conversely if $PQ = QP$ then as above we see that $PQ$ is self-adjoint and $$(PQ)^2 = PQPQ = PPQQ = P^2Q^2 = PQ$$ so $PQ$ is an orthogonal projection.

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  • $\begingroup$ So it can be indeed be as starightforward and correct without my thorough elaboration. $\endgroup$ – Rebellos Mar 17 '19 at 9:08
  • $\begingroup$ @Rebellos Yes, no need to consider kernels and images. A purely $*$-algebraical approach suffices. $\endgroup$ – mechanodroid Mar 17 '19 at 9:09
  • $\begingroup$ Thanks for clearing the simpleness up ! $\endgroup$ – Rebellos Mar 17 '19 at 9:18

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