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

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.

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.
• @Rebellos Yes, no need to consider kernels and images. A purely $*$-algebraical approach suffices. – mechanodroid Mar 17 '19 at 9:09