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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.