Given a Hilbert space $H$ and two closed linear subspace $L, M\subset H$. $P_L$ is the projection operator onto $L$ and $P_M$ onto $M$ similarly.
Prove that $P_LP_M=P_MP_L\Leftrightarrow P_LP_M=P_{L\cap M}$
My attempts:
$\Rightarrow$: $P_LP_M$ is idempotent and self-adjoint, which implies that $P_LP_M$ is a projection $P_{N}$. Clearly, $P_LP_M(H)\subset M\cap L$ and for all $x\in M\cap L$, we have $P_LP_M(x)=x$ so that $M\cap L\subset P_LP_M(H)$, which forces $N=L\cap M$.
$\Leftarrow$: My textbook just states that $P_LP_M=P_{L\cap M}=P_{M\cap L}=P_{M}P_L$. I strongly doubt the correctness of the last step and believe it needs further explanation.
THX :)
Formally the written argument seems somewhat convincing based on symmetry, however it doesn't match the exact conditions.
Rather, simply use that $P_LP_M$ is a projection by hypothesis so in particular self adjoint: $$P_LP_M=(P_LP_M)^*={P_M}^*{P_L}^*=P_MP_L\,.$$