How to show $P_MP_N=0$ if $M \perp N$?

Question

Let $M$ and $N$ be two subspaces of a vector space $V$, and consider the associated orthogonal projectors $P_M$ and $P_N$.

Proof if $M \perp N$ ($M$ is perpendicular to $N$), $P_MP_N=0$? In other words, the product of orthogonal projectors associated to the orthogonal subspaces is zero.

Answer 1

Any projector $P_M$ is $0$ on $M^{\perp}$ (and identity on $M$). Since $P_N(x) \in N \subset M^{\perp}$ it follows that $P_M(P_Nx)=0$ for all $x$.

Answer 2

Suppose $M \perp N$. Fix $v \in V$, and suppose $w \in V$ too. Recall that $P_M$ is self-adjoint. Then, $$\langle P_M P_N v, w \rangle = \langle P_N v, P_M w \rangle = 0$$ since $P_N v \in N$ and $P_M w \in M$. Since this holds for any $w$, it holds for $w = P_M P_N v$, which implies that $$\langle P_M P_N v, P_M P_N v \rangle = 0 \implies P_M P_N v = 0.$$