Does $P_2P_1 = P_2$ hold when $P_2$ and $P_1$ are orthogonal projections onto $M_2\subseteq M_1$

Question

Suppose $P_1$ and $P_2$ are the orthogonal projections onto the closed subspaces $M_1$ and $M_2$ of a Hilbert space $X$.

If $M_2 \subseteq M_1$, is it always the case that $P_2P_1 = P_2$?

Answer 1

Yes. For any $x\in X$, $$P_1P_2x=P_2x,$$ since $P_2x\in M_1$. Since orthogonal projections are selfadjoint, $$ P_2=P_2^*=(P_1P_2)^*=P_2P_1. $$

Answer 2

Yes.

Let $X \ni x = u_1 + v_1$ where $u_1 \in M_1$, $v_1 \in M^\perp$.

$M_1$ is also a Hilbert space and $M_2$ is a closed subspace of $M_1$ so $u_1$ admits the decomposition $$u_1 = w_2 + z_2$$ with $w_2 \in M_2$ and $z_2 \in M_2^{\perp_{M_1}}$ (the orthogonal complement is taken in $M_1$).

We have $x = w_2 + z_2 + v_1$. Clearly $z_2 \in M_2^{\perp_{M_1}} \subseteq M_2^\perp$, and also $v_1 \in M_1^\perp \subseteq M_2^\perp$.

Therefore $z_2 + v_1 \in M_2^\perp$ so $$x = \underbrace{w_2}_{\in M_2} + \underbrace{z_2 + v_1}_{\in M_2^\perp}$$

Now we have: $$P_2(P_1x)= P_2(u_1) = w_2 = P_2(x)$$