Orthogonal projections on $M$, $L$ and $M\cap L$ in Hilbert space

Question

I'm dealing with this functional analysis exercise:

Let $H$ be a Hilbert space and $M,L$ are two closed subspaces of $H$. $P_M$ is the orthogonal projection operator. Prove $P_MP_L=P_{M\cap L} \iff P_MP_L=P_LP_M$.

"$\Longrightarrow$" part is easy: $P_LP_M=P_{L\cap M}=P_{M\cap L}=P_MP_L$.

But for the "$\Longleftarrow$" part, denote $P=P_MP_L=P_LP_M$. Then $Px=P_M(P_Lx)\in M$ and similarly in $L$. Therefore $Px\in L\cap M$. But how can I prove it is an orthogonal projection on $L\cap M$?

Answer 1

You are almost there. To show it's an orthogonal projection, note the following fact:

If $E$ is a closed subspace of a Hilbert space $H$, then for any vector $x \in H$, the projection $P_E(x)$ is the unique vector in $E$ such that $\langle v, x\rangle = \langle v, P_E(x)\rangle$ for all $v\in E$.

Now take any $v\in L\cap M$ and we have for any vector $x \in H$: $$\langle v, x\rangle = \langle v, P_M(x)\rangle = \langle v, P_LP_M(x)\rangle.$$ Since you already know that $P_LP_M(x)\in L\cap M$, this identity means that $P_LP_M(x)$ is exactly the projection $P_{L\cap M}(x)$.

Answer 2

First note that $L\cap M=P_{L\cap M}(H).$

Now, if $x\in H,$ then $P_MP_L=P_LP_M$ implies directly that $P_MP_Lx\in L\cap M$ because $P_MP_Lx\in M$ and $P_LP_Mx\in L.$ Thus, $P_MP_L(H)\subseteq L\cap M.$

On the other hand, if $x\in L\cap M,$ then $x=P_Lx=P_Mx$ so $x=P_MP_Lx$, which implies that $L\cap M\subseteq P_MP_L(H)$ and therefore $L\cap M=P_{L\cap M}(H)= P_MP_L(H).$