Commuting Orthogonal Projections on Hilbert Space Let $\mathscr{E}$ be an orthonormal set in an Hilbert space $H$ and suppose that $F,G\subset\mathscr{E}$. Let $P_{F}\colon H\to H$ be the orthogonal projection (o.p.) on $\overline{\text{span}(F)}$, $P_{G}$ the o.p. on $\overline{\text{span}(G)}$ and $P_{F\cap G}$ the o.p. on $\overline{\text{span}(F\cap G)}$. I want to show that $P_{F}P_{G}=P_{G}P_{F}=P_{F\cap G}$.
MY ATTEMPT: Let $h\in H$ and define $x:=P_{G}h$. Since $P_{F}x$ is the unique element in $\overline{\text{span}(F)}$ that satisfies $x-P_{F}x\perp\overline{\text{span}(F)}$, I think that it suffices to prove that $x-P_{F\cap G}h\perp \overline{\text{span}(F)}$. Then, by uniqueness, we must have that $P_{F\cap G}h=P_{F}x=P_{F}P_{G}h$. The same argument shows that $P_{G}P_{F}=P_{F\cap G}$.
How do I prove that $x-P_{F\cap G}h\perp\overline{\text{span}(F)}$.
 A: It all follows easily if you use the actual formulas for the OG projections.
$$P_F (x) = \sum_{f\in F} \langle f,x \rangle f, \;\;\;\; P_G (x) = \sum_{f\in G} \langle f,x \rangle f, \;\;\;\; P_{F\cap G} (x) = \sum_{f\in F\cap G} \langle f,x \rangle f.$$
Then with your notation
$$x = P_G(h) = \sum_{f\in G} \langle f,h \rangle f, $$ $$P_{F\cap G} (h) = \sum_{f\in F\cap G} \langle f,h \rangle f,$$ $$ x - P_{F\cap G} (h) = \sum_{f \in G \setminus F} \langle f,h \rangle f.$$
To prove that it is orthogonal with $\overline{\operatorname{span} (F)}$ it is enough to show it is orthogonal to each element $f_0 \in F$:
$$\langle x - P_{F\cap G} (h), f_0 \rangle = \sum_{f \in G \setminus F} \langle f,h \rangle \langle f, f_0 \rangle = \sum_{f \in G \setminus F} \langle f,h \rangle 0 = 0.$$
Hence, $x - P_{F \cap G} \perp \overline{\operatorname{span} (F)}$.
It is worth mentioning that using the formulas above you can easily show that $P_F P_G = P_{F \cap G} = P_G P_F$ directly.
A: Let $f\in F$. 
If $f$ is also in $G$, then $\langle P_Gh, f\rangle =\langle h, f\rangle =\langle P_{F\cap G}h, f\rangle$. 
If $f$ is not in $G$, then it's orthogonal to $G$, so also to $F\cap G$, so $\langle P_Gh, f\rangle =0=\langle P_{F\cap G}h, f\rangle$.
(All these equations can also be proved e.g. by the self-adjoint property of an orthogonal projection.)
In any case, we get $P_Gh-P_{F\cap G}h\perp f$. 
