How do the products of two orthogonal projection matrices relate to their respective column spaces? I'm trying to determine the whether or not the following statements are true. I've been working on them for a while now and I think I'm mixing them up.
i. if (PA)(PB)=(PB)(PA), then C(A)=C(B). Where PA is the orthogonal projection matrix onto the column space of A. Similarly for PB.
ii. if C(A)=C(B), then (PA)(PB)=(PB)(PA).
iii. if (PA)(PB)=PA, then C(A) is a sub set or equal to C(B)
iv. if C(A) is a sub set or equal to C(B), then (PA)(PB)=PA
I feel like ii. should be true and i. is conditionally true but I think I might be misinterpreting how multiplying various projection matrices relate back to the original column spaces. For iii and iv, it seems like they are only true if the order of PA and PB are reversed. So, if PAPB=PB, then C(A)is a subset of C(B). Any help would be appreciated. -Thanks.
 A: It makes not much sense in this generality to carry the matrices for their column spaces, we should just use orthogonal projection onto a subspace.
So, let now $A$ and $B$ rather denote subspaces (what you denoted by $C(A)$ and $C(B)$). 
[The properties that $P_A^2=P_A^*=P_A$ and ${\rm ran\,}P_A=C(P_A)=C(A)$ already uniquely determine the projection $P_A$, no need to use its explicit form composed of variants of matrix $A$.]
ii. Of course, if $A=B$, then $P_A=P_B$, and $P_A$ commutes with $P_B$. Also, if $A\subseteq B$ or $B\subseteq A$, or if $A\perp B$.
i. This is not true as it stands, e.g. if $A\perp B$ then $P_AP_B=P_BP_A=0$. 
Moreover, if $A=U\oplus A'$ and $B=U\oplus B'$ with $U=A\cap B$ and $A'\perp B'$, then we again have $P_AP_B=P_BP_A=P_U$.
iii. I think you are right in that we should consider the other direction: if $P_AP_B=P_B$, then with $b\in B$ we have 
$b=P_Bb=P_AP_Bb=P_Ab$, so $b\in{\rm ran\,}P_A\,=A$. 
The other equation $P_AP_B=P_A$ would mean similar relation for the row spaces of the original matrices, I guess.
iv. Yes, the converse of iii. also holds: if $B\subseteq A$, then with arbitrary $x\in B\oplus B^\perp,\ x=b+y$, we have $$P_AP_Bx=P_Ab=b=P_Bx\,.$$
