Collection of projection operators in finite dimension and algebraic techinques Consider a set of linearly independent vectors $\{x_1,\dots,x_n\}$ in some finite-dimensional Hilbert space $H$. For any subset $S \subset [n]$, let $P_S$ be the (orthogonal) projection (operator) onto the span of $x_S := \{x_i, \;i \in S\}$. Let us also write $P_j = P_{\{j\}}$.
We would like to study the collection of projections $\{P_S : \; S \subset [n]\}$. We also have some extra information which can be encoded in the form of a graph $G = ([n], E)$ such that for any $(i,j) \notin E$
\begin{align*}
   P_S^\perp P_i \perp P_S^\perp P_j, \quad S=[n]\setminus \{i,j\}.  
\end{align*}
In other words, the residual errors after projecting $x_i$ on $x_S$ and $x_j$ on $x_S$ are orthogonal for any two nodes $i,j$ not connected with an edge.
My question is: Are there known algebraic techniques that help study these projections? Searching around, it seems that there is some connection to (finite-dimensional) von Neumann algebras, but I don't know much about them to see the link. 
As a concrete question consider this: Fix $j \in [n]$ and $S \subset [n]\setminus\{j\}$ and consider
$$
 \mathcal{T}_j(S) := \{ T \subset [n]\setminus\{j\}:\; P_T P_j = P_S P_j\}. 
$$
I believe $\mathcal{T}_j(S)$ is a complete lattice (and the minimum and maximum elements can be read from the graph $G$ ...). Does this follow easily from a more general result? 
EDIT: Concrete question 2:
Consider $A,B,C \subset [n]$ such that $C$ separates $A$ and $B$ in graph $G$, i.e., there is no path in $G$ from $A$ to $B$ that does not share a node with $C$. Then, do we have:
$$
  P_C^\perp P_A \perp P_C^\perp P_B?
$$
 A: I don't know about question 2, but question 1 can indeed be answered using a general result about the projection lattice $P$ (ordered by $p\leq q\Leftrightarrow p=pq$) of a von Neumann algebra $A$.

$Q=\{q\in P:pa=qa\}$ is a complete sublattice of $P$, for any $a\in A$ and $p\in P$

Proof: Let $[b]$ denote the range projection of any $b\in A$.  If $R\subseteq Q$ then, for all $q\in R$, $[pa]=[qa]\leq q$ so $r=\bigwedge R$ satisfies $[qa]\leq r\leq q$ and hence $ra=rqa=r[qa]qa=[qa]qa=qa=pa$, i.e. $r\in Q$.  So $Q$ is closed under taking infimums and, as $pa=qa\Leftrightarrow p^\perp a=q^\perp a$ (where $p^\perp=1-p$), the same applies to $Q^\perp=\{q^\perp:q\in Q\}$.  But $p\leq q\Leftrightarrow q^\perp\leq p^\perp$ so this is saying $Q$ is closed under taking supremums, i.e. $Q$ is a complete sublattice of $P$. $\Box$
In fact, the above proof works more generally for any Baer *-ring $A$ (see Berberian's book "Baer *-rings"), or even Rickart *-ring $A$ (where $Q$ is a complete sublattice of $P$ means that $Q$ is closed under infimums and supremums whenever they exist).
A: This is an expanded version of Tristan Bice's argument above, as far as I understand. Please feel free to correct. (For example, is it also true that $p \le q \iff p = qp$?)
Let $[b]$ be the range projection of any $b \in A$, i.e., projection onto the closure of the range of $b$. For any $q \in P$ and $a \in A$, we have $[qa] \le q$ (since the range of $qa$ is included in the range of $q$). Also note the identity (2) $b = [b] b, \; \forall b \in A$.
If $R \subset Q$, then for all $q \in R$, we have (1) $[pa] = [qa] \le q$, hence $[pa]$ is lower bound on $R$. Letting $r := \bigwedge R$, by definition of infimum, $ [pa] \le r \le q, \; \forall q \in R$, hence $[qa] \le r \le q, \forall q \in R$ by (1).  Hence,
    \begin{align*}
  ra &= r qa & (\text{By} \; r \le q \iff r = rq) \\
  & = r[qa] qa & (\text{By (2) with $b = qa$}) \\
  & = [qa] qa & (\text{By}\; [qa] \le r \iff [qa] = r [qa] ?) \\
  & = qa & (\text{By (2) with $b = qa$)}\\
  &= pa,
 \end{align*}
    showing that $r \in Q$. So, $Q$ is closed under infimums. 
