Let $\mathcal{H}$ denote an arbitrary finite dimensional Hilbert space. Associated to any subspace $V$ of $\mathcal{H}$ is a projection operator $P_V$ that projects vectors in $\mathcal{H}$ to vectors in $V$. Moreover, associated to any projector $P$ is a subspace $V_P$ of $\mathcal{H}$, defined to be the $1-$eigenspace of $P$. Thus, projectors and subspaces of $\mathcal{H}$ are in one-to-one correspondence.
I'm interested in understanding the correspondence between algebraic properties of the projectors and properties of the subspaces. For example, if $V$ and $W$ are subspaces such that $[P_V, P_W] = 0$, then $P_V P_W$ projects onto the intersection subspace $V \cap W$, and $P_V + P_W - P_V P_W$ projects onto the internal direct sum $V + W$.
My question is as follows: how does the above generalize for spaces $V$ and $W$ for which $[P_V, P_W] \not= 0$? In this case, $P_V P_W$ is not a projection operator at all. Is there a way to construct the projection operator onto $V \cap W$ and $V + W$ using projection operators onto $V$ and $W$ in the non-commuting case? It's also not clear to me geometrically why the product $P_V P_W$ doesn't work as well.
About the projection onto $V\cap W$:
To see this, let $T_n := (P_VP_W)^nP_V$ and $P := P_{V\cap W}$. With $S := P_VP_W$ and $S_* := S^* = P_WP_V$ we note that $$ T_{2n} = S^nP_VS_*^n\qquad\text{and}\qquad T_{2n+1} = S^nP_VP_WP_VS_*^n. $$ Thus, $T_n$ is self-adjoint for every $n$. We have $$ P\,\le\,T_{n+1}\,\le\,T_n $$ for all $n\in\mathbb N$. Indeed, as $P_VP = PP_V = P_WP = PP_W = P$, \begin{align} T_{2n} - P &= S^nP_VS_*^n - S^nPS_*^n = S^n(P_V-P)S_*^n = S^nP_V(I-P)P_VS_*^n\,\ge\,0 \end{align} and \begin{align} T_{2n} - T_{2n+1} &= S^n(P_V - P_VP_WP_V)S_*^n = S^nP_V(I - P_W)P_VS_*^n\,\ge\,0 \end{align} which similarly holds for odd exponents. By a theorem of Riesz, the non-increasing sequence $(T_n)$ of self-adjoint operators has a strong limit $T\in L(H)$, i.e., $T_nf\to Tf$ for all $f\in H$. We shall show that $T = P$. Surely, $P\le T$. Note that $T_n^2 = T_{2n}$, hence $T_n^2f = T_{2n}f\to Tf$ for all $f$. And since the $T_n$ are bounded (by $1$), we also have $T_n^2f\to T^2f$, which implies $T^2 = T$. Hence, $T$ is an orthogonal projection. Note furthermore that $T_nP_WT_n = T_{2n+1}$ so that $TP_WT = T$. Now, for all $f\in H$ we have $$ 0 = (Tf,f) - (TP_WTf,f) = \|Tf\|^2 - \|P_WTf\|^2 = \|(I-P_W)Tf\|^2 $$ which implies $(I-P_W)T=0$. Therefore, $T = P_WT$. But from $T_n = P_VT_n$ we also conclude that $T = P_VT$ so that $\operatorname{ran}T\subset V\cap W$, i.e., $T\le P$. From $P\le T\le T$ we indeed conclude that $T = P$.