Intersection of subspace which projector do not commute I am a little bit confused by the following.
Consider two subspace $U$ and $V$ of a given vector space. I associate to them $\Pi_U$ and $\Pi_V$ their projectors.
In the case $[\Pi_U,\Pi_V]=0$; the projector on $U \cap V$ is simply $\Pi_{U \cap V} = \Pi_U \circ \Pi_V = \Pi_V \circ \Pi_U$
But when the two projector do not commute, $\Pi_U \circ \Pi_V$ might not even be a projector.
My question is:
Is there a physical meaning for intersection of subspaces which projector do not commute ? Does that mean that the intersection is necesseraly $\{0\}$ ? If so I couldn't manage to show it.
 A: As far as I know, if the projectors do not commute there is not necessarily a significance of $U \cap V$. We may have cases where the projectors do not commute where $U \cap V \neq \{0\}$. For example, let
$$U = \text{span}\left(\hat{e}_1, \frac{1}{\sqrt 2}(\hat{e}_2 + \hat{e}_3)\right), ~ V = \text{span}\left(\hat{e}_1, \hat{e}_2\right)$$
The projectors do not commute, but clearly $\hat{e}_1 \in U \cap V$.
If they do not commute, we can say that there cannot be a complete basis of simultaneous eigenvectors of $\Pi_U$ and $\Pi_V$. Namely, there must be some vector $\vec{v}$ in either $U$ or $U^\perp$ that is not in $V$ or $V^\perp$ (and vice-versa), but as far as I know we can't really say more.
A: The projector onto $U$ is not unique!  It depends on the complement of $U$ chosen to be the kernel of $\Pi_U$.
However, given any $U$ and $V$, there are always choices of $\Pi_U$ and $\Pi_V$ which commute!

EDIT.  Here is how to get commuting $\Pi_U$ and $\Pi_V$.

*

*Find a subspace $U_1⊆U$ such that $U=(U∩V) ⊕ U_1$,


*Find a subspace $V_1⊆V$ such that $V=(U∩V) ⊕ V_1$,


*Prove that $U+V = U_1 ⊕ (U∩V) ⊕ V_1$,


*Denoting by $E$ the ambient space, find a subspace $W⊆E$ such that $E=(U+V) ⊕ W$,


*Prove that $E = U_1 ⊕ (U∩V) ⊕ V_1 ⊕ W = U ⊕ V_1 ⊕ W = U_1 ⊕ V ⊕ W$,


*Let  $\Pi_U$ be the projection onto  $U$  along (i.e. with kernel) $V_1 ⊕ W$,


*Let  $\Pi_V$ be the projection onto  $V$  along  $U_1 ⊕ W$.
Check that  $\Pi_U$ and $\Pi_V$ commute!
