The generalization of commutative property of orthogonal projectors on a subspace to the whole space

Question

Let for $i\in [n]$, $P_i$ be some orthogonal projectors defined on the Hilbert space $W$ such that they commute on subspace $V < W$ (i.e, for any $i, j \in [n]$ and $v \in V$: $P_iP_j(v) = P_jP_i(v)$). Are there $n$ projectors $\hat{P}_i$ such that they operate the same as $P_i$ on $V$ (i.e, for any $i\in [n]$ and $v\in V$, $\hat{P}_i(v)= P_i(v)$) and they commute on the Hilbert space $W$?

Answer 1

This works for two projections, and then it follows from Halmos' two projections theorem. I originally thought the same argument would address the general case, but as Dominique pointed out, that's not so clear.

This says that if $P$ and $Q$ project onto $M$ and $N$, respectively, then we can decompose the Hilbert space as $$ W = (M\cap N) \oplus (M\cap N^{\perp}) \oplus (M^{\perp}\cap N) \oplus (M^{\perp}\cap N^{\perp}) \oplus (S\oplus T) , $$ and this reduces both $P$ and $Q$, and after a unitary transformation, the action on the last two components (if present) is $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 1-H & W \\ W & H \end{pmatrix} , $$ with $0<H<1$ and $W=(H(1-H))^{1/2}$. (This of course generalizes the trivial fact that a projection in $\mathbb C^2$ looks like $\left( \begin{smallmatrix} \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha\end{smallmatrix}\right)$.)

Now if $PQv=QPv$ and $v$ has a component $(v_1,v_2)\in S\oplus T$, then it follows that $Wv_1=Wv_2=0$, so $v_1=v_2=0$. In other words, $V$ must be contained in the sum of the first four summands, and thus the desired modifications can be obtained by simply replacing the part on $S\oplus T$ by zero.