TITLE: Non-orthogonal projections summing to 1 in infinite-dimensional space
QUESTION [8 upvotes]: Consider projection operators $\rho_1,\ldots,\rho_k$ defined on vector space $V$ over field of characteristic $0$, such that
$$
\rho_1+\cdots+\rho_k = 1
$$
Projections $\rho, \pi$ are said to be orthogonal, if $\rho\circ\pi=\pi\circ\rho=0$.
Question: Are  $\rho_1,\ldots,\rho_k$ necessarily pairwise orthogonal?
If $V$ has finite dimension, the answer is yes. I expect it not to be the case in general, but I can't seem to come up with a counterexample.
Bonus question:
What about possibly infinite set of projections, where 
$$
\sum_{\rho\in P} \rho = 1
$$
is understood as "for each $v\in V$, only finitely many of $\rho v$ are nonzero, and their sum is $v$" ?

REPLY [6 votes]: That isn't true in general, so if we take our space $V=\mathbb{R}^{\mathbb{Z}}_0$ (that mean the space of sequences $(u_n)_{n\in \mathbb{Z}}$ with finitely nonzero sequences ) and we define the canonical algebraic bases $e_n$ and the projections :
$$
\left\{\begin{array}{l}
P_1(e_n)=\frac{n}{2} (e_n+e_{2-n})\\
P_2(e_n)=\frac{n}{2} (e_n-e_{2-n})\\
P_3(e_n)=\frac{-n}{2} (e_n+e_{-2-n})\\
P_4(e_n)=\frac{-n}{2} (e_n-e_{-2-n})\\
P_5(e_n)=e_n
\end{array}\right.
$$
So you can easily verify that $P_1+P_2+P_3+P_4+P_5=1$  but there are not orthogonal.