Let $P_i$ be projections from a Hilbert space $\cal{H}$ to its closed subspace $\cal{H}_i$, $i=1,2,\cdots,n$, such that $\sum^n_{i=1} P_i$ is also a projection. And let $P$ be a projection from $\cal{H}$ to $[\bigcup \cal{H}_i]$ the closed subspace spaned by $\bigcup \cal{H}_i$.
Is $\sum P_i$ justly $P$?
When $n=2$, it is true and easy to prove. But when $n>2$, there is some difficult for me to prove that $\sum^n_{i=1} P_i=P$.
Thanks a lot to anyone who can give me a hint.
Hint: the following statement solves your problem:
Statement: If $P_1,\dots,P_n\in\mathcal H$ are orthogonal projections such that $\sum_{i=1}^n P_i$ is again an orthogonal projection, then $P_i$ are pairwise orthogonal, i.e. $P_iP_j=0,\ i\neq j.$
Edit.
Proof. If $Q,R$ are projections such that $Q\geq R$ then $Q\supseteq R.$ Indeed, let $x\in Ran R.$ Then $||Qx||^2=\langle Qx,x\rangle\geq \langle Rx,x\rangle=||x||^2,$ hence $x\in Ran Q.$ In particular, $Q-R$ is also a projection and $Q-R\perp R.$
Since $\sum_{i=1}^nP_i\geq P_n$ are projections, $\sum_{i=1}^{n-1}P_i$ is a projection and $\sum_{i=1}^{n-1}P_i\perp P_n.$ Induction over $n$ completes the proof. $\Box$