Show that $P_i$ and $\sum_i P_i$ being idempotent implies $P_i P_j=\delta_{ij}$ Let $X$ be a finite dimensional real linear space, or more generally a finite dimensional vector space over a field of characteristic $0$. 
Let $(P_i)_{i=1}^n$ be a finite sequence of linear mappings $P_i :X\rightarrow X$ such that 


*

*$P_i^2=P_i$ for $i=1,...,n$, 

*$(P_1+...+P_n)^2=P_1+...+P_n$.


I wish to show that 
$$P_i\circ P_j=0 \textrm{ for } i \neq j.$$
I know how to prove it only for $n=2$:
if
$(P_1+P_2)^2=P_1+P_2$
then 
$$P_1P_2+P_2P_1=0. \tag{$\ast$}$$
By multiplying both sides this equality by $P_1$ from left and by $P_1$ by right and obtain two equalities:
$P_1 P_2+P_1P_2P_1=0$ and $P_1P_2P_1+P_2P_1=0$.
By subtracting: 
$$P_1P_2-P_2P_1=0. \tag{$\ast\ast$}$$
From $(\ast)$, $({\ast}\ast)$, we get $P_1P_2=0$, $P_2P_1=0$.
 A: First, note that the condition that $X$ has finite dimension seems actually important (see linked question in comment below).
An important property of projectors in finite dimensional spaces (over fields with characteristic $0$) is that their trace coincides with their rank (indeed, since $0$ and $1$ are the only eigenvalues, both the rank and the trace count the number of ones).
Let $p=p_1+p_2+\ldots +p_n$, $K={\sf Ker}(p),A={\sf Im}(p)$ and $A_i={\sf Im}(p_i)$. By the remark made just above,
$${\sf dim}(A)=\sum_{k=1}^n {\sf dim}(A_k), \ A \subseteq \sum_{k=1}^n A_k \tag{1}$$
The two facts above imply that $A$ is the direct sum of the $A_k$. Since $p$ is a projector, $X=K \oplus A$, and hence
$$
X=K \oplus \bigoplus_{k=1}^n A_k \tag{2}
$$
For $k\in K$, one has $0=pk=\sum_{j=1}^n p_jk$. By the unicity in decomposition
(2) above, we deduce that 
$$
p_j \ \text{is zero on} \ K \ (1\leq j\leq n) \tag{3}
$$
Now, let $q_{ij}$ be the unique endomorphism of $X$ that coincides with $p_i$
on $A_j$, and is zero on $K$ and $\bigoplus_{k\leq j}A_k$. By contruction, those
$n^2$ endomorphisms $q_{ij} (1\leq i,j \leq n)$ are linearly independent, and we have
$p_i=\displaystyle\sum_{j}q_{ij}$ for every $i$, so
$$
p=\sum_{i,j} q_{ij} \tag{4}
$$
On the other hand, since $p_i$ is a projector it is the identity on its image $A_i$, $q_{ii}$ must be the identity on $A_i$, so
$$
p=\sum_{i} q_{ii} \tag{5}
$$
Combining (4) with (5), we see that $\sum_{i\neq j}q_{ij}=0$. By the linear 
independence of the $q_{ij}$, we deduce $q_{ij}=0$ for any $i\neq j$. So each $p_i$
reduces to $q_{ii}$, and $p_i$ is the projector onto $A_i$ according to 
$K\oplus \bigoplus_{k\leq j}A_k$. The claimed property is now clear.
