Given $p\geq 2$ linear maps $f_i:E\to E$ such that $f_1+\ldots+f_p=id_E$ and $f_i^2=f_i,\forall i$. Prove that $f_j\circ f_i = 0,\forall i\neq j$ The full problem statement is as follows,
Assume that vector space $E$ is finite dimensional, and let $f_i:E \to E$ be any $p \geq 2$ linear maps such that $f_1 + \ldots + f_p = \operatorname{id}_E$. Prove that $f_i^2 = f_i, 1 \leq i \leq p$ implies $f_j \circ f_i = 0$, for all $i\neq j, 1\leq i,j \leq p$.
I have tried to apply $f_j$ on both sides,
$$f_j \circ (f_1 + \ldots +f_p) = f_j \circ \operatorname{id}_E $$
$$f_j \circ f_1 + \ldots + f_j \circ f_{j-1} + f_j \circ f_{j+1} + \ldots + f_j \circ f_p = 0$$
but don't know how to make every term go to zero. 
How can I proceed from here? Any help is highly appreciated!
 A: We have $$f_1+\ldots+f_p=Id_E$$
Then, in one hand
$$\forall x\in E,f_1(x)+\ldots+f_p(x)=x$$
which implies that $$Im(f_1)+\ldots+Im(f_p)=E.$$
In the other hand
$$trace(f_1+\ldots+f_p)=trace(Id_E)=dim(E)$$
Then by linearity of $trace$
$$trace(f_1)+\ldots+trace(f_p)=dim(E)$$
But for $i\in\{1,\ldots,p\}$.  $f_i\circ f_i=f_i$ means that $f_i$ is a projection.
So, for $i\in\{1,\ldots,p\}$, $trace(f_i)=rank(f_i)=dim(Im(f_i))$.
Then
$$dim(Im(f_1))+\ldots+dim(Im(f_p))=dim(E)$$
Hence
$$Im(f_1)\oplus\ldots\oplus Im(f_p)=E$$
Now, let $i\in\{1,\ldots,p\}$ and let $x\in E$, we have 
$$x\in\ker(f_i)\Rightarrow x=\sum_{k=1,k\neq i}^pf_k(x)$$
So
$$ker f_i\subset\oplus_{k=1,k\neq i}^pIm(f_k)$$
By applying the rank theorem we obtain 
$$dim ker f_i=dim E-dim Im f_i=\sum_{k=1,k\neq i}^pdim Im(f_k)$$
Thus
$$ker f_i=\oplus_{k=1,k\neq i}^pIm(f_k)$$
Finally,let $i,j\in\{1,\ldots,p\}$ such that $i\neq j$ and let $x\in E$, we have 
$$f_j(x)\in\oplus_{k=1,k\neq i}^pIm(f_k)=ker f_i$$
Then
$$f_i\circ f_j(x)=f_i( f_j(x))=0$$
Which implies that $f_i\circ f_j=0$
