# 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!

• This may help. May 9, 2020 at 11:30

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 $$f_i\circ f_j=0$$