Sum of Idempotent Transformation 
Let $V$ be a $n$ dimensional linear space on the number field $K$. Let $A_1,A_2,\cdots,A_s$ be idempotent transformations(or matrices) on $V$ ($A_i^2=A_i$, $i=1,2,\cdots,s$). If $A=A_1+A_2+\cdots+A_s$ is also an idempotent transformation, Prove $A_iA_j=0$ and $A_jA_i=0$ for $1\leqslant i<j\leqslant s$.

Advanced Algebra, Tsinghua University Press, Page 259
It's not hard to prove it if $s=2$. But I can't even prove the case when $s=3$. The book gives a pretty complicated method by setting
$$G=\begin{pmatrix}
A_1^2&A_1A_2&\cdots&A_1A_s\\
A_2A_1&A_2^2&\cdots&A_2A_s\\
\vdots&\vdots&&\vdots\\
A_sA_1&A_sA_2&\cdots&A_s^2
\end{pmatrix}$$
Any advice or other method appreciated.
 A: Answer inspired by the one given at the link in the comment under the question.
Note that a linear operator being idempotent means that it is diagonalisable with eigenvalues $0$ and $1$ only: the space $V$ is a direct sum of the eigenspaces for $0$ (its kernel) and the eigenspace for$~1$ (its image). 
Among other things this means that the trace of $A$ is equal to its rank (the dimension of the eigenspace for$~1$).
Our argument hinges on the additivity of the rank, and on the fact that we are considering the sum (rather than some other linear combination) of idempotents. Since the image of a linear combination of linear operators is always contained in the sum of their images, the rank of $A=A_1+\cdots+A_s$ is at most the sum of the ranks of the $A_i$; on the other hand the trace of $A$ equals the sum of the traces of the$~A_i$. If both the individual $A_i$ and their sum$~A$ are idempotent, this means that the image of$~A$, having as dimension the sum of the images of the$~A_i$, is in fact their direct sum (a sum of finite dimensional subspaces is a direct sum if and only if its dimension is the sum of their dimensions).
Thus each vector $v$ in the image of $A$ can be uniquely written as sum of vectors in the respective images of the$~A_i$, the components of $v$ in the summands of the direct sum. But one has
$$
  v=A(v)=A_1(v)+\cdots+A_s(v),
$$
so these components are simply the vectors $A_i(v)$ for $i=1,\ldots,s$. For a vector in one summand of a direct sum, it components in the other summands are clearly equal to the zero vector; this means that $A_j(A_i(v))=0$ whenever $j\neq i$. This holds for all$~v$ in the image of$~A$; on the other had for $w$ in the kernel of $A$ one has $A(w)=0=A_1(w)+\cdots+A_s(w)$ and by the directness of the sum this implies $A_i(w)=0$ for each$~i$; so $A_j(A_i(w))=0$ trivially in this case. Since for any $i\neq j$ one has that $A_jA_i$ vanishes on both eigenspaces of$~A$, and these eigenspaces span$~V$, it follows that $A_jA_i=0$ as desired.
