Sum of idempotent matrices is Identity 
[Ciarlet, Problem $1.1-10$] Let $A_k$, $1 \leq k\leq m$, be matrices of order $n$ satisfaying
  $$\sum_{k=1}^mA_k\ =\ I.$$
  Show that the following conditions are equivalent.
  
  
*
  
*$A_k = (A_k)^2$, $1  \leq k \leq m$,
  
*$A_kA_l=0$, for $k\neq l$, $1\leq k,l\leq m$
  
*$\sum\limits_{k=1}^m\mbox{rank}(A_k) = n$.
  

Remark: We already proved $(2)\to(1)$ and $(1)\to(3)$, but we couldn't prove $(3)\to(2)$ or $\{(3)\to(1)\;\; and\;\; (1)\to(2)\}$. Please help us, thanks so much.
 A: Condition 3 implies that your vector space is a direct sum of the images of the operators, i.e.
$$V = \bigoplus_{k=1}^m\mathrm{Im}(A_k)$$
Without loss of generality, let us focus on $A_1$. If $\mathbf{x}\in \ker(A_1)$ then
$$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x} = A_2\mathbf{x} + \cdots + A_m\mathbf{x}$$
Therefore it follows that $\ker(A)\subseteq \mathrm{Im}(A_2)\oplus \cdots \oplus \mathrm{Im}(A_m)$. On the other hand, if $\mathbf{x} \in \mathrm{Im}(A_2)\oplus \cdots \oplus \mathrm{Im}(A_m)$ then
$$\mathbf{x} = A_2\mathbf{x_2} + \cdots + A_m\mathbf{x_m}$$
for some $\mathbf{x_i}\in V$. But we also have
$$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x}$$
By uniqueness of representation, we must have $A_1\mathbf{x} = \mathbf{0}$ so $\mathbf{x}\in\ker(A_1)$. This shows
$$\ker(A_1) = \bigoplus_{k=2}^m\mathrm{Im}(A_k)$$ 
And in particular, we have
$$V = \mathrm{Im}(A_1) \oplus \ker(A_1)$$
For $\mathbf{x}\in \mathrm{Im}(A_1)$ we also have 
$$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x}= A\mathbf{x}$$
since by uniqueness of representation, we get that $A_k\mathbf{x} = \mathbf{0}$ for $k>1$. The two properties above characterizes $A_1$ as a projection. That's 3 implies 1. 
The above argument also shows 2, since if we have
$$\ker(A_1) = \bigoplus_{k=2}^m\mathrm{Im}(A_k)$$
then this trivially implies that for $k\neq 1$
$$A_k\mathbf{x} \in \mathrm{Im}(A_k) \subseteq \ker(A_1)$$
so that we get $A_1A_k\mathbf{x} = \mathbf{0}$ for all $\mathbf{x}\in V$ and all $k\neq 1$.
Finally as a note, 1 to 2 is probably most easily done through 1 to 3 to 2. Condition 1 very easily implies condition 3 since for projections, the trace is equal to the rank. This means
$$n = \mathrm{tr}(I) = \sum_{k=1}^m\mathrm{tr}(A_k) = \sum_{k=1}^m\mathrm{rank}(A_k)$$
Edit: I apologize for leaving out a key condition above. Hopefully this will make more sense. Let $A$ be a linear operator on an $n$-dimensional vector space $V$. The two conditions


*

*$\ker(A)\oplus\mathrm{Im}(A) = V$

*$A^2\mathbf{x} = A\mathbf{x}$ for $\mathbf{x}\in\mathrm{Im}(A)$


together characterize $A$ as an idempotent operator. I use the term "projection" synonymously with "idempotent linear transformation". This is the standard definition to my knowledge. For example, wikipedia uses it.
To see this, note that for any $\mathbf{v}\in V$ we can write $\mathbf{v}$ uniquely as
$$\mathbf{v} = \mathbf{x} + \mathbf{y},\ \ \ \ \mathbf{x}\in\mathrm{Im}(A),\ \ \mathbf{y}\in\ker(A)$$
Then this of course means that
$$A^2\mathbf{v} = A^2(\mathbf{x}+\mathbf{y}) = A^2\mathbf{x} = A\mathbf{x} = A(\mathbf{x}+\mathbf{y}) = A\mathbf{v}$$
so that $A$ is indeed idempotent on $V$.
