Sum of idempotent matrices is Identity
Solution 1:
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 $V$.