Let $A$ be a square matrix of order $n$ such that $A^{2} = A$. Prove that every $v \in \mathbb{R}^{n}$ can be decomposed as $v = v_{1} + v_{2}$ 
Let $A$ be a square matrix of order $n$ such that $A^{2} = A$. Prove that every $v \in \mathbb{R}^{n}$ can be decomposed as $v = v_{1} + v_{2}$, where $v_{1}$ is in the nullspace of $A$ and $v_{2}$ is in the column space of $A$.

My reasoning:


*

*I would first consider the matrix equation $Ax = b$.   

*The vector $b$ is in $\mathbb{R}^{n}$.  

*$b$ is either in the column  space of $A$ or not in the column space of $A$.  

*$b$ is either in the column space of $A$ or in the nullspace of $A$.  

*Thus, $b$ = $v$ can be expressed as the linear combination of $v_{1} + v_{2}$.


My reasoning is extremely fuzzy and this is the best I could think of. I personally am not convinced by Step 2 onwards. I would appreciate it if someone could strengthen my reasoning, or point out why it is wrong, and offer a better proof.
 A: Credit for your effort. Here's a better proof, with a trick involved that you would do well to remember.
Let $x \in \mathbb R^n$.Then, we write $x = (Ax) + (x - Ax)$. We know that $Ax \in \operatorname{im} A$ (where $\operatorname{im}$ denotes column space or image). The key point here is:
$$
A(x - Ax) = Ax - A(Ax) = Ax - A^2x = 0
$$
Which means that $x-Ax \in \ker A$, where $\ker$ denotes the kernel or null space.
However, we have written $x$  as a sum of an element in the column space ($Ax$) and the null space ($x - Ax$).
This completes the proof. The trick, of course, was in realizing that $x-Ax$ is actually in the kernel of $A$. 
In fact the decomposition is unique. To do this, show that the intersection of the column and null space is actually trivial (only contains zero). This is a very strong statement, which even has repercussions in infinite dimensions (there are similar statements).
A: Write 
$$v=\underbrace{(I-A)v}_{v_1}+\underbrace{Av}_{v_2}$$


*

*$v_1 \in Ker A$ because its image by $A$ is $A(I-A)(v_1)=(A-A^2)(v_1)=0.$

*$v_2$ is in the column space of $A$ because of its very form.
