Let $A$ be a square matrix of order $n$ such that $A^2 = A$. Prove that every $v \in \Bbb R^n$ can be decomposed as $v = v_1 +v_2$, where $v_1$ is in the null space of $A$ and $v_2$ is in the column space of $A$.
The decomposition is unique. How do we go about proving this?
A vector $v \in \mathbb{R}^n $ can be written uniquely as
$ v = v - A v + A v = (I - A) v + A v = v_1 + v_2 \hspace{15pt} (1) $
Now since $A^2 = A$ , then $ A - A^2 = A (I - A) = 0 $
$A v_1 = A (I - A) v = 0 $, so $v_1$ is in the null space of $A$
and $v_2 = A v $ so $v_2$ is in the column space of $A$.
Now I show that if $x$ is the null space then it cannot be in the column space unless $x = 0$, thus proving that the null space of $A$ and the columns space are disjoint.
Let $x$ be in the null space of $A$, then
$ A x = 0 $,
If $x$ is the column space of $A$ then, there exists a vector $y$ such that
$ x = A y $
Now pre-multiply both sides by $A$,
$ A x = A^2 y = A y$
But $A x = 0 $ by our assumption, then
$ A y = 0 $
Therefore,
$ x = 0 $
This proves that the null space of $A$ and its column space are disjoint.
Now suppose vector $v = u_1 + u_2$ where $u_1 $ is in the null space and $u_2$ is in the column space then
$ v = v_1 + v_2 = u_1 + u_2 $
where now we have $v_1, u_1$ are in the null space, and $v_2 , u_2$ are the column space. Then we can write
$ u_1 = v_1 + x $ implying that $u_2 = v_2 - x $
Since $u_1$ and $v_1$ are in the null space then $x$ is in the null space, and since $u_2$ and $v_2$ are in the column space then $x$ is also in the column space, thus $x$ is in both the null space and the column space, so $x$ must be the zero vector $0$. Hence $u_1 = v_1$ and $u_2 = v_2$
Thus the decomposition of a vector $v$ given above is unique.
Thanks to a comment by @Weltschmerz in this post for the decomposition that I based my answer on.