I cannot find the relation between $(I-A)x=v$ and $((I-A)^2)x=0$. This question from my textbook:

Solve $(I-A)x=v$ where $v$ is some $4\times 1$ matrix, $I$ is the identity matrix, and $A$ is some $4\times 4$ matrix and hence solve $((I-A)^2)x=0$.

I cannot find the relation between these two. How am I supposed to use the previous result when I can just do it directly?
 A: Let's look at an example: 
$$
A = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}
$$
so that 
$$
Q = I - A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}
$$
Now solve $Qx = v$:
When $v = \begin{bmatrix} 1  \\ 0 \end{bmatrix}$, for instance, we find that 
$x = \begin{bmatrix} p  \\ 1 \end{bmatrix}$ for any value of $p$, and when 
$v = \begin{bmatrix} 0  \\ 1 \end{bmatrix}$,  we find that 
there is no solution. When $v =  \begin{bmatrix} 0  \\ 0 \end{bmatrix}$, you get that $x = \begin{bmatrix} p  \\ 0 \end{bmatrix}$ for any $p$. 
You can, with some effort, let $v = \begin{bmatrix} a  \\ b \end{bmatrix}$, and write out a general solution in terms of $a$ and $b$. 
Now let's try to solve 
$$
Q (Qx) = 0
$$
If we write $v$ for $Qx$ for a moment, then 
we need to find values of $v$ that $Q$ sends to zero...and we discover from our previous work that these are exactly of the form
$$
v = \begin{bmatrix} p  \\ 0 \end{bmatrix}.
$$
Now to find an actual solution, we need an $x$ with $v = Qx$. So we again look at our earlier solutions, and say "what vectors $x$ are sent to vectors of te form $\begin{bmatrix} p  \\ 0 \end{bmatrix}$?", and the answer turns out to be "all of them!". So the solution to the problem, in this case, is "the solution set is all of $\mathbb R^2$.
