Linear algebra question: Is my approach valid in saying that let $x$ be an element of $Col(A)$ and $Null(A)$, then there exists a $y$ element of $\Bbb{R}^n$ such that 
$Ay = x$
and $Ax = 0$? and then show that $Ay = 0 = x$?
attached question
Could really use some help thanks! 
 A: I've looked at the image linked in the original post. It says that the original problem is:

Let $A\in M_{n\times n}(\Bbb{R})$ such that $A^2=I$. Prove that $\text{Col}(A)\cap\text{Null}(A)=\left\{0\right\}$.

You're off to a good start. You've let $x\in\text{Col}(A)\cap\text{Null}(A)$. From the fact that $x\in\text{Null}(A)$ we have that
$$Ax=0.$$
From the fact that $x\in\text{Col}(A)$ we have that
$$x=Ay$$
for some $y\in\Bbb{R}^n$. You'll be done if you can use these facts to show that $x=\vec{0}$.
In general, if you get stuck solving a problem, one way to get unstuck is to ask, "What was I given that I haven't used yet?" In this case, you haven't yet used the fact $A^2=I$. If you haven't already done so, I think it would be a good idea to take some time to see if you can prove $x=0$ using the fact that $A^2=I$.
In case you're still stuck, here's one way to do it:

 Note that $$y=Iy=A^2y=A(Ay)=Ax=0.$$ And since $y=0$, it follows that $$x=Ay=A0=0.$$

A: The attached question has probably a typo. If $A^2=I$, then $A$ is invertible and therefore $\operatorname{Null}(A)=\{0\}$, so clearly $\operatorname{Col}(A)\cap\operatorname{Null}(A)=\{0\}$ as well.
Indeed, if $x\in\operatorname{Null}(A)$, then $Ax=0$, so $A(Ax)=A0=0$; but $A^2=I$ implies $Ix=0$, that is, $x=0$.
A more sensible question would be:

prove that if $A^2=A$, then $\operatorname{Col}(A)\cap\operatorname{Null}(A)=\{0\}$

In order to prove it, let $x\in\operatorname{Col}(A)\cap\operatorname{Null}(A)=\{0\}$. Then $x=Ay$ and $Ax=0$, so
$$
0=Ax=A^2x=A(Ax)=Ay=x
$$
