If $A$ is an $n$ $\times$ $n$ matrix, show that $A^2=0$ $\implies$ $Im(A) \subseteq Ker(A)$. And prove/refute the converse.
I know that I need to show that any vector in the image (range) must also be in the kernel (nullspace) of $A$.
I don't understand. If $A^2=0,$ what are the possible values of $A$? 
Regarding the image, if I know A, I can just take linear combinations of of its columns.
Regarding the kernel, if I know these values, then I can just solve $A\vec x=\vec 0$ for $x \in \mathbb{R}^n$.
So you can see that my main problem is finding the possible matrices $A$ which satisfy $A^2=0$.
Could anyone offer any hints?
Thanks

Let $x\in\mathrm{Im} A$. Then there exists $y\in\mathbb{R}^n$ such that $Ay=x$. Applying $A$ on that equation gives
$$Ax = A^2 y = 0$$
That means $x\in\mathrm{Ker} A$.
So we have shown $\mathrm{Im} A\subseteq \mathrm{Ker} A$.
Note:
The converse is also true: 
If $\mathrm{Im} A\subseteq \mathrm{Ker} A$, let $x\in\mathbb{R}^n$. 
Now $Ax\in \mathrm{Im} A$, therefore also $Ax\in\mathrm{Ker} A$, i.e. $A^2 x = A(Ax)=0$.
