Show that for a $2\times 2$ matrix $A^2=0$ Assuming that given a $2\times2$ matrix $A$ with the property $A^3=0$ show that $A^2=0$.
Okay so this question is messing with me. When it says $A^3=0$ do you think it means the determinant? How would one approach this? Any help would be much appreciated

It does not mean the determinant. It means that if you multiply $A$ with itself three times, you get the zero matrix. The nullspace of $A^2$ is contained in the nullspace of $A^3$, i.e. if $A^2 x = 0$, then clearly $A^3 x = A(A^2 x) = 0$. Particularly, this means that the nullspace can only increase as you apply more powers of $A$. The nullspace of $A^2$ can be zero dimensional (just the zero vector), one dimensional or two dimensional since $A$ is a $2\times 2$ matrix.
$A^2$ cannot have trivial nullspace (consisting of just the zero vector) because this would imply that $A$ is invertible and so $A^3$ is invertible, particularly $AA^2$ couldn't be the zero matrix since the zero matrix is not invertible. This gives a contradiction.
If $A^2$ has a one dimensional nullspace, $A$ would also have a one dimensional nullspace and the nullspaces would be the same. Since they have the same nullspaces, it follows that $A^2A$ has only a one dimensional nullspace, which is again a contradiction since we need that $A^3$ is the zero matrix (that is, $A^3$ has a two dimensional nullspace).
Thus we must conclude that $A^2$ has a two dimensional nullspace, i.e. $A$ is the zero matrix.

From @amcalde's comment, I decided to work out the exact expression for $A^3$ to argue that $A^2$ must be the zero matrix. Suppose $A = \left(\begin{array}{cc} a & b \\ c & d\end{array}\right)$, then
$$A^2 = \left(\begin{array}{cc} a^2 + bc & ab + bd \\ ac + cd & bc + d^2\end{array}\right)$$
and additionally
$$A^3 = \left(\begin{array}{cc} a(a^2+bc)+b(ac+cd) & a(ab+bd)+b(bc+d^2) \\ c(a^2+bc)+d(ac+cd) & c(ab+bd)+d(bc+d^2)\end{array}\right)$$
Since $A^3 = 0$, we have the following equations:
$$a(a^2+bc) + b(ac+cd) = 0\tag{1}$$
$$a(ab+bd)+b(bc+d^2) = 0\tag{2}$$
$$c(a^2+bc)+d(ac+cd) = 0\tag{3}$$
$$c(ab+bd) + d(bc+d^2) = 0\tag{4}$$
Suppose now that not all of $a^2+bc$, $ac+cd$, $ab+bd$ and $bc+d^2$ are zero. Particularly, suppose $a^2+bc\neq 0$. If $ac+cd = 0$, then $a=0$ by equation $1$ and so by equation $2$, since $b\neq 0$, $bc+d^2 = 0$. Thus by equation $4$, we must have that $ab+bd = 0$ (since $c\neq 0$). However since $a=0$ and $b\neq 0$, we conclude that $d=0$. Since $bc+d^2=0$ and $d=0$, we conclude that $bc=0$ which gives us a contradiction.
Suppose again that $a^2+bc\neq 0$. If $ab+bd = 0$, then either $b=0$ or $a=-d$. In the first case, $a\neq 0$. By equation $4$, we get that $d=0$ and then by equation $1$, we get $a=0$ which is a contradiction since $a^2+bc=0$. In the second case, we get that $bc+d^2\neq 0$. By equation $4$, it follows that $d=0$. By equation $2$, it follows that $b=0$ which in turn gives us that $a=0$ by equation $1$. This is again a contradiction since $a^2+bc\neq 0$.
Repeating this rather yucky process, you can see that all of $a^2+bc$, $ac+cd$, $ab+bd$ and $bc+d^2$ are zero. That is to say that if $A^3 = 0$, $A^2$ must also be the zero matrix. This is a much more tedious approach as you see.
