Let $F$ be a field, and let $A$ $\in$ $F_{2}$ then the following are equivalent:
1.) $A^{2}= 0$
2.) $\mid A\mid = 0$ and $ tr A = 0$
3.) The characteristic polynomial of $A$ is $x^2$
4.) $ A$ is similar to a strictly upper triangular matrix
I have done most of the problem. However, I still don't know how to show that how any combination of $(1)$, $(2)$, and $(3)$ will imply $(4)$. I have got it down to this: $(1)$,$(2)$, and $(3)$ imply that $A$ will be of the form
$\begin{pmatrix} a & b \\ -\frac{a^{2}}{b} & -a \end{pmatrix}$
Can someone give me some guidance on how to find a matrix $C$ such that $C^{-1}AC$ is strictly upper triangular?
Since $\det A=0$, you know that $A$ has non-trivial kernel. So take $v$ with $Av=0$. Choose $w$ linearly independent with $v$, so that $v,w$ is a basis. Now think of the for of $A$ under this basis: the columns will be $Av$ and $Ax$. So in this basis $A$ is $$ A'=\begin{bmatrix} 0&r_1\\0&r_2\end{bmatrix}. $$ As the trace is independent of the basis, you have $0=\operatorname{Tr}(A)=r_2$. So $A$ is strictly triangular. We also conclude that $Aw=r_1v$. So now you take $C=\begin{bmatrix} v&w\end{bmatrix}$. Then $$ AC=\begin{bmatrix} Av&Aw\end{bmatrix} = \begin{bmatrix}0&r_1v\end{bmatrix} =\begin{bmatrix} v&w\end{bmatrix} A'=CA'. $$ From $v,w$ linearly independent you get that $C$ is invertible, so $C^{-1}AC=A'$.
All the above we can do explicitly: we can take $$v=\begin{bmatrix} b\\-a\end{bmatrix},\ \ \ w=\begin{bmatrix} a\\b\end{bmatrix}.$$ That gives you $$ C=\begin{bmatrix} b&a\\-a&b\end{bmatrix}. $$