Characteristic polynomials for $2 \times 2$ matrices 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?
 A: 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 
$$
C=\begin{bmatrix} b&a\\-a&b\end{bmatrix}.
$$
A: Since $A$ is $2\times 2$, if it has two distinct eigenvalues, then it is diagnosable so similar to a diagonal matrix. 
So we can assume that $A$ is not diagnosable and that it has one real eigenvalue (complex ones come in pairs).
Now you can use the fact that the determinant of $A$ is the product of eigenvalues of $A$ and that the trace is the sum of these.
Edit: This gives that the eigenvalue is zero and the result follows from the Jordan canonical form of $A$.
