Show that A,B are similar matrices in S, where S is the set of 2x2 matrices whose square is the zero matrix Let S be the set of 2x2 matrices whose square is the 0 matrix, i.e.
S = { A $\in$ $M_{2x2}$ | $A^2$ = 0}
Let A and B be non-zero matrices in S. Show that A and B are similar matrices.
I was able to prove the converse (essentially) of this statement:
Let A be in S and A similar to B. Show with justification that S is closed under similarity.
So, A$^2$ = $0$, B = P$^{−1}$AP, and so B$^2$ = P$^{−1}$A$^2$P = P$^{−1} 0$ P = $0$ and so B is also in S.
Thus S is closed under similarity.
But I am having trouble proving the second statement.
 A: Jordan decomposition of any matrix $A$ is $A = S J S^{-1}$, where $J$ is a Jordan matrix:
$$J = \operatorname{diag}(J_1, J_2, \dots), \quad J_k = \begin{bmatrix} \lambda_k & 1 \\ & \lambda_k & 1 \\ & & \ddots & \ddots \\ & & & \lambda_k & 1 \\ & & & & \lambda_k\end{bmatrix}.$$
Now, Jordan matrix of order $2$ can be either
$$\begin{bmatrix} \lambda_1 \\ & \lambda_2 \end{bmatrix} \quad \text{or} \quad \begin{bmatrix} \lambda & 1 \\ & \lambda \end{bmatrix}.$$
It is easy to see that $A^2 = S J^2 S^{-1}$. Now, if $J = \operatorname{diag}(\lambda_1, \lambda_2)$ and $A^2 = 0$, then $J = 0$, hence $A = 0$. Therefore, our nonzero $A$ must be nondiagonalizable with $\lambda = 0$, which means that all such matrices are similar to the Jordan matrix
$$J = \begin{bmatrix} 0 & 1 \\ & 0 \end{bmatrix},$$
and, by transitivity, to each other.
You need to fill some points here, but this is pretty much the way to go.
A: Hints: Two matrices are similar if and only if they have the same Jordan Canonical Form.
If a $2 \times 2$ matrix $A$ satisfies $A^2 = 0$, then both eigenvalues of $A$ are $0$, so the Jordan form of $A$ is either $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ or $\begin{bmatrix}0&1\\0&0\end{bmatrix}$. 
Can a non-zero matrix have a Jordan form of $\begin{bmatrix}0&0\\0&0\end{bmatrix}$?
