$A$ is a nonzero $n \times n$ matrix such that $A^2=0$. If $n=2$, is it possible that I can show that there exists an invertible $2 \times 2$ matrix $S$ such that $S^{-1}AS=\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$?
My first approach to this was to rearrange the equation so that it appears as $A=S\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}S^{-1}$ but it is kind of weird because this would mean $\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$ is the eigenvalue matrix. And this isn't a diagonal matrix.
I got no more idea and so I looked at the hint and it says I could let $\vec{u} \in \mathbb{R}^n$ such that $A\vec{u}\neq0$ and then make $S=\begin{bmatrix} \vec{u} & A\vec{u} \end{bmatrix}$ to continue from here. But I don't understand how can I just anyhow throw in a vector in $\mathbb{R}^n$ into the eigenspace matrix $S$ without even confirming that $\vec{u}$ is a eigenvector of matrix $A$?