I have the foreknowledge that the following matrix has the same minimal and characteristic polynomial:
$$A=\begin{pmatrix}1 & 1 & 0 & 0\\ -1 & -1 & 0 & 0\\ -2 & -2 & 2 & 1\\1 & 1 & -1 &0\end{pmatrix}$$
which is $x^2(x-1)^2.$
$A-xI=\begin{pmatrix}1-x & 1 & 0 & 0\\ -1 & -1-x & 0 & 0\\ -2 & -2 & 2-x & 1\\1 & 1 & -1 &-x\end{pmatrix}$
The first series of matrix operations are $R_1+(1-x)R_2, R_2+R_4,R_3+2R_4$ which yield $$\begin{pmatrix}0 & x^2 & 0 & 0\\ 0 & -x & -1 & -x\\ 0 & 0 & -x & 1-2x\\1 & 1 & -1 &-x\end{pmatrix}$$ Swap $R_4 \leftrightarrow R_1$.
$$\begin{pmatrix}1 & 1 & -1 &-x\\ 0 & -x & -1 & -x\\ 0 & 0 & -x & 1-2x\\0 & x^2 & 0 & 0\end{pmatrix}$$
The next series are column operations. I believe some are extra, but I needed them to keep it straight on paper: $C_2+C_3,C_3+C_1,C_4+xC_1,C_3 \leftrightarrow C_2, C_3-C_2, C_3\leftrightarrow C_4, C_3-xC_2$
This yields:
$$\begin{pmatrix}1 & 0 & 0 &0\\0 & -1& 0 & -x\\0 & -x & 1-2x+x^2 & 0\\0 & 0& 0& x^2\end{pmatrix} $$
Finally $R_3-xR_2$,$C_4-xC_2$, and $R_3-R_4$ yields
$$\begin{pmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\ 0 & 0& (x-1)^2 & 0 \\ 0 & 0 & 0 & x^2\end{pmatrix}$$
This is where I'm stuck since there should only be one element on the diagonal that is not a $1$.