Find the Jordan Canonical Form from c(x) and m(x) Given the matrix B:
$$
\begin{pmatrix}
2 & 1 & -2 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{pmatrix}
$$
I have found that the characteristic polynomial is:
$
c(x)=(x-1)^{3}(x+1)
$
and then found that the minimal polynomial is:
$
m(x)=c(x)
$
so then can I conclude that  is similar to the Jordan Canonical matrix:
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}
$$

From the comment of N.S., and CW. 
$$
\left(\begin{matrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & -1\\
\end{matrix}\right)
$$ 
Note that, despite the multiple eigenvalue, any matrix that commutes with $B$ can be written as a polynomial $p(B).$ 
