<s>[INST] How to find the minimal polynomial when given only the characteristic polynomial? I am trying to find all possible Jordan forms of a transformation with Characteristic Polynomial $(x-1)^2(x+2)^2$. How can I find its minimal polynomial? Or do I just assume the $2$ (minimal and characteristic pol.) coincide?
 [/INST] Actually, in the general case to obtain the Jordan canonical form you need to find monic, non-constant polynomial $d_{i}$ such that $$d_{1} \mid d_{2} \mid \dots \mid d_{n},$$ with $$d_{1} d_{2} \cdots d_{n} = \text{the characteristic polynomial},$$ and then $d_{n}$ is the minimal polynomial. (See Smith normal form.) I am assuming all eigenvalues are in the underlying field.
In this case there aren't many possibilities, it's either $n = 1$ and $d_{1} = (x-1)^2(x+2)^2$, or $n = 2$ and then you have three possibilities
$$
\begin{align}
&d_{1} = (x-1) (x+2) = d_{2}\\
&d_{1} = x-1, d_{2} = (x-1)(x+2)^2\\
&d_{1} = x+2, d_{2} = (x-1)^2(x+2)\\
\end{align}
$$
The corresponding JCF are
$$
\begin{bmatrix}1&1&&\\&1&&\\&&2&1\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&2&\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&2&1\\&&&2\end{bmatrix},
\quad
\begin{bmatrix}1&1&&\\&1&&\\&&2&\\&&&2\end{bmatrix}.
$$
Since $n \le 2$ in this case, the characteristic polynomial and the minimal polynomial together determine the JCF. But for instance if you take the characteristic polynomial to be $(x-1)^4$ and the minimal polynomial to be $(x-1)^{2}$, then there are two possible sequences of $d_{i}$
$$
d_{1} = (x-1)^{2} = d_{2},
\quad\text{and}\quad
d_{1} = d_{2} = (x-1) , d_{3} = (x-1)^{2}.
$$
The corresponding JCF are
$$
\begin{bmatrix}1&1&&\\&1&&\\&&1&1\\&&&1\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&1&1\\&&&1\end{bmatrix}.
$$
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