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?
 A: Two important facts relating the minimal and characteristic polynomials are


*

*Both have exactly the same set of roots (in an algebraic closure of the ground field) namely that of the eigenvalues of the matrix. (Without going to the algebraic closure this means that they have the same set of irreducible polynomial factors, but in your case one of the polynomials, and therefore the other as well, is already split over $\Bbb Q$.)

*The minimal polynomial divides the characteristic polynomial (this is the Cayley-Hamilton theorem).


So if you know the characteristic polynomial $P$, the minimal polynomial must be obtained by taking every distinct factor of $P$ at least once, and at most as many times as it occurs as factor of $P$. Any polynomial so obtained (in your case there are $4$ of them) can be the minimal polynomial.
A: 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
$$
\begin{bmatrix}1&1&&\\&1&&\\&&1&1\\&&&1\end{bmatrix},
\quad
\begin{bmatrix}1&&&\\&1&&\\&&1&1\\&&&1\end{bmatrix}.
$$
A: All polynomials $(x-1)^k(x+2)^m$ with $1\le k,m\le 2$ are possible minimal polynomials for different transformations, depending on the nilpotent part per Jordan block.
A: Hints: How does the Jordan Canonical Form deal with the two relatively prime factors $(x-1)^2$ and $(x-2)^2$? What happens if the eigenvectors span, and what happens if they do not span? Write down one possible Jordan Canonical Form matrix and then ask what changes to that matrix will enable you to find all. How do you read off the minimal  polynomial from the Jordan Canonical Form? 
