TITLE: How many Jordan normal forms are there for this characteristic polynomial?
QUESTION [7 upvotes]: Given the characteristic polynomial of a matrix $A \in \mathbb{C}^{6x6}$ with $p(A)=(\lambda-2)^2(\lambda-1)^4$, we were supposed to determine all Jordan normal forms that have this characteristic polynomial.
I determined 10 (is this correct?) and was wondering whether this is a general way to compute the number of them for an arbitrary characteristic polynomial.

REPLY [9 votes]: Let's see: up to order of the Jordan Blocks and the eigenvalues, we have, with $\,m_A(x)=$ the matrix's minimal polynomial:
$$m_A(x)=(x-1)(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&0&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&2\end{pmatrix}$$
$$m_A(x)=(x-1)(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}1&0&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&1\\
0&0&0&0&0&2\end{pmatrix}$$
$$(1)\;\;\;m_A(x)=(x-1)^2(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&2\end{pmatrix}$$
$$(2)\;\;\;m_A(x)=(x-1)^2(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&2\end{pmatrix}$$
$$m_A(x)=(x-1)^3(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0&0\\
0&1&1&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&2\end{pmatrix}$$
$$(1)\;\;\;m_A(x)=(x-1)^2(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}
1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&1\\
0&0&0&0&0&2\end{pmatrix}$$
$$(2)\;\;\;m_A(x)=(x-1)^2(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}
1&1&0&0&0&0\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&1\\
0&0&0&0&0&2\end{pmatrix}$$
$$m_A(x)=(x-1)^3(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}
1&1&0&0&0&0\\
0&1&1&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&1\\
0&0&0&0&0&2\end{pmatrix}$$
$$m_A(x)=(x-1)^4(x-2)\;\;\;:\;\;\;\;\begin{pmatrix}
1&1&0&0&0&0\\
0&1&1&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&0\\
0&0&0&0&0&2\end{pmatrix}$$
$$m_A(x)=(x-1)^4(x-2)^2\;\;\;:\;\;\;\;\begin{pmatrix}
1&1&0&0&0&0\\
0&1&1&0&0&0\\
0&0&1&1&0&0\\
0&0&0&1&0&0\\
0&0&0&0&2&1\\
0&0&0&0&0&2\end{pmatrix}$$
Yes, I also get $\,10\,$ different (i.e., non-similar) JCF's for that matrix, check we agree (I, you or both could be wrong), and I don't think there's a general method to come up with the different JCF's . I, for example, try to do it by checking the different possibilities for the minimal polynomial, according to the its different possbile degrees...