Determining possible rational canonical forms based on the characteristic polynomial It's been a while since I've studied linear algebra, so I'm hoping to refresh myself on this by going through some problems found here.
I wanted to see a concrete example before attempting those to jog my memory. For instance, suppose you have a linear transformation $T$ over $\mathbb{C}$ with characteristic polynomial $\chi(X)=X^2(X^2+1)^2$. Based on $\chi(X)$, how can you figure out the possible rational canonical forms of $T$?
I tried to read up a little further on this. The eigenvalues of $T$ are $0$ and $\pm 1$ all of multiplicity 2. Does this just mean that the Jordan canonical form of $T$ as a matrix will be the matrices with diagonals of blocks


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*$\text{diag}\{\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} i & 1 \\ 0 & i\end{bmatrix},\begin{bmatrix} -i & 1 \\ 0 & -i\end{bmatrix}\}$

*$\text{diag}\{\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} i & 1 \\ 0 & i\end{bmatrix},\begin{bmatrix} -i \end{bmatrix},\begin{bmatrix} -i\end{bmatrix}\}$

*$\text{diag}\{\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} -i & 1 \\ 0 & -i\end{bmatrix},\begin{bmatrix} i \end{bmatrix},\begin{bmatrix} i\end{bmatrix}\}$

*$\vdots$

*$\text{diag}\{\begin{bmatrix} i \end{bmatrix},\begin{bmatrix} i \end{bmatrix},\begin{bmatrix} -i \end{bmatrix},\begin{bmatrix} -i \end{bmatrix},\begin{bmatrix} 0 \end{bmatrix},\begin{bmatrix} 0 \end{bmatrix}\}$
For a total of $8$ Jordan normal form decompositions. Have I understood this correctly?
 A: The problem is easily reduced to the case $\chi=X^n$, that is $T$ is a nilpotent endomorphism of an $n$-dimensional vector space. Then the conjugacy classes of such endomorphisms correspond to the partitions of $n$. 
EDIT 1. More precisely, if $m_\lambda$ is the multiplicity of $\lambda$ as a root of $\chi$, and if $p(m)$ is the number of partitions of $m$, then the number of possible conjugacy classes is the product of the $p(m_\lambda)$. 
In the example, as we have three roots of multiplicity two, the number is indeed $$p(2)^3=2^3=8.$$ 
Two references: 


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*Wikipedia entry. 

*Sequence A000041 of The On-Line Encyclopedia of Integer Sequences. 
EDIT 2. You can also find the number of partitions of $n$ by typing Partitions[n] on WolframAlpha. For instance, Partitions[9] gives the number of partitions of $9$.
EDIT 3. One can also go to Keith Matthews's page Computing the n-th partition number p(n), a subpage of Number-theoretic function BCMATH programs, itself a subpage of Some BCMath/PHP number theory programs.
