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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

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

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

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

  • $\vdots$

  • $\text{diag}\{\begin{bmatrix} i \end{bmatrix},\begin{bmatrix} i \end{bmatrix},\begin{bmatrix} -i \end{bmatrix},\begin{bmatrix} -i \end{bmatrix},\begin{bmatrix} 0 \end{bmatrix},\begin{bmatrix} 0 \end{bmatrix}\}$

For a total of $8$ Jordan normal form decompositions. Have I understood this correctly?

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    Yes; however, your title is about *rational* canonical forms, not Jordan canonical forms. They are slightly different, though you will have the same total number. (The rational canonical form of a matrix whose characteristic polynomial splits has the same number of blocks associated to an eigenvalue as its Jordan canonical form has).2011-10-23
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    @ArturoMagidin Oops, I mixed them up, it's been a while since I went over them. Would the rational canonical forms just be the matrices formed by the companion matrices corresponding to the combinations $x^2$, $(x-i)^2$, $(x+i)^2$; $x^2$, $(x-i)^2$, $x+i$, $x+i$; $x^2$, $x-i$, $x-i$, $(x+i)^2$; $x$, $x$, $(x-i)^2$, $(x+i)^2$; $x^2$, $x-i$, $x-i$, $x+i$, $x+i$; $x$, $x$, $(x-i)^2$, $x+i$, $x+i$; $x$, $x$, $x-i$, $x-i$, $(x+i)^2$; $x$, $x$, $x-i$, $x-i$, $x+i$, $x+i$?2011-10-23
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    For the rational canonical form, every time you have a block of size $k$ corresponding to $\lambda$ in the Jordan canonical form, you get the companion matrix of $(x-\lambda)^k$ in the rational canonical form. So, yes.2011-10-23
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    Ah, it's coming back to me now. Thanks!2011-10-23

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