Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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). –  Arturo Magidin Oct 23 '11 at 5:01
    
@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 $x^2$, $(x-i)^2$, $(x+i)^2$; $x^2$, $(x-i)^2$, $x+i$, $x+i$; $x^2$, $x-i$, $x-i$, $(x+i)^2$; $x$, $x$, $(x-i)^2$, $(x+i)^2$; $x^2$, $x-i$, $x-i$, $x+i$, $x+i$; $x$, $x$, $(x-i)^2$, $x+i$, $x+i$; $x$, $x$, $x-i$, $x-i$, $(x+i)^2$; $x$, $x$, $x-i$, $x-i$, $x+i$, $x+i$? –  Danielle Intal Oct 23 '11 at 5:12
    
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. –  Arturo Magidin Oct 23 '11 at 5:23
    
Ah, it's coming back to me now. Thanks! –  Danielle Intal Oct 23 '11 at 5:35
add comment

1 Answer

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:

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.

share|improve this answer
    
Thanks Pierre-Yves. –  Danielle Intal Oct 23 '11 at 5:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.