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I am working through the problem below

What are the possible Jordan Canonical forms for a matrix $A \in M_n$ with characteristic polynomial $p_A(t)=(t+3)^4(t-4)^2$? Give reason for your answer.

Everything I have found regarding this question has the minimal polynomial given already. Since the minimal polynomial is not given, do I need to consider the various possibilities of the minimal polynomial as below?

  • $(t+3)^4(t-4)^2$
  • $(t+3)^3(t-4)^2$
  • $(t+3)^2(t-4)^2$
  • $(t+3)(t-4)^2$
  • $(t+3)^4(t-4)$
  • $(t+3)^3(t-4)$
  • $(t+3)^2(t-4)$
  • $(t+3)(t-4)$

Then I would need to consider the various Jordan compositions for each? For instance, I believe for minimal polynomial $(t+3)^4(t-4)^2$, we would have $\begin{pmatrix} -3 & 1 & 0 & 0 & 0 & 0\\ 0 & -3 & 1 & 0 & 0 & 0\\ 0 & 0 & -3 & 1 & 0 & 0\\ 0 & 0 & 0 & -3 & 0 & 0\\ 0 & 0 & 0 & 0 & 4 & 1\\ 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix}$.

I am not sure if any of this is correct or not, so any guidance would be much appreciated!

Tyler Clark
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    The example is fine, but check that for the same min. pol. you may have more than one possibility, for example for $;(t+3)^2(t-4)^2;$ you have two options: one with two Jordan blocks (=J.B.) of order $;2;$ each for $;-3;$ , and another one with one J.B. of order two and two of order one... – DonAntonio Dec 21 '13 at 01:24
  • @DonAntonio, thanks, so for $(t+3)^2(t-4)^2$, we would have $\begin{pmatrix} -3 & 1 & 0 & 0 & 0 & 0\ 0 & -3 & 0 & 0 & 0 & 0 \ 0 & 0 & -3 & 1 & 0 & 0 \ 0 & 0 & 0 & -3 & 0 & 0\ 0 & 0 & 0 & 0 & 4 & 1\ 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix}$ or $\begin{pmatrix} -3 & 1 & 0 & 0 & 0 & 0\ 0 & -3 & 0 & 0 & 0 & 0 \ 0 & 0 & -3 & 0 & 0 & 0 \ 0 & 0 & 0 & -3 & 0 & 0\ 0 & 0 & 0 & 0 & 4 & 1\ 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix}$? – Tyler Clark Dec 21 '13 at 01:39

1 Answers1

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I read the question as asking what forms of the Jordan canonical form, $J$, of the matrix $A$ are possible given that we know that $p_A(t) = (t + 3)^4(t-4)^3$. Assuming I've read it right, the answer I'm suspecting they're looking for is a matrix like the one you've provided above with each element of each Jordan block on the superdiagonal either a $0$ or a $1$ (which then gives $2^{n-2}$ possibilities here). The reason for this is that since $J$ is upper triangular, only the diagonal elements matter when computing the determinant.

yoknapatawpha
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  • Oh, okay! Yes, I thought it would be rather cumbersome to write all of them! – Tyler Clark Dec 21 '13 at 01:23
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    Yeah, that would be a (huge) pain. I think the intent of the question is to show that the characteristic polynomial doesn't fully specify the Jordan form, rather it only specifies the diagonal entries. – yoknapatawpha Dec 21 '13 at 01:25
  • For something like the last minimal polynomial $(t+3)(t-4)$, how would that Jordan canonical form look?

    Thanks for your help!

     – Tyler Clark Dec 21 '13 at 01:32
  • So that block would just be repeated down the diagonal three times? – Tyler Clark Dec 21 '13 at 01:49