3

Does there exists a matrix $M \in \mathrm{SL}(5, \mathbb{Z})$, such that:

  • $M$ is not diagonalisable;
  • Let $J$ be the Jordan normal form of $M$, $J$ has one Jordan block of size $3$ and one Jordan block of size $2$, and none of the Jordan blocks have a determinant with an absolute value of $1$? So $J$ looks like this,
    $$ J = \left( \begin{array}{ccc} \lambda_1 & 1 & 0 & 0& 0\\ 0 & \lambda_1 & 0 & 0& 0 \\ 0 & 0 & \lambda_2 & 1& 0 \\ 0 & 0 & 0 & \lambda_2& 1 \\ 0 & 0 & 0 & 0& \lambda_2 \\ \end{array} \right). $$ with $|\lambda_1|^2 \ne 1$ and $|\lambda_2|^3 \ne 1$?

Where I got this question from: This is a follow-up question to the one I ask here. (Thank you, TheSilverDoe, who answered my previous questions). So far, all the examples are made up of "smaller" $2$ by $2$ submatrices. I was wondering if there are matrices that are not of this form, e.g. a matrix with an odd degree.

Thanks for your time reading and help in advance.

ghc1997
  • 1,431

1 Answers1

6

5 does not seem workable. However, 9 by 9

$$ A = \left( \begin{array}{ccc|ccc|ccc} 0&1&0& 1&0&0& 0&0&0\\ 0&0&1& 0&1&0& 0&0&0\\ 1&3&0& 0&0&1& 0&0&0\\ \hline 0&0&0& 0&1&0& 1&0&0\\ 0&0&0& 0&0&1& 0&1&0\\ 0&0&0& 1&3&0& 0&0&1\\ \hline 0&0&0& 0&0&0& 0&1&0\\ 0&0&0& 0&0&0& 0&0&1\\ 0&0&0& 0&0&0& 1&3&0 \\ \end{array} \right) $$

Nice answer by Arturo about: a matrix is diagonalizable if and only if the minimal polynomial is squarefree ( and the roots are in the field being used) Minimal polynomial and diagonalizable matrix

For the 9 by 9 matrix above, the characteristic and minimal polynomials coincide, each is $(x^3 -3x-1)^3 ,$ thus repeated roots.

$$ A^3 - 3A - I = \left( \begin{array}{rrr|rrr|rrr} 0&0&0& -3&0&3& 0&3&0\\ 0&0&0& 3&6&0& 0&0&3\\ 0&0&0& 0&3&6& 3&9&0\\ \hline 0&0&0& 0&0&0& -3&0&3\\ 0&0&0& 0&0&0& 3&6&0\\ 0&0&0& 0&0&0& 0&3&6\\ \hline 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0 \\ \end{array} \right) $$

$$ (A^3 - 3A - I)^2 = \left( \begin{array}{ccc|ccc|ccc} 0&0&0& 0&0&0& 9&9&9\\ 0&0&0& 0&0&0& 9&36&9\\ 0&0&0& 0&0&0& 9&36&36\\ \hline 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0\\ \hline 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0\\ 0&0&0& 0&0&0& 0&0&0 \\ \end{array} \right) $$

Will Jagy
  • 139,541
  • But what about its Jordan blocks? – Ruy Jun 06 '22 at 23:15
  • @Ruy three real irrational eigenvalues, one full size (3 by 3) block for each – Will Jagy Jun 06 '22 at 23:29
  • Ok, I see you read it off from the minimal polynomial. However I'm not yet convinced all three roots have geometric multiplicity 1. How do you know the minimal polynomial isn't e.g $$(x-a_1)^3(x-a_2)(x-a_3),$$ where the $a_i$ are the eigenvalues? – Ruy Jun 07 '22 at 00:00
  • @Ruy the characteristic and minimal polynomials coincide here. The minimal polynomial has integer coefficients and divides the characteristic. The candidates are $x^3 - 3x-1, (x^3 - 3x - 1)^2, (x^3-3x-1)^3$ – Will Jagy Jun 07 '22 at 00:04
  • 1
    Ok, so it all follows from the fact that $x^3-3x-1$ is irreducible over $\mathbb Z$. – Ruy Jun 07 '22 at 00:07
  • @Ruy yes, the roots are $ -2 \cos \frac{2\pi}{9} , ; -2 \cos \frac{4\pi}{9} , ; -2 \cos \frac{8\pi}{9} . ; ; ; $ – Will Jagy Jun 07 '22 at 00:10