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

  • $M$ is not diagonalisable;
  • when we put $M$ in its Jordan normal form, none of the Jordan blocks have a determinant with an absolute value of $1$?

Where I got this question from: This is a follow-up question to the one I ask here. Thank you to those who provided the examples. So far, the examples are all made up of submatrices with determinant $1$. I was wondering if there are matrices that are not of this form. So maybe a matrix with its Jordan normal form looks like this, with $|\lambda_1|^2 \ne 1$ and $|\lambda_2|^3 \ne 1$.

$$ M = \left( \begin{array}{ccc} \lambda_1 & 1 & 0 & 0& 0\\ 0 & \lambda_1 & 0 & 0& 0 \\ 0 & 0 & \lambda_2 & 1& 0 \\ 0 & 0 & 0 & \lambda_2& 1 \\ 0 & 0 & 0 & 0& \lambda_2 \\ \end{array} \right). $$

ghc1997
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1 Answers1

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Here is an example :

$$A = \begin{pmatrix} 0&1&1&1\\ 1 & -6&1&1 \\ 0& 0 & 0 & 1 \\ 0& 0 & 1 & -6 \end{pmatrix}$$

Indeed, the Jordan normal form of $A$ is $$ J=\begin{pmatrix} -3-\sqrt{10}&1&0&0\\ 0 & -3-\sqrt{10}&0&0 \\ 0& 0 & -3+\sqrt{10} & 1 \\ 0& 0 & 0 & -3+\sqrt{10} \end{pmatrix}$$

TheSilverDoe
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  • on his more recent question I put a 9 by 9 example; you had deleted your answer there. For prime $p,$ matrix size $p$ by $p$ seems unworkable. https://math.stackexchange.com/questions/4466904/an-example-of-a-non-diagonalisable-matrix-in-mathrmsl5-mathbbz-whose/4466978#4466978 – Will Jagy Jun 06 '22 at 19:05
  • @WillJagy Yes, nice $9 \times 9$ example. My answer made no sense at all, but indeed, I think you are right to say that $5 \times 5$ does not work. – TheSilverDoe Jun 06 '22 at 19:32