An example of a non-diagonalisable matrix in $\mathrm{SL}(n, \mathbb{Z})$ whose eigenvalues don't all have absolute value $1$

Question

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

$M$ is not diagonalisable;
$M$ does not have all eigenvalues with absolute value $1$.

Thoughts: The only non-diagonalisable matrix in $\mathrm{SL}(n, \mathbb{Z})$ I can think of are the ones consisting of Jordan block with $\pm 1$ on the diagonal.

Any hint on how to construct such a matrix would be really appreciated.

score 5 · Answer 1 · edited Jun 05 '22 at 12:02

5

The matrix $A$ has no eigenvalues of modulus $1$ and is not diagonalizable over $\mathbb{C}$: $$A= \begin{pmatrix} 1 & 1 & 1 & 0\\ 2 & 3 & 0 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 2 & 3\\ \end{pmatrix}. $$

edited Jun 05 '22 at 12:02

David C. Ullrich

89,985

answered Jun 05 '22 at 11:22

kabenyuk

10,712

Thank you, I was looking for such matrices! – ghc1997 Jun 05 '22 at 11:29

score 3 · Accepted Answer · edited Jun 05 '22 at 12:37

3

What about $$A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 7 & 2\end{pmatrix}?$$

edited Jun 05 '22 at 12:37

Shaun

44,997

answered Jun 05 '22 at 10:46

TheSilverDoe

29,720

This matrix is not in $\operatorname{SL}(n,\mathbb{Z})$. – Branimir Ćaćić Jun 05 '22 at 10:55
2

@BranimirĆaćić Thanks for pointing that. I edited. – TheSilverDoe Jun 05 '22 at 11:01

An example of a non-diagonalisable matrix in $\mathrm{SL}(n, \mathbb{Z})$ whose eigenvalues don't all have absolute value $1$

2 Answers2

Linked