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Let $M\in \mathrm{SL}(4, \mathbb{Z})$ with all eigenvalues equal to $1$ (i.e. $M$ is a unipotent matrix).

Write $M=\begin{bmatrix} A_1&A_2\\ A_3&A_4 \end{bmatrix}, $ where each $A_i$ is a $2$ by $2$ sumbatrix of $M$.

Let $a_i = \mathrm{det}(A_i)$.

Consider the matrix $A = \begin{bmatrix} a_1&a_2\\ a_3&a_4 \end{bmatrix}$,

Question: Is it possible that

  1. All $a_i$'s are non-zero?
  2. The matrix $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, and has one eigenvalue with an absolute value not equal to $1$?

Question 1 has been answered by Dietrich Burde, any hint with question 2 would be really appreciated.

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  • $\begingroup$ ah, the fact that all elements have to be integers makes this a lot harder. $\endgroup$ Commented Jun 22, 2022 at 13:56
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    $\begingroup$ For context see this question. @BenjaminWang Actually, we still have many integral solutions, so it is not too difficult. We only need to find some solutions. $\endgroup$ Commented Jun 22, 2022 at 13:57
  • $\begingroup$ And indeed, there are many integral matrices of this type. I suppose, the next question about this you can answer yourself without posting it. $\endgroup$ Commented Jun 22, 2022 at 14:12
  • $\begingroup$ Thank you @DietrichBurde, my final question is to find $M$ such that $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, and has one eigenvalue with an absolute value not equal to $1$. I haven't been able to come up with an example for a while. If I can't find one at the end, would it be okay for me to post it? $\endgroup$
    – ghc1997
    Commented Jun 22, 2022 at 14:48
  • $\begingroup$ But the matrix $A$ in my answer has characteristic polynomial $t^2-6t+1$, so $\det(A)=1$ and the eigenvalues are $3\pm 2\sqrt{2}$. Really, try it yourself next time. You can do it. $\endgroup$ Commented Jun 22, 2022 at 15:17

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With the ideas from the previous answer we immediately find examples, e.g., $$ M=\begin{pmatrix} 0 & 4 & 1 & 0 \cr 36 & 0 & 49 & 1 \cr 2 & -3 & 2 & 0 \cr 1 & -1 & 0 & 2 \end{pmatrix}. $$ Here $M$ has characteristic polynomial $(t-1)^4$. The matrix of determinants is given by $$ A=\begin{pmatrix} -144 & 1 \cr 1 & 4 \end{pmatrix}. $$ For the second question, take $$ M=\begin{pmatrix} 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 0 & 1 & -3 & 4 \end{pmatrix}. $$ Then the matrix $A$ of block determinants is given by $$ A=\begin{pmatrix} 0 & -1 \cr 1 & 6 \end{pmatrix}\in SL_2(\Bbb Z). $$ The eigenvalues do not have absolute value $1$.

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