# Unipotent matrix with all block matrices having non-zero determinant

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.

• ah, the fact that all elements have to be integers makes this a lot harder. Commented Jun 22, 2022 at 13:56
• 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. Commented Jun 22, 2022 at 13:57
• And indeed, there are many integral matrices of this type. I suppose, the next question about this you can answer yourself without posting it. Commented Jun 22, 2022 at 14:12
• 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? Commented Jun 22, 2022 at 14:48
• 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. Commented Jun 22, 2022 at 15:17

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 $$1$$.