Submatrices of matrices in $\mathrm{SL}(4, \mathbb{Z})$ with all eigenvalues equal to $1$

Question

This is a follow-up question to my question from Math Stackexchange (Thank you Dietrich Burde and Michael Burr for the help).

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 the $A_i$ are $2$ by $2$ sumbatrix of $M$.

Let $d_i$ be the dot product of two rows of $A_i$, i.e. if $A_i = \begin{bmatrix} a&b\\ c&d \end{bmatrix}$, then $d_i = ac +bd$.

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

For example if $A_1 =\begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}$, then $d_1 = 11$ and $a_1 = - 2 - 11 = -13$.

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

Question: Suppose $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, can one of eigenvalues of $A$ have absolute value bigger than $1$?

Thoughts so far: It is clear that the matrix $A$ should have all entries being non-zero. I find this question hard because two similar matrices might have differnt "behaviour", e.g. if $ M_1=\Tiny\begin{pmatrix} 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 0 & 1 & -3 & 4 \end{pmatrix} $, then $ A=\begin{pmatrix} 0 & -1 \cr 1 & 2 \end{pmatrix}\in \mathrm{SL}_2(\Bbb Z); $ while if $ M=\Tiny\begin{pmatrix} 1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \end{pmatrix} $, we have $A=\begin{pmatrix} 0 & 0 \cr 0 & 0 \end{pmatrix}$, even though $M_1$ is simliar to $M_2$.

I also look up the Block matrix determinant, but I don't think we can use the identities since our block matrices don't satisfy their assumptions.

Thank you for reading, any reference for this would be really appreciated.

Answer 1

It is not a solution since $A$ is not in $GL_2(\mathbb{Z})$, I will try to correct it later.

Let $a,b,c,d$ be in $\mathbb{Z}$ such that $ad-bc=1$. Set $$P = \left(\begin{array}{cc} aI_2 & bI_2 \\ cI_2 & dI_2 \end{array}\right) \text{ and } T = \left(\begin{array}{cc} I_2 & S \\ 0 & I_2 \end{array}\right),$$ where $S \in \mathcal{M}_2(\mathbb{Z})$ Let $$M = PTP^{-1} = \left(\begin{array}{cc} aI_2 & bI_2 \\ cI_2 & dI_2 \end{array}\right)\left(\begin{array}{cc} I_2 & S \\ 0 & I_2 \end{array}\right) \left(\begin{array}{cc} dI_2 & -bI_2 \\ -cI_2 & aI_2 \end{array}\right),$$ Then, $$M = \left(\begin{array}{cc} aI_2 & aS+bI_2 \\ cI_2 & cS+dI_2 \end{array}\right) \left(\begin{array}{cc} dI_2 & -bI_2 \\ -cI_2 & aI_2 \end{array}\right),$$ Since $ad-bc=1$, $$M = \left(\begin{array}{cc} I_2-acS & a^2S \\ -c^2S & I_2+acS \end{array}\right).$$ Now, choose $S=I_2$. Then $$M = \left(\begin{array}{cc} (1-ac)I_2 & a^2I_2 \\ -c^2I_2 & (1+ac)I_2 \end{array}\right).$$ Hence $$A = \left(\begin{array}{cc} (1-ac)^2 & a^4 \\ c^4 & (1+ac)^2 \end{array}\right).$$ and $\det(A) = (1-ac)^2(1+ac)^2 - a^4c^4 = (1-a^2c^2)^2 - a^4c^4 = 1-2a^2c^2$ and $\mathrm{Tr}(A) = (1-ac)^2+(1+ac)^2 = 2+2a^2c^2$. If we choose $a=b=c=1$ and $d=2$, then we actually have $ad-bc=1$, $\det(A)=-1$ so $A \in GL_2(\mathbb{Z})$ and since $\mathrm{Tr}(A)=4$, the characteristic polynomial of $A$ is $X^2-4X-1$, so the eigenvalues of $A$ are $2 \pm \sqrt{5}$.