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

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.

• This question has been answered in Stackexchange Jun 22, 2022 at 9:36

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 $$2 \pm \sqrt{5}$$.
• Thanks, but $A$ is not in $\mathrm{GL}(2, \mathbb{Z})$? Jun 22, 2022 at 9:35
• @ghc1997 I had omitted this condition. I have modified the example, and I hope that it is right. I have a question: where does your problem and this transformation which replaces $2 \times 2$ blocks with entries a,b,c,d by a single entry $ad-bc-(ac+bd)$ come from? Jun 22, 2022 at 18:05
• Thank you for your time @ChristopheLeuridan, I also asked a similar question about replacing the 2 × 2 blocks by $ad-bc$. I thought I found some properties about the unipotent matrix, but people have provided counter-examples. Jun 22, 2022 at 18:13