I am looking for a matrix $M\in \mathrm{SL}(4, \mathbb{Z})$, with all eigenvalues equal to $1$, and with the following properties:
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 $A = \begin{bmatrix}
a_1&a_2\\
a_3&a_4
\end{bmatrix}$, I would like to find $M$ such that $A$ is in $\mathrm{GL}(2, \mathbb{Z})$, and has one eigenvalue with absolute value not equal to $1$.
The matrices I have tried so far: since $M$ needs to have eigenvalues all equal to $1$, I tried the matrices consisting of just Jordan blocks, unfortunately, they didn't work. I also tried the matrix in the form $M=\begin{bmatrix}
A_1&A_2\\
0&A_4
\end{bmatrix},
$ where $A_1$ and $A_4$ have all eigenvalues all equal to $1$, but in this case $A$ will have all eigenvalues with absolute values $1$. I am a bit stuck as I can't think of other $4$ by $4$ matrices that have all eigenvalues equal to $1$.
Any idea to construct such matrices will be really appreciated.
Best Answer
I have no difficulties to come up with several types of matrices $A\in SL_4(\Bbb Z)$ having all eigenvalues equal to $1$. For example, 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 of determinants minus dot products is $$ A=\begin{pmatrix} 0 & -1 \cr 1 & 2 \end{pmatrix}\in SL_2(\Bbb Z). $$ However, this one has also only eigenvalues $1$.
The polynomial system for finding $M$ is indeed solvable with Groebner, however adding the condition $\det(A)=\pm 1$ is far too much for Groebner in my case. So one has to choose a certain form to reduce the complexity.
Edit: I found solutions over $\Bbb Q$, namely $$ M=\begin{pmatrix} 0 & \frac{3}{2} & -1 & 0 \cr 0 & 0 & \frac{1}{3} & 0 \cr 2 & -3 & 3 & 0 \cr -5\eta-3 & \eta & 0 & 1 \end{pmatrix} \in SL_4(\Bbb Q). $$ for all $\eta\in \Bbb Q$. Then the matrix $A$ is given by $$ A=\begin{pmatrix} 0 & \frac{1}{3} \cr -3 & 3 \end{pmatrix}\in SL_2(\Bbb Q). $$ and has characteristic polynomial $t^2-3t+1$.
Edit: Finally, by the same procedure, I found infinitely many integral solutions:
$$ M=\begin{pmatrix} 0 & 2 & -1 & 0 \cr 1 & 0 & 1 & 0 \cr 2 & -3 & 3 & 0 \cr -5\eta-7 & \eta & 0 & 1 \end{pmatrix} \in SL_4(\Bbb Z). $$ Then we obtain $$ A=\begin{pmatrix} -2 & 1 \cr -7 & 3 \end{pmatrix}\in SL_2(\Bbb Z), $$ which has characteristic polynomial $t^2-t+1$.