Matrix that is conjugate to its own inverse in $GL(4,\mathbb{Z})$

Question

I am looking for examples of matrices that are conjugate to their own inverses in $GL(4,\mathbb{Z})$, i.e. $A=B^{-1}A^{-1}B$ for $A,B\in GL(4,\mathbb{Z}),\ A\neq I,\ B\neq I$.

I couldn't find it either by myself or Sage (it runs for a very long time and never gives an output).

I wonder if there are such matrices. If they exist, how can I find an example? I kind of want to see how they look like and be able to find a lot of examples...

Thanks in advance!

Edit: Also, I hope it is not in $GL(2,\mathbb{Z})\times GL(2,\mathbb{Z})$ or $GL(3,\mathbb{Z})$.

Edit 2: I apologize... I further hope the matrices are not finite order (better all the eigenvalues are not on the unit circle, i.e., hyperbolic matrices).

Answer 1

$$A=\left(\begin{array}{cccc}0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\end{array}\right).$$ Finding a matching $B$ is left as an exercise to the reader with the hint: the permutation $(1234)$ and its inverse $(4321)$ are conjugates in $S_4$.

That matrix does have $\lambda=1$ as an eigenvalue, so it is conjugate to a matrix in $GL_1\times GL_3$. Because $\lambda=-1$ is also an eigenvalue, it also stabilizes two 2-dimensional subspaces over $\Bbb{Q}$ and is thus also conjugate to a matrix from $GL_2\times GL_2$. Apparently this is undesirable.

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Matching the updated version of the question:

The palindromic polynomial $$f(x)=x^4+x^3+3x^2+x+1$$ is irreducible over $\Bbb{Z}$ because it remains irreducible after reduction modulo two. Because it is palindromic, $x^4 f(1/x)=f(x)$, its zeros have the property that $f(\rho)=0$ if and only if $f(1/\rho)=0$. Its companion matrix is $$ A=\left( \begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & -1 \\ \end{array} \right), $$ when the inverse is $$ A^{-1}=\left( \begin{array}{cccc} -1 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ \end{array} \right).$$

The palindromic property of $f(x)$ then tells us that $A$ and $A^{-1}$ share the same set of simple eigenvalues. This is because the eigenvalues of a companion matrix $A$ are the roots of the polynomial $f(x)$. Furthermore, the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$, which in this case are still zeros of $f(x)$, and thus the eigenvalues of $A$.

The shared simple eigenvalues already mean that the two matrices are automatically conjugate in $GL_4(\Bbb{C})$, but it turns out that they are also conjugate in $GL_4(\Bbb{Z})$. Unless I made a mistake we can use the same $B$ as above, namely $$B=\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right). $$

Irreducibility of $f(x)$ translates into the non-existence of a non-trivial subspace $V\subset \Bbb{Q}^4$ such that $AV=V$. This rules out the possibility of $A$ being conjugate to a matrix from $GL_1(\Bbb{Q})\times GL_3(\Bbb{Q})$ or $GL_2(\Bbb{Q})\times GL_2(\Bbb{Q})$.

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The same procedure works for any palindromic quartic. So it is also possible to arrange all the eigenvalues to be real. This happens for example with the companion matrix of $$g(x)=x^4 - 5 x^3 - 13 x^2 - 5 x + 1.$$

Answer 2

Using Gröbner bases, it is very easy to just come up with several types of matrices $A\in GL_4(\Bbb Z)$, where not all eigenvalues lie on the unit circle. For example, take $$ A=\begin{pmatrix} 0 & 0 & -1 & 0 \cr 1 & 0 & 0 & 1 \cr 1 & 0 & 0 & 2 \cr 1 & -1 & 1 & 1 \end{pmatrix}. $$ Now the condition $A=B^{-1}A^{-1}B$ can be rewritten as $ABA=B$, which is just a system of linear equations in the entries of $B$ together with the polynomial condition that $\det(B)=\pm 1$. In our case, we can easily find some $B\in GL_4(\Bbb Z)$ with this condition, for example $$ B=\begin{pmatrix} 1 & 0 & 0 & 2 \cr 1 & -1 & 0 & 1 \cr 0 & 0 & -1 & 0 \cr 0 & 0 & 0 & -1 \end{pmatrix}. $$ The advantage is, that you can vary the matrices as you want. First you vary $A$, which can be computed. Only afterwards you compute $B$. If you do it together, the equations are too hard. This way you may be able, perhaps, to find matrices of infinite order. Note, however, that the second part may have no solution for certain $A$.

Reference for the computations:

An example of a matrix in $\mathrm{SL}(4, \mathbb{Z})$ with the following properties

Answer 3

In any group, if $u$, $v$ are of order $2$ then

$$u(uv)u= vu = (uv)^{-1}$$

so $uv$ is conjugate to $(uv)^{-1}$.

Now, to produce elements of order $2$, start with some standard ones (like diagonal matrices) and conjugate them by other elements.