Consider the two matrices with some parameter $s \in \mathbb R$


$$A\_1= \begin{pmatrix} s& -1 &0& 0 \\1&0 &0&0 \\ 0&0&1&0 \\0&0&0&1 \end{pmatrix}$$
and
$$A\_2= \begin{pmatrix} s& -1 &-1& 0 \\1&0 &0&0 \\ -1&0&s&-1 \\0&0&1&0 \end{pmatrix}.$$


I then noticed that the eigenvalues of arbitrary products of $A\_1$ and $A\_2$, i.e. e.g. $A\_1A\_2A\_1$ and $A\_1A\_1A\_2A\_1$ etc. all have eigenvalues $\lambda\_1,1/\lambda\_1$ and $\lambda\_2, 1/\lambda\_2.$


It is clear that the product of eigenvalues is equal to one, as both matrices are in $\text{SL}(4,\mathbb R)$, but I don't see why they have to come in two pairs that multiply up to one, respectively.



Eigenvalues come in pairs
This follows from the identities
$$A\_1^{-1}=UA\_1U^{-1},\;\;A\_2^{-1}=UA\_2U^{-1},$$
$$A\_1^{\top}=VA\_1V^{-1},\;\;A\_2^{\top}=VA\_2V^{-1},$$
with
$$U=U^{-1}=\left(
\begin{array}{cccc}
 0 & 1 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 \\
\end{array}
\right),\;\;V=V^{-1}=\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & -1 \\
\end{array}
\right).$$
Hence for any string of products $M=A\_1^{n\_1}A\_2^{n\_2}A\_1^{n\_3}A\_2^{n\_4}\cdots A\_1^{n\_N-1}A\_2^{n\_N}$ of the two matrices $A\_1$ and $A\_2$ one has
$$M^\top = VU M^{-1} (VU)^{-1}.$$
It follows that if $\lambda$ is an eigenvalue of $M$, then also $1/\lambda$ is an eigenvalue:
$${\rm det}\,(\lambda-M)={\rm det}\,(\lambda-M^\top)={\rm det}\,(\lambda-VU M^{-1}(VU)^{-1})={\rm det}\,(\lambda-M^{-1}),$$
which implies that
$${\rm det}\,(\lambda-M)=0\Leftrightarrow {\rm det}\,(\lambda^{-1}-M)=0.$$
The case $\lambda=0$ is excluded because $A\_1$ and $A\_2$ are nonsingular for any $s$.


