Eigenvalues come in pairs

Question

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.

Answer 1

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 $s$.