Return to Revisions

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 $s$.