We consider a matrix
$$M_{\mu} = \begin{pmatrix} 1 & \mu & 1 & 0 \\ -\mu & 1 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \end{pmatrix}$$
One easily checks that $\operatorname{det}(M_{\mu})=1$.
I however noticed something peculiar:
Consider a sequence of real numbers $\mu_i$ then the four eigenvalues $\lambda_1,..,\lambda_4$ of
$$ A=\prod_{i=1}^n M_{\mu_i}$$
have the property that they can be chosen to satisfy $\lambda_1 = \overline{\lambda_2}$ and $\lambda_3 = 1/\lambda_1$ and consequently by the determinant $\lambda_4 = 1/\lambda_2.$
I tried to explicitly study the product and see if I can see some structure explaining all this, but so far only with limited luck.
Question: Show that the eigenvalues of $A$ for every $n$ and $\mu_i \in \mathbb R$ satisfy $$\lambda_1 = \overline{\lambda_2 } = 1/\lambda_3 = 1/\overline{\lambda_4}.$$
Update: What we know so far is that $\lambda_1 = \frac{1}{\lambda_3}$ and $\lambda_2 = \frac{1}{\lambda_4}$ and if any of the eigenvalues is non-real then we have it. However, we cannot exclude $\lambda_1>\lambda_2$ both real.
Best Answer
The explanation is pretty simple with a suitable change of basis.
Letting $$B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ i & 0 & -i & 0 \\ 0 & 1 & 0 & 1 \\ 0 & i & 0 & -i \end{pmatrix}$$ we have $$B^{-1}M_{\mu}B = \begin{pmatrix} 1+i\mu & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 1-i\mu & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}$$ Letting $N_\mu = \begin{pmatrix} 1+i\mu & 1 \\ -1 & 0 \\ \end{pmatrix}$, we thus have $$B^{-1}AB = \begin{pmatrix} \prod N_{\mu_i} & 0 \\ 0 & \overline{\prod N_{\mu_i}} \end{pmatrix}$$ where the bar denotes entry-wise complex conjugation. Thus the eigenvalues of $A$ are those of $\prod N_{\mu_i}$ plus those of $\overline{\prod N_{\mu_i}}$, which are their complex conjugates. Moreover, since $N_\mu$ has determinant $1$, so does $\prod N_{\mu_i}$, so its two eigenvalues are inverses of each other.