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.