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.



Eigenvalue pattern
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.


