Consider the matrix


$$A(\mu) = \begin{pmatrix} 0 & 1& 0 & 0 \\ -1 & -i\mu & 0 & i \\ 0 & 0 & 0 & 1 \\ 0 &i & -1 & i\mu \end{pmatrix}.$$


This matrix is for $\mu \in \mathbb R$ skew hermitian, i.e. all the eigenvalues are imaginary.


Let $(\mu\_i)\_i$ be a sequence of real numbers.


We consider the product


$$M=\prod\_{i=1}^n A(\mu\_i).$$


I claim the following two facts are true (observed numerically):


1.) If $n$ is odd, then all eigenvalues are imaginary (this is non-trivial for $n\ge 3$ since the matrix $M$ is in general not skew hermitian anymore)


2.) Show that the eigenvalues satisfy for $n \in 2\mathbb N\_0+1$ that $\lambda$ is an eigenvalue of $M$ if and only if $-\lambda$ is. If you show this for one eigenvalue it will hold for all eigenvalues of $M$.



Imaginary eigenvalues
Define the unitary and Hermitian matrices
$$U=\left(
\begin{array}{cccc}
 0 & 0 & -i & 0 \\
 0 & 0 & 0 & -i \\
 i & 0 & 0 & 0 \\
 0 & i & 0 & 0 \\
\end{array}
\right),\;\;
V=\left(
\begin{array}{cccc}
 0 & 0 & -i & 0 \\
 0 & 0 & 0 & i \\
 i & 0 & 0 & 0 \\
 0 & -i & 0 & 0 \\
\end{array}
\right),
\;\;U^2=I=V^2,$$
and note that, for $\mu\in\mathbb{R}$,
$$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu).$$


$\bullet$ Hence if $\lambda$ is an eigenvalue of $\prod\_{i=1}^n A(\mu\_i)$, then
$$0=\overline{\det\bigl(\lambda I-\prod\_i A(\mu\_i)\bigr)}=\det\bigl(\bar{\lambda} I-\prod\_i\bar{A}(\mu\_i)\bigr)=\det\bigl(\bar{\lambda} I-\prod\_i UA(\mu\_i)U\bigr)=\det\bigl(\bar{\lambda} I-\prod\_i A(\mu\_i)\bigr)=0.$$
So the eigenvalues come in complex conjugate pairs: if $\lambda$ is an eigenvalue of $\prod\_i A(\mu\_i)$, then also $\bar{\lambda}$ is an eigenvalue. (This holds irrespective of whether $n$ is even or odd.)


$\bullet$ Similarly, if $\lambda$ is an eigenvalue of $\prod\_{i=1}^n A(\mu\_i)$ and $n$ is an odd integer, then
$$0=\det\bigl(\lambda I-\prod\_{i=1}^n VA(\mu\_i)V\bigr)=\det\bigl(\lambda I-(-1)^n\prod\_{i=1}^n A(\mu\_i)\bigr)=\det\bigl(\lambda I+\prod\_{i=1}^n A(\mu\_i)\bigr)=0,$$
so the eigenvalues come in inverse pairs for odd $n$: if $\lambda$ is an eigenvalue then also $-\lambda$ is an eigenvalue. This proves property 2.


$\bullet$ Since $\det A(\mu)=1$ for any $\mu$, the product of the four eigenvalues of $\prod\_{i=1}^n A(\mu\_i)$ equals unity. This gives for odd $n$ the following three possibilities (with real $c$ and $\phi$):  

**A.** $\lambda\_1=ic$, $\lambda\_2=-ic$, $\lambda\_3=i/c$, $\lambda\_4=-i/c$ (this is property 1),  

**B.** $\lambda\_1=c$, $\lambda\_2=-c$, $\lambda\_3=1/c$, $\lambda\_4=-1/c$,  

**C.** $\lambda\_1=e^{i\phi}$, $\lambda\_2=-e^{i\phi}$, $\lambda\_3=e^{-i\phi}$, $\lambda\_4=-e^{-i\phi}$.  

The eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.


