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 $M$.