Return to Answer

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 $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu),$$ for $\mu\in\mathbb{R}$.

$\bullet$ 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 $\prod_i A(\mu_i)$ 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 three possibilities (with real $c$ and $\phi$):
$\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1),
$\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$, or
$\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$.
So the eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.

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.

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 $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu),$$ for $\mu\in\mathbb{R}$.

$\bullet$ 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 $\prod_i A(\mu_i)$ 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 three possibilities (with real $c$ and $\phi$):
$\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1),
$\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$, or
$\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$.
So all four eigenvalues on the imaginary axis, on the real axis, or on the unit circle.

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 $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu),$$ for $\mu\in\mathbb{R}$.

$\bullet$ 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 $\prod_i A(\mu_i)$ 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 three possibilities (with real $c$ and $\phi$):
$\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1),
$\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$, or
$\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$.
So the eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.

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 $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu),$$ for $\mu\in\mathbb{R}$.

$\bullet$ 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 $\prod_i A(\mu_i)$ 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.

$\bullet$ This proves property 2. For property 1 I still need to exclude four distinct eigenvalues $\lambda,-\lambda,\bar{\lambda},-\bar{\lambda}$. I note that $\det A(\mu)=1$ for any $\mu$, so if this case is realized, then the eigenvalues of $\prod_{i=1}^n A(\mu_i)$ are necessarily all four on the unit circle.
If this can be excluded, the four eigenvalues come in two imaginary reciprocal pairs $\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$, with $c>0$.

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 $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu),$$ for $\mu\in\mathbb{R}$.

$\bullet$ 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 $\prod_i A(\mu_i)$ 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 three possibilities (with real $c$ and $\phi$):
$\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1),
$\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$, or
$\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$.
So all four eigenvalues on the imaginary axis, on the real axis, or on the unit circle.