The following problem comes from a classical-mechanics treatment of lattice vibrations. It's essentially a linear algebra question, though, so I thought it would be on-topic here.
If it may be assumed that the vibrations of atoms about their equilibrium positions in a crystal lattice are small (in some sense), it can be shown the problem of finding their frequencies $\{\omega\}$ and polarizations $\{\varepsilon\}$ is equivalent to solving the eigenvalue problem
$ \begin{equation} \omega_q^2\varepsilon_q=D_q\varepsilon_q \end{equation} $
where the dynamical matrix $D$ is Hermitian and $q$ is a label which characterizes the allowed vibrations. By taking the complex conjugate of the above, replacing $q\rightarrow-q$, and exploiting the additional symmetry $D_{-q}=D_q^*$, it follows that
$ \omega_{-q}^2\varepsilon_{-q}^*=D_q\varepsilon_{-q}^* $
i.e. $\varepsilon_{-q}^*$ and $\omega_{-q}^2$ are also eigenvectors/values of $D_q$. The book I'm reading now claims we can set
$ \omega_q^2=\omega_{-q}^2 $
as well as, modulo a complex phase
$ \varepsilon_q=\varepsilon_{-q}^* $
I understand that we can make this choice and both eigenvalue problems will be satisfied, but is it necessary? Experiments can directly measure $\omega$ as a function of $q$ -- I should think that if $\omega_{q}^2\ne\omega_{-q}^2$, there should be measureable consequences.
Not sure if this is correct, but I'll post my progress on this question. If someone could let me know if it looks OK, that would be much appreciated.
Since $\{\varepsilon_{-q}^*\}$ and $\{\varepsilon_q\}$ are the eigenvectors of a Hermitian matrix $D_q$, we can expand any other vector in a series of them. In particular
$ \begin{align} \varepsilon_q&=\sum_{q'}B_{q'}\varepsilon_{-q'}^*\tag{1}\\ \varepsilon_{-q}^*&=\sum_{q'}C_{q'}\varepsilon_{q'}\tag{2} \end{align} $
Substituting $(2)$ into $(1)$
$$ \begin{equation} \displaystyle \varepsilon_q=\underbrace{\sum_{q'}B_{q'}}_{S_B}\sum_{q''}C_{q''}\varepsilon_{q''} =S_B\sum_{q'}C_{q'}\varepsilon_{q'} \end{equation} $$
Taking the dot product of both sides with $\varepsilon_{q''}^*$ gives
$$ \delta_{q,q'}=C_{q'}S_B \implies C_{q'}=\frac{\delta_{q,q'}}{S_B}\tag{3} $$
Note that $S_B\ne0$ since by summing the equation on the left-hand side above we find $S_BS_C=1$ where $S_C=\sum_qC_q$. Substituting $(3)$ into $(1)$
$$ \varepsilon_{-q}^*=\frac{1}{S_B}\times\varepsilon_q $$
The requirement that both $\{\varepsilon_{-q}^*\}$ and $\{\varepsilon_q\}$ be orthonormal bases fixes $|S_B|^2=1\implies S_B=e^{-i\phi}$. Hence $\varepsilon_{-q}^*$ are $\varepsilon_q$ are equal modulo a complex phase factor. The equality of the spectra now follows readily from the eigenvalue equations
$$ \begin{align} \omega_q^2\varepsilon_q&=D_q\varepsilon_q\\ \omega_{-q}^2\varepsilon_q&=D_q\varepsilon_q \end{align} $$
Subtracting the above two equations gives
$$ \left[\omega_q^2-\omega_{-q}^2\right]\varepsilon_q=0 $$
and hence $\omega_q^2=\omega_{-q}^2$ by the linear independence of the $\{\varepsilon_q\}$.