How do I prove a differential operator has no purely imaginary eigenvalues? Anyone who has taken a course in linear algebra knows how to prove the eigenvalues of a self-adjoint operator are real or the eigenvalues of a skew-self-adjoint operator are purely imaginary. This is easily extended to the spectrum of differential operators on Hilbert spaces. Conversely, how does one go about proving that a differential operator has no purely imaginary eigenvalues? Indeed, what condition can be imposed on a first order differential operator $L$: 
\begin{equation*}
Lu =A \frac{du}{dt} + Bu
\end{equation*}
for $u: S^1 \to \mathbb{R}^n$ that ensures the spectrum of $L$ contains no purely imaginary eigenvalues? If $n=2m$ and
\begin{equation*}
A= \left( \begin{array}{cc}
0 & -I  \\
I & 0  \end{array} \right)
\end{equation*}
then the first order part of $L$ is self-adjoint, so what condition can be imposed on $B$ so that the eigenvalues are not imaginary?
 A: Essentially, you have an ODE on $[0,2\pi]$ with a periodic endpoint condition. I'm assuming $A$ is invertible. The problem needs to be cast in $\mathbb{C}^{n}$; if you're going to require $u : S^{1}\rightarrow\mathbb{R}^{n}$, then a complex eigenvalue doesn't really make much sense. I'll use the vector inner product in $\mathbb{C}^{n}$, and constant coefficient matrices $A$ and $B$. Define $Lu$ as you have stated on the domain $\mathcal{D}(L)$ consisting of continuously differentiable vector functions $u : [0,2\pi]\rightarrow\mathbb{C}^{n}$ for which $u(0)=u(2\pi)$. Define an inner product
$$
         \langle u,v\rangle = \int_{0}^{2\pi}u(t)^{\star}v(t)dt,
$$
where $u(t)^{\star}$ is conjugate transpose of $u$. This is an inner product on continuous functions $u : [0,2\pi]\rightarrow\mathbb{C}^{n}$.
If $A^{\star}=-A$ and $B^{\star}=B$, then $L$ is symmetric on its domain $u,v\in\mathcal{D}(L)$ with respect to this inner product:
\begin{align}
   \langle Lu,v\rangle - \langle u,Lv\rangle
    & =\int_{0}^{2\pi}(A\frac{du}{dt},u)-(A^{\star}u,\frac{du}{dt})dt \\
    & = \int_{0}^{2\pi}(A\frac{du}{dt},u)+(Au,\frac{du}{dt})dt \\
    & = \int_{0}^{2\pi}\frac{d}{dt}(Au,u)dt \\
    & = (Au,u)|_{t=0}^{2\pi} = 0.
\end{align}
Symmetry is enough to force the eigenvalues to be real. It's a standard argument: if $Lf =\lambda f$ and $L$ is symmetric, then
$$
         (\lambda-\overline{\lambda})\langle f,f\rangle = \langle Lf,f\rangle-\langle f,Lf\rangle = 0,
$$
which either implies that $\lambda=\overline{\lambda}$ or $f = 0$. Note: you could also allow $B=B(t)$ if you assume that $B(t)^{\star}=B(t)$ for all $t \in [0,2\pi]$; the argument above still holds in that case.
