Suppose that I have a first-order elliptic differential operator $A: \mathrm{dom}(A) \subset L^2(E) \to L^2(E)$, where $(E,h^E) \to M$ is a hermitian vector bundle and $M$ is a compact manifold.

I know also that $A$ is $\omega$-bisectorial with $\omega < \frac{\pi}{2}$. That is to say, the spectrum $\sigma(A)$ is in a bisector of angle $\omega$ in the complex plane, and outside of this bisector, we have resolvent bounds: there exists $C > 0$ with

$$ |\zeta| ||(\zeta - A)^{-1}|| \leq C, $$

I want to be able to assert that $A$ has spectrum that goes to infinity on both sides of the imaginary axis. This should be true because you can see at the symbol level that:

$$\lambda \in \sigma( \mathrm{sym}_A(x,\xi) ) \iff -\lambda \in \sigma( \mathrm{sym}_A(x,-\xi)).$$

I have been told that this claim is true in the folklore, but I can't seem to find a reference.

Any ideas?