In the physics of open quantum systems the Lindbladian $\mathcal{L}$ governs the evolution of quantum states through the Lindblad master equation.

The Lindblad operator can be viewed as a dissipative $\langle x, \mathcal{L} x \rangle $ which has spectrum in the left half of the complex plane. If I look at the spectra of these operators, I see that oftentimes many eigenvalues lie exactly on the real axis (see the examples in this paper https://arxiv.org/pdf/2206.09879.pdf)

In a recent paper this phenomenon is explained (https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.123.234103) in Appendix $B.9$ they state that a Theorem by Carlson (https://www.semanticscholar.org/paper/On-real-eigenvalues-of-complex-matrices-Carlson/9556e7154d2cda56b4456ab23060aaa7c1946015, Theorem 1) states the following

"a necessary and sufficient condition for a complex matrix $M$ to have at least $m$ real eigenvalues is the existence of a Hermitian matrix $C$ with $\vert \sigma(C) \vert = m$, such that $MC$ is also Hermitian. Here, $σ$ denotes the signature."

If I look into the proof of this. The proof refers to the paper (https://www.sciencedirect.com/science/article/pii/0022247X65901484, Theorem 3) which again refers to the paper (https://people.math.wisc.edu/hans/paper_archive/scanned_papers/hs019.pdf, Lemma 2). Each paper with slightly new notation.

In the end, the theorem does not seem all that complicated. So does anybody know a new proof, that could make it a bit easier to shred light on the Lindbladians above?