On a theorem of Carlson on the necessary and sufficient condition for a matrix to have $m$ real eigenvalues Background: 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 is a dissipative operator $\langle x, \mathcal{L} x \rangle  \leq 0 $, so it 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/abs/2206.09879)
In particular, it has been observed that oftentimes $N$ of the $N^2$ eigenvalues of a $N^2 \times N^2$ matrix $\mathcal{L}$ lie on the real axis, see for example the picture below.

Question: 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 shed light on the Lindbladians above?
 A: I interpret the question in a bit more general terms, as a request to "shed light" on the phenomenon that "oftentimes many eigenvalues lie exactly on the real axis".
The key thing to notice is that the spectrum of the Lindbladian ${\cal L}$ is invariant under complex conjugation. So if we consider a parameter dependence ${\cal L}(t)$, an eigenvalue that is on the real axis for some $t=t_0$ cannot move into the complex plane when $t$ is increased slightly. It must first collide with another real eigenvalue, to become a complex conjugated pair.
This mechanism generically produces a number $N_{\rm real}$ of real eigenvalues that scales with the square root of the total number $N_{\rm tot}$ of eigenvalues. Precise calculations can be done in the Ginibre ensemble of random matrices, where
$$\mathbb{E}[N_{\rm real}]=\sqrt{2N_{\rm tot}/\pi}+{\cal O}(N_{\rm tot}^{-1}),$$
but the same square root law is observed in other ensembles (see, for example, appendix A.4 of arXiv: 1305.2924 ).
A heuristic argument for the square root law goes as follows:$^\ast$ The rate of change $dN_{\rm real}/dt=G-L$ contains loss terms and gain terms: A loss term $L$ appears when, upon increasing $t$, two real eigenvalues collide and move into the complex plane as a complex conjugate pair. A gain term $G$ appears when a complex conjugate pair of eigenvalues merges and splits on the real axis. It is natural to assume that $L\propto N_{\rm real}^2$, because it involves two independent real eigenvalues, while $G\propto N_{\rm tot}-N_{\rm real}$, since it involves only a single independent complex eigenvalue. A stationary state is reached when $L=G$, hence when
$$N_{\rm real}\simeq\sqrt{N_{\rm tot}},\;\;\text{for}\;\;N_{\rm tot}\gg 1.$$

$^\ast$ I am indebted to Tim Kokkeler for this argument.
