The RG transformation $R_\ell$ maps a set of coupling constants $[K]$ of a model Hamiltonian to a new set of coupling constants $[K']=R_\ell[K]$ of a coarse-grained model where the length scale is modified by a factor of $\ell>1$.
In the vicinity of a fixed point $[K^*]$, i.e. at $K_n=K_n^*+\delta K_n$, one linearizes the RG transformation, $$K_n' = K_n^*+\delta K_n' = K_n^*+\sum_m\underbrace{\left.\frac{\partial K_n'}{\partial K_m'}\right|_{[K]=[K^*]}}_{M_{nm}^{(\ell)}}\delta K_m+\mathcal{O}(\delta K^2),$$ and proceeds by relating the (right) eigenvalues of the matrix $M^{(\ell)}$ to the critical exponents.
My question is the following: the matrix $M^{(\ell)}$ is real but might not be symmetric. One thus has to distinguish left and right eigenvectors & the eigenvalues are not guaranteed to be real if they even exist. However, in the literature it is said at this point that nonetheless $M^{(\ell)}$ is generically found to be diagonalizable with real eigenvalues. Is there some (mathematical) rationalization of this statement?