N\right) }\end{array}\right) ,  \label{A3}$$ and write the decomposition of $\mathbb{M}$ and its powers as $$\mathbb{M}=\mathbb{QDQ}^{-1},\quad \text{and\quad }\mathbb{M}^{n}=\mathbb{QD}^{n}\mathbb{Q}^{-1},  \label{A8}$$ where $\mathbb{D}$ is a diagonal matrix whose entries are the eigenvalues and $\mathrm{Tr}\mathbb{M}^n = \sum_{k=1}^{N}\left(\lambda _{k}\right)^n$ since the eigenvalue equation $\mathbb{M}^{n}\mathbb{X}^{\left( k\right) }=\left( \lambda _{k}\right) ^{n}\mathbb{X}^{\left( k\right) }$ holds. The spectral decomposition of $\mathbb{M}^{n}$ in terms of the stationary state and the decaying modes is $$\mathbb{M}^{n}=\mathbb{C}_{1}+\sum_{l=2}^{N}\left( \lambda _{l}\right) ^{n}\mathbb{C}_{l}.  \label{B28}$$ The matrix $\mathbb{C}_{1}$ (associated with the PF eigenvalue) is a stationary SM; the other matrices $\mathbb{C}_{l}$, $l>1$, are not stochastic, having however the property $\mathrm{Tr}\left( \mathbb{C}_{l}\right) =1$. Since $\left\vert \lambda
_{l}\right\vert <1$, then $\lim_{n\rightarrow \infty }\mathbb{M}^{n}=\mathbb{C}_{1}$, which is asymptotically irreversible, $\mathrm{\det }\left( \mathbb{C}_{1}\right) =0$. The eigenvalues $\lambda _{l} $ can be associated with characteristic relaxation times: we write $\lambda _{l}=\mathrm{sgn}\left( \lambda _{l}\right) $ $\left\vert \lambda _{l}\right\vert 
$, where $\mathrm{sgn}\left( \cdot \right) $ is the sign of the argument. Therefore, we define the characteristic decay time as $T_{l}=-\left( \ln \left\vert \lambda _{l}\right\vert \right) ^{-1}$, such that Eq. (\[B28\]) can be written as $$\mathbb{M}^{n}=\mathbb{C}_{1}+\sum_{l=2}^{N}\left[ \mathrm{sgn}\left(
\lambda _{l}\right) \right] ^{n}\mathbb{C}_{l}\exp \left( -\frac{n}{T_{l}}\right) .  \label{B29}$$ Thereafter, one can write the evolution of the vector $\mathbb{U}^{\mathrm{T}}\left( 0\right) $, $\mathbb{U}^{\mathrm{T}}\left( n\right) =\mathbb{U}^{\mathrm{T}}\left( 0\right) \mathbb{M}^{n}$ as $$\mathbb{U}^{\mathrm{T}}\left( n\right) =\mathbb{U}^{\mathrm{T}}\left(
