About the trace operation $$\lim_{n\rightarrow \infty }\mathrm{Tr}\left[ \mathbb{M}^{n}\right]
=1+\lim_{n\rightarrow \infty }\sum_{k=2}^{N}\left( \lambda _{k}\right)
^{n}=1.  \label{B1}$$ The same holds true for doubly stochastic matrices.

**(5)** At the limit $n\rightarrow \infty $ the stationary matrix is $$\mathbb{C}_{1}:=\lim_{n\rightarrow \infty }\mathbb{M}^{n}=
\left( 
\begin{array}{ccccc}
m_{1} & m_{2} & \cdots & m_{N-1} & m_{N} \\ 
m_{1} & m_{2} & \cdots & m_{N-1} & m_{N} \\ 
\vdots & \vdots &  & \vdots & \vdots \\ 
m_{1} & m_{2} & \cdots & m_{N-1} & m_{N} \\ 
m_{1} & m_{2} & \cdots & m_{N-1} & m_{N}\end{array}\right)  \label{B2}$$

with the values of the entries depending on matrix $\mathbb{M}$ and $\sum_{j=1}^{N}m_{j}=1$; as all the rows are the same the eigenvalues are $\left\{1,0,0,...,0\right\} _{N}$ and the matrix is idempotent, $\mathbb{C}_{1}^2 = \mathbb{C}_{1}$. Although the matrix $\mathbb{M}$ determines the stationary matrix $\mathbb{C}_{1}$, the inverse is not possible, the knowledge of $\mathbb{C}_{1}$ does not permit the full determination of $\mathbb{M}$. This is the essence of the irreversible evolution. Writing one row of matrix (\[B2\]) as the vector $\mathbb{Z}^{\mathrm{T}} =\left(\begin{array}{ccccc}
m_{1} & m_{2} & \cdots & m_{N-1} & m_{N} \end{array}\right)$, it can be noted that it is a stationary distribution because $\mathbb{Z}^{\mathrm{T}}\mathbb{M}=\mathbb{Z}^{\mathrm{T}}$ (eigenvalues 1), or $\mathbb{C}_{1}\mathbb{M} =\mathbb{C}_{1}$. From Example 1, Eqs. (\[C1\]) and (\[C15\]), it 