shown in [@CRV] that if the ZF and GZ relations and are fulfilled, then ${{|\cS(z_1,z_2,\dots,z_L)\rangle}}$ satisfies $$t(z_i){{|\cS(z_1,z_2,\dots,z_L)\rangle}} = {{|\cS(z_1,z_2,\dots,z_L)\rangle}}, \quad \mbox{for} \quad 1\leq i \leq L,$$ where $t(z)$ is the inhomogeneous transfer matrix defined in with inhomogeneity parameters $z_1,\dots,z_L$. On specific models, it is even possible to prove using symmetry and degree arguments that ${{|\cS(z_1,z_2,\dots,z_L)\rangle}}$ is an eigenvector of $t(z)$ for all $z$ (but with an eigenvalue possibly different from $1$), see for instance [@CRV; @VanicatThesis].

For our process we recall that we took a specific staggered picture for the inhomogeneity parameters, see subsection \[subsec:parallel\_update\]. The steady state of the process is thus given in the staggered form (similar to the transfer matrix ) $${|{\cal S}\rangle}= \frac{1}{Z_L} {\langle\!\langle}W|\mathbf{A}(\kappa)\otimes\mathbf{A}\left(\frac{1}{\kappa}\right)\otimes\dots\otimes\mathbf{A}(\kappa)|V{\rangle\!\rangle}$$ where $Z_L= Z_L(\kappa,1/\kappa,\dots,\kappa)$. Let us briefly show that it provides the correct stationary state. The following relations indeed hold (they are a direct consequence of and ) $$U \mathbf{A}(\kappa)\otimes\mathbf{A}\left(\frac{1}{\kappa}\right)= \mathbf{A}\left(\frac{1}{\kappa}\right) \otimes \mathbf{A}(\kappa), \quad 
 B {{\langle\!\langleW|}} \mathbf{A}\left(\frac{1}{\kappa}\right)= {{\langle\!\langleW|}} \mathbf{A}(\kappa), \quad 
 \overline{B} \mathbf{A}(\kappa) {{|V\rangle\!\rangle}} =  \mathbf{A}\left(\frac{1}{\kappa}\right){{|V\rangle\!\rangle}}.$$ Using these properties, it is then easy to check that $${{\mathbb U}}^{o} {|{\cal S}\rangle}= {{|\cS'\rangle}} \quad \mbox{and} \quad {{\mathbb U}}^{e} {{|\cS'\rangle}} = {|{\cal S}\rangle},$$ where $$\label{eq:stationary_state_bis}
 {{|\cS'\rangle}} = \frac{1}{Z_L} {\langle\!\langle}W|\mathbf{A}\left(\frac{1}{\kappa}\right)\otimes \mathbf{A}(\kappa)\otimes\dots\otimes\mathbf{A}\left(\frac{1}{\kappa}\right)|V{\rangle\!\rangle}.$$ It thus implies $M{|{\cal S}\rangle}= {|{\cal S}\rangle}$. 