Consider a stochastic matrix $A=(a_{i,j})_{i,j=1}^n$ and $\omega$ its normalized left-eigenvector to the eigenvalue $1$. Define $|.|_{\omega}$ as $|x|_{\omega}=\sum_{i=1}^n|x_i|\omega_i$. For fixed $i\in\{1,...,n\}$ write $\hat{a}_i :=(a_{i,j})_{j=1}^n$ and $\bar{a}_i :=(a_{j,i})_{j=1}^n$. Assume that
- there is a partition $\mathcal{P}$ of $\{1,...,n\}$ such that for any $U,V\in\mathcal{P}$ we have for all $i,j\in U$ the identity $\sum_{k\in V} a_{i,k} = \sum_{k\in V} a_{j,k} $,
- for all $U\in\mathcal{P}$ and $i,j\in U$ there are permutations $\hat{\sigma}^{(i,j)}_U, \bar{\sigma}^{(i,j)}_U$ such that $\hat{a}_j = \hat{\sigma}^{(i,j)}_U(\hat{a}_i)$ and $\bar{a}_j = \bar{\sigma}^{(i,j)}_U(\bar{a}_i)$,
- EDIT: For all $U\in\mathcal{P}$ and all $i,j\in U$ we have for all $V\in \mathcal{P}$ that $|\{k\in V|p_{k,i}>0\}|=|\{k\in V|p_{k,j}>0\}|$
Claim: For any $U\in\mathcal{P}$ and for all $i,j\in U$ we have $|\bar{a}_i|_{\omega}=|\bar{a}_j|_{\omega}$.
My attempt
I started by using the first assumption to show that a Markov chain induced by $A$ is strongly lumpable on $\{1,...n\}$ with respect to $\mathcal{P}$ where the stationary distribution of the lumped chain is given by $\pi(U)=\sum_{i\in U}\omega _i$. Unfortunately, at this point I got stuck.
I made an attempt to use the second assumption to construct global maps $\hat{\sigma},\bar{\sigma}$ which map the vectors correctly, i.e., $\hat{a}_j = \hat{\sigma}(\hat{a}_i)$ and $\bar{a}_j = \bar{\sigma}(\bar{a}_i)$, independently of $i,j$ and $U$. My intention was to "smuggle" the $j$ inside the $|\bar{a}_i|_{\omega}$ using additionally that $\pi$ as defined above is an eigenvector of the lumped chain. But this did not lead anywhere.
Does anyone have an idea how to conclude the proof, please?
