Setup Let $\Pi$ be an $n\times n$ matrix where each row sums to one. Let $\mathcal Y$ be an eigenvector of $\Pi'$ such that \begin{equation*} \Pi' \mathcal Y=\mathcal Y \end{equation*}
Denote $\Psi$ as the diagonal matrix whose diagonal is the vector $\mathcal Y$.
Let $n\times n$ matrix $A$ satisfies the following equation \begin{equation*} (\mathbf I-\Pi) A = I-\Psi ^{-1} (\Pi')^{-1}\Psi \end{equation*} where $I$ is an identity matrix. Note that the matrix $ (\mathbf I-\Pi)$ is not full rank, and there are multiple matrices $A$ satisfying the equation above. Note, we assume $\Pi'$ is invertible and $\Pi \neq I$.
Question Let $M$ be an $n\times n$ matrix such that each row of $M$ sums to one. We want to show that for any choice of $A$, \begin{equation*} (I-M) A =D \end{equation*} where $D$ only depends on $\Pi$ and $M$, and it does not depend on the choice of $A$.
Note We have tested this claim numerically, but we are unable to find a way to prove this or characterize the matrix $D$ explicitly in terms of only $\Pi$ and $M$.