Consider a set of $n\times n$ irreducible and aperiodic Markov matrices $\{M_n\}$, we assume each row sum up to be 1 and each entry is nonnegative. Assume there exists a uniform bound $a$ for the spectrum gap, i.e,
$$ \|\lambda_2(M)\|<1-a $$
If each row of $M$ subtract the last row and we only take the first n-1 rows and n-1 columns, it generates a new matrix $J$,
$$ J=\begin{pmatrix} M_{1,1}-M_{n,1} & M_{1,2}-M_{n,2} & \cdots & M_{1,n-1}-M_{n,n-1} \\ M_{2,1}-M_{n,1} & M_{2,2}-M_{n,2} & \cdots & M_{2,n-1}-M_{n,n-1} \\ \vdots & \vdots & \ddots & \vdots \\ M_{n-1,1}-M_{n,1} & M_{n-1,2}-M_{n,2} & \cdots & M_{n-1,n-1}-M_{n,n-1} \end{pmatrix} $$ Then we have a set of matrices $\{J_n\}$, one in fact can show the eigenvalues of $J$ is the same as all eigenvalues of the corresponding $M$ that are within unit circle. So the spectrum radius of $J$ is uniformly less than 1.
I would like to show the infinite product of J from the set goes to zero, i.e, $$ \lim_{n\rightarrow \infty}\Pi_{i=1}^n J_i=0, \ \ \text{for}\ J_i\in \{J_n\} $$
The numerical simulation shows it is true. If I can find a matrix norm such that the norm of all $J$ matrix is strictly less than 1, it is also proved.