Let $A$ and $B$ be two $n\times n$ primitive row-stochastic matrices. That is, all of their entries are non-negative, all the rows sum up to $1$, and there is an integer $p\geq 1$ such that all the entries of $A^p$ and $B^p$ are strictly positive. Assume that $A_{i,j}=0\iff B_{i,j}=0$ for all $i,j$. Here is my question:
Suppose that there exists $k_0\geq 1$ such that $A^k=B^k$ for all $k\geq k_0$. Does this imply that $A=B$?
Remark: Note that if any of $A$ and $B$ is invertible, then the answer is positive. Say for instance that $B$ is invartible, then $$AB^{k_0}=AA^{k_0}=A^{k_0+1}=B^{k_0+1}=BB^{k_0}.$$ Multiplying by $B^{-k_0}$ we get that $A=B$.
The specified conditions are not sufficient to force $A=B$.
Here's an example . . .
Choose $a,b\in (0,\frac{1}{2})$ with $a\ne b$, and let $A,B$ be given by $\\[5.5pt]$ $$ A = \pmatrix { a &\frac{1}{2}-a&\frac{1}{2}\cr a &\frac{1}{2}-a&\frac{1}{2}\cr \frac{1}{2}-a&a&\frac{1}{2} } \\ $$ $$ B = \pmatrix { b &\frac{1}{2}-b&\frac{1}{2}\cr b &\frac{1}{2}-b&\frac{1}{2}\cr \frac{1}{2}-b&b&\frac{1}{2} } $$ Then $A\ne B$, but for all $k\ge 2$, we have $$ A^k=B^k= \pmatrix { {\large{\frac{1}{4}}}\;&{\large{\frac{1}{4}}}\;&{\large{\frac{1}{2}}}\cr {\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}\cr {\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}} } $$