Stochastic matrix A^n=const.

Question

Let $A\in M(n,n)$ be a doubly stochastic matrix such that $$ A^k=\begin{pmatrix} 1/n & \cdots & 1/n\\ \vdots & \ddots & \vdots\\ 1/n & \cdots & 1/n\\ \end{pmatrix} $$ for some $k>1$. Does it follow that $$ A=\begin{pmatrix} 1/n & \cdots & 1/n\\ \vdots & \ddots & \vdots\\ 1/n & \cdots & 1/n\\ \end{pmatrix}? $$

Answer 1

No, it does not follow that $$ A=\begin{pmatrix} 1/n & \cdots & 1/n\\ \vdots & \ddots & \vdots\\ 1/n & \cdots & 1/n \end{pmatrix}. $$

Here is a (minimal) counterexample. Take $$ J_3=\begin{pmatrix} 1/3 & 1/3 & 1/3\\ 1/3 & 1/3 & 1/3\\ 1/3 & 1/3 & 1/3 \end{pmatrix},\qquad N=\begin{pmatrix} 1 & 1 & -2\\ -1 & -1 & 2\\ 0 & 0 & 0 \end{pmatrix}. $$ Observe that $N^2=J_3N=NJ_3=0$. Consequently for all $\varepsilon$, $$ (J_3+\varepsilon N)^2=J_3^2=J_3. $$ When $\varepsilon\leq 1/6$, $J_3+\varepsilon N$ is doubly stochastic, producing a counterexample in the case $n=3,k=2$.