I have the following problem, for which i don't know how to start:
Prove that all Markov Chain have a unique state of equilibrium. That is, if $P$ is the transition matrix of a Markov regular chain, then there exist a unique vector $$v=(v_1,\cdots,v_n)$$ such that $v_i>0$, $\sum v_i=1$ and $Pv=v$.
Please if some one can helpme I would be grateful.
HINTS: Start by showing (using the definition of a Markov process) that $1$ is an eigenvalue of $A$. Then show that if all the entries of $A$ are positive, the vector $(1,1,\dots,1)$ is the only eigenvector (up to scalar multiples) with eigenvalue of the matrix $A^\top$. (In general, you'll need to apply this to some power of $A$. Check the definition of regular.)