Irreducible Markov chain and invariant measure We consider a Markov chain $\left(X,P\right)$
  on a finite state space $X$. 
We denote $P:=\left(p_{x,y}\right)_{x,y\in X}$
  and for $n\in\mathbb{N}$
  $P^{n}:=\left(p_{x,y}^{(n)}\right)_{x,y\in X}$.
Assume that $\left(X,P\right)$   is irreducible. Let $\pi$
  be the unique $P$-invariant measure on $X$. Prove that there exists $\rho\in]0,1[$  and $c>0$
  such that
$$\forall x,y\in X,\forall n\in\mathbb{N},\left|P^{n}(x,y)-\pi(y)\right|\leq c\rho^{n}$$
Thank you in advance.
 A: If you have an irreducible, aperiodic Markov chain on a finite state space, you can apply the Perron-Frobenius theorem to conclude that the eigenvalue of (strictly) largest magnitude is $1$ and its eigenspace is $1$-dimensional. Up to normalization this eigenspace is given by the (unique) invariant distribution $\pi$. (As an aside, Perron-Frobenius still gives you that the eigenvalue of largest magnitude is $1$ if the chain is not aperiodic, but in this case it may be only nonstrict, as $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ demonstrates. In this case the convergence may fail.)
Denote the transition matrix by $P$ (in the row-stochastic convention). So if the initial distribution is $p$, then the distribution at time $n$ is $p P^n$. If $P$ is diagonalizable, then the diagonalized form of this expression reads:
$$p P^n = \sum_{i=1}^N c_i \lambda_i^n q_i = \pi + \sum_{i=2}^N c_i \lambda_i^n q_i.$$
where $\lambda_i$ are the eigenvalues, $q_i$ are the (left) eigenvectors, and $c_i$ are coefficients depending on $p$. Take the eigenvectors to be sorted by magnitude and the eigenvectors to be unit (in whatever norm we're using). By the triangle inequality:
$$|p P^n - \pi| \leq \sum_{i=2}^N |\lambda_i|^n |c_i|$$
Replace $|\lambda_i|$ by the second largest eigenvalue in magnitude, which is $|\lambda_2|$ in this setup:
$$|p P^n - \pi| \leq |\lambda_2|^n \sum_{i=2}^N |c_i|.$$
Your result follows. In the nondiagonalizable case you can make a similar proof, but writing it out is significantly more cumbersome, because the Jordan normal form introduces some additional factors.
