TITLE: Limit of power of Markov matrix
QUESTION [16 upvotes]: Recently, a friend gave me the following problem:

Let $$M = \begin{pmatrix} p & q & r \\ q & r & p \\ r & p & q \end{pmatrix}$$ where $p, q, r > 0$ and $p + q + r = 1$. Prove that $$\lim_{n \rightarrow \infty} M^n = \begin{pmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}$$

Some of my observations include that $M^{2^n}$ remains cyclic, so I tried to bound (with no luck) the maximum and minimum to see if they both converged to $1/3$. 
It would be nice if someone could post an elementary solution to this problem.

REPLY [2 votes]: Except for the last line, this is a purely algebraic answer, as there is a lot of special structure in this problem.  (Note Gerschgorin discs are an elementary and powerful but underutilized tool to bound eigenvalues.  They also have a nice picture associated with them.)  
you have a real valued symmetric transition matrix.  Hence it is diagonalizable with real eigenvalues, and mutually orthonormal eigenvectors.  
$M = U \Lambda U^T$
It's rows and columns both sum to one i.e. it is double stochastic.  Hence we know $\lambda_1 = 1$ and $\mathbf u_1 \propto \mathbf 1$. (In fact $\mathbf u_1 = \frac{1}{\sqrt{3}}\mathbf 1$)
$trace(M) = p + q + r = 1 = \lambda_1 + \lambda_2 + \lambda_3 = 1 +\lambda_2 + \lambda_3$ 
Hence we know $\lambda_2 + \lambda_3 = 0$. Put differently: $\lambda_2 = - \lambda_3$.
Use Gerschgorin discs to find that the minimum eigenvalue $\gt -1$.  
I.e. each disc has a center point $\gt 0$ and a radius $\lt 1$.  (It's really more of a line segment than a disc since we know the eigenvalues are real, but we can still draw the disc as an upper bound.)
Hence $\vert\lambda_2\vert \lt 1$ and $\vert\lambda_3\vert \lt 1$
$M^n = U \Lambda^n U^T = \big(\lambda_1^n \mathbf u_1 \mathbf u_1^T + \lambda_2^n \mathbf u_2 \mathbf u_2^T + \lambda_3^n \mathbf u_3 \mathbf u_3^T \big) = \big(1^n \mathbf u_1 \mathbf u_1^T + \lambda_2^n \mathbf u_2 \mathbf u_2^T + \lambda_3^n \mathbf u_3 \mathbf u_3^T \big)$
now at the end, take a limit:
$\lim_{n \to \infty} M^n = \lim_{n \to \infty}\big(1^n \mathbf u_1 \mathbf u_1^T + \lambda_2^n \mathbf u_2 \mathbf u_2^T + \lambda_3^n \mathbf u_3 \mathbf u_3^T\big) = 1 \mathbf u_1 \mathbf u_1^T + 0 \mathbf u_2 \mathbf u_2^T + 0 \mathbf u_3 \mathbf u_3^T = \mathbf u_1 \mathbf u_1^T $