<s>[INST] Limit of power of Markov matrix 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.  
 [/INST] Let
$$ \sigma = p^2 + q^2 + r^2 - qr - rp - pq = \frac{1}{2} \left( (q-r)^2 + (r-p)^2 + (p-q)^2 \right) \geq 0.  $$
Note $ \sigma = (p+q+r) \sigma = p^3 + q^3 + r^3 - 3 pqr.$
Which is why the characteristic polynomial of your matrix is
$$ \lambda^3 - \lambda^2 - \sigma \lambda + \sigma = (\lambda - 1)(\lambda^2 - \sigma) $$
On the other hand
$$ 1 = p^2 + q^2 + r^2 +2 qr +2 rp +2 pq, $$ so
$$ \sigma = 1 - 3 (qr + rp + pq) < 1. $$
The eigenvalues are $1, \sqrt \sigma, -\sqrt \sigma.$ The diagonalized matrix $ D = P^T MP$ (where $P$ is orthogonal) raised to an arbitrarily large power, converges to the diagonal matrix with entries $1,0,0.$ Reverse the diagonalization, we get a symmetric rank one matrix.. give me a minute
Where was I?
$$
P =
\left(
\begin{array}{rr}
\frac{1}{\sqrt 3} & a & b \\
\frac{1}{\sqrt 3} & c & d \\
\frac{1}{\sqrt 3} & e & f 
\end{array}
\right)
$$
$$
D_\infty =
\left(
\begin{array}{rr}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 
\end{array}
\right)
$$
We then just multiply out
$$
M_\infty = P D_\infty P^T =
\left(
\begin{array}{rr}
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3} \\
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3} \\
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3} 
\end{array}
\right)
$$
</s>