Find $p_{ij}^{(n)}$ for the transition matrix 
Let
  $$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}$$
  find $p_{11}^{(n)},p_{12}^{(n)},p_{13}^{(n)}$

Since the characteristic function is $$-\mu^3+\frac{4}{3}\mu^2-\frac{1}{3}\mu=0$$
the eigenvalues are $$\mu_0=1,\mu_1=\frac{1}{3},\mu_2=0$$
Here the problem starts, from what I understood from the book, we need to found three constants such that $$p_{ij}^{(n)}=\mu_0^nA+\mu_1^n+\mu_2^nC=A+\left(\frac{1}{3}\right)^nB$$
But how can I find the values of A and B? 
 A: Hmm.  I don't know how you can directly find $A$, $B$, $C$ but I can suggest a way to go about the problem.  Now that you have found the eigenvalues $\mu_i$, solve for the corresponding normalized eigenvectors $e_i$.  Then 
$$P^n = E^{-1}\Lambda^n E$$
where $\Lambda$ is the diagonal matrix of eigenvalues and $E = \sum_{i=0}^2 {e_ie_i^T}$.
Then you can read the values off of independent entries from this result.
A: Why not just computing $P^2$, then interpolating? We have:
$$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad P^2=\begin{bmatrix}\frac{1}{3}&\frac{2}{9}&\frac{4}{9}\\\frac{1}{3}&\frac{4}{9}&\frac{2}{9}\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}$$
so $p_{11}^{(1)}=A+\frac{B}{3}=\frac{1}{3}$ and $p_{11}^{(2)}=A+\frac{B}{9}=\frac{1}{3}$ give $A=\frac{1}{3},B=0$ and $p_{11}^{(n)}=\frac{1}{3}$.
Similarly,
$$ p_{12}^{(n)}=\frac{1}{3}-\frac{1}{3^n}, \qquad p_{13}^{(n)}=\frac{1}{3}+\frac{1}{3^n}.$$
A: Hint:
You know that for $n=0$ we get
$$
A + B = I,
$$
and for $n=1$ we get
$$
A + B/3 = P.
$$
So you can express $A$ and $B$ in terms of $I$ and $P$...
