Diagonalisation of stochastic matrices

Question

Suppose that $(X_n)_{n≥0}$ is a Markov chain on a state space $I = {1, 2}$ and stochastic matrix

$$P = \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}$$

(a) Find the eigenvalues

$$|P-\lambda I| = \begin{vmatrix} \frac{1}{4}-\lambda & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3}-\lambda \end{vmatrix} = \lambda^2 - \frac{11}{12}\lambda -\frac{1}{12}$$

Then $\lambda_1 = 1, \lambda_2 = -\frac{1}{12}$

(b) You know $P$ and $P^0 = I_2$. Use this to find $a(i,j), b(i,j)$ so that you have an explicit form for $P^n$

We know by diagonalizing $P$, every entry of $P^n$ can be written as $(P^n)_{i,j}=a(i,j)\lambda_1^n+b(i,j)\lambda_2^n$

Edit:

$P^n = \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}^n = \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n \begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix}$

Edit (2):

$\begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n \begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix} = \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}^n \begin{bmatrix} 0 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n\begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix}$?

Answer 1

You're almost there. We have \begin{align} P^n &= \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}^n = \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n \begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix} \\ & = \frac 1{13} \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \left(-\frac{1}{12}\right)^n \end{bmatrix} \begin{bmatrix} 4 & 9 \\ -1 & 1 \end{bmatrix} \\ & = \frac 1{13} \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 4 & 9 \\ -\left(-\frac{1}{12}\right)^n & \left(-\frac{1}{12}\right)^n \end{bmatrix} \\ & = \frac 1{13} \begin{bmatrix} 4 + 9\left(-\frac{1}{12}\right)^n & 9 - 9\left(-\frac{1}{12}\right)^n\\ 4 - 4\left(-\frac{1}{12}\right)^n & 9 + 4\left(-\frac{1}{12}\right)^n \end{bmatrix}. \end{align}