Diagonalisation proof

Question

Suppose the nth pass through a manufacturing process is modelled by the linear equations $x_n=A^nx_0$, where $x_0$ is the initial state of the system and

$$A=\frac{1}{5} \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix}$$

Show that

$$A^n= \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}+\left( \frac{1}{5} \right)^n \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}$$

Then, with the initial state $x_0=\begin{bmatrix} p \\ 1-p \end{bmatrix}$ , calculate $\lim_{n \to \infty} x_n$.

(the original is here)

I am not sure how to do the proof part

The hint is:

First diagonalize the matrix; eigenvalues are $1, \frac{1}{5}$.

I understand the hint and have diagonalised it but I don't know how to change it into the given form? After diagonalisation, I just get 3 matrices multiplied together

Answer 1

Diagonalize the matrix $A$: $$ A=\begin{bmatrix} \frac{3}{5}&\frac{2}{5}\\ \frac{2}{5}&\frac{3}{5} \end{bmatrix}= \begin{bmatrix} -1&1\\ 1&1 \end{bmatrix} \begin{bmatrix} \frac{1}{5}&0\\ 0&1 \end{bmatrix} \begin{bmatrix} -\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix}=PDP^{-1} $$

So we have: $$ A^n=(PDP^{-1})^n=PD^nP^{-1}= \begin{bmatrix} -1&1\\ 1&1 \end{bmatrix} \begin{bmatrix} \left(\frac{1}{5}\right)^n&0\\ 0&1 \end{bmatrix} \begin{bmatrix} -\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix}= \begin{bmatrix} \frac{1}{2}\left(\frac{1}{5}\right)^n+\frac{1}{2}&-\frac{1}{2}\left(\frac{1}{5}\right)^n+ \frac{1}{2}\\ -\frac{1}{2}\left(\frac{1}{5}\right)^n+\frac{1}{2}&\frac{1}{2}\left(\frac{1}{5}\right)^n+\frac{1}{2} \end{bmatrix}= \begin{bmatrix} \frac{1}{2}&-\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2} \end{bmatrix} \left(\frac{1}{5}\right)^n+ \begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} $$

Now you can calculate the limit.

Answer 2

Hint :

$A^{n}=(P^{-1} D P)^{n}=P^{-1} D^{n} P$