Diagonalisation proof 
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
 A: Hint :
$A^{n}=(P^{-1} D P)^{n}=P^{-1} D^{n} P$
A: 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.
