Finding the limit of a sequence by diagonalising a matrix Consider the sequence  described by:
$\frac11 , \frac32 , \frac75 , ... ,\frac {a_{n}}{b_{n}}$
where $ a_{n+1} = a_n +2b_n $  and $b_{n+1} = a_n+b_n$
Find a matrix  $A$ such that 
 $$\begin{bmatrix} a_{n+1} \\b_{n+1}  \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n}  \end{bmatrix}$$
By diagonalising $A$ find explicit  formulae for $a_n \text{ and } b_n $  and  hence  show that 
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My progress so far:
$A= \begin{bmatrix} 1 & 2 \\ 1 & 1  \end{bmatrix}$ 
with eigenvalues $ \lambda_1 = 1 + \sqrt{2} \text{ and } \lambda_2 = 1 -\sqrt{2}$ that correspond to eigenvectors $(\sqrt{2},1) , (-\sqrt{2},1)$.
So the matrix  $P=\begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ 1 & 1  \end{bmatrix}$   is such that $P^{-1}A P =\begin{bmatrix} 1+ \sqrt{2} & 0 \\ 0 & 1 - \sqrt{2}  \end{bmatrix}$
I am unsure how this helps find explicit formulae for $ a_n \text{ and } b_n$.
Update:
The Solution I was  given  defined a coordinate system ,such that  $\begin{bmatrix} 1 \\1  \end{bmatrix}$ becomes  $P^{-1} \begin{bmatrix} 1 \\1  \end{bmatrix} = \begin{bmatrix} p_1 \\p_2 \end{bmatrix}$ where $$p_1 = \frac {\sqrt{2}+1} {2 \sqrt{2}} , p_2 = \frac {\sqrt{2}-
1} {2 \sqrt{2}}  $$
then
$a_n = p_1 \lambda_1^{n-1} \sqrt{2} -  p_2 \lambda_2^{n-1} \sqrt{2}$ and $b_n =  p_1 \lambda_1^{n-1}  + p_2 \lambda_2^{n-1} $
therefore $$\frac {a_n} {b_n} = \frac { p_1 \lambda_1^{n-1} \sqrt{2} -  p_2 \lambda_2^{n-1} \sqrt{2}} {p_1 \lambda_1^{n-1}  + p_2 \lambda_2^{n-1}}$$
$$ = \frac {\sqrt{2} -(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1} \sqrt{2}} {1+(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}}$$
$$= \sqrt{2} ( \frac {1-(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}  } {1 + (p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}})$$ 
Since $ 0 <|\lambda_2 /\lambda_1| <1  $ we deduce that 
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My Solution:
Upon following the advice of Git Gud and Mark Bennet, I was able to find:
$$ a_n = \frac12 (1 +\sqrt{2}) \lambda_1^{n-1}- \frac12 (\sqrt{2} -1) \lambda_2^{n-1} $$
and 
$$ b_n = \frac {(1 +\sqrt{2}) \lambda_1^{n-1}  - (\sqrt{2}-1 ) \lambda_2^{n-1}}{2\sqrt{2}}$$
so $ \frac {a_n} {b_n}$ simplifies to:
$$ \frac {a_n} {b_n} = \frac {\sqrt2 ((1 + \sqrt2)\lambda_1^{n-1} - (\sqrt2 -1 )\lambda_2^{n-1})} { (1 +\sqrt{2}) \lambda_1^{n-1}  - (\sqrt{2}-1 ) \lambda_2^{n-1}}  $$


*

*How and why are they able to define such a coordinate system?

*Why does this coordinate system simplify the question?

*Is there any discrepancy between my solution and their solution? 

 A: Hint: From $\begin{bmatrix} a_{n+1} \\b_{n+1}  \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n}  \end{bmatrix}$ find $\begin{bmatrix} a_{n+1} \\b_{n+1}  \end{bmatrix} = A^k \begin{bmatrix} a_{0} \\b_{0}  \end{bmatrix}$, for a suitable $k\in \mathbb N$.
Edit: Mark has already answered 1. and 2. As for 3., your solution is wrong:$$b_n = \frac {(1 +\sqrt{2}) \lambda_1^{n-1} \color{red}- (\sqrt{2}-1 ) \lambda_2^{n-1}}{2\sqrt{2}}.$$
The red minus sign should be a plus.
Also note that $1+\sqrt 2=\lambda _1$ and $\lambda _2=1-\sqrt 2$, this allows you to simplify the final expression a lot.
A: Hint: From $P^{-1}AP=D$ you can write $A=PDP^{-1}$ and this should enable you to compute $A^n$ 


*

*this trick of diagonalising a matrix to compute powers more easily will appear more than once, so is worth working at until you understand it.



Notice that if we have the equation $v_{n+1}=Av_n$ where $v_n$ is the obvious vector, then $$v_{n}=PDP^{-1}v_{n-1}=PD^nP^{-1}v_0$$ so that $$(P^{-1}v_n)=D^n(P^{-1}v_0)$$ and if we define $P^{-1}v_n=p_n$ we have $p_n=D^np_0$
This is the same as saying that if we transform all our vectors by $P^{-1}$ we get a diagonal transformation - another way of looking at this is to say that we are choosing a basis of eigenvectors for the underlying vector space. As we apply our transformation more times, its behaviour is dominated by what happens with the largest eigenvalue - the others soon become relatively small.
The solution you quote shows how this works in detail.

You have made an error in your calculations because $a_n$ and $b_n$ are both integers, so the ratio can't be exactly $\sqrt 2$ - however you haven't shown your working, so I can't tell you what the error is.
A: Define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z)$; multiply the recurrences by $z^n$ and sum over $n \ge 0$. Recognize some sums to get:
\begin{align}
\frac{A(z) - a_0}{z} &= A(z) + 2 B(z) \\
\frac{B(z) - b-0}{z} &= A(z) + B(z)
\end{align}
Solve the resulting linear system:
\begin{align}
A(z) &= \frac{1 + z}{1 - 2 z - z^2} \\
B(z) &= \frac{1}{1 - 2 z - z^2}
\end{align}
The zeros of the denominator are $-1 \pm \sqrt{2}$, the traditional dance with partial fractions gets a bit ugly:
\begin{align}
A(z) &= \frac{2 + \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
          - \frac{2 - \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 - \sqrt{2}) z} \\
B(z) &= \frac{1 + \sqrt{2}}{2^{3/2}}  \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
          - \frac{1 - \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 - \sqrt{2}) z}
\end{align}
This is just a pair of geometric series for each, the first term dominates in each case. As you want:
$$
\lim_{n \to \infty} \frac{a_n}{b_n}
  = \frac{2 + \sqrt{2}}{1 + \sqrt{2}}
  = \frac{3 \sqrt{2} + 4}{3}
$$
