So I have a linear equation $$ \begin{bmatrix} 1 & 2\\ 1 & 1 \end{bmatrix}\begin{bmatrix} a\\ b\\ \end{bmatrix} $$ where the first entry in the input vector represents the numerator and the second the denominator of some fraction, i.e. $$ \begin{bmatrix} a\\ b\\ \end{bmatrix} = \frac{a}{b}. $$
My initial input vector is $$ \begin{bmatrix} 1\\ 1\\ \end{bmatrix} = 1 $$ and I want to compute the limit of the resulting fraction as the applications of the matrix go to infinity. So $$ \left(\begin{bmatrix} 1 & 2\\ 1 & 1 \end{bmatrix}\right)^n \begin{bmatrix} 1\\ 1\\ \end{bmatrix} = \frac{a_n}{b_n} $$ as $n \to \infty$. The limit should be $\sqrt{2}$. I tried diagonalizing $A$ but I don't understand how to properly change the basis of the input vector so the result makes sense. Any help is greatly appreciated (I am somewhat of newbie to linear algebra so sorry if anything is unclear here!). For example,
$n = 1$ gives the vector $(3,2)$ corresponding to $\frac{3}{2}$ and so on.
A possibility to get the limit without diagonalizing the matrix:
We have $$ \frac{a_{n+1}}{b_{n+1}} = \frac{a_n+2b_n}{a_n+b_n} = 1+\frac{b_n}{a_n+b_n} =1+\frac{1}{\frac{a_n+b_n}{b_n}} = 1+\frac{1}{1+\frac{a_n}{b_n}} $$ So if we set $x_n=\frac{a_n}{b_n}$, we get $$ x_{n+1} = 1+\frac{1}{1+x_n} $$ If $x=\lim\limits_{n\to\infty}x_n$ exists, then $$ \lim\limits_{n\to\infty}x_n = \lim\limits_{n\to\infty}x_{n+1} = \lim\limits_{n\to\infty}\left(1+\frac{1}{1+x_n}\right) = 1+\frac{1}{1+\lim\limits_{n\to\infty}x_n} $$ which means $x=1+\frac{1}{1+x},$ which in turn means $x=\sqrt{2}.$ We cannot get negative values if we set $x_0=1.$
In order to show that the limit exists, we can argue that the $x_n$ are the convergents of the regular continued fraction $[1;2,2,2,2,\ldots].$ Regular continued fractions are known to converge.
Alternatively, we can use the Banach fixed-point theorem.