Let $A$ be a $2 \times 2$ matrix such that $$A\begin{pmatrix}1\\ 2\end{pmatrix} = \begin{pmatrix}1\\ 2\end{pmatrix}$$ $$A\begin{pmatrix}1\\ -1\end{pmatrix} = \begin{pmatrix}\frac{1}{2} \\ -\frac{1}{2}\end{pmatrix}$$ What is $A$? What is $\lim_{n\to \infty}A^nv$, where $v$ is an arbitrary vector $\begin{pmatrix}x\\ y\end{pmatrix}$, with $x, y > 0$? Justify your answer.
I know how to find $A$, all I had to do was augment the vectors above and solve for $A$...
$$A\begin{pmatrix} 1 & 1\\ 2 & -1\end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2}\\ -1 & -\frac{1}{2}\end{pmatrix}$$
$$A =\begin{pmatrix} 1 & \frac{1}{2}\\ -1 & -\frac{1}{2}\end{pmatrix}\begin{pmatrix} 1 & 1\\ 2 & -1\end{pmatrix} ^{-1}$$
and after finding the inverse and getting the product of the matrices above I get the following matrix :
$$A = \begin{pmatrix} \frac{2}{3} & \frac{1}{6}\\ \frac{1}{3} & \frac{5}{6}\end{pmatrix}$$
However, I am completely confused on obtaining a vector for $v$. Is $v$ supposed to be the eigenvectors for $A$?
Similar matrix: $$B=S^{-1}AS=\begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix}^{-1} \begin{pmatrix}\frac 23& \frac16 \\ \frac 13& \frac 56\end{pmatrix} \begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix}= \begin{pmatrix}1&0 \\ 0&\frac 12\end{pmatrix}.$$ Hence: $$A^n=SB^nS^{-1}=\begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix} \begin{pmatrix}1&0 \\ 0&\frac{1}{2^n}\end{pmatrix}\begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix}^{-1}$$ $$\lim_{n\to\infty}A^nv=\begin{pmatrix}\frac 13& \frac13 \\ \frac 23& \frac23\end{pmatrix}v.$$