Given eigenvectors and eigenvalues of a matrix $A$, how to find its inverse and the limit of $A^nv$?

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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 $A$?

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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: $$A^n=SB^nS^{-1}=\begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix} \begin{pmatrix}1&0 \\ 0&\frac{1}{2^n}\end{pmatrix}\begin{pmatrix}1& \frac12 \\ 2&-\frac 12\end{pmatrix}^{-1}$$ $$\lim_{n\to\infty}A^nv=\begin{pmatrix}\frac 13& \frac13 \\ \frac 23& \frac23\end{pmatrix}v.$$

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$v$ is not an eigenvector in general, but you may express it as a linear combination of two eigenvectors. (Actually, the problem already gives you all eigenvectors with eigenvalues!) Now try to investigate what happens.

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I would recommend you diagonalising the matrix, from there it’s gonna follow easily. And any term can be written out using eigenvalues.