A General Question on Powers of Matrices There is an example on positive matrices in Artin's algebra (page 113, $2^{nd}$ edition). The matrix is $A=\begin{bmatrix} 3&2\\ 1&4\end{bmatrix}$. The argument is that there must exist a positive eigen-vector for this matrix. Instead of using the standard way to show this claim, Artin argues that this matrix will send the first quadrant $S$ to itself. Therefore, we can get a nested inclusion statement like $\cdots\subset A^2S\subset AS\subset S$. Then it is intuitive to notice that the intersection of this sequence is either the smallest sector or simply a half line. By definition, it is not hard to see one eigen-vector of $A$ is $(1, 1)^T$ and alike. Alternatively, we could use limit argument to show that the limit of $A^nS$ is simply a line. In order to do this, we need to show that the limit of $A^ne_1$ and the limit of $A^ne_2$, where $e_i$ is the $i^{th}$ standard basis vector, are the same. However, it is not clear to me how to find these limits easily since the matrix $A$ is not of special pattern and its power is not easy to be expressed in some simple formula. However, I managed to do this in R as the following graph shows.  Now, my question is whether there is a simple way to deal with powers of a general matrix so that we can consider its limiting behavior. One thing I can think of is to diagonalize a matrix by its eigen-basis. But in terms of the current question, we need eigen-vectors in the first place. 
 A: Here is a simple way to characterize any power of a $2 \times 2$ matrix. Any $2 \times 2$ matrix satisfies its characteristic polynomial. Thus if 
$$ A = \begin{pmatrix}a & b\cr c & d\end{pmatrix}$$
then its characteristic polynomial is 
$$ \Delta(t) = \det  \left(  \begin{pmatrix}a & b\cr c & d\end{pmatrix} \right)=
{t}^{2}-(a+t) \,t+a\,d-b\,c$$
Hence by Cayley-Hamilton theorem we have
$$ A^2 - (a+d) A + (a\,d-b\,c) I = 0$$ or
$$A^2  = \theta A + \mu I$$ where $\theta = (a+d)$, $\mu = -( a\,d-b\,c)$.
The above condition can be used to write any power of $A$ in terms of $A$ and $I$.
There are two ways to proceed:
Method 1: Divide $t^n$ by $\Delta(t)$ and let $R(t) = \alpha_n t + \beta_n$ be the reminder (easily obtained if we know the roots of $\Delta(t)$) Then 
$$
A^n = \alpha_n A + \beta_n I
$$
Method 2: This uses the same idea but uses recursion and induction. Let
$$
A^n = \alpha_n A + \beta_n I
$$
This is clearly valid for $n=0$ with $\alpha_0=0$, $\beta_0 = 1$.
The induction/recursion step is
$$
A^{n+1} = A\, A^n = \alpha_n A^2 + \beta_n A = \alpha_n (\theta A + \mu I) + \beta_n A
= (\beta_n +\alpha_n \theta) A + \alpha_n \mu I
$$
Hence the recursive formula is
$$
\begin{align}
\alpha_{n+1} &= \theta\, \alpha_n + \beta_n \\
\beta_{n+1} &= \mu \,\alpha_n
\end{align}
$$
To be honest, the second method coverts the problem to powers of a matrix with a special structure. The above condition can be written as
$$
\begin{pmatrix}\alpha_{n+1}\cr \beta_{n+1}\end{pmatrix} = 
\begin{pmatrix}\theta & 1 \cr \mu & 0\end{pmatrix}  
\begin{pmatrix}\alpha_{n}\cr \beta_{n}\end{pmatrix}  
$$
and hence
$$
\begin{pmatrix}\alpha_{n}\cr \beta_{n}\end{pmatrix} = 
\begin{pmatrix}\theta & 1 \cr \mu & 0\end{pmatrix} ^n  
\begin{pmatrix}\alpha_{0}\cr \beta_{0}\end{pmatrix}  
$$
