Repeated Iteration of a 2x2 matrix Suppose I am given a $2$x$2$ matrix $A=$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}
And an initial vector $x_n$ = \begin{pmatrix} x_0 \\ y_0\end{pmatrix}.
Under repeated iteration $x_{n+1} = Ax_n$, we find that $x_{n+1}$ converges to a line in the direction of the $eigenvector$  of $A$ of the form $X$= $[u,mu]$ where $m$ is the slope of the line.
I know that this process somehow also involves the largest $eigenvalue, \lambda,$ of $A$, though I am unsure how it connects.
My question is: How does the largest eigenvalue connect to this repeated iteration, how do I find the slope of the line of convergence $m$, and what is meant by the "direction of the eigenvector"?
 A: 1) Let me first answer what is "$\vec{a}$ is in direction of vector $\vec{b}$". It means
$\exists c\neq0\in \mathbb{R}$, that $\vec{a}=c\vec{b}$. It can also be written in this way
$$
\cos(\theta)=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|} =1
$$
It is simpler in 2-d case, which is that $m$ is the same for $\vec{a}$ and $\vec{b}$ if you express them as $\vec{x}=(u,mu)$.
For each eigenvector, the $m$ is different. suppose you know the eigenvector you want takes the form $(u_1,u_2)$, then it is also $(u_1, (\frac{u_2}{u_1}u_1)$, so your $m=u_2/u_1$.
2)Now let me prove the statement in n-dimension.
 Suppose $n\times n$ matrix $A$ has eigenvectors $\vec{e}_1,...,\vec{e}_n$ with eigenvalues $|\lambda_1|>|\lambda_2|>...|\lambda_n|$. So we know
$$
A\vec{e}_i=\lambda_i\vec{e}_i, \quad |\vec{e}_i|=1, \quad \vec{e}_i\cdot\vec{e}_j=0 \text{ when } i\neq j
$$
$\forall \vec{v} \in \mathbb{R}^n$ is a n-d vector, then $\exists \{a_n\} \in \mathbb{R}$ s.t. $\vec{v}=\sum_{i=1}^na_i\vec{e}_i$
Now we assume $a_1\neq 0$, then
$$
A\vec{v}=\sum_{i=1}^n a_i\lambda_i\vec{e}_i, \text{ so } A^k\vec{v}=\sum_{i=1}^n a_i\lambda_i^k\vec{e}_i
$$
$$
\lim_{k\to\infty}\frac{|\vec{e}_1||A^k\vec{v}|}{\vec{e}_1\cdot A^k\vec{v}}  
=\sqrt{\sum_{i=1}^n \left( \frac{a_i}{a_1}\lim_{k\to\infty}\frac{\lambda_i^k}{\lambda_1^k}\right)^2}
=\sqrt{\left(\frac{a_1}{a_1}\cdot 1\right)^2+0^2+...+0^2} =1
$$
So  $\lim_{k\to\infty}A^k \vec{v}$ is in the same direction as $\vec{e}_i$ so long as $\vec{v}\cdot\vec{e}_1\neq0$
