Let $a,\, b,\, c,\, d, \in \mathbb{R}$. The $2\times2$ matrix acts on the pair of integers $(m_{i },\,n_{i})$ and gives a new pair of integers $(m_{j},n_{j})$:
$$\begin{align} \left[ \begin{array}{c} m_{j} \\ n_{j} \end{array} \right] = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \left[ \begin{array}{c} m_{i} \\ n_{i} \end{array} \right] \end{align} $$
The determinant is assumed to be non-vanishing. The indices, $i$ and $j$ label the pairs: $i,j=1,2,3...$
Now suppose that for all values of $i$ and $j$, the ratios are bounded and satisfy: $\epsilon \geq m_{i}/n_{i} > 1$, $\epsilon \geq m_{j}/n_{j}>1$, where $\epsilon$ is some rational number.
Given this, what can be inferred about the $2\times2$ matrix above ? Can we put any constraints on its trace, determinant etc ?
I'm going to rewrite your question slightly... You don't need to make it look as if you have a sequence of points.
Let $a,\, b,\, c,\, d, \in \mathbb{R}$. The $2\times2$ matrix acts on the pair of integers $(m,n)$ and gives a new pair of integers $(m',n')$:
$$\begin{align} \left[ \begin{array}{c} m' \\ n' \end{array} \right] = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \left[ \begin{array}{c} m \\ n \end{array} \right] \end{align} $$
The determinant is assumed to be non-vanishing.
Now suppose that for all pairs, the ratios are bounded and satisfy: $1 < m'/n' \le k$, provided we have the initial condition $1 < m/n \le k$, where $k$ is some rational number. [I don't like using $\epsilon$ for anything larger than $1$!]
Given this, what can be inferred about the $2\times2$ matrix above ? Can we put any constraints on its trace, determinant etc ?
Geometrically, you are starting with points between the lines $y=x$ and $y={x \over k}$ and you want to end up with points in the same area. You are not constraining the behaviour for points outside that area...
The initial constraint $1<m/n \le k$ can be rewritten as $m>n$ and $m \le kn$.
The constraint $1<m'/n' \le k$ can be rewritten as $m'>n'$ and $m' \le kn'$.
$m'>n' \Rightarrow am+bn>cm+dn$
$\Rightarrow am-cm>dn-bn$
$\Rightarrow (a-c)m>(d-b)n$
$\Rightarrow (a-c)>(d-b)$
$m' \le kn' \Rightarrow am+bn \le kcm+kdn$
$\Rightarrow am-kcm \le kdn-bn$
$\Rightarrow (a-kc)m \le (kd-b)n$
$\Rightarrow k(a-kc) \le (kd-b)$