The title to my question is a mouthful, but hopefully the following is clear. I am reading an article called on toral automorphisms and the authors have included a bunch of known properties with proofs to be more self-contained. One of these property proofs is the following:
Let $A \in \mathcal{GL}(2, \mathbb{Z})$. Then, assuming that $A^n - I$ is invertible, the number of solutions of $A^nx = x$ in $[0,1)^2$ is $\left|\mathcal{det}(A^n - I)\right|$.
The given proof is as follows:
Suppose that $A^n - I$ is invertible. Then note that the number of solutions to the equation $A^nx = x$ in $[0, 1)^2$ is equal to the number of integer points in the image of $[0, 1)^2$ under $A^n - I$, when $A^n - I$ is lifted to a linear map from $\mathbb{R}^2$ to $\mathbb{R}^2$. Note also that the image of $[0, 1)^2$ under $A^n - I$ is a parallelogram and hence the number of integer points in it is equal to its area, which is equal to $\left|\mathcal{det}(A^n - I)\right|$ by the following lemma.
(the following lemma)
If $T^2:\mathbb{R}^2\to \mathbb{R}^2$ is an isomorphism then for every Riemann measurable set $S\subset \mathbb{R}^2$, $T(S)$ is Riemann measurable and $\mathcal{area}(T(s)) = |\mathcal{det}(T)|\mathcal{area}(S)$
What I really don't understand/have no intuition for is why the number of integer points inside the parallelogram spanned by $A^n - I$ is equal to the area of the said parallelogram.