Cardinality of the set in $\mathbb{R^2}$ I am trying to understand the following exercise:
Let $v_1$ and $v_2$ be non-collinear vectors of $\mathbb{R}^2$. Estimate the cardinality of the set $(m,n) \in \mathbb{Z}^2$ such that $\|mv_1+nv_2\|_\infty \leq r$, as $r \rightarrow \infty$.
I don't really understand the question if someone could enlighten me please.
Thank you in advance.
 A: As $v_1,v_2$ are linearly independent, there exists a matrix $A\in\mathbb R^{2\times 2}$, such that
$$
Av_i=e_i, \quad i=1,2.
$$
Now, 
$$
\|m_1v_1+m_2v_2\|\le r\quad\Longleftrightarrow\quad \|A^{-1}(m_1e_1+m_2e_2)\|_\infty\le r,
$$
which holds iff
$$
m_1e_1+m_2e_2 \in A\big( [-r,r]\times[-r,r]\big).
$$
But $R=A \big([-r,r]\times[-r,r]\big)$ is a parallelogram of area $4r^2|\det A|$. In fact
$$
|\det A|=\frac{1}{|\det \left(v_1,\,v_2\right)|},
$$
as $A$ is the inverse of the matrix with columns $v_1$ and $v_2$.
Hence the number of such pairs is asymptotically (as $r\to\infty$) equal to $4r^2|\det A|=\dfrac{4r^2}{|\det(v_1,v_2)|}$.
A: Maybe it's worth a bit of intuitive explanation:
The points (heads of the vectors) of the set $\{mv_1+nv_2:m,n\inℤ\}$ is a lattice in $ℝ^2$, and the region $\{(x,y):\|(x,y)\|_∞<r$ is a straight square of size $2r$ centered at the origin.
For $r$ really really big, the area of each lattice cell will be very small compared to the area of the square, so you can approximate the number of lattice endpoints with the number of lattice cells that would take to form the square. The number of points that will make it wrong is small, as it's proportional to the perimeter of the square, while the number of cells is proportional to the area of the square.
Now, the area of a cell is given by the absolute value of the exterior product of the vectors: $|v_1\wedge v_2|$, that is calculated (say $v_1=(x_1,y_1)$ and $v_2=(x_2,y_2)$) as $a=|x_1y_2-x_2y_1|$. This is the ratio of growth, so the number of cells in your square of size $2r$ is $4r^2/a$.
