Lattice of integers $\mathbf{Z}$ in $\mathbb{R^2}$

Question

Lattice of integers $\mathbf{Z}$ in $\mathbb{R^2}$

The questions: Give an example of a nonempty subset of $\mathbb{R^2}$ (noted $M$) which is closed under addition and for all $m\in M$ we have $-m\in M$, but which isn't closed under scalar multiplication over $\mathbb{R^2}$ (in other words it isn't a subspace of $\mathbb{R^2}$)

After one month of thinking about this problem from time to time, I think I found the solution. It's maybe the pairs $(x,y)$ such tht $x,y\in\mathbf{Z}$

But also $(x,y)$ such that $x,y\in\mathbf{Q}$

Is THIS TRUE?

Answer 1

Yes, both of them are correct, provided you want them to be a subspace of a real vector space. If $a,b,c,d$ are integers then $(a,b)+(c,d)=(a+c,b+d)$ is also an integer. So we have closure under addition. Now for the other condition, if $(a,b)\in\mathbb{Z}^2$ then $(-a,-b)\in\mathbb{Z}^2$. However if you take a scalar from $\mathbb{R}$ say $\pi^2$, then for instance $\pi^2(1,1)=(\pi^2,\pi^2)\notin\mathbb{Z}^2$.

Intuitively speaking, $\mathbb{Z}^2$ are all of those vectors that go from the origin to those lattice points. (see some examples I've drawn in the diagram below) Now if a certain vector $\vec{v}$ goes from the origin $(0,0)$ to $(2,1)$, then it may not be the case that $\lambda\vec{v}$ will reach another point of the lattice. So it is certainly not closed under scalar multiplication.

$\,$

Answer 2

Take any $2\times2$ non-singular matrix $A$ with real entries. Regard the lattice point $(a\ b)\in\mathbf{Z}\times \mathbf{Z}$ as a vector $v$. Now take all the transformed vectors $vA$ (matrix multiplication of row vector $v$ by $A$) as $v$ varies over lattice of integers. This is also another example; choosing $A$ to be diagonal matrix with entries, say $\sqrt2$ and $\pi$ will give example of a Lattice of 'non-integer' points (excepting the origin).

In fact, every example you want arises this way (possibly allowing $A$ to be singular matrix giving rise to degenerate examples).