*$E \subset {\mathbb R}^2$ is an ellipse of area $1$ centered at the origin that contains no other point with integer coordinates. Is there a matrix $A \in SL(2,{\mathbb Z})$ such that the ellipse $A(E)$ is contained in a disc of radius 10?*


Hopefully, this is really easy and it is only my ignorance in "reduction theory" (or other similar techniques) that is at the heart of my difficulties.


More generally, given an ellipsoid $E$ of unit volume in ${\mathbb R}^n$ centered at the origin and containing no other point with integer coordinates, I'm interested in a good upper estimate of the size of the ball in which I can enclose $A(E)$ for some matrix $A \in SL(n,{\mathbb Z})$. 



Ellipsoids and lattices: an enclosure problem.
Yes, such $A$ exists, and the radius can be much smaller than $10$; indeed radius $2^{1/2}$ suffices regardless of the area of $E$.


Write the ellipse $A(E)$ as $ax^2+bxy+cy^2 \leq 1$. By reduction theory of binary quadratic forms, we can choose the ${\rm SL}\_2({\bf Z})$ transformation $A$ to obtain coefficients satisfying $|b| \leq a \leq c$. But $a \geq 1$ because $(x,y)=(1,0)$ is not in the interior of the ellipse. Hence
$$
Q(x,y) \geq a(x^2 - |xy| + y^2) \geq x^2 - |xy| + y^2 \geq \frac12(x^2+y^2)
$$
for all $x,y \in \bf R$. Therefore $A(E)$ is contained in the disc $\frac12(x^2+y^2) \leq 1$ of radius $2^{1/2}$, **QED**.


Equality holds only for $a=|b|=c=1$, in which case the ellipse has area $2\pi/\sqrt{3}$; for an ellipse of area $1$ one can improve the bound slightly using the condition $4ac-b^2=4\pi$, but the best $r$ is still somewhat larger than $1$.


There exist similar bounds in higher dimension $n$, but naturally it gets harder to compute or estimate them as $n$ grows, especially once $n>8$ and the shape of the cone of (say) Minkowski-reduced forms isn't known. (NB Minkowski reduction must give *some* bound for each $n$, but probably not the best bound once we go far enough beyond $n=2$.)


