On existence of a nonzero integer vector $x$ such that $\|Bx\|_1\leq \sqrt[n]{n!\,|\det(B)|}$

Solution 1:

This follows from Minkowski's theorem: https://en.wikipedia.org/wiki/Geometry_of_numbers

Let $r > 0$ and $K_r$ be the compact set $\{x \in \mathbb R^n : \lVert B x \rVert_1 \leq r\}$. We have $\operatorname{Vol}(K_r) = \frac{(2r)^n}{n! \cdot\lvert\det(B)\rvert}$.

Because $K_r$ is closed, by Minkowski's theorem it contains a non-zero integer vector as soon as $\operatorname{Vol}(K_r) \geq 2^n$. That is, as soon as $$r \geq \sqrt[n]{n! \cdot \lvert \det(B) \rvert} \,.$$

To compute $\operatorname{Vol}(K_r)$, it suffices to do this when $r = 1$ and $B$ is the identity matrix. Then you want to compute $2^n$ times the volume of the set $$\{(x_i) \in [0, 1]^n : x_1 + \cdots + x_n \leq 1\} \,.$$ Using Fubini's theorem you can show by induction that it is $\frac1{n!}$.