lattice orthogonal complement

Question

Let $A\in \mathbb{Z}^{m\times n}$ ($m<n$) be a matrix with orthogonal rows. Further assume that the gcd of the coefficients in each row of $A$ is $1$.

Consider $\ker A\cap \mathbb{Z}^n = \{x\in\mathbb{Z}^n: Ax = 0\}$. How is $\det(\ker A\cap \mathbb{Z}^n)$ related to $A$? I tried a few small examples and it seems that $\det(\ker A\cap \mathbb{Z}^n)\cdot\det(A\mathbb{Z}^n) = \sqrt{\det(AA^T)}$. Is this generally true?

($\det(A\mathbb{Z}^n)$ is the determinant of $A\mathbb{Z}^n = \text{im} A \cap \mathbb{Z}^m$, an $m$-dimensional sublattice in $\mathbb{Z}^m$)

Answer 1

This does not have much to do with integrality (except that you want the "maximal" lattice), and, while classical (probably in Siegel, though I can't find it right now), it is discussed in my paper Surface area and other measures of ellipsoids. (English summary) Adv. in Appl. Math. 39 (2007), no. 4, 409–427 (also on arxiv.org) section 10 - it follows from fairly simple exterior algebra.