Why does the set of integer vectors in the nullspace of $\mathbf{A}$ constitute a lattice?

Question

I can almost see that this is true for the following reason:

Let $\{\mathbf{x}_i\}$ be the set of all integer vectors in the nullspace of $\mathbf{A}$. Then any finite linear combination with integer coefficients of these vectors is again an integer vector in the nullspace of $\mathbf{A}$. This is almost a lattice, except I'm using too many "basis vectors". And I don't see how I can trim the number of $\mathbf{x}_i$ to just the dimension of the nullspace of $\mathbf{A}$.

Answer 1

Following the comments, a highly general result is as follows:

Theorem. Let $n\gt 0$ be a positive integer, and let $H$ be a subgroup of $\mathbb{Z}^n$. Then there exists a basis $a_1,\ldots,a_n$ of $\mathbb{Z}^n$, an integer $d$ with $0\leqslant d\leqslant n$, and positive integers $m_1,\ldots,m_d$ such that $m_1|m_2$, $m_2|m_3,\ldots,m_{d-1}|m_d$ and such that $m_1a_1,\ldots,m_da_d$ is a basis for $H$. In particular, $H$ is free and finitely generated.

In particular, $H$ is isomorphic to $\mathbb Z^d$.
You can see a proof this in this previous answer, with an explanation of an unclear part in this other answer.

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While the idea that $H\sim \mathbb Z^d$ for some $d$ sounds fairly intuitive, that one can produce a basis with the divisibility property is not so obvious. You might have heard of the Smith normal form for matrices, which provides an algorithm for finding exemplars of the theorem given generators for $H$.