TITLE: Lattice generated by vectors orthogonal to an integer vector
QUESTION [9 upvotes]: Given a non-zero vector $\boldsymbol{v}$ composed of integers, imagine the set of all non-zero integer vectors $\boldsymbol{u}$, such that $\boldsymbol{u} \cdot \boldsymbol{v} = 0$, i.e., the integer vectors orthogonal the original vector.  The set $S = \{\boldsymbol{u} : \boldsymbol{u} \cdot \boldsymbol{v} = 0\}$ seems to form a $dim(\boldsymbol{v})-1$ dimensional lattice.  Specifically, it's clear that for any two elements of $S$, their linear combination is also in $S$.  However, because S is a subset of the lattice of all integer vectors, it's also a lattice. It there a name for this lattice?  Additionally, how can it's basis vectors be computed?

REPLY [2 votes]: I know that is an old question and already have a beautiful answer given by Florian Bourse, but I am adding the following just for future references.
That lattice has a name: it is called orthogonal lattice. In general, we define it regarding a set of vectors $a_1, ..., a_d \in \mathbb{Z}^n$ as follows:
$$L^\bot := \left\{ u\in \mathbb{Z}^n : \langle a_i, u \rangle = 0 \text{ for }  1 \le i \le d \right\}.$$
Or equivalently we define the lattice $L$ generated by the integer linear combinations of $a_i$'s and $L^\bot$ is then the lattice whose vectors are orthogonal to all the vectors of $L$.
Your case is just a particular one where $d=1$. And you are right about the dimension, it is $n - d$.
There are a lot of other known properties of these lattices (for instance, $\det(L^\bot) \le \det(L)$ and $(L^\bot)^\bot = span(L)\cap\mathbb{Z}^n$).
Section 2 of  Merkle-Hellman Revisited: A Cryptanalysis of the Qu-Vanstone by Phong Nguyen and Jacques Stern present some discussion about this type of lattice and a polynomial-time algorithm to compute a LLL-reduced basis to them, which I've implemented in SAGE and published in this Github repository.