# Finding a basis for the solution space of a system of Diophantine equations

Let $m$, $n$, and $q$ be positive integers, with $m \ge n$. Let $\mathbf{A} \in \mathbb{Z}^{n \times m}_q$ be a matrix.

Consider the following set: $S = \big\{ \mathbf{y} \in \mathbb{Z}^m \mid \mathbf{Ay} \equiv \mathbf{0} \pmod q \big\}$.

It can be easily shown that $S$ constitutes a lattice, because it is a discrete additive subgroup of $\mathbb{R}^m$.

I want to find the basis of this lattice. In other words, I want to find a matrix $\mathbf{B} \in \mathbb{Z}^{m \times m}$, such that the following holds: $S = \{\mathbf{Bx} \mid \mathbf{x} \in \mathbb{Z}^m \}$.

Let me give some examples:

1. $q=2$, and $\mathbf{A} = [1,1]$ $\quad \xrightarrow{\qquad}\quad$ $\mathbf{B} = \left[ \begin{array}{cc} 2&1 \\ 0&1 \end{array} \right]$

2. $q=3$, and $\mathbf{A} = \left[ \begin{array}{ccc} 0&1&2 \\ 2&0&1 \end{array} \right]$ $\quad \xrightarrow{\qquad}\quad$ $\mathbf{B} = \left[ \begin{array}{ccc} 3&0&1 \\ 0&3&1 \\ 0&0&1 \end{array} \right]$

3. $q=4$, and $\mathbf{A} = \left[ \begin{array}{ccc} 0&2&3 \\ 3&1&2 \end{array} \right]$ $\quad \xrightarrow{\qquad}\quad$ $\mathbf{B} = \left[ \begin{array}{ccc} 4&2&1 \\ 0&2&1 \\ 0&0&2 \end{array} \right]$

Note that in all cases, $\mathbf{AB} =\mathbf{0} \pmod q$. However, $\mathbf{B}$ is not an arbitrary solution to this equivalence, since it must span $S$. For instance, in the example 1 above, matrix $\mathbf{\hat B} = \left[ \begin{array}{cc} 2&0\\ 0&2 \end{array} \right]$ satisfies $\mathbf{A \hat B} =\mathbf{0} \pmod 2$, but generates a sub-lattice of $S$.

Also note that if $\mathbf{B}$ is a basis of $S$, any other basis $\mathbf{\bar B}$ is also a basis of $S$, provided that there exists a unimodular matrix $\mathbf{U}$, for which $\mathbf{\bar B} = \mathbf{BU}$.

My questions:

1. How to obtain $\mathbf{B}$ from $\mathbf{A}$ and $q$?
2. Is it possible that $\mathbf{B}$ is not full rank, i.e. $\text{Rank}(\mathbf{B})< m$?
3. Is there any difference between the case where $q$ is prime and the case where it is composite?

Side note: As far as I understood, $S$ is the solution space of a system of linear Diophantine equations. The solution has something to do with Hermite normal forms, but I can't figure out how.

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There should be a way to work this out with ordinary linear algebra when $q$ is prime, as you are dealing with a field, every nonzero number has a multiplicative inverse. Other than that, i will need to think about it. The one article that comes to mind is George Leo Watson, Transformations of a Quadratic Form which do not increase the Class-Number, Proc. London Math. Soc. vol. 12 (1962) pages 577-587. Formulas 2.4 and 2.5, then Lemma 2 (ii) are related to your questions but may not be enough. Alright, I wrote "Smith Normal Form" at a few points on my copy, perhaps you should look that up. –  Will Jagy Aug 16 '12 at 23:13
@Will: Thanks for the answer. I found this paper: explicit hard instances of the shortest vector problem, which is trying to solve the same problem. Specially, section 4 describes how B is obtained from A (called Y and X in the paper). But I'm totally confused by their construction, since I can't apply it to any of my numerical examples. Could you please check that out? –  Sadeq Dousti Aug 17 '12 at 13:18
@Will: Moreover, if you have any strategy to obtain B from A, given q is a prime, please write it out as an answer. It will be invaluable to me! –  Sadeq Dousti Aug 17 '12 at 13:27
When you say that $B$ is not an arbitrary solution to $AB \equiv 0 \pmod{q},$ and then show an example of $\hat{B},$ did you notice that the example diag(2, 2) is $\hat{B} \equiv 0 \pmod{q}$? Does all non-zero (modulo $q$) solutions generate only a sub-lattice of S? –  user2468 Aug 17 '12 at 22:09
@Jennifer: Thanks for the answer. I checked your idea, but it apparently things are more complicated. Consider, for instance, the matrix $\mathbf{B'} = \left[ \begin{array}{cc} 3&2\\ 3&4 \end{array} \right]$. It is a solution for $\mathbf{AB'} \equiv 0 \pmod{2}$, and it's non-zero modulo 2. However, it generates a sub-lattice of $S$. (For example, the point $(1,1)$ is in the lattice $S$, but cannot be generated by $\mathbf{B'}$.) –  Sadeq Dousti Aug 17 '12 at 23:18

Well, here is how it works over a field. We take $q=5.$ We will start with $A$ as a 2 by 4, $$A \; = \; \left( \begin{array}{cccc} 2 & 3 & 4 & 1 \\ 3 & 4 & 0 & 1 \end{array} \right)$$ We begin a sequence of elementary row operations, first multiply the first row times 3 and the second by 2, $$\left( \begin{array}{cccc} 1 & 4 & 2 & 3 \\ 1 & 3 & 0 & 2 \end{array} \right)$$ Next subtract row 1 from row 2. $$\left( \begin{array}{cccc} 1 & 4 & 2 & 3 \\ 0 & 4 & 3 & 4 \end{array} \right)$$ then multiply the second row by 4, $$\left( \begin{array}{cccc} 1 & 4 & 2 & 3 \\ 0 & 1 & 2 & 1 \end{array} \right)$$ Finally add row 2 to row 1, $$\left( \begin{array}{cccc} 1 & 0 & 4 & 4 \\ 0 & 1 & 2 & 1 \end{array} \right).$$ This is the most favorable case. It allows us to place a little 2 by 2 identity matrix at the bottom when writing the null space we need $$\left( \begin{array}{cc} ? & ? \\ ? & ? \\ 1 & 0 \\ 0 & 1 \end{array} \right)$$ which is then forced to become $$\left( \begin{array}{cc} 1 & 1 \\ 3 & 4 \\ 1 & 0 \\ 0 & 1 \end{array} \right)$$ This can be readily filled in the way Buchmann, Lindner, Ruckert, Schneider demand at the bottom of page 2, $$B \; = \; \left( \begin{array}{cccc} 5 & 0 & 1 & 1 \\ 0 & 5 & 3 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right).$$ You can check that $AB \equiv 0 \pmod 5$ as the matrix of appropriate size.

What happens instead if the row echelon form comes out with staggered nontrivial 1's? Let us begin again with $$A \; = \; \left( \begin{array}{cccc} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 1 \end{array} \right)$$

It is first necessary to stagger the little 2 by 2 identity matrix in the same way, as in $$\left( \begin{array}{cc} ? & ? \\ 1 & 0 \\ ? & ? \\ 0 & 1 \end{array} \right)$$ and forces us to the penultimate $$\left( \begin{array}{cc} 3 & 2 \\ 1 & 0 \\ 0 & 4 \\ 0 & 1 \end{array} \right).$$ Note that it is impossible to just place this 4 by 2 as the final two columns of a 4 by 4, BLRS demand nonzero entries on the main diagonal. So what we do is simply stagger the columns with 5's in the same way, giving $$B \; = \; \left( \begin{array}{cccc} 5 & 3 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & 4 \\ 0 & 0 & 0 & 1 \end{array} \right).$$

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Even over a field, a fair amount goes into this. Here are two pages from Linear Algebra and Matrix Theory by Evar D. Nering, second edition:

From a row-echelon form for your data matrix $A,$ one can readily find the null space as a certain number of columns by placing $1$'s in certain "free" positions and back-substituting to get the other positions.

Meanwhile, once you have that information, the square matrices you display above are not quite what Buchmann, Lindner, Ruckert, and Schneider want. At the bottom of page 2 they define their HNF for $B$ as being upper triangular as well, where the first several columns have merely a single $q$ on the diagonal element and otherwise $0$'s. So you were close.

Note that, over a field ($q$ prime) there is a well-defined notion of the rank of $A.$ The number of non-trivial columns of $B$ is $m-n,$ as you have illustrated above.

Anyway, I think everything about your problem is in ordinary linear algebra books for fields, then I recommend INTEGRAL MATRICES by Morris Newman.

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> "the square matrices you display above are not quite what Buchmann, Lindner, Ruckert, and Schneider want." I converted my B matrices to HNF. Still, I don't see how Buchmann et al. algorithm can obtain these matrices. –  Sadeq Dousti Aug 18 '12 at 1:04
BTW, Buchmann and Vollmer have a book called " Binary Quadratic Forms: An Algorithmic Approach." I'm reading page 203, which seems quite related... –  Sadeq Dousti Aug 18 '12 at 2:09