
4
value by r!mr. A r×rsquare matrix is said to be non-singular if det(M) is nonzero. Notice
that in this case there exists a unique r×rmatrix M−1such that M−1M=M M−1=Ir
where Iris the identity r×rsquare matrix, i.e. the matrix with zero coefficients except on
the main diagonal where coefficients are one. This matrix can be computed by introducing
the comatrice. Let us recall that the comatrix of any r×rsquare matrix Mis the r×r
square rational matrix com(M) satisfying (com(M))ij is det(Mij ) where Mij is the matrix
obtained from Mby replacing the jth column by a column of zeros, except on the ith line in
which we put a one. Let us recall that M⊤com(M) = det(M)Irwhere M⊤is the transpose
of M. It follows that if Mis non-singular then M−1=1
det(M)com(M)⊤. Notice that if the
coefficients of Mare integers bounded in absolute value by some m∈N, then the coefficients
of com(M) are integers bounded in absolute value by (r−1)!mr−1.
Let Mbe a r×kmatrix. The rank of Mis the maximal n∈ {0,...,min{r, k}} such
that Madmits a n×nnon-singular sub-matrix. Such a matrix Mis said to be full row
rank if its rank is equal to r. We say that such a matrix Mhas an Hermite normal form if
there exists a uni-modular matrix U(i.e. a matrix of integers with a +1 or −1 determinant)
such that M U = [H0] where His a non-singular, lower triangular, non-negative matrix, in
which each row has a unique maximum entry, which is located on the main diagonal of H.
Let us recall that every full row rank matrix Mhas a unique Hermite normal form [H0]
and moreover the uni-modular matrix Usuch that M U = [H0] is unique [21, Corollary
4.3b]. Additionally, if Mis an integral matrix then His a matrix of natural numbers.
Let us recall from [21, Section 10] that from a complexity point of view, the matrices U
and Hare computable in polynomial time and moreover det(H) divides the determinant of
any r×rnon-singular square sub-matrix of M. In particular, if the coefficients of Mare
integers bounded in absolute value by some m∈N, then det(H), which is the product of
the diagonal elements of His bounded by r!mr. Let us recall that (com(H))ax = det(Hij )
where Hij is the matrix obtained from Mby replacing the jth column by a column of
zeros, except on the ith line in which we put a one. Now, let σbe a permutation of
{1,...,r}and observe that (Hij )kσ(k)≤Hkk for every k∈ {1, . . . , r}. It follows that
Qr
k=1(Hij )kσ(k)≤Qr
k=1 Hkk =δ(H). Therefore the coefficients of com(H) are integers
bounded in absolute value by r! det(H)≤(r!)2mr.
Now, let us prove Theorem 1. We consider a lattice Lspanned by a sequence v1,...,vk
of vectors in {−m, . . . , m}d. We introduce the d×kmatrix Lobtained from this sequence
by considering vjas the jth column of L, i.e. Li,j =vj(i) for every 1 ≤i≤dand
1≤j≤k. We denote by rthe rank of L. Recall that r≤min{d, k}. By reordering columns
of L(which corresponds to a permutation of v1,...,vk) and by reordering the lines of L,
(which corresponds to a permutation of the components of L) we can assume without loss
of generality that Mcan be decomposed as follows where Ais a r×rnon-singular matrix,
Bis a (d−r)×rmatrix, A′is a r×(k−r) matrix, and B′is a (d−r)×(k−r) matrix.
L=A A′
B B′
Let us introduce the r×kmatrix Mdef
= [AA′]. Notice that Mis full row rank. It follows
that there exists a uni-modular matrix Usuch that M U is in Hermite normal form [H0].
We are ready for proving the following lemma.
◮Lemma 2. The lattice Lis the set of vectors x∈Zdsuch that:
(i) det(H)divides the coefficients of [x(1) ...x(r)] com(H), and