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Let $B=(b_1,b_2)$ be the linearly independent vectors and generate a lattice $ L(B)=\{xb_1+yb_2:x,y \in \mathbb{Z}\}$. If any two linearly independent vectors, $b_1^\prime,b_2^\prime$ are taken from the lattice $L(B)$, then $L(B^\prime)$ need not be equal to $L(B)$. For example,

$b_1=[1,2],b_2=[1,-1]$ and generate the lattice $L(b_1,b_2)$ Take two lattice vectors $b_1^\prime=b_1+b_2,b_2^\prime=b_1-b_2$.

Clearly, these are linearly independent but it doesn't form a basis of lattice $L(b_1,b_2)$($b_1$ cannot be generated using $b_1^\prime,b_2^\prime$). This clearly indicates that any set of $n$($n=2$ in the example) independent lattice vectors may not be the basis of the lattice.

Given a lattice vectors, how can we generate a basis of the lattice?

qiuxicj
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preethi
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2 Answers2

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There is a general answer for modules over any commutative ring:

Let $L$ be a finitely generated free $R$-module with basis $\mathcal B=(b_1,\dots b_n)$, and $b'_1, \dots, b'_n$ be $ n$ vectors in $L$. Then $b'_1, \dots, b'_n$ are a basis of $L$ if and only if $\;\det_\mathcal{B}(b'_1, \dots, b'_n)$ is a unit in $R$.

In the specific case, as a lattice is just a free $\mathbf Z$-module of rank $2$, this means $$\det\nolimits_{\{b_1,b_2\}}(b'_1,b'_2)=\pm 1.$$

Bernard
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Let $\mathbb{R}^n$ be the n-dimensional real vector space. A lattice is a discrete, additive subgroup $L \subset \mathbb{R}^n$.By theorem 7.1 in GTM211, L is a freedom Z-module, its dimension m is less than or equal to n.when m = 1, there exist a non-zero,shortest vector $b_1$ in L, $b_1$ is the basis of L.

There exist shortest linear independent vectors ${b_i},i=1,...,m$.They can be choosen in this way: first find the shorest vector in L, saying $b_1$; then, find the shortest vector in $L\backslash L(b_1)$($L(b_1)$ is the integer linear expansion of $b_1$), saying $b_2$;then, find the shortest vector in $L\backslash L(b_1,b_2)$($L(b_1,b_2)$ is the integer linear expansion of $b_1$, $b_2$), saying $b_3$;and so on.

Suppose when m = n-1, ${b_i},i=1,...,n-1$ is the basis of $L$, we want to prove : when m = n,${b_i},i=1,...,n$ is the basis of $L$.Let$L^{\prime}=\{y-x_nb_n\,\vert\, y=\sum_{i=1}^{n}x_ib_i\in L, x_i\in \mathbb{R}\}$, then $L^{\prime}$ is a lattice, and its dimension is n - 1，its basis is ${b_i},i=1,...,n-1$.So $\forall y=\sum_{i=1}^{n}x_ib_i\in L$, we have $x_1, ... x_{n-1} \in \mathbb{Z}$, so $x_nb_n\in L$. Since $b_n$ is Shortest in $L\backslash L^{\prime}$, $x_n \in \mathbb{Z}$. So ${b_i},i=1,...,n$ is the basis of $L$.

qiuxicj
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