Finding the basis of a lattice

Question

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?

Answer 1

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 $$\det\nolimits_{\{b_1,b_2\}}(b'_1,b'_2)=\pm 1.$$