The Art Gallery Guardian

Basis of the module \Z^n

A student who is taking linear algebra asked me the following problem.

If we consider the field \R restricted to \Z, and create a "vector space" on \Z. How do we know if v,u\in \Z^2 "spans" \Z^2?

Formally, what can we say about v and u if for every w\in \Z^2, there exist n,m\in \Z, such that nv + mu = w.

We can generalize it and put it in terms of modules, as \Z is only a ring but not a field.


v_1,\ldots,v_n is a basis for the module \Z^n iff the matrix M formed by the vectors is a unimodular matrix.


\Rightarrow If \det(M)=0, then v_1,\ldots,v_n are not linearly independent. If |\det(M)|\geq 2, then the parallelepiped formed by v_1,\ldots,v_n has volume \geq 2. If there is any integer point not on the corners of the parallelepiped, then that point can't be written as linear combination of v_1,\ldots,v_n. Notice that it must contain some lattice points not on the corners of the parallelepiped. One can see why by consider a large box that contain volume of m such parallelepiped, but contain at least 2m lattice points.

This shows if M is not unimodular, then v_1,\ldots,v_n can't be a basis.

Alternative proof: M is not unimodular then M^{-1} contain a non-integer entry. This shows there exist a b, such that the solution x to Mx=b contain a non-integer entry. (proposed by Thao Do)

\Leftarrow |\det(M)|=1 implies it has a inverse over \Z, thus Mx = b for any b\in \Z^n always has a solution.

Posted by Chao Xu on 2012-08-30.
Tags: math.