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Let $ W $ be a sub vector space of $ \mathbb{R}^n $. How can we determine if $ W $ admits an integer basis?

This is equivalent to asking how to determine if $ W \cap \mathbb{Z}^n $ spans $ W $.

Obviously if $ W=\mathbb{R}^n $ then there is always an integer basis. For example the standard basis $ e_1,\dots e_n $, just the columns of the identity matrix.

$ W $ a $ 1 $ dimensional subspace of $ \mathbb{R}^2 $ is the first nontrivial case. In this case
$$ W=Span \{ (a,b) \} $$ and $ W $ has an integer basis if and only if $ a/b $ or $ b/a $ (check both in case $ a $ or $ b $ is $ 0 $) is rational.

Is there some general criterion that allows us to look at certain spanning sets for $ W $ and perhaps look at ratios of coefficients like this to solve the problem?

Ian Gershon Teixeira
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    It's suffices to solve the problem with $\mathbb{Q}$ instead of $\mathbb{Z}$. – ronno Mar 21 '23 at 23:02
  • Let $w_1,w_2,\cdots,w_k$ be an $\mathbb R$-basis for $W$, and take the exterior product $P=w_1\wedge w_2\wedge\cdots\wedge w_k$. I suspect that $W$ has a rational basis if and only if $P$ (or some multiple of $P$; e.g. divide $P$ by one of its coefficients) has rational coefficients with respect to the standard basis. – mr_e_man Mar 22 '23 at 20:16

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I think the following could be consider as an idea. Sorry, I am new here, so commenting is not an option for me, hence I post this as an answer.

Assume we live in $\mathbb{R}^{n}$. Let $W=\text{span}\{{\bf u}_1,\ldots , {\bf u}_k\}$. We can construct a basis matrix $A$ of size $k\times n$ by using vectors ${\bf u}_1,\ldots , {\bf u}_k$ as rows. Now, if $R$ is the RREF (reduced row echelon form) of $A$, then rows of $R$ still form a basis for $W$. I think that the criterion for $W$ to have a rational (and hence integer) basis is that all entries in $R$ are rational.

In this question it was "proved" that for every subspace $W$,the matrix $R=R(W)$ is unique and does not depend on the choice of the basis. The idea floating there is that if $R_1$ and $R_2$ are two different RREFs of $W$, then we consider the first row where they differ and find a vector that is in the span of one row space (say $R_1$) but not in another.

If we believe the ideas behind that question, then any integer basis of $W$ would produce a rational RREF, but that RREF is also $R$. So $R$ has to have only rational entries.

C614
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