Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $L$ is a lattice subspace and $v$ is the closest lattice point to $L$, why does dist($av$, $L$) = $a \cdot $ dist($v$, $L$)?

I saw this step in a proof that every lattice has a lattice basis. (I followed the proof on page 6 of


share|cite|improve this question

1 Answer 1

up vote 1 down vote accepted

This has nothing to do with lattices. For any vector subspace $L$ of $\mathbb{R}^n$ and any $v\in\mathbb{R}^n$, the distance between $v$ and $L$, as defined in (2.1) of the cited paper, is equal to the length of the part of $v$ orthogonal to $L$. That is, if $v=y_0+w$, where $y_0\in L$ and $w\in L^\perp$, then $\mathrm{dist}(v,L)=\|w\|$ (proof: for any $y\in L$, $\|y-v\|^2=\|(y-y_0)-w\|^2=\|(y-y_0)\|^2+\|w\|^2\ge\|w\|^2=\|y_0-v\|^2$; therefore $\inf_{y\in L}\|y-v\|=\|y_0-v\|=\|w\|$). Therefore for every $a\ge0$, we have $$\mathrm{dist}(av,L)=\mathrm{dist}(aw,L)=a\|w\|=a\,\mathrm{dist}(w,L)=a\,\mathrm{dist}(v,L).$$

share|cite|improve this answer
Thanks for your help! – badatmath Jan 16 '13 at 5:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.