# dist($av$, $L$) = $a \cdot$ dist($v$, $L$)?

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 http://www.math.lsa.umich.edu/~barvinok/latticenotes669.pdf)

Thanks!

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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 $$\mathrm{dist}(av,L)=\mathrm{dist}(aw,L)=a\|w\|=a\,\mathrm{dist}(w,L)=a\,\mathrm{dist}(v,L).$$

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Thanks for your help! –  badatmath Jan 16 '13 at 5:10