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Let $V$ be a finite set consist of nonzero vectors $v=(v_1, \cdots, v_n)\in \mathbb{Z}^n$, how to find a vector $w\in \mathbb{Z}^n$ such that the inner product $<w,v>$ is nonzero for any $v\in V$?

zhichengzhang
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  • You should be more explicit. For example, should $w$ also belong to $V$ ? Or can $w$ be an arbitrary vector from $\mathbb N^n$? – xyz Mar 05 '24 at 14:01
  • Do you mean $\langle w,v\rangle\ne 0$ for all $v\in V$? – John Douma Mar 05 '24 at 14:02
  • @xyz Well, in my case I need $w\in \mathbb{Z}^n$ – zhichengzhang Mar 05 '24 at 14:15
  • @JohnDouma Yes! for all vectors in $V$. – zhichengzhang Mar 05 '24 at 14:16
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    In some sense, "most" $w$ will have this property. It follows from this, for example (at least assuming that $0 \notin V$!) – Izaak van Dongen Mar 05 '24 at 15:19
  • Wouldn't $[-1 \quad {-}1 \quad \cdots \quad {-}1]$ work? Or $[1 \quad 1 \quad \cdots \quad 1]$ for that matter? – Brian Tung Mar 06 '24 at 00:50
  • @IzaakvanDongen Thank you, but I don't see how my question relate to your link? could you please give me some hint? – zhichengzhang Mar 06 '24 at 00:52
  • @BrianTung I'm sorry that my statement is not correct, $v\in \mathbb{Z}^n$ and in my case I don't know what these vectors look like. So I only need the existences of $w$. – zhichengzhang Mar 06 '24 at 00:58
  • As I understand it, you're asking "given a finite set of nonzero vectors in $\Bbb Z^n$, must there always be some vector which is not orthogonal to any of them?" You should show that this is equivalent to the same statement over $\Bbb Q^n$, and that if $v$ is a nonzero vector in $\Bbb Q^n$, then the set of vectors orthogonal to $v$ is a proper $\Bbb Q$-linear subspace. – Izaak van Dongen Mar 06 '24 at 02:04
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    @IzaakvanDongen Thank you! – zhichengzhang Mar 06 '24 at 03:02

1 Answers1

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For the case $n=1$, we can choose $w=(1)$. We build up to any $n$ from there.

Consider $V'$, formed by discarding the last element of each element of $V$; eg given $v \in V$, $v' \in V'$ is $(v_1, v_2, \dots, v_{n-1})$.

Suppose you have $w' \in \mathbb R^{n-1}$ that satisfies $\left<w',v'\right> \ne 0 \forall v' \in V'$. Let $m=\max_{v' \in V'} |\left<w',v'\right>|$ (the biggest inner product).

Let $w_n=m+1$. We then have $\left<w,v\right>=\left<w',v'\right> + (m+1) v_n$.

For $v$ with $v_n=0$, the inner product is unchanged: $\left<w,v\right>=\left<w',v'\right>\ne 0$.

Otherwise, $|v_n| \geq 1$; therefore, $|(m+1) v_n| > m$. Since $|\left<w',v'\right>| \leq m$, the sum $\left<w,v\right>=\left<w',v'\right> + (m+1) v_n \ne 0$.

Alex K
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