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Suppose we have a finite $s$-dimensional grid $J\subset\mathbb{Z}^{s}$ containing $0_{s}$.

Let $n_{i}\in\mathbb{Z}^{s}$, $i=1,\ldots,N$ be the vectors with ending points the points of the grid.

Can we always find a vector $u\in\mathbb{R}^{s}$ such that all dot products $n_{i}\cdot u$, $i=1,\ldots, N$ are distinct?

My intuition leads me to believe the answer is yes, since I tried to find counterexamples in $1$ and $2$ dimensions but failed. I haven't come up with a solid proof though.

A re-formulation of the problem would be proving that there exists a vector $u\in\mathbb{R}^{s}$ such that $$(n_{i}-n_{j})\cdot u\neq 0$$ for all $i\neq j$. Given that the grid contains $N$ vectors, the number of vectors $n_{i}-n_{j}$ for $i\neq j$ is $(N-1)!$.

Any pointing to the right direction would be welcome.

Nikon
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    What informs your intuition? Can you say a bit about why you intuitively feel that the answer is yes? I'm not trying to interrogate. For example, adding context like "I tried to find a counter-example, but every example I tried seemed to support the claim, so I'm inclined to think the statement is true." Even if you have only the start of a proof, or the outline of a proof, that will help potential answerers better understand how to help you best. And it also shows that you really have spent time on the problem. It doesn't matter if what you've tried is wrong. We've all been there! – amWhy Jul 26 '17 at 15:17
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    Thanks, more added. – Nikon Jul 26 '17 at 16:03
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    You're welcome, and good job with the edit to include a bit more! +1 – amWhy Jul 26 '17 at 16:04
  • Can we assume that if $(a,b)$ belongs to the grid then also $(-b,a)$ and $(b,-a)$ belong to the grid? If so then we have always two identical dot products equal to 0.. – Widawensen Jul 26 '17 at 16:57
  • @Nikon I see. The searched vector doesn't belong to the grid. Interesting problem. – Widawensen Jul 27 '17 at 11:30
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    Note that each constraint $(n_i - n_j)\cdot u \neq 0$ forbids a hyperplane. Since $\mathbb{R}^s$ is not the union of finitely (or even countably) many hyperplanes, it follows that there is a $u\in \mathbb{R}^s$ such that all the dot products $n_i \cdot u$ are distinct. It remains to prove that $\mathbb{R}^s$ is not the union of finitely many hyperplanes. If you can use a bit of measure theory or topology, the proof is very short. If the proof must use only linear algebra and a little set theory, it's going to be longer. – Daniel Fischer Jul 30 '17 at 18:13

2 Answers2

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Let a finite (or even countable) number of pairwise different vectors ${\bf n}_i$ be given. For $i\ne j$ one has ${\bf n}_i\cdot{\bf u}={\bf n}_j\cdot{\bf u}$ iff ${\bf u}$ is lying in the hyperplane $\>H_{ij}\!:\>({\bf n}_i-{\bf n}_j)\cdot{\bf u}=0$. There are at most countable many forbidden hyperplanes $H_{ij}$, and these hyperplanes do not fill all of space. It follows that there are plenty of vectors ${\bf u}$ satisfying your desires.

Christian Blatter
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After rethinking the problem I'm returning with a new proposition.

One of the solutions could be if you select components of this vector from the set of values being square roots of primes $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \dots$ etc $\dots$ let name this set $\mathbb{S}$.

From this question and comments with links provided by Jyrki Lahtonen follows that there is no possibility that any vector with components from $\mathbb{S}$ and vector with components from $\mathbb{Z}$ can have their dot product equal $0$ unless components of integer vector are equal $0$.

Note also that you can use values from the sets which look like, for example $(\sqrt{3}- \sqrt{2}), (\sqrt{13}- \sqrt{11}),\dots,(\sqrt{73}- \sqrt{71}),\dots,(\sqrt{313}- \sqrt{311}) \dots $ and other appropriate combinations with rational coefficients. Selecting appropriately them you can obtain a vector which is very close (but not equal) to any predefined form, for example to vector $(1,1,1,..,1)^T$.

Widawensen
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  • I came up with a similar solution in $2$-dimensions but was yet to generalize. Thanks, this was most useful – Nikon Jul 31 '17 at 18:42
  • @Nikon For me your question was some kind of discovery that we can have a grid (like your grid generated by integers) and through appropriate stretching it ( for example by multiplying values on axis by primes) we can obtain new grid and vectors taken from these two grids are never orthogonal to each other. The proposition makes even a greater impression for original "grids" ( which are much more dense than integer) made from rational values and stretched by a similar method. Thank you for the question. – Widawensen Aug 01 '17 at 11:15
  • That's a very nice observation. My original problem was creating a invertible Vandermonde matrix produced by the powers of dots products between an unknown vector $u$ and vectors with integer components. Your thought generalizes the problem. Well done! – Nikon Aug 02 '17 at 00:46