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Let $V$ be an $n$ dimensional vector space, let $R$ be a finite set of vectors.

  1. Will there exist a hyperplane which does not contain any of the vectros from $R$?

  2. How to construct such a hyperplane?

  3. Do I need the vectors linearly indepenedent?

I need to prove this result to show the existance of weyl chambers. I understand that there will be such hyperplane as baire category theorem says a complete metric space can not be union of no where dense sets.

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What field is $V$ over? – Chris Eagle Oct 5 '12 at 8:19
By hyperplane do you mean an affine space or a subspace? – Christopher A. Wong Oct 5 '12 at 8:32

Yes, if $0\notin R$, and the base field $K$ has infinite many elements. By arbitrarily fixing a basis, we can equip $V$ with an inner product $\langle,\rangle$.

  1. All we have to do is to find a normalvector $n$ of the desired hyperpane, that is not orthogonal to any $r\in R$.

  2. Start with an arbitrary vector $n_0$, and assume $R$ is enumerated: $R=\{r_1,r_2,\dots\}$. If $n_0$ is not yet good, there is a smallest index $k$ such that $n_0\perp r_k$. Then, let $\alpha_i:=\langle n_0, r_i\rangle\ne 0$, for $i<k$. Since $r_i\ne 0$ and $\langle r_k,r_i\rangle$ range over a finite numbers in $K$ (for $i<k$), we can find a $0\ne \beta\in K$, such that still $\langle n_0+\beta r_k ,r_i\rangle\ne 0$. Then, let $n_1:=n_0+\beta r_k $ and carry on this procedure..

  3. No.

For affine hyperplane one can omit the $0\notin R$ condition, and first find any vector $s_0$ not in $R$, and then consider $R-s_0$.

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You can actually edit your answer instead of deleting and re-submitting a new answer. – Marc van Leeuwen Oct 5 '12 at 11:52
I did so, but I got an 'Oops' window then somehow could get back the textbox and clicked on the button and it just duplicated it.. – Berci Oct 5 '12 at 13:43

I think in the following case the first could be happened:

Let your hyperplan is in $\mathbb R^n$ so it is a set of all elements in $\mathbb R^n$ which satisfy the equation $$c_1x_1+c_2x_2+...+c_nx_n=k$$ where in $u=(c_1,c_2,...c_n)\neq0$ and $u\in\mathbb R^n$. Obviously, the segment $\vec{PQ}$, associated to points $P,Q$, in the hyperplan is normal to $u$. Now I consider $R$ such that $u\in R$.

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