Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $R-s_0$.

share|cite|improve this answer
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 $u\in R$.

share|cite|improve this answer

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.