Say I have a set of $(n-1)$ linearly independent vectors $\mathbf{v}\_i$ of dimension $n$ with entries $\pm1$. I am interested in finding the $n-$dimensional vector $\mathbf{u} $which is normal to the hyperplane spanned by the $\mathbf{v}\_i$. In other words, $\mathbf{u}$ is orthogonal to each of the $\mathbf{v}\_i$, where $\mathbf{u}$ is unique up to multiplicative constant. 


I have two questions:


1) Is it true that $\mathbf{u}$ is also a vector with entries $(0, \pm 1)$ with an arbitrary pre-factor (e.g. $\mathbf{u}=\alpha[-1,1,0]$)? Can this be proven?


2) I know that $\mathbf{u}$ can be calculated with standard linear algebraic techniques (e.g. Gaussian elimination), but if property 1) is true, is there a computationally faster trick to getting the answer?


Thank you for your help!



fast way to calculate normal to set of vectors with $\pm$1 entries
To recap: we are given $n-1$ linearly independent vectors in ${\bf R}^n$ with $\pm 1$ entries. The original Q1 asked, in effect, if there is always a vector with $\pm 1$ entries that is orthogonal to all of the given ones. A counter examples is provided for $n=3$ by taking
$$
v\_1 = e\_1+e\_2+e\_3\quad,\quad v\_2=e\_1+e\_2-e\_3
$$
since the normal vector to span($v\_1$,$v\_2$) must be a multiple of $e\_1-e\_2$. (Here I am using $e\_1,e\_2,\dots$ to denote the standard o.n. basis vectors of ${\bf R}^n$.)


A modified version of Q1 now relaxes the requirement on the normal vector, so that one merely requires it to have entries in $\{-1,0,1\}$. This also fails (Adam Goucher pointed our a counterexample while I was writing this answer) and in fact we can generate a family of examples for all $n\geq 4$ as follows. Let $u= e\_1+\dots + e\_n$ iinside ${\bf R}^n$ and consider the vectors $u-2e\_1,\dots, u-2e\_{n-1}$. A little bit of manipulation shows that a normal vector to this family must have the form $ae\_1+\dots + ae\_{n-1}+be\_n$ where $(n-2)a+b=0$, and so it is not possible for such a vector to have entries in $\{-1,0,1\}$ when $n\geq 4$.


It's not clear whether the OP is interested in Q2 even when Q1 has a negative answer, and it's not something I have any immediate intuition for.


