# Maximum number of pairwise non-acute vectors in R^n

It is said to be well-known, but perhaps I've never seen it before: there are at most $$2n$$ non-zero vectors in $$R^n$$ such that the inner product of any two of them is not positive.

Proof:
We could always change the basis such that one of the vectors is $$(\delta,0,\ldots,0)$$ where $$\delta>0$$. Obviously there could not be any other vector with positive first coordinate, i.e. the rest of vectors are grouped into $$A$$ and $$B$$, where $$A$$ consists of vectors with negative first coordinate, and $$B$$ zero first coordinate. Express each vector in $$(x,v)$$ where $$x\in R$$. The $$v$$s in $$A$$ do not have duplicate, because that would violate the assumption. Similary the $$v$$s in $$B$$ are distinct. Moreover $$v$$s from $$A$$ and $$B$$ are distinct as well. This means that these $$v$$s in $$A\cup B$$ are all different and possess the property of pairwise non-positive inner product, except $$A$$ could have one zero vector. By induction this finishes the proof.
Q.E.D.

The induction proof also leads to another interesting and intuitive result: for $$n>1$$ optimality happens only with $$n$$ mutual orthogonal vectors and their scaled negatives.