TITLE: Largest number of vectors with pairwise negative dot product
QUESTION [22 upvotes]: What is the largest $m$ such that there exist  $v_1,\dots,v_m \in \mathbb{R}^n$  such that for all $i$ and $j$, $1\leq i< j\leq m$, we have  $v_i \cdot v_j < 0$.
Also, the preview screen is not displaying set braces for LaTeX.  Is that just the preview, or does it mean the site wouldn't display them after the question had been posted either?  (I formatted this question without the braces in case it's the latter.)

REPLY [4 votes]: A proof from matrix theory
Let $V=(v_1,...,v_m)\in\Bbb R^{n\times m}$ be the matrix with the $v_i$ as columns (and let us assume that $\|v_i\|=1$).
The matrix $X:=V^\top V\in\Bbb R^{m\times m}$ has the following properties:

$X$ is positive semi-definite, that is, all eigenvalues are real and $\ge 0$. Furthermore, the multiplicity of the eigenvalue zero is equal $m-d$, wehre $d$ is the rank of $X$ and equals $d:=\dim \mathrm{span}\{v_1,...,v_m\}\le n$.
$X_{ij}=v_i\cdot v_j$, in particular, the diagonal of $X$ consists of 1-s, and the off-diagonal entries of $X$ are negative.

The matrix $Y:=2I-X$ has only positive entries, and by Perron-Frobenius, its maximal eigenvalue has therefore multiplicity one.
Hence, the minimal eigenvalue of $X$ has multiplicity one.
Thus, the multiplicity of zero as eigenvalue of $X$ is at most one (it could be that zero is not an eigenvalue at all, but if it is one, it is the smallest).
Consequently, $m-d\le1\implies m\le d+1\le n+1$.