At most $n+1$ vectors, the angle between which $>\pi/2$.

Question

In a $n$ dimensional Euclidean space $V$, there exists at most $n+1$ vectors, each pair has inner product $<0$.

This is geometrically obvious in $3$ dimensions...But how can we prove it analytically?

Answer 1

(Abridged from Robin Chapman's answer in a MO thread.)

You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin.

You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$\sum_{i=1}^{n+2} a_i v_i=0.$$ This gives enough room for manoeuvre to ensure some $a_i>0$ and some $a_j<0$. Thus we get some nontrivial relation $$\sum_{i\in I}a_i v_i=\sum_{j\in J}b_j v_j\qquad\qquad(*)$$ where all the $a_i>0$ and $b_j>0$ and $I$ and $J$ are disjoint non-empty sets of indices. It follows that the dot product of the two sides of $(*)$ is negative, but that contradicts it being the square of the length of the left side.

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(Abridged from Benoît Kloeckner's answer in the same MO thread.)

Let us prove by induction that this number is $n+1$. The result is obvious for $n=1$. Assume it for some $n$ and consider a set of mutually negative dot product vectors $v_0,v_1,\ldots, v_k$ in $\mathbb{R}^{n+1}$. Then all of $v_1,\ldots,v_k$ lie in the open half-space $\{\,v\mid v_0\cdot v<0\,\}$.

Now the orthogonal projections $v_i'$ of $v_i$ ($1\leqslant i\leqslant k$) on the hyperplane $\{\,v\mid v_0\cdot v=0\,\}$ satisfy $v_i'\cdot v_j'\leqslant v_i\cdot v_j$ by a direct computation (assuming $v_i$ are unitary, one has $v_i'=v_i-(v_i\cdot v_0)v_0$ so that $v_i'\cdot v_j' = v_i\cdot v_j-(v_i\cdot v_0)(v_j\cdot v_0)$). By induction $k$ is at most $n+1$ and we are done.