Linear independence and pairwise angles Given $p+1$ vectors with pairwise angles $> \pi/2$ in $\mathbb{R}^n$ then every $p$ of them are linearly independent. How to prove this fact?
The motivation for this question is the idea of extending the notion of "linear independence" for points on tangent cones on some metric spaces (Alexandrov spaces), on which we do not have a vector space structure, but have a good notion of angle.
 A: Suppose this is true for p. Let's prove this for p+1
$\sum_\alpha\alpha v_\alpha=0$. Separate nonnegative $\alpha$'s from negative ones
$\sum_{\alpha^+} \alpha^+ v_{\alpha^+}=\sum_{\alpha^-} -\alpha^- v_{\alpha^-}$
$(\sum \alpha^+ v_{\alpha^+})^2=\sum_{\alpha^+, \alpha^-}-\alpha^+\alpha^-\langle v_{\alpha^-}v_{\alpha^+}\rangle\le 0$ and $\ge 0$ hence 
$(\sum \alpha^+ v_{\alpha^+})^2=0$ and all $\alpha^+=0$ since otherwise we would have $n\le p$ linearly dependent vectors, contradicting the induction assumption.
The scalar square of $(\sum \alpha^- v_{\alpha^-})$ gives the same for $\alpha_-$'s
A: The same idea as Dmitry K's answer yields the following non-inductive argument. Suppose you had a linear dependence among $p$ of the given $p+1$ vectors. As in Dmitry's answer, transpose all the terms with negative coefficients, so the dependence relation looks like 
$\sum_{v\in B}\beta_v v=\sum_{w\in C}\gamma_w w$, where $B$ and $C$ are disjoint subsets of your set of $p$ dependent vectors, and where all the $\beta$'s and $\gamma$'s are positive.  Write $x$ for the common value of $\sum_{v\in B}\beta_v v$ and $\sum_{w\in C}\gamma_w w$.  On the one hand, the inner product $x\cdot x$ of a vector with itself is non-negative.  On the other hand, this inner product is $\sum_{v\in B, w\in C}\beta_v\gamma_w(v\cdot w)$, in which every term is strictly negative, because the $\beta$'s and $\gamma$'s are positive while the inner products $(v\cdot w)$ are negative. The only way for such a sum to be non-negative is for it to be the empty sum.  That is, $B$ or $C$ must be empty, and our original linear dependence relation had coefficients of only one sign; without loss of generality, suppose that sign is positive, so $C$ is empty and $\sum_{v\in B}\beta_v v=0$.  Now use the assumption that there is a $(p+1)$-st vector $y$ whose inner product with all $v\in B$ is negative.  Then since $\sum_{v\in B}\beta_v v=0$, we have $0=\sum_{v\in B}\beta_v (y\cdot v)$.  But every term in this sum is strictly negative, so $B$ has to be empty too, and our original alleged linear dependence relation has disappeared.
