TITLE: Prove that vectors in Euclidean space are linearly independent
QUESTION [6 upvotes]: Could you help me with an idea of solving the following problem? I think that proof involves the positive definiteness of Gram matrix, but I don't know how.
Consider a system of vectors $e_1, e_2, ..., e_n, e_{n+1}$ in some Euclidean space such that dot product $(e_i, e_j) <0$ for all $i \neq j.$ Prove that any $n$ vectors of the system are linearly independent.

REPLY [2 votes]: The result follows from the following Lemma, which is used in the theory of root systems to study bases of root systems.

Lemma: Let $(V, \langle\,{,}\,\rangle)$ be a Euclidean vector space. Let $\ell\in V^*$ a linear form and $A\subseteq V$ a subset such that:
  (1) $\ell(x)>0$ for all $x\in A$.
  (2) $\langle x,y\rangle \le 0$ for all $x,y\in A$, $x\neq y$.
Then $A$ is linearly independent.

Apply this Lemma to $\ell = \langle\,\cdot\,,-e_{i_0}\rangle$ and $A = \{e_1,\dotsc,\widehat{e_{i_0}},\dotsc,e_{n+1}\}$ (where $\widehat{e_{i_0}}$ denotes omission) for some fixed $i_0$.
Proof of the Lemma: Let $\sum_{x\in A}\lambda_xx = 0$ be a (finite) linear combination. Now, consider
$$
v:= \sum_{x\in A_1}\lambda_xx = \sum_{x\in A_2}\mu_xx,
$$
where $A = A_1\sqcup A_2$ and $\lambda_x\ge0$ for all $x\in A_1$, $\mu_x := -\lambda_x >0$ for all $x\in A_2$. Then
$$
0\le \langle v,v\rangle = \sum_{x\in A_1}\sum_{y\in A_2}\lambda_x \mu_x \langle x,y\rangle \le 0
$$
where we have used (2) for the last inequality. It follows that $v = 0$. Hence,
$$
0 = \ell(v) = \sum_{x\in A_1}\lambda_x \ell(x) = \sum_{x\in A_2} \mu_x \ell(x).
$$
Since (1) says that $\ell(x)>0$ for all $x\in A$, it follows that $\lambda_x = 0$ for all $x\in A_1$ and $-\lambda_x = \mu_x = 0$ for all $x\in A_2$. Thus, $A$ is linearly independent.