Pairwise negative dot product implies linear independence Let $v_1, ..., v_{m+1} \in \mathbb{R}^n$ such that $v_i \cdot v_j < 0$ if $i \ne j$.
Show that $v_1, ..., v_m$ are linearly independent.
My attempt:
Assume $v_1, ..., v_m \in \mathbb{R}^n$ are linearly dependent with pairwise negative dot product. Let
$$S = \text{span}({v_1, ..., v_m})$$
I want to show that for any $v \in \mathbb{R}^n$ there exists some $v_i$ such that $v \cdot v_i \ge 0$.
Since $v = u + w$ where $u \in S$ and $w \in S^\perp$, it suffices to show that there exists some $v_i$ such that $u\cdot v_i \ge 0$. But I am stuck trying to prove this.
 A: It suffices to prove the following

Claim 1. For a linearly independent list $v_1, \ldots, v_m$ in inner product space $V$ so $\langle v_i,v_j \rangle<0$ for any $i \ne j$. Then there does not exists $u,w \in S=\text{span }(v_1, \ldots, v_m)$ so $\left \langle w,v_i \right \rangle<0, \left \langle u,v_i \right \rangle<0$ for all $i=1, \ldots, m$ and $\left \langle u,w \right \rangle<0$.

In other words, for any $m+1$ vectors $v_1, \ldots, v_{m+1}$ so $\left\langle v_i,v_j \right \rangle<0$ then any $m$ vectors among $m+1$ of them are linearly independent.
Proof. By Gram–Schmidt procedure, there exists orthonormal list $e_1, \ldots, e_m$ in $V$ such that $\text{span }(e_1, \ldots, e_j)=\text{span }(v_1, \ldots, v_j)$ for all $j=1, \ldots, m$. This follows $v_j=\sum_{i=1}^j \left \langle v_j,e_i \right \rangle e_i$ and therefore $\left \langle v_j,v_i \right \rangle = \sum_{k=1}^i \left \langle v_j,e_k \right \rangle \overline{\left \langle v_i,e_k \right \rangle}$ for any $j \ge i$.
First, we will show that $\left \langle v_i,e_i \right \rangle$ has the opposite sign with $\left \langle v_j,e_i \right \rangle$ for $j>i$. We prove this by induction on $i$. For $i=1$, since $\left \langle v_j,v_1 \right \rangle<0$ so $\left \langle v_j,e_1 \right \rangle \left \langle v_1,e_1 \right \rangle<0$, as desired. If it's true for $i$. Consider $i+1$, for $j>i+1$, we have
$$ \left \langle v_j,v_{i+1}\right \rangle<0 \implies \sum_{k=1}^{i+1}\left \langle v_j,e_k \right \rangle \left \langle v_{i+1},e_k \right \rangle<0.$$
From our inductive hypothesis, we find $\left \langle v_j,e_k \right \rangle \left \langle v_{i+1},e_k \right \rangle>0$ for $k \le i$. Therefore, $\left \langle v_j,e_{i+1}\right \rangle \left \langle v_{i+1},e_{i+1} \right \rangle<0$, as desired.
Suppose there exists such $u,w$. We will show that $\left \langle u,e_i \right \rangle$ and $\left \langle w,e_i \right \rangle$ both have the opposite sign with $\left \langle v_i,e_i \right\rangle$. We proceed by induction on $i$. For $i=1$, since $\left \langle w,v_1\right\rangle<0$ so $\left\langle w,e_1\right\rangle \left \langle v_1,e_1\right\rangle<0$. Similarly, $\left \langle u,e_1 \right\rangle \left \langle v_1,e_1\right\rangle<0$, so case $i=1$ is true. If the condition is true for $i$, consider $i+1$, we have
$$\left \langle w,v_{i+1}\right \rangle<0 \implies \sum_{k=1}^{i+1}\left \langle w,e_k \right \rangle \left \langle v_{i+1},e_k \right \rangle<0.$$
From our inductive hypothesis, we know for $k\le i$ then $\left \langle w,e_k \right\rangle \left\langle v_k,e_k \right \rangle <0$ and $\left \langle v_{i+1},e_k \right \rangle \left \langle v_k,e_k \right \rangle<0$ (from previous paragraph). Therefore, $\left \langle w,e_k \right \rangle \left\langle v_{i+1},e_k \right \rangle>0$. Combining with above inequality, we obtain $\left \langle w,e_{i+1}\right \rangle \left\langle v_{i+1},e_{i+1} \right \rangle<0$, as desired. We prove similarly for $u$.
Now, we've shown that $\left\langle w,e_i\right\rangle$ and $\left \langle u,e_i \right \rangle$ have the same sign, or $\left \langle w,e_i \right \rangle \left \langle u,e_i \right \rangle>0$. Therefore, $\left \langle w,u \right \rangle = \sum_{i=1}^m \left \langle w,e_i \right \rangle \left \langle u,e_i \right \rangle >0$, which contradicts to our condition. $\blacksquare$
