TITLE: Prove that vectors of a real inner product space are linearly independent.
QUESTION [6 upvotes]: Let $V$ be a real inner product space. Let $u, v_1, ..., v_m \in V$ such that
$$\langle u, v_i \rangle > 0, \ \forall{i}$$
$$ \langle v_i, v_j \rangle \leq 0, \ \forall i \neq j.$$
Prove that vectors $v_1, ..., v_m$ are linearly independent.
There is a hint which says if a linear combination of the vectors with nonnegative coefficients is equal to zero, then all the coefficients must be equal to zero. I'm not sure how to carry out this proof using the hint.

REPLY [2 votes]: Suppose the contrary that $v_i$s are linearly dependent. Then some non-trivial linear combination of them is zero. Thus there exist two disjoint subsets $\mathcal I,\mathcal J\subseteq\{1,2,\ldots,m\}$ such that $\mathcal I\cup\mathcal J$ is non-empty, $\sum_{i\in\mathcal I}c_iv_i-\sum_{j\in\mathcal J}c_jv_j=0$ and $c_k>0$ for all $k\in\mathcal I\cup\mathcal J$. It follows that
\begin{align}
0=\langle0,0\rangle
&=\left\langle\sum_{i\in\mathcal I}c_iv_i-\sum_{j\in\mathcal J}c_jv_j,
\ \sum_{i\in\mathcal I}c_iv_i-\sum_{j\in\mathcal J}c_jv_j\right\rangle\\
&=\left\|\sum_{i\in\mathcal I}c_iv_i\right\|^2
-2\sum_{i\in\mathcal I,\ j\in\mathcal J}c_ic_j\langle v_i,v_j\rangle
+\left\|\sum_{j\in\mathcal J}c_jv_j\right\|^2.
\end{align}
As $-2\sum_{i\in\mathcal I,\ j\in\mathcal J}c_ic_j\langle v_i,v_j\rangle\ge0$, it must be zero. Thus $\sum_{i\in\mathcal I}c_iv_i=\sum_{j\in\mathcal J}c_jv_j=0$. Since $\mathcal I\cup\mathcal J$ is non-empty, we may assume that $\mathcal I$ is non-empty. But then
$$
0=\langle u,0\rangle=\left\langle u,
\ \sum_{i\in\mathcal I}c_iv_i\right\rangle
=\sum_{i\in\mathcal I}c_i\langle u, v_i\rangle>0,
$$
which is a contradiction. Hence the $v_i$s must be linear independent at the beginning.