Lemma 5

Let $e_1, \ldots, e_n$ be vectors generating T such that: a) $(e_i|e_j) \leq 0$ for $i \neq j$; b) there does not exist a partition of $\{1, \ldots, n\}$ into two non-empty subsets I and J such that $(e_i|e_j) = 0$ for $i \in I$ and $j \in J$.

Then there are two possibilities: 1) $(e_1, \ldots, e_n)$ is a basis of T; 2) $n = \dim T + 1$; there exists a family $(c_i)_{1 \leq i \leq n}$ of real numbers $> 0$ such that $\sum_i c_i e_i = 0$, and any family $(c'_i)_{1 \leq i \leq n}$ of real numbers such that $\sum_i c'_i e_i = 0$ is proportional to $(c_i)_{1 \leq i \leq n}$.

Put $q_{ij} = (e_i|e_j)$. The matrix $Q = (q_{ij})$ then satisfies the hypotheses of Lemma 4: conditions a) and b) of Lemma 4 are the same as conditions a) and b) above, and c) is satisfied since $\sum_{i,j} q_{ij} x_i x_j = \| \sum_i x_i e_i \|_2^2$. The kernel N of the quadratic form $q$ on $\mathbf{R}^n$, with matrix $Q$, is the set of $(c_1, \ldots, c_n) \in \mathbf{R}^n$ such that $\sum_i c_i e_i = 0$. If $N = \{0\}$, the $e_i$ are linearly dependent and we are in case 1). If $\dim N > 0$, Lemma 4 (i) shows that we are in case 2).