Lemma 6

Let $(e_1, \ldots, e_n)$ be a basis of $T$ such that $(e_i|e_j) \leq 0$ for $i \neq j$.

(i) If $x = \sum_i c_i e_i \in T$ is such that $(x|e_i) \geq 0$ for all $i$, then $c_i \geq 0$ for all $i$.

(ii) If $x$ and $y$ are two elements of $T$ such that $(x|e_i) \geq 0$ and $(y|e_i) \geq 0$ for all $i$, then $(x|y) \geq 0$. If $(x|e_i) > 0$ and $(y|e_i) > 0$ for all $i$, then $(x|y) > 0$.

Under the hypotheses of (i), assume that $c_i < 0$ for some $i$. Let $f$ be the linear form on $T$ defined by $f(e_i) = 1$ and

The vectors $-x, e_1, \ldots, e_n$ then satisfy the hypotheses of Lemma 3 (ii) (taking for $q$ the metric form on $T$). We conclude that they are linearly independent, which is absurd. Hence (i). Moreover, if $x = \sum_i c_i e_i \in T$ and if $y \in T$, then $(x|y) = \sum_i c_i (e_i|y)$, so (ii) follows immediately from (i).