This question relates to the proof of theorem 11.1 of Chaper IV on page 168 of Evar Nering's "Linear Algebra and Matrix Theory", second edition. See e.g. here (page 182 of the pdf): https://download.tuxfamily.org/openmathdep/algebra_linear/Linear_Algebra_and_Matrix_Theory-Nering.pdf
If I understand the author correctly, he is saying that for a given quadratic form q over the real numbers and a set S of vectors such that, for all s in S, q(s) > 0, then for any linear combination t of elements of S, q(t) > 0. (His wording is slightly different; in particular he doesn't use the symbols S, s and t.)
Is the statement correct, and if it is, how would you prove it? The author's wording would suggest it is obvious, but to me it is anything but.
First, the author's claim is not that $S$ is a subspace (which would require $0 \in S$, which is not the case here). Rather the author claims that the value of the quadratic form at any nonzero linear combination of the basis elements $\alpha_1, \ldots, a_P$ is positive. (Incidentally, it is the case that $\overline S = \{s \in \Bbb R^n : q(s) \geq 0\}$ is a generalized cone, i.e., invariant under dilations.)
In any case, the inequality follows from the definitions: Since the basis $A = (\alpha_i)$ diagonalizes $q$, the matrix representation $[q]$ is $\operatorname{diag}(q_1, \dotsc, q_n)$, and by hypothesis $q_1, \dotsc, q_P$ are positive. Now given $\alpha \in \langle \alpha_1, \dotsc, \alpha_P \rangle$, say, $\alpha = c_1 \alpha_1 + \cdots + c_P \alpha_P$, we have $$q(\alpha) = [\alpha]^\top [q] [\alpha] = \sum_{i = 1}^P c_i^2 q_i .$$ Since all of $q_1, \dotsc, q_P$ are positive and at least one $c_i$ is nonzero, $q(\alpha) > 0$.