This question is related to this one.
Let $V$ be a $d$-dimensional real vector space, and let $g:V \times V \to \mathbb{R}$ be a symmetric bilinear form.
Suppose that for every sequence of vectors $v_1,\dots,v_{d-1} \in V$
$$ \det \big(g (v_i,v_j )\big) \ge 0,$$ and equality holds if and only if the $v_i$ are linearly independent.
Is it true that $g$ is positive? i.e. $g(v,v) \ge 0$ for every $v \in V$ with equality only if $v=0$?
Does the answer change if instead of testing sequences of $d-1$ vectors, we test $k$ tuples, when $2 \le k \le d-2$?
$g$ is either negative definite or positive definite, and if $k$ is odd then $g$ must be positive definite. A negative definite form will give positive determinants for $k$-tuples with $k$ even.
We can assume $g$ is given by a diagonal matrix. If $g$ is indefinite or $g$ is negative definite and $k$ is odd, we can pick $k$ basis vectors with an odd number of negative vectors ($g(e_i,e_i)\leq 0$), which gives a negative determinant.
In fact to prove definiteness we only need that the determinant is non-zero for linearly independent tuples. For an indefinite form, for any $1\leq k \leq d-1$ there is a set of $k$ orthogonal vectors including at least one isotropic vector. This will give a zero determinant.