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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 $2 \le k \le d-2$?

Asaf Shachar
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1 Answers1

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$g$ 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.

Dap
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  • Yes, you are right. I also just thought about something like that, but I am not sure where continuity comes in. Also, are you sure $g$ can really be negative-definite? I am not sure this is really possible. – Asaf Shachar Jan 05 '18 at 13:44
  • @AsafShachar: yeah, you don't really need continuity. Take $g$ to be the negative of the usual inner product on $\mathbb R^n.$ Your determinant will be $(-1)^k$ multiplied by the determinant for the usual inner product, which gives a non-negative determinant for $k$ even. – Dap Jan 05 '18 at 13:46