Prove that for a bilinear symmetric form $B$ there's a vector v$≠$0 that B(v,v)=0 iff $B$ and $-B$ are not positive

Question

a bit of a messy question:

Suppose there's a Bilinear symmetric form $B$ on vector space $V$ above $R$.

how can I prove that there's a $v ∈ V$ that sustains $B(v,v)=0$ iff $B$ and $-B$ are not positive?

thanks.

Answer 1

For $u\in V$ set $q(u):=B(u,u)$.

The "$\Rightarrow$" follows directly from definition of positive bilinear form.

For "$\Leftarrow$" pick $u, v\in V$ such that $q(u)=B(u,u)\geq 0$ and $q(v)=B(v,v)\leq 0$. Note that $$f:\mathbb{R}\rightarrow \mathbb{R}, t \mapsto f(t):=q(tu+(1-t)v)$$ is a continuous map (check it! Use that $B$ is bilinear and show that this function is a quadratic polynomial in $t$). Now use the intermediate value theorem to get a zero of $f$ and conclude.