Let $f : U × U → \mathbb{R}$ be a bilinear form such that $f (u, u) > 0$ and $f (v, v) < 0$ for some $u, v \in U$. I would like to show that $u, v$ are linearly independent.
We have that $u , v$ are L.I. iff $au+bv = 0$ only when $a = b = 0$. We evaluate: $$f(au+bv , au+bv) = a^2f(u,u) + b^2f(v,v) + 2ab f(u,v)$$
$u,v$ to be Linearly Independent does not imply $f(u,v) = 0$. From here I am not quite sure how to demonstrate that a and b must be 0 using the knowledge that $f(u,u) >0$ and $f(v,v)<0$
Moreover would this imply that there is some $w \neq 0$ such that $f(w,w) = 0$. This $w$ would it be of the form: $xu+yv = w$? for $x,y$ some scalars.
Thank you for the insight!
If they are not independent, then $u=av$ for some $a$. But $f(av, av) =a^2f(v,v)<0$, so this is impossible.