Jtdylktuy

Let $f$ be a definite, symmetric bilinear form.

Show that $f$ is positive or negative.

Consider the quadratic form $q(x,y,z)=x^2+y^2-z^2$, then the polar form assocaited to $q$ is symmetric, definite but negative and positive.
Is there a problem with the statement of the problem ?

EDIT:
I attempt to prove this result:

Let $x,tilde y$ such that $q(x)>0$ and $q(tilde y)<0$ where $q$ is the quadratic form associated to $f$.

Let $lambda$ such that $y=lambda tilde y$ is such that $f(x,y-x)=0$.
Note that $q(y)<0$.

We have for $tin mathbb{R}$,
$$
q(x+t(y-x))=q(x)+t^2q(y-x)
$$
and we obtain:
$q(y-x)<-q(x)<0$

By continuity of $f$, there exists $tin (0,1)$ such that $q(x+t(y-x))=0$.
(This $t$ could be such that $x+t(y-x)=0$).

But, when $tto infty$, $q(x+t(y-x))to -infty$, so it means that $q$ has to for two different value of $t$ which is absurd.

Finally $q$ is of constant sign.

Any toughts about this proof?

DO you have simpler alternative proof for this statement ?

I found what seems to be the definition at Associated Bilinear Form to Q (Quadratic Form)

The short version: given a quadratic form, create the matrix $H$ which is the Hessian matrix of second partial derivatives.

Given column vectors $x,y:$ the original quadratic form applied to $x$ is given by
$$ (1/2) x^T H x $$
The bilinear form applied to $x,y$ is
$$ (1/2) x^T H y = (1/2) y^T H x $$