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 $t\in \mathbb{R}$, $$ q(x+t(y-x))=q(x)+t^2q(y-x) $$ and we obtain: $q(y-x)<-q(x)<0$
By continuity of $f$, there exists $t\in (0,1)$ such that $q(x+t(y-x))=0$. (This $t$ could be such that $x+t(y-x)=0$).
But, when $t\to \infty$, $q(x+t(y-x))\to -\infty$, 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 $$ (1/2) x^T H y = (1/2) y^T H x $$