Linear Algebra, Quadric Form, Bilinear Form

Question

I have a question from an exercise.

So $V$ is a vector space with quadric form $q:V\to \mathbb{R}$ . I have to prove that if the exists $u$, $v$ in $V$ such that $q(v)>0$ and $q(u)<0$ then there exists $w$ in $V$ such that $q(w)=0$.

So what I did is , since we have a quadric form, which belongs to a symetric bilinear form $g(- ,- ):V \times V\to V$ we can find an orthogonal basis for it. Lets say the basis is $B=\{b_1,...,b_n\}$

Then there exists $1\leq i\leq n$ such that $q(v_i)>0$ and $1\leq j \leq n$ such that $q(v_j)<0$ (otherwise we couldn't find two vectors that has outputs on $q$ with different signs.)

Let's assume $q(v_i)=b>0$ and $q(v_j)=c<0$. We can find a vector $v$ and $u$ such that $q(v)=1$ and $q(u)=-1$ because we can use the linearity of the bilinear form

We know that for every $u$ and $v$ in $V$ $g(u,v)=\frac{1}{2}(q(u+v)-q(v)-q(u))$ we will get

$0=g(v,u)=\frac{1}{2}(q(v+u)-q(v)-q(u))$ -> $q(v+u)=q(v)+q(u)=1-1=0$ where $w=v+u$ as desired.

Is this solution correct? Can I assume that $q$ belongs to a syymetric bilinear form ? Because when we learned about quadric form, it is defined on a symetric bilinear form.

Appreciate all feedbacks, thanks!

Answer 1

Yes, over fields not characteristic $2$, each quadratic form is associated with the symmetric bilinear form you used.

The idea is indeed to produce $u,v$ such that $B(u,u)=1$, $B(v,v)=-1$ and $B(u,v)=0$, but so far in your solution you haven't explained how to get $B(u,v)=0$. In general it's possible for this to be nonzero. Accomplish this and your solution is completed.

Answer 2

Another way ... The line segment from $u$ to $v$ defined by $F(t)=(1-t)u+tv$ for $0 \le t \le 1$ is a continuous path which does not pass through the origin because that would imply $u=\alpha v$ for some real $\alpha$ and would give the contradiction that $-1=q(u)=q(\alpha v)=\alpha^{2}q(v)=\alpha^{2}$. Therefore, $q(F(t))$ is a continuous function which is -1 at $t=0$ and +1 at $t=1$. Now apply the intermediate value theorem for real functions to get what you want.