Prove that this bilinear form is a multiple of another

Question

Let $V$ be a vector space, and suppose that $g:V\times V\to\mathbb{R}$ is a symmetric bilinear, non-degenerate form on $V$. Furthermore, suppose that $g$ is negative definite on a $1$-dimensional subspace, but not on any subspace of higher dimension than $1$. Now suppose that $b: V\times V\to\mathbb{R}$ is a symmetric bilinear form, and that for every $v\in V$, $g(v,v)=0$ implies $b(v,v)=0$. Then I want to prove that $b$ is a multiple of $g$, i.e. that $b=rg$ for some $r\in\mathbb{R}$. How does one do that?

Answer 1

One can choose a basis of $V$ such that $g(v_1,v_1)=-1$, $g(v_i,v_i)=1$ for $i>1$ and $g(v_i,v_j)=0$, $i\ne j$. Then $g(v,v)=0$ iff $v=\sum a_i v_i$ with $a_1^2=a_2^2+\cdots+a_n^2$.

Let $b(v_i,v_j)=b_{i,j}$. Considering $v=v_1\pm v_i$, $i>1$, we have $b_{1,1}\pm 2b_{1,i}+b_{i,i}=0$. So $b_{1,i}=0$ and $b_{1,1}=-b_{i,i}$. Now consider $v=\sqrt 2v_1+v_i+v_j$, $1<i<j$. We get $2b_{i,j}=0$ (bearing in mind our previous identities). This is enough.