Let $l1: R^n -> R$ , $l2:R^n -> R$ be linear transformations. We assume $n\ge2$, $l1\not=0$ and $l2\not=0$. Let's define quadratic form $q: R^n -> R$ such that $q(x)=l1(x)*l2(x)$ for every $x\in R^n$.
- Prove there is a subspace of dimension $n-1$ of $R^n$ on which $q=0$.
- Prove that the signature of $q$ is either $0, 1$ or $-1$.
So for the first part I know that since $n = dimIml1 + dimKerl1$ then $dimKerl1 = n-1$ (since $Iml1$ is not zero, otherwise $l1=0$). So there is a subspace $V$ of dimension $n-1$ such that $l1=0$. I can use the same subspace to show that $q(x)=l1(x)*l2(x) =0$ for every $x\in\ V$.
- This where things gets complicated for me. I tried to use the associated bilinear form $f(u,v)=(1/2)*(q(u+v)-q(u)-q(v)) $ to prove the rank of the representing matrix of $q$ is at most 1 (hence in the diagonal form there is at most one term, whose coefficient can be either 0, positive or negative so the signature is 0, 1, -1 accordingly)
But I get a matrix of the following nature: $$ \begin{matrix} 0 & 0 & ... &0& a_1 \\ 0 & 0 &... &0& a_2 \\ . &.&. &.& .\\ .&.&.&.&.\\ 0 & 0 & ... & 0&a_{n-1} \\ a_1 & a_2 & ... & a_{n-1} & a_n \\ \end{matrix} $$
And I know that $a_n\not=0$. What can I do next?
Is there any other way to solve this?