Does the set of null vectors of an indefinite scalar product determine the product up to scale? If $V$ is a vector space with index 1, let $g$ and $\hat{g}$ be scalar products, and denote
\begin{align}\Lambda &= \{v \in V \mid g(v,v)=0\} \\
\hat{\Lambda} &=\{v \in V \mid \hat{g}(v,v)=0\} .
\end{align}
I have to show that if $\Lambda = \hat{\Lambda}$, then $g=k\hat{g}$, where $k$ is a constant. 
Other than making good sense intuitively, I'm difficulty on where to begin. Can anyone point me in the right direction?
 A: This is a well-known fact among researchers in conformal geometry (and presumably relativity), but it's slightly tricky to find a proof in the literature. $\newcommand{\bfE}{{\bf E}}\newcommand{\bfF}{{\bf F}}$
Hint Denote $n := \dim V$. Since the nondegenerate bilinear symmetric form $g$ has index $1$, it has signature $(n - 1, 1)$, and hence there is an orthonormal basis $(\bfE_1, \ldots, \bfE_{n - 1}, \bfE_-)$ such that $g_{aa} = 1$ for $1 \leq a \leq n - 1$ and $g_{--} = -1$. (As usual, for any bilinear form $h$ on $V$, we denote $h_{AB} := h(\bfE_A, \bfE_B)$ for any symbols $A, B$.) For each $a, b$ such that $1 \leq a < b \leq n - 1$, consider the family
$$\bfF_{\theta} := \bfE_- + \cos \theta \, \bfE_a + \sin \theta \, \bfE_b$$ of vectors, which we've chosen to be $g$-null, and hence which (by hypothesis) are $\hat{g}$-null.
(NB that taking pairs $(a, b)$ as above requires that $\dim V > 2$. It's not much trouble, though, to modify this proof for the special case $\dim V = 2$.)

Expanding gives $$0 = \hat{g}(\bfF_{\theta}, \bfF_{\theta}) = \hat{g}_{--} + 2 \hat{g}_{-a} \cos \theta + 2 \hat{g}_{-b} \sin \theta + \hat{g}_{ab} \sin 2\theta + \hat{g}_{aa} \cos^2 \theta + \hat{g}_{bb} \sin^2 \theta. \quad (\ast)$$ Specializing $\theta$ separately to $0, \pi$ gives us two equations, and together they imply that $$\hat{g}_{-a} = 0 \qquad \textrm{and} \qquad \hat{g}_{aa} = -\hat{g}_{--}$$ for all $a$. Substituting in $(\ast)$ leaves $$0 = \hat{g}_{ab} \sin 2 \theta,$$ and substituting, e.g., $\theta = \frac{\pi}{4}$, gives $$\hat{g}_{ab} = 0.$$ We have thus determined all of the components of $\hat{g}$, and together they yield the desired result: $$\hat{g} = k g,$$ (for $k = \hat{g}_{11}$).

This proof is adapted from the one in

Ehrlich, Paul. "Null cones and pseudo-Riemannian metrics," Semigroup Forum. 43 (1991), 337-343.

One can also find the result in

Hawking, S. W.; Ellis, G. F. R. (1973). "The Large Scale Structure of Space-Time". Cambridge University Press.

but this is evidently not freely available online. At any rate, the result is of course much older than this, but it would probably be difficult to determine where it first occurs in the literature.
Incidentally, the claim is true for nondegenerate bilinear symmetric forms of arbitrary positive index (that is, arbitrary indefinite signature), and one can modify this proof (as Ehrlich does) to show this.
