Quadratic Forms on a (finite dimensional real) vector space with same zero set are scalar multiples? Let $V$ be a finite-dimensional vector space over $\Bbb{R}$, let $g,h:V \times V \to \Bbb{R}$ be bilinear symmetric functionals, and let $Q_g,Q_h:V \to \Bbb{R}$ be the associated quadratic forms. Suppose that $Q_g, Q_h$ have the same zero set; $Q_g^{-1}(\{0\}) = Q_h^{-1}(\{0\})$.

Questions:

*

*Is it then true that $Q_g$ and $Q_h$ are scalar multiples of each other?

*If (1) isn't true as stated, could it perhaps be made true by making certain restrictions on the quadratic forms, such as restricting them to have Lorentz signature (i.e with the notation below, $p=1, n$ arbitrary and $k=0$). This is the case I'm mainly interested in, but I'd of course like to know it in the more general case if it's true.

*Can this result (if true) be generalized to arbitrary multilinear, symmetric functionals on $V$ and their associated homogeneous polynomials (as opposed to bilinear functionals and their quadratic forms). If so could you outline such a proof/ provide a reference.


My Attempt
This reminds me of a similar result in linear algebra, namely that if $\phi,\psi \in V^*$ have the same kernel then they are scalar multiples of each other. So, my attempt at "proving" (1) was to mimic that proof as much as possible. I know that every quadratic form over $\Bbb{R}$ can be "diagonalized" (Sylvester's Law?), in the sense that we can find a basis $\beta$ for $V$ such that the matrix representation of $g$ is of the type
\begin{align}
[g]_{\beta} &=
\begin{pmatrix}
I_p &  & \\
 & -I_n & \\
 & & 0_k
\end{pmatrix}
\end{align}
such that $p+n+k = \dim V$. But from here, I'm not sure how to proceed. Any help is appreciated.

Partial answer:
As can be seen in the positive definite case, it does not generally hold that two bilinear forms are scalar multiplies if they have the same zero set.
However, the statement becomes true for the case of $\dim V \leq 3$ if we extend the bilinear form to a $\Bbb C$-bilinear form over $\Bbb C$.  Without loss of generality, take $V = \Bbb R^3$. Plugging in $v_1 = (1,x,x^2)$, we see that
$$
Q_g(v_1) = g_{11} + 2g_{12} x + 2(g_{22} + g_{13}) x^2 + 2g_{23} x^3 + g_{33}x^4.
$$
Plugging in $v_2 = (-1,x,x^2)$ yields
$$
Q_g(v_2) = -g_{11} - 2g_{12} + 2(g_{22} - g_{13}) x^2 + 2g_{23} x^3 + g_{33} x^4.
$$
If two polynomials over $\Bbb C$ have the same zero-sets, then those polynomials must be multiples. Thus, if $h$ is such that $Q_h(v_1) = Q_h(v_2) = 0$ for all $x \in \Bbb C$, then it must hold that $h_{ij}$ is a multiple of $g_{ij}$ for all $j$, and hence $h$ is a multiple of $g$.
