There exists $\lambda > 0 $ with $s_1=\lambda s_2$ if $s_1(v,w)=0$ and $s_2(v,w)=0$.

Question

I am studying for an exam and do not have solutions for this one:

Let $V$ be a finit $R$ or $C$ vector space. Let $s_1$ and $s_2$ be scalar products on $V$ with the following characteristic: If $v,w \in V$ with $s_1(v,w)=0$, then $s_2(v,w)=0$.

Prove or disprove: There exists $\lambda > 0 $ with $s_1=\lambda s_2$.

Thank you for any hint!

Answer 1

Consider $s_1$ as the 'original' scalar product $\def\z{\langle} \def\x{\rangle} \z,\x$ on $V$, and fix an orthonormal basis $e_1,\dots,e_n$.
Then $s_2$ is determined by the values $s_2(e_i,e_j)$ which give an $n\times n$ diagonal matrix by assumption.
Now what happens if some diagonal elements, say $d_1:=s_2(e_1,e_1)$ and $d_2:=s_2(e_2,e_2)$ are different?

Then, though $e_1-e_2\,\perp\,e_1+e_2$, we would have $$s_2(e_1-e_2,\,e_1+e_2)=d_1-d_2\ \ne 0$$