If $G(\cdot,\cdot)$ is a symmetric bilinear form, and we know that $G(X,X)<0$ and $G(Y,X)=0$ then it must be the case that $G(Y,Y)>0$?

Question

I think the statement above should be true, and I want it to be; I tried using the polarisation identity but I ended up with a tautology (0 equals 0)...are there any easy elementary arguments that show that the statement holds?

Answer 1

On $M_2(\mathbb R)$, let $$ G(X,Y)= -X_{11}Y_{11}-X_{22}Y_{22}. $$ Now let $$ X=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\ \ \ Y=\begin{bmatrix} 0&0\\0&1\end{bmatrix}. $$ Then $$ G(X,X)=-1, \ \ G(Y,Y)=-1,\ \ G(X,Y)=0. $$

Answer 2

If $G$ has a negative index of 1, then $G(Y,Y) \geq 0$.

Otherwise, if $G$ has a negative index of 2 or more, then $G(Y,Y) < 0$ is possible, as there are at least 2 orthogonal vectors (with respect to $G$) with $G(X,X)<0$ and $G(Y,Y)<0$ and $G(X,Y)=0$.