I am not used to working with quadratics forms and their isotropy vectors, and I have trouble with the following question:
Let $x,y\in\mathbb{R}^n$ both different from $0^n$ and define $A=\begin{pmatrix}-\frac{1}{2} & \frac{1}{2}x^\top\\\frac{1}{2}x & 0^{n\times n}\end{pmatrix}$ and $B= \begin{pmatrix}0&\frac{1}{2}y^\top\\\frac{1}{2}y&I_n\end{pmatrix}$, where $I_n$ is the identity matrix of dimension $n\times n$.
Does there exist a vector $c\in\mathbb{R}^{n+1}$ such that $c^\top Ac=c^\top Bc=0$? If the answer is "no" in the general case, can we add conditions on the choice of $x$ and $y$ so that it becomes "yes"?
Here are my thoughts: First, the null cones of these forms are not subspaces in the general case. But we can still try to look at the isotropic index of these forms. For any vector $c\in\mathbb{R}^{n+1}$ whose first coefficient is $0$, we have $c^\top Ac=0$ but $c^\top Bc=c^\top c\neq 0$ (except if $c=0^{n+1}$). This shows that the isotropic index of $A$ is exactly $n$ (as it cannot be $n+1$ since $A\neq 0^{n+1\times n+1}$). But this also shows that the intersection between the $n$-dimensional subspace $\{(0~~c)\in\mathbb{R}^{n+1}| c\in\mathbb{R}^n\}$ and any maximal isotropic subspace of $B$ is empty: the isotropic index of $B$ is $\le 1$. It is, in fact, exactly $1$ as $\begin{pmatrix}1\\-y\end{pmatrix}$ is an isotropic vector of $A$.
And I do not know where to go from here. Is there any generic method for computing null cones intersection that I am unaware of? Thanks in advance for your answers.