Suppose $A=(a\_{jk})\_{j,k=1}^n$ is a symmetric complex valued matrix, that is to say, $a\_{jk}=a\_{kj}$ for all $j,k=1,\dotsc,n$. Suppose that given any two **linearly independent** vectors $\alpha=(\alpha^j)\_{j=1}^n, \beta=(\beta^j)\_{j=1}^n \in \mathbb C^n$ that satisfy $$\sum\_{j=1}^n(\alpha^j)^2=\sum\_{j=1}^n(\beta^j)^2 =0,$$
there holds
$$ A(\alpha,\beta)=0.$$
Does it follow that $A=0$?



A matrix identity
For $n = 1$, the conclusion that $A = 0$ trivially fails, as noted in the [comments](https://mathoverflow.net/questions/383252/a-matrix-identity#comment974839_383252).


For $n = 2$, the conclusion that $A = 0$ seems also to fail. We have that any vector $(v\_1, v\_2)$ of ‘norm’ $v\_1^2 + v\_2^2 = 0$ is a multiple of $(1, i)$ where $i^2 = -1$, so that, up to changing which square root of $-1$ we call $i$, we have that $\alpha = \alpha^1(1, i)$ and $\beta = \beta^1(1, -i)$. Then
$$
\alpha^{\mathsf T}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\beta
= \alpha^1\beta^1\begin{pmatrix} 1 & i \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ - i \end{pmatrix} = 0.
$$


