For a symmetric or antisymmetric bilinear form $\varphi$ on a vector space $V$, if $\varphi(x,y)=0$ then also $\varphi(y,x)=0$ ($x,y\in V$).
I was wondering if this is also a necessary condition for this to happen, that is whether a form with this property must be either symmetric or antisymmetric.
Assume we are not in characteristic 2, then this is true. Let $x_0 \in V$. Then the functional $\phi(x_0,y)$ of $y$ and the functional $\phi(y,x_0)$ have the same zero set. It follows that they are proporstional: $\phi(x_0,y) = \lambda(x_0)(y,x_0)$ (or the opposite, but it will not change anything for the proof). Substitute $y = x_0$ we get $\lambda(x_0)\phi(x_0,x_0) = \phi(x_0,x_0)$ so if $\phi(x_0,x_0) \ne 0$ then $\lambda(x_0) = 1$ and the form satisfies $\phi(x_0,y) = \phi(y,x_0)$ for every $y$.
If $\phi(x_0,x_0)$ is identically 0, then the form is anti-symmetric and we are done. Otherwise, the set of $x_0$ for which $\phi(x_0,x_0) \ne 0$ span $V$. Indeed, for each $x_0$ for which $\phi(x_0,x_0) \ne 0$ and each $y$ for each it does, we get $\phi(x_0 + \lambda y, x_0 + \lambda y ) = \phi(x_0,x_0) + \lambda (\phi(x_0,y) + \phi(y,x_0))$ and we can choose $\lambda \ne 0$ such that this expression is non-zero, since the field have more than 2 elements. But then, since the equality $\phi(x_0,y) = \phi(y,x_0)$ holds for all $y$ and all $x_0$ in a spanning set, it is true for all $x_0,y$. So the form is symmetric.
