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.



Vanishing bilinear forms
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. 


