There's this bilinear form $f$ such that $\forall x, y \in V \ \ f(x, y) = 0 \implies f(y, x) = 0$. What I need to prove is that this form is symmetric or skew-symmetric.
I don't understand how to approach this. I've tried adding something like $f(x - x, y)$ (which equals to $0$) and such to $f(x, y)$, and then applying the property above to the result, but didn't get anything. I've also tried looking at $f(x, y) - f(y, x)$ and something like $f(x + y - y, y)$, hoping I'd get some revelation if I try to transform these expressions in various ways, but to no avail.
This is not true. Take, for instance $f\colon\mathbb{R}^2\times\mathbb{R}^2\longrightarrow\mathbb R$ defined by$$f\bigl((x_1,x_2),(y_1,y_2)\bigr)=x_1y_2-x_2y_1.$$Then $f(v,w)=0\iff f(w,v)=0$, but $f$ is not symmetric.