Find all vectors $x+iy$ for which the complex bilinear form is alternating

Question

I have an alternating bilinear form $f:V\times V\rightarrow \mathbb R $ , with $V$ being a real vector space. Secondly, I have the skew symmetric hermitian linear form: $$\overline f_\mathbb C(x+iy,x'+iy')= f(x,x') +f(y,y') +if(x,y')-if(y,x')$$

I would now like to find all vectors $x+iy$ (with $x,y \in V$) for which $\overline f_\mathbb C(x+iy,x+iy) = 0$.

Since I know that $f_\mathbb C(x+iy,x+iy) = if(x,y)-if(y,x)$ (because $f(x,x)$ and $f(y,y)$ equal zero), it follows that for the function to be zero, I have the condition: $f(x,y)=f(y,x)$.

Clearly, for $x=y$ this is true. However, I am struggling to find all vectors, for which this is true. Are there any other vectors, which meet this requirement and if not, how do I know that I have already found all?

Answer 1

We have $$ 0 = f(x+y,x+y) = f(x,x+y)+f(y,x+y) = f(x,x) + f(x,y) + f(y,x) + f(y,y) \\ = f(x,y)+f(y,x) $$ since $f$ is alternating and bilinear. So $f(x,y)+f(y,x)=0$ for any $x,y \in V$. Thus you have to have $f(x,y)=0$ to satisfy your condition (since now we have both $f(x,y)=f(y,x)=-f(x,y)$).