Let $k$ be an arbitrary field and $n >1$. We define a bilinear form on $k^n$ by $\langle x,y \rangle := x^t y \in k$. Assume $v \in k^n$ with $\langle v,v \rangle=1$.
My question: Is there $w \in k^n$ with $\langle w,w \rangle=1$ and $\langle v,w \rangle=0$?
My answer: If $n$ is even then yes since we can take $w=(v_2,-v_1 , v_4,-v_3 , \dots)$. And clearly, it works for $k = \mathbb R$.
But I don't know we this should be true for $n$ odd. Are there counter-examples?