Suppose we have a vector space $V$ over some field $\mathbb F$ together with a skew-symmetric form $\langle\space ,\space \rangle$, i.e., $\langle\space ,\space \rangle$ is bilinear and $\langle v,v\rangle=0$ for all $v\in V$. Then when do we have an orthogonal bases?
Unlike the symmetric form over a field with $\mathbb F$ with $\text{char}(\mathbb F)\ne 2$ or Hermitian form over the complex field $\mathbb C$, $\langle v,v\rangle=0$ for all $v\in V$, so we can not find orthogonal bases like what we did for the above two cases.
Are there classifications for vector spaces equipped with a skew-symmetric form such that there exist orthogonal bases? Thank you!
To expand on Qiaochu's comment, suppose we have a basis $(v_1, \dots, v_k)$ for $V$ such that $\langle v_i, v_j \rangle = 0$ for all $i \neq j$. Well, for $i = j$, we also have $\langle v_i, v_j \rangle = 0$ by the alternating hypothesis. Now, in this basis, the matrix representing the bilinear form is the zero matrix! Thus, we cannot have such a basis unless the bilinear form is $0$.
Edit: Qiaochu edited his comment with more details, so nevermind :)