Orthogonal Bases w.r.t a given Bilinear Form. Under what conditions there exists an orthogonal basis? Or even better, is there a characterization of the existence of an orthogonal basis in terms of a given bilinear form and/or the base field?
For instance, if the characteristic of the field is not 2 and the bilinear form is symmetric, then there exists an orthogonal basis and we can as well extend to an orthogonal basis. Now, the proof is based on the fact that if $B$ is a nonzero symmetric bilinear form, then there exists $v \neq 0$ such that $B(v,v)\neq 0$.
So, is the above possible if the bilinear form is alternating, i.e., $B(v,v) = 0$ for all $v$? Or skew-symmetric, i.e., $B(v,w) = -B(w,v)$ for all $v,w$?
I am primarily interested on non-degenerate bilinear forms, but I am curious about the degenerate case as well. 
 A: Let's take $B$ to be a bilinear form $B : V \times V \rightarrow K$ on a finite-dimensional $K$-vector space $V$; where $K$ is a field of characteristic $p$.
Take an orthogonal basis, i.e. a basis $e_1, \ldots, e_n$ such that $B(e_i, e_j) = 0$ if $i \neq j$.


*

*If $B$ is alternating, then $B(e_i, e_i) = 0$, so in fact $B$ is zero.

*If $B$ is skew-symmetric and we have $p \neq 2$, then $B$ is alternating and thus again $B$ is zero.

*When $p = 2$, the bilinear form $B$ is skew-symmetric iff it is symmetric.
So the only interesting case is the one where $B$ is symmetric. We can reduce to the case where $B$ is non-degenerate: write $V = W \oplus V^\perp$, here $V^\perp$ is the radical of $B$. You can see that $V$ has a orthogonal basis if and only if $W$ has a orthogonal basis, and that the restriction of $B$ to $W$ is non-degenerate. 
So we might as well assume that $B$ is a non-degenerate symmetric bilinear form.
Assume that $p \neq 2$ and that $B$ is nonzero. As mentioned in your question, then $B$ does have an orthogonal basis.
We still have the situation where $p = 2$ and $B$ is a non-degenerate symmetric bilinear form. In this case you can write $V = W \oplus W'$, where $W$ has an orthogonal basis, and where restriction of $B$ to $W'$ is alternating. So $V$ has an orthogonal basis if and only if $W' = 0$.
