For each alternating bilinear form $f$ over a finite dimensional vector space $V$, there exists a basis of $V$ such that the matrix of $f$ is given by $$ \begin{pmatrix} 0&1& & & & & & & \\ -1&0& & & & & & & \\ & &\ddots & & & & & \\ & & &0&1& & & \\ & & & -1& 0& & & \\ & & & & &0& & \\ & & & & & & \ddots& \\ &&&&&&&0\end{pmatrix}.$$ (all other elements are zeroes!)
My question:
I understand where the blocks $\begin{pmatrix} 0&1\\-1&0\end{pmatrix}$ come from, but the sequence of $0$'s on the diagonal after the last block confuses me. Where does it come from?
Edit: I think it has to do with the radical of $f$ being trivial or not, but I can't seem to find a straightforward explanation.
In order to find such a basis, you need to do something like Gram-Schmidt, but with an alternating form rather than a symmetric inner product. In particular, what goes wrong with alternating forms is that they might have a radical (with inner products, this never happens). The first step in the "orthogonalisation" process is to get rid of the radical.
Let $f: V \times V \to \mathbb{F}$ be the alternating form. Let $$ R = \{v \in V \mid f(v, w) = 0 \text{ for all } w \in V\}$$ be the radical of $V$, and choose some complementary subspace $W$ such that $V = R \oplus W$. Now $f$ is a nondegenerate bilinear form on $W$, so for any subspace $U \subseteq W$, we may define the orthogonal complement $$ U^f = \{ w \in W \mid f(w, u) = 0 \text{ for all } u \in U\}$$ and the orthgonal complement now has nice properties. In order to start orthogonalising, choose a nonzero vector $e \in W$, and take any $g \in W$ such that $f(e, g) = 1$. Now, show that $W = \langle e, g \rangle \oplus \langle e, g \rangle^f$, and that $f$ is again nondegenerate on $\langle e, g \rangle^f$, and by induction you get the basis you seek.