I have a bilinear form $\phi$ and a basis $\{e_1,e_2,\dots, e_n\}.$ Using this I construct a matrix $A$ such that $A_{i,j} = \phi(e_i,e_j)$ for $1\leq i,j\leq n.$
This gives me a matrix that looks like this:
$$A = \left(\begin{array}{cccccc} \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right] & & & & & 0\\ & \ddots\\ & & \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right]\\ & & & 0\\ & & & & \ddots\\ 0 & & & & & 0 \end{array}\right).$$
I want to show that if we reorder the basis then we can get, $$B =\begin{bmatrix} 0 & -I_k & 0 \\ I_k & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$$
Here $I$ is the identity matrix. I tried a bunch of examples, but I could not see any special pattern. Any ideas will be much appreciated.
Notations: Rewrite your basis as: $(f_1,f_{-1},f_2,f_{-2},\ldots,f_k,f_{-k},g_1,\ldots,g_s)$. Under this basis you obtain the matrix $A$. Now reorder the basis as: $(f_{-1},f_{-2},\ldots,f_{-k},f_1,f_2,\ldots,f_k,g_1,\ldots,g_s)$, you will get $B$.