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So I have a bilinear form $\phi$ and a basis $\{e_1,e_2,\cdots, e_n\}.$ 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 something like this:

$$A = \left(\begin{array}{cccccc} \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right] & & & & & 0\\ & \ddots\\ & & \left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right]_k\\ & & & 0\\ & & & & \ddots\\ 0 & & & & & 0 \end{array}\right).$$

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.

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  • $\begingroup$ The diagonal dots represent 0 terms? $\endgroup$ – Bernard Mar 31 at 16:38
  • $\begingroup$ In the matrix $A$ the dots on the diagonal represent the same block being continued $k$ times. $\endgroup$ – model_checker Mar 31 at 16:40
  • $\begingroup$ Even in the lower part of the diagonal? $\endgroup$ – Bernard Mar 31 at 16:41
  • $\begingroup$ No, not in the lower part. That will be all zeros. $\endgroup$ – model_checker Mar 31 at 16:42
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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 $B$.

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  • $\begingroup$ Could you give some justification, if possible? $\endgroup$ – model_checker Mar 31 at 17:00
  • $\begingroup$ This puts $I_k$ on the top right and $-I_k$ on the bottom left. I want the reverse form? $\endgroup$ – model_checker Mar 31 at 18:07

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