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Let $A$ and $B$ be two skew symmetric matrices of order $n$ over a finite field with characteristic not equal $2.$ Also let $\mathrm{Rank}(A)=\mathrm{Rank}(B)$ (which is even, we know). Then I want to show that there is an invertible matrix $P$ such that $P^tAP=B$. I need some idea to prove this.

Thanks

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Before solving the problem, let us ponder a little. If the problem statement is true, then you can pick, from each equivalence class of skew symmetric matrices of the same ranks, a representative as a canonical form.

What canonical form would have a nice structure? Diagonal matrix would be nice, but if a matrix is both skew symmetric and diagonal, it must be zero. So, this in general doesn't work. What about a block-diagonal matrix? Perhaps a direct sum of $2\times2$ skew symmetric matrices and a zero sub-block? Can it be $\pmatrix{0&-1\\ 1&0}\oplus\cdots\oplus\pmatrix{0&-1\\ 1&0}\oplus0$?

We shall prove that, when $A$ is a skew symmetric matrix over a field of characteristic $\ne2$, its rank must be $2k$ for some integer $k$ (hence the rank is even) and it is congruent to a direct sum of $k$ sub-blocks of the form $K=\pmatrix{0&-1\\ 1&0}$ and a zero sub-block of size $(n-2k)\times(n-2k)$.

We can prove this by mathematical induction. The cases $n=1$ is trivial. If $n=2$ and $A\ne0$, since it is skew symmetric and the field has characteristic $\ne2$, the matrix must have a zero diagonal. That is, $A=aK$ for some $a\ne0$. Hence $A$ is congruent to $$ \pmatrix{1&0\\ 0&\frac1a}\pmatrix{0&-a\\ a&0}\pmatrix{1&0\\ 0&\frac1a}=K. $$ Now suppose $n>2$. If $A\ne0$, then some principal $2\times2$ submatrix of $A$ is equal to $aK$ for some $a\ne0$. Thus $A$ is congruent (actually permutation-similar) to a matrix of the form $\pmatrix{aK&-V^T\\ V&M}$, which in turn is congruent to $$ \pmatrix{I_2&0\\ \frac1aVK&I_{n-2}}\pmatrix{aK&-V^T\\ V&M}\pmatrix{I_2&-\frac1aKV^T\\ 0&I_{n-2}}=\pmatrix{aK&0\\ 0&M-\frac1aVKV^T}, $$ which is a direct sum of two smaller skew symmetric matrices. Therefore, the conclusion follows by mathematical induction.

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  • $\begingroup$ Although I up-voted the answer, I think it needs to be complete. Actually, when I was reading your answer, I was expecting to see how you find the matrix $P$ while you proved some fact that was already a theorem in Hoffman's book. Could you please let me know how we can conclude that there are such a matrix $P$ when we know that both $A$ and $B$ are similar to some same matrix? $\endgroup$
    – Majid
    Commented May 26, 2018 at 8:58

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