# Two skew symmetric matrices of same rank are congruent.

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

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.
• 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? Commented May 26, 2018 at 8:58