I have no idea how to prove/disprove this Statement.
Statement:
Let $K$ be a field of arbitrary characteristics. Let $X$ be a skew-symmetric $(n \times n)-$ matrix over $K$ and $H=(h_{ij})$ be skew-symmetric with
$$h_{ij} =
\begin{cases}
0, & i=j \\[2ex]
1, & i >j \\[2ex]
-1 & i<j \\
\end{cases}$$
Then there is $Q$ such that $X=QHQ^T.$
Any help is appreciated. Thank you!
Best Answer
Note that $\operatorname{rank}(H)=n$ when $n$ is even and $\operatorname{rank}(H)=n-1$ when $n$ is odd. To see this, add a subscript to the symbol $H$ to emphasise its size. Since the leading principal $2\times2$ submatrix of $H_n$ is $K$, by blockwise Gaussian elimination (see Wikipedia for instance), $H_n$ is congruent to $K\oplus(H_n/K)$, where $H_n/K$ denotes the Schur complement of $K$ in $H_n$. Direct calculation shows that $H_n/K=H_{n-2}$. Therefore $H_n$ is congruent to $K\oplus H_{n-2}$ and the result about $\operatorname{rank}(H_n)$ follows inductively.
Now suppose $X$ is a skew-symmetric matrix. By "skew-symmetric", I suppose that $X$ has a zero diagonal and $X^T=-X$. (The zero-ness of the diagonal makes a difference only when the characteristic of the underlying field is $2$.) With this definition of skew-symmetry, a very basic fact is that $X$ is congruent to $K\oplus K\oplus\cdots\oplus K\oplus0_{k\times k}$, where $K=\pmatrix{0&-1\\ 1&0}$ and $k$ has the same parity as $n$. This fact can be proved by a similar Schur complement trick to the above. You can also find a proof in any linear algebra textbook that covers bilinear forms (see e.g. theorem 19 of Irving Kaplansky's Linear Algebra and Geometry; if I remember correctly, this has also been proved in Adrian A. Albert's Modern Higher Algebra).
It follows that the rank of a skew-symmetric matrix is always even and the rank of $H_n$ is maximal among all skew-symmetric matrices of the same sizes. Let $H_n=P^T(K\oplus K\oplus\cdots\oplus K\oplus 0_{m\times m})P$ (where $m=1$ when $n$ is odd and $m=0$ when $n$ is even) and $X=R^T(K\oplus K\oplus\cdots\oplus K\oplus 0_{k\times k})R$, where $P$ and $R$ are invertible. Since the rank of $H_n$ is maximal, the number of $K$s in the decomposition of $H_n$ is greater than or equal to the number of $K$s in the decomposition of $X$. Hence $X=Q^TH_nQ$ where $Q=P^{-1}(I_{n-k}\oplus0)R$.