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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!

Manish Saini
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    In the link above a slightly different question was discussed. Besides, there it was about complex numbers, and here it is an arbitrary field. It seems to me that the question is closed without a solid argument. – kabenyuk Sep 24 '21 at 08:43
  • This post addresses a similar question in the case that $K = \Bbb C$. – Ben Grossmann Sep 24 '21 at 13:59
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    The statement is false over a field of characteristic $2$ if it doesn't require skew-symmetric matrices to be alternate (i.e. hollow). In general, it suffices to consider the case where $X$ is nonsingular and $n$ is even. In this case, since all nonsingular alternate matrices are congruent to $K=\pmatrix{0&-I\ I&0}$, you may show that $H$ is nonsingular or use elementary row/column operations to directly show that $H$ is congruent to $K$. – user1551 Sep 24 '21 at 15:57
  • @user1551 okay, But as we know the field of characteristic $2$ has special properties like in this link https://math.stackexchange.com/questions/1573308/whats-so-special-about-characteristic-2#:~:text=In%20a%20field%20of%20characteristic,to%20its%20own%20additive%20inverse. But for all other characteristics except $2$, How we can proceed? – Manish Saini Sep 25 '21 at 10:04
  • Can we prove the given statement by using basics stuff like characteristic polynomial, annihilating polynomial, matrix manipulation etc? – Manish Saini Sep 25 '21 at 10:10
  • This link is also addressed the same problem actually we both are friends working on the same problem, https://math.stackexchange.com/questions/4278098/a-problem-from-research-paper-on-skew-symmetric-matrices – Manish Saini Oct 17 '21 at 10:29
  • To calculate $\det\big(H\big)$ see https://math.stackexchange.com/questions/3453851/how-to-calculate-the-determinant-of-a-fully-negative-symmetric-matrix – user8675309 Oct 20 '21 at 19:22

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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 $Q=P^{-1}(I_{n-k}\oplus0)R$.

user1551
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