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Given statement is

Note: In this research paper all calculation is under field of characteristic $3$ unless specified.

$\pmb{proposition:}$ 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 $X=QHQ^T.$

Lemma: Let $K$ be a field of arbitrary characterstic. Let S be a skew-symmetric $(n \times n)-$ matrix over $K$ and $\hat H=(\hat h_{ij})$ with

$$\hat h_{ij} = \begin{cases} 0, & i=j \\[2ex] 1, & i =j+1 , i \leq n-1 \\[2ex] -1 & i=j-1, i\geq2 \\ \end{cases}$$

Then there is $P$ such that $S=P\hat HP^T.$

Proof: We apply above lemma twice. first $S=H$ and get $P$ such that $$S=H=P\hat HP^T$$ from this we get $$\hat H= P^{-1} H(P^{-1})^T.$$ Next with $S=X$ and get $P_1$ such that

$$S=X=P_1 \hat H (P_1)^T=P_1P^{-1}H(P^{-1})^T(P_1)^T $$

Now we put $$Q=P_1P^{-1}.$$

My doubts (1) In the given lemma I did't get all the entries of any order matrix. That is the matrix in the lemma is not well defined I think!

(2) As all the calculation in this paper is over modulo $3$ So can I take for particular $n=3$ numbering of rows as $0, 1 ,2 $ and same for column $0,1,2$ if yes then also i did't get all entries of the matrix $H=\hat h_{ij}$

So, If matrix in the lemma is not correct the how we can define it properly??

and then we will be able of understand the proof(then it seems easy!)

Also, I read this answer for the proof Whether a nondegenerate skew-symmetric matrix is congruent to the matrix $\begin{bmatrix} 0 & I_{\ell} \\ -I_{\ell} & 0 \end{bmatrix}$

but I want to to understand the above procedure.

So please any help is well appreciated

Mathematics learner
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  • How is the lemma proved? It seems indeed a little strange that $\hat H$ is not fully defined. About the characteristic, is it $3$ or is it arbitrary? Please clarify. – Berci Oct 16 '21 at 07:53
  • @Berci This particular lemma/propostions is defined on field of arbitrary characteristic but in the starting of the paper he write all computations are on modulo 3 unless stated – Mathematics learner Oct 16 '21 at 08:45
  • @Berci I don't know how to proved lemma, In fact that's is my main doubt – Mathematics learner Oct 16 '21 at 08:47
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    As Berci has pointed out above, the definition of $\hat{H}$ doesn't seem correct. It is not skew-symmetric and the entries on or above the second super-diagonal and those on or below the second sub-diagonal are unspecified. Perhaps you meant the tridiagonal matrix $$\hat{h}_{ij}=\begin{cases}1,&i=j+1,\ -1,&i=j-1,\ 0,&\text{otherwise}.\end{cases}$$ – user1551 Oct 18 '21 at 20:23
  • @user1551 (i) Is $\hat h_{ij}$ works for a field of arbitrary characteristic to prove the given lemma? if yes then by using this $\hat h_{ij}$ the rest proof is correct or not? (ii) How we can relate this answer with answer in this link https://math.stackexchange.com/questions/1682385/whether-a-nondegenerate-skew-symmetric-matrix-is-congruent-to-the-matrix-begin – Manish Saini Oct 19 '21 at 07:25
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    @ManishSaini As mentioned in a comment under my answer to your question, the proof of the proposition here using the lemma is incomplete, because the author has not shown that the $P$ on the line $H=P\hat{H}P^T$ can be chosen to be invertible. And as mentioned in my answer, the proposition itself is valid provided that "skew-symmetric" matrices are required to have a zero diagonal. This requirement is redundant if the characteristic of the field is not 2, but necessary otherwise. – user1551 Oct 19 '21 at 10:12
  • @user1551 Great, Thanks! – Manish Saini Oct 19 '21 at 10:16
  • @ManishSaini And the answer you mentioned clearly works only in inner product spaces, because it uses inner products explicitly. – user1551 Oct 19 '21 at 10:51
  • @user1551 Okay Thanks! – Mathematics learner Oct 20 '21 at 07:13

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