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Let $N=\{1,\dots,n\}$ and $A,B$ be $n\times n$ skew symmetric matrices such that it is possible to permute some rows and some columns from $A$ to get $B$. In other words, for some permutations $g,h: N\rightarrow N$, $$A_{i,j}=B_{g(i),h(j)}$$ for all $1\leq i,j\leq n$. Must there exist a permutation $f:N\rightarrow N$ such that $$A_{i,j}=B_{f(i),f(j)}$$for all $1\leq i,j\leq n?$

For example, let $$A=\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix} , B=\begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix} $$ If we switch the rows and also switch the columns, we get from $A$ to $B$. And there exists a permutation $f$ with $f(1)=2,f(2)=1$ such that $A_{i,j}=B_{f(i),f(j)}$ for all $1\leq i,j\leq 2$.

There exists an example with $g\neq h$. Let $$A=\begin{pmatrix} 0 & 0 & 2 & -2 \\ 0 & 0 & -2 & 2 \\ -2 & 2 & 0 & 0 \\ 2 & -2 & 0 & 0 \end{pmatrix} , B=\begin{pmatrix} 0 & 0 & -2 & 2 \\ 0 & 0 & 2 & -2 \\ 2 & -2 & 0 & 0 \\ -2 & 2 & 0 & 0 \end{pmatrix} $$

One possibility for $g,h$ is $g(i)=i$ for all $i$, $h(1)=2,h(2)=1,h(3)=4,h(4)=3$. In this case we can let $f(1)=2,f(2)=1,f(3)=3,f(4)=4$.

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  • $\begingroup$ Do you know any examples with $g \neq h$? $\endgroup$
    – uranix
    Aug 16 '16 at 7:07
  • $\begingroup$ I've added such an example $\endgroup$
    – Karo
    Aug 16 '16 at 7:13
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This is a counterexample. Let $O,I,J$ be the $2\times 2$ matrices $$\begin{pmatrix}0&0\\0&0\end{pmatrix},\,\,\begin{pmatrix}1&0\\0&1\end{pmatrix},\,\,\begin{pmatrix}0&1\\1&0\end{pmatrix},$$ respectively. Define the two $6\times6$ matrices $$A=\begin{pmatrix}O&I&I\\-I&O&I\\-I&-I&O\end{pmatrix},\,\,\,\,\, B=\begin{pmatrix}O&J&J\\-J&O&J\\-J&-J&O\end{pmatrix}.$$ We see that $B$ can be obtained by applying the permutation $(12)(34)(56)$ to the rows of $A$. On the other hand, there is no permutation $\sigma$ such that $A_{ij}=B_{\sigma(i)\sigma(j)}$. Unfortunately, I have no intuitive explanation why there is no such $\sigma$, but you can check this fact here (this is A) and here (this is B) --- there is no $\sigma$ sending the eigenvectors of $A$ to the eigenvectors of $B$.

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  • $\begingroup$ @EwanDelanoy I think that we get $J$ applying $(12)$ to the rows of $I$, so $A$ turns into $B$ with $(12)(34)(56)$. Did not you miss that I apply it only to the rows, not to both rows and columns? $\endgroup$
    – heptagon
    Aug 16 '16 at 14:15
  • $\begingroup$ Indeed I missed that. By the way, you have to edit your answer for me to be able to undo my downvote. $\endgroup$ Aug 16 '16 at 14:17
  • $\begingroup$ @EwanDelanoy 1 is not an eigenvalue of $B$. More than that, the reasoning with eigenvalues is not going to work because $A$ and $B$ are similar according to the WolframAlpha computation available via the links. $\endgroup$
    – heptagon
    Aug 16 '16 at 14:33
  • $\begingroup$ @EwanDelanoy I have edited the question, so you are able to undo the downvote if you want. $\endgroup$
    – heptagon
    Aug 16 '16 at 18:36
  • $\begingroup$ Done! you're right, my second comment was wrong too, I deleted it. I wasn't in top form yesterday. $\endgroup$ Aug 17 '16 at 7:02
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Here is a simple explanation of why the matrices $A$ and $B$ in heptagon's answer are not permutation-similar. Suppose the contrary. Then their positive parts must be permuation-similar too, i.e. $$ A_+=\pmatrix{0&I&I\\ 0&0&I\\ 0&0&0} \ \sim\ B_+=\pmatrix{0&J&J\\ 0&0&J\\ 0&0&0} $$ via some permutation. This is impossible, because the directed graph represented by the adjacency matrix $A_+$ is comprised of the paths $1\to3\to5$, $1\to5$, $2\to4\to6$ and $2\to6$, so that a node can reach at most two other different nodes, while $B_+$ is comprised of the paths $1\to4\to5$, $1\to6$, $2\to3\to6$ and $2\to5$, so that each of node 1 and node 2 can reach three other different nodes.

Alternatively, if $A_+$ is permuataion similar to $B_+$, the directed graphs represented by these two adjacency matrices must be isomorphic. So, if we turn each edge to be an undirected one, the two undirected graphs must be isomorphic too. Yet this is impossible, because the undirected graph represented by $A_++A_+^T$ has two disjoint connected components $\{1,3,5\}$ and $\{2,4,6\}$, but $B_++B_+^T$ is a connected graph.

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  • $\begingroup$ Thanks for the clarification of the argument! I have a minor comment. I think the graph of $A_+$ contains the edge $1\to5$, too. Also, I guess it should be $2\to3\to6$ in the penultimate row. $\endgroup$
    – heptagon
    Aug 16 '16 at 22:53
  • $\begingroup$ An easier explanation can be that the (undirected) graph of $B$ is connected while the graph of $A$ is not. $\endgroup$
    – heptagon
    Aug 16 '16 at 22:53
  • $\begingroup$ @heptagon 1) Thanks for the catch. Fixed now. 2) I've actually thought of an explanation using undirected graph, but there is a minor technical issue: the adjacency matrix of an undirected graph must be symmetric. $\endgroup$
    – user1551
    Aug 17 '16 at 8:46
  • $\begingroup$ I mean the graphs of $A$ and $B$, not those of $A_+$ and $B_+$. I suppose there is an edge between $i$ and $j$ in $G_A$ if and only if $A_{ij}\neq0$ (or $A_{ji}\neq0$ which is equivalent because $A$ is skew-symmetric). $\endgroup$
    – heptagon
    Aug 17 '16 at 8:49
  • $\begingroup$ @heptagon Yes, that would work too. $\endgroup$
    – user1551
    Aug 17 '16 at 8:55

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