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Introduction: Given a matrix A of a $k$ regular graph G. The matrix A can be divided into 4 sub matrices based on adjacency of vertex $x \in G$. $A_x$ is the symmetric matrix of the graph $(G-x)$, where $C$ is the symmetric matrix of the graph created by vertices of $(G-x)$ which are adjacent to $x$ and $D$ is the symmetric matrix of the graph created by vertices of $(G-x)$ which are not adjacent to $x$. For example,
$$ A_x = \left(\begin{array}{cccccc|ccc|c} 0&1&0&0&1&0&1&0&0&0\\ 1&0&1&0&0&0&0&1&0&0\\ 0&1&0&1&0&0&0&0&1&0\\ 0&0&1&0&0&1&1&0&0&0\\ 1&0&0&0&0&1&0&0&1&0\\ 0&0&0&1&1&0&0&1&0&0\\ \hline 1&0&0&1&0&0&0&0&0&1\\ 0&1&0&0&0&1&0&0&0&1\\ 0&0&1&0&1&0&0&0&0&1\\ \hline 0&0&0&0&0&0&1&1&1&0\\ \end{array}\right) = \left( \begin{array}{ccc} C & E & 0 \\ E^{T} & D & 1\\ 0 & 1 & 0\\ \end{array} \right) $$
It should be noted that
Interchanging/swapping any two rows (or columns) of $C$ does not affect matrix $D$ (and vice versa).
Any change in $C$ or $D$ or both $C$ and $D$ changes matrix $E$.
If some vertices of $G$ is rearranged (i.e., permuted), $A$ will be different, say, this new matrix is $B$. Again, matrix $B$ can be divided into 4 sub matrices based on adjacency of vertex $x \in G$ and $ B_x$ can be obtained.
$ B_x= \left( \begin{array}{cc}
S & R \\
R^{T} & Q\\
\end{array} \right) $
Given: Both matrices $A,B$ are divided based on same vertex $x$, and $D=Q$ (so that $A_x=B_x $ happens, because for all $D=Q$,$A_x=B_x $ may not happen ).
Claim: Since $D=Q$, To get $A=B$, only $S$ matrix needs to be reordered(so that $S=C$ which will imply $A=B$) . To get $A=B$, so there will be no change/permutation/ swapping in columns of $R$ to get $E=R$.
Question: Is this claim correct?
This question is related to this post.
