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I have a set of $n\times n$ matrices with entries on $\mathbb{F}_2$, given by $$\mathcal{A}=\left\{A\in\mathcal{M}_{n\times n}(\mathbb{F}_2):A= \left( \begin{array}{ccc} I_{k_1} & 0 & 0 \\ 0 & J & 0 \\ 0 & 0 & I_{k_3} \end{array} \right) \right\} $$ where $J=\left( \begin{array}{cccc} 0 & \ldots & 0 & 1 \\ 0 & \ldots & 1 & 0\\ \vdots & \vdots & \vdots & \vdots\\ 1 & \ldots & 0 & 0\end{array} \right)_{b\times b}$ is a square matrix of size $b$ and $I_{k_1},I_{k_3}$ are identity matrices such that $k_1+k_3+b=n$

This set acts a flip block set for the column vectors $X$ such that $X^T=(x_1,x_2,\ldots,x_n)\in\mathbb{F}_2^n$.

I want to break the set of those vectors in some kind of equivalence classes such that every reversed vector $Y$ can be identified with just one of those equivalence classes.

For example, in the case of $n=5$ and $b=3$ the vectors with Hamming weight $1$ are divided in $2$ sets: $\left\{(1,0,0,0,0),(0,0,1,0,0),(0,0,0,0,1)\right\}$ and $\left\{(0,1,0,0,0),(0,0,0,1,0)\right\}$

Because no matter what $A\in\mathcal{A}$ you multiply any of these vectors they will stay in their sets.

I want to get this for any case but I don't know how to express it.

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    $\begingroup$ You mean $I_3$ is identity matrix? You typed $I_2$. $\endgroup$ – edm Sep 28 '18 at 10:30
  • $\begingroup$ yes, I am sorry. Already corrected, thaks for the comment! $\endgroup$ – Luis GC Sep 28 '18 at 13:11
  • $\begingroup$ I'm a bit unsure of the definition of your set of matrices $\mathcal{A}$. Can you be a bit more precise, for example do you fix $(n, b)$ from the outset and take all $\mathrm{diag}(I_a, J_b, I_c)$ such that $a + b + c = n$, or are $a$ and $c$ fixed as well? $\endgroup$ – Joppy Oct 1 '18 at 14:05
  • $\begingroup$ @Joppy Yes, that is exactly what I mean, just after the description of $J$ I said, "$I_{k_1},I_{k_3}$ are identity matrices such that $k_1+k_3+b=n$" Please note that if $k_1,k_3$ ($a,c$ in your notation) are fixed as well then $\mathcal{A}$ contains just 1 matrix. $\endgroup$ – Luis GC Oct 1 '18 at 21:37

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