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In this post, we construct a set of matrices with the following properties

Given $M$ comprised of $n\times n$ matrices, which satisfies

  1. $I_n \in M$ and $0_{n} \not\in M$
  2. If $A,B \in M$, then $AB \in M$ or $-AB \in M$
  3. If $A,B \in M$, then $AB = BA $ or $AB = -BA$
  4. If $A\in M$ and $A\ne I_n$, then there exists $B \in M$ such that $AB=-BA$

And in this post we've already proved that the number of elements in $M$ is less than $2n^2$.

What's more if we put $G=M \cup (-M) $, $G$ will become a group.

Now I want to ask another question:

If we classify the sets $M$ by their structures, how many classes can we get?

Thanks in advance!

Added:

$\forall A\in M $ we have $A^2=\pm I$ which leads to $A$ is similar to $\mathrm{diag}(1,\dots,1,-1,\dots,-1)$ or $\mathrm{diag}(i,\dots,i,-i,\dots,-i)$.

Maybe that helps.

Added(2):

I would like to attach something intereting to this post to inspire more people to have a look at this question.

I am going to show you an elementary proof of the question mentioned above, proving the number of elements in $M$ is less than $2n^2$

Considering that the number $n^2$ relates to the dimension of $M_n(\mathbb{C})$, we aim to solve the problem via linear independent terms.

First of all we rewrite the proposition as

Given $M$ comprised of $n\times n$ matrices, which satisfies

  1. $I_n \in M$ and $0_{n} \not\in M$
  2. If $A,B \in M$, then just one of $AB$ and $-AB$ is in $M$
  3. If $A,B \in M$, then $AB = BA $ or $AB = -BA$
  4. If $A\in M$ and $A\ne I_n$, then there exists $B \in M$ such that $AB=-BA$

Prove that the number of elements in the set $M$ is less than $n^2$.

If we remove the element $m \in M $ if $-m\in M$ ( $m\ne I_n$ ) , we'll get a new set which satisfies the new rules. It's not hard to see the number of elements in the new set is larger than the original set's half, and thus we arrive at the original proposition via the new.

Proof: We fist show that if $A\in M$, then $A^2= \pm I_n $

Take any $B\in M$ then $AB=BA $ or $AB=-BA$, and thus $A^2B=BA^2$. So $A^2$ and $-A^2$ commutes with every element in $M$. Thus $A^2= I_n$ or $A^2=-I_n$ via condition 4.

If there are more than $n^2$ elements, then there exists a set $M_1=\{ A_1,A_2,\cdots,A_r\} \subset M$ s.t. $\sum_{i=1}^r \lambda_i A_i=0$ where $\sum_{i=1}^r \lambda_i^2 \ne 0$. WlOG, we assume that $\lambda_1 \ne 0$.

From $A^2 = \pm I_n$ we know $\det A \ne 0$, and thus $A_1 A_k \ne \pm A_1 A_m$, otherwise we get $A_k = \pm A_m$ which contradicts the condition 2.

Thus we have that $\tilde{M_1}=\{\tilde{A_1},\tilde{A_2},\cdots,\tilde{A_r}\}$ is linear dependent where $\tilde{A_i}=A_1A_i\text{ or } -A_1A_i$ is in $M$ and $$ \sum_{i=1}^r \tilde{\lambda_i} \tilde{A_i} =0 \tag{1} $$ where $\tilde{\lambda_1} \ne 0$ and $\tilde{A_1}=I_n$.

Then we can find a $B\in M$ s.t. $\tilde{A_2}B=-B\tilde{A_2}$.

And then we multiply $B$ in $(1)$, $$ \sum_{i=1}^r \tilde{\lambda_i} \tilde{A_i} B = \sum_{i=1}^r \pm \tilde{\lambda_i} B \tilde{A_i} = B \sum_{i=1}^r \pm \tilde{\lambda_i} \tilde{A_i} = 0 $$ which implies that $$ \sum_{i=1}^r \pm \tilde{\lambda_i} \tilde{A_i} = 0 \tag{2} $$ where the coefficient of $\tilde{A_1}$ is $\tilde{\lambda_1}$ and the coefficient of $\tilde{A_2}$ is $-\tilde{\lambda_2}$

Add $(1)$ and $(2)$ we get $$ I_n + \sum_{i=3}^r \mu_i \tilde{A_i} =0 $$

Repeating this operation we get $I_n=0$, a contradiction.

Now I am amazed at the proof in this post, which shows that it relates some profound theories from my point of view, and shows a beautiful structure. And thus I'm quite curious about the structure of the set $M$ and want to classify it just by structure.

As pointed out in the comment, it seems like that I didn't describe my question clearly. What I mean should be that we say two sets belong to the class if we can find a bijective map between them and the map maintains the given operational properties. And I would like to show you an example.

Eg:

We say that $$ \begin{gathered} I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad A=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad B=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad C=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{gathered} $$ and $$ \begin{gathered} I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \tilde{A}=\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \quad \tilde{B}=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \tilde{C}=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{gathered} $$ belong to the same class. Because the bijection to match them in order maintains the given four operational properties. For instance $AB=-BA=C$ corresponds to $\tilde{A}\tilde{B}=-\tilde{B}\tilde{A}=\tilde{C}$

Any helps or instruction would be highly appreciated! I accept the high-powered argument and I'll try to understand that.

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  • 1
    $\begingroup$ I think what you wrote is a bit ambiguous. What do you mean by 'classify the sets $M$ by their structures' or 'classifications'? How your expected answer is like, say when $n=2$? $\endgroup$ – Orat Apr 4 at 9:49
  • $\begingroup$ @Orat If we can find a bijective map between them and the map maintains the given operational properties, then we say they are equivalent and are regarded as belonging to the same class. $\endgroup$ – Zero Apr 4 at 10:53

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