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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 $AB=-BA$

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

Some thoughts

For the condition 4 , we can say the corresponding $B\ne I_n,A$.

Because if $B=I_n$ , we get $A=0$ , a contradiction.

If $B= A$ then $AB=0$ , which contradicts the condition 2.

Thus we can consider $M$ as the set of such pairs $(A,B)$.

But how to move on? Any hints? Thank you in advance!

Added

It's easy to see for all $A \in M$, $A^2$ and $-A^2$ commute with all the matrices in $M$.

Thus from conditions 2 and 4 we get $A^2 = I$ or $-I$, which might help.

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I showed my advisor this and he came up with the following argument.

First I assume you're working over $\mathbb{C}$.

Let $G:=M\cup -M = \{A\in M_n(\mathbb{C}) : \pm A \in M\}$. Then $G$ is a group: it's clearly closed under multiplications, so you just have to show it's closed under inverses. If $A\in G$ then you can show that $A^2\in G$ commutes with every element of $M$ and hence $A^2=\pm I$ since otherwise we could find an element in $M$ that skew-commutes with it. Thus $G$ is a group and every element of $G$ has square $\pm I$.

Now let $A$ denote the $\mathbb{C}$-span of elements of $G$ in $M_n(\mathbb{C})$. Then by Maschke's theorem, $A\cong \prod_{i=1}^d M_{n_i}(\mathbb{C})$ with $\sum n_i = n$. Now since $\sum 2n_i^2 \le 2n^2$, and by looking at projections and using induction, it suffices to consider the case when $A=M_n(\mathbb{C})$. In this case, there is a basis $A_1,\ldots ,A_{n^2}$ of $M_n(\mathbb{C})$ consisting of elements of $G$.

Claim: if $X\in G$ is not $\pm I$ then $X$ has trace zero.

Proof: This can be seen by noting that there is some $Y$ in $G$ with $XY=-YX$ and since $Y$ is invertible, this gives that $X$ is similar to $-X$. QED

To finish it off, we claim that $G$ must be equal to $S=\{\pm A_1,\ldots ,\pm A_{n^2}\}$ and so it has size at most $2n^2$. If not, there is some $X\in G$ that is not in $S$. Then $XA_i \neq \pm I$ for $i=1,\ldots, n^2$ since $A_i^{-1}=\pm A_i$ and so ${\rm Tr}(XA_i)=0$ for $1\le i\le n^2$. But since the $A_i$ span $M_n(\mathbb{C})$ that gives that ${\rm Tr}(XY)=0$ for all matrices $Y$ and so $X=0$, a contradiction, since $X$ is in $G$. So we're done.

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  • $\begingroup$ Well, thank you very much for this nice argument. But I looked through Maschke's theorem, it involves a lot of concepts that I'm not familiar with. Can you give me some hints to point out $A\cong \prod_{i=1}^d M_{n_i}(\mathbb{C})$ in this particular qustion? Thanks in advance! $\endgroup$ – Zero Apr 3 at 7:06
  • $\begingroup$ You should look up the Artin--Wedderburn Theorem. To be fair, this is a somewhat high-powered argument, so be careful if this is meant to be a homework exercise. $\endgroup$ – Ehsaan Apr 3 at 13:23

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