# The number of elements in a set of matrices with some 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 $$AB=-BA$$

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!

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.

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.

• 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! – Zero Apr 3 at 7:06
• 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. – Ehsaan Apr 3 at 13:23