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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.