Dgdrxrt

Given $M$ comprised of $ntimes n$ matrices, which satisfies

1. 
   $I_n in M$ and $0_n notin M$
   


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 $Ain M$ and $Ane I_n$, 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 $Bne I_n,A$.

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.

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

First I assume you're working over $mathbbC$.

Let $G:=Mcup -M = Ain M_n(mathbbC) : pm A in M$. Then $G$ is a group: it's clearly closed under multiplications, so you just have to show it's closed under inverses. If $Ain G$ then you can show that $A^2in G$ 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 $mathbbC$-span of elements of $G$ in $M_n(mathbbC)$. Then by Maschke's theorem, $Acong prod_i=1^d M_n_i(mathbbC)$ 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 $A=M_n(mathbbC)$. In this case, there is a basis $A_1,ldots ,A_n^2$ of $M_n(mathbbC)$ consisting of elements of $G$.

Claim: if $Xin G$ 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 $Xin G$ 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 $1le ile n^2$. But since the $A_i$ span $M_n(mathbbC)$ 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.