If $C$ commutes with certain matrices $A$ and $B$, why is $C$ a scalar multiple of the identity?
Solution 1:
Note that $B$ must have both primitive cube roots of unity as eigenvalues, as $B$ is conjugate to its inverse and has order $3$. Then note that $A$ and $B$ can have no common eigenvector, since if $Av = \lambda v$ and $Bv = \omega v,$ then $\lambda^{2} = 1$, so $ABAv = \lambda^{2}\omega v = \omega v,$ while $ABAv = \omega^{-1}v,$ a contradiction.
This means that there is no $1$-dimensional subspace which is left invariant by both $A$ and $B.$ But if $\mu$ is an eigenvalue of $C,$ and $A$ and $B$ commute with $C,$ then the $\mu$-eigenspace of $C$ is invariant under both $A$ and $B,$ so must be two-dimensional, and $C = \mu I.$
This is really an instance of Schur's Lemma from representation theory.
Solution 2:
This might be a more elementary approach. You only need to show that the power of $A,B$ spans $M_{2}(\mathbb{C})$. Then you can use Schur's lemma to this. From what you have you can establish the following:
- $A^{2}=I$.
- $(BA)^{2}=I$.
- $B^{3}=I$.
Since neither of $A,B$ is in $I$, we claim $A,B,B^{2},BA$ linearly independent from each other. We prove this inductively, $A\not=cB$ is clear. If $A=cB+dB^{2}$, then the fact $A^{2}=I$ would imply $c^{2}B^{2}+d^{2}B+(2cd-1)I=0$. But we know $B^{2}+B+I=0$. This force $A=B+B^{2}$, which implies $A=-I$ and $B^{2}=I$, which is impossible. If $A=cB+dB^{2}+eBA$, then similar manipulations showed this is impossible as well. So $A,B,B^{2},BA$ spans $M_{2}(\mathbb{C})$ as claimed.