If $C$ commutes with certain matrices $A$ and $B$, why is $C$ a scalar multiple of the identity?

I'm self studying Steven Roman's Advanced Linear Algebra, and this is problem 10 of Chapter 8.

Let $A,B\in M_2(\mathbb{C})$, $A^2=B^3=I$, $ABA=B^{-1}$, but $A\neq I$ and $B\neq I$. If $C\in M_2(\mathbb{C})$ commutes with $A$ and $B$, then $C=rI$ for some $r\in\mathbb{C}$.

Is there a way to solve this without writing out arbitrary matrices and attempting to solve a huge system of equations? The only thing I observe is that $A=A^{-1}$, so $B\sim B^{-1}$, so $B$ and $B^{-1}$ have the same characteristic polynomial. I'm stymied trying to show $C$ is diagonal, let alone a multiple of $I$. Thanks for any ideas.

I should add that I know that the center $Z(M_n(\mathbb{C}))$ consists of scalar multiples of $I$, but I don't see any reason to assume or prove $C$ commutes with everything.

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.

• Thanks Geoff, I can follow this. Commented Mar 11, 2013 at 7:39
• @TiffanyHwang: $B^{-1}=B^{2}$, so if $B$ has eigenvalue $\omega$, then $B^{-1}$ must have eigenvalue $\omega^{2}$. We know $B$ must have either eigenvalue $\omega$ or $\omega^{2}$ because $B\not=I$. Now use $B\cong B^{-1}$. Commented Mar 11, 2013 at 7:40

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:

1. $A^{2}=I$.
2. $(BA)^{2}=I$.
3. $B^{3}=I$.

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.

• @Geoff's beautiful solution helps me to understand linear algebra better as well. I have not touched the subject for a while. Commented Mar 11, 2013 at 7:48

As $B\sim B^{-1}$ and $B^3=I\neq B$, the eigenvalues of $B$ are the two primitive cube roots of unity, $\omega$ and $\omega^2$. Hence we may assume WLOG that $B=\operatorname{diag}(\omega,\omega^2)$. To commute with it, $C$ must be a diagonal matrix.

Now, $B$ does not commute with $A$, or else $A^2=B^3=I$ and $ABA=B^{-1}$ would imply that $B^2=B^3=I$, i.e. $B=I$. It follows that $A$ is not a diagonal matrix. Since $C$ is $2\times2$ diagonal matrix that commutes with this non-diagonal matrix $A$, its two diagonal entries must be identical and hence it is a scalar multiple of $I$.