Prove $(AB-BA)^2$ and $C$ commute in two dimensional space? Suppose $V$ is a two dimensional vector space,$A,B,C$ are linear transformations on $V$, how to prove $(AB-BA)^2$ and $C$ commute?
I see the proof using the content of trace, how can we prove without it?
 A: Proposition: Let $R$ be a commutative ring with unity and $A,B\in M_2(R)$. Then $(AB-BA)^2$ is a scalar matrix.
Proof. case 1. $R$ is a field. Let $\bar{R}$ be its algebraic closure and $B=[b_{i,j}]$.
Method 1. (mini calculation with hand) Case 1.1. $A$ is diagonalizable over $\bar{R}$; we may assume $A=diag(u,v)$ and $(AB-BA)^2=-b_{1,2}b_{2,1}(u-v)^2I_2$. Case 1.2. $A$ is not and we may assume $A=\begin{pmatrix}u&1\\0&u\end{pmatrix}$ and $(AB-BA)^2={b_{2,1}}^2I_2$.
Method 2. (The best one) $trace(AB-BA)=0$ implies that $spectrum(AB-BA)=\{\lambda,-\lambda\}$. If $\lambda=0$, then $AB-BA$ is nilpotent and $(AB-BA)^2=0$. Otherwise $AB-BA$ is diagonalizable, then $spectrum((AB-BA)^2)=\{\lambda^2,\lambda^2\}$, $(AB-BA)^2$ is diagonalizable and we are done.
Case 2. $R$ is a ring. $(AB-BA)^2=f(A,B)I_2$ is a FORMAL equality that uses only additions and multiplications concerning the $a_{i,j},b_{i,j}$. It is valid over any fielk $K$; then it is valid over any ring.
A: The only linear transformations that commute with all linear transformations are multiples of the identity, so you want to show that $(AB - BA)^2$ is a multiple of the identity.  If you don't want to use trace at all, you could just brute-force multiply it out (easier if you have a CAS).
