Prove that three $2\times2$ matrices that commute are linearly dependent

Statement:

Suppose that $A$, $B$ and $C$ are complex $2\times2$ matrices, any two of which commute under matrix multiplication. Show that $A$, $B$ and $C$ are linearly dependent.

I think one method is to show the existence of $a,b,c\in\mathbb C$, such that $aA+bB+cC=0$ while $a$, $b$, $c$ are not all zero. I'm not sure how to proceed with this.

I observed that if we add an assumption that $A$, $B$ and $C$ are diagonalizable, then they are simultaneously diagonalizable since they all commute. I think this implies that there exists a common $P$ such that $A=PD_1P^{-1}$, $B=PD_2P^{-1}$, $C=PD_3P^{-1}$, where the $D_i$ are diagonal matrices. Any three $2\times2$ diagonal matrices must be linearly dependent because they each have two non-zero entries only. As a consequence, $A$, $B$ and $C$ are linearly dependent.

Unfortunately, not all matrices are diagonalizable. I also tried to use Jordan canonical forms, but all I can see is that three $2\times2$ upper-triangular matrices may not be linearly dependent and that this line of reasoning might lead to a dead end.

Therefore, how to prove the original statement?

Solution 1:

Commuting matrices are simultaneously triangularisable. We may assume $A$, $B$ and $C$ are upper triangular. If they are linearly independent, they span the three-dimensional space $T$ of upper triangular matrices. Therefore all elements of $T$ commute. But they don't.

Solution 2:

It is hard to beat Lord Shark's answer, but we may prove the statement without knowing that $A,B,C$ can be simultaneously triangularised.

Proof. If they are linearly independent, we can find two linearly independent traceless matrices $X$ and $Y$ in their linear span. Hence we may assume that $X$ is either $\operatorname{diag}(-x,x)\ (x\ne0)$ or a nilpotent Jordan block. Any traceless matrix that commutes with $X$ is thus a scalar multiple of $X$. This contradicts the property of $Y$.