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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?

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  • $\begingroup$ If $W=\{AB-BA\mid A,B\in M_{n\times n}[F]\}$ then $dim(W)=n^2-1$ and $W=\{c\mid tr(c)=0\, , \, c\in M_{n \times n}[F]\}$. Maybe useful. $\endgroup$ – Amin235 Jul 14 '17 at 11:42
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    $\begingroup$ It means in particular that for example any rotation matrices which always commute for dim. $2$ are linearly dependent? Suprising.. $\endgroup$ – Widawensen Jul 14 '17 at 11:56
  • $\begingroup$ I don't know where mistake lies: Take $n=2$, $V=M_{2\times 2}/W$, then $dim(V)=1$? However, $\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$ and $\left(\begin{matrix}1 & 0 \\ 0 & 2\end{matrix}\right)$ seems to be linear independent?! $\endgroup$ – Rofl Ukulus Jul 14 '17 at 12:06
  • $\begingroup$ Suppose that $V$ be a vector space of dimension $n$ and $L(V,V)$ be the space of all linear transformation over $V$. Now let $W$ be a subspace of $V$ that is generated by all linear transformation as the following form $f=T_1T_2-T_2T_1$ then $dim(W)=n^2-1$. For more details you can study chapter of linear transformation of every linear algebra book. tr $\endgroup$ – Amin235 Jul 14 '17 at 12:31
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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.

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  • $\begingroup$ This is wonderful. $\endgroup$ – Frenzy Li Jul 14 '17 at 12:05
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    $\begingroup$ Is there a quick and easy proof of simultaneous triangularisability? (Man, is that word hard to type!) Somehow in my education I never really picked up the simultaneous diagonalization/triangularization stuff, and I'm sure one of my Linear Algebra books at work covers it nicely...but if there's a quick and dirty proof, I'd love to see it. $\endgroup$ – John Hughes Jul 14 '17 at 12:36
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    $\begingroup$ @JohnHughes In the 2-by-2 case, all you need is a simultaneous eigenvector. Unless $A$ is a scalar matrix, it will have some 1-dimensional eigenspace, and that will be stable under $B$ and $C$ by the commuting. $\endgroup$ – Lord Shark the Unknown Jul 14 '17 at 12:38
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    $\begingroup$ Ah...got it. And from there, it's probably not too tough to generalize inductively or something (esp. over $\Bbb C$ rather than $\Bbb R$). $\endgroup$ – John Hughes Jul 14 '17 at 13:58
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    $\begingroup$ @FrenzyLi No, it means there is an invertible $M$ with $MAM^{-1}$ and $MBM^{-1}$ upper triangular. $\endgroup$ – Lord Shark the Unknown Sep 12 '17 at 16:26
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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 $Y$.

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  • $\begingroup$ Could you elaborate on why "if they are linearly independent, we can find two linearly independent traceless matrices $X$ and $Y$ in their linear span."? $\endgroup$ – Frenzy Li Jul 14 '17 at 15:15
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    $\begingroup$ @FrenzyLi If $A,B,C$ are traceless, pick any two of them. If $C$ has a nonzero trace instead, then $X=A-(tr(A)/tr(C))C$ and $Y=B-(tr(B)/tr(C))C$ are linearly independent and traceless. $\endgroup$ – user1551 Jul 14 '17 at 15:21

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