Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$?

Question

I would like to verify the claim:

$$\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$$

I tried verifying through an example:

Given the following $3$ different matrices:

\begin{align} A & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{pmatrix}\\[2ex] B & = \begin{pmatrix} 4 & 3 & 2 & 1 \\ 3 & 6 & 4 & 2 \\ 2 & 4 & 6 & 3 \\ 1 & 2 & 3 & 4 \end{pmatrix}\\[2ex] C & = \begin{pmatrix} 4 & 3 & 2 & 1 \\ 5 & 6 & 7 & 8 \\ 8 & 7 & 6 & 5 \\ 4 & 3 & 2 & 1 \end{pmatrix} \end{align}

I calculated $\operatorname{Tr}(ABC) = 7930$ and $\operatorname{Tr}(CBA) = 7510$.

Is there any thing wrong in my calculation, or does this prove that $\operatorname{Tr}(ABC) \neq \operatorname{Tr}(CBA)$?

Many thanks!

Answer 1

As suggested in the comments, the statement $\text{Tr}(ABC)=\text{Tr}(CBA)$ is not true in general, so you should not be concerned that you found a counterexample. Actually, there is a simpler counterexample. Take $$ A= \begin{bmatrix} 5&2\\ 0&0 \end{bmatrix},\:\:\: B= \begin{bmatrix} 2&7\\ 5&3 \end{bmatrix},\:\:\: C= \begin{bmatrix} 1&1\\ 1&1 \end{bmatrix}. $$ Then $\text{Tr}(ABC)=49$ while $\text{Tr}(CBA)=61$.

Answer 2

In fact, $\mathrm{Tr}\,(ABC)=\mathrm{Tr}\,(CBA)$ cannot hold for all $A$, $B$, $C$ so long as the ring of scalars is nontrivial. For let \begin{align} A &= \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \\[2ex] B &= \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \\[2ex] C &= \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix} \\[2ex] \end{align} Then $ABC = C$, $CBA = 0$, so $\mathrm{Tr}\,(ABC) = 1$, $\mathrm{Tr}\,(CBA)=0$.