<s>[INST] Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$? 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!
 [/INST] 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 $\mathrm{Tr}\,(ABC) =  1$, $\mathrm{Tr}\,(CBA)=0$.
</s>