Cyclic property of the trace ($\mathrm{tr}(AB)=\mathrm{tr}(BA)$)

If $A$ is an $m$-by-$n$ matrix and $B$ is an $n$-by-$m$ matrix over the same field, both products are square matrices andThe products can even have different sizes: for $A=(12)$ and $B=(34{)}^T$, $AB=(11)$ and $BA=\left(\begin{array}{cc}3& 6\\ 4& 8\end{array}\right)$, but both have trace $11$.^[ref]

Grouping all but one factor proves that $\mathrm{tr}(A_1{A}_2\cdots A_k)$ is unchanged by a cyclic permutation whenever the cyclic products are defined and square. In particular, every commutator has trace zero, since $\mathrm{tr}(AB-BA)=0$ by linearity.

Cyclic rotation does not allow arbitrary reordering. Forone has $ABC=A$, so $\mathrm{tr}(ABC)=1$, but $AC=0$, so $\mathrm{tr}(ACB)=0$. Rotating $ABC$ to $BCA$ or $CAB$ preserves its trace; swapping just $B$ and $C$ need not.