Similar to a previous question here, I wonder if cyclic permutations are the only relations amongst traces of (non-commutative) monomials. Since the evaluations $\operatorname{tr}:k\langle x,y,\dots \rangle \to k$ take an infinite dimensional vector space to a one-dimensional vector space there must be quite a few relations, but I wonder if any of them are on binomials other than the cyclic permutations.
At any rate, for small dimensions, we probably get some extra relations.
It appears that $\operatorname{tr}(AABABB−AABBAB) = 0$ for all $2×2$ matrices. Is this true? How does one prove it?
First if $N$ is $2\times2$ matrix with $\operatorname{tr}(N)=0$, then $N^2$ is scalar (either $N$ is nilpotent or its eigenvalues are opposite and have same square). This implies that $\operatorname{tr}(N^3)=0$.
And $$\begin{eqnarray} (AB-BA)^3 &=& (ABABAB - BABABA) + (ABBABA + BABAAB + BAABBA) \\ && - (ABABBA + ABBAAB + BAABAB) \end{eqnarray} $$
taking the trace this gives
$$0 = 0 + 3\operatorname{tr}(ABBABA) - 3 \operatorname{tr}(ABABBA) = 3 \operatorname{tr}(AABBAB - AABABB).$$