Consider two matrices $A$ and $B$ that do not commute, so that the binomial theorem does not apply. However, one of them (say $B$) is an involutory matrix, meaning that $B^2 = I$. I am wondering whether some closed form can be given for the following quantity:
$$\mathrm{Tr}\,(A+B)^n\,.$$
Independently of the properties of $B$, it can be readily shown that for $n \leq 3$, the binomial theorem effectively applies due to the trace cyclicity, namely:
$$\mathrm{Tr}\,(A+B)^2 = \mathrm{Tr}\,(A^2 + A B + B A + B^2) = \mathrm{Tr}\,(A^2 + 2 AB + B^2)\,,$$
\begin{align} \mathrm{Tr}\,(A+B)^3 &= \mathrm{Tr}\,(A^3 + A^2 B + A B A + B A^2 + A B^2 + B A B + B^2 A + B^3) \\ &= \mathrm{Tr}\,(A^3 + 3 A^2 B + 3 A B^2+ B^2)\,. \end{align}
For higher powers, some mixed terms will appear. For instance, for $n=4$, one has:
$$\mathrm{Tr}\,(A+B)^4 \supset 4 \mathrm{Tr}\,(A^2 B^2) + 2 \mathrm{Tr}(A B A B)\,.$$
My question can be split in two parts:
- Can the trace cyclicity help get a closed form for all $n\geq4$? Such an expression would certainly contain products and powers of $A B$ and/or of $[A,B]$.
- Otherwise, can the $B^2 = I$ identity further help achieve such a closed form? It seems to me that some useful commutation relations of the form $[A^p, B^q]$ can be inferred from this property that could help in this regard.