# Trace of iterated commutators and binomial coefficients

Let $$A$$ and $$B$$ be two matrices.

I am trying to prove the following formula (and also find the conditions on $$A$$ and $$B$$ for it to work, if it is not true for any $$A$$ and $$B$$) :

$$\underset{n=0}{\sum^j} \binom{j+1}{n}\text{Tr}\left(\left[(A)^n, B\right]\left[(A)^{j-n}, B\right]\right) = \text{Tr}\left(B{\left[(A)^{j}, B\right]}\right)$$

$$\left[(A)^n, B\right]$$ is the iterated application of the adjoint of $$A$$, aka the iterated commutator $$m$$ times: $$(ad_X)^{(n)}(Y) = [(X)^n,Y] \equiv \underbrace{[X,\dotsb[X,[X}_{n \text {times }}, Y]] \dotsb],\quad [(X)^0,Y] \equiv Y$$.

I must admit I am a bit stumped, I only managed to get to:

$$\underset{n=0}{\sum^j} \binom{j}{n}\left[(A)^n, B\right]\left[(A)^{j-n}, B\right] = \left[(A)^{j}, B^2\right]$$

I would be grateful for any help one might provide,

• Yes, sorry, I used a notation that might be confusing, as defined $[(A)^2,X]=[A,[A,X]]$, while $[A^2,X] = [A.A,X]$. Commented Dec 20, 2023 at 13:38
• Usually people write $ad(A)$ for the linear map defined by $ad(A)(X)=[A,X]$, and then $ad(A)^2$ is the map with $ad(A)^2(X)=[A,[A,X]]$. In these terms books on Lie algebras have similar formulas like yours. Commented Dec 20, 2023 at 14:15