Commutator of a matrix to the power of k

Question

The question asks me to show that $$[A,B^k] = \Sigma_{r=1}^k B^{r-1}[A,B]B^{k-r}$$ (A, B are nxn matrices) but I can't get even close. I suspect there's some definition of $B^k$ that I don't know but is required. I've tried expanding the RHS, to get $$\Sigma_{r=1}^k B^{r-1}(AB-BA)B^{k-r}$$ $$= \Sigma_{r=1}^k B^{r-1}ABB^{k-r} - B^{r-1}BAB^{k-r}$$

So starting from the LHS, what can I do to $B^k$? Substituting in the diagonalised matrix such that $B^k = P\Lambda^k P^{-1}$ didn't get me anywhere and I'm not sure where the sum comes into it. The only sums I've seen in matrix calculations come from exp(B), but I don't think that's related. Any help or hints are much appreciated!

Answer 1

It's much cleaner to prove this by first proving that

$$[A, BC] = [A, B] C + B [A, C]$$

which asserts that the function $[A, -]$ given by taking the commutator with a fixed matrix is a derivation. From here, you can use induction. For example,

$$[A, B^2] = [A, B] B + B [A, B]$$ $$[A, B^3] = [A, B] B^2 + B [A, B^2] = [A, B] B^2 + B [A, B] B + B^2 [A, B].$$

It shouldn't be too hard to see how the pattern generalizes from here. One of the several advantages of doing things this way is that you don't need to know what the RHS looks like; it just falls out of the calculation automatically.

Note that, as a special case, if $B$ happens to commute with $[A, B]$, then the formula simplifies to

$$[A, B^n] = n B^{n-1} [A, B]$$

which may look familiar from calculus; taking derivatives is also a derivation.

Answer 2

You are close. \begin{align} \sum_{r=1}^k B^{r-1}(AB-BA)B^{k-r}&=\sum_{r=1}^k (B^{r-1}AB^{k-r+1} - B^{r}AB^{k-r}) \\ &=\sum_{r=1}^k B^{r-1}AB^{k-r+1}-\sum_{r=1}^k B^{r}AB^{k-r} \\ &=\sum_{r=0}^{k-1} B^{r}AB^{k-r}-\sum_{r=1}^k B^{r}AB^{k-r} \\ &=AB^{k}+\sum_{r=1}^{k-1} B^{r}AB^{k-r}-\sum_{r=1}^{k-1} B^{r}AB^{k-r}-B^kA \\ &=AB^{k}-B^kA \\ &=[A,B^k] \end{align} As @Andreas comment, this holds not only for matrix, but for general ring.