1. Sep 21, 2009

### ice109

i've never really done a proof by induction but i would like to prove a statement about commutator relations so can you please check my proof:

claim: $[A,B^n]=nB^{n-1}[A,B]$ if $[A,B]=k\cdot I$ where A,B are operators, I is the identity and k is any scalar.

proof: $$[A,B^2] = [A,B]B+B[A,B] = 2B^{2-1}[A,B]$$ where the B is brought out of the commutator by almost obvious identity and the last equality follows from the commutativity of the identity with any operator.

so for n=2 the equality is proven. we now assume that the equality holds for n-2 case and proceed to prove it holds for n case:

$$[A,B^n]=[A,B^2B^{n-2}]=[A,B^{2}]B^{n-2}+B^2[A,B^{n-2}]$$
$$=2B[A,B]B^{n-2}+B^2(n-2)B^{n-3}[A,B]$$
$$=2B^{n-1}[A,B]+(n-2)B^{n-1}[A,B]$$
$$=nB^{n-1}[A,B]$$

so how 'bout it? was successful?

2. Sep 21, 2009

### waht

I think proof by induction goes something like this:

Show that the equality is true for n = 1

Assume the equality is true for n

Then show it is true for n+1

3. Sep 22, 2009

### ice109

it's the same thing?