# Commutator Algebra proof

1. Dec 29, 2011

### carvas

1. Prove that $$[A,B^n] = nB^{n-1}[A,B]$$

Given that: $$[[A,B],B] = 0$$

My Atempt to resolution

We can write that:
$$[[A,B],B] = [A,B]B-B[A,B] = 0$$

So we get that: $$[A,B]B = B[A,B]$$

After some working several expansions, and considering that $$[X,YZ] = Y[X,Z] + [X,Y]Z$$

I arrived at this expression:

$$[A,B^n] = B^{n-1}[A,B]+B^{n-2}[A,B]+[A,B^{n-2}]B^2$$

But from here I'm a bit lost on how to get the desired result.
So, could anyone help me?

Thanks a lot!

2. Dec 30, 2011

### vela

Staff Emeritus
Prove it using induction.

3. Jan 1, 2012

### carvas

yes, i've tried that, but i cant get to the desired result...

could you help me?

thx again

4. Jan 1, 2012

### vela

Staff Emeritus
Show us what you have so far.

5. Jan 1, 2012

### carvas

what i have is the last expression in my first post.
so, by induction, and starting from this expression, i cannot get what i want to prove.

6. Jan 1, 2012

### vela

Staff Emeritus
Do you know how to do a proof by induction?

7. Jan 1, 2012

### Fredrik

Staff Emeritus
This result contradicts the formula you say that you're using. (Think X=A, Y=Bn-2, Z=B2).