# Simplify both sides of matrix

1. Jan 27, 2009

### gtfitzpatrick

1. The problem statement, all variables and given/known data
let A=B+C where B and C are nxn matrices such that C$$^{2}$$ = 0 and BC=CB show that for p>0, A$$^{p+1}$$ = B$$^{p}$$[B+(p+1)C]

I started trying to simplify both sides,

(B+C)$$^{p+1}$$ = B$$^{p+1}$$ + B$$^{p}$$C(p+1)

wanted to get rid of p so multiplied both sides by C

which gives

C(B+C)$$^{p+1}$$ =CB$$^{p+1}$$

i'm not sure if i've done this right or if im doing it the right way or where to go from here...

2. Jan 27, 2009

### Dick

Re: matrix

Since B and C commute, it's pretty easy to apply the binomial theorem to (B+C)^(p+1). Try it.

3. Jan 27, 2009

### gtfitzpatrick

Re: matrix

im not sure about the binomial therom,
do you mean (B+C)$$^{(p+1)}$$ = B(B+C)$$^{p}$$ + C(B+C)$$^{p}$$ ???

4. Jan 27, 2009

### Dick

5. Jan 27, 2009

### gtfitzpatrick

Re: matrix

still not sure...you mean (B+C)$$^{p+1}$$ = (p+1)!B$$^{p}$$C$$^{p+1}$$

6. Jan 27, 2009

### Dick

Re: matrix

That's NOT the binomial theorem. The binomial theorem is (a+b)^n=C(n,0)*a^n+C(n,1)*a^(n-1)*b+...+C(n,n-1)*b*c^(n-1)+C(n,n)*b^n, where the C(n,i) are the binomial coefficients. Look it up. Apply that with a=B, b=C and n=p+1. Why can you ignore every term after the first two? What are C(p+1,0) and C(p+1,1)? You could also prove this by induction. Would you rather do that?

7. Jan 27, 2009

### gtfitzpatrick

Re: matrix

wait maybe thats B$$^{p}$$C$$^{p+1}$$

8. Jan 27, 2009

### gtfitzpatrick

Re: matrix

sorry didnt see your reply,thanks a mill i'll look that up now, thanks again

9. Jan 27, 2009

### gtfitzpatrick

Re: matrix

think i got it (B+C)$$^{p+1}$$ = B$$^{p+1}$$ + (p+1)B$$^{p}$$C

we're only concerned with first 2 terms because C$$^{2}$$ = 0?

10. Jan 27, 2009

### Dick

Re: matrix

That's right. And remember we can only do this since BC=CB and we can rearrange the products.

11. Jan 27, 2009

### gtfitzpatrick

Re: matrix

thanks a mill for the help