Matrix Simplification with C^2 = 0 and BC = CB

[gtfitzpatrick]

Homework Statement

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 I am doing it the right way or where to go from here...

 

[Dick]

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

 

[gtfitzpatrick]

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

 

[Dick]

No. I mean the "binomial theorem". http://en.wikipedia.org/wiki/Binomial_theorem

 

[gtfitzpatrick]

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

 

[Dick]

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?

 

[gtfitzpatrick]

wait maybe that's B^{p}C^{p+1}

 

[gtfitzpatrick]

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

 

[gtfitzpatrick]

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?

 

[Dick]

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?

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

 

[gtfitzpatrick]

thanks a mill for the help