1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simplify both sides of matrix

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    let A=B+C where B and C are nxn matrices such that C[tex]^{2}[/tex] = 0 and BC=CB show that for p>0, A[tex]^{p+1}[/tex] = B[tex]^{p}[/tex][B+(p+1)C]


    I started trying to simplify both sides,

    (B+C)[tex]^{p+1}[/tex] = B[tex]^{p+1}[/tex] + B[tex]^{p}[/tex]C(p+1)

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

    which gives

    C(B+C)[tex]^{p+1}[/tex] =CB[tex]^{p+1}[/tex]

    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. jcsd
  3. Jan 27, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: matrix

    Since B and C commute, it's pretty easy to apply the binomial theorem to (B+C)^(p+1). Try it.
     
  4. Jan 27, 2009 #3
    Re: matrix

    im not sure about the binomial therom,
    do you mean (B+C)[tex]^{(p+1)}[/tex] = B(B+C)[tex]^{p}[/tex] + C(B+C)[tex]^{p}[/tex] ???
     
  5. Jan 27, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

  6. Jan 27, 2009 #5
    Re: matrix

    still not sure...you mean (B+C)[tex]^{p+1}[/tex] = (p+1)!B[tex]^{p}[/tex]C[tex]^{p+1}[/tex]
     
  7. Jan 27, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  8. Jan 27, 2009 #7
    Re: matrix

    wait maybe thats B[tex]^{p}[/tex]C[tex]^{p+1}[/tex]
     
  9. Jan 27, 2009 #8
    Re: matrix

    sorry didnt see your reply,thanks a mill i'll look that up now, thanks again
     
  10. Jan 27, 2009 #9
    Re: matrix

    think i got it (B+C)[tex]^{p+1}[/tex] = B[tex]^{p+1}[/tex] + (p+1)B[tex]^{p}[/tex]C

    we're only concerned with first 2 terms because C[tex]^{2}[/tex] = 0?
     
  11. Jan 27, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: matrix

    That's right. And remember we can only do this since BC=CB and we can rearrange the products.
     
  12. Jan 27, 2009 #11
    Re: matrix

    thanks a mill for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simplify both sides of matrix
Loading...