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Nilpotent matrices

  1. Apr 25, 2008 #1
    [SOLVED] Nilpotent matrices

    1. The problem statement, all variables and given/known data


    Use the McLaurin series for 1/(1+x) to show that I + N is invertible where N is a nilpotent matrix.



    2. Relevant equations
    n/a



    3. The attempt at a solution
    It has something to do with the inverse of 1+x is 1/(1+x), but i'm lost. I'm not sure where to start nor what exactly I need to do.
     
  2. jcsd
  3. Apr 25, 2008 #2

    Dick

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    You might want to start with the hint and find the McLaurin series for 1/(1+x).
     
  4. Apr 25, 2008 #3
    Ok, I understand that I need to start with the maclaurin series, and I've done that, but I'm not sure what that shows me.
     
  5. Apr 25, 2008 #4

    Dick

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    You might want to show us what you found for the MacLaurin series so we can make sure you are finding the correct thing. Then substitute N for x. What else? What does the nilpotency of N tell you about the infinite MacLaurin series?
     
  6. Apr 25, 2008 #5
    ok well we know:
    P(n) = 1 + x + ... + x^n
    xP(n) = x + x^2 + ... + x^n + x^(n+1)

    (x-1)P(n)=x^(n+1)-1

    P(n) = x^(n+1)-1/(x-1) goes to -1/(x-1) = 1/(1-x) <== close to 1/(1+x)

    if you use -x, P(n)= 1/(1+x)


    because the MacLaurin series for 1/(1+x)
    f(x) = f(0) + f'(0)x + f''(0)x^2 + ...
    f(x) = 1 + x + x^2 + ...
     
  7. Apr 25, 2008 #6

    Dick

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    The MacLaurin series for 1/(1-x)=1+x+x^2+x^3+x^4+..., yes. Doesn't that make the series for 1/(1+x)=1-x+x^2-x^3+x^4-...?? Taking x->-x, just as you said?
     
  8. Apr 25, 2008 #7
    yes...
    Let P(n) = 1-x+x^2-x^3+x^4-...+x^n
    xP(n) = x-x^2+.................-x^n+x^(n+1)

    add them
    (x+1)P(n)=1+x^(n+1)

    therefore P(n)=1+x^(n+1)/(x+1) which goes to 1/(1+x)
     
  9. Apr 25, 2008 #8

    Dick

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    Ok, so what happens when x is the nilpotent matrix N??
     
  10. Apr 25, 2008 #9
    P(n) approaches 1/(1+0) = 1
     
  11. Apr 25, 2008 #10
    so how does this show that I + N is invertible?
     
  12. Apr 25, 2008 #11

    Dick

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    The inverse of I+N is 1/(I+N). Does the MacLaurin expansion converge??
     
  13. Apr 25, 2008 #12
    Thanks!!!!
     
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