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Math Help - Nilpotent Matrix

  1. #1
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    Nilpotent Matrix

    Hey everyone i have no clue how to solve this one, please help.

    If B is any nilpotent matrix prove that I-B is invertible and find a formula for (I-B)^-1 in terms of powers of B?
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  2. #2
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    Quote Originally Posted by TrevK View Post
    Hey everyone i have no clue how to solve this one, please help.

    If B is any nilpotent matrix prove that I-B is invertible and find a formula for (I-B)^-1 in terms of powers of B?
    use the geometric series (1-x)^{-1} = 1+x+x^2+x^3+\ldots. If you substitute x=B then the series becomes a finite sum, because of the nilpotency.
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    Thanks for your help but i still don't really understand how to find a formula for (I-B)^-1 in terms of powers of B. What do you mean by the series is a finite sum because of the nilpotency??

    Thanks trev
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  4. #4
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    Quote Originally Posted by TrevK View Post
    Thanks for your help but i still don't really understand how to find a formula for (I-B)^-1 in terms of powers of B. What do you mean by the series is a finite sum because of the nilpotency??
    If B is nilpotent then B^N=0 for some positive integer N. Thus the series I+B+B^2+B^3+\ldots finishes with the term B^{N1}, since all the subsequent terms are zero. That suggests that (I-B)^{-1} = I+B+B^2+\ldots+B^{n-1}, a conjecture which you can verify by checking that (I-B)(I+B+B^2+\ldots+B^{n-1}) = (I+B+B^2+\ldots+B^{n-1})(I-B) = I.
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