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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- Mar 26th 2008, 08:48 PMTrevKNilpotent 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? - Mar 27th 2008, 01:38 AMOpalg
- Mar 28th 2008, 07:07 PMTrevK
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 - Mar 29th 2008, 01:26 AMOpalg
If B is nilpotent then B^N=0 for some positive integer N. Thus the series $\displaystyle I+B+B^2+B^3+\ldots$ finishes with the term B^{N–1}, since all the subsequent terms are zero. That suggests that $\displaystyle (I-B)^{-1} = I+B+B^2+\ldots+B^{n-1}$, a conjecture which you can verify by checking that $\displaystyle (I-B)(I+B+B^2+\ldots+B^{n-1}) = (I+B+B^2+\ldots+B^{n-1})(I-B) = I$.