# Thread: Nilpotent Matrix

1. ## 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?

2. Originally Posted by TrevK
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 $\displaystyle (1-x)^{-1} = 1+x+x^2+x^3+\ldots$. If you substitute x=B then the series becomes a finite sum, because of the nilpotency.

3. 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

4. Originally Posted by TrevK
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 $\displaystyle I+B+B^2+B^3+\ldots$ 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 $\displaystyle (I-B)(I+B+B^2+\ldots+B^{n-1}) = (I+B+B^2+\ldots+B^{n-1})(I-B) = I$.