Thread: 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?

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 . If you substitute x=B then the series becomes a finite sum, because of the nilpotency.

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

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 finishes with the term B^{N–1}, since all the subsequent terms are zero. That suggests that , a conjecture which you can verify by checking that .