# Nilpotent Matrix

• Mar 26th 2008, 08:48 PM
TrevK
Nilpotent Matrix

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 AM
Opalg
Quote:

Originally Posted by TrevK

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.
• Mar 28th 2008, 07:07 PM
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??

Thanks trev
• Mar 29th 2008, 01:26 AM
Opalg
Quote:

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\$.