Conversation started Jun 14, 2023 at 2:00.
D.C. the III
D.C. the III
Jun 14, 2023 02:00
So from 6(a) I got that $(I-H)^{-1} = \sum_{k = 0}^\infty H^k$ and from 7(b) I had gotten $(I- A^{-1}H)^{-1} = \sum_{k=0}^\infty (A^{-1}H)^k$.
copper.hat
copper.hat
Jun 14, 2023 02:26
Any nilpotent $H$ will also converge, regardless of norm.
Ted Shifrin
Ted Shifrin
We’re trying to get an open ball, @copper.
So it’s $A^{-1}H$ whose norm you need to control. Whence ….
copper.hat
copper.hat
An open ball around $I$ so that $I+H$ is invertible?
Ted Shifrin
Ted Shifrin
Or, more generally, around an invertible $A$.
copper.hat
copper.hat
Useful result.
D.C. the III
D.C. the III
Jun 14, 2023 02:51
That was the second part of my question too.....everything was going to be dependent on the invertible matrix $A$.
 
Conversation ended Jun 14, 2023 at 2:51.