-1

Let's consider the algebra of the square matrixes endowed with the topology of the operator norm $$ |A|:=\sup_{|v|=1}|Av| $$ let $I$ be the identity matrix and suppose that $|A-I| < r < 1$, what can be said about the distance $|A^{-1}-I|$?

flag
8 
 Do you want any more than is given by this standard argument? Let $B=I-A$ then $$A^{-1}-I=\sum_{n=1}^\infty B^n$$ and so $$\|A^{-1}-I\|\le\sum_{n=1}^\infty\|B^n\|\le\sum_{n=1}^\infty\|B\|^n < \sum_{n=1}^\infty r^n=\frac{r}{1-r}.$$ – Robin Chapman May 31 2010 at 15:38
Well, this estimate seems great even if I don't know wheter it is sharp. – Marco May 31 2010 at 16:03
4 
 It is sharp, as one sees by considering $B=bI$ and letting $b\to r$ from below. – Robin Chapman May 31 2010 at 16:32
2 
 Evidence of prior calculation/reflection would have been good in the original question... – Yemon Choi May 31 2010 at 16:33

Your Answer

Get an OpenID
or

Browse other questions tagged or ask your own question.