I'm stucked in this problem:
Prove that if $||AB-I||=\epsilon<1$ then $||A^{-1}-B||<||B||(\frac{\epsilon}{1-\epsilon})$
My process: 1) We have that $||AB-I||=||-(AB-I)||=||I-AB||$.
2) Since $||AB-I||=||I-AB||<1$ then $(AB)^{-1}=\sum_{k=0}^{\infty}{(I-AB)^k}$ then $||(AB)^{-1}||=||\sum_{k=0}^{\infty}{(I-AB)^k}||<\sum_{k=0}^{\infty}||{(I-AB)^k}||$ (this by the triangular inequality) $\leq\sum_{k=0}^{\infty}{||(I-AB)||^k}$ (this by submultiplicative property and induction: $||M^k||=||M M^{k-1}||\leq||M|| ||M^{k-1}||\leq||M||^2||M^{k-2}||\leq\cdots\leq||M||^k$).
3) Since $||I-AB||<1$ then $\sum_{k=0}^{\infty}{||I-AB||^k}$ converges to $\frac{1}{1-\epsilon}$, i.e., $||(AB)^{-1}||<\frac{1}{1-\epsilon}$.
4) Since the condition number $\kappa(A)=||A||||A^{-1}||\geq1$ for all A, then $||(AB)||||(AB)^{-1}||\geq1$ then $\frac{1}{||AB||}\leq||(AB)^{-1}||\leq\frac{1}{1-\epsilon}$.
5)By the submultiplicative property $||AB||\leq||A||||B||$ then $\frac{1}{||A||||B||}\leq\frac{1}{||AB||}$, i.e., $\frac{1}{||A||}\leq||B||(\frac{1}{1-\epsilon})$
6)Here is my problem: i want to have that $||A^{-1}||\leq\frac{1}{||A||}$ for have $||A^{-1}||||AB-I||\leq||B||(\frac{\epsilon}{1-\epsilon})$ where $||A^{-1}-B||=||(A^{-1}A)B-A^{-1}||\leq||A^{-1}||||AB-I||$.
and i have what i want...
In adition i don't understad why $A^{-1}$ and $B^{-1}$ exist.
Any hint is helpful.
Hint: To show that $A, B$ are invertible, observe \begin{align} (1-\varepsilon)\|x\|\leq \|x\|-\|ABx\|\leq \|ABx\|\leq \|x\|+\|(AB-I)x\|\leq (1+\varepsilon)\|x\| \end{align} which means $AB$ is injective and surjective (why?). Hence $AB$ is invertible which means $A$ and $B$ are invertible.
Moreover, observe \begin{align} \|Bx-A^{-1}x\| \leq \|B\|_\text{op}\|x-B^{-1}A^{-1}x\|\leq \|B\|_\text{op}\|I-(AB)^{-1}\|_\text{op}\|x\|. \end{align}
Applying your argument, then you have the result. More precisely, observe that \begin{align} (AB)^{-1} = [I-(I-AB)]^{-1} = I+(I-AB)+(I-AB)^2+\ldots \end{align} which means \begin{align} (AB)^{-1}-I = (I-AB)+(I-AB)^2+\ldots. \end{align} Then it follows \begin{align} \|I-(AB)^{-1}\|_\text{op}\leq \varepsilon +\varepsilon^2+\ldots = \frac{\varepsilon}{1-\varepsilon}. \end{align}