# Math Help - [SOLVED] Theorem on Invertible Matrices

1. ## [SOLVED] Theorem on Invertible Matrices

I am studying for finals and am looking over previous problems that I missed significant points on. My instructor doesn't comment on why things are incorrect and he doesn't discuss homework, so I am on my own to figure out why I am wrong. Anyway, I would like some help figuring out where I went wrong.

Theorem:
If $A \text{ and } B \text{ are } n \times n$ matrices such that $\|I - AB\| < 1$, then $A \text{ and }$ are invertible. Furthermore, we have
$A^{-1} = B \sum_{k=0}^{\infty}(I - AB)^k \text{ and } B^{-1} = \sum_{k=0}^{\infty}(I - AB)^k A$

Problem:
Show that if $\|AB - I\| = \varepsilon < 1,$ then
$
\|A^{-1} - B \| \le \|B\|\left(\frac{\varepsilon}{1-\varepsilon}\right)
$

Proof:
By the theorem above we have
$\|A^{-1}\| = \|B \sum_{k=0}^{\infty}(I - AB)^k\|$
$\|A^{-1}\| \le \|B\| \sum_{k=0}^{\infty}\|I - AB\|^k$
$\|A^{-1}\| \le \|B\| \sum_{k=0}^{\infty}\varepsilon^k$
$\|A^{-1}\| \le \|B\| \frac{1}{1 - \varepsilon}$
$\|A^{-1}\| + \|-B\| \le\|B\| \frac{1}{1 - \varepsilon} + \|B\|$
$\|A^{-1} - B\| \le \|B\|\frac{2 - \varepsilon}{1 - \varepsilon}$

Which doesn't seem to get me anywhere. Any help is appreciated. Thank you in advance.

2. $\Vert A^{-1} - B\Vert =\Vert B(I - \sum_{k=0}^{\infty } (I-AB)^k \Vert =\Vert B\sum_{k=1}^{\infty } (I-AB)^k \Vert \leq \Vert B\Vert \sum_{k=1}^{\infty } \epsilon ^k$ $= \Vert B\Vert \left( \frac{1}{1-\epsilon } -1\right) =\Vert B \Vert \left( \frac{\epsilon }{1- \epsilon } \right)$

3. What you did looks correct; however it is not sharp.

The solution is a little simpler: we have $A^{-1}=B\sum_{k=0}^\infty (I-AB)^k=B+B\sum_{k=1}^\infty (I-AB)^k$, hence $\|A^{-1}-B\|=\|B\sum_{k=1}^\infty (I-AB)^k\|\leq \|B\|\sum_{k=1}^\infty \varepsilon^k$, which will give you the bound you want.