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Math Help - [SOLVED] Theorem on Invertible Matrices

  1. #1
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    [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
    <br />
 \|A^{-1} - B \| \le \|B\|\left(\frac{\varepsilon}{1-\varepsilon}\right)<br />

    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.
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  2. #2
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    \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)
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  3. #3
    MHF Contributor

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    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.
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