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Prove that if A invertible and $\left\lVert A-B\right\rVert <\left\lVert A^{-1}\right\rVert ^{-1}$ then $$\left\lVert A^{-1}-B^{-1}\right\rVert \leq \left\lVert A^{-1}\right\rVert \frac{\lVert I-A^{-1}B\rVert}{1-\left\lVert I-A^{-1}B\right\rVert }$$

I am trying and trying but can't find the right solution. What I get is: $$ \left\lVert A^{-1}-B^{-1}\right\rVert = \left\lVert A^{-1}-\sum\limits_{k=0}^\infty (I-A^{-1}B)^k A^{-1}\right\rVert \leq \left\lVert A^{-1}\right\rVert + \left\lVert \sum\limits_{k=0}^\infty (I-A^{-1}B)^k A^{-1}\right\rVert \leq \\ \left\lVert A^{-1}\right\rVert + \left\lVert \sum\limits_{k=0}^\infty (I-A^{-1}B)^k \right\rVert \cdot \left\lVert A^{-1}\right\rVert \leq \left\lVert A^{-1}\right\rVert \left[ 1 + \frac{1}{1-\left\lVert I-A^{-1}B \right\rVert} \right] = \left\lVert A^{-1}\right\rVert \left[ \frac{2-\left\lVert I-A^{-1}B \right\rVert}{1-\left\lVert I-A^{-1}B \right\rVert} \right] \stackrel{?}{\le} \left\lVert A^{-1}\right\rVert \frac{\lVert I-A^{-1}B\rVert}{1-\left\lVert I-A^{-1}B\right\rVert } $$

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First, $B$ is invertible as $$ B = A- (A-B)= A( I-A^{-1}(A-B)) , $$ where $I-A^{-1}(A-B)$ is invertible since by the assumption $\|A^{-1}(A-B)\|<1$. Then we get (Neumann series) $$ B^{-1} = \sum_{k=0}^\infty (A^{-1}(A-B))^k A^{-1} = \sum_{k=0}^\infty (I-A^{-1}B)^k A^{-1} $$ and $$ \|B^{-1}\| \le\frac{ \|A^{-1} \|}{1-\|I-A^{-1}B\|}. $$ Factoring out $B^{-1}$ yields $$ A^{-1}-B^{-1} = (A^{-1}B-I)B^{-1} $$ then $$ \|A^{-1}-B^{-1}\|\le \|I-A^{-1}B\|\cdot \|B^{-1}\| \le \frac{ \|A^{-1} \| \|I-A^{-1}B\|}{1-\|I-A^{-1}B\|} $$ which is the claim.

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  • $\begingroup$ Awesome, very nice! The way you prove B to be invertible seems unfamiliar to me. How does what you do imply invertibility of B? $\endgroup$
    – Flaudre
    Nov 17, 2016 at 9:10
  • $\begingroup$ I prove that B exists by doing $\sum\limits_{k=0}^\infty \left[ A^{-1}(A-B) \right]^k=(I-A^{-1}A+A^{-1}B)^{-1}=B^{-1}A$. $\endgroup$
    – Flaudre
    Nov 17, 2016 at 9:16
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    $\begingroup$ This is the same idea: you have to argue convergence of the series. $\endgroup$
    – daw
    Nov 17, 2016 at 9:17
  • $\begingroup$ Ah okay, and so the product of two invertible matrices is also invertible. Thanks for your help! (Btw: comment above I meant $B^{-1}$, ofc) $\endgroup$
    – Flaudre
    Nov 17, 2016 at 9:27

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