# Prove that if A invertible and $\left\lVert A-B\right\rVert <\left\lVert A^{-1}\right\rVert ^{-1}$ then the following inequality holds

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: $$\left\lVert A^{-1}-B^{-1}\right\rVert = \left\lVert A^{-1}-\sum\limits_{k=0}^\infty (I-A^{-1}B)^k A^{-1}\right\rVert \leq \left\lVert A^{-1}\right\rVert + \left\lVert \sum\limits_{k=0}^\infty (I-A^{-1}B)^k A^{-1}\right\rVert \leq \\ \left\lVert A^{-1}\right\rVert + \left\lVert \sum\limits_{k=0}^\infty (I-A^{-1}B)^k \right\rVert \cdot \left\lVert A^{-1}\right\rVert \leq \left\lVert A^{-1}\right\rVert \left[ 1 + \frac{1}{1-\left\lVert I-A^{-1}B \right\rVert} \right] = \left\lVert A^{-1}\right\rVert \left[ \frac{2-\left\lVert I-A^{-1}B \right\rVert}{1-\left\lVert I-A^{-1}B \right\rVert} \right] \stackrel{?}{\le} \left\lVert A^{-1}\right\rVert \frac{\lVert I-A^{-1}B\rVert}{1-\left\lVert I-A^{-1}B\right\rVert }$$

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.
• I prove that B exists by doing $\sum\limits_{k=0}^\infty \left[ A^{-1}(A-B) \right]^k=(I-A^{-1}A+A^{-1}B)^{-1}=B^{-1}A$. Nov 17, 2016 at 9:16
• 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) Nov 17, 2016 at 9:27