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Let $A,B$ are two $p\times p$ positive definite matrices such that $0<\delta_0\leq \min\{\lambda_{\min}(A), \lambda_{\min}(B)\}\leq \max\{\lambda_{\max}(A), \lambda_{\max}(B)\}\leq \delta_1$. Also assume that $\Vert A-B\Vert_{op}\leq \varepsilon$. Can we upper bound $\Vert A^{-1} - B^{-1} \Vert_{op}$ in terms of $(\delta_0, \delta_1, \varepsilon)$. I tried several things with the definition $$ \Vert A^{-1} - B^{-1} \Vert_{op} = \max_{u,v\in \mathbb{S}^{p-1}} u^\top (A^{-1}-B^{-1})v. $$ One easy observation is that $\Vert A^{-1}-B^{-1}\Vert_{op}\leq \vert 1/\lambda_{\min}(A) - 1/\lambda_{\max}(B)\vert$. But I think this does not help me in any way. Any help will be appreciated.

Edit: It could be possible that one can not get such bound. For a related question see here.

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$A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$ so $\|A^{-1}-B^{-1}\| \le \delta_0^{-2} \epsilon $. This is the best possible bound in terms of the given parameters, as you can see by considering 2 by 2 diagonal matrices: Consider $A=$diag$(\delta_0,\delta_1)$ and $B=$diag$(\delta_0+\epsilon,\delta_1+\epsilon)$ where $\delta_1$ is large and $\epsilon \to 0$.

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  • $\begingroup$ Thanks again for your help. Yeah, you are right. This is the tightest possible bound. $\endgroup$
    – De vinci
    Commented Aug 5, 2021 at 2:00

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