# If $\Vert A-B\Vert_{op}\leq \varepsilon$ then $A^{-1}$ and $B^{-1}$ are uniformly close

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.

$$A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$$ 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 $$\epsilon \to 0$$.