I have asked the same question before but the contents were not complete, and I propose it again.
Assume that A, B are both symmetrical positive definite matrices, and $$ 0 \prec aI \preceq A \preceq bI $$ $$ A \preceq B \preceq (1+ \epsilon) A, \epsilon > 0$$ $$\gamma \in [0,1].$$
The background of this question is that $B$ is a matrix approximation of $A$ (Hessian of a strongly convex and smooth function) and the approximation error can be measured as the above.
Now we want an upper bound of $\Vert I - \gamma A B^{-1}\Vert_2 $ which can be seen as another measure of the approximation error.
Intuitively, since $B \succeq A$, this norm should be less than 1(I am not sure about this). And also, when $\gamma =0$, the corresponding norm should be $1$.
The main difficulty is that $A B^{-1}$ is not symmetrical, and thus we can not derive it in the following sense: $$ \dfrac{1}{1 + \epsilon}A^{-1} \preceq B^{-1} \preceq A^{-1}\\ \dfrac{1}{1 + \epsilon}I \preceq AB^{-1} \preceq I. $$ Any help will be appreciated.
We have the following current results: $$ I - \gamma A B^{-1} = (1 - \gamma)I + \gamma (B- A)B^{-1}, $$
and thus all the eigenvalues of $I - \gamma A B^{-1}$ are non-negative and even positive when $\gamma < 1$. And also the maximum eigenvalue of $I - \gamma A B^{-1}$ can be bounded: $$\lambda_{max}(I - \gamma A B^{-1}) \leq 1 - \gamma \lambda_{min}(A B^{-1}) \leq 1- \gamma \dfrac{a}{b(1 + \epsilon)}.$$ But the spectral norm is the maximum singular value instead of the maximum eigenvalue.