I am trying to determine whether it is possible to select $\epsilon$ such that $\|(I - \epsilon MK)^{-1}\|_2 < 1$ where $M$ is symmetric positive definite and $K$ is symmetric negative definite and $\epsilon > 0$.
As an attempt, we know that the 2-norm of a matrix is the maximum of its singular values which is related to the eigenvalues of a corresponding matrix. If $\sigma_{max},\lambda_{max}$ represent the maximum singular value and eigenvalue, respectively, then $\|(I-\epsilon MK)^{-1}\|_2 = \sigma_{max}((I-\epsilon MK)^{-1}) = \lambda_{max}([(I-\epsilon KM)(I- \epsilon MK)]^{-1})$ where I have used the symmetry of the matrices.
This means I only need to look at the minimum eigenvalue of $(I-\epsilon KM)(I- \epsilon MK) = I -\epsilon MK - \epsilon KM + \epsilon^2 KMMK $ and show that there exists $\epsilon >0$ such that the minimum eigenvalue is $> 1$. Note that this is now a symmetric matrix and so the eigenvalues are real.
I am stuck at this point and looking for suggestions.
Denote $A= M$, $B= -K$, both positive definite. It is possible that $AB+BA$ has a negative eigenvalue, see example.Then $\epsilon (AB + BA + \epsilon BAAB)$ will have a negative eigenvalue for all sufficiently small $\epsilon>0$ ( by continuity).
However, for $\epsilon>0$ large, we can write $\epsilon (AB+ BA)+ \epsilon^2 BAAB = \epsilon^2( BAAB + \frac{1}{\epsilon} (AB+BA))$. Now, the positive definite $BAAB$ dominates, so for $\epsilon $ large we get $\epsilon (AB+ BA)+ \epsilon^2 BAAB$ positive definite.
Therefore, you can always find a corresponding $\epsilon$; in some cases $\epsilon$ has to be large, rather than small.