# Showing that $\|I-\epsilon MK\|_2 < 1$ for some $\epsilon$

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.