Subtracting a positive semidefinite matrix by a hermitian matrix

Question

I have a positive definite matrix $M$ that satisfies $M \succeq \epsilon \mathbb{1}$. I also have a hermitian operator $A$ where $\operatorname{Tr}(A)=1$. I am attempting to determine a possibly loose bound on the largest $\epsilon' > 0$ such that $M+\epsilon'A$ is positive definite. I can suppose that the spectrum of $A$, $\{\lambda_i\}$ is known. Is it possible to determine how large $\epsilon'$ can be?

A sufficient but not necessary criteria may be acceptable for my application.

Answer 1

A reasonable bound is that for $\epsilon' < \epsilon(\max_{i} |\lambda_i|)^{-1}$, we can guarantee that $M + \epsilon'A$ is positive definite. Note also that because $A$ is symmetric, we have $\max_{i} |\lambda_i| = \|A\|$, the "spectral norm" of $A$.

A better bound, if we have the signs of $\lambda_i$, is $\epsilon' < \epsilon(\max_{i:\lambda_i < 0} |\lambda_i|)^{-1}$. So, if all eigenvalues of $A$ are non-negative, there is no upper bound for $\epsilon'$.