# Minimum eigenvalue of the difference of two Hermitian matrices

Consider Hermitian matrices $A$ and $B$.

Weyl's inequality tells us that $$\lambda_{\min}(A + B) \ge \lambda_{\min}(A) + \lambda_{\min}(B)$$
See this link for proof: Smallest eigenvalues of Sum of Two Positive Matrices

How can we bound $\lambda_{\min}(A - B)$, given the the eigenvalues of $A$ and $B$?

$\lambda_{min}(A) - \lambda_{max}(B)$?!
• @ssk08 it suffices to note that $\lambda_{min}(-B) = -\lambda_{max}(B)$ – Omnomnomnom Nov 9 '16 at 11:47