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 $B$?


1 Answer 1


$\lambda_{min}(A) - \lambda_{max}(B)$?!

  • $\begingroup$ That's what I thought. But just wanted to get to verify it. $\endgroup$
    – ssk08
    Nov 9, 2016 at 11:46
  • 2
    $\begingroup$ @ssk08 it suffices to note that $\lambda_{min}(-B) = -\lambda_{max}(B)$ $\endgroup$ Nov 9, 2016 at 11:47

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