2
$\begingroup$

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 $B$?

$\endgroup$
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ That's what I thought. But just wanted to get to verify it. $\endgroup$ – ssk08 Nov 9 '16 at 11:46
  • 2
    $\begingroup$ @ssk08 it suffices to note that $\lambda_{min}(-B) = -\lambda_{max}(B)$ $\endgroup$ – Omnomnomnom Nov 9 '16 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.