0
$\begingroup$

Let $A$ and $B$ be $n\times n$ Hermitian matrices and $\{\lambda_{A1},\ldots,\lambda_{An}\}$ and $\{\lambda_{B1},\ldots,\lambda_{Bn}\}$ represent their eigenvalues, respectively. If $A+B = \Lambda$ where $\Lambda$ is a diagonal matix with diagonal elements $\{\lambda_1,\ldots,\lambda_n\}$ and $\lambda_{Bn}=0$, can we find $\lambda_{An}$?

$\endgroup$
  • $\begingroup$ Do you mean that $\lambda_{Bj} = 0$ for $j = 1,\dots,n$, or do you specifically mean that $\lambda_{Bn} = 0$ (and that the other eigenvalues are unknown)? Are these eigenvalues ordered in any particular way? $\endgroup$ – Omnomnomnom Aug 8 '17 at 21:55
  • $\begingroup$ One or some eigenvalues of $B$ are zero, not all of them. I assume all eigenvalues are non-negative. Thus, $\lambda_{Bn}$ is the smallest eigenvalue. $\endgroup$ – Bob Aug 8 '17 at 21:58
1
$\begingroup$

It follows from the Rayleigh-Ritz theorem (or as an instance of Weyl's inequality) that $$ \lambda_{min}(\Lambda) = \lambda_{min}(A + B) \leq \lambda_{min}(A) + \lambda_{min}(B) = \lambda_{min}(A) $$ On the other hand, by another instance of Weyl's inequality, we can get $$ \lambda_{min}(\Lambda) = \lambda_{min}(A + B) \geq \lambda_{max}(A) + \lambda_{min}(B) = \lambda_{max}(A) $$ We can't say anything more substantial than that. Both of these inequalities are sharp, and equality is attained with diagonal matrices $A,B$.

| cite | improve this answer | |
$\endgroup$

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.