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 $\lambda_{An}$?

  • $\begingroup$ 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

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 $A,B$.

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