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Given a Hermitian matrix $A$ with largest eigenvalue $\lambda$, can we show that if $\lambda$ lies on the diagonal of $A$ (say at $a_{ii}$) that $a_{ij} = a_{ji} = 0 $ whenever $i \neq j$?

If we order the eigenvalues of $A$ as $\lambda \geq \lambda_2 \geq \lambda_3......$, then I know that we have $ a_{11} + a_{22} + \cdots+ a_{ii} \leq \lambda+\lambda_2 + \lambda_3 +\cdots+ \lambda_i$.

Is there a way to apply this to get that the $i$-th column and row are all $0$ except for possibly at $a_{ii}$?

I have thought about doing something with the diagonalized form of $A$, but that has taken me nowhere.

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    $\begingroup$ Have you heard about spectral radius? Perhaps you may want to check Section 3 here: math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect6.pdf $\endgroup$ – Lord Soth Feb 21 '14 at 3:07
  • $\begingroup$ I am familiar with spectral radius. But I am not sure how that really helps here. Are you suggesting using the spectral radius because you know of a proof using such? $\endgroup$ – Vladhagen Feb 21 '14 at 4:12
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We may assume $A>0$ (eventually changing $A$ with $A+\alpha I$). If $(e_i)_i$ is the canonical basis, then $(Ae_i,e_i)=a_{i,i}=\lambda\leq ||Ae_i||.||e_i||\leq ||A||_2=\lambda$. Thus $Ae_i=\lambda e_i$ and we are done.

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