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Under which condition is the largest eigenvalue of a positive semi-definite matrix strictly larger than the largest of the matrix's diagonal entries?

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I doubt that there's a nice necessary and sufficient condition, but a sufficient condition is that at least one off-diagonal entry in the row with the largest diagonal entry is nonzero.

EDIT: Let your matrix by $A$. Suppose the largest diagonal entry is $d_{ii}$ and $d_{ij} \ne 0$. Consider the $2 \times 2$ submatrix consisting of the $i$'th and $j$'th row and column. This is also positive semidefinite; if $p(\lambda)$ is its characteristic polynomial, $p(d_{ii}) = -d_{ij}^2 < 0$, while of course $p(\lambda) \to +\infty$, so there must be a root $\mu$ of $p$ greater than $d_{ii}$. This translates to there being a nonzero vector $u$ having only its $i$'th and $j$'th entries nonzero such that $\langle A u, u \rangle = \mu \langle u, u \rangle$, and then the largest eigenvalue of $A$ is at least $\mu$.

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  • $\begingroup$ Could you provide a reference for that? $\endgroup$ – Eigenvalue Apr 28 '15 at 19:23
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Hint: $$\lambda_{max} = \max_{|v|=1} \langle Av, v\rangle $$

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  • $\begingroup$ But this only gives me an inequality. How do I get that the largest eigenvalue is strictly larger than the largest variance entry of the matrix? $\endgroup$ – Eigenvalue Apr 28 '15 at 19:19
  • $\begingroup$ it is not strictly larger. consider a diagonal matrix. $\endgroup$ – mookid Apr 28 '15 at 19:20
  • $\begingroup$ Yes, generally not. Robert Israel gave one condition on when we have a strict inequality. $\endgroup$ – Eigenvalue Apr 28 '15 at 19:21

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