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By the Schur-Horn inequality I am thinking of the statement that for any Hermitian matrix $H$ its diagonal n-tuple $(H_{11},H_{22},..,H_{nn})$ for any choice of basis lies in the convex hull of the $n!$ permutations of its eigenvalue n-tuple.

Or any further improvement if one assumes not just the p.s.d structure but also that, $H_{ii} = \sum_{j \neq i} H_{ij}$?

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    $\begingroup$ For the first one, can't you just add a multiple of the identity to embed the usual problem into the semidefinite one, i.e. they're the same level of difficulty? $\endgroup$
    – Allen Knutson
    Commented Nov 25, 2014 at 2:49
  • $\begingroup$ @AllenKnutson I am not getting you - are you saying that any semi-definite matrix is also Hermitian and decomposable in some particular way such that the Schur-Horn can be used on the parts individually? Can you kindly elaborate? $\endgroup$
    – Student
    Commented Nov 25, 2014 at 6:17
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    $\begingroup$ If you're asking what I think you're asking, the answer is "Very obviously yes," enough so that I wonder if I'm misunderstanding the question. $\endgroup$
    – Allen Knutson
    Commented Nov 25, 2014 at 18:10
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    $\begingroup$ Let $\vec v = \sum_j c_j \vec e_j$, where $\vec v$ is the $i$th vector in your orthonormal basis, and the $(e_j)$ are an orthonormal eigenbasis (with eigenvalues $(\lambda_i))$, and $\sum_i |c_i|^2=1$. Then $A_{ii} = \langle \vec v,A \vec v\rangle = \sum_i |c_i|^2 \lambda_i \in [\lambda_{min},\lambda_{max}]$. It's a little weird to credit Schur and Horn with this. $\endgroup$
    – Allen Knutson
    Commented Nov 25, 2014 at 21:22
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    $\begingroup$ If your matrix is special then it's unlikely that Schur-Horn gives the best possible results. $\endgroup$
    – Allen Knutson
    Commented Nov 26, 2014 at 20:26

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