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I have arrived at this result from a very different perspective (quantum operations) but, being a completely algebraic result, I was hoping that there would be a simple algebraic way of looking at it too.

Let $P$ be a positive semidefinite matrix. Let $E$ be a diagonal matrix with real entries such that -

  1. Tr$(E)=0$
  2. Diag$(P+E) \succeq 0 $ [That is, for all $i$ , $P_{ii}+E_{ii}\geq 0$]

Prove that $P+E$ is positive semidefinite.

Thanks in advance!

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    I was mislead by my intuitions to believe that the claim was true. I still don't think that this was a bad question. Why the downvotes?2013-08-24

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Unless I'm missing something, the result is not true. Take $P:=\pmatrix{2&1\\1&1}$, and $E=\pmatrix{1&0\\0&-1}$. Then $P$ is symmetric, positive definite, $E$ is diagonal, $P+E=\pmatrix{3&1\\ 1&0}$. Each diagonal entry of $P+E$ is non-negative, and the trace of $E$ is $0$ but $P+E$ is not positive semi-definite (consider $x=\pmatrix{1\\-3}$).

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    So the counter-example shows that you can't get more in general about the positive definiteness of $P+E$.2012-06-23