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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? – Shitikanth Aug 24 '13 at 0:39
up vote 2 down vote accepted

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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Thanks for the counter-example. I will see where I have gone wrong. – Shitikanth Jun 21 '12 at 22:01
Maybe you missed an additional assumption. – Davide Giraudo Jun 21 '12 at 22:03
Actually I only need that Diag(P)+E is positive semidefinite and that is of course trivially true under the assumptions. – Shitikanth Jun 23 '12 at 4:10
So the counter-example shows that you can't get more in general about the positive definiteness of $P+E$. – Davide Giraudo Jun 23 '12 at 8:06

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