Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

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 $x=\pmatrix{1\\-3}$).

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.