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A question on PSD (positive semidefinite) completion

  1. Apr 4, 2010 #1
    Please help me to solve this question:

    Let [tex] A=\left[\begin{array}{ccccccc}
    |a_0|^2 &a_0\bar{a}_1 &a_0\bar{a}_2 &0 &a_0\bar{a}_3 &0&0\\
    \bar{a}_0a_1 &|a_1|^2 & c &0 &a_1\bar{a}_3 &0&0\\
    \bar{a}_0a_2 &\bar{c} &|a_2|^2 &0 &a_2\bar{a}_3 &0&0\\
    0 &0 &0 &0 &0 &0&0\\
    \bar{a}_0a_3&\bar{a}_1a_3 &\bar{a}_2a_3 &0 &|a_3|^2 &0&0\\
    0 &0 &0 &0 &0 &0 &0 \\
    0 &0 &0 &0 &0 &0 &0
    \end{array}\right][/tex].

    We note that if we choose [tex]c=a_1\bar{a}_2[/tex], then [tex] A=b^\dag b,\mbox{ with } b=(a_0~ a_1~ a_2~ 0~ a_3~ 0 ~0)^T[/tex] and hence A is PSD. Now my guess is there does not exists any other choice for c to make A PSD i.e., A is PSD iff [tex]c=a_1\bar{a}_2[/tex]. Can anyone help me to prove (or give a counter example to ) this?

    One can assume the additional condition Tr(A)=1. To give a counterexample, one can assume as many of [tex]a_i,\quad i=0,1,2,3[/tex] to be zero. Please help me.
     
  2. jcsd
  3. Apr 4, 2010 #2
    Ya, I got a proof. Its a straightforward calculation and my answer is correct--no other choice of c can make A PSD. Can anybody give me an illusive (I mean a theorem...etc) proof to this?
     
  4. Apr 5, 2010 #3
    Oh...amazingly I found a nice proof which is applicable to not only this problem but for all such matrices (atleast three a_i nonzero) . The PSD completion is unique. The main key to the proof is the property that every principle minor should be non negative.
     
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