A question on PSD (positive semidefinite) completion

[NaturePaper]

Please help me to solve this question:

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

We note that if we choose c=a_1\bar{a}_2, then A=b^\dag b,\mbox{ with } b=(a_0~ a_1~ a_2~ 0~ a_3~ 0 ~0)^T 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 c=a_1\bar{a}_2. 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 a_i,\quad i=0,1,2,3 to be zero. Please help me.

 

[NaturePaper]

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?

 

[NaturePaper]

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.