# A question on PSD (positive semidefinite) completion

1. Apr 4, 2010

### NaturePaper

Please help me to solve this question:

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

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.

2. Apr 4, 2010

### 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?

3. Apr 5, 2010

### 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook