- #1

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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.