This question is related to one property of rank-$1$, positive semidefinite matrices. Would be very useful in SDP problems (which is where I found it).

Consider a $3 \times 3$ positive semidefinite rank-$1$ matrix $A = xx^H$ $$ $$ \begin{bmatrix} x_1x^H_1 & x_1x^H_2 & x_1x^H_3 \\x_2x^H_1 & x_2x^H_2 & x_2x^H_3\\x_3x^H_1 & x_3x^H_2 & x_3x^H_3\end{bmatrix} $$ $$ Now let matrix $B$ as a partial matrix of $A$ defined as: $$ $$ \begin{bmatrix} x_1x^H_1 & x_1x^H_2 & x_1x^H_3 \\x_2x^H_1 & x_2x^H_2 & \\x_3x^H_1 & & x_3x^H_3\end{bmatrix} $$ $$ Is it possible that any completion of matrix $B$ be positive semidefinite but not $A$ (or, equivalently, psd but not rank-$1$)? I think it is impossible, but I don't know how to prove it. If it is possible, you can either tell me why or use real number as a counter example.

  • $\begingroup$ $2 \times 2$??? $\endgroup$ – Rodrigo de Azevedo Sep 21 '16 at 9:24
  • $\begingroup$ Matrix B is a partial matrix which is 2x2 psd rank-1 $\endgroup$ – Ben Wu Sep 21 '16 at 18:51

You may employ Sylvester's criterion (for positive semidefinite matrices).

  • $\begingroup$ Thank you. In fact just setting the determinant of B to be nonnegative will achieve the same result. $\endgroup$ – Ben Wu Sep 21 '16 at 20:13

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