In Boyd & Vandenberghe's Convex Optimization, we want to show that a $3 \times 3$ matrix is positive semidefinite:
$$\begin{pmatrix} x_1 & x_2 &x_3 \\ x_2 & x_4 & x_5 \\ x_3 & x_5 & x_6 \end{pmatrix}$$
Using Sylvester's criterion, I can find the conditions that would make the determinants of upper left $1 \times 1$, $2 \times 2$, and $3 \times 3$ non-negative. However, in the solution, it computes
$$z^TXz = x_1z_1^2 + 2x_2z_1z_2 + 2x_3z_1z_3 + x_4z_2^2 + 2x_5z_2z_3 + x_6z_3^2 \ge 0$$
and says if $x_1=0$, we must have $x_2 = x_3 = 0$ so $X \succeq 0$ if and only if
$$\begin{bmatrix} x_4 & x_5 \\ x_5 & x_6 \end{bmatrix} \succeq 0$$
What I don't understand is how does $x_1$ imply $x_2=x_3 = 0$?
And in their final conditions, i.e.,
$$x_1 \ge 0, x_4 \ge 0, x_6 \ge 0, x_1x_4 – x_2^2 \ge 0, x_4x_6 – x_5^2 \ge 0, x_1x_6 – x_3^2 \ge 0$$
and
$$x_1x_4x_6 + 2x_2x_3x_5 – x_1x_5^2 – x_6x_2^2 – x_4x_3^2 \ge 0$$
only some are included in Sylvester's criterion. I am assuming the criterion is not unique. Is this the case here?
Best Answer
Recall that a positive semidefinite matrix must have positive semidefinite principal submatrices. If $x_1 = 0$ but either $x_2$ or $x_3$ is non-zero, then there is a $2 \times 2$ principal submatrix that has a negative determinant.
Sylvester's criterion only applies to positive definite matrices. In order to extend the criterion to positive semidefinite matrices, we have to consider the determinant of every principal submatrix. For example, the diagonal matrix $$ \pmatrix{1\\&0\\&&-1} $$ has non-negative leading principal minors but fails to be positive semidefinite.