# How to find conditions for positive semidefinite matrix?

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?

• It is not clear what you mean by "the criterion is not unique". Commented Sep 23, 2020 at 7:29
• You are using the wrong Sylvester criterion. You are using the one for PD-ness, rather than PSD-ness. Commented Sep 23, 2020 at 8:55
• Which section of the book? Commented Apr 7, 2021 at 18:25

How does $$x_1 = 0$$ imply that $$x_2 = x_3 = 0$$?
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.
• Hi, thank you for your answer. I still don't get why $2 \times 2$ principal submatrix has a negative determinant if $x_2$ and $x_3$ are non-zero. If $x_2$ and $x_3$ are non-zero, then the determinant is $-x_2(x_2 x_6 - x_5 x_3) + x_3 (x_2 x_5 - x_3 x_4)$. how is this negative? Commented Sep 23, 2020 at 11:22
• @MoneyBall It sounds like you don't understand what a principal submatrix is. For example, if $x_1 = 0$ and $x_2 \neq 0$, then the matrix $$\pmatrix{x_1&x_2\\x_2&x_1} = \pmatrix{0&x_2\\x_2&0}$$ has negative determinant Commented Sep 23, 2020 at 11:50