$x \ge 0, z \ge 0, xz \ge y^2$ for symmetric $2 \times 2$ positive semidefinite matrix

Question

I am told that the positive semidefinite cone in $\mathbf{S}^2$ is

$$X = \begin{bmatrix} x & y \\ y & z \end{bmatrix} \in \mathbf{S}^2_+ \iff x \ge 0, \ \ \ z \ge 0, \ \ \ xz \ge y^2,$$

where $S^2_+$ is the set of symmetric $2 \times 2$ positive semidefinite matrices.

I suspect that the conditions $x \ge 0, z \ge 0, xz \ge y^2$ comes from the fact that the matrix is positive semidefinite, but can someone please make it clear why this is required for a positive semidefinite matrix?

This definition is the one I am using, but it doesn't say anything explicit about the values of the elements of the matrix.

Answer 1

Since $X$ is real symmetric and positive semi-definite, you can write it as

$$X = B^TB$$

(where B isn't unique -- $B^T$ could be for instance lower triangular as in Cholesky. But for any valid choice of $B$ the following arguments hold.)
Let $B= \begin{bmatrix} \mathbf b_1 \vert \mathbf b_2 \end{bmatrix}$. Then,

$$X = \begin{bmatrix} x & y \\ y & z \end{bmatrix} = \begin{bmatrix} \mathbf b_1^T \mathbf b_1 & \mathbf b_1^T \mathbf b_2 \\ \mathbf b_1^T \mathbf b_2 & \mathbf b_2^T \mathbf b_2 \end{bmatrix} = \begin{bmatrix} \big \Vert \mathbf b_1\big \Vert_2^2 & \mathbf b_1^T \mathbf b_2 \\ \mathbf b_1^T \mathbf b_2 & \big \Vert \mathbf b_2\big \Vert_2^2 \end{bmatrix}$$ By Cauchy-Schwarz, this implies
$$y^2 = (\mathbf b_1^T \mathbf b_2)^2 \leq \big \Vert \mathbf b_1\big \Vert_2^2 \cdot \big \Vert \mathbf b_2\big \Vert_2^2 = x \cdot z.$$

Answer 2

The eigenvalues of $X$ are given by $$\left|\begin{bmatrix} x-\lambda & y \\ y & z-\lambda \end{bmatrix}\right|=0\implies \lambda^2-(x+z)\lambda+(xz-y^2)=0.\\ \implies\lambda=\frac{(x+z)\pm\sqrt{(x+z)^2-4(xz-y^2)}}{2}.$$ A symmetric matrix is positive semidefinite if and only if all of its eigenvalues are nonnegative. \begin{align} \lambda\geq0 \iff & x+z\geq \sqrt{(x+z)^2-4(xz-y^2)}\geq 0\\ \iff& (x+z)^2 \geq (x+z)^2-4(xz-y^2) \text{ and } x+z\geq 0\\ \iff & xz\geq y^2\geq 0 \text{ and } x+z\geq 0\\ \iff & xz \geq y^2,\; x\geq 0 \text{ and } z\geq 0 \end{align}