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.
Best Answer
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}