$x \ge 0, z \ge 0, xz \ge y^2$ for symmetric $2 \times 2$ positive semidefinite matrix 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? https://en.wikipedia.org/wiki/Definiteness_of_a_matrix This definition is the one I'm using, but it doesn't say anything explicit about the values of the elements of the matrix. Thank you.
 A: 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}
A: 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
$$y^2 = (\mathbf b_1^T \mathbf b_2)^2 \leq \big \Vert \mathbf b_1\big \Vert_2^2  \cdot \big \Vert \mathbf b_2\big \Vert_2^2  = x \cdot z.$$ 
