Determine whether the following $2 \times 2$ matrix is positive semidefinite (PSD)
$$\begin{bmatrix}\frac{2}{x} & \frac{-2y}{x^2} \\\frac{-2y}{x^2} & \frac{2y^2}{x^3}\end{bmatrix}$$
where $x > 0$ and $y \in \mathbb R$.
A matrix is PSD if $v^T A v \geq 0$. So, do I just multiply by a vector $v = (v_1, v_2)$ and check if it is $\geq 0$? Thanks for any help.
Best Answer
The easiest way to check if a (symmetric/Hermitian) matrix is positive definite is using Sylvester's criterion. In this case, that means that it is sufficient to check that
The first statement is clearly true. For the second, we have $$ (2/x)(2y^2/x^3) - (-2y/x^2)^2 = \frac{4y^2 - 4y^2}{x^4} = 0 \geq 0 $$ So, your matrix will always be positive semidefinite (and singular).