Determine where Hessian matrix is semidefinite

Question

How do I determine if the matrix $\begin{bmatrix}2/y&-2x/y^2\\-2x/y^2&-2x^2/y\end{bmatrix}$ is semidefinite? Could someone please show me an algorithm to show this. I know how to do it when the matrix values are all scalar, but not when there are variables inside of it.

Answer 1

Let $A=\begin{bmatrix}2/y&-2x/y^2\\-2x/y^2&-2x^2/y\end{bmatrix}\in M_2(\mathbb{R})$

The characteristic polynomial $\chi_A(X)=X^2-(trA) X+detA=0$

And, $tr(A)=\frac{2}{y^2}(1-x^2)$

$\displaystyle detA=-\frac{4x^2}{y^4}\left(y^2+1\right)$

Now, note that $det A<0$ (squares are positive). In particular, the product of eigenvalues $\lambda_1\lambda_2<0$ and that implies that $\lambda_1$ and $\lambda_2$ are opposite signs.

Hence, it is indefinite. (Actually, we didn't have to calculate the trace in hindsight.)