Evaluate extreme point based on eigenvalue & eigenvector of Hess Matrix

Question

Is there a case, in which at an extreme point there are more than 1 positive EValue? This would lead to more than 1 EVector that shows the steepest increase of a function. Up till now I haven't run into a case like described, but I wonder about the possibility of having one. I think it is not possible because there is only one direction to make the function change fastest.

Answer 1

The gradient will tell you in which direction a function increases the fastest. Eigenvalues being positive at a point means the function is strictly convex at that point. These are different things.

Coming to the specific question (is there a case in which at an extreme point there are more than 1 positive eigenvalue) the answer is yes:

Consider, for instance

\begin{align} f \colon \mathbb{R}^2 &\to \mathbb{R} \nonumber \\ (x,y) &\mapsto 2x^2+y^2 \nonumber \end{align}

The Hessian matrix at extreme point $(0,0)$ has two positive eigenvalues, 2 and 4.