# Critical points characterization of real function

Providing a real multi-variable function $f(\bar{x})$ twice differentiable with respect of all its variables. Looking for critical points is equivalent to solve $\nabla f = \vec{0}$. And to characterize them you can investigate Hessian matrix $H(f)$ for positiveness (minimum) or negativeness (maximum). If Hessian matrix is not degenerate and eigenvalues of $H(f)$ have different signs then it is a saddle point. Otherwise, at least one eigenvalue is null, and then the test is inconclusive. Also eigenvalues are finite (function is continuous at critical point) and eigenvectors are orthogonal (Hessian matrix is symmetric).

My questions are:

• Are all these statements above right?
• Is there another critical point than extrema and saddle point? Wikipeda cites undulation point but I can get simple explanation of it.
• If the test is inconclusive what approach must I use to characterize the critical point?
• In this context, what magnitude of eigenvalues and direction of eigenvectors mean?

Is there another critical point than extrema and saddle point? Wikipeda cites undulation point but I can get simple explanation of it.

As to critical points: there is also the case, where $H$ is degenerate. If $H=0$, then one can do higher-order tests. If $H\ne0$ but positive semidefinite, nothing can be said about local optimality.

If the test is inconclusive what approach must I use to characterize the critical point?

If $H$ is not zero, semi-definite, but still degenerate then everything is possible: saddle point and local extreme. Check $f(x) = x^4$, $f(x_1,x_2)=(x_2-x_1^2)(x_2-2x_1)^2$. All these functions have critical point at $0$ with positive definite $H$. However, for the second function $(0,0)$ is not a local minimum.

In this context, what magnitude of eigenvalues and direction of eigenvectors mean?

If $H$ is positive definite with large eigenvalues, then the critical point is locally optimal in a large neighborhood. Directions of eigenvalues give directions of principal curvature (?) - more a differential geometric thing.

• +1 for answering and confirming my intuitions. I will check out your examples and investigate principal curvature. – jlandercy Sep 19 '14 at 8:52