Hessian matrix for extremum

Question

Is it enough to say a function doesn't have an extremum at a point if the determinant of the Hessian at that point is smaller than 0?

Answer 1

This is true in two dimensions, but not in general. In 3D for example the function $$ f(x,y,z)=-x^2-y^2-z^2 $$ has negative Hessian everywhere and clearly has a local (and global) maximum at the origin.

Answer 2

If the space has even dimension, then symmetric semidefinite matrices all have determinant $\ge0$, and therefore a singular point with Hessian of negative determinant must be a saddle point. In odd dimension consider the map $-\frac12x^\top x$ which has a maximum in $0$.