Non-Strict Saddle Point vs Local Minima While going through Escaping Saddle Points Efficiently, I came across the definition of Strict Saddle Point. They define a stationary point to be a strict saddle if at least one of the eigenvalue of the Hessian Matrix is negative. This implies a non-strict saddle point will have all eigenvalues of the Hessian Matrix greater than or equal to zero. But, isn't this a sufficient and necessary condition for local minima? What is the difference between a non-strict saddle point and a local-minima?

 A: When the Hessian is singular, i.e., some eigenvalues are $=0$,  then the character of the critical point cannot be decided from looking at the Hessian alone. The problem then becomes algebraically much more complicated.
The function $f(x,y):=x^2-y^2$ has a strict saddle point at $(0,0)$. The function $$g(x,y):=x^2+\alpha y^4\ +{\rm higher\ order\ terms}$$ has a strict local minimum at $(0,0)$ when $\alpha>0$, and a "non strict" saddle point when $\alpha<0$. When $\alpha=0$ you have to look at the higher order terms. 
A: That definition of saddle point does not agree wih my undersanding of the concept. Perhaps they should have defined "saddle point" as " critical point where the Hessian has at least one positive eigenvalue and at least one negative eigenvalue" and "local minimum" as " critical point where the Hessian has all eigenvalues positive." Another  difficulty is the scope of "non-." Is a "non-strict saddle point" a saddle point that is not strict or is it any point that is not a strict saddle point? A third problem with the approach to local extrema via eigenvalues of the Hessian is that what we are really interested in is the behaviour of the Hessian as a quadratic form.It is certatainly true that any real symmetric matrix, such as the Hessian, can be orthogonally diagonalized with the eigenvalues on the diagonal and that in any diagonalization by congruence ( Orthogonal diagonalization is a special case of diagonalization by congruence.) the number of positive, negative and 0 diagonal terms is the same as in any other orthogonal diagonalization. (Sylvester's theorem). However, finding the eigenvalues of an $n \times n$ matrix involves solving an equation of degree $n$ whereas a symmetric matrix can be easily diagonalized by congruence over any field of characteristic $\ne 2.$ 
A: For a function $f\colon\mathbb{R}^n\to\mathbb{R}$, the condition $\nabla^2f(x^*)\succeq 0$ is necessary for $x^*$ to be an unconstrained local minimum of $f$, but it is in general not sufficient. Consider $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3$. Then $\nabla f(x) = 3x^2$ and $\nabla^2 f(x) = 6x$. Therefore, at $x^*=0$, the function satisfies both the first- and second-order necessary conditions for optimality, i.e. $\nabla f(x^*) = 0$ and $\nabla^2 f(x^*) = 0 \succeq 0$. However, $x^*=0$ is clearly not a local minimum of this function. Indeed, it is a saddle point, but this information cannot be ascertained by the Hessian, since the second-order information at $x^*=0$ does not encapsulate enough information about the geometry of $f$ in that neighborhood. To determine the true optimality (more precisely, lack there of) of $x^*=0$ for this function, you could look at the third-order term in the Taylor series expansion of $f$ around $x^*=0$.
The authors' definition of strict saddle reflects the trouble we see in the above example. In particular, a strict saddle is a saddle point whose behavior can be ascertained by the second-order information given by the Hessian. Analyzing these points avoids the issues with nonstrict saddles where we would need to look beyond second-order information.
