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If the hessian evaluated at a critical point is positive (negative) definite, then we can conclude that it's a local minimum (maximum) there. If the hessian is indefinite (both negative and positive eigenvalues), then it's a saddle point.

What happens if the Hessian is positive SEMI-definite? According to https://en.wikipedia.org/wiki/Second_partial_derivative_test, it seems it is inconclusive whether the point is a max/min/saddle point, but I am wondering why this is the case. If the hessian was globally positive semi-definite, then the function is convex, and hence a local minima exists. But in the current local case, why can't we say that if the Hessian at a critical point is PSD, then it's a local minima?

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This is the same as the case in one dimensional optimization of the second derivative being 0. Namely, 0 is not a local minimum of the function $x^3$, and an analogous multivariate example to work out is the function $f(x,y)=x^3+y^2$.

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  • $\begingroup$ But what about the case where it is positive semi definite? $\endgroup$
    – David
    Commented Aug 21, 2020 at 17:27
  • $\begingroup$ @David it is positive semi definite here. Work out the hessian evaluated at 0, which will be diagonal, and you’ll see what I mean. $\endgroup$
    – Alex
    Commented Aug 21, 2020 at 17:34
  • $\begingroup$ Ah the hessian evaluated at (0,0), the critical point, is a 2x2 matrix with zeros everywhere except for the lower right element which is 2. So this is PSD. So at this point, we do not know if it is a max/min/saddle point based on the second derivative test? $\endgroup$
    – David
    Commented Aug 21, 2020 at 17:46
  • $\begingroup$ @David exactly, this works the exact same way as the one dimensional case (and reduces to the classic example when you restrict yourself to looking only at the x-axis $\endgroup$
    – Alex
    Commented Aug 21, 2020 at 17:47
  • $\begingroup$ In the 1-D case, if the second derivative is zero, isn't it conclusive that the critical point is a saddle point? I'm kind of having a hard time seeing the cross-over between 1D and multi-D. $\endgroup$
    – David
    Commented Aug 21, 2020 at 17:49

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