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Second derivative test when Hessian is Positive Semi-Definite

  1. May 15, 2010 #1
    Can someone tell me what this actually is.

    So, in the case when the Hessian is positive (or negative) semidefinite, the second derivative test is inconclusive.

    However, I think I've read that even in the case where the Hessian is positive semidefinite at a stationary point x, we can still conclude that the function at x is not a local maximum. Is that correct?

    Is that equivalent to the function at x being either a local minimum or a saddle point, since there are only 3 possibilities for x: local max, local min, or saddle (or are there more possibilties)?

    If someone has some online pdf notes with definitions and proof of the second derivative test in generality, that would be good too. I also can't seem to find an agreed on definition of a saddle point, which adds to the confusion.
     
  2. jcsd
  3. May 15, 2010 #2

    HallsofIvy

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    Science Advisor

    Yes, at any "critical point" we must have a maximum, minimum, or saddle point. The fact that the Hessian is not positive or negative means we cannot use the 'second derivative' test (local max if det(H)> 0 and the [itex]\partial^2 z/\partial x^2< 0[/itex], local min if det(H)> 0 and [itex]\partial^2 z/\partial x^2< 0[/itex] and a saddle point if det(H)< 0)but it will be one of those, none the less. That simply means that we cannot use that particular test to determine which.

    For example, the functions [itex]z= x^4+ y^4[/itex], [itex]z= -x^4- y^4[/itex] and [itex]z= x^3+ y^3[/itex] have first derivatives equal to 0 only at (0, 0) so that is the only critical point. The Hessian of all three functions is the 0 matrix at (0, 0) but it is obvious that (0, 0) is a minimum for the first function, a maximum for the second, and a saddle point for the third.
     
  4. May 15, 2010 #3
    I see, thanks for the reply. Although I'd like to know your definition for a saddle point. Is it a critical point that is neither a local min, nor a local max?
     
  5. Aug 11, 2011 #4
    This is an old post..... but whatever.

    My question is: If we are trying to do unconstrained optimization and using newton's method, can we adjust the hessian matrix by shifting it (adding a diagonal matrix to it) to push its eigenvales to be sufficiently negative (or positive) to force it towards a maximum (or minimum) while avoiding the saddle points; I would like to do this shift before the newton step.
    many thanks
     
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