# Local maximum when Hessian is negative semi-definite?

If it possible to have a local maximum when the Hessian is only negative semi-definite (i.e., there is one zero eigenvalue and all other eigenvalues are negative).

If not, what it the ultimate recourse to prove that a point is a local maximum is such cases?

• If it's NSD then it's inconclusive :calculus.subwiki.org/wiki/… You could try to visually look at it. – VCG Aug 23 '16 at 12:54
• Useful pointer. Thanks. Looks like I can rule out it being a local minimum, but what is the next course of action? Suppose that I am dealing with a firm's profit maximization problem, it seems like the eigen-vector associated with the zero eigen-value can be interpreted as a direction along which the firm would be indifferent as marginal step in that direction does not impact the value of the objective. Does this intuition generalize? Has some mathematician formalized it? – davidrpugh Aug 23 '16 at 16:43
• Can you give me such a profit function? Looking at it may help me get an idea. – VCG Aug 23 '16 at 17:47