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I understand that if a hessian is positive semidefinite for all x, the function is convex (though not strictly convex), so the stationary point is a local and global minimum.

However, what if a stationary point does not have a positive semi-definite hessian matrix? Then the point cannot be a local minimizer?

It is quite confusing for me, because what I think is:

If a stationary point does not have a positive semi-definite hessian matrix, then the hessian matrix is negative definite, meaning the point must be a local max.

However, I am not 100% certain about this.

Jayden Rice
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  • the function $f(x,y) = x^2 - y^2$ has a critical point at the origin, the Hessian has one positive eigenvalue and one negative. – Will Jagy Jul 09 '21 at 02:09
  • @WillJagy, can you elaborate further? Yes, $f(x,y)=x^2-y^2$ has a critical point $(0,0)$ and the Hessian has eigenvalue 2 and -2. $(0,0)$ is neither local min nor local max. – Jayden Rice Jul 09 '21 at 02:28
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    You wrote "does not have a positive semi-definite hessian matrix, then the hessian matrix is negative definite," I gave an example that shows your idea is wrong: this is usually called a saddle point – Will Jagy Jul 09 '21 at 02:31
  • Matrices do not have to be either positive or negative semi-definite, they can be indefinite like $\begin{pmatrix}1&0\0&-1\end{pmatrix}$, so your argument is invalid. But the title claim is true: at a local minimizer the Hessian is positive semi-definite (if it exists). If not, there is a negative eigenvalue and the function values get smaller along the corresponding eigenvector. – Conifold Jul 09 '21 at 02:35
  • Okay, that makes sense. Then if a stationary point does not have a positive semi-definite hessian matrix, then the point could be a saddle point or local max? Therefore "what if a stationary point does not have a positive semi-definite hessian matrix? Then the point cannot be a local minimizer?" is true? – Jayden Rice Jul 09 '21 at 02:37
  • That's the contrapositive, so yes. – Conifold Jul 09 '21 at 02:39

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