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I want the geometrical interpretation of the following:

If $f_{xx}f_{yy} < f_{xy}^2$ and $f_{xx}$ has the same sign as $f_{yy}$ at a point, then why is that point a saddle point? Because,in the case that they have the same sign,one would expect that point to be a minimum or maximum point,not a saddle point.

Bear in mind,that i have checked the intuitive explanation that is available in wikipedia,but did not understand it.

Also, i am studying physics and not math,so please don't complicate things with symbolisms that i might not understand.

Also, i want an intuitive answer,because i know the maths behind categorizing critical points.

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  • $\begingroup$ I guess $f \in C^2(\mathbb{R}^2, \mathbb{R})$? $\endgroup$
    – Keba
    Aug 6, 2015 at 14:21
  • $\begingroup$ I don't understand this notation.I am reading from the calculus book of Anton,which does not have notation for mathematicians. $\endgroup$ Aug 6, 2015 at 14:23
  • $\begingroup$ @LandosAdam, Keba meant that $f$ is twice differentiable real function of two variables $\endgroup$ Aug 6, 2015 at 14:24
  • $\begingroup$ Oh,yes it is..Thanks for the clarification $\endgroup$ Aug 6, 2015 at 14:25
  • $\begingroup$ @LandosAdam, what's your definition of saddle point? $\endgroup$ Aug 6, 2015 at 14:26

2 Answers 2

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Consider $f:\mathbb R^2 \to \mathbb R$, twice continuously differentiable in a neighborhood of $0$ with $f(0)=0$ and $f'(0)=0$. (We discuss how the value of $f$ changes near the critical point $0$)

Assume the determinant of Hessian $H=f''(0)$ of $f$ at $0$ is negative. Since $f$ is twice continuously differentiable, $H$ is symmetric. Hence, both eigenvalues $\mu_1 \le \mu_2$ of $H$ are real. From $\det H = \mu_1 \mu_2 < 0$ follows that $\mu_1 < 0 < \mu_2$.

Now, let $u$ be the eigenvector to $\mu_1$. By Taylor theorem we have $$ f(u) \approx \frac12 u^T H u = \frac12 \mu_1 u^T u < 0 $$ for $u$ sufficiently small. Similarly, for a sufficiently small eigenvector $v$ to $\mu_2$ we have $$ f(v) \approx \frac12 v^T H v = \frac12 \mu_2 v^T v > 0. $$ Thus, $0$ is a saddle point of $f$.

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    $\begingroup$ This is an answer to a very similar question.You describe the mathematical proof of what i am asking. But,i have already stated that i already know the mathematics behind it. What i need is a geometrical interpretation of it. I want to know how to describe it visually/graphycally. Like the explanation in the wiki article in the link en.wikipedia.org/wiki/Second_partial_derivative_test under "Geometric interpretation in the two-variable case". Thank you for your answer, but i am looking for a more visual/intuitive explanation $\endgroup$ Aug 6, 2015 at 18:17
  • $\begingroup$ the axes of the "saddle" are given by the eigenvectors. I don't see how it could get more graphical. for which part do you want a more graphical interpretation? $\endgroup$
    – user251257
    Aug 6, 2015 at 18:21
  • $\begingroup$ In more plain words. Like in the above article. $\endgroup$ Aug 6, 2015 at 18:23
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I found a very good answer at Geometric interpretation of $\frac {\partial^2} {\partial x \partial y} f(x,y)$ Its the last answer by @harvard man
also,a link to see before reading it is: http://www.math.harvard.edu/archive/21a_fall_08/exhibits/fxy/index.html posted by @ user26439
(sorry,i cant tag you guys)

If any of the other users have anything else to say in the lines of the answer in the link,feel free to do so.

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