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I saw the answer to this question on the nature of saddles, and it contains the reasoning:

Perhaps it suffices to note that saddles should remain saddles even when we change coordinates. For example, note that $f(x,y)=x^2+6xy+y^2$ can be rewritten in the form $f(x,y)=2(x+y)^2−(x−y)^2$. From this, we should expect that f has a saddle at zero if and only if g(x,y)=$2x^2−y^2$ has one as well.

But why should the saddle (or any critical point) remain the same exact type of critical point even after the change of coordinates? It is not entirely evident to me that it should be so.

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  • $\begingroup$ Is it entirely evident to you that a sphere is still a locus of points equidistant from its center even if you change coordinate systems? If so, then what's the difference between spheres and saddles in your mind? $\endgroup$ – Eric Lippert Jan 15 '18 at 15:15
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    $\begingroup$ @EricLippert Not all coordinate transformations preserve Euclidean distance. The coordinate change $(x,y,z) \mapsto (x/a,y/b,z/c)$, for example, deforms a sphere centered on the origin to an ellipsoid. Different coordinate transformations (e.g. smooth, linear, orthogonal linear) may preserve more or less properties of the original representation (i.e. orthogonal linear transformations preserve Euclidean distance and smooth and linear transformations may not). Certain properties are intrinsic to the object itself but it is not necessarily immediately obvious that being a saddle is one of them. $\endgroup$ – eepperly16 Jan 15 '18 at 17:54
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Your question is at least partly rather philosophical. I think you're thinking too much about the algebra and too little about the geometry.

The mathematical object you are trying to reason about is geometrical, When you see a picture like this one from http://mathworld.wolfram.com/HyperbolicParaboloid.html

enter image description here

you can recognize the saddle point by looking at the geometry. (That's essentially what @TedShifrin 's answer says.) You don't need a coordinate system or an equation.

In fact, every point on this surface is a saddle point in the geometric sense. If you imagine it in the coordinate system suggested by the lines on the surface, the middle point is where the tangent plane is horizontal, the gradient is zero but you have neither a local maximum or minimum. If you rotate the coordinate system and look at it from another angle that point will still be a saddle point, as will all the others.

If you want to reason about this figure algebraically then you choose a coordinate system and find the function that describes the surface in that system. The choice of coordinates does not determine the object. But in different coordinate systems the object will be described by different functions. Sometimes some of those functions are easier to work with than others.

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You already got a linear algebra answer, so I won't write another.

The geometric answer is that you can detect a saddle point by noting that the surface is negatively curved there (curving inward in one direction and outward in another) ... and that geometry persists when we make a linear change of coordinates.

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    $\begingroup$ Even if the change of coordinates is non-linear, it still preserves the shape of extrema; it just has to be smooth & invertible in a neighborhood of the extremum. (I think.) $\endgroup$ – Michael Seifert Jan 15 '18 at 14:58
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    $\begingroup$ @MichaelSeifert: Yes, this is right. It follows from the chain rule and Taylor's Theorem. But I was trying to keep to the spirit of the original question. $\endgroup$ – Ted Shifrin Jan 15 '18 at 15:31
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This may or may not be helpful to you depending on your background, but hopefully it will.

A point $(x,y)$ is a saddle point of $f$ if $\nabla f(x,y) = (0,0)$ and the Hessian matrix of $f$ at $(x,y)$ has both positive and negative eigenvalues. Under a change of coordinates, the (symmetric) Hessian matrix is transformed by a congruence transformation (i.e. if the Hessian was $H$ it will be transformed to $P^THP$ for some nonsingular matrix $P$) which preserves the number of positive and negative eigenvalues of the Hessian by Sylvester's Law of Inertia.

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Through a saddle point, there is (at least) one path for which the saddle point is a local maximum and (at least) one path for which the saddle point is a local minimum. If you change coordinates, the paths will be altered, but the heights will not -- there will still be a path with the (image of the) saddle point a local minimum and another with it a local maximum.

(Remember a change of coordinates changes the input coordinates, not the outputs, so the heights, minima, and maxima are unchanged.)

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  • $\begingroup$ even if the outputs are transformed, at worst the role of max and min are reversed $\endgroup$ – Hagen von Eitzen Jan 15 '18 at 14:55
  • $\begingroup$ @HagenvonEitzen : Well, ..., the constant map is continuous, and if applied to the outputs, it turns all points into mins/maxs/saddles. Not all changes of variables are injective (and this sometimes causes complications). $\endgroup$ – Eric Towers Jan 15 '18 at 17:02

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