User:IssaRice/Linear algebra/Determinant as signed volume of transformation: Difference between revisions

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Let <math>f : \mathbf R^n \to \mathbf R^n</math> be some function (not necessarily linear) and let <math>\Omega \subseteq \mathbf R^n</math> be some region in space. we will assume we can assign some "volume" to <math>\Omega</math>, e.g. by cutting it up into little cubes and adding up the volumes of the cubes (the volume of a cube is just the product of its edge lengths).
Let <math>f : \mathbf R^n \to \mathbf R^n</math> be some function (not necessarily linear) and let <math>\Omega \subseteq \mathbf R^n</math> be some region in space. we will assume we can assign some "volume" to <math>\Omega</math>, e.g. by cutting it up into little cubes and adding up the volumes of the cubes (the volume of a cube is just the product of its edge lengths).


since f takes this space to itself, the image of <math>\Omega</math> under f, denoted <math>f(\Omega)</math>, is another region in space. let's assume f is nice enough that we can assign a volume to <math>f(\Omega)</math>. we can now ask, what is the volume of <math>f(\Omega)</math>? is it related to the volume of <math>\Omega</math> somehow? does the volume change if we translate <math>\Omega</math>, stretch it, rotate it, etc.?
since f takes this space to itself, the image of <math>\Omega</math> under f, denoted <math>f(\Omega)</math>, is another region in space. let's assume f is nice enough that we can assign a volume to <math>f(\Omega)</math>. we can now ask, what is the volume of <math>f(\Omega)</math>? is it related to the volume of <math>\Omega</math> somehow? does the volume change if we translate <math>\Omega</math>, stretch it, rotate it, etc.?
in general, i don't think we can say anything too interesting here. (consider a quadratic function, where distance to the origin changes how much the volume changes.) but if we restrict attention to the following functions: for all <math>x \in \mathbf R^n</math> and all <math>\Omega \subseteq \mathbf R^n</math>, <math>\operatorname{vol} f(x+\Omega) = \operatorname{vol} f(\Omega)</math>. in other words, f alters volume "globally" in the sense that no matter where you place <math>\Omega</math> in space, the deformed volume is the same.


==References==
==References==

Revision as of 02:43, 28 December 2019

Let f:RnRn be some function (not necessarily linear) and let ΩRn be some region in space. we will assume we can assign some "volume" to Ω, e.g. by cutting it up into little cubes and adding up the volumes of the cubes (the volume of a cube is just the product of its edge lengths).

since f takes this space to itself, the image of Ω under f, denoted f(Ω), is another region in space. let's assume f is nice enough that we can assign a volume to f(Ω). we can now ask, what is the volume of f(Ω)? is it related to the volume of Ω somehow? does the volume change if we translate Ω, stretch it, rotate it, etc.?

in general, i don't think we can say anything too interesting here. (consider a quadratic function, where distance to the origin changes how much the volume changes.) but if we restrict attention to the following functions: for all xRn and all ΩRn, volf(x+Ω)=volf(Ω). in other words, f alters volume "globally" in the sense that no matter where you place Ω in space, the deformed volume is the same.

References

  • https://www.youtube.com/watch?v=xX7qBVa9cQU -- this is probably the best explanation of the determinant i have ever seen
  • sergei treil's linear algebra done wrong has a pretty good explanation. in particular, i like how he first defines determinant for a list of vectors.