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

Let $f : R^{n} \to R^{n}$ be some function (not necessarily linear) and let $Ω \subseteq R^{n}$ 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 we can restrict attention to the following functions: for all $x \in R^{n}$ and all $Ω \subseteq R^{n}$ , $v o l f (x + Ω) = v o l f (Ω)$ . 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. now write $Ω$ as a disjoint union of tiny cubes + some even smaller almost-cubes. by the property of f, these tiny cubes and almost-cubes also get deformed in such a way that their deformed volumes are the same wherever they are. but... the different cubes are simply translations of each other! so we really just need to know how f acts on one of the cubes. then we move it around inside $Ω$ -- each of these are just translations of the original tiny cube. by the property of f, the images of these translated cubes will have the same volume as the image of the original cube. so we can summarize f by assigning it a number that tells us by what factor it deforms a cube. let's call this the volume of f, $v o l f$ . to be concrete, we can take the unit cube, ${(x_{1}, \dots, x_{n}) : 0 \leq x_{j} \leq 1 for each j}$ , and say that $v o l f$ is the volume of the image of it under f. (there is no risk of confusion, since we've previously only assigned volumes to regions of space; so we are just overloading the notion for a different kind of object.)

for such a function f, we can write $v o l f (Ω) = (v o l f) (v o l Ω)$ for every $Ω$ .

now let $f, g : R^{n} \to R^{n}$ both have this property. can we assign a volume to $f \circ g$ ? well, $v o l f (Ω) = (v o l f) (v o l Ω)$ for every $Ω$ . in particular, it's true for the region $g (Ω)$ . so we get $v o l f (g (Ω)) = (v o l f) (v o l g (Ω))$ . and we know $v o l g (Ω) = (v o l g) (v o l Ω)$ since g also has this property. so we end up with $v o l (f \circ g) (Ω) = (v o l f) (v o l g) (v o l Ω)$ . so $v o l (f \circ g) = (v o l f) (v o l g)$ .

(actually, we should also check that $f \circ g$ has the volume-invariance-under-translations property)

now let's investigate what kinds of functions have the property for f mentioned above. one important class of functions that transform space is linear maps. let $T : R^{n} \to R^{n}$ be a linear map (operator). if we assume volume is invariant under transformations, then for any $x$ and $Ω$ we have $\begin{array}{r} T (x + Ω) & = T ({x + y : y \in Ω}) \\ = {T (x + y) : y \in Ω} \\ = {T x + T y : y \in Ω} \\ = T x + T (Ω) \end{array}$ Thus $v o l T (x + Ω) = v o l T (Ω)$ .

the interesting thing about linear maps is that we can summarize their entire behavior using an array of numbers. and volume is also a summary of the map, which must be some function of the array of numbers. so we can find a "volume formula" as a function of the numbers which make up the array.

References

https://www.youtube.com/watch?v=xX7qBVa9cQU -- this is probably the best explanation of the determinant i have ever seen
http://math.unt.edu/~tushar/project%20ideas/monthly%2096%20hannah%20geometric%20determinant.pdf
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.