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

Let $f:{\mathbf{R}}^n\to {\mathbf{R}}^n$ be some function (not necessarily linear) and let $\mathrm{\Omega }\subseteq {\mathbf{R}}^n$ be some region in space. we will assume we can assign some "volume" to $\mathrm{\Omega }$, 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 $\mathrm{\Omega }$ under f, denoted $f(\mathrm{\Omega })$, is another region in space. let's assume f is nice enough that we can assign a volume to $f(\mathrm{\Omega })$. we can now ask, what is the volume of $f(\mathrm{\Omega })$? is it related to the volume of $\mathrm{\Omega }$ somehow? does the volume change if we translate $\mathrm{\Omega }$, stretch it, rotate it, etc.?

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.