Let $f:E\to E$ be a linear map between an $n$-dimensional $\mathbb{K}$-vector space and itself.
If $\mathcal{B}:=\{\mathbf{e}_1,\dotsc,\mathbf{e}_n\}$ is a basis of $E$, we will say that a linear map $g:E\to E$ is a type-1 automorphism if there is a $1\leq i\leq n$ and a $\lambda\in\mathbb{K}\setminus\{0\}$ such that $g(\mathbf{e}_j)=\mathbf{e}_j$ if $j\not=i$ and $g(\mathbf{e}_i)=\lambda\mathbf{e}_i$. Similarly, we will say that $h$ is a type-2 automorphism if there are $i$,$j\in\{1,\dotsc,n\}$ such that $f(\mathbf{e}_i)=\mathbf{e}_i+\mathbf{e}_j$ and $f(\mathbf{e}_k)=\mathbf{e}_k$ otherwise.
It is easy to proof that every automorphism can be writen as a composition of several type-1 and type-2 automorphisms with respect to the same basis $\mathcal{B}$. So, lets define $\det(f)$ as the product of the determinants of its decomposition, where $\det(g):=\lambda$ for every type-1 automorphism and $\det(h):=1$ for every type-2 automorphism.
Question
How does one prove that the determinant is well defined this way? That is, give two different decompositions of $f$, are both products of determinant the same?
Given that, proving that the determinant has geometric meaning, that is, it does not depend in the chosen basis $\mathcal{B}$ is straightforward.
Motivation
This definition of determinant is somewhat intuitive, and taylormade to proof a weaker form of the change of variables theorem, which relates the idea of the determinant with the concept of volume. This weaker statement says
If $A\subset\mathbb{R}^n$ is a Lebesgue-measurable set and $f$ is a linear automorphism, the $f(A)$ is Lebesgue-measurable and $\mu(f(A))=\left\lvert\det(f)\right\rvert\mu(A)$.