Showing that Determinant is a Volume Multiplier I want to show using the change of change of variables theorem for (Riemann) integration that the determinant of a linear transformation $T$ is a scaling factor for the volume of a space.
If $1_A$ is the characteristic function for $A\subseteq\mathbb{R}^n$, define the "Riemann measure" of $A$ as the Riemann integral $\int_A 1_A$, if this integral exists (i.e. if the boundary of $A$ is a zero set) and denote it $|A|$. Now for a $C^1$ diffeomorphism $h$, we have the change of variables formula $\int_{h(A)}f=\int_Af\circ h|det(Dh)|.$ Now if $h=T:\mathbb{R}^n\to\mathbb{R}^n$ is a vector space isomorphism (which is automatically a $C^1$ diffeomorphism since $T^{-1}$ is linear, since any linear transformation $L$ is differentiable with $DL=L$, and $L$ is a homeomorphism), $A=\mathbb{R}^n$, and $f$ is the constant function $f=1$, we have that $|T(A)|=|det(B)||A|$, where $B$ is the matrix of $DT=T$ with respect to fixed bases of $\mathbb{R}^n$. Further, since similar matrices have the same determinant, this holds regardless of the matrix representation of $T$.
Does this look ok or is there anything missing?
Thanks 

Some hints:
This would be a circular argument, since the change of variables formula has been proven knowing that a linear transformation $T$ scales volumes with the factor $|\det(T)|$. 
Your problem is in reality one of linear algebra and elementary geometry. You could use the fact that any $T$ can be written as a product of (a) diagonal matrices and (b) so-called elementary matrices of the form "identity matrix plus a single nonzero entry off the diagonal".
Diagonal matrices obviously scale volumes in the described way. For the matrices of type (b) check what the matrix
$$\left[\matrix{1&\lambda \cr0&1\cr}\right]$$
does to the square $[0,1]^2$.
