It is well-known, that given a linear transformation $f \colon \mathbb R^n \rightarrow \mathbb R^m$, where $m \ge n$, the $m$-dimensional volume of an image of any measurable subset $S \subseteq \mathbb R^n$ under the transformation can be expressed as:
$$Volume(f(S)) = \sqrt{\det(A^TA)} Volume(S),$$
where $A \in \mathbb R^{m \times n}$ is a matrix of the linear transform.


If $m < n$, then this relation no longer holds (the mapping is not invertible and $\det(A^TA) = 0$). Is there any characterization of a volume change under linear transformation in this case? I am specifically interested in the case, when $S$ is a unit $L\_1$-ball in $\mathbb R^n$.



Volume change under linear transformation
The image of the $L\_1$-ball is the convex hull of the images $f(\pm e\_i)$ of the basis vectors $e\_1,\dots,e\_n$ and their opposite ones. So you are given $n$ pairs of opposite points in $\mathbb R^m$ and want to compute the volume of their convex hull (which can be an arbitrary symmetric polytope with at most $2n$ vertices).


There is no algebraic formula for this, because the answer depends on the combinatorial structure of the convex hull. For example, if one of the point happens to be inside the convex hull of the others, the volume does not depend on its location, but it starts depending on it as soon as the point moves outside the others' convex hull.


Within each combinatorial type, the volume is given by a polynomial formula. In general position, all faces are $(m-1)$-dimensional simplices, and the volume is the sum of volumes of corresponding $m$-simplices. It can be computed as twice the sum of appropriate minors of the matrix divided by $m!$. So in principle one can write down a single formula with lots of max(), min(), and abs() that separate all the cases.


On the other hand, the volume of the image of the unit cube is a simple formula: it equals the sum of absolute values of all $m\times m$ minors of the matrix.


