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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 $\mathbb R^n$.

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I wonder about the condition $m \ge n$: Take $n=1, m=2$. A line, say of lenght 1, in $\mathbb{R}$ has volume 0 in $\mathbb{R}^2$, while the projection of a unit square $S \subseteq \mathbb{R}^2$ onto $\mathbb{R}$ gives a line of length (volume) 1. – Ralph Nov 9 '11 at 20:11
It is a misprint - you should take $n$-dimensional Hausdorff meausre on the image. See "area formula" in lecture notes I've linked to in my answer. – Vít Tuček Nov 9 '11 at 20:16
up vote 10 down vote accepted

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.

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Thank you, Sergei! There is a specific reason, why the $L_1$-balls are important, rather than $L_\infty$-balls. However, because I am ultimately interested in some specific class of linear mappings, the combinatorial type is fixed and one can get a closed formula. – Grigory Yaroslavtsev Nov 9 '11 at 23:49

There is a formula that computes the integral of Hasdorff measures of level sets of a given function. It goes under the name "coarea formula". It is a generalization of a change of variables formula from multivariable calculus and of course the proof of the general formula proceeds by first showing it for linear functions. You can find all details for example in these lecture notes.

As you can see on page 13. the formula for linear mapping $A$ reads $$\int_{\mathbb{R}^m} \mathcal{H}^{n-m}(S \cap A^{-1}(y)) \mathrm{d}y = \sqrt{\mathrm{det}(LL^T)}\mathcal{L}^n(S),$$ where $\mathcal{H}^{n-m}$ denotes the $(n-m)$-dimensional Hausdorff measure, $\mathcal{L}^n$ denotes the $n$-dimensional Lebesgeue measure, $A^{-1}(y)$ denotes the set $\{x\in\mathbb{R}^n|Ax=y\}$ and $L$ is given by factoring the mapping $A$ as $A = LPQ$ where $Q$ is orthogonal, $P$ is projection onto the first $m$ coordinates and $L$ is a regular mapping $\mathbb{R}^m\to\mathbb{R}^m$.

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Thank you, but I am not sure I understand how can this be used to compute the $\mathcal{L}^m(f(S))$, which I am interested in. Also, shouldn't the formula have $\mathcal{H}^{n - m}$, rather than $\mathcal{H}^{m - n}$ on the left-hand side? – Grigory Yaroslavtsev Nov 9 '11 at 21:33
Yes, you are right. I've corrected the formula. I do not claim that this formula gives you what you want. It just popped up in my head when I read your question and I thought it might help you in your case. Anyway - from the decomposition of $A$ it is clear that what you really care about is how the projection $P$ of your measurable set $S$ changes its volume. – Vít Tuček Nov 9 '11 at 21:48

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