Measure of Image of Linear Map between Different Dimensional Space If $L \in {\mathbb R}^{m \times n}$ ($m < n$) is a linear map from ${\mathbb R}^n$ onto ${\mathbb R}^m$ (onto means $L$ has full row rank). Given a compact set ${\mathcal A} \subset {\mathbb R}^n$ (compact set in Euclidean space is Lebesgue measurable), how to calculate the measure $\mu_{m}(L({\mathcal A}))$, where $\mu_{m}(\cdot)$ is the Lebesgue measure in ${\mathbb R}^m$?
 A: When $m=n$ the answer is, of course, $\mu_m(L(\mathcal{A})) = |\det(L)| \mu_n(\mathcal{A})$.  So it's natural to ask what the analog of this formula is when $m<n$.  There is no simple formula, but we can at least understand the geometry of the problem by considering the SVD (Singular Value Decomposition) $L = U \Sigma V^T$.  Here $U$ and $V$ are orthogonal matrices (in $m$- and $n$-d, respectively, with $\det(V) > 0$, w.l.o.g.), and $\Sigma = \Sigma^{m \times m}[I^{m \times m} \,|\, 0^{m \times d}]$, where $\Sigma^{m \times m}$ is a diagonal matrix with positive diagonal elements (the singular values of $L^{m\times n}$), and $0^{m \times d}$ comprises $d = n-m$ columns of zeroes.
Now consider the effect of $U \Sigma V^T$ on $\mathcal{A}$.  The matrix $V^T$ simply rotates $\mathcal{A}$ into a special alignment with the axes of $\mathbb{R}^n$. The matrix $P = [I^{m \times m} \,|\, 0^{m \times d}]$ projects $V^T\mathcal{A}$ onto the first $m$ dimensions of $\mathbb{R}^n$.  The orthogonally projected image $\mathcal{A}_m =PV^T\mathcal{A}$ is the key to determining $\mu_m(L(\mathcal{A}))$.  If we can determine $\mu_m(\mathcal{A}_m)$, then the rest is easy.  The remaining matrices are $\Sigma^{m \times m}$, whose determinant is the product of the singular values, and $U$, whose determinant is $\pm 1$.  Thus we arrive at the formula
$$\mu_m(L(\mathcal{A})) = \det(\Sigma^{m \times m}) \mu_m(\mathcal{A}_m).$$
Unfortunately, there is no simple way to determine $\mu_m(\mathcal{A}_m)$ in general.  In particular, we cannot obtain it directly from $\mu_n(\mathcal{A})$.  If we try, what we find instead is that
$$
\mu_n(\mathcal{A}) = \mu^*_m(\mathcal{A}_m),
$$
where $\mu^*_m$ is a measure on $\mathbb{R}^m$ weighted by the "amount of $\mathcal{A}$" that was mapped to $x$, i.e.,
$$
\frac{d \mu^*_m}{d\mu_m}(x) = \mu_d(P^{-1}x \cap \mathcal{A}).
$$
For example, if $n=3$ and $m=1$, then $\mu^*_m(x)$ would be the area of a cross-section of $\mathcal{A}$ (in a plane containing $x$).  In the case $m = n$, $\mu^*_m$ reduces to $\mu_m$, but otherwise the fact different cross-sections of $\mathcal{A}$ have different areas makes it difficult to compute $\mu_m(\mathcal{A}_m)$ and $\mu_m(L(\mathcal{A}))$.
