Density of linear transformation of random vector Let $x \in \mathbb{R}^n$ be a random vector with density $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Let $A \in \mathbb{R}^{m \times n}$ with $m < n$. What can we say about the density of $Ax$?
If $A \in \mathbb{R}^{n \times n}$ is invertible then by a change of variables, we know that the density of $Ax$ is given by $f(A^{-1}x)/|\text{det} A|$. However, I can't seem to find a concrete reference for a general case.
 A: I will focus on the case where $A$ has full row rank. The key is figuring out how to handle the case where $A$ has a non-trivial kernel.
First, consider the case where $A$ is "diagonal". If we have $A = [\Sigma\ \ 0]$ with
$$
\Sigma = \pmatrix{\sigma_1\\ & \ddots\\ && \sigma_m},
$$
then for any set $S \subset \Bbb R^m$, the vector $y = Ax$ is an element of $S$ if and only if the truncated vector $\hat x = (x_1,\dots,x_m)$ is such that $\hat x = \Sigma^{-1} y$. With that, we deduce that the distribution of $y = Ax$ is given by
$$
f(y) = f_{1,\dots,m}(\Sigma^{-1} y)/|\det(\Sigma)| = ,
$$
where $f_{1,\dots,m}$ denotes the marginal distribution over $x_1,\dots,x_m$.
For an arbitrary full row-rank matrix $A$, $A$ necessarily has a singular value decomposition
$$
A = U [\Sigma \ \ 0] V^T,
$$
where both $U$ and $V$ are orthogonal matrices (which in particular implies that $|\det(U)| = |\det(V)| = 1$). The density of $V^Tx$ is given by $g(x) = f(Vx)$, the density of $[\Sigma \ \ 0] V^T x$ is given by $h(y) = g_{1,\dots,m}(\Sigma^{-1}y)/|\det\Sigma|$, and the density of $U[\Sigma \ \ 0]V^Tx = Ax$ is therefore given by
$$
p(y) = h(U^Ty).
$$
If we take $R$ to denote the row-space of $A$ and $N$ to denote the nullspace, then this amounts to the statement
$$
p(y) = f_R(A^+y)/\sqrt{\det(AA^T)},
$$
where we define
$$
f_R(\vec x) = \int_{\vec z \in N} f(\vec x + \vec z)\,d\vec z.
$$
