The standard $n$-dim Gaussian $\epsilon$ is defined through the characteristic function $\Bbb E(\exp(i(\lambda\cdot \epsilon))) = \exp(-\frac12 |\lambda|^2)$, and its PDF is given by $$f_\epsilon(x):=\frac1{(2\pi)^{n/2}}\exp(-\frac12 |x|^2),\quad \forall x\in\Bbb R^n.$$ A general $m$-dim Gaussian $Y$ arises from $\epsilon$ through an affine map: $$Y:=\mu + A\epsilon$$ where $\mu$ is a const vector and $A$ a const $m\times n$ matrix. Is there any slick way to derive the PDF for $Y$, based on the relation $Y=\mu + A\epsilon$?
PS: Assume the Cov matrix of $Y$, i.e. $AA^T$ is positive definite, so that $Y$ has a density. For the case $A$ is a non-singular square matrix I think an easy way is transformation of variables. What if $m<n$ though?
I eventually figured out a reverse engineering approach. Because I remember that the general case PDF should be compatible with the non-singular $A$ case, I know the PDF should be $$f_Y(x)=\frac1{\sqrt{(2\pi)^n|\Sigma|}}\exp(-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu))$$ where $\Sigma=AA^T$ is the Cov matrix.
So I started deducing backwards. Now I define a random vector $Y'$ with PDF $f_{Y'}(x)$ as given above, and then I am gonna compute $Y'$'s CF is exactly $\exp(i\lambda^T\mu-\frac12\lambda^T\Sigma\lambda)$ which coincides with that of the distribution $Y:=\mu+A\epsilon$, and we're done. Computations are not hard, the crucial step is to Cholesky decompose $\Sigma=CC^T$. In particular, when $A$ is non singular we may just choose $C=A$.