$\sqrt{\det(A^TA)}$ for transformations into lower dimensions If Matrix $A$ is square, then $\sqrt{\det(A^TA)}$ is the same as |$\det(A)$|. If Matrix $A$ has more rows than columns, i.e., let's say $A$ is $3\times2$, then $\sqrt{\det(A^TA)}$ returns the factor by which a $2$-dimensional subspace of $3$-dimensional space is scaled during $A$'s transformation. Is it ever appropriate to apply $\sqrt{\det(A^TA)}$ to a matrix with more columns than rows, and if so, what would this measure geometrically?
 A: If $A$ is an $m \times n$ matrix with $m < n$, then the rank of $A$ is at most $m$. As a result, $A^{\mathsf T} A$ will be an $n \times n$ matrix with rank at most $m$, so it will be singular, and $\det(A^{\mathsf T}A) = 0$. This is a very useful quantity in general, but not one that tells us very much about $A$.
We can consider $\det(AA^{\mathsf T})$ for such matrices instead, which has similar interpretations. The idea, based on the singular-value decomposition of $A$, is that $\mathbb R^n$ can be given an orthonormal basis such that $m$ of its vectors are scaled by $A$'s singular values, then mapped to some orthonormal basis of $\mathbb R^m$, while the remaining $n-m$ vectors are sent to $0$. The quantity $\sqrt{\det(AA^{\mathsf T})}$ will be the product of these $m$ scaling factors.
So we can interpret $\det(AA^{\mathsf T})$ as follows: if $A$ has rank exactly $m$, then we can restrict $A$'s transformation to an $m$-dimensional subspace of $\mathbb R^n$ which $A$ maps invertibly to $\mathbb R^m$. Then $\sqrt{\det(AA^{\mathsf T})}$ is the scaling factor of this restricted transformation. (If $A$ has rank strictly less than $m$, then this quantity is $0$.)
