I am sure the answer to this question is out there, but I cannot find it, maybe because I don't know the correct torm for 'matrix multiplication from left and right'.
Consider a matrix $W \in \mathbb{R}^{m \times n}$ and a square matrix $A \in \mathbb{R}^{m \times m}$. What can we say about the determinant
$$\det (W^TAW)?$$
In the case that $W$ is also square, since the determinant then commutes, we have
$$\det (W^TAW) = \det (A WW^T) = \det (A) \det (WW^T) = \det(A) \det(W^TW).$$
Do we have a similar result when $W$ is not square? We can not expect the two last equalities on the left generally hold, since one of $\det (WW^T)$ and $ \det(W^TW)$ will be equal to $0$, but maybe it holds for the respective one that might have full rank?
Edit:
Maybe this works indeed, using singular value decomposition:
Let $W = U \Sigma V$ and assume $W$ has full rank, and assume that $m<n$, such that $U$ is an orthogonal $m\times m$ matrix, $V$ is an orthogonal $n \times n$ matrix, and $\Sigma \in \mathbb{R}^{m\times n}$ is of the form $\begin{pmatrix}D & 0\end{pmatrix}$ with $D$ diagonal.
Then $$\det (W^T AW) = \det (V^T \Sigma^T U^T A U \Sigma V) = \det ( \Sigma^T U^T A U \Sigma VV^T) = \det ( \Sigma^T U^T A U\Sigma)$$
Bot now if we denote $U^TAU=B$, then $$\Sigma^T B \Sigma = \begin{pmatrix}D \\ 0\end{pmatrix} B \begin{pmatrix}D & 0\end{pmatrix} = DBD $$. But then
$$\det(\Sigma^T U^T A U\Sigma) = \det (\Sigma^TB\Sigma) = \det (\Sigma^T\Sigma B) = \det(D^2) \det(B) = \det(D^2) \det(U^TAU) = \det(D^2) \det(A) =\det(WW^T)\det(A).$$
Is this correct?
second edit: the 'solution' above must contain an error, since we obviously have that the determinant is zero in the case that $m<n$…
Best Answer
An analysis similar to yours works for the case where $m>n$. Let $U \Sigma V^T$ be the SVD of $W$, where $\Sigma$ has the form $$ \Sigma = \pmatrix{D\\0}. $$ Denote $B = U^TAU$. Partition $B$ into $$ B = \pmatrix{B_{11} & B_{12}\\ B_{21} & B_{22}}, $$ where $B$ is square of size $n$. We have $$ \Sigma^TB\Sigma = \pmatrix{D & 0}\pmatrix{B_{11} & B_{12}\\ B_{21} & B_{22}} \pmatrix{D \\ 0} = DB_{11}D. $$ Thus, we have $$ \det(W^TAW) = \det(\Sigma^TB\Sigma) = \det(DB_{11}D) = \det(B_{11})\det(D)^2\\ = \det(B_{11}) \det(W^TW). $$ This isn't as "nice" as the formula for the case of $m = n$, but perhaps you will find it interesting nevertheless.