How to simplify $\det(M)=\det(A^T A)$ for rectangular $A=BC$, with square diagonal $B$ and rectangular $C$ with orthonormal columns? Assume a real, square, symmetric, invertible $n \times n$ matrix $M$ and a real, rectangular $m \times n$ matrix $A$ such that $m \geq n$ and $M = A^T A$. Also assume that $A = B C$, where $B$ is diagonal ($m \times m$) and $C$ is rectangular ($m \times n$) with orthonormal columns. Thus:
\begin{align}
M
&= A^T A \\
&= (BC)^T BC \\
&= C^T B^T B C \\
&= C^T B^2 C
\end{align}
What are some possible strategies for simplifying $\det(M)$ here, ideally in terms of $A$, $B$, and/or $C$? (Note: I'm looking for a closed-form expression, not a numerical approach.)
Here are a few initial thoughts:


*

*It seems that there should be a simple solution here, given that $C$ has orthonormal columns and $B$ is diagonal.

*The fact that $C$ is rectangular complicates things a bit: e.g. if $C$ were square, $C^T B^2 C$ would provide a direct eigendecomposition of $M$.

*Singular value decomposition (SVD) appears useful here. For example, the non-zero singular values of $A$ are square roots of the eigenvalues of both $A^T A$ and $A A^T$, and $\det(M)$ is just the product of these eigenvalues; however, applying SVD to $A$ would produce the factorization: $A = U \Sigma V^T$, requiring square orthogonal $U$ and $V$ and rectangular diagonal $\Sigma$, which doesn't quite map onto the $A = BC$ above, where B is square diagonal and $C$ is rectangular with orthonormal columns.

*Maybe there is a way to apply QR decomposition? (Then $\det(M)$ is simple based on the diagonal elements of $R$.) But again, there is no clear mapping from the $B C$ above onto $Q$ (square orthogonal) and $R$ (rectangular triangular).

*Maybe some other matrix decomposition method would help? Or maybe there's something really simple that I am overlooking?


The simplest thing I can think of is to take the QR decomposition of $A=BC$, then $\det(M)$ is simply the square of the product of the diagonal elements of $R$.
