Determinant of a matrix in the form of $X'\Sigma X$

Question

Let $X \in \Bbb R^{n \times p}$, where $n > p$, be a tall matrix of full column rank. Let $\Sigma = \text{diag}\{\sigma_1, \ldots, \sigma_n\}$ be a diagonal matrix and $\sigma_i > 0$, $i = 1,\ldots,n$ are not all identical. Is there a way to calculate the determinant $|X'\Sigma X|$ without the need to compute the matrix multiplication?

An initial thought was to link the eigenvalues of $X'\Sigma X$ to $\sigma_i$ and SVD (or some decomposition, e.g. QR) of $X$ but it didn't yield anything useful.

Answer 1

One approach is to use the Cauchy Binet formula. In the notation of the linked page, we have $$ \det(X'\Sigma X) = \det([\Sigma^{1/2}X]'[\Sigma^{1/2}X]) = \sum_{S \in \binom{[n]}{p}} \det((\Sigma^{1/2} X)_{S,[p]})^2 \\ = \sum_{S \in \binom{[n]}{p}} \det(\Sigma_{S,S}) \det(X_{S,[p]})^2 \\ = \sum_{S \in \binom{[n]}{p}} \det(X_{S,[p]})^2\prod_{k \in S}\sigma_k. $$ This avoids the matrix computation, but I'm not sure if it ends up being any easier for your purposes.