Let $K = K1 \otimes K2$ where $K1$ and $K2$ are positive semidefinite matrices. Let $W$ be a diagonal matrix with positive entries. (Everything is real-valued.)
I want to calculate or bound $\det (KW + I)$ without forming $K$ or $KW$.
My current approach:
$$det(KW + I) = det((K + W^{-1})W) = det(K+W^{-1})det(W)$$
Let $a_i$ be the eigenvalues of $K$ and $b_i$ be the eigenvalues of $W^{-1}$ sorted $a_1 \geq a_2 \geq \ldots \geq a_N$ and $b_1 \geq b_2 \geq \ldots \geq b_N$. Notice that these are readily available quantities because $K$ has Kronecker structure$\dagger$ and $W$ is diagonal. Fiedler 1971 says: $$\prod_i(a_i + b_i) \leq det(K + W^{-1}) \leq \prod_i(a_i + b_{N-i+1})$$
And then I just multiply those bounds by $det(W)$.
My hope was that I could use the Kronecker times diagonal structure to solve this exactly, i.e. Subquestion: Is it possible to explicitly calculate the eigenvalues of $(K1 \otimes K2) W$? It's possible to calculate $det(KW)$ and $det(K+I)$ without forming $K$ or $KW$, which gives me some hope, but I'm inclined to think the answer is no.
Thus, I'm looking for a tighter bound than Fiedler. It seems like there's been lots of progress in this area. Relevant literature: Eigenvalues, Invariant Factors, Highest Weights, and Schubert Calculus [Fulton, 2000], Bhatia's article, and Bhatia's Matrix Analysis.
$\dagger$ Let $e_i$ be the $n$ eigenvalues of $K_1$ and $f_j$ be the $m$ eigenvalues of $K_2$. Then $det(K1 \otimes K2) = \prod_{ij} (e_i)^m (e_j)^n$.
