The well-known *matrix determinant lemma* states that for an invertible square matrix $A$ and column vectors $u,v$ one has
$$
\det(A + uv^T) = \det(A)(1 + v^T A^{-1} u).
$$
Is there any analogous formula in case we are adding to $A$ a matrix which is not rank-one, but still has some special structure? The case I have in mind is when $A,B$ are square matrices, $I$ is the identity matrix, $v$ is a vector and we are dealing with tensor (or Kronecker, if you wish) products:
$$
\det(A \otimes I + I \otimes B + (v^T v) \otimes I) = ?
$$
Eventually I'd like to work with more components of the tensor product, but for simplicity let me just state the question for two.



Matrix determinant lemma for non-rank-one updates
If $A,B$ are square matrices, then $A\otimes I+I\otimes B$ is commonly known as the Kronecker sum of $A$ and $B$. If $A$ has eigenvalues $\mu\_1,\dots,\mu\_m$ and $B$ has eigenvalues $\nu\_1,\dots,\nu\_n$, then $A\otimes I-I\otimes B$ has eigenvalues
$\mu\_i-\nu\_j$ for $1\leq i\leq m,1\leq j\leq n$. Therefore, $\det(A\otimes I-I\otimes B)=\text{res}(\chi(A),\chi(B))$ (which denotes the resultant of the characteristic polynomial of $A$ and $B$).


Therefore (Peter Taylor originally mentioned that we can do something like this), by the matrix determinant lemma, we have


$$\text{Det}(A\otimes I+I\otimes B+pq^T\otimes I+I\otimes uv^T)$$


$$=\text{res}\big(\chi(A+pq^T),\chi(B+uv^T))=\text{res}(\det(xI-A-pq^T),\det(xI-B-uv^T)\big)$$


$$=\text{res}\big(\det(xI-A)\cdot(1-q^T(xI-A)^{-1}p),\det(xI-B)\cdot(1-v^T(xI-B)^{-1}u)\big)$$


$$=\text{res}\big(\chi(A)\cdot(1-q^T(xI-A)^{-1}p),\chi(B)\cdot(1-v^T(xI-B)^{-1}u)\big).$$


