The Kronecker sum of two matrices $A \in M(n \times n, \mathbb{R})$ and $B \in M(m \times m,\mathbb{R})$ is defined by the matrix

$$A \oplus B = A \otimes I_m + I_n \otimes B \in M(nm \times nm, \mathbb{R}),$$

where $\otimes$ denotes the Kronecker product (http://en.wikipedia.org/wiki/Kronecker_product) and $I_m,I_n$ are the identity matrices of size $m$ resp. $n$. If we are given $x_A \in \ker(A)$ and $x_B \in \ker(B)$ we obtain with $x_A \otimes x_B$ an element of $\ker(A \oplus B)$. We denote now by $E_{r,j,m}$ the $m \times m$ Matrix, where at the $r$-th row and $j$-th column there is a $1$ and every other entry is $0$. If we consider another matrix $A_0 \in M(n \times n, \mathbb{R})$ we may form

$$C = A \otimes (I_m-E_{1,1,m})+A_0 \otimes E_{1,1,m} + I_n \otimes B \in M(nm \times nm, \mathbb{R}),$$ which should be interpreted as a slight modification of $A$ by $A_0$ at a certain position in the Kronecker sum.

The question is now: If we are given $x_A \in \ker(A)$ and $x_B \in \ker(B)$ (and perhaps $x_{A_0} \in \ker(A_0)$), can we calculate and element of $\ker(C)$ with an easy expression, which is hopefully only slightly more difficult than $x_A \otimes x_B$ for $\ker(A \oplus B)$?