Determinant of a Sum of Certain Kronecker Products

Dwight Nwaigwe

We compute the determinant of ${\sum }_{n=1}^N{\vec{A}}^{(n)}\otimes {\vec{B}}^{(n)}$, where ${\vec{A}}^{(n)}$ is square and ${\vec{B}}^{(n)}={\vec{x}}^{(n)}{{\vec{y}}^{(n)}}^T$ where $\vec…