Definition. Let $A=(a_{ij})$ be a $n\times n$ matrix and let $B$ be a $m\times m$ matrix. Then the Kronecker product of $A$ and $B$ is the $mn\times mn$ block matrix \begin{eqnarray*} A\otimes B &=&\left( \begin{array}{ccc} a_{11} B & \cdots & a_{1n} B \\ \vdots & \ddots & \vdots \\ a_{n1} B & \cdots & a_{nn} B \\ \end{array} \right). \end{eqnarray*}The Kronecker product is also known as the direct product or the tensor product [1].

Fundamental properties [1,2]

1. The product is bilinear. If $k$ is a scalar, and $A,B$ and $C$ are square matrices, such that $B$ and $C$ are of the same order, then \begin{eqnarray*} A\otimes (B+C) &=& A\otimes B + A\otimes C,\\ (B+C)\otimes A &=& B\otimes A + C\otimes A,\\ k(A\otimes B) &=& (kA)\otimes B = A\otimes (kB). \end{eqnarray*}
2. If $A,B,C,D$ are square matrices such that the products $AC$ and $BD$ exist, then $(A\otimes B)(C\otimes D)$ exists and \begin{eqnarray*} (A\otimes B)(C\otimes D) &=& AC\otimes BD. \end{eqnarray*}If $A$ and $B$ are invertible matrices, then \begin{eqnarray*} (A\otimes B)^{-1} &=& A^{-1} \otimes B^{-1}. \end{eqnarray*}
3. If $A$ and $B$ are square matrices, then for the transpose ($A^T$ we have \begin{eqnarray*} (A\otimes B)^{T} &=& A^{T} \otimes B^{T}. \end{eqnarray*}
4. Let $A$ and $B$ be square matrices of orders $n$ and $m$ respectively. If $\{\lambda_i | i=1,\ldots,n \}$ are the eigenvalues of $A$ and $\{\mu_j | j=1,\ldots, m \}$ are the eigenvalues of $B$ then $\{\lambda_i \mu_j | i=1,\ldots, n, \, j=1,\ldots, m \}$ are the eigenvalues of $A\otimes B$ Also, \begin{eqnarray*} \det (A\otimes B) &=& (\det A)^m (\det B)^n, \\ \rank (A\otimes B) &=& \rank A\, \rank B, \\ \trace (A\otimes B) &=& \trace A\, \trace B, \\ \end{eqnarray*}

Bibliography

1

H. Eves, Elementary Matrix Theory, Dover publications, 1980.

2

T. Kailath, A.H. Sayed, B. Hassibi, Linear estimation, Prentice Hall, 2000