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

The Kronecker product is also known as the direct product or the tensor product [1].

Fundamental properties [1, 2]

1. 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

   	$\displaystyle A\otimes(B+C)$	$\displaystyle=$	$\displaystyle A\otimes B+A\otimes C,$
   	$\displaystyle(B+C)\otimes A$	$\displaystyle=$	$\displaystyle B\otimes A+C\otimes A,$
   	$\displaystyle k(A\otimes B)$	$\displaystyle=$	$\displaystyle(kA)\otimes B=A\otimes(kB).$

2. 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

   $\displaystyle(A\otimes B)(C\otimes D)$

   $\displaystyle=$

   $\displaystyle AC\otimes BD.$

   If $A$ and $B$ are invertible matrices, then

   $\displaystyle(A\otimes B)^{{-1}}$

   $\displaystyle=$

   $\displaystyle A^{{-1}}\otimes B^{{-1}}.$

3. 3.

   If $A$ and $B$ are square matrices, then for the transpose ($A^{T}$) we have

   $\displaystyle(A\otimes B)^{{T}}$

   $\displaystyle=$

   $\displaystyle A^{{T}}\otimes B^{{T}}.$

4. 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,

   	$\displaystyle\det(A\otimes B)$	$\displaystyle=$	$\displaystyle(\det A)^{m}(\det B)^{n},$
   	$\displaystyle\mathop{\mathrm{rank}}(A\otimes B)$	$\displaystyle=$	$\displaystyle\mathop{\mathrm{rank}}A\,\mathop{\mathrm{rank}}B,$
   	$\displaystyle\mathop{\mathrm{trace}}(A\otimes B)$	$\displaystyle=$	$\displaystyle\mathop{\mathrm{trace}}A\,\mathop{\mathrm{trace}}B,$