Kronecker product

Definition.Let $A=(a_{ij})$ be a $n\times n$ matrix andlet $B$ be a $m\times m$ matrix. Then theKronecker product of $A$ and $B$ isthe $mn\times mn$ block matrix

$A\otimes B$$=$

The Kronecker product is also known as the direct productor 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

   $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).$

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

   $(A\otimes B)(C\otimes D)$$=$$AC\otimes BD.$

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

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

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

   ${(A\otimes B)}^T$$=$$A^T\otimes B^T.$

4. Let $A$ and $B$ be square matrices of orders $n$ and $m$, respectively.If $\{{\lambda }_i|i=1,\mathrm{\dots },n\}$ are the eigenvalues of $A$ and$\{{\mu }_j|j=1,\mathrm{\dots },m\}$ are the eigenvalues of $B$, then$\{{\lambda }_i{\mu }_j|i=1,\mathrm{\dots },n,j=1,\mathrm{\dots },m\}$ are the eigenvalues of$A\otimes B$. Also,

   $det(A\otimes B)$$=$${(detA)}^m{(detB)}^n,$

   $rank(A\otimes B)$$=$$rankArankB,$

   $trace(A\otimes B)$$=$$traceAtraceB,$