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<record version="4" id="4163">
 <title>Kronecker product</title>
 <name>KroneckerProduct</name>
 <created>2003-04-06 16:27:26</created>
 <modified>2006-08-08 14:00:38</modified>
 <type>Definition</type>
 <creator id="13753" name="Mathprof"/>
 <author id="13753" name="Mathprof"/>
 <author id="988" name="bwebste"/>
 <author id="1858" name="matte"/>
 <classification>
	<category scheme="msc" code="15-00"/>
 </classification>
 <synonyms>
	<synonym concept="Kronecker product" alias="tensor product (for matrices)"/>
	<synonym concept="Kronecker product" alias="direct product"/>
 </synonyms>
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{\bf Definition.}
Let $A=(a_{ij})$ be a $n\times n$ matrix  and 
let $B$ be a $m\times m$ matrix. Then the
\emph{Kronecker product} of $A$ and $B$ is 
the $mn\times mn$ block matrix
\begin{eqnarray*}
A\otimes B &amp;=&amp;\left( \begin{array}{ccc} 
a_{11} B &amp; \cdots &amp; a_{1n} B \\
\vdots &amp; \ddots &amp; \vdots \\
a_{n1} B &amp; \cdots &amp; a_{nn} B \\
\end{array} \right).
\end{eqnarray*}
The Kronecker product is also known as the \emph{direct product}
or the \emph{tensor product} \cite{eves}. 


{\bf Fundamental properties} \cite{eves, kailath}
\begin{enumerate}
\item 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) &amp;=&amp; A\otimes B + A\otimes C,\\ 
(B+C)\otimes A &amp;=&amp; B\otimes A + C\otimes A,\\
k(A\otimes B) &amp;=&amp; (kA)\otimes B = A\otimes (kB).
\end{eqnarray*}
\item 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) &amp;=&amp; AC\otimes BD.
\end{eqnarray*}
If $A$ and $B$ are invertible matrices, then
\begin{eqnarray*}
(A\otimes B)^{-1} &amp;=&amp; A^{-1} \otimes B^{-1}.
\end{eqnarray*}
\item If $A$ and $B$ are square matrices, then for the transpose ($A^T$) we have 
\begin{eqnarray*}
(A\otimes B)^{T} &amp;=&amp; A^{T} \otimes B^{T}.
\end{eqnarray*}
\item 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) &amp;=&amp; (\det A)^m (\det B)^n, \\
\rank (A\otimes B) &amp;=&amp; \rank A\, \rank B, \\
\trace (A\otimes B) &amp;=&amp; \trace A\, \trace B, \\
\end{eqnarray*}
\end{enumerate}

\begin{thebibliography}{9}
\bibitem {eves} H. Eves,
        \emph{Elementary Matrix Theory},
        Dover publications, 1980.
\bibitem {kailath} T. Kailath, A.H. Sayed, B. Hassibi,
        \emph{Linear estimation},
        Prentice Hall, 2000
\end{thebibliography}</content>
</record>
