Definition and Example
Let \(\,\boldsymbol{A}\ \text{and}\,\boldsymbol{B}\,\)
be arbitrary matrices over a field \(\,K:\)
\(\,\boldsymbol{A}\,=\,[a_{ij}]_{m\times n}\,,\)
\(\,\boldsymbol{B}\,=\,[b_{ij}]_{p\times q}\in M(K).\)
Their \(\,\) tensor product \(\,\) (Kronecker product)
written in the block notation reads
\[\begin{split}\boldsymbol{A}\otimes\boldsymbol{B}\ :\,=\
\left[\begin{array}{cccc}
a_{11}\,\boldsymbol{B} & a_{12}\,\boldsymbol{B} &
\ldots & a_{1n}\,\boldsymbol{B} \\
a_{21}\,\boldsymbol{B} & a_{22}\,\boldsymbol{B} &
\ldots & a_{2n}\,\boldsymbol{B} \\
\ldots & \ldots &
\ldots & \ldots \\
a_{m1}\,\boldsymbol{B} & a_{m2}\,\boldsymbol{B} &
\ldots & a_{mn}\,\boldsymbol{B}
\end{array}\right]\ \in\ M_{mp\times nq}(K).\end{split}\]
For instance, \(\,\) if
\(\ \,\boldsymbol{A}\,=\,\left[\begin{array}{rc}
3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right],\ \)
\(\ \boldsymbol{B}\,=\,\left[\begin{array}{rc}
2 & -1 \\ 0 & 4 \end{array}\right]\,,\ \,\) then
\[\begin{split}\boldsymbol{A}\otimes\boldsymbol{B}\,=\,
\left[\begin{array}{rr}
3\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right] &
2\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right] \\[8pt]
-1\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right] &
1\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right] \\[10pt]
-2\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right] &
0\ \left[\begin{array}{rr} 2 & -1 \\ 0 & 4 \end{array}\right]
\end{array}\right]\ =\
\left[\begin{array}{rrrr}
6 & -3 & 4 & -2 \\ 0 & 12 & 0 & 8 \\
-2 & 1 & 2 & -1 \\ 0 & -4 & 0 & 4 \\
-4 & 2 & 0 & 0 \\ 0 & -8 & 0 & 0
\end{array}\right],\end{split}\]
\[\begin{split}\boldsymbol{B}\otimes\boldsymbol{A}\,=\,
\left[\begin{array}{rr}
2\ \left[\begin{array}{rr} 3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right] &
-1\ \left[\begin{array}{rr} 3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right]
\\[16pt]
0\ \left[\begin{array}{rr} 3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right] &
4\ \left[\begin{array}{rr} 3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right]
\end{array}\right]\ =\
\left[\begin{array}{rrrr}
6 & 4 & -3 & -2 \\
-2 & 2 & 1 & -1 \\
-4 & 0 & 2 & 0 \\
0 & 0 & 12 & 8 \\
0 & 0 & -4 & 4 \\
0 & 0 & -8 & 0
\end{array}\right].\end{split}\]
The entries of a Kronecker product may be numbered by double indices
\(\ ij\,\) and \(\,kl\,:\)
\[\begin{split}\begin{array}{lr}
(\boldsymbol{A}\otimes\boldsymbol{B})_{\,ij,\,kl}\,:\,=\
(\boldsymbol{A}\otimes\boldsymbol{B})_{\ (i-1)\,p\,+\,j,\ (k-1)\,q\,+\,l}\ =\
a_{ik}\,b_{jl}, &
\begin{array}{ll}
i=1,\ldots,m; & k=1,\ldots,n; \\
j=1,\ldots,p; & l=1,\ldots,q.
\end{array}
\end{array}\end{split}\]
The indices \(\ i\ \) and \(\ k\ \) relate to block rows
and block columns, whereas \(\\\)
the indices \(\ j\ \) and \(\ l\ \) designate elementary
rows and columns, respectively.
Properties of the Kronecker Product
0.) \(\,\) The tensor product of matrices is non-commutative:
\(\ \boldsymbol{A}\otimes\boldsymbol{B}
\neq\boldsymbol{B}\otimes\boldsymbol{A}.\)
However, matrices \(\ \boldsymbol{A}\otimes\boldsymbol{B}\ \) and
\(\ \boldsymbol{B}\otimes\boldsymbol{A}\ \) are permutation equivalent,
meaning that \(\\\)
there exist permutation matrices \(\ \boldsymbol{P}\ \) and
\(\ \boldsymbol{Q}\ \) such that
\(\ \boldsymbol{B}\otimes\boldsymbol{A} \ =\
\boldsymbol{P}\,(\boldsymbol{A}\otimes\boldsymbol{B})\,\boldsymbol{Q}.\)
If \(\ \boldsymbol{A}\ \) and \(\ \boldsymbol{B}\ \) are square
matrices, then the products \(\ \boldsymbol{A}\otimes\boldsymbol{B}\ \)
and \(\ \boldsymbol{B}\otimes\boldsymbol{A}\ \) are even permutation
similar: \(\ \boldsymbol{Q}\,=\,\boldsymbol{P}^{\,T}=\,
\boldsymbol{P}^{-1}.\ \) That is to say, the product
\(\ \boldsymbol{A}\otimes\boldsymbol{B}\ \) can be transformed into
\(\ \boldsymbol{B}\otimes\boldsymbol{A}\ \) by means of a permutation
of rows followed by the same permutation of columns.
1.) \(\,\) The Kronecker product is associative and
distributive over addition of matrices:
\[ \begin{align}\begin{aligned}(\boldsymbol{A}\otimes\boldsymbol{B})\otimes\boldsymbol{C}\ =\
\boldsymbol{A}\otimes(\boldsymbol{B}\otimes\boldsymbol{C})\\(\boldsymbol{A}_1\pm\boldsymbol{A}_2)\otimes\boldsymbol{B}\ =\
(\boldsymbol{A}_1\otimes\boldsymbol{B})\pm
(\boldsymbol{A}_2\otimes\boldsymbol{B})\\\boldsymbol{A}\otimes(\boldsymbol{B}_1\pm\boldsymbol{B}_2)\ =\
(\boldsymbol{A}\otimes\boldsymbol{B}_1)\pm
(\boldsymbol{A}\otimes\boldsymbol{B}_2)\end{aligned}\end{align} \]
and is compatible with the scalar multiplication of matrices:
\[(\gamma\,\boldsymbol{A})\otimes\boldsymbol{B}\ =\
\boldsymbol{A}\otimes(\gamma\,\boldsymbol{B})\ =\
\gamma\ (\boldsymbol{A}\otimes\boldsymbol{B}),\quad\gamma\in K.\]
2.) \(\,\) If sizes of matrices
\(\ \boldsymbol{A},\boldsymbol{B},\boldsymbol{C},\boldsymbol{D}\ \)
are such that there exist products \(\ \boldsymbol{A}\boldsymbol{C}\ \)
and \(\ \boldsymbol{B}\boldsymbol{D},\ \) then
(1)\[\blacktriangleright\quad
(\boldsymbol{A}\otimes\boldsymbol{B})\,(\boldsymbol{C}\otimes\boldsymbol{D})
\ =\ (\boldsymbol{A}\boldsymbol{C})\otimes(\boldsymbol{B}\boldsymbol{D}).\]
Proof. \(\ \) Suppose the matrices
\(\ \boldsymbol{A},\,\boldsymbol{B},\,\boldsymbol{C},\,\boldsymbol{D}\,\)
are given by
\[\begin{split}\begin{array}{lr}
\boldsymbol{A}\,=\,[a_{ij}]_{m\times r}\,, & \quad
\boldsymbol{B}\,=\,[b_{ij}]_{p\times s}\,, \\
\boldsymbol{C}\,=\,[c_{ij}]_{r\times n}\,, & \quad
\boldsymbol{D}\,=\,[d_{ij}]_{s\times q}\,.
\end{array}\end{split}\]
We shall verify that matrices on both sides of Eq. (1)
have equal dimensions and are composed of the same elements.
a.) \(\ \) comparison of dimensions of matrices:
\[\begin{split}\begin{array}{rr}
\begin{array}{rr}
\boldsymbol{A}\otimes\boldsymbol{B}\ : & mp\times rs \\[4pt]
\boldsymbol{C}\otimes\boldsymbol{D}\ : & rs\times nq \\[4pt]
(\boldsymbol{A}\otimes\boldsymbol{B})
(\boldsymbol{C}\otimes\boldsymbol{D})\ : & mp\times nq
\end{array} &
\begin{array}{rr}
\boldsymbol{A}\boldsymbol{C}\ : & m\times n \\[4pt]
\boldsymbol{B}\boldsymbol{D}\ : & p\times q \\[4pt]
\qquad (\boldsymbol{A}\boldsymbol{C})\otimes
(\boldsymbol{B}\boldsymbol{D})\ : & mp\times nq
\end{array}
\end{array}\end{split}\]
b.) \(\ \) comparison of corresponding elements:
\[\begin{split}\begin{array}{l}
(\boldsymbol{A}\otimes\boldsymbol{B})
(\boldsymbol{C}\otimes\boldsymbol{D})_{\ |\ ij,\,kl}\ \ = \
\displaystyle\sum_{v=1}^r\ \sum_{w=1}^s\
(\boldsymbol{A}\otimes\boldsymbol{B})_{\ |\ ij,\,vw}\
(\boldsymbol{C}\otimes\boldsymbol{D})_{\ |\ vw,\,kl}\ \ = \\[16pt]
=\ \ \displaystyle\sum_{v=1}^r\ \sum_{w=1}^s\
a_{iv}\ b_{jw}\ c_{vk}\ d_{wl}\ \ = \
\left(\displaystyle\sum_{v=1}^r\ a_{iv}\ c_{vk}\ \right)
\left(\displaystyle\sum_{w=1}^s\ b_{jw}\ d_{wl}\ \right)\ \ = \\[26pt]
=\ \ (\boldsymbol{A}\boldsymbol{C})_{\,|\,ik}\ \cdot\
(\boldsymbol{B}\boldsymbol{D})_{\,|\,jl}\ \ = \
(\boldsymbol{A}\boldsymbol{C})\otimes
(\boldsymbol{B}\boldsymbol{D})_{\ |\ ij,\,kl}\,;
\end{array}
\\[8pt]
\begin{array}{ll}
\text{where} &
\begin{array}{ll}
i=1,2,\ldots,m; & j=1,2,\ldots,p; \\
k=1,2,\ldots,n; & l=1,2,\ldots,q.
\end{array}
\end{array}\quad\bullet\end{split}\]
It’s worthwhile to write down a special case of Eq. (1),
with \(\,\boldsymbol{A}\in M_m(C),\ \) \(\,\boldsymbol{B}\in M_p(C),\ \)
\(\,\boldsymbol{x}\in C^m\sim M_{m\times 1}(C),\ \)
\(\,\boldsymbol{y}\in C^p\sim M_{p\times 1}(C)\,:\)
\[(\boldsymbol{A}\otimes\boldsymbol{B})
(\boldsymbol{x}\otimes\boldsymbol{y})\ =\
\boldsymbol{A}\boldsymbol{x}\otimes\boldsymbol{B}\boldsymbol{y}.\]
This formula is useful for a mathematical description
of a bipartite quantum system. \(\\\)
3.) \(\,\) If
\(\ \boldsymbol{A}\,=\,[a_{ij}]_{m\times m}\in M_m(K),\
\boldsymbol{B}\,=\,[b_{ij}]_{n\times n}\in M_n(K),\ \) then
i.) \(\quad\text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B})\ =\
\text{Tr}\,\boldsymbol{A}\ \cdot\ \text{Tr}\,\boldsymbol{B}.\)
ii.) \(\quad\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\
(\det{\boldsymbol{A}})^n\ \cdot\ (\det{\boldsymbol{B}})^m.\)
iii.) \(\ \ \) If additionally \(\ \det{\boldsymbol{A}}\neq 0,\ \)
\(\ \det{\boldsymbol{B}}\neq 0,\quad\) then
\(\quad (\boldsymbol{A}\otimes\boldsymbol{B})^{-1}\ =\ \,
\boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1}.\)
Proof.
\(\begin{array}{ll}
i.) \quad\text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B}) &
= \ \ \displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^n\
(\boldsymbol{A}\otimes\boldsymbol{B})_{\ |\ ij,\,ij}\ \ = \ \
\displaystyle\sum_{i=1}^m \displaystyle\sum_{j=1}^n\ a_{ii}\ b_{jj}\ \ = \\
& = \ \ \left(\displaystyle\sum_{i=1}^m a_{ii}\right)\
\left(\displaystyle\sum_{j=1}^n b_{jj}\right)\ \ = \ \
\text{Tr}\,\boldsymbol{A}\ \cdot\ \text{Tr}\,\boldsymbol{B}\,.\quad\bullet
\end{array}\)
ii.) \(\,\) We shall use Eq. (1)
and the remarks to the item 0.) of the present discussion.
\[ \begin{align}\begin{aligned}\boldsymbol{A}\otimes\boldsymbol{B}\ =\
(\boldsymbol{A}\,\boldsymbol{I}_m)\otimes
(\boldsymbol{I}_n\,\boldsymbol{B})\ =\
(\boldsymbol{A}\otimes\boldsymbol{I}_n)\,
(\boldsymbol{I}_m\otimes\boldsymbol{B})\,;\\\boldsymbol{A}\otimes\boldsymbol{I}_n\ \, = \ \,
\boldsymbol{P}\ (\boldsymbol{I}_n\otimes
\boldsymbol{A})\,\boldsymbol{P}^{-1}.\end{aligned}\end{align} \]
\(\ \boldsymbol{I}_m\ \) and \(\ \boldsymbol{I}_n\ \)
are identity matrices of size \(\,m\,\) and \(\,n,\ \)
whereas \(\ \boldsymbol{P}\ \) is a permutation matrix.
Using the theorem on a determinant of a product of matrices, we get
\[ \begin{align}\begin{aligned}\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\
\det{(\boldsymbol{A}\otimes\boldsymbol{I}_n)}\,\cdot\,
\det{(\boldsymbol{I}_m\otimes\boldsymbol{B})},\\\begin{split}\begin{array}{lll}
\det{(\boldsymbol{A}\otimes\boldsymbol{I}_n)} &
=\ \ \det{\left[\,\boldsymbol{P}\,
(\boldsymbol{I}_n\otimes\boldsymbol{A})\,
\boldsymbol{P}^{-1}\right]}\ \ = & \\
& =\ \ \det{\boldsymbol{P}}\,\cdot\,
\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\,
\det{(\boldsymbol{P}^{-1})}\ \ = & \\
& =\ \ \det{\boldsymbol{P}}\,\cdot\,\
\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\,
(\det{\boldsymbol{P}})^{-1}\ \ = &
\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,.
\end{array}\end{split}\end{aligned}\end{align} \]
Therefore the determinant of a tensor product of two matrices,
\(\,\boldsymbol{A}\,\) and \(\,\boldsymbol{B},\ \)
is given by
(2)\[\qquad\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\
\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\,
\det{(\boldsymbol{I}_m\otimes\boldsymbol{B})}\,.\]
The matrices \(\ \boldsymbol{I}_n\otimes\boldsymbol{A}\ \) and
\(\ \boldsymbol{I}_m\otimes\boldsymbol{B}\ \) have block-diagonal
structure:
\[\begin{split}\boldsymbol{I}_n\otimes\boldsymbol{A}\ =\
\underbrace{
\left[\begin{array}{cccc}
\boldsymbol{A} & \boldsymbol{0} & \cdots & \boldsymbol{0} \\
\boldsymbol{0} & \boldsymbol{A} & \cdots & \boldsymbol{0} \\
\cdots & \cdots & \cdots & \cdots \\
\boldsymbol{0} & \boldsymbol{0} & \cdots & \boldsymbol{A}
\end{array}\right]}_{n\ \text{blocks}}\,,
\qquad
\boldsymbol{I}_m\otimes\boldsymbol{B}\ =\
\underbrace{
\left[\begin{array}{cccc}
\boldsymbol{B} & \boldsymbol{0} & \cdots & \boldsymbol{0} \\
\boldsymbol{0} & \boldsymbol{B} & \cdots & \boldsymbol{0} \\
\cdots & \cdots & \cdots & \cdots \\
\boldsymbol{0} & \boldsymbol{0} & \cdots & \boldsymbol{B}
\end{array}\right]}_{m\ \text{blocks}} \,,\end{split}\]
whereby their determinants read
(3)\[\begin{array}{ll}
\det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\ =\
(\det{\boldsymbol{A}})^n \,, & \qquad
\det{(\boldsymbol{I}_m\otimes\boldsymbol{B})}\ =\
(\det{\boldsymbol{B}})^m\,.
\end{array}\]
Inserting (3) to (2) yields the desired relation:
\(\ \det{(\boldsymbol{A}\otimes\boldsymbol{B})}\,=\,
(\det{\boldsymbol{A}})^n\,\cdot\,(\det{\boldsymbol{B}})^m\,.\ \ \bullet\)
iii.) \(\,\)
First, we note that the tensor product of two invertible matrices
is invertible as well:
\[\left(\ \det{\boldsymbol{A}}\neq 0\,,\ \det{\boldsymbol{B}}\neq 0\ \right)
\quad\Rightarrow\quad
\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ \equiv\
(\det{\boldsymbol{A}})^n\,\cdot\,(\det{\boldsymbol{B}})^m\ \neq\ 0\,.\]
Next, making use of Eq. (1), we obtain
\[ \begin{align}\begin{aligned}(\boldsymbol{A}\otimes\boldsymbol{B})\,
(\boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1})\ =\
(\boldsymbol{A}\boldsymbol{A}^{-1})\otimes
(\boldsymbol{B}\boldsymbol{B}^{-1})\ =\
\boldsymbol{I}_m\otimes\boldsymbol{I}_n\ =\
\boldsymbol{I}_{mn}\,,\\(\boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1})\,
(\boldsymbol{A}\otimes\boldsymbol{B})\ =\
(\boldsymbol{A}^{-1}\boldsymbol{A})\otimes
(\boldsymbol{B}^{-1}\boldsymbol{B})\ =\
\boldsymbol{I}_m\otimes\boldsymbol{I}_n\ =\
\boldsymbol{I}_{mn}\,.\end{aligned}\end{align} \]
This means that \(\quad (\boldsymbol{A}\otimes\boldsymbol{B})^{-1}\,=\
\boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1}\,.\quad\bullet\)
4.) \(\,\) If
\(\ \boldsymbol{A}\,=\,[a_{ij}]_{m\times n}\in M_{m\times n}(K),\
\boldsymbol{B}\,=\,[b_{ij}]_{p\times q}\in M_{p\times q}(K),\ \) then
i.) \(\quad(\boldsymbol{A}\otimes\boldsymbol{B})^T\ =\
\boldsymbol{A}^T\ \otimes\ \boldsymbol{B}^{\,T}.\)
For complex matrices (\(K=C\))
the rules for complex or Hermitian conjugate are:
ii.) \(\quad(\boldsymbol{A}\otimes\boldsymbol{B})^{\,\ast}\ =\
\boldsymbol{A}^{\ast}\otimes\ \boldsymbol{B}^{\,\ast}.\)
iii.) \(\quad(\boldsymbol{A}\otimes\boldsymbol{B})^+\ =\
\boldsymbol{A}^+\otimes\ \boldsymbol{B}^+.\)
Proof.
i.) \(\,\)
To be equal, two matrices should have the same sizes
and the same corresponding entries.
a.) \(\ \) comparison of dimensions of matrices:
\[\begin{split}\begin{array}{lcr}
\begin{array}{rr}
\boldsymbol{A} \ : & m\times n \\[4pt]
\boldsymbol{B} \ : & p\times q \\[4pt]
\boldsymbol{A}\otimes\boldsymbol{B} \ : & mp\times nq \\[4pt]
(\boldsymbol{A}\otimes\boldsymbol{B})^T \ : & nq\times mp
\end{array}
&
\begin{array}{c}
\qquad
\end{array}
&
\begin{array}{rr}
\boldsymbol{A}^T \ : & n\times m \\[4pt]
\boldsymbol{B}^T \ : & q\times p \\[4pt]
\boldsymbol{A}^T\otimes\boldsymbol{B}^T \ : & nq\times mp
\end{array}
\end{array}\end{split}\]
b.) \(\ \) comparison of corresponding elements:
\[\begin{split}\begin{array}{l}
(\boldsymbol{A}\otimes\boldsymbol{B})^T_{\ |\ ij,\,kl}\ \ =\ \
(\boldsymbol{A}\otimes\boldsymbol{B})_{\ |\ kl,\,ij}\ \ =\ \
a_{ki}\,b_{lj}
\\[4pt]
(\boldsymbol{A}^T\otimes\boldsymbol{B}^T)_{\ |\ ij,\,kl}\ \ =\ \
a^T_{ik}\,b^T_{jl}\ =\ a_{ki}\,b_{lj}
\end{array}
\\[8pt]
\begin{array}{ll}
\text{where} & \quad
\begin{array}{ll}
i=1,2,\ldots,n; & j=1,2,\ldots,q; \\
k=1,2,\ldots,m; & l=1,2,\ldots,p.\qquad\bullet
\end{array}
\end{array}\end{split}\]
Therefore, the transpose of a tensor product of two matrices
is equal to the tensor product of the transposed matrices,
the order of factors being preserved.
ii.) \(\,\)
Matrices \(\ (\boldsymbol{A}\otimes\boldsymbol{B})^*\ \) and
\(\ \boldsymbol{A}^*\otimes\boldsymbol{B}^*\ \) are of the same size
and have the same entries:
\[\begin{split}(\boldsymbol{A}\otimes\boldsymbol{B})^{*}_{\ |\ ij,\,kl}\ \ =\ \
(a_{ik}\,b_{jl})^*\,=\ \,a^*_{ik}\ b^*_{jl}\ \,=\ \,
(\boldsymbol{A}^*\otimes\,\boldsymbol{B}^*)_{\ |\ ij,\,kl}\,,
\\[8pt]
\begin{array}{ll}
\text{where} & \quad
\begin{array}{ll}
i=1,2,\ldots,m; & j=1,2,\ldots,p; \\
k=1,2,\ldots,n; & l=1,2,\ldots,q.\qquad\bullet
\end{array}
\end{array}\end{split}\]
iii.) \(\,\)
The Hermitian conjugate being composed of complex conjugate and transpose,
we obtain
\[(\boldsymbol{A}\otimes\boldsymbol{B})^+\,=\
\left[\,(\boldsymbol{A}\otimes\boldsymbol{B})^T\right]^* =\ \,
\left(\boldsymbol{A}^T\otimes\,\boldsymbol{B}^T\right)^*\ =\ \,
(\boldsymbol{A}^T)^*\otimes\,(\boldsymbol{B}^T)^*\ =\ \,
\boldsymbol{A}^+\otimes\,\boldsymbol{B}^+.\ \bullet\]
So the complex or Hermitian conjugate of a tensor product of two matrices
is equal to the tensor product of the conjugated matrices, the order of factors
being preserved.
