Definition and Properties ------------------------- Definition and Example ~~~~~~~~~~~~~~~~~~~~~~ Let :math:`\,\boldsymbol{A}\ \text{and}\,\boldsymbol{B}\,` be arbitrary matrices over a field :math:`\,K:` :math:`\,\boldsymbol{A}\,=\,[a_{ij}]_{m\times n}\,,` :math:`\,\boldsymbol{B}\,=\,[b_{ij}]_{p\times q}\in M(K).` Their :math:`\,` *tensor product* :math:`\,` (*Kronecker product*) written in the block notation reads .. :math:`\ \boldsymbol{A}\otimes\boldsymbol{B}\ ` is the block matrix .. math:: \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). For instance, :math:`\,` if :math:`\ \,\boldsymbol{A}\,=\,\left[\begin{array}{rc} 3 & 2 \\ -1 & 1 \\ -2 & 0 \end{array}\right],\ ` :math:`\ \boldsymbol{B}\,=\,\left[\begin{array}{rc} 2 & -1 \\ 0 & 4 \end{array}\right]\,,\ \,` then .. math:: \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], .. math:: \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]. The entries of a Kronecker product may be numbered by double indices :math:`\ ij\,` and :math:`\,kl\,:` .. math: \begin{array}{lr} (\boldsymbol{A}\otimes\boldsymbol{B})_{ij,kl}\,=\ a_{ik}\,b_{jl}, & \begin{array}{ll} i=1,2,\ldots,m; & k=1,2,\ldots,n; \\ j=1,2,\ldots,p; & l=1,2,\ldots,q. \end{array} \end{array} .. math:: \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} The indices :math:`\ i\ ` and :math:`\ k\ ` relate to block rows and block columns, whereas :math:`\\` the indices :math:`\ j\ ` and :math:`\ l\ ` designate elementary rows and columns, respectively. Properties of the Kronecker Product ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ **0.)** :math:`\,` The tensor product of matrices is non-commutative: :math:`\ \boldsymbol{A}\otimes\boldsymbol{B} \neq\boldsymbol{B}\otimes\boldsymbol{A}.` However, matrices :math:`\ \boldsymbol{A}\otimes\boldsymbol{B}\ ` and :math:`\ \boldsymbol{B}\otimes\boldsymbol{A}\ ` are permutation equivalent, meaning that :math:`\\` there exist permutation matrices :math:`\ \boldsymbol{P}\ ` and :math:`\ \boldsymbol{Q}\ ` such that :math:`\ \boldsymbol{B}\otimes\boldsymbol{A} \ =\ \boldsymbol{P}\,(\boldsymbol{A}\otimes\boldsymbol{B})\,\boldsymbol{Q}.` If :math:`\ \boldsymbol{A}\ ` and :math:`\ \boldsymbol{B}\ ` are square matrices, then the products :math:`\ \boldsymbol{A}\otimes\boldsymbol{B}\ ` and :math:`\ \boldsymbol{B}\otimes\boldsymbol{A}\ ` are even permutation similar: :math:`\ \boldsymbol{Q}\,=\,\boldsymbol{P}^{\,T}=\, \boldsymbol{P}^{-1}.\ ` That is to say, the product :math:`\ \boldsymbol{A}\otimes\boldsymbol{B}\ ` can be transformed into :math:`\ \boldsymbol{B}\otimes\boldsymbol{A}\ ` by means of a permutation of rows followed by the same permutation of columns. [1]_ **1.)** :math:`\,` The Kronecker product is associative and distributive over addition of matrices: .. math:: (\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) and is compatible with the scalar multiplication of matrices: .. math:: (\gamma\,\boldsymbol{A})\otimes\boldsymbol{B}\ =\ \boldsymbol{A}\otimes(\gamma\,\boldsymbol{B})\ =\ \gamma\ (\boldsymbol{A}\otimes\boldsymbol{B}),\quad\gamma\in K. **2.)** :math:`\,` If sizes of matrices :math:`\ \boldsymbol{A},\boldsymbol{B},\boldsymbol{C},\boldsymbol{D}\ ` are such that there exist products :math:`\ \boldsymbol{A}\boldsymbol{C}\ ` and :math:`\ \boldsymbol{B}\boldsymbol{D},\ ` then .. math:: :label: mixed-product \blacktriangleright\quad (\boldsymbol{A}\otimes\boldsymbol{B})\,(\boldsymbol{C}\otimes\boldsymbol{D}) \ =\ (\boldsymbol{A}\boldsymbol{C})\otimes(\boldsymbol{B}\boldsymbol{D}). **Proof.** :math:`\ ` Suppose the matrices :math:`\ \boldsymbol{A},\,\boldsymbol{B},\,\boldsymbol{C},\,\boldsymbol{D}\,` are given by .. math:: \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} We shall verify that matrices on both sides of Eq. :eq:`mixed-product` have equal dimensions and are composed of the same elements. a.) :math:`\ ` comparison of dimensions of matrices: .. math:: \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} b.) :math:`\ ` comparison of corresponding elements: .. math:: \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 It's worthwhile to write down a special case of Eq. :eq:`mixed-product`, with :math:`\,\boldsymbol{A}\in M_m(C),\ ` :math:`\,\boldsymbol{B}\in M_p(C),\ ` :math:`\,\boldsymbol{x}\in C^m\sim M_{m\times 1}(C),\ ` :math:`\,\boldsymbol{y}\in C^p\sim M_{p\times 1}(C)\,:` .. math: \boldsymbol{A}\ =\ \left[\begin{array}{ccc} a_{11} & \ldots & a_{1m} \\ \ldots & \ldots & \ldots \\ a_{m1} & \ldots & a_{mm} \end{array}\right],\quad \boldsymbol{B}\ =\ \left[\begin{array}{ccc} b_{11} & \ldots & b_{1p} \\ \ldots & \ldots & \ldots \\ b_{p1} & \ldots & b_{pp} \end{array}\right],\quad \boldsymbol{x}\ =\ \left[\begin{array}{c} x_1 \\ \ldots \\ x_m \end{array}\right],\quad \boldsymbol{y}\ =\ \left[\begin{array}{c} y_1 \\ \ldots \\ y_p \end{array}\right]: .. math:: (\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. :math:`\\` **3.)** :math:`\,` If :math:`\ \boldsymbol{A}\,=\,[a_{ij}]_{m\times m}\in M_m(K),\ \boldsymbol{B}\,=\,[b_{ij}]_{n\times n}\in M_n(K),\ ` then *i*.) :math:`\quad\text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B})\ =\ \text{Tr}\,\boldsymbol{A}\ \cdot\ \text{Tr}\,\boldsymbol{B}.` *ii*.) :math:`\quad\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\ (\det{\boldsymbol{A}})^n\ \cdot\ (\det{\boldsymbol{B}})^m.` *iii*.) :math:`\ \ ` If additionally :math:`\ \det{\boldsymbol{A}}\neq 0,\ ` :math:`\ \det{\boldsymbol{B}}\neq 0,\quad` then :math:`\quad (\boldsymbol{A}\otimes\boldsymbol{B})^{-1}\ =\ \, \boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1}.` **Proof.** .. math: \blacktriangleright\quad \text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B})\ =\ \text{Tr}\,\boldsymbol{A}\ \cdot\ \text{Tr}\,\boldsymbol{B}. .. math: \begin{array}{lll} 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}. \end{array} .. math: \begin{array}{rl} \text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B}) & =\ \ \displaystyle\sum_{i=1}^m\sum_{j=1}^n\ (\boldsymbol{A}\otimes\boldsymbol{B})_{\ |\ ij,\,ij}\ \ =\ \displaystyle\sum_{i=1}^m\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}. \end{array} \begin{array}{rll} \text{bo}\quad\text{Tr}\ (\boldsymbol{A}\otimes\boldsymbol{B}) & =\ \ \displaystyle\sum_{i=1}^m\sum_{j=1}^n\ (\boldsymbol{A}\otimes\boldsymbol{B})_{\ |\ ij,\,ij}\ \ = & \\[16pt] & =\ \ \displaystyle\sum_{i=1}^m\sum_{j=1}^n\ a_{ii}\ b_{jj}\ \ = & \\[20pt] & =\ \ \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}. \end{array} .. :math:`\begin{array}{lll} 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}. \end{array}` :math:`\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*.) :math:`\,` We shall use Eq. :eq:`mixed-product` and the remarks to the item **0.)** of the present discussion. .. math:: \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}. :math:`\ \boldsymbol{I}_m\ ` and :math:`\ \boldsymbol{I}_n\ ` are identity matrices of size :math:`\,m\,` and :math:`\,n,\ ` whereas :math:`\ \boldsymbol{P}\ ` is a permutation matrix. Using the theorem on a determinant of a product of matrices, we get .. math: \det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\ \det{(\boldsymbol{A}\otimes\boldsymbol{I}_n)}\,\cdot\, \det{(\boldsymbol{I}_m\otimes\boldsymbol{B})}, \det{(\boldsymbol{A}\otimes\boldsymbol{I}_n)}\ =\ \det{\left(\boldsymbol{P}^{-1}(\boldsymbol{I}_n\otimes\boldsymbol{A})\, \boldsymbol{P}\right)}\ =\ \det{(\boldsymbol{P}^{-1})}\cdot\,\ \det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\, \det{\boldsymbol{P}}\ = =\ (\det{\boldsymbol{P}})^{-1}\cdot\,\ \det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\, \det{\boldsymbol{P}}\ =\ \det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,. .. math:: \det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\ \det{(\boldsymbol{A}\otimes\boldsymbol{I}_n)}\,\cdot\, \det{(\boldsymbol{I}_m\otimes\boldsymbol{B})}, \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} Therefore the determinant of a tensor product of two matrices, :math:`\,\boldsymbol{A}\,` and :math:`\,\boldsymbol{B},\ ` is given by .. math:: :label: det_AxB \qquad\det{(\boldsymbol{A}\otimes\boldsymbol{B})}\ =\ \det{(\boldsymbol{I}_n\otimes\boldsymbol{A})}\,\cdot\, \det{(\boldsymbol{I}_m\otimes\boldsymbol{B})}\,. The matrices :math:`\ \boldsymbol{I}_n\otimes\boldsymbol{A}\ ` and :math:`\ \boldsymbol{I}_m\otimes\boldsymbol{B}\ ` have block-diagonal structure: .. math:: \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}} \,, whereby their determinants read .. math:: :label: I_AB \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 :eq:`I_AB` to :eq:`det_AxB` yields the desired relation: :math:`\ \det{(\boldsymbol{A}\otimes\boldsymbol{B})}\,=\, (\det{\boldsymbol{A}})^n\,\cdot\,(\det{\boldsymbol{B}})^m\,.\ \ \bullet` *iii*.) :math:`\,` First, we note that the tensor product of two invertible matrices is invertible as well: .. math:: \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. :eq:`mixed-product`, we obtain .. math:: (\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}\,. This means that :math:`\quad (\boldsymbol{A}\otimes\boldsymbol{B})^{-1}\,=\ \boldsymbol{A}^{-1}\otimes\,\boldsymbol{B}^{-1}\,.\quad\bullet` **4.)** :math:`\,` If :math:`\ \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*.) :math:`\quad(\boldsymbol{A}\otimes\boldsymbol{B})^T\ =\ \boldsymbol{A}^T\ \otimes\ \boldsymbol{B}^{\,T}.` For complex matrices (:math:`K=C`) the rules for complex or Hermitian conjugate are: .. the two additional relations hold: *ii*.) :math:`\quad(\boldsymbol{A}\otimes\boldsymbol{B})^{\,\ast}\ =\ \boldsymbol{A}^{\ast}\otimes\ \boldsymbol{B}^{\,\ast}.` *iii*.) :math:`\quad(\boldsymbol{A}\otimes\boldsymbol{B})^+\ =\ \boldsymbol{A}^+\otimes\ \boldsymbol{B}^+.` **Proof.** *i*.) :math:`\,` To be equal, two matrices should have the same sizes and the same corresponding entries. a.) :math:`\ ` comparison of dimensions of matrices: .. math:: \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} b.) :math:`\ ` comparison of corresponding elements: .. math:: \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} 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*.) :math:`\,` Matrices :math:`\ (\boldsymbol{A}\otimes\boldsymbol{B})^*\ ` and :math:`\ \boldsymbol{A}^*\otimes\boldsymbol{B}^*\ ` are of the same size and have the same entries: .. Furthermore, .. math:: (\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} *iii*.) :math:`\,` The Hermitian conjugate being composed of complex conjugate and transpose, we obtain .. math:: (\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. .. [1] H. V. Henderson; S. R. Searle (1980). "The vec-permutation matrix, the vec operator and Kronecker products: a review". LINEAR AND MULTILINEAR ALGEBRA. 9 (4): 271–288. https://dx.doi.org/10.1080%2F03081088108817379