Kronecker Product as a Transformation Matrix

Suppose we are given matrices \(\ \boldsymbol{A} = [a_{ij}]_{m\times n}\in M_{m\times n}(K)\ \) and \(\ \boldsymbol{B} = [b_{ij}]_{p\times q}\in M_{p\times q}(K).\ \) \(\\\) Then \(\,\) for every \(\ \boldsymbol{G} = [g_{ij}]_{n\times q}\in M_{n\times q}(K)\ \) there exists the product \(\ \boldsymbol{A}\,\boldsymbol{G}\,\boldsymbol{B}^T\in M_{m\times p}(K).\)

Due to the properties of matrix multiplication, the mapping

(1)\[F_{AB}\,:\qquad M_{n\times q}(K)\ni\boldsymbol{G}\ \ \mapsto\ \ F_{AB}(\boldsymbol{G}) :\,= \boldsymbol{A}\,\boldsymbol{G}\,\boldsymbol{B}^T\in M_{m\times p}(K)\]

is linear, \(\,\) i.e. \(\,\) \(\,F_{AB}\,\) is a homomorphism of vector spaces \(\,M_{n\times q}(K)\ \) and \(\ M_{m\times p}(K) :\)

\[F_{AB}\ \in\ \text{Hom}\left(M_{n\times q}(K),\,M_{m\times p}(K)\right).\]

Accepting the notation \(\,\) \(\boldsymbol{H}\,\equiv\,[h_{ij}]_{m\times p}\ :\,=\ \boldsymbol{A}\,\boldsymbol{G}\,\boldsymbol{B}^T,\ \) we obtain

\[ \begin{align}\begin{aligned}h_{ij} \ =\ \displaystyle\sum_{v=1}^n\sum_{w=1}^q a_{iv}\ g_{vw}\ b^T_{wj} \ =\ \displaystyle\sum_{v=1}^n\sum_{w=1}^q (a_{iv}\,b_{jw})\ g_{vw} \ =\ \displaystyle\sum_{v=1}^n\sum_{w=1}^q (\boldsymbol{A}\otimes\boldsymbol{B})_{\,|\,ij,\,vw}\ g_{vw}\\\text{where}\quad i=1,2,\ldots,m;\ j=1,2,\ldots,p.\end{aligned}\end{align} \]

Thus, for arbitrary matrices \(\ \boldsymbol{G}\in M_{n\times q}(K),\,\) \(\,\boldsymbol{H}\in M_{m\times p}(K):\)

(2)\[\boldsymbol{H}\ =\ F_{AB}(\boldsymbol{G}) \quad\Leftrightarrow\quad \boldsymbol{\Lambda}^{mp}(\boldsymbol{H})\ \,=\ \, (\boldsymbol{A}\otimes\boldsymbol{B})\ \cdot\ \boldsymbol{\Lambda}^{nq}(\boldsymbol{G}),\]

Relation (2) means that \(\ \boldsymbol{A}\otimes\boldsymbol{B}\ \) is the matrix of the homomorphism \(\ F_{AB}\ \) in the standard bases of the spaces of matrices. The resulting Theorem stated below shall be afterwards used for independent proofs of some properties of the Kronecker product.

Theorem. \(\\\)

Let \(\ \boldsymbol{A}\in M_{m\times n}(K),\) \(\ \boldsymbol{B}\in M_{p\times q}(K).\ \) Then the tensor product \(\ \boldsymbol{A}\otimes\boldsymbol{B}\ \) is the matrix of the homomorphism

\[F_{AB}\,:\qquad M_{n\times q}(K)\ni\boldsymbol{G} \ \ \mapsto\ \ F_{AB}(\boldsymbol{G}) :\,= \boldsymbol{A}\,\boldsymbol{G}\boldsymbol{B}^{\,T}\in M_{m\times p}(K)\]

in the standard bases \(\ \mathcal{E}_{n\times q}\ \) and \(\ \ \mathcal{E}_{m\times p}\ \) of vector spaces \(\ M_{n\times q}(K)\ \) and \(\ M_{m\times p}(K).\ \) Consequently

\[\boldsymbol{\Lambda}^{mp} (\boldsymbol{A}\,\boldsymbol{G}\,\boldsymbol{B}^{\,T}) \ \,=\ \, (\boldsymbol{A}\otimes\boldsymbol{B})\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G})\,, \qquad\forall\ \ \boldsymbol{G}\in M_{n\times q}(K)\,.\]

Here \(\ \boldsymbol{\Lambda}^{rs}(\boldsymbol{X})\ \) is a column of coordinates of matrix \(\ \boldsymbol{X}\in M_{r\times s}(K)\ \) in the basis \(\ \mathcal{E}_{r\times s}\,.\)

Let \(\ \boldsymbol{A}\in M_{m\times n}(K),\ \) \(\ \boldsymbol{B}\in M_{p\times q}(K).\ \ \) Equation (2) can be rewritten as

(3)\[\blacktriangleright\quad (\boldsymbol{A}\otimes\boldsymbol{B})\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G})\ \,=\ \, \boldsymbol{\Lambda}^{mp} (\boldsymbol{A}\,\boldsymbol{G}\,\boldsymbol{B}^T) \qquad\forall\ \ \boldsymbol{G}\in M_{n\times q}(K).\]

Substituting in (3) \(\ \boldsymbol{A}\to\boldsymbol{A}_1 + \boldsymbol{A}_2\,,\ \) where \(\ \boldsymbol{A}_1,\ \boldsymbol{A}_2 \in M_{m\times n}(K),\ \) we get

\[\begin{split}\begin{array}{ll} \left[\,(\boldsymbol{A}_1 + \boldsymbol{A}_2)\otimes\boldsymbol{B}\,\right] \,\cdot\,\boldsymbol{\Lambda}^{nq}(\boldsymbol{G}) & =\ \ \boldsymbol{\Lambda}^{mp} \left[\,(\boldsymbol{A}_1 + \boldsymbol{A}_2)\ \boldsymbol{G}\,\boldsymbol{B}^T\,\right]\ = \\[6pt] & =\ \ \boldsymbol{\Lambda}^{mp} \left(\boldsymbol{A}_1\,\boldsymbol{G}\,\boldsymbol{B}^T + \, \boldsymbol{A}_2\,\boldsymbol{G}\,\boldsymbol{B}^T\right)\ = \\[6pt] & =\ \ \boldsymbol{\Lambda}^{mp} \left(\boldsymbol{A}_1\,\boldsymbol{G}\,\boldsymbol{B}^T\right)\ +\ \boldsymbol{\Lambda}^{mp} \left(\boldsymbol{A}_2\,\boldsymbol{G}\,\boldsymbol{B}^T\right)\ = \\[6pt] & =\ \ (\boldsymbol{A}_1\otimes\boldsymbol{B})\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G})\ +\ (\boldsymbol{A}_2\otimes\boldsymbol{B})\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G})\ = \\[6pt] & =\ \ \left[\,(\boldsymbol{A}_1\otimes\boldsymbol{B})\ +\ (\boldsymbol{A}_2\otimes\boldsymbol{B})\,\right]\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G}) \end{array}\end{split}\]

for arbitrary matrix \(\ \boldsymbol{G}\in M_{n\times q}(K).\ \) Inserting, in place of \(\ \boldsymbol{G},\ \) the consecutive elements of the standard basis \(\ \mathcal{E}_{n\times q}:\ \) \(\ \boldsymbol{G} = \boldsymbol{E}_{11},\ \boldsymbol{E}_{12},\ \ldots,\ \boldsymbol{E}_{nq}\,,\ \) we ascertain the equality of the corresponding columns of matrices \(\ (\boldsymbol{A}_1 + \boldsymbol{A}_2)\otimes\boldsymbol{B}\ \,\) and \(\ \,(\boldsymbol{A}_1\otimes\boldsymbol{B})\ +\ (\boldsymbol{A}_2\otimes\boldsymbol{B})\,,\ \) which is equivalent to the equality of the matrices themselves:

\[(\boldsymbol{A}_1 + \boldsymbol{A}_2)\otimes\boldsymbol{B}\ \,=\ \, (\boldsymbol{A}_1\otimes\boldsymbol{B})\ +\ (\boldsymbol{A}_2\otimes\boldsymbol{B})\,.\quad\bullet\]

To prove the \(\,\) “mixed-product property” of the Kronecker product, we substitute in Eq. (3):

\[\begin{split}\begin{array}{lr} \boldsymbol{A}\to\boldsymbol{A}\,\boldsymbol{C},\ \ & \begin{array}{r} \boldsymbol{A}\ :\ m \times r \\ \boldsymbol{C}\ :\ r \times n \end{array}; \end{array} \qquad \begin{array}{ll} \boldsymbol{B}\to\boldsymbol{B}\,\boldsymbol{D},\ \ & \begin{array}{l} \boldsymbol{B}\ :\ p \times s \\ \boldsymbol{D}\ :\ s \times q \end{array}: \end{array} \\[8pt] \begin{array}{ll} \left[\,(\boldsymbol{A}\boldsymbol{C})\otimes (\boldsymbol{B}\boldsymbol{D})\,\right]\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G}) & =\ \ \boldsymbol{\Lambda}^{mp} \left[\,(\boldsymbol{A}\boldsymbol{C})\ \boldsymbol{G}\ (\boldsymbol{B}\boldsymbol{D})^T\,\right]\ \ =\ \\[6pt] & =\ \ \boldsymbol{\Lambda}^{mp} \left[\,\boldsymbol{A}\ (\boldsymbol{C}\boldsymbol{G}\boldsymbol{D}^T)\ \boldsymbol{B}^T\,\right]\ \ = \\[6pt] & =\ \ (\boldsymbol{A}\otimes\boldsymbol{B})\,\cdot\, \boldsymbol{\Lambda}^{rs} (\boldsymbol{C}\boldsymbol{G}\boldsymbol{D}^T)\ \ = \\[6pt] & =\ \ (\boldsymbol{A}\otimes\boldsymbol{B})\,\cdot\, \left[\,(\boldsymbol{C}\otimes\boldsymbol{D})\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G})\,\right]\ \ = \\[6pt] & =\ \ \left[\,(\boldsymbol{A}\otimes\boldsymbol{B})\cdot (\boldsymbol{C}\otimes\boldsymbol{D})\,\right]\,\cdot\, \boldsymbol{\Lambda}^{nq}(\boldsymbol{G}). \end{array}\end{split}\]

The matrix \(\ \boldsymbol{G}\in M_{n\times q}(K)\ \) being arbitrary, we come up with the matrix equality to be proved:

\[(\boldsymbol{A}\boldsymbol{C})\otimes (\boldsymbol{B}\boldsymbol{D})\ =\ (\boldsymbol{A}\otimes\boldsymbol{B})\ (\boldsymbol{C}\otimes\boldsymbol{D})\,.\quad\bullet\]