克罗内克积（Kronecker）

克罗内克积定义

克罗内克积是两个任意大小的矩阵间的运算，它是张量积的特殊形式。给定两个矩阵A∈Rm×n\boldsymbol{A} \in \mathbb{R}^{m \times n}A∈Rm×n和B∈Rp×q\boldsymbol{B} \in \mathbb{R}^{p \times q}B∈Rp×q，则这两个矩阵的克罗内克积是一个在空间Rmp×nq\mathbb{R}^{mp \times nq}Rmp×nq的分块矩阵A⊗B=[a11B⋯a1nB⋮⋱⋮am1B⋯amnB]\boldsymbol{A}\otimes\boldsymbol{B}=\left[\begin{array}{ccc}a_{11}\boldsymbol{B}&\cdots & a_{1n}\boldsymbol{B}\\\vdots&\ddots&\vdots\\a_{m1}\boldsymbol{B}&\cdots&a_{mn}\boldsymbol{B}\end{array}\right]A⊗B=⎣⎢⎡a11B⋮am1B⋯⋱⋯a1nB⋮amnB⎦⎥⎤更具体的可以表示为A⊗B=[a11b11a11b12⋯a11b1q⋯⋯a1nb11a1nb12⋯a1nb1qa11b21a11b22⋯a11b2q⋯⋯a1nb21a1nb22⋯a1nb2q⋮⋮⋱⋮⋮⋮⋱⋮a11bp1a11bp2⋯a11bpq⋯⋯a1nbp1a1nbp2⋯a1nbpq⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋱⋮⋮⋮am1b11am1b12⋯am1b1q⋯⋯amnb11amnb12⋯amnb1qam1b21am1b22⋯am1b2q⋯⋯amnb21amnb22⋯amnb2q⋮⋮⋱⋮⋮⋮⋱⋮am1bp1am1bp2⋯am1bpq⋯⋯amnbp1amnbp2⋯amnbpq]\boldsymbol{A} \otimes \boldsymbol{B}=\left[\begin{array}{ccccccccc} a_{11} b_{11} & a_{11} b_{12} & \cdots & a_{11} b_{1 q} & \cdots & \cdots & a_{1 n} b_{11} & a_{1 n} b_{12} & \cdots & a_{1 n} b_{1 q} \\ a_{11} b_{21} & a_{11} b_{22} & \cdots & a_{11} b_{2 q} & \cdots & \cdots & a_{1 n} b_{21} & a_{1 n} b_{22} & \cdots & a_{1 n} b_{2 q} \\ \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\ a_{11} b_{p 1} & a_{11} b_{p 2} & \cdots & a_{11} b_{p q} & \cdots & \cdots & a_{1 n} b_{p 1} & a_{1 n} b_{p 2} & \cdots & a_{1 n} b_{p q} \\ \vdots & \vdots & & \vdots & \ddots & & \vdots & \vdots & & \vdots \\ \vdots & \vdots & & \vdots & & \ddots & \vdots & \vdots & & \vdots \\ a_{m 1} b_{11} & a_{m 1} b_{12} & \cdots & a_{m 1} b_{1 q} & \cdots & \cdots & a_{m n} b_{11} & a_{m n} b_{12} & \cdots & a_{m n} b_{1 q} \\ a_{m 1} b_{21} & a_{m 1} b_{22} & \cdots & a_{m 1} b_{2 q} & \cdots & \cdots & a_{m n} b_{21} & a_{m n} b_{22} & \cdots & a_{m n} b_{2 q} \\ \vdots & \vdots & \ddots & \vdots & & & \vdots & \vdots & \ddots & \vdots \\ a_{m 1} b_{p 1} & a_{m 1} b_{p 2} & \cdots & a_{m 1} b_{p q} & \cdots & \cdots & a_{m n} b_{p 1} & a_{m n} b_{p 2} & \cdots & a_{m n} b_{p q} \end{array}\right] A⊗B=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡a11b11a11b21⋮a11bp1⋮⋮am1b11am1b21⋮am1bp1a11b12a11b22⋮a11bp2⋮⋮am1b12am1b22⋮am1bp2⋯⋯⋱⋯⋯⋯⋱⋯a11b1qa11b2q⋮a11bpq⋮⋮am1b1qam1b2q⋮am1bpq⋯⋯⋯⋱⋯⋯⋯⋯⋯⋯⋱⋯⋯⋯a1nb11a1nb21⋮a1nbp1⋮⋮amnb11amnb21⋮amnbp1a1nb12a1nb22⋮a1nbp2⋮⋮amnb12amnb22⋮amnbp2⋯⋯⋱⋯⋯⋯⋱⋯a1nb1qa1nb2q⋮a1nbpq⋮⋮amnb1qamnb2q⋮amnbpq⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤

克罗内克积性质

克罗内克积满足如下性质：

结合律：(A⊗B)⊗C=A⊗(B⊗C)(\boldsymbol{A}\otimes\boldsymbol{B})\otimes\boldsymbol{C}=\boldsymbol{A}\otimes(\boldsymbol{B}\otimes\boldsymbol{C})(A⊗B)⊗C=A⊗(B⊗C)
分配律：A⊗(B+C)=A⊗B+A⊗C\boldsymbol{A}\otimes(\boldsymbol{B}+\boldsymbol{C})=\boldsymbol{A}\otimes \boldsymbol{B}+\boldsymbol{A}\otimes\boldsymbol{C}\quad\quadA⊗(B+C)=A⊗B+A⊗C (A+B)⊗C=A⊗C+B⊗C(\boldsymbol{A}+\boldsymbol{B})\otimes\boldsymbol{C}=\boldsymbol{A}\otimes\boldsymbol{C}+\boldsymbol{B}\otimes\boldsymbol{C}(A+B)⊗C=A⊗C+B⊗C
双线性：(kA)⊗B=A⊗(kB)=k(A⊗B)(k\boldsymbol{A})\otimes\boldsymbol{B}=\boldsymbol{A}\otimes(k\boldsymbol{B})=k(\boldsymbol{A}\otimes\boldsymbol{B})(kA)⊗B=A⊗(kB)=k(A⊗B)
转置：(A⊗B)⊤=A⊤⊗B⊤(\boldsymbol{A} \otimes \boldsymbol{B})^{\top}=\boldsymbol{A}^{\top} \otimes \boldsymbol{B}^{\top}(A⊗B)⊤=A⊤⊗B⊤
混合乘积性：如果A,B,C,D\boldsymbol{A,B,C,D}A,B,C,D是四个矩阵，且矩阵乘积AC\boldsymbol{AC}AC和BD\boldsymbol{BD}BD存在，那么(A⊗B)(C⊗D)=AC⊗BD(\boldsymbol{A}\otimes\boldsymbol{B})(\boldsymbol{C}\otimes\boldsymbol{D})=\boldsymbol{AC}\otimes\boldsymbol{BD}(A⊗B)(C⊗D)=AC⊗BD这个性质称为“混合乘积性质”，因为它混合了矩阵乘积和克罗内克积。可以推出，A⊗B\boldsymbol{A}\otimes\boldsymbol{B}A⊗B是可逆的当且仅当A\boldsymbol{A}A和B\boldsymbol{B}B是可逆的，其逆矩阵为(A⊗B)−1=A−1⊗B−1(\boldsymbol{A}\otimes\boldsymbol{B})^{-1}=\boldsymbol{A}^{-1}\otimes\boldsymbol{B}^{-1}(A⊗B)−1=A−1⊗B−1
vec(ab⊤)=b⊗avec(\bm{ab}^{\top})=\bm{b} \otimes \bm{a}vec(ab⊤)=b⊗a

克罗内克积实例

实例1

[1231]⊗[0321]=[1⋅01⋅32⋅02⋅31⋅21⋅12⋅22⋅13⋅03⋅31⋅01⋅33⋅23⋅11⋅21⋅1]=[0306214209036321]\left[\begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array}\right] \otimes\left[\begin{array}{ll} 0 & 3 \\ 2 & 1 \end{array}\right]=\left[\begin{array}{llll} 1 \cdot 0 & 1 \cdot 3 & 2 \cdot 0 & 2 \cdot 3 \\ 1 \cdot 2 & 1 \cdot 1 & 2 \cdot 2 & 2 \cdot 1 \\ 3 \cdot 0 & 3 \cdot 3 & 1 \cdot 0 & 1 \cdot 3 \\ 3 \cdot 2 & 3 \cdot 1 & 1 \cdot 2 & 1 \cdot 1 \end{array}\right]=\left[\begin{array}{llll} 0 & 3 & 0 & 6 \\ 2 & 1 & 4 & 2 \\ 0 & 9 & 0 & 3 \\ 6 & 3 & 2 & 1 \end{array}\right] [1321]⊗[0231]=⎣⎢⎢⎡1⋅01⋅23⋅03⋅21⋅31⋅13⋅33⋅12⋅02⋅21⋅01⋅22⋅32⋅11⋅31⋅1⎦⎥⎥⎤=⎣⎢⎢⎡0206319304026231⎦⎥⎥⎤

实例2

[a11a12a21a22a31a32]⊗[b11b12b13b21b22b23]=[a11b11a11b12a11b13a12b11a12b12a12b13a11b21a11b22a11b23a12b21a12b22a12b23a21b11a21b12a21b13a22b11a22b12a22b13a21b21a21b22a21b23a22b21a22b22a22b23a31b11a31b12a31b13a32b11a32b12a32b13a31b21a31b22a31b23a32b21a32b22a32b23]\left[\begin{array}{l} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \otimes\left[\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{array}\right]=\left[\begin{array}{llllll} a_{11} b_{11} & a_{11} b_{12} & a_{11} b_{13} & a_{12} b_{11} & a_{12} b_{12} & a_{12} b_{13} \\ a_{11} b_{21} & a_{11} b_{22} & a_{11} b_{23} & a_{12} b_{21} & a_{12} b_{22} & a_{12} b_{23} \\ a_{21} b_{11} & a_{21} b_{12} & a_{21} b_{13} & a_{22} b_{11} & a_{22} b_{12} & a_{22} b_{13} \\ a_{21} b_{21} & a_{21} b_{22} & a_{21} b_{23} & a_{22} b_{21} & a_{22} b_{22} & a_{22} b_{23} \\ a_{31} b_{11} & a_{31} b_{12} & a_{31} b_{13} & a_{32} b_{11} & a_{32} b_{12} & a_{32} b_{13} \\ a_{31} b_{21} & a_{31} b_{22} & a_{31} b_{23} & a_{32} b_{21} & a_{32} b_{22} & a_{32} b_{23} \end{array}\right] ⎣⎡a11a21a31a12a22a32⎦⎤⊗[b11b21b12b22b13b23]=⎣⎢⎢⎢⎢⎢⎢⎡a11b11a11b21a21b11a21b21a31b11a31b21a11b12a11b22a21b12a21b22a31b12a31b22a11b13a11b23a21b13a21b23a31b13a31b23a12b11a12b21a22b11a22b21a32b11a32b21a12b12a12b22a22b12a22b22a32b12a32b22a12b13a12b23a22b13a22b23a32b13a32b23⎦⎥⎥⎥⎥⎥⎥⎤