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 ∈ R m × n \boldsymbol{A} \in \mathbb{R}^{m \times n} A∈Rm×n和 B ∈ R p × 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 R m p × n q \mathbb{R}^{mp \times nq} Rmp×nq的分块矩阵 A ⊗ B = [ a 11 B ⋯ a 1 n B ⋮ ⋱ ⋮ a m 1 B ⋯ a m n B ] \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 = [ a 11 b 11 a 11 b 12 ⋯ a 11 b 1 q ⋯ ⋯ a 1 n b 11 a 1 n b 12 ⋯ a 1 n b 1 q a 11 b 21 a 11 b 22 ⋯ a 11 b 2 q ⋯ ⋯ a 1 n b 21 a 1 n b 22 ⋯ a 1 n b 2 q ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋱ ⋮ a 11 b p 1 a 11 b p 2 ⋯ a 11 b p q ⋯ ⋯ a 1 n b p 1 a 1 n b p 2 ⋯ a 1 n b p q ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ a m 1 b 11 a m 1 b 12 ⋯ a m 1 b 1 q ⋯ ⋯ a m n b 11 a m n b 12 ⋯ a m n b 1 q a m 1 b 21 a m 1 b 22 ⋯ a m 1 b 2 q ⋯ ⋯ a m n b 21 a m n b 22 ⋯ a m n b 2 q ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋱ ⋮ a m 1 b p 1 a m 1 b p 2 ⋯ a m 1 b p q ⋯ ⋯ a m n b p 1 a m n b p 2 ⋯ a m n b p q ] \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\quad A⊗(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: ( k A ) ⊗ B = A ⊗ ( k B ) = 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 A C \boldsymbol{AC} AC和 B D \boldsymbol{BD} BD存在,那么 ( A ⊗ B ) ( C ⊗ D ) = A C ⊗ B D (\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
- v e c ( a b ⊤ ) = b ⊗ a vec(\bm{ab}^{\top})=\bm{b} \otimes \bm{a} vec(ab⊤)=b⊗a
克罗内克积实例
实例1
[ 1 2 3 1 ] ⊗ [ 0 3 2 1 ] = [ 1 ⋅ 0 1 ⋅ 3 2 ⋅ 0 2 ⋅ 3 1 ⋅ 2 1 ⋅ 1 2 ⋅ 2 2 ⋅ 1 3 ⋅ 0 3 ⋅ 3 1 ⋅ 0 1 ⋅ 3 3 ⋅ 2 3 ⋅ 1 1 ⋅ 2 1 ⋅ 1 ] = [ 0 3 0 6 2 1 4 2 0 9 0 3 6 3 2 1 ] \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
[ a 11 a 12 a 21 a 22 a 31 a 32 ] ⊗ [ b 11 b 12 b 13 b 21 b 22 b 23 ] = [ 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 ] \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⎦⎥⎥⎥⎥⎥⎥⎤