设A=(aij)m×n,B=(bij)p×q,则称如下分块矩阵(a11Ba12B⋯a1nBa21Ba22B⋯a2nB⋮⋮⋱⋮an1Ban2⋯ann)为A与B的张量积记作A⊗B=(aijB)mp×nq\begin{aligned} &设A=(a_{ij})_{m\times n},B=(b_{ij})_{p\times q},则称如下分块矩阵\left( \begin{matrix} a_{11}B&a_{12}B&\cdots&a_{1n}B\\ a_{21}B&a_{22}B&\cdots&a_{2n}B\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}B&a_{n2}&\cdots&a_{nn} \end{matrix} \right)为A与B的张量积\\ &记作A\otimes B=(a_{ij}B)_{mp\times nq} \end{aligned} 设A=(aij)m×n,B=(bij)p×q,则称如下分块矩阵⎝⎜⎜⎜⎛a11Ba21B⋮an1Ba12Ba22B⋮an2⋯⋯⋱⋯a1nBa2nB⋮ann⎠⎟⎟⎟⎞为A与B的张量积记作A⊗B=(aijB)mp×nq
eg
A=(abcd),B=(23)A=\left( \begin{matrix} a&b\\c&d \end{matrix} \right),B=\left( \begin{matrix} 2\\3 \end{matrix} \right) A=(acbd),B=(23)
A⊗B=(aBbBcBdB)=(2a2b3a3b2c2d3c3d),B⊗A=(2A3B)=(2a2b2c2d3a3b3c3d)\begin{aligned} &A\otimes B=\left( \begin{matrix} aB&bB\\cB&dB \end{matrix} \right)=\left( \begin{matrix} 2a&2b\\3a&3b\\2c&2d\\3c&3d \end{matrix} \right),B\otimes A=\left( \begin{matrix} 2A\\3B \end{matrix} \right)=\left( \begin{matrix} 2a&2b\\2c&2d\\3a&3b\\3c&3d \end{matrix} \right) \end{aligned} A⊗B=(aBcBbBdB)=⎝⎜⎜⎛2a3a2c3c2b3b2d3d⎠⎟⎟⎞,B⊗A=(2A3B)=⎝⎜⎜⎛2a2c3a3c2b2d3b3d⎠⎟⎟⎞
右进右出
(ABCD)⊗F=(A⊗FB⊗FC⊗FD⊗F)(AB)⊗F=(A⊗FB⊗F)\left( \begin{matrix} A&B\\C&D \end{matrix} \right)\otimes F=\left( \begin{matrix} A\otimes F&B\otimes F\\C\otimes F&D\otimes F \end{matrix} \right) (A\quad B)\otimes F=(A\otimes F\quad B\otimes F) (ACBD)⊗F=(A⊗FC⊗FB⊗FD⊗F)(AB)⊗F=(A⊗FB⊗F)
一般情况下:(AB)⊗F≠(A⊗FB⊗F)(A\quad B) \otimes F\neq (A\otimes F\quad B\otimes F)(AB)⊗F=(A⊗FB⊗F)
列向量α=(a1a2⋮an)β=(b1b2⋮bq),则α⊗β=(a1⊗βa2⊗β⋮an⊗β)nq×1=(a1b1a1b2⋮a1bq⋮anb1anb2⋮anbq)nq×1列向量\alpha=\left( \begin{matrix} a_1\\a_2\\\vdots\\a_n \end{matrix} \right)\beta=\left( \begin{matrix} b_1\\b_2\\\vdots\\b_q \end{matrix} \right),则\alpha \otimes \beta=\left( \begin{matrix} a_1\otimes \beta\\a_2\otimes \beta\\\vdots\\a_n\otimes \beta \end{matrix} \right)_{nq\times 1}=\left( \begin{matrix} a_1b_1\\a_1b_2\\\vdots\\a_1b_q\\\vdots \\a_nb_1\\a_nb_2\\\vdots\\a_nb_q \end{matrix} \right)_{nq\times 1} 列向量α=⎝⎜⎜⎜⎛a1a2⋮an⎠⎟⎟⎟⎞β=⎝⎜⎜⎜⎛b1b2⋮bq⎠⎟⎟⎟⎞,则α⊗β=⎝⎜⎜⎜⎛a1⊗βa2⊗β⋮an⊗β⎠⎟⎟⎟⎞nq×1=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛a1b1a1b2⋮a1bq⋮anb1anb2⋮anbq⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞nq×1
列向量α=(a1a2⋮am),Bn×q=(β1,β2,⋯,βq),α⊗B=(a1Ba2B⋮amB)=(a1(β1,β2,⋯,βq)a2(β1,β2,⋯,βq)⋮am(β1,β2,⋯,βq))=(a1b11a1b12⋯a1b1qa1b21a1b22⋯a1b2q⋮⋮⋱⋮a1bp1a1bp2⋯a1bpqa2b11a2b12⋯a2b1qa2b21a2b22⋯a2b2q⋮⋮⋱⋮a2bp1a2bp2⋯a2bpq⋮⋮⋮⋮amb11amb12⋯amb1qamb21amb22⋯amb2q⋮⋮⋱⋮ambp1ambp2⋯ambpq)mp×q\begin{aligned} &列向量\alpha=\left( \begin{matrix} a_1\\a_2\\\vdots\\a_m \end{matrix} \right),B_{n\times q}=(\beta_1,\beta_2,\cdots,\beta_q),\\ &\alpha\otimes B=\left( \begin{matrix} a_1B\\a_2B\\\vdots\\a_mB \end{matrix} \right)=\left( \begin{matrix} a_1(\beta_1,\beta_2,\cdots,\beta_q)\\ a_2(\beta_1,\beta_2,\cdots,\beta_q)\\ \vdots\\ a_m(\beta_1,\beta_2,\cdots,\beta_q)\\ \end{matrix} \right)=\left( \begin{matrix} a_1b_{11}&a_1b_{12}&\cdots&a_1b_{1q}\\ a_1b_{21}&a_1b_{22}&\cdots&a_1b_{2q}\\ \vdots&\vdots&\ddots&\vdots\\ a_1b_{p1}&a_1b_{p2}&\cdots&a_1b_{pq}\\ a_2b_{11}&a_2b_{12}&\cdots&a_2b_{1q}\\ a_2b_{21}&a_2b_{22}&\cdots&a_2b_{2q}\\ \vdots&\vdots&\ddots&\vdots\\ a_2b_{p1}&a_2b_{p2}&\cdots&a_2b_{pq}\\ \vdots&\vdots&\vdots&\vdots\\ a_mb_{11}&a_mb_{12}&\cdots&a_mb_{1q}\\ a_mb_{21}&a_mb_{22}&\cdots&a_mb_{2q}\\ \vdots&\vdots&\ddots&\vdots\\ a_mb_{p1}&a_mb_{p2}&\cdots&a_mb_{pq}\\ \end{matrix} \right)_{mp\times q} \end{aligned} 列向量α=⎝⎜⎜⎜⎛a1a2⋮am⎠⎟⎟⎟⎞,Bn×q=(β1,β2,⋯,βq),α⊗B=⎝⎜⎜⎜⎛a1Ba2B⋮amB⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎛a1(β1,β2,⋯,βq)a2(β1,β2,⋯,βq)⋮am(β1,β2,⋯,βq)⎠⎟⎟⎟⎞=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛a1b11a1b21⋮a1bp1a2b11a2b21⋮a2bp1⋮amb11amb21⋮ambp1a1b12a1b22⋮a1bp2a2b12a2b22⋮a2bp2⋮amb12amb22⋮ambp2⋯⋯⋱⋯⋯⋯⋱⋯⋮⋯⋯⋱⋯a1b1qa1b2q⋮a1bpqa2b1qa2b2q⋮a2bpq⋮amb1qamb2q⋮ambpq⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞mp×q
k(A⊗B)=(kA)⊗B=A⊗(kB)k(A\otimes B)=(kA)\otimes B=A\otimes (kB)k(A⊗B)=(kA)⊗B=A⊗(kB)
分配律:(A+B)⊗C=A⊗C+B⊗C(A+B)\otimes C=A\otimes C+B\otimes C(A+B)⊗C=A⊗C+B⊗C ,C⊗(A+B)=C⊗A+C⊗BC\otimes(A+B)=C\otimes A+C\otimes BC⊗(A+B)=C⊗A+C⊗B
结合律:(A⊗B)⊗C=A⊗(B⊗C)(A\otimes B)\otimes C=A\otimes (B\otimes C)(A⊗B)⊗C=A⊗(B⊗C)
吸收律:(A⊗B)(C⊗D)=(AC)⊗(BD)(A\otimes B)(C\otimes D)=(AC)\otimes (BD)(A⊗B)(C⊗D)=(AC)⊗(BD)
推论:
若A=Am×m为m阶方阵,B=Bn×n为n阶方阵,则(A⊗B)k=Ak⊗Bk(A⊗In)(Im⊗B)=A⊗B\begin{aligned} &若A=A_{m\times m}为m阶方阵,B=B_{n\times n}为n阶方阵,则\\ &(A\otimes B)^k=A^k\otimes B^k\\ &(A\otimes I_n)(I_m\otimes B)=A\otimes B \end{aligned} 若A=Am×m为m阶方阵,B=Bn×n为n阶方阵,则(A⊗B)k=Ak⊗Bk(A⊗In)(Im⊗B)=A⊗B
(A1⊗B1)(A2⊗B2)⋯(Ak⊗Bk)=(A1A2⋯Ak)⊗(B1⊗B2⋯Bk)(A1⊗A2⊗⋯⊗Ak)(B1⊗B2⊗⋯⊗Bk)=(A1B1)⊗(A2⊗⋯⊗Ak)(B2⊗⋯⊗Bk)=(A1B1)⊗(A2B2)⊗⋯⊗(AkBk)\begin{aligned} &(A_1\otimes B_1)(A_2\otimes B_2)\cdots(A_k\otimes B_k)=(A_1A_2\cdots A_k)\otimes (B_1\otimes B_2\cdots B_k)\\ &(A_1\otimes A_2\otimes \cdots\otimes A_k)(B_1\otimes B_2\otimes \cdots\otimes B_k)=(A_1B_1)\otimes(A_2\otimes \cdots\otimes A_k)(B_2\otimes \cdots\otimes B_k)=(A_1B_1)\otimes (A_2B_2)\otimes \cdots \otimes (A_kB_k) \end{aligned} (A1⊗B1)(A2⊗B2)⋯(Ak⊗Bk)=(A1A2⋯Ak)⊗(B1⊗B2⋯Bk)(A1⊗A2⊗⋯⊗Ak)(B1⊗B2⊗⋯⊗Bk)=(A1B1)⊗(A2⊗⋯⊗Ak)(B2⊗⋯⊗Bk)=(A1B1)⊗(A2B2)⊗⋯⊗(AkBk)
eg:
A=Am×m,证明eA⊗In=eA⊗InA=A_{m\times m},证明e^{A\otimes I_n}=e^A\otimes I_n A=Am×m,证明eA⊗In=eA⊗In
eA⊗In=∑k=1∞1k!(A⊗I)k=∑k=1∞1k!(Ak⊗Ik)=(∑k=1∞1k!Ak)⊗Ik=eA⊗I\begin{aligned} &e^{A\otimes I_n}=\sum_{k=1}\limits^\infty\frac{1}{k!}(A\otimes I)^k=\sum_{k=1}\limits^\infty\frac{1}{k!}(A^k\otimes I^k)=(\sum_{k=1}\limits^\infty\frac{1}{k!}A^k)\otimes I^k=e^A\otimes I \end{aligned} eA⊗In=k=1∑∞k!1(A⊗I)k=k=1∑∞k!1(Ak⊗Ik)=(k=1∑∞k!1Ak)⊗Ik=eA⊗I
转置公式:
(A⊗B)H=AH⊗BH(A⊗B)−1=A−1⊗B−1\begin{aligned} &(A\otimes B)^H=A^H\otimes B^H\\ &(A\otimes B)^{-1}= A^{-1}\otimes B^{-1} \end{aligned} (A⊗B)H=AH⊗BH(A⊗B)−1=A−1⊗B−1
若A与B都是U阵,则 A⊗BA\otimes BA⊗B U阵
秩公式:r(A⊗B)=r(A)r(B)r(A\otimes B)=r(A)r(B)r(A⊗B)=r(A)r(B)
推论:
若X1、⋯、Xp为Cm中p个线性无关的列向量,Y1、⋯、Yq为Cn中q个线性无关列向量,则则pq个列向量(Xi⊗Yj)线性无关\begin{aligned} &若X_1、\cdots、X_p为C^m中p个线性无关的列向量,Y_1、\cdots、Y_q为C^n中q个线性无关列向量,则\\ &则pq个列向量(X_i\otimes Y_j)线性无关 \end{aligned} 若X1、⋯、Xp为Cm中p个线性无关的列向量,Y1、⋯、Yq为Cn中q个线性无关列向量,则则pq个列向量(Xi⊗Yj)线性无关
由于 r({X⊗Y})=r({X})r({Y})r(\{X\otimes Y\})=r(\{X\})r(\{Y\})r({X⊗Y})=r({X})r({Y}) ,张量积的秩等于两个向量组的秩的乘积,所以张量积线性无关
设 A=(aij)∈Cm×m,B=(bij)∈Cn×nA=(a_{ij}) \in C^{m\times m} ,B=(b_{ij})\in C^{n\times n}A=(aij)∈Cm×m,B=(bij)∈Cn×n ,则
tr(A⊗B)=tr(A)∗tr(B)∣A⊗B∣=∣A∣n∣B∣m\begin{aligned} &tr(A\otimes B)=tr(A)*tr(B)\\ &\vert A\otimes B \vert=\vert A \vert^n\vert B \vert^m \end{aligned} tr(A⊗B)=tr(A)∗tr(B)∣A⊗B∣=∣A∣n∣B∣m
证明:
由许尔公式,存在可逆阵P使P−1AP=(λ1∗⋱0λm)为上三角,且P⊗I为可逆阵,构造一个新公式(P⊗I)−1A⊗B(P⊗I)=(P−1AP)⊗B=A1⊗B=(λ1B(∗B)λ2B⋱0λmB)故∣A⊗B∣=∣λ1nB∣∣λ2nB∣⋯∣λmnB∣m=∣λ1nλ2n⋯λmn∣∣B∣m=∣A∣n∣B∣m\begin{aligned} &由许尔公式,存在可逆阵P使P^{-1}AP=\left( \begin{matrix} \lambda_1&&*\\ &\ddots&\\ 0&&\lambda_m \end{matrix} \right)为上三角,且P\otimes I为可逆阵,构造一个新公式\\ &(P\otimes I)^{-1}A\otimes B(P\otimes I)=(P^{-1}AP)\otimes B=A_1\otimes B=\left( \begin{matrix} \lambda_1B&&&(*B)\\ &\lambda_2B&&\\ &&\ddots&\\ 0&&&\lambda_mB \end{matrix} \right)\\ &故\vert A\otimes B \vert=\vert \lambda_1^nB\vert\vert \lambda_2^nB\vert\cdots\vert \lambda_m^nB\vert^m=\vert \lambda_1^n\lambda_2^n\cdots\lambda_m^n\vert\vert B \vert^m = \vert A \vert^n\vert B\vert ^m \end{aligned} 由许尔公式,存在可逆阵P使P−1AP=⎝⎛λ10⋱∗λm⎠⎞为上三角,且P⊗I为可逆阵,构造一个新公式(P⊗I)−1A⊗B(P⊗I)=(P−1AP)⊗B=A1⊗B=⎝⎜⎜⎛λ1B0λ2B⋱(∗B)λmB⎠⎟⎟⎞故∣A⊗B∣=∣λ1nB∣∣λ2nB∣⋯∣λmnB∣m=∣λ1nλ2n⋯λmn∣∣B∣m=∣A∣n∣B∣m
若A=Am×m的特征根为λ1,λ2,⋯,λm,B=Bn×n的特征根为t1,t2,⋯,tn,则A⊗B的全体特征根为mn个数{λktj},(k=1,2,⋯,mj=1,2,⋯,n)\begin{aligned} &若A=A_{m\times m} 的特征根为\lambda_1,\lambda_2,\cdots,\lambda_m,B=B_{n\times n} 的特征根为t_1,t_2,\cdots,t_n,则\\ &A\otimes B的全体特征根为mn个数\{\lambda_kt_j\},(k=1,2,\cdots,m\quad j=1,2,\cdots,n) \end{aligned} 若A=Am×m的特征根为λ1,λ2,⋯,λm,B=Bn×n的特征根为t1,t2,⋯,tn,则A⊗B的全体特征根为mn个数{λktj},(k=1,2,⋯,mj=1,2,⋯,n)
设{X1,⋯,Xp}是A∈Cm×m关于λ的线性无关的特征向量,{Y1,⋯,Yq}是B∈Cn×n关于t的线性无关的特征向量,则pq个向量{Xi⊗Yj}是A⊗B关于λt的特征向量\begin{aligned} &设\{X_1,\cdots,X_p\}是A\in C^{m\times m}关于\lambda的线性无关的特征向量,\{Y_1,\cdots,Y_q\}是B\in C^{n\times n}关于t的线性\\ &无关的特征向量,则pq个向量 \{X_i\otimes Y_j\} 是A\otimes B关于\lambda t的特征向量 \end{aligned} 设{X1,⋯,Xp}是A∈Cm×m关于λ的线性无关的特征向量,{Y1,⋯,Yq}是B∈Cn×n关于t的线性无关的特征向量,则pq个向量{Xi⊗Yj}是A⊗B关于λt的特征向量
eg
A=(2213)⊗(1−101)=B⊗D,且B为行和等矩阵,B为对角阵,则λ(A)={4,tr(A)−4}={4,1},λ(B)={1,1}∴A⊗B=∏λAλB={4,4,1,1},特式∣λI−A∣=(λ−4)2(λ−1)2可知,(11)(−21)是B的一个特征向量,(10)是D的特征值,故A⊗B的特征向量为(1010),(−2010)\begin{aligned} &A=\left( \begin{matrix} 2&2\\1&3 \end{matrix} \right)\otimes \left( \begin{matrix} 1&-1\\0&1 \end{matrix} \right)=B\otimes D,且B为行和等矩阵,B为对角阵,则\lambda(A)=\{4,tr(A)-4\}=\{4,1\},\lambda(B)=\{1,1\}\\ &\therefore A\otimes B=\prod\lambda_A\lambda_B=\{4,4,1,1\},特式\vert \lambda I-A\vert=(\lambda-4)^2(\lambda-1)^2\\ &可知,\left( \begin{matrix} 1\\1 \end{matrix} \right)\left( \begin{matrix} -2\\1 \end{matrix} \right)是B的一个特征向量,\left( \begin{matrix} 1\\0 \end{matrix} \right)是D的特征值,故A\otimes B的特征向量为\left( \begin{matrix} 1\\0\\1\\0 \end{matrix} \right),\left( \begin{matrix} -2\\0\\1\\0 \end{matrix} \right) \end{aligned} A=(2123)⊗(10−11)=B⊗D,且B为行和等矩阵,B为对角阵,则λ(A)={4,tr(A)−4}={4,1},λ(B)={1,1}∴A⊗B=∏λAλB={4,4,1,1},特式∣λI−A∣=(λ−4)2(λ−1)2可知,(11)(−21)是B的一个特征向量,(10)是D的特征值,故A⊗B的特征向量为⎝⎜⎜⎛1010⎠⎟⎟⎞,⎝⎜⎜⎛−2010⎠⎟⎟⎞
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