矩阵的Kronecker积的相关结论 |
您所在的位置:网站首页 › a的转置等于a的性质 › 矩阵的Kronecker积的相关结论 |
矩阵的Kronecker积的相关结论
矩阵的Kronecker积矩阵的Kronecker积的定义矩阵的Kronecker积的性质
矩阵的拉直矩阵的拉直的定义矩阵的拉直的性质
矩阵的方程
矩阵的Kronecker积
矩阵的Kronecker积的定义
设矩阵 A = ( a i j ) ∈ M m , n , B ∈ M s , t . A=(a_{ij})\in M_{m,n},B\in M_{s,t}. A=(aij)∈Mm,n,B∈Ms,t. A A A和 B B B的 K r o n e c k e r Kronecker Kronecker积(也称为张量积)记作 A ⊗ B A\otimes B A⊗B,定义为下面的分块矩阵: A ⊗ B = ( a 11 B a 12 B ⋯ a 1 n B a 21 B a 22 B ⋯ a 2 n B ⋮ ⋮ ⋮ a m 1 B a m 2 B ⋯ a m n B ) ∈ M m s , n t . A\otimes B=\begin{pmatrix} a_{11}B & a_{12}B & \cdots & a_{1n}B\\ a_{21}B & a_{22}B & \cdots & a_{2n}B\\ \vdots & \vdots & & \vdots\\ a_{m1}B & a_{m2}B & \cdots & a_{mn}B \end{pmatrix} \in M_{ms,nt}. A⊗B=⎝⎜⎜⎜⎛a11Ba21B⋮am1Ba12Ba22B⋮am2B⋯⋯⋯a1nBa2nB⋮amnB⎠⎟⎟⎟⎞∈Mms,nt. 矩阵的Kronecker积的性质A ∈ M m , n , B ∈ M s , t , C ∈ M p , q , D ∈ M t , r A\in M_{m,n},B\in M_{s,t},C\in M_{p,q},D\in M_{t,r} A∈Mm,n,B∈Ms,t,C∈Mp,q,D∈Mt,r ( α A ) ⊗ B = A ⊗ ( α B ) = α ( A ⊗ B ) , α ∈ C (\alpha A)\otimes B=A\otimes (\alpha B)=\alpha(A\otimes B),\alpha\in \mathbf{C} (αA)⊗B=A⊗(αB)=α(A⊗B),α∈C ( A ⊗ B ) T = A T ⊗ B T (A \otimes B)^T=A^T\otimes B^T (A⊗B)T=AT⊗BT ( A ⊗ B ) H = A H ⊗ B H (A \otimes B)^H=A^H\otimes B^H (A⊗B)H=AH⊗BH ( A ⊗ B ) ⊗ C = A ⊗ ( B ⊗ C ) (A \otimes B)\otimes C=A\otimes (B\otimes C) (A⊗B)⊗C=A⊗(B⊗C) A ⊗ ( B + C ) = ( A ⊗ B ) + ( A ⊗ C ) A\otimes (B+C)=(A\otimes B)+(A\otimes C) A⊗(B+C)=(A⊗B)+(A⊗C) ( A + B ) ⊗ C = ( A ⊗ C ) + ( B ⊗ C ) (A+B)\otimes C=(A\otimes C)+(B\otimes C) (A+B)⊗C=(A⊗C)+(B⊗C) A ⊗ B = 0 A \otimes B=0 A⊗B=0当且仅当 A = 0 A=0 A=0或 B = 0 B=0 B=0 ( A ⊗ B ) ( C ⊗ D ) = ( A C ) ⊗ ( B D ) (A\otimes B)(C\otimes D)=(AC)\otimes(BD) (A⊗B)(C⊗D)=(AC)⊗(BD)设 A ∈ M m , B ∈ M n A\in M_{m},B\in M_{n} A∈Mm,B∈Mn 若 A , B A,B A,B对称,则 A ⊗ B A \otimes B A⊗B对称若 A , B A,B A,B为 H e r m i t Hermit Hermit矩阵,则 A ⊗ B A\otimes B A⊗B是 H e r m i t Hermit Hermit矩阵若 A , B A,B A,B可逆,则 A ⊗ B A\otimes B A⊗B也可逆,且 ( A ⊗ B ) − 1 = A − 1 ⊗ B − 1 (A\otimes B)^{-1}=A^{-1}\otimes B^{-1} (A⊗B)−1=A−1⊗B−1若 A , B A,B A,B为正规矩阵,则 A ⊗ B A\otimes B A⊗B是正规矩阵若 A , B A,B A,B为酉矩阵,则 A ⊗ B A\otimes B A⊗B是酉矩阵若 λ ∈ σ ( A ) \lambda\in\sigma(A) λ∈σ(A), x x x是对应的特征向量, μ ∈ σ ( B ) \mu\in\sigma(B) μ∈σ(B), y y y是对应的特征向量,则 λ μ ∈ σ ( A ⊗ B ) \lambda\mu\in\sigma(A\otimes B) λμ∈σ(A⊗B), x ⊗ y x\otimes y x⊗y是对应的特征向量若 σ ( A ) = { λ 1 , ⋯ , λ m } , σ ( B ) = { μ 1 , ⋯ , μ n } \sigma(A)= \{ \lambda_1,\cdots,\lambda_{m}\},\sigma(B)= \{ \mu_1,\cdots,\mu_{n}\} σ(A)={λ1,⋯,λm},σ(B)={μ1,⋯,μn},则 σ ( A ⊗ B ) = { λ i μ j ∣ i = 1 , ⋯ , m , j = 1 , ⋯ , n } \sigma(A\otimes B)=\{\lambda_{i}\mu_{j} | i=1,\cdots,m,j=1,\cdots,n\} σ(A⊗B)={λiμj∣i=1,⋯,m,j=1,⋯,n} d e t ( A ⊗ B ) = ( d e t A ) n ( d e t B ) m det(A\otimes B)=(detA)^n(detB)^m det(A⊗B)=(detA)n(detB)m若 s v ( A ) = { s 1 , ⋯ , s m } , s v ( B ) = { t 1 , ⋯ , t n } sv(A)= \{ s_1,\cdots,s_{m}\},sv(B)= \{ t_1,\cdots,t_{n}\} sv(A)={s1,⋯,sm},sv(B)={t1,⋯,tn},则 s v ( A ⊗ B ) = { s i t j ∣ i = 1 , ⋯ , m , j = 1 , ⋯ , n } sv(A\otimes B)=\{s_{i}t_{j} | i=1,\cdots,m,j=1,\cdots,n\} sv(A⊗B)={sitj∣i=1,⋯,m,j=1,⋯,n} r a n k ( A ⊗ B ) = ( r a n k A ) ( r a n k B ) rank(A\otimes B)=(rankA)(rankB) rank(A⊗B)=(rankA)(rankB) 矩阵的拉直 矩阵的拉直的定义设 A = ( a 1 , a 2 , ⋯ , a n ) ∈ M m , n A=(a_1,a_2,\cdots,a_n)\in M_{m,n} A=(a1,a2,⋯,an)∈Mm,n,则 v e c ( A ) = ( a 1 a 2 ⋮ a n ) . vec(A)=\begin{pmatrix} a_{1}\\ a_{2}\\ \vdots \\ a_{n} \end{pmatrix}. vec(A)=⎝⎜⎜⎜⎛a1a2⋮an⎠⎟⎟⎟⎞. 矩阵的拉直的性质1、设 A ∈ M m , n , B ∈ M n , k , C ∈ M k , t , A\in M_{m,n},B\in M_{n,k},C\in M_{k,t}, A∈Mm,n,B∈Mn,k,C∈Mk,t,则 v e c ( A B C ) = ( C T ⊗ A ) v e c B . vec(ABC)=(C^T\otimes A)vecB. vec(ABC)=(CT⊗A)vecB. 2、存在一个只依赖于 m , n m,n m,n的 m n mn mn阶置换矩阵 P ( m , n ) P(m,n) P(m,n)使得 v e c X T = P ( m , n ) v e c X , vecX^T=P(m,n)vecX, vecXT=P(m,n)vecX,对任何 X ∈ M m , n X\in M_{m,n} X∈Mm,n成立. 矩阵的方程定理1: 矩阵方程 A X − X B = C ( 称 为 S y l v e s t e r 方 程 ) , A ∈ M m , B ∈ M n , C ∈ M m , n AX-XB=C(称为Sylvester方程), A\in M_m,B\in M_n,C\in M_{m,n} AX−XB=C(称为Sylvester方程),A∈Mm,B∈Mn,C∈Mm,n有唯一解当且仅当 A A A和 B B B没有公共特征值 定理2: 矩阵方程 A X − X B = C ( 称 为 S y l v e s t e r 方 程 ) , A ∈ M m , B ∈ M n , C ∈ M m , n AX-XB=C(称为Sylvester方程), A\in M_m,B\in M_n,C\in M_{m,n} AX−XB=C(称为Sylvester方程),A∈Mm,B∈Mn,C∈Mm,n有解当且仅当 ( A 0 0 B ) \begin{pmatrix} A&0\\ 0&B \end{pmatrix} (A00B)和 ( A C 0 B ) \begin{pmatrix} A&C\\ 0&B \end{pmatrix} (A0CB)相似 例1、 设 A ∈ C m × m , B ∈ C n × n , X ( t ) ∈ C m × n A\in \mathbf{C}^{m\times m},B\in \mathbf{C}^{n\times n},X(t)\in \mathbf{C}^{m\times n} A∈Cm×m,B∈Cn×n,X(t)∈Cm×n,求下列微分方程初值问题的解: { d X ( t ) d t = A X ( t ) + X ( t ) B X ( 0 ) = X 0 \begin{cases}\dfrac{dX(t)}{dt}=AX(t)+X(t)B\\ X(0)=X_0 \end{cases} ⎩⎨⎧dtdX(t)=AX(t)+X(t)BX(0)=X0 引理: 设矩阵 A ∈ C m × m , B ∈ C n × n A\in \mathbf{C}^{m\times m},B\in\mathbf{C}^{n\times n} A∈Cm×m,B∈Cn×n,则 e A × I n = e A × I n e^{A\times I_n}=e^A\times I_n eA×In=eA×In, e I m × B = I m × B e^{I_m\times B}=I_m\times B eIm×B=Im×B. p r o o f : proof: proof: e A × I n = ∑ k = 1 ∞ 1 k ! ( A ⊗ I ) k = ∑ k = 1 ∞ 1 k ! ( A k ⊗ I k ) = ( ∑ k = 1 ∞ 1 k ! A k ) ⊗ I = e A × I n e^{A\times I_n}=\sum\limits_{k=1}^{\infty}\frac{1}{k!}(A\otimes I)^k=\sum\limits_{k=1}^{\infty}\frac{1}{k!}(A^k\otimes I^k)=(\sum\limits_{k=1}^{\infty}\frac{1}{k!}A^k)\otimes I=e^A\times I_n eA×In=k=1∑∞k!1(A⊗I)k=k=1∑∞k!1(Ak⊗Ik)=(k=1∑∞k!1Ak)⊗I=eA×In 同理可得: e I m × B = I m × B e^{I_m\times B}=I_m\times B eIm×B=Im×B 解、 对微分方程两边拉直,易得: { d v e c X ( t ) d t = ( I n ⊗ A + B T ⊗ I m ) v e c X ( t ) v e c X ( 0 ) = v e c X 0 \begin{cases}\dfrac{dvecX(t)}{dt}=(I_n\otimes A+B^T\otimes I_m)vecX(t)\\ vecX(0)=vecX_0 \end{cases} ⎩⎨⎧dtdvecX(t)=(In⊗A+BT⊗Im)vecX(t)vecX(0)=vecX0 由引理可得: v e c X ( t ) = e t ( I n ⊗ A + B T ⊗ I m ) v e c X 0 = ( e B T t ⊗ e A t ) v e c X 0 = v e c ( e A t X 0 ( e B T t ) T ) = v e c ( e A t X 0 e B t ) vecX(t)=e^{t(I_n\otimes A+B^T\otimes I_m)}vecX_0=(e^{B^Tt}\otimes e^{At})vecX_0=vec(e^{At}X_0(e^{B^Tt})^T)=vec(e^{At}X_0e^{B^t}) vecX(t)=et(In⊗A+BT⊗Im)vecX0=(eBTt⊗eAt)vecX0=vec(eAtX0(eBTt)T)=vec(eAtX0eBt) 于是 X ( t ) = e A t X 0 e B t X(t)=e^{At}X_0e^{B^t} X(t)=eAtX0eBt为微分方程的解 |
今日新闻 |
点击排行 |
|
推荐新闻 |
图片新闻 |
|
专题文章 |
CopyRight 2018-2019 实验室设备网 版权所有 win10的实时保护怎么永久关闭 |