【矩阵论】3. 矩阵运算与函数——张量积

矩阵论
1. 准备知识——复数域上的矩阵与换位公式)
1. 准备知识——复数域上的内积域正交阵
1. 准备知识——相似对角化与合同&正定阵
2. 矩阵分解—— SVD准备知识——奇异值
2. 矩阵分解——SVD
2. 矩阵分解——QR分解
2. 矩阵分解——乔利斯分解&平方根公式
2. 矩阵分解——正规谱分解——正规阵
2. 矩阵分解——正规分解
2. 矩阵分解——单阵及特征值特征向量一些求法
2. 矩阵分解——高低分解
3. 矩阵运算与函数——张量积

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9.1 张量积

9.1.1 定义

$$\begin{array}{rl}& 设A=(a_{ij}{)}_{m\times n},B=(b_{ij}{)}_{p\times q},则称如下分块矩阵\left(\begin{array}{cccc}{a}_{11}B& a_{12}B& \cdots & a_{1n}B\\ a_{21}B& a_{22}B& \cdots & a_{2n}B\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}B& a_{n2}& \cdots & a_{nn}\end{array}\right)为A与B的张量积\\ & 记作A\otimes B=(a_{ij}B{)}_{mp\times nq}\end{array}$$

eg
$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),B=\left(\begin{array}{c}2\\ 3\end{array}\right)$$

$$\begin{array}{rl}& A\otimes B=\left(\begin{array}{cc}aB& bB\\ cB& dB\end{array}\right)=\left(\begin{array}{cc}2a& 2b\\ 3a& 3b\\ 2c& 2d\\ 3c& 3d\end{array}\right),B\otimes A=\left(\begin{array}{c}2A\\ 3B\end{array}\right)=\left(\begin{array}{cc}2a& 2b\\ 2c& 2d\\ 3a& 3b\\ 3c& 3d\end{array}\right)\end{array}$$

• 张量积不满足交换律

定理

1. 两个上三角的张量积也是上三角
2. 两个对角阵的张量积是对角阵
3. $I_n\otimes I_m=I_m\otimes I_n=I_{m\times n}$

9.1.2 计算

a. 分块法

右进右出
$$\begin{gathered} \left(\begin{array}{cc}A& B\\ C& D\end{array}\right)\otimes F=\left(\begin{array}{cc}A\otimes F& B\otimes F\\ C\otimes F& D\otimes F\end{array}\right),\left(\begin{array}{c}A\\ C\end{array}\right)\otimes F=\left(\begin{array}{c}A\otimes F\\ C\otimes F\end{array}\right) \\ (AB)\otimes F=(A\otimes FB\otimes F) \end{gathered}$$
一般情况下：$(AB)\otimes F\ne (A\otimes FB\otimes F)$

b. 向量与向量张量积

$$列向量\alpha =\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_n\end{array}\right)\beta =\left(\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_q\end{array}\right),则\alpha \otimes \beta ={\left(\begin{array}{c}{a}_1\otimes \beta \\ a_2\otimes \beta \\ ⋮\\ a_n\otimes \beta \end{array}\right)}_{nq\times 1}={\left(\begin{array}{c}{a}_1{b}_1\\ a_1{b}_2\\ ⋮\\ a_1{b}_q\\ ⋮\\ a_n{b}_1\\ a_n{b}_2\\ ⋮\\ a_n{b}_q\end{array}\right)}_{nq\times 1}$$

向量与矩阵张量积

$$列向量\alpha =\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_m\end{array}\right),B=({\beta }_1,{\beta }_2,\cdots ,{\beta }_q)，\alpha \otimes B=\left(\begin{array}{c}{a}_{1}B\\ a_{2}B\\ ⋮\\ a_{m}B\end{array}\right)=\left(\begin{array}{c}{a}_1({\beta }_1,{\beta }_2,\cdots ,{\beta }_q)\\ a_2({\beta }_1,{\beta }_2,\cdots ,{\beta }_q)\\ ⋮\\ a_m({\beta }_1,{\beta }_2,\cdots ,{\beta }_q)\end{array}\right)={\left(\begin{array}{cccc}{a}_1{b}_{11}& a_1{b}_{12}& \cdots & a_1{b}_{1q}\\ a_1{b}_{21}& a_1{b}_{22}& \cdots & a_1{b}_{2q}\\ ⋮& ⋮& \ddots & ⋮\\ a_1{b}_{p1}& a_1{b}_{p2}& \cdots & a_1{b}_{pq}\\ a_2{b}_{11}& a_2{b}_{12}& \cdots & a_2{b}_{1q}\\ a_2{b}_{21}& a_2{b}_{22}& \cdots & a_2{b}_{2q}\\ ⋮& ⋮& \ddots & ⋮\\ a_2{b}_{p1}& a_2{b}_{p2}& \cdots & a_2{b}_{pq}\\ ⋮& ⋮& ⋮& ⋮\\ a_m{b}_{11}& a_m{b}_{12}& \cdots & a_m{b}_{1q}\\ a_m{b}_{21}& a_m{b}_{22}& \cdots & a_m{b}_{2q}\\ ⋮& ⋮& \ddots & ⋮\\ a_m{b}_{p1}& a_m{b}_{p2}& \cdots & a_m{b}_{pq}\end{array}\right)}_{mp\times q}$$

c. 性质

1. $k(A\otimes B)=(kA)\otimes B=A\otimes (kB)$

2. 分配律：$(A+B)\otimes C=A\otimes C+B\otimes C$ ，$C\otimes (A+B)=C\otimes A+C\otimes B$

3. 结合律：$(A\otimes B)\otimes C=A\otimes (B\otimes C)$

4. 吸收律：$(A\otimes B)(C\otimes D)=(AC)\otimes (BD)$

   

   推论：
   $$\begin{array}{rl}& 若A=A_{m\times m}为m阶方阵,B=B_{n\times n}为n阶方阵，则\\ & (A\otimes B{)}^k=A^k\otimes B^k\\ & (A\otimes I_n)(I_m\otimes B)=A\otimes B\end{array}$$

   $$\begin{array}{rl}& (A_1\otimes B_1)(A_2\otimes B_2)\cdots (A_k\otimes B_k)=(A_1{A}_2\cdots A_k)\otimes (B_1\otimes B_2\cdots B_k)\\ & (A_1\otimes A_2\otimes \cdots \otimes A_k)(B_1\otimes B_2\otimes \cdots \otimes B_k)=(A_1{B}_1)\otimes (A_2\otimes \cdots \otimes A_k)(B_2\otimes \cdots \otimes B_k)=(A_1{B}_1)\otimes (A_2{B}_2)\otimes \cdots \otimes (A_k{B}_k)\end{array}$$

   eg：
   $$A=A_{m\times m}，证明e^{A\otimes I_n}=e^A\otimes I_n$$

   $$\begin{array}{rl}& e^{A\otimes I_n}=\sum _{k=1}^{\infty }\frac{1}{k!}(A\otimes I{)}^k=\sum _{k=1}^{\infty }\frac{1}{k!}(A^k\otimes I^k)=(\sum _{k=1}^{\infty }\frac{1}{k!}{A}^k)\otimes I^k=e^A\otimes I\end{array}$$

5. 转置公式：
   $$\begin{array}{rl}& (A\otimes B{)}^H=A^H\otimes B^H\\ & (A\otimes B{)}^{-1}=A^{-1}\otimes B^{-1}\end{array}$$

6. 若A与B都是U阵，则 $A\otimes B$ U阵

7. 秩公式：$r(A\otimes B)=r(A)r(B)$

   推论：
   $$\begin{array}{rl}& 若X_1、\cdots 、X_p为C^m中p个线性无关的列向量，Y_1、\cdots 、Y_q为C^n中q个线性无关列向量，则\\ & 则pq个列向量(X_i\otimes Y_j)线性无关\end{array}$$
   由于 $r(\{X\otimes Y\})=r(\{X\})r(\{Y\})$ ，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

9.1.3 张量积行列式

设 $A=(a_{ij})\in C^{m\times m}，B=(b_{ij})\in C^{n\times n}$ ，则
$$\begin{array}{rl}& tr(A\otimes B)=tr(A)\ast tr(B)\\ & |A\otimes B|=|A{|}^n|B{|}^m\end{array}$$
证明：

$$\begin{array}{rl}& 由许尔公式，存在可逆阵P使P^{-1}AP=\left(\begin{array}{ccc}{\lambda }_1& & \ast \\ & \ddots & \\ 0& & {\lambda }_m\end{array}\right)为上三角，且P\otimes I为可逆阵，构造一个新公式\\ & (P\otimes I{)}^{-1}A\otimes B(P\otimes I)=(P^{-1}AP)\otimes B=A_1\otimes B=\left(\begin{array}{cccc}{\lambda }_{1}B& & & (\ast B)\\ & {\lambda }_{2}B& & \\ & & \ddots & \\ 0& & & {\lambda }_{m}B\end{array}\right)\\ & 故|A\otimes B|=|{\lambda }_1^{n}B||{\lambda }_2^{n}B|\cdots |{\lambda }_m^{n}B{|}^m=|{\lambda }_1^n{\lambda }_2^n\cdots {\lambda }_m^n||B{|}^m=|A{|}^n|B{|}^m\end{array}$$

9.1.4 张量积特征值特征向量

特征值

$$\begin{array}{rl}& 若A=A_{m\times m}的特征根为{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_m，B=B_{n\times n}的特征根为t_1,t_2,\cdots ,t_n，则\\ & A\otimes B的全体特征根为mn个数\{{\lambda }_k{t}_j\}，(k=1,2,\cdots ,mj=1,2,\cdots ,n)\end{array}$$

特征向量

$$\begin{array}{rl}& 设\{X_1,\cdots ,X_p\}是A\in C^{m\times m}关于\lambda 的线性无关的特征向量,\{Y_1,\cdots ,Y_q\}是B\in C^{n\times n}关于t的线性\\ & 无关的特征向量，则pq个向量\{X_i\otimes Y_j\}是A\otimes B关于\lambda t的特征向量\end{array}$$

eg

$$\begin{array}{rl}& A=\left(\begin{array}{cc}2& 2\\ 1& 3\end{array}\right)\otimes \left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)=B\otimes D,且B为行和等矩阵，B为对角阵，则\lambda (A)=\{4,tr(A)-4\}=\{4,1\},\lambda (B)=\{1,1\}\\ & \therefore A\otimes B=\prod {\lambda }_A{\lambda }_B=\{4,4,1,1\},特式|\lambda I-A|=(\lambda -4{)}^2(\lambda -1{)}^2\\ & 可知，\left(\begin{array}{c}1\\ 1\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)是B的一个特征向量，\left(\begin{array}{c}1\\ 0\end{array}\right)是D的特征值,故A\otimes B的特征向量为\left(\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}-2\\ 0\\ 1\\ 0\end{array}\right)\end{array}$$

$A\otimes I\pm I\otimes B$