（八）二阶张量与矩阵（二）

1. 二阶张量的代数运算与矩阵的代数运算

4.1. 二阶张量的相等、加（减）、数乘

二阶张量的相等、加（减）、数乘运算与矩阵相等、加（减）、数乘运算一 一对应；

1.2. 二阶张量的缩并

与二阶张量的缩并相关的运算为求二阶张量的迹$tr(\mathbf{T})$：
$$tr(\mathbf{T})=T^{ij}{g}_{ij}=T_i^{\bullet i}=T_1^{\bullet 1}+T_2^{\bullet 2}+T_3^{\bullet 3}=T_{\bullet i}^i=T_{\bullet 1}^1+T_{\bullet 2}^2+T_{\bullet 3}^3=T_{ij}{g}^{ij}$$显然，与之相关的矩阵运算为求方阵 ${\tau }_2、{\tau }_3$ 的迹。

二阶张量迹的运算性质：

若 $\mathbf{A},\mathbf{B},\mathbf{C}\in {\mathcal{T}}_2(V)$，则：
$$\begin{array}{rl}& (1)tr(\mathbf{A}+\mathbf{B})=tr(\mathbf{A})+tr(\mathbf{B})\\ & \\ & (2)tr(\mathbf{A}\cdot \mathbf{B})=tr(\mathbf{B}\cdot \mathbf{A})\\ & \\ & (3)tr(\mathbf{A}\cdot \mathbf{B}\cdot \mathbf{C})=tr(\mathbf{B}\cdot \mathbf{C}\cdot \mathbf{A})=tr(\mathbf{C}\cdot \mathbf{A}\cdot \mathbf{B})\\ & \\ & (4)\mathbf{A}:\mathbf{B}=tr({\mathbf{A}}^T\cdot \mathbf{B})=tr(\mathbf{A}\cdot {\mathbf{B}}^T)=tr({\mathbf{B}}^T\cdot \mathbf{A})=tr(\mathbf{B}\cdot {\mathbf{A}}^T)\end{array}$$
证明：
$$\begin{array}{rl}& (1)tr(\mathbf{A}+\mathbf{B})=A_{\bullet i}^i+B_{\bullet i}^i=tr(\mathbf{A})+tr(\mathbf{B})\\ & \\ & (2)tr(\mathbf{A}\cdot \mathbf{B})=A_{\bullet j}^i{B}_{\bullet i}^j=B_{\bullet i}^j{A}_{\bullet j}^i=tr(\mathbf{B}\cdot \mathbf{A})\\ & \\ & (3)tr(\mathbf{A}\cdot \mathbf{B}\cdot \mathbf{C})=A_{\bullet j}^i{B}_{\bullet k}^j{C}_{\bullet i}^k=B_{\bullet k}^j{C}_{\bullet i}^k{A}_{\bullet j}^i=tr(\mathbf{B}\cdot \mathbf{C}\cdot \mathbf{A})\\ & \\ & =C_{\bullet i}^k{A}_{\bullet j}^i{B}_{\bullet k}^j=tr(\mathbf{C}\cdot \mathbf{A}\cdot \mathbf{B})\\ & \\ & (4)\mathbf{A}:\mathbf{B}=A_{\bullet j}^i{B}_i^{\bullet j}=(B^T{)}_{\bullet i}^j{A}_{\bullet j}^i=tr({\mathbf{B}}^T\cdot \mathbf{A})\\ & \\ & =(A^T{)}_j^{\bullet i}{B}_i^{\bullet j}=tr({\mathbf{A}}^T\cdot \mathbf{B})\end{array}$$

1.3. 二阶张量与矢量的点积 —— 线性变换

w ⃗ = T ∙ u ⃗ ⟺ w i = T ∙ j i u j ⟺ [ w 1 w 2 w 3 ] = [ T ∙ 1 1 T ∙ 2 1 T ∙ 3 1 T ∙ 1 2 T ∙ 2 2 T ∙ 3 2 T ∙ 1 3 T ∙ 2 3 T ∙ 3 3 ] [ u 1 u 2 u 3 ] ⟺ w ⃗ = τ 3 u ⃗   t ⃗ = u ⃗ ∙ T ⟺ t i = u j T j ∙ i ⟺ [ t 1 t 2 t 3 ] = [ T 1 ∙ 1 T 2 ∙ 1 T 3 ∙ 1 T 1 ∙ 2 T 2 ∙ 2 T 3 ∙ 2 T 1 ∙ 3 T 2 ∙ 3 T 3 ∙ 3 ] [ u 1 u 2 u 3 ] ⟺ t ⃗ = τ 2 u ⃗ \vec{w}=\bold{T}\bullet\vec{u} \Longleftrightarrow w^i=T^{i}_{\bullet j}u^j \Longleftrightarrow \begin{bmatrix}w^1\\\\w^2\\\\w^3\end{bmatrix} =\begin{bmatrix} T^{1}_{\bullet 1} & T^{1}_{\bullet 2} & T^{1}_{\bullet 3} \\ \\ T^{2}_{\bullet 1} & T^{2}_{\bullet 2} & T^{2}_{\bullet 3} \\ \\ T^{3}_{\bullet 1} & T^{3}_{\bullet 2} & T^{3}_{\bullet 3} \end{bmatrix} \begin{bmatrix}u^1\\\\u^2\\\\u^3\end{bmatrix} \Longleftrightarrow \vec{w}=\tau_{3}\vec{u} \\\ \\ \vec{t}=\vec{u}\bullet\bold{T} \Longleftrightarrow t^i=u^jT^{\bullet i}_{j} \Longleftrightarrow \begin{bmatrix}t^1\\\\t^2\\\\t^3\end{bmatrix} =\begin{bmatrix} T_{1}^{\bullet 1} & T_{2}^{\bullet 1} & T_{3}^{\bullet 1} \\ \\ T_{1}^{\bullet 2} & T_{2}^{\bullet 2} & T_{3}^{\bullet 2} \\ \\ T_{1}^{\bullet 3} & T_{2}^{\bullet 3} & T_{3}^{\bullet 3} \end{bmatrix} \begin{bmatrix}u^1\\\\u^2\\\\u^3\end{bmatrix} \Longleftrightarrow \vec{t}=\tau_{2}\vec{u} w =T∙u ⟺wi=T∙ji​uj⟺ ​w1w2w3​ ​= ​T∙11​T∙12​T∙13​​T∙21​T∙22​T∙23​​T∙31​T∙32​T∙33​​ ​ ​u1u2u3​ ​⟺w =τ3​u  t =u ∙T⟺ti=ujTj∙i​⟺ ​t1t2t3​ ​= ​T1∙1​T1∙2​T1∙3​​T2∙1​T2∙2​T2∙3​​T3∙1​T3∙2​T3∙3​​ ​ ​u1u2u3​ ​⟺t =τ2​u 显然，二阶张量与矢量的左、右点积一般不等：$$\mathbf{T}\bullet u\ne u\bullet \mathbf{T}$$且有：
$$({\tau }^T{)}_{3}u={\tau }_{2}u⟺{\mathbf{T}}^T\bullet u=u\bullet \mathbf{T}$$那么有：
$$\begin{gathered} \mathbf{N}\bullet u=u\bullet \mathbf{N}(\mathbf{N}为对称二阶张量) \\ \mathbf{\Omega }\bullet u=-u\bullet \mathbf{\Omega }(\mathbf{\Omega }为反对称二阶张量) \end{gathered}$$与矩阵与列向量的乘法相同，二阶张量可将任意向量映射为其它的向量，故也将二阶张量与矢量的点积称作线性变换。另外，任意对称二阶张量也对应着一个二次型，即：
x ⃗ ∙ N ∙ x ⃗ = N : x ⃗ x ⃗ = N i j x i x j = [ x 1 x 2 x 3 ] [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] [ x 1 x 2 x 3 ] = x ⃗ T N 1 x ⃗ \vec{x}\bullet\bold{N}\bullet\vec{x} =\bold{N}:\vec{x}\vec{x} =N_{ij}x^ix^j =\begin{bmatrix}x^1 & x^2 & x^3\end{bmatrix} \begin{bmatrix} N_{11} & N_{12} & N_{13} \\ \\ N_{21} & N_{22} & N_{23} \\ \\ N_{31} & N_{32} & N_{33} \end{bmatrix} \begin{bmatrix}x^1 \\\\ x^2 \\\\ x^3\end{bmatrix} =\vec{x}^TN_{1}\vec{x} x ∙N∙x =N:x x =Nij​xixj=[x1​x2​x3​] ​N11​N21​N31​​N12​N22​N32​​N13​N23​N33​​ ​ ​x1x2x3​ ​=x TN1​x

1.4. 二阶张量与二阶张量的点积

二阶张量的点积采用分量形式有：
C i j = A i ∙ k B k j = A i k B ∙ j k ⟺ C 1 = [ C i j ] = [ A 1 ∙ 1 A 2 ∙ 1 A 3 ∙ 1 A 1 ∙ 2 A 2 ∙ 2 A 3 ∙ 2 A 1 ∙ 3 A 2 ∙ 3 A 3 ∙ 3 ] T [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 2 T B 1 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B ∙ 1 1 B ∙ 2 1 B ∙ 3 1 B ∙ 1 2 B ∙ 2 2 B ∙ 3 2 B ∙ 1 3 B ∙ 2 3 B ∙ 3 3 ] = A 1 B 3   C i ∙ j = A i ∙ k B k ∙ j = A i k B k j ⟺ C 2 = [ C i ∙ j ] = [ A 1 ∙ 1 A 2 ∙ 1 A 3 ∙ 1 A 1 ∙ 2 A 2 ∙ 2 A 3 ∙ 2 A 1 ∙ 3 A 2 ∙ 3 A 3 ∙ 3 ] [ B 1 ∙ 1 B 2 ∙ 1 B 3 ∙ 1 B 1 ∙ 2 B 2 ∙ 2 B 3 ∙ 2 B 1 ∙ 3 B 2 ∙ 3 B 3 ∙ 3 ] = A 2 B 2 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 1 B 4   C ∙ j i = A ∙ k i B ∙ j k = A i k B k j ⟺ C 3 = [ C ∙ j i ] = [ A ∙ 1 1 A ∙ 2 1 A ∙ 3 1 A ∙ 1 2 A ∙ 2 2 A ∙ 3 2 A ∙ 1 3 A ∙ 2 3 A ∙ 3 3 ] [ B ∙ 1 1 B ∙ 2 1 B ∙ 3 1 B ∙ 1 2 B ∙ 2 2 B ∙ 3 2 B ∙ 1 3 B ∙ 2 3 B ∙ 3 3 ] = A 3 B 3 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 4 B 1   C i j = A ∙ k i B k j = A i k B k ∙ j ⟺ C 2 = [ C i j ] = [ A ∙ 1 1 A ∙ 2 1 A ∙ 3 1 A ∙ 1 2 A ∙ 2 2 A ∙ 3 2 A ∙ 1 3 A ∙ 2 3 A ∙ 3 3 ] [ B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33 ] = A 3 B 4 = [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] [ B 1 ∙ 1 B 2 ∙ 1 B 3 ∙ 1 B 1 ∙ 2 B 2 ∙ 2 B 3 ∙ 2 B 1 ∙ 3 B 2 ∙ 3 B 3 ∙ 3 ] T = A 4 B 2 T C_{ij}=A_{i}^{\bullet k}B_{k j}=A_{ik}B^k_{\bullet j} \Longleftrightarrow C_{1}=[C_{ij}] =\begin{bmatrix} A_{1}^{\bullet 1} & A_{2}^{\bullet 1} & A_{3}^{\bullet 1} \\\\ A_{1}^{\bullet 2} & A_{2}^{\bullet 2} & A_{3}^{\bullet 2} \\\\ A_{1}^{\bullet 3} & A_{2}^{\bullet 3} & A_{3}^{\bullet 3} \end{bmatrix}^T \begin{bmatrix} B_{11} & B_{12} & B_{13} \\\\ B_{21} & B_{22} & B_{23} \\\\ B_{31} & B_{32} & B_{33} \end{bmatrix} =A_{2}^TB_{1} =\begin{bmatrix} A_{11} & A_{12} & A_{13} \\\\ A_{21} & A_{22} & A_{23} \\\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} B^1_{\bullet 1} & B^1_{\bullet 2} & B^1_{\bullet 3} \\\\ B^2_{\bullet 1} & B^2_{\bullet 2} & B^2_{\bullet 3} \\\\ B^3_{\bullet 1} & B^3_{\bullet 2} & B^3_{\bullet 3} \end{bmatrix} =A_{1}B_{3}\\\ \\ %%%%%%%%%%%%%%%% C_{i}^{\bullet j}=A_{i}^{\bullet k}B_k^{\bullet j}=A_{ik}B^{kj} \Longleftrightarrow C_{2}=[C_{i}^{\bullet j}] =\begin{bmatrix} A_{1}^{\bullet 1} & A_{2}^{\bullet 1} & A_{3}^{\bullet 1} \\\\ A_{1}^{\bullet 2} & A_{2}^{\bullet 2} & A_{3}^{\bullet 2} \\\\ A_{1}^{\bullet 3} & A_{2}^{\bullet 3} & A_{3}^{\bullet 3} \end{bmatrix} \begin{bmatrix} B_{1}^{\bullet 1} & B_{2}^{\bullet 1} & B_{3}^{\bullet 1} \\\\ B_{1}^{\bullet 2} & B_{2}^{\bullet 2} & B_{3}^{\bullet 2} \\\\ B_{1}^{\bullet 3} & B_{2}^{\bullet 3} & B_{3}^{\bullet 3} \end{bmatrix} =A_{2}B_{2} =\begin{bmatrix} A_{11} & A_{12} & A_{13} \\\\ A_{21} & A_{22} & A_{23} \\\\ A_{31} & A_{32} & A_{33} \end{bmatrix} \begin{bmatrix} B^{11} & B^{12} & B^{13} \\\\ B^{21} & B^{22} & B^{23} \\\\ B^{31} & B^{32} & B^{33} \end{bmatrix} =A_{1}B_{4}\\\ \\ %%%%%%%%%%%%%%%% C_{\bullet j}^{i}=A^{i}_{\bullet k}B^k_{\bullet j}=A^{ik}B_{kj} \Longleftrightarrow C_{3}=[C_{\bullet j}^{i}] =\begin{bmatrix} A^{1}_{\bullet 1} & A^{1}_{\bullet 2}& A^{1}_{\bullet 3} \\\\ A^{2}_{\bullet 1} & A^{2}_{\bullet 2}& A^{2}_{\bullet 3} \\\\ A^{3}_{\bullet 1} & A^{3}_{\bullet 2}& A^{3}_{\bullet 3} \end{bmatrix} \begin{bmatrix} B^{1}_{\bullet 1} & B^{1}_{\bullet 2}& B^{1}_{\bullet 3} \\\\ B^{2}_{\bullet 1} & B^{2}_{\bullet 2}& B^{2}_{\bullet 3} \\\\ B^{3}_{\bullet 1} & B^{3}_{\bullet 2}& B^{3}_{\bullet 3} \end{bmatrix} =A_{3}B_{3} =\begin{bmatrix} A^{11} & A^{12} & A^{13} \\\\ A^{21} & A^{22} & A^{23} \\\\ A^{31} & A^{32} & A^{33} \end{bmatrix} \begin{bmatrix} B_{11} & B_{12} & B_{13} \\\\ B_{21} & B_{22} & B_{23} \\\\ B_{31} & B_{32} & B_{33} \end{bmatrix} =A_{4}B_{1}\\\ \\ %%%%%%%%%%%%%%%% C^{ij}=A^{i}_{\bullet k}B^{kj}=A^{ik}B_k^{\bullet j} \Longleftrightarrow C_{2}=[C^{ij}] =\begin{bmatrix} A^{1}_{\bullet 1} & A^{1}_{\bullet 2}& A^{1}_{\bullet 3} \\\\ A^{2}_{\bullet 1} & A^{2}_{\bullet 2}& A^{2}_{\bullet 3} \\\\ A^{3}_{\bullet 1} & A^{3}_{\bullet 2}& A^{3}_{\bullet 3} \end{bmatrix} \begin{bmatrix} B^{11} & B^{12} & B^{13} \\\\ B^{21} & B^{22} & B^{23} \\\\ B^{31} & B^{32} & B^{33} \end{bmatrix} =A_{3}B_{4} =\begin{bmatrix} A^{11} & A^{12} & A^{13} \\\\ A^{21} & A^{22} & A^{23} \\\\ A^{31} & A^{32} & A^{33} \end{bmatrix} \begin{bmatrix} B_{1}^{\bullet 1} & B_{2}^{\bullet 1} & B_{3}^{\bullet 1} \\\\ B_{1}^{\bullet 2} & B_{2}^{\bullet 2} & B_{3}^{\bullet 2} \\\\ B_{1}^{\bullet 3} & B_{2}^{\bullet 3} & B_{3}^{\bullet 3} \end{bmatrix}^T =A_{4}B_{2}^T %%%%%%%%%%%%%%%% Cij​=Ai∙k​Bkj​=Aik​B∙jk​⟺C1​=[Cij​]= ​A1∙1​A1∙2​A1∙3​​A2∙1​A2∙2​A2∙3​​A3∙1​A3∙2​A3∙3​​ ​T ​B11​B21​B31​​B12​B22​B32​​B13​B23​B33​​ ​=A2T​B1​= ​A11​A21​A31​​A12​A22​A32​​A13​A23​A33​​ ​ ​B∙11​B∙12​B∙13​​B∙21​B∙22​B∙23​​B∙31​B∙32​B∙33​​ ​=A1​B3​ Ci∙j​=Ai∙k​Bk∙j​=Aik​Bkj⟺C2​=[Ci∙j​]= ​A1∙1​A1∙2​A1∙3​​A2∙1​A2∙2​A2∙3​​A3∙1​A3∙2​A3∙3​​ ​ ​B1∙1​B1∙2​B1∙3​​B2∙1​B2∙2​B2∙3​​B3∙1​B3∙2​B3∙3​​ ​=A2​B2​= ​A11​A21​A31​​A12​A22​A32​​A13​A23​A33​​ ​ ​B11B21B31​B12B22B32​B13B23B33​ ​=A1​B4​ C∙ji​=A∙ki​B∙jk​=AikBkj​⟺C3​=[C∙ji​]= ​A∙11​A∙12​A∙13​​A∙21​A∙22​A∙23​​A∙31​A∙32​A∙33​​ ​ ​B∙11​B∙12​B∙13​​B∙21​B∙22​B∙23​​B∙31​B∙32​B∙33​​ ​=A3​B3​= ​A11A21A31​A12A22A32​A13A23A33​ ​ ​B11​B21​B31​​B12​B22​B32​​B13​B23​B33​​ ​=A4​B1​ Cij=A∙ki​Bkj=AikBk∙j​⟺C2​=[Cij]= ​A∙11​A∙12​A∙13​​A∙21​A∙22​A∙23​​A∙31​A∙32​A∙33​​ ​ ​B11B21B31​B12B22B32​B13B23B33​ ​=A3​B4​= ​A11A21A31​A12A22A32​A13A23A33​ ​ ​B1∙1​B1∙2​B1∙3​​B2∙1​B2∙2​B2∙3​​B3∙1​B3∙2​B3∙3​​ ​T=A4​B2T​
由两个二阶张量点乘与矩阵乘法的对应关系也可得到：张量点积不可随意更换次序，即
$$\mathbf{A}\bullet \mathbf{B}\ne \mathbf{B}\bullet \mathbf{A}$$另外，可借助张量点积与矩阵乘法的对应关系推导出如下类似于矩阵乘法与矩阵转置关系的张量点积与张量转置的关系式：
$$(A_3{B}_3{)}^T=(B_3{)}^T(A_3{)}^T=(B^T{)}_2(A^T{)}_2⟺(\mathbf{A}\bullet \mathbf{B}{)}^T={\mathbf{B}}^T\bullet {\mathbf{A}}^T$$张量点积的行列式与张量行列式的关系式：$$det(A_3{B}_3)=det(A_3)det(B_3)⟺det(\mathbf{A}\bullet \mathbf{B})=det(\mathbf{A})det(\mathbf{B})$$

2. 正则二阶张量与可逆矩阵

若二阶张量 $\mathbf{T}$ 的矩阵可逆，则称二阶张量正则，反之称作退化的二阶张量。
显然，二阶张量正则的充要条件为二阶张量的行列式不为零
则有：$\mathbf{T}$如果正则，那么 ${\mathbf{T}}^T$也正则。
另外，根据正则张量、张量点积与可逆矩阵、矩阵乘法的对应关系可知：对于正则的二阶张量$\mathbf{T}$，必唯一存在正则的二阶张量${\mathbf{T}}^{-1}$，使得：
$$\mathbf{T}\bullet {\mathbf{T}}^{-1}={\mathbf{T}}^{-1}\bullet \mathbf{T}=\mathbf{G}(1)$$将${\mathbf{T}}^{-1}$称作正则张量$\mathbf{T}$的逆张量，且逆张量的矩阵等于原张量矩阵的逆，即
$$({\tau }^{-1}{)}_3=({\tau }_3{)}^{-1}$$则，
$$det[({\tau }^{-1}{)}_3]=det[({\tau }_3{)}^{-1}]=\frac{1}{det({\tau }_3)}⟺det({\mathbf{T}}^{-1})=\frac{1}{det(\mathbf{T})}$$此外，进一步根据（1）式可得：
$$\begin{gathered} ({\mathbf{T}}^{-1}{)}^{-1}=\mathbf{T} \\ {\mathbf{G}}^T=(\mathbf{T}\bullet {\mathbf{T}}^{-1}{)}^T=({\mathbf{T}}^{-1}{)}^T\bullet {\mathbf{T}}^T=\mathbf{G}=({\mathbf{T}}^T{)}^{-1}\bullet {\mathbf{T}}^T \end{gathered}$$由逆张量的唯一性可知：
$$({\mathbf{T}}^T{)}^{-1}=({\mathbf{T}}^{-1}{)}^T$$若二阶张量线性变换可逆，则：
$$\begin{gathered} w={\tau }_{3}u⟺w=T\bullet u \\ u=({\tau }_3{)}^{-1}w=({\tau }^{-1}{)}_{3}w⟺u=T^{-1}\bullet w \end{gathered}$$文章来源地址https://www.toymoban.com/news/detail-444642.html

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