本文主要内容如下：

1. 二阶张量的矩阵表示

任意二阶张量含有九个分量，正好可以通过三阶方阵进行表示，但二阶张量具有四种不同的分量形式，不同的分量对应于不同的方阵：

τ 1 = [ T i j ] = [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ]   τ 2 = [ T i ∙ j ] = [ 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 ]   τ 3 = [ T ∙ j i ] = [ 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 ]   τ 4 = [ T i j ] = [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] %%%%%%%%%%% \tau_1=[T_{ij}]= \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ \\ T_{21} & T_{22} & T_{23} \\ \\ T_{31} & T_{32} & T_{33} \end{bmatrix} \\\ \\ %%%%%%%%%%%% \tau_2=[T_{i}^{\bullet j}]= \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} \\\ \\ %%%%%%%%%%%% \tau_3=[T^{i}_{\bullet j}]= \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} \\\ \\ %%%%%%%%%%%% \tau_4=[T^{ij}]= \begin{bmatrix} T^{11} & T^{12} & T^{13} \\ \\ T^{21} & T^{22} & T^{23} \\ \\ T^{31} & T^{32} & T^{33} \end{bmatrix} τ1​=[Tij​]=⎣ ⎡​T11​T21​T31​​T12​T22​T32​​T13​T23​T33​​⎦ ⎤​ τ2​=[Ti∙j​]=⎣ ⎡​T1∙1​T1∙2​T1∙3​​T2∙1​T2∙2​T2∙3​​T3∙1​T3∙2​T3∙3​​⎦ ⎤​ τ3​=[T∙ji​]=⎣ ⎡​T∙11​T∙12​T∙13​​T∙21​T∙22​T∙23​​T∙31​T∙32​T∙33​​⎦ ⎤​ τ4​=[Tij]=⎣ ⎡​T11T21T31​T12T22T32​T13T23T33​⎦ ⎤​

注：本文采用前（上）指标为行编号，后（下）指标为列编号的对应规则，而在《张量分析》-黄克智中采用前指标为行编号的对应规则，两种规则的区别仅在于 τ 2 \tau_{2} τ2​的不同，两种规则得到的 τ 2 \tau_{2} τ2​互为矩阵转置的关系。

特别地，度量张量也是二阶张量，其矩阵形式为:
G 1 = [ g i j ] = [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ]   G 2 = G 3 = [ δ j i ] = [ 1 0 0 0 1 0 0 0 1 ] = E   G 4 = [ g i j ] = [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] = G 1 − 1 \mathscr{G}_{1}= [g_{ij}]= \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix}\\\ \\ %%%%%%%%%%%% \mathscr{G}_{2}= \mathscr{G}_{3}= [\delta^i_j]= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =E\\\ \\ %%%%%%%%%%%% \mathscr{G}_{4}= [g^{ij}]= \begin{bmatrix} g^{11} & g^{12} & g^{13} \\\ \\ g^{21} & g^{22} & g^{23} \\\ \\ g^{31} & g^{32} & g^{33} \end{bmatrix} =\mathscr{G}_{1}^{-1} G1​=[gij​]=⎣ ⎡​g11​ g21​ g31​​g12​g22​g32​​g13​g23​g33​​⎦ ⎤​ G2​=G3​=[δji​]=⎣ ⎡​100​010​001​⎦ ⎤​=E G4​=[gij]=⎣ ⎡​g11 g21 g31​g12g22g32​g13g23g33​⎦ ⎤​=G1−1​注意到，在一般坐标系下，二阶张量对应的四个矩阵是不相等的，它们之间的转换通过指标升降关系的矩阵形式来实现：
[ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] =   [ 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 ] T [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] =   [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] [ 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 ] =   [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] [ T 11 T 12 T 13 T 21 T 22 T 23 T 31 T 32 T 33 ] [ g 11 g 12 g 13   g 21 g 22 g 23   g 31 g 32 g 33 ] \begin{bmatrix} T_{11} & T_{12} & T_{13} \\ \\ T_{21} & T_{22} & T_{23} \\ \\ T_{31} & T_{32} & T_{33} \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}^T \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} =\ \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \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} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} \begin{bmatrix} T^{11} & T^{12} & T^{13} \\ \\ T^{21} & T^{22} & T^{23} \\ \\ T^{31} & T^{32} & T^{33} \end{bmatrix} \begin{bmatrix} g_{11} & g_{12} & g_{13} \\\ \\ g_{21} & g_{22} & g_{23} \\\ \\ g_{31} & g_{32} & g_{33} \end{bmatrix} ⎣ ⎡​T11​T21​T31​​T12​T22​T32​​T13​T23​T33​​⎦ ⎤​= ⎣ ⎡​T1∙1​T1∙2​T1∙3​​T2∙1​T2∙2​T2∙3​​T3∙1​T3∙2​T3∙3​​⎦ ⎤​T⎣ ⎡​g11​ g21​ g31​​g12​g22​g32​​g13​g23​g33​​⎦ ⎤​= ⎣ ⎡​g11​ g21​ g31​​g12​g22​g32​​g13​g23​g33​​⎦ ⎤​⎣ ⎡​T∙11​T∙12​T∙13​​T∙21​T∙22​T∙23​​T∙31​T∙32​T∙33​​⎦ ⎤​= ⎣ ⎡​g11​ g21​ g31​​g12​g22​g32​​g13​g23​g33​​⎦ ⎤​⎣ ⎡​T11T21T31​T12T22T32​T13T23T33​⎦ ⎤​⎣ ⎡​g11​ g21​ g31​​g12​g22​g32​​g13​g23​g33​​⎦ ⎤​即:
τ 1 = τ 2 T G 1 = G 1 τ 3 = G 1 τ 4 G 1 \tau_{1} =\tau_{2}^T\mathscr{G}_{1} =\mathscr{G}_{1}\tau_{3} =\mathscr{G}_{1}\tau_{4}\mathscr{G}_{1} τ1​=τ2T​G1​=G1​τ3​=G1​τ4​G1​那么：
G 1 − 1 τ 2 T G 1 = τ 3 ⟹ τ 2 T ∼ τ 3 \mathscr{G}_{1}^{-1}\tau_{2}^T\mathscr{G}_{1}=\tau_3\Longrightarrow \tau_{2}^T\sim\tau_{3} G1−1​τ2T​G1​=τ3​⟹τ2T​∼τ3​在笛卡尔坐标系中，由于 G 1 = E \mathscr{G}_{1}=E G1​=E 故在笛卡尔坐标系中有： τ 1 = τ 2 T = τ 3 = τ 4 \tau_{1}=\tau_{2}^T=\tau_{3}=\tau_{4} τ1​=τ2T​=τ3​=τ4​通常如不加说明，定义 τ 3 \tau_{3} τ3​为张量的矩阵。

2. 二阶张量转置与矩阵转置

二阶转置张量的分量形式为：
( T T ) i j = T j i ⟹ ( τ T ) 1 = [ ( T T ) i j ] = [ ( T T ) 11 ( T T ) 12 ( T T ) 13 ( T T ) 21 ( T T ) 22 ( T T ) 23 ( T T ) 31 ( T T ) 32 ( T T ) 33 ] = [ T 11 T 21 T 31 T 12 T 22 T 32 T 13 T 23 T 33 ] = ( τ 1 ) T   ( T T ) i ∙ j = T ∙ i j ⟹ ( τ T ) 2 = [ ( T T ) i ∙ j ] = [ ( T T ) 1 ∙ 1 ( T T ) 2 ∙ 1 ( T T ) 3 ∙ 1 ( T T ) 1 ∙ 2 ( T T ) 2 ∙ 2 ( T T ) 3 ∙ 2 ( T T ) 1 ∙ 3 ( T T ) 2 ∙ 3 ( T T ) 3 ∙ 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 ] = τ 3   ( T T ) ∙ j i = T j ∙ i ⟹ ( τ T ) 3 = [ ( T T ) ∙ j i ] = [ ( T T ) ∙ 1 1 ( T T ) ∙ 2 1 ( T T ) ∙ 3 1 ( T T ) ∙ 1 2 ( T T ) ∙ 2 2 ( T T ) ∙ 3 2 ( T T ) ∙ 2 3 ( T T ) ∙ 2 3 ( T T ) ∙ 3 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 ] = τ 2   ( T T ) i j = T j i ⟹ ( τ T ) 4 = [ ( T T ) i j ] = [ ( T T ) 11 ( T T ) 12 ( T T ) 13 ( T T ) 21 ( T T ) 22 ( T T ) 23 ( T T ) 31 ( T T ) 32 ( T T ) 33 ] = [ T 11 T 21 T 31 T 12 T 22 T 32 T 13 T 23 T 33 ] = ( τ 4 ) T (T^{T})_{ij}=T_{ji} \Longrightarrow (\tau^T)_{1}=[(T^{T})_{ij}] =\begin{bmatrix} (T^{T})_{11} & (T^{T})_{12} & (T^{T})_{13} \\ \\ (T^{T})_{21} & (T^{T})_{22} & (T^{T})_{23} \\ \\ (T^{T})_{31} & (T^{T})_{32} & (T^{T})_{33} \end{bmatrix} =\begin{bmatrix} T_{11} & T_{21} & T_{31} \\ \\ T_{12} & T_{22} & T_{32} \\ \\ T_{13} & T_{23} & T_{33} \end{bmatrix} =(\tau_{1})^T\\\ \\ %%%%%%%%%%%% (T^{T})_{i}^{\bullet j}=T^{j}_{\bullet i} \Longrightarrow (\tau^T)_{2}=[(T^{T})_{i}^{\bullet j}] =\begin{bmatrix} (T^{T})_{1}^{\bullet 1} & (T^{T})_{2}^{\bullet 1} & (T^{T})_{3}^{\bullet 1} \\ \\ (T^{T})_{1}^{\bullet 2} & (T^{T})_{2}^{\bullet 2} & (T^{T})_{3}^{\bullet 2} \\ \\ (T^{T})_{1}^{\bullet 3} & (T^{T})_{2}^{\bullet 3} & (T^{T})_{3}^{\bullet 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} =\tau_{3}\\\ \\ %%%%%%%%%%%% (T^{T})^{i}_{\bullet j}=T_{j}^{\bullet i} \Longrightarrow (\tau^T)_{3}=[(T^{T})^{i}_{\bullet j}] =\begin{bmatrix} (T^{T})^{1}_{\bullet 1} & (T^{T})^{1}_{\bullet 2} & (T^{T})^{1}_{\bullet 3} \\ \\ (T^{T})^{2}_{\bullet 1} & (T^{T})^{2}_{\bullet 2} & (T^{T})^{2}_{\bullet 3} \\ \\ (T^{T})^{3}_{\bullet 2} & (T^{T})^{3}_{\bullet 2} & (T^{T})^{3}_{\bullet 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} =\tau_{2} \\\ \\ %%%%%%%%%%%% (T^{T})^{ij}=T^{ji} \Longrightarrow (\tau^T)_{4}=[(T^{T})^{ij}] =\begin{bmatrix} (T^{T})^{11} & (T^{T})^{12} & (T^{T})^{13} \\ \\ (T^{T})^{21} & (T^{T})^{22} & (T^{T})^{23} \\ \\ (T^{T})^{31} & (T^{T})^{32} & (T^{T})^{33} \end{bmatrix} =\begin{bmatrix} T^{11} & T^{21} & T^{31} \\ \\ T^{12} & T^{22} & T^{32} \\ \\ T^{13} & T^{23} & T^{33} \end{bmatrix} =(\tau_{4})^T (TT)ij​=Tji​⟹(τT)1​=[(TT)ij​]=⎣ ⎡​(TT)11​(TT)21​(TT)31​​(TT)12​(TT)22​(TT)32​​(TT)13​(TT)23​(TT)33​​⎦ ⎤​=⎣ ⎡​T11​T12​T13​​T21​T22​T23​​T31​T32​T33​​⎦ ⎤​=(τ1​)T (TT)i∙j​=T∙ij​⟹(τT)2​=[(TT)i∙j​]=⎣ ⎡​(TT)1∙1​(TT)1∙2​(TT)1∙3​​(TT)2∙1​(TT)2∙2​(TT)2∙3​​(TT)3∙1​(TT)3∙2​(TT)3∙3​​⎦ ⎤​=⎣ ⎡​T∙11​T∙12​T∙13​​T∙21​T∙22​T∙23​​T∙31​T∙32​T∙33​​⎦ ⎤​=τ3​ (TT)∙ji​=Tj∙i​⟹(τT)3​=[(TT)∙ji​]=⎣ ⎡​(TT)∙11​(TT)∙12​(TT)∙23​​(TT)∙21​(TT)∙22​(TT)∙23​​(TT)∙31​(TT)∙32​(TT)∙33​​⎦ ⎤​=⎣ ⎡​T1∙1​T1∙2​T1∙3​​T2∙1​T2∙2​T2∙3​​T3∙1​T3∙2​T3∙3​​⎦ ⎤​=τ2​ (TT)ij=Tji⟹(τT)4​=[(TT)ij]=⎣ ⎡​(TT)11(TT)21(TT)31​(TT)12(TT)22(TT)32​(TT)13(TT)23(TT)33​⎦ ⎤​=⎣ ⎡​T11T12T13​T21T22T23​T31T32T33​⎦ ⎤​=(τ4​)T这说明若两个张量互为转置，则它们的 τ 1 、 τ 4 \tau_1、\tau_4 τ1​、τ4​ 矩阵也互为转置，而转置张量的 τ 2 \tau_2 τ2​ 矩阵与原张量的 τ 3 \tau_3 τ3​ 矩阵相等，转置张量的 τ 3 \tau_3 τ3​ 矩阵与原张量的 τ 2 \tau_2 τ2​ 矩阵相等。

  • 更特别地，对于对称张量 N \bold{N} N有：
    ( N T ) i j = N i j ⟹ ( N 1 ) T = ( N T ) 1 = [ ( N T ) i j ] = [ ( N T ) 11 ( N T ) 12 ( N T ) 13 ( N T ) 21 ( N T ) 22 ( N T ) 23 ( N T ) 31 ( N T ) 32 ( N T ) 33 ] = [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] = N 1   ( N T ) i ∙ j = N i ∙ j ⟹ N 3 = ( N T ) 2 = [ ( N T ) i ∙ j ] = [ ( N T ) 1 ∙ 1 ( N T ) 2 ∙ 1 ( N T ) 3 ∙ 1 ( N T ) 1 ∙ 2 ( N T ) 2 ∙ 2 ( N T ) 3 ∙ 2 ( N T ) 1 ∙ 3 ( N T ) 2 ∙ 3 ( N T ) 3 ∙ 3 ] = [ N 1 ∙ 1 N 2 ∙ 1 N 3 ∙ 1 N 1 ∙ 2 N 2 ∙ 2 N 3 ∙ 2 N 1 ∙ 3 N 2 ∙ 3 N 3 ∙ 3 ] = N 2   ( N T ) ∙ j i = N ∙ j i ⟹ N 2 = ( N T ) 3 = [ ( N T ) ∙ j i ] = [ ( N T ) ∙ 1 1 ( N T ) ∙ 2 1 ( N T ) ∙ 3 1 ( N T ) ∙ 1 2 ( N T ) ∙ 2 2 ( N T ) ∙ 3 2 ( N T ) ∙ 2 3 ( N T ) ∙ 2 3 ( N T ) ∙ 3 3 ] = [ N ∙ 1 1 N ∙ 2 1 N ∙ 3 1 N ∙ 1 2 N ∙ 2 2 N ∙ 3 2 N ∙ 2 3 N ∙ 2 3 N ∙ 3 3 ] = N 3   ( N T ) i j = N i j ⟹ ( N 4 ) T = ( N T ) 4 = [ ( N T ) i j ] = [ ( N T ) 11 ( N T ) 12 ( N T ) 13 ( N T ) 21 ( N T ) 22 ( N T ) 23 ( N T ) 31 ( N T ) 32 ( N T ) 33 ] = [ N 11 N 12 N 13 N 21 N 22 N 23 N 31 N 32 N 33 ] = N 4 (N^T)_{ij}=N_{ij} \Longrightarrow (N_{1})^T=(N^T)_{1}=[(N^{T})_{ij}] =\begin{bmatrix} (N^{T})_{11} & (N^{T})_{12} & (N^{T})_{13} \\ \\ (N^{T})_{21} & (N^{T})_{22} & (N^{T})_{23} \\ \\ (N^{T})_{31} & (N^{T})_{32} & (N^{T})_{33} \end{bmatrix} =\begin{bmatrix} N_{11} & N_{12} & N_{13} \\ \\ N_{21} & N_{22} & N_{23} \\ \\ N_{31} & N_{32} & N_{33} \end{bmatrix} =N_1\\\ \\ %%%%%%%%%%%% (N^{T})_{i}^{\bullet j}=N_{i}^{\bullet j} \Longrightarrow N_{3}=(N^T)_{2}=[(N^{T})_{i}^{\bullet j}] =\begin{bmatrix} (N^{T})_{1}^{\bullet 1} & (N^{T})_{2}^{\bullet 1} & (N^{T})_{3}^{\bullet 1} \\ \\ (N^{T})_{1}^{\bullet 2} & (N^{T})_{2}^{\bullet 2} & (N^{T})_{3}^{\bullet 2} \\ \\ (N^{T})_{1}^{\bullet 3} & (N^{T})_{2}^{\bullet 3} & (N^{T})_{3}^{\bullet 3} \end{bmatrix} =\begin{bmatrix} N_{1}^{\bullet 1} & N_{2}^{\bullet 1} & N_{3}^{\bullet 1} \\ \\ N_{1}^{\bullet 2} & N_{2}^{\bullet 2} & N_{3}^{\bullet 2} \\ \\ N_{1}^{\bullet 3} & N_{2}^{\bullet 3} & N_{3}^{\bullet 3} \end{bmatrix} =N_2 \\\ \\ %%%%%%%%%%%% (N^{T})^{i}_{\bullet j}=N^{i}_{\bullet j} \Longrightarrow N_2=(N^T)_{3}=[(N^{T})^{i}_{\bullet j}] =\begin{bmatrix} (N^{T})^{1}_{\bullet 1} & (N^{T})^{1}_{\bullet 2} & (N^{T})^{1}_{\bullet 3} \\ \\ (N^{T})^{2}_{\bullet 1} & (N^{T})^{2}_{\bullet 2} & (N^{T})^{2}_{\bullet 3} \\ \\ (N^{T})^{3}_{\bullet 2} & (N^{T})^{3}_{\bullet 2} & (N^{T})^{3}_{\bullet 3} \end{bmatrix} =\begin{bmatrix} N^{1}_{\bullet 1} & N^{1}_{\bullet 2} & N^{1}_{\bullet 3} \\ \\ N^{2}_{\bullet 1} & N^{2}_{\bullet 2} & N^{2}_{\bullet 3} \\ \\ N^{3}_{\bullet 2} & N^{3}_{\bullet 2} & N^{3}_{\bullet 3} \end{bmatrix} =N_{3} \\\ \\ %%%%%%%%%%%% (N^{T})^{ij}=N^{ij} \Longrightarrow (N_4)^T=(N^T)_{4}=[(N^{T})^{ij}] =\begin{bmatrix} (N^{T})^{11} & (N^{T})^{12} & (N^{T})^{13} \\ \\ (N^{T})^{21} & (N^{T})^{22} & (N^{T})^{23} \\ \\ (N^{T})^{31} & (N^{T})^{32} & (N^{T})^{33} \end{bmatrix} =\begin{bmatrix} N^{11} & N^{12} & N^{13} \\ \\ N^{21} & N^{22} & N^{23} \\ \\ N^{31} & N^{32} & N^{33} \end{bmatrix} =N_{4} (NT)ij​=Nij​⟹(N1​)T=(NT)1​=[(NT)ij​]=⎣ ⎡​(NT)11​(NT)21​(NT)31​​(NT)12​(NT)22​(NT)32​​(NT)13​(NT)23​(NT)33​​⎦ ⎤​=⎣ ⎡​N11​N21​N31​​N12​N22​N32​​N13​N23​N33​​⎦ ⎤​=N1​ (NT)i∙j​=Ni∙j​⟹N3​=(NT)2​=[(NT)i∙j​]=⎣ ⎡​(NT)1∙1​(NT)1∙2​(NT)1∙3​​(NT)2∙1​(NT)2∙2​(NT)2∙3​​(NT)3∙1​(NT)3∙2​(NT)3∙3​​⎦ ⎤​=⎣ ⎡​N1∙1​N1∙2​N1∙3​​N2∙1​N2∙2​N2∙3​​N3∙1​N3∙2​N3∙3​​⎦ ⎤​=N2​ (NT)∙ji​=N∙ji​⟹N2​=(NT)3​=[(NT)∙ji​]=⎣ ⎡​(NT)∙11​(NT)∙12​(NT)∙23​​(NT)∙21​(NT)∙22​(NT)∙23​​(NT)∙31​(NT)∙32​(NT)∙33​​⎦ ⎤​=⎣ ⎡​N∙11​N∙12​N∙23​​N∙21​N∙22​N∙23​​N∙31​N∙32​N∙33​​⎦ ⎤​=N3​ (NT)ij=Nij⟹(N4​)T=(NT)4​=[(NT)ij]=⎣ ⎡​(NT)11(NT)21(NT)31​(NT)12(NT)22(NT)32​(NT)13(NT)23(NT)33​⎦ ⎤​=⎣ ⎡​N11N21N31​N12N22N32​N13N23N33​⎦ ⎤​=N4​可见，对称二阶张量 N \bold{N} N的 N 1 、 N 4 N_1、N_4 N1​、N4​ 矩阵为对称矩阵且 N 2 = N 3 N_2=N_3 N2​=N3​，但 N 2 、 N 3 N_2、N_3 N2​、N3​ 一般并不是对称矩阵。
  • 对于反对称张量 Ω \bold{\Omega} Ω有：
    ( Ω T ) i j = − Ω i j ⟹ ( Ω 1 ) T = ( Ω T ) 1 = [ ( Ω T ) i j ] = [ ( Ω T ) 11 ( Ω T ) 12 ( Ω T ) 13 ( Ω T ) 21 ( Ω T ) 22 ( Ω T ) 23 ( Ω T ) 31 ( Ω T ) 32 ( Ω T ) 33 ] = − [ Ω 11 Ω 12 Ω 13 Ω 21 Ω 22 Ω 23 Ω 31 Ω 32 Ω 33 ] = − Ω 1   ( Ω T ) i ∙ j = − Ω i ∙ j ⟹ Ω 3 = ( Ω T ) 2 = [ ( Ω T ) i ∙ j ] = [ ( Ω T ) 1 ∙ 1 ( Ω T ) 2 ∙ 1 ( Ω T ) 3 ∙ 1 ( Ω T ) 1 ∙ 2 ( Ω T ) 2 ∙ 2 ( Ω T ) 3 ∙ 2 ( Ω T ) 1 ∙ 3 ( Ω T ) 2 ∙ 3 ( Ω T ) 3 ∙ 3 ] = − [ Ω 1 ∙ 1 Ω 2 ∙ 1 Ω 3 ∙ 1 Ω 1 ∙ 2 Ω 2 ∙ 2 Ω 3 ∙ 2 Ω 1 ∙ 3 Ω 2 ∙ 3 Ω 3 ∙ 3 ] = − Ω 2   ( Ω T ) ∙ j i = − Ω ∙ j i ⟹ Ω 2 = ( Ω T ) 3 = [ ( Ω T ) ∙ j i ] = [ ( Ω T ) ∙ 1 1 ( Ω T ) ∙ 2 1 ( Ω T ) ∙ 3 1 ( Ω T ) ∙ 1 2 ( Ω T ) ∙ 2 2 ( Ω T ) ∙ 3 2 ( Ω T ) ∙ 2 3 ( Ω T ) ∙ 2 3 ( Ω T ) ∙ 3 3 ] = − [ Ω ∙ 1 1 Ω ∙ 2 1 Ω ∙ 3 1 Ω ∙ 1 2 Ω ∙ 2 2 Ω ∙ 3 2 Ω ∙ 2 3 Ω ∙ 2 3 Ω ∙ 3 3 ] = − Ω 3   ( Ω T ) i j = − Ω i j ⟹ ( Ω 4 ) T = ( Ω T ) 4 = [ ( Ω T ) i j ] = [ ( Ω T ) 11 ( Ω T ) 12 ( Ω T ) 13 ( Ω T ) 21 ( Ω T ) 22 ( Ω T ) 23 ( Ω T ) 31 ( Ω T ) 32 ( Ω T ) 33 ] = − [ Ω 11 Ω 12 Ω 13 Ω 21 Ω 22 Ω 23 Ω 31 Ω 32 Ω 33 ] = − Ω 4 ({\Omega}^T)_{ij}=-{\Omega}_{ij} \Longrightarrow ({\Omega}_{1})^T=({\Omega}^T)_{1}=[({\Omega}^{T})_{ij}] =\begin{bmatrix} ({\Omega}^{T})_{11} & ({\Omega}^{T})_{12} & ({\Omega}^{T})_{13} \\ \\ ({\Omega}^{T})_{21} & ({\Omega}^{T})_{22} & ({\Omega}^{T})_{23} \\ \\ ({\Omega}^{T})_{31} & ({\Omega}^{T})_{32} & ({\Omega}^{T})_{33} \end{bmatrix} =-\begin{bmatrix} {\Omega}_{11} & {\Omega}_{12} & {\Omega}_{13} \\ \\ {\Omega}_{21} & {\Omega}_{22} & {\Omega}_{23} \\ \\ {\Omega}_{31} & {\Omega}_{32} & {\Omega}_{33} \end{bmatrix} =-{\Omega}_1\\\ \\ %%%%%%%%%%%% ({\Omega}^{T})_{i}^{\bullet j}=-{\Omega}_{i}^{\bullet j} \Longrightarrow {\Omega}_{3}=({\Omega}^T)_{2}=[({\Omega}^{T})_{i}^{\bullet j}] =\begin{bmatrix} ({\Omega}^{T})_{1}^{\bullet 1} & ({\Omega}^{T})_{2}^{\bullet 1} & ({\Omega}^{T})_{3}^{\bullet 1} \\ \\ ({\Omega}^{T})_{1}^{\bullet 2} & ({\Omega}^{T})_{2}^{\bullet 2} & ({\Omega}^{T})_{3}^{\bullet 2} \\ \\ ({\Omega}^{T})_{1}^{\bullet 3} & ({\Omega}^{T})_{2}^{\bullet 3} & ({\Omega}^{T})_{3}^{\bullet 3} \end{bmatrix} =-\begin{bmatrix} {\Omega}_{1}^{\bullet 1} & {\Omega}_{2}^{\bullet 1} & {\Omega}_{3}^{\bullet 1} \\ \\ {\Omega}_{1}^{\bullet 2} & {\Omega}_{2}^{\bullet 2} & {\Omega}_{3}^{\bullet 2} \\ \\ {\Omega}_{1}^{\bullet 3} & {\Omega}_{2}^{\bullet 3} & {\Omega}_{3}^{\bullet 3} \end{bmatrix} =-{\Omega}_2 \\\ \\ %%%%%%%%%%%% ({\Omega}^{T})^{i}_{\bullet j}=-{\Omega}^{i}_{\bullet j} \Longrightarrow {\Omega}_2=({\Omega}^T)_{3}=[({\Omega}^{T})^{i}_{\bullet j}] =\begin{bmatrix} ({\Omega}^{T})^{1}_{\bullet 1} & ({\Omega}^{T})^{1}_{\bullet 2} & ({\Omega}^{T})^{1}_{\bullet 3} \\ \\ ({\Omega}^{T})^{2}_{\bullet 1} & ({\Omega}^{T})^{2}_{\bullet 2} & ({\Omega}^{T})^{2}_{\bullet 3} \\ \\ ({\Omega}^{T})^{3}_{\bullet 2} & ({\Omega}^{T})^{3}_{\bullet 2} & ({\Omega}^{T})^{3}_{\bullet 3} \end{bmatrix} =-\begin{bmatrix} {\Omega}^{1}_{\bullet 1} & {\Omega}^{1}_{\bullet 2} & {\Omega}^{1}_{\bullet 3} \\ \\ {\Omega}^{2}_{\bullet 1} & {\Omega}^{2}_{\bullet 2} & {\Omega}^{2}_{\bullet 3} \\ \\ {\Omega}^{3}_{\bullet 2} & {\Omega}^{3}_{\bullet 2} & {\Omega}^{3}_{\bullet 3} \end{bmatrix} =-{\Omega}_{3} \\\ \\ %%%%%%%%%%%% ({\Omega}^{T})^{ij}=-{\Omega}^{ij} \Longrightarrow ({\Omega}_4)^T=({\Omega}^T)_{4}=[({\Omega}^{T})^{ij}] =\begin{bmatrix} ({\Omega}^{T})^{11} & ({\Omega}^{T})^{12} & ({\Omega}^{T})^{13} \\ \\ ({\Omega}^{T})^{21} & ({\Omega}^{T})^{22} & ({\Omega}^{T})^{23} \\ \\ ({\Omega}^{T})^{31} & ({\Omega}^{T})^{32} & ({\Omega}^{T})^{33} \end{bmatrix} =-\begin{bmatrix} {\Omega}^{11} & {\Omega}^{12} & {\Omega}^{13} \\ \\ {\Omega}^{21} & {\Omega}^{22} & {\Omega}^{23} \\ \\ {\Omega}^{31} & {\Omega}^{32} & {\Omega}^{33} \end{bmatrix} =-{\Omega}_{4} (ΩT)ij​=−Ωij​⟹(Ω1​)T=(ΩT)1​=[(ΩT)ij​]=⎣ ⎡​(ΩT)11​(ΩT)21​(ΩT)31​​(ΩT)12​(ΩT)22​(ΩT)32​​(ΩT)13​(ΩT)23​(ΩT)33​​⎦ ⎤​=−⎣ ⎡​Ω11​Ω21​Ω31​​Ω12​Ω22​Ω32​​Ω13​Ω23​Ω33​​⎦ ⎤​=−Ω1​ (ΩT)i∙j​=−Ωi∙j​⟹Ω3​=(ΩT)2​=[(ΩT)i∙j​]=⎣ ⎡​(ΩT)1∙1​(ΩT)1∙2​(ΩT)1∙3​​(ΩT)2∙1​(ΩT)2∙2​(ΩT)2∙3​​(ΩT)3∙1​(ΩT)3∙2​(ΩT)3∙3​​⎦ ⎤​=−⎣ ⎡​Ω1∙1​Ω1∙2​Ω1∙3​​Ω2∙1​Ω2∙2​Ω2∙3​​Ω3∙1​Ω3∙2​Ω3∙3​​⎦ ⎤​=−Ω2​ (ΩT)∙ji​=−Ω∙ji​⟹Ω2​=(ΩT)3​=[(ΩT)∙ji​]=⎣ ⎡​(ΩT)∙11​(ΩT)∙12​(ΩT)∙23​​(ΩT)∙21​(ΩT)∙22​(ΩT)∙23​​(ΩT)∙31​(ΩT)∙32​(ΩT)∙33​​⎦ ⎤​=−⎣ ⎡​Ω∙11​Ω∙12​Ω∙23​​Ω∙21​Ω∙22​Ω∙23​​Ω∙31​Ω∙32​Ω∙33​​⎦ ⎤​=−Ω3​ (ΩT)ij=−Ωij⟹(Ω4​)T=(ΩT)4​=[(ΩT)ij]=⎣ ⎡​(ΩT)11(ΩT)21(ΩT)31​(ΩT)12(ΩT)22(ΩT)32​(ΩT)13(ΩT)23(ΩT)33​⎦ ⎤​=−⎣ ⎡​Ω11Ω21Ω31​Ω12Ω22Ω32​Ω13Ω23Ω33​⎦ ⎤​=−Ω4​可见，反对称二阶张量 Ω \bold{{\Omega}} Ω的 Ω 1 、 Ω 4 {\Omega}_1、{\Omega}_4 Ω1​、Ω4​ 矩阵为反对称矩阵且 Ω 2 = − Ω 3 {\Omega}_2=-{\Omega}_3 Ω2​=−Ω3​，但 Ω 2 、 Ω 3 {\Omega}_2、{\Omega}_3 Ω2​、Ω3​ 一般并不是反对称矩阵。

3. 二阶张量的行列式与矩阵的行列式

二阶张量对应的四种矩阵具有不同的行列式值，根据四种矩阵间的转换关系，有：
d e t ( τ 1 ) = g   d e t ( τ 2 ) = g   d e t ( τ 3 ) = g 2   d e t ( τ 4 ) ( 1 ) det(\tau_{1})=g\ det(\tau_{2})=g\ det(\tau_{3})=g^2\ det(\tau_{4})\qquad(1) det(τ1​)=g det(τ2​)=g det(τ3​)=g2 det(τ4​)(1)通常，定义张量 T \bold{T} T的行列式为 d e t ( T ) = d e t ( τ 3 ) det(\bold{T})=det(\tau_3) det(T)=det(τ3​)，根据行列式的定义及(1)式可知：互为转置的张量行列式相等，即 d e t ( T T ) = d e t ( T ) det(\bold{T^T})=det(\bold{T}) det(TT)=det(T)
二阶张量行列式的几何意义：通过 T 对矢量组进行映射，映射前后矢量组所构成的平行六面体的体积比为 d e t ( T ) det(\bold T) det(T)。证明如下：

由行列式交换行/列与行列式的置换符号表述关系知：
T ∙ p i T ∙ q j T ∙ l k e i j k = T ∙ 1 i T ∙ 2 j T ∙ 3 k e i j k e p q l = d e t ( [ T ∙ j i ] ) e p q l T^{i}_{\bullet p}T^{j}_{\bullet q}T^{k}_{\bullet l}e_{ijk} =T^{i}_{\bullet 1}T^{j}_{\bullet 2}T^{k}_{\bullet 3}e_{ijk}e_{pql} =det([T^{i}_{\bullet j}])e_{pql} T∙pi​T∙qj​T∙lk​eijk​=T∙1i​T∙2j​T∙3k​eijk​epql​=det([T∙ji​])epql​那么，
[ T ∙ u ⃗ T ∙ v ⃗ T ∙ w ⃗ ]   = ( T ∙ p i u p ) ( T ∙ q j v q ) ( T ∙ l k w l ) ϵ i j k   = ( T ∙ p i T ∙ q j T ∙ l k e i j k ) ( u p v q w l ) g   = d e t ( [ T ∙ j i ] ) ( u p v q w l g e p q l )   = d e t ( [ T ∙ j i ] ) ( u p v q w l ϵ p q l )   = d e t ( τ 3 ) [ u ⃗ v ⃗ w ⃗ ] \begin{aligned} &\quad\begin{bmatrix}T\bullet\vec{u} & T\bullet\vec{v} & T\bullet\vec{w}\end{bmatrix}\\\ \\ &=(T^{i}_{\bullet p}u^p)(T^{j}_{\bullet q}v^q)(T^{k}_{\bullet l}w^l)\epsilon_{ijk}\\\ \\ &=(T^{i}_{\bullet p}T^{j}_{\bullet q}T^{k}_{\bullet l}e_{ijk})(u^pv^qw^l)\sqrt{g}\\\ \\ &=det([T^{i}_{\bullet j}])(u^pv^qw^l\sqrt{g}e_{pql})\\\ \\ &=det([T^{i}_{\bullet j}])(u^pv^qw^l\epsilon_{pql})\\\ \\ &=det(\tau_3)\begin{bmatrix}\vec{u} & \vec{v} & \vec{w}\end{bmatrix} \end{aligned}      ​[T∙u ​T∙v ​T∙w ​]=(T∙pi​up)(T∙qj​vq)(T∙lk​wl)ϵijk​=(T∙pi​T∙qj​T∙lk​eijk​)(upvqwl)g ​=det([T∙ji​])(upvqwlg ​epql​)=det([T∙ji​])(upvqwlϵpql​)=det(τ3​)[u ​v ​w ​]​

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