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

$$\begin{gathered} \vec{w}=\mathbf{T}\bullet \vec{u}⟺w^i=T_{\bullet j}^i{u}^j⟺\left[\begin{array}{c}{w}^1\\ \\ w^2\\ \\ w^3\end{array}\right]=\left[\begin{array}{ccc}{T}_{\bullet 1}^1& T_{\bullet 2}^1& T_{\bullet 3}^1\\ & & \\ T_{\bullet 1}^2& T_{\bullet 2}^2& T_{\bullet 3}^2\\ & & \\ T_{\bullet 1}^3& T_{\bullet 2}^3& T_{\bullet 3}^3\end{array}\right]\left[\begin{array}{c}{u}^1\\ \\ u^2\\ \\ u^3\end{array}\right]⟺\vec{w}={\tau }_3\vec{u} \\ \vec{t}=\vec{u}\bullet \mathbf{T}⟺t^i=u^j{T}_j^{\bullet i}⟺\left[\begin{array}{c}{t}^1\\ \\ t^2\\ \\ t^3\end{array}\right]=\left[\begin{array}{ccc}{T}_1^{\bullet 1}& T_2^{\bullet 1}& T_3^{\bullet 1}\\ & & \\ T_1^{\bullet 2}& T_2^{\bullet 2}& T_3^{\bullet 2}\\ & & \\ T_1^{\bullet 3}& T_2^{\bullet 3}& T_3^{\bullet 3}\end{array}\right]\left[\begin{array}{c}{u}^1\\ \\ u^2\\ \\ u^3\end{array}\right]⟺\vec{t}={\tau }_2\vec{u} \end{gathered}$$显然，二阶张量与矢量的左、右点积一般不等：$$\mathbf{T}\bullet \vec{u}\ne \vec{u}\bullet \mathbf{T}$$且有：
$$({\tau }^T{)}_3\vec{u}={\tau }_2\vec{u}⟺{\mathbf{T}}^T\bullet \vec{u}=\vec{u}\bullet \mathbf{T}$$那么有：
$$\begin{gathered} \mathbf{N}\bullet \vec{u}=\vec{u}\bullet \mathbf{N}(\mathbf{N}为对称二阶张量) \\ \mathbf{\Omega }\bullet \vec{u}=-\vec{u}\bullet \mathbf{\Omega }(\mathbf{\Omega }为反对称二阶张量) \end{gathered}$$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：
$$\vec{x}\bullet \mathbf{N}\bullet \vec{x}=\mathbf{N}:\vec{x}\vec{x}=N_{ij}{x}^i{x}^j=\left[\begin{array}{ccc}{x}^1& x^2& x^3\end{array}\right]\left[\begin{array}{ccc}{N}_{11}& N_{12}& N_{13}\\ & & \\ N_{21}& N_{22}& N_{23}\\ & & \\ N_{31}& N_{32}& N_{33}\end{array}\right]\left[\begin{array}{c}{x}^1\\ \\ x^2\\ \\ x^3\end{array}\right]={\vec{x}}^T{N}_1\vec{x}$$

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：
$$\begin{gathered} C_{ij}=A_i^{\bullet k}{B}_{kj}=A_{ik}{B}_{\bullet j}^k⟺C_1=[C_{ij}]={\left[\begin{array}{ccc}{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{array}\right]}^T\left[\begin{array}{ccc}{B}_{11}& B_{12}& B_{13}\\ & & \\ B_{21}& B_{22}& B_{23}\\ & & \\ B_{31}& B_{32}& B_{33}\end{array}\right]=A_2^T{B}_1=\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ & & \\ A_{21}& A_{22}& A_{23}\\ & & \\ A_{31}& A_{32}& A_{33}\end{array}\right]\left[\begin{array}{ccc}{B}_{\bullet 1}^1& B_{\bullet 2}^1& B_{\bullet 3}^1\\ & & \\ B_{\bullet 1}^2& B_{\bullet 2}^2& B_{\bullet 3}^2\\ & & \\ B_{\bullet 1}^3& B_{\bullet 2}^3& B_{\bullet 3}^3\end{array}\right]=A_1{B}_3 \\ {C}_i^{\bullet j}=A_i^{\bullet k}{B}_k^{\bullet j}=A_{ik}{B}^{kj}⟺C_2=[C_i^{\bullet j}]=\left[\begin{array}{ccc}{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{array}\right]\left[\begin{array}{ccc}{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{array}\right]=A_2{B}_2=\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ & & \\ A_{21}& A_{22}& A_{23}\\ & & \\ A_{31}& A_{32}& A_{33}\end{array}\right]\left[\begin{array}{ccc}{B}^{11}& B^{12}& B^{13}\\ & & \\ B^{21}& B^{22}& B^{23}\\ & & \\ B^{31}& B^{32}& B^{33}\end{array}\right]=A_1{B}_4 \\ {C}_{\bullet j}^i=A_{\bullet k}^i{B}_{\bullet j}^k=A^{ik}{B}_{kj}⟺C_3=[C_{\bullet j}^i]=\left[\begin{array}{ccc}{A}_{\bullet 1}^1& A_{\bullet 2}^1& A_{\bullet 3}^1\\ & & \\ A_{\bullet 1}^2& A_{\bullet 2}^2& A_{\bullet 3}^2\\ & & \\ A_{\bullet 1}^3& A_{\bullet 2}^3& A_{\bullet 3}^3\end{array}\right]\left[\begin{array}{ccc}{B}_{\bullet 1}^1& B_{\bullet 2}^1& B_{\bullet 3}^1\\ & & \\ B_{\bullet 1}^2& B_{\bullet 2}^2& B_{\bullet 3}^2\\ & & \\ B_{\bullet 1}^3& B_{\bullet 2}^3& B_{\bullet 3}^3\end{array}\right]=A_3{B}_3=\left[\begin{array}{ccc}{A}^{11}& A^{12}& A^{13}\\ & & \\ A^{21}& A^{22}& A^{23}\\ & & \\ A^{31}& A^{32}& A^{33}\end{array}\right]\left[\begin{array}{ccc}{B}_{11}& B_{12}& B_{13}\\ & & \\ B_{21}& B_{22}& B_{23}\\ & & \\ B_{31}& B_{32}& B_{33}\end{array}\right]=A_4{B}_1 \\ {C}^{ij}=A_{\bullet k}^i{B}^{kj}=A^{ik}{B}_k^{\bullet j}⟺C_2=[C^{ij}]=\left[\begin{array}{ccc}{A}_{\bullet 1}^1& A_{\bullet 2}^1& A_{\bullet 3}^1\\ & & \\ A_{\bullet 1}^2& A_{\bullet 2}^2& A_{\bullet 3}^2\\ & & \\ A_{\bullet 1}^3& A_{\bullet 2}^3& A_{\bullet 3}^3\end{array}\right]\left[\begin{array}{ccc}{B}^{11}& B^{12}& B^{13}\\ & & \\ B^{21}& B^{22}& B^{23}\\ & & \\ B^{31}& B^{32}& B^{33}\end{array}\right]=A_3{B}_4=\left[\begin{array}{ccc}{A}^{11}& A^{12}& A^{13}\\ & & \\ A^{21}& A^{22}& A^{23}\\ & & \\ A^{31}& A^{32}& A^{33}\end{array}\right]{\left[\begin{array}{ccc}{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{array}\right]}^T=A_4{B}_2^T \end{gathered}$$
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：
$$\begin{gathered} \vec{w}={\tau }_3\vec{u}⟺\vec{w}=T\bullet \vec{u} \\ \vec{u}=({\tau }_3{)}^{-1}\vec{w}=({\tau }^{-1}{)}_3\vec{w}⟺\vec{u}=T^{-1}\bullet \vec{w} \end{gathered}$$