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(八)二阶张量与矩阵(二)_二阶张量的点乘推导

二阶张量的点乘推导

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 t r ( T ) tr(\bold{T}) tr(T)
t r ( T ) = T i j g i j = T i ∙ i = T 1 ∙ 1 + T 2 ∙ 2 + T 3 ∙ 3 = T ∙ i i = T ∙ 1 1 + T ∙ 2 2 + T ∙ 3 3 = T i j g i j tr(\bold{T}) =T^{ij}g_{ij} =T_i^{\bullet i} =T_1^{\bullet 1}+T_2^{\bullet 2}+T_3^{\bullet 3} =T^i_{\bullet i} =T^1_{\bullet 1}+T^2_{\bullet 2}+T^3_{\bullet 3} =T_{ij}g^{ij} tr(T)=Tijgij=Tii=T11+T22+T33=Tii=T11+T22+T33=Tijgij显然,与之相关的矩阵运算为求方阵 τ 2 、 τ 3 \tau_2、\tau_3 τ2τ3 的迹

二阶张量迹的运算性质:

A , B , C ∈ T 2 ( V ) \bold{A},\bold{B},\bold{C}\in\mathscr{T}_2(V) A,B,CT2(V),则:
( 1 )   t r ( A + B ) = t r ( A ) + t r ( B ) ( 2 )   t r ( A ⋅ B ) = t r ( B ⋅ A ) ( 3 )   t r ( A ⋅ B ⋅ C ) = t r ( B ⋅ C ⋅ A ) = t r ( C ⋅ A ⋅ B ) ( 4 )   A : B = t r ( A T ⋅ B ) = t r ( A ⋅ B T ) = t r ( B T ⋅ A ) = t r ( B ⋅ A T ) (1) tr(\boldA+\boldB)=tr(\boldA)+tr(\boldB)(2) tr(\boldA\boldB)=tr(\boldB\boldA)(3) tr(\boldA\boldB\boldC)=tr(\boldB\boldC\boldA)=tr(\boldC\boldA\boldB)(4) \boldA:\boldB=tr(\boldAT\boldB)=tr(\boldA\boldBT)=tr(\boldBT\boldA)=tr(\boldB\boldAT) (1) tr(A+B)=tr(A)+tr(B)(2) tr(AB)=tr(BA)(3) tr(ABC)=tr(BCA)=tr(CAB)(4) A:B=tr(ATB)=tr(ABT)=tr(BTA)=tr(BAT)
证明:
( 1 )   t r ( A + B ) = A ∙ i i + B ∙ i i = t r ( A ) + t r ( B ) ( 2 )   t r ( A ⋅ B ) = A ∙ j i B ∙ i j = B ∙ i j A ∙ j i = t r ( B ⋅ A ) ( 3 )   t r ( A ⋅ B ⋅ C ) = A ∙ j i B ∙ k j C ∙ i k = B ∙ k j C ∙ i k A ∙ j i = t r ( B ⋅ C ⋅ A )    = C ∙ i k A ∙ j i B ∙ k j = t r ( C ⋅ A ⋅ B ) ( 4 )   A : B = A ∙ j i B i ∙ j = ( B T ) ∙ i j A ∙ j i = t r ( B T ⋅ A )   = ( A T ) j ∙ i B i ∙ j = t r ( A T ⋅ B ) (1) tr(\boldA+\boldB)=Aii+Bii=tr(\boldA)+tr(\boldB)(2) tr(\boldA\boldB)=AijBji=BjiAij=tr(\boldB\boldA)(3) tr(\boldA\boldB\boldC)=AijBjkCki=BjkCkiAij=tr(\boldB\boldC\boldA)  =CkiAijBjk=tr(\boldC\boldA\boldB)(4) \boldA:\boldB=AijBji=(BT)jiAij=tr(\boldBT\boldA) =(AT)ijBji=tr(\boldAT\boldB) (1) tr(A+B)=Aii+Bii=tr(A)+tr(B)(2) tr(AB)=AjiBij=BijAji=tr(BA)(3) tr(ABC)=AjiBkjCik=BkjCikAji=tr(BCA)  =CikAjiBkj=tr(CAB)(4) A:B=AjiBij=(BT)ijAji=tr(BTA) =(AT)jiBij=tr(ATB)

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 [w1w2w3] =[T11T12T13T21T22T23T31T32T33] [u1u2u3] \Longleftrightarrow \vec{w}=\tau_{3}\vec{u} \\\ \\ \vec{t}=\vec{u}\bullet\bold{T} \Longleftrightarrow t^i=u^jT^{\bullet i}_{j} \Longleftrightarrow [t1t2t3] =[T11T12T13T21T22T23T31T32T33] [u1u2u3] \Longleftrightarrow \vec{t}=\tau_{2}\vec{u} w =Tu wi=Tjiuj w1w2w3 = T11T12T13T21T22T23T31T32T33 u1u2u3 w =τ3u  t =u Tti=ujTji t1t2t3 = T11T12T13T21T22T23T31T32T33 u1u2u3 t =τ2u 显然,二阶张量与矢量的左、右点积一般不等: T ∙ u ⃗ ≠ u ⃗ ∙ T \bold{T}\bullet\vec{u}\ne\vec{u}\bullet\bold{T} Tu =u T且有:
( τ T ) 3 u ⃗ = τ 2 u ⃗ ⟺ T T ∙ u ⃗ = u ⃗ ∙ T (\tau^T)_3\vec{u}=\tau_{2}\vec{u} \Longleftrightarrow \bold{T}^T\bullet\vec{u}=\vec{u}\bullet\bold{T} (τT)3u =τ2u TTu =u T那么有:
N ∙ u ⃗ = u ⃗ ∙ N ( N 为对称二阶张量 )   Ω ∙ u ⃗ = − u ⃗ ∙ Ω ( Ω 为反对称二阶张量 ) \bold{N}\bullet\vec{u}=\vec{u}\bullet\bold{N} \qquad(\bold{N}为对称二阶张量)\\\ \\ \bold{\Omega}\bullet\vec{u}=-\vec{u}\bullet\bold{\Omega} \qquad(\bold{\Omega}为反对称二阶张量) Nu =u N(N为对称二阶张量) Ωu =u Ω(Ω为反对称二阶张量)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 =[x1x2x3] [N11N12N13N21N22N23N31N32N33] [x1x2x3] =\vec{x}^TN_{1}\vec{x} x Nx =N:x x =Nijxixj=[x1x2x3] N11N21N31N12N22N32N13N23N33 x1x2x3 =x TN1x

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}] =[A11A12A13A21A22A23A31A32A33]^T [B11B12B13B21B22B23B31B32B33] =A_{2}^TB_{1} =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =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}] =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =A_{2}B_{2} =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =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}] =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =A_{3}B_{3} =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =A_{4}B_{1}\\\ \\ %%%%%%%%%%%%%%%% C^{ij}=A^{i}_{\bullet k}B^{kj}=A^{ik}B_k^{\bullet j} \Longleftrightarrow C_{2}=[C^{ij}] =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33] =A_{3}B_{4} =[A11A12A13A21A22A23A31A32A33] [B11B12B13B21B22B23B31B32B33]^T =A_{4}B_{2}^T %%%%%%%%%%%%%%%% Cij=AikBkj=AikBjkC1=[Cij]= A11A12A13A21A22A23A31A32A33 T B11B21B31B12B22B32B13B23B33 =A2TB1= A11A21A31A12A22A32A13A23A33 B11B12B13B21B22B23B31B32B33 =A1B3 Cij=AikBkj=AikBkjC2=[Cij]= A11A12A13A21A22A23A31A32A33 B11B12B13B21B22B23B31B32B33 =A2B2= A11A21A31A12A22A32A13A23A33 B11B21B31B12B22B32B13B23B33 =A1B4 Cji=AkiBjk=AikBkjC3=[Cji]= A11A12A13A21A22A23A31A32A33 B11B12B13B21B22B23B31B32B33 =A3B3= A11A21A31A12A22A32A13A23A33 B11B21B31B12B22B32B13B23B33 =A4B1 Cij=AkiBkj=AikBkjC2=[Cij]= A11A12A13A21A22A23A31A32A33 B11B21B31B12B22B32B13B23B33 =A3B4= A11A21A31A12A22A32A13A23A33 B11B12B13B21B22B23B31B32B33 T=A4B2T
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
A ∙ B ≠ B ∙ A \bold{A}\bullet\bold{B}\ne\bold{B}\bullet\bold{A} AB=BA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 ⟺ ( A ∙ B ) T = B T ∙ A T (A_3B_3)^T=(B_3)^T(A_3)^T=(B^T)_2(A^T)_2 \Longleftrightarrow (\bold{A}\bullet\bold{B})^T=\bold{B}^T\bullet\bold{A}^T (A3B3)T=(B3)T(A3)T=(BT)2(AT)2(AB)T=BTAT张量点积的行列式与张量行列式的关系式 d e t ( A 3 B 3 ) = d e t ( A 3 ) d e t ( B 3 ) ⟺ d e t ( A ∙ B ) = d e t ( A ) d e t ( B ) det(A_3B_3)=det(A_3)det(B_3) \Longleftrightarrow det(\bold{A}\bullet\bold{B})=det(\bold{A})det(\bold{B}) det(A3B3)=det(A3)det(B3)det(AB)=det(A)det(B)

2. 正则二阶张量与可逆矩阵

若二阶张量 T \bold T 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: T \bold T T如果正则,那么 T T \bold T^T TT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 T \bold{T} T,必唯一存在正则的二阶张量 T − 1 \bold{T}^{-1} T1,使得:
T ∙ T − 1 = T − 1 ∙ T = G ( 1 ) \bold{T}\bullet\bold{T}^{-1}=\bold{T}^{-1}\bullet\bold{T}=\bold{G}\qquad(1) TT1=T1T=G(1) T − 1 \bold{T}^{-1} T1称作正则张量 T \bold{T} T的逆张量,且逆张量的矩阵等于原张量矩阵的逆,即
( τ − 1 ) 3 = ( τ 3 ) − 1 (\tau^{-1})_3=(\tau_3)^{-1} (τ1)3=(τ3)1则,
d e t [ ( τ − 1 ) 3 ] = d e t [ ( τ 3 ) − 1 ] = 1 d e t ( τ 3 ) ⟺ d e t ( T − 1 ) = 1 d e t ( T ) det[(\tau^{-1})_3]=det[(\tau_3)^{-1}]=\frac{1}{det(\tau_3)} \Longleftrightarrow det(\bold{T}^{-1})=\frac{1}{det(\bold{T})} det[(τ1)3]=det[(τ3)1]=det(τ3)1det(T1)=det(T)1此外,进一步根据(1)式可得:
( T − 1 ) − 1 = T   G T = ( T ∙ T − 1 ) T = ( T − 1 ) T ∙ T T = G = ( T T ) − 1 ∙ T T (\bold{T}^{-1})^{-1}=\bold{T}\\\ \\ \bold{G}^T =(\bold{T}\bullet\bold{T}^{-1})^T =(\bold{T}^{-1})^T\bullet\bold{T}^T =\bold{G} =(\bold{T}^T)^{-1}\bullet\bold{T}^T (T1)1=T GT=(TT1)T=(T1)TTT=G=(TT)1TT由逆张量的唯一性可知:
( T T ) − 1 = ( T − 1 ) T (\bold{T}^T)^{-1}=(\bold{T}^{-1})^T (TT)1=(T1)T若二阶张量线性变换可逆,则:
w ⃗ = τ 3 u ⃗ ⟺ w ⃗ = T ∙ u ⃗   u ⃗ = ( τ 3 ) − 1 w ⃗ = ( τ − 1 ) 3 w ⃗ ⟺ u ⃗ = T − 1 ∙ w ⃗ \vec{w}=\tau_3\vec{u} \Longleftrightarrow \vec{w}=T\bullet\vec{u}\\\ \\ \vec{u}=(\tau_3)^{-1}\vec{w}=(\tau^{-1})_3\vec{w} \Longleftrightarrow \vec{u}=T^{-1}\bullet\vec{w} w =τ3u w =Tu  u =(τ3)1w =(τ1)3w u =T1w

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