定义 若二阶张量 Q \bold{Q} Q的逆张量与转置张量相等,则称 Q \bold{Q} Q为正交张量,记作, Q ∈ O 3 \bold Q\in\mathcal{O}_3 Q∈O3,即
Q − 1 = Q T \bold{Q}^{-1}=\bold{Q}^T Q−1=QT
或者说
Q ∙ Q T = Q T ∙ Q = G \bold{Q}\bullet\bold{Q}^T=\bold{Q}^T\bullet\bold{Q}=\bold{G} Q∙QT=QT∙Q=G
其对应的矩阵形式为:
Q 3 ( Q T ) 3 = ( Q T ) 3 Q 3 = E Q_3(Q^T)_3=(Q^T)_3Q_3=E Q3(QT)3=(QT)3Q3=E
由于
( Q T ) 3 = Q 2 ≠ ( Q 3 ) T (Q^T)_3=Q_2\ne (Q_3)^T (QT)3=Q2=(Q3)T
故正交张量的矩阵一般不是正交矩阵。不过在笛卡尔坐标系中
( Q T ) 3 = Q 2 = ( Q 3 ) T (Q^T)_3=Q_2=(Q_3)^T (QT)3=Q2=(Q3)T
则正交张量在笛卡尔坐标系中的分量构成的矩阵为正交矩阵(这与 τ 2 \tau_2 τ2的 定义方式无关,若采用黄克智-《张量分析》中的定义仍为正交矩阵)。由正交张量的定义可知:
d e t ( Q ∙ Q T ) = d e t ( Q ) d e t ( Q T ) = [ d e t ( Q ) ] 2 = d e t ( G ) = 1 det(\bold{Q}\bullet\bold{Q}^T)=det(\bold{Q})det(\bold{Q}^T)=[det(\bold{Q})]^2=det(G)=1 det(Q∙QT)=det(Q)det(QT)=[det(Q)]2=det(G)=1
得到
d e t ( Q ) = ± 1 det(\bold{Q})=\pm1 det(Q)=±1
将行列式为 1 1 1的正交张量称作正常正交张量,记作 Q ∈ O 3 + \bold Q\in\mathcal{O}_3^+ Q∈O3+;行列式为 − 1 -1 −1的正交张量称作反常正交张量。
将正交张量对应的线性变换称作正交变换。对于任意两向量 u ⃗ \vec{u} u、 v ⃗ \vec{v} v,有:
( Q ∙ u ⃗ ) ∙ ( Q ∙ v ⃗ ) = ( u ⃗ ∙ Q T ) ∙ ( Q ∙ v ⃗ ) = ( u ⃗ ∙ Q − 1 ) ∙ ( Q ∙ v ⃗ ) = u ⃗ ∙ ( Q − 1 ∙ Q ) ∙ v ⃗ = u ⃗ ∙ ( G ∙ v ⃗ ) = u ⃗ ∙ v ⃗ ( u ⃗ ∙ Q ) ∙ ( v ⃗ ∙ Q ) = ( u ⃗ ∙ Q ) ∙ ( Q T ∙ v ⃗ ) = ( u ⃗ ∙ Q ) ∙ ( Q − 1 ∙ v ⃗ ) = u ⃗ ∙ ( Q ∙ Q − 1 ) ∙ v ⃗ = u ⃗ ∙ ( G ∙ v ⃗ ) = u ⃗ ∙ v ⃗ (\bold{Q}\bullet\vec{u})\bullet(\bold{Q}\bullet\vec{v})= (\vec{u}\bullet\bold{Q}^{T})\bullet(\bold{Q}\bullet\vec{v})= (\vec{u}\bullet\bold{Q}^{-1})\bullet(\bold{Q}\bullet\vec{v})= \vec{u}\bullet(\bold{Q}^{-1}\bullet\bold{Q})\bullet\vec{v}= \vec{u}\bullet(\bold{G}\bullet\vec{v})= \vec{u}\bullet\vec{v} \\\ \\ (\vec{u}\bullet\bold{Q})\bullet(\vec{v}\bullet\bold{Q})= (\vec{u}\bullet\bold{Q})\bullet(\bold{Q}^T\bullet\vec{v})= (\vec{u}\bullet\bold{Q})\bullet(\bold{Q}^{-1}\bullet\vec{v})= \vec{u}\bullet(\bold{Q}\bullet\bold{Q}^{-1})\bullet\vec{v}= \vec{u}\bullet(\bold{G}\bullet\vec{v})= \vec{u}\bullet\vec{v} (Q∙u)∙(Q∙v)=(u∙QT)∙(Q∙v)=(u∙Q−1)∙(Q∙v)=u∙(Q−1∙Q)∙v=u∙(G∙v)=u∙v (u∙Q)∙(v∙Q)=(u∙Q)∙(QT∙v)=(u∙Q)∙(Q−1∙v)=u∙(Q∙Q−1)∙v=u∙(G∙v)=u∙v
故,同种正交变换具有“保内积性”。反过来,若张量的线性变换具有保内积性,那么该变换为正交变换证明如下,由于
( Q ∙ u ⃗ ) ∙ ( Q ∙ v ⃗ ) = u ⃗ ∙ ( Q T ∙ Q ) ∙ v ⃗ = u ⃗ ∙ v ⃗ ( u ⃗ ∙ Q ) ∙ ( v ⃗ ∙ Q ) = u ⃗ ∙ ( Q ∙ Q T ) ∙ v ⃗ = u ⃗ ∙ v ⃗ (\bold{Q}\bullet\vec{u})\bullet(\bold{Q}\bullet\vec{v})= \vec{u}\bullet(\bold{Q}^{T}\bullet\bold{Q})\bullet\vec{v}= \vec{u}\bullet\vec{v}\\\ \\ (\vec{u}\bullet\bold{Q})\bullet(\vec{v}\bullet\bold{Q})= \vec{u}\bullet(\bold{Q}\bullet\bold{Q}^{T})\bullet\vec{v}= \vec{u}\bullet\vec{v} (Q∙u)∙(Q∙v)=u∙(QT∙Q)∙v=u∙v (u∙Q)∙(v∙Q)=u∙(Q∙QT)∙v=u∙v
则
u ⃗ ∙ ( Q T ∙ Q − G ) ∙ v ⃗ = 0 ⟺ ( Q i ∙ k Q ∙ j i − δ j k ) u k v j = 0 u ⃗ ∙ ( Q ∙ Q T − G ) ∙ v ⃗ = 0 ⟺ ( Q i ∙ k Q ∙ j i − δ j k ) u k v j = 0 \vec{u}\bullet(\bold{Q}^{T}\bullet\bold{Q}-\bold{G})\bullet\vec{v}=0 \Longleftrightarrow (Q^{\bullet k}_iQ^{i}_{\bullet j}-\delta^{k}_{j})u_kv^j=0\\\ \\ \vec{u}\bullet(\bold{Q}\bullet\bold{Q}^{T}-\bold{G})\bullet\vec{v}=0 \Longleftrightarrow (Q^{\bullet k}_iQ^{i}_{\bullet j}-\delta^{k}_{j})u_kv^j=0 u∙(QT∙Q−G)∙v=0⟺(Qi∙kQ∙ji−δjk)ukvj=0 u∙(Q∙QT−G)∙v=0⟺(Qi∙kQ∙ji−δjk)ukvj=0
由 u k , v j ( k , j = 1 , 2 , 3 ) u_k,v^j\ (k,j=1,2,3) uk,vj (k,j=1,2,3) 的任意性知:
Q i ∙ k Q ∙ j i − δ j k = 0 ⟺ Q i ∙ k Q ∙ j i = δ j k ⟺ Q T ∙ Q = G Q^{\bullet k}_iQ^{i}_{\bullet j}-\delta^{k}_{j}=0 \Longleftrightarrow Q^{\bullet k}_iQ^{i}_{\bullet j}=\delta^{k}_{j} \Longleftrightarrow \bold{Q}^{T}\bullet\bold{Q}=\bold{G} Qi∙kQ∙ji−δjk=0⟺Qi∙kQ∙ji=δjk⟺QT∙Q=G
根据保内积的性质可以得到:正交变换将标准正交基映射为标准正交基。由于:
Q ∙ g ⃗ k = G ⃗ k = Q ∙ k i g ⃗ i Q ∙ g ⃗ k = G ⃗ k = Q i ∙ k g ⃗ i \bold{Q}\bullet\vec{g}_k=\vec{\mathscr{G}}_k=Q^i_{\bullet k}\vec{g}_i \\\ \\ \bold{Q}\bullet\vec{g}^k=\vec{\mathscr{G}}^k=Q_i^{\bullet k}\vec{g}^i Q∙gk=Gk=Q∙kigi Q∙gk=Gk=Qi∙kgi
其中, G ⃗ k \vec{\mathscr{G}}_k Gk为通过正交变换得到的新协变基矢,而 G ⃗ k \vec{\mathscr{G}}^k Gk为通过正交变换得到的新逆变基矢。则:
Q = ( Q ∙ k i g ⃗ i ) g ⃗ k = ( Q i ∙ k g ⃗ i ) g ⃗ k = G ⃗ k g ⃗ k = G ⃗ k g ⃗ k \bold{Q}=(Q^i_{\bullet k}\vec{g}_i)\vec{g}^k=(Q_i^{\bullet k}\vec{g}^i)\vec{g}_k=\vec{\mathscr{G}}_k\vec{g}^k=\vec{\mathscr{G}}^k\vec{g}_k Q=(Q∙kigi)gk=(Qi∙kgi)gk=Gkgk=Gkgk
这意味着正交张量可表示为正交变换得到的新协(逆)变基矢与原逆(协)变基矢的并矢之和,需要注意的是新基矢在前,原基矢量在后。若取原基矢为标准正交基 { e ⃗ 1 , e ⃗ 2 , e ⃗ 3 } \{\vec{e}_1,\vec{e}_2,\vec{e}_3\} {e1,e2,e3},则有: Q = E 1 ⃗ e ⃗ 1 + E 2 ⃗ e ⃗ 2 + E 3 ⃗ e ⃗ 3 \bold{Q}=\vec{\mathscr{E}_1}\vec{e}_1+\vec{\mathscr{E}_2}\vec{e}_2+\vec{\mathscr{E}_3}\vec{e}_3 Q=E1e1+E2e2+E3e3
其中, { E 1 ⃗ , E 2 ⃗ , E 3 ⃗ } \{\vec{\mathscr{E}_1},\vec{\mathscr{E}_2},\vec{\mathscr{E}_3}\} {E1,E2,E3}为通过 Q \bold{Q} Q代表的正交变换得到的新标准正交基。那么:
c o s ( E i ⃗ , e ⃗ j ) = E i ⃗ ∙ e ⃗ j = ( Q ∙ e ⃗ i ) ∙ e ⃗ j = Q i j cos(\vec{\mathscr{E}_i},\vec{e}_j)=\vec{\mathscr{E}_i}\bullet\vec{e}_j=(\bold{Q}\bullet\vec{e}_i)\bullet\vec{e}_j=Q_{ij} cos(Ei,ej)=Ei∙ej=(Q∙ei)∙ej=Qij
这说明,正交张量在原标准正交基中的分量 Q i j Q_{ij} Qij为原基矢量 e ⃗ i \vec{e}_i ei与新基矢量 E j ⃗ \vec{\mathscr{E}_j} Ej夹角的方向余弦。对正交变换的几何意义进行分析,由于
[ G ⃗ 1 G ⃗ 2 G ⃗ 3 ] = d e t ( Q ) [ g ⃗ 1 g ⃗ 2 g ⃗ 3 ] [ G ⃗ 1 G ⃗ 2 G ⃗ 3 ] = d e t ( Q ) [ g ⃗ 1 g ⃗ 2 g ⃗ 3 ] \begin{bmatrix}\vec{\mathscr{G}}_1& \vec{\mathscr{G}}_2 & \vec{\mathscr{G}}_3\end{bmatrix} =det(\bold{Q})\begin{bmatrix}\vec{g}_1& \vec{g}_2 & \vec{g}_3\end{bmatrix} \\\ \\ \begin{bmatrix}\vec{\mathscr{G}}^1& \vec{\mathscr{G}}^2 & \vec{\mathscr{G}}^3\end{bmatrix} =det(\bold{Q})\begin{bmatrix}\vec{g}^1& \vec{g}^2 & \vec{g}^3\end{bmatrix} [G1G2G3]=det(Q)[g1g2g3] [G1G2G3]=det(Q)[g1g2g3]
对于正常正交张量,混合积 [ G ⃗ 1 G ⃗ 2 G ⃗ 3 ] \begin{bmatrix}\vec{\mathscr{G}}_1& \vec{\mathscr{G}}_2 & \vec{\mathscr{G}}_3\end{bmatrix} [G1G2G3]与 [ g ⃗ 1 g ⃗ 2 g ⃗ 3 ] \begin{bmatrix}\vec{g}_1& \vec{g}_2 & \vec{g}_3\end{bmatrix} [g1g2g3]同号,而对于反常正交张量混合积 [ G ⃗ 1 G ⃗ 2 G ⃗ 3 ] \begin{bmatrix}\vec{\mathscr{G}}_1& \vec{\mathscr{G}}_2 & \vec{\mathscr{G}}_3\end{bmatrix} [G1G2G3]与 [ g ⃗ 1 g ⃗ 2 g ⃗ 3 ] \begin{bmatrix}\vec{g}_1& \vec{g}_2 & \vec{g}_3\end{bmatrix} [g1g2g3]异号。结合正交变换的保内积性知:
正常正交变换仅对应基的刚性转动,基之间的夹角与长度均不变;
反常正交变换不仅对应于基的刚性转动,而且附加一次镜面反射的操作。
不妨设正交张量的特征值分别为:
λ 1 ∈ C 、 λ 2 ∈ C 、 λ 3 ∈ R \lambda_1\in\mathbf{C}、\lambda_2\in\mathbf{C}、\lambda_3\in\mathbf{R} λ1∈C、λ2∈C、λ3∈R
而
u ⃗ 1 ∈ C 3 、 u ⃗ 2 ∈ C 3 、 u ⃗ 3 ∈ R 3 \vec{u}_1\in\mathbf{C}^3、\vec{u}_2\in\mathbf{C}^3、\vec{u}_3\in\mathbf{R}^3 u1∈C3、u2∈C3、u3∈R3
分别为其对应的特征向量,则有:
{ Q ∙ u ⃗ 1 = λ 1 u ⃗ 1 Q ∙ u ⃗ 2 = λ 2 u ⃗ 2 Q ∙ u ⃗ 3 = λ 3 u ⃗ 3 \begin{cases} \bold{Q}\bullet\vec{u}_1=\lambda_1\vec{u}_1\\ \\ \bold{Q}\bullet\vec{u}_2=\lambda_2\vec{u}_2 \\ \\ \bold{Q}\bullet\vec{u}_3=\lambda_3\vec{u}_3 \end{cases} ⎩ ⎨ ⎧Q∙u1=λ1u1Q∙u2=λ2u2Q∙u3=λ3u3
根据正交变换保内积的性质:
( Q ∙ u ⃗ 3 ) ∙ ( Q ∙ u ⃗ 3 ) = u ⃗ 3 ∙ u ⃗ 3 = ∣ u ⃗ 3 ∣ 2 = ( λ 3 u ⃗ 3 ) ∙ ( λ 3 u ⃗ 3 ) = λ 3 2 ∣ u ⃗ 3 ∣ 2 ⟹ ∣ λ 3 ∣ = 1 ( Q ∙ u ⃗ 1 ) ∙ ( Q ∙ u ⃗ 1 ) = u ⃗ 1 ∙ u ⃗ 1 = ( λ 1 u ⃗ 1 ) ∙ ( λ 1 u ⃗ 1 ) = λ 1 2 ( u ⃗ 1 ∙ u ⃗ 1 ) ⟹ ∣ λ 1 ∣ = 1 ( Q ∙ u ⃗ 2 ) ∙ ( Q ∙ u ⃗ 2 ) = u ⃗ 2 ∙ u ⃗ 2 = ( λ 2 u ⃗ 2 ) ∙ ( λ 2 u ⃗ 2 ) = λ 2 2 ( u ⃗ 2 ∙ u ⃗ 2 ) ⟹ ∣ λ 2 ∣ = 1 \begin{aligned} &(\bold{Q}\bullet\vec{u}_3)\bullet(\bold{Q}\bullet\vec{u}_3) =\vec{u}_3\bullet\vec{u}_3 =|\vec{u}_3|^2 =(\lambda_3\vec{u}_3)\bullet(\lambda_3\vec{u}_3) =\lambda_3^2|\vec{u}_3|^2 \Longrightarrow |\lambda_3|=1 \\ \\ &(\bold{Q}\bullet\vec{u}_1)\bullet(\bold{Q}\bullet\vec{u}_1) =\vec{u}_1\bullet\vec{u}_1 =(\lambda_1\vec{u}_1)\bullet(\lambda_1\vec{u}_1) =\lambda_1^2(\vec{u}_1\bullet\vec{u}_1) \Longrightarrow |\lambda_1|=1\\ \\ &(\bold{Q}\bullet\vec{u}_2)\bullet(\bold{Q}\bullet\vec{u}_2) =\vec{u}_2\bullet\vec{u}_2 =(\lambda_2\vec{u}_2)\bullet(\lambda_2\vec{u}_2) =\lambda_2^2(\vec{u}_2\bullet\vec{u}_2) \Longrightarrow |\lambda_2|=1 \end{aligned} (Q∙u3)∙(Q∙u3)=u3∙u3=∣u3∣2=(λ3u3)∙(λ3u3)=λ32∣u3∣2⟹∣λ3∣=1(Q∙u1)∙(Q∙u1)=u1∙u1=(λ1u1)∙(λ1u1)=λ12(u1∙u1)⟹∣λ1∣=1(Q∙u2)∙(Q∙u2)=u2∙u2=(λ2u2)∙(λ2u2)=λ22(u2∙u2)⟹∣λ2∣=1
说明:正交张量的特征值模为1。又因为
d e t ( Q ) = λ 1 λ 2 λ 3 = ± 1 det(\bold{Q})=\lambda_1\lambda_2\lambda_3=\pm1 det(Q)=λ1λ2λ3=±1
则正交张量的特征值包括如下情况:
由Nanson公式可知:
( Q ∙ u ⃗ ) × ( Q ∙ v ⃗ ) = d e t ( Q ) Q ∙ ( u ⃗ × v ⃗ ) ( a ) (\bold{Q}\bullet\vec{u})\times(\bold{Q}\bullet\vec{v})=det(\bold{Q})\bold{Q}\bullet(\vec{u}\times\vec{v})\qquad(a) (Q∙u)×(Q∙v)=det(Q)Q∙(u×v)(a)
综上所述,正交张量特征向量的性质包括:
(1) 共轭复特征值 λ 1 = λ ˉ 2 \lambda_1=\bar{\lambda}_2 λ1=λˉ2 + 实特征值 λ 3 \lambda_3 λ3: u ⃗ 3 ∙ u ⃗ 1 = u ⃗ 3 ∙ u ⃗ 2 = 0 \vec{u}_3\bullet\vec{u}_1=\vec{u}_3\bullet\vec{u}_2=0 u3∙u1=u3∙u2=0;
(2) 二重实特征值 λ 1 = λ 2 \lambda_1=\lambda_2 λ1=λ2 + 实特征值 λ 3 \lambda_3 λ3: u ⃗ 3 ∙ u ⃗ 1 = u ⃗ 3 ∙ u ⃗ 2 = 0 \vec{u}_3\bullet\vec{u}_1=\vec{u}_3\bullet\vec{u}_2=0 u3∙u1=u3∙u2=0;
(3) 三重实特征值 λ 1 = λ 2 = λ 3 \lambda_1=\lambda_2=\lambda_3 λ1=λ2=λ3:任意向量均为正交张量的主方向;
且正交张量的几何重数必定等于其代数重数。
定理:正交仿射量同一特征值对应的左、右特征空间相同。
证明如下:设正交仿射量 Q \bold Q Q的特征值与右特征向量分别为 λ \lambda λ 与 r ⃗ \vec{r} r,即
Q ∙ r ⃗ = λ r ⃗ \bold Q\bullet\vec{r}=\lambda\vec{r} Q∙r=λr
那么
( λ Q T ) ∙ ( Q ∙ r ⃗ ) = λ 2 ( Q T ∙ r ⃗ ) = Q T ∙ r ⃗ = r ⃗ ∙ Q = λ r ⃗ ( 证毕 ) (\lambda\bold Q^T)\bullet(\bold Q\bullet\vec{r})=\lambda^2(\bold Q^T\bullet\vec{r})=\bold Q^T\bullet\vec{r}=\vec{r}\bullet\bold Q=\lambda\vec{r}\quad(证毕) (λQT)∙(Q∙r)=λ2(QT∙r)=QT∙r=r∙Q=λr(证毕)
推论: Q \bold Q Q 与 Q T \bold Q^T QT 有相同的特征值与特征空间。
证明:由于任何互为转置的张量具有相同的特征值,且对同一特征值, Q \bold Q Q 的左特征向量为 Q T \bold Q^T QT 的右特征向量,而正交张量左右特征空间相同,故得证。