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高等代数教学辅导

8.3 酉矩阵、正交矩阵与标准型

建设中!

练习 练习

基础题.

1.
证明:酉(正交)相似关系是等价关系。
解答.
  • 自反性:对任意\(n\)阶方阵\(A\),取酉矩阵\(U = E_n\),有 \(A = U^HAU \),故\(A\)酉相似于自身。
  • 对称性:若\(A\)酉相似于\(B\),则存在酉矩阵\(U\)使得\(B = U^HAU\)。由\(U\)是酉矩阵知\(U^H\)也是酉矩阵,又\(A = (U^H)^HB U^H\) ,因此\(B\)酉相似于\(A\)
  • 传递性:若\(A\)酉相似于\(B\)\(B\)酉相似于\(C\),则存在酉矩阵 \(U,V\)使得
    \begin{equation*} B = U^H A U , C = V^H B V. \end{equation*}
    于是
    \begin{equation*} C = V^H (U^H A U) V = (V^H U^H) A (U V) = (U V)^H A (U V). \end{equation*}
    注意到\(U,V\)是酉矩阵,所以\(UV\)也是酉矩阵,因此\(A\)酉相似于\(C\)
2.
\(A\)\(n\)阶幂零矩阵,且\(A\neq {\bf 0}\),证明:\(A\)不可能酉相似于对角矩阵。
解答.
假设\(A\)酉相似于对角矩阵\({\rm diag}(\lambda_1,\dots,\lambda_n)\),则 \(\lambda_1,\dots,\lambda_n\)\(A\)的全部特征值。由于\(A\)是幂零矩阵,所以\(A\)的特征值只能为\(0\),故\(A\)酉相似于零矩阵。而酉相似于零矩阵的只能是零矩阵,这与\(A\neq {\bf 0}\)相矛盾,因此\(A\)不可能酉相似于对角矩阵。
3.
\(A\)\(n\)阶正交矩阵且\(\det A =-1\)。证明:\(-1\)\(A\)的特征值。
解答.
因为\(A\)是正交矩阵,所以存在正交矩阵\(Q\)使得
\begin{equation*} Q^{-1}AQ={\rm diag}\left( \begin{pmatrix} \cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{pmatrix}, \dots , \begin{pmatrix} \cos\theta_l & -\sin\theta_l\\ \sin\theta_l & \cos\theta_l\end{pmatrix}, E_r, -E_s\right), \end{equation*}
这里\(\cos\theta_1\pm\sin\theta_1,\dots,\cos\theta_l\pm\sin\theta_l,1,\dots,1,-1,\dots,-1\)\(A\)的全部特征值,其中\(-1\)代数重数为\(s\)。 两边同取行列式得
\begin{equation*} \det A = \prod\limits_{i=1}^l \begin{vmatrix} \cos\theta_i & -\sin\theta_i\\ \sin\theta_i & \cos\theta_i\end{vmatrix}\cdot 1^r\cdot (-1)^s=(-1)^s, \end{equation*}
由题设\(\det A=-1\)\(s > 0\),因此\(-1\)\(A\)的特征值。
4.
\(A\)\(n\)阶正交矩阵,其特征值均为实数,证明:\(A\)是实对称矩阵。
解答.
因为正交矩阵\(A\)的特征值全为实数,所以存在正交矩阵\(Q\)使得
\begin{equation} A = Q^{-1}\begin{pmatrix} E_r & {\bf 0}\\ {\bf 0} & -E_s\end{pmatrix}Q. \tag{8.3.1} \end{equation}
注意到\(Q^{-1}=Q^T\),所以(8.3.1)两边同取转置得
\begin{equation*} A^T=Q^T\begin{pmatrix} E_r & {\bf 0}\\ {\bf 0} & -E_s\end{pmatrix}Q=A, \end{equation*}
因此\(A\)是实对称矩阵。

提高题.

5.
\(\phi\)是欧氏空间\(V\)上的变换,且对任意\(\alpha,\beta\in V\),都有
\begin{equation*} \left(\phi(\alpha),\phi(\beta)\right)=\left(\alpha,\beta\right), \end{equation*}
证明:\(\phi\)是线性变换,因而是正交变换。
解答.
对任意\(\alpha,\beta\in V, a,b\in\mathbb{R}\),根据已知条件,有
\begin{equation*} \begin{array}{cl} &\left(\phi(a\alpha+b \beta)-a\phi(\alpha)-b\phi(\beta),\phi(a\alpha+b \beta)-a\phi(\alpha)-b\phi(\beta)\right)\\ =&\left(\phi(a\alpha+b \beta),\phi(a\alpha+b \beta)\right)-2a\left(\phi(\alpha),\phi(a\alpha+b \beta)\right)\\ &-2b\left(\phi(\beta),\phi(a\alpha+b \beta)\right)+a^2\left(\phi(\alpha),\phi(\alpha)\right)\\ &+2ab\left(\phi(\alpha),\phi(\beta)\right)+b^2\left(\phi(\beta),\phi(\beta)\right)\\ =&\left(a\alpha+b \beta,a\alpha+b \beta\right)-2a\left(\alpha,a\alpha+b \beta\right)-2b\left(\beta,a\alpha+b \beta\right)\\ &+a^2\left(\alpha,\alpha\right)+2ab\left(\alpha,\beta\right)+b^2\left(\beta,\beta\right)\\ =&0 \end{array} \end{equation*}
\(\phi(a\alpha+b \beta)-a\phi(\alpha)-b\phi(\beta)=0\),即
\begin{equation*} \phi(a\alpha+b \beta)=a\phi(\alpha)+b\phi(\beta). \end{equation*}
由此可知\(\phi\)是线性变换,因而是正交变换。
6.
\(\phi\)是正交变换,\(U\)\(\phi\)-子空间,证明:\(U^\perp\)也是\(\phi\)-子空间。
解答.
证法一:由\(\phi\)是正交变换可知,\(\phi^{-1}\)也是正交变换。故\(\forall\alpha\in U^\perp ,\beta\in U\)
\begin{equation*} (\phi(\alpha),\beta)=(\alpha,\phi^{-1}(\beta)), \end{equation*}
因为\(U\)\(\phi\)-不变子空间且\(\phi\)可逆,所以\(U\)也是\(\phi^{-1}\)-不变子空间,于是\(\phi^{-1}(\beta)\in U\),则
\begin{equation*} (\phi(\alpha),\beta)=(\alpha,\phi^{-1}(\beta))=0, \end{equation*}
\(\phi(\alpha)\in U^\perp\)。因此\(U^\perp\)也是\(\phi\)-不变子空间。
证法二:设\(\xi_1,\dots ,\xi_r\)\(U\)的一个标准正交基,将其扩充为\(V\)的一个标准正交基\(\xi_1,\dots ,\xi_r,\xi_{r+1},\dots ,\xi_n\),则
\begin{equation*} U^\perp =\langle \xi_{r+1},\dots ,\xi_n\rangle. \end{equation*}
\(\phi\)是正交变换,所以\(\phi(\xi_1),\dots ,\phi(\xi_r),\phi(\xi_{r+1}),\dots ,\phi(\xi_n)\)也是\(V\)的一个标准正交基。故\(\phi(\xi_1),\dots ,\phi(\xi_r)\)线性无关。又\(U\)\(\phi\)-不变子空间,所以\(\phi(\xi_1),\dots ,\phi(\xi_r)\in U\)。因此\(\phi(\xi_1),\dots ,\phi(\xi_r)\)\(U\)的一个基。
对任意\(r+1\leq i \leq n, 1\leq j\leq r\),由
\begin{equation*} (\phi(\xi_i),\phi(\xi_j))=(\xi_i,\xi_j)=0 \end{equation*}
\(\phi(\xi_i)\in U^\perp\)。因此\(U^\perp\)也是\(\phi\)-不变子空间。
证法三:设\(\xi_1,\dots ,\xi_r\)\(U\)的一个标准正交基,将其扩充为\(V\)的一个标准正交基\(\xi_1,\dots ,\xi_r,\xi_{r+1},\dots ,\xi_n\),则
\begin{equation*} U^\perp =\langle \xi_{r+1},\dots ,\xi_n\rangle. \end{equation*}
因为\(U\)\(\phi\)-不变子空间,所以
\begin{equation*} \phi(\xi_1,\dots ,\xi_n)=(\xi_1,\dots ,\xi_n)\begin{pmatrix} A&B\\{\bf 0}&C \end{pmatrix}. \end{equation*}
又因为\(\phi\)是正交变换,故\(\begin{pmatrix} A&B\\{\bf 0}&C \end{pmatrix}\)是正交矩阵,则\(\begin{pmatrix} A&B\\{\bf 0}&C \end{pmatrix}^{-1}=\begin{pmatrix} A&B\\{\bf 0}&C \end{pmatrix}^T\),即
\begin{equation*} \begin{pmatrix} A^{-1}&-A^{-1}BC^{-1}\\{\bf 0}&C^{-1} \end{pmatrix}=\begin{pmatrix} A^T&{\bf 0}\\B^T&C^T \end{pmatrix}, \end{equation*}
\(B={\bf 0}\)。从而\(U^\perp\)也是\(\phi\)-不变子空间。
7.
\(\phi\)是酉变换,证明:\(\phi\)的属于不同特征值的特征向量必正交。
解答.
\(\lambda, \mu\)\(\phi\)的两个不同的特征值,\(\alpha ,\beta\)分别是\(\phi\)的属于\(\lambda, \mu\)的特征向量,则
\begin{equation*} \phi(\alpha)=\lambda \alpha,\phi(\beta)=\mu \beta\mbox{,且}\left|\lambda\right|=\left|\mu\right|=1\mbox{。} \end{equation*}
\((\phi(\alpha),\phi(\beta))=(\alpha,\beta)\)
\begin{equation*} \lambda\overline{\mu}(\alpha,\beta)=(\alpha,\beta). \end{equation*}
注意到\(\left|\mu\right|=1\)\(\lambda\neq \mu\),所以\(\lambda\overline{\mu}\neq 1\)。因此\((\alpha,\beta)=0\)
8.
\(\xi,\eta\)\(n\)维欧氏空间\(V\)中两个不同的单位向量,证明:存在一个镜面反射\(\phi\),使得\(\phi(\xi)=\eta\)
解答.
\(\alpha=\frac{\xi- \eta}{\|\xi-\eta\|}\),则\(\alpha\)是单位向量。定义\(V\)上变换\(\phi\)如下
\begin{equation*} \phi (\beta)=\beta-2(\alpha,\beta)\alpha, \end{equation*}
\(\phi\)是镜面反射,且
\begin{equation*} \begin{array}{ccl}\phi(\xi)&=&\xi-2(\frac{\xi- \eta}{\|\xi-\eta\|},\xi)\frac{\xi- \eta}{\|\xi-\eta\|}\\ &=&\xi-2\frac{(\xi-\eta,\xi)}{(\xi-\eta,\xi-\eta)}(\xi-\eta)\\ &=&\xi-2\frac{1-(\xi,\eta)}{2-2(\xi,\eta)}(\xi-\eta)\\ &=&\eta . \end{array} \end{equation*}
9.
\(A\)\(n\)阶正交矩阵。证明:\(A\)是第一类正交矩阵的充分必要条件是存在正交矩阵\(B\),使得\(A=B^2\)
解答.
充分性:由于\(B\)是正交矩阵,所以\(\det B=\pm 1\)。又\(A=B^2\),两边同取行列式得
\begin{equation*} \det A = (\det B)^2 = (\pm 1)^2=1. \end{equation*}
必要性:因为\(A\)是正交矩阵,所以存在正交矩阵\(Q\)使得
\begin{equation*} A=Q^{-1}{\rm diag}\left( \begin{pmatrix} \cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{pmatrix}, \dots , \begin{pmatrix} \cos\theta_l & -\sin\theta_l\\ \sin\theta_l & \cos\theta_l\end{pmatrix}, E_r, -E_s\right)Q. \end{equation*}
由于\(\det A=1\),所以\(s\)为偶数,设\(s=2t\)。注意到
\begin{equation*} \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}^2, \end{equation*}
所以令 \(B=Q^{-1}{\rm diag}\left(B_1, \dots , B_l, E_r, C_1, \dots ,C_t\right)Q\), 其中\(B_i=\begin{pmatrix} \cos\frac{\theta_i}{2} & -\sin\frac{\theta_i}{2}\\ \sin\frac{\theta_i}{2} & \cos\frac{\theta_i}{2}\end{pmatrix},C_j =\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix},i=1,\dots,l,j=1,\dots,t\), 则\(B\)是正交矩阵,且\(A=B^2\)

挑战题.

10.
证明:\(n\)维欧氏空间\(V\)中任意正交变换\(\phi\)都可以表为一系列镜面反射的乘积。
解答.
证法一:设\(\xi_1,\dots ,\xi_n\)\(V\)的一个标准正交基,\(\eta_i=\phi(\xi_i),i=1,\dots ,n\)。由\(\phi\)是正交变换知,\(\eta_1,\dots ,\eta_n\)也是\(V\)的一个标准正交基。
  1. \(\xi_i=\eta_i(i=1,2,\cdots ,n)\),令\(\phi_1(\alpha)=\alpha-2(\xi_1,\alpha)\xi_1\),则
    \begin{equation*} \phi_1(\xi_1)=-\xi_1,\phi_1(\xi_j)=\xi_j(j=2,\cdots ,n), \end{equation*}
    \(\phi_1\)是镜面反射且\(\phi=\phi_1\phi_1\)
  2. \(\xi_1,\cdots ,\xi_n\)\(\eta_1,\cdots ,\eta_n\)不尽相同,不妨设\(\xi_1\neq \eta_1\),那么由上题结论,存在镜面反射\(\phi_1\),使得
    \begin{equation*} \phi_1(\xi_1)=\eta_1,\phi_1(\xi_j)=\zeta_j(j=2,\cdots ,n), \end{equation*}
    \begin{equation*} \xi_1,\xi_2,\cdots,\xi_n\stackrel{\phi_1}{\longrightarrow}\eta_1,\zeta_2,\cdots ,\zeta_n. \end{equation*}
    • \(\eta_j=\zeta_j(j=2,\cdots ,n)\),则\(\phi=\phi_1\),结论成立。
    • 否则,不妨假设\(\eta_2\neq\zeta_2\),令镜面反射
      \begin{equation*} \phi_2(\alpha)=\alpha-2(\eta,\alpha)\eta,\ \mbox{其中}\eta=\frac{\eta_2-\zeta_2}{\|\eta_2-\zeta_2\|}, \end{equation*}
      \(\phi_2(\eta_2)=\zeta_2\)。注意到\((\eta_2,\eta_1)=(\zeta_2,\eta_1)=0\),所以\(\phi_2(\eta_1)=\eta_1\)。于是
      \begin{equation*} \xi_1,\xi_2,\cdots,\xi_n\stackrel{\phi_1}{\longrightarrow}\eta_1,\zeta_2,\cdots ,\zeta_n\stackrel{\phi_2}{\longrightarrow}\eta_1,\eta_2,\chi_3\cdots ,\chi_n. \end{equation*}
      依此类推,
      \begin{equation*} \xi_1,\xi_2,\cdots,\xi_n\stackrel{\phi_1}{\longrightarrow}\eta_1,\zeta_2,\cdots ,\zeta_n\stackrel{\phi_2}{\longrightarrow}\eta_1,\eta_2,\chi_3\cdots ,\chi_n \end{equation*}
      \begin{equation*} \stackrel{\phi_3}{\longrightarrow}\cdots\stackrel{\phi_s}{\longrightarrow}\eta_1,\eta_2,\cdots ,\eta_n. \end{equation*}
      \(\phi=\phi_s\phi_{s-1}\cdots\phi_1\),其中\(\phi_i\)都是镜面反射\((i=1,2,\cdots,s)\)
证法二:对维数\(n\)用数学归纳法。
  1. \(n=1\)时,设\(\xi\)\(V\)的一个标准正交基,定义镜面反射
    \begin{equation*} \psi:V\rightarrow V,\ \alpha\mapsto \alpha-2(\xi,\alpha)\xi. \end{equation*}
    \(\phi\)是正交变换,所以\(\phi={\rm id}_V\)\(-{\rm id}_V\)
    1. \(\phi=-{\rm id}_V\) 时,则\(\phi=\psi\)是一个镜面反射。
    2. \(\phi={\rm id}_V\)时,则\(\phi=\psi\psi\)是两个镜面反射的乘积。
  2. 假设对于\(n-1\)维欧氏空间结论成立。对\(n\)维欧氏空间\(V\)上的任一正交变换\(\phi\)
    1. \(\phi={\rm id}_V\),则\(\phi=\psi\psi\),其中\(\psi\)是任一镜面反射。
    2. \(\phi\neq {\rm id}_V\),则存在\(V\)中单位向量\(\xi_1\),使得\(\phi(\xi_1)\neq \xi_1\)。令\(\eta_1=\phi(\xi_1)\),由\(\phi\)是正交变换知,\(\|\eta_1\|=\|\phi(\xi_1)\|=\|\xi_1\|=1\)。由上题结论知,存在镜面反射\(\phi_1\),使得\(\phi_1(\eta_1)=\xi_1\),则\(\phi_1\phi(\xi_1)=\xi_1\),即\(U=\langle \xi_1\rangle\)\(\phi_1\phi\)-不变子空间。由于\(\phi_1\phi\)仍是正交变换,所以由 练习 8.3.6 结论可知,\(U^\perp\)也是\(\phi_1\phi\)-不变子空间。故\(\left. (\phi_1\phi)\right|_{U^\perp}\)是正交变换,其中\(\dim U^\perp =n-1\)。由归纳假设,在\(U^\perp\)内存在单位向量\(\xi_2,\cdots,\xi_s\),它们分别决定\(U^\perp\)上的镜面反射\(\phi_2,\cdots ,\phi_s\),使得
      \begin{equation*} \left. (\phi_1\phi)\right|_{U^\perp}=\phi_2\cdots\phi_s. \end{equation*}
现将\(\phi_i(i=2,\cdots, s)\)定义范围扩大到\(V\)上,即补充定义\(\phi_i(\xi_1)=\xi_1\)
\(\forall \alpha\in V\),存在唯一的\(\alpha_1=a \xi_1\in U, \alpha_2\in U^\perp\),使得\(\alpha=\alpha_1+\alpha_2\),注意到
\begin{equation*} \phi_i(\alpha_1)=a\phi_i(\xi_1)=a \xi_1=\alpha_1, \end{equation*}
所以
\begin{equation*} \begin{array}{ccl} \phi_i(\alpha)&=&\phi_i(\alpha_1)+\phi_i(\alpha_2)=\alpha_1+\alpha_2-2(\xi_i,\alpha_2)\xi_i\\ &=&\alpha-2(\xi_i,\alpha_1+\alpha_2)\xi_i=\alpha-2(\xi_i,\alpha)\xi_i,\end{array} \end{equation*}
\(\phi_i\)\(\xi_i\)\(V\)上确定的镜面反射。又
\begin{equation*} \begin{array}{ccl} \phi_2\cdots\phi_s(\alpha)&=&\phi_2\cdots\phi_s(\alpha_1)+\phi_2\cdots\phi_s(\alpha_2)=\alpha_1+\phi_2\cdots\phi_s(\alpha_2)\\&=&\phi_1\phi(\alpha_1)+\phi_1\phi(\alpha_2)=\phi_1\phi(\alpha),\end{array} \end{equation*}
所以\(\phi_1\phi=\phi_2\cdots \phi_s\)。注意到\(\phi_1^2={\rm id}_V\),上式两边同时左乘\(\phi_1\),得
\begin{equation*} \phi=\phi_1\phi_2\cdots\phi_s \end{equation*}
是一系列镜面反射的乘积。
11.
\(\alpha_1,\dots,\alpha_m\)\(\beta_1,\dots,\beta_m\)\(n\)维欧氏空间\(V\)中的两个向量组,证明:存在\(V\)上的一个正交变换\(\phi\),使得
\begin{equation*} \phi(\alpha_i)=\beta_i,\ i=1,\dots,m \end{equation*}
的充分必要条件是
\begin{equation*} \left(\alpha_i,\alpha_j\right)=\left(\beta_i,\beta_j\right),\ i,j=1,\dots,m. \end{equation*}
解答.
必要性:因为\(\phi\)是正交变换且\(\phi(\alpha_i)=\beta_i\),所以
\begin{equation*} \left(\alpha_i,\alpha_j\right)=\left(\phi(\beta_i),\phi(\beta_j)\right)=\left(\beta_i,\beta_j\right),\ i,j=1,2,\cdots ,m. \end{equation*}
充分性:设\(\alpha_1,\dots ,\alpha_r\)\(\alpha_1,\dots ,\alpha_m\)的一个极大无关组,则
\begin{equation*} \det G(\alpha_1,\dots ,\alpha_r)\neq 0, \end{equation*}
这里\(G(\alpha_1,\dots ,\alpha_r)=\left((\alpha_i,\alpha_j)\right)_{r\times r}\)。由\(\left(\alpha_i,\alpha_j\right)=\left(\beta_i,\beta_j\right)\)
\begin{equation*} \det G(\beta_1,\dots ,\beta_r)=\det G(\alpha_1,\dots ,\alpha_r)\neq 0, \end{equation*}
\(\beta_1,\dots ,\beta_r\)线性无关。
\(\alpha_1,\dots ,\alpha_r\)\(\alpha_1,\dots ,\alpha_m\)的一个极大无关组知\(\forall 1\leq i\leq m\),存在\(a_{1i},\dots ,a_{ri}\in\mathbb{R}\),使得\(\alpha_i=a_{1i}\alpha_1+\dots +a_{ri}\alpha_r\),则
\begin{equation*} \begin{array}{ll} & \left(\beta_i-\sum\limits_{j=1}^ra_{ji}\beta_j,\beta_i-\sum\limits_{j=1}^ra_{ji}\beta_j\right)\\ =&\left(\alpha_i-\sum\limits_{j=1}^ra_{ji}\alpha_j,\alpha_i-\sum\limits_{j=1}^ra_{ji}\alpha_j\right)\\ &=0,\end{array} \end{equation*}
\(\beta_i=a_{1i}\beta_1+\cdots +a_{ri}\beta_r\),故\(\beta_1,\dots,\beta_r\)是向量组\(\beta_1,\dots ,\beta_m\)的一个极大无关组。令\(U_1=\langle \alpha_1,\dots ,\alpha_m\rangle ,\ U_2=\langle \beta_1,\dots ,\beta_m\rangle\),则\(\dim U_1=\dim U_2=r\)
定义\(U_1\)\(U_2\)的线性映射\(\phi_1\),满足\(\phi_1(\alpha_i)=\beta_i,\ i=1,\dots ,r\),则\(\phi_1\)是线性同构。由\(\left(\alpha_i,\alpha_j\right)=\left(\beta_i,\beta_j\right)\)知:\(\forall \alpha=\sum\limits_{k=1}^r a_k \alpha_k ,\beta=\sum\limits_{l=1}^r b_l \alpha_l\in U_1\),有
\begin{equation*} \begin{array}{ccl} (\phi_1(\alpha),\phi_1(\beta))&=&\left(\sum\limits_{k=1}^r a_k \phi_1(\alpha_k),\sum\limits_{l=1}^r b_l \phi_1(\alpha_l)\right)\\ &=&\sum\limits_{k=1}^r\sum\limits_{l=1}^r a_kb_l\left(\phi_1(\alpha_k),\phi_1(\alpha_l)\right)\\ &=&\sum\limits_{k=1}^r\sum\limits_{l=1}^r a_kb_l\left(\beta_k,\beta_l\right)\\ &=&\sum\limits_{k=1}^r\sum\limits_{l=1}^r a_kb_l\left(\alpha_k,\alpha_l\right)\\ &=&\left(\sum\limits_{k=1}^r a_k \alpha_k,\sum\limits_{l=1}^r b_l \alpha_l\right)\\ &=&(\alpha,\beta) \end{array} \end{equation*}
\(\phi_1\)是欧氏空间\(U_1\)\(U_2\)的同构映射。
又因为\(\dim U_1^\bot=\dim U_2^\bot=n-r\),所以存在欧氏空间同构映射
\begin{equation*} \phi_2:U_1^\bot\rightarrow U_2^\bot \end{equation*}
注意到\(V=U_1\oplus U_1^\bot\),所以可定义\(V\)上的变换\(\phi\)如下
\begin{equation*} \phi(\alpha+\beta)=\phi_1(\alpha)+\phi_2(\beta),\ \forall \alpha\in U_1,\ \beta\in U_1^\bot . \end{equation*}
不难验证,\(\phi\)\(V\)上的线性变换,下证\(\phi\)保内积。对任意\(X,Y\in V\),存在唯一的\(X_1,Y_1\in U_1, X_2,Y_2\in U_1^\bot\),使得\(X=X_1+X_2,\ Y=Y_1+Y_2\),则
\begin{equation*} \begin{array}{ccl} \left(\phi(X),\phi(Y)\right)&=&\left(\phi_1(X_1)+\phi_2(X_2),\phi_1(Y_1)+\phi_2(Y_2)\right)\\ &=&\left(\phi_1(X_1),\phi_1(Y_1)\right)+\left(\phi_1(X_1),\phi_2(Y_2)\right)\\ &&+\left(\phi_2(X_2),\phi_1(Y_1)\right)+\left(\phi_2(X_2),\phi_2(Y_2)\right)\\ &=&\left(\phi_1(X_1),\phi_1(Y_1)\right)+\left(\phi_2(X_2),\phi_2(Y_2)\right)\\ &=&(X_1,Y_1)+(X_2,Y_2)\\ &=&(X_1,Y_1)+(X_1,Y_2)+(X_2,Y_1)+(X_2,Y_2)\\ &=&(X,Y), \end{array} \end{equation*}
\(\phi\)保内积,进而\(\phi\)是正交变换,且对\(i=1,\dots,m\),有
\begin{equation*} \begin{array}{ll} \phi(\alpha_i)&=\phi_1(\alpha_i)\\ &=\phi_1(a_{1i}\alpha_1+\cdots +a_{ri}\alpha_r)\\ &=a_{1i}\beta_1+\dots +a_{ri}\beta_r\\ &=\beta_i.\end{array} \end{equation*}
12.
\begin{equation*} R_{xy}= \left\{ \left.\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\right| \theta\in \R \right\},\quad \mbox{(xoy平面上的旋转)} \end{equation*}
\begin{equation*} R_{yz}= \left\{ \left.\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}\right| \theta\in \R \right\},\quad \mbox{(yoz平面上的旋转)} \end{equation*}
证明:对欧氏空间\(\R^3\)上的第一类正交变换\(A\),存在\(B_1,B_2\in R_{xy}\)\(C\in R_{yz}\),使得
\begin{equation*} A = B_1CB_2. \end{equation*}