第四章 （四）

1.Givens旋转

[ C S − S C ] [ a b ] = [ a 2 + b 2 0 ]

= [C−S​SC​][ab​]=[a2+b2 ​0​]
G = [ C S − S C ] 为正交矩阵 G=

[CS−SC]" role="presentation" style="position: relative;">[C−SSC]$$\left[\begin{array}{cc}C& S\\ -S& C\end{array}\right]$$

为正交矩阵 G=[C−S​SC​]为正交矩阵
得
{ C a + S b = a 2 + b 2 − S a + C b = 0 C 2 + S 2 = 1 ⇒ { C = a a 2 + b 2 = c o s θ S = b a 2 + b 2 = s i n θ

{Ca+Sb=a2+b2−Sa+Cb=0C2+S2=1" role="presentation" style="position: relative;">⎧⎩⎨Ca+Sb=a2+b2−−−−−−√−Sa+Cb=0C2+S2=1$$\{\begin{array}{l}Ca+Sb=\sqrt{a^2+b^2}\\ -Sa+Cb=0\\ C^2+S^2=1\end{array}$$

\Rightarrow

{C=aa2+b2=cosθS=ba2+b2=sinθ" role="presentation" style="position: relative;">⎧⎩⎨C=aa2+b2√=cosθS=ba2+b2√=sinθ$$\{\begin{array}{l}C=\frac{a}{\sqrt{a^2+b^2}}=cos\theta \\ S=\frac{b}{\sqrt{a^2+b^2}}=sin\theta \end{array}$$

⎩ ⎨ ⎧​Ca+Sb=a2+b2 ​−Sa+Cb=0C2+S2=1​⇒{C=a2+b2 ​a​=cosθS=a2+b2 ​b​=sinθ​
这个$G$就是旋转矩阵了
两个向量模长相等
类似地
G ( i , j , θ ) = [ 1 ⋯ 0 ⋯ 0 ⋯ 0 ⋮ ⋱ ⋮ ⋮ ⋮ 0 ⋯ C ⋯ S ⋯ 0 ⋮ ⋮ ⋱ ⋮ ⋮ 0 ⋯ − S ⋯ C ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 0 ⋯ 0 ⋯ 0 ⋯ 1 ] G(i,j,\theta)=

[1⋯0⋯0⋯0⋮⋱⋮⋮⋮0⋯C⋯S⋯0⋮⋮⋱⋮⋮0⋯−S⋯C⋯0⋮⋮⋮⋱⋮0⋯0⋯0⋯1]" role="presentation" style="position: relative;">⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢1⋮0⋮0⋮0⋯⋱⋯⋯⋯0⋮C⋮−S⋮0⋯⋯⋱⋯⋯0⋮S⋮C⋮0⋯⋯⋯⋱⋯0⋮0⋮0⋮1⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥$$\left[\begin{array}{ccccccc}1& \cdots & 0& \cdots & 0& \cdots & 0\\ ⋮& \ddots & ⋮& & ⋮& & ⋮\\ 0& \cdots & C& \cdots & S& \cdots & 0\\ ⋮& & ⋮& \ddots & ⋮& & ⋮\\ 0& \cdots & -S& \cdots & C& \cdots & 0\\ ⋮& & ⋮& & ⋮& \ddots & ⋮\\ 0& \cdots & 0& \cdots & 0& \cdots & 1\end{array}\right]$$

G(i,j,θ)=⎣ ⎡​1⋮0⋮0⋮0​⋯⋱⋯⋯⋯​0⋮C⋮−S⋮0​⋯⋯⋱⋯⋯​0⋮S⋮C⋮0​⋯⋯⋯⋱⋯​0⋮0⋮0⋮1​⎦ ⎤​
其中$g_{kk}=1,k\ne i,j$
$g_{i,i}=g_{j,j}=C$
$g_{i,j}=S$
$g_{j,i}=-S$
其余元素为0
$Gx=y$,将$x$在$i,j$平面，顺时针旋转$\theta$,得到$y$

2. 采用Givens旋转的QR分解

Givens旋转也可以用来计算QR分解。
这里以$4\times 3$矩阵为例，说明Givens QR分解的思想:

其中，$\otimes$表示用Givens旋转进行变换的元素。
从上述说明中易得出结论:如果令$G_j$代表约化过程中的第$j$次Givens旋转，则$Q^{T}A=R$是上三角矩阵，其中，$Q=G_t{G}_{t-1}\dots G_1$，而$t$是总的旋转次数。

例
用Givens旋转求下列线性方程的最小二乘解。
$x_1+2x_2=5$
$2x_1+3x_2=8$
$6x_1+7x_2=21$
解构造增广矩阵
X = [ 1 2 5 2 3 8 6 7 21 ] X=

X=⎣ ⎡​126​237​5821​⎦ ⎤​
先用Givens旋转消去$X$的(3,1)元素6。为此，选择$c=-01663，s=0.9859$，则有
G 31 = [ − 0.1663 0 − 0.9859 0 1 0 0.9859 0 − 0.1663 ] → G 31 X = [ − 6.081 − 7.234 − 21.536 2 3 8 0 0.808 1.438 ] G_{31}=

[−0.16630−0.98590100.98590−0.1663]" role="presentation" style="position: relative;">⎡⎣⎢−0.166300.9859010−0.98590−0.1663⎤⎦⎥$$\left[\begin{array}{ccc}-0.1663& 0& -0.9859\\ 0& 1& 0\\ 0.9859& 0& -0.1663\end{array}\right]$$

\rightarrow G_{31}X=

[−6.081−7.234−21.53623800.8081.438]" role="presentation" style="position: relative;">⎡⎣⎢−6.08120−7.23430.808−21.53681.438⎤⎦⎥$$\left[\begin{array}{ccc}-6.081& -7.234& -21.536\\ 2& 3& 8\\ 0& 0.808& 1.438\end{array}\right]$$

G31​=⎣ ⎡​−0.166300.9859​010​−0.98590−0.1663​⎦ ⎤​→G31​X=⎣ ⎡​−6.08120​−7.23430.808​−21.53681.438​⎦ ⎤​
为了消去$G_{31}X$的(2,1)元素2，选择c=0.952,s=0.313，故
G 21 = [ 0.952 − 0.313 0 0.313 0.952 0 0 0 1 ] → G 21 G 31 X = [ − 6.081 − 7.234 − 23.008 0 0.592 0.875 0 0.808 1.438 ] G_{21}=

[0.952−0.31300.3130.9520001]" role="presentation" style="position: relative;">⎡⎣⎢0.9520.3130−0.3130.9520001⎤⎦⎥$$\left[\begin{array}{ccc}0.952& -0.313& 0\\ 0.313& 0.952& 0\\ 0& 0& 1\end{array}\right]$$

\rightarrow G_{21}G_{31}X=

[−6.081−7.234−23.00800.5920.87500.8081.438]" role="presentation" style="position: relative;">⎡⎣⎢−6.08100−7.2340.5920.808−23.0080.8751.438⎤⎦⎥$$\left[\begin{array}{ccc}-6.081& -7.234& -23.008\\ 0& 0.592& 0.875\\ 0& 0.808& 1.438\end{array}\right]$$

G21​=⎣ ⎡​0.9520.3130​−0.3130.9520​001​⎦ ⎤​→G21​G31​X=⎣ ⎡​−6.08100​−7.2340.5920.808​−23.0080.8751.438​⎦ ⎤​
在$G_{32}$中选择c=-0.6527，s=0.890，消去$G_{21}{G}_{31}X$中的(3,2)元素，得QR分解
G 32 = [ 1 0 0 0 − 0.6527 − 0.8900 0 0.8900 − 0.6527 ] → G 32 G 21 G 31 X = [ − 6.415 − 7.826 − 23.008 0 − 1.105 − 1.851 0 0 − 0.160 ] G_{32}=

[1000−0.6527−0.890000.8900−0.6527]" role="presentation" style="position: relative;">⎡⎣⎢1000−0.65270.89000−0.8900−0.6527⎤⎦⎥$$\left[\begin{array}{ccc}1& 0& 0\\ 0& -0.6527& -0.8900\\ 0& 0.8900& -0.6527\end{array}\right]$$

\rightarrow G_{32}G_{21}G_{31}X=

[−6.415−7.826−23.0080−1.105−1.85100−0.160]" role="presentation" style="position: relative;">⎡⎣⎢−6.41500−7.826−1.1050−23.008−1.851−0.160⎤⎦⎥$$\left[\begin{array}{ccc}-6.415& -7.826& -23.008\\ 0& -1.105& -1.851\\ 0& 0& -0.160\end{array}\right]$$

G32​=⎣ ⎡​100​0−0.65270.8900​0−0.8900−0.6527​⎦ ⎤​→G32​G21​G31​X=⎣ ⎡​−6.41500​−7.826−1.1050​−23.008−1.851−0.160​⎦ ⎤​
由此得线性方程组
$-6.415x_1-7.826x_2=-23.008$
$-1.105x_2=-1.851$
其解为$x_1=1.543$和$x_2=1.675$。