4.7 QR分解三：Givens旋转

Givens矩阵

  Givens矩阵有三个参数组成，$i,j,\theta$.它长这样：
G ( i , j , θ ) = ( I 0 0 0 0 0 cos ⁡ ( θ ) 0 sin ⁡ ( θ ) 0 0 0 I 0 0 0 − sin ⁡ ( θ ) 0 cos ⁡ ( θ ) 0 0 0 0 0 I ) G(i,j,\theta)=\begin{pmatrix} I & 0 & 0 & 0 & 0\\ 0 & \cos(\theta) & 0 & \sin(\theta) & 0 \\ 0 & 0 & I & 0 & 0\\ 0 & -\sin(\theta) & 0 & \cos(\theta) & 0\\ 0 & 0 & 0 & 0 & I \end{pmatrix}\\ G(i,j,θ)= ​I0000​0cos(θ)0−sin(θ)0​00I00​0sin(θ)0cos(θ)0​0000I​ ​
  如果角度不好计算也可以把$\cos \theta$和$\sin \theta$改成$c,s$，于是Given矩阵就由四个参数定义：
G ( i , j , c , s ) = ( I 0 0 0 0 0 c 0 s 0 0 0 I 0 0 0 − s 0 c 0 0 0 0 0 I ) G(i,j,c,s)=\begin{pmatrix} I & 0 & 0 & 0 & 0\\ 0 & c & 0 & s & 0 \\ 0 & 0 & I & 0 & 0\\ 0 & -s & 0 & c & 0\\ 0 & 0 & 0 & 0 & I \end{pmatrix}\\ G(i,j,c,s)= ​I0000​0c0−s0​00I00​0s0c0​0000I​ ​
  整个矩阵除了对角线和$a_{ii},a_{ij},a_{ji},a_{jj}$，其余位置都是0。对角线上除了第$i$行和第$j$行，其余位置都是1.
  Givens矩阵和Householder差不多，都可以把向量变成第一个坐标为它的长度，其余坐标变成0.不同的是这样变，只需要一个Householder矩阵，而Givens变换需要多个Givens矩阵。

旋转到标准基

  回忆利用Householder变换进行QR分解就是将向量投影到标准基上。Givens旋转也是一样，不过旋转比较麻烦。把任意向量x旋转到标准基$e_1$上，这个旋转变换是$n$个Givens矩阵的乘积。乘积的每一项定义如下：
$$T_{1i}=G(1,i,\frac{\sqrt{{\sum }_{j=1}^{i-1}{x}_j^2}}{\sqrt{{\sum }_{j=1}^i{x}_j^2}},\frac{a_i}{\sqrt{{\sum }_{j=1}^i{x}_j^2}})$$
  把所有的这些Givens矩阵倒序乘起来得到一个矩阵：
$$T=\prod _{i=n}^2{T}_{1i}$$
  这个矩阵乘以向量会把向量旋转到$e_1$方向，也就是第一个坐标为向量的长度，其余坐标为0.说得这么多不如举个例子:
x = ( 3 4 3   4 5 ) x=\begin{pmatrix}3\\ 4\\ 3\ 4\\ 5\\ \end{pmatrix} x= ​343 45​ ​
  分解步骤：
T 12 = ( 0.6 0.8 0 0 0 − 0.8 0.6 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 ) T 13 = ( 0.857 0 0.514 0 0 0 1 0 0 0 − 0.514 0 0.857 0 0 0 0 0 1 0 0 0 0 0 1 ) T 14 = ( 0.825 0 0 0.566 0 0 1 0 0 0 0 0 1 0 0 − 0.566 0 0 0.825 0 0 0 0 0 1 ) T 15 = ( 0.816 0 0 0 0.577 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 − 0.577 0 0 0 0.816 ) T = T 15 T 14 T 13 T 12 = ( 0.346 0.462 0.346 0.462 0.577 − 0.8 0.6 0 0 0 − 0.309 − 0.412 0.857 0 0 − 0.291 − 0.388 − 0.291 0.825 0 − 0.245 − 0.327 − 0.245 − 0.327 0.816 ) T x = ( 8.66 0 0 0 0 ) T_{12}=\begin{pmatrix}0.6 & 0.8 & 0 & 0 & 0\\ -0.8 & 0.6 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}\\ T_{13}=\begin{pmatrix}0.857 & 0 & 0.514 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ -0.514 & 0 & 0.857 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}\\ T_{14}=\begin{pmatrix}0.825 & 0 & 0 & 0.566 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ -0.566 & 0 & 0 & 0.825 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}\\ T_{15}= \begin{pmatrix}0.816 & 0 & 0 & 0 & 0.577\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ -0.577 & 0 & 0 & 0 & 0.816\\ \end{pmatrix}\\ T=T_{15}T_{14}T_{13}T_{12}=\begin{pmatrix}0.346 & 0.462 & 0.346 & 0.462 & 0.577\\ -0.8 & 0.6 & 0 & 0 & 0\\ -0.309 & -0.412 & 0.857 & 0 & 0\\ -0.291 & -0.388 & -0.291 & 0.825 & 0\\ -0.245 & -0.327 & -0.245 & -0.327 & 0.816\\ \end{pmatrix}\\ Tx=\begin{pmatrix}8.66\\ 0\\ 0\\ 0\\ 0\\ \end{pmatrix} T12​= ​0.6−0.8000​0.80.6000​00100​00010​00001​ ​T13​= ​0.8570−0.51400​01000​0.51400.85700​00010​00001​ ​T14​= ​0.82500−0.5660​01000​00100​0.566000.8250​00001​ ​T15​= ​0.816000−0.577​01000​00100​00010​0.5770000.816​ ​T=T15​T14​T13​T12​= ​0.346−0.8−0.309−0.291−0.245​0.4620.6−0.412−0.388−0.327​0.34600.857−0.291−0.245​0.462000.825−0.327​0.5770000.816​ ​Tx= ​8.660000​ ​

QR分解步骤

  思想和Householder变换一样，按列循环，对于第一列找到一个$T_1$把矩阵的第一列变成上三角矩阵的第一列。对于第二列，就用同样的办法，生成Givens矩阵，连乘起来变成变换阵，把第二行以下的数字变成0,但是这个时候Givens矩阵的$c,s$参数的计算方式要变了，变成这个样子：
$$T_{ki}=G(k,i,\frac{\sqrt{{\sum }_{j=k}^{i-1}{x}_j^2}}{\sqrt{{\sum }_{j=k}^i{x}_j^2}},\frac{a_i}{\sqrt{{\sum }_{j=k}^i{x}_j^2}})$$
  还是以这个向量为例子，通过一系列Givens矩阵可以将其第二行以下变成0：
T 23 = ( 1 0 0 0 0 0 0.8 0.6 0 0 0 − 0.6 0.8 0 0 0 0 0 1 0 0 0 0 0 1 ) T 24 = ( 1 0 0 0 0 0 0.781 0 0.625 0 0 0 1 0 0 0 − 0.625 0 0.781 0 0 0 0 0 1 ) T 25 = ( 1 0 0 0 0 0 0.788 0 0 0.615 0 0 1 0 0 0 0 0 1 0 0 − 0.615 0 0 0.788 ) T 2 = T 25 T 24 T 23 = ( 1 0 0 0 0 0 0.492 0.369 0.492 0.615 0 − 0.6 0.8 0 0 0 − 0.5 − 0.375 0.781 0 0 − 0.384 − 0.288 − 0.384 0.788 ) T 2 x = ( 3 8.124 0 0 0 ) T_{23}=\begin{pmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 0.8 & 0.6 & 0 & 0\\ 0 & -0.6 & 0.8 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}\\ T_{24}=\begin{pmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 0.781 & 0 & 0.625 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & -0.625 & 0 & 0.781 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{pmatrix}\\ T_{25}= \begin{pmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 0.788 & 0 & 0 & 0.615\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & -0.615 & 0 & 0 & 0.788\\ \end{pmatrix}\\ T_2=T_{25}T_{24}T_{23}=\begin{pmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 0.492 & 0.369 & 0.492 & 0.615\\ 0 & -0.6 & 0.8 & 0 & 0\\ 0 & -0.5 & -0.375 & 0.781 & 0\\ 0 & -0.384 & -0.288 & -0.384 & 0.788\\ \end{pmatrix}\\ T_2x=\begin{pmatrix}3\\ 8.124\\ 0\\ 0\\ 0\\ \end{pmatrix} T23​= ​10000​00.8−0.600​00.60.800​00010​00001​ ​T24​= ​10000​00.7810−0.6250​00100​00.62500.7810​00001​ ​T25​= ​10000​00.78800−0.615​00100​00010​00.615000.788​ ​T2​=T25​T24​T23​= ​10000​00.492−0.6−0.5−0.384​00.3690.8−0.375−0.288​00.49200.781−0.384​00.615000.788​ ​T2​x= ​38.124000​ ​
  所以完整的QR分解就是这样按列迭代。但是要注意的是这一系列迭代是这样的过程：
$$\begin{gathered} T_{n-1}\cdots T_{1}A=R \\ A=T_1^{-1}{T}_2^{-1}\cdots T_{n-1}^{-1}R=QR \end{gathered}$$
  Givens矩阵是正交阵，他们的乘积也是正交阵，所以有：
$$T_i^{-1}=T_i^T$$
  所以有：
$$Q=T_1^T{T}_2^T\cdots T_{n-1}^T$$

Python代码

  经过上述讲解，代码写起来就容易了,最终代码如下：

   @staticmethod
    def givens_matrix(n, i, j, c, s):
        array = Matrix.unit_matrix(n)
        array[i][i] = c
        array[j][j] = c

        array[i][j] = -s
        array[j][i] = s

        return Matrix(array)

    @staticmethod
    def get_givens_matrix(vector, line, i):
        sum_vector = 0
        for j in range(line, i):
            sum_vector += vector[j] * vector[j]
        numerator = math.sqrt(sum_vector)
        vector_i_pow = vector[i] * vector[i]
        denominator = math.sqrt(sum_vector + vector_i_pow)
        alpha = numerator / denominator
        beta = math.sqrt(vector_i_pow) / denominator
        return Matrix.givens_matrix(len(vector), line, i, alpha, beta)

    @staticmethod
    def get_givens_rotator(vector, line):
        rotator = Matrix(Matrix.unit_matrix(len(vector)))
        n = len(vector)
        for i in range(n - 1, line, -1):
            givens_matrix = Matrix.get_givens_matrix(vector, line, i)
            rotator = rotator * givens_matrix
        return rotator

    def givens_qr(self):
        n = len(self.__vectors)
        r = self
        q = Matrix(Matrix.unit_matrix(n))
        for i in range(n-1):
            vector = r.__vectors[i]
            givens = Matrix.get_givens_rotator(vector, i)
            # 正交矩阵的逆矩阵就是它的转置矩阵
            q = q * givens.transpose_matrix()
            r = givens * r
        return q, r