矩阵分解相关知识点总结（三）

书接上回，CSDN发不了长文，只能这样分开发了😔

  完成矩阵的QR分解有三种常用的方法：Schmidt正交化方法，Givens变换方法和Householder变换方法。下面通过一个具体实例，将矩阵$A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]$分别用上述三种方法进行QR分解。

Schmidt正交化方法：

  令${\mathbf{a}}_1=(1,2,1{)}^{\mathrm{T}},{\mathbf{a}}_2=(2,1,2{)}^{\mathrm{T}},{\mathbf{a}}_3=(2,2,1{)}^{\mathrm{T}}$，将其Schmidt正交化可得：
$$\begin{array}{l}{\mathbf{b}}_1={\mathbf{a}}_1=(1,2,1{)}^{\mathrm{T}}\\ {\mathbf{b}}_2={\mathbf{a}}_2-\frac{({\mathbf{a}}_2,{\mathbf{b}}_1)}{({\mathbf{b}}_1,{\mathbf{b}}_1)}{\mathbf{b}}_1={\mathbf{a}}_2-{\mathbf{b}}_1=(1,-1,1{)}^{\mathrm{T}}\\ {\mathbf{b}}_3={\mathbf{a}}_3-\frac{({\mathbf{a}}_3,{\mathbf{b}}_1)}{({\mathbf{b}}_1,{\mathbf{b}}_1)}{\mathbf{b}}_1-\frac{({\mathbf{a}}_3,{\mathbf{b}}_2)}{({\mathbf{b}}_2,{\mathbf{b}}_2)}{\mathbf{b}}_2={\mathbf{a}}_3-\frac{1}{3}{\mathbf{b}}_2-\frac{7}{6}{\mathbf{b}}_1=(\frac{1}{2},0,-\frac{1}{2}{)}^{\mathrm{T}}\end{array}$$

进而有：
$$({\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3)=({\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3)\cdot C=({\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3)\left[\begin{array}{ccc}1& 1& \frac{7}{6}\\ 0& 1& \frac{1}{3}\\ 0& 0& 1\end{array}\right]$$

  取：
$$\begin{array}{l}{\mathbf{q}}_1=\frac{1}{|{\mathbf{b}}_1|}|{\mathbf{b}}_1|=\frac{1}{\sqrt{6}}(1,2,1{)}^{\mathrm{T}}\\ {\mathbf{q}}_2=\frac{1}{|{\mathbf{b}}_2|}|{\mathbf{b}}_2|=\frac{1}{\sqrt{3}}(1,-1,1{)}^{\mathrm{T}}\\ {\mathbf{q}}_3=\frac{1}{|{\mathbf{b}}_3|}|{\mathbf{b}}_3|=\frac{1}{\sqrt{2}}(1,0,-1{)}^{\mathrm{T}}\end{array}$$

  令：
$$\begin{array}{l}Q=({\mathbf{q}}_1,{\mathbf{q}}_2,{\mathbf{q}}_3)=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& -\frac{1}{\sqrt{3}}& 0\\ \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{2}}\end{array}\right]\\ R=\mathrm{diag}(|{\mathbf{b}}_1|,|{\mathbf{b}}_2|,|{\mathbf{b}}_3|)\cdot \mathrm{C}=\left[\begin{array}{ccc}\sqrt{6}& 0& 0\\ 0& \sqrt{3}& 0\\ 0& 0& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}1& 1& \frac{7}{6}\\ 0& 1& \frac{1}{3}\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\sqrt{6}& \sqrt{6}& \frac{7}{\sqrt{6}}\\ 0& \sqrt{3}& \frac{1}{\sqrt{3}}\\ 0& 0& \frac{1}{\sqrt{2}}\end{array}\right]\end{array}$$

则有$A=QR$。

Givens变换方法：
  第1步，对$A^{(0)}=A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]$的第1列${\mathbf{b}}^{(1)}=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]$构造旋转矩阵$T_1$，使$T_1{\mathbf{b}}^{(1)}=|{\mathbf{b}}^{(1)}|{\mathbf{e}}_1$：

$$T_1=T_{13}{T}_{12}=\left[\begin{array}{ccc}\frac{\sqrt{5}}{\sqrt{6}}& 0& \frac{1}{\sqrt{6}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{6}}& 0& \frac{\sqrt{5}}{\sqrt{6}}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{5}}& \frac{2}{\sqrt{5}}& 0\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 0\\ -\frac{1}{\sqrt{30}}& -\frac{2}{\sqrt{30}}& \frac{\sqrt{5}}{\sqrt{6}}\end{array}\right]$$

$$T_1{\mathbf{b}}^{(1)}=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 0\\ -\frac{1}{\sqrt{30}}& -\frac{2}{\sqrt{30}}& \frac{\sqrt{5}}{\sqrt{6}}\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}\sqrt{6}\\ 0\\ 0\end{array}\right]$$

$$T_1{A}^{(0)}=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 0\\ -\frac{1}{\sqrt{30}}& -\frac{2}{\sqrt{30}}& \frac{\sqrt{5}}{\sqrt{6}}\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]=\left[\begin{array}{ccc}\sqrt{6}& \sqrt{6}& \frac{7}{\sqrt{6}}\\ 0& -\frac{3}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\\ 0& \frac{6}{\sqrt{30}}& -\frac{1}{\sqrt{30}}\end{array}\right]$$

  第2步，对$A^{(1)}=\left[\begin{array}{cc}-\frac{3}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\\ \frac{6}{\sqrt{30}}& -\frac{1}{\sqrt{30}}\end{array}\right]$的第1列${\mathbf{b}}^{(2)}=\left[\begin{array}{c}-\frac{3}{\sqrt{5}}\\ \frac{6}{\sqrt{30}}\end{array}\right]$构造旋转矩阵$T_2$，使$T_2{\mathbf{b}}^{(2)}=|{\mathbf{b}}^{(2)}|{\mathbf{e}}_1$：

$$T_2=T_{12}=\left[\begin{array}{cc}-\frac{\sqrt{3}}{\sqrt{5}}& \frac{2}{\sqrt{10}}\\ -\frac{2}{\sqrt{10}}& -\frac{\sqrt{3}}{\sqrt{5}}\end{array}\right]$$

$$T_2{\mathbf{b}}^{(2)}=\left[\begin{array}{cc}-\frac{\sqrt{3}}{\sqrt{5}}& \frac{2}{\sqrt{10}}\\ -\frac{2}{\sqrt{10}}& -\frac{\sqrt{3}}{\sqrt{5}}\end{array}\right]\left[\begin{array}{c}-\frac{3}{\sqrt{5}}\\ \frac{6}{\sqrt{30}}\end{array}\right]=\left[\begin{array}{c}\sqrt{3}\\ 0\end{array}\right]$$

$$T_2{A}^{(1)}=\left[\begin{array}{cc}-\frac{\sqrt{3}}{\sqrt{5}}& \frac{2}{\sqrt{10}}\\ -\frac{2}{\sqrt{10}}& -\frac{\sqrt{3}}{\sqrt{5}}\end{array}\right]\left[\begin{array}{cc}-\frac{3}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\\ \frac{6}{\sqrt{30}}& -\frac{1}{\sqrt{30}}\end{array}\right]=\left[\begin{array}{cc}\sqrt{3}& \frac{1}{\sqrt{3}}\\ 0& \frac{1}{\sqrt{2}}\end{array}\right]$$

  最后，令
$$T=\left[\begin{array}{cc}1& O^{\text{T}}\\ O& T_2\end{array}\right]T_1=\left[\begin{array}{ccc}1& 0& 0\\ 0& -\frac{\sqrt{3}}{\sqrt{5}}& \frac{2}{\sqrt{10}}\\ 0& -\frac{2}{\sqrt{10}}& -\frac{\sqrt{3}}{\sqrt{5}}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}& 0\\ -\frac{1}{\sqrt{30}}& -\frac{2}{\sqrt{30}}& \frac{\sqrt{5}}{\sqrt{6}}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\end{array}\right]$$

$$Q=T^{\text{T}}==\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& -\frac{1}{\sqrt{3}}& 0\\ \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{2}}\end{array}\right],R=TA=\left[\begin{array}{ccc}\sqrt{6}& \sqrt{6}& \frac{7}{\sqrt{6}}\\ 0& \sqrt{3}& \frac{1}{\sqrt{3}}\\ 0& 0& \frac{1}{\sqrt{2}}\end{array}\right]$$

则有$A=QR$。

Householder变换方法：
  第1步，对$A^{(0)}=A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]$的第1列${\mathbf{b}}^{(1)}=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]$构造Householder矩阵$H_1$，使$H_1{\mathbf{b}}^{(1)}=|{\mathbf{b}}^{(1)}|{\mathbf{e}}_1$：
$${\mathbf{b}}^{(1)}-|{\mathbf{b}}^{(1)}|{\mathbf{e}}_1=\left[\begin{array}{c}1\\ 2\\ 1\end{array}\right]-\sqrt{6}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}1-\sqrt{6}\\ 2\\ 1\end{array}\right]$$

$$\mathbf{u}=\frac{{\mathbf{b}}^{(1)}-|{\mathbf{b}}^{(1)}|{\mathbf{e}}_1}{|{\mathbf{b}}^{(1)}-|{\mathbf{b}}^{(1)}|{\mathbf{e}}_1|}=\frac{1}{\sqrt{12-2\sqrt{6}}}\left[\begin{array}{c}1-\sqrt{6}\\ 2\\ 1\end{array}\right]$$

$$H_1=I-2{\mathbf{uu}}^{\mathrm{T}}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]-\frac{1}{6-\sqrt{6}}\left[\begin{array}{c}1-\sqrt{6}\\ 2\\ 1\end{array}\right]\left[\begin{array}{ccc}1-\sqrt{6}& 2& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ \frac{2}{\sqrt{6}}& \frac{\sqrt{6}-2}{\sqrt{6}-6}& \frac{2}{\sqrt{6}-6}\\ \frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}-6}& \frac{\sqrt{6}-5}{\sqrt{6}-6}\end{array}\right]$$

$$H_1{A}^{(0)}=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ \frac{2}{\sqrt{6}}& \frac{\sqrt{6}-2}{\sqrt{6}-6}& \frac{2}{\sqrt{6}-6}\\ \frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}-6}& \frac{\sqrt{6}-5}{\sqrt{6}-6}\end{array}\right]\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]=\left[\begin{array}{ccc}\sqrt{6}& \sqrt{6}& \frac{7}{\sqrt{6}}\\ 0& \frac{3-\sqrt{6}}{\sqrt{6}-1}& \frac{2}{\sqrt{6}}\\ 0& \frac{\sqrt{6}}{\sqrt{6}-1}& \frac{1}{\sqrt{6}}\end{array}\right]$$

  第2步，对$A^{(1)}=\left[\begin{array}{cc}\frac{3-\sqrt{6}}{\sqrt{6}-1}& \frac{2}{\sqrt{6}}\\ \frac{\sqrt{6}}{\sqrt{6}-1}& \frac{1}{\sqrt{6}}\end{array}\right]$的第1列${\mathbf{b}}^{(2)}=\left[\begin{array}{c}\frac{3-\sqrt{6}}{\sqrt{6}-1}\\ \frac{\sqrt{6}}{\sqrt{6}-1}\end{array}\right]$构造Householder矩阵$H_2$，使$H_2{\mathbf{b}}^{(2)}=|{\mathbf{b}}^{(2)}|{\mathbf{e}}_1$：
$${\mathbf{b}}^{(2)}-|{\mathbf{b}}^{(2)}|{\mathbf{e}}_1=\left[\begin{array}{c}\frac{3-\sqrt{6}}{\sqrt{6}-1}\\ \frac{\sqrt{6}}{\sqrt{6}-1}\end{array}\right]-\frac{\sqrt{21-6\sqrt{6}}}{\sqrt{6}-1}\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}\frac{3-\sqrt{6}-\sqrt{21-6\sqrt{6}}}{\sqrt{6}-1}\\ \frac{\sqrt{6}}{\sqrt{6}-1}\end{array}\right]=\left[\begin{array}{c}\frac{x}{\sqrt{6}-1}\\ \frac{\sqrt{6}}{\sqrt{6}-1}\end{array}\right]$$

$$\mathbf{u}=\frac{{\mathbf{b}}^{(2)}-|{\mathbf{b}}^{(2)}|{\mathbf{e}}_1}{|{\mathbf{b}}^{(2)}-|{\mathbf{b}}^{(2)}|{\mathbf{e}}_1|}=\frac{\sqrt{6}-1}{\sqrt{x^2+6}}\left[\begin{array}{c}\frac{x}{\sqrt{6}-1}\\ \frac{\sqrt{6}}{\sqrt{6}-1}\end{array}\right]=\left[\begin{array}{c}\frac{x}{\sqrt{x^2+6}}\\ \frac{\sqrt{6}}{\sqrt{x^2+6}}\end{array}\right]$$

$$H_2=I-2{\mathbf{uu}}^{\text{T}}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-2\left[\begin{array}{c}\frac{x}{\sqrt{x^2+6}}\\ \frac{\sqrt{6}}{\sqrt{x^2+6}}\end{array}\right]\left[\begin{array}{cc}\frac{x}{\sqrt{x^2+6}}& \frac{\sqrt{6}}{\sqrt{x^2+6}}\end{array}\right]=\left[\begin{array}{cc}\frac{6-x^2}{x^2+6}& \frac{-2\sqrt{6}x}{x^2+6}\\ \frac{-2\sqrt{6}x}{x^2+6}& \frac{x^2-6}{x^2+6}\end{array}\right]$$

$$H_2{A}^{(1)}=\left[\begin{array}{cc}\frac{6-x^2}{x^2+6}& \frac{-2\sqrt{6}x}{x^2+6}\\ \frac{-2\sqrt{6}x}{x^2+6}& \frac{x^2-6}{x^2+6}\end{array}\right]\left[\begin{array}{cc}\frac{3-\sqrt{6}}{\sqrt{6}-1}& \frac{2}{\sqrt{6}}\\ \frac{\sqrt{6}}{\sqrt{6}-1}& \frac{1}{\sqrt{6}}\end{array}\right]=\left[\begin{array}{cc}\sqrt{3}& \frac{1}{\sqrt{3}}\\ 0& \frac{1}{\sqrt{2}}\end{array}\right]$$

  最后，令
$$S=\left[\begin{array}{cc}1& O^{\text{T}}\\ O& H_2\end{array}\right]H_1=\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{6-x^2}{x^2+6}& \frac{-2\sqrt{6}x}{x^2+6}\\ 0& \frac{-2\sqrt{6}x}{x^2+6}& \frac{x^2-6}{x^2+6}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ \frac{2}{\sqrt{6}}& \frac{\sqrt{6}-2}{\sqrt{6}-6}& \frac{2}{\sqrt{6}-6}\\ \frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}-6}& \frac{\sqrt{6}-5}{\sqrt{6}-6}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\end{array}\right]$$

$$Q=S^{\mathrm{T}}==\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& -\frac{1}{\sqrt{3}}& 0\\ \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{2}}\end{array}\right],R=SA=\left[\begin{array}{ccc}\sqrt{6}& \sqrt{6}& \frac{7}{\sqrt{6}}\\ 0& \sqrt{3}& \frac{1}{\sqrt{3}}\\ 0& 0& \frac{1}{\sqrt{2}}\end{array}\right]$$

则有$A=QR$。（感兴趣的读者可用MATLAB验证结果，没想到一个小小的矩阵计算这么复杂(╯﹏╰)）