【国科大矩阵论】矩阵分解复习

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前言

作者进行矩阵分解章节复习总结的几种计算。
矩阵分解主要有LDU分解、Cholesky分解、QR分解、满秩分解、奇异值分解。
本文不讲理论知识，只是理清计算过程。

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提示：以下是本篇文章正文内容，下面例题可供参考

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一、LDU分解

三角分解，其中还有上三角和下三角形式。
求$A=\left[\begin{array}{lll}2& -1& 3\\ 1& 2& 1\\ 2& 4& 2\end{array}\right]$的LDU分解

解答：

(1)根据A构造第1列的单位下三角矩阵

$$L_1=\left[\begin{array}{lll}1& & \\ \frac{1}{2}& 1& \\ 1& 0& 1\end{array}\right]$$ $${L_1}^{-1}=\left[\begin{array}{lll}1& & \\ -\frac{1}{2}& 1& \\ -1& 0& 1\end{array}\right]$$

问：怎么得来的$\frac{1}{2}和1？$
$$\begin{gathered} \frac{a_{21}}{a_{11}}=\frac{1}{2} \\ \frac{a_{31}}{a_{a11}}=\frac{2}{2}=1 \end{gathered}$$

$${L_1}^{-1}A=\left[\begin{array}{lll}2& -1& 3\\ 0& \frac{5}{2}& -\frac{1}{2}\\ 0& 5& -1\end{array}\right]=A^{(1)}$$

(2)根据$A^{(1)}$第2列构造单位下三角矩阵

$$L_2=\left[\begin{array}{lll}1& & \\ 0& 1& \\ 0& 2& 1\end{array}\right]$$ $${L_2}^{-1}=\left[\begin{array}{lll}1& & \\ 0& 1& \\ 0& -2& 1\end{array}\right]$$

$${L_2}^{-1}{A}^{(1)}=\left[\begin{array}{lll}2& -1& 3\\ 0& \frac{5}{2}& -\frac{1}{2}\\ 0& 0& 0\end{array}\right]=A^{(2)}$$

此时可以看到矩阵为上三角形式，因此可以分解成如下形式，得出对角矩阵和单位上三角矩阵

$$A^{(2)}=\left[\begin{array}{lll}2& & \\ & \frac{5}{2}& \\ & & 0\end{array}\right]\left[\begin{array}{lll}1& -\frac{1}{2}& \frac{3}{2}\\ & 1& -\frac{1}{5}\\ & & 1\end{array}\right]=DU$$

$$L=L_1{L}_2=\left[\begin{array}{lll}1& & \\ \frac{1}{2}& 1& \\ 1& 2& 1\end{array}\right]$$

因此A是经过L变换得到的DU形式

$$A=LDU$$
$$A=\left[\begin{array}{lll}1& & \\ \frac{1}{2}& 1& \\ 1& 2& 1\end{array}\right]\left[\begin{array}{lll}2& & \\ & \frac{5}{2}& \\ & & 0\end{array}\right]\left[\begin{array}{lll}1& -\frac{1}{2}& \frac{3}{2}\\ & 1& -\frac{1}{5}\\ & & 1\end{array}\right]$$

二、Cholesky分解

$求A=\left[\begin{array}{lll}2& -1& 3\\ 1& 2& 1\\ 2& 4& 2\end{array}\right]的Cholesky分解$

解答：过程与LDU一致，只是多了一个步骤

$$A=L(\sqrt{D}\sqrt{D})U=L\sqrt{D}(L\sqrt{D}{)}^T=G{G}^T$$

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三、QR分解

最熟悉的就是正交三角分解方法，但是某些时候运算比较复杂，所以了解一下后两种方法很有必要。

1. 正交三角分解

$A=\left[\begin{array}{lll}1& 2& 1\\ 2& 1& 2\\ 1& 2& 1\end{array}\right]，QR分解$

解答：

(1)取A的各列

$$a_1=(1,2,1{)}^T,a_2=(2,1,2{)}^T,a_3=(2,2,1{)}^T$$

(2)Schmide正交化

$$b_1=a_1=(1,2,1{)}^T,|b_1|=\sqrt{6}$$

$$\begin{gathered} b_2=a_2-k_{21}{b}_1=a_2-b_1(1,-1,1{)}^T \\ |b_2|=\sqrt{3},k_{21}=\frac{(a2,b1)}{(b1,b1)}=1 \end{gathered}$$

$$\begin{gathered} b_2=a_2-k_{21}{b}_1=a_2-b_1=(1,-1,1{)}^T \\ |b_2|=\sqrt{3},k_{21}=\frac{(a2,b1)}{(b1,b1)}=1 \end{gathered}$$

(3)正交矩阵

$$Q=\left[\begin{array}{lll}1& 1& \frac{1}{2}\\ 2& -1& 0\\ 1& 1& -\frac{1}{2}\end{array}\right]$$

$$C=\left[\begin{array}{lll}1& k_{21}& k_{31}\\ & 1& k_{32}\\ & & 1\end{array}\right]=\left[\begin{array}{lll}1& 1& \frac{7}{6}\\ & 1& \frac{1}{3}\\ & & 1\end{array}\right]$$

$$R=diag(|b_1|,|b_2|,|b_3|)C$$
$$R=\left[\begin{array}{lll}\sqrt{6}& & \\ & \sqrt{3}& \\ & & \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{lll}1& 1& \frac{7}{6}\\ & 1& \frac{1}{3}\\ & & 1\end{array}\right]$$

$$\therefore A=QR$$

2. Givens变换方法（初等旋转）

$$A=\left[\begin{array}{lll}0& 1& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]，QR分解$$
解答：

(1)$对A的第1列a_1=(0,1,1{)}^T，构造T_{12}$

$$T_1{a}_1=|a_1|e_1=\sqrt{2}{e}_1$$ $$T_{12}(c,s),c=0,s=1$$ $$T_{12}=\left[\begin{array}{lll}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right],T_{12}{a}_1=\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]$$

(2)$对T_1{a}_1构造T_{13}$

$$T_{13}(c,s),c=\frac{1}{\sqrt{2}},s=\frac{1}{\sqrt{2}}$$

$$T_{13}=\left[\begin{array}{lll}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$$

得到变换矩阵$T_1$

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

$A经过T_1$变换

$$T_{1}A=\left[\begin{array}{ccc}\sqrt{2}& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ 0& -1& -1\\ 0& -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$$

(3)得到新矩阵，继续构造变换矩阵

$令A^{(1)}=\left[\begin{array}{cc}-1& -1\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]，b_1={\left(-1,-\frac{1}{\sqrt{2}}\right)}^T构造T_2$

$$T_{12}=\left[\begin{array}{cc}-\frac{\sqrt{2}}{3}& -\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}& -\sqrt{\frac{2}{3}}\end{array}\right]T_{12}{b}_1=\left[\begin{array}{c}\frac{\sqrt{3}}{2}\\ 0\end{array}\right]$$

$得到变换矩阵T_2$

$$\begin{array}{rl}& T_2=T_{12}=\left[\begin{array}{cc}-\frac{\sqrt{2}}{3}& -\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}& -\frac{\sqrt{2}}{3}\end{array}\right]\end{array}$$

$A_{(1)}经由T_2变换$

$$\begin{array}{r}{T}_2{A}^{(1)}=\left[\begin{array}{cc}-\sqrt{\frac{2}{3}}& \frac{1}{\sqrt{6}}\\ 0& -\frac{2}{\sqrt{3}}\end{array}\right]\end{array}$$

(3)已经得到了上三角矩阵，得到了最后的变换矩阵T，写出分解结果

$$T=\left[\begin{array}{l}1\\ T_2\end{array}\right]T_1$$

$$Q=T^T,R=TA$$

$$A=QR$$

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3. Householder变换方法（初等反射）

$$A=\left[\begin{array}{ccc}3& 14& 9\\ 6& 43& 3\\ 6& 22& 15\end{array}\right]，QR分解$$

解答：

（1）对A的第1列$a_1=(3,6,6{)}^T$构造H变换

$$\begin{array}{rl}& a_1-\left|a_1\right|e_1=6\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]\\ & \mu =\frac{a_1-\left|a_1\right|e_1}{\left|a_1-\right|a_1\left|e_1\right|}=\frac{1}{\sqrt{3}}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]\\ & H_1=1-2\mu {\mu }^{\top }=\frac{1}{3}\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]\\ & H_{1}A=\left[\begin{array}{ccc}9& 48& 15\\ 0& 9& -3\\ 0& -12& 9\end{array}\right]\end{array}$$

$由H_1变换得到A^{(1)}$

$$A^{(1)}=\left[\begin{array}{cc}9& -3\\ -12& 9\end{array}\right]$$

（2）对$A^{(1)}$的第1列$b_1=\left[\begin{array}{c}9\\ -12\end{array}\right]$构造H变换矩阵

$$\begin{array}{rl}& b_1-\left|b_1\right|e_1=6\left[\begin{array}{c}-1\\ -2\end{array}\right]\\ & \mu =\frac{b_1-\left|b_1\right|e_1}{\left|b_1-\right|b_1\left|e_1\right|}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}-1\\ -2\end{array}\right]\\ & H_2=1-2\mu {\mu }^{\top }=\frac{1}{5}\left[\begin{array}{cc}3& -4\\ -4& -3\end{array}\right]\\ & H_2{A}^{(1)}=\left[\begin{array}{ll}15& -9\\ 0& -3\end{array}\right]\end{array}$$

(3)写出A的H变换矩阵S，得出分解结果

$$\begin{array}{rl}& S=\left[\begin{array}{ll}1& \\ & H_2\end{array}\right]H_1=\frac{1}{15}\left[\begin{array}{ccc}5& 10& 10\\ -1& 11& -10\\ -14& 2& 5\end{array}\right]\\ & R=SA=\left[\begin{array}{ccc}9& 48& 15\\ 15& -9& \\ 15& -3& \end{array}\right]\\ & Q=S^T=\frac{1}{15}\left[\begin{array}{ccc}5& -2& -14\\ 10& 11& 2\\ 10& -10& 5\end{array}\right]\end{array}$$

$$\begin{array}{r}\therefore A=QR\end{array}$$

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四、 满秩分解

两种方法：初等行变换和Hermite标准形
$$A=\left[\begin{array}{cccc}-1& 0& 1& 2\\ 1& 2& -1& 1\\ 2& 2& -2& -1\end{array}\right]，对矩阵A满秩分解FG$$

方法（1）初等行变换

$$\begin{array}{c}[A|I]=\left[\begin{array}{cccccccc}-1& 0& 1& 2& ⋮& 1& 0& 0\\ 1& 2& -1& 1& ⋮& 0& 1& 0\\ 2& 2& -2& -1& ⋮& 0& 0& 1\end{array}\right]\\ \stackrel{行变换}{⟶}\left[\begin{array}{cccccccc}-1& 0& 1& 2& ⋮& 1& 0& 0\\ 0& 2& 0& 3& ⋮& 0& 1& 0\\ 0& 0& 0& 0& ⋮& 1& -1& 1\end{array}\right]\end{array}$$

左边是一个熟悉的行最简，前两行就是G。右边为矩阵P，算出P的逆矩阵，前两列就是F

$$P=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& -1& 1\end{array}\right]，P^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -2& 1& 1\end{array}\right]$$

$$F=\left[\begin{array}{cc}-1& 0\\ -1& 1\\ -2& 1\end{array}\right],G=\left[\begin{array}{cccc}-1& 0& 1& 2\\ 0& 2& 0& 3\end{array}\right]$$

相乘得到结果

$$\therefore A=FG=\left[\begin{array}{cc}-1& 0\\ -1& 1\\ -2& 1\end{array}\right]\left[\begin{array}{cccc}-1& 0& 1& 2\\ 0& 2& 0& 3\end{array}\right]$$

方法（2）Hermite标准形：

通过行变换得到标准形

$$\begin{array}{rlr}& A=\left[\begin{array}{lll}0& 0& 1\\ 2& 1& 1\\ 2j& j& 0\end{array}\right](j=\sqrt{-1})& \stackrel{\text{ 行变换 }}{\to }\left[\begin{array}{lll}1& \frac{1}{2}& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]=G\text{ (Hermite) }\end{array}$$

$\begin{array}{rl}& \therefore c_1=1,c_2=3,F取第1和第3列\end{array}$

$$\begin{array}{r}F=\left[\begin{array}{cc}0& 1\\ 2& 1\\ 2j& 0\end{array}\right]\end{array}$$

$$\begin{array}{rl}& \therefore A=FG=\left[\begin{array}{cc}0& 1\\ 2& 1\\ 2j& 0\end{array}\right]\left[\begin{array}{lll}1& \frac{1}{2}& 0\\ 0& 0& 1\end{array}\right]\end{array}$$

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五、奇异值分解

$$奇异值分解即A=U\left[\begin{array}{ll}\sum & 0\\ 0& 0\end{array}\right]V^H$$
$$A=\left[\begin{array}{lll}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]，对矩阵A进行奇异值分解。$$
解答：
（1）首先求$A^{T}A$特征值和特征向量

$$B=A^{T}A=\left[\begin{array}{lll}1& 0& 1\\ 0& 1& 1\\ 1& 1& 2\end{array}\right]\Rightarrow \begin{array}{rl}& {\lambda }_1=3\\ & \genfrac{}{}{0ex}{}{{\lambda }_2=1}{{\lambda }_3=0}\end{array}$$

$特征值向量为$
$${\xi }_1=\left[\begin{array}{l}1\\ 1\\ 2\end{array}\right]{\xi }_2=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]{\xi }_3=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$$

$$\mathrm{rank}A=2,\Sigma =\left[\begin{array}{cc}\sqrt{3}& 0\\ 0& 1\end{array}\right]$$

（2）得到正交矩阵

$$\therefore V=\left[\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{3}}\\ \frac{2}{\sqrt{6}}& 0& \frac{1}{\sqrt{3}}\end{array}\right]（正交矩阵)$$
$$\begin{array}{rl}& V_1=\left[\begin{array}{cc}\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& 0\end{array}\right]\end{array}$$

（3）根据公式计算出U

$$\begin{array}{c}{U}_1=A{V}_1{\Sigma }^{-1}=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ 0& 0\end{array}\right]\\ U_2=\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\\ U=\left[U_1⋮U_2\right]=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}& 0\\ 0& 0& 1\end{array}\right]\end{array}$$

$$\therefore A=U\left[\begin{array}{ccc}\sqrt{3}& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]V^T$$

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总结

建议有时间的可以好好学学这一部分，这部分在编程的时候用处比较大，便于计算机计算分析矩阵。

参考资料

本文章题目以及知识点部分源于
1、《矩阵论》—张凯院，西北工业大学出版社第4版
2、《矩阵论—讲稿》—张凯院
3、《矩阵论》——国科大2021秋季叶世伟上课PPT