【矩阵论】2. 矩阵分解——正规分解

矩阵论
1. 准备知识——复数域上的矩阵与换位公式)
1. 准备知识——复数域上的内积域正交阵
1. 准备知识——相似对角化与合同&正定阵
2. 矩阵分解—— SVD准备知识——奇异值
2. 矩阵分解——SVD
2. 矩阵分解——QR分解
2. 矩阵分解——乔利斯分解&平方根公式
2. 矩阵分解——正规谱分解——正规阵
2. 矩阵分解——正规分解
2. 矩阵分解——单阵及特征值特征向量一些求法

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6.2 正规分解定理

若 $A=A_{n\times n}$ 正规，则存在U阵Q，使 $Q^{H}AQ=\Lambda =\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)$

证明
$$\begin{array}{rl}& 设A=A_{n\times n}正规，由U相似定理，Q^{H}AQ正规，由许尔公式，存在U阵Q，使\\ & Q^{H}AQ=D=\left(\begin{array}{ccc}{\lambda }_1& & \ast \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)为上三角阵\\ & D=Q^{H}AQ为正规三角阵，由"正规三角定理"可知D为对角阵\\ & 即Q^{-1}AQ=D=\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)成立\end{array}$$

6.2.1 正规阵A恰有n个正交特向

$$\begin{array}{rl}& 由Q^{-1}AQ=D=\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)\Rightarrow AQ=QD\\ & \Rightarrow A\left(q_1,\cdots ,q_n\right)=\left(q_1,\cdots ,q_n\right)D\\ & \Rightarrow \left(A{q}_1,\cdots ,A{q}_n\right)=\left({\lambda }_1{q}_1,\cdots ,{\lambda }_n{q}_n\right)\\ & U阵Q为A的特征向量组成的矩阵，且n个特征向量相互正交，q_1\perp \cdots \perp q_n\end{array}$$

6.2.2 正规分解方法

1. 先令特征根 ${\lambda }_1,\cdots ,{\lambda }_n$ ，求正交特征向量 $X_1\perp \cdots \perp X_n$

2. 令U阵 $Q=\left(q_1,\cdots ,q_n\right)=\left(\frac{X_1}{|X_1|},\cdots ,\frac{X_n}{|X_n|}\right)$

   则有U相似阵 $Q^{H}AQ=D=\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)$ 为对角阵

3. 可写正规分解 $A=QD{Q}^H$

定义法

对矩阵 $A=\left(\begin{array}{cc}1& -1\\ 1& 0\end{array}\right)$ 正规分解
$$\begin{array}{rl}& A=\left(\begin{array}{cc}1& -1\\ 1& 0\end{array}\right)为正规阵，计算可得{\lambda }_1=i,{\lambda }_2=-i，X_1=\left(\begin{array}{c}i\\ 1\end{array}\right),X_2=\left(\begin{array}{c}1\\ i\end{array}\right),\\ & 且X_1与X_2为不同特征值的特征向量，所以X_1\perp X_2\\ & 令U阵Q=\left(\begin{array}{c}\frac{X_1}{|X_1|},\frac{X_2}{|X_2|}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}i& 1\\ 1& i\end{array}\right),可得Q^{H}AQ=D=\left(\begin{array}{cc}i& \\ & -i\end{array}\right)\\ & 则有正规分解A=QD{Q}^H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}i& 1\\ 1& i\end{array}\right)\left(\begin{array}{cc}i& \\ & -i\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-i& 1\\ 1& -i\end{array}\right)\end{array}$$

平移法

$$\begin{array}{rl}& 令B=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)=I+A=\left(\begin{array}{cc}1& \\ & 1\end{array}\right)+\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),其中A为正规U阵\\ & \lambda (B)=\{{\lambda }_1,{\lambda }_2\}=\{1+i,1-i\},X_1=\left(\begin{array}{c}i\\ 1\end{array}\right),X_2=\left(\begin{array}{c}1\\ i\end{array}\right),\\ & 得U阵Q=\left(\frac{X_1}{|X_1|},\frac{X_2}{|X_2|}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}i& 1\\ 1& i\end{array}\right)，故有正规分解A=QD{Q}^H\\ & =\frac{1}{\sqrt{2}}\left(\begin{array}{cc}i& 1\\ 1& i\end{array}\right)\left(\begin{array}{cc}1+i& \\ & 1-i\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-i& 1\\ 1& -i\end{array}\right)\end{array}$$

6.3 正规谱分解

6.3.1 正规分解推导

$$\begin{array}{rl}& 若A=A_{n\times n}正规，互异根为{\lambda }_1,\cdots ,{\lambda }_k，则有Q^{H}AQ=D=\left(\begin{array}{ccc}{\lambda }_1{I}_1& & \\ & \ddots & \\ & & {\lambda }_k{I}_k\end{array}\right)\\ & 其中Q为U阵，Q^H=Q^{-1},I_1,\cdots ,I_k为单位阵\\ & \left(如D=\left(\begin{array}{cc}2\left(\begin{array}{cc}1& \\ & 1\end{array}\right)& \\ & 3\left(\begin{array}{cc}1& \\ & 1\end{array}\right)\end{array}\right)=\left(\begin{array}{cc}2I_1& \\ & 3I_2\end{array}\right)\right)\\ & 可设Q^{-1}AQ=D=\left(\begin{array}{ccc}{\lambda }_1{I}_1& & 0\\ & \ddots & \\ 0& & {\lambda }_k{I}_k\end{array}\right)(Q为U阵，Q^H=Q^{-1})\\ & 写为D={\lambda }_1\left(\begin{array}{ccc}{I}_1& & 0\\ & \ddots & \\ 0& & 0\end{array}\right)+{\lambda }_2\left(\begin{array}{ccc}0& & 0\\ & I_2& \\ 0& & \ddots \end{array}\right)+\cdots +{\lambda }_k\left(\begin{array}{ccc}0& & 0\\ & \ddots & \\ 0& & I_k\end{array}\right)\\ & 则令D_1=\left(\begin{array}{ccc}{I}_1& & 0\\ & \ddots & \\ 0& & 0\end{array}\right),D_2=\left(\begin{array}{ccc}0& & 0\\ & I_2& \\ 0& & \ddots \end{array}\right),\cdots ,D_k=\left(\begin{array}{ccc}0& & 0\\ & \ddots & \\ 0& & I_k\end{array}\right)\\ & \Rightarrow Q^{-1}AQ=D={\lambda }_1{D}_1+{\lambda }_2{D}_2+\cdots +{\lambda }_k{D}_k\end{array}$$

则可得出结论：
$$\begin{array}{rl}& ①和为单位阵：D_1+D_2+\cdots +D_k=\left(\begin{array}{ccc}{I}_1& & \\ & \ddots & \\ & & I_k\end{array}\right)=I(单位阵)，\\ & ②正交：D_1{D}_2=0,\cdots ,D_i{D}_j=0(i\ne j)\\ & ③幂等：D_1^2=D_1,\cdots ,D_k^2=D_k，且D_1^H=D_1,\cdots ,D_k^H=D_k\end{array}$$
故可等价写为：
$$\begin{array}{rl}& Q^{H}AQ=D={\lambda }_1{D}_1+{\lambda }_2{D}_2+\cdots +{\lambda }_k{D}_k\\ & \Rightarrow A=QD{Q}^H={\lambda }_{1}Q{D}_1{Q}^H+{\lambda }_{2}Q{D}_2{Q}^H+\cdots +{\lambda }_{k}Q{D}_k{Q}^H\\ & 可令G_1=Q{D}_1{Q}^H,\cdots ,G_k=Q{D}_k{Q}^H\\ & \Rightarrow A={\lambda }_1{G}_1+\cdots +{\lambda }_k{G}_k\end{array}$$

有类似推论：
$$\begin{array}{rl}& ①G_1+G_2+\cdots +G_k=I\\ & \because G_1+G_2+\cdots +G_k=Q(D_1{Q}^{-1}+D_2+\cdots +D_k)Q^{-1}=QI{Q}^{-1}=I\\ & ②G_1{G}_2=0,\cdots ,G_i{G}_j=0(i\ne j)\\ & \because G_1{G}_2=(Q{D}_1{Q}^{-1})(Q{D}_2{Q}^{-1})=0\\ & ③G_1^2=G_1,\cdots ,G_k^2=G_k,且G_1^H=G_1,\cdots ,G_k^H=G_k都是Hermite阵\end{array}$$

6.3.2 正规阵谱分解与谱阵性质

$$\begin{array}{rl}& 若A=A_{n\times n}正规，全体互异根为{\lambda }_1,\cdots ,{\lambda }_k，则有A={\lambda }_1{G}_1+{\lambda }_2{G}_2+\cdots +{\lambda }_k{G}_k\\ & 其中G_1,\cdots ,G_k为A的谱阵\end{array}$$

性质

$$\begin{array}{rl}& ①和为I：G_1+G_2+\cdots +G_k=I\\ & ②正交：G_1{G}_2=0,\cdots ,G_i{G}_j=0(i\ne j)\\ & ③幂等：G_1^2=G_1,\cdots ,G_k^2=G_k(幂等),且G_1^H=G_1,\cdots ,G_k^H=G_k\\ & ④正规阵幂次：A^p={\lambda }_1^p{G}_1+\cdots +{\lambda }_k{G}_k,p=0,1,2.\cdots \\ & ⑤正规阵函数：f(A)=f({\lambda }_1)G_1+\cdots +f({\lambda }_k)G_k,f(x)=c_0+c_1{x}_1+\cdots +c_p{x}^p\\ & \because f(A)=c_{0}I+c_{1}A+\cdots +c_k{A}^p\\ & =c_0(G_1+\cdots +G_k)+c_1({\lambda }_1{G}_1+{\lambda }_2{G}_2+\cdots +{\lambda }_k{G}_k)+\\ & \cdots +c_p({\lambda }_1^p{G}_1+{\lambda }_2^p{G}_2+\cdots +{\lambda }_p^k{G}_k)\\ & =(c_0+c_1{\lambda }_1+c_p{\lambda }_1^k)G_1+(c_0+c_1{\lambda }_2+\cdots +c_p{\lambda }_2^{p}k)G_2+\\ & \cdots +(c_0+c_1{\lambda }_k+\cdots +c_p{\lambda }_k^p)G_k\\ & =f({\lambda }_1)G_1+\cdots +f({\lambda }_k)G_k\\ & ⑥正规阵求法：设A正规，全体不同根为{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_k，则有谱阵公式G_1,G_2,\cdots ,G_k\\ & G_1=\frac{(A-{\lambda }_{2}I)\cdots (A-{\lambda }_{k}I)}{({\lambda }_1-{\lambda }_2)\cdots ({\lambda }_1-{\lambda }_k)},G_1=\frac{(A-{\lambda }_{1}I)(A-{\lambda }_{3}I)\cdots (A-{\lambda }_{k}I)}{({\lambda }_2-{\lambda }_1)({\lambda }_2-{\lambda }_3)\cdots ({\lambda }_2-{\lambda }_k)}\\ & \cdots \\ & G_k=\frac{(A-{\lambda }_{1}I)(A-{\lambda }_{2}I)\cdots (A-{\lambda }_{k-1}I)}{({\lambda }_k-{\lambda }_1)({\lambda }_k-{\lambda }_2)\cdots ({\lambda }_k-{\lambda }_{k-1})}\\ & ⑦谱阵中列是A的特征向量：A{G}_1={\lambda }_1{G}_1,A{G}_2={\lambda }_2{G}_2,\cdots ,A{G}_k={\lambda }_k{G}_k\\ & \because A{G}_1=({\lambda }_1{G}_1+\cdots +{\lambda }_k{G}_k)G_1={\lambda }_1{G}_1\end{array}$$

谱阵求法

由推论⑤可知，取k个不同的函数f(x)可求出谱阵 $G_1,G_2,\cdots .G_k$

证明：

A正规有2个不同根

$$\begin{array}{r}{G}_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2},G_2=I-G_1=\frac{A-{\lambda }_{1}I}{{\lambda }_2-{\lambda }_1}\end{array}$$

A正规且有3个不同根

$$\begin{array}{rl}& G_1=\frac{(A-{\lambda }_{2}I)(A-{\lambda }_{3}I)}{({\lambda }_1-{\lambda }_2)({\lambda }_1-{\lambda }_3)},G_2=\frac{(A-{\lambda }_{1}I)(A-{\lambda }_{3}I)}{({\lambda }_2-{\lambda }_1)({\lambda }_2-{\lambda }_3)},\\ & G_3=I-G_1-G_2=\frac{(A-{\lambda }_{1}I)(A-{\lambda }_{2}I)}{({\lambda }_3-{\lambda }_1)({\lambda }_3-{\lambda }_2)}\end{array}$$

正规分解性质的应用例

1

$$\begin{array}{r}A=\left(\begin{array}{ccc}1& 0& -2\\ 0& 0& 0\\ -2& 0& 4\end{array}\right)(实对称正规阵)，求A与f(A)的谱分解，与A^{100}\end{array}$$

$$\begin{array}{rl}& 可知A为秩1矩阵，r(A)=1,特征根为\lambda (A)=5,0,0,不同根为{\lambda }_1=5,{\lambda }_2=0\\ & G_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}=\frac{1}{5}\left(\begin{array}{ccc}1& 0& -2\\ 0& 0& 0\\ -2& 0& 4\end{array}\right),G_2=\frac{A-{\lambda }_{1}I}{{\lambda }_2-{\lambda }_1}=I-G_1=\frac{1}{5}\left(\begin{array}{ccc}4& 0& 2\\ 0& 5& 0\\ 2& 0& 1\end{array}\right)\\ & 得谱分解：A={\lambda }_1{G}_1+{\lambda }_2{G}_2,且f(A)=f({\lambda }_1)G_1+f({\lambda }_2)G_2\\ & 即f(A)=f(5)G_1+f(0)G_2\\ & 令f(x)=x^{100}\Rightarrow A^{100}=f(5)G_1+f(0)G_2=5^{100}{G}_1=5^{100}\frac{1}{5}A=5^{99}A\\ & 由于G中列向量都是A的特征向量，所以{\lambda }_1=5的特征向量{\alpha }_1=\left(\begin{array}{c}1\\ 0\\ -2\end{array}\right)\\ & {\lambda }_2=0的特征向量{\alpha }_2=\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right),{\alpha }_3=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\end{array}$$

2

3.分块法

6.3.3 新平方根公式

$$\begin{array}{rl}& \sqrt{A}=\sqrt{{\lambda }_1}{G}_1+\sqrt{{\lambda }_2}{G}_2+\cdots +\sqrt{{\lambda }_n}{G}_n\\ & 条件：A正规，且A={\lambda }_1{G}_1+\cdots +{\lambda }_k{G}_k\end{array}$$

证明：

$$\begin{array}{rl}(\sqrt{A}{)}^2& =(\sqrt{{\lambda }_1}{G}_1+\sqrt{{\lambda }_2}{G}_2+\cdots +\sqrt{{\lambda }_k}{G}_k{)}^2\\ & ={\lambda }_1{G}_1^2+{\lambda }_2{G}_2^2+\cdots +{\lambda }_k{G}_k^2\\ & ={\lambda }_1{G}_1+{\lambda }_2{G}_2+\cdots +{\lambda }_k{G}_k\end{array}$$

eg

$$A=\left(\begin{array}{cc}5& 4\\ 4& 5\end{array}\right),求\sqrt{A}$$

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$$\begin{array}{rl}& 由于A为行和相等矩阵，\therefore \lambda (A)=\{9,tr(A)-9\}=\{9,1\},令{\lambda }_1=9,{\lambda }_2=1\\ & \Rightarrow G_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),G_2=\frac{A-{\lambda }_{1}I}{{\lambda }_2-{\lambda }_1}=\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\\ & \Rightarrow \sqrt{A}=\sqrt{{\lambda }_1}{G}_1+\sqrt{{\lambda }_2}{G}_2=3G_1+G_2=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\end{array}$$