【矩阵论】1.准备知识——复数域上的内积域正交阵

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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1.7 内积

$$\begin{array}{rl}& X=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right),Y=\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_n\end{array}\right)\in C^n\end{array}$$

1.7.1 复向量内积

$$\begin{array}{rl}(X,Y)& \stackrel{\Delta }{=}{Y}^{H}X=x_1\overline{y_1}+x_2\overline{y_2}+\cdots +x_n\overline{y_n}=\sum _{i=1}^n{x}_i\overline{y_i}\\ & =tr(Y^{H}X)\\ (Y,X)& =X^{H}Y=\overline{x_1}{y}_1+\overline{x_2}{y}_2+\cdots +\overline{x_n}{y}_1=\sum _{i=1}^n\overline{x_i}{y}_i\\ & =tr(X^{H}Y)\end{array}$$

若取Y=X，则其内积
$$\begin{array}{rl}(X,X)& =X^{H}X=x_1\overline{x_1}+x_2\overline{x_2}+\cdots +x_n\overline{x_n}=\sum _{i=1}^n{x}_i\overline{x_i}\\ & =|x_1{|}^2+|x_2{|}^2+\cdots +|x_n{|}^2=|X{|}^2\\ & =tr(X^{H}X)=tr(X{X}^H)\end{array}$$

向量内积性质

1. $(X,X)\ge 0$ ；若 $X\ne 0,(X,X)\ge 0$
2. $(X,Y)=\overline{(Y,X)}$
3. $(kX,Y)=k(X,Y),(X,kY)=\overline{k}(X,Y)$
4. (X+Y,W)=(X,Y)+(X,W)
5. $|(X,Y){|}^2\le |X|\cdot |Y|$

1.7.2 复矩阵内积

定义

$$\begin{array}{rl}& (A,B)\stackrel{\Delta }{=}tr(A{B}^H)=tr(A^{H}B)=\sum a_{ij}\overline{b_{ij}},A,B\in C^{m,n}\\ & (A,A)=tr(A{A}^H)=tr(A^{H}A)=\sum a_{ij}\overline{a_{ij}}=\sum |a_{ij}{|}^2\end{array}$$

矩阵A的模长：$||A||=\sqrt{(A,A)}=\sqrt{tr(A{A}^H)}=\sqrt{\sum |a_{ij}{|}^2}$

性质

1. $(A,A)=tr(A{A}^H)=\sum |a_{ij}{|}^2\ge 0$ ；若 $A\ne 0$ ，则 $(A,A)>0$

2. $(A,B)=\overline{(B,A)}$

3. $(kA,B)=k(A,B),(A,kB)=\overline{k}(A,B)$

4. (A+B,D)=(A,D)+(B,D)，(D,A+B)=(D,A)+(D,B)

5. $|(A,B){|}^2\le |A|\cdot |B|$

矩阵的内积形式

列分块

$$\begin{array}{rl}A& =\left(\begin{array}{cccc}& a_{11}& \cdots & a_{1p}\\ & ⋮& \ddots & ⋮\\ & a_{n1}& \cdots & a_{np}\end{array}\right)\in C^{n\times p}\\ & =({\alpha }_1,\cdots ,{\alpha }_p),其中{\alpha }_i为n维列向量(n\times 1阶矩阵)\\ & \\ A^H& =\left(\begin{array}{cccc}& \overline{a_{11}}& \cdots & \overline{a_{n1}}\\ & ⋮& \ddots & ⋮\\ & \overline{a_{1p}}& \cdots & \overline{a_{np}}\end{array}\right)\in C^{p\times n}\\ & =\left(\begin{array}{c}{\overline{{\alpha }_1}}^T\\ ⋮\\ {\overline{{\alpha }_p}}^T\end{array}\right)，其中{\overline{{\alpha }_1}}^T是n维行向量(1\times n阶矩阵)\\ & \end{array}$$

$$\begin{array}{rl}{A}^{H}A& =\left(\begin{array}{c}{\overline{{\alpha }_1}}^T\\ ⋮\\ {\overline{{\alpha }_p}}^T\end{array}\right)({\alpha }_1,\cdots ,{\alpha }_p)\\ & \\ & =\left(\begin{array}{ccccc}& {\overline{{\alpha }_1}}^T{\alpha }_1& {\overline{{\alpha }_1}}^T{\alpha }_2& \cdots & {\overline{{\alpha }_1}}^T{\alpha }_p\\ & {\overline{{\alpha }_2}}^T{\alpha }_1& {\overline{{\alpha }_2}}^T{\alpha }_2& \cdots & {\overline{{\alpha }_2}}^T{\alpha }_p\\ & ⋮& ⋮& \ddots & ⋮\\ & {\overline{{\alpha }_p}}^T{\alpha }_1& {\overline{{\alpha }_p}}^T{\alpha }_2& \cdots & {\overline{{\alpha }_p}}^T{\alpha }_p\end{array}\right)\\ & \\ & =\left(\begin{array}{ccccc}& ({\alpha }_1,{\alpha }_1)& ({\alpha }_2,{\alpha }_1)& \cdots & ({\alpha }_p,{\alpha }_1)\\ & ({\alpha }_1,{\alpha }_2)& ({\alpha }_2,{\alpha }_2)& \cdots & ({\alpha }_p,{\alpha }_2)\\ & ⋮& ⋮& \ddots & ⋮\\ & ({\alpha }_1,{\alpha }_p)& ({\alpha }_2,{\alpha }_p)& \cdots & ({\alpha }_p,{\alpha }_p)\end{array}\right)\\ & \\ & =\left(\begin{array}{ccccc}& \overline{({\alpha }_1,{\alpha }_1)}& \overline{({\alpha }_1,{\alpha }_2)}& \cdots & \overline{({\alpha }_1,{\alpha }_p)}\\ & \overline{({\alpha }_2,{\alpha }_1)}& \overline{({\alpha }_2,{\alpha }_2)}& \cdots & \overline{({\alpha }_2,{\alpha }_p)}\\ & ⋮& ⋮& \ddots & ⋮\\ & \overline{({\alpha }_p,{\alpha }_1)}& \overline{({\alpha }_p,{\alpha }_2)}& \cdots & \overline{({\alpha }_p,{\alpha }_p)}\end{array}\right)\end{array}$$

行分块

$$\begin{array}{rl}A& =\left(\begin{array}{cccc}& a_{11}& \cdots & a_{1p}\\ & ⋮& \ddots & ⋮\\ & a_{n1}& \cdots & a_{np}\end{array}\right)\in C^{n\times p}\\ & =\left(\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ ⋮\\ {\alpha }_n\end{array}\right),其中{\alpha }_i为p维行向量(1\times p阶矩阵)\\ & \\ A^H& =\left(\begin{array}{c}{\overline{{\alpha }_1}}^T{\overline{{\alpha }_2}}^T\cdots {\overline{{\alpha }_n}}^T\end{array}\right),其中{\overline{{\alpha }_i}}^T为p维列向量(p\times 1阶矩阵)\end{array}$$

$$\begin{array}{rl}A{A}^H& =\left(\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ ⋮\\ {\alpha }_n\end{array}\right)\left(\begin{array}{c}{\overline{{\alpha }_1}}^T{\overline{{\alpha }_2}}^T\cdots {\overline{{\alpha }_n}}^T\end{array}\right)\\ & \\ & =\left(\begin{array}{ccccc}& {\alpha }_1{\overline{{\alpha }_1}}^T& {\alpha }_1{\overline{{\alpha }_2}}^T& \cdots & {\alpha }_1{\overline{{\alpha }_n}}^T\\ & {\alpha }_2{\overline{{\alpha }_1}}^T& {\alpha }_2{\overline{{\alpha }_2}}^T& \cdots & {\alpha }_2{\overline{{\alpha }_n}}^T\\ & ⋮& ⋮& \ddots & ⋮\\ & {\alpha }_n{\overline{{\alpha }_1}}^T& {\alpha }_n{\overline{{\alpha }_2}}^T& \cdots & {\alpha }_n{\overline{{\alpha }_n}}^T\end{array}\right)\\ & \\ & =\left(\begin{array}{ccccc}& ({\alpha }_1,{\alpha }_1)& ({\alpha }_1,{\alpha }_2)& \cdots & ({\alpha }_1,{\alpha }_n)\\ & ({\alpha }_2,{\alpha }_1)& ({\alpha }_2,{\alpha }_2)& \cdots & ({\alpha }_2,{\alpha }_n)\\ & ⋮& ⋮& \ddots & ⋮\\ & ({\alpha }_n,{\alpha }_1)& ({\alpha }_n,{\alpha }_2)& \cdots & ({\alpha }_n,{\alpha }_n)\end{array}\right)\end{array}$$

1.8 正交

1.8.1 向量正交

$$\begin{array}{rl}X& =\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right),Y=\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_n\end{array}\right)\in C^n\end{array}$$

$$\begin{array}{rl}X\perp Y⟺(X,Y)=0& =x_1\overline{y_1}+x_2\overline{y_2}+\cdots +x_n\overline{y_n}\\ & =\overline{\overline{x_1}{y}_1+\overline{x_2}{y}_2+\cdots +\overline{x_n}{y}_n}\\ =(Y,X)& \end{array}$$

正交性质

1. $X\perp Y\Rightarrow aX\perp bY$

   证：$(aX,bY)=\overline{b}{Y}^{H}aX=a\overline{b}{Y}^{H}X=a\overline{b}(X,Y)=0$

2. 勾股定理：$X_1\perp X_2\perp \cdots \pm X_n\Rightarrow |c_1{X}_1\pm c_2{X}_2\pm \cdots \pm c_n{X}_n{|}^2=|c_1{X}_1{|}^2+|c_2{X}_2{|}^2+\cdots +|c_n{X}_n{|}^2$

   此时，$X_1,X_2,\cdots ,x_n$ 称为一个正交组

1.8.2 单位向量

若 $X\ne 0$ ，$\frac{X}{|X|}$ 是一个单位向量（$|\frac{X}{|X|}|=1$）

1.8.3 U阵(正交阵)

预：非单位列向量

半：p个n维列向量(p

预-半U阵(预-半正交阵)

$$\begin{gathered} {\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p是n维列向量，且p\le n，且{\alpha }_1\perp {\alpha }_2\perp \cdots \perp {\alpha }_p \\ 则称A=({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p)为预半U阵 \end{gathered}$$

判定

$$\begin{array}{rl}A& =({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p)是预半U阵\\ & ⟺A^{H}A=\left(\begin{array}{cccc}& ({\alpha }_1,{\alpha }_1)& \cdots & 0\\ & ⋮& \ddots & 0\\ & 0& \cdots & ({\alpha }_p,{\alpha }_p)\end{array}\right)是对角阵\\ & 其中，{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p是n维列向量\end{array}$$

证明：

区分 ： $A^{H}A$ 是 $p\times p$ 阶满秩方阵，而 $A{A}^H$ 是 $n\times n$ 不满秩方阵

半U阵(半正交阵)

$$\begin{array}{rl}& A=({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p)是预半U阵，其中{\alpha }_i是n维列向量，若满足\\ & |{\alpha }_1|=|{\alpha }_2|=\cdots =|{\alpha }_p|=1,则A为半U阵\end{array}$$

判定

$$\begin{array}{rl}A=({\alpha }_1,\cdots ,{\alpha }_p)是半U阵& ⟺{\alpha }_1\perp \cdots \perp {\alpha }_p,且|{\alpha }_1|=\cdots =|{\alpha }_p|=1\\ & ⟺A^{H}A=I_p\end{array}$$

性质

1. 保模长 A为半U阵，则 $|Ax{|}^2=|x{|}^2$

   $|Ax{|}^2=(Ax{)}^H(Ax)=x^H{A}^{H}Ax=|X{|}^2$

2. 保正交 A为半U阵，$x\perp y$ ，则$Ax\perp Ay$

预-U阵(预-单位正交阵)

$$\begin{array}{rl}& {\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n是n维列向量，且{\alpha }_1\perp {\alpha }_2\perp \cdots \perp {\alpha }_n,\\ & 则A=({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)是预U阵\end{array}$$

eg
$$\begin{array}{rl}& X_1=\left(\begin{array}{c}1\\ i\\ i\end{array}\right),X_2=\left(\begin{array}{c}2i\\ 1\\ 1\end{array}\right),X_3=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)\\ & \\ & (X_1,X_2)=0,(X_2,X_3)=0,(X_1,X_3)=0,\\ & \\ & \therefore X_1\perp X_2\perp X_3,A=(X_1,X_2,X_3)是预-U阵\end{array}$$

判定

$$\begin{array}{rl}A& =({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)\\ & ⟺A^{H}A=\left(\begin{array}{cccc}& ({\alpha }_1,{\alpha }_1)& \cdots & 0\\ & ⋮& \ddots & 0\\ & 0& \cdots & ({\alpha }_n,{\alpha }_n)\end{array}\right)是对角阵\\ & 其中，{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_p是n维列向量\end{array}$$

等价判定

• $A=A_{n\times n}$ 为U阵($A^{H}A=I$)，即A的列向量 ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$ 互正交，且长度为1
• $A^{-1}=A^H$
• $A^{H}A=I,且A{A}^H=I$

U阵(A正交阵)

$$\begin{array}{r}A是预U阵且|{\alpha }_1|=\cdots =|{\alpha }_n|=1,则A是一个U阵(正交阵)\end{array}$$

判定

$$\begin{array}{rl}A& =({\alpha }_1,\cdots ,{\alpha }_n)\\ & ⟺A^{H}A=\left(\begin{array}{ccccc}& 1& 0& \cdots & 0\\ & 0& 1& \cdots & 0\\ & ⋮& ⋮& \ddots & ⋮\\ & 0& 0& \cdots & 1\end{array}\right)=I是单位阵\end{array}$$

性质

$$\begin{array}{rl}& 1.A是U阵⟺A^{H}A=I⟺A^{-1}A=I⟺A{A}^H=I\\ & ⟺A=({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)，且{\alpha }_1\perp {\alpha }_2,\cdots \perp {\alpha }_n,且|{\alpha }_1|=\cdots =|{\alpha }_n|=1\\ & \\ & 2.|Ax{|}^2=|x{|}^2，A是U阵\\ & \because |Ax{|}^2=(Ax{)}^H(Ax)=x^H{A}^{H}Ax=x^{H}Ix=(x,x)=|x{|}^2\\ & \\ & 3.x\perp y\Rightarrow Ax\perp Ay，A是U阵\\ & \because (Ax,Ay)=(Ay{)}^{H}Ax=y^H{A}^{H}Ax=(x,y)=0⟺Ax\perp Ay\\ & \\ & 4.(Ax,Ay)=(x,y)，A是U阵\end{array}$$

预U阵与U阵

$$\begin{array}{r}A=({\alpha }_1,\cdots ,{\alpha }_n)是预U阵\Rightarrow A=(\frac{{\alpha }_1}{|{\alpha }_1|},\frac{{\alpha }_2}{|{\alpha }_2|},\cdots ,\frac{{\alpha }_n}{|{\alpha }_n|})是U阵\end{array}$$

U阵构造新U阵

$$\begin{array}{rl}若A& =({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)为U阵，则\\ & 1.k=\pm 1,kA=(k{\alpha }_1,k{\alpha }_2,\cdots ,k{\alpha }_n)为U阵\\ & 2.B=({\beta }_1,{\beta }_2,\cdots ,{\beta }_n)为U阵，其中\beta 组为\alpha 组的重排\\ & 3.(封闭性)若A、B为同阶U阵，则AB也为U阵\end{array}$$

向量构造U阵

$$\begin{array}{rl}若\alpha & =\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_n\end{array}\right)\in C,A=I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2}是一个U阵\\ & \\ & 1.A^H=A且A^2=I(A^{-1}=A)\\ & 2.A为U阵(A^{H}A=I)\\ & \\ 证明:1.A^2& =(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})\\ & =I_n^2-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4(\alpha {\alpha }^H)(\alpha {\alpha }^H)}{|\alpha {|}^4}=I_n-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4\alpha ({\alpha }^H\alpha ){\alpha }^H}{|\alpha {|}^4}\\ & =I_n-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4\alpha (|\alpha {|}^2){\alpha }^H}{|\alpha {|}^4}=I_n-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}=I_n\\ 2.A^{H}A& =(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2}{)}^H(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})\\ & =(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})=A^2=I\\ & \therefore A是U阵\end{array}$$

eg ：
$$\begin{array}{rl}\alpha =\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),其U阵为I_3-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2}& =\left(\begin{array}{cccc}& 1& 0& 0\\ & 0& 1& 0\\ & 0& 0& 1\end{array}\right)-\frac{2}{3}\left(\begin{array}{cccc}& 1& 1& 1\\ & 1& 1& 1\\ & 1& 1& 1\end{array}\right)\\ & =\left(\begin{array}{cccc}& \frac{1}{3}& -\frac{2}{3}& -\frac{2}{3}\\ & -\frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ & -\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right)\end{array}$$