8.4 镜面阵

法向量确定一个镜面

8.4.1 镜面阵的作用

对法向量

• $A\alpha =-\alpha$
• $A(A\alpha )=A^2\alpha =\alpha$

对镜面上向量

• $AY=Y$

8.4.2 镜面阵表示

$$A=I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2},其中\alpha =\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)\in C^n,且\alpha \ne 0$$

令 $ϵ=\frac{\alpha }{|\alpha |}$ ，镜面阵 $A=I-2ϵ{ϵ}^H$ ，满足 ${ϵ}^Hϵ=|ϵ{|}^2=1$

验证：
$$\begin{array}{rl}& 对于法向量\alpha ，A\alpha =(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})\alpha =\alpha -\frac{2\alpha {\alpha }^H\alpha }{|\alpha {|}^2}=\alpha -\frac{2\alpha |\alpha {|}^2}{|\alpha {|}^2}=-\alpha \\ & 对于Y\in 镜面，有\alpha \perp Y,即内积(\alpha ,Y)=(Y,\alpha )=0,且(Y,\alpha )={\alpha }^{H}Y=0\\ & AY=(I_n-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2})Y=Y-\frac{2\alpha {\alpha }^{H}Y}{|\alpha {|}^2}=Y\end{array}$$

8.4.3 性质

a. 特根

镜面阵至少有两个特征向量 $\alpha$ 与 $Y$

• $A\alpha =-\alpha$ ，特根为 $-1$
• $AY=Y$ ，特根为 $1$

代数方法求特根
平移法： A − I = − 2 α α H ∣ α ∣ 2 为秩 1 矩阵 , 特根 λ ( A − I ) = { − 2 ∣ α ∣ 2 t r ( α α H ) , 0 , ⋯   , 0 ⏟ n − 1 } = = t r ( X X H ) = X H X = ∣ X ∣ 2 { − 2 , 0 , ⋯   , 0 } ⇒ λ ( A ) = λ ( − 2 α α H ∣ α ∣ 2 ) + 1 = { − 1 , 1 , ⋯   , 1 ⏟ n − 1 } \begin{aligned} &平移法：\\ &A-I=-2\frac{\alpha\alpha^H}{\vert \alpha\vert^2}为秩1矩阵,\\ &特根\lambda(A-I)=\{\frac{-2}{\vert \alpha\vert^2}tr(\alpha\alpha^H),\underbrace{0,\cdots,0}_{n-1}\}=\xlongequal{tr(XX^H)=X^HX=\vert X\vert^2}\{-2,0,\cdots,0\}\\ &\Rightarrow \lambda(A)=\lambda\left(-2\frac{\alpha\alpha^H}{\vert \alpha\vert^2}\right)+1=\{-1,\underbrace{1,\cdots,1}_{n-1}\} \end{aligned} ​平移法：A−I=−2∣α∣2ααH​为秩1矩阵,特根λ(A−I)={∣α∣2−2​tr(ααH),n−1 0,⋯,0​​}=tr(XXH)=XHX=∣X∣2 {−2,0,⋯,0}⇒λ(A)=λ(−2∣α∣2ααH​)+1={−1,n−1 1,⋯,1​​}​
特征多项式 $|\lambda I-A|=(\lambda +1)(\lambda -1{)}^{n-1}$

b. 特向

镜面上有 $n-1$ 个独立的(线性无关)向量，$Y_1,Y_2,\cdots ,Y_{n-1}$ $(A{Y}_1=Y_1,A{Y}_2=Y_2,\cdots ,A{Y}_{n-1}=Y_{n-1})$ 都是 $1$ 的特向

• $Y_1,Y_2,\cdots ,Y_{n-1}$ 为 ${\alpha }^{H}X=0$ 的 $n-1$ 个线性无关的解

  证明：
  $$\begin{array}{rl}& A-I=\frac{-2}{|\alpha {|}^2}\alpha {\alpha }^H为秩1矩阵，故\alpha 为A-I的特向，r({\alpha }^H)=1,{\alpha }^{H}X=0\\ & 且n-r({\alpha }^H)=n-1,故{\alpha }^{H}X=0有n-1个线性无关解Y_1,\cdots ,Y_{n-1}\\ & (A-I)X=(\frac{-2}{|\alpha {|}^2}\alpha {\alpha }^H)X=(\frac{-2}{|\alpha {|}^2}\alpha ){\alpha }^{H}X=0\Rightarrow Y_1,\cdots ,Y_{n-1}为(A-I)特根为0的根\\ & 由平移矩阵性质，A-I与A有相同的特根，\alpha ,Y_1,\cdots ,Y_{n-1}为A的特征向量\\ & (A-I)\alpha =-2\alpha \Rightarrow A\alpha =-\alpha \\ & (A-I)Y_1=0,\cdots ,(A-I)Y_{n-1}=0\Rightarrow A{Y}_1=Y_1,\cdots ,A{Y}_{n-1}=Y_{n-1}\end{array}$$

• 镜面阵内 $n-1$ 个特征向量有不同选法，可取 $Y_1\perp Y_2\perp \cdots \perp Y_{n-1}$ (互正交)，故可对镜面阵有 $n$ 个正交特向 { X ⊥ Y 1 ⊥ Y 2 ⊥ ⋯ ⊥ Y n − 1 } \{X\bot Y_1\bot Y_2\bot \cdots \bot Y_{n-1}\} {X⊥Y1​⊥Y2​⊥⋯⊥Yn−1​}

  正规阵，Hermite阵有 $n$ 个相互正交的向量

$A$ 的全体特根为 λ ( A ) = { − 1 , 1 , ⋯   , 1 ⏟ n − 1 个 1 } \lambda(A)=\{-1,\underbrace{1,\cdots,1}_{n-1个1}\} λ(A)={−1,n−1个1 1,⋯,1​​}

c. 镜面阵为 $Hermite$ 阵

$A=I_n-\frac{2(\alpha {\alpha }^H)}{|\alpha {|}^2}$ Hermite阵的差

d. $A^2=I$

$A^2\alpha =A(A\alpha )=A(-\alpha )=-(A\alpha )=-(-\alpha )=\alpha$

即证明： $A^2=I$

$$\begin{array}{rl}& 任取C^n中向量W,可分解为W=Y+k\alpha (直角分解)\\ & AW=A(Y+k\alpha )=AY+k(A\alpha )=Y+k(-\alpha )=Y-k\alpha =\tilde{W}\\ & A^{2}W=A(AW)=A\tilde{W}=A(Y-k\alpha )=AY-kA\alpha =Y+k\alpha =W\\ & \forall W\in C^n，有A^{2}W=W,故A^2=I\end{array}$$

---

$$\begin{array}{rl}& A^2={\left(I-\frac{2\alpha {\alpha }^H}{|\alpha {|}^2}\right)}^2=I^2-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4}{|\alpha {|}^4}{\left(\alpha {\alpha }^H\right)}^2\\ & 其中(\alpha {\alpha }^H{)}^2=(\alpha {\alpha }^H)(\alpha {\alpha }^H)=\alpha ({\alpha }^H\alpha ){\alpha }^H=|\alpha {|}^2\alpha {\alpha }^H\\ & \Rightarrow A^2=I^2-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4}{|\alpha {|}^4}{\left(\alpha {\alpha }^H\right)}^2=I^2-\frac{4\alpha {\alpha }^H}{|\alpha {|}^2}+\frac{4}{|\alpha {|}^4}|\alpha {|}^2\alpha {\alpha }^H=I\end{array}$$

d. $A^{-1}=A$

e. $A$ 为优阵

由于 $A^H=A$ ，且 $A^{-1}=A$ ，故 $A^{H}A=A^{-1}A=I$ ，$A$ 为优阵

特根 λ ( A ) = { − 1 , 1 , ⋯   , 1 ⏟ n − 1 } \lambda(A)=\{-1,\underbrace{1,\cdots,1}_{n-1}\} λ(A)={−1,n−1 1,⋯,1​​} ，特向 $\alpha \perp Y_1\perp \cdots \perp Y_{n-1}$ ，优阵 $Q=\left(\frac{\alpha }{|\alpha |},\frac{Y_1}{|Y_1|},\cdots ,\frac{Y_{n-1}}{|Y_{n-1}|}\right)$ ，$Q^{-1}=Q^H$

$\Rightarrow Q^{-1}AQ=Q^{H}AQ=D=\left(\begin{array}{cccc}-1& & & \\ & 1& & \\ & & \ddots & \\ & & & 1\end{array}\right)$ 为对角阵

$D=\left(\begin{array}{cccc}-1& & & \\ & 1& & \\ & & \ddots & \\ & & & 1\end{array}\right)$ 为一个特殊镜面阵，法向量指向 $x$ 轴方向

8.4.3 向量构造镜面阵

设 $R^n$ 中2个 实向量，$\alpha =\left(\begin{array}{c}{a}_1\\ ⋮\\ a_n\end{array}\right),\beta =\left(\begin{array}{c}{b}_1\\ ⋮\\ b_n\end{array}\right)\in R^n$ ，且 $\{\begin{array}{rl}& |\alpha |=|\beta |\\ & \alpha \ne \beta \end{array}$ ，则有镜面阵 $P=I-\frac{2(\alpha -\beta )(\alpha -\beta {)}^H}{|(\alpha -\beta ){|}^2}$ ，使得 $P\alpha =\beta ,P\beta =\alpha$

证明：
$$\begin{array}{rl}& 令P=I-\frac{2(\alpha -\beta )(\alpha -\beta {)}^H}{|(\alpha -\beta ){|}^2}\\ & \Rightarrow PX=P(\alpha -\beta )=(\alpha -\beta )-2\frac{(\alpha -\beta )(\alpha -\beta {)}^H(\alpha -\beta )}{|(\alpha -\beta ){|}^2}=-(\alpha -\beta )=-X\\ & (\alpha +\beta ,X)=(\alpha +\beta ,\alpha -\beta )=(\alpha ,\alpha )-(\alpha ,\beta )+(\beta ,\alpha )-(\beta ,\beta )=|\alpha {|}^2-|\beta {|}^2=0\\ & \Rightarrow 法向量(\alpha -\beta )\perp (\alpha +\beta ),其中(\alpha +\beta )\in 镜面\\ & 由于\{\begin{array}{rl}& P(\alpha -\beta )=-(\alpha -\beta )\\ & P(\alpha +\beta )=(\alpha +\beta )\end{array}，两式相加得2P\alpha =2\beta \Rightarrow P\alpha =\beta ,P^{-1}\beta =P\beta =\alpha \end{array}$$
eg

8.4.4 镜面阵与QR分解

$P_x$ 表示 $x$ 阶镜面阵 ，$A_x$ 表示 $x$ 阶方阵
$$\begin{array}{rl}{P}_n{A}_n& =P_n({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n)\stackrel{P_n{\alpha }_1={\left(\begin{array}{c}{\mu }_1\\ 0\\ ⋮\\ 0\end{array}\right)}_n}{=}\left(\begin{array}{cc}{\mu }_1& \ast \\ O& A_{n-1}\end{array}\right)\\ & \\ \left(\begin{array}{cc}1& 0\\ 0& P_{n-1}\end{array}\right)P_n{A}_n& =\left(\begin{array}{cc}1& 0\\ 0& P_{n-1}\end{array}\right)\left(\begin{array}{cc}{\mu }_1& \ast \\ 0& A_{n-1}\end{array}\right)=\left(\begin{array}{cc}{\mu }_1& \ast \\ 0& P_{n-1}{A}_{n-1}\end{array}\right)\\ & =\left(\begin{array}{cc}{\mu }_1& \ast \\ 0& P_{n-1}({\beta }_1,\cdots ,{\beta }_{n-1}{)}_{n-1\times n-1}\end{array}\right)\\ & \stackrel{P_2{\beta }_1={\left(\begin{array}{c}{\mu }_2\\ 0\\ ⋮\\ 0\end{array}\right)}_{n-1}}{=}\left(\begin{array}{ccc}{\mu }_1& \ast & \ast \\ 0& {\mu }_2& \ast \\ 0& 0& A_{n-2}\end{array}\right)\end{array}$$

$$\begin{array}{rl}\left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & P_2\end{array}\right)\cdots \left(\begin{array}{cc}1& 0\\ 0& P_{n-1}\end{array}\right)P_{n}A& =\left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & P_2\end{array}\right)\left(\begin{array}{ccccc}{\mu }_1& \ast & \ast & \cdots & \ast \\ 0& {\mu }_2& \ast & \cdots & \ast \\ 0& 0& {\mu }_3& \cdots & \ast \\ ⋮& ⋮& ⋮& \ddots & \ast \\ 0& 0& 0& \cdots & A_2\end{array}\right)\\ & =\left(\begin{array}{ccccc}{\mu }_1& \ast & \ast & \cdots & \ast \\ 0& {\mu }_2& \ast & \cdots & \ast \\ 0& 0& {\mu }_3& \cdots & \ast \\ ⋮& ⋮& ⋮& \ddots & \ast \\ 0& 0& 0& \cdots & P_2{A}_2\end{array}\right)\\ & =\left(\begin{array}{ccccc}{\mu }_1& \ast & \ast & \cdots & \ast \\ 0& {\mu }_2& \ast & \cdots & \ast \\ 0& 0& {\mu }_3& \cdots & \ast \\ ⋮& ⋮& ⋮& \ddots & \ast \\ 0& 0& 0& \cdots & P_2({\gamma }_1,{\gamma }_2{)}_{2\times 2}\end{array}\right)\\ & \stackrel{P_2{\gamma }_1=\left(\begin{array}{c}{\mu }_{n-1}\\ 0\end{array}\right)}{=}\left(\begin{array}{cccccc}{\mu }_1& \ast & \ast & \cdots & \ast & \ast \\ 0& {\mu }_2& \ast & \cdots & \ast & \ast \\ 0& 0& {\mu }_3& \cdots & \ast & \ast \\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & {\mu }_{n-1}& ⋮\\ 0& 0& 0& \cdots & 0& \ast \end{array}\right)=R\end{array}$$

故有镜面阵 $Q=P_n\left(\begin{array}{cc}1& 0\\ 0& P_{n-1}\end{array}\right)\cdots \left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & P_2\end{array}\right)$ ，使 $QA=R$ ，由于镜面阵为优H阵 $Q^H=Q^{-1}=Q$ ，故有 $QR$ 分解 $A=QR$

2阶例题

$$\begin{array}{rl}& 令\alpha =\left(\begin{array}{c}1\\ i\end{array}\right),|\alpha |=\sqrt{2}，令\beta =\left(\begin{array}{c}\sqrt{2}\\ 0\end{array}\right),令镜面法向量X=\alpha -\beta =\left(\begin{array}{c}1-\sqrt{2}\\ i\end{array}\right)\\ & X{X}^H=\left(\begin{array}{c}1-\sqrt{2}\\ i\end{array}\right)\left(1-\sqrt{2}-i\right)=\left(\begin{array}{cc}(1-\sqrt{2}{)}^2& -i(1-\sqrt{2})\\ i(1-\sqrt{2})& 1\end{array}\right)=\left(\begin{array}{cc}3-\sqrt{2}& -i(1-\sqrt{2})\\ i(1-\sqrt{2})& 1\end{array}\right)\\ & |X{|}^2=X^{H}X=\left(\begin{array}{c}1-\sqrt{2}-i\end{array}\right)\left(\begin{array}{c}1-\sqrt{2}\\ i\end{array}\right)=2(2-\sqrt{2})\\ & 镜面阵Q=I_2-2\frac{X{X}^H}{|X{|}^2}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-\frac{2}{2(2-\sqrt{2})}\left(\begin{array}{cc}3-2\sqrt{2}& -i(1-\sqrt{2})\\ i(1-\sqrt{2})& 1\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -i\\ i& -1\end{array}\right)\\ & 使得Q\left(\begin{array}{c}1\\ i\end{array}\right)=\left(\begin{array}{c}\sqrt{2}\\ 0\end{array}\right)\Rightarrow QA=Q({\alpha }_1,{\alpha }_2)=\left(Q{\alpha }_1,Q{\alpha }_2\right)=\left(\begin{array}{cc}\sqrt{2}& \frac{i}{\sqrt{2}}\\ 0& -\frac{3}{\sqrt{2}}\end{array}\right)=R\\ & 解出A的QR分解A=QR,Q=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -i\\ i& -1\end{array}\right)\end{array}$$

3阶例题

$$\begin{array}{rl}& 令P_3=I,P_3\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 0\\ 0\end{array}\right),使得P_{3}A=\left(\begin{array}{cc}2& \ast \\ 0& A_2\end{array}\right)\\ & 令{\alpha }_2=\left(\begin{array}{c}3\\ 4\end{array}\right),{\beta }_2=\left(\begin{array}{c}|{\alpha }_2|\\ 0\end{array}\right)=\left(\begin{array}{c}5\\ 0\end{array}\right),二维镜面法向量X_2={\alpha }_2-{\beta }_2=\left(\begin{array}{c}-2\\ 4\end{array}\right),|X_2{|}^2=20,X{X}^H=\left(\begin{array}{cc}5& -8\\ -8& 16\end{array}\right)\\ & P_2=I_2-\frac{2}{|X_2{|}^2}X{X}^H=\left(\begin{array}{cc}\frac{3}{5}& \frac{4}{5}\\ \frac{4}{5}& -\frac{3}{5}\end{array}\right),使得P_2{A}_2=\left(\begin{array}{cc}5& -\frac{2}{5}\\ 0& -\frac{11}{5}\end{array}\right)\\ & \Rightarrow \left(\begin{array}{cc}1& 0\\ 0& P_2\end{array}\right)P_{3}A=\left(\begin{array}{ccc}1& 0& 0\\ 0& \frac{3}{5}& \frac{4}{5}\\ 0& \frac{4}{5}& -\frac{3}{5}\end{array}\right)A=R=\left(\begin{array}{ccc}2& -1& 3\\ 0& \frac{3}{5}& -\frac{2}{5}\\ 0& 0& -\frac{11}{5}\end{array}\right)\\ & 故有Q=\left(\begin{array}{cc}1& 0\\ 0& P_2\end{array}\right)P_3=\left(\begin{array}{ccc}1& 0& 0\\ 0& \frac{3}{5}& \frac{4}{5}\\ 0& \frac{4}{5}& -\frac{3}{5}\end{array}\right)，R=\left(\begin{array}{ccc}2& -1& 3\\ 0& \frac{3}{5}& -\frac{2}{5}\\ 0& 0& -\frac{11}{5}\end{array}\right),使得A=QR\end{array}$$

8.5 正定阵

8.5.1 二次型

a. 二次型定义

令 $A$ 为 $Hermite$ 阵 $(A^H=A\in C^{n\times n})$， $X=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)$ ，称 $X^{H}AX=\left(\overline{x_1},\overline{x_2},\cdots ,\overline{x_n}\right)A\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)$ ，为矩阵 $A$ 产生的二次型，记为 $f(x)=X^{H}AX$

b. 正定二次型与正定阵定义

若 $A^H=A$ ,对一切 $X\ne 0$ ,有 $X^{H}AX>0$ ,则 $f(x)=X^{H}AX$ 为正定二次型，A为正定阵，记为 $A>0$
若 $A^H=A$ ,对一切 $X\ne 0$ ，有 $X^{H}AX\ge 0$ ,则 $f(x)=X^{H}AX$ 为半正定二次型，A为半正定阵，记为 $A\ge 0$

8.5.2 正定阵的定理

$A>0⟺$ $A$ 为Hermite阵，且 ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n>0$
$A\ge 0⟺$ $A$ 为Hermite阵，且 ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n\ge 0$

---

证明

$\Rightarrow$

若 $A$ 为正定阵，则 $A$ 生成的二次型 $f(x)=X^{H}AX>0$ ，$\therefore {\lambda }_i=\frac{X^{H}AX}{|X{|}^2}>0$

$\Leftarrow$
$$\begin{array}{rl}& 由Hermite分解定理，A为Hermite阵，则存在U阵Q，\\ & 使得Q^{H}AQ=\Lambda =\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)\\ & \therefore A\stackrel{\Delta }{=}\Lambda ,而{\lambda }_i>0，\Lambda 为正定阵，故A为正定阵\end{array}$$
单位阵是正定阵 ：${\lambda }_i=1$ 显然大于0

8.5.3 正定阵间必合同

1. $A>0（正定阵）⟺A\stackrel{\Delta }{=}\Lambda$
2. $\Lambda \stackrel{\Delta }{=}I$ 对角阵一定合同于单位阵
3. 若A,B为同阶正定阵，则 $A\stackrel{\Delta }{=}B$

证明1：
$$\begin{array}{rl}& 若A正定，则有A^H=A\\ & 由Hermite分解定理，必\exists 优阵Q使得Q^{H}AQ=\Lambda ,且{\lambda }_i>0\end{array}$$
证明2：
$$\begin{array}{rl}若\Lambda & =\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right),其中{\lambda }_i>0\\ & \Rightarrow f(x)=X^H\Lambda X={\lambda }_1|x_1{|}^2+{\lambda }_2|x_2{|}^2+\cdots +{\lambda }_n|x_n{|}^2>0\\ & \Rightarrow 可分解为\Lambda =\left(\begin{array}{ccc}\sqrt{{\lambda }_1}& & \\ & \ddots & \\ & & \sqrt{{\lambda }_n}\end{array}\right)I\left(\begin{array}{ccc}\sqrt{{\lambda }_1}& & \\ & \ddots & \\ & & \sqrt{{\lambda }_n}\end{array}\right)\\ & =PIP\\ & 可知P可逆，且P^H=P,故P^{H}IP=\Lambda \\ & 即对角正定阵合同于单位阵，记为\Lambda \stackrel{\Delta }{=}I\end{array}$$
证明3：
$$\begin{array}{rl}& 由A>0,B>0,则A\stackrel{\Delta }{=}I,B\stackrel{\Delta }{=}I\Rightarrow A\stackrel{\Delta }{=}B\end{array}$$

8.5.4 乘积形式的正定阵

1. 对一切矩阵 $A=A_{n\times p}$ 且 $n\ge p$ ，$A^{H}A$ 与 $A{A}^H$ 都是Hermite阵
2. $A^{H}A$ 与 $A{A}^H$ 只相差 $n-p$ 个0根
3. $A^{H}A\ge 0$ ，$A^{H}A\ge 0$
4. $r(A^{H}A)=r(A{A}^H)=r(A)$

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a. $A^{H}A$ 为Hermite阵

$$\begin{array}{r}(A^{H}A{)}^H=A^{H}A，且(A{A}^H{)}^H=A{A}^H，则A^{H}A与A{A}^H为Hermite阵\end{array}$$

b. $A^{H}A与A{A}^H$ 相差n-p个0根

$$\begin{array}{rl}& A=A_{n\times p},B=p\times n,且n\ge p\\ & 由换位公式|\lambda I-AB|={\lambda }^{n-p}|\lambda I-BA|,\\ & 则AB与BA必有相同的非零根，故A^{H}A与A{A}^H只相差n-p个零根\end{array}$$

c. $A^{H}A与A{A}^H$ 是半正定阵（不是方阵的正定阵）

$$\begin{array}{rl}& 对任意非零向量X，有二次型f(x)=X^H{A}^{H}AX=(AX{)}^H(AX)=|AX{|}^2\ge 0,\\ & 可知f(x)为半正定二次型，A^{H}A为半正定阵\end{array}$$

d. $r(A{A}^H)=r(A^{H}A)=r(A)$

由 A H A 为半正定阵，则 A H A 与 A A H 都只有非负根 可写为 λ ( A H A ) = { λ 1 , λ 2 , . . . λ p } ≥ 0 由换位公式，知 λ A H A 与 λ A A H 只相差 n − p 个零根 ∴ λ ( A A H ) = { λ 1 , λ 2 , . . . , λ p , 0 , 0 , . . . , 0 } ≥ 0 r ( A H A ) = r ( A A H ) = p = r ( A ) = r ( A H ) \begin{aligned} &由A^HA为半正定阵，则A^HA与AA^H都只有非负根\\ &可写为\lambda(A^HA)=\{\lambda_1,\lambda_2,...\lambda_p\}\ge 0\\ &由换位公式，知\lambda{A^HA}与\lambda{AA^H}只相差n-p个零根\\ &\therefore \lambda(AA^H)=\{\lambda_1,\lambda_2,...,\lambda_p,0,0,...,0\}\ge 0\\ &r(A^HA)=r(AA^H)=p=r(A)=r(A^H) \end{aligned} ​由AHA为半正定阵，则AHA与AAH都只有非负根可写为λ(AHA)={λ1​,λ2​,...λp​}≥0由换位公式，知λAHA与λAAH只相差n−p个零根∴λ(AAH)={λ1​,λ2​,...,λp​,0,0,...,0}≥0r(AHA)=r(AAH)=p=r(A)=r(AH)​

$$\begin{array}{rl}& 齐次方程组AX=0,A^{H}AX=0，解集相同(同解)\\ & 若(A^{H}A)X=0成立，则|AX{|}^2=(AX{)}^H(AX)=X^H{A}^{H}AX=(AX{)}^2=0\\ & \therefore AX=0,r(A^{H}A)=r(A)\end{array}$$