【矩阵论】4. 矩阵运算——广义逆——广义逆的计算

4.3.2 $A^{+}$ 计算

a. 秩1公式

$$\begin{array}{c}若A=(a_{ij}{)}_{m\times n},r(A)=1,则A^{+}=\frac{1}{\sum |a_{ij}{|}^2}{A}^H=\frac{1}{tr(A^{H}A)}{A}^H\end{array}$$

$A^{H}A$ 的相关结论：

• $tr(A^{H}A)=t(A{A}^H)=\sum |a_{ij}{|}^2=\Vert A{\Vert }_F^2$
• $r(A^{H}A)=r(A{A}^H)=r(A)$
• $A^{H}A,A{A}^H$ 为半正定阵

证明：
$$\begin{array}{rl}& 由于r(A)=1,则A^{H}A优相似与对角阵D=\left(\begin{array}{cccc}{\lambda }_1& & & \\ & 0& & \\ & & \ddots & \\ & & & 0\end{array}\right)，其中\sqrt{{\lambda }_1}为正奇值\Rightarrow \\ & tr(A^{H}A)=tr(A{A}^H)={\lambda }_1=\sum |a_{ij}{|}^2,则A的正SVD为A=P\Delta Q^H=P\left(\sqrt{{\lambda }_1}\right)Q^H,\\ & 其中P^{H}P=Q^{H}Q=I\\ & A^H=Q\Delta P^H=Q(\sqrt{{\lambda }_1})P^H,A^{+}=Q{\Delta }^{-1}{P}^H=Q(\frac{1}{\sqrt{{\lambda }_1}})P^H\\ & \Rightarrow A^{+}=\frac{1}{(\sqrt{{\lambda }_1}{)}^2}Q(\sqrt{{\lambda }_1})P^H=\frac{1}{(\sqrt{{\lambda }_1}{)}^2}Q\Delta P^H=\frac{1}{(\sqrt{{\lambda }_1}{)}^2}{A}^H=\frac{1}{tr(A^{H}A)}{A}^H=\frac{1}{\sum |a_{ij}{|}^2}{A}^H\end{array}$$
eg：

b. 正奇公式(第一公式)

$$\begin{array}{rl}& 若A=A_{m\times n}有正SVD，A=P\Delta Q^H,\Delta =\left(\begin{array}{ccc}{s}_1& & \\ & \ddots & \\ & & s_r\end{array}\right)>0,\\ & 其中P，Q为半U阵，P^{H}P=I_r=Q^{H}Q\\ & 则有公式A^{+}=Q{\Delta }^{-1}{P}^H,其中{\Delta }^{-1}={\Delta }^{+}\end{array}$$

由 $A^{+}$ 的正奇公式，可证 逆的H穿脱公式 $(A^H{)}^{+}=(A^{+}{)}^H$

• $A^{+}$ 的正奇值为 $\{s_1^{-1},s_2^{-1},\cdots ,s_r^{-1}\}$ ，且有公式 $\Vert A^{+}{\Vert }_F=\sqrt{s_1^{-2},\cdots ,s_r^{-2}}$ ，$\Vert A^{+}{\Vert }_2=s_r^{-1}$

证明：
$$\begin{array}{rl}& A=P\Delta Q^H\Rightarrow A^H=Q\Delta P^H为A^H的正SVD\\ & A^{+}=Q{\Delta }^{-1}{P}^H,(A^H{)}^{+}=P{\Delta }^{-1}{Q}^H\Rightarrow (A^{+}{)}^H=P{\Delta }^{-1}{Q}^H=(A^H{)}^{+}\end{array}$$
eg

$$\begin{array}{rl}& (1)求正奇值，A^{H}A=\left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\\ 2& 0\end{array}\right)=\left(\begin{array}{cc}5& 0\\ 0& 1\end{array}\right),\lambda (A)=\{5,1\},正奇值为\sqrt{5},1,令\Delta =\left(\begin{array}{cc}\sqrt{5}& 0\\ 0& 1\end{array}\right)\\ & (2)SVD，Q=(q_1,q_2)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),P=\left(\begin{array}{c}\frac{A{q}_1}{|A{q}_1|},\frac{A{q}_2}{|A{q}_2|}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& 0\\ 0& 1\\ \frac{2}{\sqrt{5}}& 0\end{array}\right),A=P\Delta Q^H=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& 0\\ 0& 1\\ \frac{2}{\sqrt{5}}& 0\end{array}\right)\left(\begin{array}{cc}\sqrt{5}& 0\\ 0& 1\end{array}\right){\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)}^H\\ & (3)A^{+}=Q{\Delta }^{-1}{P}^H=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 0\\ 0& \sqrt{5}\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{5}}& 0& \frac{2}{\sqrt{5}}\\ 0& 1& 0\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{5}}& 0& \frac{2}{\sqrt{5}}\\ 0& 1& 0\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{5}& 0& \frac{2}{5}\\ 0& 1& 0\end{array}\right)\\ & (4)\because B=A^H,则B^{+}=(A^H{)}^{+}=(A^{+}{)}^H=\left(\begin{array}{cc}\frac{1}{5}& 0\\ 0& 1\\ \frac{2}{5}& 0\end{array}\right)\end{array}$$

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$$\begin{array}{rl}& A^{H}A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right),\lambda (A^{H}A)=\{3,1\},正奇值S^{+}=\{\sqrt{3},1\},故A^{H}A优相似与\Delta =\left(\begin{array}{cc}\sqrt{3}& 0\\ 0& 1\end{array}\right),\\ & {\lambda }_1=3的特根q_1=\left(\begin{array}{c}1\\ 1\end{array}\right),{\lambda }_2=1的特根为q_2=\left(\begin{array}{c}1\\ -1\end{array}\right),\\ & Q=(q_1,q_2)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right),P=\left(\frac{A{q}_1}{|A{q}_1|},\frac{A{q}_2}{|A{q}_2|}\right)=\left(\begin{array}{c}\frac{1}{\sqrt{6}}\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& 0\end{array}\right)\\ & A=P\Delta Q^H=\left(\begin{array}{cc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{6}}& -\frac{1}{\sqrt{2}}\\ \frac{2}{\sqrt{6}}& 0\end{array}\right)\left(\begin{array}{cc}\sqrt{3}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\\ & A^{+}=Q{\Delta }^{-1}{P}^H=\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{3}}& 0\\ 0& 1\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{6}}& \frac{1}{\sqrt{6}}& \frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}& 0\end{array}\right)=\frac{1}{6}=\left(\begin{array}{ccc}4& -2& 2\\ -2& 4& 2\end{array}\right)\end{array}$$

c. 优分解公式

$$\begin{array}{rl}& 设P,Q为优阵，则(PDQ{)}^{+}=Q^H{D}^{+}{P}^H=Q^{H}D{P}^H\end{array}$$

• 引理：若 $D=\left(\begin{array}{ccc}{\lambda }_1& & \\ & \ddots & \\ & & {\lambda }_n\end{array}\right)$ 为对角阵，则 $D^{+}D=D{D}^{+}$ ，且 $(D^{+}{)}^k=(D^k{)}^{+}$ ，(k=1,2,…)

d. QR分解公式

$$\begin{array}{rl}& 设A=A_{m\times p}为高阵，且有QR分解，A=QR，Q为列U阵，R=R_{p\times p},则A^{+}=R^{-1}{Q}^H\end{array}$$

eg

$$\begin{array}{rl}& (1)求A的QR分解，令{\beta }_1=\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right),{\beta }_2={\alpha }_2-\frac{({\alpha }_2,{\beta }_1)}{({\beta }_1,{\beta }_1)}{\beta }_1=\left(\begin{array}{c}2\\ 0\\ -1\end{array}\right),Q=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ 0& 0\\ 2& -1\end{array}\right)\\ & R=Q^{H}A=\sqrt{5}\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\Rightarrow A=QR=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ 0& 0\\ 2& -1\end{array}\right)\sqrt{5}\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\\ & (2)A^{+}=(QR{)}^{+}=R^{+}{Q}^{+}=R^{-1}{Q}^H=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)\frac{1}{\sqrt{5}}\left(\begin{array}{ccc}1& 0& 2\\ 2& 0& -1\end{array}\right)=\frac{1}{5}\left(\begin{array}{ccc}-1& 0& 3\\ 2& 0& -1\end{array}\right)\end{array}$$

e. 单阵谱分解

若A为单阵不可逆，则不能用谱分解方式求逆

$$\begin{array}{rl}& tr(A)=1,故A^{+}=\frac{1}{2}{A}^H=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)\\ & A的特征多项式为|A-\lambda I|=\lambda (1-\lambda )\Rightarrow \lambda (A)=\{1,0\},G_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\\ & A有谱分解A=G_1,A^{+}\ne A,故对于不可逆的单阵，A^{+}\ne {\lambda }_1^{-1}{G}_1+\cdots +{\lambda }_k^{-1}{G}_k\end{array}$$

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若A为单阵且可逆，则有谱分解 $A={\lambda }_1{G}_1+\cdots +{\lambda }_k{G}_k$ 且 $A^{+}={\lambda }_1^{-1}{G}_1+\cdots +{\lambda }_k^{-1}{G}_k$

$$\begin{array}{rl}& A为行和阵，\lambda (A)=\{4,1\},G_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}=\frac{1}{3}\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right),G_2=\frac{1}{3}\left(\begin{array}{cc}2& -2\\ -1& 1\end{array}\right),A=4G_1+G_2\\ & r(A)=2，则A可逆，故\Rightarrow A^{+}=A^{-1}=\frac{1}{3}{G}_1+G_2=\frac{1}{4}\left(\begin{array}{cc}3& -2\\ -1& 2\end{array}\right)\end{array}$$

正规阵谱分解

$$\begin{array}{rl}& 若A为正规阵，且有谱分解：A={\lambda }_1{G}_1+\cdots +{\lambda }_k{G}_k,则有A^{+}={\lambda }_1^{+}{G}_1+{\lambda }_2^{+}{G}_2+\cdots +{\lambda }_k^{+}{G}_k\end{array}$$

eg

$$\begin{array}{rl}& (1)求A的谱分解：\\ & A有特征多项式|A-\lambda I|=\left|\begin{array}{cc}2-\lambda & 1-i\\ 1+i& 1-\lambda \end{array}\right|=(2-\lambda )(1-\lambda )-2=\lambda (\lambda -3)\Rightarrow \lambda (A)=\{3,0\}\\ & G_1=\frac{A-{\lambda }_{2}I}{{\lambda }_1-{\lambda }_2}=\frac{1}{3}\left(\begin{array}{cc}2& 1-i\\ 1+i& 1\end{array}\right),G_2=I-G_1=\frac{1}{3}\left(\begin{array}{cc}1& 1-i\\ -1-i& 2\end{array}\right)\\ & 可得谱分解A=3G_1+0G_2\\ & (2)A^{+}=3^{+}{G}_1=\frac{1}{3}{G}_1=\frac{1}{9}\left(\begin{array}{cc}2& 1-i\\ 1+i& 1\end{array}\right)\\ & 验证：\\ & (1-i)r_2=r_1，即A为秩1公式，故有A^{+}=\frac{1}{\sum |a_{ij}{|}^2}\left(\begin{array}{cc}2& 1+i\\ 1-i& 1\end{array}\right)=\frac{1}{9}\left(\begin{array}{cc}2& 1-i\\ 1+i& 1\end{array}\right)\end{array}$$

g. 高低分解公式(第二公式)

引理

高低阵引理：

高低阵消去性质：

Hermite性质：

高低广逆公式

$$\begin{array}{rl}& 高阵公式：若A=A_{m\times r}为列满秩阵(高阵)，则有加号逆A^{+}=(A^{H}A{)}^{-1}{A}^H,且A^{+}A=I\\ & 低阵公式：若A=A_{r\times n}为行满秩阵(低阵)，则有加号逆A^{+}=A^H(A{A}^H{)}^{-1},且A{A}^{+}=I\end{array}$$

eg

$$\begin{array}{rl}& 由高阵公式：A^{+}=(A^{H}A{)}^{-1}{A}^H,计算可得A^{H}A=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right),(A^{H}A{)}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right),\\ & 可得A^{+}=(A^{H}A{)}^{-1}{A}^H=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right)=\frac{1}{2}\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\end{array}\right),可验证\\ & A^{+}A=I\end{array}$$

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$$\begin{array}{rl}& 由于B=A^H,则B^{+}=(A^H{)}^{+}=(A^{+}{)}^H=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 2\\ 1& 0\end{array}\right)\end{array}$$

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$$\begin{array}{rl}& 由低阵公式A^{+}=A^H(A{A}^H{)}^{-1},A{A}^H=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right),(A{A}^H{)}^{-1}=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)\\ & A^{+}=A^H(A{A}^H{)}^{-1}=\frac{1}{3}\left(\begin{array}{cc}1& 1\\ -1& 2\\ 2& -1\end{array}\right)\\ & 由于B=A^H,B^{+}=(A^H{)}^{+}=(A^{+}{)}^H=\frac{1}{3}\left(\begin{array}{ccc}1& -1& 2\\ 1& 2& -1\end{array}\right)\end{array}$$

$A^{+}$ 第二公式（高低分解公式）

$$\begin{array}{rl}& 若有高低分解A=BC，则有公式：A^{+}=C^{+}{B}^{+},其中B^{+}=(B^{H}B{)}^{-1}{B}^H,C^{+}=A^H(A{A}^H{)}^{-1}\\ & 且有B^{+}B=I,C{C}^{+}=I\end{array}$$

注

$(BC{)}^{+}$ 一般不成立，只有在高低分解时才成立

eg

$$\begin{array}{rl}& \because A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),故选第1,2列作为高阵，B=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 0\\ 2& 1\end{array}\right),C=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& -1\end{array}\right),A=BC(高低分解)\\ & (B^{H}B{)}^{-1}=\frac{1}{4}\left(\begin{array}{cc}1& -1\\ -1& 3\end{array}\right),(C{C}^H{)}^{-1}=\frac{1}{2}\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right),\\ & B^{+}=(B^{H}B{)}^{-1}{B}^H=\frac{1}{4}\left(\begin{array}{cc}1& -1\\ -1& 3\end{array}\right)\left(\begin{array}{cccc}1& 0& 1& 2\\ 0& 1& 0& 1\end{array}\right)=\frac{1}{4}\left(\begin{array}{cccc}1& 0& 1& 1\\ -1& 3& -1& 1\end{array}\right),\\ & C^{+}=C^H(C{C}^H{)}^{-1}=\frac{1}{2}\left(\begin{array}{cc}2& 0\\ 0& 1\\ 0& -1\end{array}\right),\\ & A=C^{+}{B}^{+}=\frac{1}{8}\left(\begin{array}{cccc}2& -2& 2& 2\\ -1& 3& -1& 1\\ 1& -3& 1& -1\end{array}\right)\end{array}$$

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h. 第三公式

$$A^{+}=(A^{H}A{)}^{+}{A}^H$$

应用

eg

i. 张量分解公式

$$(A\otimes B{)}^{+}=A^{+}\otimes B^{+}$$

j. 分块公式

$$\begin{gathered} {\left(\begin{array}{c}A\\ 0\end{array}\right)}^{+}=\left(A^{+},0\right),(A,0{)}^{+}=\left(\begin{array}{c}{A}^{+},0\end{array}\right) \\ {\left(\begin{array}{cc}{A}_1& 0\\ 0& A_2\end{array}\right)}^{+}=\left(\begin{array}{cc}{A}_1^{+}& 0\\ 0& A_2^{+}\end{array}\right) \\ {\left(\begin{array}{cc}0& A_1\\ A_2& 0\end{array}\right)}^{+}=\left(\begin{array}{cc}0& A_1^{+}\\ A_2^{+}& 0\end{array}\right) \\ {\left(\begin{array}{c}A\\ A\\ A\end{array}\right)}^{+}=\frac{1}{3}\left(A^{+},A^{+},A^{+}\right) \\ (A,A,A{)}^{+}=\frac{1}{3}\left(\begin{array}{c}{A}^{+}\\ A^{+}\\ A^{+}\end{array}\right) \\ {\left(\begin{array}{cc}A& A\\ A& A\end{array}\right)}^{+}=\frac{1}{4}\left(\begin{array}{cc}{A}^{+}& A^{+}\\ A^{+}& A^{+}\end{array}\right) \end{gathered}$$

eg

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$$\begin{array}{rl}& A=\left(\begin{array}{c}D\\ 0\end{array}\right),D为低阵，D^{+}=D^H(D{D}^H{)}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 2\\ 1& 0\end{array}\right),A^{+}=\left(\begin{array}{cc}{D}^{+}& 0\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ \frac{1}{2}& 0& 0\end{array}\right)\end{array}$$

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eg

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