酉阵和QR分解

 酉阵（Unitary Matrix）：复数域上的矩阵

 

内积：$X,Y \in C^{n}, k \in C$ 内积$(X,Y)$定义为:

\[(X,Y) = {Y^H}X = {x_1}{\overline y _1} + {x_2}{\overline y _2} + {x_3}{\overline y _3} + ... + {x_n}{\overline y _n}\]

有四个性质

(1)正性:$X \not\equiv 0$，则$(X,X) \ge 0$

 

(2)共轭对称

\[(Y,X) = {{\rm{X}}^H}Y = \overline {{x_1}} {y_1} + \overline {{x_2}} {y_2} + ... + \overline {{x_n}} {y_n} = \overline {{x_1}\overline {{y_1}}  + {x_2}\overline {{y_2}}  + ... + {x_n}\overline {{y_n}} }  = \overline {(X,Y)} \]

(3)齐性

\[\begin{array}{l}
(kX,Y) = k(X,Y)\\
(X,kY) = \overline k (X,Y)
\end{array}\]

(4)可加性

\[\begin{array}{l}
(X + Y,Z) = (X,Z) + (Y,Z)\\
(X,Y + Z) = (X,Y) + (X,Z)
\end{array}\]

酉阵的其他性质：

(1)$f(x)$为任意多项式(包括$e^x,sinx,cosx$)，则$f(A^H)=f(A)^H$

(2)根据公式$\det ({A^H}) = \overline {\det (A)} $，若$P$为酉阵，则$\det (P) =  \pm 1$

证明：

\[P{P^H} = I \Rightarrow \det (P)\det ({P^H}) = \det (P)\overline {\det (P)}  = {\left| {\det (P)} \right|^2} = \det (I) \Rightarrow \det (P) =  \pm 1\]

勾股定理:

\[{\left| {X + Y} \right|^2} = (X + Y,X + Y) = (X,X) + (X,Y) + (Y,X) + (Y,Y) = (X,X) + (Y,Y) = {\left| X \right|^2} + {\left| Y \right|^2}\]

酉阵（Unitary Matrix）定义:

方阵$A \in {C^{n \times n}}$:若方阵$A=[\alpha_{1},\alpha_{2},...,\alpha_{n}]$且$\left| {{\alpha _1}} \right| = \left| {{\alpha _2}} \right| = ... = \left| {{\alpha _n}} \right| = 1$,则A为酉阵

A为酉阵等价条件

(1)A为酉阵$ \Leftrightarrow {A^{ - 1}} = {A^H}$

(2)A为酉阵$ \Leftrightarrow A{A^H} = {A^H}A = {I_n}$

$A \in {C^{m \times n}}$, 满足以下条件，则A为行半酉阵:

\[A = \left[ {\begin{array}{*{20}{c}}
{{\alpha _1}}&{{\alpha _2}}&{{\alpha _k}}&{{\alpha _n}}
\end{array}} \right]\]

\[A{A^H} = \left[ {\begin{array}{*{20}{c}}
{{\beta _1}\beta _1^H}&0&0&0\\
0&{{\beta _2}\beta _2^H}&0&0\\
0&0&{{\beta _k}\beta _k^H}&0\\
0&0&0&{{\beta _m}\beta _m^H}
\end{array}} \right] = I\]

$A \in {C^{m \times n}}$, 满足以下条件，则A为列半酉阵:

\[A = \left[ {\begin{array}{*{20}{c}}
{{\beta _1}}\\
{{\beta _2}}\\
{{\beta _k}}\\
{{\beta _m}}
\end{array}} \right]\]

\[{A^H}A = \left[ {\begin{array}{*{20}{c}}
{\alpha _1^H{\alpha _1}}&0&0&0\\
0&{\alpha _2^H{\alpha _2}}&0&0\\
0&0&{\alpha _k^H{\alpha _k}}&0\\
0&0&0&{\alpha _n^H{\alpha _n}}
\end{array}} \right] = I\]

QR分解:设A$ \in {C^{m \times n}}$且列满秩($rank(A)=p$)，则有$A=QR$

求解Q:

\[\begin{array}{l}
{Y_1} = {X_1}\\
{Y_1} = {X_2} - \frac{{({X_2},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1}\\
{Y_3} = {X_3} - \frac{{({X_3},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1} - \frac{{({X_3},{Y_2})}}{{{{\left| {{Y_2}} \right|}^2}}}{{\rm{Y}}_2}\\
{Y_p} = {X_p} - \frac{{({X_p},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1} - \frac{{({X_p},{Y_2})}}{{{{\left| {{Y_2}} \right|}^2}}}{{\rm{Y}}_2} - ... - \frac{{({X_p},{Y_{p - 1}})}}{{{{\left| {{Y_{p - 1}}} \right|}^2}}}{{\rm{Y}}_{p - 1}}
\end{array}\]

则${Y_1} \bot {{\rm{Y}}_2} \bot ... \bot {{\rm{Y}}_p}$

令${\varepsilon _1} = \frac{{{Y_1}}}{{\left| {{Y_1}} \right|}},{\varepsilon _2} = \frac{{{Y_2}}}{{\left| {{Y_2}} \right|}},...,{\varepsilon _p} = \frac{{{Y_p}}}{{\left| {{Y_p}} \right|}}$

最终$Q = (\begin{array}{*{20}{c}}{{\varepsilon _1}}&{{\varepsilon _2}}&{...}&{{\varepsilon _p}}\end{array})$, Q为列半酉阵($Q^HQ=I$)

求解R:

\[A = QR \Rightarrow {Q^H}A = {Q^H}QR \Rightarrow {Q^H}A{\rm{ = R}}\]

所以$R=Q^HA$

\[R = \left[ {\begin{array}{*{20}{c}}
{\left| {{Y_1}} \right|}& \otimes & \otimes & \otimes \\
0&{\left| {{Y_2}} \right|}& \otimes & \otimes \\
0&0&{...}& \otimes \\
0&0&0&{\left| {{Y_p}} \right|}
\end{array}} \right](\left| {{Y_1}} \right| > 0,\left| {{Y_2}} \right| > 0,...,\left| {{Y_p}} \right| > 0)\]

R为正线上三角(对角线元素大于0)，且对角线元素为$|Y_k|$

证明对角线元素为$|Y_k|$:

\[R = {Q^H}A = \left[ {\begin{array}{*{20}{c}}
{{\varepsilon _1}^H}\\
\begin{array}{l}
{\varepsilon _2}^H\\
...
\end{array}\\
{{\varepsilon _p}^H}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{X_1}}&{{X_2}}&{...}&{{X_p}}
\end{array}} \right]\]

 

第k个对角线元素：

\[dia{g_k} = {\varepsilon _k}^H{X_k} = \frac{{{\rm{Y}}_k^H}}{{\left| {{Y_k}} \right|}}{X_k}\]

因为Schmidt正交方法有:

\[\begin{array}{l}
{Y_k} = {X_k} - \frac{{({X_k},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1} - \frac{{({X_k},{Y_2})}}{{{{\left| {{Y_2}} \right|}^2}}}{{\rm{Y}}_2} - ... - \frac{{({X_k},{Y_{k - 1}})}}{{{{\left| {{Y_{k - 1}}} \right|}^2}}}{{\rm{Y}}_{k - 1}}\\
{X_k} = {Y_k} + \frac{{({X_k},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1} + \frac{{({X_k},{Y_2})}}{{{{\left| {{Y_2}} \right|}^2}}}{{\rm{Y}}_2} + ... + \frac{{({X_k},{Y_{k - 1}})}}{{{{\left| {{Y_{k - 1}}} \right|}^2}}}{{\rm{Y}}_{k - 1}}
\end{array}\]

所以：

\[{\varepsilon _k}^H{X_k} = \frac{{Y_k^H}}{{\left| {{Y_k}} \right|}}{X_k} = \frac{{Y_k^H}}{{\left| {{Y_k}} \right|}}({Y_k} + \frac{{({X_k},{Y_1})}}{{{{\left| {{Y_1}} \right|}^2}}}{{\rm{Y}}_1} + ... + \frac{{({X_k},{Y_{k - 1}})}}{{{{\left| {{Y_{k - 1}}} \right|}^2}}}{{\rm{Y}}_{k - 1}}) = \frac{{Y_k^H}}{{\left| {{Y_k}} \right|}}{Y_k} = \frac{{{{\left| {{Y^k}} \right|}^2}}}{{\left| {{Y_k}} \right|}} = \left| {{Y_k}} \right|\]

镜面阵公式:

\[A = I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}}(X \in {C^n})\]

(1)$A^2=I$

\[{A^2} = {(I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}})^2} = {I^2} - \frac{{4X{X^H}}}{{{{\left| X \right|}^2}}}{\rm{ + (}}\frac{{2X{X^H}}}{{{{\left| X \right|}^4}}}{)^2} = {I^2} - \frac{{4X{X^H}}}{{{{\left| X \right|}^2}}} + \frac{{4X({X^H}X){X^H}}}{{{{\left| X \right|}^4}}} = {I^2} - \frac{{4X{X^H}}}{{{{\left| X \right|}^2}}} + \frac{{4X{X^H}}}{{{{\left| X \right|}^2}}} = I\]

(2)$A^H=A$

\[{A^H} = {(I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}})^H} = {I^H} - \frac{{2{{(X{X^H})}^H}}}{{{{\left| X \right|}^2}}} = I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}} = A\]

(3)A为酉阵

\[{A^H}A = A{A^H} = {A^2} = I\]

(4)$AX=-X$

\[AX = (I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}})X = X - \frac{{2X{X^H}{\rm{X}}}}{{{{\left| X \right|}^2}}} = X - 2X =  - X\]

(5)若$X \bot Y$,则$AY=Y$

\[AY = (I - \frac{{2X{X^H}}}{{{{\left| X \right|}^2}}})Y = Y - \frac{{2X{X^H}Y}}{{{{\left| X \right|}^2}}} = Y - \frac{{2X(Y,X)}}{{{{\left| X \right|}^2}}} = Y\]

(6)$\lambda (A) = \{  - 1,1,...,1\} $(n-1个1)

向量空间$C^n$中，除了X，平面中恰有n-1个$Y(Y_1,Y_2,...,Y_{n-1})$,(X,Y_1,Y_2,...,Y_{n-1})$之间两两正交

\[\begin{array}{l}
AX = - X\\
A{Y_i} = Y
\end{array}\]

(7)$det(A)=|A|=(-1)*1*...*1=-1$

(8)$A=A^{-1}=A^{H}$

\[AA = {A^{ - 1}}A = I \Rightarrow AAA = {A^{ - 1}}AA \Rightarrow A = {A^{ - 1}}\]

引理：在$C^n$中，若$|\alpha|=|\beta|$, 且内积$ \left( {\alpha ,\beta } \right) = {\beta ^H}\alpha $为实数，则$ \left( {\alpha  + \beta } \right) \bot \left( {\alpha  - \beta } \right) $

\[\left( {\alpha ,\beta } \right) = \overline {\left( {\beta ,\alpha } \right)}  = \left( {\beta ,\alpha } \right)\]

\[(\alpha  + \beta ,\alpha  - \beta ) = \left( {\alpha ,\alpha } \right) + \left( {\beta ,\alpha } \right) - \left( {\alpha ,\beta } \right) - \left( {\beta ,\beta } \right) = {\left| \alpha  \right|^2} - {\left| \beta  \right|^2} = 0 \Rightarrow (\alpha  + \beta ) \bot (\alpha  - \beta )\]

定理：

$\alpha ,\beta \in {C^n}$，且$|\alpha|=|\beta|$，且内积$ \left( {\alpha ,\beta } \right) = {\beta ^H}\alpha $为实数，则存在镜面阵A（酉阵），使得$A \alpha=\beta$,且${\rm{A = I - }}\frac{{2(\alpha  - \beta ){{(\alpha  - \beta )}^H}}}{{\left| {\alpha  - \beta } \right|}}$

\[\begin{array}{l}
X = \alpha - \beta \Rightarrow A(\alpha - \beta ) = - (\alpha - \beta )\\
\left( {\alpha + \beta } \right) \bot \left( {\alpha - \beta } \right) \Rightarrow A(\alpha + \beta ) = (\alpha + \beta )\\
A(\alpha - \beta ) + A(\alpha + \beta ) = - (\alpha - \beta ) + (\alpha + \beta ) \Rightarrow A\alpha = \beta
\end{array}\]

引理：

$\alpha \in {C^n},\forall \alpha = \left[ {\begin{array}{*{20}{c}}
{{\alpha _1}}\\
{...}\\
{{\alpha _n}}
\end{array}} \right]$令$\beta = \left[ {\begin{array}{*{20}{c}}{\lambda |\alpha| }\\{...}\\0\end{array}} \right]$,当$a_1 \ne 0$时，$\lambda  = \frac{{{a_1}}}{{\left| {{a_1}} \right|}}$当$a_1 = 0$时，$\lambda  = 1$,则存在镜面A，使得$A \alpha= \beta$

证明:

满足两个条件$(\alpha,\beta)$为实数和$|\alpha|=|\beta|$

验第一个条件：$(\alpha,\beta)$为实数

\[\begin{array}{l}
\left( {\alpha ,\beta } \right) = \overline \lambda {a_1}\left| \alpha \right|\\
{a_1} = 0 \Rightarrow \left( {\alpha ,\beta } \right) = 0\\
{a_1} \ne 0 \Rightarrow \overline \lambda {a_1}\left| \alpha \right| = \overline {\frac{{{a_1}}}{{\left| {{a_1}} \right|}}} {a_1}\left| \alpha \right| = \frac{{{{\left| {{a_1}} \right|}^2}}}{{\left| {{a_1}} \right|}}\left| \alpha \right| = \left| {{a_1}} \right|\left| \alpha \right|
\end{array}\]

验第二个条件：$|\alpha|=|\beta|$

\[\left| \beta  \right|{\rm{ = }}\left| {\lambda \left| \alpha  \right|} \right| = \left| \lambda  \right|\left| \alpha  \right| = \left| \alpha  \right|\]

镜面阵做QR分解：设$A=[\alpha_1,\alpha_2,...,\alpha_n]$(按列分块)：

(1)将第一列对角线以下元素清零

令${\alpha _1} = \left[ {\begin{array}{*{20}{c}}{{a_1}}\\{...}\\{{a_n}}\end{array}} \right]$和${\beta _1} = \left[ {\begin{array}{*{20}{c}}{\lambda \left| {{\alpha _1}} \right|}\\{...}\\0\end{array}} \right]$

计算镜面$\widetilde {{P_1}}$：

\[\widetilde {{P_1}} = I - \frac{{2({\alpha _1} - {\beta _1}){{({\alpha _1} + {\beta _1})}^H}}}{{{{\left| {{\alpha _1} - {\beta _1}} \right|}^2}}}\]

那么$\widetilde {{P_1}}A$有：

\[\widetilde {{P_1}}A = {P_1}\left[ {\begin{array}{*{20}{c}}
{{\alpha _1}}&{{\alpha _2}}&{...}&{{\alpha _n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{\beta _1}}&{{{\overline P }_1}{\alpha _2}}&{...}&{{P_1}{\alpha _n}}
\end{array}} \right]\]

\[{P_1} = \widetilde {{P_1}}\]

第一列除了第一个元素外其他为零：

\[\widetilde {{P_1}}A = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes & \otimes & \otimes \\
0& \otimes & \otimes & \otimes \\
0& \otimes &{...}& \otimes \\
0& \otimes & \otimes & \otimes
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes \\
0&{{A^{(1)}}}
\end{array}} \right]\]

(2)继续将第二列对角线以下元素清零：

由(1)得到$A^{(1)}$，并按列分块${A^{(1)}} = \left[ {\begin{array}{*{20}{c}}{\alpha _1^{(1)}}&{\alpha _2^{(1)}}&{...}&{\alpha _{n - 1}^{(1)}}\end{array}} \right]$

得到$\alpha^{(1)}_1$和$\beta^{(1)}_1$：

\[\alpha _1^{(1)} = \left[ {\begin{array}{*{20}{c}}
{a_1^{(1)}}\\
{...}\\
{a_n^{(1)}}
\end{array}} \right]\]

\[\beta _1^{(1)} = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {\alpha _1^{(1)}} \right|}\\
{...}\\
0
\end{array}} \right]\]

计算镜面：

\[\widetilde {{P_2}} = I - \frac{{2(\alpha _1^{(1)} - \beta _1^{(1)}){{(\alpha _1^{(1)} + \beta _1^{(1)})}^H}}}{{{{\left| {\alpha _1^{(1)} - \beta _1^{(1)}} \right|}^2}}}\]

计算$\widetilde {{P_2}}{A^{(1)}}$：

\[\widetilde {{P_2}}{A^{(1)}} = \left[ {\begin{array}{*{20}{c}}
{\alpha _1^{(1)}}&{\alpha _1^{(1)}}&{...}&{\alpha _{n - 1}^{(1)}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\beta _{_1}^{(1)}}&{{P_2}\alpha _{_2}^{(1)}}&{...}&{{P_2}\alpha _{n - 1}^{(1)}}
\end{array}} \right]\]

第二列也被清零：

\[\widetilde {{P_2}}{A^{(1)}} = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {\alpha _1^{(1)}} \right|}& \otimes & \otimes & \otimes \\
0& \otimes & \otimes & \otimes \\
0& \otimes &{...}& \otimes \\
0& \otimes & \otimes & \otimes
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {\alpha _1^{(1)}} \right|}& \otimes \\
0&{{A^{(2)}}}
\end{array}} \right]\]

得到$P_2$：

\[{P_2} = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&{\widetilde {{P_2}}}
\end{array}} \right]\]

由(1)(2)可以得到：

\[{P_2}{P_1}A = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&{\overline {{P_2}} }
\end{array}} \right]{P_1}A = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&{\overline {{P_2}} }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes \\
0&{{A^{(1)}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes \\
0&{\overline {{P_2}} {A^{(1)}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes & \otimes \\
0&{\lambda \left| {\alpha _1^{(1)}} \right|}& \otimes \\
0&0&{{A^{(2)}}}
\end{array}} \right]\]

前两列被清零。

(3)继续执行直到n-1次：

\[\begin{array}{l}
{P_{n - 1}}{P_{n - 2}}...{P_2}{P_1}A = R\\
R = \left[ {\begin{array}{*{20}{c}}
{\lambda \left| {{\alpha _1}} \right|}& \otimes & \otimes & \otimes \\
0&{\lambda \left| {\alpha _1^{(1)}} \right|}& \otimes & \otimes \\
0&0&{...}& \otimes \\
0&0&0&{\lambda \left| {\alpha _1^{(n - 1)}} \right|}
\end{array}} \right]\\
A = {({P_{n - 1}}{P_{n - 2}}...{P_2}{P_1})^{ - 1}}R = P_1^{ - 1}P_2^{ - 1}...P_{n - 2}^{ - 1}P_{n - 1}^{ - 1}R = ({P_1}{P_2}...{P_{n - 2}}{P_{n - 1}})R\\
Q = {P_1}{P_2}...{P_{n - 2}}{P_{n - 1}}
\end{array}\]

 

镜面阵QR分解举例：

\[A = \left[ {\begin{array}{*{20}{c}}
0&3&1\\
0&4&{ - 2}\\
2&1&1
\end{array}} \right]\]

(1)将第一列对角线以下元素化零：

得到$\alpha$：

\[\alpha = \left[ {\begin{array}{*{20}{c}}
0\\
0\\
2
\end{array}} \right]\]

 计算$\beta$：

\[\beta = \left[ {\begin{array}{*{20}{c}}
2\\
0\\
0
\end{array}} \right]\]

计算镜面阵：

\[{P_1} = I - \frac{{2(\alpha - \beta ){{({\alpha _1} + \beta )}^H}}}{{{{\left| {\alpha - \beta } \right|}^2}}} = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]\]

$P_1 A$：

\[{P_1}A = \left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&3&1\\
0&4&{ - 2}\\
2&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&4&{ - 2}\\
0&3&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
2& \otimes \\
0&{{A^{(1)}}}
\end{array}} \right]\]

得到$A^{(1)}$

(2)将第二列对角线以下元素清零：

根据$A^{(1)}$取第一列得到：

\[\alpha = \left[ {\begin{array}{*{20}{c}}
4\\
3
\end{array}} \right]\]

计算$\beta$：

\[\beta = \left[ {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right]\]

计算镜面：

\[\widetilde {{P_2}} = I - \frac{{2(\alpha - \beta ){{({\alpha _1} + \beta )}^H}}}{{{{\left| {\alpha - \beta } \right|}^2}}} = \frac{1}{5}\left[ {\begin{array}{*{20}{c}}
4&3\\
{{\rm{ - }}3}&4
\end{array}} \right]\]

\[{P_2}{\rm{ = }}\left[ {\begin{array}{*{20}{c}}
1&0\\
0&{\frac{1}{5}\left[ {\begin{array}{*{20}{c}}
4&3\\
{{\rm{ - }}3}&4
\end{array}} \right]}
\end{array}} \right]{\rm{ = }}\frac{1}{5}\left[ {\begin{array}{*{20}{c}}
5&0&0\\
0&4&3\\
0&{{\rm{ - }}3}&4
\end{array}} \right]\]

\[{P_2}{\rm{ = }}\left[ {\begin{array}{*{20}{c}}
1&0\\
0&{\frac{1}{5}\left[ {\begin{array}{*{20}{c}}
4&3\\
{{\rm{ - }}3}&4
\end{array}} \right]}
\end{array}} \right]{\rm{ = }}\frac{1}{5}\left[ {\begin{array}{*{20}{c}}
5&0&0\\
0&4&3\\
0&{{\rm{ - }}3}&4
\end{array}} \right]\]

$P_2 P_1 A$为：

\[{P_2}{P_1}A = \frac{1}{5}\left[ {\begin{array}{*{20}{c}}
5&0&0\\
0&4&3\\
0&{{\rm{ - }}3}&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&0&1\\
0&1&0\\
1&0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&3&1\\
0&4&{ - 2}\\
2&1&1
\end{array}} \right]{\rm{ = }}\frac{1}{5}\left[ {\begin{array}{*{20}{c}}
5&0&0\\
0&4&3\\
0&{{\rm{ - }}3}&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&4&{ - 2}\\
0&3&1
\end{array}} \right]{\rm{ = }}\left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&5&{ - 1}\\
0&0&2
\end{array}} \right]\]

\[A = {({P_2}{P_1})^{ - 1}}\left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&5&{ - 1}\\
0&0&2
\end{array}} \right] = {P_1}^{ - 1}{P_2}^{ - 1}\left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&5&{ - 1}\\
0&0&2
\end{array}} \right] = {P_1}{P_2}\left[ {\begin{array}{*{20}{c}}
2&1&1\\
0&5&{ - 1}\\
0&0&2
\end{array}} \right]\]

令$Q=P_1P_2$，得到$A=QR$

 

 

参考PPT：https://wenku.baidu.com/view/1b482f39f68a6529647d27284b73f242336c31b4.html