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矩阵分析(十一)酉矩阵、正交矩阵

November 23, 2020 • Read: 3055 • 数学阅读设置

酉矩阵

若$n$阶复矩阵$A$满足

$$ A^HA=AA^H=E $$

则称$A$是酉矩阵,记为$A\in U^{n\times n}$

设$A\in C^{n\times n}$,则$A$是酉矩阵的充要条件是$A$的$n$个列(或行)向量是标准正交向量组

酉矩阵的性质
  1. $A^{-1}=A^H\in U^{n \times n}$
  2. $\mid \det A\mid=1$
  3. $A^T\in U^{n\times n}$
  4. $AB, BA\in U^{n\times n}$,其中$B\in U^{n\times n}$
  5. 酉矩阵的特征值的模为1
  6. 标准正交基到标准正交基的过渡矩阵是酉矩阵
酉变换

设$V$是$n$维酉空间,$\mathscr{A}$是$V$的线性变换,若$\forall \alpha, \beta \in V$都有

$$ (\mathscr{A}(\alpha), \mathscr{A}(\beta))=(\alpha,\beta) $$


正交矩阵

若$n$阶实矩阵$A$满足

$$ A^TA=A^HA=E $$

则称$A$是正交矩阵,记为$A\in E^{n\times n}$

设$A\in R^{n\times n}$,则$A$是正交矩阵的充要条件是$A$的$n$个列(或行)向量是标准正交向量组

正交矩阵的性质
  1. $A^{-1}=A^T\in E^{n\times n}$
  2. $\det A=±1$
  3. $AB,BA\in E^{n\times n}$
正交变换

设$V$是$n$维欧氏空间,若线性变换$\mathscr{B}$满足$\forall \alpha,\beta\in V$都有

$$ (\mathscr{B}(\alpha), \mathscr{B}(\beta))=(\alpha, \beta) $$

设$\mathscr{A}$是酉空间(或欧式空间)$V$的线性变换,则下列命题等价:

  1. $\mathscr{A}$是酉变换(或正交变换)
  2. $||\mathscr{A}(\alpha)||=||\alpha||$,其中$\forall \alpha \in V$
  3. $\sigma$将$V$的标准正交基变到标准正交基
  4. 酉变换(或正交变换)在标准正交基下的矩阵表示是酉矩阵(或正交矩阵)

满秩矩阵的QR分解

若$n$阶实矩阵$A\in \mathbb{C}^{n\times n}$满秩,且

$$ A = [\alpha_1,...,\alpha_n] $$

其中$\alpha_1,...,\alpha_n$是$\mathbb{C}^{n\times n}$中线性无关向量组

正交化

$$ \begin{aligned} \beta_1&=\alpha_1\\ \beta_2&=\alpha_2 - \frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}\beta_1\\ \vdots\\ \beta_n &= \alpha_n - \frac{\left<\beta_n,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}\beta_1-\cdots-\frac{\left<\beta_{n-1},\alpha_n\right>}{\left<\beta_{n-1},\alpha_{n-1}\right>}\beta_{n-1} \end{aligned} $$

变形得

$$ \begin{aligned} \alpha_1 &= \beta_1\\ \alpha_2 &= \frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}\beta_1 + \beta_2\\ \vdots \\ \alpha_n &= \frac{\left<\beta_n,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}\beta_1-\cdots-\frac{\left<\beta_{n-1},\alpha_n\right>}{\left<\beta_{n-1},\alpha_{n-1}\right>}\beta_{n-1} + \beta_n \end{aligned} $$

写成矩阵形式

$$ \begin{aligned} \begin{bmatrix}\alpha_1,\alpha_2,...,\alpha_n\end{bmatrix} &= \begin{bmatrix}\beta_1,\beta_2,...,\beta_n\end{bmatrix}\begin{bmatrix}1&\frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}&\cdots &\frac{\left<\beta_1,\alpha_n\right>}{\left<\beta_1,\beta_1\right>}\\0&1&\cdots&\frac{\left<\beta_2,\alpha_n\right>}{\left<\beta_2,\beta_2\right>}\\ &&\ddots\\0&&\cdots& 1\end{bmatrix}\\ &\triangleq B\begin{bmatrix}1&\frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}&\cdots &\frac{\left<\beta_1,\alpha_n\right>}{\left<\beta_1,\beta_1\right>}\\0&1&\cdots&\frac{\left<\beta_2,\alpha_n\right>}{\left<\beta_2,\beta_2\right>}\\ &&\ddots\\0&&\cdots& 1\end{bmatrix} \end{aligned} $$

单位化

$$ q_1=\frac{\beta_1}{||\beta_1||}\\ \vdots \\ q_n = \frac{\beta_n}{||\beta_n||} $$

变形得

$$ \beta_1 = q_1||\beta_1||\\ \vdots \\ \beta_n = q_n ||\beta_n|| $$

写成矩阵形式

$$ \begin{aligned} \begin{bmatrix}\beta_1,\beta_2,...,\beta_n\end{bmatrix} &= \begin{bmatrix}q_1,q_2,...,q_n\end{bmatrix}\begin{bmatrix}||\beta_1||&0&\cdots &0\\0&||\beta_2||&\cdots&0\\ &&\ddots\\0&&\cdots& ||\beta_n||\end{bmatrix}\\ &\triangleq Q\begin{bmatrix}||\beta_1||&0&\cdots &0\\0&||\beta_2||&\cdots&0\\ &&\ddots\\0&&\cdots& ||\beta_n||\end{bmatrix} \end{aligned} $$

综上,结合正交化和单位化可得

$$ \begin{aligned} A &= B\begin{bmatrix}1&\frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}&\cdots &\frac{\left<\beta_1,\alpha_n\right>}{\left<\beta_1,\beta_1\right>}\\0&1&\cdots&\frac{\left<\beta_2,\alpha_n\right>}{\left<\beta_2,\beta_2\right>}\\ &&\ddots\\0&&\cdots& 1\end{bmatrix}\\ &=Q\begin{bmatrix}||\beta_1||&0&\cdots &0\\0&||\beta_2||&\cdots&0\\ &&\ddots\\0&&\cdots& ||\beta_n||\end{bmatrix}\begin{bmatrix}1&\frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}&\cdots &\frac{\left<\beta_1,\alpha_n\right>}{\left<\beta_1,\beta_1\right>}\\0&1&\cdots&\frac{\left<\beta_2,\alpha_n\right>}{\left<\beta_2,\beta_2\right>}\\ &&\ddots\\0&&\cdots& 1\end{bmatrix}\\ &=Q\begin{bmatrix}||\beta_1||&\frac{\left<\beta_1,\alpha_1\right>}{\left<\beta_1,\beta_1\right>}&\cdots &\frac{\left<\beta_1,\alpha_n\right>}{\left<\beta_1,\beta_1\right>}\\0&||\beta_2||&\cdots&\frac{\left<\beta_2,\alpha_n\right>}{\left<\beta_2,\beta_2\right>}\\ &&\ddots\\0&&\cdots& ||\beta_n||\end{bmatrix}\\ &\triangleq QR \end{aligned} $$

QR分解定理:$A\in \mathbb{C}^{n\times n}$,则存在酉矩阵$Q$和正线上三角阵$R$,使

$$ A=QR $$

且分解唯一

QR分解的求法

  1. 取矩阵$A=(A_1,A_2,...,A_n)$的列向量,进行Schmidt标准正交化,得$v_1,v_2,...,v_n$,有$Q=(v_1,v_2,...,v_n)$
  2. 再由$R=Q^HA$得到$R$,于是$A=QR$
Last Modified: December 10, 2020
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