矩阵分析（十一）酉矩阵、正交矩阵

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

令

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

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

令

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

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$A=QR$