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矩阵分析笔记(六)矩阵等价与线性映射的最简表示

October 9, 2020 • Read: 98 • 数学阅读设置

矩阵等价

矩阵$A\cong B$的充分必要条件是存在$m$阶可逆矩阵$P$及$n$阶可逆矩阵$Q$,使$PAQ=B$


线性映射的最简表示

在指定了空间$V_1$与$V_2$的基之后,便可以求得线性映射$\mathscr{A}:V_1\to V_2$在指定一对基下的矩阵表示。但是空间基是不唯一的,自然应该考虑以下两个问题:

  1. 线性映射在不同对基下的矩阵表示之间有什么关系?
  2. 对一个线性映射,能否选择一对基,使它的矩阵表示最简单(零多)?

先回答第一个问题

设$\mathscr{A}$是$V_1\to V_2$的一个线性映射,$\alpha_1,\alpha_2,...,\alpha_n$与$\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n$是$V_1$的两组基,由$\alpha_i$到$\alpha^{'}_i$的过渡矩阵为$P$。设$\beta_1,\beta_2,...,\beta_m$与$\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m$是$V_2$的两组基,由$\beta_j$到$\beta^{'}_j$的过渡矩阵为$Q$。线性映射$\mathscr{A}$在基$\alpha_1,\alpha_2,...,\alpha_n$与$\beta_1,\beta_2,...,\beta_m$下的矩阵表示为$A$,在基$\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n$与$\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m$下的矩阵表示为$B$,则

$$ B = Q^{-1}AP $$

证明:

由假设条件知

$$ \begin{gather} \mathscr{A}(\alpha_1,\alpha_2,...,\alpha_n)=(\beta_1,\beta_2,...,\beta_m)A \tag{1}\\ \mathscr{A}(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)=(\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m)B \tag{2}\\ (\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)=(\alpha_1,\alpha_2,...,\alpha_n)P \tag{3}\\ (\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m)=(\beta_1,\beta_2,...,\beta_m)Q \tag{4} \end{gather} $$

将式$(3)$和式$(4)$带入式$(2)$得

$$ \begin{gather} \mathscr{A}(\alpha_1,\alpha_2,...,\alpha_n)P=(\beta_1,\beta_2,..,\beta_m)QB \tag{5} \end{gather} $$

将式$(1)$带入式$(5)$得

$$ \begin{aligned} (\beta_1,\beta_2,...,\beta_m)AP=(\beta_1,\beta_2,...,\beta_m)QB \end{aligned} $$

因为$(\beta_1,\beta_2,...,\beta_m)$线性无关,故

$$ AP=QB $$

由于$Q$是满秩方阵(因为过渡矩阵都是满秩的),所以

$$ B = Q^{-1}AP \tag{6} $$

回答第二个问题

大学线性代数中有这么一个结论:对于$m\times n$矩阵$A$,总可经过初等变换(行变换和列变换)把它化为标准形

$$ Q^{-1}AP=\begin{bmatrix}E_r& 0_{r\times (n-r)}\\0_{(m-r)\times r}&0_{(m-r)\times (n-r)}\end{bmatrix}_{m\times n},\ r = rank(A) $$

将式$(6)$带入式$(2)$得

$$ \begin{aligned} \mathscr{A}(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)&=(\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m)Q^{-1}AP\\ &=(\beta^{'}_1,\beta^{'}_2,...,\beta^{'}_m)\begin{bmatrix}E_r& 0_{r\times (n-r)}\\0_{(m-r)\times r}&0_{(m-r)\times (n-r)}\end{bmatrix}_{m\times n} \end{aligned} $$

所以,对于一个线性映射,一定可以找到一对基,使得线性映射对应的矩阵最简单


线性变换

接下来的线性映射$\mathscr{A}$都是指线性空间$V$到$V$的映射,特称这样的$\mathscr{A}$为线性空间$V$的线性变换。由于线性变换时线性空间$V$到它自身的映射,所以只需取$V$的一组基$\alpha_1,\alpha_2,...,\alpha_n$即可

设$\alpha = \begin{bmatrix}\alpha_1,\alpha_2,...,\alpha_n\end{bmatrix}\begin{bmatrix}x_1\\x_2\\ \vdots x_n\end{bmatrix} \in V$,若

$$ \mathscr{A}(\alpha) = \begin{bmatrix}\alpha_1, \alpha_2,...,\alpha_n\end{bmatrix}\begin{bmatrix}y_1\\ y_2 \\ \vdots \\ y_n\end{bmatrix} $$

则原像$\alpha$与像$\mathscr{A}(\alpha)$的坐标变换公式为

$$ \begin{bmatrix}y_1\\y_2\\ \vdots \\ y_n\end{bmatrix}=A\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix} \tag{7} $$

例1

设$\mathbb{R}^3$中线性变换$\mathscr{A}$将基

$$ \alpha_1 = \begin{bmatrix}1\\1\\-1\end{bmatrix}, \alpha_2=\begin{bmatrix}0\\2\\-1\end{bmatrix},\alpha_3=\begin{bmatrix}1\\0\\-1\end{bmatrix} $$

变为基

$$ \alpha^{'}_1 = \begin{bmatrix}1\\-1\\0\end{bmatrix}, \alpha^{'}_2=\begin{bmatrix}0\\1\\-1\end{bmatrix},\alpha^{'}_3=\begin{bmatrix}0\\3\\-2\end{bmatrix} $$

  1. 求$\mathscr{A}$在基$\alpha_1,\alpha_2,\alpha_3$下的矩阵表示$A$
  2. 求向量$\xi=(1,2,3)^T$及$\mathscr{A}(\xi)$在基$\alpha_1, \alpha_2,\alpha_3$下的坐标
  3. 求向量$\xi$及$\mathscr{A}(\xi)$在基$\alpha^{'}_1,\alpha^{'}_2,\alpha^{'}_3$下的坐标

解:(1)

$$ \because \mathscr{A}(\alpha_1,\alpha_2,\alpha_3)=(\alpha_1,\alpha_2,\alpha_3)A=(\alpha^{'}_1,\alpha^{'}_2,\alpha^{'}_3)\\ \therefore A = (\alpha_1,\alpha_2,\alpha_3)^{-1}(\alpha^{'}_1,\alpha^{'}_2,\alpha^{'}_3)=\begin{bmatrix}1&-1&-1\\-1&1&2\\0&1&1\end{bmatrix} $$

(2)

设$\xi=\begin{bmatrix}\alpha_1,\alpha_2,\alpha_3\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}$,即

$$ \begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}1&0&1\\1&2&0\\-1&-1&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix} $$

解得

$$ k_1 = 10, k_2=-4,k_3=-9 $$

所以$\xi$在基$\alpha_1,\alpha_2,\alpha_3$下的坐标为$(10,-4,-9)^T$

$\mathscr{A}(\xi)$在基$\alpha_1,\alpha_2,\alpha_3$下的坐标可由公式$(7)$得

$$ \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}=\begin{bmatrix}1&-1&-1\\-1&1&2\\0&1&1\end{bmatrix}\begin{bmatrix}10\\-4\\-9\end{bmatrix}=\begin{bmatrix}23\\-32\\-13\end{bmatrix} $$

(3)

设$\xi=[\alpha_1,\alpha_2,\alpha_3]\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$,即

$$ \begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}1&0&0\\-1&1&3\\0&-1&-2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\Rightarrow \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}1\\-15\\6\end{bmatrix} $$

$\mathscr{A}(\xi)$在基$\alpha_1,\alpha_2,\alpha_3$下的坐标可由公式$(7)$得

$$ \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}=\begin{bmatrix}1&-1&-1\\-1&1&2\\0&1&1\end{bmatrix}\begin{bmatrix}1\\-15\\6\end{bmatrix}=\begin{bmatrix}10\\-4\\-9\end{bmatrix} $$

例2

求线性空间$\mathbb{R}^3$绕指定了正方向的固定轴旋转角度$\theta$的变换$\mathscr{A}$的矩阵表示

解:以$O$为起点沿旋转轴正方向取单位长有向线段,记为$e_z$,再取以$O$为起点的另两单位长有向线段$e_x,e_y$,使得$e_x,e_y,e_z$构成线性空间$V$中的右手直角坐标系。入口基和出口基都选为$e_x,e_y,e_z$

$$ \mathscr{A}(e_x,e_y,e_z)=\begin{bmatrix}e_x&e_y&e_z\end{bmatrix}\begin{bmatrix}cos\theta&-sin\theta&0\\sin\theta&cos\theta&0\\0&0&1\end{bmatrix} $$

因此$\mathscr{A}$的矩阵表示为$\begin{bmatrix}cos\theta&-sin\theta&0\\sin\theta&cos\theta&0\\0&0&1\end{bmatrix}$

例3

求几何空间中以$XOY$面为镜面反射变换$\mathscr{B}$的矩阵表示

解:

$$ \begin{aligned} \mathscr{B}(e_x,e_y,e_z)&=\begin{bmatrix}e_x&e_y&-e_z\end{bmatrix}\\ &=\begin{bmatrix}e_x&e_y&e_z\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix} \end{aligned} $$

因此$\mathscr{B}$的矩阵表示为$\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$

线性变换的运算

设$\mathscr{A},\mathscr{B}$是线性空间$V$的两个线性变换,$\lambda \in \mathbb{F}$

  1. 加法:$(\mathscr{A}+\mathscr{B})(\alpha)=\mathscr{A}(\alpha)+\mathscr{B}(\alpha)$
  2. 乘法:$\mathscr{AB}(\alpha)=\mathscr{A}(\mathscr{B}(\alpha))$
  3. 数乘:$(\lambda\mathscr{A})(\alpha)=\lambda \mathscr{A}(\alpha)$
  4. 可逆:设$\mathscr{A}\mathscr{B}=\mathscr{B}\mathscr{A}=E$,其中$E$为恒等变换,这时变换$\mathscr{B}$称为$\mathscr{A}$的拟变换,记为$\mathscr{A}^{-1}$

问题:对于一般的线性映射,能否定义加法、乘法、数乘?

很明显加法和数乘都可以,乘法不行(维度不匹配)


不同基下的矩阵关系

设$\mathscr{A}$为线性空间$V$上的线性变换,设$\alpha_1,\alpha_2,...,\alpha_n$与$\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n$为$V$的基且过渡矩阵为$P$。若$\mathscr{A}$在基$\alpha_1,\alpha_2,...,\alpha_n$下的矩阵表示为$A$,在基$\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n$下的矩阵表示为$B$,则

$$ B=P^{-1}AP $$

证明:

由已知得

$$ \begin{aligned} \mathscr{A}(\alpha_1,\alpha_2,...,\alpha_n)=(\alpha_1,\alpha_2,...,\alpha_n)A \\ \mathscr{A}(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)=(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)B \\ (\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)=(\alpha_1,\alpha_2,...,\alpha_n)P \end{aligned} $$

于是有

$$ \begin{aligned} \because (\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)&=(\alpha_1,\alpha_2,...,\alpha_n)P\\ \therefore \mathscr{A}(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)&=\mathscr{A}(\alpha_1,\alpha_2,...,\alpha_n)P\\ &=(\alpha_1,\alpha_2,...,\alpha_n)AP\\ \because \mathscr{A}(\alpha^{'}_1,\alpha^{'}_2,...,\alpha^{'}_n)&=((\alpha_1,\alpha_2,...,\alpha_n)P)B\\ &=(\alpha_1,\alpha_2,...,\alpha_n)PB\\ \therefore AP=PB &\Rightarrow B=P^{-1}AP \end{aligned} $$

相似

设$A,B\in \mathbb{F}^{m\times n}$,若存在$P\in \mathbb{F}^{n\times n}$,满足

$$ B=P^{-1}AP $$

则称$B$与$A$相似,记为$B\sim A$


同构

设$V_1,V_2$为线性空间,若存在一一映射$\sigma:V_1\to V_2$满足$\forall \alpha, \beta \in V_1, \lambda \in \mathbb{F}$,有

$$ \begin{aligned} \sigma(\alpha+\beta)&=\sigma(\alpha)+\sigma(\beta)\\ \sigma(\lambda \alpha)&=\lambda \sigma(\alpha) \end{aligned} $$

则称$V_1$与$V_2$同构,$\sigma$称为同构映射

同构的充要条件

数域$\mathbb{F}$上两个有限维线性空间$V_1,V_2$同构的充要条件是$\dim (V_1)=\dim (V_2)$

同构的性质

同构映射具有以下四个基本性质

  1. $\sigma(0)=0, \sigma(\alpha)=-\sigma(\alpha)$
  2. $\sigma(k_1\alpha_1+k_2\alpha_2+···+k_s\alpha_s)=k_1\sigma(\alpha_1)+k_2\sigma(\alpha_2)+···k_s\sigma(\alpha_s)$
  3. $V$中向量组$\alpha_1,\alpha_2,...,\alpha_s$线性相(无)关$\Longleftrightarrow$像$\sigma(\alpha_1),\sigma(\alpha_2),...,\sigma(\alpha_s)$线性相(无)关
  4. 如果$V_1$是$V$的一个子空间,则$V_1$在$\sigma$下的像集合$\sigma(V_1)=\{\sigma(\alpha)\mid \alpha \in V_1\}$是$\sigma(V)$的子空间,并且$V_1$与$\sigma(V_1)$维数相同

例4

设$\mathscr{A}$是线性空间$\mathbb{R}^3$的线性变换,它在$\mathbb{R}^3$中基$\alpha_1, \alpha_2,\alpha_3$下的矩阵表示为

$$ A = \begin{bmatrix}1&2&3\\-1&0&3\\2&1&5\end{bmatrix} $$

(1)求$\mathscr{A}$在基$\beta_1=\alpha_1,\beta_2=\alpha_1+\alpha_2,\beta_3=\alpha_1+\alpha_2+\alpha_3$下的矩阵表示

(2)求$\mathscr{A}$在基$\alpha_1,\alpha_2,\alpha_3$下的核与值域

解:

(1)

$$ \begin{aligned} \because [\beta_1,\beta_2,\beta_3]&=[\alpha_1,\alpha_2,\alpha_3]\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\\ &=[\alpha_1,\alpha_2,\alpha_3]P \end{aligned} $$

设$\mathscr{A}$在基$\beta_1,\beta_2,\beta_3$下的矩阵表示为$B$

$$ \begin{aligned} B &= P^{-1}AP\\ &=\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&-1\end{bmatrix}\begin{bmatrix}1&2&3\\-1&0&3\\2&1&5\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\\ &=\begin{bmatrix}4&4&4\\-3&-4&-6\\2&3&8\end{bmatrix} \end{aligned} $$

(2)

由于方程组$|A|\neq 0$,故$AX=0$只有零解,所以$\mathscr{A}$的核是零空间。由维数定理可知,$\mathscr{A}$的值域是线性空间$\mathbb{R}^3$

例5

设线性变换$\mathscr{A}$在基$\alpha_1=(-1,1,1)^T,\alpha_2=(1,0,-1)^T,\alpha_3=(0,1,1)^T$下的矩阵表示是

$$ A = \begin{bmatrix}1&0&-1\\1&1&0\\-1&2&3\end{bmatrix} $$

(1)求$\mathscr{A}$在基$\varepsilon_1=(1,0,0)^T,\varepsilon_2=(0,1,0)^T,\varepsilon_3=(0,1,1)^T$下的矩阵表示

(2)求$\mathscr{A}$的核与值域

解:

(1)

$$ \because [\varepsilon_1,\varepsilon_2,\varepsilon_3] = [\alpha_1,\alpha_2,\alpha_3]P\\ \Rightarrow P = [\alpha_1,\alpha_2,\alpha_3]^{-1}·[\varepsilon_1,\varepsilon_2,\varepsilon_3]\\ \Rightarrow P=\begin{bmatrix}-1&1&-1\\0&1&-1\\1&0&1\end{bmatrix} $$

设$\mathscr{A}$在基$\varepsilon_1,\varepsilon_2,\varepsilon_3$下的矩阵表示为为$B$

$$ \begin{aligned} B &= P^{-1}AP\\ &=\begin{bmatrix}-1&1&0\\1&-1&1\\1&0&1\end{bmatrix}\begin{bmatrix}1&0&-1\\1&1&0\\-1&2&3\end{bmatrix}\begin{bmatrix}-1&1&-1\\0&1&-1\\1&0&1\end{bmatrix}\\ &=\begin{bmatrix}1&1&0\\2&2&0\\3&0&2\end{bmatrix} \end{aligned} $$

(2)

由于方程组$AX=0$的基础解系是$[1,-1,1]^T$,所以$\mathscr{A}$的核子空间

$$ \mathcal{N}(\mathscr{A})=span\{\alpha_1-\alpha_2+\alpha_3\}=span\{[0,2,3]^T\} $$

$\mathscr{A}$的值域

$$ \begin{aligned} \mathcal{R}(\mathscr{A})&=span\{\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),\mathscr{A}(\alpha_3)\}\\ &=span\{\alpha_1+\alpha_2-\alpha_3,\alpha_2+2\alpha_3,-\alpha_1+3\alpha_3\}\\ &=span\{[0,0,-1]^T,[1,2,1]^T,[1,2,2]^T\}\\ &=span\{[0,0,1]^T,[1,2,0]^T\} \end{aligned} $$

Last Modified: October 12, 2020
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