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矩阵分析(五)线性映射

October 5, 2020 • Read: 1305 • 数学阅读设置

线性映射定义

设$V_1,V_2$是数域$\mathbb{F}$上两个线性空间,映射$\mathscr{A}:V_1→V_2$,如果它保持加法和数乘法:

  1. $\forall \alpha_1, \alpha_2\in V_1$有$\mathscr{A}(\alpha_1+\alpha_2)=\mathscr{A}(\alpha_1)+\mathscr{A}(\alpha_2)$
  2. $\forall \alpha \in V_1, k\in \mathbb{F}$有$\mathscr{A}(\alpha ·k)=\mathscr{A}(\alpha)·k$

则称$\mathscr{A}$为从线性空间$V_1$到$V_2$的线性映射,如果$V_1$和$V_2$是同一个线性空间,则称$\mathscr{A}$为线性变换。从$V_1$到$V_2$的所有线性映射构成的集合记为$\mathscr{L}(V_1, V_2)$

例如,映射

$$ \mathscr{A}:\mathbb{R}^2→\mathbb{R}^2\\ \begin{bmatrix} x_1\\ x_2 \end{bmatrix}\mapsto \begin{bmatrix}x_1+x_2 \\ x_1x_2\end{bmatrix} $$

不是线性映射;映射

$$ \mathscr{B}:\mathbb{R}^3 \to \mathbb{R}^2\\ \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}\mapsto \begin{bmatrix}x_1+x_2-x_3 \\ 2x_1+3x_2\end{bmatrix} $$

是线性映射

线性映射的简单性质

  1. $\mathscr{A}(0)=0$
  2. $\mathscr{A}(\sum_{i=1}^sk_i\alpha_i)=\sum_{i=1}^sk_i\mathscr{A}(\alpha_i)$,其中$\alpha_i \in V, k_i\in \mathbb{F}$
  3. 设$\alpha_1, \alpha_2,...,\alpha_s \in V_1 ,\alpha_1, \alpha_2,...,\alpha_s$线性相关,则$\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_s)$也线性相关
  4. 若$V=span\{\alpha_1,\alpha_2,...,\alpha_s\}$,则$\mathscr{A}$的值域$\mathscr{A}(V)=span\{\mathscr{A}(\alpha_1),...,\mathscr{A}(\alpha_s)\}$

现证明性质(3)

事实上,由于$\alpha_1,\alpha_2,...,\alpha_s$线性相关,不失一般性,不妨设

$$ \alpha_s = k_1\alpha_1 + k_2\alpha_2+···+k_{s-1}\alpha_{s-1} $$

于是

$$ \mathscr{\alpha_s} = k_1\mathscr{\alpha_1}+k_2\mathscr{\alpha_2}+···+k_{s-1}\alpha_{\alpha_{s-1}} $$

因此,$\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_s)$线性相关

特别要注意,若$\alpha_1, \alpha_2, ..., \alpha_s \in V_1$线性无关,则$\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2), ..., \mathscr{A}(\alpha_s)$不一定线性无关

线性变换的简单性质

设$\mathscr{A,B,C}\in \mathscr{L}(V,V)$,且$x\in V$,则

  1. (乘法的可结合性)$((\mathscr{A}\mathscr{B})\mathscr{C})(x)=(\mathscr{A}(\mathscr{B}\mathscr{C}))(x)$
  2. (线性映射的左分配律)$\mathscr{A}(\mathscr{B}+\mathscr{C})=\mathscr{A}\mathscr{B}+\mathscr{A}\mathscr{C}$
  3. (线性映射的右分配律)$(\mathscr{A}+\mathscr{B})\mathscr{C}=\mathscr{A}\mathscr{C}+\mathscr{B}\mathscr{C}$

线性映射的乘法一般是不满足交换律的

现证明性质(1)

$$ \because ((\mathscr{A}\mathscr{B})\mathscr{C})(x)=(\mathscr{A}\mathscr{B})(\mathscr{C}(x))=\mathscr{A}(\mathscr{B}(\mathscr{C}(x))) $$

$$ \because (\mathscr{A}(\mathscr{B}\mathscr{C}))(x)=\mathscr{A}((\mathscr{B}\mathscr{C})(x))=\mathscr{A}(\mathscr{B}(\mathscr{C}(x))) $$

所以$((\mathscr{A}\mathscr{B})\mathscr{C})(x)=(\mathscr{A}(\mathscr{B}\mathscr{C}))(x)$


线性映射的矩阵表示

给定$\mathbb{F}$上的线性空间$V_1, V_2$,及线性映射$\mathscr{A}:V_1\to V_2$。设$\dim(V_1)=n, \dim(V_2)=m$,并设$\varepsilon_1,\varepsilon_2,...,\varepsilon_n$为$V_1$的一个基(称为入口基);$\eta_1, \eta_2,...,\eta_m$为$V_2$的一个基(称为出口基)。记第$j$个入口基向量$\varepsilon_j \in V_1$在$\mathscr{A}$下的像$\mathscr{A}(\varepsilon_j)\in V_2$在出口基$\eta_1, \eta_2,...,\eta_n$下的坐标为

$$ a_j = \begin{bmatrix}a_{1j} \\ a_{2j}\\ \vdots \\ a_{mj}\end{bmatrix}\in \mathbb{F}^m $$

$$ \mathscr{A}(\varepsilon_j)=\begin{bmatrix}\eta_1 & \eta_2 & \cdots & \eta_m\end{bmatrix}\begin{bmatrix}a_{1j}\\ a_{2j}\\ \vdots \\ a_{mj}\end{bmatrix},\ j=1,2,...,n $$

则由$\mathbb{F}^m$中的向量组$a_1, a_2,...,a_n$拼成的矩阵

$$ A = \begin{bmatrix}a_1 & a_2 & \cdots & a_n\end{bmatrix} = [a_{ij}]_{m\times n} $$

称为$\mathscr{A}$在相应的入口基和出口基下的表示

定义记号

$$ \mathscr{A}\begin{bmatrix}\varepsilon_1 & \varepsilon_2 & \cdots & \varepsilon_n\end{bmatrix}=\begin{bmatrix}\mathscr{A}(\varepsilon_1) & \mathscr{A}(\varepsilon_2) & \cdots & \mathscr{A}(\varepsilon_n)\end{bmatrix} $$

即线性映射作用在向量组拼成的矩阵上,定义为向量组中每个向量的像按原顺序所成的向量组(简称为向量组的像)拼成的矩阵。则有下面的公式

$$ {\color{red} {\mathscr{A}\begin{bmatrix}\varepsilon_1 & \varepsilon_2 & \cdots & \varepsilon_n\end{bmatrix}=\begin{bmatrix}\eta_1 & \eta_2 & \cdots & \eta_m\end{bmatrix}A}} $$

用文字表示,读作

$$ [线性映射][入口基矩阵]=[出口基矩阵][表示矩阵] $$

事实上,只要确定了线性映射两个空间的(例如$(\varepsilon_1,\cdots,\varepsilon_n)$和$(\beta_1,\cdots,\beta_m)$),就有唯一确定的一个矩阵$A$与之对应,而且矩阵$A$的每一个列向量就是对应的原基向量映射后的坐标;反之,如果基确定,任何一个矩阵都唯一确定了一个线性映射

我个人理解,线性映射其实就是将一个$m$维的矩阵,转换为$n$维的矩阵,而在转换过程中,需要一个$m\times n$的矩阵$A$,这类似于PyTorch中的nn.Linear(m, n, bias=False)函数


用坐标计算线性映射

设线性映射$\mathscr{A}:V_1\to V_2$在入口基$\varepsilon_1, \varepsilon_2,...,\varepsilon_n$和出口基$\eta_1, \eta_2,...,\eta_m$下的矩阵表示为$A$,又设$\alpha \in V_1$在入口基下的坐标为$X \in \mathbb{F}^n$,则$\mathscr{A}(\alpha)\in V_2$在出口基下的坐标为$AX\in \mathbb{F}^m$

证明:也就是要证$\mathscr{A}(\alpha)=\begin{bmatrix}\eta_1 & \eta_2 & \cdots & \eta_m\end{bmatrix}(AX)$

因为

$$ \begin{aligned} \begin{bmatrix}\eta_1 & \eta_2 & \cdots & \eta_m\end{bmatrix}(AX)&= (\begin{bmatrix}\eta_1 & \eta_2 & \cdots & \eta_m\end{bmatrix}A)X\\ &=(\mathscr{A}\begin{bmatrix}\varepsilon_1 & \varepsilon_2 & \cdots & \varepsilon_n\end{bmatrix})X\\ &=\mathscr{A}(\begin{bmatrix}\varepsilon_1 & \varepsilon_2 & \cdots & \varepsilon_n\end{bmatrix}X)\\ &=\mathscr{A}(\alpha) \end{aligned} $$

证毕


例2

设$B=\begin{bmatrix}1 & 2 \\ 1 & 1 \\ 0 & 1\end{bmatrix}$,映射$\mathscr{A}:\mathbb{R}^2\to \mathbb{R}^3$由下式确定

$$ \mathscr{A}(\alpha)=B\alpha, \ \ \ \alpha \in \mathbb{R}^2 $$

试求$\mathscr{A}$在基$\alpha_1=(1, 0)^T, \alpha_2=(0,1)^T$与基$\beta_1=(1,0,0)^T,\beta_2=(0,1,0)^T,\beta_3=(0,0,1)^T$下的矩阵表示$A$

解:因为$\mathscr{A}(\alpha_1)=(1, 1, 0)^T, \mathscr{A}(\alpha_2)=(2, 1, 1)^T$

又由$\begin{bmatrix}\mathscr{A}(\alpha_1) & \mathscr{A}(\alpha_2)\end{bmatrix}=\begin{bmatrix}\beta_1 & \beta_2 & \beta_3\end{bmatrix}A$得

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


例3

求线性映射$\mathscr{A}:R[x]_{n+1}\to R[x]_n$

$$ \mathscr{A}(f(x))=\frac{d}{dx}f(x) $$

在基$1, x, x^2, ..., x^n$与基$1, x, x^2,..., x^{n-1}$下的矩阵表示$D$

解:因为$\mathscr{A}(1)=0, \mathscr{A}(x)=1,...,\mathscr{A}(x^n)=nx^{n-1}$

又由$\begin{bmatrix}\mathscr{A}(1) & \mathscr{A}(x) & \cdots & \mathscr{x^n}\end{bmatrix}=\begin{bmatrix}1 & x & \cdots & x^{n-1}\end{bmatrix}A$得

$$ A = \begin{bmatrix}0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 2 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & n\end{bmatrix}_{n\times (n+1)} $$


例4

设$\mathscr{A}$是$n$维线性空间$V$的一个线性变换,对某个$\xi\in V$有$\mathscr{A}^{k-1}(\xi)\neq 0, \mathscr{A}^k(\xi)=0$,试证:$\xi,\mathscr{A}(\xi),\mathscr{A}^2(\xi),...,\mathscr{A}^{k-1}(\xi)$线性无关

解:$\forall a_1,...,a_{k-1}$,使

$$ a_1\xi+a_2\mathscr{A}(\xi)+···+a_k\mathscr{A}^{k-1}(\xi)=0\quad(1) $$

有且仅有零解,且线性无关

用$\mathscr{A}^{k-1}$左乘$(1)$式两端,由$\mathscr{A}^{k}(\xi)=0$可得

$$ a_1\mathscr{A}^{k-1}(\xi)=0 $$

因为$\mathscr{A}^{k-1}(\xi)\neq 0$,所以$a_1=0$,带入$(1)$式得

$$ a_2\mathscr{A}(\xi)+···+a_k\mathscr{A}^{k-1}(\xi)=0\quad(2) $$

用$\mathscr{A}^{k-2}$左乘$(2)$式两端,由$\mathscr{A}^k(\xi)=0$可得$a_2=0$,以此类推,可得$a_3=···=a_k=0$

所以$\xi,\mathscr{A}(\xi),...,\mathscr{A}^{k-1}(\xi)$线性无关


例5

设$\beta_1,\beta_2,...,\beta_m$线性无关,且

$$ \begin{aligned} \xi_{i} &={a}_{1 i} {\beta}_{1}+a_{2 i}{\beta}_{2}+\cdots+a_{m i} \boldsymbol{\beta}_{m} \\ &=\left({\beta}_{1}, {\beta}_{2}, \cdots ,{\beta}_{m}\right)\left[\begin{array}{c} a_{1 i} \\ a_{2 i} \\ \vdots \\ a_{m i} \end{array}\right] \quad(i=1,2, \cdots, s) \end{aligned} $$

试证:向量组$\xi_1,\xi_2,...,\xi_s$的秩=矩阵$(a_{ij})_{m\times s}$的秩

解:令$B=(\beta_1,\beta_2,...,\beta_m)$,$A=(a_{ij})_{m\times s}$,$P=(\xi_1,...,\xi_s)$,则$P=BA$

因为$\beta_1,...,\beta_m$线性无关,所以$B$满秩

又因为$rank(P)=rank(BA)$,所以$rank(P)=rank(A)$


线性映射的值域

设$\mathscr{A}$为线性空间$V_1$到$V_2$的线性映射,则称$\{\mathscr{A}(\alpha)\mid \alpha \in V_1\}$为$\mathscr{A}$的值域,记为$\mathcal{R}(\mathscr{A})$,且$\mathcal{R}(\mathscr{A})$为$V_2$的子空间。$\dim (\mathcal{R(\mathscr{A})})$称为$\mathscr{A}$的秩,记作$rank(\mathscr{A})$

证明($\mathcal{R}(\mathscr{A})$为$V_2$的子空间):任取$\alpha, \beta \in \mathcal{R}(\mathscr{A}), \lambda \in \mathbb{F}$,则存在$\alpha_1, \beta_1 \in V_1$,使得$\mathscr{A}(\alpha_1)=\alpha, \mathscr{A}(\beta_1)=\beta$,因为

$$ \begin{aligned} \alpha + \beta = \mathscr{A}(\alpha_1) + \mathscr{A}(\beta_1)=\mathscr{A}(\alpha_1 + \beta_1) \in \mathcal{R}(\mathscr{A})\\ \lambda \alpha = \lambda \mathscr{\alpha_1}=\mathscr{\lambda \alpha_1} \in \mathcal{R}(\mathscr{A}) \end{aligned} $$

故$\mathcal{R}(\mathscr{A})$为$V_2$的子空间(保加法、保数乘)

值域与秩的求法(定理)

设$\alpha_1, \alpha_2,...,\alpha_n$为$V_1$的基,$\beta_1, \beta_2,...,\beta_m$为$V_2$的基,$\mathscr{A}$为$V_1$到$V_2$的线性映射且在基$\alpha_1, \alpha_2,...,\alpha_n$与基$\beta_1, \beta_2,...,\beta_m$下的矩阵为$A$,即

$$ (\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n))=(\beta_1,\beta_2,...,\beta_m)A $$

  • $\mathcal{R}(\mathscr{A})=span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\}$
  • $rank(\mathscr{A})=rank(A)$

证明:任取$\alpha \in \mathcal{R}(\mathscr{A})$,则存在$\beta \in V_1$,使得$\mathscr{A}(\beta)=\alpha$

$$ \begin{aligned} &\because\ \beta = k_1\alpha_1 + k_2\alpha_2 +···+k_n \alpha_n \\ &\therefore\ \alpha = \mathscr{A}(\beta) = k_1\mathscr{A}(\alpha_1) + k_2\mathscr{A}(\alpha_2) + ··· + k_n \mathscr{A}(\alpha_n) \\ &\Rightarrow \ \alpha \in span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\} \\ &\Rightarrow \ \mathcal{R}(\mathscr{A}) \subseteq span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\} \end{aligned} $$

任取$\alpha \in span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\}$,则

$$ \begin{aligned} &\alpha = k_1\mathscr{A}(\alpha_1) + k_2\mathscr{A}(\alpha_2) + ··· + k_n \mathscr{A}(\alpha_n)\\ &\because \ \mathscr{A}(\alpha_i)\in \mathcal{R}(\mathscr{A}),\mathcal{R}(\mathscr{A})为子空间\\ &\therefore \ \alpha \in \mathcal{R}(\mathscr{A})\\ &\Rightarrow \ span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\} \subseteq \mathcal{R}(\mathscr{A}) \end{aligned} $$

故$\mathcal{R}(\mathscr{A})=span\{\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2),...,\mathscr{A}(\alpha_n)\}$


线性映射的核

设$\mathscr{A}$为$V_1$到$V_2$的线性映射,则称$\{\alpha \mid \mathscr{A}(\alpha)=0, \alpha \in V_1\}$为$\mathscr{A}$的核,记为$\mathcal{N}(\mathscr{A})$,且$\mathcal{N}(\mathscr{A})$为$V_1$的子空间。$\dim (\mathcal{N}(\mathscr{A}))$称为$\mathscr{A}$的零度,记为$null(\mathscr{A})$

证明($\mathcal{N}(\mathscr{A})$为$V_1$的子空间):任取$\alpha, \beta \in \mathcal{N}(\mathscr{A}), \lambda \in \mathbb{F}$,则$\mathscr{A}(\alpha) = 0, \mathscr{A}(\beta)=0$,因为

$$ \mathscr{A}(\alpha + \beta) = \mathscr{A}(\alpha) + \mathscr{A}(\beta)=0 \Rightarrow \ \alpha+\beta \in \mathcal{N}(\mathscr{A})\\ \mathscr{A}(\lambda \alpha) = \lambda \mathscr{A}(\alpha)=0 \Rightarrow \ \lambda \alpha \in \mathcal{N}(\mathscr{A}) $$

故$\mathcal{N}(\mathscr{A})$为$V_1$的子空间(保加法、保数乘)

核与零度的求法(定理)

设$\alpha_1, \alpha_2, ..., \alpha_n$为$V_1$的基,$\beta_1, \beta_2,...,\beta_m$为$V_2$的基,$\mathscr{A}$为$V_1$到$V_2$的线性映射且在$\alpha_1, \alpha_2,...,\alpha_n$与基$\beta_1, \beta_2, ...,\beta_m$下的矩阵为$A$,且$Ax=0$的基础解系为$\xi_1, \xi_2,..,\xi_{n-r}$,则

  • $\mathcal{N}(\mathscr{A})=span\{\gamma_1, \gamma_2, ..., \gamma_{n-r}\}$,其中$\gamma_i=(\alpha_1, \alpha_2,...,\alpha_n)\xi_i$
  • $null(\mathscr{A})=n-rank(\mathscr{A})=n-rank(A)$
  • $\dim(\mathcal{R}(\mathscr{A}))+\dim(\mathcal{N}(\mathscr{A}))=\dim(V)$

证明:任取$\alpha \in \mathcal{N}(\mathscr{A})$,且$\alpha$在基$\alpha_1, \alpha_2,...,\alpha_n$下的坐标为$\xi$

则$\mathscr{A}(\alpha)=0$,且$\mathscr{A}(\alpha)$在基$\beta_1, \beta_2,...,\beta_m$下的坐标为$A\xi$

$\Rightarrow A\xi=0$,即$\xi$为$Ax=0$的解,则有

$$ \begin{aligned} &\xi = k_1\xi_1 + k_2\xi_2 +···+k_{n-r}\xi_{n-r}\\\\ &\Rightarrow \alpha = (\alpha_1,\alpha_2,...,\alpha_n)\xi=k_1[(\alpha_1,\alpha_2,...,\alpha_n)\xi_1]+k_2[(\alpha_1,\alpha_2,...,\alpha_n)\xi_2]\\ &+···+k_{n-r}[(\alpha_1,\alpha_2,...,\alpha_n)\xi_1]+k_2[(\alpha_1,\alpha_2,...,\alpha_n)\xi_{n-r}]\\ &=k_1\gamma_1+k_2\gamma_2+···+k_{n-r}\gamma_{n-r}\\ \\ &\Rightarrow \alpha \in span\{\gamma_1, \gamma_2,...,\gamma_{n-r}\}\Rightarrow \mathcal{N}(\mathscr{A})\subseteq span\{\gamma_1, \gamma_2,...,\gamma_{n-r}\} \end{aligned} $$

任取$\alpha \in span\{\gamma_1, \gamma_2,...,\gamma_{n-r}\}$,则

$$ \begin{aligned} \alpha&=k_1\gamma_1 + k_2\gamma_2+···+k_{n-r}\gamma_{n-r}\\ \Rightarrow \mathscr{A}(\alpha)&=k_1\mathscr{A}(\gamma_1)+k_2\mathscr{A}(\gamma_2)+···+k_{n-r}\mathscr{A}(\gamma_{n-r})\\ \because \mathscr{A}(\gamma_i)&=\mathscr{A}[(\alpha_1, \alpha_2,...,\alpha_n)\xi_i]=[\mathscr{A}(\alpha_1,\alpha_2,...,\alpha_n)]\xi_i\\ &=[(\beta_1,\beta_2,...,\beta_m)A]\xi_i=(\beta_1,\beta_2,...,\beta_m)(A\xi_i)\\ &=(\beta_1, \beta_2,...,\beta_m)0=0\\ \therefore \mathscr{A}(\alpha)&=0 \Rightarrow \in \mathcal{N}(\mathscr{A}) \Rightarrow span\{\gamma_1, \gamma_2, ..., \gamma_{n-r}\}\subseteq \mathcal{N}(\mathscr{A}) \end{aligned} $$

故$\mathcal{N}(\mathscr{A})=span\{\gamma_1, \gamma_2,...,\gamma_{n-r}\}$


*矩阵的列空间

对于一个$m\times n$的矩阵$A$,它的列空间就是$A$中列向量$(A_1,A_2,...,A_n)$的所有线性组合,即

$$ span\{A_1,A_2,...,A_n\} $$


例6

设线性变换$\mathscr{A}: V\to V$在基$\alpha_1,\alpha_2,\alpha_3$下的矩阵表示为$A=\begin{bmatrix}1&2&0\\0&-1&1\\1&1&1\end{bmatrix}$,求$\mathcal{R}(\mathscr{A}),\mathcal{N}(\mathscr{A})$的基与维数

解:对矩阵$A$做初等行变换得

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

因为$\mathcal{R}(\mathscr{A})=span\{\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),\mathscr{A}(\alpha_3)\}$,所以$\mathcal{R}(\mathscr{A})$的基是$(\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),\mathscr{A}(\alpha_3))$的极大线性无关组

又因为

$$ (\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),\mathscr{A}(\alpha_3))=(\alpha_1,\alpha_2,\alpha_3)A\\ $$

所以$\mathcal{R}(\mathscr{A})$的基为$(\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2))=(\alpha_1+\alpha_3,2\alpha_1-\alpha_2+\alpha_3)$,$\dim(\mathcal{R}(\mathscr{A}))=2$

$A$的核空间为$AX=0$的解空间,即

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

其基础解系为$[-2,1,1]^T$,所以$\mathcal{N}(\mathscr{A})$的基为$-2\alpha_1+\alpha_2+\alpha_3$,且$\dim(\mathcal{N}(\mathscr{A}))=1$


例7

设线性变换$\mathscr{A}:\mathbb{F}^{2\times 2}\to \mathbb{F}^{2\times 2}$在基$E_{11},E_{12},E_{21},E_{22}$下的定义为:

$$ \forall X=\begin{bmatrix}a&b\\c&d\end{bmatrix},\mathscr{A}(X)=\begin{bmatrix}a+b&b+c\\c+d&d+a\end{bmatrix} $$

求$\mathcal{R}(\mathscr{A}),\mathcal(N)(\mathscr{A})$的一组基与维数

解:

因为$\mathcal{R}(\mathscr{A})=span\{\mathscr{A}(E_{11}),\mathscr{A}(E_{12}),\mathscr{A}(E_{21}),\mathscr{A}(E_{22})\}$,所以$\mathcal{R}(\mathscr{A})$的基是$(\mathscr{A}(E_{11}),\mathscr{A}(E_{12}),\mathscr{A}(E_{21}),\mathscr{A}(E_{22}))$的极大线性无关组

又因为

$$ \begin{aligned} (\mathscr{A}(E_{11}),\mathscr{A}(E_{12}),\mathscr{A}(E_{21}),\mathscr{A}(E_{22}))&=(E_{11},E_{12},E_{21},E_{22})A\\ \Rightarrow A&= \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{bmatrix}\\ \to &\begin{bmatrix}1&0&0&1\\0&1&0&-1\\0&0&1&1\\0&0&0&0\end{bmatrix} \end{aligned} $$

所以$\mathcal{R}(\mathscr{A})$的基为$(\mathscr{A}(E_{11}),\mathscr{A}(E_{12}),\mathscr{A}(E_{21}))=(\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix})$,$\dim(\mathcal{R}(\mathscr{A}))=3$

注意:最后找基的时候,要借助原来$A$矩阵的各个列向量来看,不能使用经过初等行变换后的矩阵

$A$的核空间为$AX=0$的解空间,即

$$ \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=0\\ \Rightarrow \begin{cases}x_1=-x_4\\x_2=x_4\\x_3=-x_4\end{cases} $$

其基础解系为$[-1,1,-1,1]^T$,所以$\mathcal{N}(\mathscr{A})$的基为$-E_{11}+E_{12}-E_{21}+E_{22}$,且$\dim(\mathcal{N}(\mathscr{A}))=1$


例8

求矩阵$A=\begin{bmatrix}1&1&6\\0&4&2\\1&1&6\end{bmatrix}$的值域$\mathcal{R}(A)$与核空间$\mathcal{N}(A)$

解:$A$的值域$\mathcal{R}(A)$为

$$ \begin{aligned} \mathcal{R}(A) &=\operatorname{span}\left\{[1,0,1]^{T} ,[1,4,1]^{T} ,[6,2,6]^{T}\right\} \\ &=\operatorname{span}\left\{[1,0,1]^{T} ,[1,4,1]^{T}\right\} \\ &=\operatorname{span}\left\{[1,0,1]^{T},[0,1,0]^{T}\right\} \end{aligned} $$

又因为$A$的核空间为$AX=0$的解空间,其基础解系为$[11, 1, -2]^T$,所以

$$ \mathcal{N}(A)=span\{[11, 1, -2]^T\} $$


例9

设$B=\begin{bmatrix}1& 2& 3 \\ 1 & 1 & 2 \\ 2& 0 & 2\\ 0 & 1& 1\end{bmatrix}$,定义$\mathscr{A}:\mathbb{R}^3\to \mathbb{R}^4, \mathscr{A}(\alpha)=B\alpha$,$\alpha_1=\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, \alpha_2=\begin{bmatrix}0 \\ 1\\ 1\end{bmatrix}, \alpha_3 = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$为$\mathbb{R}^3$的基,$\beta_1 = \begin{bmatrix}1 \\ 1 \\ 0 \\ 0\end{bmatrix}, \beta_2 = \begin{bmatrix}1 \\ 0\\ 1\\ 0\end{bmatrix}, \beta_3=\begin{bmatrix}1\\ 0\\ 0\\ 1\end{bmatrix}, \beta_4 = \begin{bmatrix}0 \\ 1\\ 1\\ 0\end{bmatrix}$为$\mathbb{R}^4$的基

求:(1)$\mathscr{A}$在基$\alpha_1, \alpha_2, \alpha_3$与基$\beta_1, \beta_2, \beta_3, \beta_4$下的矩阵$A$;(2)$\mathcal{R}(\mathscr{A})$与$\mathcal{N}(\mathscr{A})$的基

解:

(1)

$$ \because \mathscr{A}(\alpha_1)=\begin{bmatrix}1&2&3\\1&1&2\\2&0&2\\0&1&1\end{bmatrix}\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}3\\2\\2\\1\end{bmatrix}\\ \mathscr{A}(\alpha_2)=\begin{bmatrix}5\\3\\2\\2\end{bmatrix}\\ \mathscr{A}(\alpha_3)=\begin{bmatrix}4\\3\\4\\1\end{bmatrix}\\ \therefore \mathscr{A}(\alpha_1, \alpha_2, \alpha_3)=(\mathscr{A}(\alpha_1), \mathscr{A}(\alpha_2), \mathscr{A}(\alpha_3))=\begin{bmatrix}3&5&4\\2&3&3\\2&2&4\\1&2&1\end{bmatrix} $$

$$ \mathscr{A}(\alpha_1, \alpha_2, \alpha_3)=(\beta_1, \beta_2,\beta_3,\beta_4)A\\ \therefore \begin{bmatrix}3&5&4\\2&3&3\\2&2&4\\1&2&1\end{bmatrix}=\begin{bmatrix}1&1&1&0\\1&0&0&1\\0&1&0&1\\0&0&1&0\end{bmatrix}A \Rightarrow A = \begin{bmatrix}1&2&1\\1&1&2\\1&2&1\\1&1&2\end{bmatrix} $$

(2)

由于$rank(A)=2$,所以齐次线性方程组$AX=0$只有零解,故线性变换的核子空间$\mathcal{N}(\mathscr{A})$为零空间,维数为0

根据维数定理可知$\mathscr{A}$的值域$\mathcal{R}(\mathscr{A})$的维数为3,又

$$ \begin{aligned} \mathcal{R}(\mathscr{A})&=span\{\mathscr{A}(\alpha_1),\mathscr{A}(\alpha_2),\mathscr{A}(\alpha_3)\}\\ &=span\{\beta_1+\beta_2+\beta_3+\beta_4,2\beta_1+\beta_2+2\beta_3+\beta_4,\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \beta_1+2\beta_2+\beta_3+2\beta_4\}\\ &=span\{[3,2,2,1]^T,[5,3,2,2]^T,[4,3,4,1]^T\} \end{aligned} $$

所以$\mathcal{R}(\mathscr{A})$的基为$[3,2,2,1]^T,[5,3,2,2]^T,[4,3,4,1]^T$


线性映射是单射、满射的充要条件

单射:原像空间中任意两个不同的向量经过线性映射后得到的向量也是不同的

满射:像空间中的任意一个向量,都能在原像空间中找到原向量

  • $\mathscr{A}$是单射$\Leftrightarrow \mathcal{N}(\mathscr{A})={0}$
  • $\mathscr{A}$是满射$\Leftrightarrow \mathcal{R}(\mathscr{A})=V_2$
  • 降维不单,升维不满
Last Modified: June 14, 2021
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2 Comments
  1. Mask Mask

    最后一道例题(1)中的A求错了吧?

    1. mathor mathor

      @Mask用软件验证了下确实求错了,感谢提醒,第一问和第二问我都改了,麻烦帮忙检验下还有什么问题