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高等代数教学辅导

6.6 线性映射的表示矩阵

子节 6.6.1 基础知识回顾

同构交换关系图
6.6.2. 同构交换关系图

练习 6.6.2 练习

基础题.

1.
\(n \geq 1\),定义\(\F_{n-1}[x]\)\(\F_{n}[x]\)的映射\(\varphi\)如下:
\begin{equation*} \varphi(a_{0} + a_{1} x + \cdots + a_{n-1}x^{n-1}) = a_{0} x + \frac{1}{2} a_{1} x^{2} + \frac{1}{3} a_{2} x^{3} + \cdots + \frac{1}{n}a_{n-1}x^{n}, \end{equation*}
其中\(a_{0},a_{1},\ldots,a_{n-1}\in \F\)
  1. 证明:\(\varphi\)是线性映射;
  2. \(\varphi\)\(\F_{n-1}[x]\)的基\((1,x,\ldots,x^{n-1})\)\(\F_{n}[x]\)的基\((1,x,\ldots,x^{n-1},x^{n})\)下的矩阵。
解答.
  1. 对任意\(f,g \in \F_{n-1}[x]\),设\(f(x)=\sum_{i=0}^{n-1}a_{i} x^{i}\)\(g(x)=\sum_{i=0}^{n-1}b_{i} x^{i}\),则
    \begin{align*} \varphi(f+g)\amp = \varphi\left( \sum_{i=0}^{n-1} (a_i+b_i) x^i \right) = \sum_{i=0}^{n-1} \frac{a_i+b_i}{i+1} x^{i+1} \\ \amp = \sum_{i=0}^{n-1} \frac{a_i}{i+1} x^{i+1} + \sum_{i=0}^{n-1} \frac{b_i}{i+1} x^{i+1} = \varphi(f) + \varphi(g). \end{align*}
    对任意\(c \in \F\)
    \begin{align*} \varphi(cf) \amp = \varphi\left( \sum_{i=0}^{n-1}c a_{i} x^{i} \right) \\ \amp = \sum_{i=0}^{n-1}\frac{c a_{i}}{i+1}x^{i+1}\\ \amp = c \sum_{i=0}^{n-1}\frac{a_{i}}{i+1}x^{i+1} \\ \amp = c \varphi(f). \end{align*}
    所以\(\varphi\)是线性映射。
  2. 对每个\(k = 0,1,\ldots,n-1\),有
    \begin{equation*} \varphi(x^{k}) = \frac{1}{k+1}x^{k+1}. \end{equation*}
    所以\(\varphi(x^{k})\)在基\((1, x, x^{2}, \ldots, x^{n-1})\)下的坐标为:第\(k+2\)个分量为\(\frac{1}{k+1}\),其余分量为0,即\((0,\ldots,0, \frac{1}{k+1}, 0, \ldots, 0)^{T}\)。因此,\(\varphi\)的表示矩阵\(A\)是一个\((n+1) \times n\)矩阵,其第\(j\)列(对应基向量\(x^{j-1}\)\(j=1,\ldots,n\))的第\(i\)行(对应基向量\(x^{i-1}\)\(i=1,\ldots,n+1\))为:
    \begin{equation*} a_{ij}= \begin{cases}\frac{1}{j}, & \text{若 } i = j+1, \\ 0, & \text{否则}.\end{cases} \end{equation*}
    \begin{equation*} A = \begin{pmatrix} 0 & 0 & \cdots & 0 \\ 1 & 0 & \cdots & 0 \\ 0 & \frac{1}{2}& \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{1}{n}\end{pmatrix}, \end{equation*}
    其中第一行全为零,第\(i\)行(\(i \geq 2\))只有第\(i-1\)列非零,值为\(\frac{1}{i-1}\)
2.
\(V,U,W\)为线性空间,\((\xi_{1}, \ldots, \xi_{n})\)\(V\)的基,\((\eta_{1}, \ldots, \eta_{m})\)\(U\)的基,\((\zeta_{1}, \ldots, \zeta_{p})\)\(W\)的基,又设线性映射\(\varphi \in \mathcal{L}(V,U)\)\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的表示矩阵为\(A\),线性映射\(\psi \in \mathcal{L}(U,W)\)\((\eta_{1}, \ldots, \eta_{m})\)\((\zeta_{1}, \ldots, \zeta_{p})\)下的表示矩阵为\(B\)。证明:\(\psi \varphi\)在基\((\xi_{1}, \ldots, \xi_{n})\)\((\zeta_{1}, \ldots, \zeta_{p})\)下的表示矩阵为\(BA\)
解答.
由表示矩阵的定义,有
\begin{equation*} \varphi(\xi_{1}, \ldots, \xi_{n}) = (\eta_{1}, \ldots, \eta_{m}) A, \end{equation*}
即对每个\(j\)\(\varphi(\xi_{j}) = \sum_{i=1}^{m} a_{ij}\eta_{i}\)
同理,
\begin{equation*} \psi(\eta_{1}, \ldots, \eta_{m}) = (\zeta_{1}, \ldots, \zeta_{p}) B, \end{equation*}
\(\psi(\eta_{i}) = \sum_{k=1}^{p} b_{ki}\zeta_{k}\)
考虑乘积映射\(\psi \varphi: V \to W\),计算\(\psi\varphi(\xi_{j})\)
\begin{align*} (\psi \varphi)(\xi_j)\amp = \psi(\varphi(\xi_j)) = \psi\left( \sum_{i=1}^m a_{ij} \eta_i \right) = \sum_{i=1}^m a_{ij} \psi(\eta_i) \\ \amp = \sum_{i=1}^m a_{ij} \left( \sum_{k=1}^p b_{ki} \zeta_k \right) = \sum_{k=1}^p \left( \sum_{i=1}^m b_{ki} a_{ij} \right) \zeta_k. \end{align*}
所以\((\psi \varphi)(\xi_{j})\)在基\((\zeta_{1}, \ldots, \zeta_{p})\)下的坐标是\(\left( \sum_{i=1}^{m} b_{1i}a_{ij}, \ldots, \sum_{i=1}^{m} b_{pi}a_{ij}\right)^{\top}\),这正是矩阵\(BA\)的第\(j\)列。因此,\(\psi \varphi\)在基\((\xi_{1}, \ldots, \xi_{n})\)\((\zeta_{1}, \ldots, \zeta_{p})\)下的表示矩阵为\(BA\)
3.
\(V\)\(U\)为线性空间,\((\xi_{1}, \ldots, \xi_{n})\)\(V\)的基,\((\eta_{1}, \ldots, \eta_{m})\)\(U\)的基,\(\varphi\)\(V\)\(U\)的线性映射,且在\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的表示矩阵为\(A\)。若矩阵\(B\)\(A\)相抵,证明:\(B\)\(\varphi\)在另一对\(V\)\(U\)的基下的表示矩阵。
解答.
矩阵\(A\)\(B\)相抵,即存在可逆矩阵\(P \in \F^{m \times m}\)\(Q \in \F^{n \times n}\)使得\(B = P A Q\)
我们构造\(V\)的新基和\(U\)的新基如下: 令\(V\)的新基为\((\xi'_{1}, \ldots, \xi'_{n})\),满足
\begin{equation*} (\xi'_{1}, \ldots, \xi'_{n}) = (\xi_{1}, \ldots, \xi_{n}) Q. \end{equation*}
\(U\)的新基为\((\eta'_{1}, \ldots, \eta'_{m})\),满足
\begin{equation*} (\eta'_{1}, \ldots, \eta'_{m}) = (\eta_{1}, \ldots, \eta_{m}) P^{-1}. \end{equation*}
由于\(P\)\(Q\)可逆,\(P^{-1}\)也可逆,由命题6.2.2\((\xi'_{1}, \ldots, \xi'_{n})\)\((\eta'_{1}, \ldots, \eta'_{m})\)确实是基。
另一方面,
\begin{align*} \varphi(\xi'_{1}, \ldots, \xi'_{n}) \amp = \varphi\left( (\xi_{1}, \ldots, \xi_{n}) Q \right) \\ \amp = \varphi(\xi_{1}, \ldots, \xi_{n}) Q \\ \amp = (\eta_{1}, \ldots, \eta_{m}) A Q. \end{align*}
\((\eta_{1}, \ldots, \eta_{m}) = (\eta'_{1}, \ldots, \eta'_{m}) P\),代入得
\begin{equation*} \varphi(\xi'_{1}, \ldots, \xi'_{n}) = (\eta'_{1}, \ldots, \eta'_{m}) P A Q. \end{equation*}
因此,\(P A Q = B\)\(\varphi\)在新基\((\xi'_{1}, \ldots, \xi'_{n})\)\((\eta'_{1}, \ldots, \eta'_{m})\)下的表示矩阵。
4.
\(V\)是数域\(\F\)上的\(n\)维线性空间,证明:\(V\)上所有线性函数构成的线性空间,即\(\mathcal{L}(V, \F)\),同构于\(\F^{n}\)
解答.
\(V\)的一组基\((\xi_{1}, \ldots, \xi_{n})\)。定义映射\(\Phi: \mathcal{L}(V, \F) \to \F^{n}\)如下:对任意\(f \in \mathcal{L}(V, \F)\),令
\begin{equation*} \Phi(f) = (f(\xi_{1}), f(\xi_{2}), \ldots, f(\xi_{n}))^{\top}. \end{equation*}
首先,\(\Phi\)是线性映射:对任意\(f,g \in \mathcal{L}(V, \F)\)\(c \in \F\),有
\begin{align*} \Phi(f+g) \amp = ((f+g)(\xi_1), \ldots, (f+g)(\xi_n))^\top \\ \amp = (f(\xi_1)+g(\xi_1), \ldots, f(\xi_n)+g(\xi_n))^\top\\ \amp= (f(\xi_1), \ldots, f(\xi_n))^\top + (g(\xi_1), \ldots, g(\xi_n))^\top \\ \amp = \Phi(f) + \Phi(g),\\ \Phi(cf) \amp = ((cf)(\xi_1), \ldots, (cf)(\xi_n))^\top \\ \amp = (c f(\xi_1), \ldots, c f(\xi_n))^\top\\ \amp= c \Phi(f). \end{align*}
其次,\(\Phi\)是单射:若\(\Phi(f)=0\),则\(f(\xi_{i})=0\)对所有\(i\)成立。由于\((\xi_{1}, \ldots, \xi_{n})\)是基,对任意\(\alpha \in V\)\(\alpha = \sum_{i=1}^{n} a_{i} \xi_{i}\),则\(f(\alpha) = \sum a_{i} f(\xi_{i})=0\),所以\(f=0\)。根据命题6.3.5\(\Phi\)是单射。
最后,\(\Phi\)是满射:对任意\((c_{1}, \ldots, c_{n})^{\top} \in \F^{n}\),定义线性函数\(f: V \to \F\)使得\(f(\xi_{i})=c_{i}\),并由线性扩张到整个\(V\)(命题6.6.1)。则\(\Phi(f) = (c_{1}, \ldots, c_{n})^{\top}\)
因此,\(\Phi\)是线性双射,即\(\mathcal{L}(V, \F) \cong \F^{n}\)

提高题.

5.
\(V\)\(n\)维线性空间,\(U\)\(m\)维线性空间,\(\varphi\)是从\(V\)\(U\)的线性映射,证明:存在\(V\)的一个基\((\xi_{1}, \ldots, \xi_{n})\)\(U\)的一个基\((\eta_{1}, \ldots, \eta_{m})\)使得\(\varphi\)在这两个基下的表示矩阵形如
\begin{equation*} \begin{pmatrix}E_{r}&0 \\ 0&0\end{pmatrix}_{m \times n}. \end{equation*}
解答.
\(V\)的一组基\((\varepsilon_{1}, \ldots, \varepsilon_{n})\)。考虑\(U\)中向量组\(\varphi(\varepsilon_{1}), \ldots, \varphi(\varepsilon_{n})\),设其极大线性无关组包含\(r\)个向量,不妨设为\(\varphi(\varepsilon_{1}), \ldots, \varphi(\varepsilon_{r})\)(若不然,可通过重排基向量的顺序实现)。则对任意\(j = r+1, \ldots, n\),存在标量\(a_{1j}, \ldots, a_{rj}\in \F\)使得
\begin{equation*} \varphi(\varepsilon_{j}) = \sum_{i=1}^{r} a_{ij}\varphi(\varepsilon_{i}). \end{equation*}
定义\(V\)的新向量组\((\xi_{1}, \ldots, \xi_{n})\)如下:
\begin{equation*} \xi_{i} = \varepsilon_{i} \quad (i=1,\ldots,r), \qquad \xi_{j} = \varepsilon_{j} - \sum_{i=1}^{r} a_{ij}\varepsilon_{i} \quad (j=r+1,\ldots,n). \end{equation*}
\((\varepsilon_{1}, \ldots, \varepsilon_{n})\)\((\xi_{1}, \ldots, \xi_{n})\)的变换矩阵为
\begin{equation*} P = \begin{pmatrix}E_{r}&-A \\ 0&E_{n-r}\end{pmatrix}, \end{equation*}
其中\(A = (a_{ij})_{r \times (n-r)}\)\(E_{r}\)\(E_{n-r}\)分别为\(r\)阶和\(n-r\)阶单位矩阵。因为\(\det P = 1 \neq 0\),所以\(P\)可逆,从而\((\xi_{1}, \ldots, \xi_{n})\)也是\(V\)的一组基。
由于\(\varphi(\varepsilon_{1}), \ldots, \varphi(\varepsilon_{r})\)线性无关,可将它们扩充为\(U\)的一组基。令
\begin{equation*} \eta_{i} = \varphi(\varepsilon_{i}) \quad (i=1,\ldots,r), \end{equation*}
再选取\(\eta_{r+1}, \ldots, \eta_{m} \in U\)使得\((\eta_{1}, \ldots, \eta_{m})\)构成\(U\)的一组基。
现在计算\(\varphi\)在基\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的表示矩阵。
  • \(i=1,\ldots,r\),有\(\varphi(\xi_{i}) = \varphi(\varepsilon_{i}) = \eta_{i}\),故其坐标为第\(i\)个分量为\(1\)、其余分量为\(0\)的列向量。
  • \(j=r+1,\ldots,n\),有
    \begin{align*} \varphi(\xi_{j}) \amp = \varphi\!\left(\varepsilon_{j} - \sum_{i=1}^{r} a_{ij}\varepsilon_{i}\right)\\ \amp= \varphi(\varepsilon_{j}) - \sum_{i=1}^{r} a_{ij}\varphi(\varepsilon_{i}) \\ \amp = \sum_{i=1}^{r} a_{ij}\eta_{i} - \sum_{i=1}^{r} a_{ij}\eta_{i} = 0, \end{align*}
    故其坐标为零向量。
因此,\(\varphi\)的表示矩阵为
\begin{equation*} \begin{pmatrix}E_{r}&0 \\ 0&0\end{pmatrix}_{m \times n}, \end{equation*}
其中\(E_{r}\)\(r\)阶单位矩阵,左上角块为\(r \times r\),其余块为零矩阵。
6.
\((\xi_{1}, \ldots, \xi_{n})\)是线性空间\(V\)的一个基,\(\eta_{1}, \ldots, \eta_{n}\)是有限维线性空间\(U\)中的\(n\)个向量,命题6.6.1表明存在唯一的从\(V\)\(U\)的线性映射使得\(\varphi(\xi_{i}) = \eta_{i}, \forall i \in [n]\)。证明:
  1. \(\varphi\)是单射当且仅当\(\eta_{1}, \ldots, \eta_{n}\)线性无关;
  2. \(\varphi\)是满射当且仅当\(U = \langle \eta_{1}, \ldots, \eta_{n} \rangle\),即向量\(\eta_{1}, \ldots, \eta_{n}\)可以生成线性空间\(U\)
解答.
  1. 必要性:若\(\varphi\)是单射,设\(\sum_{i=1}^{n} c_{i} \eta_{i} = 0\),则\(\varphi\left( \sum_{i=1}^{n} c_{i} \xi_{i} \right) = \sum_{i=1}^{n} c_{i} \varphi(\xi_{i}) = \sum_{i=1}^{n} c_{i} \eta_{i} = 0\)。因为\(\varphi\)单射,所以\(\sum_{i=1}^{n} c_{i} \xi_{i} = 0\)。而\(\xi_{1}, \ldots, \xi_{n}\)是基,故线性无关,于是所有\(c_{i} = 0\)。因此\(\eta_{1}, \ldots, \eta_{n}\)线性无关。
    充分性:若\(\eta_{1}, \ldots, \eta_{n}\)线性无关,设\(\varphi(\alpha) = 0\),将\(\alpha\)表示为\(\alpha = \sum_{i=1}^{n} c_{i} \xi_{i}\)。则\(0 = \varphi(\alpha) = \sum_{i=1}^{n} c_{i} \eta_{i}\),由线性无关性得所有\(c_{i} = 0\),故\(\alpha = 0\)。由命题6.3.5\(\varphi\)是单射。
  2. 必要性:若\(\varphi\)是满射,则对任意\(\beta \in U\),存在\(\alpha \in V\)使得\(\varphi(\alpha) = \beta\)。设\(\alpha = \sum_{i=1}^{n} c_{i} \xi_{i}\),则\(\beta = \varphi(\alpha) = \sum_{i=1}^{n} c_{i} \eta_{i}\),所以\(\beta \in \langle \eta_{1}, \ldots, \eta_{n} \rangle\)。故\(U \subseteq \langle \eta_{1}, \ldots, \eta_{n} \rangle\),显然\(\langle \eta_{1}, \ldots, \eta_{n} \rangle \subseteq U\),所以\(U = \langle \eta_{1}, \ldots, \eta_{n} \rangle\)
    充分性:若\(U = \langle \eta_{1}, \ldots, \eta_{n} \rangle\),则对任意\(\beta \in U\),存在\(c_{1}, \ldots, c_{n}\)使得\(\beta = \sum_{i=1}^{n} c_{i} \eta_{i} = \varphi( \sum_{i=1}^{n} c_{i} \xi_{i} )\),故存在\(\beta\)\(\varphi\)下的原像,因此\(\varphi\)是满射。
7.
\((\xi_{1}, \ldots, \xi_{n})\)是线性空间\(V\)的一个基,\((\eta_{1}, \ldots, \eta_{m})\)是线性空间\(U\)的一个基,\(\varphi\)是从\(V\)\(U\)的线性映射且\(\varphi\)\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的表示矩阵为\(A \in \F^{m \times n}\),证明:
  1. \(\varphi\)是单射当且仅当\(A\)是列满秩矩阵;
  2. \(\varphi\)是满射当且仅当\(A\)是行满秩矩阵。
解答.
  1. \(A = (a_{ij})_{m \times n}\),即\(\varphi(\xi_{j}) = \sum_{i=1}^{m} a_{ij}\eta_{i}\)\(j=1,\ldots,n\)。定义坐标映射\(\sigma_{V}: V \to \F^{n}\)\(\sigma_{V}(\alpha) = (x_{1}, \ldots, x_{n})^{\top}\),其中\(\alpha = \sum_{j=1}^{n} x_{j} \xi_{j}\);类似定义\(\sigma_{U}: U \to \F^{m}\)。则线性映射\(\varphi\)对应于矩阵乘法\(L_{A}: \F^{n} \to \F^{m}\)\(L_{A}(x) = Ax\)。由同构交换关系(见图6.6.2),即命题6.6.3,有\(\varphi = \sigma_{U}^{-1}L_{A} \sigma_{V}\)
    因为\(\sigma_{V}\)\(\sigma_{U}\)是同构映射,所以\(\varphi\)是单射当且仅当\(L_{A}\)是单射。而\(L_{A}\)是单射当且仅当\(AX=0\)只有零解,即\(A\)的列向量线性无关,亦即\(A\)的列秩为\(n\)(列满秩)。
  2. 同上,\(\varphi\)是满射当且仅当\(L_{A}\)是满射。\(L_{A}\)是满射当且仅当\(A\)的列向量张成\(\F^{m}\),即\(A\)的列空间等于\(\F^{m}\),这等价于\(A\)的行秩为\(m\)(因为行秩等于列秩),即\(A\)是行满秩矩阵。