定理 6.7.1. 线性映射基本定理.
设\(V\)为有限维线性空间，\(\phi\)是从\(V\)到线性空间\(U\)的线性映射，则\(\Ima \phi\)也是有限维空间，且满足
\begin{equation*}
\dim \Ima \phi + \dim \Ker \phi = \dim V.
\end{equation*}
节 6.7 线性映射的像与核
子节 6.7.1 基础知识回顾
命题 6.7.2.
设\(\phi \in \mathcal{L}(V,U)\)为从数域\(\F\)上的线性空间\(V\)到\(U\)的线性映射，则\(\phi\)是单射的充分必要条件是\(\Ker \phi = 0\)。
练习 6.7.2 练习
基础题.
1.
对线性映射\(\phi: \F^{n \times n}\to \F, A \mapsto {\rm tr } A\)，求\(\Ker \phi\)和\(\Ima \phi\)，并求它们的一个基和维数。
解答.
-
\(\Ima \phi = \F\)，因为对任意\(c \in \F\)，取矩阵\(A = c E_{11}\)，则\({\rm tr }A = c\)，所以\(\phi\)是满射。从而\(\dim \Ima \phi = 1\)，一个基为\((1)\)（将\(\F\)视为\(\F\)上的线性空间）。
-
\(\Ker \phi = \{ A \in \F^{n \times n}\mid {\rm tr } A = 0 \}\)。由维数公式：\begin{equation*} \dim \Ker \phi = \dim \F^{n \times n}- \dim \Ima \phi = n^{2} - 1. \end{equation*}取\(\F^{n \times n}\)的标准基\(\{ E_{ij}\mid 1 \leq i,j \leq n \}\)，其中\(E_{ij}\)是\((i,j)\)位置为1，其余为0的矩阵。则\(\Ker \phi\)的一组基为：\begin{equation*} \{ E_{ij}\mid i \neq j \} \cup \{ E_{11}- E_{ii}\mid i = 2,3,\ldots,n \}. \end{equation*}这些矩阵共\(n(n-1) + (n-1) = n^{2} - 1\)个，且线性无关，它们都满足迹为零。
2.
\begin{equation*}
\phi(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\(a_{0},a_{1},\ldots,a_{n-1}\in \F\)。 求\(\Ker \phi\)和\(\Ima \phi\)以及它们的维数。
解答.
-
\begin{equation*} a_{0} x + \frac{1}{2} a_{1} x^{2} + \cdots + \frac{1}{n}a_{n-1}x^{n} = 0. \end{equation*}比较系数得\(a_{i} = 0\)对所有\(i=0,\ldots,n-1\)成立，所以\(f=0\)。因此\(\Ker \phi = \{0\}\)，\(\dim \Ker \phi = 0\)。
-
\(\Ima \phi\)：由定义，\(\Ima \phi\)由所有形如\(a_{0} x + \frac{1}{2} a_{1} x^{2} + \cdots + \frac{1}{n}a_{n-1}x^{n}\)的多项式组成，其中\(a_{i} \in \F\)。因为系数\(a_{i}\)可任意选取，所以\(\Ima \phi\)是\(\F_{n}[x]\)中由\(x, x^{2}, \ldots, x^{n}\)张成的子空间，即所有常数项为零的多项式构成的子空间。故\begin{equation*} \Ima \phi = \{ p(x) \in \F_{n}[x] \mid p(0)=0 \}. \end{equation*}
3.
设\(V,U,W\)是有限维线性空间，\(\phi \in \mathcal{L}(V,U), \psi \in \mathcal{L}(U,W)\)，证明：
\begin{equation*}
\dim \Ker \psi \phi \leq \dim \Ker \phi + \dim \Ker \psi.
\end{equation*}
解答.
考虑\(\phi\)在子空间\(K = \Ker \psi \phi\)上的限制\(\phi|_{K}: K \to U\)。对任意\(\alpha \in K\)，有\(\psi(\phi(\alpha)) = 0\)，故\(\phi(\alpha) \in \Ker \psi\)，即\(\Ima \phi|_{K} \subseteq \Ker \psi\)。因此
\begin{equation}
\dim \phi(K) \leq \dim \Ker \psi.\tag{6.7.1}
\end{equation}
另一方面，将线性映射基本定理应用于\(\phi|_{K}\)，有
\begin{equation*}
\dim K = \dim \Ker(\phi|_{K}) + \dim \Ima (\phi|_{K}).
\end{equation*}
注意到\(\Ker(\phi|_{K}) = \{ \alpha \in K \mid \phi(\alpha)=0 \} = K \cap \Ker \phi \subseteq \Ker \phi\)，所以
\begin{equation}
\dim \Ker(\phi|_{K}) \leq \dim \Ker \phi.\tag{6.7.2}
\end{equation}
\begin{equation*}
\dim K \leq \dim \Ker \phi + \dim \Ker \psi,
\end{equation*}
即\(\dim \Ker \psi \phi \leq \dim \Ker \phi + \dim \Ker \psi\)。
4.
设\(V,U\)是有限维线性空间，\(\phi \in \mathcal{L}(V,U)\)，证明：存在\(V\)的子空间\(V'\)使得\(V' \cap \Ker \phi = 0\)且\(\Ima \phi = \{ \phi(\alpha) \mid \alpha \in V'\}\)。
解答.
取\(\Ker \phi\)的一个基\((\alpha_{1}, \ldots, \alpha_{k})\)，将其扩充为\(V\)的一个基\((\alpha_{1}, \ldots, \alpha_{k}, \beta_{1}, \ldots, \beta_{r})\)。令\(V' = \langle \beta_{1}, \ldots, \beta_{r} \rangle\)，即由\(\beta_{1}, \ldots, \beta_{r}\)张成的子空间。则显然\(V' \cap \Ker \phi = 0\)（因为\(\alpha_{1}, \ldots, \alpha_{k}, \beta_{1}, \ldots, \beta_{r}\)线性无关）。
对任意\(\alpha \in V\)，可唯一表示为\(\alpha = \sum_{i=1}^{k} a_{i} \alpha_{i} + \sum_{j=1}^{r} b_{j} \beta_{j}\)，则
\begin{equation*}
\phi(\alpha) = \sum_{i=1}^{k} a_{i} \phi(\alpha_{i}) + \sum_{j=1}^{r} b_{j} \phi(\beta_{j}) = \sum_{j=1}^{r} b_{j} \phi(\beta_{j}) \in \phi(V').
\end{equation*}
所以\(\Ima \phi \subseteq \phi(V')\)。反之，显然\(\phi(V') \subseteq \Ima \phi\)。故\(\Ima \phi = \phi(V')\)。
提高题.
5.
设\(V\)是数域\(\F\)上的有限维线性空间，\(\phi \in \mathcal{L}(V, \F)\)，证明：若存在\(\alpha \in V\backslash \Ker \phi\)，则
\begin{equation*}
V = \Ker \phi \oplus \langle \alpha \rangle.
\end{equation*}
解答.
由于\(\phi \neq 0\)（因为存在\(\alpha \notin \Ker \phi\)），所以\(\phi\)是满射（因为\(\Ima \phi\)是\(\F\)的子空间，且包含非零元，故为整个\(\F\)）。由维数公式，
\begin{equation*}
\dim V = \dim \Ker \phi + \dim \Ima \phi = \dim \Ker \phi + 1.
\end{equation*}
设\(k = \dim \Ker \phi\)，取\(\Ker \phi\)的一个基\((\alpha_{1}, \ldots, \alpha_{k})\)。由于\(\alpha \notin \Ker \phi\)，我们断言\(\alpha_{1}, \ldots, \alpha_{k}, \alpha\)线性无关。事实上，若有线性组合
\begin{equation*}
c_{1} \alpha_{1} + \cdots + c_{k} \alpha_{k} + c \alpha = 0,
\end{equation*}
两边用\(\phi\)作用得
\begin{equation*}
c \phi(\alpha) = 0.
\end{equation*}
因为\(\phi(\alpha) \neq 0\)，所以\(c=0\)，于是\(c_{1} \alpha_{1} + \cdots + c_{k} \alpha_{k} = 0\)，由基的线性无关性得\(c_{1} = \cdots = c_{k} = 0\)。因此\(\alpha_{1}, \ldots, \alpha_{k}, \alpha\)线性无关，且个数为\(k+1 = \dim V\)，故构成\(V\)的一组基。于是\(V = \langle \alpha_{1}, \ldots, \alpha_{k}, \alpha \rangle = \Ker \phi + \langle \alpha \rangle\)。又因为\(\Ker \phi \cap \langle \alpha \rangle = \{0\}\)（若有\(c\alpha \in \Ker \phi\)，则\(\phi(c\alpha)=c\phi(\alpha)=0\)，得\(c=0\)），所以和为直和，即\(V = \Ker \phi \oplus \langle \alpha \rangle\)。
6.
设\(V,U\)是数域\(\F\)上的有限维线性空间，\(\phi \in \mathcal{L}(V,U)\)，\((\eta_{1}, \ldots, \eta_{r})\)是\(\Ima \phi\)的一个基。证明：存在\(\psi_{1}, \ldots, \psi_{r} \in \mathcal{L}(V,\F)\)使得
\begin{equation*}
\phi(\alpha) = \psi_{1}(\alpha) \eta_{1} + \cdots + \psi_{r}(\alpha) \eta_{r}, \quad \forall \alpha \in V.
\end{equation*}
解答.
对任意\(\alpha \in V\)，\(\phi(\alpha) \in \Ima \phi\)，故存在唯一的标量\(\psi_{1}(\alpha), \ldots, \psi_{r}(\alpha) \in \F\)使得
\begin{equation*}
\phi(\alpha) = \psi_{1}(\alpha) \eta_{1} + \cdots + \psi_{r}(\alpha) \eta_{r}.
\end{equation*}
这定义了映射\(\psi_{i}: V \to \F\)。下证每个\(\psi_{i}\)是线性的。对任意\(\alpha, \beta \in V\)和\(c \in \F\)，有
\begin{align*}
\phi(\alpha + \beta)\amp = \phi(\alpha) + \phi(\beta)\\
\amp = \sum_{i=1}^{r} \psi_{i}(\alpha) \eta_{i} + \sum_{i=1}^{r} \psi_{i}(\beta) \eta_{i}\\
\amp = \sum_{i=1}^{r} (\psi_{i}(\alpha) + \psi_{i}(\beta)) \eta_{i}, \\
\phi(c\alpha) \amp = c \phi(\alpha) \\
\amp = c \sum_{i=1}^{r} \psi_{i}(\alpha) \eta_{i} \\
\amp = \sum_{i=1}^{r} (c \psi_{i}(\alpha)) \eta_{i}.
\end{align*}
另一方面，由定义，
\begin{equation*}
\phi(\alpha + \beta) = \sum_{i=1}^{r} \psi_{i}(\alpha + \beta) \eta_{i}, \quad \phi(c\alpha) = \sum_{i=1}^{r} \psi_{i}(c\alpha) \eta_{i}.
\end{equation*}
由于\(\eta_{1}, \ldots, \eta_{r}\)线性无关，比较系数得
\begin{equation*}
\psi_{i}(\alpha + \beta) = \psi_{i}(\alpha) + \psi_{i}(\beta), \quad \psi_{i}(c\alpha) = c \psi_{i}(\alpha),
\end{equation*}
所以\(\psi_{i}\)是线性函数，即\(\psi_{i} \in \mathcal{L}(V, \F)\)。
7.
设\(V\)是\(n\)维线性空间，\(U\)是\(m\)维线性空间，\(\phi \in \mathcal{L}(V,U)\)是从\(V\)到\(U\)的线性映射且是单射，证明：存在\(U\)到\(V\)的满射\(\psi\)使得\(\psi \phi ={\rm id}_{V}\)。
解答.
取\(V\)的一个基\((\xi_{1}, \ldots, \xi_{n})\)，因\(\phi\)是单射，由定理6.7.1的证明可知，\((\eta_{1} = \phi(\xi_{1}), \ldots, \eta_{n} = \phi(\xi_{n}))\)构成\(\Ima \phi\)的一个基。将其扩充为\(U\)的一个基\((\eta_{1}, \ldots, \eta_{n}, \eta_{n+1}, \ldots, \eta_{m})\)。定义线性映射\(\psi: U \to V\)为：在基\(\eta_{1}, \ldots, \eta_{m}\)上规定
\begin{equation*}
\psi(\eta_{i}) = \begin{cases}\xi_{i},&i=1,\ldots,n, \\ 0,&i=n+1,\ldots,m,\end{cases}
\end{equation*}
然后线性扩张到整个\(U\)。则对任意\(\alpha \in V\)，设\(\alpha = \sum_{i=1}^{n} a_{i} \xi_{i}\)，有
\begin{equation*}
\phi(\alpha) = \sum_{i=1}^{n} a_{i} \eta_{i}.
\end{equation*}
于是
\begin{equation*}
\psi(\phi(\alpha)) = \psi\left( \sum_{i=1}^{n} a_{i} \eta_{i} \right) = \sum_{i=1}^{n} a_{i} \xi_{i} = \alpha,
\end{equation*}
所以\(\psi \phi = \operatorname{id}_{V}\)。由此可得\(\psi\)是满射：因为对任意\(\beta \in V\)，有\(\beta = \psi(\phi(\beta))\)，即\(\beta \in \Ima \psi\)，故\(\psi\)是满射。
8.
设\(V,U\)是有限维线性空间，\(\phi\)是从\(V\)到\(U\)的线性映射。证明下列三个命题是等价的。
-
\(\phi\)是单射；
解答.
(1) \(\Rightarrow\) (2): 假设\(\phi\)是单射，且\(\phi \psi = \phi \psi'\)。则对任意\(\beta \in U\)，有\(\phi(\psi(\beta)) = \phi(\psi'(\beta))\)，由单射性得\(\psi(\beta) = \psi'(\beta)\)，所以\(\psi = \psi'\)。
(2) \(\Rightarrow\) (1): 假设(2)成立。若\(\phi\)不是单射，由命题6.7.2则\(\Ker \phi \neq 0\)。设\(\alpha \neq 0 \in \Ker \phi\)。取\(U\)的一个基\((\xi_{1}, \ldots, \xi_{k})\)，定义两个线性映射\(\psi, \psi': U \to V\)使得
\begin{equation*}
\psi(\xi_{1}) = 0, \quad \psi'(\xi_{1}) = \alpha, \quad \psi(\xi_{i}) = \psi'(\xi_{i}) = 0, ~~\forall i = 2, \ldots, k.
\end{equation*}
则对任意\(\beta = \sum c_{i} \xi_{i} \in U\)，有\(\phi(\psi(\beta)) = \phi(c_{1} \cdot 0) = 0\)，\(\phi(\psi'(\beta)) = \phi(c_{1} \alpha) = c_{1} \phi(\alpha)=0\)，所以\(\phi \psi = \phi \psi'\)，但\(\psi \neq \psi'\)，与(2)矛盾。故\(\phi\)是单射。
(1) \(\Rightarrow\) (3): 由\(\phi\)是单射，知\(\dim V = \dim \Ima \phi \leq \dim U\)。取\(V\)的一个基\((\xi_{1}, \ldots, \xi_{n})\)，则由命题6.5.4的证明过程知\(\phi(\xi_{1}), \ldots, \phi(\xi_{n})\)线性无关。将其扩充为\(U\)的一个基\((\phi(\xi_{1}), \ldots, \phi(\xi_{n}), \eta_{n+1}, \ldots, \eta_{m})\)。定义\(\psi: U \to V\)为\(\psi(\phi(\xi_{i})) = \xi_{i}, i=1,\ldots, n\)，\(\psi(\eta_{j})=0, j=n+1, \ldots, m\)，则\(\psi\)线性且满足\(\psi \phi ={\rm id}_{V}\)。
(3) \(\Rightarrow\) (1): 假设存在\(\psi\)使得\(\psi \phi = \operatorname{id}_{V}\)。若\(\phi(\alpha)=0\)，则\(\alpha = \psi(\phi(\alpha)) = \psi(0)=0\)，所以\(\phi\)是单射。
9.
设\(V,U\)是有限维线性空间，\(\phi\)是从\(V\)到\(U\)的线性映射。证明下列三个命题是等价的。
-
\(\phi\)是满射；
解答.
(1) \(\Rightarrow\) (2): 假设\(\phi\)是满射，且\(\psi \phi = \psi' \phi\)。对任意\(\beta \in U\)，由满射性，存在\(\alpha \in V\)使得\(\beta = \phi(\alpha)\)。则\(\psi(\beta) = \psi(\phi(\alpha)) = \psi'(\phi(\alpha)) = \psi'(\beta)\)，所以\(\psi = \psi'\)。
(2) \(\Rightarrow\) (1): 假设(2)成立。若\(\phi\)不是满射，则\(\Ima \phi\)是\(U\)的真子空间。设\(\dim U = m\)，\(\dim \Ima \phi = r < m\)。取\(\Ima \phi\)的一个基\((\eta_{1}, \ldots, \eta_{r})\)，扩充为\(U\)的基\((\eta_{1}, \ldots, \eta_{r}, \eta_{r+1}, \ldots, \eta_{m})\)。定义两个线性映射\(\psi, \psi': U \to V\)满足：
\begin{equation*}
\psi(\eta_{r+1}) = 0, \quad \psi'(\eta_{r+1}) = \beta \neq 0 \in V, \quad \psi(\eta_{i}) = \psi'(\eta_{i}) = 0, ~~\forall i \neq r+1.
\end{equation*}
则对任意\(\alpha \in V\)，\(\phi(\alpha)\)可表示为\(\eta_{1}, \ldots, \eta_{r}\)的线性组合，故\(\psi(\phi(\alpha))=0\)，\(\psi'(\phi(\alpha))=0\)，所以\(\psi \phi = \psi' \phi\)，但\(\psi \neq \psi'\)，与(2)矛盾。故\(\phi\)是满射。
(1) \(\Rightarrow\) (3): 由\(\phi\)是满射，知\(\dim U = \dim \Ima \phi \leq \dim V\)。取\(U\)的一个基\((\eta_{1}, \ldots, \eta_{m})\)，对每个\(\eta_{i}\)，选取\(\xi_{i} \in V\)使得\(\phi(\xi_{i}) = \eta_{i}\)。定义\(\psi: U \to V\)为\(\psi(\eta_{i}) = \xi_{i}\)，则\(\phi \psi = \operatorname{id}_{U}\)。
(3) \(\Rightarrow\) (1): 假设存在\(\psi\)使得\(\phi \psi = \operatorname{id}_{U}\)。对任意\(\beta \in U\)，有\(\beta = \phi(\psi(\beta)) \in \Ima \phi\)，所以\(\phi\)是满射。
10.
设\(\phi\)是\(n\)维线性空间\(V\)上的线性变换，\(i\)是任意正整数，证明：
\begin{equation*}
\dim (\Ima \phi^{i-1}\cap \Ker \phi) = \dim \Ker \phi^{i} - \dim \Ker \phi^{i-1}.
\end{equation*}
解答.
考虑映射\(\phi\)在子空间\(\Ima \phi^{i-1}\)上的限制，记为\(\phi|_{\Ima \phi^{i-1}}: \Ima \phi^{i-1}\to V\)。显然，\(\Ker (\phi|_{\Ima \phi^{i-1}}) = \Ima \phi^{i-1}\cap \Ker \phi\)。而\(\Ima (\phi|_{\Ima \phi^{i-1}}) = \phi(\Ima \phi^{i-1}) = \Ima \phi^{i}\)。由线性映射基本定理，
\begin{equation*}
\dim \Ima \phi^{i-1}= \dim \Ker (\phi|_{\Ima \phi^{i-1}}) + \dim \Ima (\phi|_{\Ima \phi^{i-1}}),
\end{equation*}
即
\begin{equation*}
\dim \Ima \phi^{i-1}= \dim (\Ima \phi^{i-1}\cap \Ker \phi) + \dim \Ima \phi^{i}.
\end{equation*}
移项得
\begin{equation*}
\dim (\Ima \phi^{i-1}\cap \Ker \phi) = \dim \Ima \phi^{i-1}- \dim \Ima \phi^{i}.
\end{equation*}
对线性变换\(\phi^{j}\)应用维数公式：\(\dim V = \dim \Ker \phi^{j} + \dim \Ima \phi^{j}\)，所以\(\dim \Ima \phi^{j} = \dim V - \dim \Ker \phi^{j}\)。代入上式得
\begin{align*}
\dim (\Ima \phi^{i-1}\cap \Ker \phi)\amp = (\dim V - \dim \Ker \phi^{i-1}) - (\dim V - \dim \Ker \phi^{i}) \\
\amp = \dim \Ker \phi^{i} - \dim \Ker \phi^{i-1}.
\end{align*}
证毕。
11.
从线性映射的观点证明Sylvester不等式：
\begin{equation*}
r(AB) \geq r(A) + r(B) - n,
\end{equation*}
其中，\(A \in \F^{m \times n}, B \in \F^{n \times p}\)。
解答.
设\(A\)对应于线性映射\(\phi_{A}: \F^{n} \to \F^{m}\)，\(B\)对应于线性映射\(\phi_{B}: \F^{p} \to \F^{n}\)，则\(AB\)对应于\(\phi_{A} \phi_{B}: \F^{p} \to \F^{m}\)。我们需要证明
\begin{equation*}
r(AB) \geq r(A) + r(B) - n.
\end{equation*}
即
\begin{equation*}
\dim \Ima (\phi_{A} \phi_{B}) \geq \dim \Ima \phi_{A} + \dim \Ima \phi_{B} - n.
\end{equation*}
由于\(\Ima (\phi_{A} \phi_{B}) = \phi_{A}(\Ima \phi_{B})\)，且\(\Ima \phi_{B} \subseteq \F^{n}\)，考虑\(\phi_{A}\)在子空间\(\Ima \phi_{B}\)上的限制\(\phi_{A}|_{\Ima \phi_B}: \Ima \phi_{B} \to \F^{m}\)。则
\begin{equation*}
\Ima (\phi_{A}|_{\Ima \phi_B}) = \phi_{A}(\Ima \phi_{B}) = \Ima (\phi_{A} \phi_{B}).
\end{equation*}
由线性映射基本定理，
\begin{equation*}
\dim \Ima \phi_{B} = \dim \Ker (\phi_{A}|_{\Ima \phi_B}) + \dim \Ima (\phi_{A}|_{\Ima \phi_B}).
\end{equation*}
而\(\Ker (\phi_{A}|_{\Ima \phi_B}) = \Ima \phi_{B} \cap \Ker \phi_{A}\)，所以
\begin{equation*}
\dim \Ima (\phi_{A} \phi_{B}) = \dim \Ima \phi_{B} - \dim (\Ima \phi_{B} \cap \Ker \phi_{A}).
\end{equation*}
又因为\(\Ima \phi_{B} \cap \Ker \phi_{A} \subseteq \Ker \phi_{A}\)，故
\begin{equation*}
\dim (\Ima \phi_{B} \cap \Ker \phi_{A}) \leq \dim \Ker \phi_{A} = n - \dim \Ima \phi_{A}.
\end{equation*}
代入得
\begin{align*}
\dim \Ima (\phi_{A} \phi_{B})\amp \geq \dim \Ima \phi_{B} - (n - \dim \Ima \phi_{A}) \\
\amp = \dim \Ima \phi_{A} + \dim \Ima \phi_{B} - n.
\end{align*}
即\(r(AB) \geq r(A) + r(B) - n\)。
挑战题.
12.
从线性映射的观点证明：
\begin{equation*}
r(ABC) \geq r(AB) + r(BC) - r(B),
\end{equation*}
其中，矩阵\(A,B,C\)的维数使\(ABC\)相乘合法。
