主要内容

高等代数 多项式与线性代数

6.7 线性映射的像与核

在本节中,我们将讨论与一个线性映射紧密关联的两个线性子空间:像空间与核空间。

子节 6.7.1 像空间与核空间

我们首先讨论线性映射的像相关的概念。

定义 6.7.1.

\(V\)\(U\)为两个线性空间,\(\varphi\)\(V\)\(U\)的线性映射。 对于\(V\)中向量\(\alpha\),我们称\(\varphi(\alpha)\)\(\alpha\)\(\varphi\)下的像。 对于\(U\)中向量\(\beta\),我们称像是\(\beta\)\(V\)中所有向量的集合为\(\beta\)\(\varphi\)下的原像,记为\({ \varphi^{-1}(\beta)} := \{ \alpha \in V \mid \varphi(\alpha) = \beta\}\)

备注 6.7.2.

附录中针对一般映射的相关定义B.2.26与这里的定义是一致的。

定义 6.7.3.

\(V\)\(U\)为两个线性空间,\(\varphi\)\(V\)\(U\)的线性映射。 称\(V\)中所有向量在\(\varphi\)下的像的集合为\(\varphi\)的像,记作
\begin{equation*} \Ima \varphi := \varphi(V) = \{ \varphi(\alpha) \mid \alpha \in V \} \subseteq U. \end{equation*}
\(U\)中零向量\(0_{U}\)\(\varphi\)下的原像集合为线性映射\(\varphi\),记作
\begin{equation*} \Ker \varphi := \varphi^{-1}(0_{U}) = \{ \alpha \in V \mid \varphi(\alpha) = 0_{U}\}. \end{equation*}
线性映射的像与核不仅仅是两个集合,它们还具有更多的代数性质。特别地,这两个集合结合相应线性空间\(U\)\(V\)的加法和数乘构成子空间(见定义4.4.1)。

证明.

要证明\(\Ima \varphi\)\(U\)的子空间,只需要验证集合\(\Ima \varphi\)关于\(U\)上定义的加法和数乘封闭。
\(\beta_{1}, \beta_{2} \in \Ima \varphi\),则存在\(\alpha_{1}, \alpha_{2} \in V\)使得\(\varphi(\alpha_{1}) = \beta_{1}, \varphi(\alpha_{2}) = \beta_{2}\)。由于\(\varphi\)保持线性运算,我们有
\begin{equation*} \beta_{1} + \beta_{2} = \varphi(\alpha_{1}) + \varphi(\alpha_{2}) = \varphi(\alpha_{1} + \alpha_{2}) \in \Ima \varphi. \end{equation*}
所以,\(\Ima \varphi\)关于\(U\)上的加法封闭。
\(\beta \in \Ima \varphi\),则存在\(\alpha \in V\)使得\(\varphi(\alpha) = \beta\)。对于任意的\(c \in \F\),有\(c \beta = c \varphi(\alpha) = \varphi(c \alpha) \in \Ima \varphi\),即\(\Ima \varphi\)关于\(U\)上的数乘封闭。
接下来,为了证明\(\Ker \varphi\)\(V\)的子空间,我们验证\(\Ker \varphi\)关于\(V\)上的加法和数乘封闭。
\(\alpha_{1}, \alpha_{2} \in \Ker \varphi\),则\(\varphi(\alpha_{1}) = \varphi(\alpha_{2}) = 0_{U}\)。由于\(\varphi\)保持线性运算,我们有
\begin{equation*} \varphi(\alpha_{1} + \alpha_{2}) = 0_{U}. \end{equation*}
由线性映射核的定义,\(\alpha_{1} + \alpha_{2} \in \Ker \varphi\)。所以,\(\Ker \varphi\)关于\(V\)上的加法封闭。
\(\alpha \in \Ker \varphi\),则\(\varphi(\alpha) = 0_{U}\)。对于任意的\(c \in \F\),有\(\varphi(c \alpha) = c \varphi(\alpha) = 0_{U}\),即\(c \alpha \in \Ker \varphi\)。因此,\(\Ker \varphi\)关于\(V\)上的数乘封闭。
既然\(\Ima \varphi\)\(\Ker \varphi\)是子空间,我们可以讨论它的维数。我们称\(\dim (\Ima \varphi)\)为线性映射\(\varphi\),称\(\dim (\Ker \varphi)\)为线性映射\(\varphi\)零度

备注 6.7.5.

稍后在第6.6节中我们将看到,线性映射秩的概念与矩阵秩的概念密切相关。
下面是一些线性映射像空间与核空间的例子。

6.7.6.

考虑\(V\)\(U\)的零映射\(\varphi: \alpha \mapsto 0_{U}\)(见例子6.3.3),则\(\Ima \varphi = \{0_{U}\}, \Ker \varphi = V\)。零映射的秩\(\dim (\Ima \varphi) = 0\),零度\(\dim (\Ker \varphi) = \dim V\)

6.7.7.

考虑\(n\)维线性空间\(V\)上的恒等映射\({\rm id}_{V}\),则\(\Ima{\rm id}_{V} = V, \Ker{\rm id}_{V} = 0\)。恒等映射的秩\(\dim{\rm id}_{V} = n\),零度\(\dim \Ker{\rm id}_{V} = 0\)

6.7.8.

考虑映射\(\varphi_{A}: \F^{n} \to \F^{m}, \alpha \mapsto A \alpha\),其中\(A \in \F^{m \times n}\)(见例子6.3.10)。不难看出,像空间\(\Ima \varphi_{A}\)就是矩阵\(A\)的列向量空间\(\Ima A\),而核空间\(\Ker \varphi_{A}\)就是齐次线性方程组\(AX = 0\)的解空间\(\Ker A\)。此时,\(\varphi_{A}\)的秩等于矩阵\(A\)的秩,\(\varphi_{A}\)的零度等于\(n - r(A)\)
我们最后举一个无穷维空间的例子。

6.7.9.

考虑求导映射\(D \in \mathcal{L}(D[a,b], D[a,b])\)(见例子6.3.7)。对于任意\(f \in D[a,b]\)\(D f = f' \in D[a,b]\),所以\(\Ima D \subseteq D[a,b]\)。同时有界闭区间\([a,b]\)上的可微函数必然可积,所以对于任意可微函数\(f \in D[a,b]\),可以找到\(g \in D[a,b]\)使得\(D g = g' = f\)。因此,\(\Ima D = D[a,b]\)\(D\)的秩为无穷。
另一方面,满足\(D f = f' = 0\)的函数\(f\)\([a,b]\)上的常值函数,因此\(\Ker D = \{ f \in D[a,b] \mid f(x) = C, \forall x \in [a,b]\}\)\(D\)的零度为\(\dim \Ker D = 1\)
本小节的最后,我们看看核空间的一个应用:判断线性映射是否为单射。

证明.

必要性:假设\(\varphi\)是单射。若存在\(\alpha \neq 0_{V} \in \Ker \varphi\),则有\(\varphi(\alpha) = \varphi(0) = 0\)\(\alpha \neq 0\),与\(\varphi\)是单射矛盾。
充分性:设\(\Ker \varphi = 0\)。对于任意\(\alpha_{1}, \alpha_{2} \in V\)满足\(\varphi(\alpha_{1}) = \varphi(\alpha_{2})\),由\(\varphi\)保持线性运算有\(\varphi(\alpha_{1} - \alpha_{2}) = 0_{U}\),即\(\alpha_{1} - \alpha_{2} \in \Ker \varphi\)。又因为\(\Ker \varphi\)仅包含零向量,所以\(\alpha_{1} = \alpha_{2}\)。因此,\(\varphi\)是单射。

子节 6.7.2 线性映射基本定理

从上述有限维线性空间上线性映射的像空间与核空间的例子中,我们看出线性空间\(V\)\(U\)的线性映射\(\varphi\)的秩与零度满足关系
\begin{equation*} \dim \Ima \varphi + \dim \Ker \varphi = \dim V. \end{equation*}
上述关系具有一般性,是线性映射的重要基本性质。

备注 6.7.12.

虽然线性映射基本定理表明\(\dim \Ima \varphi + \dim \Ker \varphi\)等于全空间\(V\)的维数。但是我们应该注意到\(\Ima \varphi + \Ker \varphi = V\)并不成立。首先\(\Ima \varphi\)\(U\)的子空间而不是\(V\)的子空间,因此像空间与核空间未必能相加。即是对于线性变换\(\varphi\),在像空间与核空间可以相加的情况下,\(\Ima \varphi + \Ker \varphi = V\)依然未必成立。思考题:举出具体的反例。
下面我们首先使用第4.4.3节中介绍过的扩基法证明线性映射基本定理。

证法一.

\(n = \dim V, r = \dim \Ker \varphi\)\((\xi_{1}, \ldots, \xi_{r})\)为核空间\(\Ker \varphi\)的一个基。依据扩基定理4.4.17,可通过添加向量\(\xi_{r+1}, \ldots, \xi_{n} \in V\)构成\(V\)的一组基\((\xi_{1}, \ldots, \xi_{r}, \xi_{r+1}, \ldots, \xi_{n})\)
要完成线性映射基本定理的证明,我们只需要证明\((\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n}))\)\(\Ima \varphi\)的一个基,即\(\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n})\)线性无关且可以线性表出\(\Ima \varphi\)中的所有向量(见定义4.4.11)。
我们首先说明\(\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n}) \in U\)是线性无关向量组。设
\begin{equation*} c_{r+1}\varphi(\xi_{r+1}) + \cdots + c_{n} \varphi(\xi_{n}) = 0, \end{equation*}
则由\(\varphi\)保持线性运算可得
\begin{equation*} \varphi(c_{r+1}\xi_{r+1}+ \cdots + c_{n} \xi_{n}) = 0, \end{equation*}
\(c_{r+1}\xi_{r+1}+ \cdots + c_{n} \xi_{n} \in \Ker \varphi\)。由于\((\xi_{1}, \ldots, \xi_{r})\)\(\Ker \varphi\)的基,所以存在\(c_{1}, \ldots, c_{r} \in \F\)使得
\begin{equation*} c_{1} \xi_{1} + \cdots + c_{r} \xi_{r} = c_{r+1}\xi_{r+1}+ \cdots + c_{n} \xi_{n}. \end{equation*}
又因为\(\xi_{1}, \ldots, \xi_{n}\)线性无关(是\(V\)的基),我们可知\(c_{r+1}= \cdots = c_{n} = 0 = c_{1} = \cdots = c_{r}\)。因此\(\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n})\)线性无关。
下面证明任意\(\beta \in \Ima \varphi\)可以由\(\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n})\)线性表出。由像空间的定义,存在\(\alpha \in V\)使得\(\beta = \varphi(\alpha)\)。又因为\((\xi_{1}, \ldots, \xi_{n})\)\(V\)的基,\(\alpha\)可由\(\xi_{1}, \ldots, \xi_{n}\)线性表出,即存在\(a_{1}, \ldots, a_{n}\)使得
\begin{equation*} \alpha = a_{1} \xi_{1} + \cdots + a_{r} \xi_{r} + a_{r+1}\xi_{r+1}+ \cdots + a_{n} \xi_{n}. \end{equation*}
等式两边同时作用线性映射\(\varphi\)
\begin{align*} \beta = \varphi(\alpha) \amp = \varphi(a_{1} \xi_{1} + \cdots + a_{r} \xi_{r} + a_{r+1}\xi_{r+1}+ \cdots + a_{n} \xi_{n})\\ \amp = a_{1} \varphi(\xi_{1}) + \cdots + a_{r} \varphi(\xi_{r}) + a_{r+1}\varphi(\xi_{r+1}) + \cdots + a_{n} \varphi(\xi_{n}), \end{align*}
其中最后一个等式是因为\(\varphi\)保持线性运算。注意到\(\xi_{1}, \ldots, \xi_{r} \in \Ker \varphi\),所以我们有
\begin{equation*} \beta = a_{r+1}\varphi(\xi_{r+1}) + \cdots + a_{n} \varphi(\xi_{n}), \end{equation*}
\(\beta\)可由\(\varphi(\xi_{r+1}), \ldots, \varphi(\xi_{n})\)线性表出。
接下来,我们利用线性映射空间与矩阵空间的同构关系给出线性映射基本定理的第二种证明方法。
\((\xi_{1}, \ldots, \xi_{n})\)\(V\)的基,\((\eta_{1}, \ldots, \eta_{m})\)\(U\)的基,且\(\varphi\)\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的矩阵为\(A \in \F^{m \times n}\)。 在第6.6.2节中,我们已经看到\(\varphi\)与线性映射\(\varphi_{A}: \F^{n} \times \F^{m}, x \mapsto Ax\)密切相关。下面,我们进一步证明两者的像空间与核空间同构。
我们先证明一个引理。

证明.

首先验证\(\varphi_{V'}\)是双射。根据构造\(\varphi_{V'}\)是满射,仅需验证其为单射。对于任意\(\alpha_{1}, \alpha_{2} \in V'\),若\(\varphi_{V'}(\alpha_{1}) = \varphi_{V'}(\alpha_{2})\),则\(\varphi(\alpha_{1}) = \varphi(\alpha_{2})\)。由于\(\varphi\)是单射,所以\(\alpha_{1} = \alpha_{2}\),即\(\varphi_{V'}\)也是单射。
再验证\(\varphi_{V'}\)是线性映射。因\(V'\)是子空间,对于任意\(\alpha_{1}, \alpha_{2} \in V'\)\(c_{1}, c_{2} \in \F\),我们有\(c_{1} \alpha_{1} + c_{2} \alpha_{2} \in V'\)。因此,结合\(\varphi\)是线性映射,有
\begin{align*} \varphi_{V'}(c_{1} \alpha_{1} + c_{2} \alpha_{2})\amp = \varphi(c_{1} \alpha_{1} + c_{2} \alpha_{2})\\ \amp = c_{1} \varphi(\alpha_{1}) + c_{2} \varphi(\alpha_{2})\\ \amp = c_{1} \varphi_{V'}(\alpha_{1}) + c_{2} \varphi_{V'}(\alpha_{2}). \end{align*}
所以,\(\varphi_{V'}\)保持线性运算,是线性映射。

证明.

首先证明\(\Ima \varphi\)\(\Ima \varphi_{A}\)同构。设\(\sigma_{U}: U \to \F^{m}\)\(U\)中向量到其在\((\eta_{1}, \ldots, \eta_{m})\)下坐标的映射(见第6.6.2节)。由例6.5.2\(\sigma_{U}\)\(U\)\(\F^{m}\)的同构映射。根据引理6.7.13,将\(\sigma_{U}\)的作用范围限制到像空间\(\Ima \varphi \subseteq U\)之后所得的映射\(\sigma_{\Ima \varphi}\)依然是\(\Ima \varphi\)\(\sigma_{U}(\Ima \varphi)\)的同构映射。因此,完成证明,仅需说明\(\sigma_{U}(\Ima \varphi) = \Ima \varphi_{A}\)
一方面,对于任意\(\beta \in \Ima \varphi\),存在\(\alpha \in V\)使得\(\varphi(\alpha) = \beta\)。设\(\alpha\)\((\xi_{1}, \ldots, \xi_{n})\)下的坐标为\(x \in \F^{n}\),由于\(\varphi\)\((\xi_{1}, \ldots, \xi_{n})\)\((\eta_{1}, \ldots, \eta_{m})\)下的矩阵为\(A\),则根据命题6.6.5\(\varphi(\alpha)\)\((\eta_{1}, \ldots, \eta_{m})\)下的坐标为\(A x\)。根据\(\sigma_{U}\)的定义,
\begin{equation*} \sigma_{U}(\beta) = \sigma_{U}(\varphi(\alpha)) = Ax \in \Ima \varphi_{A}. \end{equation*}
因此,\(\sigma_{U}(\Ima \varphi) \subseteq \Ima \varphi_{A}\)
另一方面,对于任意\(y \in \Ima \varphi_{A}\),存在\(x \in \F^{m}\)使得\(y = Ax\)。设\(\alpha \in V\)是以\(x\)\((\xi_{1}, \ldots, \xi_{n})\)下坐标的向量,则再次根据根据命题6.6.5\(\varphi(\alpha)\)\((\eta_{1}, \ldots, \eta_{m})\)下的坐标为\(A x = y\),即\(y = \sigma_{U}(\varphi(\alpha)) \in \sigma_{U}(\Ima \varphi)\)。因此,\(\Ima \varphi_{A} \subseteq \sigma_{U}(\Ima \varphi)\)
综上\(\sigma_{U}(\Ima \varphi) = \Ima \varphi_{A}\),所以\(\Ima \varphi\)\(\Ima \varphi_{A}\)同构。
接下来,证明\(\Ker \varphi\)\(\Ker \varphi_{A}\)同构。设\(\sigma_{V}: V \to \F^{m}\)\(V\)中向量到其在\((\xi_{1}, \ldots, \xi_{n})\)下坐标的映射(见第6.6.2节)。类似于像空间的证明,我们仅须说明\(\sigma_{V}(\Ker \varphi) = \Ker \varphi_{A}\)即可。
一方面,对于任意\(\alpha \in \Ker \varphi\),设\(\sigma_{V}(\alpha) = x \in \F^{n}\)是它的坐标。根据根据命题6.6.5\(\varphi(\alpha)=0\)\((\eta_{1}, \ldots, \eta_{m})\)下的坐标为\(A x\)。因为\(0 \in U\)的坐标就是\(0 \in \F^{m}\),由坐标的唯一性知\(Ax = 0\),即\(x = \sigma_{V}(\alpha) \in \Ker \varphi_{A}\)。所以,\(\sigma_{V}(\Ker \varphi) \subseteq \Ker \varphi_{A}\)
另一方面,对于任意\(x \in \Ker \varphi_{A}\)\(Ax = 0\)。取坐标为\(x\)的向量\(\alpha \in V\),再次根据根据命题6.6.5\(\varphi(\alpha)\)\((\eta_{1}, \ldots, \eta_{m})\)下的坐标为\(A x = 0\),所以\(\varphi(\alpha) = 0\),即\(\alpha \in \Ker \varphi\)。所以,\(\Ker \varphi_{A} \subseteq \sigma_{V}(\Ker \varphi)\)
综上,\(\sigma_{V}(\Ker \varphi) = \Ker \varphi_{A}\),得证。
接着,我们建立\(\varphi_{A}\)像空间的秩与核空间的零度与矩阵\(A\)的秩和解空间维数之间的关系。

证明.

结合命题6.7.14和命题6.7.15可以得到线性映射基本定理的另一种证明方法。

定理6.7.11 证法二.

结合定理6.5.9、命题6.7.14和命题6.7.15
\begin{equation*} \dim \Ima \varphi = \dim \Ima \varphi_{A} = r(A) \quad \text{且}\quad \dim \Ker \varphi = \dim \Ker \varphi_{A} = n - r(A). \end{equation*}
因此,\(\dim \Ima \varphi + \dim \Ker \varphi = n = \dim V\)
下面给出线性映射基本定理的两个简单推论:维数不同的线性空间之间的线性映射不可能同时为单射和双射。

证明.

\(\varphi \in \mathcal{L}(V,U)\),由定理6.7.11
\begin{equation*} \dim \Ker \varphi = \dim V - \dim \Ima \varphi \geq \dim V - \dim W > 0. \end{equation*}
因此\(\Ker \varphi \neq 0\)。由命题6.7.10\(\varphi\)不是单射。

证明.

\(\varphi \in \mathcal{L}(V,U)\),由定理6.7.11
\begin{equation*} \dim \Ima \varphi = \dim V - \dim \Ker \varphi \leq \dim V < \dim U. \end{equation*}
因此,存在\(\beta \in \dim U\)\(\beta \not\in \Ima \varphi\)\(\varphi\)不是满射。
关于线性映射基本定理的两种证明方法:利用扩基法证明以及利用同构关系证明都具有普遍性,下面我们再看一个例子(该例子与第5.2.4节中介绍的Morre-Penrose广义逆的存在性有关)。

证法一.

\(n = \dim V, m = \dim U, r = \dim \Ima \varphi\)。由定理6.7.11\(\dim \Ker \varphi = n-r\)
\(\xi_{r+1}, \ldots, \xi_{n}\)\(\Ker \varphi\)的基,将其扩为\(V\)的基\(\xi_{1}, \ldots, \xi_{r}, \xi_{r+1}, \ldots, \xi_{n}\)。则由定理6.7.11的证明过程知\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{r}))\)\(\Ima \varphi\)的基。
将其扩为\(U\)的基\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{r}), \beta_{r+1}, \ldots, \beta_{m})\)。由命题6.6.1知,为了构造\(\psi \in \mathcal{L}(U,V)\),只需要指定基向量的像即可。我们构造\(\psi\)满足
\begin{equation*} \psi(\varphi(\xi_{i})) = \xi_{i}, \forall i=1,\ldots,r, \quad \psi(\beta_{j}) = 0, \forall j=r+1, \ldots, m. \end{equation*}
此时,对于任意\(\alpha = \sum_{i=1}^{n} c_{i} \xi_{i} \in V\),我们有
\begin{equation*} \varphi \psi \varphi(\alpha) = \sum_{i=1}^{r} c_{i} \varphi \psi (\varphi)(\xi_{i})) = \sum_{i=1}^{r} c_{i} \varphi(\xi_{i}) = \varphi(\alpha), \end{equation*}
其中,第一个和最后一个等式是由于\(\varphi \psi \varphi\)\(\varphi\)是线性映射且\(\xi_{j} \in \Ker \varphi, j > r\),第二个等式是根据\(\psi\)的构造。所以,\(\varphi \psi \varphi = \varphi\)
类似地,对于任意\(\beta = \sum_{i=1}^{r} c_{i} \varphi(\xi_{i}) + \sum_{j=r+1}^{n} b_{j} \beta_{j} \in U\),又由\(\psi\)的构造方式,我们有
\begin{equation*} \psi \varphi \psi(\beta) = \sum_{i=1}^{r} c_{i} \psi \varphi \psi( \varphi(\xi_{i})) + \sum_{j=r+1}^{n} c_{i} \psi \varphi \psi( \beta_{j} ) = \sum_{i=1}^{r} c_{i} \xi_{i} = \psi(\beta). \end{equation*}
所以,\(\psi \varphi \psi = \psi\)

证法二.

\(V\)的基\((\xi_{1}, \ldots, \xi_{n})\)\(U\)的基\((\eta_{1}, \ldots, \eta_{m})\),设\(\varphi\)在这两组基下的矩阵为\(A \in \F^{m \times n}\)。将\(A\)化为相抵标准型,即存在可逆矩阵\(P \in \F^{m \times m}, Q \in \F^{n \times n}\)使得
\begin{equation*} A = P \begin{pmatrix}E_{r}&0 \\ 0&0\end{pmatrix}_{m \times n}Q. \end{equation*}
取矩阵\(B \in \F^{n \times m}\)满足
\begin{equation*} B = Q^{-1}\begin{pmatrix}E_{r}&0 \\ 0&0\end{pmatrix}_{n \times m}P^{-1}, \end{equation*}
\(ABA = A\)\(BAB = B\)
\(\psi \in \mathcal{L}(U,V)\)为在\((\eta_{1}, \ldots, \eta_{m})\)\((\xi_{1}, \ldots, \xi_{n})\)下矩阵为\(B\)的线性映射,即
\begin{equation*} \psi(\eta_{1}, \ldots, \eta_{m}) = (\xi_{1}, \ldots, \xi_{n}) B. \end{equation*}
\(\sigma_{V}: V \to \F^{n}\)\(V\)中向量到其在\((\xi_{1}, \ldots, \xi_{n})\)下坐标的映射,\(\sigma_{U}: U \to \F^{m}\)\(U\)中向量到其在\((\eta_{1}, \ldots, \eta_{m})\)下坐标的映射。又令\(\varphi_{A}: x \in \F^{n} \mapsto Ax \in \F^{m}\)\(\psi_{B}: y \in \F^{m} \mapsto By \in \F^{n}\)。则由命题6.6.11可知
\begin{equation*} \varphi \psi \varphi = \sigma_{U}^{-1}\varphi_{A} \sigma_{V} \sigma_{V}^{-1}\psi_{B} \sigma_{U} \sigma_{U}^{-1}\varphi_{A} \sigma_{V} = \sigma_{U}^{-1}\varphi_{A} \psi_{B} \varphi_{A} \sigma_{V}. \end{equation*}
因此,对于\(\alpha = \sum_{i=1}^{n} x_{i} \xi_{i} \in V\),即在\((\xi_{1}, \ldots, \xi_{n})\)下坐标为\(x \in \F^{n}\)的向量\(\alpha\),有
\begin{equation*} \varphi \psi \varphi (\alpha) = \sigma_{U}^{-1}\varphi_{A} \psi_{B} \varphi_{A} \sigma_{V} (\alpha) = \sigma_{U}^{-1}ABAx. \end{equation*}
由于\(ABA = A\)
\begin{equation*} \varphi \psi \varphi (\alpha) = \sigma_{U}^{-1}A x = \sigma_{U}^{-1}\varphi_{A} \sigma_{V} (\alpha) = \varphi(\alpha), \quad \forall \alpha \in V. \end{equation*}
所以,\(\varphi \psi \varphi = \varphi\)
关于\(\psi \varphi \psi = \psi\)的证明可通过类似的方式完成。
本节的最后,我们尝试从线性映射的观点证明一个关于秩的不等式。

6.7.19.

\(A \in \F^{m \times n}, B \in \F^{n \times p}\),考虑线性映射\(\varphi_{A}: \F^{n} \to \F^{m}, \alpha \mapsto A\alpha\)\(\varphi_{B}: \F^{p} \to \F^{n}, \beta \mapsto B\beta\)\(\varphi_{AB}: \F^{p} \to \F^{m}, \beta \mapsto AB \beta\)
注意到\(\varphi_{AB}= \varphi_{A} \varphi_{B}\),因此\(\Ima \varphi_{AB}= \varphi_{A}(\Ima \varphi_{B})\)。又由于\(\Ima \varphi_{B}\)\(\F^{n}\)的子空间,因而
\begin{equation*} \Ima \varphi_{AB}\subseteq \varphi_{A}(\F^{n}) = \Ima \varphi_{A}. \end{equation*}
所以\(r(AB) = \dim \Ima \varphi_{AB}\leq \dim \Ima \varphi_{A} = r(A)\)
类似地,我们可以证明\(r(B^{T} A^{T}) \leq r(B^{T})\)。又由于\(r(AB) = r((AB)^{T}) = r(B^{T} A^{T})\)\(r(B) = r(B^{T})\),所以\(r(AB) \leq r(B)\)
综上\(r(AB) \leq \min\{ r(A), r(B) \}\)

练习 6.7.3

基础题.

1.
对线性映射\(\varphi: \F^{n \times n}\to \F, A \mapsto {\rm tr } A\),求\(\Ker \varphi\)\(\Ima \varphi\),并求它们的一个基和维数。
解答.
  • \(\Ima \varphi = \F\),因为对任意\(c \in \F\),取矩阵\(A = c E_{11}\),则\({\rm tr }A = c\),所以\(\varphi\)是满射。从而\(\dim \Ima \varphi = 1\),一个基为\((1)\)(将\(\F\)视为\(\F\)上的线性空间)。
  • \(\Ker \varphi = \{ A \in \F^{n \times n}\mid {\rm tr } A = 0 \}\)。由维数公式:
    \begin{equation*} \dim \Ker \varphi = \dim \F^{n \times n}- \dim \Ima \varphi = n^{2} - 1. \end{equation*}
    \(\F^{n \times n}\)的标准基\(\{ E_{ij}\mid 1 \leq i,j \leq n \}\),其中\(E_{ij}\)\((i,j)\)位置为1,其余为0的矩阵。则\(\Ker \varphi\)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.
考虑第6.6.4节习题6.6.4.1中的从\(\F_{n-1}[x]\)\(\F_{n}[x]\)的线性映射\(\varphi\)
\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}, \quad \forall a_{0},a_{1},\ldots,a_{n-1}\in \F. \end{equation*}
\(\Ker \varphi\)\(\Ima \varphi\)以及它们的维数。
解答.
  • \(\Ker \varphi\):设\(f = a_{0} + a_{1} x + \cdots + a_{n-1}x^{n-1}\in \F_{n-1}[x]\),使得\(\varphi(f)=0\)。则
    \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\(\Ker \varphi = \{0\}\)\(\dim \Ker \varphi = 0\)
  • \(\Ima \varphi\):由定义,\(\Ima \varphi\)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\(\Ima \varphi\)\(\F_{n}[x]\)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 \varphi = \{ p(x) \in \F_{n}[x] \mid p(0)=0 \}. \end{equation*}
    一组基为\(\{ x, x^{2}, \ldots, x^{n} \}\),维数为\(n\)
3.
\(V,U,W\)是有限维线性空间,\(\varphi \in \mathcal{L}(V,U), \psi \in \mathcal{L}(U,W)\),证明:
\begin{equation*} \dim \Ker \psi \varphi \leq \dim \Ker \varphi + \dim \Ker \psi. \end{equation*}
解答.
考虑\(\varphi\)在子空间\(K = \Ker \psi \varphi\)上的限制\(\varphi|_{K}: K \to U\)。对任意\(\alpha \in K\),有\(\psi(\varphi(\alpha)) = 0\),故\(\varphi(\alpha) \in \Ker \psi\),即\(\Ima \varphi|_{K} \subseteq \Ker \psi\)。因此
\begin{equation} \dim \varphi(K) \leq \dim \Ker \psi.\tag{6.7.1} \end{equation}
另一方面,将线性映射基本定理定理应用于\(\varphi|_{K}\),有
\begin{equation*} \dim K = \dim \Ker(\varphi|_{K}) + \dim \Ima (\varphi|_{K}). \end{equation*}
注意到\(\Ker(\varphi|_{K}) = \{ \alpha \in K \mid \varphi(\alpha)=0 \} = K \cap \Ker \varphi \subseteq \Ker \varphi\),所以
\begin{equation} \dim \Ker(\varphi|_{K}) \leq \dim \Ker \varphi.\tag{6.7.2} \end{equation}
结合((6.7.1))和((6.7.2)),得
\begin{equation*} \dim K \leq \dim \Ker \varphi + \dim \Ker \psi, \end{equation*}
\(\dim \Ker \psi \varphi \leq \dim \Ker \varphi + \dim \Ker \psi\)
4.
\(V,U\)是有限维线性空间,\(\varphi \in \mathcal{L}(V,U)\),证明:存在\(V\)的子空间\(V'\)使得\(V' \cap \Ker \varphi = 0\)\(\Ima \varphi = \{ \varphi(\alpha) \mid \alpha \in V'\}\)
解答.
\(\Ker \varphi\)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\(V' \cap \Ker \varphi = 0\)(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*} \varphi(\alpha) = \sum_{i=1}^{k} a_{i} \varphi(\alpha_{i}) + \sum_{j=1}^{r} b_{j} \varphi(\beta_{j}) = \sum_{j=1}^{r} b_{j} \varphi(\beta_{j}) \in \varphi(V'). \end{equation*}
所以\(\Ima \varphi \subseteq \varphi(V')\)。反之,显然\(\varphi(V') \subseteq \Ima \varphi\)。故\(\Ima \varphi = \varphi(V')\)

提高题.

5.
\(V\)是数域\(\F\)上的有限维线性空间,\(\varphi \in \mathcal{L}(V, \F)\),证明:若存在\(\alpha \not\in \Ker \varphi\),则
\begin{equation*} V = \Ker \varphi \oplus \langle \alpha \rangle. \end{equation*}
解答.
由于\(\varphi \neq 0\)(因为存在\(\alpha \notin \Ker \varphi\)),所以\(\varphi\)是满射(因为\(\Ima \varphi\)\(\F\)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 \varphi + \dim \Ima \varphi = \dim \Ker \varphi + 1. \end{equation*}
\(k = \dim \Ker \varphi\),取\(\Ker \varphi\)的一个基\((\alpha_{1}, \ldots, \alpha_{k})\)。由于\(\alpha \notin \Ker \varphi\),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\(\varphi\)作用得
\begin{equation*} c \varphi(\alpha) = 0. \end{equation*}
因为\(\varphi(\alpha) \neq 0\),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\(V = \langle \alpha_{1}, \ldots, \alpha_{k}, \alpha \rangle = \Ker \varphi + \langle \alpha \rangle\)。又因为\(\Ker \varphi \cap \langle \alpha \rangle = \{0\}\)(若有\(c\alpha \in \Ker \varphi\),则\(\varphi(c\alpha)=c\varphi(\alpha)=0\),得\(c=0\)),所以和为直和,即\(V = \Ker \varphi \oplus \langle \alpha \rangle\)
6.
\(V,U\)是数域\(\F\)上的有限维线性空间,\(\varphi \in \mathcal{L}(V,U)\)\((\eta_{1}, \ldots, \eta_{r})\)\(\Ima \varphi\)的一个基。证明:存在\(\psi_{1}, \ldots, \psi_{r} \in \mathcal{L}(V,\F)\)使得
\begin{equation*} \varphi(\alpha) = \psi_{1}(\alpha) \eta_{1} + \cdots + \psi_{r}(\alpha) \eta_{r}, \quad \forall \alpha \in V. \end{equation*}
解答.
对任意\(\alpha \in V\)\(\varphi(\alpha) \in \Ima \varphi\),故存在唯一的标量\(\psi_{1}(\alpha), \ldots, \psi_{r}(\alpha) \in \F\)使得
\begin{equation*} \varphi(\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\(\alpha, \beta \in V\)\(c \in \F\),有
\begin{align*} \varphi(\alpha + \beta)\amp = \varphi(\alpha) + \varphi(\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}, \\ \varphi(c\alpha) \amp = c \varphi(\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*} \varphi(\alpha + \beta) = \sum_{i=1}^{r} \psi_{i}(\alpha + \beta) \eta_{i}, \quad \varphi(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.
\(V\)\(n\)维线性空间,\(U\)\(m\)维线性空间,\(\varphi \in \mathcal{L}(V,U)\)是从\(V\)\(U\)的线性映射且是单射,证明:存在\(U\)\(V\)的满射\(\psi\)使得\(\psi \varphi ={\rm id}_{V}\)
解答.
\(V\)的一个基\((\xi_{1}, \ldots, \xi_{n})\),因\(\varphi\)是单射,由定理6.7.11的证明可知,\((\eta_{1} = \varphi(\xi_{1}), \ldots, \eta_{n} = \varphi(\xi_{n}))\)构成\(\Ima \varphi\)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*} \varphi(\alpha) = \sum_{i=1}^{n} a_{i} \eta_{i}. \end{equation*}
于是
\begin{equation*} \psi(\varphi(\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\(\psi \varphi = \operatorname{id}_{V}\)。由此可得\(\psi\)是满射:因为对任意\(\beta \in V\),有\(\beta = \psi(\varphi(\beta))\),即\(\beta \in \Ima \psi\),故\(\psi\)是满射。
8.
\(V,U\)是有限维线性空间,\(\varphi\)是从\(V\)\(U\)的线性映射。证明下列三个命题是等价的。
  1. \(\varphi\)是单射;
  2. 对于任意从\(U\)\(V\)的线性映射\(\psi, \psi'\),若\(\varphi \psi = \varphi \psi'\),则\(\psi = \psi'\)
  3. 存在从\(U\)\(V\)的线性\(\psi\)使得\(\psi \varphi ={\rm id}_{V}\)
解答.
(1) \(\Rightarrow\) (2): 假设\(\varphi\)是单射,且\(\varphi \psi = \varphi \psi'\)。则对任意\(\beta \in U\),有\(\varphi(\psi(\beta)) = \varphi(\psi'(\beta))\),由单射性得\(\psi(\beta) = \psi'(\beta)\),所以\(\psi = \psi'\)
(2) \(\Rightarrow\) (1): 假设(2)成立。若\(\varphi\)不是单射,由命题6.7.10\(\Ker \varphi \neq 0\)。设\(\alpha \neq 0 \in \Ker \varphi\)。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\(\varphi(\psi(\beta)) = \varphi(c_{1} \cdot 0) = 0\)\(\varphi(\psi'(\beta)) = \varphi(c_{1} \alpha) = c_{1} \varphi(\alpha)=0\),所以\(\varphi \psi = \varphi \psi'\),但\(\psi \neq \psi'\),与(2)矛盾。故\(\varphi\)是单射。
(1) \(\Rightarrow\) (3): 由\(\varphi\)是单射,知\(\dim V = \dim \Ima \varphi \leq \dim U\)。取\(V\)的一个基\((\xi_{1}, \ldots, \xi_{n})\),则由命题6.5.4的证明过程知\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)线性无关。将其扩充为\(U\)的一个基\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{n}), \eta_{n+1}, \ldots, \eta_{m})\)。定义\(\psi: U \to V\)\(\psi(\varphi(\xi_{i})) = \xi_{i}, i=1,\ldots, n\)\(\psi(\eta_{j})=0, j=n+1, \ldots, m\),则\(\psi\)线性且满足\(\psi \varphi ={\rm id}_{V}\)
(3) \(\Rightarrow\) (1): 假设存在\(\psi\)使得\(\psi \varphi = \operatorname{id}_{V}\)。若\(\varphi(\alpha)=0\),则\(\alpha = \psi(\varphi(\alpha)) = \psi(0)=0\),所以\(\varphi\)是单射。
9.
\(V,U\)是有限维线性空间,\(\varphi\)是从\(V\)\(U\)的线性映射。证明下列三个命题是等价的。
  1. \(\varphi\)是满射;
  2. 对于任意从\(U\)\(V\)的线性映射\(\psi, \psi'\),若\(\psi \varphi = \psi' \varphi\),则\(\psi = \psi'\)
  3. 存在从\(U\)\(V\)的线性\(\psi\)使得\(\varphi \psi ={\rm id}_{U}\)
解答.
(1) \(\Rightarrow\) (2): 假设\(\varphi\)是满射,且\(\psi \varphi = \psi' \varphi\)。对任意\(\beta \in U\),由满射性,存在\(\alpha \in V\)使得\(\beta = \varphi(\alpha)\)。则\(\psi(\beta) = \psi(\varphi(\alpha)) = \psi'(\varphi(\alpha)) = \psi'(\beta)\),所以\(\psi = \psi'\)
(2) \(\Rightarrow\) (1): 假设(2)成立。若\(\varphi\)不是满射,则\(\Ima \varphi\)\(U\)的真子空间。设\(\dim U = m\)\(\dim \Ima \varphi = r < m\)。取\(\Ima \varphi\)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\(\alpha \in V\)\(\varphi(\alpha)\)可表示为\(\eta_{1}, \ldots, \eta_{r}\)的线性组合,故\(\psi(\varphi(\alpha))=0\)\(\psi'(\varphi(\alpha))=0\),所以\(\psi \varphi = \psi' \varphi\),但\(\psi \neq \psi'\),与(2)矛盾。故\(\varphi\)是满射。
(1) \(\Rightarrow\) (2): 由\(\varphi\)是满射,知\(\dim U = \dim \Ima \varphi \leq \dim V\)。取\(U\)的一个基\((\eta_{1}, \ldots, \eta_{m})\),对每个\(\eta_{i}\),选取\(\xi_{i} \in V\)使得\(\varphi(\xi_{i}) = \eta_{i}\)。定义\(\psi: U \to V\)\(\psi(\eta_{i}) = \xi_{i}\),则\(\varphi \psi = \operatorname{id}_{U}\)
(3) \(\Rightarrow\) (1): 假设存在\(\psi\)使得\(\varphi \psi = \operatorname{id}_{U}\)。对任意\(\beta \in U\),有\(\beta = \varphi(\psi(\beta)) \in \Ima \varphi\),所以\(\varphi\)是满射。
10.
\(\varphi\)\(n\)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 \varphi^{i-1}\cap \Ker \varphi) = \dim \Ker \varphi^{i} - \dim \Ker \varphi^{i-1}. \end{equation*}
解答.
考虑映射\(\varphi\)在子空间\(\Ima \varphi^{i-1}\)上的限制,记为\(\varphi|_{\Ima \varphi^{i-1}}: \Ima \varphi^{i-1}\to V\)。显然,\(\Ker (\varphi|_{\Ima \varphi^{i-1}}) = \Ima \varphi^{i-1}\cap \Ker \varphi\)。而\(\Ima (\varphi|_{\Ima \varphi^{i-1}}) = \varphi(\Ima \varphi^{i-1}) = \Ima \varphi^{i}\)。由线性映射基本定理,
\begin{equation*} \dim \Ima \varphi^{i-1}= \dim \Ker (\varphi|_{\Ima \varphi^{i-1}}) + \dim \Ima (\varphi|_{\Ima \varphi^{i-1}}), \end{equation*}
\begin{equation*} \dim \Ima \varphi^{i-1}= \dim (\Ima \varphi^{i-1}\cap \Ker \varphi) + \dim \Ima \varphi^{i}. \end{equation*}
移项得
\begin{equation*} \dim (\Ima \varphi^{i-1}\cap \Ker \varphi) = \dim \Ima \varphi^{i-1}- \dim \Ima \varphi^{i}. \end{equation*}
对线性变换\(\varphi^{j}\)应用维数公式:\(\dim V = \dim \Ker \varphi^{j} + \dim \Ima \varphi^{j}\),所以\(\dim \Ima \varphi^{j} = \dim V - \dim \Ker \varphi^{j}\)。代入上式得
\begin{align*} \dim (\Ima \varphi^{i-1}\cap \Ker \varphi)\amp = (\dim V - \dim \Ker \varphi^{i-1}) - (\dim V - \dim \Ker \varphi^{i}) \\ \amp = \dim \Ker \varphi^{i} - \dim \Ker \varphi^{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.
\(A\)对应于线性映射\(\varphi_{A}: \F^{n} \to \F^{m}\)\(B\)对应于线性映射\(\varphi_{B}: \F^{p} \to \F^{n}\),则\(AB\)对应于\(\varphi_{A} \varphi_{B}: \F^{p} \to \F^{m}\)。我们需要证明
\begin{equation*} r(AB) \geq r(A) + r(B) - n. \end{equation*}
\begin{equation*} \dim \Ima (\varphi_{A} \varphi_{B}) \geq \dim \Ima \varphi_{A} + \dim \Ima \varphi_{B} - n. \end{equation*}
由于\(\Ima (\varphi_{A} \varphi_{B}) = \varphi_{A}(\Ima \varphi_{B})\),且\(\Ima \varphi_{B} \subseteq \F^{n}\),考虑\(\varphi_{A}\)在子空间\(\Ima \varphi_{B}\)上的限制\(\varphi_{A}|_{\Ima \varphi_B}: \Ima \varphi_{B} \to \F^{m}\)。则
\begin{equation*} \Ima (\varphi_{A}|_{\Ima \varphi_B}) = \varphi_{A}(\Ima \varphi_{B}) = \Ima (\varphi_{A} \varphi_{B}). \end{equation*}
由线性映射基本定理,
\begin{equation*} \dim \Ima \varphi_{B} = \dim \Ker (\varphi_{A}|_{\Ima \varphi_B}) + \dim \Ima (\varphi_{A}|_{\Ima \varphi_B}). \end{equation*}
\(\Ker (\varphi_{A}|_{\Ima \varphi_B}) = \Ima \varphi_{B} \cap \Ker \varphi_{A}\),所以
\begin{equation*} \dim \Ima (\varphi_{A} \varphi_{B}) = \dim \Ima \varphi_{B} - \dim (\Ima \varphi_{B} \cap \Ker \varphi_{A}). \end{equation*}
又因为\(\Ima \varphi_{B} \cap \Ker \varphi_{A} \subseteq \Ker \varphi_{A}\),故
\begin{equation*} \dim (\Ima \varphi_{B} \cap \Ker \varphi_{A}) \leq \dim \Ker \varphi_{A} = n - \dim \Ima \varphi_{A}. \end{equation*}
代入得
\begin{align*} \dim \Ima (\varphi_{A} \varphi_{B})\amp \geq \dim \Ima \varphi_{B} - (n - \dim \Ima \varphi_{A}) \\ \amp = \dim \Ima \varphi_{A} + \dim \Ima \varphi_{B} - n. \end{align*}
\(r(AB) \geq r(A) + r(B) - n\)

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。