定义 6.5.1.
设\(V\)和\(U\)为数域\(\F\)上的两个线性空间，若存在映射\(\varphi : V \to U\)满足
-
\(\varphi\)是双射，
-
\(\varphi\)是线性映射，
则称\(\varphi\)为\(V\)到\(U\)的同构映射。
节 6.5 同构映射与线性空间同构
子节 6.5.1 同构映射
例 6.5.2. 坐标映射是同构映射.
命题 6.5.3.
设\(V\)和\(U\)为数域\(\F\)上的两个线性空间，若\(\varphi : V \to U\)为\(V\)到\(U\)的同构映射，则向量\(\alpha \in V\)在\(V\)中可由向量组\(\alpha_{1}, \ldots, \alpha_{n} \in V\)线性表出当且仅当\(\varphi(\alpha)\)在\(U\)中可由向量组\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)线性表出。
证明.
必要性可由命题6.3.16直接得到。 考虑充分性，设
\begin{equation*}
\varphi(\alpha) = c_{1} \varphi(\alpha_{1}) + \cdots + c_{n} \varphi(\alpha_{n}), c_{1}, \ldots, c_{n} \in \F.
\end{equation*}
由于同构映射\(\varphi\)保持线性运算，有
\begin{equation*}
\varphi(\alpha) = \varphi(c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n}).
\end{equation*}
又因为同构映射\(\varphi\)是单射，必有\(\alpha = c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n}\)。
命题 6.5.4.
设\(V\)和\(U\)为数域\(\F\)上的两个线性空间，若\(\varphi : V \to U\)为\(V\)到\(U\)的同构映射，则向量组\(\alpha_{1}, \ldots, \alpha_{n}\)在\(V\)中线性相关当且仅当\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)在\(U\)中线性相关。
证明.
必要性可由命题6.3.17直接得到。 考虑充分性，设存在非全零系数\(c_{1}, \ldots, c_{n} \in \F\)使得
\begin{equation*}
c_{1} \varphi(\alpha_{1}) + \cdots + c_{n} \varphi(\alpha_{n}) = 0_{U}.
\end{equation*}
应用\(\varphi\)保持线性运算的性质和命题6.3.14可得
\begin{equation*}
\varphi(c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n}) = \varphi(0_{V}).
\end{equation*}
又因为同构映射\(\varphi\)是单射，所以\(c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n} = 0_{V}\)。
此外，子空间直和结构在同构映射下也是等价的。
命题 6.5.5.
设\(V\)和\(U\)为数域\(\F\)上的两个线性空间，\(V_{1}, V_{2} \subseteq V\)分别为\(V\)的两个子空间，若\(\varphi : V \to U\)为\(V\)到\(U\)的同构映射，则\(V_{1} \oplus V_{2}\)当且仅当\(\varphi(V_{1}) \oplus \varphi(V_{2})\)。
证明.
充分性：由命题4.4.22，我们仅需证明\(V_{1} \cap V_{2} = 0\)（此处\(0\)表示零子空间）。设\(\alpha \in V_{1} \cap V_{2}\)，则\(\varphi(\alpha) \in \varphi(V_{1}) \cap \varphi(V_{2})\)。由已知条件\(\varphi(V_{1}) \oplus \varphi(V_{2})\)，我们有\(\varphi(\alpha) = 0_{U} = \varphi(0_{V})\)，其中最后一个等式根据命题6.3.14得到。又因为\(\varphi\)是单射，所以\(\alpha = 0_{V}\)。因此，\(V_{1} \cap V_{2} = 0\)。
必要性：仅需证明\(\varphi(V_{1}) \cap \varphi(V_{2}) = 0\)。设\(\beta \in \varphi(V_{1}) \cap \varphi(V_{2})\)，则存在\(\alpha_{1} \in V_{1}, \alpha_{2} \in V_{2}\)使得\(\varphi(\alpha_{1}) = \varphi(\alpha_{2}) = \beta\)。由于\(\varphi\)保持线性运算，\(\varphi(\alpha_{1} - \alpha_{2}) = 0_{U} = \varphi(0_{V})\)。又因为\(\varphi\)是单射，所以\(\alpha_{1} - \alpha_{2} = 0_{V}\)，即\(\alpha_{1} = \alpha_{2} \in V_{1} \cap V_{2}\)。由已知条件\(V_{1} \oplus V_{2}\)，我们有\(\alpha_{1} = \alpha_{2} = 0_{V}\)，因此\(\beta = \varphi(0_{V}) = 0_{U}\)。
我们注意到，在上述三个命题的证明中，我们其实并未完全用到\(\varphi\)是双射这一强有力的保证，而是仅仅用到了\(\varphi\)是单射。 若是加上\(\varphi\)是满射这一性质，我们可以得到更有趣的结果：线性空间的基在同构映射下也是等价的。
命题 6.5.6.
设\(V\)和\(U\)为数域\(\F\)上的两个线性空间，若\(\varphi : V \to U\)为\(V\)到\(U\)的同构映射，则向量组\((\alpha_{1}, \ldots, \alpha_{n})\)是\(V\)的基当且仅当\((\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n}))\)是\(U\)的基。
证明.
由命题6.5.4可知，\(\alpha_{1}, \ldots, \alpha_{n} \)线性无关当且仅当\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)线性无关。 因此，由基的定义4.4.11，仅需证明\(\alpha_{1}, \ldots, \alpha_{n}\)可线性表出\(V\)中所有向量当且仅当\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)可线性表出\(U\)中所有向量。
必要性：设\(\beta \in U\)，由于\(\varphi\)是\(V\)到\(U\)的满射，\(\beta\)在\(V\)中存在原像\(\alpha \in V\)使得\(\varphi(\alpha) = \beta\)。又因为\(\alpha_{1}, \ldots, \alpha_{n}\)可线性表出\(V\)中所有向量，所以存在\(c_{1}, \ldots, c_{n} \in \F\)使得
\begin{equation*}
c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n} = \alpha.
\end{equation*}
将保持线性运算的同构映射同时作用于等式两边的向量得
\begin{equation*}
\varphi(c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n}) = c_{1} \varphi(\alpha_{1}) + \cdots + c_{n} \varphi(\alpha_{n}) = \varphi(\alpha) = \beta,
\end{equation*}
即\(\beta\)可由\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)线性表出。
充分性：设\(\alpha \in V\)，由于\(\varphi(\alpha)\)可在\(U\)中由\(\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n})\)线性表出，存在\(c_{1}, \ldots, c_{n} \in \F\)使得
\begin{equation*}
c_{1} \varphi(\alpha_{1}) + \cdots + c_{n} \varphi(\alpha_{n}) = \varphi(\alpha).
\end{equation*}
由\(\varphi\)保持线性运算的性质及其单射的性质，我们有\(c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n} = \alpha\)。
一个简单的推论是子空间的基和维数在同构映射下也等价。
推论 6.5.7.
设\(\varphi : V \to U\)为线性空间\(V\)到\(U\)的线性映射，\(V' \subseteq V\)是\(V\)的任意子空间，则\(\dim V' = \dim \varphi(V')\)，且\((\alpha_{1}, \ldots, \alpha_{n})\)是\(V'\)的基当且仅当\((\varphi(\alpha_{1}), \ldots, \varphi(\alpha_{n}))\)是\(U\)的子空间\(\varphi(V')\)的基。
子节 6.5.2 线性空间同构
同构映射可以保持线性空间几乎所有的性质。特别地，当两个线性空间之间存在同构映射时，它们从线性代数的本质上来说是一样的。
定义 6.5.8.
可以验证线性空间的同构关系是一个等价关系，满足：
-
自反性：\(V \cong V\)；
因此，同构关系将线性空间划分为不同的等价类。同一个等价类中的线性空间有什么共同特点呢？ 下面的定理是本节中关于线性空间同构理论的最主要结论，它表明线性空间的维数是同构等价类中最核心的等价不变量。
定理 6.5.9.
证明.
必要性：若\(V\)与\(U\)同构，则存在从\(V\)到\(U\)的同构映射\(\varphi\)，由命题6.5.6可知，\(\varphi\)将\(V\)的一组基映射成\(U\)的一组基，因此\(V\)与\(U\)维数相同。
充分性：设\(\dim V = \dim U = n\)，令\((\alpha_{1}, \ldots, \alpha_{n}) \in V\)为\(V\)的基，\((\beta_{1}, \ldots, \beta_{n}) \in U\)为\(U\)的基。由于\((\alpha_{1}, \ldots, \alpha_{n})\)是\(V\)的基，任意\(\alpha \in V\)可由改组基唯一地线性表示为\(\alpha = c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n}\)，其中\(c_{1}, \ldots, c_{n} \in \F\)。我们定义\(V\)到\(U\)的映射\(\varphi\)如下：
\begin{equation*}
\varphi(\alpha) := c_{1} \beta_{1} + \cdots + c_{n} \beta_{n}, \quad \forall \alpha = c_{1} \alpha_{1} + \cdots + c_{n} \alpha_{n} \in V.
\end{equation*}
下面证明\(\varphi\)是\(V\)到\(U\)的同构映射。首先说明\(\varphi\)是单射。对于任意\(\gamma_{1}, \gamma_{2} \in V\)，设
\begin{equation*}
\gamma_{1} = a_{1} \alpha_{1} + \cdots + a_{n} \alpha_{n},
\end{equation*}
\begin{equation*}
\gamma_{2} = b_{1} \alpha_{1} + \cdots + b_{n} \alpha_{n},
\end{equation*}
其中\(a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n} \in \F\)。 若\(\varphi(\gamma_{1}) = \varphi(\gamma_{2})\)，由\(\varphi\)的定义有
\begin{align*}
\amp a_{1} \beta_{1} + \cdots + a_{n} \beta_{n} = b_{1} \beta_{1} + \cdots + b_{n} \beta_{n}\\
\quad \iff \quad \amp (a_{1} - b_{1}) \beta_{1} + \cdots + (a_{n} - b_{n}) \beta_{n} = 0.
\end{align*}
又因为\(\beta_{1}, \ldots, \beta_{n}\)线性无关，所以\(a_{i} = b_{i}, \forall i=1,\ldots,n\)，即\(\gamma_{1}=\gamma_{2}\)。因此，\(\varphi\)是单射。
接下来，说明\(\varphi\)是满射。设\(\beta\)为\(U\)中任一向量，由于\((\beta_{1}, \ldots, \beta_{n})\)是\(U\)的基，存在唯一一组\(d_{1}, \ldots, d_{n} \in \F\)使得
\begin{equation*}
\beta = d_{1} \beta_{1} + \cdots + d_{n} \beta_{n}.
\end{equation*}
我们令\(\alpha= d_{1} \alpha_{1} + \cdots + d_{n} \alpha_{n} \in V\)，由\(\varphi\)的定义可知\(\varphi(\alpha) = \beta\)，即\(\beta\)可在\(V\)中找到\(\varphi\)下的原像。由\(\beta\)的任意性，\(\varphi\)是满射。
最后，我们说明\(\varphi\)保持线性运算。对于任意\(\alpha_{1}, \alpha_{2} \in V\)，设
\begin{equation*}
\alpha_{1} = a_{1} \alpha_{1} + \cdots + a_{n} \alpha_{n}, \quad \alpha_{2} = a'_{1} \alpha_{1} + \cdots + a'_{n} \alpha_{n}.
\end{equation*}
考虑任意\(c_{1}, c_{2} \in \F\)，根据\(\varphi\)的定义，我们有
\begin{align*}
\varphi(c_{1} \alpha_{1} + c_{2} \alpha_{2}) \amp =\varphi( (c_{1} a_{1} + c_{2} a'_{1}) \alpha_{1} + \cdots + (c_{1} a_{n} + c_{2} a'_{n}) \alpha_{n} ) \\
\amp = (c_{1} a_{1} + c_{2} a'_{1}) \beta_{1} + \cdots + (c_{1} a_{n} + c_{2} a'_{n}) \beta_{n}.
\end{align*}
结合\(U\)中加法和数乘的性质以及\(\varphi\)的定义可得，
\begin{align*}
\varphi(c_{1} \alpha_{1} + c_{2} \alpha_{2})\amp = c_{1} (a_{1} \beta_{1} + \cdots + a_{n} \beta_{n}) + c_{2} (a'_{1} \beta_{1} + \cdots + a'_{n} \beta_{n}) \\
\amp = c_{1} \varphi(\alpha_{1}) + c_{2} \varphi(\alpha_{2}).
\end{align*}
根据命题6.3.12，\(\varphi\)保持线性运算。
推论 6.5.10.
练习 6.5.3 练习
基础题.
1.
设\((\xi_{1}, \ldots, \xi_{n})\)是线性空间\(V\)的一个基，\(\varphi: V \to V\)是\(V\)上的线性变换，证明：\(\varphi\)是双射当且仅当\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)线性无关。
解答.
先证必要性：设\(\varphi\)是双射，则\(\varphi\)是可逆线性变换。假设存在标量\(c_{1}, \ldots, c_{n}\)使得
\begin{equation*}
c_{1} \varphi(\xi_{1}) + \cdots + c_{n} \varphi(\xi_{n}) = 0.
\end{equation*}
由于\(\varphi\)是线性的，上式等价于\(\varphi(c_{1} \xi_{1} + \cdots + c_{n} \xi_{n}) = 0\)。因为\(\varphi\)是单射（双射蕴含单射），所以\(c_{1} \xi_{1} + \cdots + c_{n} \xi_{n} = 0\)。而\(\xi_{1}, \ldots, \xi_{n}\)是基，故线性无关，于是\(c_{1} = \cdots = c_{n} = 0\)。因此\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)线性无关。
再证充分性：设\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)线性无关。因为\(\dim V = n\)，所以\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)也是\(V\)的一个基。对任意\(\alpha \in V\)，存在唯一标量\(a_{1}, \ldots, a_{n}\)使得\(\alpha = a_{1} \varphi(\xi_{1}) + \cdots + a_{n} \varphi(\xi_{n})\)。定义映射\(\psi: V \to V\)为\(\psi(\alpha) = a_{1} \xi_{1} + \cdots + a_{n} \xi_{n}\)，则易验证\(\psi\)是线性映射，且\(\psi \varphi = \varphi \psi ={\rm id}_{V}\)，故\(\varphi\)可逆，从而是双射。
2.
设\(\varphi\)是从线性空间\(V\)到\(U\)的同构映射，\(V_{1}, V_{2}\)是\(V\)的子空间，证明：
\begin{equation*}
\varphi(V_{1} + V_{2}) = \varphi(V_{1}) + \varphi(V_{2}) \quad \text{且}\quad \varphi(V_{1} \cap V_{2}) = \varphi(V_{1}) \cap \varphi(V_{2}).
\end{equation*}
解答.
先证第一个等式。任取\(\beta \in \varphi(V_{1} + V_{2})\)，则存在\(\alpha \in V_{1} + V_{2}\)使得\(\beta = \varphi(\alpha)\)。由\(V_{1} + V_{2}\)的定义，存在\(\alpha_{1} \in V_{1}, \alpha_{2} \in V_{2}\)使得\(\alpha = \alpha_{1} + \alpha_{2}\)。于是\(\beta = \varphi(\alpha_{1} + \alpha_{2}) = \varphi(\alpha_{1}) + \varphi(\alpha_{2}) \in \varphi(V_{1}) + \varphi(V_{2})\)。
反之，任取\(\beta \in \varphi(V_{1}) + \varphi(V_{2})\)，则存在\(\beta_{1} \in \varphi(V_{1}), \beta_{2} \in \varphi(V_{2})\)使得\(\beta = \beta_{1} + \beta_{2}\)。由定义，存在\(\alpha_{1} \in V_{1}, \alpha_{2} \in V_{2}\)使得\(\beta_{1} = \varphi(\alpha_{1}), \beta_{2} = \varphi(\alpha_{2})\)。于是\(\beta = \varphi(\alpha_{1}) + \varphi(\alpha_{2}) = \varphi(\alpha_{1} + \alpha_{2})\)，而\(\alpha_{1} + \alpha_{2} \in V_{1} + V_{2}\)，故\(\beta \in \varphi(V_{1} + V_{2})\)。所以\(\varphi(V_{1} + V_{2}) = \varphi(V_{1}) + \varphi(V_{2})\)。
再证第二个等式。任取\(\beta \in \varphi(V_{1} \cap V_{2})\)，则存在\(\alpha \in V_{1} \cap V_{2}\)使得\(\beta = \varphi(\alpha)\)。由于\(\alpha \in V_{1}\)且\(\alpha \in V_{2}\)，所以\(\beta = \varphi(\alpha) \in \varphi(V_{1})\)且\(\beta \in \varphi(V_{2})\)，即\(\beta \in \varphi(V_{1}) \cap \varphi(V_{2})\)。
反之，任取\(\beta \in \varphi(V_{1}) \cap \varphi(V_{2})\)，则\(\beta \in \varphi(V_{1})\)且\(\beta \in \varphi(V_{2})\)。存在\(\alpha_{1} \in V_{1}\)使得\(\beta = \varphi(\alpha_{1})\)，也存在\(\alpha_{2} \in V_{2}\)使得\(\beta = \varphi(\alpha_{2})\)。于是\(\varphi(\alpha_{1}) = \varphi(\alpha_{2})\)，由\(\varphi\)是单射（同构映射是双射）得\(\alpha_{1} = \alpha_{2}\)，记\(\alpha = \alpha_{1} = \alpha_{2}\)，则\(\alpha \in V_{1} \cap V_{2}\)，且\(\beta = \varphi(\alpha) \in \varphi(V_{1} \cap V_{2})\)。所以\(\varphi(V_{1} \cap V_{2}) = \varphi(V_{1}) \cap \varphi(V_{2})\)。
3.
设\(A \in \F^{n \times n}\)是可逆矩阵，定义\(\varphi_{A}: \F^{n} \to \F^{n}, \alpha \mapsto A \alpha\)，证明：\(\varphi_{A}\)是同构映射。
解答.
首先，\(\varphi_{A}\)是线性映射：对任意\(\alpha, \beta \in \F^{n}\)和\(c \in \F\)，有
\begin{equation*}
\varphi_{A}(\alpha + \beta) = A(\alpha + \beta) = A\alpha + A\beta = \varphi_{A}(\alpha) + \varphi_{A}(\beta),
\end{equation*}
\begin{equation*}
\varphi_{A}(c\alpha) = A(c\alpha) = k(A\alpha) = c \varphi_{A}(\alpha).
\end{equation*}
其次，由于\(A\)可逆，定义映射\(\psi: \F^{n} \to \F^{n}\)为\(\psi(\alpha) = A^{-1}\alpha\)，则\(\psi\)也是线性映射，且对任意\(\alpha \in \F^{n}\)，
\begin{equation*}
(\psi \varphi_{A})(\alpha) = \psi(A\alpha) = A^{-1}(A\alpha) = \alpha,
\end{equation*}
\begin{equation*}
(\varphi_{A} \psi)(\alpha) = \varphi_{A}(A^{-1}\alpha) = A(A^{-1}\alpha) = \alpha.
\end{equation*}
所以，\(\psi \varphi_{A} = \varphi_{A} \psi ={\rm id}_{\F^n}\)。因此\(\varphi_{A}\)是可逆线性映射，从而是同构映射。
4.
\begin{equation*}
H = \left\{ \begin{pmatrix}\alpha & \beta \\ -\overline{\beta} & \overline{\alpha}\end{pmatrix} \middle| \alpha, \beta \in \C \right\}.
\end{equation*}
证明：\(H\)与\(\R^{4}\)同构，并写出\(H\)到\(\R^{4}\)的一个同构映射。
解答.
设\(\alpha = a + b {\rm i}, \beta = c + d {\rm i}\)，其中\(a,b,c,d \in \R\)，则
\begin{align*}
\begin{pmatrix}\alpha & \beta \\ -\overline{\beta} & \overline{\alpha}\end{pmatrix} \amp = \begin{pmatrix}a+b{\rm i}&c+d{\rm i} \\ -c+d{\rm i}&a-b{\rm i}\end{pmatrix} \\
\amp = a \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + b \begin{pmatrix}{\rm i} & 0 \\ 0 & -{\rm i}\end{pmatrix} + c \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} + d \begin{pmatrix}0 & {\rm i} \\ {\rm i} & 0\end{pmatrix}.
\end{align*}
\ 记
\begin{equation*}
I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\quad J = \begin{pmatrix}{\rm i} & 0 \\ 0 & -{\rm i}\end{pmatrix},\quad K = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix},\quad L = \begin{pmatrix}0 & {\rm i} \\ {\rm i} & 0\end{pmatrix}.
\end{equation*}
定义映射\(\varphi: H \to \R^{4}\)为
\begin{equation*}
\varphi\left( aI + bJ + cK + dL \right) = (a, b, c, d)^{\top}.
\end{equation*}
由之前的讨论\((I,J,K,L)\)是\(H\)的一个基，因此\(\varphi\)为\(H\)中向量到其在\((I,J,K,L)\)下坐标的映射。由例6.5.2知\(\varphi\)是同构映射。
提高题.
5.
设有矩阵\(A \in \F^{n \times n}\)，令\(M = \{ A B \mid B \in \F^{n \times n}\} \subseteq \F^{n \times n}\)。
解答.
-
显然\(0 = A0 \in M\)，故\(M\)非空。对任意\(X,Y \in M\)，存在\(B,C \in \F^{n \times n}\)使得\(X = AB\)，\(Y = AC\)。则\(X+Y = A(B+C) \in M\)。对任意\(c \in \F\)，\(cX = A(cB) \in M\)。所以\(M\)对加法和数乘封闭，从而是\(\F^{n \times n}\)的子空间。
-
考虑满秩分解：由于\(r(A)=r\)，存在列满秩矩阵\(L \in \F^{n \times r}\)和行满秩矩阵\(R \in \F^{r \times n}\)使得\(A = LR\)。定义映射\(\varphi: \F^{r \times n}\to M\)为\(\Phi(C) = LC\)。容易验证\(\varphi\)是线性映射。下面证明\(\varphi\)是同构映射。\(\varphi\)是满射：对任意\(X \in M\)，存在\(B \in \F^{n \times n}\)使得\(X = AB = LRB\)。令\(C = RB \in \F^{r \times n}\)，则\(\varphi(C) = LC = LRB = X\)。\(\varphi\)是单射：假设存在\(C_{1}, C_{2} \in \F^{r \times n}\)使得\(\varphi(C_{1})= \varphi(C_{2})\)，则由\(\varphi\)的线性性有\(\varphi(C_{1} - C_{2}) = L(C_{1} - C_{2}) = 0\)。由于\(L\)列满秩，\(L\)的列线性无关，故\(L(C_{1} - C_{2})=0\)蕴含\(C_{1} - C_{2} = 0\)（以\(C_{1} - C_{2}\)的每一列作为系数对\(L\)的列向量进行线性组合都得到零向量）。所以，\(C_{1} = C_{2}\)，即\(\varphi\)是单射。
6.
设\(A,B \in F^{m \times n}\)满足\(r(A) = r(B)\)，设\(U\)是\(AX=0\)的解空间，\(W\)是\(BX=0\)的解空间，证明：\(U \cong W\)，并给出\(U\)到\(W\)的一个同构映射。
解答.
设\(r(A)=r(B)=r\)。由于秩相等，存在可逆矩阵\(P \in \F^{m \times m}\)和\(Q \in \F^{n \times n}\)使得\(B = PAQ\)（\(A\)与\(B\)相抵）。定义映射\(\varphi: U \to W\)为\(\varphi(\alpha) = Q^{-1}\alpha, \forall \alpha \in U\)。
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首先验证\(\varphi\)良定义：若\(\alpha \in U\)，则\(A \alpha=0\)。计算\(B\varphi(\alpha) = B Q^{-1}\alpha = PAQ Q^{-1}\alpha = PA\alpha = P0 = 0\)，所以\(\varphi(\alpha) \in W\)。
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\(\varphi\)是线性映射：对任意\(\alpha, \beta \in U\)和\(c \in \F\)，有\(\varphi(\alpha+\beta)=Q^{-1}(\alpha+\beta)=Q^{-1}\alpha+Q^{-1}\beta=\varphi(\alpha)+\varphi(\beta)\)，\(\varphi(c \alpha)=Q^{-1}(c \alpha)=c Q^{-1}\alpha = c\varphi(\alpha)\)。
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\(\varphi\)是单射：若存在\(\alpha, \alpha' \in U\)使得\(\varphi(\alpha)=\varphi(\alpha')\)，则\(Q^{-1}(\alpha - \alpha')=0\)，故\(\alpha=\alpha'\)。所以\(\varphi\)是单射。
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\(\varphi\)是满射：对任意\(\beta \in W\)，有\(B \beta =0\)。令\(\alpha = Q \beta\)，则\(A\alpha = AQ\beta\)。由\(B=PAQ\)且\(P\)可逆，\(B\beta=0\)等价于\(AQ\beta=0\)，所以\(A\alpha=0\)，即\(\alpha \in U\)。且\(\varphi(\alpha)=Q^{-1}\alpha = Q^{-1}Q\beta = \beta\)。所以\(\varphi\)是满射。
7.
设\(V,V'\)都是\(n\)维线性空间，\(\varphi\)是\(V\)到\(V'\)的线性映射，证明：\(\varphi\)是单射当且仅当\(\varphi\)是满射。（换言之，对于两个维数相同的线性空间之间的线性映射，我们仅需要单射或双射条件就能保证它是同构映射）
解答.
先证必要性：设\(\varphi\)是单射。 设\((\xi_{1}, \ldots, \xi_{n})\)是\(V\)的一个基，则基向量在\(V\)中线性无关。注意到命题6.5.4的证明中，仅用到了\(\varphi\)是单射的性质，所以其结论在此也使用，即\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)在\(V'\)中也线性无关。又因为\(V'\)的维数是\(n\)，所以\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{n}))\)是\(V'\)的基。
再证充分性：设\(\varphi\)是满射。设\((\xi_{1}, \ldots, \xi_{n})\)是\(V\)的一个基，我们首先证明\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{n}))\)是\(V'\)的一个基。
因\(\varphi\)是满射，对于任意\(\beta \in V'\)，可以找到\(\alpha \in V\)使得\(\varphi(\alpha) = \beta\)。设\(\alpha = c_{1} \xi_{1} + \cdots + c_{n} \xi_{n}\)。则由\(\varphi\)是线性映射可得
\begin{equation*}
\beta = \varphi(\alpha) = c_{1} \varphi(\xi_{1}) + \cdots + c_{n} \varphi(\xi_{n}),
\end{equation*}
即，任意\(\beta \in V'\)可由\(\varphi(\xi_{1}), \ldots, \varphi(\xi_{n})\)线性表出。因为\(V'\)的维数是\(n\)，根据定理4.4.16可知\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{n}))\)是\(V'\)的基。
因此，对于任意\(\alpha = \sum_{i=1}^{n} c_{i} \xi_{i}\)和\(\alpha' = \sum_{i=1}^{n} c'_{i} \xi_{i}\)，若\(\varphi(\alpha) = \varphi(\alpha')\)，则因\(\varphi\)是线性映射，所以
\begin{equation*}
\sum_{i=1}^{n} c_{i} \varphi(\xi_{i}) = \sum_{i=1}^{n} c'_{i} \varphi(\xi_{i}).
\end{equation*}
由于\((\varphi(\xi_{1}), \ldots, \varphi(\xi_{n}))\)是\(V'\)的基，任意\(V'\)中向量的表出方式唯一，故\(c_{i} = c'_{i}, \forall i \in [n]\)。所以\(\alpha = \alpha'\)，\(\varphi\)是单射。
