# 矩阵表示让我们从一类线性映射出发m × n m \times n m × n A A A
f A : R n → R m , f A ( x ) = A x f_A: \mathbb R^n \to \mathbb R^m, \quad f_A(\boldsymbol x) = A \boldsymbol x
f A : R n → R m , f A ( x ) = A x
则容易验证 f A f_A f A 由矩阵诱导的线性映射
也非常容易验证的是，f A f_A f A A x = 0 A \boldsymbol x = \boldsymbol 0 A x = 0 A A A
K e r f A = N ( A ) , dim ( K e r f A ) = n − r a n k ( A ) \mathrm{Ker} f_A = N(A),\quad \dim(\mathrm{Ker} f_A) = n - \mathrm{rank}(A) K e r f A = N ( A ) , dim ( K e r f A ) = n − r a n k ( A ) I m f A = C ( A ) , dim ( I m f A ) = r a n k ( A ) \mathrm{Im} f_A = C(A),\quad \dim(\mathrm{Im} f_A) = \mathrm{rank}(A) I m f A = C ( A ) , dim ( I m f A ) = r a n k ( A )
那么代入到单射与满射的判定，可以得到
f A f_A f A ⟺ r ( A ) = n \iff r(A) = n ⟺ r ( A ) = n f A f_A f A ⟺ r ( A ) = m \iff r(A) = m ⟺ r ( A ) = m 若 A A A ⟺ \iff ⟺ ⟺ \iff ⟺ ⟺ \iff ⟺ A A A
也就是说，通过分析矩阵 A A A f A f_A f A
令 V , W V, W V , W F \mathbb F F dim V = n , dim W = m \dim V = n, \dim W = m dim V = n , dim W = m f : V → W f: V \to W f : V → W
A = ( a 1 a 2 ⋯ a n ) \mathscr A = \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n \end{pmatrix} A = ( a 1 a 2 ⋯ a n ) V V V B = ( b 1 b 2 ⋯ b m ) \mathscr B = \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_m \end{pmatrix} B = ( b 1 b 2 ⋯ b m ) W W W
显然，对于 V V V a j \boldsymbol a_j a j W W W W W W
f ( a j ) = k 1 j b 1 + k 2 j b 2 + ⋯ + k m j b m , j = 1 , 2 , … , n f(\boldsymbol a_j) = k_{1j} \boldsymbol b_1 + k_{2j} \boldsymbol b_2 + \cdots + k_{mj} \boldsymbol b_m, \quad j = 1, 2, \ldots, n
f ( a j ) = k 1 j b 1 + k 2 j b 2 + ⋯ + k m j b m , j = 1 , 2 , … , n
逐一计算所有 V V V A \mathscr A A
{ f ( a 1 ) = k 11 b 1 + k 21 b 2 + ⋯ + k m 1 b m f ( a 2 ) = k 12 b 1 + k 22 b 2 + ⋯ + k m 2 b m ⋮ f ( a n ) = k 1 n b 1 + k 2 n b 2 + ⋯ + k m n b m \begin{cases}
f(\boldsymbol a_1) = k_{11} \boldsymbol b_1 + k_{21} \boldsymbol b_2 + \cdots + k_{m1} \boldsymbol b_m \\
f(\boldsymbol a_2) = k_{12} \boldsymbol b_1 + k_{22} \boldsymbol b_2 + \cdots + k_{m2} \boldsymbol b_m \\
\quad \vdots \\
f(\boldsymbol a_n) = k_{1n} \boldsymbol b_1 + k_{2n} \boldsymbol b_2 + \cdots + k_{mn} \boldsymbol b_m
\end{cases} ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ f ( a 1 ) = k 1 1 b 1 + k 2 1 b 2 + ⋯ + k m 1 b m f ( a 2 ) = k 1 2 b 1 + k 2 2 b 2 + ⋯ + k m 2 b m ⋮ f ( a n ) = k 1 n b 1 + k 2 n b 2 + ⋯ + k m n b m
将上述等式整理为矩阵形式
( f ( a 1 ) f ( a 2 ) ⋮ f ( a n ) ) 1 × n = ( k 11 k 21 ⋯ k m 1 k 12 k 22 ⋯ k m 2 ⋮ ⋮ ⋱ ⋮ k 1 n k 2 n ⋯ k m n ) n × m ( b 1 b 2 ⋮ b m ) m × 1 \begin{pmatrix}
f(\boldsymbol a_1) \\
f(\boldsymbol a_2) \\
\vdots \\
f(\boldsymbol a_n)
\end{pmatrix}_{1 \times n}
= \begin{pmatrix}
k_{11} & k_{21} & \cdots & k_{m1} \\
k_{12} & k_{22} & \cdots & k_{m2} \\
\vdots & \vdots & \ddots & \vdots \\
k_{1n} & k_{2n} & \cdots & k_{mn}
\end{pmatrix}_{n \times m}
\begin{pmatrix}
\boldsymbol b_1 \\
\boldsymbol b_2 \\
\vdots \\
\boldsymbol b_m
\end{pmatrix}_{m \times 1} ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ f ( a 1 ) f ( a 2 ) ⋮ f ( a n ) ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ 1 × n = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ k 1 1 k 1 2 ⋮ k 1 n k 2 1 k 2 2 ⋮ k 2 n ⋯ ⋯ ⋱ ⋯ k m 1 k m 2 ⋮ k m n ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ n × m ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ b 1 b 2 ⋮ b m ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ m × 1
对两边取转置，得到基的格式
( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) = ( b 1 b 2 ⋯ b m ) ( k 11 k 12 ⋯ k 1 n k 21 k 22 ⋯ k 2 n ⋮ ⋮ ⋱ ⋮ k m 1 k m 2 ⋯ k m n ) \begin{pmatrix} f(\boldsymbol a_1) & f(\boldsymbol a_2) & \cdots & f(\boldsymbol a_n) \end{pmatrix}
= \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_m \end{pmatrix}
\begin{pmatrix}
k_{11} & k_{12} & \cdots & k_{1n} \\
k_{21} & k_{22} & \cdots & k_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
k_{m1} & k_{m2} & \cdots & k_{mn}
\end{pmatrix} ( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) = ( b 1 b 2 ⋯ b m ) ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ k 1 1 k 2 1 ⋮ k m 1 k 1 2 k 2 2 ⋮ k m 2 ⋯ ⋯ ⋱ ⋯ k 1 n k 2 n ⋮ k m n ⎠ ⎟ ⎟ ⎟ ⎟ ⎞
其中 k i j k_{ij} k i j f f f V V V j j j W W W i i i
称此处出现的矩阵
T = ( k 11 k 12 ⋯ k 1 n k 21 k 22 ⋯ k 2 n ⋮ ⋮ ⋱ ⋮ k m 1 k m 2 ⋯ k m n ) T = \begin{pmatrix}
k_{11} & k_{12} & \cdots & k_{1n} \\
k_{21} & k_{22} & \cdots & k_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
k_{m1} & k_{m2} & \cdots & k_{mn}
\end{pmatrix} T = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ k 1 1 k 2 1 ⋮ k m 1 k 1 2 k 2 2 ⋮ k m 2 ⋯ ⋯ ⋱ ⋯ k 1 n k 2 n ⋮ k m n ⎠ ⎟ ⎟ ⎟ ⎟ ⎞
为线性映射 f f f A , B \mathscr A, \mathscr B A , B 矩阵表示 (Matrix Representation)「表現行列」
线性映射的矩阵表示如同线性映射的身份信息。完整描述 了所有与基底有关的变化情况。而任意的向量都可以被基底线性表示出来。T T T 所有 元的映射规则
不难看出，矩阵表示依赖于两个线性空间的基的选择，不同的基底会导致不同的矩阵表示F n \mathbb F^n F n F m \mathbb F^m F m f f f 标准矩阵表示
示例 R 2 \mathbb R^2 R 2 R 2 \mathbb R^2 R 2 R R R E = ( e 1 e 2 ) \mathscr E = \begin{pmatrix} \boldsymbol e_1 & \boldsymbol e_2 \end{pmatrix} E = ( e 1 e 2 )
R ( θ ) = ( sin θ − cos θ cos θ sin θ ) R(\theta) = \begin{pmatrix}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{pmatrix} R ( θ ) = ( sin θ cos θ − cos θ sin θ )
此线性变换表示将 R 2 \mathbb R^2 R 2 θ \theta θ
一般地，对于由 A A A f A : V → W f_A: V \to W f A : V → W
取 V V V
取 W W W
则
( f A ( e 1 ) f A ( e 2 ) ⋯ f A ( e n ) ) = ( A e 1 A e 2 ⋯ A e n ) = ( e 1 ′ e 2 ′ ⋯ e m ′ ) ( A e 1 A e 2 ⋯ A e n ) = ( e 1 ′ e 2 ′ ⋯ e m ′ ) A \begin{aligned}
\begin{pmatrix} f_A(\boldsymbol e_1) & f_A(\boldsymbol e_2) & \cdots & f_A(\boldsymbol e_n) \end{pmatrix}
&= \begin{pmatrix} A \boldsymbol e_1 & A \boldsymbol e_2 & \cdots & A \boldsymbol e_n \end{pmatrix} \\
&= \begin{pmatrix} \boldsymbol e_1' & \boldsymbol e_2' & \cdots & \boldsymbol e_m' \end{pmatrix}
\begin{pmatrix} A \boldsymbol e_1 & A \boldsymbol e_2 & \cdots & A \boldsymbol e_n \end{pmatrix} \\
&= \begin{pmatrix} \boldsymbol e_1' & \boldsymbol e_2' & \cdots & \boldsymbol e_m' \end{pmatrix} A
\end{aligned} ( f A ( e 1 ) f A ( e 2 ) ⋯ f A ( e n ) ) = ( A e 1 A e 2 ⋯ A e n ) = ( e 1 ′ e 2 ′ ⋯ e m ′ ) ( A e 1 A e 2 ⋯ A e n ) = ( e 1 ′ e 2 ′ ⋯ e m ′ ) A
因此，A A A f A f_A f A 标准基底下的矩阵表示
矩阵表示方便的一点是
可以直接用矩阵积来计算线性映射的复合
可以直接用逆矩阵来计算线性映射的逆映射
命题 f : U → V , g : V → W f: U \to V, g: V \to W f : U → V , g : V → W A , B , C \mathscr A, \mathscr B, \mathscr C A , B , C f f f A , B \mathscr A, \mathscr B A , B T f T_f T f g g g B , C \mathscr B, \mathscr C B , C T g T_g T g g ∘ f : U → W g \circ f: U \to W g ∘ f : U → W A , C \mathscr A, \mathscr C A , C T g ∘ f = T g T f T_{g \circ f} = T_g T_f T g ∘ f = T g T f
证明
令基
A = ( a 1 a 2 ⋯ a n ) B = ( b 1 b 2 ⋯ b p ) C = ( c 1 c 2 ⋯ c m ) \begin{aligned}
\mathscr A &= \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n \end{pmatrix} \\
\mathscr B &= \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_p \end{pmatrix} \\
\mathscr C &= \begin{pmatrix} \boldsymbol c_1 & \boldsymbol c_2 & \cdots & \boldsymbol c_m \end{pmatrix}
\end{aligned} A B C = ( a 1 a 2 ⋯ a n ) = ( b 1 b 2 ⋯ b p ) = ( c 1 c 2 ⋯ c m )
从矩阵表示的定义出发得到
( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) = ( b 1 b 2 ⋯ b p ) T f \begin{pmatrix} f(\boldsymbol a_1) & f(\boldsymbol a_2) & \cdots & f(\boldsymbol a_n) \end{pmatrix}
= \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_p \end{pmatrix} T_f ( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) = ( b 1 b 2 ⋯ b p ) T f
两边同时经由 g g g
( g ( f ( a 1 ) ) g ( f ( a 2 ) ) ⋯ g ( f ( a n ) ) ) = ( g ( b 1 ) g ( b 2 ) ⋯ g ( b p ) ) T f \begin{pmatrix} g(f(\boldsymbol a_1)) & g(f(\boldsymbol a_2)) & \cdots & g(f(\boldsymbol a_n)) \end{pmatrix}
= \begin{pmatrix} g(\boldsymbol b_1) & g(\boldsymbol b_2) & \cdots & g(\boldsymbol b_p) \end{pmatrix} T_f ( g ( f ( a 1 ) ) g ( f ( a 2 ) ) ⋯ g ( f ( a n ) ) ) = ( g ( b 1 ) g ( b 2 ) ⋯ g ( b p ) ) T f
又因为根据 g g g
( g ( b 1 ) g ( b 2 ) ⋯ g ( b p ) ) = ( c 1 c 2 ⋯ c m ) T g \begin{pmatrix} g(\boldsymbol b_1) & g(\boldsymbol b_2) & \cdots & g(\boldsymbol b_p) \end{pmatrix}
= \begin{pmatrix} \boldsymbol c_1 & \boldsymbol c_2 & \cdots & \boldsymbol c_m \end{pmatrix} T_g ( g ( b 1 ) g ( b 2 ) ⋯ g ( b p ) ) = ( c 1 c 2 ⋯ c m ) T g
将其代入上式，得到
( g ( f ( a 1 ) ) g ( f ( a 2 ) ) ⋯ g ( f ( a n ) ) ) = ( c 1 c 2 ⋯ c m ) T g T f \begin{pmatrix} g(f(\boldsymbol a_1)) & g(f(\boldsymbol a_2)) & \cdots & g(f(\boldsymbol a_n)) \end{pmatrix}
= \begin{pmatrix} \boldsymbol c_1 & \boldsymbol c_2 & \cdots & \boldsymbol c_m \end{pmatrix} T_g T_f ( g ( f ( a 1 ) ) g ( f ( a 2 ) ) ⋯ g ( f ( a n ) ) ) = ( c 1 c 2 ⋯ c m ) T g T f
根据矩阵表示的定义，g ∘ f g \circ f g ∘ f A , C \mathscr A, \mathscr C A , C T g ∘ f = T g T f T_{g \circ f} = T_g T_f T g ∘ f = T g T f □ \square □
命题 f : V → W f: V \to W f : V → W A , B \mathscr A, \mathscr B A , B f f f f − 1 : W → V f^{-1}: W \to V f − 1 : W → V B , A \mathscr B, \mathscr A B , A
证明
因为逆映射的定义得到
f ∘ f − 1 = i d W , f − 1 ∘ f = i d V f \circ f^{-1} = \mathrm{id}_W, \quad f^{-1} \circ f = \mathrm{id}_V
f ∘ f − 1 = i d W , f − 1 ∘ f = i d V
所以应用复合映射的结论，得到
T f T f − 1 = E m , T f − 1 T f = E n T_f T_{f^{-1}} = E_m, \quad T_{f^{-1}} T_f = E_n
T f T f − 1 = E m , T f − 1 T f = E n
因此 T f − 1 = T f − 1 T_{f^{-1}} = T_f^{-1} T f − 1 = T f − 1 □ \square □
注意可逆等价于双射，此时同构给出 dim V = dim W \dim V = \dim W dim V = dim W
现在，让我们来讨论一下元素在经由线性映射后，坐标会发生什么样的变化
命题 V , W V, W V , W A = ( a 1 a 2 ⋯ a n ) , B = ( b 1 b 2 ⋯ b m ) \mathscr A = \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n \end{pmatrix}, \quad \mathscr B = \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_m \end{pmatrix} A = ( a 1 a 2 ⋯ a n ) , B = ( b 1 b 2 ⋯ b m ) f : V → W f: V \to W f : V → W A , B \mathscr A, \mathscr B A , B T T T v ∈ V \boldsymbol v \in V v ∈ V
[ f ( v ) ] B = T [ v ] A [f(\boldsymbol v)]_{\mathscr B} = T [\boldsymbol v]_{\mathscr A}
[ f ( v ) ] B = T [ v ] A
证明
由坐标的定义
v = ( a 1 a 2 ⋯ a n ) [ v ] A \boldsymbol v = \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n \end{pmatrix} [\boldsymbol v]_{\mathscr A}
v = ( a 1 a 2 ⋯ a n ) [ v ] A
将等式两边通过线性映射 f f f
f ( v ) = ( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) [ v ] A = ( b 1 b 2 ⋯ b m ) T ⋅ [ v ] A f(\boldsymbol v) = \begin{pmatrix} f(\boldsymbol a_1) & f(\boldsymbol a_2) & \cdots & f(\boldsymbol a_n) \end{pmatrix} [\boldsymbol v]_{\mathscr A}
= \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_m \end{pmatrix} T \cdot [\boldsymbol v]_{\mathscr A} f ( v ) = ( f ( a 1 ) f ( a 2 ) ⋯ f ( a n ) ) [ v ] A = ( b 1 b 2 ⋯ b m ) T ⋅ [ v ] A
同时，f ( v ) f(\boldsymbol v) f ( v ) B \mathscr B B
f ( v ) = ( b 1 b 2 ⋯ b m ) [ f ( v ) ] B f(\boldsymbol v) = \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_m \end{pmatrix} [f(\boldsymbol v)]_{\mathscr B}
f ( v ) = ( b 1 b 2 ⋯ b m ) [ f ( v ) ] B
即得
[ f ( v ) ] B = T [ v ] A [f(\boldsymbol v)]_{\mathscr B} = T [\boldsymbol v]_{\mathscr A}
[ f ( v ) ] B = T [ v ] A
□ \square □
注意由 A \mathscr A A B \mathscr B B T T T
但是由 A \mathscr A A B \mathscr B B T T T
# 基底变换令 V V V
A = ( a 1 a 2 ⋯ a n ) \mathscr A = \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n \end{pmatrix} \quad A = ( a 1 a 2 ⋯ a n ) B = ( b 1 b 2 ⋯ b n ) \mathscr B = \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_n \end{pmatrix} \quad B = ( b 1 b 2 ⋯ b n )
由于各自都是各自的基底，所以可以用 A \mathscr A A B \mathscr B B
b j = k 1 j a 1 + k 2 j a 2 + ⋯ + k n j a n , j = 1 , 2 , … , n \boldsymbol b_j = k_{1j} \boldsymbol a_1 + k_{2j} \boldsymbol a_2 + \cdots + k_{nj} \boldsymbol a_n, \quad j = 1, 2, \ldots, n
b j = k 1 j a 1 + k 2 j a 2 + ⋯ + k n j a n , j = 1 , 2 , … , n
将上述等式整理为矩阵形式
( b 1 b 2 ⋯ b n ) = ( a 1 a 2 ⋯ a n ) ( k 11 k 12 ⋯ k 1 n k 21 k 22 ⋯ k 2 n ⋮ ⋮ ⋱ ⋮ k n 1 k n 2 ⋯ k n n ) \begin{pmatrix}
\boldsymbol b_1 & \boldsymbol b_2 & \cdots & \boldsymbol b_n
\end{pmatrix}
= \begin{pmatrix}
\boldsymbol a_1 & \boldsymbol a_2 & \cdots & \boldsymbol a_n
\end{pmatrix}
\begin{pmatrix}
k_{11} & k_{12} & \cdots & k_{1n} \\
k_{21} & k_{22} & \cdots & k_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
k_{n1} & k_{n2} & \cdots & k_{nn}
\end{pmatrix} ( b 1 b 2 ⋯ b n ) = ( a 1 a 2 ⋯ a n ) ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ k 1 1 k 2 1 ⋮ k n 1 k 1 2 k 2 2 ⋮ k n 2 ⋯ ⋯ ⋱ ⋯ k 1 n k 2 n ⋮ k n n ⎠ ⎟ ⎟ ⎟ ⎟ ⎞
从线性映射的角度来说，这无非是一个矩阵表示，令
P = ( k 11 k 12 ⋯ k 1 n k 21 k 22 ⋯ k 2 n ⋮ ⋮ ⋱ ⋮ k n 1 k n 2 ⋯ k n n ) P = \begin{pmatrix}
k_{11} & k_{12} & \cdots & k_{1n} \\
k_{21} & k_{22} & \cdots & k_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
k_{n1} & k_{n2} & \cdots & k_{nn}
\end{pmatrix} P = ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ k 1 1 k 2 1 ⋮ k n 1 k 1 2 k 2 2 ⋮ k n 2 ⋯ ⋯ ⋱ ⋯ k 1 n k 2 n ⋮ k n n ⎠ ⎟ ⎟ ⎟ ⎟ ⎞
那么由 P P P f P : V → V f_P: V \to V f P : V → V 基底变换 (Change of Basis)「基底変換」 P P P A \mathscr A A B \mathscr B B 过渡矩阵 (Transition Matrix)「変換行列」 P − 1 P^{-1} P − 1
对于任意给出的向量 v ∈ V \boldsymbol v \in V v ∈ V
v = A [ v ] A = B [ v ] B \boldsymbol v = \mathscr A [\boldsymbol v]_{\mathscr A} = \mathscr B [\boldsymbol v]_{\mathscr B}
v = A [ v ] A = B [ v ] B
将 B = A P \mathscr B = \mathscr A P B = A P
v = A P [ v ] B \boldsymbol v = \mathscr A P [\boldsymbol v]_{\mathscr B}
v = A P [ v ] B
所以
[ v ] A = P [ v ] B [\boldsymbol v]_{\mathscr A} = P [\boldsymbol v]_{\mathscr B}
[ v ] A = P [ v ] B
基底变换中的结构是 新基底 = 旧基底 × 过渡矩阵
坐标变换中的结构是 旧坐标 = 过渡矩阵 × 新坐标，这个方向一定要注意
请注意：给出线性映射 f : V → W f: V \to W f : V → W A , B \mathscr A, \mathscr B A , B T T T T T T W ( B ) W(\mathscr B) W ( B ) V ( A ) V(\mathscr A) V ( A ) T T T
V → f W ⟹ V ( A ) ← T W ( B ) V \xrightarrow{f} W \implies V(\mathscr A) \xleftarrow{T} W(\mathscr B)
V f W ⟹ V ( A ) T W ( B )
示例 R 3 \mathbb R^3 R 3 子空间 的两组基底
A = { a 1 = ( 0 4 − 1 ) , a 2 = ( 2 1 1 ) , } , B = { b 1 = ( 4 − 2 3 ) , b 2 = ( 6 − 1 4 ) , } \begin{aligned}
\mathscr A &= \left\{
\boldsymbol a_1 = \begin{pmatrix} 0 \\ 4 \\ -1 \end{pmatrix},
\boldsymbol a_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix},
\right\}, \\
\mathscr B &= \left\{
\boldsymbol b_1 = \begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix},
\boldsymbol b_2 = \begin{pmatrix} 6 \\ -1 \\ 4 \end{pmatrix},
\right\}
\end{aligned} A B = ⎩ ⎪ ⎨ ⎪ ⎧ a 1 = ⎝ ⎛ 0 4 − 1 ⎠ ⎞ , a 2 = ⎝ ⎛ 2 1 1 ⎠ ⎞ , ⎭ ⎪ ⎬ ⎪ ⎫ , = ⎩ ⎪ ⎨ ⎪ ⎧ b 1 = ⎝ ⎛ 4 − 2 3 ⎠ ⎞ , b 2 = ⎝ ⎛ 6 − 1 4 ⎠ ⎞ , ⎭ ⎪ ⎬ ⎪ ⎫
求从基底 A \mathscr A A B \mathscr B B
解
求过渡矩阵，等价于求出 B \mathscr B B A \mathscr A A
( a 1 a 2 ) x = b j , j = 1 , 2 \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 \end{pmatrix}
\boldsymbol x = \boldsymbol b_j, \quad j = 1, 2 ( a 1 a 2 ) x = b j , j = 1 , 2
构造增广矩阵
P ~ : = [ a 1 a 2 b 1 b 2 ] = ( 0 2 4 6 4 1 − 2 − 1 − 1 1 3 4 ) \widetilde P := [ \begin{array}{cc|c|c} \boldsymbol a_1 & \boldsymbol a_2 & \boldsymbol b_1 & \boldsymbol b_2 \end{array} ]
= \begin{pmatrix}
0 & 2 & 4 & 6 \\
4 & 1 & -2 & -1 \\
-1 & 1 & 3 & 4
\end{pmatrix} P : = [ a 1 a 2 b 1 b 2 ] = ⎝ ⎛ 0 4 − 1 2 1 1 4 − 2 3 6 − 1 4 ⎠ ⎞
行化简后得到
→ R r e f ( 1 0 − 1 − 1 0 1 2 3 0 0 0 0 ) \xrightarrow{Rref}
\begin{pmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0
\end{pmatrix} R r e f ⎝ ⎛ 1 0 0 0 1 0 − 1 2 0 − 1 3 0 ⎠ ⎞
因此解得
{ b 1 = − 1 a 1 + 2 a 2 b 2 = − 1 a 1 + 3 a 2 \begin{cases}
\boldsymbol b_1 = -1 \boldsymbol a_1 + 2 \boldsymbol a_2 \\
\boldsymbol b_2 = -1 \boldsymbol a_1 + 3 \boldsymbol a_2
\end{cases} { b 1 = − 1 a 1 + 2 a 2 b 2 = − 1 a 1 + 3 a 2
这等价于（一定要注意转置）
( b 1 b 2 ) = ( a 1 a 2 ) ( − 1 − 1 2 3 ) \begin{pmatrix} \boldsymbol b_1 & \boldsymbol b_2 \end{pmatrix}
= \begin{pmatrix} \boldsymbol a_1 & \boldsymbol a_2 \end{pmatrix}
\begin{pmatrix}
-1 & -1 \\
2 & 3
\end{pmatrix} ( b 1 b 2 ) = ( a 1 a 2 ) ( − 1 2 − 1 3 )
所以从基底 A \mathscr A A B \mathscr B B
P = ( − 1 − 1 2 3 ) P = \begin{pmatrix}
-1 & -1 \\
2 & 3
\end{pmatrix} P = ( − 1 2 − 1 3 )
□ \square □
现在让我们分析一个稍微复杂一些的情况V V V
A = { v 1 , v 2 , … , v n } \mathscr A = \{\boldsymbol v_1, \boldsymbol v_2, \ldots, \boldsymbol v_n\} \quad A = { v 1 , v 2 , … , v n } A ′ = { v 1 ′ , v 2 ′ , … , v n ′ } \mathscr A' = \{\boldsymbol v_1', \boldsymbol v_2', \ldots, \boldsymbol v_n'\} \quad A ′ = { v 1 ′ , v 2 ′ , … , v n ′ }
在 W W W
B = { w 1 , w 2 , … , w m } \mathscr B = \{\boldsymbol w_1, \boldsymbol w_2, \ldots, \boldsymbol w_m\} \quad B = { w 1 , w 2 , … , w m } B ′ = { w 1 ′ , w 2 ′ , … , w m ′ } \mathscr B' = \{\boldsymbol w_1', \boldsymbol w_2', \ldots, \boldsymbol w_m'\} \quad B ′ = { w 1 ′ , w 2 ′ , … , w m ′ }
分别令
P P P A \mathscr A A A ′ \mathscr A' A ′ Q Q Q B \mathscr B B B ′ \mathscr B' B ′
取线性映射 T : V → W T: V \to W T : V → W T T T A , B \mathscr A, \mathscr B A , B
V ( A ) ← T W ( B ) ↓ P ↓ Q V ( A ′ ) ← ? W ( B ′ ) \begin{array}{ccc}
V(\mathscr A) & \xleftarrow{T} & W(\mathscr B) \\
\downarrow P & & \downarrow Q \\
V(\mathscr A') & \xleftarrow{?} & W(\mathscr B')
\end{array} V ( A ) ↓ P V ( A ′ ) T ? W ( B ) ↓ Q W ( B ′ )
问题：? ? ?
假设 X X X ? ? ? A ′ \mathscr A' A ′ B ′ \mathscr B' B ′
( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( w 1 ′ , w 2 ′ , … , w m ′ ) X (T(\boldsymbol v_1'), T(\boldsymbol v_2'), \ldots, T(\boldsymbol v_n'))
= (\boldsymbol w_1', \boldsymbol w_2', \ldots, \boldsymbol w_m') X ( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( w 1 ′ , w 2 ′ , … , w m ′ ) X
整理以下已经有的关系式
( v 1 ′ , v 2 ′ , … , v n ′ ) = ( v 1 , v 2 , … , v n ) P ( w 1 ′ , w 2 ′ , … , w m ′ ) = ( w 1 , w 2 , … , w m ) Q ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) = ( w 1 , w 2 , … , w m ) T \begin{aligned}
(\boldsymbol v_1', \boldsymbol v_2', \ldots, \boldsymbol v_n') &= (\boldsymbol v_1, \boldsymbol v_2, \ldots, \boldsymbol v_n) P \\
(\boldsymbol w_1', \boldsymbol w_2', \ldots, \boldsymbol w_m') &= (\boldsymbol w_1, \boldsymbol w_2, \ldots, \boldsymbol w_m) Q \\
(T(\boldsymbol v_1), T(\boldsymbol v_2), \ldots, T(\boldsymbol v_n)) &= (\boldsymbol w_1, \boldsymbol w_2, \ldots, \boldsymbol w_m) T
\end{aligned} ( v 1 ′ , v 2 ′ , … , v n ′ ) ( w 1 ′ , w 2 ′ , … , w m ′ ) ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) = ( v 1 , v 2 , … , v n ) P = ( w 1 , w 2 , … , w m ) Q = ( w 1 , w 2 , … , w m ) T
由于将线性映射同时作用于线性变换的两边，可以得到
( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) P (T(\boldsymbol v_1'), T(\boldsymbol v_2'), \ldots, T(\boldsymbol v_n'))
= (T(\boldsymbol v_1), T(\boldsymbol v_2), \ldots, T(\boldsymbol v_n)) P ( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) P
注意关系式 ( A B ) − 1 = B − 1 A − 1 (AB)^{-1} = B^{-1} A^{-1} ( A B ) − 1 = B − 1 A − 1
( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) P = ( w 1 , w 2 , … , w m ) T P = ( w 1 ′ , w 2 ′ , … , w m ′ ) Q − 1 T P \begin{aligned}
(T(\boldsymbol v_1'), T(\boldsymbol v_2'), \ldots, T(\boldsymbol v_n'))
&= (T(\boldsymbol v_1), T(\boldsymbol v_2), \ldots, T(\boldsymbol v_n)) P \\
&= (\boldsymbol w_1, \boldsymbol w_2, \ldots, \boldsymbol w_m) T P \\
&= (\boldsymbol w_1', \boldsymbol w_2', \ldots, \boldsymbol w_m') Q^{-1} T P
\end{aligned} ( T ( v 1 ′ ) , T ( v 2 ′ ) , … , T ( v n ′ ) ) = ( T ( v 1 ) , T ( v 2 ) , … , T ( v n ) ) P = ( w 1 , w 2 , … , w m ) T P = ( w 1 ′ , w 2 ′ , … , w m ′ ) Q − 1 T P
因此可以得到
X = Q − 1 T P X = Q^{-1} T P
X = Q − 1 T P
命题 f : V → W f: V \to W f : V → W V V V A , A ′ \mathscr A, \mathscr A' A , A ′ W W W B , B ′ \mathscr B, \mathscr B' B , B ′ P P P A \mathscr A A A ′ \mathscr A' A ′ Q Q Q B \mathscr B B B ′ \mathscr B' B ′ f f f A , B \mathscr A, \mathscr B A , B T T T f f f A ′ , B ′ \mathscr A', \mathscr B' A ′ , B ′
T ′ = Q − 1 T P T' = Q^{-1} T P
T ′ = Q − 1 T P
证明
