Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
§ 8。决定因素
1. AN 自同态的决定因素
设 M M M A A A n n n u u u M M M A A A ⋀ n ( M ) \bigwedge^n(M) ⋀ n ( M ) A A A ⋀ n ( u ) \bigwedge^n(u) ⋀ n ( u ) λ ∈ A \lambda \in A λ ∈ A z ↦ λ z z \mapsto \lambda z z ↦ λ z
通过 § 7, no. 2 的公式 (4)，det u \det u det u
(1)
u ( x 1 ) ∧ u ( x 2 ) ∧ ⋯ ∧ u ( x n ) = ( det u ) ⋅ x 1 ∧ x 2 ∧ ⋯ ∧ x n
u(x_1) \wedge u(x_2) \wedge \cdots \wedge u(x_n) = (\det u) \cdot x_1 \wedge x_2 \wedge \cdots \wedge x_n
u ( x 1 ) ∧ u ( x 2 ) ∧ ⋯ ∧ u ( x n ) = ( det u ) ⋅ x 1 ∧ x 2 ∧ ⋯ ∧ x n ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n n n n M M M det ( u ) = 1 \det(u) = 1 det ( u ) = 1 u u u
(2)
det ( u ∘ v ) = ( det u ) ( det v ) .
\det(u \circ v) = (\det u)(\det v).
det ( u ∘ v ) = ( det u ) ( det v ) . (ii) det ( 1 M ) = 1 \det(1_M) = 1 det ( 1 M ) = 1 对于 M M M det u \det u det u 中的每个自同构 u u u 可逆于 A A A 和
(3)
det ( u − 1 ) = ( det u ) − 1 .
\det(u^{-1}) = (\det u)^{-1}.
det ( u − 1 ) = ( det u ) − 1 . 如果 n n n M M M ⋀ n ( u ∘ v ) = ( ⋀ n ( u ) ) ∘ ( ⋀ n ( v ) ) \bigwedge^n(u \circ v) = (\bigwedge^n(u)) \circ (\bigwedge^n(v)) ⋀ n ( u ∘ v ) = ( ⋀ n ( u )) ∘ ( ⋀ n ( v ))
设 M M M A A A ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n n n n ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n M M M 相对于 给定的基 ( e i ) (e_i) ( e i ) det ( x 1 , x 2 , … , x n ) \det(x_1, x_2, \ldots, x_n) det ( x 1 , x 2 , … , x n ) u u u M M M u ( e i ) = x i u(e_i) = x_i u ( e i ) = x i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
x 1 ∧ x 2 ∧ ⋯ ∧ x n = det ( x 1 , x 2 , … , x n ) e 1 ∧ e 2 ∧ ⋯ ∧ e n
x_1 \wedge x_2 \wedge \cdots \wedge x_n = \det(x_1, x_2, \ldots, x_n) \ e_1 \wedge e_2 \wedge \cdots \wedge e_n
x 1 ∧ x 2 ∧ ⋯ ∧ x n = det ( x 1 , x 2 , … , x n ) e 1 ∧ e 2 ∧ ⋯ ∧ e n 这个关系将M n M^n M n ( x i ) ↦ det ( x 1 , x 2 , … , x n ) (x_i) \mapsto \det(x_1, x_2, \ldots, x_n) ( x i ) ↦ det ( x 1 , x 2 , … , x n ) A A A 交错的 n 线性形式 。由于，凭借§ 7, no. 4，命题 7，交错的n n n A A A ⋀ n ( M ) \bigwedge^n(M) ⋀ n ( M ) ⋀ n ( M ) \bigwedge^n(M) ⋀ n ( M ) A A A M n M^n M n
( x 1 , … , x n ) ↦ α det ( x 1 , x 2 , … , x n ) (x_1, \ldots, x_n) \mapsto \alpha \det(x_1, x_2, \ldots, x_n) ( x 1 , … , x n ) ↦ α det ( x 1 , x 2 , … , x n ) 对于一些α ∈ A \alpha \in A α ∈ A
设 M M M A A A ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n v v v M M M n n n M M M ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n
det ( v ( x 1 ) , … , v ( x n ) ) = ( det v ) det ( x 1 , … , x n ) .
\det(v(x_1), \ldots, v(x_n)) = (\det v) \det(x_1, \ldots, x_n).
det ( v ( x 1 ) , … , v ( x n )) = ( det v ) det ( x 1 , … , x n ) . 如果 u u u M M M u ( e i ) = x i u(e_i) = x_i u ( e i ) = x i i i i
v ( x i ) = ( v ∘ u ) ( e i )
v(x_i) = (v \circ u)(e_i)
v ( x i ) = ( v ∘ u ) ( e i ) 因此，(5) 由(2) 得出。
2. 有限维自由模的刻画自同构
令M M M A A A u u u M M M u u u u u u u u u u u u det u \det u det u A A A
引用了 5 · 被以下引用 5 引用了 推论 1, II § 1 , 命题 15 , 推论 2, II § 1 , 命题 3, III § 7 , 定理 2, III § 7
被以下引用 § 8. 。决定因素 , 注 , 命题 11 , 命题 3 , 命题 3, III § 9
设( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n M M M x i = u ( e i ) x_i = u(e_i) x i = u ( e i ) 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det u ) e 1 ∧ e 2 ∧ ⋯ ∧ e n .
x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det u) e_1 \wedge e_2 \wedge \cdots \wedge e_n.
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det u ) e 1 ∧ e 2 ∧ ⋯ ∧ e n . 由§ 7, no. 9，定理 2 M M M x i x_i x i det u \det u det u A A A M M M v v v v ∘ u = 1 M v \circ u = 1_M v ∘ u = 1 M u ∘ v = 1 M u \circ v = 1_M u ∘ v = 1 M ( det v ) ( det u ) = 1 (\det v)(\det u) = 1 ( det v ) ( det u ) = 1 det u \det u det u A A A u u u ⋀ n ( u ) \bigwedge^n(u) ⋀ n ( u ) A A A u u u u u u
与定理 1 的证明相同的记号，对于u u u x i x_i x i λ x 1 ∧ x 2 ∧ ⋯ ∧ x n = 0 \lambda x_1 \wedge x_2 \wedge \cdots \wedge x_n = 0 λ x 1 ∧ x 2 ∧ ⋯ ∧ x n = 0 λ ∈ A \lambda \in A λ ∈ A λ = 0 \lambda = 0 λ = 0 λ ( det u ) = 0 \lambda (\det u) = 0 λ ( det u ) = 0 e 1 ∧ ⋯ ∧ e n e_1 \wedge \cdots \wedge e_n e 1 ∧ ⋯ ∧ e n Λ n ( M ) \Lambda^n(M) Λ n ( M )
3. 平方矩阵的行列式
令I I I ( I , I ) (I, I) ( I , I ) A A A A A A X X X A A A A I A^I A I u u u A I A^I A I X X X X X X det X \det X det X
引用了 1 引用了 II § 10. 。 矩阵
如果X = ( ξ i j ) ( i , j ) ∈ I × I X = (\xi_{ij})_{(i,j) \in I \times I} X = ( ξ ij ) ( i , j ) ∈ I × I ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I A I A^I A I u u u
u ( e i ) = ∑ j ∈ J ξ j i e j .
u(e_i) = \sum_{j \in J} \xi_{ji} e_j.
u ( e i ) = j ∈ J ∑ ξ j i e j . 当I = { 1 , n } ⊂ N I = \{1, n\} \subset \mathbf{N} I = { 1 , n } ⊂ N i ∈ I i \in I i ∈ I x i = u ( e i ) x_i = u(e_i) x i = u ( e i ) X X X
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det X ) e 1 ∧ e 2 ∧ ⋯ ∧ e n
x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det X) e_1 \wedge e_2 \wedge \cdots \wedge e_n
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det X ) e 1 ∧ e 2 ∧ ⋯ ∧ e n det X \det X det X A n A^n A n det ( x 1 , x 2 , … , x n ) \det(x_1, x_2, \ldots, x_n) det ( x 1 , x 2 , … , x n )
对于 A n A^n A n n n n x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n X ( x 1 , … , x n ) X(x_1, \ldots, x_n) X ( x 1 , … , x n ) n n n i i i x i x_i x i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
( x 1 , … , x n ) ↦ det ( X ( x 1 , … , x n ) )
(x_1, \ldots, x_n) \mapsto \det(X(x_1, \ldots, x_n))
( x 1 , … , x n ) ↦ det ( X ( x 1 , … , x n )) *
( A n ) n (A^n)^n ( A n ) n 到
A A A 的交错的和
n n n 是线性的。
特别地，其中两个列相等的矩阵的行列式是零。如果对矩阵的列执行置换，则新矩阵的行列式等于旧矩阵的行列式乘以ε σ \varepsilon_{\sigma} ε σ
更一般地，令M M M n n n A A A ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I M M M M M M u u u ( e i ) (e_i) ( e i ) X X X u u u
det ( u ) = det ( X )
\det(u) = \det(X)
det ( u ) = det ( X ) 从定义开始如下。
当I = { 1 , n } I = \{1, n\} I = { 1 , n } X X X
det ( ξ i j ) 1 ≤ i ≤ n , 1 ≤ j ≤ n
\det(\xi_{ij})_{1 \leq i \leq n, 1 \leq j \leq n}
det ( ξ ij ) 1 ≤ i ≤ n , 1 ≤ j ≤ n 或者简单地 det ( ξ i j ) \det(\xi_{ij}) det ( ξ ij )
∣ ξ 11 ξ 12 … ξ 1 n ξ 21 ξ 22 … ξ 2 n … … … … ξ n 1 ξ n 2 … ξ n n ∣
\begin{vmatrix}
\xi_{11} & \xi_{12} & \ldots & \xi_{1n} \\
\xi_{21} & \xi_{22} & \ldots & \xi_{2n} \\
\ldots & \ldots & \ldots & \ldots \\
\xi_{n1} & \xi_{n2} & \ldots & \xi_{nn}
\end{vmatrix}
ξ 11 ξ 21 … ξ n 1 ξ 12 ξ 22 … ξ n 2 … … … … ξ 1 n ξ 2 n … ξ nn 当X = 1 X = 1 X = 1 X X X
(1)空矩阵的行列式等于1； 阶 1 的平方矩阵的行列式等于这个矩阵的唯一元素。对于阶 2 中的矩阵
( ξ 11 ξ 12 ξ 21 ξ 22 )
\begin{pmatrix}
\xi_{11} & \xi_{12} \\
\xi_{21} & \xi_{22}
\end{pmatrix}
( ξ 11 ξ 21 ξ 12 ξ 22 ) 那么，在上面的记号中，
x 1 ∧ x 2 = ( ξ 11 e 1 + ξ 21 e 2 ) ∧ ( ξ 12 e 1 + ξ 22 e 2 ) = ξ 11 ξ 22 e 1 ′ ∧ e 2 + ξ 21 ξ 12 e 2 ∧ e 1
x_1 \wedge x_2 = (\xi_{11} e_1 + \xi_{21} e_2) \wedge (\xi_{12} e_1 + \xi_{22} e_2) = \xi_{11} \xi_{22} e_1' \wedge e_2 + \xi_{21} \xi_{12} e_2 \wedge e_1
x 1 ∧ x 2 = ( ξ 11 e 1 + ξ 21 e 2 ) ∧ ( ξ 12 e 1 + ξ 22 e 2 ) = ξ 11 ξ 22 e 1 ′ ∧ e 2 + ξ 21 ξ 12 e 2 ∧ e 1 whence
∣ ξ 11 ξ 12 ξ 21 ξ 22 ∣ = ξ 11 ξ 22 − ξ 12 ξ 21 .
\begin{vmatrix}
\xi_{11} & \xi_{12} \\
\xi_{21} & \xi_{22}
\end{vmatrix} = \xi_{11} \xi_{22} - \xi_{12} \xi_{21}.
ξ 11 ξ 21 ξ 12 ξ 22 = ξ 11 ξ 22 − ξ 12 ξ 21 . 我们将 nos 的一些结果翻译成矩阵的语言。 1 和 2：
如果X X X Y Y Y A A A
引用了 1 · 被以下引用 1 引用了 命题 1
被以下引用 § 8. 。决定因素
det ( X Y ) = ( det X ) ( det Y ) .
\det(XY) = (\det X)(\det Y).
det ( X Y ) = ( det X ) ( det Y ) . 为了使 X 可逆，det X 是 A 的可逆元是充分必要的，然后
(9)
det ( X − 1 ) = ( det X ) − 1 .
\det(X^{-1}) = (\det X)^{-1}.
det ( X − 1 ) = ( det X ) − 1 . 这紧接着no. 1、命题 1 和no. 2、定理 1。
如果 P 是可逆的平方矩阵，则 det ( P X P − 1 ) = det X \det(PXP^{-1}) = \det X det ( P X P − 1 ) = det X
For the columns of a square matrix X of finite order to be linearly independent, it is necessary and sufficient that det X be not a divisor of zero in A.
引用了 1 · 被以下引用 1 引用了 命题 3
被以下引用 推论 2
4. 行列式的计算
设 A 为交换环，M 自由A-模以及基 ( e j ) j ∈ J (e_j)_{j \in J} ( e j ) j ∈ J p ≤ C a r d ( J ) p \leq \mathrm{Card}(J) p ≤ Card ( J ) f : M p → N f : M^p \to N f : M p → N ( 1 ≤ i ≤ p ) (1 \leq i \leq p) ( 1 ≤ i ≤ p ) x i = ∑ j ∈ J ξ j i e j x_i = \sum_{j \in J} \xi_{ji} e_j x i = ∑ j ∈ J ξ j i e j
f ( x 1 , x 2 , … , x p ) = ∑ j 1 < j 2 < ⋯ < j p ( ∑ σ ∈ S p ε σ ⋅ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( p ) , p ) f ( e j 1 , … , e j p )
f(x_1, x_2, \ldots, x_p)
= \sum_{j_1 < j_2 < \cdots < j_p} \left( \sum_{\sigma \in S_p} \varepsilon_\sigma \cdot \xi_{j_{\sigma(1)}, 1} \xi_{j_{\sigma(2)}, 2} \cdots \xi_{j_{\sigma(p)}, p} \right) f(e_{j_1}, \ldots, e_{j_p})
f ( x 1 , x 2 , … , x p ) = j 1 < j 2 < ⋯ < j p ∑ σ ∈ S p ∑ ε σ ⋅ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( p ) , p f ( e j 1 , … , e j p ) 其中* ( j k ) 1 ≤ k ≤ p (j_k)_{1 \leq k \leq p} ( j k ) 1 ≤ k ≤ p
引用了 1 引用了 推论 1, III § 7
现在
f ( x 1 , … , x p ) = ∑ ( j k ) ξ j 1 , 1 ξ j 2 , 2 ⋯ ξ j p , p f ( e j 1 , e j 2 , … , e j p )
f(x_1, \ldots, x_p) = \sum_{(j_k)} \xi_{j_1, 1} \xi_{j_2, 2} \cdots \xi_{j_p, p} f(e_{j_1}, e_{j_2}, \ldots, e_{j_p})
f ( x 1 , … , x p ) = ( j k ) ∑ ξ j 1 , 1 ξ j 2 , 2 ⋯ ξ j p , p f ( e j 1 , e j 2 , … , e j p ) 其中 ( j k ) 1 ≤ k ≤ p (j_k)_{1 \leq k \leq p} ( j k ) 1 ≤ k ≤ p
特别地，如果 J 是有限的并且有 n 元素和 x i = ∑ j ∈ J ξ j i e j x_i = \sum_{j \in J} \xi_{ji} e_j x i = ∑ j ∈ J ξ j i e j ( 1 ≤ i ≤ n ) (1 \leq i \leq n) ( 1 ≤ i ≤ n )
(11)
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( ∑ σ ∈ S n ε σ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( n ) , n ) e j 1 ∧ e j 2 ∧ ⋯ ∧ e j n
x_1 \wedge x_2 \wedge \cdots \wedge x_n
= \left( \sum_{\sigma \in S_n} \varepsilon_\sigma \xi_{j_{\sigma(1)}, 1} \xi_{j_{\sigma(2)}, 2} \cdots \xi_{j_{\sigma(n)}, n} \right) e_{j_1} \wedge e_{j_2} \wedge \cdots \wedge e_{j_n}
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( σ ∈ S n ∑ ε σ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( n ) , n ) e j 1 ∧ e j 2 ∧ ⋯ ∧ e j n 其中( j k ) 1 ≤ k ≤ n (j_k)_{1 \leq k \leq n} ( j k ) 1 ≤ k ≤ n
(12)
det ( x 1 , x 2 , … , x n ) = ∑ σ ∈ S ε σ ⋅ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( n ) , n .
\det(x_1, x_2, \ldots, x_n) = \sum_{\sigma \in S} \varepsilon_\sigma \cdot \xi_{j_{\sigma(1)}, 1} \xi_{j_{\sigma(2)}, 2} \cdots \xi_{j_{\sigma(n)}, n}.
det ( x 1 , x 2 , … , x n ) = σ ∈ S ∑ ε σ ⋅ ξ j σ ( 1 ) , 1 ξ j σ ( 2 ) , 2 ⋯ ξ j σ ( n ) , n . 当记号为引理 1 时，比较公式 (10) 和 (12) 得出
x 1 ∧ x 2 ∧ ⋯ ∧ x p = ∑ H ∈ F p ( J ) det ( x H , 1 , x H , 2 , … , x H , p ) e H
x_1 \wedge x_2 \wedge \cdots \wedge x_p = \sum_{H \in \mathscr{F}_p(J)} \det(x_{H,1}, x_{H,2}, \ldots, x_{H,p}) e_H
x 1 ∧ x 2 ∧ ⋯ ∧ x p = H ∈ F p ( J ) ∑ det ( x H , 1 , x H , 2 , … , x H , p ) e H 其中F p ( J ) \mathscr{F}_p(J) F p ( J ) J J J p p p H ∈ F p ( J ) H \in \mathscr{F}_p(J) H ∈ F p ( J ) x H , i = ∑ j ∈ H ξ j i e j x_{H,i} = \sum_{j \in H} \xi_{ji} e_j x H , i = ∑ j ∈ H ξ j i e j e H = e j 1 ∧ e j 2 ∧ ⋯ ∧ e j p e_H = e_{j_1} \wedge e_{j_2} \wedge \cdots \wedge e_{j_p} e H = e j 1 ∧ e j 2 ∧ ⋯ ∧ e j p ( j k ) 1 ≤ k ≤ p (j_k)_{1 \leq k \leq p} ( j k ) 1 ≤ k ≤ p H H H det ( x H , 1 , … , x H , p ) \det(x_{H,1}, \ldots, x_{H,p}) det ( x H , 1 , … , x H , p ) ( e j k ) 1 ≤ k ≤ p (e_{jk})_{1 \leq k \leq p} ( e j k ) 1 ≤ k ≤ p
det X = ∑ σ ∈ S I ε σ ( ∏ i ∈ I ξ σ ( i ) , i )
\det X = \sum_{\sigma \in S_I} \varepsilon_\sigma \left( \prod_{i \in I} \xi_{\sigma(i), i} \right)
det X = σ ∈ S I ∑ ε σ ( i ∈ I ∏ ξ σ ( i ) , i ) 其中σ \sigma σ I I I S I S_I S I ε σ \varepsilon_\sigma ε σ σ \sigma σ （I, § 5, no. 7）的签名。
注意可能仅限于I = { 1 , n } ⊂ N I = \{1, n\} \subset \mathbf{N} I = { 1 , n } ⊂ N ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n A n A^n A n x i x_i x i X X X
特别是，对于阶 3 的矩阵行列式
X = ( ξ 11 ξ 12 ξ 13 ξ 21 ξ 22 ξ 23 ξ 31 ξ 32 ξ 33 )
X = \begin{pmatrix}
\xi_{11} & \xi_{12} & \xi_{13} \\
\xi_{21} & \xi_{22} & \xi_{23} \\
\xi_{31} & \xi_{32} & \xi_{33}
\end{pmatrix}
X = ξ 11 ξ 21 ξ 31 ξ 12 ξ 22 ξ 32 ξ 13 ξ 23 ξ 33 我们有
det ( X ) = ξ 11 ξ 22 ξ 33 + ξ 12 ξ 23 ξ 31 + ξ 21 ξ 32 ξ 13 − ξ 13 ξ 22 ξ 31 − ξ 12 ξ 21 ξ 33 − ξ 11 ξ 23 ξ 32 .
\det(X) = \xi_{11} \xi_{22} \xi_{33} + \xi_{12} \xi_{23} \xi_{31} + \xi_{21} \xi_{32} \xi_{13} - \xi_{13} \xi_{22} \xi_{31} - \xi_{12} \xi_{21} \xi_{33} - \xi_{11} \xi_{23} \xi_{32}.
det ( X ) = ξ 11 ξ 22 ξ 33 + ξ 12 ξ 23 ξ 31 + ξ 21 ξ 32 ξ 13 − ξ 13 ξ 22 ξ 31 − ξ 12 ξ 21 ξ 33 − ξ 11 ξ 23 ξ 32 .
假设X X X σ , τ \sigma, \tau σ , τ S I S_I S I
∏ i ∈ I ξ σ ( i ) , i = ∏ j ∈ I ξ σ ( τ ( j ) ) , τ ( j ) .
\prod_{i \in I} \xi_{\sigma(i), i} = \prod_{j \in I} \xi_{\sigma(\tau(j)), \tau(j)}.
i ∈ I ∏ ξ σ ( i ) , i = j ∈ I ∏ ξ σ ( τ ( j )) , τ ( j ) . 特别是取τ = σ − 1 \tau = \sigma^{-1} τ = σ − 1 ε σ − 1 = ε σ \varepsilon_{\sigma^{-1}} = \varepsilon_\sigma ε σ − 1 = ε σ
det X = ∑ σ ∈ S I ε σ ( ∏ i ∈ I , i ξ i , σ ( i ) ) ,
\det X = \sum_{\sigma \in S_I} \varepsilon_\sigma \left( \prod_{i \in I, i} \xi_{i, \sigma(i)} \right),
det X = σ ∈ S I ∑ ε σ i ∈ I , i ∏ ξ i , σ ( i ) , 这就证明了命题。
对于 A n \mathbf{A}^n A n x 1 , … , x n x_1, \ldots, x_n x 1 , … , x n Y ( x 1 , … , x n ) Y(x_1, \ldots, x_n) Y ( x 1 , … , x n ) x i x_i x i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
( x 1 , … , x n ) ↦ det ( Y ( x 1 , … , x n ) )
(x_1, \ldots, x_n) \mapsto \det(Y(x_1, \ldots, x_n))
( x 1 , … , x n ) ↦ det ( Y ( x 1 , … , x n )) 从
( A n ) n (\mathbf{A}^n)^n ( A n ) n 到
A \mathbf{A} A 是交错的并且是 n 线性的。
被以下引用 2 被以下引用 § 8. 。决定因素 , 命题 11
对于平方矩阵 X X X A \mathbf{A} A X X X X X X det X \det X det X A \mathbf{A} A
引用了 2 引用了 命题 6 , 命题 8
设 u u u A \mathbf{A} A M M M M ∗ M^* M ∗ t u ^t u t u
det ( t u ) = det ( u ) .
\det(^t u) = \det(u).
det ( t u ) = det ( u ) . 如果
X X X 相对于
M M M 的基是
u u u 的矩阵，则
t X ^t X t X 相对于对偶基是
t u ^t u t u 的矩阵（II, § 10, no. 4， 命题3);如
det ( u ) = det ( X ) \det(u) = \det(X) det ( u ) = det ( X ) 和
det ( t u ) = det ( t X ) \det(^t u) = \det(^t X) det ( t u ) = det ( t X ) ，由命题8得出结论。
引用了 3 · 被以下引用 1 引用了 定义 5, II § 2 , 命题 3, II § 10 , 命题 8
被以下引用 推论 3, VIII § 17
5. 矩阵的未成年人
令X X X ( ξ i j ) ( i , j ) ∈ I × J (\xi_{ij})_{(i,j) \in I \times J} ( ξ ij ) ( i , j ) ∈ I × J H ⊂ I H \subset I H ⊂ I K ⊂ J K \subset J K ⊂ J p p p ϕ : H → K \phi : H \to K ϕ : H → K X H , K X_{H,K} X H , K ( ξ i , ϕ ( j ) ) ( i , j ) ∈ H × H (\xi_{i,\phi(j)})_{(i,j) \in H \times H} ( ξ i , ϕ ( j ) ) ( i , j ) ∈ H × H X X X A \mathbf{A} A det ( X H , K ) \det(X_{H,K}) det ( X H , K ) X X X p p p p p p X X X
令 M M M ( e i ) i ∈ J (e_i)_{i \in J} ( e i ) i ∈ J A \mathbf{A} A J J J p > 0 p > 0 p > 0 ( e H ) H ∈ S p ( J ) (e_H)_{H \in \mathfrak{S}_p(J)} ( e H ) H ∈ S p ( J ) ∧ p ( M ) \wedge^p(M) ∧ p ( M ) ( x i ) 1 ≤ i ≤ p (x_i)_{1 \leq i \leq p} ( x i ) 1 ≤ i ≤ p p p p M M M
x i = ∑ j ∈ J ξ j i e j for i ∈ I = { 1 , p }
x_i = \sum_{j \in J} \xi_{ji} e_j \quad \text{for } i \in I = \{1, p\}
x i = j ∈ J ∑ ξ j i e j for i ∈ I = { 1 , p } 引用了 1 · 被以下引用 2 引用了 III § 7. 。外观代数
被以下引用 命题 10 , 命题 4, VII § 4
并令 X X X ( J , I ) (J, I) ( J , I ) ( ξ j i ) (\xi_{ji}) ( ξ j i )
x 1 ∧ x 2 ∧ ⋯ ∧ x p = ∑ H ∈ F p ( J ) ( det X H , I ) e H .
x_1 \wedge x_2 \wedge \cdots \wedge x_p = \sum_{H \in \mathfrak{F}_p(J)} (\det X_{H,I}) e_H.
x 1 ∧ x 2 ∧ ⋯ ∧ x p = H ∈ F p ( J ) ∑ ( det X H , I ) e H . 其中H H H J J J F p ( J ) \mathfrak{F}_p(J) F p ( J ) p p p
这紧接着是no. 4的公式(12)和no. 3的公式(6)。
令M M M N N N u u u m m m n n n u : M → N u : M \to N u : M → N X X X A A A M M M ( e i ) 1 ≤ i ≤ m (e_i)_{1 \leq i \leq m} ( e i ) 1 ≤ i ≤ m N N N ( f j ) 1 ≤ j ≤ n (f_j)_{1 \leq j \leq n} ( f j ) 1 ≤ j ≤ n p ≤ inf ( m , n ) p \leq \inf(m, n) p ≤ inf ( m , n ) ∧ p ( u ) \wedge^p(u) ∧ p ( u ) ∧ p ( M ) \wedge^p(M) ∧ p ( M ) ( e K ) K ∈ F p ( I ) (e_K)_{K \in \mathfrak{F}_p(I)} ( e K ) K ∈ F p ( I ) ∧ p ( N ) \wedge^p(N) ∧ p ( N ) ( f H ) H ∈ F p ( J ) (f_H)_{H \in \mathfrak{F}_p(J)} ( f H ) H ∈ F p ( J ) I = { 1 , m } I = \{1, m\} I = { 1 , m } J = { 1 , n } J = \{1, n\} J = { 1 , n } ( F p ( J ) , F p ( I ) ) (\mathfrak{F}_p(J), \mathfrak{F}_p(I)) ( F p ( J ) , F p ( I )) ( det ( X H , K ) ) (\det(X_{H,K})) ( det ( X H , K )) ( n p ) \binom{n}{p} ( p n ) ( m p ) \binom{m}{p} ( p m )
引用了 2 · 被以下引用 3 引用了 III § 7. 。外观代数 , 命题 9
被以下引用 § 8. 。决定因素 , 命题 11 , VIII § 17. 。 简化范数与迹
对于子集K ⊂ J K \subset J K ⊂ J p p p ( j k ) 1 ≤ k ≤ p (j_k)_{1 \leq k \leq p} ( j k ) 1 ≤ k ≤ p K K K ∧ p ( u ) \wedge^p(u) ∧ p ( u )
∧ p ( u ) ( e K ) = u ( e j 1 ) ∧ u ( e j 2 ) ∧ ⋯ ∧ u ( e j p ) .
\wedge^p(u)(e_K) = u(e_{j_1}) \wedge u(e_{j_2}) \wedge \cdots \wedge u(e_{j_p}).
∧ p ( u ) ( e K ) = u ( e j 1 ) ∧ u ( e j 2 ) ∧ ⋯ ∧ u ( e j p ) . 因此，位于指标H H H ∧ p ( u ) \wedge^p(u) ∧ p ( u ) K K K H H H u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) u(e_{j_1}) \wedge \cdots \wedge u(e_{j_p}) u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) det ( X H , K ) \det(X_{H,K}) det ( X H , K )
矩阵( det ( X H , K ) ) (\det(X_{H,K})) ( det ( X H , K )) X X X p p p ∧ p ( X ) \wedge^p(X) ∧ p ( X ) p = m = n p = m = n p = m = n ∧ n ( X ) \wedge^n(X) ∧ n ( X ) det ( X ) \det(X) det ( X )
令M M M n n n A A A M M M u u u A A A ξ , η \xi, \eta ξ , η
det ( ξ 1 M + η u ) = ∑ k ≥ 0 Tr ( ∧ k ( u ) ) ξ n − k η k .
\det(\xi 1_M + \eta u) = \sum_{k \geq 0} \operatorname{Tr}(\wedge^k(u)) \xi^{n-k} \eta^k.
det ( ξ 1 M + η u ) = k ≥ 0 ∑ Tr ( ∧ k ( u )) ξ n − k η k . 引用了 3 · 被以下引用 2 引用了 推论 1 , 定理 1 , 命题 10
被以下引用 § 8. 。决定因素 , 推论
令( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n M M M I = { 1 , n } I = \{1, n\} I = { 1 , n }
( ξ e 1 + η u ( e 1 ) ) ∧ ( ξ e 2 + η u ( e 2 ) ) ∧ ⋯ ∧ ( ξ e n + η u ( e n ) )
(\xi e_1 + \eta u(e_1)) \wedge (\xi e_2 + \eta u(e_2)) \wedge \cdots \wedge (\xi e_n + \eta u(e_n))
( ξ e 1 + η u ( e 1 )) ∧ ( ξ e 2 + η u ( e 2 )) ∧ ⋯ ∧ ( ξ e n + η u ( e n )) 等于项 ξ n − p η p z K \xi^{n-p} \eta^p z_K ξ n − p η p z K
z K = x 1 ∧ x 2 ∧ ⋯ ∧ x n
z_K = x_1 \wedge x_2 \wedge \cdots \wedge x_n
z K = x 1 ∧ x 2 ∧ ⋯ ∧ x n 其中x i = u ( e i ) x_i = u(e_i) x i = u ( e i ) i ∈ K i \in K i ∈ K x i = e i x_i = e_i x i = e i i ∈ H = I − K i \in H = I - K i ∈ H = I − K p p p [ 0 , n ] [0, n] [ 0 , n ] p p p K K K p p p i 1 < i 2 < ⋯ < i n − p i_1 < i_2 < \cdots < i_{n-p} i 1 < i 2 < ⋯ < i n − p j 1 < j 2 < ⋯ < j p j_1 < j_2 < \cdots < j_p j 1 < j 2 < ⋯ < j p § 7 § 7 §7
z K = ρ H , K e i 1 ∧ e i 2 ∧ ⋯ ∧ e i n − p ∧ u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) .
z_K = \rho_{H, K} e_{i_1} \wedge e_{i_2} \wedge \cdots \wedge e_{i_{n-p}} \wedge u(e_{j_1}) \wedge \cdots \wedge u(e_{j_p}).
z K = ρ H , K e i 1 ∧ e i 2 ∧ ⋯ ∧ e i n − p ∧ u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) . 但如果X X X ( e i ) (e_i) ( e i ) u u u
u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) = ∑ L ∈ F p ( I ) ( det X L , K ) e L
u(e_{j_1}) \wedge \cdots \wedge u(e_{j_p}) = \sum_{L \in \mathfrak{F}_p(I)} (\det X_{L, K}) e_L
u ( e j 1 ) ∧ ⋯ ∧ u ( e j p ) = L ∈ F p ( I ) ∑ ( det X L , K ) e L 因此
z K = ρ H , K ∑ L ∈ F p ( I ) ( det X L , K ) e H ∧ e L .
z_K = \rho_{H, K} \sum_{L \in \mathfrak{F}_p(I)} (\det X_{L, K}) e_H \wedge e_L.
z K = ρ H , K L ∈ F p ( I ) ∑ ( det X L , K ) e H ∧ e L . 现在H ∩ L ≠ ∅ H \cap L \neq \varnothing H ∩ L = ∅ L = K L = K L = K § 7 § 7 §7 z K = ( det X K , K ) e 1 ∧ e 2 ∧ ⋯ ∧ e n z_K = (\det X_{K, K}) e_1 \wedge e_2 \wedge \cdots \wedge e_n z K = ( det X K , K ) e 1 ∧ e 2 ∧ ⋯ ∧ e n § 10 § 10 §10
在与命题 11相同的假设下，对于自同态⋀ ( u ) \bigwedge(u) ⋀ ( u ) ⋀ ( M ) \bigwedge(M) ⋀ ( M )
引用了 1 引用了 命题 11
Tr ( ⋀ ( u ) ) = det ( 1 M + u ) .
\operatorname{Tr}(\bigwedge(u)) = \det(1_M + u).
Tr ( ⋀ ( u )) = det ( 1 M + u ) . 在(18)中，替换ξ \xi ξ η \eta η ⋀ ( u ) \bigwedge(u) ⋀ ( u ) e H e_H e H H ∈ F ( I ) H \in \mathfrak{F}(I) H ∈ F ( I ) ⋀ k ( u ) \bigwedge^k(u) ⋀ k ( u ) ≥ k \geq k ≥ k § 10 § 10 §10
6. 行列式的展开
令I为全序有限索引集合。对于I的每个子集 H，令H'表示补集I - H。令X = ( ξ i j ) X = (\xi_{ij}) X = ( ξ ij ) M = A I M = A^I M = A I u u u ( e i ) i ∈ I (e_i)_{i \in I} ( e i ) i ∈ I n = Card ( I ) n = \operatorname{Card}(I) n = Card ( I ) q ≤ n q \leq n q ≤ n n − q n - q n − q
( ⋀ q ( u ) ) ( e H ) = ∑ R det ( X R , H ) e R
(\bigwedge^q(u))(e_H) = \sum_R \det(X_{R, H}) e_R
( ⋀ q ( u )) ( e H ) = R ∑ det ( X R , H ) e R ( ⋀ n − q ( u ) ) ( e K ) = ∑ S det ( X S , K ) e S
(\bigwedge^{n-q}(u))(e_K) = \sum_S \det(X_{S, K}) e_S
( ⋀ n − q ( u )) ( e K ) = S ∑ det ( X S , K ) e S 其中 R（或 S）穿过 I 的子集和集合，以及 q q q n − q n - q n − q § 7 § 7 §7
e R ∧ e S = 0
e_R \wedge e_S = 0
e R ∧ e S = 0 S = R ′ S = R' S = R ′
( 20 ) ( ∧ q ( u ) ( e H ) ) ∧ ( ∧ n − q ( u ) ( e K ) ) = ∑ R ρ R , R ′ det ( X R , H ) det ( X R ′ , K ) e I
(20) \quad (\wedge^q(u)(e_H)) \wedge (\wedge^{n-q}(u)(e_K)) = \sum_R \rho_{R, R'} \det(X_{R, H}) \det(X_{R', K}) e_I
( 20 ) ( ∧ q ( u ) ( e H )) ∧ ( ∧ n − q ( u ) ( e K )) = R ∑ ρ R , R ′ det ( X R , H ) det ( X R ′ , K ) e I 其中R R R I I I q q q F q ( I ) \mathscr{F}_q(I) F q ( I )
如果我们取K = H ′ K = H' K = H ′ ∧ n ( u ) \wedge^n(u) ∧ n ( u ) ρ H , H ′ ∧ n ( u ) ( e I ) \rho_{H, H'} \wedge^n(u)(e_I) ρ H , H ′ ∧ n ( u ) ( e I )
( 21 ) det ( X ) = ρ H , H ′ ∑ R ∈ F q ; R ′ ≠ R ρ R , R ′ det ( X R , H ) det ( X R ′ , H ′ ) .
(21) \quad \det(X) = \rho_{H, H'} \sum_{R \in \mathscr{F}_q; R' \neq R} \rho_{R, R'} \det(X_{R, H}) \det(X_{R', H'}).
( 21 ) det ( X ) = ρ H , H ′ R ∈ F q ; R ′ = R ∑ ρ R , R ′ det ( X R , H ) det ( X R ′ , H ′ ) . 另一方面，如果K ≠ H ′ K \neq H' K = H ′ H ∩ K ≠ ∅ H \cap K \neq \varnothing H ∩ K = ∅ ± ∧ n ( u ) ( e H ∧ e K ) \pm \wedge^n(u)(e_H \wedge e_K) ± ∧ n ( u ) ( e H ∧ e K )
( 22 ) ∑ R ρ R , R ′ det ( X R , H ) det ( X R ′ , K ) = 0 for K ≠ H ′ .
(22) \quad \sum_R \rho_{R, R'} \det(X_{R, H}) \det(X_{R', K}) = 0 \quad \text{for } K \neq H'.
( 22 ) R ∑ ρ R , R ′ det ( X R , H ) det ( X R ′ , K ) = 0 for K = H ′ . (21) 的右侧被称为矩阵 X X X 拉普拉斯展开 由 q 列（其指标属于到 H）和 n - q 列（其指标） 属于到 H 的补码 H'。次要的 det ( X R , H ) \det(X_{R, H}) det ( X R , H ) det ( X R ′ , H ′ ) \det(X_{R', H'}) det ( X R ′ , H ′ )
拉普拉斯展开的一个重要简单情况是I = { 1 , n } I = \{1, n\} I = { 1 , n } q = 1 q = 1 q = 1 H = { i } H = \{i\} H = { i } I I I R = { j } R = \{j\} R = { j } det X R , H = ξ j i \det X_{R, H} = \xi_{ji} det X R , H = ξ j i det X R ′ , H ′ \det X_{R', H'} det X R ′ , H ′ X X X j j j i i i X j i X^{ji} X j i ρ H , H ′ = ( − 1 ) i − 1 \rho_{H, H'} = (-1)^{i-1} ρ H , H ′ = ( − 1 ) i − 1 ρ R , R ′ = ( − 1 ) j − 1 \rho_{R, R'} = (-1)^{j-1} ρ R , R ′ = ( − 1 ) j − 1
( 23 ) det X = ∑ j = 1 n ( − 1 ) i + j ξ j i det ( X j i )
(23) \quad \det X = \sum_{j=1}^n (-1)^{i+j} \xi_{ji} \det(X^{ji})
( 23 ) det X = j = 1 ∑ n ( − 1 ) i + j ξ j i det ( X j i ) 我们从 (22) 中得到类似的结果
( 24 ) ∑ j = 1 n ( − 1 ) j i ξ j i det ( X j k ) = 0 for k ≠ i .
(24) \quad \sum_{j=1}^n (-1)^{ji} \xi_{ji} \det(X^{jk}) = 0 \quad \text{for } k \neq i.
( 24 ) j = 1 ∑ n ( − 1 ) j i ξ j i det ( X j k ) = 0 for k = i . 公式 (23) 被称为 X 与指标 i* 的列的行列式 展开。标量( − 1 ) i + j det ( X j i ) (-1)^{i+j} \det(X^{ji}) ( − 1 ) i + j det ( X j i ) j j j i i i 辅因子*（或者，滥用语言，ξ j i \xi^{ji} ξ j i X X X
X X X
( 25 ) Y = ( ( − 1 ) i + j det ( X j i ) )
(25) \quad Y = ((-1)^{i+j} \det(X^{ji}))
( 25 ) Y = (( − 1 ) i + j det ( X j i )) 其中第j j j i i i j j j i i i
t Y . X = ( det X ) I n .
{}^tY . X = (\det X) I_n.
t Y . X = ( det X ) I n . 所以：
X − 1 = ( det X ) − 1 t Y
X^{-1} = (\det X)^{-1} {}^t Y
X − 1 = ( det X ) − 1 t Y 其中Y Y Y X X X
通过考虑X X X det X \det X det X
X . t Y = ( det X ) I n ,
X . {}^t Y = (\det X) I_n,
X . t Y = ( det X ) I n , 在上面的记号中。
容易验证，如果X X X n n n A \mathbf{A} A M M M u u u ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n t Y {}^t Y t Y u ~ \tilde{u} u ~ M M M n n n x , y 2 , … , y n x, y_2, \ldots, y_n x , y 2 , … , y n M M M
u ~ ( x ) ∧ y 2 ∧ ⋯ ∧ y n = x ∧ u ( y 2 ) ∧ ⋯ ∧ u ( y n ) .
\tilde{u}(x) \wedge y_2 \wedge \cdots \wedge y_n = x \wedge u(y_2) \wedge \cdots \wedge u(y_n).
u ~ ( x ) ∧ y 2 ∧ ⋯ ∧ y n = x ∧ u ( y 2 ) ∧ ⋯ ∧ u ( y n ) . u ~ \tilde{u} u ~ u u u
(1) Vandermonde 行列式。 给定 A \mathbf{A} A n n n ( ζ i ) 1 ≤ i ≤ n (\zeta_i)_{1 \leq i \leq n} ( ζ i ) 1 ≤ i ≤ n Vandermonde 行列式 是行列式
V ( ζ 1 , ζ 2 , … , ζ n ) = ∣ 1 1 … 1 ζ 1 ζ 2 … ζ n ζ 1 2 ζ 2 2 … ζ n 2 ⋮ ⋮ ⋱ ⋮ ζ 1 n − 1 ζ 2 n − 1 … ζ n n − 1 ∣
V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \begin{vmatrix}
1 & 1 & \ldots & 1 \\
\zeta_1 & \zeta_2 & \ldots & \zeta_n \\
\zeta_1^2 & \zeta_2^2 & \ldots & \zeta_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
\zeta_1^{n-1} & \zeta_2^{n-1} & \ldots & \zeta_n^{n-1}
\end{vmatrix}
V ( ζ 1 , ζ 2 , … , ζ n ) = 1 ζ 1 ζ 1 2 ⋮ ζ 1 n − 1 1 ζ 2 ζ 2 2 ⋮ ζ 2 n − 1 … … … ⋱ … 1 ζ n ζ n 2 ⋮ ζ n n − 1 我们将证明
V ( ζ 1 , ζ 2 , … , ζ n ) = ∏ i < j ( ζ j − ζ i ) .
V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \prod_{i < j} (\zeta_j - \zeta_i).
V ( ζ 1 , ζ 2 , … , ζ n ) = i < j ∏ ( ζ j − ζ i ) . 由于命题对于n = 1 n = 1 n = 1 n n n
对于每个指标k ≥ 2 k \geq 2 k ≥ 2 k k k k − 1 k - 1 k − 1 ζ 1 \zeta_1 ζ 1
V ( ζ 1 , ζ 2 , … , ζ n ) = ∣ 1 1 ⋯ 1 0 ζ 2 − ζ 1 ⋯ ζ n − ζ 1 0 ζ 2 ( ζ 2 − ζ 1 ) ⋯ ζ n ( ζ n − ζ 1 ) ⋯ ⋯ ⋯ ⋯ 0 ζ 2 n − 2 ( ζ 2 − ζ 1 ) ⋯ ζ n n − 2 ( ζ n − ζ 1 ) ∣
V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \begin{vmatrix}
1 & 1 & \cdots & 1 \\
0 & \zeta_2 - \zeta_1 & \cdots & \zeta_n - \zeta_1 \\
0 & \zeta_2(\zeta_2 - \zeta_1) & \cdots & \zeta_n(\zeta_n - \zeta_1) \\
\cdots & \cdots & \cdots & \cdots \\
0 & \zeta_2^{n-2}(\zeta_2 - \zeta_1) & \cdots & \zeta_n^{n-2}(\zeta_n - \zeta_1)
\end{vmatrix}
V ( ζ 1 , ζ 2 , … , ζ n ) = 1 0 0 ⋯ 0 1 ζ 2 − ζ 1 ζ 2 ( ζ 2 − ζ 1 ) ⋯ ζ 2 n − 2 ( ζ 2 − ζ 1 ) ⋯ ⋯ ⋯ ⋯ ⋯ 1 ζ n − ζ 1 ζ n ( ζ n − ζ 1 ) ⋯ ζ n n − 2 ( ζ n − ζ 1 ) 从而，通过第一个列展开，然后从指标k − 1 k - 1 k − 1 ζ k − ζ 1 \zeta_k - \zeta_1 ζ k − ζ 1 2 ≤ k ≤ n 2 \leq k \leq n 2 ≤ k ≤ n
V ( ζ 1 , … , ζ n ) = ( ζ 2 − ζ 1 ) ( ζ 3 − ζ 1 ) ⋯ ( ζ n − ζ 1 ) V ( ζ 2 , … , ζ n )
V(\zeta_1, \ldots, \zeta_n) = (\zeta_2 - \zeta_1)(\zeta_3 - \zeta_1) \cdots (\zeta_n - \zeta_1) V(\zeta_2, \ldots, \zeta_n)
V ( ζ 1 , … , ζ n ) = ( ζ 2 − ζ 1 ) ( ζ 3 − ζ 1 ) ⋯ ( ζ n − ζ 1 ) V ( ζ 2 , … , ζ n ) 由归纳成立 (29)。
(2) 考虑阶n n n
X = ( Y T 0 Z )
X = \begin{pmatrix} Y & T \\ 0 & Z \end{pmatrix}
X = ( Y 0 T Z ) 我们表明
det X = ( det Y ) ( det Z ) .
\det X = (\det Y)(\det Z).
det X = ( det Y ) ( det Z ) . 令n n n X X X Y Y Y h h h A n \mathbf{A}^n A n ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n x i x_i x i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n X X X A n \mathbf{A}^n A n x 1 , … , x h x_1, \ldots, x_h x 1 , … , x h e 1 , … , e h e_1, \ldots, e_h e 1 , … , e h
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det Y ) e 1 ∧ e 2 ∧ ⋯ ∧ e h .
x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det Y)e_1 \wedge e_2 \wedge \cdots \wedge e_h.
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det Y ) e 1 ∧ e 2 ∧ ⋯ ∧ e h . 另一方面，对于每个指标i > h i > h i > h x i = y i + z i x_i = y_i + z_i x i = y i + z i y i y_i y i e 1 , e 2 , … , e h e_1, e_2, \ldots, e_h e 1 , e 2 , … , e h z i z_i z i e h + 1 , … , e n e_{h+1}, \ldots, e_n e h + 1 , … , e n x 1 ∧ x 2 ∧ ⋯ ∧ x n ∧ y i = 0 x_1 \wedge x_2 \wedge \cdots \wedge x_n \wedge y_i = 0 x 1 ∧ x 2 ∧ ⋯ ∧ x n ∧ y i = 0 i > h i > h i > h
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det Y ) e 1 ∧ e 2 ∧ ⋯ ∧ e h ∧ z h + 1 ∧ ⋯ ∧ z n .
x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det Y)e_1 \wedge e_2 \wedge \cdots \wedge e_h \wedge z_{h+1} \wedge \cdots \wedge z_n.
x 1 ∧ x 2 ∧ ⋯ ∧ x n = ( det Y ) e 1 ∧ e 2 ∧ ⋯ ∧ e h ∧ z h + 1 ∧ ⋯ ∧ z n . 但到定义
z h + 1 ∧ z h + 2 ∧ ⋯ ∧ z n = ( det Z ) e h + 1 ∧ e h + 2 ∧ ⋯ ∧ e n
z_{h+1} \wedge z_{h+2} \wedge \cdots \wedge z_n = (\det Z)e_{h+1} \wedge e_{h+2} \wedge \cdots \wedge e_n
z h + 1 ∧ z h + 2 ∧ ⋯ ∧ z n = ( det Z ) e h + 1 ∧ e h + 2 ∧ ⋯ ∧ e n 从而公式 (30)。
根据p p p X X X
X = ( X 11 X 12 ⋯ X 1 p 0 X 22 ⋯ X 2 p ⋯ ⋯ ⋯ ⋯ 0 0 ⋯ X p p )
X = \begin{pmatrix}
X_{11} & X_{12} & \cdots & X_{1p} \\
0 & X_{22} & \cdots & X_{2p} \\
\cdots & \cdots & \cdots & \cdots \\
0 & 0 & \cdots & X_{pp}
\end{pmatrix}
X = X 11 0 ⋯ 0 X 12 X 22 ⋯ 0 ⋯ ⋯ ⋯ ⋯ X 1 p X 2 p ⋯ X pp (31)
det X = ( det X 11 ) ( det X 22 ) … ( det X p p ) .
\det X = (\det X_{11})(\det X_{22}) \ldots (\det X_{pp}).
det X = ( det X 11 ) ( det X 22 ) … ( det X pp ) . 这尤其适用于三角形矩阵（其中所有 X i i X_{ii} X ii
(32)
det ( diag ( α 1 , α 2 , … , α n ) ) = α 1 α 2 … α n .
\det(\operatorname{diag}(\alpha_1, \alpha_2, \ldots, \alpha_n)) = \alpha_1 \alpha_2 \ldots \alpha_n.
det ( diag ( α 1 , α 2 , … , α n )) = α 1 α 2 … α n . (3) 令M , M ′ M, M' M , M ′ n , n ′ n, n' n , n ′ M M M u u u M ′ M' M ′ u ′ u' u ′ A A A
(33)
det ( u ⊗ u ′ ) = ( det u ) n ′ ( det u ′ ) n .
\det(u \otimes u') = (\det u)^{n'} (\det u')^n.
det ( u ⊗ u ′ ) = ( det u ) n ′ ( det u ′ ) n . 因为我们可以写 u ⊗ u ′ = ( u ⊗ 1 M ′ ) ∘ ( 1 M ⊗ u ′ ) u \otimes u' = (u \otimes 1_{M'}) \circ (1_M \otimes u') u ⊗ u ′ = ( u ⊗ 1 M ′ ) ∘ ( 1 M ⊗ u ′ ) u , u ′ u, u' u , u ′ u ′ = 1 M ′ u' = 1_{M'} u ′ = 1 M ′ X X X M M M ( e i ) (e_i) ( e i ) u u u u ⊗ 1 M ′ u \otimes 1_{M'} u ⊗ 1 M ′ M ′ M' M ′ ( e i ) (e_i) ( e i ) n ′ n' n ′ n ′ n' n ′ n n n n n n
( X 0 … 0 0 X … 0 … … … … 0 0 … X )
\begin{pmatrix}
X & 0 & \ldots & 0 \\
0 & X & \ldots & 0 \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0 & \ldots & X
\end{pmatrix}
X 0 … 0 0 X … 0 … … … … 0 0 … X 从而，凭借*例2*，
det ( u ⊗ 1 M ′ ) = ( det X ) n ′ = ( det u ) n ′
\det(u \otimes 1_{M'}) = (\det X)^{n'} = (\det u)^{n'}
det ( u ⊗ 1 M ′ ) = ( det X ) n ′ = ( det u ) n ′ 这立即给出公式 (33)。
引用了 1 引用了 II § 10. 。 矩阵
7. 在线性方程中的应用
考虑在 (可交换 ) 环 A A A n n n n n n
(34)
∑ j = 1 n λ i j x j = η i ( 1 ≤ i ≤ n ) .
\sum_{j=1}^n \lambda_{ij} x_j = \eta_i \quad (1 \leq i \leq n).
j = 1 ∑ n λ ij x j = η i ( 1 ≤ i ≤ n ) . 令L L L ( λ i j ) (\lambda_{ij}) ( λ ij ) n n n ξ i \xi_i ξ i η i \eta_i η i A n \mathbf{A}^n A n x = ( ξ i ) x = (\xi_i) x = ( ξ i ) A n \mathbf{A}^n A n y = ( η i ) y = (\eta_i) y = ( η i )
(35)
令u u u A n \mathbf{A}^n A n x ↦ L . x x \mapsto L.x x ↦ L . x L L L y ∈ A n y \in \mathbf{A}^n y ∈ A n u u u
对于交换环上n n n n n n
引用了 1 · 被以下引用 1 引用了 命题 3
被以下引用 命题 16
若A \mathbf{A} A det L \det L det L
( det L ) L . x = ( det L ) y .
(\det L)L.x = (\det L)y.
( det L ) L . x = ( det L ) y . 如果
M M M 是
L L L 的辅因子矩阵，我们可以从no. 6关系的(34)和公式(26)导出 (36)
( det L ) x = t M . y
(\det L)x = {}^tM.y
( det L ) x = t M . y 也可以写成 (37)
( det L ) ξ i = ∑ j = 1 n ( − 1 ) i + j ( det L i j ) η j = det L i ( 1 ≤ i ≤ n )
(\det L)\xi_i = \sum_{j=1}^n (-1)^{i+j}(\det L^{ij})\eta_j = \det L_i \quad (1 \leq i \leq n)
( det L ) ξ i = j = 1 ∑ n ( − 1 ) i + j ( det L ij ) η j = det L i ( 1 ≤ i ≤ n ) 其中，
L i L_i L i 表示矩阵由替换列、指标
i i i 的
L L L 通过
y y y 获得。 公式 (37) 被称为
Cramer 的公式 对于系统 (34)； (34) 的每个解也是 (37) 的解。相反，我们从 (36) 推导，考虑到 no. 6 的公式 (28)， (38)
( det L ) ( L . x − y ) = 0
(\det L)(L.x - y) = 0
( det L ) ( L . x − y ) = 0 因此，如果
det L \det L det L 不是
A \mathbf{A} A 中零的除子，系统 (34) 和 (37)
等价 ；如果
det L \det L det L 可逆，则 (34) 的唯一解由下式给出 (39)
ξ i = ( det L ) − 1 ( det L i ) ( 1 ≤ i ≤ n ) .
\xi_i = (\det L)^{-1}(\det L_i) \quad (1 \leq i \leq n).
ξ i = ( det L ) − 1 ( det L i ) ( 1 ≤ i ≤ n ) . 系统 (34) 使得
det L \det L det L 可逆，也称为
Cramer 系统 。
特别是令 y = 0 y = 0 y = 0
对于交换环上n n n n n n
8. 交换域案例
当环A \mathbf{A} A
因此，在这种情况下，§ 7, no. 9的命题12可以表述如下：
令E \mathbf{E} E p p p x i ∈ E x_i \in \mathbf{E} x i ∈ E 1 ≤ i ≤ p 1 \leq i \leq p 1 ≤ i ≤ p x 1 ∧ x 2 ∧ ⋯ ∧ x p ≠ 0 x_1 \wedge x_2 \wedge \cdots \wedge x_p \neq 0 x 1 ∧ x 2 ∧ ⋯ ∧ x p = 0
被以下引用 3 被以下引用 推论 , 命题 16 , 定理 1
令X X X ( m , n ) (m, n) ( m , n ) X X X p p p p p p X X X ≠ 0 \neq 0 = 0
引用了 2 · 被以下引用 1 引用了 定义 7, II § 10 , 命题 15
被以下引用 命题 16
X X X X X X
现在考虑交换域 K \mathbf{K} K n n n m m m
∑ j = 1 n λ i j ξ j = η i ( 1 ≤ i ≤ m ) .
\sum_{j=1}^n \lambda_{ij} \xi_j = \eta_i \quad (1 \leq i \leq m).
j = 1 ∑ n λ ij ξ j = η i ( 1 ≤ i ≤ m ) . (41)
令 L = ( λ i j ) L = (\lambda_{ij}) L = ( λ ij ) ( m , n ) (m, n) ( m , n ) M M M ( m , n + 1 ) (m, n+1) ( m , n + 1 ) L L L ( n + 1 ) (n+1) ( n + 1 ) ( η i ) (\eta_i) ( η i ) p p p L L L L L L Δ \Delta Δ L L L ≥ p + 1 \geq p + 1 ≥ p + 1 ≠ 0 \neq 0 = 0 L L L L L L M M M p + 1 p + 1 p + 1 p + 1 p + 1 p + 1 M M M 1 , 2 , … , p 1, 2, \ldots, p 1 , 2 , … , p n + 1 n+1 n + 1 p p p
∑ j = 1 p λ i j ξ j = η i − ∑ k = p + 1 n λ i k ξ k ( 1 ≤ i ≤ p )
\sum_{j=1}^p \lambda_{ij} \xi_j = \eta_i - \sum_{k=p+1}^n \lambda_{ik} \xi_k \quad (1 \leq i \leq p)
j = 1 ∑ p λ ij ξ j = η i − k = p + 1 ∑ n λ ik ξ k ( 1 ≤ i ≤ p ) 引用了 5 引用了 II § 10. 。 矩阵 , 推论 , 命题 12, II § 10 , 命题 13 , 命题 15
这个系统的所有解都是通过取指标k > p k > p k > p ξ k \xi_k ξ k ξ j \xi_j ξ j j ≤ p j \leq p j ≤ p
我们知道（II, § 10, no. 12，命题 12），系统（41）至少有一个解，矩阵L L L M M M L L L a i a_i a i 1 ≤ i ≤ p 1 \leq i \leq p 1 ≤ i ≤ p L L L p p p y = ( η i ) y = (\eta_i) y = ( η i ) M M M ( n + 1 ) (n+1) ( n + 1 ) L L L a i a_i a i M M M L L L p p p y y y a i a_i a i a 1 ∧ ⋯ ∧ a p ∧ y = 0 a_1 \wedge \cdots \wedge a_p \wedge y = 0 a 1 ∧ ⋯ ∧ a p ∧ y = 0 M M M p p p > p > p > p
9. 幺模群S L ( n , A ) \mathbf{SL}(n, A) SL ( n , A )
让M n ( A ) \mathbf{M}_n(A) M n ( A ) A A A n n n det : M n ( A ) → A \det : \mathbf{M}_n(A) \to A det : M n ( A ) → A M n ( A ) \mathbf{M}_n(A) M n ( A ) G L ( n , A ) \mathbf{GL}(n, A) GL ( n , A ) A A A A n A^n A n A A A A ∗ A^* A ∗ det : G L ( n , A ) → A ∗ \det : \mathbf{GL}(n, A) \to A^* det : GL ( n , A ) → A ∗
此外，映射det : M n ( A ) → A \det : \mathbf{M}_n(A) \to A det : M n ( A ) → A det : G L ( n , A ) → A ∗ \det : \mathbf{GL}(n, A) \to A^* det : GL ( n , A ) → A ∗ λ ∈ A \lambda \in A λ ∈ A
det ( diag ( λ , 1 , … , 1 ) ) = λ
\det(\operatorname{diag}(\lambda, 1, \ldots, 1)) = \lambda
det ( diag ( λ , 1 , … , 1 )) = λ 凭借no. 6的公式(32)。
满同态 det : G L ( n , A ) → A ∗ \det : \mathbf{GL}(n, A) \to A^* det : GL ( n , A ) → A ∗ G L ( n , A ) \mathbf{GL}(n, A) GL ( n , A ) S L n ( A ) \mathbf{SL}_n(A) SL n ( A ) S L ( n , A ) \mathbf{SL}(n, A) SL ( n , A ) n n n A A A
在这个没有。我们将检查 A A A 域 的情况。回想一下，对于1 ≤ i ≤ n , 1 ≤ j ≤ n 1 \leq i \leq n, 1 \leq j \leq n 1 ≤ i ≤ n , 1 ≤ j ≤ n E i j E_{ij} E ij n n n i i i j j j I n I_n I n n n n 不同 指标 i , j i, j i , j λ ∈ A \lambda \in A λ ∈ A B i j ( λ ) = I n + λ E i j B_{ij}(\lambda) = I_n + \lambda E_{ij} B ij ( λ ) = I n + λ E ij
令K K K S L ( n , K ) \mathbf{SL}(n, K) SL ( n , K ) B i j ( λ ) B_{ij}(\lambda) B ij ( λ ) i ≠ j i \neq j i = j λ ∈ K \lambda \in K λ ∈ K
引用了 1 · 被以下引用 4 引用了 命题 14, II § 10
被以下引用 推论 , 注 2, LIE VII § 3 , 定理 2, LIE VIII § 1 , 命题 3, LIE VIII § 11
由II, § 10, no. 13、命题 14，我们知道G L ( n , K ) \mathbf{GL}(n, K) GL ( n , K ) B i j ( λ ) B_{ij}(\lambda) B ij ( λ ) diag ( 1 , 1 , … , 1 , α ) \operatorname{diag}(1, 1, \ldots, 1, \alpha) diag ( 1 , 1 , … , 1 , α ) α ∈ K ∗ \alpha \in K^* α ∈ K ∗ det ( B i j ( λ ) ) = 1 \det(B_{ij}(\lambda)) = 1 det ( B ij ( λ )) = 1 det ( diag ( 1 , … , 1 , α ) ) = α \det(\operatorname{diag}(1, \ldots, 1, \alpha)) = \alpha det ( diag ( 1 , … , 1 , α )) = α
因为S L ( n , K ) \mathbf{SL}(n, K) SL ( n , K ) G L ( n , K ) \mathbf{GL}(n, K) GL ( n , K ) K ∗ K^* K ∗ S L ( n , K ) \mathbf{SL}(n, K) SL ( n , K ) Γ \Gamma Γ G L ( n , K ) \mathbf{GL}(n, K) GL ( n , K ) S L ( n , K ) = Γ \mathbf{SL}(n, K) = \Gamma SL ( n , K ) = Γ λ ∈ K ∗ \lambda \in K^* λ ∈ K ∗ B i j ( λ ) B_{ij}(\lambda) B ij ( λ ) Γ \Gamma Γ B i j ( λ ) = Q . B i j ( 1 ) . Q − 1 B_{ij}(\lambda) = Q . B_{ij}(1) . Q^{-1} B ij ( λ ) = Q . B ij ( 1 ) . Q − 1 B i j ( λ ) B_{ij}(\lambda) B ij ( λ ) G L ( n , K ) \mathbf{GL}(n, K) GL ( n , K ) B i j ( 1 ) B_{ij}(1) B ij ( 1 ) Q Q Q K n K^n K n v v v ( e i ) (e_i) ( e i ) v ( e i ) = λ e i , v ( e k ) = e k v(e_i) = \lambda e_i, v(e_k) = e_k v ( e i ) = λ e i , v ( e k ) = e k k ≠ i k \neq i k = i u i j u_{ij} u ij i ≠ j i \neq j i = j K n K^n K n u i j ( e i ) = − e j , u i j ( e j ) = e i , u i j ( e k ) = e k u_{ij}(e_i) = -e_j, \ u_{ij}(e_j) = e_i, \ u_{ij}(e_k) = e_k u ij ( e i ) = − e j , u ij ( e j ) = e i , u ij ( e k ) = e k k ∉ { i , j } k \notin \{i, j\} k ∈ / { i , j } S L ( n , K ) \mathbf{SL}(n, K) SL ( n , K ) B j i ( λ ) = U i j B i j ( − λ ) U i j − 1 B_{ji}(\lambda) = U_{ij} B_{ij}(-\lambda) U_{ij}^{-1} B j i ( λ ) = U ij B ij ( − λ ) U ij − 1 U i j U_{ij} U ij u i j u_{ij} u ij 1 < i < j 1 < i < j 1 < i < j B 1 j ( λ ) = U 1 i B i j ( λ ) U 1 i − 1 B_{1j}(\lambda) = U_{1i} B_{ij}(\lambda) U_{1i}^{-1} B 1 j ( λ ) = U 1 i B ij ( λ ) U 1 i − 1 2 < j 2 < j 2 < j B 12 ( λ ) = U 2 j B 1 j ( λ ) U 2 j − 1 B_{12}(\lambda) = U_{2j} B_{1j}(\lambda) U_{2j}^{-1} B 12 ( λ ) = U 2 j B 1 j ( λ ) U 2 j − 1 G L ( n , K ) / Γ \mathbf{GL}(n, K)/\Gamma GL ( n , K ) /Γ B i j ( λ ) B_{ij}(\lambda) B ij ( λ ) s s s s s s
首先假设K K K λ \lambda λ 1 = λ + ( 1 − λ ) 1 = \lambda + (1 - \lambda) 1 = λ + ( 1 − λ ) ≠ 0 \neq 0 = 0 B 12 ( 1 ) = B 12 ( λ ) B 12 ( 1 − λ ) B_{12}(1) = B_{12}(\lambda) B_{12}(1 - \lambda) B 12 ( 1 ) = B 12 ( λ ) B 12 ( 1 − λ ) s 2 = s s^2 = s s 2 = s s s s
现在假设n ≥ 3 n \geq 3 n ≥ 3 B 21 ( 1 ) B 31 ( 1 ) B_{21}(1) B_{31}(1) B 21 ( 1 ) B 31 ( 1 ) u u u K n K^n K n u ( e 1 ) = e 1 + e 2 + e 3 , u ( e i ) = e i u(e_1) = e_1 + e_2 + e_3, u(e_i) = e_i u ( e 1 ) = e 1 + e 2 + e 3 , u ( e i ) = e i i ≠ 1 i \neq 1 i = 1 S S S u ′ u' u ′ K n K^n K n u ′ ( e 2 ) = e 2 + e 3 , u ′ ( e i ) = e i u'(e_2) = e_2 + e_3, u'(e_i) = e_i u ′ ( e 2 ) = e 2 + e 3 , u ′ ( e i ) = e i i ≠ 2 i \neq 2 i = 2 S . B 21 ( 1 ) B 31 ( 1 ) . S − 1 = B 21 ( 1 ) S . B_{21}(1) B_{31}(1) . S^{-1} = B_{21}(1) S . B 21 ( 1 ) B 31 ( 1 ) . S − 1 = B 21 ( 1 ) s 2 = s s^2 = s s 2 = s
10. 与 A \mathbf{A} A A [ X ] \mathbf{A}[X] A [ X ]
令M M M A \mathbf{A} A u u u M M M A [ X ] \mathbf{A}[X] A [ X ] A \mathbf{A} A X X X p ∈ A [ X ] p \in \mathbf{A}[X] p ∈ A [ X ] x ∈ M x \in M x ∈ M
p . x = p ( u ) ( x ) .
p . x = p(u)(x).
p . x = p ( u ) ( x ) . 作为A [ X ] \mathbf{A}[X] A [ X ] p , q p, q p , q ( p q ) ( u ) = p ( y ) ∘ q ( u ) (pq)(u) = p(y) \circ q(u) ( pq ) ( u ) = p ( y ) ∘ q ( u ) A [ X ] \mathbf{A}[X] A [ X ] M M M M M M M u M_u M u M M M A \mathbf{A} A M u M_u M u A \mathbf{A} A M u M_u M u M M M u u u
由于A [ X ] × M A[X] \times M A [ X ] × M M M M ( p , x ) ↦ p . x (p, x) \mapsto p.x ( p , x ) ↦ p . x A A A A A A ϕ : A [ X ] ⊗ A M → M \phi : A[X] \otimes_A M \to M ϕ : A [ X ] ⊗ A M → M
ϕ ( p ⊗ x ) = p . x = p ( u ) ( x ) .
\phi(p \otimes x) = p.x = p(u)(x).
ϕ ( p ⊗ x ) = p . x = p ( u ) ( x ) . A [ X ] ⊗ A M A[X] \otimes_A M A [ X ] ⊗ A M A [ X ] A[X] A [ X ] M [ X ] M[X] M [ X ] A [ X ] A[X] A [ X ] ϕ : M [ X ] → M u \phi : M[X] \to M_u ϕ : M [ X ] → M u A [ X ] A[X] A [ X ] A [ X ] A[X] A [ X ] x ∈ M x \in M x ∈ M p , q p, q p , q ϕ ( q ( p ⊗ x ) ) = ϕ ( ( q p ) ⊗ x ) = ( q p ) . x = q ( u ) ( p ( u ) ( x ) ) = q . ϕ ( p ⊗ x ) .
\phi(q(p \otimes x)) = \phi((qp) \otimes x) = (qp).x = q(u)(p(u)(x)) = q.\phi(p \otimes x).
ϕ ( q ( p ⊗ x )) = ϕ (( q p ) ⊗ x ) = ( q p ) . x = q ( u ) ( p ( u ) ( x )) = q . ϕ ( p ⊗ x ) . u u u M u M_u M u A [ X ] A[X] A [ X ] u ( p . x ) = u ( p ( u ) ( x ) ) = ( u p ( x ) ) ( x ) = p . u ( x ) .
u(p.x) = u(p(u)(x)) = (up(x))(x) = p.u(x).
u ( p . x ) = u ( p ( u ) ( x )) = ( u p ( x )) ( x ) = p . u ( x ) . M [ X ] M[X] M [ X ] A [ X ] A[X] A [ X ] u ˉ \bar{u} u ˉ u ˉ ( p ⊗ x ) = p ⊗ u ( x ) .
\bar{u}(p \otimes x) = p \otimes u(x).
u ˉ ( p ⊗ x ) = p ⊗ u ( x ) . A [ X ] A[X] A [ X ] u , u ˉ u, \bar{u} u , u ˉ ϕ \phi ϕ ϕ ∘ u ˉ = u ∘ ϕ .
\phi \circ \bar{u} = u \circ \phi.
ϕ ∘ u ˉ = u ∘ ϕ . ψ \psi ψ M [ X ] M[X] M [ X ] A [ X ] A[X] A [ X ] X − u ˉ X - \bar{u} X − u ˉ ψ ( p ⊗ x ) = ( X p ) ⊗ x − p ⊗ u ( x ) \psi(p \otimes x) = (Xp) \otimes x - p \otimes u(x) ψ ( p ⊗ x ) = ( X p ) ⊗ x − p ⊗ u ( x )
由于 ϕ ( 1 ⊗ x ) = x \phi(1 \otimes x) = x ϕ ( 1 ⊗ x ) = x x ∈ M x \in M x ∈ M ϕ \phi ϕ
ϕ ( X ( p ⊗ x ) ) = X . ϕ ( p ⊗ x ) = u ( ϕ ( p ⊗ x ) ) ,
\phi(X(p \otimes x)) = X.\phi(p \otimes x) = u(\phi(p \otimes x)),
ϕ ( X ( p ⊗ x )) = X . ϕ ( p ⊗ x ) = u ( ϕ ( p ⊗ x )) , 其他字，
ϕ ∘ X = u ∘ ϕ = ϕ ∘ u ˉ \phi \circ X = u \circ \phi = \phi \circ \bar{u} ϕ ∘ X = u ∘ ϕ = ϕ ∘ u ˉ 由(46)；这证明了
ϕ ∘ ψ = 0 \phi \circ \psi = 0 ϕ ∘ ψ = 0 。仍有待验证
Ker ϕ ⊂ Im ψ \operatorname{Ker} \phi \subset \operatorname{Im} \psi Ker ϕ ⊂ Im ψ 。对于这个注，由于
A A A 模
A [ X ] A[X] A [ X ] 的单项式
X k X^k X k (
k ≥ 0 k \geq 0 k ≥ 0 )形式基，每个元素
z ∈ M [ X ] z \in M[X] z ∈ M [ X ] 都可以唯一地写成形式
z = ∑ k X k ⊗ x k z = \sum_k X^k \otimes x_k z = ∑ k X k ⊗ x k ，其中
( x k ) (x_k) ( x k ) 是元素、
M M M 、有限支撑的家庭。如果
z ∈ Ker ϕ z \in \operatorname{Ker} \phi z ∈ Ker ϕ ，那么
ϕ ( z ) = ∑ k u k ( x k ) = 0 \phi(z) = \sum_k u^k(x_k) = 0 ϕ ( z ) = ∑ k u k ( x k ) = 0 ，我们可以写
z = ∑ k ( X k ⊗ x k − 1 ⊗ u k ( x k ) ) = ∑ k ( X k − u ˉ k ) ( 1 ⊗ x k ) .
z = \sum_k (X^k \otimes x_k - 1 \otimes u^k(x_k)) = \sum_k (X^k - \bar{u}^k)(1 \otimes x_k).
z = k ∑ ( X k ⊗ x k − 1 ⊗ u k ( x k )) = k ∑ ( X k − u ˉ k ) ( 1 ⊗ x k ) . 但由于
M [ X ] M[X] M [ X ] 的
A [ X ] A[X] A [ X ] 自同态
X X X 和
u ˉ \bar{u} u ˉ 是可置换的，那么
X k − u ˉ k = ( X − u ˉ ) ∘ ( ∑ j = 0 k − 1 X j u ˉ k − j − 1 )
X^k - \bar{u}^k = (X - \bar{u}) \circ \left( \sum_{j=0}^{k-1} X^j \bar{u}^{k-j-1} \right)
X k − u ˉ k = ( X − u ˉ ) ∘ ( j = 0 ∑ k − 1 X j u ˉ k − j − 1 ) 这证明存在一个y ∈ M [ X ] y \in M[X] y ∈ M [ X ] z = ψ ( y ) z = \psi(y) z = ψ ( y )
现在让M ′ M' M ′ A A A u ′ u' u ′ M ′ M' M ′ M u ′ ′ M'_{u'} M u ′ ′ ϕ ′ \phi' ϕ ′ u ˉ ′ \bar{u}' u ˉ ′ ψ ′ \psi' ψ ′ M ′ M' M ′ u ′ u' u ′ M u M_u M u ϕ \phi ϕ u ˉ \bar{u} u ˉ ψ \psi ψ M M M u u u
对于 M 到 M' 的映射 g 是 M_u 到 M'_u 的 A[X]-同态，g 是 M 到 M' 的 A-同态使得 g ∘ u = u ′ ∘ g g \circ u = u' \circ g g ∘ u = u ′ ∘ g g ˉ \bar{g} g ˉ 1 A [ x ] ⊗ g 1_{A[x]} \otimes g 1 A [ x ] ⊗ g
引用了 1 · 被以下引用 2 引用了 II § 5. 。标量的环的扩张
被以下引用 习题 26, § 8 , VII § 5. 。 自同态 / 向量空间
M [ X ] → ψ M [ X ] → ϕ M u ⟶ 0 ↓ g ˉ ↓ g ˉ ↓ g M ′ [ X ] → ψ ′ M ′ [ X ] → ϕ ′ M u ′ ′ ⟶ 0
\begin{array}{cccccc}
M[X] & \xrightarrow{\psi} & M[X] & \xrightarrow{\phi} & M_u & \longrightarrow 0 \\
\downarrow \bar{g} & & \downarrow \bar{g} & & \downarrow g & \\
M'[X] & \xrightarrow{\psi'} & M'[X] & \xrightarrow{\phi'} & M'_{u'} & \longrightarrow 0
\end{array}
M [ X ] ↓ g ˉ M ′ [ X ] ψ ψ ′ M [ X ] ↓ g ˉ M ′ [ X ] ϕ ϕ ′ M u ↓ g M u ′ ′ ⟶ 0 ⟶ 0 是*可交换的*。
条件 g ∘ u = u ′ ∘ g g \circ u = u' \circ g g ∘ u = u ′ ∘ g g g g n > 0 n > 0 n > 0 g ∘ u n = u ′ n ∘ g g \circ u^n = {u'}^n \circ g g ∘ u n = u ′ n ∘ g x ∈ M x \in M x ∈ M p ∈ A [ X ] p \in A[X] p ∈ A [ X ]
ϕ ′ ( g ˉ ( p ⊗ x ) ) = ϕ ′ ( p ⊗ g ( x ) ) = p ( u ′ ) ( g ( x ) ) = g ( p ( u ) ( x ) ) = g ( ϕ ( p ⊗ x ) )
\phi'(\bar{g}(p \otimes x)) = \phi'(p \otimes g(x)) = p(u')(g(x)) = g(p(u)(x)) = g(\phi(p \otimes x))
ϕ ′ ( g ˉ ( p ⊗ x )) = ϕ ′ ( p ⊗ g ( x )) = p ( u ′ ) ( g ( x )) = g ( p ( u ) ( x )) = g ( ϕ ( p ⊗ x )) 和
u ˉ ′ ( g ( p ⊗ x ) ) = u ˉ ′ ( p ⊗ g ( x ) ) = p ⊗ u ′ ( g ( x ) ) = p ⊗ g ( u ( x ) ) = g ˉ ( u ˉ ( p ⊗ x ) )
\bar{u}'(g(p \otimes x)) = \bar{u}'(p \otimes g(x)) = p \otimes u'(g(x)) = p \otimes g(u(x)) = \bar{g}(\bar{u}(p \otimes x))
u ˉ ′ ( g ( p ⊗ x )) = u ˉ ′ ( p ⊗ g ( x )) = p ⊗ u ′ ( g ( x )) = p ⊗ g ( u ( x )) = g ˉ ( u ˉ ( p ⊗ x )) 这就证明了图表 (48) 的交换律。
11. AN 自同态的特征多项式
令M M M n n n u u u A A A M M M A [ X , Y ] A[X, Y] A [ X , Y ] A [ X , Y ] A[X, Y] A [ X , Y ] M [ X , Y ] = A [ X , Y ] ⊗ A M M[X, Y] = A[X, Y] \otimes_A M M [ X , Y ] = A [ X , Y ] ⊗ A M u ˉ \bar{u} u ˉ A [ X , Y ] A[X, Y] A [ X , Y ] M [ X , Y ] M[X, Y] M [ X , Y ] u u u
det ( X − Y u ˉ ) = ∑ j = 0 n ( − 1 ) j T r ( ∧ j ( u ) ) X n − j Y j
\det(X - Y \bar{u}) = \sum_{j=0}^n (-1)^j \mathrm{Tr}(\wedge^j(u)) X^{n-j} Y^j
det ( X − Y u ˉ ) = j = 0 ∑ n ( − 1 ) j Tr ( ∧ j ( u )) X n − j Y j 因为如果相对于M M M ( e i ) 1 ≤ i ≤ n (e_i)_{1 \leq i \leq n} ( e i ) 1 ≤ i ≤ n U U U u u u M [ X , Y ] M[X, Y] M [ X , Y ] ( 1 ⊗ e i ) 1 ≤ i ≤ n (1 \otimes e_i)_{1 \leq i \leq n} ( 1 ⊗ e i ) 1 ≤ i ≤ n U U U u ˉ \bar{u} u ˉ
T r ( ∧ j ( u ˉ ) ) = T r ( ∧ j ( u ) ) .
\mathrm{Tr}(\wedge^j(\bar{u})) = \mathrm{Tr}(\wedge^j(u)).
Tr ( ∧ j ( u ˉ )) = Tr ( ∧ j ( u )) .
令 M M M A A A M M M u u u A [ X ] A[X] A [ X ] M [ X ] M[X] M [ X ] X − u ˉ X - \bar{u} X − u ˉ u u u χ u ( X ) \chi_u(X) χ u ( X )
被以下引用 1 被以下引用 VIII § 17. 。 简化范数与迹
如果 M M M n n n
χ u ( X ) = ∑ j = 0 n ( − 1 ) j Tr ( ∧ j ( u ) ) X n − j
\chi_u(X) = \sum_{j=0}^n (-1)^j \operatorname{Tr}(\wedge^j(u)) X^{n-j}
χ u ( X ) = j = 0 ∑ n ( − 1 ) j Tr ( ∧ j ( u )) X n − j 对于 det ( X − Y u ˉ ) = det ( X . I n + Y U ) \det(X - Y \bar{u}) = \det(X . I_n + Y U) det ( X − Y u ˉ ) = det ( X . I n + Y U ) det ( X − u ˉ ) = det ( X . I n − U ) \det(X - \bar{u}) = \det(X . I_n - U) det ( X − u ˉ ) = det ( X . I n − U ) χ u ( X ) \chi_u(X) χ u ( X ) n n n X n − 1 X^{n-1} X n − 1 − Tr ( u ) -\operatorname{Tr}(u) − Tr ( u ) ( − 1 ) n det ( u ) (-1)^n \det(u) ( − 1 ) n det ( u )
在命题 18 的记号（§ 3 § 3 §3 x ∈ M x \in M x ∈ M χ u ( u ) ( x ) \chi_u(u)(x) χ u ( u ) ( x ) χ u ( X ) ⊗ x \chi_u(X) \otimes x χ u ( X ) ⊗ x ϕ \phi ϕ v v v M [ X ] M[X] M [ X ] X − u ˉ X - \bar{u} X − u ˉ
χ u ( X ) ⊗ x = χ u ( X ) ( 1 ⊗ x ) = ( X − u ˉ ) ( v ( 1 ⊗ x ) )
\chi_u(X) \otimes x = \chi_u(X)(1 \otimes x) = (X - \bar{u})(v(1 \otimes x))
χ u ( X ) ⊗ x = χ u ( X ) ( 1 ⊗ x ) = ( X − u ˉ ) ( v ( 1 ⊗ x )) 结论由命题18 of no. 10得出。
习题
参见习题为 § 8 。