Jordan标准形知识梳理补充证明
证明: 充分性: 因为 ( λ I − B ) ≃ ( λ I − A ) (\lambda \bm{I}-\bm{B}) \simeq (\lambda \bm{I}-\bm{A}) (λI−B)≃(λI−A), 所以存在 n n n 阶可逆 λ \lambda λ-矩阵 U ( λ ) \bm{U}(\lambda) U(λ), V ( λ ) \bm{V}(\lambda) V(λ), 使得 λ I − A = U ( λ ) ( λ I − B ) V ( λ ) (1) \lambda \bm{I}- \bm{A}=\bm{U}(\lambda)(\lambda \bm{I}-\bm{B})\bm{V}(\lambda) \tag{1} λI−A=U(λ)(λI−B)V(λ)(1)
由于 λ I − A \lambda \bm I-\bm{A} λI−A 展开式的首项系数矩阵可逆, 由带余除法的可行性定理, 存在 n n n 阶 λ \lambda λ-矩阵 Q ( λ ) \bm{Q}(\lambda) Q(λ), U 0 ( λ ) \bm{U}_0(\lambda) U0(λ), deg U 0 ( λ ) < 1 \deg{\bm{U}_0}(\lambda)<1 degU0(λ)<1, (即 U 0 ( λ ) \bm{U}_0(\lambda) U0(λ) 为 0 0 0 次, 下文简记为 U 0 \bm{U}_0 U0), 使得 U ( λ ) = ( λ I − A ) Q ( λ ) + U 0 (U) \bm{U}(\lambda) = (\lambda \bm{I}-\bm{A}) \bm{Q}(\lambda) + \bm{U}_0 \tag{U} U(λ)=(λI−A)Q(λ)+U0(U)
存在 n n n 阶 λ \lambda λ-矩阵 R ( λ ) \bm{R}(\lambda) R(λ), V 0 ( λ ) \bm{V}_0(\lambda) V0(λ), deg V 0 ( λ ) < 1 \deg{\bm{V}_0}(\lambda)<1 degV0(λ)<1 (即 V 0 ( λ ) \bm{V}_0(\lambda) V0(λ) 为 0 0 0 次, 下文简记为 V 0 \bm{V}_0 V0), 使得 V ( λ ) = R ( λ ) ( λ I − A ) + V 0 (V) \bm{V}(\lambda) = \bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) + \bm{V}_0 \tag{V} V(λ)=R(λ)(λI−A)+V0(V)
下面证明: λ I − A = U 0 ( λ I − B ) V 0 (2) \lambda \bm{I}-\bm{A}=\bm{U}_0 (\lambda \bm{I}-\bm{B}) \bm{V}_0 \tag{2} λI−A=U0(λI−B)V0(2)
将 (2) 代入 (1) 可知, U ( λ ) ( λ I − B ) V ( λ ) = U 0 ( λ I − B ) V 0 \bm{U}(\lambda) (\lambda \bm{I}-\bm{B}) \bm{V}(\lambda) =\bm{U}_0 (\lambda \bm{I}-\bm{B}) \bm{V}_0 U(λ)(λI−B)V(λ)=U0(λI−B)V0
下面对此式进行变换, 左边减去右边:
U ( λ ) ( λ I − A ) ( V ( λ ) − V 0 ) + ( U ( λ ) − U 0 ) ( λ I − B ) V 0 = 0 \bm{U}(\lambda) (\lambda \bm{I}-\bm{A}) (\bm{V}(\lambda)-\bm{V}_0)+(\bm{U}(\lambda)-\bm{U}_0)(\lambda \bm{I}-\bm{B})\bm{V}_0 = \bm{0} U(λ)(λI−A)(V(λ)−V0)+(U(λ)−U0)(λI−B)V0=0
U ( λ ) ( λ I − B ) R ( λ ) ( λ I − A ) + ( U ( λ ) − U 0 ) ( λ I − B ) V 0 = 0 \bm{U}(\lambda) (\lambda \bm{I}-\bm{B}) \bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) + (\bm{U}(\lambda)-\bm{U}_0) (\lambda \bm{I}-\bm{B})\bm{V}_0 = \bm{0} U(λ)(λI−B)R(λ)(λI−A)+(U(λ)−U0)(λI−B)V0=0
左右同乘 U − 1 ( λ ) \bm{U}^{-1}(\lambda) U−1(λ), ( λ I − B ) R ( λ ) ( λ I − A ) − U − 1 ( λ ) U 0 ( λ I − B ) V 0 + ( λ I − B ) V 0 = 0 (\lambda \bm{I}-\bm{B}) \bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) - \bm{U}^{-1}(\lambda)\bm{U}_0 (\lambda \bm{I}-\bm{B})\bm{V}_0 + (\lambda \bm{I}-\bm{B})\bm{V}_0=\bm{0} (λI−B)R(λ)(λI−A)−U−1(λ)U0(λI−B)V0+(λI−B)V0=0
整理化简: ( λ I − B ) R ( λ ) ( λ I − A ) − U − 1 ( λ ) ( U 0 ( λ I − B ) V 0 ) = − ( λ I − B ) V 0 (\lambda \bm{I}-\bm{B}) \bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) - \bm{U}^{-1}(\lambda)(\bm{U}_0 (\lambda \bm{I}-\bm{B})\bm{V}_0) = - (\lambda \bm{I}-\bm{B})\bm{V}_0 (λI−B)R(λ)(λI−A)−U−1(λ)(U0(λI−B)V0)=−(λI−B)V0
( λ I − B ) R ( λ ) ( λ I − A ) − U − 1 ( λ I − A ) = − ( λ I − B ) V 0 (\lambda \bm{I}-\bm{B}) \bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) - \bm{U}^{-1} (\lambda \bm{I}-\bm{A}) = - (\lambda \bm{I}-\bm{B})\bm{V}_0 (λI−B)R(λ)(λI−A)−U−1(λI−A)=−(λI−B)V0
下面得到 (2) 成立的一个必要条件:
( U − 1 ( λ ) − ( λ I − B ) R ( λ ) ) ( λ I − A ) = ( λ I − B ) V 0 (\bm{U}^{-1}(\lambda) - (\lambda \bm{I}-\bm{B})\bm{R}(\lambda))(\lambda \bm{I}-\bm{A}) = (\lambda \bm{I}-\bm{B})\bm{V}_0 (U−1(λ)−(λI−B)R(λ))(λI−A)=(λI−B)V0
为便于书写, 记 C ( λ ) = U − 1 ( λ ) − ( λ I − B ) R ( λ ) \bm{C}(\lambda) = \bm{U}^{-1}(\lambda) - (\lambda \bm{I}-\bm{B})\bm{R}(\lambda) C(λ)=U−1(λ)−(λI−B)R(λ), 将上式写为 C ( λ ) ( λ I − A ) = ( λ I − B ) V 0 (3) \bm{C}(\lambda)(\lambda \bm{I}-\bm{A}) = (\lambda \bm{I}-\bm{B})\bm{V}_0\tag{3} C(λ)(λI−A)=(λI−B)V0(3)
由 (2) 还可以推出: λ I − A = λ U 0 V 0 − U 0 B V 0 \lambda \bm{I}-\bm{A} = \lambda \bm{U}_0\bm{V}_0 - \bm{U}_0\bm{B}\bm{V}_0 λI−A=λU0V0−U0BV0.
比较系数可知: I = U 0 V 0 , A = U 0 B V 0 (4) \bm{I}=\bm{U}_0\bm{V}_0, \bm{A}=\bm{U}_0\bm{B}\bm{V}_0 \tag{4} I=U0V0,A=U0BV0(4)
由左式可知: U 0 \bm{U}_0 U0 和 V 0 \bm{V}_0 V0 可逆, U 0 − 1 = V 0 (5) \bm{U}_0^{-1} = \bm{V}_0 \tag{5} U0−1=V0(5)
将 (5) 代入 (2) 可得 ( λ I − B ) V 0 = U 0 − 1 ( λ I − A ) (\lambda \bm{I}-\bm{B})\bm{V}_0 = \bm{U}_0^{-1} (\lambda \bm{I}-\bm{A}) (λI−B)V0=U0−1(λI−A), 再将此式代入 (3), 得到:
( C ( λ ) − U 0 − 1 ) ( λ I − A ) = 0 (\bm{C}(\lambda) - \bm{U}_0^{-1})(\lambda \bm{I}-\bm{A})=\bm{0} (C(λ)−U0−1)(λI−A)=0
比较系数, 有 C ( λ ) − U 0 − 1 = 0 \bm{C}(\lambda) - \bm{U}_0^{-1} = \bm{0} C(λ)−U0−1=0, 即 C ( λ ) = U 0 − 1 (6) \bm{C}(\lambda) = \bm{U}_0^{-1} \tag{6} C(λ)=U0−1(6)
接下来验证必要条件 (3), (4), (5), (6) 的成立:
下面验证 (3) 成立.
将 (V) 代入 (1):
λ I − A = U ( λ ) ( λ I − B ) ( R ( λ ) ( λ I − A ) + V 0 ) \lambda\bm{I}-\bm{A}=\bm{U}(\lambda)(\lambda\bm{I}-\bm{B})(\bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) + \bm{V}_0) λI−A=U(λ)(λI−B)(R(λ)(λI−A)+V0)
左右同乘 U − 1 ( λ ) \bm{U}^{-1}(\lambda) U−1(λ): U − 1 ( λ ) ( λ I − A ) = ( λ I − B ) ( R ( λ ) ( λ I − A ) + V 0 ) \bm{U}^{-1}(\lambda) (\lambda\bm{I}-\bm{A}) = (\lambda\bm{I}-\bm{B})(\bm{R}(\lambda) (\lambda \bm{I}-\bm{A}) + \bm{V}_0) U−1(λ)(λI−A)=(λI−B)(R(λ)(λI−A)+V0)
整理得: ( U − 1 ( λ ) − ( λ I − B ) R ( λ ) ) ( λ I − A ) = ( λ I − B ) V 0 (7) (\bm{U}^{-1}(\lambda) - (\lambda\bm{I}-\bm{B})\bm{R}(\lambda)) (\lambda\bm{I}-\bm{A}) = (\lambda\bm{I}-\bm{B})\bm{V}_0 \tag{7} (U−1(λ)−(λI−B)R(λ))(λI−A)=(λI−B)V0(7)
此条件成立.
下面验证 U 0 \bm{U}_0 U0 可逆且 (6) 成立.
在 (7) 中通过比较系数还可以得到: C ( λ ) \bm{C}(\lambda) C(λ) 是数字矩阵, 为简便起见, 记为 C \bm{C} C.
U ( λ ) C = I − U ( λ ) ( λ I − B ) R ( λ ) = I − ( λ I − A ) V − 1 ( λ ) R ( λ ) (9) \bm{U}(\lambda)\bm{C} = \bm{I} - \bm{U}(\lambda)(\lambda \bm{I}-\bm{B})\bm{R}(\lambda) = \bm{I} - (\lambda \bm{I}-\bm{A})\bm{V}^{-1}(\lambda)\bm{R}(\lambda) \tag{9} U(λ)C=I−U(λ)(λI−B)R(λ)=I−(λI−A)V−1(λ)R(λ)(9)
代入 (U), 得到: U ( λ ) C = [ ( λ I − A ) Q ( λ ) + U 0 ] C (10) \bm{U}(\lambda)\bm{C} = [(\lambda \bm{I}-\bm{A}) \bm{Q}(\lambda) + \bm{U}_0]\bm{C} \tag{10} U(λ)C=[(λI−A)Q(λ)+U0]C(10)
联立 (9) 和 (10), 有： I − U 0 C = ( λ I − A ) [ V − 1 ( λ ) R ( λ ) + Q ( λ ) ] \bm{I} - \bm{U}_0\bm{C}= (\lambda \bm{I}-\bm{A})[\bm{V}^{-1}(\lambda)\bm{R}(\lambda)+\bm{Q}(\lambda)] I−U0C=(λI−A)[V−1(λ)R(λ)+Q(λ)]
比较系数得: V − 1 ( λ ) R ( λ ) + Q ( λ ) = 0 \bm{V}^{-1}(\lambda)\bm{R}(\lambda)+\bm{Q}(\lambda) = \bm 0 V−1(λ)R(λ)+Q(λ)=0, I − U 0 C = − A [ V − 1 ( λ ) R ( λ ) + Q ( λ ) ] = 0 \bm{I} - \bm{U}_0\bm{C} = -\bm{A}[\bm{V}^{-1}(\lambda)\bm{R}(\lambda)+\bm{Q}(\lambda)] = \bm{0} I−U0C=−A[V−1(λ)R(λ)+Q(λ)]=0, 因此 C \bm{C} C 和 U 0 \bm{U}_0 U0 可逆, 且 C = U 0 − 1 \bm{C} = \bm{U}_0^{-1} C=U0−1.
下面验证 (4), (5) 成立:
至此, 我们已经证明出: U 0 − 1 ( λ I − A ) = ( λ I − B ) V 0 \bm{U}_0^{-1}(\lambda \bm{I}-\bm{A})=(\lambda \bm{I}-\bm{B})\bm{V}_0 U0−1(λI−A)=(λI−B)V0
因此 ( λ I − A ) = U 0 ( λ I − B ) V 0 (\lambda \bm{I}-\bm{A})=\bm{U}_0(\lambda \bm{I}-\bm{B})\bm{V}_0 (λI−A)=U0(λI−B)V0, 比较系数可知 (4),(5) 成立.
进而 A = U 0 B U 0 − 1 \bm{A}=\bm{U}_0 \bm{B} \bm{U}_0^{-1} A=U0BU0−1, A ∼ B \bm{A} \sim \bm{B} A∼B.
推论. A ∼ B \mathbf{A} \sim \mathbf{B} A∼B 的另一个充要条件是 A \mathbf{A} A 与 B \mathbf{B} B 有相同的不变因子, 各阶行列式因子, 初等因子.
必要性: 若 A ∼ B \mathbf{A}\sim \mathbf{B} A∼B, 则存在 n n n 阶可逆矩阵 P \mathbf{P} P, P − 1 A P = B \mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{B} P−1AP=B, 进而有 P − 1 ( λ I − A ) P = λ I − B \mathbf{P}^{-1}(\lambda\mathbf{I}-\mathbf{A})\mathbf{P}=\lambda\mathbf{I}-\mathbf{B} P−1(λI−A)P=λI−B, 因此 ( λ I − A ) ≃ ( λ I − B ) (λ\mathbf{I}-\mathbf{A}) \simeq (λ\mathbf{I}-\mathbf{B}) (λI−A)≃(λI−B).