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多元多项式的特征列与零点的关系定理

2024/12/23 7:14:34 来源:https://blog.csdn.net/tianwangwxm/article/details/139755853  浏览:    关键词:多元多项式的特征列与零点的关系定理

下面这个定理来自《计算机代数》6.1三角列与特征列(王东明、夏壁灿著)

【定理】

C = [ C 1 , … , C r ] \mathbb{C =}\left\lbrack C_{1},\ldots,C_{r} \right\rbrack C=[C1,,Cr]为多项式组 P ⊂ K [ x ] \mathbb{P \subset}\mathcal{K\lbrack}\mathbf{x}\rbrack PK[x]的特征列,且命

I i = i n i ( C i ) P i = P ∪ { I i } i = 1 , … , r I_{i} = ini\left( C_{i} \right)\ \ \ \ \ \ \mathbb{P}_{i}\mathbb{= P \cup}\left\{ I_{i} \right\}\ \ \ \ \ i = 1,\ldots,r Ii=ini(Ci)      Pi=P{Ii}     i=1,,r

I = i n i ( C ) = { I 1 , … , I r } \mathbb{I =}ini\left( \mathbb{C} \right) = \left\{ I_{1},\ldots,I_{r} \right\} I=ini(C)={I1,,Ir}

Z e r o ( C \ I ) ⊂ Z e r o ( P ) ⊂ Z e r o ( C ) Zero\left( \mathbb{C\backslash I} \right) \subset Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C} \right) Zero(C\I)Zero(P)Zero(C)

Z e r o ( P ) = Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) Zero\left( \mathbb{P} \right) = Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} Zero(P)=Zero(C\I)i=1rZero(Pi)

K \mathcal{K} K以及 K \mathcal{K} K的任意扩域中成立

【证明】

  1. Z e r o ( C \ I ) ⊂ Z e r o ( P ) Zero\left( \mathbb{C\backslash I} \right) \subset Zero\left( \mathbb{P} \right) Zero(C\I)Zero(P)

由于 C = [ C 1 , … , C r ] \mathbb{C =}\left\lbrack C_{1},\ldots,C_{r} \right\rbrack C=[C1,,Cr]为多项式组 P ⊂ K [ x ] \mathbb{P \subset}\mathcal{K\lbrack}\mathbf{x}\rbrack PK[x]的特征列,所以 p r e m ( P , C ) = { 0 } prem\left( \mathbb{P,C} \right) = \left\{ 0 \right\} prem(P,C)={0},也就是说对于任意 P ∈ P P \in \mathbb{P} PP,都有

I 1 q 1 … I r q r P = ∑ i = 1 r C i I_{1}^{q_{1}}\ldots I_{r}^{q_{r}}P = \sum_{i = 1}^{r}C_{i} I1q1IrqrP=i=1rCi

而对于任意的 x ∈ Z e r o ( C \ I ) x \in Zero\left( \mathbb{C\backslash I} \right) xZero(C\I),都有 x ∉ Z e r o ( I 1 q 1 … I r q r ) x \notin Zero\left( I_{1}^{q_{1}}\ldots I_{r}^{q_{r}} \right) x/Zero(I1q1Irqr) x ∈ Z e r o ( C i ) x \in Zero\left( C_{i} \right) xZero(Ci),那么 P = 0 P = 0 P=0,可得 x ∈ Z e r o ( P ) x \in Zero\left( \mathbb{P} \right) xZero(P),即 Z e r o ( C \ I ) ⊂ Z e r o ( P ) Zero\left( \mathbb{C\backslash I} \right) \subset Zero\left( \mathbb{P} \right) Zero(C\I)Zero(P)

  1. Z e r o ( P ) ⊂ Z e r o ( C ) Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C} \right) Zero(P)Zero(C)

根据特征列的定义,有 C ⊂ ⟨ P ⟩ \mathbb{C \subset}\left\langle \mathbb{P} \right\rangle CP,也就是

C i = ∑ P ∈ P k P P C_{i} = \sum_{P \in \mathbb{P}}^{}{k_{P}P} Ci=PPkPP

所以,当多项式 P P P的值为 0 0 0时, C i C_{i} Ci必为 0 0 0,即 Z e r o ( P ) ⊂ Z e r o ( C ) Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C} \right) Zero(P)Zero(C)

  1. Z e r o ( P ) ⊂ Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} Zero(P)Zero(C\I)i=1rZero(Pi)

x ∈ Z e r o ( P ) x \in Zero\left( \mathbb{P} \right) xZero(P),根据2,那么有 x ∈ Z e r o ( C ) x \in Zero\left( \mathbb{C} \right) xZero(C)

x ∈ Z e r o ( I ) x \in Zero\left( \mathbb{I} \right) xZero(I),则 x ∈ ⋃ i = 1 r Z e r o ( I i ) x \in \bigcup_{i = 1}^{r}{Zero\left( I_{i} \right)} xi=1rZero(Ii),又因为 x ∈ Z e r o ( P ) x \in Zero\left( \mathbb{P} \right) xZero(P),所以 x ∈ ⋃ i = 1 r Z e r o ( P i ) x \in \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} xi=1rZero(Pi)

x ∉ Z e r o ( I ) x \notin Zero\left( \mathbb{I} \right) x/Zero(I),结合 x ∈ Z e r o ( C ) x \in Zero\left( \mathbb{C} \right) xZero(C),可得 x ∈ Z e r o ( C \ I ) x \in Zero\left( \mathbb{C\backslash I} \right) xZero(C\I)

结合上述两种情况的讨论,可得 Z e r o ( P ) ⊂ Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} Zero(P)Zero(C\I)i=1rZero(Pi)

  1. Z e r o ( P ) ⊃ Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) Zero\left( \mathbb{P} \right) \supset Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} Zero(P)Zero(C\I)i=1rZero(Pi)

根据1, Z e r o ( C \ I ) ⊂ Z e r o ( P ) Zero\left( \mathbb{C\backslash I} \right) \subset Zero\left( \mathbb{P} \right) Zero(C\I)Zero(P)

因为 Z e r o ( P i ) ⊂ Z e r o ( P ) Zero\left( \mathbb{P}_{i} \right) \subset Zero\left( \mathbb{P} \right) Zero(Pi)Zero(P),所以 ⋃ i = 1 r Z e r o ( P i ) ⊂ Z e r o ( P ) \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} \subset Zero\left( \mathbb{P} \right) i=1rZero(Pi)Zero(P)

综合可得 Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) ⊂ Z e r o ( P ) Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} \subset Zero\left( \mathbb{P} \right) Zero(C\I)i=1rZero(Pi)Zero(P)

综合1、2可得
Z e r o ( C \ I ) ⊂ Z e r o ( P ) ⊂ Z e r o ( C ) Zero\left( \mathbb{C\backslash I} \right) \subset Zero\left( \mathbb{P} \right) \subset Zero\left( \mathbb{C} \right) Zero(C\I)Zero(P)Zero(C)

综合3、4可得
Z e r o ( P ) = Z e r o ( C \ I ) ∪ ⋃ i = 1 r Z e r o ( P i ) Zero\left( \mathbb{P} \right) = Zero\left( \mathbb{C\backslash I} \right) \cup \bigcup_{i = 1}^{r}{Zero\left( \mathbb{P}_{i} \right)} Zero(P)=Zero(C\I)i=1rZero(Pi)

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