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2023-10-30
数学
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目录

题目1

题目1

定义:

σL(V1,V2)\sigma \in \mathcal{L}(V_1,V_2)σ\sigma是线性空间V1V_1V2V_2的一个线性映射)

Im(σ){σ(α)αV1}I_m(\sigma)\triangleq\{\sigma(\alpha)\mid\alpha\in V_1\}ImI_mσ\sigma的像/值域)

Ker(σ){αV1σ(α)=0}\operatorname{Ker}(\sigma)\triangleq\{\alpha\in V_1\mid\sigma(\alpha)=0\}Ker(σ)\operatorname{Ker}(\sigma)σ\sigma的核)

证明:

  1. Im(σ)I_m(\sigma)V2V_2的子空间,Ker(σ)\operatorname{Ker}(\sigma)V1V_1的子空间
  2. Im(σ)=span(σ(α1),σ(α2),,σ(αn))I_m(\sigma)=\operatorname{span}(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_n)),其中α1,α2,,αn\alpha_1,\alpha_2,\cdots,\alpha_nV1V_1的基
  3. dim(Im(σ))=rank(A)\operatorname{dim}(I_m(\sigma))=\operatorname{rank}(A),其中AAσ\sigma的矩阵
  4. dim(Im(σ))+dim(Ker(σ))=n\operatorname{dim}(I_m(\sigma))+\operatorname{dim}(\operatorname{Ker}(\sigma))=n

解:

(1) 证明子空间只需要验证加法与数乘封闭性

α,βV1,kP\forall \alpha,\beta \in V_1, k \in \mathbb{P}

由线性映射的定义有,σ(α)+σ(β)=σ(α+β)Im(σ)\sigma(\alpha)+\sigma(\beta)=\sigma(\alpha+\beta)\in I_m(\sigma)kσ(α)=σ(kα)Im(σ)k\sigma(\alpha)=\sigma(k\alpha)\in I_m(\sigma),故Im(σ)I_m(\sigma)V2V_2的子空间

α,βKer(σ),kP\forall \alpha,\beta \in \operatorname{Ker}(\sigma), k \in \mathbb{P}

有,σ(α)+σ(β)=σ(α+β)=0\sigma(\alpha)+\sigma(\beta)=\sigma(\alpha+\beta)=0σ(kα)=kσ(α)=0\sigma(k\alpha)=k\sigma(\alpha)=0,故Ker(σ)\operatorname{Ker}(\sigma)V1V_1的子空间

(2)αV1\forall \alpha \in V_1有,

α=i=1nxiαi\alpha=\sum_{i=1}^{n}x_i\alpha_i

β=σ(α)=i=1nxiσ(αi)\beta=\sigma(\alpha)=\sum_{i=1}^{n}x_i\sigma(\alpha_i)

也即βIm(σ),β=i=1nxiσ(αi)\forall \beta \in I_m(\sigma),\beta=\sum_{i=1}^{n}x_i\sigma(\alpha_i),故Im(σ)=span(σ(α1),σ(α2),,σ(αn))I_m(\sigma)=\operatorname{span}(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_n))

(3)由于dim(Im(σ))=rank(σ(α1),σ(α2),,σ(αn))\operatorname{dim}(I_m(\sigma))=\operatorname{rank}(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_n))rank(σ(α1),σ(α2),,σ(αn))=rank(A)\operatorname{rank}(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_n))=\operatorname{rank}(A),故dim(Im(σ))=rank(A)\operatorname{dim}(I_m(\sigma))=\operatorname{rank}(A)

(4)设dim(Ker(σ))=r\operatorname{dim}(\operatorname{Ker}(\sigma))=r,在Ker(σ)\operatorname{Ker}(\sigma)中取一组基α1,α2,,αr\alpha_1,\alpha_2,\cdots,\alpha_r并将其扩充到V1V_1的基α1,α2,,αr,αr+1,,αn\alpha_1,\alpha_2,\cdots,\alpha_r,\alpha_{r+1},\cdots,\alpha_n 则有

Im(σ)=span(σ(α1),σ(α2),,σ(αr),σ(αr+1),,σ(αn))I_m(\sigma)=\operatorname{span}(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_r),\sigma(\alpha_{r+1}),\cdots,\sigma(\alpha_n))

由于σ(α1),σ(α2),,σ(αr)=0\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_r)=0,则有

Im(σ)=span(σ(αr+1),,σ(αn))I_m(\sigma)=\operatorname{span}(\sigma(\alpha_{r+1}),\cdots,\sigma(\alpha_n))

现在证明σ(αr+1),,σ(αn)\sigma(\alpha_{r+1}),\cdots,\sigma(\alpha_n)是线性无关的

σ(αr+1),,σ(αn)\sigma(\alpha_{r+1}),\cdots,\sigma(\alpha_n)是线性相关的,则

j=r+1nkjσ(αj)=0\sum_{j=r+1}^{n}k_j\sigma(\alpha_j)=0

则有

σ(j=r+1nkjαj)=0\sigma(\sum_{j=r+1}^{n}k_j\alpha_j)=0

也就是说j=r+1nkjαjKer(σ)\sum_{j=r+1}^{n}k_j\alpha_j\in \operatorname{Ker}(\sigma),而α1,α2,,αr\alpha_1,\alpha_2,\cdots,\alpha_rKer(σ)\operatorname{Ker}(\sigma)的一组基,故

j=r+1nkjαj=j=1rkjαj\sum_{j=r+1}^{n}k_j\alpha_j=\sum_{j=1}^{r}k_j\alpha_j

α1,α2,,αr,αr+1,,αn\alpha_1,\alpha_2,\cdots,\alpha_r,\alpha_{r+1},\cdots,\alpha_n线性无关,故只有kj=0k_j=0,与σ(αr+1),,σ(αn)\sigma(\alpha_{r+1}),\cdots,\sigma(\alpha_n)是线性相关的矛盾

因此dim(Im(σ))=nr\operatorname{dim}(I_m(\sigma))=n-r,也就是有dim(Im(σ))+dim(Ker(σ))=n\operatorname{dim}(I_m(\sigma))+\operatorname{dim}(\operatorname{Ker}(\sigma))=n

本文作者:insomnia

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