类不变子空间与不变子空间的关系

古雯; 倪军娜

doi:10.6054/j.jscnun.2019072

类不变子空间与不变子空间的关系

古雯,
倪军娜

华南师范大学数学科学学院，广州 510631

基金项目:

广东省自然科学基金项目 2016A030313850

详细信息

通讯作者:
倪军娜，副教授，Email:nijunna@ 126.com

中图分类号: O15
计量
- 文章访问数: 2051
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出版历程
- 收稿日期: 2018-09-19
- 网络出版日期: 2021-03-21
- 刊出日期: 2019-08-24

The Relationship between Similar Invariant Subspaces and Invariant Subspaces

GU Wen,
NI Junna

School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China

摘要

摘要: 给出了“类不变子空间”的定义，研究了可逆线性变换和一般线性变换的类不变子空间与不变子空间的关系：利用向量空间的理论，证明了对于可逆线性变换，类不变子空间与不变子空间是等价的；进一步证明对于非可逆的线性变换，类不变子空间是不变子空间，反之不成立.
- 类不变子空间 /
- 不变子空间 /
- 线性变换 /
- 向量空间
Abstract: The concept of "similar invariant subspace" is defined and the relationship between similar invariant subspace and invariant subspace under the conditions of reversible linear transformation and general linear transformation is discussed. Using the theory of vector space, it is proved that similar invariant subspace is equivalent to invariant subspace under the condition of reversible linear transformation. Furthermore, it is proved that for a linear transformation σ of vector space V, if W is a similar invariant subspace, W must be an invariant subspace.
- similar invariant subspace /
- invariant subspace /
- linear transformation /
- vector space

HTML全文

近30余年，学者们讨论了特殊空间的不变子空间^[1-3]、特殊方程的不变子空间^[4-6]和特殊矩阵的不变子空间^[7-8]等问题，如：研究多圆盘的加权Bergman空间上的不变子空间和约化子空间，给出了一类解析Toeplitz算子T_{z_i} (1≤i≤n)的不变子空间的Beurling型定理^[1]; 结合压力变换和不变子空间方法中的等价变换，给出了一般非齐次非线性扩散方程的等价方程的高维不变子空间^[4].另外，“不变子空间问题”作为算子理论中的重要问题，许多学者做了相关研究^[9-17].

本文基于不变子空间的定义，给出了“类不变子空间”的定义，并利用向量空间有关理论，探讨类不变子空间的性质，研究类不变子空间与不变子空间的关系.

1. 预备知识

定义1^[18] 令V是数域F上的向量空间，σ是V的线性变换，W是V的子空间，当σ(W)⊆W, 称W是σ的不变子空间，也就是说，若α?W, 则σ(α)?W, 称W是σ的不变子空间.

引理1^[18] 若α₁, α₂, …, α_r是n维向量空间V中一组线性无关的向量，则总可以添加n－r个向量α_r+1, …, α_n, 使得{α₁, α₂, …, α_r, α_r+1, …, α_n}作为V的一个基.特别地，n维向量空间中任意n个线性无关的向量都可以作为V的基.

引理2^[18] 若W₁和W₂都是数域F上向量空间V的有限维子空间，则W₁+W₂也是有限维的，且

${\rm{dim}}\left( {{W_1} + {W_2}} \right) = {\rm{dim}}{W_1} + {\rm{dim}}{W_2}{\rm{dim}}\left( {{W_1} \cap {W_2}} \right).$

引理3^[18] 一个齐次线性方程组有非零解的充要条件是其系数矩阵的秩r小于其未知量的个数n.

2. 主要结果

首先，在定义1的基础上给出类不变子空间的定义:

定义2 令V是数域F上的向量空间，σ是V的线性变换，W是V的子空间.如果满足：对于任意的α?V，若σ(α)?W，有α?W，则称W是σ的类不变子空间.

下面探讨类不变子空间的性质，研究不同情形下“类不变子空间”与不变子空间的关系.

2.1 可逆线性变换的类不变子空间

定理1 令V是数域F上的向量空间，σ是V的可逆的线性变换，W是V的子空间，则W是σ的类不变子空间当且仅当W是σ的不变子空间.

证明先证必要性.设W是σ的类不变子空间，即∀α?W, σ^-1(α)?W, 则W是σ^-1的不变子空间，因此σ^-1(W) ⊆W，即W⊆σ(W).下证:σ(W)=W.

设{α₁, α₂，…, α_r}是向量空间W的一个基，ξ=k₁α₁+k₂α₂+…+k_rα_r?W.因为σ(ξ)=k₁σ(α₁)+k₂σ(α₂)+…+k_rσ(α_r), 所以 $\sigma \left( W \right) = {\cal L}\left( {\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right),\sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right), \cdots \sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right)} \right)$ ，故dim σ(W)≤r.又因为σ(W)⊇W，dimσ(W)≥dim W=r，所以dim σ(W)=dimW=r.又W⊆σ(W)，有σ(W)=W，则σ(W)⊆W.故W是σ的不变子空间.

再证充分性.设W是σ的不变子空间，设{α₁, α₂，…, α_r}是向量空间W的基，则易证σ(α₁), σ(α₂), …, σ(α_r)线性无关.又由于

$\left. {\sigma \left( W \right) = {\cal L}\left( {\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right)} \right), \sigma \left( {\mathit{\boldsymbol{\alpha }}{_2}} \right), \ldots , \sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right)} \right)$

则dimσ(W)=r=dim W.因为W是σ的不变子空间，σ(W)⊆W，所以σ(W)=W. ∀α?W，由于σ(W)=W，故α?σ(W).因此，存在β?W, 使得σ(β)=α，从而有σ^－1(α)=σ^－1(σ(β))=β?W, 故W是σ的类不变子空间.证毕.

2.2 一般线性变换的类不变子空间

对于双射情形下，类不变子空间与不变子空间是等价的，下面研究一般情形下的类不变子空间.设V是数域F上的向量空间，σ是V的线性变换，W是V的子空间.由于W中一定包含零向量，则由定义2可知Ker(σ)⊆W.因此，在一般线性变换的情形下，有以下定理.

定理2 设V是数域F上的向量空间，且dimV=n，σ是V的线性变换，若W是σ的类不变子空间，则W一定是σ的不变子空间.

证明 (1)若Ker(σ)={0}，即σ是可逆的线性变换.则由定理1，可得W是不变子空间.

(2) 若Ker(σ)≠{0}，设dim Ker(σ)=m，分以下2种情形讨论.

① 若dim W=dim Ker(σ)=m，由于Ker(σ)⊆W, 则有W=Ker(σ)，显然Ker(σ)是V的不变子空间，因此W是不变子空间.

② 若dim W＞dim Ker(σ)=m，由引理1，Ker(σ)的一个基{α₁, α₂, …, α_m}可以扩充为W的基{α₁, α₂, …, α_m, α_m+1, …, α_q}, 并进一步扩充到V的基{α₁, α₂, …, α_m, α_m+1, …, α_q, α_q+1, …, α_n}.

任取ξ?W, 设

$\mathit{\boldsymbol{\xi }} = {k_1}\mathit{\boldsymbol{\alpha }}{_1} + {k_2}\mathit{\boldsymbol{\alpha }}{_2} + \ldots + {k_m}\mathit{\boldsymbol{\alpha }}{_m} + {k_{m + 1}}\mathit{\boldsymbol{\alpha }}{_{m + 1}} + \ldots + {k_q}\mathit{\boldsymbol{\alpha }}{_q}$

因为σ是线性变换，所以

$\begin{array}{l} \sigma \left( \mathit{\boldsymbol{\xi }} \right) = {k_1}\sigma \left( {\mathit{\boldsymbol{\alpha }}{_1}} \right) + {k_2}\sigma \left( {\mathit{\boldsymbol{\alpha }}{_2}} \right) + \ldots + {k_m}\sigma \left( {\mathit{\boldsymbol{\alpha }}{_m}} \right) + \\ \;\;\;\;\;\;\;{k_{m + 1}}\sigma \left( {\mathit{\boldsymbol{\alpha }}{_{m + 1}}} \right) + \ldots + {k_q}\sigma (\mathit{\boldsymbol{\alpha }}{_q}). \end{array}$

由于{α₁, α₂，…, α_m}是Ker(σ)的一个基，故σ(α₁)=σ(α₂)=…=σ(α_m)=0.

若要证W是不变子空间，即要证σ(ξ)?W, 即证k_m+1σ(α_m+1)+…+k_qσ(α_q)?W, 故只需证σ(α_m+1), σ(α_m+2), …,σ(α_q)?W.首先证明σ(α_m+1)?W(用反证法).

若σ(α_i)?W (i＞q)，由于W是类不变子空间，故可得α_i?W (i＞q).显然α_i?W (i＞q)，矛盾.故σ(α_i)?W (i＞q).从而可得σ(α_i)?W (i=q+1, q+2, …, n).

若σ(α_m+1)?W, 由于W是类不变子空间，故对于V中的每个向量ξ，设

$\begin{array}{l} \mathit{\boldsymbol{\xi }}{\rm{ }} = {k_1}\mathit{\boldsymbol{\alpha }}{_1} + {k_2}\mathit{\boldsymbol{\alpha }}{_2} + \ldots + {k_m}\mathit{\boldsymbol{\alpha }}{_m} + {k_{m + 1}}\mathit{\boldsymbol{\alpha }}{_{m + 1}} + \ldots + {k_q}\mathit{\boldsymbol{\alpha }}{_q} + \\ \;\;\;\;\;{k_{q + 1}}\mathit{\boldsymbol{\alpha }}{_{q + 1}} + \ldots + {k_n}\mathit{\boldsymbol{\alpha }}{_n}. \end{array}$

若满足σ(ξ)=t₁α₁+t₂α₂+…+t_mα_m+t_m+1α_m+1+…+t_qα_q, 必然可得到ξ?W，即k_q+1=k_q+2=…=k_n=0.

令 $W\prime = {\cal L}\left( {\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{m + 1}}} \right)} \right), \sigma \left( {\mathit{\boldsymbol{\alpha }}{_{q + 1}}} \right), \sigma \left( {\mathit{\boldsymbol{\alpha }}{_{q + 2}}} \right), \ldots , \sigma (\mathit{\boldsymbol{\alpha }}{_n}))$ , 设η?W′∩W, 由于η?W′, 设

$\begin{array}{l} \mathit{\boldsymbol{\eta }} = {k_{m + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{m + 1}}} \right) + {k_{q + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{q + 1}}} \right) + \ldots + {k_{q + 2}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{q + 2}}} \right) + \ldots + \\ \;\;\;\;\;{k_n}\sigma ({\mathit{\boldsymbol{\alpha }}_n})?W \end{array}$

其中, k_m+1, k_q+1, k_q+2, …, k_n?F，可得k_q+1=k_q+2=…=k_n=0.否则，若k_i≠0 (i≥q+1)，此时存在

$\begin{array}{l} \mathit{\boldsymbol{\xi }} = {k_1}{\mathit{\boldsymbol{\alpha }}_1} + \ldots + {k_m}{\mathit{\boldsymbol{\alpha }}_m} + {k_{m + 1}}{\mathit{\boldsymbol{\alpha }}_{m + 1}} + 0{\mathit{\boldsymbol{\alpha }}_{m + 2}} + \ldots + 0{\mathit{\boldsymbol{\alpha }}_q} + \\ \;\;\;\;\;{k_{q + 1}}{\mathit{\boldsymbol{\alpha }}_{q + 1}} + \ldots + {k_i}{\mathit{\boldsymbol{\alpha }}_i} + \ldots + {k_n}{\mathit{\boldsymbol{\alpha }}_n},\\ \begin{array}{*{20}{l}} {\sigma \left( \mathit{\boldsymbol{\xi }} \right) = {k_{m + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{m + 1}}} \right) + {k_{q + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{q + 1}}} \right) + \ldots + {k_i}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_i}} \right) + \ldots + }\\ {\;\;\;\;\;{k_n}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_n}} \right)?W.} \end{array} \end{array}$

但由于k_i≠0 (i≥q+1)，不满足k_q+1=k_q+2=…=k_n=0, 与定义1矛盾，则k_q+1=k_q+2=…=k_n=0, 从而有k_m+1σ(α_m+1)?W.而σ(α_m+1)?W，则k_m+1=0，因此η=0.故W′∩W={0}.

而对于子空间W′，设

${k_{m + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{m + 1}}} \right) + {k_{q + 1}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{q + 1}}} \right) + {k_{q + 2}}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_{q + 2}}} \right) + \ldots + {k_n}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_n}} \right) = 0$

即σ(k_m+1α_m+1+k_q+1α_q+1+…+k_nα_n)=0.由于{α₁, α₂，…, α_m}是Ker(σ)的基，故可设

${k_{m + 1}}{\mathit{\boldsymbol{\alpha }}_{m + 1}} + {k_{q + 1}}{\mathit{\boldsymbol{\alpha }}_{q + 1}} + \ldots + {k_n}{\mathit{\boldsymbol{\alpha }}_n} = {k_1}{\alpha _1} + {k_2}{\mathit{\boldsymbol{\alpha }}_2} + \ldots {k_m}{\mathit{\boldsymbol{\alpha }}_m}.$

由于{α₁, α₂, …, α_n}是V的基，则k₁=k₂=…=k_m=k_m+1=k_q+1=…=k_n=0，故σ(α_m+1), σ(α_q+1), …σ(α_n)线性无关，即dim W′=n－q+1.

根据前面的证明可得：W+W′⊆V，且W∩W′={0}.由引理2，有

$\begin{array}{*{20}{l}} {{\rm{dim}}\left( {W + W'} \right) = {\rm{dim}}W + {\rm{dim}}W' - {\rm{dim}}\left( {W \cap W'} \right) = }\\ {\;\;\;\;\;\;q + n - q + 1 - 0 = n + 1n = {\rm{dim}}V} \end{array}$

这与W+W′⊆V矛盾.故σ(α_m+1)?W.

同理可证，σ(α_m+2)?W, σ(α_m+3)?W, …, σ(α_q)?W.证毕.

由定理2可知：对于一般的线性变换，类不变子空间一定是不变子空间.但反之不成立.例如：令F_n(x)表示数域F上一切次数≤n的多项式连同零多项式所组成的向量空间，σ:f(x)→f′(x).易证σ是F_n(x)的线性变换，而且不是可逆线性变换.由定义2可知，若W是V的类不变子空间，由于0?W, 则Ker(σ)⊆W, 从而有

$\begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ {f\left( x \right) = {a_0}|{a_0}?F} \right\} \subseteq W, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ {f\left( x \right) = {a_1}x + {a_0}|{a_1}, {a_0}?F} \right\} \subseteq W, \\ \;\;\;\;\;\;\left\{ {f\left( x \right) = {a_2}{x^2} + {a_1}x + {a_0}|{a_2}, {a_1}, {a_0}?F\} \subseteq W, } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \\ \left\{ {f\left( x \right) = {a_n}{x^n} + {a_{n1}}{x^{n1}} + \ldots + {a_2}{x^2} + {a_1}x + {a_0}|{a_i}?F, } \right.\\ \;\;\;\;\;\;\;\;\;\;\left. {i = 0, 1, 2, \ldots , n} \right\} \subseteq W, \end{array}$

则W=F_n(x), 故F_n(x)只有1个类不变子空间，也就是它本身，显然F_n(x)是它本身的不变子空间.但是，一般情形下，类不变子空间与不变子空间并不是等价的.例如F_n－1(x)作为F_n(x)的子空间，根据不变子空间与类不变子空间的定义，易发现F_n－1(x)是F_n(x)的不变子空间，但不是F_n(x)的类不变子空间.

下面着重研究σ不是V的可逆的线性变换的情况.

定理3 设V是数域F上的向量空间，且dimV=n，σ是V的线性变换，且不可逆，若W是V的类不变子空间，则σ(W)≠W.

证明根据定理2，若W是V的类不变子空间，则W是V的不变子空间，即σ(W)⊆W，下证σ(W)≠W.

由引理1，W的基{α₁, α₂, …, α_r}可以扩充为V的基{α₁, α₂, …, α_n}，令(σ(α₁), σ(α₂), …, σ(α_n))= (α₁, α₂, …, α_n)A, 其中A为线性变换σ在V中的基{α₁, α₂, …, α_n}下的矩阵.设α=m₁α₁+m₂α₂+…+m_nα_n，令σ(α)?W，不妨设σ(α)=k₁α₁+k₂α₂+…+k_rα_r.由定义2可得α?W，即m_r+1=m_r+2=…=m_n=0.

而由于

$\begin{array}{l} \sigma \left( \mathit{\boldsymbol{\alpha }} \right) = {m_1}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right) + {m_2}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right) + \ldots + {m_n}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_n}} \right) = \\ \;\;\;\;\;\left( {\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right), \sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right), \ldots , \sigma \left( {{\mathit{\boldsymbol{\alpha }}_n}} \right)} \right){\left( {{m_1}, {m_2}, \ldots , {m_n}} \right)^{\rm{T}}} = \\ \;\;\;\;\;\left( {{\mathit{\boldsymbol{\alpha }}_1}, {\rm{ }}{\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\mathit{\boldsymbol{\alpha }}_n}} \right){\rm{ }}\mathit{\boldsymbol{A}}{\left( {{m_1}, {m_2}, \ldots , {m_n}} \right)^{\rm{T}}}, \end{array}$

则类不变子空间的定义可转变一种表达形式：由

$\begin{array}{l} \left( {{\mathit{\boldsymbol{\alpha }}_1}, {\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\mathit{\boldsymbol{\alpha }}_n}} \right)\mathit{\boldsymbol{A}}{\rm{ }}{\left( {{m_1}, {m_2}, \ldots , {m_n}} \right)^{\rm{T}}} = \\ \;\;\;\;\;\left( {{\mathit{\boldsymbol{\alpha }}_1}, {\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\mathit{\boldsymbol{\alpha }}_n}} \right){\left( {{k_1}, {k_2}, \ldots , {k_r}, 0, 0, \ldots , 0} \right)^{\rm{T}}} \end{array}$

可推出m_r+1=m_r+2=…=m_n=0, 其中k₁α₁+k₂α₂+…+k_rα_r?σ(V)∩W.即由

$\mathit{\boldsymbol{A}}{\left( {{m_1}, {m_2}, \ldots , {m_n}} \right)^{\rm{T}}} = {\left( {{k_1}, {k_2}, \ldots , {k_r}, 0, 0, \ldots , 0} \right)^{\rm{T}}}$

可推出m_r+1=m_r+2=…=m_n=0, k₁α₁+k₂α₂+…+k_rα_r?σ(V)∩W.

由于此时研究的σ是不可逆的线性变换，所以σ在V中的基{α₁, α₂, …, α_n}下的矩阵A不可逆.设

$\mathit{\boldsymbol{A}} = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1r}}}&{{a_{1, r + 1}}}& \cdots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2r}}}&{{a_{2, r + 1}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots &{}& \vdots & \vdots &{}& \vdots \\ {{a_{r1}}}&{{a_{r2}}}& \cdots &{{a_{rr}}}&{{a_{r, r + 1}}}& \cdots &{{a_{r, n}}}\\ {{a_{r + 1, 1}}}&{{a_{r + 1, 2}}}& \cdots &{{a_{r + 1, r}}}&{{a_{r + 1, r + 1}}}& \cdots &{{a_{r + 1, n}}}\\ \vdots & \vdots &{}& \vdots & \vdots &{}& \vdots \\ {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nr}}}&{{a_{n, r + 1}}}& \cdots &{{a_{nn}}} \end{array}} \right),$

即对于线性方程组

$\left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1r}}}&{{a_{1, r + 1}}}& \cdots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2r}}}&{{a_{2, r + 1}}}& \cdots &{{a_{2n}}}\\ \vdots & \vdots &{}& \vdots & \vdots &{}& \vdots \\ {{a_{r1}}}&{{a_{r2}}}& \cdots &{{a_{rr}}}&{{a_{r, r + 1}}}& \cdots &{{a_{r, n}}}\\ {{a_{r + 1, 1}}}&{{a_{r + 1, 2}}}& \cdots &{{a_{r + 1, r}}}&{{a_{r + 1, r + 1}}}& \cdots &{{a_{r + 1, n}}}\\ \vdots & \vdots &{}& \vdots & \vdots &{}& \vdots \\ {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nr}}}&{{a_{n, r + 1}}}& \cdots &{{a_{nn}}} \end{array}} \right)\left( \begin{array}{l} {m_1}\\ {m_2}\\ \; \vdots \\ {m_r}\\ {m_{r + 1}}\\ \; \vdots \\ {m_n} \end{array} \right) = \left( \begin{array}{l} {k_1}\\ {k_2}\\ \; \vdots \\ {k_r}\\ 0\\ \; \vdots \\ 0 \end{array} \right),$

(1)

将m₁, m₂, …, m_n当作自变量，当满足k₁α₁+k₂α₂+…+k_rα_r?σ(V)∩F的k_i (i=1, 2, …, r)选取后，方程组的解均形如(m₁, m₂, …, m_r, 0, 0, …, 0)^T.那么，对于k₁= k₂=…=k_r=0，其解也均形如(m₁, m₂, …, m_r, 0, 0, …, 0)^T.由于A不可逆，故方程组(1)一定有非零解.

设其非零解为(m₁, m₂, …, m_r, 0, 0, …, 0)^T，其中m_i不全为0.即方程组

$\left\{ \begin{array}{l} {a_{11}}{m_1} + {a_{12}}{m_2} + \ldots + {a_{1r}}{m_r} = 0, \\ {a_{21}}{m_1} + {a_{22}}{m_2} + \ldots + {a_{2r}}{m_r} = 0, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \\ {a_{r1}}{m_1} + {a_{r2}}{m_2} + \ldots + {a_{rr}}{m_r} = 0, \\ {a_{r + 1, 1}}{m_1} + {a_{r + 1, 2}}{m_2} + \ldots + {a_{r + 1, r}}{m_r} = 0, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \\ {a_{n1}}{m_1} + {a_{n2}}{m_2} + \ldots + {a_{nr}}{m_r} = 0 \end{array} \right.$

(2)

一定有非零解.

由引理3可知系数矩阵的秩小于未知量的个数r.令β_i (i=1, 2, …, r)表示线性方程组(2)的系数矩阵的第i个列向量，则 ${\rm{dim}}\left( {{\cal L}\left( {{\mathit{\boldsymbol{\beta }}_1}, {\rm{ }}{\mathit{\boldsymbol{\beta }}_2}, \ldots , {\rm{ }}{\mathit{\boldsymbol{\beta }}_r}} \right)} \right) < r$ ，即β₁, β₂, …, β_r线性相关.因此，存在t₁, t₂, …, t_r不全为0，使得

${t_1}{\mathit{\boldsymbol{\beta }}_1} + {t_2}{\mathit{\boldsymbol{\beta }}_2} + \ldots + {t_r}{\mathit{\boldsymbol{\beta }}_r} = \mathit{\pmb{0}}.$

由于

$\begin{array}{l} \sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right) = {a_{11}}{\mathit{\boldsymbol{\alpha }}_1} + {a_{21}}{\mathit{\boldsymbol{\alpha }}_2} + \ldots + {a_{n1}}{\mathit{\boldsymbol{\alpha }}_n} = \left( {{\rm{ }}{\mathit{\boldsymbol{\alpha }}_1}, {\rm{ }}{\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\rm{ }}{\mathit{\boldsymbol{\alpha }}_n}} \right){\rm{ }}{\mathit{\boldsymbol{\beta }}_1}, \\ \sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right) = {a_{12}}{\mathit{\boldsymbol{\alpha }}_1} + {a_{22}}{\mathit{\boldsymbol{\alpha }}_2} + \ldots + {a_{n2}}{\mathit{\boldsymbol{\alpha }}_n} = \left( {{\rm{ }}{\mathit{\boldsymbol{\alpha }}_1}, {\rm{ }}{\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\rm{ }}{\mathit{\boldsymbol{\alpha }}_n}} \right){\rm{ }}{\mathit{\boldsymbol{\beta }}_2}, \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \\ \sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right) = {a_{1r}}{\mathit{\boldsymbol{\alpha }}_1} + {a_{2r}}{\mathit{\boldsymbol{\alpha }}_2} + \ldots + {a_{nr}}{\mathit{\boldsymbol{\alpha }}_n} = \left( {{\rm{ }}{\mathit{\boldsymbol{\alpha }}_1}, {\rm{ }}{\mathit{\boldsymbol{\alpha }}_2}, \ldots , {\rm{ }}{\mathit{\boldsymbol{\alpha }}_n}} \right){\rm{ }}{\mathit{\boldsymbol{\beta }}_r}, \end{array}$

则

$\begin{array}{*{20}{l}} {{t_1}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right) + {t_2}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right) + \ldots + {t_r}\sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right) = }\\ {\;\;\;\;\;\left( {{\mathit{\boldsymbol{\alpha }}_1},{\mathit{\boldsymbol{\alpha }}_2}, \ldots ,{\mathit{\boldsymbol{\alpha }}_n}} \right)\left( {{t_1}{\mathit{\boldsymbol{\beta }}_1} + {t_2}{\mathit{\boldsymbol{\beta }}_2} + \ldots + {t_r}{\mathit{\boldsymbol{\beta }}_r}} \right) = }\\ {\;\;\;\;\;\left( {{\mathit{\boldsymbol{\alpha }}_1},{\mathit{\boldsymbol{\alpha }}_2}, \ldots ,{\mathit{\boldsymbol{\alpha }}_n}} \right)\mathit{\pmb{0}} = \mathit{\pmb{0}},} \end{array}$

故σ(α₁), σ(α₂), …, σ(α_r)线性相关.而 $\sigma \left( W \right) = {\cal L}\left( {\sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right), \sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right), \cdots \sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right)} \right), \sigma \left( {{\mathit{\boldsymbol{\alpha }}_1}} \right), \sigma \left( {{\mathit{\boldsymbol{\alpha }}_2}} \right), \cdots , \sigma \left( {{\mathit{\boldsymbol{\alpha }}_r}} \right)$ 线性相关，则dimσ (W)＜r=dimW.因此，σ(W)≠W.证毕.

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