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Latex源代码推荐例子

时间:2009-11-14 17:22:56  来源:互联网  作者:未知
/documentclass[twoside,11pt,openany]{cctbook}
/usepackage{amsfonts}
/usepackage{mathrsfs}
/usepackage{amsmath}
/usepackage{amssymb}
/usepackage{fancyhdr}
/usepackage{graphicx}
%/usepackage{float} %不想浮动可用 float 包中的 H 参数/usepackage{float}[H]
/usepackage{cite}%产生如(文[2,4-8])的效果
/usepackage[nooneline,flushleft]{caption2}
/newtheorem{thm}{定理}[chapter]
/newtheorem{defn}{定义}[chapter]
/newtheorem{lt}{例}[chapter]
/newtheorem{tl}{推论}[chapter]
/newtheorem{mt}{命题}[chapter]
/newtheorem{yl}{引理}[chapter]
/newtheorem{jx}{假设}[chapter]
/input vatola.sty 
/textwidth 159mm /textheight 251mm /topmargin -5mm /oddsidemargin
5mm /evensidemargin 5mm
/renewcommand/baselinestretch{1.5}
/addtocounter{page}{0}/parindent=2/ccwd


/catcode`@=11

/@matchccfont{/itshape}{/oitshape}{/songti }


% Change definition of /thebibliography environment to use smaller font.
/renewenvironment{thebibliography}[1]
%org     {/chapter*{/bibname
%org    /@mkboth{/MakeUppercase/bibname}{/MakeUppercase/bibname}}%
%%%     {/def/chaptername{}/chapter{/bibname}%                            !!!
     {/def/chaptername{}/chapter{/bibname}%                            !!!
      /list{/@biblabel{/@arabic/c@enumiv}}%
           {/settowidth/labelwidth{/@biblabel{#1}}%
            /leftmargin/labelwidth
            /advance/leftmargin/labelsep
            /@openbib@code
            /usecounter{enumiv}%
            /let/p@enumiv/@empty
            /renewcommand/theenumiv{/@arabic/c@enumiv}}%
      /small%                                               !!!
      /sloppy
      /clubpenalty4000
      /@clubpenalty /clubpenalty
      /widowpenalty4000%
      /sfcode`/./@m}
     {/def/@noitemerr
       {/@latex@warning{Empty `thebibliography' environment}}%
      /endlist}


%%%%%%%%%%%%%%%%%%%%%%%%%%%开始章的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define chapter
/renewcommand/chaptername{第/thechapter章}
/def/@chapter[#1]#2{/ifnum /c@secnumdepth >/m@ne
                           /thispagestyle{headings}%                       !!!
                       /if@mainmatter
                           /refstepcounter{chapter}%
                          /protected@xdef/@currentlabel{/chaptername}%  !!!
                          /typeout{/huge /@chapapp /space /thechapter.}%
                          /addcontentsline{toc}{chapter}{/protect/numberline{/heiti/zihao{-4}/chaptername} {/heiti/zihao{-4}#1}}%  !!!
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                          /addcontentsline{toc}{chapter}{/heiti/zihao{-4}#1}%
                       /fi
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                       /addcontentsline{toc}{chapter}{/zihao{-4}#1}%
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                    /chaptermark{#1}%
                    /addtocontents{lof}{/protect/addvspace{10/p@}}%
                    /addtocontents{lot}{/protect/addvspace{10/p@}}%
                    /if@twocolumn
                       /@topnewpage[/@makechapterhead{#2}]%
                    /else
                       /@makechapterhead{#2}%
                       /@afterheading
                    /fi
                    }
/def/@chapapp{Chapter}%                   !!!
/def/chapterformat{/huge/bfseries/centering}%      |||
/def/@makechapterhead#1{%

                /if@mainmatter
                          /renewcommand/leftmark{/zihao{5}第/thechapter 章/ / / #1}
                       /else
                          /renewcommand/leftmark{/zihao{5} #1}%
                       /fi
                         /vspace*{-/headsep}/vspace*{-/headheight}/vspace*{15/p@}%      !!!
                         {/chapterformat%                       !!!
                         /ifnum /c@secnumdepth >/m@ne%                                !!!
                             /if@mainmatter%                                            !!!
                                /chaptername /quad #1 /par/nobreak%      !!!
                             /else%                                                     !!!
                                #1 /par/nobreak%                         !!!
                             /fi%                                                       !!!
                         /fi%                                                         !!!
                         /vskip 15/p@%                                                !!!
                         }
                        }
/def/@makeschapterhead#1{%
                            /vspace*{-/headsep}/vspace*{-/headheight}/vspace*{15/p@}%   !!!
                            {/chapterformat%                        !!!
                             /interlinepenalty/@M%                                     !!!
                             #1/par/nobreak%                                           !!!
                             /vskip 15/p@%                                             !!!
                            }
                         }
%%%%%%%%%%%%%%%%%%%%%%%%%%%结束章的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%开始节的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
/renewcommand/section{
/renewcommand/sectionname{/hspace*{1.6/ccwd}/thechapter./arabic{section}}
/@startsection
            {section}{0}{/z@}%   {name}{depth}{}
                {3.5ex /@plus 1ex /@minus .2ex}% positive->leave parindent
                {2.3ex /@plus.2ex}%
                {/protected@xdef/@currentlabel{/sectionname}/reset@font/Large/bfseries/sectionformat}
                }
/renewcommand/sectionformat{}

%%%%%%%%%%%%%%%%%%%%%%%%%%%结束节的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%开始目录的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
/def/@pnumwidth{1.55em}
/def/@tocrmarg {2.55em}
/def/@dotsep{4.5}
/setcounter{tocdepth}{2}
/def/tableofcontents{/@restonecolfalse
/thispagestyle{plain}%
/if@twocolumn/@restonecoltrue/onecolumn /fi
/chapter*{目录}
/@starttoc{toc}/if@restonecol/twocolumn/fi}
/def/l@chapter{/@dottedtocline{0}{1.5em}{2.3em}}
/renewcommand/sectionname{/zihao{-4}/large/thechapter./thesection}
/def/l@section{/@dottedtocline{1}{-0.3em}{3.2em}}
/def/l@subsection{/@dottedtocline{2}{7.0em}{4.1em}}
/renewcommand/subsectionname{/zihao{-4}/thechapter./arabic{section}./arabic{subsection}}
/def/l@subsubsection{/@dottedtocline{3}{10.0em}{5.0em}}
/def/l@paragraph{/@dottedtocline{4}{12.0em}{6.0em}}
/def/l@subparagraph{/@dottedtocline{5}{12em}{6em}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%结束目录的定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%开始页眉和页脚定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
/def/ps@headings{%
/def/@evenfoot{/ /hfill /centering{/small/zihao{5}/ /thepage/ /hfill / }}
/def/@oddfoot{/ /hfill /centering{/small/zihao{5} / /thepage/ /hfill / }}
/def/@evenhead{/vbox{/centering{/small/kaishu /zihao{5}满足秩条件的矩阵对的标准形}//[2pt]
/rule{/textwidth}{1mm}
//[2pt]
/rule{/textwidth}{.5pt}}}
/def/@oddhead{/vbox{/centering{/small/zihao{5}/kaishu/protect满足秩条件的矩阵对的标准形}//[2pt]
/rule{/textwidth}{1mm}//[2pt]
/rule{/textwidth}{.5pt} }} } /ps@headings
%%%%%%%%%%%%%%%%%%%%%%%%%%%结束页眉和页脚定义%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
/catcode`@=12


/def/proof{/indent{/heiti 证明/ / } /ignorespaces}
/def/endproof{/vbox{/hrule height0.6pt/hbox{/vrule height1.3ex width0.6pt/hskip0.8ex/vrule width0.6pt}
               /hrule height0.6pt}/vskip 3mm}




/begin{document}
%############封面###############
%/baselineskip=0.80cm
/thispagestyle{empty} /vspace{6cm} /LARGE/zihao{3}
%/pagestyle{empty}
{/ / / } /vspace{3cm}

/begin{center}
{/huge/zihao{2}/bf/heiti/ziju{0.1} 满足秩条件的矩阵对的标准形}
/end{center}
/vspace{4cm}

$$/begin{array}{ll}
/mbox{{/heiti 学/hspace{8mm}院:}}&/mbox{/heiti 学院}//[5mm]
/mbox{{/heiti 年/hspace{8mm}级:}}&/mbox{/heiti 2004级}//[5mm]
/mbox{{/heiti 专/hspace{8mm}业:}}&/mbox{/heiti 数学与应用数学}//[5mm]
/mbox{{/heiti 姓/hspace{8mm}名:}}&/mbox{/heiti  姓名}//[5mm]
/mbox{{/heiti 学/hspace{8mm}号:}}&/mbox{/heiti 888}//[5mm]
/mbox{{/heiti 指导教师:}}&/mbox{/heiti 老师名字}
/end{array}$$
%##############封面结束##############

%/frontmatter
/chapter*{摘/ / 要}
/renewcommand{/thepage}{/Roman{page}}
/setcounter{page}{1} /headsep=6mm
/headheight=15.24pt%5mm
%摘要
%/addcontentsline{toc}{chapter}{/heiti 摘要} /label{abstract}

/zihao{-4}/large
矩阵的秩是矩阵代数中一个基本而深刻的概念,它贯穿于整个矩阵代数的理论,在代数研究中有着非常重要的作用。并且关于矩阵的
秩早已有很多经典的不等式被证明,例如:Sylvester不等式与Frobenius不等式。但是我们有时更需要知道这些不等式
等号成立的充分必要条件。关于这些不等式的等号成立问题早已有人关注,并获得了一些成果。本文用矩阵的等价分解和分块技术给出了几个矩阵秩的
不等式等号成立的充要条件。

/vspace{2cm}
/begin{center}
{/heiti/zihao{2} 关键词}
/end{center}

秩;矩阵;标准形。

%英文摘要
/chapter*{Abstract}
%/addcontentsline{toc}{chapter}{/bf Abstract}
Rank of the matrices is a basic and profound concept of the matrix algebra, which runs through 
the whole theory of matrix algebra and plays a very important role in the study of the algebra. And many classic inequalities
about the rank of the matrices have been proved, for example, the Sylvester and Frobenius inequalities. But sometimes we more often want to know the equivalent conditions
under which the equality holds. Inequalities on these issues have long been concerned, and some people have
achieved some results. In this paper we will give some necessary and sufficient conditions under which several inequalities
about the rank of matrices became equalities by using matrix equivalent decomposition and block technology of matrices.



/vspace{2cm}
/begin{center}
{/bf/huge Key words}
/end{center}

Rank;/ / Matrix;/ / Standard Form.
/newpage
%摘要完毕
%目录
%/makeatletter
%/renewcommand*/l@chapter{/@dottedtocline{0}{0em}{1.5em}}
%/makeatother%这三行命令会让目录中的章也有.....连接
%/tableofcontents
/vspace*{0.8cm} /centerline{/zihao{2} /heiti 目录} /vspace{0.6cm}
/thispagestyle{empty}
%/pagestyle{empty}
/zihao{-4}/large {/heiti  摘要/dotfill I}

{/bf Abstract}/dotfill II//


{/heiti 第1章/ / 前言/dotfill 1}

{/heiti 第2章/ / 预备知识/dotfill 2}

{/heiti 第3章/ / 满足秩条件的矩阵对的标准形/dotfill 3}

/hspace{0.9cm}3.1 矩阵乘积的秩等于因子秩的矩阵对的标准形/dotfill 3

/hspace{0.9cm}3.2 满足秩可加性条件的矩阵对的标准形/dotfill 6

/hspace{0.9cm}3.3 满足Sylvester秩等式的矩阵对的标准形/dotfill 8

{/heiti 结论 /dotfill 11}

{/heiti 参考文献 /dotfill 12}

{/heiti 致谢 /dotfill 13}

%目录结束
/newpage
/chapter{前言}
%/addcontentsline{toc}{chapter}{/heiti 前言} 
/renewcommand{/thepage}{/arabic{page}}
/setcounter{page}{1}
矩阵的秩是矩阵的非常重要的数字特征,它反应了矩阵的本质特性。它贯穿于整个矩阵代数的理论,它与线性方程组、线性空间等都有着非常密切的联系。并且关于矩阵的
秩早已有很多经典的不等式被证明,例如:

(i)/ / 秩($AB$)$/leq$$/rm{min}$/{秩$(A)$,秩($B$)/}$^{[2]}$;

(ii)秩($A+B$)$/leq$秩($A$)+秩($B$)$^{[2]}$;

(iii)/ / (Sylvester不等式$^{[4]}$)设$A$和$B$分别是$m/times n$和$n/times l$矩阵,
则/begin{center}秩($AB$)$/geq$秩($A$)+秩($B$)$-$$n$;/end{center}

(iv)/ / (Frobenius不等式$^{[5]}$)
设$A$、$B$和$C$分别是$m/times n $、$n/times l$和$l/times s$矩阵,则/begin{center}秩($ABC$)$/geq$秩($AB$)+秩($BC$)$-$秩($B$)。/end{center}并由此延伸出许多的关于矩阵秩的不等式。

但是我们有时更需要知道这些不等式什么时候取等
号,即想知道这些不等式的等号成立的充分必要条件是什么。这些问题已有一些结果,例如:

秩($AB$)=秩($A$)的充要条件是存在矩阵$C$使得$A=ABC$$^{[3]}$;

秩($AB$)=秩($B$)的充要条件是$ABx=0$和$Bx=0$同解$^{[1]}$;

秩($AB$)$=$秩($A$)+秩($B$)$-$$n$的充要条件是存在矩阵$X,Y$使得$XA+BY=I_n$$^{[5]}$;

虽然找出了一些等价条件,但它们的形式和所使用方法并不统一,各个等价条件的相关程度并不大。本文要做的工作就是
用比较统一的标准形方法来推出几个矩阵秩的不等式等号成立的充要条件,从矩阵的秩和矩阵标准形作为出发点来描述这些充分必要条件。
/chapter{预备知识}
本章介绍的几个基本概念和后面将使用的几个基本结论均摘自文献[2]。
/begin{thm}/label{t1.1.1}
对于矩阵$A$,存在可逆矩阵$P$/mbox{和}$Q$使
/begin{equation}A=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q/label{1.1}
/end{equation}
/end{thm}

/begin{defn}
称$(/ref{1.1})$式为矩阵$A$的等价分解,称$/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )$为矩阵$A$的等价标准形。
/end{defn}

/begin{defn}/label{d1.1.2}
称矩阵$A$的等价标准形中的$r$为矩阵$A$的秩。
/end{defn}

/begin{thm}/label{t1.1.1}
若$B$可逆,则秩$(AB)$=秩$(A)$=秩$(BA)$。
/end{thm}

/begin{thm}/label{t1.1.1}
矩阵$A$左乘可逆矩阵$B$相当于对矩阵$A$进行初等行变换,矩阵$A$右乘可逆矩阵$B$相当于对矩阵$A$进行初等列变换。
/end{thm}

/chapter{满足秩条件的矩阵对的标准形}

本章给出几类满足秩条件的矩阵对的标准形,先在$/S3.1$中给出矩阵乘积的秩等于因子的秩的矩阵对的标准形,然后在$/S3.2$中给出
满足秩可加条件的矩阵对的标准形,最后在$/S3.3$中给出满足Sylvester秩等式的矩阵对的标准形。
/section{矩阵乘积的秩等于因子秩的矩阵对的标准形}
/begin{thm}/label{t1.1.1}
设$A/in F^{m/times n}$,$B/in F^{n/times p}$,则秩$(AB)$=秩$(A)$的充要条件是存在可逆阵$P/in F^{m/times n}$、$Q/in F^{n/times n}$和$R/in F^{p/times p}$使得
$$
A=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q
$$$$
B=Q^{-1} /left (/begin{array}{cc}
I_s&0//
0&0
/end{array}
/right )R
$$
其中$r/leq s$。
/end{thm}
{/heiti 证明}/ /充分性。当存在可逆阵$P/in F^{m/times n}$、$Q/in F^{n/times n}$和$R/in F^{p/times p}$使得
$$
A=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q/mbox{,}
/hspace{5mm}
B=Q^{-1} /left (/begin{array}{cc}
I_s&0//
0&0
/end{array}
/right )R
$$
时,其中$r/leq s$,由分块矩阵的乘法知$$AB=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )R$$注意到$P,R$可逆,使用定理2.2推出秩($AB$)=$r$,即秩($AB$)=秩($A$)。

必要性。由定理2.1可知存在可逆阵$P$和$Q_1$使得$$A=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q_1$$注意到$Q_1B$可经初等列变换化为
/begin{center}
$/left( /begin{array}{cc}
0&B_1//
0&B_2
/end{array} /right )$/hspace{3mm}((0/hspace{2mm}$B_1$)是前$r$行)/end{center}
于是由定理2.3知有可逆阵$R_1$使/begin{center}$B=Q_1^{-1}/left( /begin{array}{cc}
0&B_1//
0&B_2
/end{array} /right )R_1$/end{center}此时$$AB=P/left( /begin{array}{cc}
0&B_1//
0&0
/end{array} /right )R_1$$由已知条件秩($AB$)=秩($A$)及$P$、$R_1$可逆,推出秩$B_1$=$r$,即$B_1$为行满秩,从而有可逆阵$N$使得$B_1=(I_r/hspace{2mm}0)N$。
此时$$
B=Q_1^{-1}/left( /begin{array}{cc}
0&(I_r/hspace{2mm}0)N//
0&B_2
/end{array} /right )R_1=Q_1^{-1}/left ( /begin {array}{ccc}
0&I_r&0//
0&B_3&B_4
/end{array} /right )/left ( /begin {array}{cc}
I&0//
0&N
/end{array} /right ) R_1%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$
其中$(B_3/hspace {2mm} B_4)$是$B_2N^{-1}$的分块写法。故
$$
B=Q_1^{-1}/left ( /begin {array}{cc}
I&0//
B_3&I
/end{array} /right )
/left ( /begin {array}{ccc}
0&I_r&0//
0&0&B_4
/end{array} /right )/left ( /begin {array}{cc}
I&0//
0&N
/end{array} /right ) R_1
$$
对于$B_4$,有可逆阵$X$和$Y$使$$B_4=X /left ( /begin {array}{cc}
I_l&0//
0&0
/end{array} /right ) Y$$设$$
Q^{-1}=Q_1^{-1}/left ( /begin {array}{cc}
I&0//
B_3&I
/end{array} /right )
/left ( /begin{array}{cc}
I&0//
0&X
/end{array}/right )
$$
$$R_2=/left(/begin{array}{ccc}
I&0&0//
0&I&0//
0&0&Y
/end{array}/right )
/left ( /begin {array}{cc}
I&0//
0&N
/end{array} /right ) R_1$$
于是
$$
B=Q^{-1}
/left ( /begin {array}{cccc}
0&I_r&0&0//
0&0&I_l&0//
0&0&0&0
/end{array} /right )R_2
$$
又因为矩阵$$/left ( /begin {array}{cccc}
0&I_r&0&0//
0&0&I_l&0//
0&0&0&0
/end{array} /right )$$可以经列变换化为
$$/left( /begin {array}{cc}
I_s&0//
0&0
/end{array}/right )$$其中$s=r+l$,则有可逆阵$S$使
$$
/left ( /begin {array}{cccc}
0&I_r&0&0//
0&0&I_l&0//
0&0&0&0
/end{array} /right )=/left( /begin {array}{cc}
I_s&0//
0&0
/end{array}/right )S
$$
此时/begin{equation}
B=Q^{-1} 
/left( /begin {array}{cc}
I_s&0//
0&0
/end{array}/right )
R /label{3.1}
/end{equation}
其中$R=SR_2$。并且
$$/left./begin{array}{rl}
&P/left ( /begin{array}{cc} 
I_r&0//
0&0 /end{array} /right )Q//
=&P/left ( /begin{array}{cc} 
I_r&0//
0&0 /end{array} /right )/left ( /begin{array}{cc} 
I&0//
0&X^{-1} /end{array} /right )/left ( /begin{array}{cc} 
I&0//
-B_3&I /end{array} /right )Q_1//
=&P/left ( /begin{array}{cc} 
I_r&0//
0&0 /end{array} /right )Q_1//
=&A
/end{array}
/right.$$
这和(3.1)一起完成了定理3.1的证明。
/begin{thm}/label{t1.1.2}
设$A/in F^{m/times n}$,$B/in F^{n/times p}$,则秩$(AB)$=秩$(B)$的充要条件是存在可逆阵$P$、$Q$和$R$使$$A=P/left ( /begin{array}{cc} 
I_r&0//
0&0 /end{array} /right )Q^{-1}$$$$B=Q/left ( /begin{array}{cc} 
I_s&0//
0&0 /end{array} /right )R$$其中$r/geq s$。
/end{thm}
{ /heiti 证明}由定理3.1易见

/hspace{8mm}秩$(AB)$=秩$(B)$

$/Longleftrightarrow$秩$(B^TA^T)$=秩$(B^T)$

$/Longleftrightarrow$存在可逆阵/~{P}、/~{Q}和/~{R}使得
$$
B^T=/mbox{/~{P}}/left( /begin{array}{cc}
I_s&0//
0&0
/end{array} /right)/mbox{/~{Q}}
$$
$$
A^T={/mbox{/~{Q}}}^{-1}/left( /begin{array}{cc}
I_r&0//
0&0
/end{array} /right)/mbox{/~{R}}
$$
其中$s /leq r$。

$/Longleftrightarrow$存在可逆阵$P$、$Q$和$R$使得
$$
A=P/left ( /begin{array}{cc} 
I_r&0//
0&0 /end{array} /right )Q^{-1}
$$
$$
B=Q/left ( /begin{array}{cc} 
I_s&0//
0&0 /end{array} /right )R
$$
其中$r/geq s$。
/section{满足秩可加性条件的矩阵对的标准形}
/begin{thm}/label{t1.1.3}
设$A,B/in F^{m/times n}$,则秩$(A+B)$=秩$(A)$$+$秩$(B)$的充要条件是存在可逆矩阵$P/in F^{m/times m}$和$Q/in F^{n/times n }$使得$$A=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q$$$$B=P/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )Q$$且$r+s/leq /rm{ min}/{m,n/}$。
/end{thm}
{/heiti 证明}充分性。由已知易见$$A+B=P/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q+P/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )Q=P/left (/begin{array}{ccc}
I_r&0&0//
0&0&0//
0&0&I_s
/end{array}
/right )Q$$($r+s/leq /rm{min}/{m,n/}$),于是秩$(A+B)$=秩$(A)$+秩$(B)$。

必要性。由定理2.1知存在可逆矩阵$P_1/mbox{、}P_2/in F^{m/times m}/mbox{及}Q_1/mbox{、}Q_2/in F^{n/times n}$使得$$A=P_1/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q_1$$$$B=P_2/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )Q_2$$
令$$P_2^{-1}P_1=/left (/begin{array}{cc}
A_1&A_3//
A_2&A_4
/end{array}
/right )$$$$Q_2Q_1^{-1}=/left (/begin{array}{cc}
B_1&B_3//
B_2&B_4
/end{array}
/right )$$其中$A_2,B_2/in F^{s/times r}$,则
$$
A=P_1/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q_1=P_2P_2^{-1}P_1/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q_1$$
/hspace{3.9cm}
$
=P_2/left (/begin{array}{cc}
A_1&A_3//
A_2&A_4
/end{array}
/right )/left (/begin{array}{cc}
I_r&0//
0&0
/end{array}
/right )Q_1=P_2/left (/begin{array}{cc}
A_1&0//
A_2&0
/end{array}
/right )Q_1
$
$$
B=P_2/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )Q_2=P_2/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )Q_2Q_1^{-1}Q_1$$
/hspace{4.05cm}$=P_2/left (/begin{array}{cc}
0&0//
0&I_s
/end{array}
/right )/left (/begin{array}{cc}
B_1&B_3//
B_2&B_4
/end{array}
/right )Q_1=P_2/left (/begin{array}{cc}
0&0//
B_2&B_4
/end{array}
/right )Q_1
$//从而$$A+B=P_2/left (/begin{array}{cc}
A_1&0//
A_2+B_2&B_4
/end{array}
/right )Q_1$$
因为/begin{center}秩($A+B$)$/leq$秩$/left (/begin{array}{ccc}
A_1&A_1&0//
0&A_2+B_2&B_4
/end{array}
/right )$=秩$/left (/begin{array}{ccc}
A_1&0&0//
0&A_2+B_2&B_4
/end{array}
/right )$/end{center}
/begin{center}=秩($A_1$)+秩($A_2+B_2/hspace{2mm}B_4$)$/leq$秩($A_1$)+$s$/end{center}再由条件秩($A+B$)=秩($A$)+秩($B$)=$r+s$可知秩($A_1$)$/geq$$r$,于是$A_1$是列满秩矩阵,
故存在可逆矩阵
$P_3/in F^{(m-s)/times (m-s)}$使$A_1=P_3/left (/begin{array}{c}
I_r//
0
/end{array}
/right )$。同理存在可逆阵$Q_3/in F^{(n-r)/times(n-r)}$,使得$B_4=(0/hspace{2mm}I_s)Q_3$。

(1)当$r+s< /rm{min}/{m,n/}$时,易见
$$
A=P_2/left (/begin{array}{cc}
A_1&0//
A_2&0
/end{array}
/right )Q_1=P_2/left (/begin{array}{cc}
P_3/left (/begin{array}{c}
I_r//
0
/end{array}
/right )&0//
A_2&0
/end{array}
/right )Q_1 //
$$
/hspace{3.8cm}$
=P_2/left (/begin{array}{cc}
P_3&0//
(A_2 /hspace{2mm}0)&I_s
/end{array}
/right )/left (/begin{array}{cc}
/left (/begin{array}{c}
I_r//
0
/end{array}
/right )&0//
0&0
/end{array}
/right )Q_1
$

/hspace{3.0cm}$
=P/left(/begin{array}{cc}
I_r&0//
0&0
/end{array}/right )Q
$/hspace{7.8cm}(3.3)
$$
B=P_2/left(/begin{array}{cc}
0&0//
B_2&B_4
/end{array}/right )Q_1=P_2/left(/begin{array}{cc}
0&0//
B_2&(0/hspace{2mm}I_s)Q_3
/end{array}/right )Q_1
$$
/hspace{3.7cm}$
=P_2/left(/begin{array}{cc}
0&0//
0&(0 /hspace{2mm}I_s)
/end{array}/right )/left(/begin{array}{cc}
I_r&0//
/left(/begin{array}{c}
0//
B_3
/end{array}/right )&Q_3
/end{array}/right )Q_1$

/hspace{2.9cm}$=P/left(/begin{array}{cc}
0&0//
0&I_s
/end{array}/right )Q
$/hspace{7.8cm}(3.4)
//其中$$P=P_2/left (/begin{array}{cc}
P_3&0//
(A_2 /hspace{2mm}0)&I_s
/end{array}
/right )$$$$Q=/left(/begin{array}{cc}
I_r&0//
/left(/begin{array}{c}
0//
B_2
/end{array}/right )&Q_3
/end{array}/right )Q_1$$
(2)当$r+s=/rm{min}/{m,n/}$时,证明类似于情形(1)。

/section{满足Sylvester秩等式的矩阵对的标准形}
/begin{thm}/label{t1.1.4}
设$A/in F^{m/times n}$、$B/in F^{n/times p}$,则秩$(AB)$=秩$(A)$$+$秩$(B)$$-$$n$的充要条件是存在可逆阵$P/in F^{m/times m}$、$Q/in F^{n/times n}$和$R/in F^{p/times p}$
使得$$A=P/left ( /begin{array}{cc} 
I_r&0//
0&0/end{array} /right )Q$$$$B=Q^{-1}/left ( /begin{array}{cc} 
0&0//
0&I_s/end{array} /right )R$$且$r+s/geq n$。
/end{thm}
{/heiti 证明}充分性。如果存在这样的矩阵$P$、$Q$和$R$,为了使矩阵满足分块相乘把$B$分解成
$$B=Q^{-1}/left ( /begin{array}{ccc} 
0&0&0//
0&I_{s-(n-r)}&0//
0&0&I_{n-r}/end{array} /right )R$$则$$AB=P/left ( /begin{array}{ccc} 
0&0&0//
0&I_{s-(n-r)}&0//
0&0&0/end{array} /right )R$$故秩$(AB)$=$s-(n-r)=s+r-n$=秩($A$)+秩($B$)$-$$n$

必要性。由定理2.1存在可逆矩阵$P_1/in F^{m/times m}$和$Q_1/in F^{n/times n}$使$$A=P_1/left ( /begin{array}{cc} 
I_r&0//
0&0/end{array} /right )Q_1$$因$Q_1B$可经列变换化为$$/left ( /begin{array}{cc} 
0&B_1//
0&B_2/end{array} /right )$$其中(0/hspace{2mm}$B_1$)是前$r$行,$/left ( /begin{array}{c} 
B_1//
B_2/end{array} /right )$为列满秩,且它的秩是$s$,则有可逆矩阵$R_1/in F^{p/times p}$使$$B=Q_1^{-1}/left ( /begin{array}{cc} 
0&B_1//
0&B_2/end{array} /right )R_1$$于是
$$
AB=P_1/left ( /begin{array}{cc} 
0&B_1//
0&0/end{array} /right )R_1
$$设$AB$的秩是$k$,则存在可逆矩阵$M/in F^{r/times r}$,$N/in F^{s/times s}$使$$B_1=M/left ( /begin{array}{cc} 
0&0//
I_k&0/end{array} /right )N$$从而有

/hspace{1.5cm}$
B=Q_1^{-1}/left ( /begin{array}{cc} 
0&M/left ( /begin{array}{cc} 
0&0//
I_k&0/end{array} /right )N//
0&B_2/end{array} /right )R_1$
$$=Q_1^{-1}/left ( /begin{array}{cc} 
M&0//
0&I_{n-r}/end{array} /right )/left ( /begin{array}{ccc} 
0&0&0//
0&I_k&0//
0&B_3&B_4  /end{array} /right )/left ( /begin{array}{cc} 
I_{p-s}&0//
0&N/end{array} /right )R_1
$$其中$(B_3/hspace{2mm} B_4)$是$B_2N^{-1}$的分块写法,且$B_4$的列数是$s-k$,行数是$n-r$。使用已知条件秩($AB$)=秩($A$)+秩($B$)$-$$n$
(即$k=r+s-n$)推出$s-k=n-r$,即$B_4$是一$(s-k)/times (s-k)$方阵,且$B_4$可逆。故
$$
B=Q^{-1}/left ( /begin{array}{cc} 
0&0//
0&I_s/end{array} /right )R$$其中
$$
Q=/left ( /begin{array}{ccc} 
I_{n-s}&0&0//
0&I_k&0//
0&-B_3&I_{s-k} /end{array} /right )/left ( /begin{array}{cc} 
M^{-1}&0//
0&I_{n-r}/end{array} /right )Q_1
$$
$$
R=/left ( /begin{array}{cc} 
I_{p-(s-k)}&0//
0&B_4/end{array} /right )/left ( /begin{array}{cc} 
I_{p-s}&0//
0&N/end{array} /right )R_1
$$
设$$
P=P_1/left ( /begin{array}{cc} 
M&0//
0&I_{m-r}/end{array} /right )
$$

$$/left./begin{array}{rl}
&P/left ( /begin{array}{cc} 
I_r&0//
0&0/end{array} /right )Q//
=&P_1/left ( /begin{array}{cc} 
M&0//
0&I_{m-r}/end{array} /right )/left ( /begin{array}{cc} 
I_r&0//
0&0/end{array} /right )/left ( /begin{array}{ccc} 
I_{n-s}&0&0//
0&I_k&0//
0&-B_3&I_{s-k} /end{array} /right )/left ( /begin{array}{cc} 
M^{-1}&0//
0&I_{n-r}/end{array} /right )Q_1//
=&P_1/left ( /begin{array}{cc} 
I_r&0//
0&0/end{array} /right )Q_1//
=&A
/end{array}/right.
$$
证完。
/backmatter
/chapter{结论}
本文给出了几个著名的矩阵秩的不等式等号成立的充要条件,问题从满足秩条件的矩阵对的标准形角度出发,用比较统一的
方法处理了几个问题,使用的方法基于矩阵的等价分解和分块技术。

$/S3.2$节中关于满足秩可加条件的矩阵对的标准形已经被用于解决秩可加性的加法保持问题$^{[6]}$,寻找本文取得的结论的其它应用,是继续要研究的课题。
%/addcontentsline{toc}{chapter}{/heiti 结论} 
/renewcommand{/thepage}{/arabic{page}}


/begin{thebibliography}{0}
/bibitem{bei}北京大学数学系几何与代数教研室前代数小组.高等代数(第三版)[M].北京:高等教育出版社,2003.
/bibitem{cao}曹重光.线性代数[M].赤峰:内蒙古科学技术出版社,1999.
/bibitem{yang}杨子胥.高等代数习题解(修订版)[M].济南:山东科技出版社,2004.
/bibitem{li}李书超,蒋君.一类矩阵的秩的恒等式及其推广[J]. 武汉科技大学学报(自然科学版). 2004,27(1):96-98.
/bibitem{hu}胡付高.关于一类矩阵秩的恒等式注记[J]. 武汉科技大学学报(自然科学版). 2004,27(3):322-323.
/bibitem{hu}X.Zhang.Linear operators that preserve pairs of matrices which satisfy extreme rank properties —a supplymentary version[J]. Linear Algebra Appl. 2003,375:283-290.
/end{thebibliography}
/chapter{致谢}
%/addcontentsline{toc}{chapter}{/heiti 致谢}

经过两个多月的学习与研究,本次毕业论文的撰写就要画上一个句号了。可是,对我来说,这次毕业论文设计的本身所产生的影响,还远远没有结束,
我不仅对学到的知识进行了巩固,而且从中学到了许多课本上没有的知识,也为以后从事研究工作积累了一些经验。
在这里我要感谢我的指导教师张显老师,在整个设计过程中,张老师给了我很大的帮助,使我的毕业设计能顺利的完成。

最后我还要感谢所有帮助过我的老师和同学们,谢谢你们一直对我的支持和帮助, 感谢大家。
/end{document}
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