latex1.ctx

毕业论文的源代码

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\begin{document}
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{\ \ \ } \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}
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%\frontmatter
\chapter*{摘\ \ 要}
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%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
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%\makeatother%这三行命令会让目录中的章也有.....连接
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\thispagestyle{empty}
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\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\\