#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
\save_transient_properties true
\origin unavailable
\textclass beamer
\begin_preamble
% 如果没有这一句命令，XeTeX会出错，原因参见
% http://bbs.ctex.org/viewthread.php?tid=60547
\DeclareRobustCommand\nobreakspace{\leavevmode\nobreak\ }
% \usepackage{tkz-euclide}
% \usetkzobj{all}
\usepackage{multicol}
\usepackage[define-L-C-R]{nicematrix}

\usetheme[lw]{uantwerpen}

\AtBeginDocument{
\renewcommand\logopos{111.png} 
\renewcommand\logoneg{111.png} 
\renewcommand\logomonowhite{111.png} 
\renewcommand\iconfile{111.png} 
}

\setbeamertemplate{theorems}[numbered]

\AtBeginSection[]
{
	\begin{frame}{章节内容}
		\transfade%淡入淡出效果
		\begin{multicols}{2}
		\tableofcontents[sectionstyle=show/shaded,subsectionstyle=show/shaded/hide]
		\end{multicols}
		\addtocounter{framenumber}{-1}  %目录页不计算页码
	\end{frame}
}

\usepackage{amsmath, amsfonts, amssymb, mathtools, yhmath, mathrsfs}
% http://ctan.org/pkg/extarrows
% long equal sign
\usepackage{extarrows}

\DeclareMathOperator{\sech}{sech}
\DeclareMathOperator{\curl}{curl}

%\everymath{\color{blue}\everymath{}}
%\everymath\expandafter{\color{blue}\displaystyle}
%\everydisplay\expandafter{\the\everydisplay \color{red}}

\def\degree{^\circ}
\def\bt{\begin{theorem}}
\def\et{\end{theorem}}
\def\bl{\begin{lemma}}
\def\el{\end{lemma}}
\def\bc{\begin{corrolary}}
\def\ec{\end{corrolary}}
\def\ba{\begin{proof}[解]}
\def\ea{\end{proof}}
\def\ue{\mathrm{e}}
\def\ud{\,\mathrm{d}}
\def\GF{\mathrm{GF}}
\def\ui{\mathrm{i}}
\def\Re{\mathrm{Re}}
\def\Im{\mathrm{Im}}
\def\uRes{\mathrm{Res}}
\def\diag{\,\mathrm{diag}\,}
\def\be{\begin{equation}}
\def\ee{\end{equation}}
\def\bee{\begin{equation*}}
\def\eee{\end{equation*}}
\def\sumcyc{\sum\limits_{cyc}}
\def\prodcyc{\prod\limits_{cyc}}
\def\i{\infty}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\l{\lambda}
\def\m{\mu}
\def\t{\theta}
\def\p{\partial}
\def\wc{\rightharpoonup}
\def\udiv{\mathrm{div}}
\def\diam{\mathrm{diam}}
\def\dist{\mathrm{dist}}
\def\uloc{\mathrm{loc}}
\def\uLip{\mathrm{Lip}}
\def\ucurl{\mathrm{curl}}
\def\usupp{\mathrm{supp}}
\def\uspt{\mathrm{spt}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\providecommand{\abs}[1]{\left\lvert#1\right\rvert}
\providecommand{\norm}[1]{\left\Vert#1\right\Vert}
\providecommand{\paren}[1]{\left(#1\right)}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\FF}{\mathbb{F}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\WW}{\mathbb{W}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\PP}{\mathbb{P}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\TT}{\mathbb{T}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\pNN}{\mathbb{N}_{+}}
\newcommand{\cZ}{\mathcal{Z}}
\newcommand{\cM}{\mathcal{M}}
\newcommand{\cS}{\mathcal{S}}
\newcommand{\cX}{\mathcal{X}}
\newcommand{\cW}{\mathcal{W}}

\newcommand{\eqdef}{\xlongequal{\text{def}}}%
\newcommand{\eqexdef}{\xlongequal[\text{存在}]{\text{记为}}}%
\end_preamble
\options aspectratio = 1610, 11pt, UTF8
\use_default_options true
\begin_modules
theorems-ams
theorems-sec
\end_modules
\maintain_unincluded_children false
\language chinese-simplified
\language_package default
\inputencoding utf8-cjk
\fontencoding global
\font_roman "default" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\font_cjk gbsn
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format pdf2
\output_sync 0
\bibtex_command default
\index_command default
\float_placement H
\paperfontsize default
\spacing single
\use_hyperref true
\pdf_bookmarks true
\pdf_bookmarksnumbered false
\pdf_bookmarksopen false
\pdf_bookmarksopenlevel 1
\pdf_breaklinks true
\pdf_pdfborder true
\pdf_colorlinks true
\pdf_backref false
\pdf_pdfusetitle true
\papersize default
\use_geometry true
\use_package amsmath 2
\use_package amssymb 2
\use_package cancel 1
\use_package esint 2
\use_package mathdots 1
\use_package mathtools 2
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_minted 0
\index Index
\shortcut idx
\color #008000
\end_index
\leftmargin 2cm
\topmargin 2cm
\rightmargin 2cm
\bottommargin 2cm
\secnumdepth 3
\tocdepth 2
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tracking_changes false
\output_changes false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\end_header

\begin_body

\begin_layout Section
化二次型为标准形
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
引言
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
若二次型 
\begin_inset Formula $f\left(x_{1},x_{2},\cdots,x_{n}\right)$
\end_inset

 经可逆线性变换化为只含平方项的形式 
\begin_inset Formula 
\begin{equation}
b_{1}y_{1}^{2}+b_{2}y_{2}^{2}+\cdots+b_{n}y_{n}^{2},\label{eq:5.2-1}
\end{equation}

\end_inset

则称 
\series bold
(
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5.2-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 为二次型 
\begin_inset Formula $f\left(x_{1},x_{2},\cdots,x_{n}\right)$
\end_inset

 的标准形
\series default
.
 由上节讨论知, 二次型 
\begin_inset Formula $f\left(x_{1},x_{2},\cdots,x_{n}\right)=X^{T}AX$
\end_inset

 在线性变换 
\begin_inset Formula $X=CY$
\end_inset

 下, 可化为 
\begin_inset Formula $Y^{T}\left(C^{T}AC\right)Y$
\end_inset

.
 如果 
\begin_inset Formula $C^{T}AC$
\end_inset

 为对角矩阵 
\begin_inset Formula 
\[
B=\begin{bmatrix}b_{1}\\
 & b_{2}\\
 &  & \ddots\\
 &  &  & b_{n}
\end{bmatrix},
\]

\end_inset

则 
\begin_inset Formula $f\left(x_{1},x_{2},\cdots,x_{n}\right)$
\end_inset

 就可化为标准形 
\begin_inset Formula $b_{1}y_{1}^{2}+b_{2}y_{2}^{2}+\cdots+b_{n}y_{n}^{2}$
\end_inset

, 其标准形中的系数恰好为对角阵 
\begin_inset Formula $B$
\end_inset

 的对角线上的元素, 因此上面的问题归结为 
\begin_inset Formula $A$
\end_inset

 能否合同于一个对角矩阵.
\end_layout

\end_deeper
\begin_layout Subsection
用配方法化二次型为标准形
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用配方法化二次型为标准形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:5.2-1"

\end_inset

 任何二次型都可以通过可逆线性变换化为标准形.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
拉格朗日配方法
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
拉格朗日配方法的步骤:
\end_layout

\begin_layout Standard
(1) 若二次型含有 
\begin_inset Formula $x_{i}$
\end_inset

 的平方项, 则先把含有 
\begin_inset Formula $x_{i}$
\end_inset

 的项按 
\begin_inset Formula $x_{i}$
\end_inset

 的幂降次排列, 然后配方.
 再对其余的变量进行同样过程的操作, 直到所有的变量都配成平方项为止.
 经过可逆线性变换, 就得到标准形;
\end_layout

\begin_layout Standard
(2) 若二次型中不含有平方项, 但是 
\begin_inset Formula $a_{ij}\neq0$
\end_inset

, (
\begin_inset Formula $i\neq j$
\end_inset

), 则先作可逆变换
\begin_inset Formula 
\[
\begin{cases}
x_{i}=y_{i}-y_{j},\\
x_{j}=y_{i}+y_{j}, & (k=1,2,\cdots,n\text{ 且 }k\neq i,j)\\
x_{k}=y_{k},
\end{cases}
\]

\end_inset

化二次型为含有平方项的二次型, 然后再按 (1) 中方法配方.
\end_layout

\begin_layout Remark*
配方法是一种可逆线性变换, 但平方项的系数与 
\begin_inset Formula $A$
\end_inset

 的特征值无关.
\end_layout

\begin_layout Standard
因为二次型 
\begin_inset Formula $f$
\end_inset

 与它的对称矩阵 
\begin_inset Formula $A$
\end_inset

 有一一对应的关系, 由定理 
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:5.2-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

 即得:
\end_layout

\begin_layout Theorem
对任一实对称矩阵 
\begin_inset Formula $A$
\end_inset

, 存在非奇异矩阵 
\begin_inset Formula $C$
\end_inset

, 使 
\begin_inset Formula $B=C^{T}AC$
\end_inset

 为对角矩阵.
 即任一实对称矩阵都与一个对角矩阵合同.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
将二次型化为标准形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
E01
\end_layout

\end_inset

 将 
\begin_inset Formula $x_{1}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}^{2}+4x_{2}x_{3}+x_{3}^{2}$
\end_inset

 化为标准形.
\end_layout

\begin_layout Solution*
因标准形是平方项的代数和, 可利用配方法解之.
\begin_inset Formula 
\begin{equation}
\begin{aligned}x_{1}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}^{2}+4x_{2}x_{3}+x_{3}^{2} & =x_{1}^{2}+2x_{1}\left(x_{2}+x_{3}\right)+\left(x_{2}+x_{3}\right)^{2}-\left(x_{2}+x_{3}\right)^{2}+2x_{2}^{2}+4x_{2}x_{3}+x_{3}^{2}\\
 & =\left(x_{1}+x_{2}+x_{3}\right)^{2}+x_{2}^{2}+2x_{2}x_{3}\\
 & =\left(x_{1}+x_{2}+x_{3}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}-x_{3}^{2},
\end{aligned}
\label{eq:5.2-2}
\end{equation}

\end_inset

其线性变换矩阵的行列式 
\begin_inset Formula $\mathrm{det}(C)=\begin{vmatrix}1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 & 1
\end{vmatrix}=1\neq0$
\end_inset

, 将 
\begin_inset Formula $X=CY$
\end_inset

 代入 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5.2-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 式, 得二次型的标准形 
\begin_inset Formula $y_{1}^{2}+y_{2}^{2}-y_{3}^{2}$
\end_inset

, 该二次型的矩阵为 
\begin_inset Formula $B=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{bmatrix}$
\end_inset

, 而原二次型的矩阵为 
\begin_inset Formula $A=\begin{bmatrix}1 & 1 & 1\\
1 & 2 & 2\\
1 & 2 & 1
\end{bmatrix}$
\end_inset

, 线性替换的矩阵为 
\begin_inset Formula $C=\begin{bmatrix}1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 & 1
\end{bmatrix}$
\end_inset

, 易验证 
\begin_inset Formula $C^{T}AC=B=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{bmatrix}$
\end_inset

 是对角矩阵, 且 
\begin_inset Formula $y^{T}By=y_{1}^{2}+y_{2}^{2}-y_{3}^{2}$
\end_inset

.
 
\end_layout

\begin_layout Remark*
可见, 要把二次型化为标准形, 关键在于求出一个非奇异矩阵 
\begin_inset Formula $C$
\end_inset

, 使得 
\begin_inset Formula $C^{T}AC$
\end_inset

 是对角矩阵.
 而这需要在 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5.2-2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 式中令 
\begin_inset Formula $y_{1}=x_{1}+x_{2}+x_{3}$
\end_inset

, 
\begin_inset Formula $y_{2}=x_{2}+x_{3}$
\end_inset

, 
\begin_inset Formula $y_{3}=x_{3}$
\end_inset

 反解出 
\begin_inset Formula $x_{1},x_{2},x_{3}$
\end_inset

, 用矩阵的语言来表述就是从 
\begin_inset Formula 
\[
\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix}=\begin{bmatrix}1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}\Longleftrightarrow\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}1 & -1\\
 & 1 & -1\\
 &  & 1
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix}\Longleftrightarrow X=CY.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
用配方法化二次型为标准形并求变换矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
化二次型 
\begin_inset Formula $f=x_{1}^{2}+2x_{2}^{2}+5x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+6x_{2}x_{3}$
\end_inset

 为标准形, 并求所用的变换矩阵.
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\begin{align*}
f & =x_{1}^{2}+2x_{2}^{2}+5x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+6x_{2}x_{3}=x_{1}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}^{2}+5x_{3}^{2}+6x_{2}x_{3}\\
 & =\left(x_{1}+x_{2}+x_{3}\right)^{2}-x_{2}^{2}-x_{3}^{2}-2x_{2}x_{3}+2x_{2}^{2}+5x_{3}^{2}+6x_{2}x_{3}\\
 & =\left(x_{1}+x_{2}+x_{3}\right)^{2}+x_{2}^{2}+4x_{3}^{2}+4x_{2}x_{3}\\
 & =\left(x_{1}+x_{2}+x_{3}\right)^{2}+\left(x_{2}+2x_{3}\right)^{2}.
\end{align*}

\end_inset

令
\begin_inset Formula 
\[
\begin{cases}
y_{1}=x_{1}+x_{2}+x_{3}\\
y_{2}=x_{2}+2x_{3}\\
y_{3}=x_{3}
\end{cases}\Longrightarrow\begin{cases}
x_{1}=y_{1}-y_{2}+y_{3}\\
x_{2}=y_{2}-2y_{3}\\
x_{3}=y_{3}
\end{cases},
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
令
\begin_inset Formula 
\[
\begin{cases}
y_{1}=x_{1}+x_{2}+x_{3}\\
y_{2}=x_{2}+2x_{3}\\
y_{3}=x_{3}
\end{cases}\Longrightarrow\begin{cases}
x_{1}=y_{1}-y_{2}+y_{3}\\
x_{2}=y_{2}-2y_{3}\\
x_{3}=y_{3}
\end{cases},
\]

\end_inset

也即
\begin_inset Formula 
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}1 & -1 & 1\\
0 & 1 & -2\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix}.
\]

\end_inset

因此二次型 
\begin_inset Formula 
\[
f=x_{1}^{2}+2x_{2}^{2}+5x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+6x_{2}x_{3}=y_{1}^{2}+y_{2}^{2}.
\]

\end_inset

所用变换矩阵为 
\begin_inset Formula $C=\begin{bmatrix}1 & -1 & 1\\
0 & 1 & -2\\
0 & 0 & 1
\end{bmatrix}$
\end_inset

, (
\begin_inset Formula $|C|=1\neq0$
\end_inset

).
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
另一个例子 (但需要拉格朗日配方法的 (2))
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:5.2-3"

\end_inset


\begin_inset Argument 1
status open

\begin_layout Plain Layout
E02
\end_layout

\end_inset

 化二次型 
\begin_inset Formula $f=2x_{1}x_{2}+2x_{1}x_{3}-6x_{2}x_{3}$
\end_inset

 成标准形, 并求所用的变换矩阵.
\end_layout

\begin_layout Solution*
由于所给二次型中无平方项, 所以先令
\begin_inset Formula 
\[
\begin{cases}
x_{1}=y_{1}+y_{2}\\
x_{2}=y_{1}-y_{2},\\
x_{3}=y_{3}
\end{cases}\Longleftrightarrow\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}1 & 1 & 0\\
1 & -1 & 0\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix},
\]

\end_inset

代入原二次型得 
\begin_inset Formula $f=2y_{1}^{2}-2y_{2}^{2}-4y_{1}y_{3}+8y_{2}y_{3}$
\end_inset

.
 再配方得 
\begin_inset Formula $f=2\left(y_{1}-y_{3}\right)^{2}-2\left(y_{2}-2y_{3}\right)^{2}+6y_{3}^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
由 
\begin_inset Formula $f=2\left(y_{1}-y_{3}\right)^{2}-2\left(y_{2}-2y_{3}\right)^{2}+6y_{3}^{2}$
\end_inset

.
 令
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
\begin{cases}
z_{1}=y_{1}-y_{3}\\
z_{2}=y_{2}-2y_{3}\\
z_{3}=y_{3}
\end{cases}\Longrightarrow\begin{cases}
y_{1}=z_{1}+z_{3}\\
y_{2}=z_{2}+2z_{3}\text{, }\\
y_{3}=z_{3}
\end{cases}\Longrightarrow\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix}=\begin{bmatrix}1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}z_{1}\\
z_{2}\\
z_{3}
\end{bmatrix}.
\]

\end_inset

代入原二次型得标准形 
\begin_inset Formula $f=2z_{1}^{2}-2z_{2}^{2}+6z_{3}^{2}$
\end_inset

.
 所用变换矩阵为
\begin_inset Formula 
\[
C=\begin{bmatrix}1 & 1 & 0\\
1 & -1 & 0\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}1 & 1 & 3\\
1 & -1 & -1\\
0 & 0 & 1
\end{bmatrix},\quad(|C|=-2\neq0).
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
用配方法化二次型为标准形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
用配方法将下列二次型化为标准形 
\begin_inset Formula 
\[
f\left(x_{1},x_{2},x_{3},x_{4}\right)=2x_{1}x_{2}-x_{1}x_{3}+x_{1}x_{4}-x_{2}x_{3}+x_{2}x_{4}-2x_{3}x_{4}.
\]

\end_inset


\end_layout

\begin_layout Solution*
因二次型 
\begin_inset Formula $f$
\end_inset

 缺少 
\begin_inset Formula $x_{1}^{2}$
\end_inset

 (
\begin_inset Formula $i=1,2,3,4$
\end_inset

) 项, 无法配方.
 作变换: 
\begin_inset Formula 
\begin{equation}
\begin{cases}
x_{1}=y_{1}+y_{2}\\
x_{2}=y_{1}-y_{2}\\
x_{3}=y_{3}\\
x_{4}=y_{4}
\end{cases},\label{eq:5.2-3}
\end{equation}

\end_inset

代入原二次型得关于 
\begin_inset Formula $y_{i}$
\end_inset

 的二次型:
\begin_inset Formula 
\[
f=2y_{1}^{2}-2y_{2}^{2}-2y_{1}y_{3}+2y_{1}y_{4}-2y_{3}y_{4},
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
由于
\begin_inset Formula 
\[
f=2y_{1}^{2}-2y_{2}^{2}-2y_{1}y_{3}+2y_{1}y_{4}-2y_{3}y_{4},
\]

\end_inset

中 
\begin_inset Formula $y_{1}^{2}$
\end_inset

 项的系数不为零, 配方得
\begin_inset Formula 
\[
\begin{aligned}f & =\left(2y_{1}^{2}-2y_{1}y_{3}+2y_{1}y_{4}\right)-2y_{2}^{2}-2y_{3}y_{4}\\
 & =2\left[\left(y_{1}-\frac{1}{2}y_{3}+\frac{1}{2}y_{4}\right)^{2}-\frac{1}{4}y_{3}^{2}-\frac{1}{4}y_{4}^{2}+\frac{1}{2}y_{3}y_{4}\right]-2y_{2}^{2}-2y_{3}y_{4}\\
 & =2\left(y_{1}-\frac{1}{2}y_{3}+\frac{1}{2}y_{4}\right)^{2}-2y_{2}^{2}-\frac{1}{2}y_{3}^{2}-y_{3}y_{4}-\frac{1}{2}y_{4}^{2}\\
 & =2\left(y_{1}-\frac{1}{2}y_{3}+\frac{1}{2}y_{4}\right)^{2}-2y_{2}^{2}-\frac{1}{2}\left(y_{3}+y_{4}\right)^{2}.
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
故 
\begin_inset Formula $f=2\left(y_{1}-\frac{1}{2}y_{3}+\frac{1}{2}y_{4}\right)^{2}-2y_{2}^{2}-\frac{1}{2}\left(y_{3}+y_{4}\right)^{2}$
\end_inset

, 令
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\begin{equation}
\begin{cases}
z_{1}=y_{1}-\frac{1}{2}y_{3}+\frac{1}{2}y_{4}\\
z_{2}=y_{2}\\
z_{3}=y_{3}+y_{4}\\
z_{4}=y_{4}
\end{cases}\label{eq:5.2-4}
\end{equation}

\end_inset

故标准形为 
\begin_inset Formula $2z_{1}^{2}-2z_{2}^{2}-\frac{1}{2}z_{3}^{2}$
\end_inset

.
 为求变换矩阵 
\begin_inset Formula $C$
\end_inset

, 从 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5.2-4"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 解出 
\begin_inset Formula $y_{i}$
\end_inset

: 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{cases}
y_{1}=z_{1}+\frac{1}{2}z_{3}-z_{4}\\
y_{2}=z_{2}\\
y_{3}=z_{3}-z_{4}\\
y_{4}=z_{4}
\end{cases}\xrightarrow{\text{ 代入 }(\ref{eq:5.2-3})}\begin{cases}
x_{1}=z_{1}+z_{2}+\frac{1}{3}z_{3}-z_{4}\\
x_{2}=z_{1}-z_{2}+\frac{1}{2}z_{3}-z_{4}.\\
x_{3}=z_{3}-z_{4}\\
x_{4}=z_{4}
\end{cases}
\]

\end_inset

于是 
\begin_inset Formula $C=\begin{bmatrix}1 & 1 & 1/2 & -1\\
1 & -1 & 1/2 & -1\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 1
\end{bmatrix}$
\end_inset

, 所用线性变换为 
\begin_inset Formula $x=Cz$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
用初等变换化二次型为标准形
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用初等变换化二次型为标准形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
设有可逆线性变换为 
\begin_inset Formula $X=CY$
\end_inset

, 它把二次型 
\begin_inset Formula $X^{T}AX$
\end_inset

 化为标准形 
\begin_inset Formula $Y^{T}BY$
\end_inset

, 则 
\begin_inset Formula $C^{T}AC=B$
\end_inset

.
 已知任一非奇异矩阵均可表示为若干个初等矩阵的乘积, 故存在初等矩阵 
\begin_inset Formula $P_{1},P_{2},\cdots,P_{s}$
\end_inset

, 使 
\begin_inset Formula $C=P_{1}P_{2}\cdots P_{s}$
\end_inset

, 于是
\begin_inset Formula 
\begin{align*}
C & =EP_{1}P_{2}\cdots P_{s}\\
C^{T}AC & =P_{s}^{T}\cdots P_{2}^{T}P_{1}^{T}AP_{1}P_{2}\cdots P_{s}\\
 & =\mathrm{diag}\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}\right)\eqqcolon\Lambda.
\end{align*}

\end_inset

由此可见, 对 
\begin_inset Formula $2n\times n$
\end_inset

 矩阵 
\begin_inset Formula $\begin{bmatrix}A\\
E
\end{bmatrix}$
\end_inset

 施以相应于右乘 
\begin_inset Formula $P_{1}P_{2}\cdots P_{s}$
\end_inset

 的初等列变换
\begin_inset Note Note
status open

\begin_layout Plain Layout
初等变换细节解析
\end_layout

\end_inset

, 再对 
\begin_inset Formula $A$
\end_inset

 施以相应于左乘 
\begin_inset Formula $P_{1}^{T},P_{2}^{T},\cdots,P_{s}^{T}$
\end_inset

 的初等行变换, 则矩阵 
\begin_inset Formula $A$
\end_inset

 变为对角矩阵 
\begin_inset Formula $B$
\end_inset

, 而单位矩阵 
\begin_inset Formula $E$
\end_inset

 就变为所要求的可逆矩阵 
\begin_inset Formula $C$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用初等变换化二次型为标准形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
E03
\end_layout

\end_inset

 求一可逆线性变换将 
\begin_inset Formula 
\[
x_{1}^{2}+2x_{2}^{2}+x_{3}^{2}+2x_{1}x_{2}+2x_{1}x_{3}+4x_{2}x_{3}
\]

\end_inset

化为标准形.
\end_layout

\begin_layout Solution*
二次型对应的矩阵为 
\begin_inset Formula $A=\begin{bmatrix}1 & 1 & 1\\
1 & 2 & 2\\
1 & 2 & 1
\end{bmatrix}$
\end_inset

, 利用初等变换, 有
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-4mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{bmatrix}A\\
E
\end{bmatrix}=\begin{bmatrix}1 & 1 & 1\\
1 & 2 & 2\\
1 & 2 & 1\\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\boxed{\xrightarrow[c_{3}-c_{1}]{c_{2}-c_{1}}\begin{bmatrix}1 & 0 & 0\\
1 & 1 & 1\\
1 & 1 & 0\\
1 & -1 & -1\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\xrightarrow[r_{3}-r_{1}]{r_{2}-r_{1}}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 1\\
0 & 1 & 0\\
1 & -1 & -1\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}}\xrightarrow[r_{3}-r_{2}]{c_{3}-c_{2}}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1\\
1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 & 1
\end{bmatrix},
\]

\end_inset

因此, 
\begin_inset Formula $C=\begin{bmatrix}1 & -1 & 0\\
0 & 1 & -1\\
0 & 0 & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $|C|=1\neq0$
\end_inset

.
 令 
\begin_inset Formula $\begin{cases}
x_{1}=z_{1}-z_{2}\\
x_{2}=z_{2}-z_{3}\\
x_{3}=z_{3}
\end{cases}$
\end_inset

, 代入原二次型可得标准形 
\begin_inset Formula $z_{1}^{2}+z_{2}^{2}-z_{3}^{2}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
一个特殊的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
求一可逆线性变换化 
\begin_inset Formula $2x_{1}x_{2}+2x_{1}x_{3}-4x_{2}x_{3}$
\end_inset

 为标准形.
\end_layout

\begin_layout Solution*
此二次型对应的矩阵为 
\begin_inset Formula $A=\begin{bmatrix}0 & 1 & 1\\
1 & 0 & -2\\
1 & -2 & 0
\end{bmatrix}$
\end_inset

, 对矩阵 
\begin_inset Formula $\begin{bmatrix}A\\
E
\end{bmatrix}$
\end_inset

 做合同变换, 有
\begin_inset Formula 
\begin{align*}
\begin{bmatrix}A\\
E
\end{bmatrix} & =\begin{bmatrix}0 & 1 & 1\\
1 & 0 & -2\\
1 & -2 & 0\\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\boxed{\xrightarrow{c_{1}+c_{2}}\begin{bmatrix}1 & 1 & 1\\
1 & 0 & -2\\
-1 & -2 & 0\\
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\xrightarrow{r_{1}+r_{2}}\begin{bmatrix}2 & 1 & -1\\
1 & 0 & -2\\
-1 & -2 & 0\\
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}}\\
 & \qquad\boxed{\xrightarrow[c_{3}-c_{1}/2]{c_{2}-c_{1}/2}\begin{bmatrix}2 & 0 & 0\\
1 & -1/2 & -3/2\\
-1 & -3/2 & -1/2\\
1 & -1/2 & 1/2\\
1 & 1/2 & 1/2\\
0 & 0 & 1
\end{bmatrix}\xrightarrow[r_{3}-r_{1}/2]{r_{2}-r_{1}/2}\begin{bmatrix}2 & 0 & 0\\
0 & -1/2 & -3/2\\
0 & -3/2 & -1/2\\
1 & -1/2 & 1/2\\
1 & 1/2 & 1/2\\
0 & 0 & 1
\end{bmatrix}}
\end{align*}

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula 
\begin{align*}
\begin{bmatrix}A\\
E
\end{bmatrix} & {\color{gray}\xrightarrow[r_{3}-r_{1}/2]{r_{2}-r_{1}/2}\begin{bmatrix}2 & 0 & 0\\
0 & -1/2 & -3/2\\
0 & -3/2 & -1/2\\
1 & -1/2 & 1/2\\
1 & 1/2 & 1/2\\
0 & 0 & 1
\end{bmatrix}}\xrightarrow{c_{3}-3c_{2}}\begin{bmatrix}2 & 0 & 0\\
0 & -1/2 & 0\\
0 & -3/2 & 4\\
1 & -1/2 & 2\\
1 & 1/2 & -1\\
0 & 0 & 1
\end{bmatrix}\xrightarrow{r_{3}-3r_{2}}\begin{bmatrix}2 & 0 & 0\\
0 & -1/2 & 0\\
0 & 0 & 4\\
1 & -1/2 & 2\\
1 & 1/2 & -1\\
0 & 0 & 1
\end{bmatrix}.
\end{align*}

\end_inset

所以 
\begin_inset Formula $C=\begin{bmatrix}1 & -1/2 & 2\\
1 & 1/2 & -1\\
0 & 0 & 1
\end{bmatrix}$
\end_inset

, 
\begin_inset Formula $|C|=1\neq0$
\end_inset

.
 令 
\begin_inset Formula $\begin{cases}
x_{1}=z_{1}-(1/2)z_{2}+2z_{3}\\
x_{2}=z_{1}+(1/2)z_{2}-z_{3}\\
x_{3}=z_{3}
\end{cases}$
\end_inset

, 代入原二次型可得标准形 
\begin_inset Formula 
\[
2z_{1}^{2}-(1/2)z_{2}^{2}+4z_{3}^{2}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Subsection
用正交变换化二次型为标准型
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用正交变换化二次型为标准型
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Theorem
若 
\begin_inset Formula $A$
\end_inset

 为对称矩阵, 
\begin_inset Formula $C$
\end_inset

 为任一可逆矩阵, 令 
\begin_inset Formula $B=C^{T}AC$
\end_inset

, 则 
\begin_inset Formula $B$
\end_inset

 也为对称矩阵, 且 
\begin_inset Formula $r(B)=r(A)$
\end_inset

.
\end_layout

\begin_layout Remark*
(1) 二次型经可逆变换 
\begin_inset Formula $X=CY$
\end_inset

 后, 其秩不变, 但 
\begin_inset Formula $f$
\end_inset

 的矩阵由 
\begin_inset Formula $A$
\end_inset

 变为 
\begin_inset Formula $B=C^{T}AC$
\end_inset

;
\end_layout

\begin_layout Remark*
(2) 要使二次型 
\begin_inset Formula $f$
\end_inset

 经可逆变换 
\begin_inset Formula $X=CY$
\end_inset

 变成标准形, 即要使 
\begin_inset Formula $C^{T}AC$
\end_inset

 成为对角矩阵, 即
\begin_inset Formula 
\[
Y^{T}C^{T}ACY=\begin{bmatrix}y_{1} & y_{2} & \cdots & y_{n}\end{bmatrix}\begin{bmatrix}b_{1}\\
 & b_{2}\\
 &  & \ddots\\
 &  &  & b_{n}
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
\vdots\\
y_{n}
\end{bmatrix}=b_{1}y_{1}^{2}+b_{2}y_{2}^{2}+\cdots+b_{n}y_{n}^{2}\text{. }
\]

\end_inset


\end_layout

\begin_layout Theorem
任给二次型 
\begin_inset Formula $f=\sum_{i,j=1}^{n}a_{ij}x_{i}x_{j}$
\end_inset

, (
\begin_inset Formula $a_{ji}=a_{ij}$
\end_inset

), 总有正交变换 
\begin_inset Formula $X=PY$
\end_inset

, 使 
\begin_inset Formula $f$
\end_inset

 化为标准形
\begin_inset Formula 
\[
f=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\cdots+\lambda_{n}y_{n}^{2},
\]

\end_inset

其中 
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$
\end_inset

 是 
\begin_inset Formula $f$
\end_inset

 的矩阵 
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset

 的特征值.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
用正交变换化二次型为标准形的步骤
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
用正交变换化二次型为标准形的步骤:
\end_layout

\begin_layout Standard
(1) 将二次型表成矩阵形式 
\begin_inset Formula $f=X^{T}AX$
\end_inset

, 求出 
\begin_inset Formula $A$
\end_inset

;
\end_layout

\begin_layout Standard
(2) 求出 
\begin_inset Formula $A$
\end_inset

 的所有特征值 
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$
\end_inset

;
\end_layout

\begin_layout Standard
(3) 求出对应于特征值的特征向量 
\begin_inset Formula $\xi_{1},\xi_{2},\cdots,\xi_{n}$
\end_inset

;
\end_layout

\begin_layout Standard
(4) 将特征向量 
\begin_inset Formula $\xi_{1},\xi_{2},\cdots,\xi_{n}$
\end_inset

 正交化, 单位化, 得 
\begin_inset Formula $\eta_{1},\eta_{2},\cdots,\eta_{n}$
\end_inset

, 记 
\begin_inset Formula $C=\left(\eta_{1},\eta_{2},\cdots,\eta_{n}\right)$
\end_inset

;
\end_layout

\begin_layout Standard
(5) 作正交变换 
\begin_inset Formula $X=CY$
\end_inset

, 则得 
\begin_inset Formula $f$
\end_inset

 的标准形
\begin_inset Formula 
\[
f=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\cdots+\lambda_{n}y_{n}^{2}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
用正交变换将二次型化为标准形的例
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
E04
\end_layout

\end_inset

 将二次型 
\begin_inset Formula $f=17x_{1}^{2}+14x_{2}^{2}+14x_{3}^{2}-4x_{1}x_{2}-4x_{1}x_{3}-8x_{2}x_{3}$
\end_inset

 通过正交变换 
\begin_inset Formula $x=PY$
\end_inset

, 化成标准形.
\end_layout

\begin_layout Solution*
(1) 写出二次型矩阵: 
\begin_inset Formula $A=\begin{bmatrix}17 & -2 & -2\\
-2 & 14 & -4\\
-2 & -4 & 14
\end{bmatrix}$
\end_inset

,
\end_layout

\begin_layout Solution*
(2) 求其特征值: 由
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
|\lambda E-A|=\begin{vmatrix}\lambda-17 & 2 & 2\\
2 & \lambda-14 & 4\\
2 & 4 & \lambda-14
\end{vmatrix}=(\lambda-18)^{2}(\lambda-9)\Longrightarrow\lambda_{1}=9,\ \lambda_{2}=\lambda_{3}=18.
\]

\end_inset


\end_layout

\begin_layout Solution*
(3) 求特征向量:
\end_layout

\begin_layout Solution*
将 
\begin_inset Formula $\lambda_{1}=9$
\end_inset

 代入 
\begin_inset Formula $(\lambda E-A)x=0$
\end_inset

, 得基础解系 
\begin_inset Formula $\xi_{1}=(1/2,1,1)^{T}$
\end_inset

.
\end_layout

\begin_layout Solution*
将 
\begin_inset Formula $\lambda_{2}=\lambda_{3}=18$
\end_inset

 代入 
\begin_inset Formula $(\lambda E-A)x=0$
\end_inset

, 得基础解系 
\begin_inset Formula $\xi_{2}=(-2,1,0)^{T}$
\end_inset

, 
\begin_inset Formula $\xi_{3}=(-2,0,1)^{T}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
(4) 将特征向量正交化
\end_layout

\begin_layout Solution*
取 
\begin_inset Formula $\alpha_{1}=\xi_{1}$
\end_inset

, 
\begin_inset Formula $\alpha_{2}=\xi_{2}$
\end_inset

, 
\begin_inset Formula $\alpha_{3}=\xi_{3}-\frac{\left[\alpha_{2},\xi_{3}\right]}{\left[\alpha_{2},\alpha_{2}\right]}\alpha_{2}$
\end_inset

, 得正交向量组: 
\begin_inset Formula $\alpha_{1}=(1/2,1,1)^{T}$
\end_inset

, 
\begin_inset Formula $\alpha_{2}=(-2,1,0)^{T}$
\end_inset

, 
\begin_inset Formula $\alpha_{3}=(-2/5,-4/5,1)^{T}$
\end_inset

.
 将其单位化得:
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
\eta_{1}=\begin{bmatrix}1/3\\
2/3\\
2/3
\end{bmatrix},\ \eta_{2}=\begin{bmatrix}-2/\sqrt{5}\\
1/\sqrt{5}\\
0
\end{bmatrix},\ \eta_{3}=\begin{bmatrix}-2/\sqrt{45}\\
-4/\sqrt{45}\\
5/\sqrt{45}
\end{bmatrix}.
\]

\end_inset

作正交矩阵: 
\begin_inset Formula $P=\begin{bmatrix}1/3 & -2/\sqrt{5} & -2/\sqrt{45}\\
2/3 & 1/\sqrt{5} & -4/\sqrt{45}\\
2/3 & 0 & 5/\sqrt{45}
\end{bmatrix}$
\end_inset

.
\end_layout

\begin_layout Solution*
(5) 故所求正交变换为 
\begin_inset Formula $\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}1/3 & -2/\sqrt{5} & -2/\sqrt{45}\\
2/3 & 1/\sqrt{5} & -4/\sqrt{45}\\
2/3 & 0 & 5/\sqrt{45}
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}
\end{bmatrix}$
\end_inset

, 在此变换下原二次型化为标准形: 
\begin_inset Formula $f=9y_{1}^{2}+18y_{2}^{2}+18y_{3}^{2}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
用正交变换将二次型化为标准形的例
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设 
\begin_inset Formula $f=2x_{1}x_{2}+2x_{1}x_{3}-2x_{1}x_{4}-2x_{2}x_{3}+2x_{2}x_{4}+2x_{3}x_{4}$
\end_inset

, 求一个正交变换 
\begin_inset Formula $X=PY$
\end_inset

, 把该二次型化为标准形.
 
\end_layout

\begin_layout Solution*
二次型的矩阵为 
\begin_inset Formula $A=\begin{bmatrix}0 & 1 & 1 & -1\\
1 & 0 & -1 & 1\\
1 & -1 & 0 & 1\\
-1 & 1 & 1 & 0
\end{bmatrix}$
\end_inset

, 其特征多项式为
\begin_inset Formula 
\[
\begin{aligned}|A-\lambda E| & =\begin{vmatrix}-\lambda & 1 & 1 & -1\\
1 & -\lambda & -1 & 1\\
1 & -1 & -\lambda & 1\\
-1 & 1 & 1 & -\lambda
\end{vmatrix}=(-\lambda+1)\begin{vmatrix}1 & 1 & 1 & -1\\
1 & -\lambda & -1 & 1\\
1 & -1 & -\lambda & 1\\
1 & 1 & 1 & -\lambda
\end{vmatrix}=(-\lambda+1)\begin{vmatrix}\boxed{1} & 1 & 1 & -1\\
0 & -\lambda-1 & -2 & 2\\
0 & -2 & -\lambda-1 & 2\\
0 & 0 & 0 & \boxed{-\lambda+1}
\end{vmatrix}\\
 & =(-\lambda+1)^{2}\begin{vmatrix}-\lambda-1 & -2\\
-2 & -\lambda-1
\end{vmatrix}=(-\lambda+1)^{2}\left(\lambda^{2}+2\lambda-3\right)=(\lambda+3)(\lambda-1)^{3}.
\end{aligned}
\]

\end_inset

故 
\begin_inset Formula $A$
\end_inset

 的特征值 
\begin_inset Formula $\lambda_{1}=-3$
\end_inset

, 
\begin_inset Formula $\lambda_{2}=\lambda_{3}=\lambda_{4}=1$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
当 
\begin_inset Formula $\lambda_{1}=-3$
\end_inset

 时, 解方程 
\begin_inset Formula $(A+3E)x=0$
\end_inset

, 得基础解系 
\begin_inset Formula $\xi_{1}=\begin{bmatrix}1\\
-1\\
-1\\
1
\end{bmatrix}$
\end_inset

,
\end_layout

\begin_layout Solution*
当 
\begin_inset Formula $\lambda_{2}=\lambda_{3}=\lambda_{4}=1$
\end_inset

 时, 解方程 
\begin_inset Formula $(A-E)x=0$
\end_inset

, 可得正交的基础解系 
\begin_inset Formula 
\[
\xi_{2}=\begin{bmatrix}1\\
1\\
0\\
0
\end{bmatrix},\ \xi_{3}=\begin{bmatrix}0\\
0\\
1\\
1
\end{bmatrix},\ \xi_{4}=\begin{bmatrix}1\\
-1\\
1\\
-1
\end{bmatrix}.
\]

\end_inset

单位化得
\begin_inset Formula 
\[
P_{1}=\frac{1}{2}\begin{bmatrix}1\\
-1\\
-1\\
1
\end{bmatrix},\ P_{2}=\begin{bmatrix}1/\sqrt{2}\\
1/\sqrt{2}\\
0\\
0
\end{bmatrix},\ P_{3}=\begin{bmatrix}0\\
0\\
1/\sqrt{2}\\
1/\sqrt{2}
\end{bmatrix},\ P_{4}=\begin{bmatrix}1/2\\
-1/2\\
1/2\\
-1/2
\end{bmatrix}.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
由
\begin_inset Formula 
\[
P_{1}=\frac{1}{2}\begin{bmatrix}1\\
-1\\
-1\\
1
\end{bmatrix},\ P_{2}=\begin{bmatrix}1/\sqrt{2}\\
1/\sqrt{2}\\
0\\
0
\end{bmatrix},\ P_{3}=\begin{bmatrix}0\\
0\\
1/\sqrt{2}\\
1/\sqrt{2}
\end{bmatrix},\ P_{4}=\begin{bmatrix}1/2\\
-1/2\\
1/2\\
-1/2
\end{bmatrix}.
\]

\end_inset

构成的正交变换为
\begin_inset Formula 
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}\\
x_{4}
\end{bmatrix}=\begin{bmatrix}1/2 & 1/\sqrt{2} & 0 & 1/2\\
-1/2 & 1/\sqrt{2} & 0 & -1/2\\
-1/2 & 0 & 1/\sqrt{2} & 1/2\\
1/2 & 0 & 1/\sqrt{2} & -1/2
\end{bmatrix}\begin{bmatrix}y_{1}\\
y_{2}\\
y_{3}\\
y_{4}
\end{bmatrix}.
\]

\end_inset

在此变换下原二次型化为标准形 
\begin_inset Formula $f=-3y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
二次型与对称矩阵的规范形
\end_layout

\begin_layout Frame
\begin_inset Argument 3
status open

\begin_layout Plain Layout
allowframebreaks
\end_layout

\end_inset


\begin_inset Argument 4
status open

\begin_layout Plain Layout
二次型与对称矩阵的规范形
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
将二次型化为平方项的代数和形式后, 如有必要可重新安排各项的次序 (相当于作一次可逆线性变换), 使这个标准形为
\begin_inset Formula 
\[
d_{1}x_{1}^{2}+\cdots+d_{p}x_{p}^{2}-d_{p+1}x_{p+1}^{2}-\cdots-d_{r}x_{r}^{2},
\]

\end_inset

其中 
\begin_inset Formula $d_{i}>0$
\end_inset

, (
\begin_inset Formula $i=1,2,\cdots,r$
\end_inset

).
\end_layout

\begin_layout Theorem
任何二次型都可通过可逆线性变换化为规范形, 且规范形是由二次型本身唯一决定的形式, 与所作的可逆线性变换无关.
\end_layout

\begin_layout Remark*
把
\bar under
规范形中的正项个
\series bold
数 
\begin_inset Formula $p$
\end_inset

 称为二次型的正惯性指数
\series default
\bar default
, 
\bar under
负项个
\series bold
数 
\begin_inset Formula $r-p$
\end_inset

 称为二次型的负惯性指数
\series default
\bar default
, 
\series bold
\bar under

\begin_inset Formula $r$
\end_inset

 是二次型的秩
\series default
\bar default
.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
任何合同的对称矩阵具有相同的规范形 
\begin_inset Formula $\begin{bmatrix}E_{p} & 0 & 0\\
0 & -E_{r-p} & 0\\
0 & 0 & 0
\end{bmatrix}$
\end_inset

.
\end_layout

\begin_layout Theorem
设 
\begin_inset Formula $A$
\end_inset

 为任意对称矩阵, 如果存在可逆矩阵 
\begin_inset Formula $C,Q$
\end_inset

, 且 
\begin_inset Formula $C\neq Q$
\end_inset

, 使得
\begin_inset Formula 
\[
C^{T}AC=\begin{bmatrix}E_{p} & 0 & 0\\
0 & -E_{r-p} & 0\\
0 & 0 & 0
\end{bmatrix},\quad Q^{T}AQ=\begin{bmatrix}E_{q} & 0 & 0\\
0 & -E_{r-q} & 0\\
0 & 0 & 0
\end{bmatrix}.
\]

\end_inset

则 
\begin_inset Formula $p=q$
\end_inset

.
\end_layout

\begin_layout Remark*
说明二次型的正惯性指数、负惯性指数是被二次型本身唯一确定的.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
将标准型 
\begin_inset Formula $2y_{1}^{2}-2y_{2}^{2}-\frac{1}{2}y_{3}^{2}$
\end_inset

 规范化.
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
2y_{1}^{2}-2y_{2}^{2}-\frac{1}{2}y_{3}^{2}=\left(\sqrt{2}y_{1}\right)^{2}-\left(\sqrt{2}y_{2}\right)^{2}-\left(\frac{1}{\sqrt{2}}y_{3}\right)^{2},
\]

\end_inset

假如做变换: 
\begin_inset Formula $\begin{cases}
w_{1}=\sqrt{2}y_{1}\\
w_{2}=\sqrt{2}y_{2}\\
w_{3}=\frac{1}{\sqrt{2}}y_{3}
\end{cases}$
\end_inset

, 则原二次型就成为 
\begin_inset Formula $w_{1}^{2}-w_{2}^{2}-w_{3}^{2}$
\end_inset

, 就为所求规范标准形.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
E05
\end_layout

\end_inset

 化二次型 
\begin_inset Formula $f=2x_{1}x_{2}+2x_{1}x_{3}-6x_{2}x_{3}$
\end_inset

 为规范形, 并求其正惯性指数.
\end_layout

\begin_layout Solution*
由例 
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:5.2-3"
plural "false"
caps "false"
noprefix "false"

\end_inset

 知, 
\begin_inset Formula $f$
\end_inset

 经线性变换 
\begin_inset Formula $\begin{cases}
x_{1}=z_{1}+z_{2}+3z_{3}\\
x_{2}=z_{1}-z_{2}-z_{3}\\
x_{3}=y_{3}
\end{cases}$
\end_inset

, 化为标准形 
\begin_inset Formula $f=2z_{1}^{2}-2z_{2}^{2}+6z_{3}^{2}$
\end_inset

.
 令 
\begin_inset Formula $\begin{cases}
w_{1}=\sqrt{2}z_{1}\\
w_{3}=\sqrt{2}z_{2},\\
w_{2}=\sqrt{6}z_{3}
\end{cases}$
\end_inset

 即 
\begin_inset Formula $\begin{cases}
z_{1}=\frac{1}{\sqrt{2}}w_{1}\\
z_{2}=\frac{1}{\sqrt{2}}w_{3}\\
z_{3}=\frac{1}{\sqrt{6}}w_{2}
\end{cases}$
\end_inset

, 就把 
\begin_inset Formula $f$
\end_inset

 化成规范形 
\begin_inset Formula $f=w_{1}^{2}+w_{2}^{2}-w_{3}^{2}$
\end_inset

, 且 
\begin_inset Formula $f$
\end_inset

 的正惯性指数为 
\begin_inset Formula $2$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
作业
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
作业
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Solution
求一正交变换, 将二次型
\begin_inset Formula 
\[
f\left(x_{1},x_{2},x_{3}\right)=5x_{1}^{2}+5x_{2}^{2}+3x_{3}^{2}-2x_{1}x_{2}+6x_{1}x_{3}-6x_{2}x_{3}
\]

\end_inset

化为标准型, 并指出 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=1$
\end_inset

 表示何种二次曲面.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\end_body
\end_document