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\end_header

\begin_body

\begin_layout Section
二次型及其矩阵
\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
在解析几何中, 为了便于研究二次曲线
\begin_inset Formula 
\[
ax^{2}+bxy+cy^{2}=1
\]

\end_inset

的几何性质, 可以选择适当的坐标旋转变换
\begin_inset Formula 
\[
\begin{cases}
x=x^{\prime}\cos\theta-y^{\prime}\sin\theta,\\
y=x^{\prime}\sin\theta+y^{\prime}\cos\theta,
\end{cases}
\]

\end_inset

把方程化为标准形式
\begin_inset Formula 
\[
mx^{\prime2}+cy^{\prime2}=1.
\]

\end_inset


\end_layout

\begin_layout Standard
这类问题具有普遍性, 在许多理论问题和实际问题中常会遇到, 本章将把这类问题一般化, 讨论 
\begin_inset Formula $n$
\end_inset

 个变量的二次多项式的化简问题.
\end_layout

\end_deeper
\begin_layout Subsection
二次型的概念
\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
二次型的概念
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
含有 
\begin_inset Formula $n$
\end_inset

 个变量 
\begin_inset Formula $x_{1},x_{2},\cdots,x_{n}$
\end_inset

 的二次齐次函数
\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO: 齐次函数的定义
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{aligned}f\left(x_{1},x_{2},\cdots,x_{n}\right) & =a_{11}x_{1}^{2}+a_{22}x_{2}^{2}+a_{33}x_{3}^{2}+\cdots+a_{nn}x_{n}^{2}\\
 & \quad+2a_{12}x_{1}x_{2}+2a_{13}x_{1}x_{3}+\cdots+2a_{1n}x_{1}x_{n}\\
 & \quad\enskip\phantom{+2a_{12}x_{1}x_{2}}+2a_{23}x_{2}x_{3}+\cdots+2a_{2n}x_{2}x_{n}\\
 & \quad\enskip\phantom{+2a_{12}x_{1}x_{2}+2a_{13}x_{1}x_{3}}+\cdots+2a_{n-1,n}x_{n-1}x_{n},
\end{aligned}
\]

\end_inset

称为
\series bold
二次型
\series default
.
 当 
\begin_inset Formula $a_{ij}$
\end_inset

 为复数时, 
\begin_inset Formula $f$
\end_inset

 称为
\series bold
复二次型
\series default
; 当 
\begin_inset Formula $a_{ij}$
\end_inset

 为实数时, 
\begin_inset Formula $f$
\end_inset

 称为
\series bold
实二次型
\series default
.
 在本章中只讨论实二次型.
\end_layout

\begin_layout Definition
只含有平方项的二次型 
\begin_inset Formula $f=k_{1}y_{1}^{2}+k_{2}y_{2}^{2}+\cdots+k_{n}y_{n}^{2}$
\end_inset

 称为
\series bold
二次型的标准型
\series default
(或
\series bold
二次型的法式
\series default
).
\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
二次型的例子
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example

\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $f(x,y)=x^{2}+3xy+y^{2}$
\end_inset

 是一个含有 
\begin_inset Formula $2$
\end_inset

 个变量的实二次型.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f(x,y,z)=3x^{2}+2xy+\sqrt{2}xz-y^{2}-4yz+5z^{2}$
\end_inset

 是一个含有 
\begin_inset Formula $3$
\end_inset

 个变量的实二次型.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f\left(x_{1},x_{2},x_{3},x_{4}\right)=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{4}^{2}$
\end_inset

 是一个含有 
\begin_inset Formula $4$
\end_inset

 个变量的实二次型.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f\left(x_{1},x_{2},x_{3},x_{4}\right)=x_{1}x_{2}+2x_{1}x_{3}-4x_{1}x_{4}+3x_{2}x_{4}$
\end_inset

 是一个含有 
\begin_inset Formula $4$
\end_inset

 个变量的实二次型.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f(x,y)=x^{2}+xy-y^{2}+5x+1$
\end_inset

 不是一个实二次型, 因为它含有一次项 
\begin_inset Formula $5x$
\end_inset

 及常数项 
\begin_inset Formula $1$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=x_{1}^{3}+x_{1}x_{2}+x_{1}x_{3}$
\end_inset

 不是一个实二次型, 因为它含有 
\begin_inset Formula $3$
\end_inset

 次项 
\begin_inset Formula $x_{1}^{3}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f(x,y)=x^{2}+iy^{2}$
\end_inset

, (
\begin_inset Formula $i=\sqrt{-1}$
\end_inset

), 不是一个实二次型, 因为 
\begin_inset Formula $i$
\end_inset

 是虚数, 但它是一个复二次型.
\end_layout

\end_deeper
\end_deeper
\begin_layout Subsection
二次型的矩阵
\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
二次型的矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Fact*
取 
\begin_inset Formula $a_{ji}=a_{ij}$
\end_inset

, 则 
\begin_inset Formula $2a_{ij}x_{i}x_{j}=a_{ij}x_{i}x_{j}+a_{ji}x_{j}x_{i}$
\end_inset

, 于是
\end_layout

\begin_layout Fact*
\begin_inset Formula 
\[
\begin{aligned}f\left(x_{1},x_{2},\cdots,x_{n}\right) & =a_{11}x_{1}^{2}+a_{12}x_{1}x_{2}+\cdots+a_{1n}x_{1}x_{n}\\
 & \quad+a_{21}x_{2}x_{1}+a_{22}x_{2}^{2}+\cdots+a_{2n}x_{2}x_{n}\\
 & \quad+\cdots\cdots\cdots\cdots\cdots\cdots+\\
 & \quad+a_{n1}x_{n}x_{1}+a_{n2}x_{n}x_{2}+\cdots+a_{nn}x_{n}^{2}\\
 & =\sum_{i,j=1}^{n}a_{ij}x_{i}x_{j}.
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Fact*
仍旧取
\begin_inset Formula $a_{ji}=a_{ij}$
\end_inset

, 但对上面的函数采取矩阵运算的语言来表达如下
\begin_inset Formula 
\[
\begin{aligned}f\left(x_{1},x_{2},\cdots,x_{n}\right) & =x_{1}\left(a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}\right)\\
 & \quad+x_{2}\left(a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}\right)\\
 & \quad+\cdots\cdots\cdots\cdots\cdots\cdots+\\
 & \quad+x_{n}\left(a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}\right)\\
 & =\begin{bmatrix}x_{1} & x_{2} & \cdots & x_{n}\end{bmatrix}\begin{bmatrix}a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n}
\end{bmatrix}\\
 & =\begin{bmatrix}x_{1} & x_{2} & \cdots & x_{n}\end{bmatrix}\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix}\\
 & =X^{T}AX,
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Definition
简而言之, 二次型
\begin_inset Formula 
\begin{align*}
f\left(x_{1},x_{2},\cdots,x_{n}\right) & =\begin{bmatrix}x_{1} & x_{2} & \cdots & x_{n}\end{bmatrix}\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix}=X^{T}AX,
\end{align*}

\end_inset

其中 
\begin_inset Formula 
\[
X=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix},\quad A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{bmatrix}.
\]

\end_inset

称 
\series bold
\color red

\begin_inset Formula $f(x)=X^{T}AX$
\end_inset

 为二次型的矩阵形式
\series default
\color inherit
, 其中
\series bold
\color red
实对称矩阵 
\begin_inset Formula $A$
\end_inset

 称为该二次型的矩阵
\series default
\color inherit
, 
\series bold
\color red
二次型 
\begin_inset Formula $f$
\end_inset

 称为实对称矩阵 
\begin_inset Formula $A$
\end_inset

 的二次型
\series default
\color inherit
, 
\series bold
\color red
实对称矩阵 
\begin_inset Formula $A$
\end_inset

 的秩称为二次型的秩
\series default
\color inherit
.
 于是, 二次型 
\begin_inset Formula $f$
\end_inset

 与其实对称矩阵 
\begin_inset Formula $A$
\end_inset

 之间有一一对应关系.
\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
给出二次型的矩阵
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
写出下列是二次型相应的对称阵.
\end_layout

\begin_layout Example
(1) 
\begin_inset Formula $f(x,y)=x^{2}+3xy+y^{2}=x^{2}+\frac{3}{2}xy+\frac{3}{2}xy+y^{2}$
\end_inset

, 其矩阵为
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{bmatrix}1 & 3/2\\
3/2 & 1
\end{bmatrix}.
\]

\end_inset


\end_layout

\begin_layout Example
(2) 
\begin_inset Formula $f(x,y,z)=3x^{2}+2xy+\sqrt{2}xz-y^{2}-4yz+5z^{2}=3x^{2}+xy+\frac{\sqrt{2}}{2}xz+xy-y^{2}-2xy+\frac{\sqrt{2}}{2}xz-2yz+5z^{2}$
\end_inset

, 其相应的实对称阵为
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset

 
\begin_inset Formula 
\[
\begin{bmatrix}3 & 1 & \sqrt{2}/2\\
1 & -1 & -2\\
\sqrt{2}/2 & -2 & 5
\end{bmatrix}.
\]

\end_inset


\end_layout

\begin_layout Example
(3) 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3},x_{4}\right)=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{4}^{2}$
\end_inset

, 相应的实对称阵是一个对角阵:
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{bmatrix}.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
(4) 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3},x_{4}\right)=x_{1}x_{2}+2x_{1}x_{3}-4x_{1}x_{4}+3x_{2}x_{4}$
\end_inset

 相应的对称阵为
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
vspace{-3mm}
\end_layout

\end_inset


\begin_inset Formula 
\[
\begin{bmatrix}0 & 1/2 & 1 & -2\\
1/2 & 0 & 0 & 3/2\\
1 & 0 & 0 & 0\\
-2 & 3/2 & 0 & 0
\end{bmatrix}.
\]

\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
给出实对称阵对应的二次型
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设有实对称矩阵 
\begin_inset Formula $A=\begin{bmatrix}-1 & 1 & 0\\
1 & 0 & -1/2\\
0 & -1/2 & \sqrt{2}
\end{bmatrix}$
\end_inset

, 求 
\begin_inset Formula $A$
\end_inset

 对应的实二次型.
\end_layout

\begin_layout Solution*
由 
\begin_inset Formula $A$
\end_inset

 是三阶阵, 故有 
\begin_inset Formula $3$
\end_inset

 个变量, 则实二次型为
\begin_inset Formula 
\[
f\left(x_{1},x_{2},x_{3}\right)=\begin{bmatrix}x_{1} & x_{2} & x_{3}\end{bmatrix}\begin{bmatrix}-1 & 1 & 0\\
1 & 0 & -1/2\\
0 & -1/2 & \sqrt{2}
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=-x_{1}^{2}+2x_{1}x_{2}-x_{2}x_{3}+\sqrt{2}x_{3}^{2}.
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Frame
\begin_inset Argument 4
status open

\begin_layout Plain Layout
实对称阵与二次型的相互表示
\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

 二次型 
\begin_inset Formula $x_{1}x_{2}+x_{1}x_{3}+2x_{2}^{2}-3x_{2}x_{3}$
\end_inset

 的矩阵
\begin_inset Formula 
\[
A=\begin{bmatrix}0 & 1/2 & 1/2\\
1/2 & 2 & -3/2\\
1/2 & -3/2 & 0
\end{bmatrix}.
\]

\end_inset

反之, 对称矩阵 
\begin_inset Formula $A=\begin{bmatrix}0 & 1/2 & 1/2\\
1/2 & 2 & -3/2\\
1/2 & -3/2 & 0
\end{bmatrix}$
\end_inset

 所对应的二次型是
\begin_inset Formula 
\[
x^{T}Ax=\begin{bmatrix}x_{1} & x_{2} & x_{3}\end{bmatrix}\begin{bmatrix}0 & 1/2 & 1/2\\
1/2 & 2 & -3/2\\
1/2 & -3/2 & 0
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=x_{1}x_{2}+x_{1}x_{3}+2x_{2}^{2}-3x_{2}x_{3}.
\]

\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
求二次型的秩
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open

\begin_layout Plain Layout
E02
\end_layout

\end_inset

 求二次型 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=x_{1}^{2}-4x_{1}x_{2}+2x_{1}x_{3}-2x_{2}^{2}+6x_{3}^{2}$
\end_inset

 的秩.
\end_layout

\begin_layout Solution*
先求二次型的矩阵,
\begin_inset Formula 
\[
f\left(x_{1},x_{2},x_{3}\right)=x_{1}^{2}-2x_{1}x_{2}+x_{1}x_{3}-2x_{2}x_{1}-2x_{2}^{2}+0x_{2}x_{3}+x_{3}x_{1}+0x_{3}x_{2}+6x_{3}^{2}.
\]

\end_inset

所以 
\begin_inset Formula $A=\begin{bmatrix}1 & -2 & 1\\
-2 & -2 & 0\\
1 & 0 & 6
\end{bmatrix}$
\end_inset

, 对 
\begin_inset Formula $A$
\end_inset

 作初等变换
\begin_inset Formula 
\[
A\longrightarrow\begin{bmatrix}1 & -2 & 1\\
0 & -6 & 2\\
0 & 2 & 5
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & -2 & 1\\
0 & 2 & 5\\
0 & 0 & 17
\end{bmatrix},
\]

\end_inset

即 
\begin_inset Formula $r(A)=3$
\end_inset

, 所以二次型的秩为 
\begin_inset Formula $3$
\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
矩阵的合同
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Definition
设 
\begin_inset Formula $A,B$
\end_inset

 为两个 
\begin_inset Formula $n$
\end_inset

 阶矩阵, 如果存在 
\begin_inset Formula $n$
\end_inset

 阶非奇异矩阵 
\begin_inset Formula $C$
\end_inset

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

, 则称
\series bold
矩阵 
\begin_inset Formula $A$
\end_inset

 合同于矩阵 
\begin_inset Formula $B$
\end_inset


\series default
, 或 
\series bold

\begin_inset Formula $A$
\end_inset

 与 
\begin_inset Formula $B$
\end_inset

 合同
\series default
, 记为 
\color red

\begin_inset Formula $A\cong B$
\end_inset


\color inherit
.
\end_layout

\begin_layout Standard
易见, 二次型 
\begin_inset Formula $f\left(x_{1},x_{2},\cdots,x_{n}\right)=X^{T}AX$
\end_inset

 的矩阵 
\begin_inset Formula $A$
\end_inset

 与经过非退化线性变换 
\begin_inset Formula $X=CY$
\end_inset

 得到的二次型的矩阵 
\begin_inset Formula $B=C^{T}AC$
\end_inset

 是合同的.
\end_layout

\begin_layout Fact*
矩阵的合同关系基本性质:
\end_layout

\begin_layout Fact*
(1) 反身性: 对任意方阵 
\begin_inset Formula $A$
\end_inset

, 
\begin_inset Formula $A\cong A$
\end_inset

; (因为 
\begin_inset Formula $E^{T}AE=A$
\end_inset

);
\end_layout

\begin_layout Fact*
(2) 对称性: 若 
\begin_inset Formula $A\cong B$
\end_inset

, 则 
\begin_inset Formula $B\cong A$
\end_inset

;
\end_layout

\begin_layout Fact*
(3) 传递性 若 
\begin_inset Formula $A\cong B$
\end_inset

, 
\begin_inset Formula $B\cong C$
\end_inset

, 则 
\begin_inset Formula $A\cong C$
\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
对二次型做线性变换后的新二次型
\end_layout

\end_inset


\end_layout

\begin_deeper
\begin_layout Example
设二次型 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=2x_{1}x_{2}-4x_{1}x_{3}+10x_{2}x_{3}$
\end_inset

, 且
\begin_inset Formula 
\begin{equation}
\begin{cases}
x_{1}=y_{1}-y_{2}-5y_{3},\\
x_{2}=y_{1}+y_{2}+2y_{3},\\
x_{3}=y_{3}.
\end{cases}\label{eq:5.1-1}
\end{equation}

\end_inset

求经过上述线性变换后新的二次型.
\end_layout

\begin_layout Solution*
因 
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)$
\end_inset

 对应的矩阵为 
\begin_inset Formula $A=\begin{bmatrix}0 & 1 & -2\\
1 & 0 & 5\\
-2 & 5 & 0
\end{bmatrix}$
\end_inset

.
 而变换 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:5.1-1"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 所决定的变换矩阵 
\begin_inset Formula $C=\begin{bmatrix}1 & -1 & -5\\
1 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}$
\end_inset

,
\begin_inset Formula 
\[
C^{T}AC=\begin{bmatrix}1 & 1 & 0\\
-1 & 1 & 0\\
-5 & 2 & 1
\end{bmatrix}\begin{bmatrix}0 & 1 & -2\\
1 & 0 & 5\\
-2 & 5 & 0
\end{bmatrix}\begin{bmatrix}1 & -1 & -5\\
1 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}2 & 0 & 0\\
0 & -2 & 0\\
0 & 0 & 20
\end{bmatrix}.
\]

\end_inset

于是新的二次型为 
\begin_inset Formula $2y_{1}^{2}-2y_{2}^{2}+20y_{3}^{2}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Frame

\end_layout

\end_body
\end_document