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\begin_body
\begin_layout Section
正定二次型
\end_layout
\begin_layout Subsection
二次型的定性理论
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
二次型的定性理论
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设对称矩阵
\begin_inset Formula $A$
\end_inset
对应的二次型为
\begin_inset Formula $f=X^{T}AX$
\end_inset
,
\end_layout
\begin_layout Definition
(1) 如果对任何非零向量
\begin_inset Formula $X$
\end_inset
, 都有
\begin_inset Formula
\[
X^{T}AX>0,\quad(\text{或 }X^{T}AX<0)
\]
\end_inset
成立, 则称
\series bold
\begin_inset Formula $f=X^{T}AX$
\end_inset
为正定 (负定) 二次型
\series default
,
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
称为正定矩阵 (负定矩阵)
\series default
.
\end_layout
\begin_layout Definition
(2) 如果对任何非零向量
\begin_inset Formula $X$
\end_inset
, 都有
\begin_inset Formula
\[
X^{T}AX\geq0,\quad(\text{或 }X^{T}AX\leq0)
\]
\end_inset
\end_layout
\begin_layout Definition
成立, 且有非零向量
\begin_inset Formula $X_{0}$
\end_inset
, 使
\begin_inset Formula $X_{0}{}^{T}AX_{0}=0$
\end_inset
, 则称
\series bold
\begin_inset Formula $f=X^{T}AX$
\end_inset
为半正定 (半负定) 二次型
\series default
,
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
称为半正定矩阵 (半负定矩阵)
\series default
.
\end_layout
\begin_layout Remark
二次型的正定 (负定)、半正定 (半负定) 统称为
\series bold
二次型及其矩阵的有定性
\series default
.
不具备有定性的二次型及其矩阵称为
\series bold
不定的
\series default
.
\end_layout
\begin_layout Remark
二次型的有定性与其矩阵的有定性之间具有一一对应关系.
因此, 二次型的正定性判别可转化为对称矩阵的正定性判别.
\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
判断二次型的正定性
\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 $f\left(x_{1},x_{2},\cdots,x_{n}\right)=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}$
\end_inset
, 当
\begin_inset Formula $X=\left(x_{1},x_{2},\cdots,x_{n}\right)^{T}\neq0$
\end_inset
时, 显然有
\begin_inset Formula
\[
f\left(x_{1},x_{2},\cdots,x_{n}\right)>0,
\]
\end_inset
所以这个二次型是正定的, 其矩阵
\begin_inset Formula $E_{n}$
\end_inset
是正定矩阵.
\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
半负定矩阵的例
\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
二次型
\begin_inset Formula $f=-x_{1}^{2}-2x_{1}x_{2}+4x_{1}x_{3}-x_{2}^{2}+4x_{2}x_{3}-4x_{3}^{2}$
\end_inset
, 将其改写成
\begin_inset Formula
\[
f\left(x_{1},x_{2},x_{3}\right)=-\left(x_{1}+x_{2}-2x_{3}\right)^{2}\leq0,
\]
\end_inset
当
\begin_inset Formula $x_{1}+x_{2}-2x_{3}=0$
\end_inset
时,
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=0$
\end_inset
, 故
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)$
\end_inset
是半负定二次型, 其对应的矩阵
\begin_inset Formula $\begin{bmatrix}-1 & -1 & 2\\
-1 & -1 & 2\\
2 & 2 & -4
\end{bmatrix}$
\end_inset
是半负定矩阵.
\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
不定二次型的例
\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
二次型
\begin_inset Formula $f\left(x_{1},x_{2}\right)=x_{1}^{2}-2x_{2}^{2}$
\end_inset
是不定二次型, 因其符号有时正有时负, 如
\begin_inset Formula
\[
f(1,1)=-1<0,\quad f(2,1)=2>0.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsection
正定矩阵的判别法
\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
正定矩阵的判别法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
设
\begin_inset Formula $A$
\end_inset
为正定矩阵, 若
\begin_inset Formula $A\cong B$
\end_inset
, (
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
合同), 则
\begin_inset Formula $B$
\end_inset
也是正定矩阵.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
对角矩阵
\begin_inset Formula $D=\mathrm{diag}\left(d_{1},d_{2},\cdots,d_{n}\right)$
\end_inset
正定的充分必要条件是
\begin_inset Formula $d_{i}>0$
\end_inset
, (
\begin_inset Formula $i=1,2,\cdots,n$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
对称矩阵
\begin_inset Formula $A$
\end_inset
为正定的充分必要条件是它的特征值全大于零.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
矩阵
\begin_inset Formula $A$
\end_inset
为正定矩阵的充分必要条件是
\begin_inset Formula $A$
\end_inset
的正惯性指数
\begin_inset Formula $p=n$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
矩阵
\begin_inset Formula $A$
\end_inset
为正定矩阵的充分必要条件矩阵是: 存在非奇异矩阵
\begin_inset Formula $C$
\end_inset
, 使
\begin_inset Formula $A=C^{T}C$
\end_inset
.
即
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $E$
\end_inset
合同.
\end_layout
\begin_layout Corollary
若
\begin_inset Formula $A$
\end_inset
为正定矩阵, 则
\begin_inset Formula $|A|>0$
\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
判断正定二次型的例
\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
当
\begin_inset Formula $\lambda$
\end_inset
取何值时, 二次型
\begin_inset Formula
\[
f\left(x_{1},x_{2},x_{3}\right)=x_{1}^{2}+2x_{1}x_{2}+4x_{1}x_{3}+2x_{2}^{2}+6x_{2}x_{3}+\lambda x_{3}^{2}
\]
\end_inset
为正定二次型.
\end_layout
\begin_layout Solution*
由题设给出二次型的矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 1 & 2\\
1 & 2 & 3\\
2 & 3 & \lambda
\end{bmatrix}$
\end_inset
, 由于
\begin_inset Formula $\left|A_{1}\right|=1>0$
\end_inset
,
\begin_inset Formula $\left|A_{2}\right|=\begin{vmatrix}1 & 1\\
1 & 2
\end{vmatrix}=1>0$
\end_inset
,
\begin_inset Formula $\left|A_{3}\right|=|A|=\lambda-5>0$
\end_inset
, 所以
\begin_inset Formula $\lambda>5$
\end_inset
时, 二次型
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)$
\end_inset
正定二次型.
\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
惯性定理
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
秩为
\begin_inset Formula $r$
\end_inset
的
\begin_inset Formula $n$
\end_inset
元实二次型
\begin_inset Formula $f=X^{T}AX$
\end_inset
, 设其规范形为
\begin_inset Formula
\[
z_{1}^{2}+z_{2}^{2}+\cdots+z_{p}^{2}-z_{p+1}^{2}-\cdots-z_{r}^{2},
\]
\end_inset
则
\end_layout
\begin_layout Theorem
(1)
\begin_inset Formula $f$
\end_inset
负定的充分必要条件是
\begin_inset Formula $p=0$
\end_inset
, 且
\begin_inset Formula $r=n$
\end_inset
, (即负定二次型, 其规范形为
\begin_inset Formula $f=-z_{1}^{2}-z_{2}^{2}-\cdots-z_{n}^{2}$
\end_inset
).
\end_layout
\begin_layout Theorem
(2)
\begin_inset Formula $f$
\end_inset
半正定的充分必要条件是
\begin_inset Formula $p=r<n$
\end_inset
.
(即半正定二次型的规范形为
\begin_inset Formula $f=z_{1}^{2}+z_{2}^{2}+\cdots+z_{r}^{2}$
\end_inset
,
\begin_inset Formula $r<n$
\end_inset
).
\end_layout
\begin_layout Theorem
(3)
\begin_inset Formula $f$
\end_inset
半负定的充分必要条件是
\begin_inset Formula $p=0$
\end_inset
,
\begin_inset Formula $r<n$
\end_inset
.
(即
\begin_inset Formula $f=-z_{1}^{2}-z_{2}^{2}-\cdots-z_{r}^{2}$
\end_inset
,
\begin_inset Formula $r<n$
\end_inset
).
\end_layout
\begin_layout Theorem
(4)
\begin_inset Formula $f$
\end_inset
不定的充分必要条件是
\begin_inset Formula $0<p<r\leq n$
\end_inset
.
(即
\begin_inset Formula $f=z_{1}^{2}+z_{2}^{2}+\cdots+z_{p}^{2}-z_{p+1}^{2}-\cdots-z_{r}^{2}$
\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
子式与主子式的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
的
\begin_inset Formula $k$
\end_inset
个行标和列标相同的子式
\begin_inset Formula
\[
\begin{vmatrix}a_{i_{1}i_{1}} & a_{i_{1}i_{2}} & \cdots & a_{i_{1}i_{k}}\\
a_{i_{2}i_{1}} & a_{i_{2}i_{2}} & \cdots & a_{i_{2}i_{k}}\\
\vdots & \vdots & \ddots & \vdots\\
a_{i_{k}i_{1}} & a_{i_{k}i_{2}} & \cdots & a_{i_{k}i_{k}}
\end{vmatrix},\quad\left(1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n\right),
\]
\end_inset
称为
\begin_inset Formula $A$
\end_inset
的一个
\begin_inset Formula $k$
\end_inset
阶主子式.
而子式
\begin_inset Formula
\[
\left|A_{k}\right|=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1k}\\
a_{21} & a_{22} & \cdots & a_{2k}\\
\vdots & \vdots & \ddots & \vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk}
\end{vmatrix},\quad(k=1,2,\cdots,n),
\]
\end_inset
称为
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $k$
\end_inset
阶顺序主子式.
\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
用顺序主子式判别矩阵的正 (负) 定性
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
为正定矩阵的充分必要条件是
\begin_inset Formula $A$
\end_inset
的所有顺序主子式
\begin_inset Formula $\left|A_{k}\right|>0$
\end_inset
, (
\begin_inset Formula $k=1,2,\cdots,n$
\end_inset
).
\end_layout
\begin_layout Remark
\end_layout
\begin_deeper
\begin_layout Enumerate
若
\begin_inset Formula $A$
\end_inset
是负定矩阵, 则
\begin_inset Formula $-A$
\end_inset
为正定矩阵.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
是负定矩阵的充要条件是:
\begin_inset Formula $(-1)^{k}\left|A_{k}\right|>0$
\end_inset
, (
\begin_inset Formula $k=1,2,\cdots,n$
\end_inset
), 其中
\begin_inset Formula $A_{k}$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $k$
\end_inset
阶顺序主子式.
\end_layout
\begin_layout Enumerate
对半正定 (半负定) 矩阵可证明以下三个结论等价:
\end_layout
\begin_deeper
\begin_layout Enumerate
对称矩阵
\begin_inset Formula $A$
\end_inset
是半正定 (半负定) 的;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
的所有主子式大于 (小于) 或等于零;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A$
\end_inset
的全部特征值大于 (小于) 或等于零.
\end_layout
\end_deeper
\end_deeper
\begin_layout Remark
\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
判断负定二次型的例
\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
判别二次型
\begin_inset Formula
\[
f(x,y,z)=-5x^{2}-6y^{2}-4z^{2}+4xy+4xz,
\]
\end_inset
为负定二次型.
\end_layout
\begin_layout Solution*
由题设给出二次型的矩阵
\begin_inset Formula $A=\begin{bmatrix}-5 & 2 & 2\\
2 & -6 & 0\\
2 & 0 & -4
\end{bmatrix}$
\end_inset
, 由于
\begin_inset Formula $\left|A_{1}\right|=-5<0$
\end_inset
,
\begin_inset Formula $\left|A_{2}\right|=\begin{vmatrix}-5 & 2\\
2 & -6
\end{vmatrix}=26>0$
\end_inset
,
\begin_inset Formula $\left|A_{3}\right|=|A|=-80<0$
\end_inset
, 所以
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)$
\end_inset
为负定二次型.
\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
正定矩阵的逆也正定
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
E06
\end_layout
\end_inset
证明: 如果
\begin_inset Formula $A$
\end_inset
为正定矩阵, 则
\begin_inset Formula $A^{-1}$
\end_inset
也是正定矩阵.
\end_layout
\begin_layout Proof
由于
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
正定, 则存在非奇异矩阵
\begin_inset Formula $C$
\end_inset
, 使
\begin_inset Formula $C^{T}AC=E_{n}$
\end_inset
, 两边取逆得:
\begin_inset Formula $C^{-1}A^{-1}\left(C^{T}\right)^{-1}=E_{n}$
\end_inset
.
\end_layout
\begin_layout Proof
又因为
\begin_inset Formula $\left(C^{T}\right)^{-1}=\left(C^{-1}\right)^{T}$
\end_inset
,
\begin_inset Formula $\left(\left(C^{-1}\right)^{T}\right)^{T}=C^{-1}$
\end_inset
, 因此
\begin_inset Formula $\left(\left(C^{-1}\right)^{T}\right)^{T}A^{-1}\left(C^{-1}\right)^{T}=E_{n}$
\end_inset
,
\begin_inset Formula $\left|\left(C^{-1}\right)^{T}\right|=|C|^{-1}\neq0$
\end_inset
, 故
\begin_inset Formula $A^{-1}$
\end_inset
与
\begin_inset Formula $E_{n}$
\end_inset
合同, 即
\begin_inset Formula $A^{-1}$
\end_inset
为正定矩阵.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\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 Problem
设二次型
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=x_{1}^{2}+x_{2}^{2}+2x_{3}^{2}+2tx_{1}x_{2}-2x_{1}x_{3}$
\end_inset
, 试确定当
\begin_inset Formula $t$
\end_inset
取何值时,
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)$
\end_inset
为正定二次型.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
判别二次型
\begin_inset Formula $f\left(x_{1},x_{2},x_{3}\right)=2x_{1}^{2}+4x_{2}^{2}+5x_{3}^{2}-4x_{1}x_{3}$
\end_inset
是否正定.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $A,B$
\end_inset
分别为
\begin_inset Formula $m$
\end_inset
阶,
\begin_inset Formula $n$
\end_inset
阶正定矩阵, 试判定分块矩阵
\begin_inset Formula $C=\begin{bmatrix}A & 0\\
0 & B
\end{bmatrix}$
\end_inset
是否为正定矩阵.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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