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\end_header
\begin_body
\begin_layout Section
矩阵的特征值与特征向量
\end_layout
\begin_layout Subsection
特征值与特征向量
\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
特征值与特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $A$
\end_inset
是
\begin_inset Formula $n$
\end_inset
阶方阵, 如果数
\begin_inset Formula $\lambda$
\end_inset
和
\begin_inset Formula $n$
\end_inset
维非零向量
\begin_inset Formula $X$
\end_inset
使
\begin_inset Formula
\[
AX=\lambda X
\]
\end_inset
成立, 则称数
\begin_inset Formula $\lambda$
\end_inset
为方阵
\begin_inset Formula $A$
\end_inset
的特征值, 非零向量
\begin_inset Formula $X$
\end_inset
称为
\begin_inset Formula $A$
\end_inset
的对应于特征值
\begin_inset Formula $\lambda$
\end_inset
的特征向量 (或称为
\begin_inset Formula $A$
\end_inset
的属于特征值
\begin_inset Formula $\lambda$
\end_inset
的特征向量).
\end_layout
\begin_layout Remark
\begin_inset Formula $n$
\end_inset
阶方阵
\begin_inset Formula $A$
\end_inset
的特征值
\begin_inset Formula $\lambda$
\end_inset
, 就是使齐次线性方程组
\begin_inset Formula
\[
(\lambda E-A)X=0
\]
\end_inset
有非零解的值, 即满足方程
\begin_inset Formula
\[
|\lambda E-A|=0
\]
\end_inset
的
\begin_inset Formula $\lambda$
\end_inset
都是矩阵
\begin_inset Formula $A$
\end_inset
的特征值.
\end_layout
\begin_layout Definition
称关于
\begin_inset Formula $\lambda$
\end_inset
的一元
\begin_inset Formula $n$
\end_inset
次方程
\begin_inset Formula $|\lambda E-A|=0$
\end_inset
为矩阵
\begin_inset Formula $A$
\end_inset
的特征方程, 称
\begin_inset Formula $\lambda$
\end_inset
的一元
\begin_inset Formula $n$
\end_inset
次多项式
\begin_inset Formula
\[
f(\lambda)=|\lambda E-A|
\]
\end_inset
为矩阵
\begin_inset Formula $A$
\end_inset
的特征多项式.
\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
特征向量的求法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
根据上述定义, 即可给出特征向量的求法:
\end_layout
\begin_layout Standard
设
\begin_inset Formula $\lambda=\lambda_{i}$
\end_inset
为方阵
\begin_inset Formula $A$
\end_inset
的一个特征值, 则由齐次线性方程组
\begin_inset Formula
\[
\left(\lambda_{i}E-A\right)X=0
\]
\end_inset
可求得非零解
\begin_inset Formula $p_{i}$
\end_inset
, 那么
\begin_inset Formula $p_{i}$
\end_inset
就是
\begin_inset Formula $A$
\end_inset
的对应于特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
的特征向量, 且
\begin_inset Formula $A$
\end_inset
的对应于特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
的特征向量全体是方程组
\begin_inset Formula $\left(\lambda_{i}E-A\right)X=0$
\end_inset
的全体非零解.
即设
\begin_inset Formula $p_{1},p_{2},\cdots,p_{s}$
\end_inset
为
\begin_inset Formula $\left(\lambda_{i}E-A\right)X=0$
\end_inset
的基础解系, 则
\begin_inset Formula $A$
\end_inset
的对应于特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
的特征向量全体是
\begin_inset Formula
\[
p=k_{1}p_{1}+k_{2}p_{2}+\cdots+k_{s}p_{s},\quad\left(k_{1},\cdots,k_{s}\text{ 不同时 }0\right).
\]
\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
求矩阵的特征值和特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:4.2-1"
\end_inset
求矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 1\\
5 & -1
\end{bmatrix}$
\end_inset
的特征值和特征向量.
\end_layout
\begin_layout Solution*
矩阵
\begin_inset Formula $A$
\end_inset
的特征方程为
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda-3 & -1\\
-5 & \lambda+1
\end{vmatrix}=0\Longrightarrow(\lambda-4)(\lambda+2)=0,
\]
\end_inset
所以
\begin_inset Formula $\lambda_{1}=4$
\end_inset
,
\begin_inset Formula $\lambda_{2}=-2$
\end_inset
是矩阵
\begin_inset Formula $A$
\end_inset
的两个不同的特征值.
\end_layout
\begin_layout Solution*
以
\begin_inset Formula $\lambda_{1}=4$
\end_inset
代入与特征方程对应的齐次线性方程组, 得
\begin_inset Formula
\[
\begin{cases}
x_{1}-x_{2}=0\\
-5x_{1}+5x_{2}=0
\end{cases}\Longrightarrow\text{ 基础解系是 }\begin{bmatrix}1\\
1
\end{bmatrix},
\]
\end_inset
故
\begin_inset Formula $k_{1}\begin{bmatrix}1\\
1
\end{bmatrix}$
\end_inset
, (
\begin_inset Formula $k_{1}\neq0$
\end_inset
) 是矩阵
\begin_inset Formula $A$
\end_inset
对应于
\begin_inset Formula $\lambda_{1}=4$
\end_inset
的全部特征向量.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
矩阵
\begin_inset Formula
\[
\lambda E-A=\begin{bmatrix}\lambda-3 & -1\\
-5 & \lambda+1
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Solution*
以
\begin_inset Formula $\lambda_{2}=-2$
\end_inset
代入与特征方程对应的齐次线性方程组, 得
\begin_inset Formula
\[
\begin{cases}
-5x_{1}-x_{2}=0\\
-5x_{1}-x_{2}=0
\end{cases}\Longrightarrow\text{ 基础解系是 }\begin{bmatrix}1\\
-5
\end{bmatrix},
\]
\end_inset
故
\begin_inset Formula $k_{2}\begin{bmatrix}1\\
-5
\end{bmatrix}$
\end_inset
, (
\begin_inset Formula $k_{2}\neq0$
\end_inset
) 是矩阵
\begin_inset Formula $A$
\end_inset
对应于
\begin_inset Formula $\lambda_{1}=-2$
\end_inset
的全部特征向量.
\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
求矩阵的特征值和特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}-2 & 1 & 1\\
0 & 2 & 0\\
-4 & 1 & 3
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A$
\end_inset
的特征值与特征向量.
\end_layout
\begin_layout Solution*
\begin_inset Formula $|\lambda E-A|=\begin{vmatrix}\lambda+2 & -1 & -1\\
0 & \lambda-2 & 0\\
4 & -1 & \lambda-3
\end{vmatrix}=(\lambda+1)(\lambda-2)^{2},$
\end_inset
解得特征值
\begin_inset Formula $\lambda_{1}=-1$
\end_inset
,
\begin_inset Formula $\lambda_{2}=\lambda_{3}=2$
\end_inset
.
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $\lambda_{1}=-1$
\end_inset
时, 解方程
\begin_inset Formula $(-A-E)x=0$
\end_inset
.
由
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
-A-E=\begin{bmatrix}1 & -1 & -1\\
0 & -3 & 0\\
4 & -1 & -4
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & -1\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix},
\]
\end_inset
这个矩阵对应的线性方程组的基础解系
\begin_inset Formula $p_{1}=\begin{bmatrix}1\\
0\\
1
\end{bmatrix}$
\end_inset
, 故对应于
\begin_inset Formula $\lambda_{1}=-1$
\end_inset
的全体特征向量为
\begin_inset Formula $k_{1}p_{1}$
\end_inset
, (
\begin_inset Formula $k_{1}\neq0$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $\lambda_{2}=\lambda_{3}=2$
\end_inset
时, 解方程
\begin_inset Formula $(2E-A)x=0$
\end_inset
.
由
\begin_inset Formula
\[
2E-A=\begin{bmatrix}4 & -1 & -1\\
0 & 0 & 0\\
4 & -1 & -1
\end{bmatrix}\longrightarrow\begin{bmatrix}4 & -1 & -1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix},
\]
\end_inset
类似地解得基础解系
\begin_inset Formula $p_{2}=\begin{bmatrix}1\\
4\\
0
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $p_{3}=\begin{bmatrix}1\\
0\\
4
\end{bmatrix}$
\end_inset
, 故对应于
\begin_inset Formula $\lambda_{2}=\lambda_{3}=2$
\end_inset
的全部特征向量为:
\begin_inset Formula
\[
k_{2}p_{2}+k_{3}p_{3},\quad\left(k_{2},k_{3}\text{ 不同时为 }0\right).
\]
\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
求矩阵的特征值和特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求
\begin_inset Formula $3$
\end_inset
阶矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & -1 & 1\\
1 & 3 & -1\\
1 & 1 & 1
\end{bmatrix}$
\end_inset
的特征值以及相应的线性无关的特征向量组.
\end_layout
\begin_layout Solution*
\begin_inset Formula $A$
\end_inset
的特征多项式为
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda-1 & 1 & -1\\
-1 & \lambda-3 & 1\\
-1 & -1 & \lambda-1
\end{vmatrix}=\lambda^{3}-5\lambda^{2}+8\lambda-4=(\lambda-2)^{2}(\lambda-1).
\]
\end_inset
这个多项式的根为
\begin_inset Formula $\lambda_{1}=1$
\end_inset
,
\begin_inset Formula $\lambda_{2}=\lambda_{3}=2$
\end_inset
, 因此
\begin_inset Formula $A$
\end_inset
的特征值等于
\begin_inset Formula $1,2,2$
\end_inset
.
\end_layout
\begin_layout Solution*
接下来求特征向量: 对
\begin_inset Formula $\lambda_{1}=1$
\end_inset
, 将
\begin_inset Formula $\lambda=1$
\end_inset
代入
\begin_inset Formula $(\lambda E-A)x=0$
\end_inset
, 得
\begin_inset Formula
\begin{equation}
(E-A)x=\begin{bmatrix}0 & 1 & -1\\
-1 & -2 & 1\\
-1 & -1 & 0
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}0\\
0\\
0
\end{bmatrix}.\label{eq:4.2-1}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
(E-A)x=\begin{bmatrix}0 & 1 & -1\\
-1 & -2 & 1\\
-1 & -1 & 0
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}0\\
0\\
0
\end{bmatrix}.
\]
\end_inset
容易算出这个方程组的系数矩阵等于
\begin_inset Formula $2$
\end_inset
, 因此齐次线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的基础解系只有一个线性无关的向量, 不难求出为
\begin_inset Formula
\[
\eta_{1}=(-1,1,1)^{T}.
\]
\end_inset
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{2}=\lambda_{3}=2$
\end_inset
, 将
\begin_inset Formula $\lambda=2$
\end_inset
代入可得齐次方程组:
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{bmatrix}1 & 1 & -1\\
-1 & -1 & 1\\
-1 & -1 & 1
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\begin{bmatrix}0\\
0\\
0
\end{bmatrix}.
\]
\end_inset
求出这个齐次线性方程组的基础解系为
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\eta_{2}=(1,0,1)^{T},\quad\eta_{3}=(0,1,1)^{T}.
\]
\end_inset
因此
\begin_inset Formula $A$
\end_inset
的相应于特征值
\begin_inset Formula $1$
\end_inset
的线性无关的特征向量有
\begin_inset Formula $1$
\end_inset
个, 而相应于特征值
\begin_inset Formula $2$
\end_inset
的线性无关的特征向量有
\begin_inset Formula $2$
\end_inset
个, 于是
\begin_inset Formula $A$
\end_inset
的线性无关的特征向量有
\begin_inset Formula $3$
\end_inset
个, 正好等于
\begin_inset Formula $A$
\end_inset
的阶数
\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
求解特征向量问题过程中可能出现的问题
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $\lambda_{1}$
\end_inset
和
\begin_inset Formula $\lambda_{2}$
\end_inset
是矩阵
\begin_inset Formula $A$
\end_inset
的两个不同的特征值, 对应的特征向量依次为
\begin_inset Formula $p_{1}$
\end_inset
和
\begin_inset Formula $p_{2}$
\end_inset
, 证明
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
不是
\begin_inset Formula $A$
\end_inset
的特征向量.
\end_layout
\begin_layout Proof
按题设, 有
\begin_inset Formula $Ap_{1}=\lambda_{1}p_{1}$
\end_inset
,
\begin_inset Formula $Ap_{2}=\lambda_{2}p_{2}$
\end_inset
, 故
\begin_inset Formula $A\left(p_{1}+p_{2}\right)=\lambda_{1}p_{1}+\lambda_{2}p_{2}$
\end_inset
.
\end_layout
\begin_layout Proof
用
\series bold
反证法
\series default
, 设
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的特征向量, 则应存在数
\begin_inset Formula $\lambda$
\end_inset
, 使
\begin_inset Formula
\[
A\left(p_{1}+p_{2}\right)=\lambda\left(p_{1}+p_{2}\right),
\]
\end_inset
于是
\begin_inset Formula $\lambda\left(p_{1}+p_{2}\right)=\lambda_{1}p_{1}+\lambda_{2}p_{2}$
\end_inset
, 即
\begin_inset Formula $\left(\lambda_{1}-\lambda\right)p_{1}+\left(\lambda_{2}-\lambda\right)p_{2}=0$
\end_inset
.
\end_layout
\begin_layout Proof
因
\begin_inset Formula $\lambda_{1}\neq\lambda_{2}$
\end_inset
, 由本节定理知
\begin_inset Formula $p_{1},p_{2}$
\end_inset
线性无关, 故由上式得
\begin_inset Formula
\[
\lambda_{1}-\lambda=\lambda_{2}-\lambda=0,
\]
\end_inset
即
\begin_inset Formula $\lambda_{1}=\lambda_{2}$
\end_inset
, 与题设矛盾.
因此
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
不是
\begin_inset Formula $A$
\end_inset
的特征向量.
\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
求矩阵的特征值和特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求
\begin_inset Formula $n$
\end_inset
阶数量矩阵
\begin_inset Formula $A=\begin{bmatrix}a & 0 & \cdots & 0\\
0 & a & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a
\end{bmatrix}$
\end_inset
的特征值与特征向量.
\end_layout
\begin_layout Solution*
\begin_inset Formula $|\lambda E-A|=\begin{vmatrix}\lambda-a & 0 & \cdots & 0\\
0 & \lambda-a & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda-a
\end{vmatrix}=(\lambda-a)^{n}=0$
\end_inset
, 故
\begin_inset Formula $A$
\end_inset
的特征值为
\begin_inset Formula $\lambda_{1}=\lambda_{2}=\cdots=\lambda_{n}=a$
\end_inset
.
\end_layout
\begin_layout Solution*
把
\begin_inset Formula $\lambda=a$
\end_inset
代入
\begin_inset Formula $(\lambda E-A)x=0$
\end_inset
得
\begin_inset Formula $0\cdot x_{1}=0$
\end_inset
,
\begin_inset Formula $0\cdot x_{2}=0$
\end_inset
,
\begin_inset Formula $\cdots$
\end_inset
,
\begin_inset Formula $0\cdot x_{n}=0$
\end_inset
.
这个方程组的系数矩阵是零矩阵, 所以任意
\begin_inset Formula $n$
\end_inset
个线性无关的向量都是它的基础解系, 取单位向量组
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\varepsilon_{1}=\begin{bmatrix}1\\
0\\
\vdots\\
0
\end{bmatrix},\ \varepsilon_{n}=\begin{bmatrix}0\\
1\\
\vdots\\
0
\end{bmatrix},\ \cdots,\ \varepsilon_{n}=\begin{bmatrix}0\\
0\\
\vdots\\
1
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Solution*
作为基础解系, 于是,
\begin_inset Formula $A$
\end_inset
的全部特征向量为
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
c_{1}\varepsilon_{1}+c_{2}\varepsilon_{2}+\cdots+c_{n}\varepsilon_{n},\quad\left(c_{1},c_{2},\cdots,c_{n}\text{ 不全为零}\right).
\]
\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
求上三角矩阵的特征值
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
试求上三角阵
\begin_inset Formula $A$
\end_inset
的特征值:
\begin_inset Formula $A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
0 & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a_{nn}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $|\lambda E-A|=\begin{vmatrix}\lambda-a_{11} & -a_{12} & \cdots & -a_{1m}\\
0 & \lambda-a_{22} & \cdots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda-a_{nn}
\end{vmatrix}$
\end_inset
, 这是一个上三角行列式, 因此,
\begin_inset Formula
\[
|\lambda E-A|=\left(\lambda-a_{11}\right)\left(\lambda-a_{22}\right)\cdots\left(\lambda-a_{nn}\right).
\]
\end_inset
因此
\begin_inset Formula $A$
\end_inset
的特征值等于
\begin_inset Formula $a_{11},a_{22},\cdots,a_{nn}$
\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
特征值相关反例
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
令
\begin_inset Formula $A=\begin{bmatrix}1 & 1\\
0 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 2\\
2 & 1
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula $A+B=\begin{bmatrix}2 & 3\\
2 & 2
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $AB=\begin{bmatrix}3 & 3\\
2 & 1
\end{bmatrix}$
\end_inset
, 不难看出
\begin_inset Formula $1$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的一个特征值,
\begin_inset Formula $-1$
\end_inset
是
\begin_inset Formula $B$
\end_inset
的一个特征值, 但
\begin_inset Formula $1+(-1)$
\end_inset
不是
\begin_inset Formula $A+B$
\end_inset
的特征值, 因为
\begin_inset Formula $A+B$
\end_inset
的特征值是
\begin_inset Formula $2+\sqrt{6}$
\end_inset
,
\begin_inset Formula $2-\sqrt{6}$
\end_inset
.
又
\begin_inset Formula $AB$
\end_inset
的特征值是
\begin_inset Formula $2+\sqrt{7}$
\end_inset
,
\begin_inset Formula $2-\sqrt{7}$
\end_inset
.
因此
\begin_inset Formula $1\times(-1)=-1$
\end_inset
也不是
\begin_inset Formula $AB$
\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
奇异矩阵的特征值
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
试证:
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
是奇异矩阵的充分必要条件是
\begin_inset Formula $A$
\end_inset
有一个特征值为零.
\end_layout
\begin_layout Proof
必要性: 若
\begin_inset Formula $A$
\end_inset
是奇异矩阵, 则
\begin_inset Formula $|A|=0$
\end_inset
.
于是
\begin_inset Formula
\[
|0E-A|=|-A|=(-1)^{n}|A|=0,
\]
\end_inset
即
\begin_inset Formula $0$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的一个特征值.
\end_layout
\begin_layout Proof
充分性: 设
\begin_inset Formula $A$
\end_inset
有一个特征值为
\begin_inset Formula $0$
\end_inset
, 对应的特征向量为
\begin_inset Formula $p$
\end_inset
, 由特征值的定义, 有
\begin_inset Formula
\[
Ap=0p=0,\quad(p\neq0),
\]
\end_inset
所以齐次线性方程组
\begin_inset Formula $Ax=0$
\end_inset
有非零解
\begin_inset Formula $p$
\end_inset
.
由此可知
\begin_inset Formula $|A|=0$
\end_inset
, 即
\begin_inset Formula $A$
\end_inset
为奇异矩阵.
\end_layout
\begin_layout Remark
此例也可以叙述为:
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
可逆
\begin_inset Formula $\Leftrightarrow$
\end_inset
它的任一特征值不为零.
\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
矩阵函数的特征值
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $\lambda$
\end_inset
是方阵
\begin_inset Formula $A$
\end_inset
的特征值, 证明
\end_layout
\begin_layout Example
(1)
\begin_inset Formula $\lambda^{2}$
\end_inset
是
\begin_inset Formula $A^{2}$
\end_inset
的特征值;
\end_layout
\begin_layout Example
(2) 当
\begin_inset Formula $A$
\end_inset
可逆时,
\begin_inset Formula $\frac{1}{\lambda}$
\end_inset
是
\begin_inset Formula $A^{-1}$
\end_inset
的特征值.
\end_layout
\begin_layout Proof
(2) 因
\begin_inset Formula $\lambda$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的特征值, 故有
\begin_inset Formula $p\neq0$
\end_inset
使
\begin_inset Formula $Ap=\lambda p$
\end_inset
, 有
\begin_inset Formula $p=\lambda A^{-1}p$
\end_inset
, 因
\begin_inset Formula $p\neq0$
\end_inset
, 知
\begin_inset Formula $\lambda\neq0$
\end_inset
, 故
\begin_inset Formula $A^{-1}p=\frac{1}{\lambda}p$
\end_inset
, 即
\begin_inset Formula $\frac{1}{\lambda}$
\end_inset
是
\begin_inset Formula $A^{-1}$
\end_inset
的特征值.
证毕.
\end_layout
\begin_layout Remark
易进一步证明: 若
\begin_inset Formula $\lambda$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的特征值, 则
\begin_inset Formula $\lambda^{k}$
\end_inset
是
\begin_inset Formula $A^{k}$
\end_inset
的特征值,
\begin_inset Formula $\varphi(\lambda)$
\end_inset
是
\begin_inset Formula $\varphi(A)$
\end_inset
的特征值, 其中
\begin_inset Formula $\varphi(x)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}$
\end_inset
.
特别地, 设特征多项式
\begin_inset Formula $f(\lambda)=|\lambda E-A|$
\end_inset
, 则
\begin_inset Formula $f(\lambda)$
\end_inset
是
\begin_inset Formula $f(A)$
\end_inset
的特征值, 且
\begin_inset Formula
\[
A^{n}-\left(a_{11}+a_{22}+\cdots+a_{nn}\right)A^{n-1}+\cdots+(-1)^{n}|A|E=0.
\]
\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
通过特征值求行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $3$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
的特征值为
\begin_inset Formula $1,-1,2$
\end_inset
, 求
\begin_inset Formula $\left|A^{*}+3A-2E\right|$
\end_inset
.
\end_layout
\begin_layout Solution*
因
\begin_inset Formula $A$
\end_inset
的特征值全不为
\begin_inset Formula $0$
\end_inset
, 知
\begin_inset Formula $A$
\end_inset
可逆, 故
\begin_inset Formula $A^{*}=|A|A^{-1}$
\end_inset
.
而
\begin_inset Formula $|A|=\lambda_{1}\lambda_{2}\lambda_{3}=-2$
\end_inset
, 所以
\begin_inset Formula
\[
A^{*}+3A-2E=-2A^{-1}+3A-2E.
\]
\end_inset
把上式记作
\begin_inset Formula $\varphi(A)$
\end_inset
, 有
\begin_inset Formula $\varphi(\lambda)=-\frac{2}{\lambda}+3\lambda-2$
\end_inset
, 故
\begin_inset Formula $\varphi(A)$
\end_inset
的特征值为
\begin_inset Formula
\[
\varphi(1)=-1,\quad\varphi(-1)=-3,\quad\varphi(2)=3.
\]
\end_inset
于是
\begin_inset Formula
\[
\left|A^{*}+3A-2E\right|=(-1)\cdot(-3)\cdot3=9.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsection
特征值与特征向量的性质
\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
特征值与特征向量的性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
与它的转置矩阵
\begin_inset Formula $A^{T}$
\end_inset
有相同的特征值.
\end_layout
\begin_layout Theorem
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset Formula $\triangle\triangle\triangle$
\end_inset
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "thm:4.2-1"
\end_inset
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
的互不相等的特征值
\begin_inset Formula $\lambda_{1},\cdots,\lambda_{m}$
\end_inset
对应的特征向量
\begin_inset Formula $p_{1},p_{2},\cdots,p_{m}$
\end_inset
线性无关.
\end_layout
\begin_layout Remark
1.
属于不同特征值的特征向量是线性无关的;
\end_layout
\begin_layout Remark
2.
属于同一特征值的特征向量的非零线性组合仍是属于这个特征值的特征向量;
\end_layout
\begin_layout Remark
3.
矩阵的特征向量总是相对于矩阵的特征值而言的, 一个特征值具有的特征向量不唯一; 一个特征向量不能属于不同的特征值.
\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
特征向量相关问题
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $\lambda_{1}$
\end_inset
和
\begin_inset Formula $\lambda_{2}$
\end_inset
是矩阵
\begin_inset Formula $A$
\end_inset
的两个不同的特征值, 对应的特征向量依次为
\begin_inset Formula $p_{1}$
\end_inset
和
\begin_inset Formula $p_{2}$
\end_inset
, 证明
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
不是
\begin_inset Formula $A$
\end_inset
的特征向量.
\end_layout
\begin_layout Proof
按题设, 有
\begin_inset Formula $Ap_{1}=\lambda_{1}p_{1}$
\end_inset
,
\begin_inset Formula $Ap_{2}=\lambda_{2}p_{2}$
\end_inset
, 故
\begin_inset Formula $A\left(p_{1}+p_{2}\right)=\lambda_{1}p_{1}+\lambda_{2}p_{2}$
\end_inset
.
\end_layout
\begin_layout Proof
用
\series bold
反证法
\series default
, 设
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的特征向量, 则应存在数
\begin_inset Formula $\lambda$
\end_inset
, 使
\begin_inset Formula
\[
A\left(p_{1}+p_{2}\right)=\lambda\left(p_{1}+p_{2}\right),
\]
\end_inset
于是
\begin_inset Formula $\lambda\left(p_{1}+p_{2}\right)=\lambda_{1}p_{1}+\lambda_{2}p_{2}$
\end_inset
, 即
\begin_inset Formula $\left(\lambda_{1}-\lambda\right)p_{1}+\left(\lambda_{2}-\lambda\right)p_{2}=0$
\end_inset
.
\end_layout
\begin_layout Proof
因
\begin_inset Formula $\lambda_{1}\neq\lambda_{2}$
\end_inset
, 由本节定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
知
\begin_inset Formula $p_{1},p_{2}$
\end_inset
线性无关, 故由上式得
\begin_inset Formula
\[
\lambda_{1}-\lambda=\lambda_{2}-\lambda=0,
\]
\end_inset
即
\begin_inset Formula $\lambda_{1}=\lambda_{2}$
\end_inset
, 与题设矛盾.
因此
\begin_inset Formula $p_{1}+p_{2}$
\end_inset
不是
\begin_inset Formula $A$
\end_inset
的特征向量.
\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
矩阵的特征方程与迹的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
设
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
是
\begin_inset Formula $n$
\end_inset
阶矩阵, 则
\begin_inset Formula
\[
\begin{aligned}f(\lambda) & =|\lambda E-A|=\left|\begin{array}{cccc}
\lambda-a_{11} & -a_{12} & \cdots & -a_{1n}\\
-a_{21} & \lambda-a_{22} & \cdots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & -a_{n2} & \cdots & \lambda-a_{nn}
\end{array}\right|\\
& =\lambda^{n}-\left(\sum_{i=1}^{n}a_{ii}\right)\lambda^{n-1}+\cdots+(-1)^{k}S_{k}\lambda^{n-k}+\cdots+(-1)^{n}|A|.
\end{aligned}
\]
\end_inset
其中
\begin_inset Formula $S_{k}$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的全体
\begin_inset Formula $k$
\end_inset
阶主子式的和.
设
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $n$
\end_inset
个特征值, 则由
\begin_inset Formula $n$
\end_inset
次代数方程的根与系数的关系知, 有
\end_layout
\begin_layout Proposition
(1)
\begin_inset Formula $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}=a_{11}+a_{22}+\cdots+a_{nn}$
\end_inset
;
\end_layout
\begin_layout Proposition
(2)
\begin_inset Formula $\lambda_{1}\lambda_{2}\cdots\lambda_{n}=|A|$
\end_inset
.
\end_layout
\begin_layout Proposition
其中
\begin_inset Formula $A$
\end_inset
的全体特征值的和
\begin_inset Formula $a_{11}+a_{22}+\cdots+a_{nn}$
\end_inset
称为矩阵
\begin_inset Formula $A$
\end_inset
的迹, 记为
\begin_inset Formula $\mathrm{tr}(A)$
\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
正交矩阵的特征值
\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
正交矩阵的实特征值的绝对值为
\begin_inset Formula $1$
\end_inset
.
\end_layout
\begin_layout Proof
设
\begin_inset Formula $A$
\end_inset
为正交矩阵,
\begin_inset Formula $p$
\end_inset
是方阵
\begin_inset Formula $A$
\end_inset
的对应于特征值
\begin_inset Formula $\lambda$
\end_inset
的特征向量, 则
\begin_inset Formula $Ap=\lambda p$
\end_inset
.
因
\begin_inset Formula
\begin{align}
(Ap)^{T}Ap & =p^{T}A^{T}Ap^{T}=p^{T}p=\|p\|^{2},\label{eq:4.2-2}\\
(Ap)^{T}Ap & =(\lambda p)^{T}(\lambda p)=\lambda^{2}p^{T}p=\lambda^{2}\|p\|^{2},\label{eq:4.2-3}
\end{align}
\end_inset
又
\begin_inset Formula $p\neq0$
\end_inset
, 所以
\begin_inset Formula $\|p\|>0$
\end_inset
, 式 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:4.2-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
)-式 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:4.2-3"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 得
\begin_inset Formula $\lambda^{2}=1$
\end_inset
, 即
\begin_inset Formula $|\lambda|=1$
\end_inset
.
\end_layout
\begin_layout Remark
\begin_inset Formula $A$
\end_inset
的特征值
\begin_inset Formula $\lambda$
\end_inset
是特征方程
\begin_inset Formula $|\lambda E-A|=0$
\end_inset
的根, 也是
\begin_inset Formula $|A-\lambda E|=0$
\end_inset
的根.
\begin_inset Formula $A$
\end_inset
的对应特征值
\begin_inset Formula $\lambda$
\end_inset
的特征向量是齐次方程组
\begin_inset Formula $(\lambda E-A)X=0$
\end_inset
的非零解, 也是
\begin_inset Formula $(A-\lambda E)X=0$
\end_inset
的非零解.
\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
矩阵特征值的估计
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
设
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
是
\begin_inset Formula $n$
\end_inset
阶矩阵, 如果
\end_layout
\begin_layout Proposition
\begin_inset Formula
\[
\sum_{j=1}^{n}\left|a_{ij}\right|<1,\qquad(i=1,2,\cdots,n)
\]
\end_inset
或
\end_layout
\begin_layout Proposition
\begin_inset Formula
\[
\sum_{i=1}^{n}\left|a_{ij}\right|<1,\qquad(j=1,2,\cdots,n)
\]
\end_inset
有一个成立, 则矩阵
\begin_inset Formula $A$
\end_inset
的所有特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
的模小于 1 , 即
\begin_inset Formula $\left|\lambda_{i}\right|<1$
\end_inset
, (
\begin_inset Formula $i=1,2,\cdots,n$
\end_inset
).
\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
求矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & -1\\
-1 & 3
\end{bmatrix}$
\end_inset
的特征值和特征向量.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset Formula $\checkmark$
\end_inset
\end_layout
\end_inset
\begin_inset CommandInset label
LatexCommand label
name "prob:4.2-2"
\end_inset
求矩阵
\begin_inset Formula $A=\begin{bmatrix}4 & 6 & 0\\
-3 & -5 & 0\\
-3 & -6 & 1
\end{bmatrix}$
\end_inset
的特征值与特征向量.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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