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\end_header
\begin_body
\begin_layout Section
实对称矩阵的对角化
\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
一个
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
具备什么条件才能对角化? 这是一个比较复杂的问题.
本节我们仅对
\begin_inset Formula $A$
\end_inset
为实对称矩阵的情况进行讨论.
实对称矩阵具有许多一般矩阵所没有的特殊性质.
\end_layout
\end_deeper
\begin_layout Subsection
实对称矩阵的特征值问题
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
实对称矩阵的特征值问题
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
实对称矩阵的特征值都为实数.
\end_layout
\begin_layout Remark
对实对称矩阵
\begin_inset Formula $A$
\end_inset
, 因其特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
为实数, 故方程组
\begin_inset Formula
\[
\left(A-\lambda_{i}E\right)X=0
\]
\end_inset
是实系数方程组, 由
\begin_inset Formula $\left|A-\lambda_{i}E\right|=0$
\end_inset
知它必有实的基础解系, 所以
\begin_inset Formula $A$
\end_inset
的特征向量可以取实向量.
\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
实对称矩阵的特征向量
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
设
\begin_inset Formula $\lambda_{1},\lambda_{2}$
\end_inset
是对称矩阵
\begin_inset Formula $A$
\end_inset
的两个特征值,
\begin_inset Formula $p_{1},p_{2}$
\end_inset
是对应的特征向量.
若
\begin_inset Formula $\lambda_{1}\neq\lambda_{2}$
\end_inset
, 则
\begin_inset Formula $p_{1}$
\end_inset
与
\begin_inset Formula $p_{2}$
\end_inset
正交.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶实对称矩阵,
\begin_inset Formula $\lambda$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的特征方程的
\begin_inset Formula $k$
\end_inset
重根, 则矩阵
\begin_inset Formula $A-\lambda E$
\end_inset
的秩
\begin_inset Formula $r(A-\lambda E)=n-k$
\end_inset
, 从而对应特征值
\begin_inset Formula $\lambda$
\end_inset
恰有
\begin_inset Formula $k$
\end_inset
个线性无关的特征向量.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶实对称矩阵, 则必有正交矩阵
\begin_inset Formula $P$
\end_inset
,使
\begin_inset Formula
\[
P^{-1}AP=\Lambda,
\]
\end_inset
其中
\begin_inset Formula $\Lambda$
\end_inset
是以
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $n$
\end_inset
个特征值为对角元素的对角矩阵.
\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
将实对称矩阵对角化的步骤
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
与上节
\bar under
将一般矩阵对角化的方法
\bar default
类似, 根据上述结论, 可求正交变换矩阵
\begin_inset Formula $P$
\end_inset
将实对称矩阵
\begin_inset Formula $A$
\end_inset
对角化的步骤为:
\end_layout
\begin_layout Standard
(1) 求出
\begin_inset Formula $A$
\end_inset
的全部特征值
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{s}$
\end_inset
;
\end_layout
\begin_layout Standard
(2) 对每一个特征值
\begin_inset Formula $\lambda_{i}$
\end_inset
, 由
\begin_inset Formula $\left(\lambda_{i}E-A\right)X=0$
\end_inset
求出基础解系 (特征向量);
\end_layout
\begin_layout Standard
(3) 将基础解系 (特征向量) 正交化; 再单位化;
\end_layout
\begin_layout Standard
(4) 以这些单位向量作为列向量构成一个正交矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula
\[
P^{-1}AP=\Lambda.
\]
\end_inset
\end_layout
\begin_layout Remark
\begin_inset Formula $P$
\end_inset
中列向量的次序与矩阵
\begin_inset Formula $\Lambda$
\end_inset
对角线上的特征值的次序相对应.
\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
将实对称阵对角化
\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
设实对称矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & -2 & 0\\
-2 & 2 & -2\\
0 & -2 & 3
\end{bmatrix}$
\end_inset
, 求正交矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula $P^{-1}AP$
\end_inset
为对角矩阵.
\end_layout
\begin_layout Solution*
矩阵
\begin_inset Formula $A$
\end_inset
的特征方程为
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda-1 & 2 & 0\\
2 & \lambda-2 & 2\\
0 & 2 & \lambda-3
\end{vmatrix}=0\Longrightarrow(\lambda+1)(\lambda-2)(\lambda-5)=0\Longrightarrow\lambda_{1}=-1,\ \lambda_{2}=2,\ \lambda_{3}=5.
\]
\end_inset
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $\lambda_{1}=-1$
\end_inset
时, 由
\begin_inset Formula $(-E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{1}=(2,2,1)^{T}$
\end_inset
;
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $\lambda_{2}=2$
\end_inset
时, 由
\begin_inset Formula $(2E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{2}=(2,-1,-2)^{T}$
\end_inset
;
\end_layout
\begin_layout Solution*
当
\begin_inset Formula $\lambda_{3}=5$
\end_inset
时, 由
\begin_inset Formula $(5E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{3}=(1,-2,2)^{T}$
\end_inset
.
\end_layout
\begin_layout Solution*
不难验证
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
是正交向量组, 把
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
单位化, 得
\begin_inset Formula
\[
\eta_{1}=\frac{p_{1}}{\left\Vert p_{1}\right\Vert }=\begin{bmatrix}2/3\\
2/3\\
1/3
\end{bmatrix},\ \eta_{2}=\frac{p_{2}}{\left\Vert p_{2}\right\Vert }=\begin{bmatrix}2/3\\
-1/3\\
-2/3
\end{bmatrix},\ \eta_{3}=\frac{p_{3}}{\left\Vert p_{3}\right\Vert }=\begin{bmatrix}1/3\\
-2/3\\
2/3
\end{bmatrix}.
\]
\end_inset
令
\begin_inset Formula $P=\left(\eta_{1},\eta_{2},\eta_{3}\right)=\begin{bmatrix}2/3 & 2/3 & 1/3\\
2/3 & -1/3 & -2/3\\
1/3 & -2/3 & 2/3
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula $P^{-1}AP=P^{T}AP=\begin{bmatrix}-1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 5
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
将实对称阵对角化
\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
设有对称矩阵
\begin_inset Formula $A=\begin{bmatrix}4 & 0 & 0\\
0 & 3 & 1\\
0 & 1 & 3
\end{bmatrix}$
\end_inset
, 试求出正交矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula $P^{-1}AP$
\end_inset
为对角阵.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda-4 & 0 & 0\\
0 & \lambda-3 & -1\\
0 & -1 & \lambda-3
\end{vmatrix}=(\lambda-2)(4-\lambda)^{2}\Longrightarrow\lambda_{1}=2,\ \lambda_{2}=\lambda_{3}=4,
\]
\end_inset
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{1}=2$
\end_inset
, 由
\begin_inset Formula $(2E-A)x=0\Longrightarrow$
\end_inset
基础解系
\begin_inset Formula $p_{1}=\begin{bmatrix}0\\
1\\
-1
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{2}=\lambda_{3}=4$
\end_inset
, 由
\begin_inset Formula $(4E-A)x=0\Longrightarrow$
\end_inset
基础解系
\begin_inset Formula $p_{2}=\begin{bmatrix}1\\
0\\
0
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $p_{3}=\begin{bmatrix}0\\
1\\
1
\end{bmatrix}$
\end_inset
.
\begin_inset Formula $p_{2}$
\end_inset
与
\begin_inset Formula $p_{3}$
\end_inset
恰好正交, 所以
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
两两正交.
再将
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
单位化, 令
\begin_inset Formula $\eta_{i}=p_{i}/\left\Vert p_{i}\right\Vert $
\end_inset
, (
\begin_inset Formula $i=1,2,3$
\end_inset
), 得
\begin_inset Formula
\[
\eta_{1}=\begin{bmatrix}0\\
1/\sqrt{2}\\
-1/\sqrt{2}
\end{bmatrix},\ \eta_{2}=\begin{bmatrix}1\\
0\\
0
\end{bmatrix},\ \eta_{3}=\begin{bmatrix}0\\
1/\sqrt{2}\\
1/\sqrt{2}
\end{bmatrix},
\]
\end_inset
故所求正交矩阵
\begin_inset Formula
\[
P=\left(\eta_{1},\eta_{2},\eta_{3}\right)=\begin{bmatrix}0 & 1 & 0\\
1/\sqrt{2} & 0 & 1/\sqrt{2}\\
-1/\sqrt{2} & 0 & 1/\sqrt{2}
\end{bmatrix}\text{ 且 }P^{-1}AP=\begin{bmatrix}2 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 4
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
将实对称矩阵对角化
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知
\begin_inset Formula $A=\begin{bmatrix}2 & 0 & 0\\
0 & a & 2\\
0 & 2 & a
\end{bmatrix}$
\end_inset
, (其中
\begin_inset Formula $a>0$
\end_inset
), 有一特征值为
\begin_inset Formula $1$
\end_inset
, 求正交矩阵
\begin_inset Formula $P$
\end_inset
使得
\begin_inset Formula $P^{-1}AP$
\end_inset
为对角矩阵.
\end_layout
\begin_layout Solution*
\begin_inset Formula $A$
\end_inset
的特征多项式为
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda-2 & 0 & 0\\
0 & \lambda-a & -2\\
0 & -2 & \lambda-a
\end{vmatrix}=(\lambda-2)(\lambda-a+2)(\lambda-a-2),
\]
\end_inset
由于
\begin_inset Formula $A$
\end_inset
有特征值
\begin_inset Formula $1$
\end_inset
, 故有两种情形:
\end_layout
\begin_layout Solution*
若
\begin_inset Formula $a-2=1$
\end_inset
, 则
\begin_inset Formula $a=3$
\end_inset
; 若
\begin_inset Formula $a+2=1$
\end_inset
, 则
\begin_inset Formula $a=-1$
\end_inset
.
但
\begin_inset Formula $a>0$
\end_inset
, 所以只能是
\begin_inset Formula $a=3$
\end_inset
.
从而得
\begin_inset Formula $A$
\end_inset
的特征值为
\begin_inset Formula $2,1,5$
\end_inset
.
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{1}=2$
\end_inset
, 由
\begin_inset Formula $(2E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{1}=(1,0,0)^{T}$
\end_inset
;
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{2}=1$
\end_inset
, 由
\begin_inset Formula $(E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{2}=(0,1,-1)^{T}$
\end_inset
;
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $\lambda_{3}=5$
\end_inset
, 由
\begin_inset Formula $(5E-A)x=0$
\end_inset
, 得基础解系
\begin_inset Formula $p_{3}=(1,0,0)^{T}$
\end_inset
;
\end_layout
\begin_layout Solution*
因实对称矩阵的
\bar under
属于不同特征值的特征向量
\bar default
必相互正交, 故特征向量
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
已是正交向量组, 因此只需将其单位化:
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\eta_{1}=(1,0,0)^{T};\ \eta_{2}=\left(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)^{T};\ \eta_{3}=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)^{T};
\]
\end_inset
令
\begin_inset Formula $P=\left(\eta_{1},\eta_{2},\eta_{3}\right)=\begin{bmatrix}1 & 0 & 0\\
0 & 1/\sqrt{2} & 1/\sqrt{2}\\
0 & -1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula $P^{-1}AP=\begin{bmatrix}2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 5
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求矩阵的幂
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}2 & -1\\
-1 & 2
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A^{n}$
\end_inset
.
\end_layout
\begin_layout Solution*
因
\begin_inset Formula $A$
\end_inset
对称, 故
\begin_inset Formula $A$
\end_inset
可对角化, 即有可逆矩阵
\begin_inset Formula $P$
\end_inset
及对角阵
\begin_inset Formula $\Lambda$
\end_inset
, 使
\begin_inset Formula $P^{-1}AP=\Lambda$
\end_inset
.
于是
\begin_inset Formula
\[
A=P\Lambda P^{-1}\Longrightarrow A^{n}=P\Lambda^{n}P^{-1}.
\]
\end_inset
由
\begin_inset Formula
\[
|A-\lambda E|=\begin{vmatrix}2-\lambda & -1\\
-1 & 2-\lambda
\end{vmatrix}=\lambda^{2}-4\lambda+3=(\lambda-1)(\lambda-3),
\]
\end_inset
得
\begin_inset Formula $A$
\end_inset
的特征值
\begin_inset Formula $\lambda_{1}=1$
\end_inset
,
\begin_inset Formula $\lambda_{2}=3$
\end_inset
.
于是
\begin_inset Formula
\[
\Lambda=\begin{bmatrix}1 & 0\\
0 & 3
\end{bmatrix},\ \Lambda^{n}=\begin{bmatrix}1 & 0\\
0 & 3^{n}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Solution*
对应
\begin_inset Formula $\lambda_{1}=1$
\end_inset
, 由
\begin_inset Formula $(A-E)x=0$
\end_inset
, 解得对应特征向量
\begin_inset Formula $P_{1}=\begin{bmatrix}1\\
1
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Solution*
对应
\begin_inset Formula $\lambda_{2}=3$
\end_inset
, 由
\begin_inset Formula $(A-3E)x=0$
\end_inset
, 解得对应特征向量
\begin_inset Formula $P_{2}=\begin{bmatrix}1\\
-1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
令
\begin_inset Formula $P=\left(p_{1},p_{2}\right)=\begin{bmatrix}1 & 1\\
1 & -1
\end{bmatrix}$
\end_inset
, 求出
\begin_inset Formula $P^{-1}=\frac{1}{2}\begin{bmatrix}1 & 1\\
1 & -1
\end{bmatrix}$
\end_inset
.
于是
\begin_inset Formula
\[
A^{n}=P\Lambda^{n}P^{-1}=\frac{1}{2}\begin{bmatrix}1 & 1\\
1 & -1
\end{bmatrix}\begin{bmatrix}1 & 0\\
0 & 3^{n}
\end{bmatrix}\begin{bmatrix}1 & 1\\
1 & -1
\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1+3^{n} & 1-3^{n}\\
1-3^{n} & 1+3^{n}
\end{bmatrix}.
\]
\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 Problem
设实对称矩阵
\begin_inset Formula $A=\begin{bmatrix}2 & -2 & 0\\
-2 & 1 & -2\\
0 & -2 & 0
\end{bmatrix}$
\end_inset
, 试求出正交矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula $P^{-1}AP$
\end_inset
为对角阵.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $n$
\end_inset
阶实对称矩阵
\begin_inset Formula $A$
\end_inset
满足
\begin_inset Formula $A^{2}=A$
\end_inset
, 且
\begin_inset Formula $A$
\end_inset
的秩为
\begin_inset Formula $r$
\end_inset
, 试求行列式
\begin_inset Formula $\mathrm{det}\left(2E-A\right)$
\end_inset
的值.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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