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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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\end_header
\begin_body
\begin_layout Section
相似矩阵
\end_layout
\begin_layout Subsection
相似矩阵的概念
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
相似矩阵的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $A,B$
\end_inset
都是
\begin_inset Formula $n$
\end_inset
阶矩阵, 若存在可逆矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula
\[
P^{-1}AP=B,
\]
\end_inset
则称
\series bold
\begin_inset Formula $B$
\end_inset
是
\begin_inset Formula $A$
\end_inset
的相似矩阵
\series default
, 并称
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似
\series default
.
记为
\begin_inset Formula $A\sim B$
\end_inset
.
\end_layout
\begin_layout Definition
对
\begin_inset Formula $A$
\end_inset
进行运算
\begin_inset Formula $P^{-1}AP$
\end_inset
称为
\series bold
对
\begin_inset Formula $A$
\end_inset
进行相似变换
\series default
, 称
\series bold
可逆矩阵
\begin_inset Formula $P$
\end_inset
为相似变换矩阵
\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
验证两个矩阵相似
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
能否避免求
\begin_inset Formula $P^{-1}$
\end_inset
?
\end_layout
\end_inset
设有矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 1\\
5 & -1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}4 & 0\\
0 & -2
\end{bmatrix}$
\end_inset
, 试验证存在可逆矩阵
\begin_inset Formula $P=\begin{bmatrix}1 & 1\\
1 & -5
\end{bmatrix}$
\end_inset
, 使得
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似.
\end_layout
\begin_layout Proof
易见
\begin_inset Formula $P$
\end_inset
可逆, 且
\begin_inset Formula
\[
P^{-1}=\begin{bmatrix}\frac{5}{6} & \frac{1}{6}\\
\frac{1}{6} & -\frac{1}{6}
\end{bmatrix},
\]
\end_inset
由
\begin_inset Formula
\[
P^{-1}AP=\begin{bmatrix}\frac{5}{6} & \frac{1}{6}\\
\frac{1}{6} & -\frac{1}{6}
\end{bmatrix}\begin{bmatrix}3 & 1\\
5 & -1
\end{bmatrix}\begin{bmatrix}1 & 1\\
1 & -5
\end{bmatrix}=\begin{bmatrix}4 & 0\\
0 & -2
\end{bmatrix}=B.
\]
\end_inset
故
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似.
\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
验证两个矩阵相似
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\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 "exa:4.3-1"
\end_inset
\begin_inset Formula $A=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 1\\
0 & 1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
容易算出
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
的特征多项式均为
\begin_inset Formula $(\lambda-1)^{2}$
\end_inset
.
但可以证明
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
不相似.
事实上,
\begin_inset Formula $A$
\end_inset
是一个单位阵, 对任意的非奇异阵
\begin_inset Formula $P$
\end_inset
有
\begin_inset Formula
\[
P^{-1}AP=P^{-1}IP=P^{-1}P=I.
\]
\end_inset
因此若
\begin_inset Formula $B$
\end_inset
与
\begin_inset Formula $A$
\end_inset
相似,
\begin_inset Formula $B$
\end_inset
也必须是单位阵, 而现在
\begin_inset Formula $B$
\end_inset
不是单位阵.
所以
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
不相似.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
相似矩阵的性质
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
相似矩阵之间的关系
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
矩阵的相似关系是一种
\color red
等价关系
\color inherit
, 满足:
\end_layout
\begin_layout Standard
(1) 反身性: 对任意
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
, 有
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $A$
\end_inset
相似;
\end_layout
\begin_layout Standard
(2) 对称性: 若
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似, 则
\begin_inset Formula $B$
\end_inset
与
\begin_inset Formula $A$
\end_inset
相似;
\end_layout
\begin_layout Standard
(3) 传递性: 若
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似, 且
\begin_inset Formula $B$
\end_inset
与
\begin_inset Formula $C$
\end_inset
相似, 则
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $C$
\end_inset
相似.
\end_layout
\begin_layout Standard
两个常用运算表达式:
\end_layout
\begin_layout Standard
(1)
\begin_inset Formula $P^{-1}ABP=\left(P^{-1}AP\right)\left(P^{-1}BP\right)$
\end_inset
;
\end_layout
\begin_layout Standard
(2)
\begin_inset Formula $P^{-1}(kA+lB)P=kP^{-1}AP+lP^{-1}BP$
\end_inset
, 其中
\begin_inset Formula $k,l$
\end_inset
为任意实数.
\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
相似矩阵的性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
若
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
相似, 则
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
的特征多项式相同, 从而
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
的特征值亦相同.
\end_layout
\begin_layout Remark*
由例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:4.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
知, 上述定理的逆命题不正确, 也即特征值相同的两个同型矩阵可以不相似.
\end_layout
\begin_layout Standard
相似矩阵的其它性质:
\end_layout
\begin_layout Enumerate
相似矩阵的秩相等;
\end_layout
\begin_layout Enumerate
相似矩阵的行列式相等;
\end_layout
\begin_layout Enumerate
相似矩阵具有相同的可逆性, 当它们可逆时, 则它们的逆矩阵也相似.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
矩阵与对角矩阵相似的条件
\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
矩阵与对角矩阵相似的条件
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:4.3-2"
\end_inset
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
与对角矩阵
\begin_inset Formula $\Lambda=\begin{bmatrix}\lambda_{1}\\
& \lambda_{2}\\
& & \ddots\\
& & & \lambda_{n}
\end{bmatrix}$
\end_inset
相似的充分必要条件为矩阵
\begin_inset Formula $A$
\end_inset
有
\begin_inset Formula $n$
\end_inset
个线性无关的特征向量.
\end_layout
\begin_layout Remark
定理的证明过程实际上已经给出了把方阵对角化的方法.
\end_layout
\begin_layout Corollary
若
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
有
\begin_inset Formula $n$
\end_inset
个相异的特征值
\begin_inset Formula $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$
\end_inset
, 则
\begin_inset Formula $A$
\end_inset
与对角矩阵
\begin_inset Formula
\[
\Lambda=\begin{bmatrix}\lambda_{1}\\
& \lambda_{2}\\
& & \ddots\\
& & & \lambda_{n}
\end{bmatrix}
\]
\end_inset
相似.
\end_layout
\begin_layout Standard
对于
\begin_inset Formula $n$
\end_inset
阶方阵
\begin_inset Formula $A$
\end_inset
, 若存在可逆矩阵
\begin_inset Formula $P$
\end_inset
, 使
\begin_inset Formula $P^{-1}AP=\Lambda$
\end_inset
为对角阵, 则称
\series bold
方阵
\begin_inset Formula $A$
\end_inset
可对角化
\series default
.
\end_layout
\begin_layout Theorem
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
可对角化的充要条件是对应于
\begin_inset Formula $A$
\end_inset
的每个特征值的线性无关的特征向量的个数恰好等于该特征值的重数.
即设
\begin_inset Formula $\lambda_{i}$
\end_inset
是矩阵
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $n_{i}$
\end_inset
重特征值, 则
\begin_inset Formula
\[
A\text{ 与 }\Lambda\text{ 相似 }\Leftrightarrow r\left(A-\lambda_{i}E\right)=n-n_{i},\quad(i=1,2,\cdots,n).
\]
\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
验证矩阵可以对角化
\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
试对矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 1\\
5 & -1
\end{bmatrix}$
\end_inset
验证前述定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:4.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
的结论.
\end_layout
\begin_layout Solution*
由本章第
\begin_inset CommandInset ref
LatexCommand pageref
reference "exa:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
页例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
知
\begin_inset Foot
status open
\begin_layout Plain Layout
第
\begin_inset CommandInset ref
LatexCommand pageref
reference "exa:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
页例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
所提供的信息仅仅是: 矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 1\\
5 & -1
\end{bmatrix}$
\end_inset
的特征值为
\begin_inset Formula $4$
\end_inset
和
\begin_inset Formula $-2$
\end_inset
, 对应的特征向量分别为
\begin_inset Formula $\begin{bmatrix}1\\
1
\end{bmatrix}$
\end_inset
和
\begin_inset Formula $\begin{bmatrix}1\\
-5
\end{bmatrix}$
\end_inset
.
\end_layout
\end_inset
, 题设矩阵
\begin_inset Formula $A$
\end_inset
有两个互不相同的特征值
\begin_inset Formula $\lambda_{1}=4$
\end_inset
,
\begin_inset Formula $\lambda_{2}=-2$
\end_inset
, 其对应特征向量分别为:
\begin_inset Formula
\[
p_{1}=\begin{bmatrix}1\\
1
\end{bmatrix},\ p_{2}=\begin{bmatrix}1\\
-5
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Solution*
如果取
\begin_inset Formula $\Lambda_{1}=\begin{bmatrix}4 & 0\\
0 & -2
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $P=\left(p_{1},p_{2}\right)=\begin{bmatrix}1 & 1\\
1 & -5
\end{bmatrix}$
\end_inset
, 则有
\begin_inset Formula $P^{-1}AP=\Lambda_{1}$
\end_inset
, 即
\begin_inset Formula $A\sim\Lambda_{1}$
\end_inset
.
\end_layout
\begin_layout Solution*
如果取
\begin_inset Formula $\Lambda_{2}=\begin{bmatrix}-2 & 0\\
0 & 4
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $P=\left(p_{2},p_{1}\right)=\begin{bmatrix}1 & 1\\
-5 & 1
\end{bmatrix}$
\end_inset
, 则亦有
\begin_inset Formula $P^{-1}AP=\Lambda_{2}$
\end_inset
, 即
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $\Lambda_{2}$
\end_inset
相似.
\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
矩阵特征值有重数的情况
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
试对矩阵
\begin_inset Formula $A=\begin{bmatrix}4 & 6 & 0\\
-3 & -5 & 0\\
-3 & -6 & 1
\end{bmatrix}$
\end_inset
验证定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:4.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
的结论.
\end_layout
\begin_layout Solution*
由上节课堂练习
\begin_inset CommandInset ref
LatexCommand ref
reference "prob:4.2-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
知, 题设矩阵
\begin_inset Formula $A$
\end_inset
有两个互不相同的特征值
\begin_inset Formula $\lambda_{1}=-2$
\end_inset
,
\begin_inset Formula $\lambda_{2}=\lambda_{3}=1$
\end_inset
.
其对应特征向量分别为:
\begin_inset Formula
\[
p_{1}=\begin{bmatrix}-1\\
1\\
1
\end{bmatrix},\ p_{2}=\begin{bmatrix}-2\\
1\\
0
\end{bmatrix},\ p_{3}=\begin{bmatrix}0\\
0\\
1
\end{bmatrix}.
\]
\end_inset
容易验证
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
线性无关.
若取
\begin_inset Formula
\[
P=\left(p_{1},p_{2},p_{3}\right)=\begin{bmatrix}-1 & -2 & 0\\
1 & 1 & 0\\
1 & 0 & 1
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
取
\begin_inset Formula
\[
P=\left(p_{1},p_{2},p_{3}\right)=\begin{bmatrix}-1 & -2 & 0\\
1 & 1 & 0\\
1 & 0 & 1
\end{bmatrix},
\]
\end_inset
则
\begin_inset Formula
\[
P^{-1}=\begin{bmatrix}1 & 2 & 0\\
-1 & -1 & 0\\
-1 & -2 & 1
\end{bmatrix}\Longrightarrow P^{-1}AP=\begin{bmatrix}-2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Remark*
本例子说明了
\begin_inset Formula $A$
\end_inset
的特征值不全互异时,
\begin_inset Formula $A$
\end_inset
也可能化为对角矩阵.
\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
矩阵特征值有重数的情况 - 2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
判断矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & -2 & 2\\
-2 & -2 & 4\\
2 & 4 & -2
\end{bmatrix}$
\end_inset
能否化为对角阵.
\end_layout
\begin_layout Solution*
求解特征方程
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
|A-\lambda E|=\begin{vmatrix}1-\lambda & -2 & 2\\
-2 & -2-\lambda & 4\\
2 & 4 & -2-\lambda
\end{vmatrix}=-(\lambda-2)^{2}(\lambda+7)=0\Longrightarrow\lambda_{1}=\lambda_{2}=2,\lambda_{3}=-7.
\]
\end_inset
将
\begin_inset Formula $\lambda_{1}=\lambda_{2}=2$
\end_inset
代入
\begin_inset Formula $(A-\lambda E)x=0$
\end_inset
, 得方程组
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\left\{ \begin{array}{l}
-x_{1}-2x_{2}+2x_{3}=0\\
-2x_{1}-4x_{2}+4x_{3}=0\\
2x_{1}+4x_{2}-4x_{3}=0
\end{array}\Longrightarrow\text{ 基础解系 }p_{1}=\begin{bmatrix}2\\
0\\
1
\end{bmatrix},\ p_{2}=\begin{bmatrix}0\\
1\\
1
\end{bmatrix}\right..
\]
\end_inset
同理, 对
\begin_inset Formula $\lambda_{3}=-7$
\end_inset
, 由
\begin_inset Formula $\left(A-\lambda_{3}E\right)x=0\Longrightarrow$
\end_inset
基础解系
\begin_inset Formula $p_{3}=(1,2,-2)^{T}$
\end_inset
.
\end_layout
\begin_layout Solution*
由于
\begin_inset Formula $\begin{vmatrix}2 & 0 & 1\\
0 & 1 & 2\\
1 & 1 & -2
\end{vmatrix}\neq0$
\end_inset
, 所以
\begin_inset Formula $p_{1},p_{2},p_{3}$
\end_inset
线性无关.
即
\begin_inset Formula $A$
\end_inset
有
\begin_inset Formula $3$
\end_inset
个线性无关的特征向量, 因而
\begin_inset Formula $A$
\end_inset
可对角化.
\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
矩阵特征值有重数的情况 - 3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}0 & 0 & 1\\
1 & 1 & a\\
1 & 0 & 0
\end{bmatrix}$
\end_inset
, 问
\begin_inset Formula $a$
\end_inset
为何值时, 矩阵
\begin_inset Formula $A$
\end_inset
能对角化?
\end_layout
\begin_layout Solution*
求解特征方程
\begin_inset Formula
\[
|\lambda E-A|=\begin{vmatrix}\lambda & 0 & -1\\
-1 & \lambda-1 & -a\\
-1 & 0 & \lambda
\end{vmatrix}=(\lambda-1)^{2}(\lambda+1)\Longleftrightarrow\lambda_{1}=-1,\ \lambda_{2}=\lambda_{3}=1.
\]
\end_inset
\end_layout
\begin_layout Solution*
对于单根
\begin_inset Formula $\lambda_{1}=-1$
\end_inset
, 可求得线性无关的特征向量恰有
\begin_inset Formula $1$
\end_inset
个, 而对应重根
\begin_inset Formula $\lambda_{2}=\lambda_{3}=1$
\end_inset
, 欲使矩阵
\begin_inset Formula $A$
\end_inset
能对角化, 应有
\begin_inset Formula $2$
\end_inset
个线性无关的特征向量, 即方程组
\begin_inset Formula $(E-A)x=0$
\end_inset
有
\begin_inset Formula $2$
\end_inset
个线性无关的解, 亦即系数矩阵
\begin_inset Formula $E-A$
\end_inset
的秩
\begin_inset Formula $r(E-A)=1$
\end_inset
, 而
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
E-A=\begin{bmatrix}1 & 0 & -1\\
-1 & 0 & -a\\
-1 & 0 & 1
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & -1\\
0 & 0 & a+1\\
0 & 0 & 0
\end{bmatrix},
\]
\end_inset
要
\begin_inset Formula $r(E-A)=1$
\end_inset
, 得
\begin_inset Formula $a+1=0$
\end_inset
, 即
\begin_inset Formula $a=-1$
\end_inset
.
因此, 当
\begin_inset Formula $a=-1$
\end_inset
时, 矩阵
\begin_inset Formula $A$
\end_inset
能对角化.
\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
判断矩阵
\begin_inset Formula $A=\begin{bmatrix}-2 & 1 & -2\\
-5 & 3 & -3\\
1 & 0 & 2
\end{bmatrix}$
\end_inset
能否化为对角阵.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
判断下列两矩阵
\begin_inset Formula $A,B$
\end_inset
是否相似.
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 1 & \cdots & 1\\
1 & 1 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & 1
\end{bmatrix},\ B=\begin{bmatrix}n & 0 & \cdots & 0\\
1 & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
1 & 0 & \cdots & 0
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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