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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
矩阵的运算
\end_layout
\begin_layout Subsection
矩阵的线性运算
\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
矩阵的线性运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设有两个
\begin_inset Formula $m\times n$
\end_inset
矩阵
\begin_inset Formula $A=(a_{ij})$
\end_inset
和
\begin_inset Formula $B=(b_{ij})$
\end_inset
, 矩阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
的
\series bold
和
\series default
记作
\begin_inset Formula $A+B$
\end_inset
, 规定为
\begin_inset Formula
\[
A+B=(a_{ij}+b_{ij})_{m\times n}=\begin{bmatrix}a_{11}+b_{11} & a_{12}+b_{12} & \cdots & a_{1n}+b_{1n}\\
a_{21}+b_{21} & a_{22}+b_{22} & \cdots & a_{2n}+b_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1}+b_{m1} & a_{m2}+b_{m2} & \cdots & a_{mn}+b_{mn}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Remark*
只有两个矩阵是同型矩阵时, 才能进行矩阵的加法运算.
两个同型矩阵的和, 即为两个矩阵对应位置元素相加得到的矩阵.
\end_layout
\begin_layout Definition
设矩阵
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
, 记
\begin_inset Formula
\[
-A=\left(-a_{ij}\right),
\]
\end_inset
称
\begin_inset Formula $-A$
\end_inset
为矩阵
\begin_inset Formula $A$
\end_inset
的
\series bold
负矩阵
\series default
, 显然有
\begin_inset Formula
\[
A+(-A)=O\text{. }
\]
\end_inset
由此规定矩阵的减法为
\begin_inset Formula
\[
A-B=A+(-B).
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Definition
数
\begin_inset Formula $k$
\end_inset
与矩阵
\begin_inset Formula $A$
\end_inset
的乘积记作
\begin_inset Formula $kA$
\end_inset
或
\begin_inset Formula $Ak$
\end_inset
, 规定为
\begin_inset Formula
\[
kA=Ak=(ka_{ij})=\begin{bmatrix}ka_{11} & ka_{12} & \cdots & ka_{1n}\\
ka_{21} & ka_{22} & \cdots & ka_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
ka_{m1} & ka_{m2} & \cdots & ka_{mn}
\end{bmatrix}.
\]
\end_inset
数与矩阵的乘积运算称为
\series bold
数乘运算
\series default
.
\end_layout
\begin_layout Definition
\end_layout
\begin_layout Standard
矩阵的加法与矩阵的数乘两种运算统称为
\series bold
矩阵的线性运算
\series default
.
它满足下列运算规律:
\end_layout
\begin_layout Standard
设
\begin_inset Formula $A,B,C,O$
\end_inset
都是同型矩阵,
\begin_inset Formula $k,l$
\end_inset
是常数, 则
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A+B=B+A$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(A+B)+C=A+(B+C)$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A+O=A$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A+(-A)=O$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $1A=A$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(kl)A=k(lA)$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(k+l)A=kA+lA$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $k(A+B)=kA+kB$
\end_inset
.
\end_layout
\begin_layout Remark*
在数学中, 把满足上述八条规律的运算称为
\series bold
线性运算
\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
矩阵的线性运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知
\begin_inset Formula $A=\begin{bmatrix}-1 & 2 & 3 & 1\\
0 & 3 & -2 & 1\\
4 & 0 & 3 & 2
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}4 & 3 & 2 & -1\\
5 & -3 & 0 & 1\\
1 & 2 & -5 & 0
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $3A-2B$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
3A-2B & =3\begin{bmatrix}-1 & 2 & 3 & 1\\
0 & 3 & -2 & 1\\
4 & 0 & 3 & 2
\end{bmatrix}-2\begin{bmatrix}4 & 3 & 2 & -1\\
5 & -3 & 0 & 1\\
1 & 2 & -5 & 0
\end{bmatrix}=\begin{bmatrix}-3-8 & 6-6 & 9-4 & 3+2\\
0-10 & 9+6 & -6-0 & 3-2\\
12-2 & 0-4 & 9+10 & 6-0
\end{bmatrix}\\
& =\begin{bmatrix}-11 & 0 & 5 & 5\\
-10 & 15 & -6 & 1\\
10 & -4 & 19 & 6
\end{bmatrix}.
\end{align*}
\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
矩阵的线性运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知
\begin_inset Formula $A=\begin{bmatrix}3 & -1 & 2 & 0\\
1 & 5 & 7 & 9\\
2 & 4 & 6 & 8
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}7 & 5 & -2 & 4\\
5 & 1 & 9 & 7\\
3 & 2 & -1 & 6
\end{bmatrix}$
\end_inset
, 且
\begin_inset Formula $A+2X=B$
\end_inset
, 求
\begin_inset Formula $X$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $X=\frac{1}{2}(B-A)=\frac{1}{2}\begin{bmatrix}4 & 6 & -4 & 4\\
4 & -4 & 2 & -2\\
1 & -2 & -7 & -2
\end{bmatrix}=\begin{bmatrix}2 & 3 & -2 & 2\\
2 & -2 & 1 & -1\\
\frac{1}{2} & -1 & -\frac{7}{2} & -1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Remark*
\begin_inset Formula $n$
\end_inset
阶数量矩阵
\begin_inset Formula $A=\begin{bmatrix}a & 0 & \cdots & 0\\
0 & a & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & a
\end{bmatrix}=aE_{n}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
矩阵的相乘
\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
矩阵的相乘
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\[
A=\left(a_{ij}\right)_{m\times s}=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1s}\\
a_{2s} & a_{2s} & \cdots & a_{2s}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{ms}
\end{bmatrix},\quad B=\left(b_{ij}\right)_{s\times n}=\begin{bmatrix}b_{11} & b_{12} & \cdots & b_{1n}\\
b_{21} & b_{22} & \cdots & b_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
b_{s1} & b_{s2} & \cdots & b_{sn}
\end{bmatrix}
\]
\end_inset
矩阵
\begin_inset Formula $A$
\end_inset
与矩阵
\begin_inset Formula $B$
\end_inset
的
\color red
乘积
\color inherit
记作
\begin_inset Formula $AB$
\end_inset
, 规定为
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\[
AB=\left(c_{ij}\right)_{m\times n}=\begin{bmatrix}c_{11} & c_{12} & \cdots & c_{1n}\\
c_{21} & c_{22} & \cdots & c_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
c_{m1} & c_{m2} & \cdots & c_{mn}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\[
c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{is}b_{sj}={\color{gray}\sum_{k=1}^{s}}{\color{red}a_{ik}b_{kj}},\quad(i=1,2,\cdots,m;\ j=1,2,\cdots,n).
\]
\end_inset
记号
\begin_inset Formula $AB$
\end_inset
常读作
\begin_inset Formula $A$
\end_inset
左乘
\begin_inset Formula $B$
\end_inset
或
\begin_inset Formula $B$
\end_inset
右乘
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Remark*
只有当
\color magenta
左边矩阵的列数等于右边矩阵的行数
\color inherit
时, 两个矩阵才能进行乘法运算.
\end_layout
\begin_layout Standard
若
\begin_inset Formula $C=AB$
\end_inset
, 则矩阵
\begin_inset Formula $C$
\end_inset
的元素
\begin_inset Formula $c_{ij}$
\end_inset
即为矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行元素与矩阵
\begin_inset Formula $B$
\end_inset
的第
\begin_inset Formula $j$
\end_inset
列对应元素乘积的和.
即
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-10mm}
\end_layout
\end_inset
\end_layout
\begin_layout ColumnsCenterAligned
\end_layout
\begin_deeper
\begin_layout Column
6cm
\end_layout
\end_deeper
\begin_layout ColumnsCenterAligned
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
hspace{24mm}
\end_layout
\end_inset
\begin_inset Formula $c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{is}b_{sj}.$
\end_inset
\end_layout
\begin_deeper
\begin_layout Column
7cm
\end_layout
\end_deeper
\begin_layout ColumnsCenterAligned
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
usetikzlibrary{fit}
\end_layout
\begin_layout Plain Layout
\backslash
tikzset{highlight/.style={rectangle,
\end_layout
\begin_layout Plain Layout
fill=red!15,
\end_layout
\begin_layout Plain Layout
blend mode = multiply,
\end_layout
\begin_layout Plain Layout
rounded corners = 0.5 mm,
\end_layout
\begin_layout Plain Layout
inner sep=1pt,
\end_layout
\begin_layout Plain Layout
fit = #1}}
\end_layout
\begin_layout Plain Layout
\backslash
tikzset{mes-options/.style={remember picture,
\end_layout
\begin_layout Plain Layout
overlay,
\end_layout
\begin_layout Plain Layout
name prefix = exemple-,
\end_layout
\begin_layout Plain Layout
highlight/.style = {fill = red!15,
\end_layout
\begin_layout Plain Layout
blend mode = multiply,
\end_layout
\begin_layout Plain Layout
inner sep = 0pt,
\end_layout
\begin_layout Plain Layout
fit = #1}}}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
\backslash
begin{NiceArray}{*{6}{c}@{
\backslash
hspace{6mm}}*{5}{c}}[nullify-dots]
\end_layout
\begin_layout Plain Layout
\backslash
CodeBefore
\end_layout
\begin_layout Plain Layout
\backslash
SubMatrix[{2-7}{6-11}]
\end_layout
\begin_layout Plain Layout
\backslash
SubMatrix[{7-2}{11-6}]
\end_layout
\begin_layout Plain Layout
\backslash
SubMatrix[{7-7}{11-11}]
\end_layout
\begin_layout Plain Layout
\backslash
Body
\end_layout
\begin_layout Plain Layout
& & & & & & & &
\backslash
color{blue}
\backslash
scriptstyle C_j
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & b_{11} &
\backslash
Cdots & b_{1j} &
\backslash
Cdots & b_{1n}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & &
\backslash
Vdots & &
\backslash
Vdots & &
\backslash
Vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & & & b_{kj}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & & &
\backslash
Vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & b_{n1} &
\backslash
Cdots & b_{nj} &
\backslash
Cdots & b_{nn}
\backslash
\backslash
[3mm]
\end_layout
\begin_layout Plain Layout
& a_{11} &
\backslash
Cdots & & & a_{1n}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
Vdots & & & &
\backslash
Vdots & & &
\backslash
Vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
color{blue}
\backslash
scriptstyle L_i & a_{i1} &
\backslash
Cdots & a_{ik} &
\backslash
Cdots & a_{in} &
\backslash
Cdots & & c_{ij}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
Vdots & & & &
\backslash
Vdots
\backslash
\backslash
& a_{n1} &
\backslash
Cdots & & & a_{nn}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
CodeAfter
\end_layout
\begin_layout Plain Layout
\backslash
begin{tikzpicture}
\end_layout
\begin_layout Plain Layout
\backslash
node [highlight = (9-2) (9-6)] { } ;
\end_layout
\begin_layout Plain Layout
\backslash
node [highlight = (2-9) (6-9)] { } ;
\end_layout
\begin_layout Plain Layout
\backslash
draw [gray,shorten > = 1mm, shorten < = 1mm] (9-4.north) to [bend left] (4-9.west)
;
\end_layout
\begin_layout Plain Layout
\backslash
end{tikzpicture}
\end_layout
\begin_layout Plain Layout
\backslash
end{NiceArray}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\begin_layout Standard
矩阵的乘法满足下列运算规律 (假定运算都是可行的):
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(AB)C=A(BC)$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(A+B)C=AC+BC$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $C(A+B)=CA+CB$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $k(AB)=(kA)B=A(kB)$
\end_inset
.
\end_layout
\begin_layout Remark*
矩阵的乘法一般不满足交换律, 即
\begin_inset Formula $AB\neq BA$
\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
矩阵乘法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}-2 & 4\\
1 & -2
\end{bmatrix},B=\begin{bmatrix}2 & 4\\
-3 & -6
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula
\[
\begin{aligned}AB & =\begin{bmatrix}-2 & 4\\
1 & -2
\end{bmatrix}\begin{bmatrix}2 & 4\\
-3 & -6
\end{bmatrix}=\begin{bmatrix}-16 & -32\\
8 & 16
\end{bmatrix}\text{, }\\
BA & =\begin{bmatrix}2 & 4\\
-3 & -6
\end{bmatrix}\begin{bmatrix}-2 & 4\\
1 & -2
\end{bmatrix}=\begin{bmatrix}0 & 0\\
0 & 0
\end{bmatrix},
\end{aligned}
\]
\end_inset
于是
\begin_inset Formula $AB\neq BA;$
\end_inset
且
\begin_inset Formula $BA=O$
\end_inset
, 而
\begin_inset Formula $A\ne O$
\end_inset
,
\begin_inset Formula $B\ne O$
\end_inset
.
\end_layout
\begin_layout Standard
从上例还可看出:
\color red
两个非零矩阵相乘, 可能是零矩阵, 故不能从
\begin_inset Formula $AB=O$
\end_inset
必然推出
\begin_inset Formula $A=O$
\end_inset
或
\begin_inset Formula $B=O$
\end_inset
.
\end_layout
\begin_layout Standard
此外, 矩阵乘法一般也不满足消去律, 即不能从
\begin_inset Formula $AC=BC$
\end_inset
必然推出
\begin_inset Formula $A=B$
\end_inset
.
例如,
\end_layout
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
矩阵乘法不满足消去率的反例
\end_layout
\end_inset
设
\begin_inset Formula
\[
\begin{gathered}A=\begin{bmatrix}1 & 2\\
0 & 3
\end{bmatrix},\ B=\begin{bmatrix}1 & 0\\
0 & 4
\end{bmatrix},\ C=\begin{bmatrix}1 & 1\\
0 & 0
\end{bmatrix},\\
AC=\begin{bmatrix}1 & 2\\
0 & 3
\end{bmatrix}\begin{bmatrix}1 & 1\\
0 & 0
\end{bmatrix}=\begin{bmatrix}1 & 1\\
0 & 0
\end{bmatrix}=\begin{bmatrix}1 & 0\\
0 & 4
\end{bmatrix}\begin{bmatrix}1 & 1\\
0 & 0
\end{bmatrix}=BC,
\end{gathered}
\]
\end_inset
但
\begin_inset Formula $A\neq B$
\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
乘法可交换的矩阵乘积性质
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
如果两矩阵相乘, 有
\begin_inset Formula
\[
AB=BA
\]
\end_inset
则称矩阵
\begin_inset Formula $A$
\end_inset
与矩阵
\begin_inset Formula $B$
\end_inset
\series bold
可交换
\series default
.
简称
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
\series bold
可换
\series default
.
\end_layout
\begin_layout Remark
对于单位矩阵
\begin_inset Formula $E$
\end_inset
, 容易证明
\begin_inset Formula
\[
E_{m}A_{m\times n}=A_{m\times n},\quad A_{m\times n}E_{n}=A_{m\times n}.
\]
\end_inset
当
\begin_inset Formula $m\ne n$
\end_inset
时, 我们不能说
\begin_inset Formula $A_{m\times n}$
\end_inset
与
\begin_inset Formula $E_{m}$
\end_inset
可交换, 或
\begin_inset Formula $A_{m\times n}$
\end_inset
与
\begin_inset Formula $E_{n}$
\end_inset
可交换.
特别地, 当
\begin_inset Formula $m=n$
\end_inset
时, 有
\begin_inset Formula
\[
EA=AE=A.
\]
\end_inset
此时称
\begin_inset Formula $n$
\end_inset
阶方阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $n$
\end_inset
阶单位阵
\begin_inset Formula $E_{n}$
\end_inset
可交换, 且乘积
\begin_inset Formula $AE$
\end_inset
,
\begin_inset Formula $EA$
\end_inset
等于矩阵
\begin_inset Formula $A$
\end_inset
自身.
\end_layout
\begin_layout Standard
可见单位矩阵
\begin_inset Formula $E$
\end_inset
在矩阵乘法中的作用类似于数
\begin_inset Formula $1$
\end_inset
.
\end_layout
\begin_layout Standard
更进一步我们有
\end_layout
\begin_layout Proposition
设
\begin_inset Formula $B$
\end_inset
是一个
\begin_inset Formula $n$
\end_inset
阶矩阵, 则
\begin_inset Formula $B$
\end_inset
是一个数量矩阵的充分必要条件是
\begin_inset Formula $B$
\end_inset
与任何
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
可换.
\end_layout
\begin_layout Solution*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
Hint
\end_layout
\end_inset
取矩阵
\begin_inset ERT
status open
\begin_layout Plain Layout
$A=
\backslash
begin{bNiceMatrix}[last-col,last-row]
\end_layout
\begin_layout Plain Layout
0&
\backslash
cdots&0&
\backslash
cdots&0&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
vdots&
\backslash
ddots&
\backslash
vdots&
\backslash
ddots&
\backslash
vdots&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&
\backslash
cdots&1&
\backslash
cdots&0&
\backslash
leftarrow p
\backslash
text{ 行}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
vdots&
\backslash
ddots&
\backslash
vdots&
\backslash
ddots&
\backslash
vdots&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0&
\backslash
cdots&0&
\backslash
cdots&0&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&
\backslash
overset{
\backslash
uparrow}{q
\backslash
text{ 列}}&&&
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\end_layout
\begin_layout Plain Layout
$,
\end_layout
\end_inset
所以若矩阵
\begin_inset Formula $B=(b_{ij})_{n\times n}$
\end_inset
与矩阵
\begin_inset Formula $A$
\end_inset
交换, 则必然有
\begin_inset Formula
\[
b_{pp}=b_{qq},\ \begin{cases}
b_{pk}\equiv0, & k\ne p\\
b_{kq}\equiv0, & k\ne q
\end{cases}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proposition
设
\begin_inset Formula $A,B$
\end_inset
均为
\begin_inset Formula $n$
\end_inset
阶矩阵, 则下列命题等价:
\end_layout
\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $AB=BA$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(A+B)^{2}=A^{2}+2AB+B^{2}$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(A-B)^{2}=A^{2}-2AB+B^{2}$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $(A+B)(A-B)=(A-B)(A+B)=A^{2}-B^{2}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Proposition
\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
乘法可交换的矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
证明: 如果
\begin_inset Formula $CA=AC$
\end_inset
,
\begin_inset Formula $CB=BC$
\end_inset
, 则有
\end_layout
\begin_layout Example
\begin_inset Formula
\[
(A+B)C=C(A+B);\quad(AB)C=C(AB).
\]
\end_inset
\end_layout
\begin_layout Proof
由于
\begin_inset Formula $CA=AC$
\end_inset
,
\begin_inset Formula $CB=BC$
\end_inset
, 所以
\begin_inset Formula
\[
(A+B)C=AC+BC=CA+CB=C(A+B);
\]
\end_inset
\begin_inset Formula
\[
(AB)C=A(BC)=A(CB)=(AC)B=(CA)B=C(AB).
\]
\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
矩阵乘法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
若
\begin_inset Formula $A=\begin{bmatrix}2 & 3\\
1 & -2\\
3 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & -2 & -3\\
2 & -1 & 0
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $AB$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
AB & =\begin{bmatrix}2 & 3\\
1 & -2\\
3 & 1
\end{bmatrix}\begin{bmatrix}1 & -2 & -3\\
2 & -1 & 0
\end{bmatrix}\\
& =\begin{bmatrix}2\times1+3\times2 & 2\times(-2)+3\times(-1) & 2\times(-3)+3\times0\\
1\times1+(-2)\times2 & 1\times(-2)+(-2)\times(-1) & 1\times(-3)+(-2)\times0\\
3\times1+1\times2 & 3\times(-2)+1\times(-1) & 3\times(-3)+1\times0
\end{bmatrix}\\
& =\begin{bmatrix}8 & -7 & -6\\
-3 & 0 & -3\\
5 & -7 & -9
\end{bmatrix}.
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
就此例顺便求一下
\begin_inset Formula $BA$
\end_inset
.
\begin_inset Formula
\begin{align*}
BA & =\begin{bmatrix}1 & -2 & -3\\
2 & -1 & 0
\end{bmatrix}\begin{bmatrix}2 & 3\\
1 & -2\\
3 & 1
\end{bmatrix}\\
& =\begin{bmatrix}1\times2+(-2)\times1+(-3)\times3 & 1\times3+(-2)\times(-2)+(-3)\times1\\
2\times2+(-1)\times1+0\times3 & 2\times3+(-1)\times(-2)+0\times1
\end{bmatrix}\\
& =\begin{bmatrix}-9 & 4\\
3 & 8
\end{bmatrix}.
\end{align*}
\end_inset
显然
\begin_inset Formula $AB\neq BA$
\end_inset
.
\end_layout
\begin_layout Example
设
\begin_inset Formula $A=(1,0,4)$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1\\
1\\
0
\end{bmatrix}$
\end_inset
.
\begin_inset Formula $A$
\end_inset
是一个
\begin_inset Formula $1\times3$
\end_inset
矩阵,
\begin_inset Formula $B$
\end_inset
是
\begin_inset Formula $3\times1$
\end_inset
矩阵, 因此
\begin_inset Formula $AB$
\end_inset
有意义,
\begin_inset Formula $BA$
\end_inset
也有意义; 但
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
AB & =(1,0,4)\begin{bmatrix}1\\
1\\
0
\end{bmatrix}=1\times1+0\times1+4\times0=1,\\
BA & =\begin{bmatrix}1\\
1\\
0
\end{bmatrix}(1,0,4)=\begin{bmatrix}1\times1 & 1\times0 & 1\times4\\
1\times1 & 1\times0 & 1\times4\\
0\times1 & 0\times0 & 0\times4
\end{bmatrix}=\begin{bmatrix}1 & 0 & 4\\
1 & 0 & 4\\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}a_{1}\\
& a_{2}\\
& & \ddots\\
& & & a_{n}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}b_{1}\\
& b_{2}\\
& & \ddots\\
& & & b_{n}
\end{bmatrix}$
\end_inset
.
(这种记法表示主对角线以外没有注明的元素均为零), 则
\end_layout
\begin_layout Example
(1)
\begin_inset Formula $k\begin{bmatrix}a_{1}\\
& a_{2}\\
& & \ddots\\
& & & a_{n}
\end{bmatrix}=\begin{bmatrix}ka_{1}\\
& ka_{2}\\
& & \ddots\\
& & & ka_{n}
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Example
(2)
\begin_inset Formula $\begin{bmatrix}a_{1}\\
& a_{2}\\
& & \ddots\\
& & & a_{n}
\end{bmatrix}+\begin{bmatrix}b_{1}\\
& b_{2}\\
& & \ddots\\
& & & b_{n}
\end{bmatrix}=\begin{bmatrix}a_{1}+b_{1}\\
& a_{2}+b_{2}\\
\\
& & & a_{n}+b_{n}
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Example
(3)
\begin_inset Formula $\begin{bmatrix}a_{1}\\
& a_{2}\\
& & \ddots\\
& & & a_{n}
\end{bmatrix}\begin{bmatrix}b_{1}\\
& b_{2}\\
& & \ddots\\
& & & b_{n}
\end{bmatrix}=\begin{bmatrix}a_{1}b_{1}\\
& a_{2}b_{2}\\
& & \ddots\\
& & & a_{n}b_{n}
\end{bmatrix};$
\end_inset
\end_layout
\begin_layout Example
可见, 如果
\begin_inset Formula $A,B$
\end_inset
为
\series bold
同阶对角矩阵
\series default
, 则
\begin_inset Formula $kA$
\end_inset
,
\begin_inset Formula $A+B$
\end_inset
,
\begin_inset Formula $A\times B$
\end_inset
仍为
\series bold
同阶对角矩阵
\series default
.
\end_layout
\begin_layout Example
\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
乘法可交换的矩阵 (中心化子)
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求与矩阵
\begin_inset Formula $A=\begin{bmatrix}0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}$
\end_inset
可交换的一切矩阵.
\end_layout
\begin_layout Solution*
设与
\begin_inset Formula $A$
\end_inset
可交换的矩阵为
\begin_inset Formula $B=\begin{bmatrix}a & b & c & d\\
a_{1} & b_{1} & c_{1} & d_{1}\\
a_{2} & b_{2} & c_{2} & d_{2}\\
a_{3} & b_{3} & c_{3} & d_{3}
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula
\[
AB=\begin{bmatrix}0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}a & b & c & d\\
a_{1} & b_{1} & c_{1} & d_{1}\\
a_{2} & b_{2} & c_{2} & d_{2}\\
a_{3} & b_{3} & c_{3} & d_{3}
\end{bmatrix}=\begin{bmatrix}a_{1} & b_{1} & c_{1} & d_{1}\\
a_{2} & b_{2} & c_{2} & d_{2}\\
a_{3} & b_{3} & c_{3} & d_{3}\\
0 & 0 & 0 & 0
\end{bmatrix};
\]
\end_inset
\begin_inset Formula
\[
BA=\begin{bmatrix}a & b & c & d\\
a_{1} & b_{1} & c_{1} & d_{1}\\
a_{2} & b_{2} & c_{2} & d_{2}\\
a_{3} & b_{3} & c_{3} & d_{3}
\end{bmatrix}\begin{bmatrix}0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}=\begin{bmatrix}0 & a & b & c\\
0 & a_{1} & b_{1} & c_{1}\\
0 & a_{2} & b_{2} & c_{2}\\
0 & a_{3} & b_{3} & c_{3}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
由
\begin_inset Formula
\begin{align*}
AB=BA & \Longrightarrow a_{1}=0,\ b_{1}=a,\ c_{1}=b,\ d_{1}=c,\\
& \qquad a_{2}=0,\ b_{2}=a_{1}=0,\ c_{2}=b_{1}=a,\ d_{2}=c_{1}=b,\\
& \qquad a_{3}=0,\ b_{3}=a_{2}=0,\ c_{3}=b_{2}=0,\ d_{3}=c_{2}=a.
\end{align*}
\end_inset
于是可得
\begin_inset Formula $B=\begin{bmatrix}a & b & c & d\\
0 & a & b & c\\
0 & 0 & a & b\\
0 & 0 & 0 & a
\end{bmatrix}$
\end_inset
, 其中
\begin_inset Formula $a,b,c$
\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 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵乘法定义的来源
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
关于矩阵乘法的定义的来源, 一种解释是源于变量替换, 对于从
\begin_inset Formula $x=(x_{1},x_{2},\cdots,x_{n})$
\end_inset
到
\begin_inset Formula $y=(y_{1},y_{2},\cdots,y_{m})$
\end_inset
的变量替换
\begin_inset Formula
\begin{equation}
\begin{cases}
y_{1}=b_{11}x_{1}+b_{12}x_{2}+\cdots+b_{1n}x_{n},\\
y_{2}=b_{21}x_{1}+b_{22}x_{2}+\cdots+b_{2n}x_{n},\\
\vdots\\
y_{m}=b_{m1}x_{1}+b_{m2}x_{2}+\cdots+b_{mn}x_{n},
\end{cases}\Longleftrightarrow y_{k}=\sum_{j=1}^{n}b_{kj}x_{j}.\label{eq:x2y}
\end{equation}
\end_inset
一般简写为
\begin_inset Formula $y=Bx$
\end_inset
.
而从
\begin_inset Formula $y$
\end_inset
到
\begin_inset Formula $z=(z_{1},z_{2},\cdots,z_{t})$
\end_inset
的变量替换
\begin_inset Formula
\begin{equation}
\begin{cases}
z_{1}=a_{11}y_{1}+a_{12}y_{2}+\cdots+a_{1m}y_{m},\\
z_{2}=a_{21}y_{1}+a_{22}y_{2}+\cdots+a_{2m}y_{m},\\
\vdots\\
z_{t}=a_{t1}y_{1}+a_{t2}y_{2}+\cdots+a_{tm}y_{m},
\end{cases}\Longleftrightarrow z_{i}=\sum_{k=1}^{m}a_{ik}y_{k}.\label{eq:y2z}
\end{equation}
\end_inset
一般简写为
\begin_inset Formula $z=Ay$
\end_inset
, 于是从变量
\begin_inset Formula $x$
\end_inset
直接变换到
\begin_inset Formula $z$
\end_inset
的变量替换为
\begin_inset Formula
\[
z=Ay=A(Bx).
\]
\end_inset
\end_layout
\begin_layout Standard
为此希望记直接从
\begin_inset Formula $x$
\end_inset
到
\begin_inset Formula $z$
\end_inset
的变换为
\begin_inset Formula $z=Cx\coloneqq(AB)\cdot x$
\end_inset
, 也即
\begin_inset Formula
\[
\begin{cases}
z_{1}=c_{11}x_{1}+c_{12}x_{2}+\cdots+c_{1n}x_{n},\\
z_{2}=c_{21}x_{1}+c_{22}x_{2}+\cdots+c_{2n}x_{n},\\
\vdots\\
z_{t}=c_{t1}x_{1}+c_{t2}x_{2}+\cdots+c_{tn}x_{n},
\end{cases}\Longleftrightarrow z_{i}=\sum_{j=1}^{n}c_{ij}x_{j}.
\]
\end_inset
为了得到
\begin_inset Formula $c_{ij}$
\end_inset
的具体表达式, 这相当于直接将 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:x2y"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 代入 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:y2z"
plural "false"
caps "false"
noprefix "false"
\end_inset
), 所以有
\begin_inset Formula
\[
z_{i}=\sum_{k=1}^{m}a_{ik}y_{k}=\sum_{k=1}^{m}a_{ik}\sum_{j=1}^{n}b_{kj}x_{j}=\sum_{j=1}^{n}\boxed{{\color{red}\sum_{k=1}^{m}a_{ik}b_{kj}}}x_{j}=\sum_{j=1}^{n}\boxed{{\color{red}c_{ij}}}x_{j}.
\]
\end_inset
因此为了能使变量替换与通常实数时的情况一致, 矩阵乘法中的元素
\begin_inset Formula $c_{ij}$
\end_inset
是由矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行元素与矩阵
\begin_inset Formula $B$
\end_inset
的第
\begin_inset Formula $j$
\end_inset
列元素对应相乘的累和.
\begin_inset Foot
status open
\begin_layout Plain Layout
矩阵乘法的定义归功于Cayley, 参考: 莫里斯·克莱因,《古今数学思想》, vol.
3, p.207-216.
\end_layout
\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
矩阵乘法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
某地区有四个工厂 I、II、III、IV, 生产甲、乙、丙三种产品, 矩阵
\begin_inset Formula $A$
\end_inset
表示一年中各工厂生产各种产品的数量, 矩阵
\begin_inset Formula $B$
\end_inset
表示各种产品的单位价格 (元) 及单位利润 (元), 矩阵
\begin_inset Formula $C$
\end_inset
表示各工厂的总收入及总利润.
\end_layout
\begin_layout Example
\begin_inset ERT
status open
\begin_layout Plain Layout
$$A=
\backslash
begin{bNiceArray}{ccc}[last-row,last-col]
\end_layout
\begin_layout Plain Layout
a_{11} & a_{12} & a_{13} &
\backslash
text{I}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{21} & a_{22} & a_{23} &
\backslash
text{II}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{31} & a_{32} & a_{33} &
\backslash
text{III}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{41} & a_{42} & a_{43} &
\backslash
text{IV}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
text{甲} &
\backslash
text{乙} &
\backslash
text{丙} &
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
;,
\backslash
\end_layout
\begin_layout Plain Layout
B=
\backslash
begin{bNiceArray}{cc}[last-row,last-col]
\end_layout
\begin_layout Plain Layout
b_{11} & b_{12} &
\backslash
text{甲}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
b_{21} & b_{22} &
\backslash
text{乙}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
b_{31} & b_{32} &
\backslash
text{丙}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
text{单位价格} &
\backslash
text{单位利润} &
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}
\backslash
\backslash
;,
\backslash
\end_layout
\begin_layout Plain Layout
C=
\backslash
begin{bNiceArray}{cc}[last-row,last-col]
\end_layout
\begin_layout Plain Layout
c_{11} & c_{12} &
\backslash
text{I}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{21} & c_{22} &
\backslash
text{II}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{31} & c_{32} &
\backslash
text{III}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{41} & c_{42} &
\backslash
text{IV}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
text{总收入} &
\backslash
text{总利润} &
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}.
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
其中,
\begin_inset Formula $a_{ik}$
\end_inset
(
\begin_inset Formula $i=1,2,3,4$
\end_inset
;
\begin_inset Formula $k=1,2,3$
\end_inset
) 是第
\begin_inset Formula $i$
\end_inset
个工厂生产第
\begin_inset Formula $k$
\end_inset
种产品的数量,
\begin_inset Formula $b_{k1}$
\end_inset
及
\begin_inset Formula $b_{k2}$
\end_inset
(
\begin_inset Formula $k=1,2,3$
\end_inset
) 分别是第
\begin_inset Formula $k$
\end_inset
种产品的单位价格及单位利润,
\begin_inset Formula $c_{i1}$
\end_inset
及
\begin_inset Formula $c_{i2}$
\end_inset
(
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
\begin_inset Formula $i=1,2,3,4$
\end_inset
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
) 分别是第
\begin_inset Formula $i$
\end_inset
个工厂生产三种产品的总收入及总利润.
则矩阵
\begin_inset Formula $A,B,C$
\end_inset
的元素之间有下列关系:
\end_layout
\begin_layout Example
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31} & a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31} & a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31} & a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{41}b_{11}+a_{42}b_{21}+a_{43}b_{31} & a_{41}b_{12}+a_{42}b_{22}+a_{43}b_{32}
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}=
\backslash
begin{bNiceArray}{cc}[last-row]
\end_layout
\begin_layout Plain Layout
c_{11} & c_{12}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{21} & c_{22}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{31} & c_{32}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
c_{41} & c_{42}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
text{总收入} &
\backslash
text{总利润}
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}.$$
\end_layout
\end_inset
其中
\begin_inset Formula $c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+a_{i3}b_{3j}$
\end_inset
(
\begin_inset Formula $i=1,2,3,4$
\end_inset
;
\begin_inset Formula $j=1,2$
\end_inset
), 即
\begin_inset Formula $C=AB$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
线性方程组的矩阵表示
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
线性方程组的矩阵表示
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设有线性方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\left\{ \begin{array}{l}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2}\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=b_{m}
\end{array}\right.\label{eq:2.3-1}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
若记
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix},\quad x=\begin{bmatrix}x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{bmatrix},\quad b=\begin{bmatrix}b_{1}\\
b_{2}\\
\vdots\\
b_{m}
\end{bmatrix},
\]
\end_inset
则利用矩阵的乘法, 线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 可表示为矩阵形式:
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
Ax=b.\label{eq:2.3-2}
\end{equation}
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
其中矩阵
\begin_inset Formula $A$
\end_inset
称为线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的
\series bold
系数矩阵
\series default
.
方程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 又称为
\series bold
矩阵方程
\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
线性方程组的矩阵表示
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
如果
\begin_inset Formula $x_{j}=c_{j}$
\end_inset
, (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset
) 是方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解, 记列矩阵
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\begin_inset Formula
\[
c=\begin{bmatrix}c_{1}\\
c_{2}\\
\vdots\\
c_{n}
\end{bmatrix},
\]
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
则
\begin_inset Formula $Ac=b$
\end_inset
.
这时也称
\begin_inset Formula $c$
\end_inset
是矩阵方程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解;
\end_layout
\begin_layout Standard
反之, 如果列矩阵
\begin_inset Formula $c$
\end_inset
是矩阵方程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解, 即有矩阵等式
\begin_inset Formula $Ac=b$
\end_inset
成立, 则
\begin_inset Formula $x=c$
\end_inset
, 即
\begin_inset Formula $x_{j}=c_{j}$
\end_inset
, (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset
) 也是线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的解.
\end_layout
\begin_layout Standard
这样, 对线性方程组 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的讨论便等价于对矩阵方程 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.3-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的讨论.
\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
齐次线性方程组的矩阵表示
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
特别地, 齐次线性方程组可以表示为
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
Ax=O.
\]
\end_inset
\end_layout
\begin_layout Standard
将线性方程组写成矩阵方程的形式, 不仅
\series bold
书写方便
\series default
, 而且可以把线性方程组的理论与矩阵理论联系起来, 这给线性方程组的讨论带来很大的便利.
\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
矩阵方程的例子
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解矩阵方程
\begin_inset Formula $\begin{bmatrix}2 & 1\\
1 & 2
\end{bmatrix}X=\begin{bmatrix}1 & 2\\
-1 & 4
\end{bmatrix}$
\end_inset
, 其中
\begin_inset Formula $X$
\end_inset
为二阶矩阵.
\end_layout
\begin_layout Solution*
设
\begin_inset Formula $X=\begin{bmatrix}x_{11} & x_{12}\\
x_{21} & x_{22}
\end{bmatrix}$
\end_inset
, 由题设, 有
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{bmatrix}2 & 1\\
1 & 2
\end{bmatrix}\begin{bmatrix}x_{11} & x_{12}\\
x_{21} & x_{22}
\end{bmatrix}=\begin{bmatrix}1 & 2\\
-1 & 4
\end{bmatrix},\quad\begin{bmatrix}2x_{11}+x_{21} & 2x_{12}+x_{22}\\
x_{11}+2x_{21} & x_{12}+2x_{22}
\end{bmatrix}=\begin{bmatrix}1 & 2\\
-1 & 4
\end{bmatrix}.
\]
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
即
\begin_inset Formula
\[
\begin{cases}
2x_{11}+x_{21}=1 & (1)\\
x_{11}+2x_{21}=-1 & (2)
\end{cases},\qquad\begin{cases}
2x_{12}+x_{22}=2 & (3)\\
x_{12}+2x_{22}=4 & (4)
\end{cases}.
\]
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
分别解 (1), (2) 和 (3), (4) 两个方程组得
\begin_inset Formula
\[
x_{11}=1,\quad x_{12}=0,\quad x_{21}=-1,\quad x_{22}=2\Longrightarrow X=\begin{bmatrix}1 & 0\\
-1 & 2
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Subsection
线性变换的概念
\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
线性变换的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
变量
\begin_inset Formula $x_{1},x_{2},\cdots,x_{n}$
\end_inset
与变量
\begin_inset Formula $y_{1},y_{2},\cdots,y_{m}$
\end_inset
之间的关系式:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\begin{cases}
y_{1}=a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}\\
y_{2}=a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
y_{m}=a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}.
\end{cases}\label{eq:2.4-2}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
称为从变量
\begin_inset Formula $x_{1},x_{2},\cdots,x_{n}$
\end_inset
到变量
\begin_inset Formula $y_{1},y_{2},\cdots,y_{m}$
\end_inset
的
\series bold
线性变换
\series default
.
其中
\begin_inset Formula $a_{ij}$
\end_inset
(
\begin_inset Formula $i=1,2,\cdots,m;\ j=1,2,\cdots,n$
\end_inset
) 为常数.
线性变换 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.4-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的系数
\begin_inset Formula $a_{ij}$
\end_inset
构成矩阵
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
, 称其为线性变换 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2.4-2"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 的
\series bold
系数矩阵
\series default
.
\end_layout
\begin_layout Standard
易见
\end_layout
\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
thickness "0.4pt"
separation "3pt"
shadowsize "4pt"
framecolor "black"
backgroundcolor "none"
status open
\begin_layout Plain Layout
线性变换与其系数矩阵之间存在一一对应关系.
\end_layout
\end_inset
\end_layout
\begin_layout Standard
因而可利用矩阵来研究线性变换, 亦可利用线性变换来研究矩阵.
\end_layout
\begin_layout Standard
线性变换
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{cases}
y_{1}=x_{1}\\
y_{2}=x_{2}\\
\vdots\\
y_{n}=x_{n}
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Standard
称为
\series bold
恒等变换
\series default
, 其系数矩阵就是
\series bold
单位矩阵
\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
线性变换的几何意义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设有线性变换
\begin_inset Formula $y=Ax$
\end_inset
, 其中
\begin_inset Formula $A=\begin{bmatrix}1 & 2\\
0 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $x=\begin{bmatrix}1\\
1
\end{bmatrix}$
\end_inset
, 试求出向量
\begin_inset Formula $y$
\end_inset
, 并指出该变换的几何意义.
\end_layout
\begin_layout Solution*
\begin_inset Formula $y=Ax=\begin{bmatrix}1 & 2\\
0 & 1
\end{bmatrix}\begin{bmatrix}1\\
1
\end{bmatrix}=\begin{bmatrix}3\\
1
\end{bmatrix}$
\end_inset
.
其几何意义是: 线性变换
\begin_inset Formula $y=Ax$
\end_inset
将平面
\begin_inset Formula $x_{1}Ox_{2}$
\end_inset
上的向量
\begin_inset Formula $x=\begin{bmatrix}1\\
1
\end{bmatrix}$
\end_inset
变换为该平面上的另一向量
\begin_inset Formula $y=\begin{bmatrix}3\\
1
\end{bmatrix}$
\end_inset
, (见下图).
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
begin{center}
\end_layout
\begin_layout Plain Layout
\backslash
begin{tikzpicture}[remember picture]
\end_layout
\begin_layout Plain Layout
\backslash
draw[help lines] (0,0) grid (3,3);
\end_layout
\begin_layout Plain Layout
\backslash
draw[red,fill=red!20] (0,0)--(2,0)--(2,2)--(0,2)--(0,0);
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
\backslash
filldraw [gray] (1.5,1.5) circle [radius=2pt]
\end_layout
\begin_layout Plain Layout
node[above] (n3) at (1.5,1.5) {$x$};
\end_layout
\begin_layout Plain Layout
\backslash
end{tikzpicture}
\end_layout
\begin_layout Plain Layout
\backslash
hspace{3mm}
\end_layout
\begin_layout Plain Layout
\backslash
begin{tikzpicture}[remember picture]
\end_layout
\begin_layout Plain Layout
\backslash
draw[help lines] (0,0) grid (7.5,3);
\end_layout
\begin_layout Plain Layout
\backslash
draw[gray,fill=gray!20] (0,0)--(2,0)--(2,2)--(0,2)--(0,0);
\end_layout
\begin_layout Plain Layout
\backslash
draw[red,fill=red!20,opacity=0.7] (0,0)--(2,0)--(6,2)--(4,2)--(0,0);
\end_layout
\begin_layout Plain Layout
\end_layout
\begin_layout Plain Layout
\backslash
filldraw [gray] (4.5,1.5) circle [radius=2pt]
\end_layout
\begin_layout Plain Layout
node[above] (n2) at (4.5,1.5) {$Ax$};
\end_layout
\begin_layout Plain Layout
\backslash
end{tikzpicture}
\end_layout
\begin_layout Plain Layout
\backslash
begin{tikzpicture}[remember picture,overlay]
\end_layout
\begin_layout Plain Layout
\backslash
draw[overlay,->,very thick,yshift=5mm] (n3) to[bend left] (n2);
\end_layout
\begin_layout Plain Layout
\backslash
end{tikzpicture}
\end_layout
\begin_layout Plain Layout
\backslash
end{center}
\end_layout
\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
线性变换的几何意义
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}$
\end_inset
, 为三维空间一向量, 试讨论矩阵变换
\begin_inset Formula $x\rightarrow Ax$
\end_inset
的几何意义.
\end_layout
\begin_layout Solution*
如图所示, 设
\begin_inset Formula $x=\overrightarrow{OP}=\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula
\[
\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}\rightarrow\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{bmatrix}=\boxed{{\color{red}\begin{bmatrix}1\\
0\\
0
\end{bmatrix}x_{1}+\begin{bmatrix}0\\
1\\
0
\end{bmatrix}x_{2}+\begin{bmatrix}0\\
0\\
0
\end{bmatrix}x_{3}}}=\begin{bmatrix}x_{1}\\
x_{2}\\
0
\end{bmatrix},
\]
\end_inset
从几何上看, 在变换
\begin_inset Formula $x\rightarrow Ax$
\end_inset
下, 空间中的点
\begin_inset Formula $P\left(x_{1},x_{2},x_{3}\right)$
\end_inset
被投影到了
\begin_inset Formula $x_{1}Ox_{2}$
\end_inset
平面上.
\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
矩阵的转置
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
把矩阵
\begin_inset Formula $A$
\end_inset
的行换成同序数的列得到的新矩阵, 称为
\begin_inset Formula $A$
\end_inset
的
\series bold
转置矩阵
\series default
, 记作
\begin_inset Formula $A^{T}$
\end_inset
(或
\begin_inset Formula $A'$
\end_inset
).
即若
\begin_inset Formula
\[
A=\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Definition
则
\end_layout
\begin_layout Definition
\begin_inset Formula
\[
A^{T}=\begin{bmatrix}a_{11} & a_{21} & \cdots & a_{m1}\\
a_{12} & a_{22} & \cdots & a_{m2}\\
\vdots & \vdots & \ddots & \vdots\\
a_{1n} & a_{2n} & \cdots & a_{mn}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Example
(1) 设
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & -1 & 0\\
-1 & 0 & 1 & 4\\
2 & 5 & -3 & 1
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula $A^{T}=\begin{bmatrix}1 & -1 & 2\\
2 & 0 & 5\\
-1 & 1 & -3\\
0 & 4 & 1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Example
(2) 设
\begin_inset Formula $A=(1,2,3,-1)$
\end_inset
, 则
\begin_inset Formula $A^{T}=\begin{bmatrix}1\\
2\\
3\\
-1
\end{bmatrix}$
\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
矩阵的转置
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
矩阵的转置满足以下运算规律 (假设运算都是可行的):
\end_layout
\begin_layout Standard
(1)
\begin_inset Formula $\left(A^{T}\right)^{T}=A$
\end_inset
;
\end_layout
\begin_layout Standard
(2)
\begin_inset Formula $(A+B)^{T}=A^{T}+B^{T}$
\end_inset
;
\end_layout
\begin_layout Standard
(3)
\begin_inset Formula $(kA)^{T}=kA^{T}$
\end_inset
;
\end_layout
\begin_layout Standard
(4)
\begin_inset Formula $(AB)^{T}=B^{T}A^{T}$
\end_inset
.
\end_layout
\begin_layout Example
已知
\begin_inset Formula $A=\begin{bmatrix}2 & 0 & -1\\
1 & 3 & 2
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 7 & -1\\
4 & 2 & 3\\
2 & 0 & 1
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $(AB)^{T}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $AB=\begin{bmatrix}2 & 0 & -1\\
1 & 3 & 2
\end{bmatrix}\begin{bmatrix}1 & 7 & -1\\
4 & 2 & 3\\
2 & 0 & 1
\end{bmatrix}=\begin{bmatrix}0 & 14 & -3\\
17 & 13 & 10
\end{bmatrix}$
\end_inset
, 所以
\begin_inset Formula
\[
(AB)^{T}=\begin{bmatrix}0 & 17\\
14 & 13\\
-3 & 10
\end{bmatrix}.
\]
\end_inset
另一方面, 也可以按下式计算
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
(AB)^{T}=B^{T}A^{T}=\begin{bmatrix}1 & 4 & 2\\
7 & 2 & 0\\
-1 & 3 & 1
\end{bmatrix}\begin{bmatrix}2 & 1\\
0 & 3\\
-1 & 2
\end{bmatrix}=\begin{bmatrix}0 & 17\\
14 & 13\\
-3 & 10
\end{bmatrix}.
\]
\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
方阵的幂
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设方阵
\begin_inset Formula $A=\left(a_{ij}\right)_{n\times n}$
\end_inset
, 规定
\begin_inset Formula
\[
A^{0}=E,\quad A^{k}=\underbrace{A\cdot A\cdot\cdots\cdot A}_{k\text{个}A},\quad k\text{ 为自然数. }
\]
\end_inset
\begin_inset Formula $A^{k}$
\end_inset
称为
\series bold
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $k$
\end_inset
次幂
\series default
.
\end_layout
\begin_layout Standard
方阵的幂满足以下运算规律 (假设运算都是可行的):
\end_layout
\begin_layout Standard
(1).
\begin_inset Formula $A^{m}A^{n}=A^{m+n}$
\end_inset
, (
\begin_inset Formula $m,n$
\end_inset
为非负整数);
\end_layout
\begin_layout Standard
(2).
\begin_inset Formula $\left(A^{m}\right)^{n}=A^{mn}$
\end_inset
.
\end_layout
\begin_layout Remark*
一般地,
\begin_inset Formula $(AB)^{m}\neq A^{m}B^{m}$
\end_inset
,
\begin_inset Formula $m$
\end_inset
为自然数.
\end_layout
\begin_layout Proposition
设
\begin_inset Formula $A,B$
\end_inset
均为
\begin_inset Formula $n$
\end_inset
阶矩阵, 且有
\begin_inset Formula $AB=BA$
\end_inset
, 则
\begin_inset Formula $(AB)^{m}=A^{m}B^{m}$
\end_inset
, 其中
\begin_inset Formula $m$
\end_inset
为自然数, 反之不成立.
\end_layout
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A^{3}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
A^{2}=\begin{bmatrix}\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{bmatrix}\begin{bmatrix}\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{bmatrix}=\begin{bmatrix}\lambda^{2} & 2\lambda & 1\\
0 & \lambda^{2} & 2\lambda\\
0 & 0 & \lambda^{2}
\end{bmatrix},
\]
\end_inset
\begin_inset Formula
\[
A^{3}=A^{2}A=\begin{bmatrix}\lambda^{2} & 2\lambda & 1\\
0 & \lambda^{2} & 2\lambda\\
0 & 0 & \lambda^{2}
\end{bmatrix}\begin{bmatrix}\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{bmatrix}=\begin{bmatrix}\lambda^{3} & 3\lambda^{2} & 3\lambda\\
0 & \lambda^{3} & 3\lambda^{2}\\
0 & 0 & \lambda^{3}
\end{bmatrix}.
\]
\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
方阵的行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
由
\begin_inset Formula $n$
\end_inset
阶方阵
\begin_inset Formula $A$
\end_inset
的元素所构成的行列式 (各元素的位置不变), 称为方阵
\begin_inset Formula $A$
\end_inset
的行列式, 记作
\begin_inset Formula $|A|$
\end_inset
或
\begin_inset Formula $\mathrm{det}A$
\end_inset
.
\end_layout
\begin_layout Remark*
方阵与行列式是两个不同的概念,
\begin_inset Formula $n$
\end_inset
阶方阵是
\begin_inset Formula $n^{2}$
\end_inset
个数按一定方式排成的数表, 而
\begin_inset Formula $n$
\end_inset
阶行列式则是这些数按一定的运算法则所确定的一个数值 (实数或复数).
\end_layout
\begin_layout Standard
方阵
\begin_inset Formula $A$
\end_inset
的行列式
\begin_inset Formula $|A|$
\end_inset
满足以下运算规律 (设
\begin_inset Formula $A,B$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶方阵,
\begin_inset Formula $k$
\end_inset
为常数):
\end_layout
\begin_layout Standard
(1)
\begin_inset Formula $\left|A^{T}\right|=|A|$
\end_inset
, (行列式性质1);
\end_layout
\begin_layout Standard
(2)
\begin_inset Formula $|kA|=k^{n}|A|$
\end_inset
;
\end_layout
\begin_layout Standard
(3)
\begin_inset Formula $|AB|=|A||B|$
\end_inset
.
进一步
\begin_inset Formula $|A||B|=|AB|=|B||A|$
\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
方阵的行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & -1\\
2 & 1 & 0\\
3 & 2 & -1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}-2 & 1 & 0\\
0 & 3 & 1\\
0 & 0 & 2
\end{bmatrix}$
\end_inset
, 则
\end_layout
\begin_layout Example
\begin_inset Formula
\[
AB=\begin{bmatrix}-2 & 1 & -2\\
-4 & 5 & 1\\
-6 & 9 & 0
\end{bmatrix},\quad|AB|=\begin{vmatrix}-2 & 1 & -2\\
-4 & 5 & 1\\
-6 & 9 & 0
\end{vmatrix}=24.
\]
\end_inset
又
\end_layout
\begin_layout Example
\begin_inset Formula
\[
|A|=\begin{vmatrix}1 & 0 & -1\\
2 & 1 & 0\\
3 & 2 & -1
\end{vmatrix}=-2,\quad|B|=\begin{vmatrix}-2 & 1 & 0\\
0 & 3 & 1\\
0 & 0 & 2
\end{vmatrix}=-12\text{, }
\]
\end_inset
因此
\begin_inset Formula $|AB|=24=(-2)(-12)=|A||B|$
\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
对称矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶方阵, 如果
\begin_inset Formula $A^{T}=A$
\end_inset
, 即
\begin_inset Formula
\[
a_{ij}=a_{ji},\quad(i,j=1,2,\cdots,n),
\]
\end_inset
则称
\begin_inset Formula $A$
\end_inset
为
\series bold
对称矩阵
\series default
.
\end_layout
\begin_layout Standard
显然, 对称矩阵
\begin_inset Formula $A$
\end_inset
的元素关于主对角线对称.
例如
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{bmatrix}0 & -1\\
-1 & 0
\end{bmatrix},\begin{bmatrix}8 & 6 & 1\\
6 & 9 & 0\\
1 & 0 & 5
\end{bmatrix}
\]
\end_inset
\end_layout
\begin_layout Standard
均为对称矩阵.
\end_layout
\begin_layout Standard
如果
\begin_inset Formula $A^{T}=-A$
\end_inset
, 则称
\begin_inset Formula $A$
\end_inset
为
\series bold
反对称矩阵
\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
对称阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设列矩阵
\begin_inset Formula $X=\left(x_{1},x_{2},\cdots,x_{n}\right)^{T}$
\end_inset
满足
\begin_inset Formula $X^{T}X=1$
\end_inset
,
\begin_inset Formula $E$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶单位矩阵,
\begin_inset Formula $H=E-2XX^{T}$
\end_inset
, 证明
\begin_inset Formula $H$
\end_inset
是对称矩阵, 且
\begin_inset Formula $HH^{T}=E$
\end_inset
.
\end_layout
\begin_layout Proof
由于
\begin_inset Formula $H^{T}=\left(E-2XX^{T}\right)^{T}=E^{T}-2\left(XX^{T}\right)^{T}=E-2XX^{T}=H$
\end_inset
, 所以
\begin_inset Formula $H$
\end_inset
是对称矩阵.
\begin_inset Formula
\[
\begin{aligned}HH^{T} & =H^{2}=\left(E-2XX^{T}\right)^{2}\\
& =E-4XX^{T}+4\left(XX^{T}\right)\left(XX^{T}\right)\\
& =E-4XX^{T}+4X\left(X^{T}X\right)X^{T}\\
& =E-4XX^{T}+4XX^{T}=E.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
由此可知: 矩阵方程
\begin_inset Formula $X^{2}=E$
\end_inset
有无穷多解, 称满足矩阵方程
\begin_inset Formula $X^{2}=E$
\end_inset
的矩阵为幂等矩阵.
\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
反对称阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
是两个
\begin_inset Formula $n$
\end_inset
阶反对称矩阵, 证明: 当且仅当
\begin_inset Formula $AB=-BA$
\end_inset
时,
\begin_inset Formula $AB$
\end_inset
是反对称矩阵.
\end_layout
\begin_layout Solution*
由于
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
是反对称矩阵
\begin_inset Formula $\Longrightarrow A=-A^{T}$
\end_inset
,
\begin_inset Formula $B=-B^{T}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Itemize
若
\begin_inset Formula $AB=-BA$
\end_inset
, 则
\begin_inset Formula $(AB)^{T}=B^{T}A^{T}=BA=-AB\Longrightarrow AB$
\end_inset
反对称.
\end_layout
\begin_layout Itemize
反之, 若
\begin_inset Formula $AB$
\end_inset
反对称, 即
\begin_inset Formula $(AB)^{T}=-AB\Longrightarrow AB=-(AB)^{T}=-B^{T}A^{T}=-(-B)(-A)=-BA$
\end_inset
.
\end_layout
\begin_layout Standard
证毕.
\end_layout
\end_deeper
\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
共轭矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
设
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
为复 (数) 矩阵, 记
\begin_inset Formula
\[
\overline{A}=\left(\overline{a_{ij}}\right),
\]
\end_inset
其中
\begin_inset Formula $\overline{a_{ij}}$
\end_inset
表示
\begin_inset Formula $a_{ij}$
\end_inset
的共轭复数, 称
\begin_inset Formula $\overline{A}$
\end_inset
为
\begin_inset Formula $A$
\end_inset
的
\series bold
共轭矩阵
\series default
.
\end_layout
\begin_layout Standard
共轭矩阵满足以下运算规律 (设
\begin_inset Formula $A,B$
\end_inset
为复矩阵,
\begin_inset Formula $k$
\end_inset
为复数, 且运算都是可行的):
\end_layout
\begin_layout Standard
(1)
\begin_inset Formula $\overline{A+B}=\overline{A}+\overline{B}$
\end_inset
;
\end_layout
\begin_layout Standard
(2)
\begin_inset Formula $\overline{\lambda A}=\overline{\lambda A}$
\end_inset
;
\end_layout
\begin_layout Standard
(3)
\begin_inset Formula $\overline{AB}=\overline{A}\overline{B}$
\end_inset
.
\end_layout
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1+i & 0 & 1-\sqrt{2}i\\
2i & -1 & -4i\\
-4-i & \sqrt{3}i & i
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $i=\sqrt{-1}$
\end_inset
, 则
\begin_inset Formula $\overline{A}=\begin{bmatrix}1-i & 0 & 1+\sqrt{2}i\\
-2i & -1 & 4i\\
-4+i & -\sqrt{3}i & -i
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
作业
\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
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
设
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
为三阶矩阵, 若已知
\begin_inset Formula $|A|=-2$
\end_inset
, 求
\begin_inset Formula $\left|\vphantom{\int}|A|\cdot A\right|$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula
\[
A=\begin{bmatrix}\lambda_{1}\\
& \ddots\\
& & \lambda_{n}
\end{bmatrix},
\]
\end_inset
\begin_inset Formula $B=\left(b_{ij}\right)$
\end_inset
, 求
\begin_inset Formula $AB$
\end_inset
,
\begin_inset Formula $BA$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula
\[
E_{23}=\begin{bmatrix}1\\
& & 1\\
& 1\\
& & & 1
\end{bmatrix},\ E_{2(\lambda)}=\begin{bmatrix}1\\
& \lambda\\
& & 1\\
& & & 1
\end{bmatrix},\ E_{24(k)}=\begin{bmatrix}1\\
& 1 & & k\\
& & 1\\
& & & 1
\end{bmatrix}.
\]
\end_inset
若记矩阵
\begin_inset Formula $A=\begin{bmatrix}a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\
a_{31} & a_{32} & a_{33} & a_{34} & a_{35}\\
a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Problem
(1).
试求
\begin_inset Formula $E_{23}A$
\end_inset
,
\begin_inset Formula $E_{2(\lambda)}A$
\end_inset
与
\begin_inset Formula $E_{24(k)}A$
\end_inset
并比较它们与矩阵
\begin_inset Formula $A$
\end_inset
的不同 (并写出不同点).
\end_layout
\begin_layout Problem
(2).
试求
\begin_inset Formula $AE_{23}$
\end_inset
,
\begin_inset Formula $AE_{2(\lambda)}$
\end_inset
与
\begin_inset Formula $AE_{24(k)}$
\end_inset
并比较它们与矩阵
\begin_inset Formula $A$
\end_inset
的不同 (并写出不同点).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $A=\begin{bmatrix}1 & 1 & 1\\
-1 & 1 & 1\\
1 & -1 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 2 & 1\\
1 & 3 & -1\\
2 & 1 & 4
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Problem
(1).
计算
\begin_inset Formula $AB-2A$
\end_inset
,
\begin_inset Formula $AB-BA$
\end_inset
;
\end_layout
\begin_layout Problem
(2).
问：
\begin_inset Formula $(A+B)(A-B)$
\end_inset
是否等于
\begin_inset Formula $A^{2}-B^{2}$
\end_inset
?
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
计算方阵
\begin_inset Formula $A$
\end_inset
的矩阵多项式
\begin_inset Formula $f(A)$
\end_inset
, 其中
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $A=\begin{bmatrix}2 & -1\\
-3 & 3
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $f(x)=x^{2}-x-1$
\end_inset
;
\end_layout
\begin_layout Problem
(2).
\begin_inset Formula $A=\begin{bmatrix}2 & 1 & 1\\
3 & 1 & 2\\
1 & -1 & 0
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $f(x)=x^{2}-5x+3$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
举例说明下列命题是错误的:
\end_layout
\begin_layout Problem
(1).
如果
\begin_inset Formula $A^{2}=O$
\end_inset
, 则
\begin_inset Formula $A=O$
\end_inset
;
\end_layout
\begin_layout Problem
(2).
如果
\begin_inset Formula $A^{2}=A$
\end_inset
, 则
\begin_inset Formula $A=O$
\end_inset
或者
\begin_inset Formula $A=E$
\end_inset
;
\end_layout
\begin_layout Problem
(3).
如果
\begin_inset Formula $AX=AY$
\end_inset
且
\begin_inset Formula $A\neq O$
\end_inset
, 则
\begin_inset Formula $X=Y$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $k$
\end_inset
为正整数, 计算：
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula
\[
\begin{bmatrix}\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}^{k};
\]
\end_inset
\end_layout
\begin_layout Problem
(2).
\begin_inset Formula
\[
\begin{bmatrix}1 & 0\\
\lambda & 1
\end{bmatrix}^{k};
\]
\end_inset
\end_layout
\begin_layout Problem
(3).
\begin_inset Formula
\[
\begin{bmatrix}\lambda & 1\\
& \lambda & 1\\
& & \lambda
\end{bmatrix}^{k}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
计算矩阵乘积
\begin_inset Formula $\begin{bmatrix}b_{1} & b_{2} & b_{3}\end{bmatrix}\begin{bmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix}\begin{bmatrix}b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
(1).
证明两个上三角形方阵的积仍为上三角形 (
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset
为上三角形是指当
\begin_inset Formula $i>j$
\end_inset
时,
\begin_inset Formula $a_{ij}=0$
\end_inset
).
\end_layout
\begin_layout Problem
(2).
证明两个下三角形方阵的积仍为下三角形 (若当
\begin_inset Formula $i<j$
\end_inset
时
\begin_inset Formula $a_{ij}=0$
\end_inset
, 则称
\begin_inset Formula $A$
\end_inset
为下三角形).
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
证明任一
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
都可表示成对称阵与反对称阵之和.
\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
***
\end_layout
\end_inset
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n\times m$
\end_inset
矩阵,
\begin_inset Formula $B$
\end_inset
为
\begin_inset Formula $m\times n$
\end_inset
矩阵, 如果
\begin_inset Formula $E_{n}-AB$
\end_inset
可逆, 证明:
\begin_inset Formula $E_{m}-BA$
\end_inset
也可逆, 并求
\begin_inset Formula $\left(E_{m}-BA\right)^{-1}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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