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\begin_body
\begin_layout Section
分块矩阵
\end_layout
\begin_layout Subsection
分块矩阵的概念
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
分块矩阵的概念
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
对于行数和列数较高的矩阵;
\end_layout
\begin_layout Itemize
为了简化运算, 经常采用分块法;
\end_layout
\begin_layout Itemize
使大矩阵的运算化成若干小矩阵间的运算;
\end_layout
\begin_layout Itemize
同时也使原矩阵的结构显得简单而清晰.
\end_layout
\begin_layout Standard
具体做法是: 将大矩阵用若干条纵线和横线分成多个小矩阵.
每个小矩阵称为
\begin_inset Formula $A$
\end_inset
的
\series bold
子块
\series default
,
\series bold
以子块为元素
\series default
的形式上的矩阵称为
\series bold
分块矩阵
\series default
.
\end_layout
\begin_layout Standard
矩阵的分块有多种方式, 可根据具体需要而定.
\end_layout
\begin_layout Remark*
一个矩阵也可看作以
\begin_inset Formula $m\times n$
\end_inset
个元素为
\begin_inset Formula $1$
\end_inset
阶子块的分块矩阵.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
分块矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset ERT
status open
\begin_layout Plain Layout
$A=
\backslash
begin{bNiceArray}{ccc:c}
\end_layout
\begin_layout Plain Layout
1 & 3 & -1 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
2 & 5 & 0 & -2
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hdottedline
\end_layout
\begin_layout Plain Layout
3 & 1 & -1 & 3
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$,
\end_layout
\end_inset
则
\begin_inset Formula $A$
\end_inset
就是一个分块矩阵.
若记
\begin_inset Formula
\[
\begin{array}{ll}
A_{11}=\begin{bmatrix}1 & 3 & -1\\
2 & 5 & 0
\end{bmatrix}, & A_{12}=\begin{bmatrix}0\\
-2
\end{bmatrix},\\
A_{21}=(3,1,-1), & A_{22}=(3),
\end{array}
\]
\end_inset
则
\begin_inset Formula $A$
\end_inset
可表示为
\begin_inset Formula
\[
A=\begin{bmatrix}A_{11} & A_{12}\\
A_{21} & A_{22}
\end{bmatrix},
\]
\end_inset
这是一个分成了 4 块的分块矩阵.
\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
分块矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset ERT
status open
\begin_layout Plain Layout
$A=
\backslash
begin{bNiceArray}{cc:cc:c}
\end_layout
\begin_layout Plain Layout
1 & 1 & 0 & 0 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
-1 & 1 & 0 & 0 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hdottedline
\end_layout
\begin_layout Plain Layout
0 & 0 & 1 & 0 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0 & 0 & 1 & 1 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hdottedline
\end_layout
\begin_layout Plain Layout
0 & 0 & 0 & 0 & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$,
\end_layout
\end_inset
则
\begin_inset Formula $A$
\end_inset
是一个分了块的矩阵, 且
\begin_inset Formula $A$
\end_inset
的分块有一个特点, 若记
\begin_inset Formula
\[
A_{1}=\begin{bmatrix}1 & 1\\
-1 & 1
\end{bmatrix},\ A_{2}=\begin{bmatrix}1 & 0\\
1 & 1
\end{bmatrix},\ A_{3}=(1),
\]
\end_inset
\begin_inset Formula
\[
A=\begin{bmatrix}A_{1} & O & O\\
O & A_{2} & O\\
O & O & A_{3}
\end{bmatrix},
\]
\end_inset
即
\begin_inset Formula $A$
\end_inset
作为分块矩阵来看, 除了主对角线上的块外, 其余各块都是零矩阵, 以后我们会看到这种分块成对角形状的矩阵在运算上是比较简便的.
\end_layout
\end_deeper
\begin_layout Subsection
分块矩阵的运算
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
分块矩阵的运算与普通矩阵的运算规则相似.
分块时要注意, 运算的两矩阵按块能运算, 并且参与运算的子块也能运算, 即, 内外都能运算.
\end_layout
\begin_layout Enumerate
设矩阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
\end_inset
的行数相同, 列数相同, 采用相同的分块法, 若
\begin_inset Formula
\[
A=\begin{bmatrix}A_{11} & \cdots & A_{1t}\\
\vdots & \ddots & \vdots\\
A_{s1} & \cdots & A_{st}
\end{bmatrix},\ B=\begin{bmatrix}B_{11} & \cdots & B_{1t}\\
\vdots & \ddots & \vdots\\
B_{s1} & \cdots & B_{st}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset Formula $A_{ij}$
\end_inset
与
\begin_inset Formula $B_{ij}$
\end_inset
的行数相同, 列数相同, 则
\begin_inset Formula
\[
A+B=\begin{bmatrix}A_{11}+B_{11} & \cdots & A_{1t}+B_{1t}\\
\vdots & \ddots & \vdots\\
A_{s1}+B_{s1} & \cdots & A_{st}+B_{st}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Enumerate
设
\begin_inset Formula $A=\begin{bmatrix}A_{11} & \cdots & A_{1t}\\
\vdots & \ddots & \vdots\\
A_{s1} & \cdots & A_{st}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $k$
\end_inset
为数, 则
\begin_inset Formula $kA=\begin{bmatrix}kA_{11} & \cdots & kA_{1t}\\
\vdots & \ddots & \vdots\\
kA_{s1} & \cdots & kA_{st}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Enumerate
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $m\times l$
\end_inset
矩阵,
\begin_inset Formula $B$
\end_inset
为
\begin_inset Formula $l\times n$
\end_inset
矩阵, 分块成
\begin_inset Formula
\[
A=\begin{bmatrix}A_{11} & \cdots & A_{1t}\\
\vdots & \ddots & \vdots\\
A_{s1} & \cdots & A_{st}
\end{bmatrix},\ B=\begin{bmatrix}B_{11} & \cdots & B_{1r}\\
\vdots & \ddots & \vdots\\
B_{t1} & \cdots & B_{tr}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset Formula $A_{p1},A_{p2},\cdots,A_{pt}$
\end_inset
的列数分别等于
\begin_inset Formula $B_{1q},B_{2q},\cdots,B_{tq}$
\end_inset
的行数, 则
\begin_inset Formula
\[
AB=\begin{bmatrix}C_{11} & \cdots & C_{1r}\\
\vdots & \ddots & \vdots\\
C_{s1} & \cdots & C_{sr}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset Formula $C_{pq}=\sum_{k=1}^{t}A_{pk}B_{kq}$
\end_inset
, (
\begin_inset Formula $p=1,2,\cdots,s;\ q=1,2,\cdots,r$
\end_inset
).
\end_layout
\begin_layout Enumerate
\series bold
分块矩阵的转置
\series default
, 设
\begin_inset Formula $A=\begin{bmatrix}A_{11} & \cdots & A_{1t}\\
\vdots & \ddots & \vdots\\
A_{s1} & \cdots & A_{st}
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula $A^{T}=\begin{bmatrix}A_{11}^{T} & \cdots & A_{s1}^{T}\\
\vdots & \ddots & \vdots\\
A_{1t}^{T} & \cdots & A_{st}^{T}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Enumerate
设
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶矩阵, 若
\begin_inset Formula $A$
\end_inset
的分块矩阵只有在对角线上有非零子块, 其余子块都为零矩阵, 且在对角线上的子块都是方阵, 即
\begin_inset Formula
\[
A=\begin{bmatrix}A_{1} & & & O\\
& A_{2}\\
& & \ddots\\
O & & & A_{s}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset Formula $A_{i}$
\end_inset
(
\begin_inset Formula $i=1,2,\cdots,s$
\end_inset
) 都是方阵, 则称
\begin_inset Formula $A$
\end_inset
为
\series bold
分块对角矩阵
\series default
.
\begin_inset Newline newline
\end_inset
分块对角矩阵具有以下性质:
\end_layout
\begin_deeper
\begin_layout Enumerate
若
\begin_inset Formula $\left|A_{i}\right|\neq0$
\end_inset
, (
\begin_inset Formula $i=1,2,\cdots,s$
\end_inset
), 则
\begin_inset Formula $|A|\neq0$
\end_inset
, 且
\begin_inset Formula $|A|=\left|A_{1}\right|\left|A_{2}\right|\cdots\left|A_{s}\right|$
\end_inset
;
\end_layout
\begin_layout Enumerate
\begin_inset Formula $A^{-1}=\begin{bmatrix}A_{1}^{-1} & & & O\\
& A_{2}^{-1}\\
& & \ddots\\
O & & & A_{s}^{-1}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Enumerate
同结构的对角分块矩阵的和、差、积、商仍是对角分块矩阵.
且运算表现为对应子块的运算.
\end_layout
\end_deeper
\begin_layout Enumerate
形如
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{bmatrix}A_{11} & A_{12} & \cdots & A_{1s}\\
0 & A_{22} & \cdots & A_{2s}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & A_{ss}
\end{bmatrix}\text{ 或 }\begin{bmatrix}A_{11} & 0 & \cdots & 0\\
A_{21} & A_{22} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
A_{s1} & A_{s2} & \cdots & A_{ss}
\end{bmatrix}
\]
\end_inset
的分块矩阵, 分别称为
\series bold
上三角分块矩阵
\series default
或
\series bold
下三角分块矩阵
\series default
, 其中
\begin_inset Formula $A_{pp}$
\end_inset
(
\begin_inset Formula $p=1,2,\cdots,s$
\end_inset
) 是方阵.
同结构的上 (下) 三角分块矩阵的和、差、积、商仍是上 (下) 三角分块矩阵.
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1 & 3\\
0 & 1 & 2 & 4\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 2 & 0 & 0\\
2 & 0 & 0 & 0\\
6 & 3 & 1 & 0\\
0 & -2 & 0 & 1
\end{bmatrix}$
\end_inset
, 用分块矩阵计算
\begin_inset Formula $kA,A+B$
\end_inset
.
\end_layout
\begin_layout Solution*
将矩阵计算
\begin_inset Formula $A,B$
\end_inset
分块如下:
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 0 & 1 & 3\\
0 & 1 & 2 & 4\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{bmatrix}=\begin{bmatrix}E & C\\
O & -E
\end{bmatrix},\quad B=\begin{bmatrix}1 & 2 & 0 & 0\\
2 & 0 & 0 & 0\\
6 & 3 & 1 & 0\\
0 & -2 & 0 & 1
\end{bmatrix}=\begin{bmatrix}D & O\\
F & E
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\color gray
将矩阵计算
\begin_inset Formula $A,B$
\end_inset
分块如下:
\end_layout
\begin_layout Solution*
\color gray
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 0 & 1 & 3\\
0 & 1 & 2 & 4\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{bmatrix}=\begin{bmatrix}E & C\\
O & -E
\end{bmatrix},\quad B=\begin{bmatrix}1 & 2 & 0 & 0\\
2 & 0 & 0 & 0\\
6 & 3 & 1 & 0\\
0 & -2 & 0 & 1
\end{bmatrix}=\begin{bmatrix}D & O\\
F & E
\end{bmatrix},
\]
\end_inset
\color inherit
则
\begin_inset Formula $A=k\begin{bmatrix}E & C\\
O & -E
\end{bmatrix}=\begin{bmatrix}kE & kC\\
O & -kE
\end{bmatrix}=\begin{bmatrix}k & 0 & k & 3k\\
0 & k & 2k & 4k\\
0 & 0 & -k & 0\\
0 & 0 & 0 & -k
\end{bmatrix}$
\end_inset
,
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
A+B=\begin{bmatrix}E & C\\
O & -E
\end{bmatrix}+\begin{bmatrix}D & O\\
F & E
\end{bmatrix}=\begin{bmatrix}E+D & C\\
F & O
\end{bmatrix}=\begin{bmatrix}2 & 2 & 1 & 3\\
2 & 1 & 2 & 4\\
6 & 3 & 0 & 0\\
0 & -2 & 0 & 0
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-1 & 2 & 1 & 0\\
1 & 1 & 0 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}1 & 0 & 1 & 0\\
-1 & 2 & 0 & 1\\
1 & 0 & 4 & 1\\
-1 & -1 & 2 & 0
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $AB$
\end_inset
.
\end_layout
\begin_layout Solution*
把
\begin_inset Formula $A,B$
\end_inset
分块成
\begin_inset Formula
\[
A=\begin{bmatrix}E & O\\
A_{1} & E
\end{bmatrix},\ B=\begin{bmatrix}B_{11} & E\\
B_{21} & B_{22}
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\color gray
把
\begin_inset Formula $A,B$
\end_inset
分块成
\begin_inset Formula
\[
A=\begin{bmatrix}E & O\\
A_{1} & E
\end{bmatrix},\ B=\begin{bmatrix}B_{11} & E\\
B_{21} & B_{22}
\end{bmatrix},
\]
\end_inset
\color inherit
则
\begin_inset Formula $AB=\begin{bmatrix}E & O\\
A_{1} & E
\end{bmatrix}\cdot\begin{bmatrix}B_{11} & E\\
B_{21} & B_{22}
\end{bmatrix}=\begin{bmatrix}B_{11} & E\\
A_{1}B_{11}+B_{21} & A_{1}+B_{22}
\end{bmatrix}$
\end_inset
.
又
\begin_inset Formula
\begin{align*}
A_{1}B_{11}+B_{21} & =\begin{bmatrix}-1 & 2\\
1 & 1
\end{bmatrix}\begin{bmatrix}1 & 0\\
-1 & 2
\end{bmatrix}+\begin{bmatrix}1 & 0\\
-1 & -1
\end{bmatrix}=\begin{bmatrix}-3 & 4\\
0 & 2
\end{bmatrix}+\begin{bmatrix}1 & 0\\
-1 & -1
\end{bmatrix}=\begin{bmatrix}-2 & 4\\
-1 & 1
\end{bmatrix}\\
A_{1}+B_{22} & =\begin{bmatrix}-1 & 2\\
1 & 1
\end{bmatrix}+\begin{bmatrix}4 & 1\\
2 & 0
\end{bmatrix}=\begin{bmatrix}3 & 3\\
3 & 1
\end{bmatrix},\\
AB & =\begin{bmatrix}B_{11} & E\\
A_{1}B_{11}+B_{21} & A_{1}+B_{22}
\end{bmatrix}=\begin{bmatrix}1 & 0 & 1 & 0\\
-1 & 2 & 0 & 1\\
-2 & 4 & 3 & 3\\
-1 & 1 & 3 & 1
\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 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset ERT
status open
\begin_layout Plain Layout
$A=
\backslash
begin{bNiceArray}{cc|c|cc}
\end_layout
\begin_layout Plain Layout
1 & 0 & 2 & -1 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0 & 1 & 1 & -2 & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hline
\end_layout
\begin_layout Plain Layout
0 & 0 & 3 & 1 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
1 & 0 & -2 & 0 & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$, $B=
\backslash
begin{bNiceArray}{cc|c}
\end_layout
\begin_layout Plain Layout
1 & 0 & 2
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0 & 1 & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hline
\end_layout
\begin_layout Plain Layout
-1 & 1 & 3
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
hline
\end_layout
\begin_layout Plain Layout
2 & 0 & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}$,
\end_layout
\end_inset
也可写为
\begin_inset Formula
\[
A=\begin{bmatrix}A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}
\end{bmatrix},\ B=\begin{bmatrix}B_{11} & B_{12}\\
B_{21} & B_{22}\\
B_{31} & B_{32}
\end{bmatrix}.
\]
\end_inset
设
\begin_inset Formula $C=AB$
\end_inset
, 易见
\begin_inset Formula $C$
\end_inset
是
\begin_inset Formula $2\times2$
\end_inset
分块矩阵, 若记
\begin_inset Formula $C=\begin{bmatrix}C_{11} & C_{12}\\
C_{21} & C_{22}
\end{bmatrix}$
\end_inset
, 则
\begin_inset Formula
\begin{align*}
C_{11} & =A_{11}B_{11}+A_{12}B_{21}+A_{13}B_{31}\\
& =\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}+\begin{bmatrix}2\\
1
\end{bmatrix}\begin{bmatrix}-1 & 1\end{bmatrix}+\begin{bmatrix}-1 & 0\\
-2 & 1
\end{bmatrix}\begin{bmatrix}0 & 1\\
2 & 0
\end{bmatrix}\\
& =\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}+\begin{bmatrix}-2 & 2\\
-1 & 1
\end{bmatrix}+\begin{bmatrix}0 & -1\\
2 & -2
\end{bmatrix}=\begin{bmatrix}-1 & 1\\
1 & 0
\end{bmatrix},
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example
\begin_inset Formula $A=\begin{bmatrix}A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}B_{11} & B_{12}\\
B_{21} & B_{22}\\
B_{31} & B_{32}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $C=\begin{bmatrix}C_{11} & C_{12}\\
C_{21} & C_{22}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula
\begin{align*}
C_{12} & =A_{11}B_{12}+A_{12}B_{22}+A_{13}B_{32}\\
& =\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}0\\
2
\end{bmatrix}+\begin{bmatrix}2\\
1
\end{bmatrix}(3)+\begin{bmatrix}-1 & 0\\
-2 & 1
\end{bmatrix}\begin{bmatrix}-1\\
1
\end{bmatrix}\\
& =\begin{bmatrix}2\\
0
\end{bmatrix}+\begin{bmatrix}6\\
3
\end{bmatrix}+\begin{bmatrix}1\\
3
\end{bmatrix}=\begin{bmatrix}9\\
6
\end{bmatrix},
\end{align*}
\end_inset
\begin_inset Formula
\begin{align*}
C_{21} & =A_{21}B_{11}+A_{22}B_{21}+A_{23}B_{31}\\
& =\begin{bmatrix}0 & 0\\
1 & 1
\end{bmatrix}\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}+\begin{bmatrix}3\\
-2
\end{bmatrix}\begin{bmatrix}-1 & 1\end{bmatrix}+\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}0 & 1\\
2 & 0
\end{bmatrix}\\
& =\begin{bmatrix}0 & 0\\
1 & 0
\end{bmatrix}+\begin{bmatrix}-3 & 3\\
2 & -2
\end{bmatrix}+\begin{bmatrix}0 & 1\\
2 & 0
\end{bmatrix}=\begin{bmatrix}-3 & 4\\
5 & -2
\end{bmatrix},
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example
\begin_inset Formula $A=\begin{bmatrix}A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}B_{11} & B_{12}\\
B_{21} & B_{22}\\
B_{31} & B_{32}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $C=\begin{bmatrix}C_{11} & C_{12}\\
C_{21} & C_{22}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula
\begin{align*}
C_{22} & =A_{21}B_{12}+A_{22}B_{22}+A_{23}B_{32}\\
& =\begin{bmatrix}0 & 0\\
1 & 0
\end{bmatrix}\begin{bmatrix}2\\
0
\end{bmatrix}+\begin{bmatrix}3\\
-2
\end{bmatrix}(3)+\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}-1\\
1
\end{bmatrix}=\begin{bmatrix}8\\
-3
\end{bmatrix},
\end{align*}
\end_inset
则
\begin_inset Formula $C_{11}=\begin{bmatrix}-1 & 1\\
1 & 0
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $C_{12}=\begin{bmatrix}9\\
6
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $C_{21}=\begin{bmatrix}-3 & 4\\
5 & -2
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $C_{22}=\begin{bmatrix}8\\
-3
\end{bmatrix}$
\end_inset
,
\end_layout
\begin_layout Example
于是
\begin_inset Formula $C=\begin{bmatrix}C_{11} & C_{12}\\
C_{21} & C_{22}
\end{bmatrix}=\begin{bmatrix}-1 & 1 & 9\\
1 & 0 & 6\\
-3 & 4 & 8\\
5 & -2 & -3
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\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 $\star\star\star\star\star$
\end_inset
\end_layout
\end_inset
如果将矩阵
\begin_inset Formula $A_{m\times n}$
\end_inset
,
\begin_inset Formula $E_{n}$
\end_inset
分块为
\begin_inset ERT
status open
\begin_layout Plain Layout
$$A=
\backslash
begin{bNiceArray}{c|c|c|c}
\end_layout
\begin_layout Plain Layout
a_{11} & a_{12} &
\backslash
cdots & a_{1n}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{21} & a_{22} &
\backslash
cdots & a_{2n}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
vdots &
\backslash
vdots &
\backslash
ddots &
\backslash
vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
a_{m1} & a_{m2} &
\backslash
cdots & a_{mn}
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}=
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
A_1 & A_2 &
\backslash
cdots & A_n
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix},
\backslash
E=
\backslash
begin{bNiceArray}{c|c|c|c}
\end_layout
\begin_layout Plain Layout
1 & 0 &
\backslash
cdots & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0 & 1 &
\backslash
cdots & 0
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
vdots &
\backslash
vdots &
\backslash
ddots &
\backslash
vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
0 & 0 &
\backslash
cdots & 1
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceArray}=
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
\backslash
varepsilon_1 &
\backslash
varepsilon_2 &
\backslash
cdots &
\backslash
varepsilon_n
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix},$$
\end_layout
\end_inset
则
\begin_inset Formula
\[
\begin{aligned}AE_{n} & =A\begin{bmatrix}\varepsilon_{1} & \varepsilon_{2} & \cdots & \varepsilon_{n}\end{bmatrix}=\begin{bmatrix}A\varepsilon_{1} & A\varepsilon_{2} & \cdots & A\varepsilon_{n}\end{bmatrix}=\begin{bmatrix}A_{1} & A_{2} & \cdots & A_{n}\end{bmatrix}\\
& \Rightarrow A\varepsilon_{j}=A_{j},\quad(j=1,2,\cdots,n).
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Remark
已知点
\begin_inset Formula $P(x,y)$
\end_inset
, 点
\begin_inset Formula $P'$
\end_inset
是点
\begin_inset Formula $P$
\end_inset
绕原点
\begin_inset Formula $O$
\end_inset
顺时针旋转
\begin_inset Formula $\theta$
\end_inset
角得到的, 求点
\begin_inset Formula $P'$
\end_inset
的坐标.
先用复数法推导, 设点
\begin_inset Formula $P$
\end_inset
的复坐标为
\begin_inset Formula $z=x+\ui y$
\end_inset
, 旋转后
\begin_inset Formula $P'$
\end_inset
的复坐标为
\begin_inset Formula $z'=x'+\ui y'$
\end_inset
, 由于
\begin_inset Formula $P'$
\end_inset
为
\begin_inset Formula $P$
\end_inset
绕原点
\begin_inset Formula $O$
\end_inset
顺时针旋转
\begin_inset Formula $\theta$
\end_inset
得到, 所以
\begin_inset Formula
\[
z'=z\ue^{-\ui\theta}\Longrightarrow x'+\ui y'=(x+\ui y)\ue^{-\ui\theta}=\left(x+\ui y\right)\left(\cos\theta-\ui\sin\theta\right),
\]
\end_inset
比较实虚部便得
\begin_inset Formula
\[
\begin{cases}
x'=x\cos\theta+y\sin\theta\\
y'=-x\sin\theta+y\cos\theta
\end{cases}\Longrightarrow\begin{bmatrix}x'\\
y'
\end{bmatrix}=\begin{bmatrix}\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}\begin{bmatrix}x\\
y
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Remark
根据前面的表述, 矩阵
\begin_inset Formula $A=\begin{bmatrix}\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}$
\end_inset
为旋转矩阵, 将其每一列进行分块
\begin_inset Formula $A=\begin{bmatrix}A_{1} & A_{2}\end{bmatrix}$
\end_inset
, 则有
\begin_inset Formula $A_{1}=A\varepsilon_{1}$
\end_inset
,
\begin_inset Formula $A_{2}=A\varepsilon_{2}$
\end_inset
.
从几何上来看, 矩阵的列
\begin_inset Formula $A_{1},A_{2}$
\end_inset
分别是两个坐标轴上的单位向量
\begin_inset Formula $\varepsilon_{1},\varepsilon_{2}$
\end_inset
经过旋转矩阵
\begin_inset Formula $A$
\end_inset
左乘得到的点, 也就是经过顺时针旋转
\begin_inset Formula $\theta$
\end_inset
角之后的位置.
使用类似的方法, 可以给出逆时针旋转
\begin_inset Formula $\theta$
\end_inset
角的旋转矩阵.
\end_layout
\begin_layout Remark*
\begin_inset ERT
status open
\begin_layout Plain Layout
{
\backslash
color{red}{
\end_layout
\end_inset
矩阵按行 (列) 分块是最常见的一种分块方法.
一般地,
\begin_inset Formula $m\times n$
\end_inset
矩阵
\begin_inset Formula $A$
\end_inset
有
\begin_inset Formula $m$
\end_inset
行, 称为
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $m$
\end_inset
个行向量
\series default
, 若第
\begin_inset Formula $i$
\end_inset
行记作
\begin_inset Formula
\[
\alpha_{i}^{T}=\left(\alpha_{i1},\alpha_{i2},\cdots,\alpha_{in}\right),
\]
\end_inset
\end_layout
\begin_layout Remark*
则矩阵
\begin_inset Formula $A$
\end_inset
就可表示为
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\end_layout
\begin_layout Remark*
\begin_inset Formula
\[
A=\begin{bmatrix}\alpha_{1}^{T}\\
\alpha_{2}^{T}\\
\vdots\\
\alpha_{m}^{T}
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Remark*
\begin_inset Formula $m\times n$
\end_inset
矩阵
\begin_inset Formula $A$
\end_inset
有
\begin_inset Formula $n$
\end_inset
列, 称为
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
的
\begin_inset Formula $n$
\end_inset
个列向量
\series default
, 若第
\begin_inset Formula $j$
\end_inset
列记作
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\end_layout
\begin_layout Remark*
\begin_inset Formula
\[
\alpha_{j}=\begin{bmatrix}\alpha_{1j}\\
\alpha_{2j}\\
\vdots\\
\alpha_{mj}
\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Remark*
则
\begin_inset Formula $A=\left(\alpha_{1},\alpha_{2},\cdots,\alpha_{n}\right)$
\end_inset
.
\begin_inset ERT
status open
\begin_layout Plain Layout
}}
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A$
\end_inset
是一个
\begin_inset Formula $m\times n$
\end_inset
矩阵,
\begin_inset Formula $B$
\end_inset
是一个
\begin_inset Formula $n\times l$
\end_inset
矩阵, 同样, 可对
\begin_inset Formula $A$
\end_inset
作行分块, 即将
\begin_inset Formula $A$
\end_inset
的每一行作为一块, 则
\begin_inset Formula
\[
A=\begin{bmatrix}\alpha_{1}\\
\alpha_{2}\\
\vdots\\
\alpha_{m}
\end{bmatrix},
\]
\end_inset
其中
\begin_inset Formula $\alpha_{i}=\left(\alpha_{i1},\alpha_{i2},\cdots,\alpha_{in}\right)$
\end_inset
, (
\begin_inset Formula $i=1,2,\cdots,m$
\end_inset
), 是
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行.
这时也将
\begin_inset Formula $B$
\end_inset
看成
\begin_inset Formula $1\times1$
\end_inset
分块矩阵, 则有
\begin_inset Formula
\[
AB=\begin{bmatrix}\alpha_{1}B\\
\alpha_{2}B\\
\vdots\\
\alpha_{4}B
\end{bmatrix}.
\]
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:4.2-8"
\end_inset
设有二个分块对角阵:
\begin_inset Formula
\[
A=\begin{bmatrix}A_{1} & & & 0\\
& A_{2}\\
& & \ddots\\
0 & & & A_{k}
\end{bmatrix},\ B=\begin{bmatrix}B_{1} & & & 0\\
& B_{2}\\
& & \ddots\\
0 & & & B_{k}
\end{bmatrix}.
\]
\end_inset
其中矩阵
\begin_inset Formula $A_{i}$
\end_inset
与
\begin_inset Formula $B_{i}$
\end_inset
都是
\begin_inset Formula $n_{i}$
\end_inset
阶方阵 (因此
\begin_inset Formula $A,B$
\end_inset
是同阶方阵), 因此
\begin_inset Formula $A_{i}$
\end_inset
与
\begin_inset Formula $B_{i}$
\end_inset
可以相乘, 用分块矩阵的乘法不难求得
\begin_inset Formula
\[
AB=\begin{bmatrix}A_{1}B_{1} & & & 0\\
& A_{2}B_{2}\\
& & \ddots\\
0 & & & A_{k}B_{k}
\end{bmatrix},
\]
\end_inset
即分块对角阵相乘时只需将主对角线上的块乘起来即可.
\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
分块矩阵的运算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A$
\end_inset
是一个分块对角矩阵:
\begin_inset Formula $A=\begin{bmatrix}A_{1} & & & O\\
& A_{2}\\
& & \ddots\\
O & & & A_{k}
\end{bmatrix}$
\end_inset
, 且每块
\begin_inset Formula $A_{i}$
\end_inset
都是非奇异方阵 (因此
\begin_inset Formula $A$
\end_inset
也是方阵), 则
\begin_inset Formula $A$
\end_inset
也是非奇异方阵且
\begin_inset Formula $A^{-1}=\begin{bmatrix}A_{1}^{-1} & & & O\\
& A_{2}^{-1}\\
& & \ddots\\
O & & & A_{k}^{-1}
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
事实是由例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:4.2-8"
plural "false"
caps "false"
noprefix "false"
\end_inset
知
\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
\begin_inset Formula
\[
A\begin{bmatrix}A_{1}^{-1} & & & 0\\
& A_{2}^{-1}\\
& & \ddots\\
0 & & & A_{k}^{-1}
\end{bmatrix}=\begin{bmatrix}A_{1}A_{1}^{-1}\\
& A_{2}A_{2}^{-1}\\
& & \ddots\\
& & & A_{k}A_{k}^{-1}
\end{bmatrix}=\begin{bmatrix}E_{n_{1}}\\
& E_{n_{2}}\\
& & \ddots\\
& & & E_{n_{k}}
\end{bmatrix}=E,
\]
\end_inset
其中
\begin_inset Formula $E_{n_{k}}$
\end_inset
表示与
\begin_inset Formula $A_{k}$
\end_inset
同阶的单位阵, 一个分块对角阵主对角线上的块都是单位阵, 则它自己也是一个单位阵, 故
\begin_inset Formula
\[
A^{-1}=\begin{bmatrix}A_{1}^{-1} & & & O\\
& A_{2}^{-1}\\
& & \ddots\\
O & & & A_{k}^{-1}
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
用分块矩阵求逆矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}5 & 0 & 0\\
0 & 3 & 1\\
0 & 2 & 1
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A^{-1}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $A=\begin{bmatrix}5 & 0 & 0\\
0 & 3 & 1\\
0 & 2 & 1
\end{bmatrix}=\begin{bmatrix}A_{1} & O\\
O & A_{2}
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $A_{1}=(5)$
\end_inset
,
\begin_inset Formula $A_{2}=\begin{bmatrix}3 & 1\\
2 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $A_{1}^{-1}=\left(\frac{1}{5}\right)$
\end_inset
,
\begin_inset Formula $A_{2}^{-1}=\begin{bmatrix}1 & -1\\
-2 & 3
\end{bmatrix}$
\end_inset
; 所以
\begin_inset Formula
\[
A^{-1}=\begin{bmatrix}A_{1}^{-1} & O\\
O & A_{2}^{-1}
\end{bmatrix}=\begin{bmatrix}\frac{1}{5} & 0 & 0\\
0 & 1 & -1\\
0 & -2 & 3
\end{bmatrix}.
\]
\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
如何证明矩阵是零矩阵 (
\strikeout on
矩阵的迹
\strikeout default
)
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知实矩阵
\begin_inset Formula $A$
\end_inset
满足
\begin_inset Formula $A^{T}A=O$
\end_inset
, 证明
\begin_inset Formula $A=O$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
设
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
, 把
\begin_inset Formula $A$
\end_inset
用列向量表示为
\begin_inset Formula $A=\left(\alpha_{1},\alpha_{2},\cdots,\alpha_{n}\right)$
\end_inset
, 则
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
A^{T}A=\begin{bmatrix}\alpha_{1}^{T}\\
\alpha_{2}^{T}\\
\vdots\\
\alpha_{n}^{T}
\end{bmatrix}\left(\alpha_{1},\alpha_{2},\cdots,\alpha_{n}\right)=\begin{bmatrix}\alpha_{1}^{T}\alpha_{1} & \alpha_{1}^{T}\alpha_{2} & \cdots & \alpha_{1}^{T}\alpha_{n}\\
\alpha_{2}^{T}\alpha_{1} & \alpha_{2}^{T}\alpha_{2} & \cdots & \alpha_{2}^{T}\alpha_{n}\\
\vdots & \vdots & \ddots & \vdots\\
\alpha_{n}^{T}\alpha_{1} & \alpha_{n}^{T}\alpha_{2} & \cdots & \alpha_{n}^{T}\alpha_{n}
\end{bmatrix},
\]
\end_inset
即
\begin_inset Formula $A^{T}A$
\end_inset
的
\begin_inset Formula $(i,j)$
\end_inset
元为
\begin_inset Formula $\alpha_{i}^{T}\alpha_{j}$
\end_inset
, 因
\begin_inset Formula $A^{T}A=O$
\end_inset
, 故
\begin_inset Formula $\alpha_{i}^{T}\alpha_{j}=0$
\end_inset
, (
\begin_inset Formula $i,j=1,2,\cdots,n$
\end_inset
).
\end_layout
\begin_layout Proof
特别地, 有
\begin_inset Formula $\alpha_{j}^{T}\alpha_{j}=0$
\end_inset
, (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset
), 而
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\alpha_{j}^{T}\alpha_{j}=\left(a_{1j},a_{2j},\cdots,a_{mj}\right)\begin{bmatrix}a_{1j}\\
a_{2j}\\
\vdots\\
a_{mj}
\end{bmatrix}=a_{1j}^{2}+a_{2j}^{2}+\cdots+a_{mj}^{2},
\]
\end_inset
由
\begin_inset Formula $a_{1j}^{2}+a_{2j}^{2}+\cdots+a_{mj}^{2}=0$
\end_inset
, (因
\begin_inset Formula $a_{ij}$
\end_inset
为实数) 得
\begin_inset Formula $a_{1j}=a_{2j}=\cdots=a_{mj}=0$
\end_inset
, (
\begin_inset Formula $j=1,2,\cdots,n$
\end_inset
), 即
\begin_inset Formula $A=O$
\end_inset
.
证毕.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\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 $A=\begin{bmatrix}a & 1 & 0 & 0\\
0 & a & 0 & 0\\
0 & 0 & b & 1\\
0 & 0 & 1 & b
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}a & 0 & 0 & 0\\
1 & a & 0 & 0\\
0 & 0 & b & 0\\
0 & 0 & 1 & b
\end{bmatrix}$
\end_inset
, 用分块矩阵的方式求
\begin_inset Formula $ABA$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $A$
\end_inset
和
\begin_inset Formula $B$
\end_inset
是
\begin_inset Formula $n$
\end_inset
阶方阵, 计算
\begin_inset Formula
\[
\begin{bmatrix}O & E_{n}\\
E_{n} & O
\end{bmatrix}\begin{bmatrix}A & O\\
O & B
\end{bmatrix}\begin{bmatrix}O & E_{n}\\
E_{n} & O
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设方阵
\begin_inset Formula $A$
\end_inset
和
\begin_inset Formula $B$
\end_inset
可交换, 求
\begin_inset Formula $\begin{bmatrix}A & B\\
O & A
\end{bmatrix}^{n}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设方阵
\begin_inset Formula $A$
\end_inset
和
\begin_inset Formula $B$
\end_inset
分别为
\begin_inset Formula $p$
\end_inset
阶方阵和
\begin_inset Formula $q$
\end_inset
阶方阵, 证明:
\begin_inset Formula
\[
\begin{bmatrix}A & E\\
E & O
\end{bmatrix}\begin{bmatrix}B & E\\
O & -A
\end{bmatrix}=\begin{bmatrix}AB & O\\
B & E
\end{bmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
分块方阵
\begin_inset Formula $D=\begin{bmatrix}A & C\\
O & B
\end{bmatrix}$
\end_inset
, 其中
\begin_inset Formula $A,B$
\end_inset
均为可逆方阵, 证明
\begin_inset Formula $D$
\end_inset
可逆, 并求
\begin_inset Formula $D^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
\begin_inset CommandInset label
LatexCommand label
name "prob:2.4-19"
\end_inset
称方阵
\begin_inset Formula
\[
C=\begin{bmatrix}0 & 1 & 0 & \cdots & 0 & 0\\
0 & 0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & 0\\
0 & 0 & 0 & \cdots & 0 & 1\\
1 & 0 & 0 & \cdots & 0 & 0
\end{bmatrix}_{n\times n}=\begin{bmatrix}0 & E_{n-1}\\
1 & 0
\end{bmatrix}
\]
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶循环移位矩阵.
证明:
\end_layout
\begin_layout Problem
(1).
用
\begin_inset Formula $C$
\end_inset
左乘矩阵, 相当于将这个矩阵的行向上移一行, 而第一行移到最后一行; 用
\begin_inset Formula $C$
\end_inset
右乘矩阵, 相当于将这个矩阵的列向右移一列, 而最后一列移到第一列;
\end_layout
\begin_layout Problem
(2).
\begin_inset Formula $C^{k}=\begin{cases}
\begin{bmatrix}O & E_{n-k}\\
E_{k} & O
\end{bmatrix}, & k=1,2,\ldots,n-1,\\
E_{n}, & k=n.
\end{cases}$
\end_inset
(接下一页.)
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "prob:2.4-19"
plural "false"
caps "false"
noprefix "false"
\end_inset
\end_layout
\end_inset
(3).
设
\begin_inset Formula $J$
\end_inset
是所有元素为
\begin_inset Formula $1$
\end_inset
的
\begin_inset Formula $n$
\end_inset
阶方阵, 则
\begin_inset Formula
\[
E+C+C^{2}+\cdots+C^{n-1}=J;
\]
\end_inset
\end_layout
\begin_layout Problem*
(4).
设
\begin_inset Formula
\[
A=\begin{bmatrix}a_{1} & a_{2} & \cdots & a_{n-1} & a_{n}\\
a_{n} & a_{1} & \cdots & a_{n-2} & a_{n-1}\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_{3} & a_{4} & \cdots & a_{1} & a_{2}\\
a_{2} & a_{3} & \cdots & a_{n} & a_{1}
\end{bmatrix}
\]
\end_inset
称
\begin_inset Formula $A$
\end_inset
是循环矩阵, 则
\begin_inset Formula
\[
A=a_{1}E+a_{2}C+a_{3}C^{2}+\cdots+a_{n}C^{n-1};
\]
\end_inset
\end_layout
\begin_layout Problem*
(5) 两个
\begin_inset Formula $n$
\end_inset
阶循环阵的乘积仍是循环阵;
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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