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% Title page details
\title{{\color{red} 1.4. 分块矩阵}}
%\subtitle{}
\author{} 
\date{}


\begin{document}
\begin{frame}
% Print the title page as the first slide
\titlepage
\end{frame}

\begin{frame}
    
给定矩阵 
\[\mA=\begin{pmatrix}
  1 & 0 & 0 &0\\
  0 & 1 & 0 & 0\\
  11 & 123 & 1& 0\\
  453 & 987 & 0 &1
\end{pmatrix},\]
如何计算$\mA^{1\,000}$
\end{frame}
 
\begin{frame}
   
\[\mA=\left(\begin{array}{cc|cc}
1 & 0 & 0 &0\\
0 & 1 & 0 & 0\\
\hline
11 & 123 & 1& 0\\
453 & 987 & 0 &1
\end{array}\right),\]
如果令
\[\mX=\begin{pmatrix}
11 & 123 \\
543 & 987
\end{pmatrix}
\]
那么$\mA$由四个矩阵拼接而成： 
\[\mA=\left(\begin{array}{cc}
  \mI & \mO\\
  \mX & \mI
\end{array}
\right).\]
\end{frame}
 
\begin{frame}
   
下面带问号的等式成立吗？

\[\begin{aligned}
    & \left(\!\begin{array}{cc}
  \mI & \mO\\
  \mX & \mI
\end{array}
\right) \left(\begin{array}{cc}
  \mI & \mO\\
  \mX & \mI
\end{array}
\right)\\
 \stackrel{?}{=} & \left(\begin{array}{cc}
  \mI\mI+\mO\mX & \mI\mO+\mO\mI\\
  \mX\mI+\mI\mX & \mX\mO+\mI\mI
\end{array}
\right)\\
  = & \left(\!\begin{array}{cc}
  \mI & \mO\\
2\mX & \mI
\end{array}
\right)
\end{aligned}\]

如果成立的话，重复这个过程可以得到 
\[\mA^{1\,000}=\begin{pmatrix}
  \mI & \mO\\
1\,000\mX & \mI
\end{pmatrix}.\]
\end{frame}

\begin{frame}

实际上，上面的等式的确是成立的！

\begin{Def}\label{def-partitioned-matrix}
在一个矩阵$\mA$的行和列之间横平竖直地加上一些线，将其分成很多小块矩阵的形式称为{\em 分块矩阵}(partitioned matrix)。 一个矩阵可以根据需要进行不同的分块。分块矩阵的一般形式如下.
\[\mA= 
  \begin{blockarray}{ccccc}
    &n_{1} &n_2&\ldots &n_r\\
    \begin{block}{c(cccc)}
      m_{1}     &\mA_{11} &  \mA_{12}  & \ldots & \mA_{1r}\\
      m_{2} & \mA_{21} &  \mA_{22} & \ldots & \mA_{2r}\\
     \vdots & \vdots & \vdots &  & \vdots\\
    m_{s}  & \mA_{s1}  &   \mA_{s2}       &\ldots & \mA_{sr}\\
    \end{block}
  \end{blockarray},\]
其中矩阵$\mA_{ij}$是$m_i$行$n_j$列矩阵。
\end{Def} 
\end{frame}


\begin{frame}{分块矩阵的乘法}
   
对于分块矩阵
\[\mA= 
\begin{blockarray}{ccccc}
  &n_{1} &n_2&\ldots &n_r\\
  \begin{block}{c(cccc)}
    m_{1}     &\mA_{11} &  \mA_{12}  & \ldots & \mA_{1r}\\
    m_{2} & \mA_{21} &  \mA_{22} & \ldots & \mA_{2r}\\
   \vdots & \vdots & \vdots & & \vdots\\
  m_{s}  & \mA_{s1}  &   \mA_{s2}       &\ldots & \mA_{sr}\\
  \end{block}
\end{blockarray}\;, \mB= 
\begin{blockarray}{ccccc}
  &l_{1} &l_2&\ldots &l_t\\
  \begin{block}{c(cccc)}
    n_{1}     &\mB_{11} &  \mB_{12}  & \ldots & \mB_{1t}\\
    n_{2} & \mB_{21} &  \mB_{22} & \ldots & \mB_{2t}\\
    \vdots & \vdots & \vdots & & \vdots\\
  n_{r}  & \mB_{r1}  &   \mB_{r2}       &\ldots & \mB_{rt}\\
  \end{block}
\end{blockarray}\]
其中$\mA$的列分法与$\mB$的行分法一致. 
\end{frame}

\begin{frame}
   
那么
$$\mA\mB= 
\begin{blockarray}{ccccc}
  &l_{1} &l_2&\ldots &l_t\\
  \begin{block}{c(cccc)}
    m_{1}     &\mC_{11} &  \mC_{12}  & \ldots & \mC_{1t}\\
    m_{2} & \mC_{21} &  \mC_{22} & \ldots & \mC_{2t}\\
   \vdots & \vdots & \vdots &  & \vdots\\
  m_{s}  & \mC_{s1}  &   \mC_{s2}       &\ldots & \mC_{st}\\
  \end{block}
\end{blockarray}\;,$$
其中
$$\mC_{ij}=\sum_{k=1}^{r}\mA_{ik}\mB_{kj}.$$
\end{frame}

\begin{frame}
   
对矩阵进行分块，
  $$\mA=\left(\begin{array}{cc|cc}
    1&0& 3&0\\
    0&1 &0&3\\
    \hline
    0&0&4&2
  \end{array}\right)=\begin{pmatrix}
    \mI_2 & 3\mI_2\\
    \mO & \mA_{22}
  \end{pmatrix}, \quad \mA_{22}=\big(4\;\; 2\big);$$
$$\mB=\left(\begin{array}{c|cc}
  1&1&2\\
  1&3& 1\\
  \hline
  0&1&0\\
  0&0&1
\end{array}\right)=\begin{pmatrix}
  \mB_{11} &\mB_{12}\\
  \mO& \mI_2
\end{pmatrix},\quad \mB_{11}=\begin{pmatrix}
  1\\1
\end{pmatrix},\quad \mB_{12}=\begin{pmatrix}
  1&2\\3&1
\end{pmatrix}.$$
\end{frame}


\begin{frame}
   
$$\begin{aligned} 
\mA\mB & =\begin{pmatrix}
  \mI_2 & 3\mI_2\\
  \mO & \mA_{22}
\end{pmatrix}\begin{pmatrix}
  \mB_{11} &\mB_{12}\\
  \mO& \mI_2
\end{pmatrix}=\begin{pmatrix}
  \mI_2\mB_{11} & \mI_2\mB_{12}+3\mI_2\mI_2\\
  \mO & \mA_{22}\mI_2
\end{pmatrix}\\
& =\begin{pmatrix}
  \mB_{11} & \mB_{12}+3\mI_2\\
  \mO & \mA_{22}
\end{pmatrix}.
\end{aligned}$$
由此得到
$$\mA\mB=\left(
\begin{array}{c|cc}
  1&4& 2\\
  1& 3& 4\\
  \hline 
  0 & 4& 2  
\end{array}
\right).$$
\end{frame}

\begin{frame}{分块矩阵的加法、数乘和转置}
   
对于两个\textbf{\color{red} 分块方式相同}的分块矩阵
$$\mA= 
\begin{blockarray}{ccccc}
  &n_{1} &n_2&\ldots &n_r\\
  \begin{block}{c(cccc)}
    m_{1}     &\mA_{11} &  \mA_{12}  & \ldots & \mA_{1r}\\
    m_{2} & \mA_{21} &  \mA_{22} & \ldots & \mA_{2r}\\
   \vdots & \vdots & \vdots &  & \vdots\\
  m_{s}  & \mA_{s1}  &   \mA_{s2}       &\ldots & \mA_{sr}\\
  \end{block}
\end{blockarray}$$
$$\mB= 
\begin{blockarray}{ccccc}
  &n_{1} &n_2&\ldots &n_r\\
  \begin{block}{c(cccc)}
    m_{1}     &\mB_{11} &  \mB_{12}  & \ldots & \mB_{1r}\\
    m_{2} & \mB_{21} &  \mB_{22} & \ldots & \mB_{2r}\\
    \vdots & \vdots & \vdots &  & \vdots\\
  m_{s}  & \mB_{s1}  &   \mB_{s2}       &\ldots & \mB_{sr}\\
  \end{block}
\end{blockarray}$$
\end{frame}

\begin{frame}{分块矩阵的加法}

$$\mA+\mB=\begin{pmatrix}
  \mA_{11}+\mB_{11} &  \mA_{12}+\mB_{12}  & \cdots & \mA_{1r}+\mB_{1r}\\
  \mA_{21}+\mB_{21} &  \mA_{22}+\mB_{22} & \cdots & \mA_{2r}+\mB_{2r}\\
     \vdots & \vdots &  & \vdots\\
     \mA_{s1}+\mB_{s1}  &   \mA_{s2}+\mB_{s2}       &\cdots & \mA_{sr}+\mB_{sr}\\
  \end{pmatrix}$$
\end{frame}

\begin{frame}{分块矩阵的数乘}

对任意数$\lambda$, 
$$\lambda \mA=\begin{pmatrix}
  \lambda \mA_{11} &  \lambda \mA_{12}  & \cdots & \lambda \mA_{1r}\\
  \lambda \mA_{21} &  \lambda \mA_{22} & \ldots & \lambda \mA_{2r}\\
  \vdots & \vdots &  & \vdots\\
  \lambda \mA_{s1}  &  \lambda \mA_{s2}       &\cdots &\lambda \mA_{sr}\\
\end{pmatrix}$$
\end{frame}


\begin{frame}{分块矩阵的转置}
   

$$\mA\trp=\begin{pmatrix}
   \mA_{11}\trp &  \mA_{21}\trp  & \cdots &  \mA_{s1}\trp\\
  \mA_{12}\trp &   \mA_{22}\trp & \cdots &  \mA_{s2}\trp\\
  \vdots & \vdots &  & \vdots\\
  \mA_{1r}\trp  &   \mA_{2r}\trp       &\cdots & \mA_{sr}\trp\\
\end{pmatrix}$$
\end{frame}


\begin{frame}{分块对角矩阵}

一个{\em 分块对角矩阵}形如
\[\left(
  \begin{array}{cccc}
    \mA_1 &  &   &    \\
     & \mA_2 &   &   \\
     &  & \ddots &   \\
    &   &   &\mA_s \\
  \end{array}
\right),\]
其中$\mA_i$是$n_i$阶方阵, $i=1, 2, \cdots, s$, 其他位置上的分块都是零矩阵.
\end{frame}


\begin{frame}{分块对角矩阵的乘法}
 如果 
\[\mB=\left(
  \begin{array}{cccc}
    \mB_1 &  &  &  \\
     & \mB_2 &   &   \\
    &   & \ddots &   \\
    &   &  &\mB_s \\
  \end{array}
\right)\]
且$\mB_i$也为$n_i$阶方阵，$i=1,2,\cdots, s$，那么
$$
\mA\mB=\left(
  \begin{array}{cccc}
    \mA_1\mB_1 &  &   &  \\
     & \mA_2\mB_2 &  &    \\
    &   &  \ddots&  \\
     &   &   &\mA_s\mB_s \\
  \end{array}
\right),
$$
   \end{frame}
\begin{frame}
   
$$
\mA+\mB=\left(
  \begin{array}{cccc}
    \mA_1+\mB_1 &  &   &  \\
     & \mA_2+\mB_2 &  &    \\
    &   &  \ddots&  \\
     &   &   &\mA_s+\mB_s \\
  \end{array}
\right).
$$
\end{frame}


\end{document}