\bordermatrix 的问题

Question 1

仍然需要一些手动工作，但我希望您的文档中不要有太多这样的对象。

\documentclass{article}
\usepackage{amsmath}
\usepackage{blkarray}

\makeatletter
\newcommand{\tranhoang@uparrowfill}{%
  \par % make sure we're in vertical mode
  \begingroup\offinterlineskip
  \sbox\z@{$\uparrow$}%
  \sbox\tw@{\hskip 0.5em \hb@xt@\wd\z@{\hss$|$\hss}}%
  \hbox{\hskip 0.5em \copy\z@}%
  \kern-4\p@
  \leaders\vbox{\kern-4\p@\copy\tw@\kern-4\p@}\vfill
  \kern-4\p@
  \copy\tw@
  \endgroup
}
\newcommand{\tranhoang@downarrowfill}{%
  \par % make sure we're in vertical mode
  \begingroup\offinterlineskip
  \sbox\z@{$\downarrow$}%
  \sbox\tw@{\hskip 0.5em \hb@xt@\wd\z@{\hss$|$\hss}}%
  \copy\tw@
  \kern-4\p@
  \leaders\vbox{\kern-4\p@\copy\tw@\kern-4\p@}\vfill
  \kern-4\p@
  \hbox{\hskip 0.5em \copy\z@}%
  \endgroup
}
\newcommand{\spanrows}[3][0pt]{%
  \dimen@=#2\dimexpr\ht\@arstrutbox+\dp\@arstrutbox\relax
  \advance\dimen@ #1\relax
  \vbox to \dimen@{
    \kern\p@
    \tranhoang@uparrowfill
    \strut#3%
    \tranhoang@downarrowfill
    \kern\p@
  }%
}
\makeatother
\newdimen\tranhoangwidth

\begin{document}

\[
\settowidth\tranhoangwidth{$n-k$ rows}
\langle k\colon e_{k + 1},\dots, e_n\rangle =
\begin{blockarray}{lllll}
\begin{block}{|llll|\BAmultirow{\tranhoangwidth}}
  x_1 & x_2 & \dots & x_n  & \spanrows{4}{$k$ rows} \\
  x_1 & x_2 & \dots & x_n \\
  \dots  & \dots  & \dots & \dots  \\
  x_1 & x_2 & \dots & x_n \\[0.5ex]
\end{block}
\begin{block}{|llll|\BAmultirow{\tranhoangwidth}}
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dots & y_n^{p^{e_{k + 1}}} &
    \spanrows{3}{$n-k$ rows} \\
  \dots & \dots & \dots  & \dots \\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dots & y_n^{p^{e_n}} \\
\end{block}
\end{blockarray}
\]

\[
\settowidth\tranhoangwidth{$n-k$ rows}% the widest label
\begin{blockarray}{lll@{\quad}llll}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
x_1 & \dots & x_n & x_1 & \dots & x_n & \spanrows{4}{$n+1$ rows} \\
y_1 & \dots & y_n & y_1 & \dots & y_n \\
\dots & \dots & \dots & \dots & \dots & \dots \\
y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} & y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} \\
\end{block}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
&&& x_1 & \dots & x_n & \spanrows{3}{$k-1$ rows} \\
&&& \dots & \dots & \dots \\
&&& x_1 & \dots & x_n \\
\end{block}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
&&& y_1 & \dots & y_n & \spanrows[10pt]{9}{$n-k$ rows} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{s_1-1}} & \dots & y_n^{p^{s_1-1}} \\
&\smash{\raisebox{1ex}{\LARGE$0$}}
  && y_1^{p^{s_1+1}} & \dots & y_n^{p^{s_1+1}} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{s_k-1}} & \dots & y_n^{p^{s_k-1}} \\
&&& y_1^{p^{s_k+1}} & \dots & y_n^{p^{s_k+1}} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} \\
\end{block}
\end{blockarray}
\]

\end{document}

该\spanrows命令有两个参数：要跨越的行数和文本。还有一个可选参数，因此调用

\spanrows[10pt]{4}{text}

如果矩阵具有特殊高度的行，则可选参数中给出的量将添加到高度中。

Answer

仍然需要一些手动工作，但我希望您的文档中不要有太多这样的对象。

\documentclass{article}
\usepackage{amsmath}
\usepackage{blkarray}

\makeatletter
\newcommand{\tranhoang@uparrowfill}{%
  \par % make sure we're in vertical mode
  \begingroup\offinterlineskip
  \sbox\z@{$\uparrow$}%
  \sbox\tw@{\hskip 0.5em \hb@xt@\wd\z@{\hss$|$\hss}}%
  \hbox{\hskip 0.5em \copy\z@}%
  \kern-4\p@
  \leaders\vbox{\kern-4\p@\copy\tw@\kern-4\p@}\vfill
  \kern-4\p@
  \copy\tw@
  \endgroup
}
\newcommand{\tranhoang@downarrowfill}{%
  \par % make sure we're in vertical mode
  \begingroup\offinterlineskip
  \sbox\z@{$\downarrow$}%
  \sbox\tw@{\hskip 0.5em \hb@xt@\wd\z@{\hss$|$\hss}}%
  \copy\tw@
  \kern-4\p@
  \leaders\vbox{\kern-4\p@\copy\tw@\kern-4\p@}\vfill
  \kern-4\p@
  \hbox{\hskip 0.5em \copy\z@}%
  \endgroup
}
\newcommand{\spanrows}[3][0pt]{%
  \dimen@=#2\dimexpr\ht\@arstrutbox+\dp\@arstrutbox\relax
  \advance\dimen@ #1\relax
  \vbox to \dimen@{
    \kern\p@
    \tranhoang@uparrowfill
    \strut#3%
    \tranhoang@downarrowfill
    \kern\p@
  }%
}
\makeatother
\newdimen\tranhoangwidth

\begin{document}

\[
\settowidth\tranhoangwidth{$n-k$ rows}
\langle k\colon e_{k + 1},\dots, e_n\rangle =
\begin{blockarray}{lllll}
\begin{block}{|llll|\BAmultirow{\tranhoangwidth}}
  x_1 & x_2 & \dots & x_n  & \spanrows{4}{$k$ rows} \\
  x_1 & x_2 & \dots & x_n \\
  \dots  & \dots  & \dots & \dots  \\
  x_1 & x_2 & \dots & x_n \\[0.5ex]
\end{block}
\begin{block}{|llll|\BAmultirow{\tranhoangwidth}}
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dots & y_n^{p^{e_{k + 1}}} &
    \spanrows{3}{$n-k$ rows} \\
  \dots & \dots & \dots  & \dots \\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dots & y_n^{p^{e_n}} \\
\end{block}
\end{blockarray}
\]

\[
\settowidth\tranhoangwidth{$n-k$ rows}% the widest label
\begin{blockarray}{lll@{\quad}llll}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
x_1 & \dots & x_n & x_1 & \dots & x_n & \spanrows{4}{$n+1$ rows} \\
y_1 & \dots & y_n & y_1 & \dots & y_n \\
\dots & \dots & \dots & \dots & \dots & \dots \\
y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} & y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} \\
\end{block}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
&&& x_1 & \dots & x_n & \spanrows{3}{$k-1$ rows} \\
&&& \dots & \dots & \dots \\
&&& x_1 & \dots & x_n \\
\end{block}
\begin{block}{|lll@{\quad}lll|\BAmultirow{\tranhoangwidth}}
&&& y_1 & \dots & y_n & \spanrows[10pt]{9}{$n-k$ rows} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{s_1-1}} & \dots & y_n^{p^{s_1-1}} \\
&\smash{\raisebox{1ex}{\LARGE$0$}}
  && y_1^{p^{s_1+1}} & \dots & y_n^{p^{s_1+1}} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{s_k-1}} & \dots & y_n^{p^{s_k-1}} \\
&&& y_1^{p^{s_k+1}} & \dots & y_n^{p^{s_k+1}} \\
&&& \dots & \dots & \dots \\
&&& y_1^{p^{n-1}} & \dots & y_n^{p^{n-1}} \\
\end{block}
\end{blockarray}
\]

\end{document}

该\spanrows命令有两个参数：要跨越的行数和文本。还有一个可选参数，因此调用

\spanrows[10pt]{4}{text}

如果矩阵具有特殊高度的行，则可选参数中给出的量将添加到高度中。

Question 2

这里有一种方法blkarray：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{blkarray, makebox}
\usepackage{mathtools}

\begin{document}

\[ \langle\mkern1mu k\colon e_{k + 1},\dotsm, e_n\mkern0.5mu\rangle =%
 \begin{blockarray}{llllc}
  \begin{block}{|llll|c\Right{\}}{$ k $ rows}}%
  x_1 & x_2 & \dotsm & x_n   & \!\!   \\
  x_1 & x_2 & \dotsm & x_n \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}  \\
  x_1 & x_2 & \dotsm & x_n \\[0.5ex]
  \end{block}
  \begin{block}{|llll|c\Right{\}}{$n- k $ rows}}%
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dotsm & y_n^{p^{e_{k + 1}}} & \!\! \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}\\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dotsm & y_n^{p^{e_n}} \\
  \end{block}
  \end{blockarray} \]    

\end{document}

如果您想要使用箭头而不是括号，可以使用以下方法获取它们pstricks（使用进行编译xelatex）：

  .......
\usepackage{pst-node} %
\begin{document}

   \[ \begin{pspicture}%
 $ \langle\mkern1mu k\colon e_{k + 1},\dotsm, e_n\mkern0.5mu\rangle =%
 \begin{blockarray}{|llll|c}
  x_1 & x_2 & \dotsm & x_n   & \pnode[0,1ex]{A}   \\
  x_1 & x_2 & \dotsm & x_n \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}  \\
  x_1 & x_2 & \dotsm & x_n & \pnode[0,-1ex]{B} \\[0.5ex]%
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dotsm & y_n^{p^{e_{k + 1}}} & \pnode[0,1ex]{C} \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}\\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dotsm & y_n^{p^{e_n}} & \pnode[0,-1ex]{D}
  \end{blockarray} $%
  \psset{arrows=<->, arrowinset=0.15}
  \ncline{A}{B}\ncput*{$ k $\rlap{\enspace rows}}
  \ncline{C}{D}\ncput*{$ n $\rlap{$- k $\enspace   \text{rows}}}
  \end{pspicture}\]

Answer

这里有一种方法blkarray：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{blkarray, makebox}
\usepackage{mathtools}

\begin{document}

\[ \langle\mkern1mu k\colon e_{k + 1},\dotsm, e_n\mkern0.5mu\rangle =%
 \begin{blockarray}{llllc}
  \begin{block}{|llll|c\Right{\}}{$ k $ rows}}%
  x_1 & x_2 & \dotsm & x_n   & \!\!   \\
  x_1 & x_2 & \dotsm & x_n \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}  \\
  x_1 & x_2 & \dotsm & x_n \\[0.5ex]
  \end{block}
  \begin{block}{|llll|c\Right{\}}{$n- k $ rows}}%
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dotsm & y_n^{p^{e_{k + 1}}} & \!\! \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}\\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dotsm & y_n^{p^{e_n}} \\
  \end{block}
  \end{blockarray} \]    

\end{document}

如果您想要使用箭头而不是括号，可以使用以下方法获取它们pstricks（使用进行编译xelatex）：

  .......
\usepackage{pst-node} %
\begin{document}

   \[ \begin{pspicture}%
 $ \langle\mkern1mu k\colon e_{k + 1},\dotsm, e_n\mkern0.5mu\rangle =%
 \begin{blockarray}{|llll|c}
  x_1 & x_2 & \dotsm & x_n   & \pnode[0,1ex]{A}   \\
  x_1 & x_2 & \dotsm & x_n \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}  \\
  x_1 & x_2 & \dotsm & x_n & \pnode[0,-1ex]{B} \\[0.5ex]%
  y_1^{p^{e_{k + 1}}} & y_2^{p^{e_{k + 1}}} & \dotsm & y_n^{p^{e_{k + 1}}} & \pnode[0,1ex]{C} \\
  \makebox*{$ x_1 $}{$\cdot$} & \makebox*{$ x_2 $}{$\cdot$}  & \dotsm  & \makebox*{$ x_n$}{$\cdot$}\\
  y_1^{p^{e_n}} & y_2^{p^{e_n}} & \dotsm & y_n^{p^{e_n}} & \pnode[0,-1ex]{D}
  \end{blockarray} $%
  \psset{arrows=<->, arrowinset=0.15}
  \ncline{A}{B}\ncput*{$ k $\rlap{\enspace rows}}
  \ncline{C}{D}\ncput*{$ n $\rlap{$- k $\enspace   \text{rows}}}
  \end{pspicture}\]

Question 3

我想在这里重申我的座右铭：有些矩阵不是用来排版的。请注意，在您的屏幕截图中，无论是谁做的，实际上都是在试图精确地描述用户根本不会真正关心的细节。Fortran 编译器不会读取矩阵的规格。他们需要知道矩阵中有什么，并且矩阵中有一个定义非常明确的结构，而您根本不会使用它，甚至没有提到它，这确实是一种浪费。

以下是我要做的事情：

\documentclass{article}
\usepackage{mathtools}

\begin{document}

Let $x,y$, be row vectors of size $n$, $\otimes$ denote the Kronecker product 
and also the notation $y(a,b,\cdots)$ denote the rows 

\[\begin{bmatrix}y_1^{p^{a}}&\cdots&y_k^{p^{a}}\\
y_1^{p^{b}}&\cdots&y_k^{p^{b}}\\
&\vdots\end{bmatrix}\]
Then, following determinant of an $n\times n$ matrix is well-defined 
\[
\langle k: e_{k+1},\cdots,e_n\rangle = 
\left|
\begin{array}{c}
\mathbf{1}_{k\times 1} \otimes x \\
y(e_{k+1},\cdots,e_n)
\end{array}
\right|
\]
also the following is well-defined too
\[
\left|
\begin{array}{cc}
x & x \\
y(0,\cdots,n-1)&y(0,\cdots,n-1)\\
0 & \mathbf{1}_{k- 1\times 1} \otimes x\\
0 & 
y(0,1,\cdots,s_k,??????)
\end{array}
\right|
\]
\end{document}

您可以看到，在您的第二个示例中，我甚至不理解这种模式，所以我留下了问号。相反，您可以继续输入不同的术语y(...)，它们会自动分隔开，而不是在内容方面意义不大的行号。

Answer

我想在这里重申我的座右铭：有些矩阵不是用来排版的。请注意，在您的屏幕截图中，无论是谁做的，实际上都是在试图精确地描述用户根本不会真正关心的细节。Fortran 编译器不会读取矩阵的规格。他们需要知道矩阵中有什么，并且矩阵中有一个定义非常明确的结构，而您根本不会使用它，甚至没有提到它，这确实是一种浪费。

以下是我要做的事情：

\documentclass{article}
\usepackage{mathtools}

\begin{document}

Let $x,y$, be row vectors of size $n$, $\otimes$ denote the Kronecker product 
and also the notation $y(a,b,\cdots)$ denote the rows 

\[\begin{bmatrix}y_1^{p^{a}}&\cdots&y_k^{p^{a}}\\
y_1^{p^{b}}&\cdots&y_k^{p^{b}}\\
&\vdots\end{bmatrix}\]
Then, following determinant of an $n\times n$ matrix is well-defined 
\[
\langle k: e_{k+1},\cdots,e_n\rangle = 
\left|
\begin{array}{c}
\mathbf{1}_{k\times 1} \otimes x \\
y(e_{k+1},\cdots,e_n)
\end{array}
\right|
\]
also the following is well-defined too
\[
\left|
\begin{array}{cc}
x & x \\
y(0,\cdots,n-1)&y(0,\cdots,n-1)\\
0 & \mathbf{1}_{k- 1\times 1} \otimes x\\
0 & 
y(0,1,\cdots,s_k,??????)
\end{array}
\right|
\]
\end{document}

您可以看到，在您的第二个示例中，我甚至不理解这种模式，所以我留下了问号。相反，您可以继续输入不同的术语y(...)，它们会自动分隔开，而不是在内容方面意义不大的行号。

Question 4

这是一个{vNiceMatrix}使用的解决方案nicematrix。

\documentclass{article}
\usepackage{nicematrix,tikz}

\begin{document}

\begin{center}
\NiceMatrixOptions{xdots={horizontal-labels,line-style=<->}}
$\langle k\colon e_{k + 1},\dots, e_n\rangle =
\begin{vNiceMatrix}[last-col=5]
  x_1              & x_2               & \dots  & x_n  
& \Vdotsfor{4}:{\makebox[4mm][l]{\small $k$ rows}} \\
  x_1              & x_2               & \dots  & x_n & \\
  \dots            & \dots             & \dots  & \dots  \\
  x_1              & x_2               & \dots  & x_n \\[1mm]
  y_1^{p^{e_{k+1}}} & y_2^{p^{e_{k + 1}}} & \dots  & y_n^{p^{e_{k + 1}}} 
& \Vdotsfor{3}:{\makebox[4mm][l]{\small $n-k$ rows}} \\
  \dots           & \dots              & \dots  & \dots \\
  y_1^{p^{e_n}} & y_2^{p^{e_n}}          & \dots  & y_n^{p^{e_n}} \\
\end{vNiceMatrix}$
\end{center}

\end{document}

您需要多次编译（因为nicematrix在后台使用 PGF/Tikz 节点）。

Answer

这是一个{vNiceMatrix}使用的解决方案nicematrix。

\documentclass{article}
\usepackage{nicematrix,tikz}

\begin{document}

\begin{center}
\NiceMatrixOptions{xdots={horizontal-labels,line-style=<->}}
$\langle k\colon e_{k + 1},\dots, e_n\rangle =
\begin{vNiceMatrix}[last-col=5]
  x_1              & x_2               & \dots  & x_n  
& \Vdotsfor{4}:{\makebox[4mm][l]{\small $k$ rows}} \\
  x_1              & x_2               & \dots  & x_n & \\
  \dots            & \dots             & \dots  & \dots  \\
  x_1              & x_2               & \dots  & x_n \\[1mm]
  y_1^{p^{e_{k+1}}} & y_2^{p^{e_{k + 1}}} & \dots  & y_n^{p^{e_{k + 1}}} 
& \Vdotsfor{3}:{\makebox[4mm][l]{\small $n-k$ rows}} \\
  \dots           & \dots              & \dots  & \dots \\
  y_1^{p^{e_n}} & y_2^{p^{e_n}}          & \dots  & y_n^{p^{e_n}} \\
\end{vNiceMatrix}$
\end{center}

\end{document}

您需要多次编译（因为nicematrix在后台使用 PGF/Tikz 节点）。

\bordermatrix 的问题

答案1

答案2

答案3

答案4

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