☰

带箭头的矩阵

tag-icon • tag-icon matrices arrows
带箭头的矩阵

有人知道如何处理垂直箭头吗？我目前的代码如下：

在此处输入图片描述

\hspace{0.75cm}
$\xleftrightarrow{\text{\hspace{4.25cm}j \hspace{4.25cm}}}$
\newline
\updownarrow{\text{i}}
\newline
\updownarrow{\text{i}}
\[
\begin{pmatrix}
y_0 & y_1 & ... & y_{j-1} \\
... & ... & ... & ... \\
y_{i-2} & y_{i-1} & ... & y_{i+j-3} \\
y_{i-i} & y_{i} & ... & y_{i+j-2} \\
\noindent\rule{2cm}{0.4pt} & \noindent\rule{2cm}{0.4pt} & \noindent\rule{2cm}{0.4pt} & \noindent\rule{2cm}{0.4pt} \\
y_{i} & y_{i+1} & ... & y_{i+j-1} \\
y_{i+1} & y_{i+2} & ... & y_{i+j} \\
... & ... & ... & ... \\
y_{2i-1} & y_{2i} & ... & y_{2i+j-2} \\
\end{pmatrix}
\]

答案1

这是一种可能的方法：标记一些特殊点（例如矩阵的角）的坐标，然后用来tikz绘制箭头。你可能需要编译几次

\documentclass[]{article}

\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{positioning}

\newcommand{\pincoord}[1]{\tikz[overlay, remember picture] \coordinate (#1);}

\begin{document}



\[
  Y_{0|2i-1} \stackrel{\rm def}{=} \quad\pincoord{A}
  \begin{pmatrix}
    \pincoord{B}  y_0 & y_1 & ... & y_{j-1} \pincoord{C} \\
    ... & ... & ... & ... \\
    y_{i-2} & y_{i-1} & ... & y_{i+j-3} \\
     \pincoord{F} y_{i-i} & y_{i} & ... & y_{i+j-2} \pincoord{G} \\
    y_{i} & y_{i+1} & ... & y_{i+j-1} \\
    y_{i+1} & y_{i+2} & ... & y_{i+j} \\
    ... & ... & ... & ... \\
     \pincoord{D}  y_{2i-1}& y_{2i} & ... & y_{2i+j-2} \pincoord{E}\\
  \end{pmatrix}
  \pincoord{H} \qquad\stackrel{\rm def}{=} \left(\frac{Y_{0|i-1}}{Y_{i|2i-1}}\right)  \quad\stackrel{\rm def}{=} 
  \left(\frac{Y_p}{Y_f}\right)
\]

\begin{tikzpicture}[overlay, remember picture, >=stealth]
  \draw[<->] ([xshift = -1ex, yshift = 2pt]A) -- ([xshift = -1ex]A |- B) node[midway, left] {i};
  \draw[<->] ([xshift = -1ex, yshift =-2pt]A) -- ([xshift = -1ex]A |- D) node[midway, left] {i};
  \draw[<->] ([yshift = 3ex]B) -- ([yshift = 3ex]B -| C) node[midway, above] {J};
  \draw[] ([yshift = -5pt]F) -- ([yshift = -5pt]G);
  \draw[<->] ([xshift = 1ex, yshift = 2pt]H) -- ([xshift = 1ex]H |- B) node[midway, right] {``past''};
  \draw[<->] ([xshift = 1ex, yshift =-2pt]H) -- ([xshift = 1ex]H |- D) node[midway, right] {``future''};
\end{tikzpicture}

\end{document}

在此处输入图片描述

答案2

对于 来说这是一份完美的工作tikz matrix！

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix, arrows.meta}
\begin{document}
    \[
    Y_{0\mid 2i-1} \stackrel{\text{def}}{=}
\begin{tikzpicture}[>=Stealth, baseline=(mybaseline)]
\matrix[
    matrix of math nodes,
    left delimiter=(,
    right delimiter=),
    inner xsep=2pt,
    ] (mymatr) {%
    y_0 & y_1 & \cdots & y_{j-1} \\[-6pt]
    \vdots & \vdots & \ddots & \vdots \\
    y_{i-2} & y_{i-1} & \cdots & y_{i+j-3} \\
    y_{i-1}% there was a little typo here 
    & y_{i} & \cdots & y_{i+j-2} \\[12pt]
    y_{i} & y_{i+1} & \cdots & y_{i+j-1} \\
    y_{i+1} & y_{i+2} & \cdots & y_{i+j} \\[-6pt]
    \vdots & \vdots & \ddots & \vdots \\
    y_{2i-1} & y_{2i} & \cdots & y_{2i+j-2} \\
    };
\coordinate (mybaseline) at ([yshift=-9pt]mymatr-4-1.south west);
\coordinate (mystart) at ([yshift=-6pt]mymatr-4-1.south west);
\coordinate (mystop)  at ([yshift=-6pt]mymatr-4-4.south east);
\coordinate (myne)  at ([xshift=16pt]mymatr.north east);
\coordinate (myse)  at ([xshift=16pt]mymatr.south east);
\coordinate (mynw)  at ([xshift=-16pt]mymatr.north west);
\coordinate (mysw)  at ([xshift=-16pt]mymatr.south west);
\draw (mystart) -- (mystop);
\begin{scope}[every node/.style={midway},
every path/.style={<->}]
\draw (myne) -- (myne |- mystop) node[right] {``past''};
\draw (myse) -- (myse |- mystop) node[right] {``future''};
\draw (mynw) -- (mynw |- mystop) node[left] {$i$};
\draw (mysw) -- (mysw |- mystop) node[left] {$i$};
\draw ([yshift=10pt]mymatr.north west) -- ([yshift=10pt]mymatr.north east) node[above] {$j$};
\end{scope}
\end{tikzpicture}
\hspace{-34pt}% <--- but I would omit this
\stackrel{\text{def}}{=}
\biggl(\frac{Y_{0\mid i-1}}{Y_{i\mid 2i-1}}\biggr)
\stackrel{\text{def}}{=}
\biggl(\frac{Y_{p}}{Y_{f}}\biggr)
\]
\end{document}

在此处输入图片描述

答案3

与。{NiceArray}​nicematrix

\documentclass{article}
\usepackage{nicematrix}
\usepackage{tikz}
\usepackage{booktabs}% for \cmidrule

\begin{document}

\NiceMatrixOptions
  {
    exterior-arraycolsep,
    xdots = 
      { 
        horizontal-labels ,
        line-style = { solid , <-> }
      }
  }

\[ 
Y_{0\mid 2i-1} \stackrel{\text{def}}{=}
\begin{NiceArray}{cccccc}[baseline = line-6]
    & \Hdotsfor{4}^{j} \\
  \Vdotsfor{4}_{i} 
    & y_0 & y_1 & \cdots & y_{j-1} & 
        \Vdotsfor{4}^{\text{``past''}} \\
    & \vdots & \vdots & \ddots & \vdots \\
    & y_{i-2} & y_{i-1} & \cdots & y_{i+j-3} \\
    & y_{i-1} & y_{i} & \cdots & y_{i+j-2} \\
    \cmidrule(lr){2-5}
  \Vdotsfor{4}_{j} 
    & y_{i} & y_{i+1} & \cdots & y_{i+j-1} & 
        \Vdotsfor{4}^{\text{``future''}} \\
    & y_{i+1} & y_{i+2} & \cdots & y_{i+j} \\
    & \vdots & \vdots & \ddots & \vdots \\
    & y_{2i-1} & y_{2i} & \cdots & y_{2i+j-2} \\
\CodeAfter
  \SubMatrix({2-2}{9-5})
\end{NiceArray}
\stackrel{\text{def}}{=}
\biggl(\frac{Y_{0\mid i-1}}{Y_{i\mid 2i-1}}\biggr)
\stackrel{\text{def}}{=}
\biggl(\frac{Y_{p}}{Y_{f}}\biggr)
\]

\end{document}

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

上述代码的输出

相关内容