☰

设置拉伸 + 数组

tag-icon • tag-icon spacing matrices
设置拉伸 + 数组

我目前正在尝试排版 Gell-Mann-Matrices：

\documentclass{article}
\usepackage{array}
\usepackage{setspace}

\newcommand\matthree[9]{\left(\begin{array}{c c c} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9\end{array}\right)}

\begin{document}

The generators are often written in a standard basis
$T^a = \frac{1}{2} \lambda^a$, with $\lambda^a$ being the Gell-Mann-Matrices

\begin{center}
  $\lambda^1 = \matthree {0}{1}{}{1}{0}{}{}{}{0}$,
  $\lambda^2 = \matthree {0}{-i}{}{i}{0}{}{}{}{0}$,
  $\lambda^3 = \matthree {1}{}{}{}{-1}{}{}{}{0}$,
  $\lambda^4 = \matthree {0}{}{1}{}{0}{}{1}{}{0}$,
  $\lambda^5 = \matthree {0}{}{-i}{}{0}{}{i}{}{0}$,
  $\lambda^6 = \matthree {0}{}{}{}{0}{1}{}{1}{0}$,
  $\lambda^7 = \matthree {0}{}{}{}{0}{-i}{}{i}{0}$,
  $\lambda^8 = \frac{1}{\sqrt{3}} \matthree {1}{}{}{}{1}{}{}{}{-2}$ 
\end{center}

\end{document}

这看起来已经相当合理了：标准

但是，行距看起来有点太小了，所以我想

\begin{center}\setstretch{1.5}\par
   ...
\end{center}

拉伸

和我预期的不太一样。现在，我想我可以像这样修复它：

\newcommand\matthree[9]{\begingroup\setstretch{1.0}\left(\begin{array}{c c c} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9\end{array}\right)\endgroup}

但这只会让一切回到我开始的地方。有什么建议可以让我增加行距而不弄乱矩阵中的行距吗？

编辑：这不是重复的减少矩阵中的行距，因为我实际上想保持矩阵中的行间距不变，同时增加周围文本的行间距。

答案1

setspace如果由于机构的具体指示而没有强制您使用它，请避免使用它。

这里您只需要一个标准amsmath环境，并借助一些帮助在行间提供更多的空间。

\documentclass{article}
\usepackage{amsmath}

\newcommand\matthree[9]{%
  \begin{pmatrix}
  #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9
  \end{pmatrix}%
}

\begin{document}

The generators are often written in a standard basis
$T^a = \frac{1}{2} \lambda^a$, with $\lambda^a$ being the Gell-Mann-Matrices
\begin{gather*}
\lambda^1 = \matthree {0}{1}{}{1}{0}{}{}{}{0},\quad
\lambda^2 = \matthree {0}{-i}{}{i}{0}{}{}{}{0},\quad
\lambda^3 = \matthree {1}{}{}{}{-1}{}{}{}{0},\\[1ex]
\lambda^4 = \matthree {0}{}{1}{}{0}{}{1}{}{0},\quad
\lambda^5 = \matthree {0}{}{-i}{}{0}{}{i}{}{0},\quad
\lambda^6 = \matthree {0}{}{}{}{0}{1}{}{1}{0},\\[1ex]
\lambda^7 = \matthree {0}{}{}{}{0}{-i}{}{i}{0},\quad
\lambda^8 = \frac{1}{\sqrt{3}} \matthree {1}{}{}{}{1}{}{}{}{-2}
\end{gather*}

\end{document}

在此处输入图片描述

只是为了好玩，这里有一个我们让 LaTeX 进行居中的解决方案：

\documentclass{article}
\usepackage{amsmath}

\newcommand\matthree[9]{%
  \begin{pmatrix}
  #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9
  \end{pmatrix}%
}

\begin{document}

The generators are often written in a standard basis
$T^a = \frac{1}{2} \lambda^a$, with $\lambda^a$ being the Gell-Mann-Matrices
\[
\begin{minipage}{.75\displaywidth}
\centering\setlength{lineskip}{2ex}
$\lambda^1 = \matthree {0}{1}{}{1}{0}{}{}{}{0}$,
$\lambda^2 = \matthree {0}{-i}{}{i}{0}{}{}{}{0}$,
$\lambda^3 = \matthree {1}{}{}{}{-1}{}{}{}{0}$,
$\lambda^4 = \matthree {0}{}{1}{}{0}{}{1}{}{0}$,
$\lambda^5 = \matthree {0}{}{-i}{}{0}{}{i}{}{0}$,
$\lambda^6 = \matthree {0}{}{}{}{0}{1}{}{1}{0}$,
$\lambda^7 = \matthree {0}{}{}{}{0}{-i}{}{i}{0}$,
$\lambda^8 = \frac{1}{\sqrt{3}} \matthree {1}{}{}{}{1}{}{}{}{-2}$
\end{minipage}
\]

\end{document}

答案2

这是一个使用环境的解决方案。还请注意，3x3 矩阵的环境align*的使用以及将三个主要行之间的垂直空间增加。pmatrix*[2ex]2ex

在此处输入图片描述

\documentclass{article}
\usepackage{mathtools} % for 'pmatrix*' env.
\newcommand\matthree[9]{%
     \begin{pmatrix*}[r]
         #1 & #2 & #3 \\ 
         #4 & #5 & #6 \\ 
         #7 & #8 & #9
     \end{pmatrix*}}

\begin{document}
\noindent
Using an \verb+align*+ environment:
\begin{align*}
  \lambda^1 &= \matthree {0}{1}{}{1}{0}{}{}{}{0} &
  \lambda^2 &= \matthree {0}{-i}{}{i}{0}{}{}{}{0}&
  \lambda^3 &= \matthree {1}{}{}{}{-1}{}{}{}{0}  \\[2ex]
  \lambda^4 &= \matthree {0}{}{1}{}{0}{}{1}{}{0} &
  \lambda^5 &= \matthree {0}{}{-i}{}{0}{}{i}{}{0}&
  \lambda^6 &= \matthree {0}{}{}{}{0}{1}{}{1}{0} \\[2ex]
  \lambda^7 &= \matthree {0}{}{}{}{0}{-i}{}{i}{0}&
  \lambda^8 &= \frac{1}{\sqrt{3}} \matthree {1}{}{}{}{1}{}{}{}{-2} 
\end{align*}
\end{document}

相关内容