如何形成这种间距?

如何形成这种间距?

嗯

$ (E\mathcal{V}_{i+1})^{\perp}=(EA^{-1}(E\mathcal{V}_i))^T \stackrel{Lem. 1.1(i)}{=} E^{-T}(A^{-1}(E\mathcal{V}_i))^{\perp}
  \stackrel{Lem. 1.1(ii)}{=} E^{-T}(A^T(E\mathcal{V}_i))^{\perp}=E^{-T}(A^T\widehat{\mathcal{W}}_{i+1})=\widehat{\mathcal{W}}_{i+2}$

我想像图片一样书写。我该怎么做?

答案1

也许是这样的?

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath} % for 'align*' environment and '\stackrel' macro
\newlength\mylength
\settowidth{\mylength}{\scriptsize Lem.\ 1.1(ii)}
\begin{document}
\begin{align*}
(E\mathcal{V}_{i+1})^{\perp}
&\stackrel{\makebox[\mylength]{}}{=}
    (EA^{-1}(E\mathcal{V}_i))^T \\
&\stackrel{\makebox[\mylength]{\scriptsize Lem.\ 1.1(i)}}{=} 
    E^{-T}(A^{-1}(E\mathcal{V}_i))^{\perp}\\
&\stackrel{\text{Lem.\ 1.1(ii)}}{=} 
    E^{-T}(A^T(E\mathcal{V}_i))^{\perp}\\
&\stackrel{\makebox[\mylength]{}}{=}E^{-T}
    (A^T\widehat{\mathcal{W}}_{i+1})
    =\widehat{\mathcal{W}}_{i+2}
\end{align*}
\end{document}

附录:如果您有很多这样的表达式,那么设置一个专用宏来表示专门的=符号可能是一个好主意,例如按照以下代码行(它生成与上面显示的示例相同的输出)。如果符号上方没有任何内容需要排版=,只需写入\myeq{}即可获得正确的间距。

\documentclass{article}
\usepackage{amsmath} % for 'align*' environment and '\stackrel' macro
\newlength\mylength
\settowidth{\mylength}{\scriptsize Lem.\ 1.1(ii)} % width of  longest label
\newcommand\myeq[1]{\stackrel{\makebox[\mylength]{\scriptsize #1}}{=}}
\begin{document}
\begin{align*}
(E\mathcal{V}_{i+1})^{\perp}
&\myeq{}              (E A^{-1}(E\mathcal{V}_i))^T \\
&\myeq{Lem.\ 1.1(i)}  E^{-T}(A^{-1}(E\mathcal{V}_i))^{\perp}\\
&\myeq{Lem.\ 1.1(ii)} E^{-T}(A^T(E\mathcal{V}_i))^{\perp}\\
&\myeq{}              E^{-T} (A^T\widehat{\mathcal{W}}_{i+1})
     =\widehat{\mathcal{W}}_{i+2}
\end{align*}
\end{document}

答案2

您可以使用eqparbox;第一个参数\widerel是任意标签(每个环境不同)。

但是,第二种解决方案似乎更好。当然,不建议对引理数进行硬编码,而\ref应该使用。

\documentclass{article}
\usepackage{amsmath}
\usepackage{eqparbox}

\DeclareRobustCommand{\widerel}[2]{%
  \mathrel{\eqmakebox[#1]{$\displaystyle#2$}}%
}

\begin{document}

\begin{align*}
(E\mathcal{V}_{i+1})^{\perp}
  &\widerel{A}{=}                                 (EA^{-1}(E\mathcal{V}_i))^T \\
  &\widerel{A}{\overset{\text{Lem.\ 1.1(i)}}{=}}  E^{-T}(A^{-1}(E\mathcal{V}_i))^{\perp}\\
  &\widerel{A}{\overset{\text{Lem.\ 1.1(ii)}}{=}} E^{-T}(A^T(E\mathcal{V}_i))^{\perp}\\
  &\widerel{A}{=}                                 E^{-T}(A^T\widehat{\mathcal{W}}_{i+1})
                                                  =\widehat{\mathcal{W}}_{i+2}
\end{align*}

\begin{align*}
(E\mathcal{V}_{i+1})^{\perp}
  &= (EA^{-1}(E\mathcal{V}_i))^T \\
  &= E^{-T}(A^{-1}(E\mathcal{V}_i))^{\perp} &&\text{Lemma 1.1(i)}\\
  &= E^{-T}(A^T(E\mathcal{V}_i))^{\perp}    &&\text{Lemma 1.1(ii)}\\
  &= E^{-T}(A^T\widehat{\mathcal{W}}_{i+1})
     =\widehat{\mathcal{W}}_{i+2}
\end{align*}
\end{document}

在此处输入图片描述

答案3

这里有两种解决方案。第一种使用array环境,将等号放在单独的列中。第二种解决方案是我推荐的,因为它更易于阅读;它将理由放在额外的列中,而不是等号上方。

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath,array}
\newcommand\VV{{\mathcal V}}
\newcommand\WW{{\mathcal W}}
\begin{document}
\[\begin{array}{r@{}>{{}}c<{{}}@{}l}
  (E\VV_{i+1})^{\perp}
  &=&(EA^{-1}(E\VV_i))^T\\
  &\stackrel{\text{Lem.1.1(i)}}{=}&E^{-T}(A^{-1}(E\VV_i))^{\perp}\\
  &\stackrel{\text{Lem.1.1(ii)}}{=}&E^{-T}(A^T(E\VV_i))^{\perp}\\
  &=&E^{-T}(A^T\widehat{\WW}_{i+1})\\
  &=&\widehat{\WW}_{i+2}
\end{array}
\]
\end{document}

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\newcommand\VV{{\mathcal V}}
\newcommand\WW{{\mathcal W}}
\begin{document}
\begin{align*}
(E\VV_{i+1})^{\perp}
&=(EA^{-1}(E\VV_i))^T\\
&=E^{-T}(A^{-1}(E\VV_i))^{\perp} && \text{by Lem.~1.1(i)}\\
&=E^{-T}(A^T(E\VV_i))^{\perp} && \text{by Lem.~1.1(ii)}\\
&=E^{-T}(A^T\widehat{\WW}_{i+1})\\
&=\widehat{\WW}_{i+2}
\end{align*}
\end{document}

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