我想删除显示的方程式和“收敛到 q”之间的大空白。我正在使用环境tabularx，因此必须\\在显示的方程式后插入一个回车符 ( ) 才能使表格正常工作。我该如何删除该空格？
由于我正在从事一个大型项目并且无法改变整个项目，因此如果有一个不改变整体结构的解决方案我将不胜感激（tabularx，\hspace{3em}）。
这是我的代码。
\documentclass{article}
\usepackage[english]{babel}
\usepackage[a4paper,top=2cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]
{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage{xcolor}
\usepackage{indentfirst}
\usepackage{tabularx}
\usepackage{bm}
\usepackage{nicematrix}
\usepackage[colorlinks=true, allcolors=teal]{hyperref}
\usepackage[T1]{fontenc}
\begin{document}
\begin{tabularx}{0.9\linewidth}{l@{}c@{}X}
\hypertarget{5.5.1}{\textbf{5.5.1}} & $\quad$ & If $P$ is the transition matrix for a
regular Markov chain, then :
\end{tabularx}
\renewcommand{\arraystretch}{1.5}
\begin{tabularx}{0.9\linewidth}{l@{}c@{}X}
& \hspace{3em} & (a) There is a unique probability vector $\mathbf{q}$ with positive
entries such that \\
& \hspace{3em} & \hspace{1.25em} $P \mathbf{p} = \mathbf{q}$. \\
& \hspace{3em} & (b) For any initial probability vector $\mathbf{x}_{0}$, the sequence
of state vectors
$$ \mathbf{x}_{0}, \quad P \mathbf{x}_{0}, \quad \cdots, \quad P^{k} \mathbf{x}_{0},
\quad \cdots $$ \\
& \hspace{3em} & \hspace{1.25em} converges to $\mathbf{q}$. \\
& \hspace{3em} & (c) The sequence $P, P^{2}, \cdots, P^{k}, \cdots$ converges to the
matrix $Q$ each of whose \\
& \hspace{3em} & \hspace{1.25em} column vectors is $\mathbf{q}$. \\
\end{tabularx}
\end{document}
答案1
我认为你严重滥用tabularx机器，并且你正在进行大量难以破译的（并且，我不想这么说，看起来相当业余的）视觉格式化。
我建议您enumerate按照以下方法切换到自定义环境。附带的好处是，您遇到的虚拟空间过大问题会自行消失。
\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage[a4paper, vmargin=2cm, hmargin=3cm,
marginparwidth=1.75cm]{geometry}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{enumitem}
\usepackage[colorlinks=true, allcolors=teal]{hyperref}
\newcommand\xzero{\mathbf{x}_{0}}
\begin{document}
\noindent
\hypertarget{5.5.1}{\textbf{5.5.1}} \quad
\begin{minipage}[t]{0.9\textwidth}
If $P$ is the transition matrix for a regular Markov chain, then:
\begin{enumerate}[label=(\alph*),left=0pt,align=left,itemsep=0pt]
\item There is a unique probability vector $\mathbf{q}$ with positive entries such that $P \mathbf{p} = \mathbf{q}$.
\item For any initial probability vector $\xzero$, the sequence of state vectors
\[
\xzero, P \xzero, \dots, P^{k} \xzero
\]
converges to $\mathbf{q}$.
\item The sequence $P, P^{2}, \dots, P^{k}, \dots$ converges to the matrix $Q$, each of whose column vectors is~$\mathbf{q}$.
\end{enumerate}
\end{minipage}
\end{document}