使用 \underbrace 进行矩阵垂直对齐

Question

我想要垂直对齐下支撑符号。

现在看起来像：

应该是

女士：

\documentclass{article}

\usepackage{mathtools}
\usepackage[no-math]{fontspec}
\usepackage{unicode-math}
\setmathfont[StylisticSet=1]{XITS Math}


\begin{document}

\begin{equation*}
\underbrace{\begin{pmatrix}  
    \ b_1 \ \null \\
    \ b_2 \ \null \\
    \ b_3 \ \null  \\
    \ \vdots\ \null \\
    \ b_n \
    \end{pmatrix}}_{\substack{\\[0.2em] \text{\normalsize\strut $\mathbf{b}$}}} =  
\underbrace{\begin{pmatrix}
    \ s_{11} & s_{12} & s_{13} & \dots & s_{1n} \ \null \\
    \ s_{21} & s_{22} & s_{23} & \dots & s_{2n} \ \null \\
    \ s_{31} & s_{32} & s_{33} & \dots & s_{3n} \ \null \\
    \ \vdots & \vdots & \vdots & \ddots& \vdots \ \null \\
    \ s_{n1} & \dots  & \dots  & \dots & s_{nn}\ \\
    \end{pmatrix}}_{\substack{\\[0.2em] \text{\normalsize $\mathbf{S}$}}} \cdot 
\underbrace{\begin{pmatrix}
    \ a_1 \ \null \\
    \ a_2 \  \null \\
    \ a_3 \  \null \\
    \ \vdots\ \null \\
    \ a_n \
    \end{pmatrix}}_{\substack{}{ \\[0.2em] \text{\normalsize\strut $\mathbf{a}$}}}
\end{equation*}

\end{document}

谢谢大家

更新：

就像@RuixiZhang 说的，问题是由 Mathfont 引起的

如果使用\,\,\,\, a_1 \,\,\,\,

矩阵只有一列，宽度为两列

Answer 1

您需要强制一个宽度，使得 XITS Math 选择更大的括号。

\documentclass{article}

\usepackage{mathtools}
%\usepackage[no-math]{fontspec}
\usepackage{unicode-math}
\setmathfont[StylisticSet=1]{XITS Math}


\begin{document}

\begin{equation*}
{\underbrace{
  \mspace{3mu}
  \left(\begin{array}{c} b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{array}\right)
  \mspace{3mu}
}_{\textstyle\mathstrut\symbf{b}}} =
{\underbrace{
  \left(\begin{array}{*{5}{c}}
    s_{11} & s_{12} & s_{13} & \dots & s_{1n} \\
    s_{21} & s_{22} & s_{23} & \dots & s_{2n} \\
    s_{31} & s_{32} & s_{33} & \dots & s_{3n} \\
    \vdots & \vdots & \vdots & \ddots& \vdots \\
    s_{n1} & \dots  & \dots  & \dots & s_{nn}
    \end{array}\right)
}_{\textstyle\mathstrut\symbf{S}}}
\mspace{3mu}\cdot 
{\underbrace{
  \mspace{3mu}
  \left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{array}\right)
  \mspace{3mu}
}_{\textstyle\mathstrut\symbf{a}}}
\end{equation*}

\end{document}

请注意 2mu 太小了。还请注意，输入可以大大简化。周围的括号\underbrace{...}_{...}是获得正确间距所必需的；前面又加了一些括号，\cdot以补偿右侧较大的空间。

Answer 2

TikZ 的解决方案：

\documentclass{article}

\usepackage{mathtools}
\usepackage[no-math]{fontspec}
\usepackage{unicode-math}
\setmathfont[StylisticSet=1]{XITS Math}

\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{decorations.pathreplacing,calligraphy}
\tikzset{
    mybrace/.style={
            line width=1.2pt,
            decorate,
            decoration={
            calligraphic brace, amplitude=4pt,
            mirror,
            }        
        }
    }
    
\begin{document}
\begin{equation*}
\begin{tikzpicture}
\node (vettb) {$\begin{pmatrix}  
    \ b_1 \ \null \\
    \ b_2 \ \null \\
    \ b_3 \ \null  \\
    \ \vdots\ \null \\
    \ b_n \
    \end{pmatrix}$};
\draw [mybrace] ([xshift=4pt]vettb.south west) -- ([xshift=-4pt]vettb.south east) 
node [pos=0.5,anchor=north,yshift=-4pt] {\strut $\mathbf{b}$};
\node[right=0pt of vettb, inner sep=0pt] (eqsign) {$=$};
\node[right=0pt of eqsign] (matrS) {$\begin{pmatrix}
    \ s_{11} & s_{12} & s_{13} & \dots & s_{1n} \ \null \\
    \ s_{21} & s_{22} & s_{23} & \dots & s_{2n} \ \null \\
    \ s_{31} & s_{32} & s_{33} & \dots & s_{3n} \ \null \\
    \ \vdots & \vdots & \vdots & \ddots& \vdots \ \null \\
    \ s_{n1} & \dots  & \dots  & \dots & s_{nn}\ \\
    \end{pmatrix}$};
\draw [mybrace] ([xshift=4pt]matrS.south west) -- ([xshift=-4pt]matrS.south east) 
node [pos=0.5,anchor=north,yshift=-4pt] {\strut $\mathbf{S}$};
\node[right=0pt of matrS, inner sep=0pt] (dotop) {$\cdot$};
\node[right=0pt of dotop] (vetta) {$\begin{pmatrix}
    \ a_1 \ \null \\
    \ a_2 \  \null \\
    \ a_3 \  \null \\
    \ \vdots\ \null \\
    \ a_n \
    \end{pmatrix}$};
\draw [mybrace] ([xshift=4pt]vetta.south west) -- ([xshift=-4pt]vetta.south east) 
node [pos=0.5,anchor=north,yshift=-4pt] {\strut $\mathbf{a}$};
\end{tikzpicture}
\end{equation*}
\end{document}