[L]
我怎样才能用一个巨大的矩阵替换原始的正常大小的矩阵,如下图所示(红色的结果是我想要的):
答案1
一种选择:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\begin{bmatrix}
\delta_{1} \\
\delta_{2} \\
\vdots \\
\delta_{M} \\
\end{bmatrix} =
\begin{bmatrix}
\qquad\qquad \\
\\
\raisebox{0pt}[0pt][0pt]{\Huge $L$} \\
\\
\end{bmatrix}_{M\times M}
\begin{bmatrix}
v_{1} \\
v_{2} \\
\vdots \\
v_{M} \\
\end{bmatrix}
\]
\end{document}
答案2
如果希望括号大小相同,请使用\vphantom
:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\begin{bmatrix}
\delta_{1} \\
\delta_{2} \\
\vdots \\
\delta_{M} \\
\end{bmatrix} =
\left[
\vphantom{%
\begin{matrix}
v_{1} \\
v_{2} \\
\vdots \\
v_{M} \\
\end{matrix}%
}
\quad\begin{gathered}\text{\Huge $L$}\end{gathered}\quad
\right]_{M\times M}
\begin{bmatrix}
v_{1} \\
v_{2} \\
\vdots \\
v_{M} \\
\end{bmatrix}
\]
\end{document}
答案3
这边走?
\documentclass{article}
\usepackage[showframe, nomarginpar, textwidth = 16cm]{geometry}
\usepackage{mathtools}
\usepackage{graphics}
\begin{document}
\[
\begin{bmatrix}
δ_1\\δ_{2} \\ \vdots \\ δ_M
\end{bmatrix} =
\begin{bmatrix}
\vphantom{\begin{matrix} δ_1 & & δ_M \\δ_{2} \\ \vdots \\ δ_M \end{matrix}}\scalebox{2.5}{\enspace$ L $\enspace}
\end{bmatrix}_{M× M}
\begin{bmatrix}
v_1\\v_{2} \\ \vdots \\ v_M
\end{bmatrix} ,
\]
\end{document}
答案4
简洁,有堆栈。
\documentclass{article}
\usepackage{stackengine}
\stackMath
\setstackgap{L}{1.3\baselineskip}
\begin{document}
\savestack{\tmp}{\Huge $L$}
\begin{equation}
\left[\addstackgap{\Centerstack{\delta_1 \delta_2 {\vdots} \delta_M}}\right]
=
\left[\addstackgap[2.5\baselineskip]{\smash{~~~~~~\raisebox{-.3\ht\tmpcontent}{\tmp}~~~~~~}}\right]_{M\times M}
\left[\addstackgap{\Centerstack{v_1 v_2 {\vdots} v_M}}\right]
\end{equation}
\end{document}