我可以在两页上显示一组特征向量吗？

Question

我想知道是否可以将数组拆分为两页，因为我的数组公式很长？我已经旋转了公式并减小了大小，但我认为我仍然必须将其拆分为两页才能使其可读。我使用了以下软件包和代码（抱歉，除了编写乳胶方程式之外，我不知道如何显示我的代码）：

\documentclass[a4paper,twoside,12pt,openany]{report}
\begin{document}
\usepackage{mathtools}
\usepackage{rotating}
\usepackage{graphicx}
\usepackage[T1]{fontenc} 
\usepackage{placeins}
\usepackage{amssymb}
\usepackage{amsmath,scalefnt}

Eigenvector (D12): \scalefont{0.3}{\begin{equation} \rotatebox{-90}{$\left(\begin{array}{ccc} \frac{6\,\left(d-2\right)\,\left(\frac{d}{3}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{6}+\frac{2^{2/3}\,\left(d^2-1\right)}{3\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d\,\left(d+2\right)}-\frac{{\left(2\,d-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2\right)}^2}{3\,d\,\left(d+2\right)}-\frac{4\,\left(d-1\right)}{d+2} & -\frac{4\,\left(d-1\right)}{d+2}-\frac{{\left(4\,d+2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-4\right)}^2}{12\,d\,\left(d+2\right)}+\frac{6\,\left(d-2\right)\,\left(\frac{d}{3}+\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{12}-\frac{2^{2/3}\,\left(d^2-1\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d\,\left(d+2\right)} & -\frac{4\,\left(d-1\right)}{d+2}-\frac{{\left(\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-4\,d+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+4\right)}^2}{12\,d\,\left(d+2\right)}-\frac{6\,\left(d-2\right)\,\left(\frac{2^{2/3}\,\left(d^2-1\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{12}-\frac{d}{3}+\frac{2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{1}{3}\right)}{d\,\left(d+2\right)}\\ \frac{6\,\left(\frac{d}{3}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{6}+\frac{2^{2/3}\,\left(d^2-1\right)}{3\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d}-\frac{{\left(2\,d-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2\right)}^2}{6\,d\,\left(d-1\right)}-3 & -\frac{{\left(4\,d+2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-4\right)}^2}{24\,d\,\left(d-1\right)}+\frac{2^{2/3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}}{4}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-d^2+2^{1/3}\,d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+1-\sqrt{3}\,1{}\mathrm{i}+\sqrt{3}\,d^2\,1{}\mathrm{i}+\frac{2^{2/3}\,\sqrt{3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}\right)}{d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-3 & -\frac{{\left(\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-4\,d+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+4\right)}^2}{24\,d\,\left(d-1\right)}+\frac{2^{2/3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}}{4}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-d^2+2^{1/3}\,d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+1+\sqrt{3}\,1{}\mathrm{i}-\sqrt{3}\,d^2\,1{}\mathrm{i}-\frac{2^{2/3}\,\sqrt{3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}\right)}{d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-3\\ 1 & 1 & 1 \end{array}\right) $}  \end{equation}}


\end{document}

Answer 1

如果您希望读者有机会了解正在发生的事情，您需要根据特征向量的共同因子重新表达其元素，可能遵循以下思路（不能保证我说的正确！）：

\documentclass[a4paper,twoside,12pt,openany]{report}
\usepackage{mathtools}
\usepackage[T1]{fontenc}
\providecommand\dsq{d^{\mkern1mu 2}}

\begin{document}
The eigenvector matrix $D$ is given by
\[
D=\begin{pmatrix}
  d_{11} & d_{12} & d_{13} \\
  d_{21} & d_{22} & d_{23} \\
  1 & 1 & 1
\end{pmatrix}
\] 
where \allowdisplaybreaks
\begin{align*} 
d_{11}
&= \frac{6(d-2)V}{d(d+2)}
  -\frac{{6V}^2}{3d(d+2)}
  -\frac{4(d-1)}{d+2} \\
d_{12}
&=-\frac{(W+X)^2}{12d(d+2)} 
  +\frac{6(d-2)(Y+Z)}{d(d+2)}
  -\frac{4(d-1)}{d+2} \\
d_{13} 
&=-\frac{(W-X)^2}{12d(d+2)}
  -\frac{6(d-2)(Y-Z)}{d(d+2)}
  -\frac{4(d-1)}{d+2} \\
d_{21}
&=\frac{6V}{d}
 -\frac{(6V)^2}{6d(d-1)}
 -3 \\
d_{22}
&=-\frac{(W+X)^2}{24d(d-1)} 
  +\frac{2^{2/3}(K-L)}{d\,U}
  -3
\\
d_{23}
&= -\frac{(W-X)^2}{24d(d-1)} 
   +\frac{2^{2/3}\left(K+L\right)}{d\,U}
   -3 
\\
\shortintertext{and}
U&= (\sqrt{\dsq (d-1)^2(17\dsq +40d+24)}-d^3-3\dsq +4)^{1/3}\\
V&= \frac{d}{3}-\frac{2^{1/3}U}{6}+\frac{2^{2/3}(\dsq -1)}{3U}-\frac{1}{3}\\
W&= 2\cdot2^{2/3}\sqrt{3}(2^{2/3}U^2 \mathrm{i}/4+(\dsq -1) \mathrm{i})/U \\
X&= 4d+2^{1/3}U-2\cdot2^{2/3}(\dsq -1)/U-4 \\
Y&= 2^{2/3}\sqrt{3}(2^{2/3}U^2 \mathrm{i}/4+(\dsq -1) \mathrm{i})/(6U) \\
Z&= \frac{d}{3}+\frac{2^{1/3}U}{12}-\frac{2^{2/3}(\dsq -1)}{6U}-\frac{1}{3}  \\
K&= 2^{2/3}U^2/4-2^{1/3}U-\dsq +2^{1/3}d\,U+1 \\
L&= \sqrt{3} \mathrm{i}-\sqrt{3}\dsq  \mathrm{i}-2^{2/3}\sqrt{3}U^2 \mathrm{i}/4 \,.
\end{align*}

\end{document}