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校样分为两页

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校样分为两页

我有一个证明，它应该只占用半页多一点的空间；但是，当我编译它时，只有前几句话出现在证明开始的那一页的下半部分，其余的都移到了下一页。我宁愿让文本只在可用的空白处被使用后才出现在下一页上（通常都是这样）。

证明如下：

\documentclass{article}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{lipsum}
\begin{document}

\lipsum[1-3]
\begin{enumerate}
        \item Show Rank($A$) = Rank($B$), for $A = {^S}B$.
            \begin{proof}
                    For $S^{-1}AS$ to be defined, $A$ must be $n \times n$ for S $n \times n$, and as $S^{-1}AS = B$, $B$ is also $n \times n$. By the Rank-Nullity theorem, Rank($A$) = n $-$ Null($A$). Hence, it suffices to show Null($A$) = Null($B$), as both $A$ and $B$ have $n$ columns. 
                    \begin{align*}
                            \mathbf{0} & = B\mathbf{x} = S^{-1}AS\mathbf{x},\ \mathbf{x} \in \mathrm{Ker}(B) \\
                            & = SS^{-1}AS\mathbf{x},\ \text{as}\ S\mathbf{0} = \mathbf{0} \\
                            & = I_{n \times n}AS\mathbf{x} \\
                            & = AS\mathbf{x} \\
                            & \Rightarrow S\mathbf{x} \in \mathrm{Ker}(A) \tag{71.a.1} \\
                            \mathbf{0} & = A\mathbf{x} = SBS^{-1}\mathbf{x},\ \mathbf{x} \in \mathrm{Ker}(A) \\
                            & = S^{-1}SBS^{-1}\mathbf{x},\ \text{as}\ S^{-1}\mathbf{0} = \mathbf{0} \\
                            & = I_{n \times n}BS^{-1}\mathbf{x} \\
                            & = BS^{-1}\mathbf{x} \\
                            & \Rightarrow S^{-1}\mathbf{x} \in \mathrm{Ker}(B) \tag{71.a.2} \\
                            \mathbf{0} & = c_1\mathbf{v_1} + \dotsb + c_p\mathbf{v_p} \Rightarrow c_i = 0\ \text{for}\ \bigl\{v_i \vert i \in \{1,\dotsc,p\}\bigr\}\ \text{a basis of $B$} \\
                            & = c_1S\mathbf{v_1} + \dotsb + c_nS\mathbf{v_p},\ \text{as $S$ is linear} \\
                            & \Rightarrow S\mathbf{v_i}\ \text{are linearly independent} \bigl(\text{in}\ \mathrm{Ker}(A)\ \text{by (71.a.1)}\bigr),\ \text{as $c_i$ are unchanged and all zero} \\
                            & \Rightarrow \mathrm{dim}\bigl(\mathrm{Ker}(A)\bigr) \geq \mathrm{dim}\bigl(\mathrm{Ker}(B)\bigr) = p \tag{71.b.1} \\
                            \mathbf{0} & = c_1\mathbf{w_1} + \dotsb + c_q\mathbf{w_q} \Rightarrow c_i = 0\ \text{for}\ \bigl\{w_i \vert i \in \{1,\dotsc,q\}\bigr\}\ \text{a basis of $A$} \\ 
                            & = c_1S^{-1}\mathbf{w_1} + \dotsb + c_qS^{-1}\mathbf{w_q},\ \text{as $S^{-1}$ is linear} \\
                            & \Rightarrow S^{-1}\mathbf{w_i}\ \text{are linearly independent}\ \bigl(\text{in}\ \mathrm{Ker}(B)\ \text{by (71.a.2)}\bigr),\ \text{as $c_i$ are unchanged and all zero} \\
                            & \Rightarrow \mathrm{dim}\bigl(\mathrm{Ker}(B)\bigr) \geq \mathrm{dim}\bigl(\mathrm{Ker}(A)\bigr) = q \tag{71.b.2}
                    \end{align*}
                    Combining (71.b.1) and (71.b.2), as $p \geq q$ and $q \geq p$, by the antisymmetric property of the order relation, $p = q$, and Nul($A$) = Nul($B$).
            \end{proof}
\end{enumerate}
\end{document}

尽管我检查过，但这是在枚举环境中与许多其他问题一起使用的，即使没有枚举环境，问题仍然存在。

感谢任何人提供的建议。

我的完整代码（使用这里建议的 afterpage 包）在这里： http://pastebin.com/HZhB5x4i

答案1

加载后在序言中添加此内容amsmath

\allowdisplaybreaks

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