6

I have a proof that should take up only just over half a page; however, when I compile it, only the first few sentences appear in the bottom half of the page where my proof starts, and the rest is moved to the next page. I rather would want text to only appear on the next page after the available white space has been used (as is usually the case).

Here is the proof:

\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}

This is being used within an enumerate environment with many other questions, though I checked, and the problem still persists even without the enumerate environment.

Thanks for any suggestions anyone can provide.

My full code (using the afterpage package suggested here) is here: http://pastebin.com/HZhB5x4i

  • 3
    Without a full compilable example it's hard to tell, but I guess your alignment is so huge that it doesn't fit onto the first page. – Hendrik Vogt Mar 3 '11 at 14:16
  • @Sara: It's recommended to post a minimal example that illustrates your problem. However, I compiled your code and didn't see any problems in the output! – Hendrik Vogt Mar 3 '11 at 15:04
  • @Hendrik I probably could make my example more minimal, but I've at least removed a section of text that didn't pertain to the issue at hand. Hope that helps. You didn't see the proof I mentioned being labeled by a letter instead of a separate number? – Sara Fauzia Mar 3 '11 at 15:15
  • @Sara: I think you just shouldn't use \afterpage and have a look at Herbert's answer. – Hendrik Vogt Mar 3 '11 at 15:19
  • 1
    As a side note, you might consider using \DeclareMathOperator\Rank{Rank} \DeclareMathOperator\Ker{Ker} and then use \Rank and \Ker and similarly for \Null (and any other operators I might have missed). \dim already exists. That first line should probably be Show $\Rank(A) = \Rank(B)$, for $A = {^S}B$. – TH. Mar 3 '11 at 17:28
16

Add this in the preamble after loading amsmath

\allowdisplaybreaks
  • This solution works perfectly and elegantly. Thanks a lot! (None of the aforementioned issues with afterpage occur with this solution). – Sara Fauzia Mar 3 '11 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.