3

I am currently working with some pretty big matrices and was wondering how I could better format them to make my final proof appear less sloppy. What formatting techniques can I use to better present these matrices? Current Mess

\documentclass[reqno]{amsart}
    \usepackage{amsmath}
    \usepackage{amssymb}

    \begin{document}
    \begin{enumerate}
    \begin{enumerate}
    \item
    \begin{proof}
        \begin{align*} A^n\begin{pmatrix} 1 \\ 0 \end{pmatrix}&=P\begin{pmatrix}
                \left(\frac{1+\sqrt{5}}{2}\right)^n & 0 \\ 0 & \left(\frac{1-\sqrt{5}}{2}\right)^n
            \end{pmatrix}P^{-1}\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
            \begin{pmatrix}
                F_{n+1} \\ F_n
            \end{pmatrix}&=\begin{pmatrix}
            \frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\[4pt] 1 & 1
        \end{pmatrix}\begin{pmatrix}
                \left(\frac{1+\sqrt{5}}{2}\right)^n & 0 \\[4pt] 0 & \left(\frac{1-\sqrt{5}}{2}\right)^n
            \end{pmatrix}P^{-1}\begin{pmatrix} 1 \\[4pt] 0 \end{pmatrix} \\
            &=\left(\frac{1}{10}\right)\begin{pmatrix}
            \left(\frac{1+\sqrt{5}}{2}\right)^{n+1} & \left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \\ \left(\frac{1+\sqrt{5}}{2}\right)^n & \left(\frac{1-\sqrt{5}}{2}\right)^n  
            \end{pmatrix}\begin{pmatrix}
            2\sqrt{5} & 5-\sqrt{5} \\[4pt]-2\sqrt{5} & 5+\sqrt{5}
        \end{pmatrix}\begin{pmatrix}
            1 \\ 0
        \end{pmatrix} \\
            &=\left(\frac{1}{10}\right)\begin{pmatrix} 2\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-2\sqrt{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} & (5-\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}+(5+\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)^{n+1} \\ 2\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^n-2\sqrt{5}\left(\frac{1-\sqrt{5}}{2}\right)^n & (5-\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)^n+(5+\sqrt{5})\left(\frac{1+\sqrt{5}}{2}\right)^n 
        \end{pmatrix}\begin{pmatrix}
            1 \\ 0
        \end{pmatrix} \\
        \begin{pmatrix}
                F_{n+1} \\ F_n
            \end{pmatrix}&=\left(\frac{1}{10}\right)\begin{pmatrix}
        2\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-2\sqrt{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \\ 2\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^n-2\sqrt{5}\left(\frac{1-\sqrt{5}}{2}\right)^n  
        \end{pmatrix}
        \end{align*}
        We see that $F_n$ must equal the bottom entry in this matrix. $$F_n=\frac{1}{10}\left(2\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^n-2\sqrt{5}\left(\frac{1-\sqrt{5}}{2}\right)^n\right)$$ Which simplifies to $$F_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n \right]$$.
    \end{proof}
        \end{enumerate}          
    \end{enumerate}
    \end{document}
  • 1
    Use a symbol, say \varphi instead of \frac{1+\sqrt{5}}{2} and, say, \bar\phi for \frac{1-\sqrt{5}}{2}. This will immediately reduce the size of your equations. – egreg Nov 20 '15 at 9:42
  • just a side node... F_n is equal to the bottom entry in the vector - after the matrix-vector multiplication you only have a vector left. – Ronny Nov 20 '15 at 10:37
3

A trick is to replace complicated objects with simpler ones.

\documentclass[reqno]{amsart}
\usepackage{amsmath}
\usepackage{amssymb}

\newcommand{\fib}{\varphi} % Fibonacci constant
\newcommand{\fibc}{\bar\varphi} % conjugate

\begin{document}
\begin{proof}
Set $\fib=(1+\sqrt{5})/2$ and $\fibc=(1-\sqrt{5})/2$. Then
\begin{align*}
A^n\begin{pmatrix} 1 \\ 0 \end{pmatrix}
  &=P\begin{pmatrix}\fib^n & 0 \\ 0 & \fibc^n \end{pmatrix}P^{-1}
    \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\\[2ex]
\begin{pmatrix}
F_{n+1} \\ F_n
\end{pmatrix}
  &=\begin{pmatrix} \fib & \fibc \\ 1 & 1 \end{pmatrix}
    \begin{pmatrix} \fib^n & 0 \\ 0 & \fibc^n \end{pmatrix}P^{-1}
    \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\\[1ex]
  &=\frac{1}{10}
    \begin{pmatrix} \fib^{n+1} & \fibc^{n+1} \\ \fib^n & \fibc^n \end{pmatrix}
    \begin{pmatrix} 2\sqrt{5} & 5-\sqrt{5} \\ -2\sqrt{5} & 5+\sqrt{5} \end{pmatrix}
    \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\\[1ex]
  &=\frac{1}{10}
    \begin{pmatrix}
    2\sqrt{5}\,\fib^{n+1}-2\sqrt{5}\,\fibc^{n+1} & (5-\sqrt{5})\fib^{n+1}+(5+\sqrt{5})\fib^{n+1}
    \\[1ex]
    2\sqrt{5}\,\fib^n-2\sqrt{5}\,\fibc^n & (5-\sqrt{5})\fib^n+(5+\sqrt{5})\fib^n 
    \end{pmatrix}
    \begin{pmatrix} 1 \\[1ex] 0 \end{pmatrix}
\\[2ex]
\begin{pmatrix} F_{n+1} \\[1ex] F_n \end{pmatrix}
  &=\frac{1}{10}
    \begin{pmatrix}
    2\sqrt{5}\,\fib^{n+1}-2\sqrt{5}\,\fibc^{n+1} \\[1ex]
    2\sqrt{5}\,\fib^n-2\sqrt{5}\,\fibc^n
    \end{pmatrix}
\end{align*}
We see that $F_n$ must equal the bottom entry in this matrix, so
\[
F_n=\frac{1}{10}\bigl(2\sqrt{5}\,\fib^n-2\sqrt{5}\,\fibc^n\bigr)
\]
which simplifies to 
\[
F_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n \right].
\qedhere
\]
\end{proof}

\end{document}

I increased the leading just in selected matrices. The parentheses around the 1/10 fraction were misleading and unnecessary. Note also that \, has been inserted between the square root and the phi.

Note that $$ should never be used in LaTeX (and I really mean it).

enter image description here

  • You beat me to this answer by a couple of minutes... (I was going to use \mu and \nu where you use \varphi and \bar{\varphi}.) – Mico Nov 20 '15 at 10:15
  • @Mico Just replace them in the definition of \fib and \fibc. ;-) – egreg Nov 20 '15 at 10:17
  • Thank you so much. Could you explain why $$should never be used? – cpage Nov 20 '15 at 11:27
  • 2
    @cpage See Why is \[ … \] preferable to $$? – egreg Nov 20 '15 at 11:33

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.