# Formatting Large Equations with Matrices

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?

\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} • 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 ## 1 Answer 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 $$should never be used in LaTeX (and I really mean it). • You beat me to this answer by a couple of minutes... (I was going to use \mu and \nu where you use \varphi and \bar{\varphi}.) – Mico Nov 20 '15 at 10:15 • @Mico Just replace them in the definition of \fib and \fibc. ;-) – egreg Nov 20 '15 at 10:17 • Thank you so much. Could you explain why$$should never be used? – cpage Nov 20 '15 at 11:27
• @cpage See Why is $…$ preferable to ? – egreg Nov 20 '15 at 11:33