Formatting Large Equations with Matrices

Question

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}

Answer 1

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).