\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item
Applying 2 fold symmetry // X$_1$ :
\\
\\
\[
a=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\]
\\
\[
\alpha =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\]
\\
Applying 3 fold symmetry // X$_3$:
\\
\\
\[
a=
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\\
\[
\alpha_1 =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\]
\\
For d$_{im}$,
\\
d' = a * d * \alpha
\[
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\end{enumerate}
\end{document}
4
14
Adaptations:
- math mode for
d' = a * d * \alpha - no linebreak inside
\[ ... \] - use
align*instead of\[ ... \] - use
\cdotinstead of* - use linebreaks instead of
\\ - use
$X_1$instead ofX$_1$ - you can use
\renewcommand{\arraystretch}{1.5}to increase the spacing of the matrices with the fractions (see How can I increase the line spacing in a matrix?) or write the fractions with$a/b$as JPi suggested.
Code:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item
Applying 2 fold symmetry // $X_1$:
\begin{align*}
a &=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\\
\alpha &=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\end{align*}
Applying 3 fold symmetry // $X_3$:
\begin{align*}
a &=
{
\renewcommand{\arraystretch}{1.5}
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
}
\\
\alpha_1 &=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\end{align*}
For $d_{im}$,
$d' = a \cdot d \cdot \alpha$
{
\renewcommand{\arraystretch}{1.5}
\begin{align*}
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\\
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\end{align*}
}
\end{enumerate}
\end{document}
Result:
-
3You've forgotten to list the adaption
$X_1$instead ofX$_1$. Nevertheless +1 for listing all adaptions! – Schweinebacke Sep 5 '17 at 9:39 -
3You may want to increase the spacing in the matrices or better yet: write $\sqrt{3}/2$ instead of $\frac{\sqrt{3}}{2}$. – JPi Sep 5 '17 at 10:03
-
Thank you Schweinebacke and JPi. I added your suggestions to the answer. – dexteritas Sep 5 '17 at 10:51
-
the width of the matrices in the first two blocks makes it easy to identify them as separate. but it wouldn't hurt to have a little vertical space between the two in the last block. add with an option after the double backslash separating them, say
\\[3pt]and be sure not to leave a (typed) space between the second backslash and the[. – barbara beeton Sep 5 '17 at 15:43
2
I did not do much, just corrected some shells
- for \alpha, it can only be used in a mathematical environment, either $\alpha$ or \[\alpha\]
it is possible to print the \alpha character in a text, but then it is \textalpha.
so I will avoid writing
X$_1$ but rather $X_1$ to leave its coherence to the mathematical text.
I also corrected some spaces generating in the mathematical formulas
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item
Applying 2 fold symmetry // X$_1$ :
\\
\\
\[
a=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{bmatrix}
\]
\\
\[
\alpha =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\]
\\
Applying 3 fold symmetry // X$_3$:
\\
\\
\[
a=
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\\
\[
\alpha_1 =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\]
\\
For d$_{im}$,
\\
$d' = a * d * \alpha$
\[
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}\\
\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\end{enumerate}
\end{document}
d' = a * d * \alphais outside math mode, but it needs math mode:$d' = a * d * \alpha$. But the whole code is ugly. You should avoid\\before and after\[…\]. You should not useX$_3$but$X_3$etc. – Schweinebacke Sep 5 '17 at 9:30$\text{x}_3, if the intent is a non-italic variable. (My memory is fuzzy on the correct command to use, though.\mathrminstead?) – chepner Sep 5 '17 at 11:34