3

I have something published by IEEE. I noticed that with the PDF generated by myself (top picture), some inline vectors look not as compact as in the PDF version produced by IEEE (bottom picture). The comparison is as follows. My question is, in general (it could be IEEEtran class, book class, or something else), how to make those inline math look more compact? What package or options should I use to achieve a similar result?

enter image description here

enter image description here

This is the code for producing those lines:

columns of $\boldsymbol{\mathcal{A}}$ identical and its rank is reduced to 2,
with the null space of $\left\{ \left[\begin{array}{ccc} -1/\sqrt{2} & 0 &
      1/\sqrt{2}\end{array}\right]^{\text{T}}\right\}$. The other part of the
condition shows that for an $\boldsymbol{\mathbf{N}}=\left[\begin{array}{ccc}
    N_{1} & N_{2} & N_{3}\end{array}\right]^{\text{T}}$ in
$\boldsymbol{\mathcal{A}}$ null space to be a reasonable solution, $N_{1}=N_{3}$
must be satisfied.
1
  • Obviously the \left and \right are adding a lot of space for one thing
    – Au101
    Sep 5 '18 at 16:48
4

I don't think the braces should go in the first instance. Anyway, inline math with \left and \right will usually disrupt the interline spacing. I suggest using simply [...].

I also suggest a few improvements:

  1. instead of the clumsy \boldsymbol{\mathcal{A}} you can define a new command, \bcal;
  2. using matrix instead of array will remove the sidebearings from the array;
  3. \text{T} for the transpose would give bad results in case the text is in a theorem statement, better using \mathrm (inside a semantic command).

Note also that \boldsymbol{\mathbf{N}} does nothing different from \mathbf{N}.

The vinculum of the square root would clash with the bracket, so I recommend \, for inserting a thin space.

I suggest two realizations for \rowvector: one with the default intercolumn space, one with half of it.

\documentclass{article}
\usepackage{amsmath,bm}

\newcommand{\bcal}[1]{\bm{\mathcal{#1}}}
\newcommand{\transpose}{^{\mathrm{T}}}

\newcommand{\rowvector}[1]{[\begin{matrix}#1\end{matrix}]}

\begin{document}

\noindent % just for this example
columns of $\bcal{A}$ identical and its rank is reduced to~$2$, with the null 
space of $\rowvector{-1/\sqrt{2} & 0 & 1/\sqrt{2}\,}\transpose$.
The other part of the condition shows that for an 
$\mathbf{N}=\rowvector{N_{1} & N_{2} & N_{3}}\transpose$ in
$\bcal{A}$ null space to be a reasonable solution, $N_{1}=N_{3}$
must be satisfied.

\bigskip

% for the second example
\renewcommand{\rowvector}[1]{%
  [\begingroup
  \setlength\arraycolsep{0.5\arraycolsep}%
  \begin{matrix}#1\end{matrix}%
  \endgroup]
}


\noindent % just for this example
columns of $\bcal{A}$ identical and its rank is reduced to~$2$, with the null 
space of $\rowvector{-1/\sqrt{2} & 0 & 1/\sqrt{2}\,}\transpose$.
The other part of the condition shows that for an 
$\mathbf{N}=\rowvector{N_{1} & N_{2} & N_{3}}\transpose$ in
$\bcal{A}$ null space to be a reasonable solution, $N_{1}=N_{3}$
must be satisfied.

\end{document}

enter image description here

1
  • As better alternatives to \boldsymbol{\mathcal{...}}, mathalpha defines \mathbcal and unicode-math defines \mathbfcal or \symbfcal.
    – Davislor
    Oct 2 '20 at 6:11
0

do not use \left and \right and no array:

columns of $\boldsymbol{\mathcal{A}}$ identical and its rank is reduced to 2, with the null space 
of $\{ [-1/\sqrt{2} 0 1/\sqrt{2}]^{\text{T}}\}$. The other part of the condition 
shows that for an $\mathbf{N}=[N_{1} N_{2} N_{3}]^{\text{T}}$ 
in $\boldsymbol{\mathcal{A}}$ null space to be 
a reasonable solution, $N_{1}=N_{3}$ must be satisfied.

enter image description here

If you want some space between the elements write N_1\, N_2,...

2
  • Somehow I checked their webpage version and it seems that the code they are using are the same as mine. They seem to have used \left \right and array so I was wondering if they use some additional package.
    – nanjun
    Sep 5 '18 at 17:02
  • You're short some space between elements within the vector...
    – Werner
    Sep 5 '18 at 17:45
0

In math mode there are plenty of options. The relevant one here might be \! (bang-escape a space).

Overleaf has a great resource on the topic. Here is an extract from their excellent page:

\begin{align*}
f(x) &= x^2\! +3x\! +2 \\
f(x) &= x^2+3x+2 \\
f(x) &= x^2\, +3x\, +2 \\
f(x) &= x^2\: +3x\: +2 \\
f(x) &= x^2\; +3x\; +2 \\
f(x) &= x^2\ +3x\ +2 \\
f(x) &= x^2\quad +3x\quad +2 \\
f(x) &= x^2\qquad +3x\qquad +2
\end{align*}

The rendering shows how the different options finely tunes spacing (result available at the linked page).

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.