2

Starting from this code

\documentclass[12pt]{book}
\usepackage{mathtools,amssymb} 
\usepackage{newtxtext}
\usepackage[lite]{mtpro2}
\begin{document}
$(L\boldsymbol{\mathcal A}')^\intercal =\boldsymbol{\mathcal A}'{}^\intercal  L^\intercal $
\end{document}

with this output

enter image description here

I'd like to move the \intercal command a little higher and to the left in the direction indicated by the arrow. But does LaTeX have a direct command to move the symbol up or down or do I need to use specific macros? What could be the ways to have an adequate and beautiful way to write this formula?

Mico
  • 506,678
Sebastiano
  • 54,118

2 Answers2

7

Combining David's suggestion with Sebastiano's requirement...

I use \mathsf{T} as the core of the revised \intercal (you can rename it something else if you don't want to overwrite it), but then to make it smaller, I use it in cramped, superscripted form (revision 1). Alternately, I just set it in \scriptscriptstyle, which lowers its elevation a bit, while still keeping it higher than the original \intercal (revision 2).

Original, then two revisions.

\documentclass[12pt]{book}
\usepackage{mathtools,amssymb} 
\usepackage{newtxtext}
\usepackage{newtxmath}%[lite]{mtpro2}
\begin{document}
$(L\boldsymbol{\mathcal A}')^\intercal =\boldsymbol{\mathcal A}'{}^\intercal  L^\intercal $

\renewcommand\intercal{{\cramped{{}^\mathsf{T}}}}
$(L\boldsymbol{\mathcal A}')^\intercal =\boldsymbol{\mathcal A}'{}^\intercal  L^\intercal $

\renewcommand\intercal{\mathsf{\scriptscriptstyle T}}
$(L\boldsymbol{\mathcal A}')^\intercal =\boldsymbol{\mathcal A}'{}^\intercal  L^\intercal $
\end{document}

enter image description here

3

Here's another variation on David's idea to use \mathsf{T} instead of either \intercal or \top. The macro \tp (short for "transpose", naturally) employs \scriptscriptstyle math mode. The \tp macro takes an optional argument: a scalar number, to indicate how many mu the "T" should be shifted to the left. The ability to fine-tune the horizontal position can come in handy if you're dealing with letters, such as L, which have no component in the upper-right quadrant. (The default amount of left-shift is 3, as in -3mu. Aside: \mkern-3mu is the same as \!.)

enter image description here

\documentclass[12pt]{book}
\usepackage{mathtools,amssymb,newtxtext}
\usepackage[lite]{mtpro2}
\newcommand\bmcA{\boldsymbol{\mathcal A}'{}}
\newcommand\tp[1][3]{^{^{\mkern-#1mu\mathsf{T}}}}
\begin{document}
$\intercal$ $\top$ $\mathsf{T}$

\medskip
\begin{tabular}{@{}ll}
$(L\bmcA)^\intercal = \bmcA{}^\intercal L^\intercal$ & \verb+\intercal+ \\
$(L\bmcA)^\top = \bmcA^\top L^\top$                  & \verb+\top+      \\
$(L\bmcA)\tp = \bmcA\tp L\tp = \bmcA{}\tp L\tp[6]$   & \verb+\tp+
\end{tabular}
\end{document}
Mico
  • 506,678
  • The answers are all very good. :-( If I don't give the green check it's because I'm good with the two answers at the same time. – Sebastiano Apr 04 '19 at 18:57