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what are your solutions to properly place the shortcut 'sym' in symmetric matrices?

Thank you

\documentclass{book}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{bmatrix}u,_1+x_3\theta_2,_1-x_2\theta_3,_1&\frac{1}{2}(v,_1-x_3
\theta_1,_1-\theta_3) & \frac{1}{2}(w,_1+x_2\theta_1,_1+\theta_2)\\
\text{Sym.} & {0} & {0} \\ {} & {} & {0} 
\end{bmatrix}
\end{equation}
\end{document}

[text edited for a more complex configuration]

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1 Answer 1

up vote 3 down vote accepted

I see two possible approaches.

\documentclass{article}
\usepackage{amsmath}
\usepackage{multirow}

\begin{document}
  \begin{equation}
    \begin{bmatrix}
      1 & 1 & 1 & 1 \\
        & 1 & 1 & 1 \\
        & \multirow{2}{*}{\makebox[0pt]{\text{sym.}}}  & 1 & 1 \\
        &   &   & 1 
    \end{bmatrix}
  \end{equation}
  \begin{equation}
    \begin{bmatrix}
      1 & 1 & 1 & 1 \\
        & 1 & 1 & 1 \\
        &   & 1 & 1 \\
        \multicolumn{2}{c}{\text{sym.}} & & 1 
    \end{bmatrix}
  \end{equation}
\end{document}

Supplement:

\documentclass{article}
\usepackage{amsmath}

\begin{document}
  \begin{equation}
    \begin{bmatrix}
      u,_1+x_3\theta_2,_1-x_2\theta_3,_1&\frac{1}{2}(v,_1-x_3\theta_1,_1-\theta_3) & \frac{1}{2}(w,_1+x_2\theta_1,_1+\theta_2) \\
      & 0 & {0} \\
      \multicolumn{2}{c}{\text{\smash{\raisebox{1.5ex}{Sym.}}}} & {0} 
    \end{bmatrix}
  \end{equation}
\end{document}
share|improve this answer
    
I just edited the initial code for a more challenging configuration. Your solution is fine but not always suitable, I think. –  pluton Oct 9 '10 at 23:59
    
It requires some creativity but is possible with simple measures (see supplement). Now we should have covered all possible cases except a matrix within a matrix. –  Thorsten Donig Oct 10 '10 at 16:15
    
correct. I was investigating the \raisebox strategy without being aware of the smash command. You are right, most of the possible cases are now covered. Thanks –  pluton Oct 10 '10 at 16:58

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