# symmetric matrices

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

Thank you

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


[text edited for a more complex configuration]

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I see two possible approaches.

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

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


Supplement:

\documentclass{article}
\usepackage{amsmath}

\begin{document}
$$\begin{bmatrix} u,_1+x_3\theta_2,_1-x_2\theta_3,_1&\frac{1}{2}(v,_1-x_3\theta_1,_1-\theta_3) & \frac{1}{2}(w,_1+x_2\theta_1,_1+\theta_2) \\ & 0 & {0} \\ \multicolumn{2}{c}{\text{\smash{\raisebox{1.5ex}{Sym.}}}} & {0} \end{bmatrix}$$
\end{document}

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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