# Offset a symbol without changing layout

I'm using a certain symbol that I would like to shift down without effecting it's layout(change the surrounding markup's positions):

a test case is:

$\begin{array}{ccc} a & b & c\\ a & b & e\\ \end{array}$


and I want to move c down just right on top of e so the bottom of c touches, or even overlap, the top of e without effecting the layout at all(except for c). The two matrices should look identical except for how c changes.

(the answer should be relatively generic since the above is just a test case)

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You can use \smash and \raisebox:

## Notes:

• The \hlines were added so as to be able to compare the before and after.
• The \raisebox is used to lower the desired character, and the \smash is used to ensure that the remainder of the environment is not affected by it.

## Code:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
$\begin{array}{ccc}\hline a & b & c\\ a & b & e\\\hline \end{array}$
$\begin{array}{ccc}\hline a & b & \smash{\raisebox{-1.75ex}{c}}\\ a & b & e\\\hline \end{array}$
\end{document}

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A nested array puts a c over e. This avoids that c and e are on different lines in the outer array and the positioning becomes independent from the contents of the other cells that might increase the distances of the rows. The inner array needs b to put the e on the baseline, see Peter Grill's comment.

Another trick is using \arraystretch with meaning 0. Then the struts are reset with nullified heights and depths that are inserted by environment array.

As requested in the question, \smash nullifies the height (and depth) of c.

\documentclass{article}
\begin{document}
$\begin{array}{ccc} a & b & \\ a & b & \renewcommand*{\arraystretch}{0}% \begin{array}[b]{@{}c@{}}\smash{c}\\e\end{array}\\ \end{array}$
\end{document}


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The bottom of the e is not quite at the baseline. Using \begin{array}[b] for the inner array appears to fix it. –  Peter Grill Nov 15 '12 at 7:41
@PeterGrill Thanks, fixed. –  Heiko Oberdiek Nov 15 '12 at 7:52