Tabular environment - text vertically positions itself by bottom of tikz picture in adjacent cell
I would like to get my text (actually a matrix) to vertically center itself in its cell. Instead, it centers itself based on where the bottom of the tikz picture in the adjacent cell lies. This adds unnecessary height to a table I would like to eliminate. Is there any way to force the text column to align with the center of the tikz picture or to ignore the tikz picture entirely?
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
tables vertical-alignment
add a comment |
I would like to get my text (actually a matrix) to vertically center itself in its cell. Instead, it centers itself based on where the bottom of the tikz picture in the adjacent cell lies. This adds unnecessary height to a table I would like to eliminate. Is there any way to force the text column to align with the center of the tikz picture or to ignore the tikz picture entirely?
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
tables vertical-alignment
add a comment |
I would like to get my text (actually a matrix) to vertically center itself in its cell. Instead, it centers itself based on where the bottom of the tikz picture in the adjacent cell lies. This adds unnecessary height to a table I would like to eliminate. Is there any way to force the text column to align with the center of the tikz picture or to ignore the tikz picture entirely?
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
tables vertical-alignment
I would like to get my text (actually a matrix) to vertically center itself in its cell. Instead, it centers itself based on where the bottom of the tikz picture in the adjacent cell lies. This adds unnecessary height to a table I would like to eliminate. Is there any way to force the text column to align with the center of the tikz picture or to ignore the tikz picture entirely?
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
tables vertical-alignment
tables vertical-alignment
asked yesterday
AlexJAlexJ
384
384
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can use baseline=(current bounding box.center)
to achieve the desired alignment as shown in the following code:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
As the border of the image now overlaps with the horizontal lines, you might want to add some extra vertical space (as done here using the cellspace
package):
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{cellspace}
setlengthcellspacetoplimit{6pt}
setlengthcellspacebottomlimit{6pt}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| Sc | Sc | Sc |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
A fix with an optional argument for the baseline of the tikzpicture:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}[1][-17pt]
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=#1]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}%
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic[-25pt] & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use baseline=(current bounding box.center)
to achieve the desired alignment as shown in the following code:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
As the border of the image now overlaps with the horizontal lines, you might want to add some extra vertical space (as done here using the cellspace
package):
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{cellspace}
setlengthcellspacetoplimit{6pt}
setlengthcellspacebottomlimit{6pt}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| Sc | Sc | Sc |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
You can use baseline=(current bounding box.center)
to achieve the desired alignment as shown in the following code:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
As the border of the image now overlaps with the horizontal lines, you might want to add some extra vertical space (as done here using the cellspace
package):
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{cellspace}
setlengthcellspacetoplimit{6pt}
setlengthcellspacebottomlimit{6pt}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| Sc | Sc | Sc |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
You can use baseline=(current bounding box.center)
to achieve the desired alignment as shown in the following code:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
As the border of the image now overlaps with the horizontal lines, you might want to add some extra vertical space (as done here using the cellspace
package):
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{cellspace}
setlengthcellspacetoplimit{6pt}
setlengthcellspacebottomlimit{6pt}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| Sc | Sc | Sc |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
You can use baseline=(current bounding box.center)
to achieve the desired alignment as shown in the following code:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
As the border of the image now overlaps with the horizontal lines, you might want to add some extra vertical space (as done here using the cellspace
package):
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{cellspace}
setlengthcellspacetoplimit{6pt}
setlengthcellspacebottomlimit{6pt}
usepackage{tikz}
newcommand{pic}{
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=(current bounding box.center)]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}}
}
begin{document}
begin{tabular}{| Sc | Sc | Sc |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
edited yesterday
answered yesterday
leandriisleandriis
9,5451530
9,5451530
add a comment |
add a comment |
A fix with an optional argument for the baseline of the tikzpicture:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}[1][-17pt]
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=#1]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}%
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic[-25pt] & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
A fix with an optional argument for the baseline of the tikzpicture:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}[1][-17pt]
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=#1]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}%
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic[-25pt] & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
add a comment |
A fix with an optional argument for the baseline of the tikzpicture:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}[1][-17pt]
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=#1]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}%
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic[-25pt] & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
A fix with an optional argument for the baseline of the tikzpicture:
documentclass[12pt]{article}
usepackage{amsmath}
usepackage{tikz}
newcommand{pic}[1][-17pt]
{centering
begin{tikzpicture}[x=1cm,y=1cm,baseline=#1]
useasboundingbox (0,.5) rectangle (3, -2);
draw (current bounding box.north east) -- (current bounding box.north west) -- (current bounding box.south west) -- (current bounding box.south east) -- cycle;
end{tikzpicture}%
}
begin{document}
begin{tabular}{| c | c | c |} hline
Initial Pic & Final Pic & U \ hline
pic & pic & \ hline
pic & pic & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
pic[-25pt] & that &
$text{U} = .5 begin{bmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
this & that & $text{U} = begin{bmatrix}
1 & i & 1 & -i \
-i & 1 & i & 1 \
1 & -i & 1 & i \
i & 1 & -i & 1 end{bmatrix}$ \ hline
end{tabular}
end{document}
answered yesterday
koleygrkoleygr
12.3k11038
12.3k11038
add a comment |
add a comment |
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