Tabular environment - text vertically positions itself by bottom of tikz picture in adjacent cell












3















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}


Table










share|improve this question



























    3















    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}


    Table










    share|improve this question

























      3












      3








      3








      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}


      Table










      share|improve this question














      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}


      Table







      tables vertical-alignment






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      AlexJAlexJ

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          4














          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}


          enter image description here



          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}


          enter image description here






          share|improve this answer

































            3














            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}


            enter image description here






            share|improve this answer























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              2 Answers
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              2 Answers
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              active

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              active

              oldest

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              active

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              4














              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}


              enter image description here



              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}


              enter image description here






              share|improve this answer






























                4














                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}


                enter image description here



                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}


                enter image description here






                share|improve this answer




























                  4












                  4








                  4







                  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}


                  enter image description here



                  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}


                  enter image description here






                  share|improve this answer















                  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}


                  enter image description here



                  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}


                  enter image description here







                  share|improve this answer














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                  edited yesterday

























                  answered yesterday









                  leandriisleandriis

                  9,5451530




                  9,5451530























                      3














                      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}


                      enter image description here






                      share|improve this answer




























                        3














                        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}


                        enter image description here






                        share|improve this answer


























                          3












                          3








                          3







                          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}


                          enter image description here






                          share|improve this answer













                          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}


                          enter image description here







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          koleygrkoleygr

                          12.3k11038




                          12.3k11038






























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