Diagonal dots spanning multiple lines/columns of a matrix

Question

I'm filling in large sparse matrices, for example

\begin{bmatrix}
0  & -2     & 1      &        & -1     & 2  \\
2  & \ddots & \ddots & \ddots &        & -1 \\
-1 & \ddots & \ddots & \ddots & \ddots &    \\
   & \ddots & \ddots & \ddots & \ddots & 1  \\
1  &        & \ddots & \ddots & \ddots & -2 \\
-2 & 1      &        & -1     & 2      & 0
\end{bmatrix}

I do not like the way the dots are spaced. I would like them to be equally spaced across the diagonals. I guess this can be done using TikZ but is there an easier way?

Answer 1

It's not difficult with TikZ. I have just removed the ddots

\documentclass{article}
\usepackage{amsmath,tikz}
\usetikzlibrary{matrix}
\begin{document}
Following quadratic form involves a skew symmetric matrix. 
\begin{equation}
\begin{pmatrix}
    a\\b\\\vdots\\z
\end{pmatrix}^T
\begin{tikzpicture}[baseline=(current bounding box.center)]
\matrix (m) [matrix of math nodes,nodes in empty cells,right delimiter={]},left delimiter={[} ]{
0  & -2 & 1  &  & -1 & 2  \\
2  & & & & & -1 \\
-1 & & & & &    \\
   & & & & & 1  \\
1  & & & & & -2 \\
-2 &1 & & -1 & 2 & 0\\
} ;
\draw[loosely dotted] (m-1-1)-- (m-6-6);
\draw[loosely dotted] (m-1-2)-- (m-5-6);
\draw[loosely dotted] (m-2-1)-- (m-6-5);
\end{tikzpicture}\begin{pmatrix}
    a\\b\\\vdots\\z
\end{pmatrix}=0
\label{eq:eqq1}
\end{equation}

\end{document}

It looks slightly faint in the picture but if you wish you can add the option thick to the draw commands.

Answer 2

Such diagonals are definitely possible using graphic packages like tikz/pgf or pstricks. However, here's one using TeX leaders:

\documentclass{article}

\usepackage{amsmath,graphicx}

\newcommand{\diagdots}[3][-25]{%
  \rotatebox{#1}{\makebox[0pt]{\makebox[#2]{\xleaders\hbox{$\cdot$\hskip#3}\hfill\kern0pt}}}%
}

\begin{document}

\[
  \begin{bmatrix}
     0 & -2 & 1 &    & -1 &  2 \\
     2 &    &   &    &    & -1 \\
    -1 &    &   &    &    &    \\
       &    &  \multicolumn{2}{c}{\smash{\raisebox{.5\normalbaselineskip}{\diagdots{8em}{.5em}}}} &    &  1 \\
     1 &    &   &    &    & -2 \\
    -2 &  1 & \phantom{-2}  & -1 &  2 &  0
  \end{bmatrix}
\]

\end{document}

The minimal example provides \diagdots[<angle>]{<len>}{<skip>} that draws a diagonal array of dots (actually \cdots) of length <len> at an angle of <angle> (default is -25). The <skip> defines the approximate length between dots.

I've placed \diagdots in the middle of your bmatrix (horizontally by using \multicolumn{2}{c}{...} and vertically by using \raisebox{.5\normalbaselineskip}{...}), and \smashed it to remove any vertical height distortion. The \diagdots output has zero width (by virtue of \makebox[0pt]).

You can play around with the lengths and angles so see what suits you.

Answer 3

Ellipses are usually used to indicate that something is repeated. In this case you want to indicate that the 0 entry is repeated, and hence should show that there is a 0 on either end of the ellipses.

So, assuming that there are a fixed number of rows (as your example seems to indicate), then I would just use \cdots in either one column or both:

If the number of rows is not fixed, then I would use one of the following. My preference would be the last one as then there is no confusion as to where the 0s are.

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\[
\begin{bmatrix}
 0 & -2 & 1 & 0      & -1 &  2 \\
 2 &  0 &   & \cdots &  0 & -1 \\
-1 &  0 &   & \cdots &  0 &  0 \\
 0 &  0 &   & \cdots &  0 &  1 \\
 1 &  0 &   & \cdots &  0 & -2 \\
-2 &  1 & 0 & -1     &  2 &  0
\end{bmatrix}
%
\begin{bmatrix}
 0 & -2 &  1     &  0     & -1 &  2 \\
 2 &  0 & \cdots & \cdots &  0 & -1 \\
-1 &  0 & \cdots & \cdots &  0 &  0 \\
 0 &  0 & \cdots & \cdots &  0 &  1 \\
 1 &  0 & \cdots & \cdots &  0 & -2 \\
-2 &  1 &  0     & -1     &  2 &  0
\end{bmatrix}
\]

If the number of rows is not fixed:
\[
\begin{bmatrix}
 0 & -2 & 1 & 0      & -1 &  2 \\
 2 &  0 &   & \cdots &  0 & -1 \\
-1 &  0 &   & \cdots &  0 &  0 \\
 0 &  0 &   & \cdots &  0 &  0 \\
 \vdots &  \vdots &   & \vdots &  \vdots &  \vdots \\
 0 &  0 &   & \cdots &  0 &  0 \\
 0 &  0 &   & \cdots &  0 &  1 \\
 1 &  0 &   & \cdots &  0 & -2 \\
-2 &  1 & 0 & -1     &  2 &  0
\end{bmatrix}
%
\begin{bmatrix}
 0      & -2     &  1     & 0      & -1     &  2 \\
 2      &  0     & \cdots & \cdots &  0     & -1 \\
-1      &  0     & \cdots & \cdots &  0     &  0 \\
 0      &  0     & \cdots & \cdots &  0     &  0 \\
 \vdots & \vdots & \ddots &        &  \vdots &  \vdots \\
 \vdots & \vdots &        & \ddots &  \vdots &  \vdots \\
 0      &  0     & \cdots & \cdots &  0     &  0 \\
 0      &  0     & \cdots & \cdots &  0     &  1 \\
 1      &  0     & \cdots & \cdots &  0     & -2 \\
-2      &  1     &  0     & -1     &  2     &  0
\end{bmatrix}
\]

\end{document}

Answer 4

I suggest to use the package nicematrix which has functionalities dedicated to his problem. The dotted lines are drawn with Tikz.

\documentclass{article}
\usepackage{nicematrix}
\NiceMatrixOptions{nullify-dots}
\begin{document}
\[\begin{bNiceMatrix}[columns-width = 5mm]
0  & -2     & 1      &        & -1 & 2  \\
2  & \Ddots & \Ddots & \Ddots &    & -1 \\
-1 & \Ddots &        &        &    &    \\
   & \Ddots &        &        &    & 1  \\
1  &        &        &        &    & -2 \\
-2 & 1      &        & -1     & 2  & 0
\end{bNiceMatrix}\]
\end{document}

Answer 5

runt it with xelatex

\documentclass{article}
\usepackage{mathtools}
\usepackage{pst-node}
\begin{document}
Following quadratic form involves a skew symmetric matrix. 
\begin{equation}\label{eq:eqq1}
\begin{pmatrix*}[r]
    a\\b\\c\\\vdots\\z
\end{pmatrix*}^T
\begin{bmatrix*}[r]
\rnode[rb]{C}{0}  & \rnode[rb]{B}{-2} & \rnode[rb]{A}{1}  &  & -1 & 2  \\
\rnode[rb]{D}{2}  & & & & & -1 \\
\rnode[rb]{E}{-1} & & & & &    \\
                  & & & & & \rnode[l]{a}{1}  \\
               1  & & & & & -\rnode[l]{b}{2} \\
              -2  &1& & -\rnode[l]{e}{1} & \rnode[rb]{d}{2} & \rnode[l]{c}{0}\\
\end{bmatrix*}
\begin{pmatrix*}[r]
    a\\b\\c\\\vdots\\z
\end{pmatrix*}=0
\psset{linestyle=dotted,nodesep=2mm}
\ncline{A}{a}\ncline{B}{b}\ncline{C}{c}\ncline{D}{d}\ncline{E}{e}
\end{equation}

\end{document}