Is there any way to put \ddots in an "x" shape?

Question

I tried using this code but ir isn't the result I wanted

\newcommand{\xdots}{\ensuremath{{}^\cdot_\cdot}\cdot^\cdot_\cdot}

It has to go into this matrix

$$B_j\rightarrow\tilde{B}_j=\begin{pmatrix}a_{11}&\dots&a_{1,j-1}&0&a_{1,j+1}&\dots&a_{1n}\\\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\a_{i-1,1}&\dots&a_{i-1,j-1}&0&a_{i-1,j+1}&\dots&a_{i-1,n}\\0&\dots&0&1&0&\dots&0\\a_{i+1,1}&\dots&a_{i+1,j-1}&0&a_{i+1,j+1}&\dots&a_{i+1,n}\\\vdots&\xdots&\vdots&\vdots&\vdots&\ddots&\vdots\\a_{n11}&\dots&a_{n,j-1}&0&a_{n,j+1}&\dots&a_{nn}\end{pmatrix}$$

Instead of \ddots but the upper and lower dots are smaller in my way, if there isn't a way i will use \ddots

Steven B. Segletes · Answer 1 · 2018-01-30T14:23:32.163

7

Here, I reflect \ddots and superimpose it to create the X shape of dots.

EDITED to address row spacing issue raised by barbara.

\documentclass{article}
\usepackage{amsmath,graphicx}
\newcommand\xdots{\ooalign{$\ddots$\cr\scalebox{-1}[1]{$\ddots$}}}
\begin{document}
\[
B_j\rightarrow\tilde{B}_j=\begin{pmatrix}
a_{11}&\dots&a_{1,j-1}&0&a_{1,j+1}&\dots&a_{1n}\\
\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\
a_{i-1,1}&\dots&a_{i-1,j-1}&0&a_{i-1,j+1}&\dots&a_{i-1,n}\\[5pt]
0&\dots&0&1&0&\dots&0\\[3pt]
a_{i+1,1}&\dots&a_{i+1,j-1}&0&a_{i+1,j+1}&\dots&a_{i+1,n}\\
\vdots&\xdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
a_{n11}&\dots&a_{n,j-1}&0&a_{n,j+1}&\dots&a_{nn}
\end{pmatrix}
\]
\end{document}

edited Jan 30 '18 at 14:23

answered Jan 30 '18 at 11:36

Steven B. Segletes

237,551

Nice! The same as I did, but with much less work :). What does \ooalign do? – Phelype Oleinik Jan 30 '18 at 11:43
2

@PhelypeOleinik \ooalign is a TeX primative for superimposing things. It is defined on p.356 of the TeXbook. – Steven B. Segletes Jan 30 '18 at 11:45
Thanks for that! I'm on page 74 yet :) – Phelype Oleinik Jan 30 '18 at 11:49
2

@PhelypeOleinik This answer from egreg contains a nice explanation of \ooalign, if you don't want to skip 300 pages ;-) – campa Jan 30 '18 at 12:09
irrelevant to answer, but the line with the zeros is really too close to the adjacent lines. – barbara beeton Jan 30 '18 at 13:46

score 3 · Accepted Answer · answered Jan 30 '18 at 12:17

3

I am inspired by the definition of \ddots macro and I suggest to define:

\def\xdots{\mathinner{\mkern1mu
   \raise.1em\hbox{.}\kern-.33em{\ddots}\kern-.33em\raise.7em\hbox{.}\mkern1mu}}

answered Jan 30 '18 at 12:17

wipet

74,238

score 2 · Answer 3 · answered Jan 30 '18 at 11:38

Here's a different approach of what you suggested.

The macro \xdots first saves a \ddots in a box. Then this box is used to place the \ddots, and then a negative \hskip of the width of the box is applied to return to the original position. Finally, a \reflectbox is used to place the other part of the X.

\documentclass{article}

\usepackage{amsmath}
\usepackage{graphicx}

\newcommand{\xdots}{%
\newbox\dotsbox%
\savebox\dotsbox{\(\ddots\)}%
\usebox\dotsbox\hskip-\wd\dotsbox\relax%
\reflectbox{\(\ddots\)}%
}

\begin{document}

$$B_j\rightarrow\tilde{B}_j=\begin{pmatrix}a_{11}&\dots&a_{1,j-1}&0&a_{1,j+1}&\dots&a_{1n}\\\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\a_{i-1,1}&\dots&a_{i-1,j-1}&0&a_{i-1,j+1}&\dots&a_{i-1,n}\\0&\dots&0&1&0&\dots&0\\a_{i+1,1}&\dots&a_{i+1,j-1}&0&a_{i+1,j+1}&\dots&a_{i+1,n}\\\vdots&\xdots&\vdots&\vdots&\vdots&\ddots&\vdots\\a_{n11}&\dots&a_{n,j-1}&0&a_{n,j+1}&\dots&a_{nn}\end{pmatrix}$$

\end{document}

score 0 · Answer 4 · answered Jan 30 '18 at 13:36

You can obtain it with the \adots glyph from yhmath, superimposed to \ddots. Here are two ways:

\documentclass{article}
\usepackage[utf8]{inputenc}%
\usepackage{xcolor} 
\usepackage{mathtools, yhmath}
\usepackage{stackengine} 
\newcommand{\xdots}{\mathrlap{\ddots}{\mkern-1mu\color{red}\adots}}
\newcommand{\sixdots}{\stackMath\stackinset{c}{-0.32ex}{c}{0pt}{ \color{blue}\adots}{ \ddots}}

\begin{document}


\[ \xdots\qquad \sixdots\]%


\end{document}

Is there any way to put \ddots in an "x" shape?

4 Answers4