Question

I can represent an upper trapezoidal matrix say as shown below. I would like to replace all those zeros with a single big zero that spans across the low triangle rows and columns, and maybe also add a delimiter along the diagonal that clearly shows it is an upper diagonal matrix. How can I do that?

\newcommand\x{\XSolid}
%\newcommand\x{\ding{53}}
\begin{equation}
  \left(
    \begin{array}{*5{c}}
    \x & \x & \x & \x & \x \\
     0 & \x & \x & \x & \x \\
     0 &  0 & \x & \x & \x \\
     0 &  0 &  0 & \x & \x \\
     0 &  0 &  0 &  0 & \x \\
  \end{array}\right)
\end{equation}

Separate question ... why the \x command I define outputs # rather than the intended cross symbol? It outputs the same symbol # no matter if I use \XSolid or \ding{53}

UPDATE: taking the answer as input, I ended doing this:

\newcommand\x{\times}
\newcommand\bigzero{\makebox(0,0){\text{\huge0}}}
\newcommand*{\bord}{\multicolumn{1}{c|}{}}
\begin{equation}
  \left(
    \begin{array}{ccccc}
    \x    & \x       & \x    & \x    & \x \\ \cline{1-1}
    \bord & \x       & \x    & \x    & \x \\ \cline{2-2}
          & \bord    & \x    & \x    & \x \\ \cline{3-3}
          & \bigzero & \bord & \x    & \x \\ \cline{4-4}
          &          &       & \bord & \x \\ \cline{5-5}
  \end{array}\right)
\end{equation}

which produces this:

enter image description here

Answer 1

\documentclass[]{article}
\usepackage{mathtools}
\begin{document}

\[
    \left(
    \begin{array}{ccccc}
    1                                    \\
      & 1             &   & \text{\huge0}\\
      &               & 1                \\
      & \text{\huge0} &   & 1            \\
      &               &   &   & 1
    \end{array}
    \right)
\]  

\end{document}

enter image description here

or \makebox(0,0){\text{\huge0}} if you want to have the same line spacing.

Answer 2

Just in case : you might need repeated dots across the diagonal. Here is a ugly way to do so.

\newcount\dotcnt\newdimen\deltay
\def\Ddot#1#2(#3,#4,#5,#6){\deltay=#6\setbox1=\hbox to0pt{\smash{\dotcnt=1
\kern#3\loop\raise\dotcnt\deltay\hbox to0pt{\hss#2}\kern#5\ifnum\dotcnt<#1
\advance\dotcnt 1\repeat}\hss}\setbox2=\vtop{\box1}\ht2=#4\box2}

And an exemple (using amsmath, of course) :

\[\begin{pmatrix}
1\Ddot{12}.(6pt,-2pt,6pt,-5pt)&1\Ddot8.(9pt,2pt,6pt,0pt)&\quad&\quad&1\\
&&&&\\
&&&&\\
&&&&\\
&\mbox{\Huge 0}&&&\\
&&&&1\\
\end{pmatrix}\]

$\Ddot in action$

Answer 3

With {pNiceMatrix} of nicematrix.

\documentclass{article}
\usepackage{nicematrix,tikz}

\begin{document}

$\begin{pNiceMatrix}[left-margin]
\times & \times & \times & \times & \times \\
       & \times & \times & \times & \times \\
       &        & \times & \times & \times \\ 
\Block{2-2}<\Huge>{0}
       &        &        & \times & \times \\
       &        &        &        & \times \\
\CodeAfter
 \tikz \draw (2-|1) -| (3-|2) -| (4-|3) -| (5-|4) -| (6-|5) ;
\end{pNiceMatrix}$

\end{document}