Typeset a matrix with crossing lines

Question

Section 5.4, "Invariant Subspaces and the Cayley–Hamilton Theorem", of Friedberg, Insel, and Spence's Linear Algebra represents the direct sum of matrices like so:

I'm interested in recreating the weird crossing lines thing, preferably with the standard amsmath matrix environments.

Answer 1

I'm interested in recreating the weird crossing lines thing, preferably with the standard amsmath matrix environments.

For the matrix at hand, a basic array environment generates a better outcome than a pmatrix environment does. By "better", I mean that the horizontal lines do not touch the tall round parentheses when using an array environment.

\documentclass{article}
\usepackage{amsmath} % for 'pmatrix' env.
\newcommand\mc[1]{\multicolumn{1}{c|}{#1}} % handy shortcut macro

\begin{document}
\[
B_1 \oplus B_2 \oplus B_3 =
\left(\begin{array}{ *{6}{c} }
1 & \mc{2} & 0 & 0 & 0 & 0 \\
1 & \mc{1} & 0 & 0 & 0 & 0 \\
\cline{1-3}
0 & \mc{0} & \mc{3} & 0 & 0 & 0 \\
\cline{3-6}
0 & 0 & \mc{0} & 1 & 2 & 1 \\
0 & 0 & \mc{0} & 1 & 2 & 3 \\
0 & 0 & \mc{0} & 1 & 1 & 1 
\end{array}\right)
\qquad
\begin{pmatrix}
1 & \mc{2} & 0 & 0 & 0 & 0 \\
1 & \mc{1} & 0 & 0 & 0 & 0 \\
\cline{1-3}
0 & \mc{0} & \mc{3} & 0 & 0 & 0 \\
\cline{3-6}
0 & 0 & \mc{0} & 1 & 2 & 1 \\
0 & 0 & \mc{0} & 1 & 2 & 3 \\
0 & 0 & \mc{0} & 1 & 1 & 1 
\end{pmatrix}
\]

\end{document} 

Answer 2

I know you asked for standard matrix environments, but I suggest nicematrix.

If you want additional space for the delimiters (so the lines don't touch them) you can include [extra-margin=3pt] in the following code:

\documentclass{article}

\usepackage{nicematrix, tikz}

\begin{document}

\[
B_1\oplus B_2\oplus B_3=
\begin{pNiceMatrix}%[extra-margin=3pt]
\CodeBefore
    \tikz \draw(1-|3) |- (4-|last) (3-|1) -| (last-|4);
\Body
1&2&0&0&0&0\\
1&1&0&0&0&0\\
0&0&3&0&0&0\\
0&0&0&1&2&1\\
0&0&0&1&2&3\\
0&0&0&1&1&1
\end{pNiceMatrix}.
\]

\end{document}

To clarify the (i-|j) notation: If tikz is loaded (note that nicematrix does not automatically load tikz) then (i-|j) is the intersection of the ith horizontal rule with the jth vertical rule (whether or not the rules are actually drawn). The 1st horizontal rule is above row 1; the 1st vertical rule is left of column 1. There is also position i.5 which is halfway between i and i+1 (useful for striking out rows or columns). The last position has the alias last.

These positions can be accessed using CodeBefore or CodeAfter (see the excellent nicematrix manual for details) just as one would use a node in TikZ. So in the CodeBefore, a tikzpicture can, for example, include \draw (1-|2) -- (3-|4); for a diagonal line, or \draw (1-|2) -| (3-|4); to draw horizontally then vertically.

Answer 3

It is possible with just the amsmath package, but the syntax is a bit convoluted and the lines are not nicely connected at the corners, therefore I am unsure whether I should recommend this:

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\[
B_{1} \oplus B_{2} \oplus B_{3} =
\left(
\begin{array}{*6{c}}
1 & 2 & \multicolumn{1}{|c}{0}  & 0 & 0 & 0 \\
1 & 1 & \multicolumn{1}{|c}{0}  & 0 & 0 & 0 \\ \cline{1-3}
0 & 0 & \multicolumn{1}{|c|}{3} & 0 & 0 & 0 \\ \cline{3-6}
0 & 0 & \multicolumn{1}{c|}{0}  & 1 & 2 & 1 \\
0 & 0 & \multicolumn{1}{c|}{0}  & 1 & 2 & 3 \\
0 & 0 & \multicolumn{1}{c|}{0}  & 1 & 1 & 1 \\
\end{array}
\right)
\]

\end{document}

Answer 4

You can do it with array and a couple of fixes.

\documentclass{article}
\usepackage{amsmath}
\usepackage{array}

\begin{document}
\[
% local command to ease input
\newcommand{\K}[1]{\multicolumn{1}{c|}{#1}}%
\newcommand{\B}{\vrule width 0pt height \dimeval{\ht\strutbox+0.5\dp\strutbox}}%
\addtolength{\arraycolsep}{-1pt}
B_1\oplus B_2\oplus B_3=
\left(\,
\begin{array}{@{}*{6}{c}@{}}
1 & \K{2} & 0     & 0 & 0 & 0 \\
1 & \K{1} & 0     & 0 & 0 & 0 \B \\
\cline{1-3}
0 & \K{0} & \K{3} & 0 & 0 & 0 \B \\
\cline{3-6}
0 & 0     & \K{0} & 1 & 2 & 1 \B \\
0 & 0     & \K{0} & 1 & 2 & 3 \B \\
0 & 0     & \K{0} & 1 & 1 & 1 \B
\end{array}
\,\right)
\]

\end{document}

The center cell seems more “square”.