Add Codespell CI job for spell checking the source

.codespell-ignorelines

@@ -0,0 +1,10 @@
+# SPDX-License-Identifier: FSFAP
+# Copyright (C) 2025 Colin B. Macdonald
+#
+# Copying and distribution of this file, with or without modification,
+# are permitted in any medium without royalty provided the copyright
+# notice and this notice are preserved.  This file is offered as-is,
+# without any warranty.
+
+# lines that codespell should ignore: whitespace matters!
+

.codespell-ignorewords

@@ -0,0 +1,13 @@
+# SPDX-License-Identifier: FSFAP
+# Copyright (C) 2025 Colin B. Macdonald
+#
+# Copying and distribution of this file, with or without modification,
+# are permitted in any medium without royalty provided the copyright
+# notice and this notice are preserved.  This file is offered as-is,
+# without any warranty.
+
+# words that codespell should not complain about
+
+co-ordinates
+co-ordinate
+ans

.github/workflows/main.yml

@@ -0,0 +1,30 @@
+# SPDX-License-Identifier: FSFAP
+# Copyright (C) 2025 Colin B. Macdonald
+#
+# Copying and distribution of this file, with or without modification,
+# are permitted in any medium without royalty provided the copyright
+# notice and this notice are preserved.  This file is offered as-is,
+# without any warranty.
+
+name: CI
+
+on:
+  push:
+  pull_request:
+  schedule:
+    - cron: '17 0 * * 0'
+  # Allows you to run this workflow manually from the Actions tab
+  workflow_dispatch:
+
+jobs:
+  codespell:
+    runs-on: ubuntu-latest
+    steps:
+      - uses: actions/checkout@v4
+      - uses: codespell-project/actions-codespell@v2
+        with:
+          ignore_words_file: .codespell-ignorewords
+          exclude_file: .codespell-ignorelines
+          check_filenames: true
+          check_hidden: true
+          skip: '*.pdf'

Chapters/chapter2.tex

@@ -1224,7 +1224,7 @@ \subsection{MATLAB: assigning matrices and {\tt det} and {\tt cross}
 
 \subsection{MATLAB: generating scripts with the MATLAB editor}
 
-Often times using the command window in MATLAB to solve a problem can be tedious, because if the need arises to redo the problem, or change a parameter, one has to rewrite it all. The editor comes in handy for such cases. The editor is a text window (accesed from the command window: {\tt File $\rightarrow$ New $\rightarrow$ Blank M-file}) where one can write commands in the same syntax as the editor, and when one runs it, the results appear in the command window exactly as if one had written them there one after the other.
+Often times using the command window in MATLAB to solve a problem can be tedious, because if the need arises to redo the problem, or change a parameter, one has to rewrite it all. The editor comes in handy for such cases. The editor is a text window (accessed from the command window: {\tt File $\rightarrow$ New $\rightarrow$ Blank M-file}) where one can write commands in the same syntax as the editor, and when one runs it, the results appear in the command window exactly as if one had written them there one after the other.
 
 For example, the code to generate three random orthogonal vectors would look something like this:
 \begin{verbatim}
@@ -1853,7 +1853,7 @@ \subsection{Problems}
 \begin{problem}
 \label{op1_32}
 Find the equation for the line through $[2,-1,-1]$ and parallel to each of the
-two planes $x_1+x_2=0$ and $x_1-x_2+2x_3=0$. Express the equation fo the line
+two planes $x_1+x_2=0$ and $x_1-x_2+2x_3=0$. Express the equation of the line
 in both parametric and equation form.
 \end{problem}
 

Chapters/chapter3.tex

@@ -1124,16 +1124,16 @@ \subsection{Connection of solutions to homogeneous and inhomogeneous systems.}
 \]
 To see the implications of this let us  suppose that $\xx=\qq$ is any particular
 solution to a (non-homogeneous) system of equations. Then if $\yy$ is any other
-solution $\yy-\xx=\zz$ is a solution of the corresponding homogenous system.
+solution $\yy-\xx=\zz$ is a solution of the corresponding homogeneous system.
 So $\yy=\qq+\zz$. In other words any solution can be written as $\qq +$ some
-solution of the corresponding homogenous system. Going the other way, if
-$\zz$ is any solution of the corresponding homogenous system, then $\qq+\zz$
+solution of the corresponding homogeneous system. Going the other way, if
+$\zz$ is any solution of the corresponding homogeneous system, then $\qq+\zz$
 solves the original system. This can be seen by plugging $\qq+\zz$ into the
 equation. So the structure of the set of solutions is
 \[
 \xx = \qq + (\mbox{{\ \bf solution to homogeneous system}})
 \]
-As you run through all solutions to the homogenous system on the right,
+As you run through all solutions to the homogeneous system on the right,
 $\xx$ runs through all solutions of the original system. Notice that it doesn't
 matter which $\qq$ you choose as the starting point. This is completely analogous
 to the parametric form for a line, where the base point can be any
@@ -2720,7 +2720,7 @@ \section{Solutions to Chapter Problems}
 \end{eqnarray*}
 Hence, we have that $\qq = (-3, -6, 1, 7)$.
 
-Now, to find $\aa$, the homogenous row echelon form is
+Now, to find $\aa$, the homogeneous row echelon form is
 $$
 \left[\begin{array}{cccc|c}
      1 & 2 & 2 & 2 & 0 \\

Chapters/chapter4.tex

@@ -3032,9 +3032,8 @@ \section{Solutions to Chapter Problems}
 0&1 + t + {{t^2}\over{2}} + {{t^3}\over{3!}} + \cdots\cr
 }\right] \\
 &=&\left[\matrix{
-e^t&
-te^t\cr
-0&e^t\cr
+  e^t & t e^t\cr
+  0   & e^t\cr
 }\right]
 \end{eqnarray*}
 \item[d,e)] We are looking for all matrices that satisfy $B^2=A$. Let 
@@ -4068,4 +4067,4 @@ \section{Solutions to Chapter Problems}
 \end{eqnarray*}
 
 
-\end{enumerate}
+\end{enumerate}

Chapters/chapter6.tex

@@ -1779,7 +1779,7 @@ \subsection{The matrix exponential and differential equations}
 cases. Consider the matrix $A=\left[\matrix{1&1\cr0&1}\right]$. This
 matrix does not have a basis of eigenvectors. So it cannot be
 diagonalized.  However, in a homework problem, you showed that $e^{tA}
-= \left[\matrix{e^t&te^t\cr 0&e^t}\right]$. Thus the solution to
+= \left[\matrix{e^t & t e^t\cr 0 & e^t}\right]$. Thus the solution to
 \[
 \yy'(t) = \left[\matrix{1&1\cr 0&1}\right]\yy(t)
 \]
@@ -1790,8 +1790,8 @@ \subsection{The matrix exponential and differential equations}
 is
 \[
 \yy(t) = e^{tA}\left[\matrix{2\cr 1\cr}\right]
-=\left[\matrix{e^t&te^t\cr 0&e^t}\right]\left[\matrix{2\cr 1\cr}\right]
-=\left[\matrix{2e^t+te^t\cr e^t\cr}\right]
+=\left[\matrix{e^t & t e^t\cr 0 & e^t}\right]\left[\matrix{2\cr 1\cr}\right]
+=\left[\matrix{2e^t + t e^t\cr e^t\cr}\right]
 \]
 Notice that this solution involves a power of $t$ in addition to
 exponentials.