\documentclass[a4paper]{article}
\usepackage{amssymb,amsmath}

\newcommand\CC{\mathbb{C}}
\newcommand\RR{\mathbb{R}}
\newcommand\ZZ{\mathbb{Z}}
\newcommand\NN{\mathbb{N}}
\newcommand\spn{\mathop{\rm span}\nolimits}

\begin{document}

\section*{Linear Algebra}

\subsection*{Richard Kaye and Robert Wilson,\\
Oxford University Press, 1988.}
\subsubsection*{ISBN 0 19 850238 9 (Hbk)\\ISBN 0 19 850237 9 (Pbk)}

This document lists all the errors (including typographical errors and misprints and two 
slightly more serious errors) that we are aware of at the current time.  The authors 
would be grateful to hear from you if you discover any others.  Please email us
at \texttt{R.W.Kaye@bham.ac.uk}, \texttt{R.A.Wilson@bham.ac.uk}, or write
to us at the School of Mathematics and Statistics, University of Birmingham,
Birmingham,~B15~2TT.
\medskip

\subsubsection*{Example 4.13, page 67.} Replace this with the following.

\paragraph{Example 4.13}
%\begin{example}
In the complex vector space $\CC^2$,  take ordered bases 
$\mathbf a_1,\mathbf a_2$ and $\mathbf b_1,\mathbf b_2$ where
\[
\mathbf a_1=\begin{pmatrix} 1\\ i \end{pmatrix},\ 
\mathbf a_2=\begin{pmatrix} 1\\ 1+i \end{pmatrix},\qquad
\mathbf b_1=\begin{pmatrix} i\\ -i \end{pmatrix},\ 
\mathbf b_2=\begin{pmatrix} 1\\ 1 \end{pmatrix}.
\]
Then the base-change matrices from the usual basis $(1,0)^T,(0,1)^T$ to
$\mathbf a_1,\mathbf a_2$ is
\[
\mathbf P = \begin{pmatrix} 1 & 1 \\ i & 1+i \end{pmatrix},
\]
and from the usual basis $(1,0)^T,(0,1)^T$ to
$\mathbf b_1,\mathbf b_2$ is
\[
\mathbf Q = \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}.
\]
This can be used to find the base-change matrix from 
$\mathbf a_1,\mathbf a_2$ to $\mathbf b_1,\mathbf b_2$ as follows.

If $(v_1,v_2)^T$ is the coordinate form  of a vector $\mathbf v$
with respect to $\mathbf b_1,\mathbf b_2$, then 
$\mathbf Q(v_1,v_2)^T$ is the coordinate form of $\mathbf v$ with
respect to the usual basis.  Then 
$\mathbf P^{-1}\mathbf Q(v_1,v_2)^T$ is the coordinate form of 
$\mathbf v$ with respect to $\mathbf a_1,\mathbf a_2$, so the 
base-change matrix from $\mathbf a_1,\mathbf a_2$ to 
$\mathbf b_1,\mathbf b_2$ is
\begin{align*}
\mathbf P^{-1}\mathbf Q &= 
\begin{pmatrix} 1 & 1 \\ i & 1+i \end{pmatrix}^{-1} 
\begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}= 
\begin{pmatrix} 1+i & -1 \\ -i & 1 \end{pmatrix} 
\begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}\\
&= \begin{pmatrix} -1+2i & i \\ 1-i & 1-i \end{pmatrix}.
\end{align*}
%WAS:
%If $(v_1,v_2)^T$ is the coordinate form  of a vector $\mathbf v$
%with respect to $\mathbf a_1,\mathbf a_2$, then 
%$\mathbf P(v_1,v_2)^T$ is the coordinate form of $\mathbf v$ with
%respect to the usual basis.  Then 
%$\mathbf Q^{-1}\mathbf P(v_1,v_2)^T$ is the coordinate form of 
%$\mathbf v$ with respect to $\mathbf b_1,\mathbf b_2$, so the 
%base-change matrix from $\mathbf a_1,\mathbf a_2$ to 
%$\mathbf b_1,\mathbf b_2$ is
%\begin{align*}
%\mathbf Q^{-1}\mathbf P &= 
%\begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}^{-1}
%\begin{pmatrix} 1 & 1 \\ i & 1+i \end{pmatrix} = 
%\frac{1}{2i} \begin{pmatrix} 1 & -1 \\ i & i \end{pmatrix}^{-1}\begin{pmatrix} 1 & 1 \\ i & 1+i \end{pmatrix} \\
%&= \frac{1}{2i}\begin{pmatrix} 1-i & -i \\ i-1 & 2i-1 \end{pmatrix} = 
%\frac{1}{2}\begin{pmatrix} -1-i & -1 \\ 1+i & 2+i \end{pmatrix}.
%\end{align*}
%\end{example}

\subsubsection*{Exercise 4.6, page 72.}

The last occurrence of $\mathbf f_1$ should read $\mathbf f_3$.

\subsubsection*{p. 117, line 15}
The second partial derivative should be 
$\partial F/\partial y$ not $\partial F/\partial x$.

\subsubsection*{p. 140, Exercise 8.4} 
Replace $-y+3z$ by $-y-2z$ in the definition of $f$.

\subsubsection*{p. 160, line 14} It is not true that 
$(\mathbf A-\lambda_n\mathbf I)\mathbf e_n = \mathbf 0$.
In fact it is a linear combination of $\mathbf e_1, \ldots, \mathbf e_{n-1}$.
Thus the induction still works, but the proof needs some slight adjustments.
For example, replace the third paragraph of the proof (beginning 'Now \ldots')
by the following:

Now
$(\mathbf A-\lambda_n\mathbf I)\mathbf e_n$ is a linear combination of 
$\mathbf e_1, \ldots, \mathbf e_{n-1}$.
 Also,
the linear transformation given by $\mathbf A$ on the subspace
$\spn(\mathbf e_1,\ldots,\mathbf e_{n-1})$
has upper triangular matrix with respect to this basis,
with diagonal entries $\lambda_1,\ldots,\lambda_{n-1}$,
so by the induction hypothesis we have
\[
(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I)
\mathbf e_i =\mathbf 0
\]
for all $i<n$.  Thus
\[
(\mathbf A-\lambda_n\mathbf I)
(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I)
\mathbf e_i =\mathbf 0
\]
for $i<n$. Moreover, 
\[
(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I)
  (\mathbf A-\lambda_n\mathbf I)\mathbf e_n
\]
is a linear combination of 
the\[
(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I)
\mathbf e_i
\]
for $i<n$, all of which are $\mathbf 0$. Therefore
\[
(\mathbf A-\lambda_1 \mathbf I)\ldots(\mathbf A-\lambda_{n-1} \mathbf I)
(\mathbf A-\lambda_n\mathbf I)
\mathbf e_i =\mathbf 0
\]
for all $i\le n$. 

\subsubsection*{Exercise 12.8(a), page 186.}

$x_{n+1} = x_n+4x_n+1$ should read $x_{n+1} = x_n+4y_n+1$.

\end{document}
