\documentclass[12pt]{article}
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\title{Algebra 1; MA20008; Sheet 2 Solutions}
\author{{\tt G.C.Smith@bath.ac.uk}}
\date{19-x-2004}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Suppose that $U$ and $V$ are vector spaces over the
same field $F$, and that
$W$ is a subspace of $V$. Let $\alpha: U \rightarrow V$
be a linear map. Show that
$Y = \{ {\bf x } \mid {\bf x } \in U,\ \alpha({\bf x}) \in V\}$
is a subspace of $U$.} \newline \noindent {\bf Solution\ }
$\alpha({\bf 0}) = {\bf 0} \in W$ so $Y \not = \emptyset.$
Suppose that ${\bf x}, {\bf y} \in Y$ and $\lambda, \mu \in F$,
then $\alpha( \lambda{\bf x}+ \mu{\bf y})
= \lambda \alpha({\bf x})+ \mu \alpha({\bf y})\in W.$
Therefore $Y \leq U$. 
\item {\em Suppose that $U$ and $V$ are vector spaces over the same
field $F$, and
that we have linear maps $\alpha:  U \longrightarrow V$
and $\beta:  U \longrightarrow V$.
Show that $Z = \{  {\bf x } \mid {\bf x } \in U,\ \alpha({\bf x })
= \beta({\bf x }) \}$ is a subspace of $U$.}\newline \noindent
{\bf Solution\ } Note that ${\bf 0} \in Z \not = \emptyset.$
Now suppose that ${\bf x}, {\bf y} \in Z$ and $\lambda, \mu \in F$,
then \[ \alpha (\lambda{\bf x} + \mu {\bf y}) 
= \lambda\alpha ({\bf x}) + \mu \alpha ({\bf y})
= \lambda\beta ({\bf x}) + \mu \beta ({\bf y})
= \beta(\lambda{\bf x} + \mu {\bf y})\]
so $Z \leq U.$
\item {\em Let $\mathbb C$ be the complex numbers, viewed
as a vector space over $\mathbb R$. We have shown that
the map $\varphi : \mathbb C  \longrightarrow \mathbb C$
defined by complex conjugation is a linear map. Let
$n$ be a natural number, and define $\theta_n$ to
be multiplication by $e^{\frac{2\pi i}{n}}$; more formally
$\theta_n : {\mathbb C} \longrightarrow {\mathbb C}$ with
$\theta_n(z) = e^{\frac{2\pi i}{n}}z$ for all $z \in {\mathbb C}.$}
\begin{enumerate}
\item[(a)] {\em Show that each map $\theta_n$ is linear.}
 \newline \noindent {\bf Solution\ } In fact multiplication by
any fixed complex number is a linear map. This is another
way of viewing the distributive law of multiplication
over addition.
\item[(b)] {\em How many different maps can you get by composing
the maps $\theta_4$ and $\theta_6$? {\rm (For example,
$\theta_4 \theta_4 \theta_6 \theta_6 \theta_4$ is one such
composition.)}}  \newline \noindent {\bf Solution\ } There are
12 maps that can be obtained. Each such map is a rotation of the complex 
plane (Argand diagram) about the origin which preserves the 
vertices of the regular 12-gon with centre 0, and one vertex at 1.
There are clearly 12 such maps, and each can be obtained since 
$\theta_4 \theta_6^5$ is rotation through $\pi/6$, and this map has 12 
different positive powers which are all the possible rotations
respecting this regular 12-gon,
\item[(c)] {\em How many different maps can you get by composing
the maps $\theta_5$ and $\varphi$?} \newline \noindent {\bf Solution\ }
The answer is 10. These are actually the rigid symmetries of
the regular pentagon (5-gon) with centre 0 and a vertex at 1 (rotations
and reflections).
\item[(d)] {\em How many different maps can you get by composing
the maps $\theta_4$, $\theta_6$ and $\varphi$?} There are 24 maps
that can be obtained.  
 \newline \noindent {\bf Solution\ } The answer is 24, the rigid symmetries
of the obvious regular dodecagon (12-gon), reflections and rotations.
\end{enumerate}
\item {\em Let $V$ we a vector space over a field $F$. We define
a {\em line} as follows. Suppose that ${\bf a }, {\bf b} \in V$
with ${\bf b } \not = {\bf 0}.$ The set
\[ L = \{ {\bf r} \mid {\bf } {\bf r}  = {\bf a} + t{\bf b},\ t \in F\}\]
is a {\em line}. Suppose that $U$ is also a vector space over $F$
and that \[\alpha : V \longrightarrow U\] is a linear map.
Show that if ${\bf b} \not \in \mbox{ Ker }\alpha$,
then \[K = \{ \alpha({\bf r}) \mid  {\bf r} \in L\}\] is a line.
What happens if ${\bf b } = {\bf 0}?$}
 \newline \noindent {\bf Solution\ } 
\[ \{ \alpha({\bf r}) \mid  {\bf r} \in L\} = 
\{ {\bf r} \mid {\bf } {\bf r}  = \alpha({\bf a}) + \alpha(t{\bf b}),
\ t \in F\}
\] \[
= \{ {\bf r} \mid {\bf } {\bf r}  = \alpha({\bf a}) + t  \alpha({\bf b}),
\ t \in F\}
\] which is a line. If ${\bf b } = {\bf 0}$ or more generally
if ${\bf b } \in  \mbox{ Ker }\alpha$, then we get a set consisting
of a single point instead.
\item {\em Regard ${\mathbb R}^n$ as a vector space over $\mathbb R$.
Define a map $\mu : {\mathbb R}^n \longrightarrow {\mathbb R}^n$
by $(x_1, x_2, \ldots, x_n) \mapsto (x_2, x_3, \ldots, x_n,0)$
for all $(x_1, x_2, \ldots, x_n) \in \mathbb R^n.$}
\begin{enumerate}
\item[(a)] {\em Show that $\mu$ is a linear map.}
 \newline \noindent {\bf Solution\ } This is entirely routine.
Suppose that $\lambda, \theta \in \mathbb R$ and 
${\bf x} = (x_1, x_2, \ldots, x_n),\ 
{\bf y}= (y_1, y_2, \ldots, y_n) \in \mathbb R^n$.
Now \[\mu(\lambda {\bf x} + \theta {\bf y}) =
\lambda (x_2,x_3, \ldots, x_n, 0) +
\theta(y_2,y_3, \ldots, y_n, 0) \]\[=
(\lambda x_2 + \theta y_2, \lambda x_3 + \theta y_3, \ldots,
\lambda x_n + \theta y_n, 0),\] whereas
\[\lambda \mu({\bf x}) + \theta \mu ({\bf y})
= \lambda (x_2,x_3, \ldots, x_n, 0) +
\theta(y_2,y_3, \ldots, y_n, 0) \]\[=
(\lambda x_2 + \theta y_2, \lambda x_3 + \theta y_3, \ldots,
\lambda x_n + \theta y_n, 0).\] We are done.
\item[(b)] {\em Show that $\mu^n$ is the zero map ($\mu^n$ denotes
the map obtained by composing $n$ copies of $\mu$).} 
\newline \noindent {\bf Solution\ }
Induct on $r$
to show that 
\[ \mbox{Im }\mu^r = \{(y_1, y_2, \ldots, y_{n-r}, 0, \ldots, 0) \mid
y_i \in \mathbb R \mbox{ for all }i\}.\]
We omit the details.
\item[(c)] {\em Show that $\mu^{n-1}$ is not the zero map.}
 \newline \noindent {\bf Solution\ }
This follows from the argument above.

\end{enumerate}
\item {\em Let $V$ be a vector spaces, and suppose that
$\alpha$ and $\beta$ are both projections onto subspaces
of $V$ with suitable kernels. Suppose also that
$\alpha \beta= \beta \alpha$. Show that $\alpha \beta$
is a projection.}  \newline \noindent {\bf Solution\ }
We are given that $\alpha, \beta : V \longrightarrow V$
are linear maps which commute and satisfy $\alpha \beta = \beta \alpha$.
Moreover $\alpha^2 = \alpha$ and $\beta^2 = \beta$.
(we allow a slight notational abuse here, and inflate the codomains
of $\alpha$ and $\beta$ to $V$ from the given subspaces of $V$).
Now $(\alpha \beta)^2 = \alpha \beta \alpha \beta = \alpha^2 \beta^2
= \alpha \beta.$ We have used the fact that $\alpha$ an $\beta$
are projections so $\alpha^2 = \alpha$ and $\beta^2 = \beta$, 
and commutativity. Now we proved in lectures that
$(\alpha \beta)^2 = \alpha \beta $  forces $\alpha \beta$ to be a projection,
so we are done.
\end{enumerate}
\end{document}



