\documentclass[12pt]{article}
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\title{Algebra 1; MA20008; Sheet 4 Solutions}
\author{{\tt G.C.Smith@bath.ac.uk}}
\date{2-xi-2004}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Suppose that $n$ is a natural number or $0$,
and $F$ is a field. Show that there is a vector
space over $F$ of dimension $n$.}\newline \noindent
{\bf Solution\ } The zero space has basis $\emptyset$ 
and so has dimension $|\emptyset| = 0$. If $n > 0$,
then $F^n$ has dimension $n$. 
\item {\em Suppose that $V$ is a vector space with subspaces
$U$ and $W$ both of dimension $n < \infty$.
Does it follow that $V$ is finite dimensional?
Does it follow that $V$ has dimension $n$?
Does it follow that $U = W$?
In each case you should supply a reason for your answer.}
\newline \noindent
{\bf Solution\ } 
The answers are no, no and no.
To illustrate all three points, let $F$ be a field
and let $V = F[X]$ be the set of polynomials in $X$ with
coefficients in $F$. This $V$ is not a vector space of finite
dimension. Let $U$ be the subset of $V$ consisting
of polynomials of degree at most 1. Let $W$  be the subset of $V$ consisting
of polynomials of degree at most 2 but which have constant term 0.
Now $U \not = W$, $\dim U = \dim W = 2$ but $\dim V \not = 2.$
\item {\em Let $V$ be a vector space of dimension $n$.
Suppose that $V_0, V_1, \ldots, V_m$ are subspaces
of $V$ with
\[ V_0 \leq V_1 \leq \cdots \leq V_m.\]}
\begin{enumerate}
\item {\em Suppose that $m > n$. Show that there 
is $i \in \{ 1,2, \ldots, m\}$ such that $V_i = V_{i-1}.$}
\newline \noindent {\bf Solution\ }
The dimensions of the spaces $V_0, V_1, \ldots, V_m$ are
weakly increasing. If $m >n$, then two spaces 
$V_i$ and $V_{i-1}$ must have the same dimension,
and therefore (theorem in lectures) $V_{i-1} = V_i.$
\item {\em Suppose that $m \leq n$. Show that it may be
that the spaces \[V_0, V_1, \ldots, V_m\] are distinct.}
\newline \noindent {\bf Solution\ }      
Let ${\bf v_1}, {\bf v_2}, \ldots , {\bf v_n}$.
Let $V_j$ be the span of ${\bf v_1}, {\bf v_2}, \ldots , {\bf v_j}.$
This does the job.
\end{enumerate}
\item Suppose that $\alpha : U \longrightarrow W$
is a linear map between vector spaces over the same field.
Let ${\bf x_1}, {\bf x_2}, \ldots, {\bf x_n}$ be vectors in $U$.
\begin{enumerate}
\item  {\em Suppose that $U = \langle {\bf x_1}, {\bf x_2}, \ldots, {\bf x_n} \rangle$ 
and $\alpha$
is surjective. Prove that $W = \langle \alpha({\bf x_1}), \alpha({\bf x_2}), 
\ldots, \alpha({\bf x_n}) \rangle.$} \newline \noindent {\bf Solution\ }
Suppose that ${\bf w} \in W$.
Since $\alpha$ is surjective there is ${\bf u} \in U$ sich that
$\alpha({\bf u}) = {\bf w}.$ Now
${\bf u} = \sum_i \lambda {\bf x_i}$ so 
${\bf w} = \alpha\left(\sum_i \lambda_i {\bf x_i}\right)
= \sum_i \lambda_i \alpha({\bf x_i}).$
\item {\em Suppose that ${\bf x_1}, {\bf x_2}, \ldots, {\bf x_n}$
are linearly independent and $\alpha$ is injective.
Show that $\alpha({\bf x_1}), \alpha({\bf x_2}),
\ldots, \alpha({\bf x_n})$ are linearly independent.}
\newline \noindent {\bf Solution\ } 
Suppose that $\sum_i \lambda_i \alpha({\bf x_i}) = {\bf 0}.$
Therefore $\alpha ( \sum_i \lambda_i {\bf x_i}) = {\bf 0}.$
Now by injectivity 
$\sum_i \lambda_i {\bf x_i} = {\bf 0}.$ However, the ${\bf x_i}$
are linearly independent, so each $\lambda_i$ is $0$.
\end{enumerate}
\item {\em Let $\zeta = e^{\frac{2\pi i}{5}} \in \mathbb C.$}
\begin{enumerate}
\item {\em Suppose that we view $\mathbb C$ as a vector space over
$\mathbb Q$. Show that $1, \zeta, \zeta^2, \zeta^3$
are linearly independent.} \newline \noindent {\bf Solution\ }
$\zeta$ is a root of $X^4 + X^3 + X^2 + X + 1$. However, $\zeta$
is not a root of any rational polynomial of smaller
degree. To see this, note that a smallest degree non-zero
rational polynomial having $\alpha$ as a root must divide 
$X^4 + X^3 + X^2 + X + 1$. First eliminate the
possibility of a linear factor, and then a quadratic factor.
\item {\em Suppose that we view $\mathbb C$ as a vector space over
$\mathbb R$. Show that $1, \zeta, \zeta^2, \zeta^3$
are linearly dependent.}  \newline \noindent {\bf Solution\ }        
$\overline \zeta = \zeta^4 = \zeta^{-1}.$ Now 
$(X - \zeta)(X - \overline \zeta)$ is a real polynomial of degree
2 which has $\zeta$ as a root.
\end{enumerate}
\item Let $V$ be a vector space with subspaces 
$U, W$ such that $U$ and $W$ are both finite 
dimensional. Let 
${\bf u_1}, {\bf u_2}, \ldots, {\bf u_m}$
be a basis of $U$ and 
${\bf w_1}, {\bf w_2}, \ldots, {\bf w_n}$ be
a basis of $W$.
\begin{enumerate}
\item {\em Show that $U + W$ is finite dimensional.} 
\newline \noindent {\bf Solution\ } Certainly
$U + W = \langle {\bf u_1}, {\bf u_2}, \ldots, {\bf u_m}, {\bf w_1}, {\bf w_2}, \ldots, {\bf w_n}
\rangle$, so $U+W$ has a finite spanning set and therefore a finite basis.
\item {\em Show that ${\bf u_1}, {\bf u_2}, \ldots, {\bf u_m},
{\bf w_1}, {\bf w_2}, \ldots, {\bf w_n}$ need not be a basis
of $U + W$.} \newline \noindent {\bf Solution\ } 
Well, perhaps ${\bf u_1} = {\bf w_1}.$ That would do. 
\item {\em Suppose that $U + W = U \oplus W$. Show that
\[{\bf u_1}, {\bf u_2}, \ldots, {\bf u_m},
{\bf w_1}, {\bf w_2}, \ldots, {\bf w_n}\]
is a basis of $U+W$.} \newline \noindent {\bf Solution\ }              
Spanning is not an issue; linear independence is.
Suppose that there are scalars $\lambda_i, \mu_i$ such that
$\sigma_i (\lambda_i  {\bf u_i}) + \sigma_j (\mu_j  {\bf w_j})  = {\bf 0}.$
Since $U + W = U \oplus W$ it follows that 
$\sigma_i (\lambda_i  {\bf u_i}) = {\bf 0}$ and 
$\sigma_i \mu_j  {\bf w_j} = {\bf 0}$. Now $\lambda_i$ and $\mu_j$ are $0$
for every $i$ and $j$ by the linear independence of the relevant sequences. 
\item {\em Suppose that ${\bf u_1}, {\bf u_2}, \ldots, {\bf u_m},
{\bf w_1}, {\bf w_2}, \ldots, {\bf w_n}$
is a basis of $U+W$. Show that $U+W = U \oplus W$. }
\newline \noindent {\bf Solution\ }    
It suffices to show that $U \cap W = 0.$ Suppose not, then there is
${\bf v} \in U \cap W$ with ${\bf v} \not = {\bf 0}$. Now 
${\bf v} \in U$ so ${\bf v} = \sum \theta_i {\bf u_i}$
and ${\bf v} = \sum \psi_j  {\bf u_j}.$  This yields 
a non-trivial linear relation $\sum \theta_i {\bf u_i} - \sum \psi_j  {\bf w_j}$
among the vectors ${\bf u_i}$ and ${\bf w_i}$, which cannot therefore be linearly
independent.
\end{enumerate}
\item {\em Let $I$ be a set. Let $V$ be the set of real
valued functions on $I$; more formally
\[ V = \left\{ f \vert f: I \longrightarrow \mathbb R\right\}.\]
Define addition on $V$ by $(f + h)(x) :=
f(x) + g(x)$ for all $x \in I$. If $\lambda \in \mathbb R$
and $f \in V$ we define $\lambda \cdot f \in V$
by $(\lambda \cdot f)(x) = (\lambda)(f(x))$ where the final
multiplication is just the product (in $\mathbb R$).}
\begin{enumerate}
\item {\em Check that $V$ is now a vector space over $\mathbb R$.}
\newline \noindent {\bf Solution } This is routine.
\item {\em For each $i \in I$, define a function $\delta_i \in V$
where $\delta_i(x) = \delta_{i,x}$ (Kr\"onecker delta).
Thus $\delta_i(i) = 1$ and $\delta_i(x) = 0$ if $x \not = i$.
Show that the vectors $\delta_i$ are linearly independent.}
\newline \noindent {\bf Solution } 
Suppose that $\lambda_i$ are scalars (all but finitely many of which
are $0$). Then $\sum_i' \lambda_i \delta_i = 0$. Choose any 
$j \in I$< then $\lambda_i$ are scalars (all but finitely many of which
are $0$). Then $\sum_i' \lambda_i \delta_i)(j) = 0$ so
$\lambda_j = 0$. Thus these maps are linearly independent.
\item {\em Let $W = \langle \delta_i : i \in I \rangle$ be the
span of all the $\delta_i$. Show that the vectors $\delta_i$
form a basis of $W$ (in that they are a linearly independent
spanning set for $W$).} Well they are linearly independent by
the previous answer, and course they span $W$ by the design of $W$.
\item {\em Show that $W = V$ if and only if $I$ is finite.}
$W$ is the subset of $V$ consisting of functions of finite support,
i.e. functions which take non-zero values at only finitely many 
elements of the domain. The two subsets co-incide if and only if 
$I$ is finite.
\item {\em Give an explicit example of a vector space with an 
clearly describable uncountable basis (no set theoretic
metaphysics allowed).}  \newline \noindent {\bf Solution}
See part (c).
\end{enumerate}
\end{enumerate}  
\end{document}