\documentclass[12pt]{article}
\usepackage{latexsym}
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\title{Algebra 1; MA20008; Sheet 3}
\author{{\tt G.C.Smith@bath.ac.uk}}
\date{25-x-2004}
\begin{document}
\maketitle
\begin{enumerate}
\item {\em Suppose that $V$ is a vector space over $F$, and
that $S \subseteq V$. Let $\overline S$ be the intersection
of those subspaces of $V$ which contain the subset $S$, 
or put formally
\[ \overline S = \bigcap \{ U \mid U \leq V, S \subseteq U\}.\]
Show that $\overline S = \langle S  \rangle.$}
\newline \noindent {\bf Solution:\ } We have
$ S \subseteq \langle S \rangle \leq V$. Now $\langle S \rangle$ is therefore
one of the sets being intersected in the definition of 
$\overline S.$ Therefore $\overline S \subseteq \langle S \rangle.$ Conversely
if $S \subseteq U \leq V$ and ${\bf v} \in \langle S\rangle,$ then  ${\bf v}
\in U$ since $U$ is closed under the formation of linear combinations.
Therefore $\langle S \rangle \subseteq \overline S.$ Since we have both
inclusions it follows that $\overline S = \langle S  \rangle.$
\item  {\em Let $V$ be a vector space over $F$
and ${\bf v_1}, {\bf v_2},\ldots, {\bf v_n} \in V$. Suppose that
whenever $\theta_1, \ldots, \theta_n, \psi_1, \ldots, \psi_n \in F$
and $\sum_{i=1}^n \theta_i {\bf v_i} = \sum_{i=i}^n \psi_i {\bf v_i}$,
then necessarily $\lambda_i = \mu_i$ for each $i$, $1 \leq i \leq n$.
Show that ${\bf v_1}, {\bf v_2},\ldots, {\bf v_n} \in V$ is
linearly independent.} \newline \noindent {\bf Solution:\ }         
It suffices to choose $\psi_i = 0$ for every $i$, and we obtain the condition
for linear independence.
\item {\em Consider $V = \mathbb R$ as a vector space over $\mathbb Q$.}
\begin{enumerate}
\item {\em Show that $1, \sqrt 2, \sqrt 3$ are linearly independent.}
\newline \noindent {\bf Solution:\ }  It is first year work to show that $\sqrt 2,\
\sqrt3$ and $\sqrt 6$ are irrational, and we assume that you can do this. 
Suppose that $a + b \sqrt 2 + c \sqrt 3 = 0$ for rational $a,b$ and $c$. 
Therefore $(b \sqrt 2 + c \sqrt 3)^2 \in \mathbb Q$, so
$bc \sqrt 6 \in  \mathbb Q.$ Now $bc =0$ else $\sqrt 6$ would be 
rational. If only one of $b$ and $c$ were $0$, then
$\sqrt 2$ or $\sqrt 3$ would be rational, which it isn't.
Therefore $b = c = 0$, so $a=0$ and we are done.
\item {\em Let $\alpha = e^{\frac{\pi i}{3}}$. Which lists
of the form $1, \alpha, \ldots, \alpha^n$ are linearly independent?
Justify your answer.} \newline \noindent {\bf Solution:\ }   
$\alpha$ is a root of $X^3+1 = (X+1)(X^2-X+1)$ but not of $X+1$
so $\alpha$ is a root of $X^2-X+1$. Thus
$1 - \alpha + \alpha^2 = 0$ so $1, \alpha, \alpha^2$ is a linearly
dependent list, as is $1, \alpha, \ldots, \alpha^n$ whenever $n \geq 2$.
Also 1 is linearly independent, and $1, \alpha$ is linearly
independent, since a non-trivial linear relation would force
$\alpha \in \mathbb R$, which is false. 
\item {\em Suppose that $1, \beta, \beta^2, \ldots, \beta^n$ are 
linearly independent. Show that
$1, (\beta +1), (\beta +1)^2, \ldots,  (\beta +1)^n$ are linearly independent.} 
\newline \noindent {\bf Solution:\ } Suppose, for contradiction,
that these powers
of $1+\beta$ are linearly dependent.
Thus there is a non-zero polynomial $f$ (or $f(X)$) with
rational coefficients so that $f(1 + \beta) = 0$. Now $\beta$ will
be a root of $h := f(X+1)$, and $\deg h = \deg f$ (and indeed the leading
coefficients co-incide). Therefore $h$ is not the zero polynomial
and the given powers of $\beta$ satisfy a non-trivial linear relation.
However, we are given that these powers of $\beta$ are linearly
independent over $\mathbb Q,$ so this is absurd. We have the required
contradiction.
\end{enumerate}
\item {\em Suppose that $X$ and $Y$ are both linearly independent subsets
of $V$. Does it follow that $X \cap Y$ is linearly independent?
What about $X \cup Y$?} \newline \noindent {\bf Solution:\ } A 
subset of a l.i. set of vectors is l.i. for formal 
reasons, and $X \cap Y \subseteq X$, so we are done. 
However, the same is not true for the formation of unions. 
Let $V = F = \mathbb R$. Let $X = \{1\},\ Y = \{2\}$ which are both
l.i., but $X \cup Y = \{1,2\}$ which is l.d. because
$2 \cdot 1 + (-1) \cdot 2 = 0.$

\item {\em Suppose that $V = U \oplus W$. We are given
a sets of vectors $X \subseteq U$
and $Y  \subseteq W$. Is $X \cup Y$ necessarily a linearly
independent set of vectors?}\newline \noindent {\bf Solution:\ }  
We have proved that a direct sum yields uniqueness of decomposition,
so if ${\bf v} \in V$ and ${\bf v} = {\bf u_1} + {\bf w_1}
= {\bf u_2} + {\bf w_2}$ for ${\bf u_1}, {\bf u_2} \in U$ and
${\bf w_1}, {\bf w_2} \in W$, then ${\bf u_1}= {\bf u_2}$
and ${\bf w_1} = {\bf w_2}$. Now suppose that we have scalars
$\lambda_i, \theta_j$ so that
\[ \sum_{i=1}^m \lambda_i {\bf u_i} +  \sum_{j=1}^n \theta_j {\bf w_j} = {\bf 0}.\]
Here ${\bf u_1}, \ldots, {\bf u_m}\in X$ and $ 
{\bf v_1}, \ldots, {\bf v_n}\in Y$. The uniqueness of expression, compared
to ${\bf 0} + {\bf 0} = {\bf 0}$, ensures that both
\[ \sum_{i=1}^m \lambda_i {\bf u_i} =  {\bf 0}\]  and 
\[ \sum_{i=1}^n \theta_j {\bf w_j} =  {\bf 0}.\]  The l.i. of both
$X$ and $Y$ forces all scalars to vanish,  and we are done.   

\item {\em Suppose that ${\bf v_1}, {\bf v_2},\ldots, {\bf v_n}$ is a linearly
independent list of vectors in the vector space $V$. We are given 
${\bf w} \in V$. Does it follow that
\[{\bf v_1}+ {\bf w},  {\bf v_2}+ {\bf w}, \ldots, {\bf v_n} + {\bf w}\]
are linearly independent?}\newline \noindent {\bf Solution:\ } No.
Choose $ {\bf w} = -{\bf v_1}$ and the zero vector occurs in the list.
\item {\em Suppose that ${\bf v_1}, {\bf v_2},\ldots, {\bf v_n}$ is a linearly
dependent list of vectors in the vector space $V$. We are given
${\bf w} \in V$. Does it follow that
\[{\bf v_1}+ {\bf w},  {\bf v_2}+ {\bf w}, \ldots, {\bf v_n} + {\bf w}\]
is linearly dependent?}\newline \noindent {\bf Solution:\ }
No. Let $V  = \mathbb R$, $F = \mathbb R$. Let $n=1$
and ${\bf v_1}= {\bf 0}.$ Let ${\bf w}=1$.
\item {\em Let $V = \mathbb R^3$ viewed as vector space over $\mathbb R$.
Let ${\bf v_1}, \ldots, {\bf v_8}$ be the position vectors of the vertices
of a cube.}
\begin{enumerate}
\item {\em Let 
\[ A = \left\{ \sum_i \lambda_i {\bf v_i} \mid 0 \leq \lambda_i \leq 1 \mbox{ for all }
i, \sum_i \lambda_i = 1\right\}. \]
Describe the set $A$, viewed as a collection of position vectors, geometrically.}
\newline \noindent {\bf Solution:\ } The given position vectors point 
to the points inside and on the surface of the cube. 
\item  {\em Let
\[ B = \left\{ \sum_i \lambda_i {\bf v_i}  \mid \lambda_i \geq 0 \mbox{ for all } i, 
\right\}. \]
Under what circumstances is $B = \mathbb R^3$? Under what circumstances
is $B$ a closed half space (i.e. one side of a plane and all the
points on that plane)? What other shapes can arise?}
\newline \noindent {\bf Solution:\ } We have $B = \mathbb R^3$ exactly
when the origin is strictly inside the cube. If the origin
is in the interior of a face, then $B$ is a half-space. If the origin
is in the interior of an edge, then $B$ is the intersection of two
half-spaces defined by perpendicular planes.  If the origin
is at a vertex of the cube, then  $B$ is the intersection of three
half-spaces defined by pairwise perpendicular planes (also known as
an octant). If  the origin
is strictly outside the cube, then $B$ will be an infinite cone with finitely
many planar faces.
\end{enumerate}
\end{enumerate}
\end{document}