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\title{Algebra 1: Math20008: Sheet 1}
\author{\tt G.C.Smith@bath.ac.uk}
\date{October 12, 2004}
\begin{document}
\maketitle

{\it The course web site is available via 
{\tt http://www.bath.ac.uk/$\sim$masgcs/}}

\begin{enumerate}

\item {\em Let $V$ be a vector space over $F$. {\rm Here is a proof that
$0 {\bf v} = {\bf 0}$ for all ${\bf v} \in V$. First note that
$0+0 = 0$ (Field Axiom 3). Therefore 
$(0+0){\bf v} = 0{\bf v}$. Now by Vector Space Axiom 7 we have
$0{\bf v} + 0{\bf v} = 0{\bf v}.$ Now, by Vector Space Axiom 4
the vector $0{\bf v}$ has an additive inverse $-0{\bf v}$.
It follows that
\[ (0{\bf v} + 0{\bf v}) + -0{\bf v} = 0{\bf v} + -0{\bf v}.\]
Now rebracket the left hand side using Vector Space Axiom 2.
Therefore
\[ 0{\bf v} + (0{\bf v} + -0{\bf v}) = 0{\bf v} + -0{\bf v}.\]
Now on both sides we have the opportunity to use the 
definition
of an additive inverse so
$0{\bf v} + {\bf 0} = {\bf 0}.$
Now by Vector Space Axiom 3 we deduce that
\[ 0{\bf v} = {\bf 0}.\]} Now, using this as a model, and 
quoting this result if necessary, prove that if ${\bf v} \in V$,
then $(-1){\bf v} = -{\bf v}.$}\newline \noindent {\bf Solution\ }
By the definition of inverses, $1 + -1 = 0.$
Therefore $(1 + -1){\bf v} = 0 {\bf v}.$ We apply Vector Space Axiom 7,
and the result justified in the above proof, to see that
$1 {\bf v} + (-1){\bf v} = {\bf 0}.$ Now $1 {\bf v} = {\bf v}$ by
Vector Space Axiom 9. We add the additive inverse of ${\bf v}$
to each side so $-{\bf v} + ({\bf v} + (-1){\bf v}) = {\bf 0} + -{\bf v}.$
We rebracket the left side using Vector Space Axiom 2, and use the
definition
of ${\bf 0}$ on the right, so
$(-{\bf v} + {\bf v}) +  (-1){\bf v}) =  -{\bf v}.$
Now by the definition of additive inverses we have
$-{\bf v} + {\bf v}  = {\bf 0}$ so
$ {\bf 0}+  (-1){\bf v}) = -{\bf v}.$    
Finally the definition of $ {\bf 0}$ yields 
$(-1){\bf v}) = -{\bf v}$ as required.

\item Let $V$ be a vector space over $F$. Suppose that $U$ is a subset 
of $V$. Show that $U$ is a subspace of $V$ if and only
if the following three conditions are satisfied:
\begin{enumerate}
\item $U \not = \emptyset$.
\item If ${\bf x} \in U$ and $\lambda \in F$, then $\lambda{\bf x} \in U$.
\item If ${\bf x},\ {\bf y} \in U,$ then ${\bf x}+{\bf y} \in U.$
\end{enumerate} \noindent {\bf Solution\ }
The two conditions for $U$ to be a suspace are (a) that $U \not 
= \emptyset$ and (b) that $U$ be closed under arbitrary linear
combinations. First assume (i), (ii) and (iii).
Now (i) is the same as (a) so (a) holds. Now suppose that
${\bf u},\ {\bf v} \in U$ and $\lambda,\ \mu \in F$. Then
$\lambda {\bf u}, \mu {\bf v} \in U$ by (ii) and 
$\lambda {\bf u} + \mu {\bf v} \in U$ by (iii). Conversely
we assume that (a) and (b) hold. Now (i) is the same as 
(a) so (i) holds. Now choosing $\lambda = 1$ and $\mu = 0$ we obtain (ii).
Finally choosing $\lambda = \mu =1$ we obtain (iii).
\item Consider the usual Cartesian description
of the plane as ${\mathbb R}^2$ (with perpendicular axes).
This collection of ordered pairs is a vector space
over $\mathbb R$ in a natural way as discussed
in lectures. In each case you should justify your answer.
\begin{enumerate}
\item[(a)] {\em Prove that the ordered pairs corresponding a straight
line through the origin form a subspace.} \newline \noindent {\bf Solution\ }
Each straight line through the origin is a set of the form
$L = \{ (x,y) \mid ax + by = 0\}$ for suitable choice of constants $a, b \in 
\mathbb R$. Note that $(0,0) \in L \not = \emptyset.$ It remains
to demonstrate closure under linear combinations.
Suppose that ${\bf u},\ {\bf v} \in L$ with 
${\bf u} = (e,f)$ and ${\bf v} = (g,h).$ 
Now if $\lambda, \mu \in \mathbf R$, then
\[ \lambda{\bf x} + \mu{\bf y} = (\lambda e + \mu g, \lambda f + \mu h)\]
and \[a(\lambda e + \mu g)+ b(\lambda f + \mu h)
= \lambda (ae + bf) + \mu (ag + bh)
= \lambda 0 + \mu 0 = 0\]
and so $\lambda{\bf x} + \mu{\bf y} \in L$. Thus $L$ is a subspace. 
\item[(b)] {\em Consider the 
straight line S = $\{(x,1) \mid x \in \mathbb R \}.$
Is this a subspace?} \newline \noindent {\bf Solution\ }
No. $(1,1) \in S$ but $2(1,1) = (2,2) \not \in S.$
\item[(c)] {\em Consider the circle $C = \{ (x, y) \mid x^2 + (y-1)^2 = 1 \}.$
Is this a subspace?} \newline \noindent {\bf Solution\ }
Certainly not. $(1,1) \in C$ but
$(2,2) = 2(1,1) \not \in C$.
\item[(d)] {\em Prove that $Z = \{ (0,0)\} $ is a subspace.} \newline
\noindent {\bf Solution\ } Suppose that ${\bf x}, {\bf y} \in Z$
so ${\bf x}={\bf y} = {\bf 0}$. If $\lambda, \mu \in \mathbb R$, then
$\lambda {\bf x} + \mu {\bf y} = {\bf 0} \in Z$ so $Z \leq {\mathbb R}^2.$
\item {\em Prove that $\emptyset$ is not a subspace.}\newline
\noindent {\bf Solution\ } This is true by definition.
\end{enumerate}
\item {\em Describe as many different subspaces of $\mathbb R^3$ 
as you can find.}\newline
\noindent {\bf Solution\ } It should be possible to spot $0$,
each line through the origin, each plane through the origin, and
the whole space. In fact there are no other subspaces, but at this
stage no proof is available.
\item Let $\mathbb R[X]$ denote the set of polynomials in $X$ which
have coefficients in $\mathbb R$. This set has a natural vector
space structure over $\mathbb R$. Which of the following are subspaces
of $\mathbb R[X]$, and why?
\begin{enumerate}
\item[(a)] {\em $\{ f \mid f \in \mathbb R[X], f(42) = 0\}$.}\newline
\noindent {\bf Solution\ }  Let the set be $S$.
The zero polynomial satisfies the
condition so $S \not = \emptyset.$ Suppose that
$f, g \in S$ and $\lambda, \mu \in \mathbb R$. Let $h
= \lambda f + \mu g$ so
\[ h(42) = \lambda f(42) + \mu g(42) = 0 + 0 = 0\]
so $S$ is a subspace.
\item[(b)] {\em $\{ f \mid f \in \mathbb R[X], f(42) = 1\}$.} \newline
\noindent {\bf Solution\ } The constant
polynomial 1 is in this set but $2 = 2 \cdot 1$ is not, so it cannot
be a subspace.             
\item[(c)] {\em $\{ f \mid f \in \mathbb R[X], 
f \mbox{ has at most two real roots }\}$. } \newline
\noindent {\bf Solution\ } This set is not a subspace
because $1, -1$ are in the set (these constant polynomials have
no rela roots) but $0 = 1 + (-1)$ has infinitely many real roots
(all real numbers are roots).
\item[(d)] {\em $\{ f \mid f \in \mathbb R[X], \deg f \leq n\}$ where $n \in \mathbb N \cup
\{0\}.$ Let the set be $S_n$ for each possible $n \in \mathbb N \cup \{0\}.$
Note that the degree of the zero polynomial is $-\infty$, a symbol
deemed to be smaller than all integers.}\newline \noindent {\bf Solution\ }
Each $S_n$ is a subspace. This is because the zero polynomial is in 
each $S_n$, and if $f, g \in S_n$ and $\lambda, \mu \in \mathbb R$,
then \[\deg (\lambda f + \mu g) \leq 
\max \{ \deg (\lambda f),\ \deg (\mu g)\} \leq \max \{ n, n\} = n.\] 
We are done.
\item[(e)] {\em $\{ f \mid f \in \mathbb R[X], f''(X) - Xf'(X) + f(X) = 0\}$
where a dash in the exponent indicates differentiation with respect to $X$.}
\newline \noindent {\bf Solution\ } Let the ste be $S$.
The zero polynomial satisfies the
differential equation so $S \not = \emptyset.$ Suppose that
$f, g \in S$ and $\lambda, \mu \in \mathbb R$. Let $h
= \lambda f + \mu g$ so
\[ h'' - Xh' + h = (\lambda f + \mu g)'' -X (\lambda f + \mu g)'
+ \lambda f + \mu g \]\[= \lambda (f'' -Xf' + f) + \mu(g'' -Xg' + g) 
= 0 + 0 = 0\]
so $S$ is a subspace.
\item[(f)] {\em $\{ f \mid f(X)^2 = f(X^2)\}.$ }\newline \noindent
{\bf Solution\ } Let the set be $S$. Observe that $X \in S$. However
$3X \not \in S$ because $(3X)^2 = 9X^2 \not = 3X^2.$ Thus
$S$ is not closed under scalar multiplication so cannot be a subspace.
\end{enumerate}
\item {\em Suppose that 
$V$ is a vector space and that ${\bf v}_1, {\bf v}_2, \ldots,
{\bf v}_m \in V.$ Let \[U = 
\left\{ \sum_{i=1}^n \lambda_i {\bf v}_i \Biggm\vert \lambda_i \in \mathbb R
\mbox{ for all } 1 \leq i \leq n\right\}.\] Show that $U$ is a subspace of $V$.
} \newline \noindent {\bf Solution\ } By choosing the scalars to be all 0, we
deduce that ${\bf 0} \in U \not = \emptyset.$ Now suppose that
${\bf x}, {\bf y} \in U$, so there are scalars $\theta_i$ and $\varphi_i$
such that 
\[ {\bf x} = \sum_{i=1}^n \theta_i {\bf v}_i\]
and 
\[ {\bf y} = \sum_{i=1}^n \varphi_i {\bf v}_i\]
so 
\[ {\bf x} +{\bf y} 
= \sum_{i=1}^n (\theta_i +\varphi_i) {\bf v}_i \in U.\]
Thus $U$ is a subspace.
 
\item {\em Suppose that $U, W$ are subspaces of $V$ and that
$U \cup W$ is also a subspace of $V$.
Prove that either $U \subseteq W$ or $W \subseteq U$.} Suppose (for contradiction)
that 
neither $U \subseteq W$ nor $W \subseteq U$. Thus there is ${\bf w} \in W - U$
and ${\bf u} \in U - W.$ Consider ${\bf x} = {\bf u} + {\bf w}.$
Now $U \cup W$ is a subspace of $V$ so ${\bf x} \in U \cup W$.
If ${\bf x} \in U$ then ${\bf w} = {\bf x} -  {\bf u} \in U$ which is false.
If ${\bf x} \in W$ then ${\bf u} = {\bf x} -  {\bf w} \in W$ which is false.
This is the required contradiction, so we are done.
\end{enumerate} 
\end{document}