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\begin{document}
\section*{Lior Silberman's Math 223: Problem Set 3 (due 1/2/2021)}
\begin{center}
\textbf{Practice problems (recommended, but do not submit)}
\par\end{center}
Section 1.6, Problems 1 (except (g)), 2-5, 7, 11,12, 22{*}, 24{*}.
\begin{center}
\textbf{Bases and dimension}
\par\end{center}
\begin{lyxlist}{10.}
\item [{1.}] (\S1.6 E8)Let $W=\left\{ \underline{x}\in\R^{5}\mid\sum_{i=1}^{5}x_{i}=0\right\} $
be the set of vectors in $\R^{5}$ whose co-ordinates sum to zero.
It is a subspace (but you don't have to check this). The following
8 vectors span $W$ (you don't have to check that either). Find a
subset of them which forms a basis for $W$. $\underline{u}_{1}=\left(2,-3,4,-5,2\right)$,
$\underline{u}_{2}=\left(-6,9,-12,15,-6\right)$, $\underline{u}_{3}=\left(3,-2,7,-9,1\right)$,
$\underline{u}_{4}=\left(2,-8,2,-2,6\right)$, $\underline{u}_{5}=\left(-1,1,2,1,-3\right)$,
$\underline{u}_{6}=\left(0,-3,-18,9,12\right)$, $\underline{u}_{7}=\left(1,0,-2,3,-2\right)$,
$\underline{u}_{8}=\left(2,-1,1,-9,7\right)$.\bigskip{}
\item [{2.}] Find a basis for the subspace $\left\{ \underline{x}\in\R^{4}\mid x_{1}+3x_{2}-x_{3}=0\right\} $
of $\R^{4}$. What is the dimension?\bigskip{}
\item [{3.}] Let $U=\left\{ p\in\R[x]^{\leq n}\mid p(-x)=p(x)\right\} $.
Find a basis for $U$ and determine its dimension.\bigskip{}
\item [{{*}4.}] Let $\R(x)$ be the space of functions of the form $\frac{f}{g}$
where $f,g\in\R[x]$ are polynomials. $\R(x)$ is called ``the field
of rational functions in one variable, and has the same relation to
the ring of polynomials $\R[x]$ that the rational numbers $\Q$ have
to the ring of integers $\Z$. We will consider $\R(x)$ as a real
vector space.
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\frac{1}{1-x}\in\R(x)$ is linearly independent
of the set $\left\{ x^{k}\right\} _{k=0}^{\infty}\subset\R(x)$.
\item [{RMK}] It's true that $\sum_{k=0}^{\infty}x^{k}=\frac{1}{1-x}$
holds on the interval $(-1,1)$, but don't forget that the summation
symbol on the left \emph{does not stand} for repeated addition. Rather,
it stands for a kind of limit.
\item [{(b)}] Show that the subset $\left\{ \frac{1}{x-a}\right\} _{a\in\R}\subset\R(x)$
is linearly independent.
\item [{RMK}] The vector space $\R[x]$ has countable dimension, but by
part (b) the dimension of $\R(x)$ as a real vector space is at least
the cardinality of the continuum (in fact there is equality, because
the cardinality of all of $\R(x)$ is that of the continuum). \bigskip{}
\end{lyxlist}
\end{lyxlist}
\begin{center}
\textbf{Linear Functionals}
\par\end{center}
Fix a vector space $V$. A \emph{linear functional} on $V$ is a map
$\varphi\colon V\to\R$ such that for all $a,b\in\R$ and $\vu,\vv\in V$,
$\varphi(a\vv+b\vu)=a\varphi(\vv)+b\varphi(\vu)$. Let $V^{*}\eqdef\left\{ \varphi\colon V\to\R\mid\varphi\textrm{ is a linear functional}\right\} $
be the set of linear functionals on $V$ (called vector\emph{ }space
\emph{dual} to $V$, in short the \emph{dual space}).
\begin{lyxlist}{10.}
\item [{5.}] (The basic example)
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\varphi\left(\left(\begin{array}{c}
x\\
y\\
z
\end{array}\right)\right)=x-2y+3z$ defines a linear functional on $\R^{3}$.
\item [{(b)}] Let $\varphi$ be a linear functional on $\R^{2}$. Show
that $\varphi\left(\left(\begin{array}{c}
x\\
y
\end{array}\right)\right)=x\cdot\varphi\left(\left(\begin{array}{c}
1\\
0
\end{array}\right)\right)+y\cdot\varphi\left(\left(\begin{array}{c}
0\\
1
\end{array}\right)\right)$ and conclude that every linear functional on $\R^{2}$ is of the
form $\varphi\left(\left(\begin{array}{c}
x\\
y
\end{array}\right)\right)=ax+by$ for some $a,b\in\R$.
\item [{SUPP}] Construct an identification of $\left(\R^{n}\right)^{*}$
with $\R^{n}$.\bigskip{}
\end{lyxlist}
\item [{6.}] Show that $V^{*}$ is a subspace of $\R^{V}$, hence a vector
space.\bigskip{}
\item [{7.}] Let $V$ be a vector space and let $\varphi\in V^{*}$ be
non-zero.
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\Ker\varphi\eqdef\left\{ \vv\in V\mid\varphi(\vv)=0\right\} $
is a subspace.
\item [{({*}b)}] Show that there is $\vv\in V$ satisfying $\varphi(\vv)=1$.
\item [{({*}{*}c)}] Let $B$ be a basis of $\Ker\varphi$, and let $\vv\in V$
be as in part (b). Show that $B\cup\left\{ \vv\right\} $ is a basis
of $V$.
\item [{RMK}] If $V$ is finite-dimensional this shows: $\dim V=\dim\Ker\varphi+1$.
In general we say that $\Ker\varphi$ is of \emph{codimension $1$.}\bigskip{}
\end{lyxlist}
\end{lyxlist}
\newpage{}
\begin{center}
\textbf{A Linear Transformation}
\par\end{center}
In this problem our choice of letters follows conventions from physics.
Thus $v$ will be a numerical parameter rather than a vector, and
we write the coordinates of a vector in $\R^{2}$ as $\left(\begin{array}{c}
x\\
t
\end{array}\right)$ rather than $\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)$.\bigskip{}
\begin{lyxlist}{10.}
\item [{8.}] In the course of his researches on electromagnetism, Henri
Poincaré wrote down the following map $L_{v}\colon\R^{2}\to\R^{2}$:
\[
L_{v}\left(\begin{array}{c}
x\\
t
\end{array}\right)\eqdef\gamma_{v}\cdot\left(\begin{array}{c}
x-vt\\
t-vx
\end{array}\right)\,.
\]
Here $v$ is a real parameter such that $\left|v\right|<1$ and $\gamma_{v}$
is also a number, defined by $\gamma_{v}=\left(1-v^{2}\right)^{-1/2}$.
\begin{lyxlist}{10.}
\item [{(a)}] Suppose $v=0.6$ so that $\gamma_{v}=(1-0.6^{2})^{-1/2}=1.25$.
Calculate $L_{v}\left(\begin{array}{c}
3\\
2
\end{array}\right)$, $L_{v}\left(\begin{array}{c}
-1\\
1
\end{array}\right)$ and $L_{v}\left(\begin{array}{c}
2\\
3
\end{array}\right)$. Check that $L_{v}\left(\begin{array}{c}
2\\
3
\end{array}\right)=L_{v}\left(\begin{array}{c}
3\\
2
\end{array}\right)+L_{v}\left(\begin{array}{c}
-1\\
1
\end{array}\right)$.
\item [{(b)}] Show that $L_{v}$ is a linear transformation.
\item [{(c)}] (``Relativistic addition of velocities'') Let $v,v'\in(-1,1)$
be two parameters. Show that $L_{v}\circ L_{v'}=L_{u}$ for $u=\frac{v+v'}{1+vv'}$.
It is a fact that if $v,v'\in(-1,1)$ then $\frac{v+v'}{1+vv'}\in(-1,1)$
as well.\\
\emph{Hint}: Start by showing $\gamma_{v}\gamma_{v'}=\frac{\gamma_{u}}{1+vv'}$.
\item [{RMK}] If $g\colon A\to B$ and $f\colon B\to C$ are functions
then$f\circ g$ denotes their \emph{composition}, the function $f\circ g\colon A\to C$
such that $\left(f\circ g\right)(a)=f\left(g(a)\right)$for all $a\in A$.
\end{lyxlist}
\end{lyxlist}
\rule[0.5ex]{1\columnwidth}{1pt}
\begin{center}
\textbf{Supplementary problems}
\par\end{center}
\begin{lyxlist}{10.}
\item [{A.}] Let $V$ be a vector space and let $W_{1},W_{2}\subset V$
be finite-dimensional subspaces.
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\dim(W_{1}+W_{2})\leq\dim W_{1}+\dim W_{2}$.
\item [{({*}{*}b)}] Show that $\dim(W_{1}+W_{2})+\dim(W_{1}\cap W_{2})=\dim W_{1}+\dim W_{2}$.
\item [{RMK}] Let $A,B$ be finite sets. Then the ``inclusion-exclusion''
formula states $\#A+\#B=\#(A\cup B)+\#(A\cap B)$
\end{lyxlist}
\item [{B.}] Let $V$ be a vector space, $W$ a subspace. Let $B\subset W$
be a basis for $W$ and let $C\subset V$ be disjoint from $B$ and
such that $B\cup C$ is a basis for $V$ (that is, we extend $B$
until we get a basis for $V$).
\begin{lyxlist}{10.}
\item [{(a)}] Show that $\left\{ \vv+W\right\} _{\vv\in C}$ is a basis
for the quotient vector space $V/W$ ($V/W$ is defined in the supplement
to PS2).
\item [{(b)}] Show that $\dim W+\dim(V/W)=\dim V$.
\end{lyxlist}
\end{lyxlist}
The following problem requires some background in set theory.
\begin{lyxlist}{10.}
\item [{C.}] Let $V$ be a vector space, and let $B,C$ be a bases of $V$.
\begin{lyxlist}{10.}
\item [{(a)}] Suppose one of $B,C$ is finite, Show that the other is finite
and that they have the same size.
\item [{--}] We may therefore assume both $B,C$ are finitely.
\item [{(b)}] For a finite subset $A\subset B$ show that $C\cap\Span(A)$
is finite.
\item [{--}] Let $\cF_{B},\cF_{C}$ be the sets of finite subsets of $B,C$
respectively, and let $f\colon\cF_{B}\to\cF_{C}$ be the function
$f(A)=C\cap\Span(A)$.
\item [{(c)}] Show that the image of $f$ covers $C$.
\item [{(d)}] Show that the cardinality of the image of $f$ is at least
that of $C$.
\item [{(e)}] Show that $\left|B\right|\geq\left|C\right|$. Conclude that
$\left|B\right|=\left|C\right|$, in other words that infinite-dimensional
vector spaces also have well-defined dimensions.
\end{lyxlist}
\end{lyxlist}
\end{document}