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\begin{document}
\centerline{Math 201 lecture for Friday, Week 2}
\bigskip

\centerline{\bf Subspaces and spanning sets I}
\medskip

{\bf Note:} Unless otherwise stated, from now on~$V$ will denote a vector space
over a field~$F$.
\medskip

The over-arching goal of the next several classes is to define the notions of
\emph{dimension} and \emph{isomorphism} and show that every finite-dimensional
vector space over~$F$ is isomorphic to the vector space of~$d$-tuples, $F^{d}$,
where~$d$ is the dimension.  Today's class lays some of the groundwork for reaching that
goal.
\medskip

{\bf Definition.} Let~$S$ be a nonempty subset of~$V$.  Then~$v\in V$ is a
\emph{linear combination} of vectors in~$S$ if there exist
$u_1,\dots,u_n\in S$ and $a_1,\dots,a_n\in F$ (for some~$n$) such that
\[
  v = \sum_{i=1}^{n}a_iu_i=a_1u_1+\dots+a_nu_n.
\]
\medskip

{\bf Example.} Let~$S=\left\{ (3,2),(2,-1) \right\}\subset\R$.  Is~$(-1,4)$ a linear
combination of vectors in~$S$?  In other words, do there exist~$a,b\in \R$ such
that
\[
  a(3,2)+b(2,-1)=(-1,4)?
\]
Since $a(3,2)+b(2,-1)=(3a+2b,2a-b)$, the above requirement is equivalent to the
existence of~$a,b\in\R$ such that
\[
  \sysdelim..
  \systeme{3a+2b=-1,2a-b=4{,}}
\]
a system of linear equations!  Apply our algorithm to look for solutions:
\begin{align*}
  &\left(\begin{array}{rr|r}
      3&2&-1\\
      2&-1&4
  \end{array} \right)
  \xrightarrow{r_1\to r_1-r_2}
  \left(\begin{array}{rr|r}
      1&3&-5\\
      2&-1&4
  \end{array} \right)
  \xrightarrow{r_2\to r_2-2r_1}
  \left(\begin{array}{rr|r}
      1&3&-5\\
      0&-7&14
  \end{array} \right)
  \xrightarrow{r_2\to -\tfrac{1}{7}r_2}\\[10pt]
  &\left(\begin{array}{rr|r}
      1&3&-5\\
      0&1&-2
  \end{array} \right)
  \xrightarrow{r_1\to r_1-3r_2}
  \left(\begin{array}{rr|r}
      1&0&1\\
      0&1&-2
  \end{array} \right).
\end{align*}
Thus,~$a=1$ and~$b=-2$.  Check:
\[
  1\cdot(3,2)-2(2,-1)=(-1,4).\quad \checkmark
\]

So~$(-1,4)$ is a linear combination of the two given
vectors.  (If it were not, we would have had an inconsistent system, i.e., a
  system with no solutions.)
\medskip

{\bf Definition.} Let~$S$ be a nonempty subset of~$V$.  The \emph{span}
of~$S$, denoted~$\Span(S)$, is the set of all linear combinations of elements
of~$S$.  By convention~$\Span\emptyset:=\left\{ 0 \right\}$, and we say
that~$0$ is the {\em empty linear combination}.
\medskip

{\bf Example.} In~$\R^2$,
\[
  \Span\left\{ (1,1) \right\}=\left\{ (a,a):a\in \R \right\}.
\]
In~$\R^3$,
\medskip
\[
  \Span\left\{ (1,0,0),(0,1,0) \right\}=\left\{ a(1,0,0)+b(0,1,0):a,b\in\R
  \right\}=\left\{ (a,b,0):a,b\in\R \right\}.
\]
Note that the same set can be spanned by different sets of vectors, for
instance,
\begin{align*}
  \Span\left\{ (1,0,0),(0,1,0) \right\}&= \Span\left\{ (1,0,0),(0,2,0)
  \right\}\\
  &= \Span\left\{ (1,0,0),(0,1,0),(2,3,0) \right\}.
\end{align*}
A point in~$\R^3$ is in any of these sets if and only if its third component
is~$0$.
\medskip

{\bf Definition.} A subset~$W\subseteq V$ is a \emph{subspace}
of~$V$ if~$W$ is a vector space itself with the operations of addition and
scalar multiplication inherited from~$V$.
\medskip

{\bf Proposition.} $W\subseteq V$ is a subspace of~$V$ if and only if
\begin{enumerate}[itemsep=0pt]
  \item $0\in W$
  \item $W$ is closed under addition ($x,y\in W\Rightarrow x+y\in W$)
  \item $W$ is closed under scalar multiplication ($c\in F$ and $w\in
    W\Rightarrow cw\in W$).
\end{enumerate}
\smallskip

{\bf Proof.} Exercise.  Part 1 is there to ensure that~$W$ is nonempty.  (Note
  that Part 2 and Part 3 are vacuously true for the empty set, and yet the
  empty set is not a subspace because of Part~1.)  \hfill$\Box$
\medskip

{\bf Examples.} 
\begin{enumerate}
  \item $W=\left\{ (a,0):a\in \R \right\}$ is a subspace of~$\R^2$.  
    \smallskip

  {\bf Proof.} Letting~$a=0$, we see~$(0,0)\in W$.  If~$(a,0),(b,0)\in W$,
  then~$(a,0)+(b,0)=(a+b,0)\in W$. If~$c\in \R$ and~$(a,0)\in W$,
  then~$c(a,0)=(ca,0)\in W$.  Thus,~$W$ is a subspace of~$\R^2$.
  \hfill$\Box$
\item Let 
  \begin{align*}
    V&= \left\{ f\colon\R\to\R: \text{$f$ is continuous} \right\},\\
  W&= \left\{ f\colon\R\to\R: \text{$f$ is differentiable} \right\}.
  \end{align*}
  Both~$V$ and~$W$ are subspaces of the vector space~$\R^{\R}$ of functions
  from~$\R$ to~$\R$ (recall our earlier notation~$F^S$ for functions from a
  set~$S$ to a field~$F$), and~$W$ is a subspace of~$V$.
\item Let~$W=\left\{ (a,b)\in\R^2:ab=0 \right\}$.  So~$W$ is the union of the
  two coordinate axes in~$\R^2$.  Each of these coordinate axes is a subspace
  of~$\R^2$, but~$W$ is not.  For instance, $(1,0),(0,1)\in W$,
  but~$(1,0)+(0,1)=(1,1)\notin W$.  So~$W$ is not closed under addition.
\item $\left\{ 0 \right\}$ and~$V$ are always subspaces of~$V$.  The empty
  set~$\emptyset$ is not a subspace (since it does not contain~$0$).
\end{enumerate}

\medskip

{\bf Proposition.} If~$W_1$ and~$W_2$ are subspaces of~$V$, so is~$W_1\cap W_2$.
\smallskip

{\bf Proof.} Since~$W_1$ and~$W_2$ are subspaces, we have~$0\in W_i$
for~$i=1,2$.  Hence, $0\in W_1\cap W_2$.  If~$u,v\in W_1\cap W_2$,
then~$u,v\in W_i$ for $i=1,2$.  Hence,~$u+v\in W_i$ for $i=1,2$.  Similarly, for
each~$\lambda\in F$, 
\begin{align*}
  u\in W_1\cap W_2
  &\quad\Rightarrow\quad 
  u\in W_1\text{ and } u\in W_2\\[5pt]
  &\quad\Rightarrow\quad 
  \lambda u\in W_1\text{ and } \lambda u\in W_2\\[5pt]\
  &\quad\Rightarrow\quad 
  \lambda u\in W_1\cap W_2.
\end{align*}
\hfill$\Box$
\medskip

{\bf Proposition.} Let~$S$ be a subset of~$V$. Then:
\begin{enumerate}[itemsep=0pt]
  \item $\Span(S)$ is a subspace of~$V$.
  \item If~$W\subseteq V$ is a subspace and~$S\subseteq W$,
    then~$\Span(S)\subseteq W$.  (In other words: a subspace is closed under the
    process of taking linear combinations of its elements.)
  \item Every subspace of~$V$ is the span of some subset of~$V$.
\end{enumerate}
\smallskip

{\bf Proof.} 1. If~$S=\emptyset$, then~$\Span(S)=\left\{ 0 \right\}$, which is
a subspace of~$V$.  Otherwise, we will show~$0\in\Span(S)$ and~$\Span(S)$ is
closed under addition and scalar multiplication.  Since~$S\neq\emptyset$, there
exists some~$u\in S$.  Then~$0\cdot u$ is a linear combination of elements
in~$S$, and~$0\cdot u=0$ (the first~$0$ in this equation is in~$F$, and the second is
in~$V$).  Hence,~$0\in\Span(S)$.  Now let~$x,y\in\Span(S)$ so that
\begin{align*}
  x&= a_1u_1+\dots+a_m u_m\\
  y&= b_1v_1+\dots+b_n v_n
\end{align*}
for some~$a_i,b_i\in F$ and~$u_i,v_i\in S$.  Then
\[
  x+y=a_1u_1+\dots+a_m u_m+b_1v_1+\dots+b_n v_n\in\Span(S)
\]
and for each~$\lambda\in F$,
\[
  \lambda x=\lambda(a_1u_1+\dots+a_m u_m)=(\lambda a_1)u_1+\dots+(\lambda_m
  a_m)u_m\in\Span(S).
\]

2. Take~$x\in\Span(S)$.  Then~$x=a_1u_1+\dots+a_mu_m$ for some~$a_i\in
F$ and~$u_i\in S$. Since~$S\subseteq W$, we have~$u_i\in W$ for all~$i$, and
since~$W$ is a subspace, it is closed under vector addition and scalar
multiplication.  Therefore,~$x\in W$.

3. $\Span(W)=W$.
\hfill$\Box$
\medskip

{\bf Definition.} A subset~$S\subseteq V$ {\em generates} a subspace~$W$
if~$\Span(S)=W$. 
\medskip

{\bf Examples.} 
\begin{enumerate}
  \item $\left\{1, x, x^2,\dots, \right\}$ generates~$P(F)$, the vector space of
    polynomials in one variable over~$F$.  More commonly, this vector space is
    denoted~$F[x]$.
  \item $\left\{ (1,0),(0,1) \right\}$ generates~$\R^2$.  So do $\left\{
    (1,0),(0,1),(3,-2) \right\}$ and~$\left\{ (1,1),(0,1) \right\}$.
  \item The \emph{$i$-the standard basis vector} for~$F^n$ is
    $e_i:=(0,\dots,0,1,0,\dots,0)$, the vector whose only nonzero entry is in
    the~$i$-th component.  We have that~$\{e_1,\dots,e_n\}$ generates~$F^n$.
\end{enumerate}
\end{document}
