% $Id: CVectorSpace.tex,v 1.6 1999/04/26 14:06:02 herman Exp $

\subsection{Vector Spaces: An external operation}

\begin{definition}
Let $F$ be a field.  An $F$-{\em vector space\/} is a 
tuple \struct{G,\cdot}, with $G$ an abelian group and $\cdot$ 
an operation from $F\times G$ to $G$ satisfying
\begin{eqnarray*}
(\alpha\beta) \cdot a &=& \alpha\cdot(\beta\cdot a),\\
1\cdot a &=& a,\\
(\alpha+\beta)\cdot a &=& (\alpha\cdot a) + (\beta\cdot a),\\
\alpha\cdot(a+b) &=& (\alpha\cdot a) + (\beta\cdot a).
%\alpha\cdot a \noto 0 &\implies& \alpha\noto 0 \wedge a\noto 0,
\end{eqnarray*}
where $\alpha,\beta$ range over $F$ (\emph{scalars}) and $a,b$ range
over $G$ (\emph{vectors}).
Where no confusion arises we will omit the operation $\cdot$,
replacing it with juxtaposition. 
\end{definition}
Note that $\cdot$ is strongly extensional, due to our Convention
\ref{convstrext}.

\begin{lemma} For all $\alpha,a$,
\begin{eqnarray*}
& \alpha 0 = 0,\\
& 0 a = 0,\\
& \alpha a \noto 0 \implies \alpha\noto 0 \wedge a\noto 0,\\
& (-\alpha)a =  -(\alpha a)  = \alpha(-a).
\end{eqnarray*}
\end{lemma}
\begin{proof}
For the first, by cancellation in the group G, it suffices to note
$(\alpha 0)+(\alpha 0)=\alpha (0+0)=\alpha 0$.
The argument for the second is similar. For the third we use the first
and the second. Suppose $\alpha a \noto 0=0 a$. By strong
extensionality of $\cdot$, $\alpha\noto 0$ or $a\noto a$; the latter is absurd.
Similarly, $a\noto 0$. \qed
\end{proof}

\begin{lemma} For all $\alpha,\beta,a,b$,
\begin{eqnarray*}
\alpha\noto 0\wedge a\noto b&\implies & \alpha\cdot a \noto \alpha\cdot b,\\
\alpha\noto 0\wedge a\noto 0&\implies & \alpha\cdot a \noto 0,\\
\alpha\noto \beta\wedge a\noto 0&\implies & \alpha\cdot a \noto \beta\cdot a.
\end{eqnarray*}
\end{lemma}
\begin{proof}
For the first we have: $\alpha^{-1}(\alpha a)=a\noto
b=\alpha^{-1}(\alpha b)$, so $\alpha a \noto\alpha b$ by strong
extensionality of $\cdot$.  The second follows by taking $b=0$.  For
the third, we have $\alpha-\beta\noto 0$ (from \ref{lemgrstrext}), and
$a\noto 0$, so $\alpha a-\beta a=(\alpha-\beta)a\noto 0$, and $\alpha
a\noto\beta a$ by \ref{lemgrstrext} again. \qed
\end{proof}

