=\,\omega(\b{e}_1)\b{e}^1+\cdots\omega(\b{e}_n)\b{e}^n
\,=\,\sum_{i=1}^n\omega(\b{e}_i)\b{e}^i$$
for any covector $\omega$.

\endSuppose $\b{l}\in
\vs
\state{The case $m=1$ and $n=1$.}

Suppose $l:{\bf R}\rightarrow{\bf R}$
(Most of the time we do not use boldface to denote a function when the function takes scalar values.)
 is linear.  Let $a=l(1)$.  Then 
$$ l(x)\,=\l(x\,1)\,=\,x\,l(1)\,=\,ax\qquad\hbox{for any $x\in{\bf R}$.}$$

If, on the other hand, $a\in{\bf R}$ and we {\it define} 
$l:{\bf R}\rightarrow{\bf R}$ by setting 
$$l(x)\,=\,ax\qquad\hbox{whenever $x\in{\bf R}$}\leqno(1)$$
then
$$l(cx)\,=\,a(cx)\,=\,c(ax)\,=\,cl(x)\qquad\hbox{whenever $c,x\in{\bf R}$}$$
and
$$ l(x+y)\,=\,a(x+y)\,=\,ax+ay\,=\,l(x)+l(y)
\qquad\hbox{whenever $x,y\in{\bf R}$}$$ so $l$ is linear.  

Thus if $l:{\bf R}\rightarrow{\bf R}$ then $l$ is linear if and only if there
is $a\in{\bf R}$ such that (1) holds in which case $a=l(1)$.
\vs
\state{The case $m=1$ and $n=2$.}
Suppose $l:{\bf R}^2\rightarrow{\bf R}$
 is linear. Let $a=l(\b{i})$ and let $b=l(\b{j})$.  Then for and $\b{x}
=(x,y)\in{\bf R}^2$ we have
$$ l(\b{x})\,=\,l(x\b{i}+y\b{j})\,=\,
l(x\b{i})+l(y\b{j})\,=\,x\,l(\b{i})+yl(\b{j})\,
=\,ax+by.$$

If, on the other $a,b\in{\bf R}$ and we {\it define} 
$l:{\bf R}^2\rightarrow{\bf R}$ at $\b{x}=(x,y)\in{\bf R}^2$ by setting 
$$l(\b{x})\,=\,ax+by\qquad\hbox{whenever $x\in{\bf R}$}\leqno(2)$$
then
$$l(c\b{x})\,=\,l(cx,cy)\,=\,a(cx)+b(cy)\,=\,c(ax+by)\,=\,cl(\b{x})
\qquad\hbox{whenever $c\in{\bf R}$ and $\b{x}=(x,y)\in{\bf R}$}$$
and
$$ l(\b{x}^1+\b{x}^2)\,=\,l(x^1+x^2,y^1+y^2)\,=\,a(x^1+x^2)+b(y^1+y^2)\,
=\,(ax^1+by^1)+(ax^2+by^2)\,=\,l(\b{x}^1)+l(\b{x}^2)
\qquad\hbox{whenever $\b{x}^1=(x^1,y^1),\b{x}^2=(x^2,y^2)\in{\bf R}$}$$ so $l$ is linear.  

Thus if $l:{\bf R}^2\rightarrow{\bf R}$ then $l$ is linear if and only if there
are $a,b\in{\bf R}$ such that (2) holds in which case $a=l(\b{i})$ and
$b=l(\b{j})$.
\vs
\state{The case $m=1$ and $n=3$.} Proceeding as above, the reader should be
able easily to verify that if $l:{\bf R}^3\rightarrow{\bf R}$
the $l$ is linear if and only if there are $a,b,c\in{\bf R}$ such that
$$l(\b{x})\,=\,ax+by+cz\qquad\hbox{whenever $\b{x}=(x,y,z)\in{\bf R}^3$}$$
in which case $a=l(\b{i}),b=l(\b{j})$ and $c=l(\b{k})$.
\vs
The reader should be able to state
\end

Thus if $l:{\bf R}\rightarrow{\bf R}$ then $l$ is linear if and only if there
is $a\in{\bf R}$ such that (1) holds in which case $a=l(1)$.
\vs
\

Let $a=l(\b{i})$ and let $b=l(\b{j}).  Suppose
$\b{x}=(x,y)\int




\state{Proposition.} Suppose $\b{l}:{\bf R}^n\rightarrow{\bf R}^m$ and
$\b{l}$ is linear and suppose, for each $i=1\ldots,m$ and $j=1\ldots,n$,
$$ l_{i,j} $$
is the $j$-th component of $\b{l}(\b{e}_j)$.  Then
$$ \b{l}(\b{x})\,=\,\sum_{i=1}^m (\sum_{j=1}^n l_{i,j}x^j)\b{e}_i\leqno(1)$$
whenever $\b{x}\in{\bf R}^n$.
\proof Suppose $\b{x}\in{\bf R}^n$. Then
$$ \b{x}\,=\,\sum_{j=1}^n x^j\b{e}_j.$$
Applying $\b{l}$ to this equation and using the fact that $\b{l}$ is linear
we obtain
$$ \b{l}(\b{x})\,=\,\sum_{j=1}^n x^j\b{l}(\b{e}_j).$$
But
$$ \b{l}(\b{e}_j)\,=\,\sum_{i=1}^m l_{i,j}\b{e}_i.$$
\endproof
\state{Remark.} The rectangular array
$$\left[\matrix{l_{1,1}&l_{1,2}&\ldots&l_{1,n}\cr
l_{2,1}&l_{2,2}&\ldots&l_{2,n}\cr
\vdots&\vdots&\ddots&\vdots\cr
l_{m,1}&l_{m,2}&\ldots&l_{m,n}}\right]$$
is called the {\bf matrix} of $\b{l}$.  If you know the matrix of $\b{l}$
you can compute $\b{l}(\b{x})$ by using the formula $(1)$ above.
In particular, if $\b{y}=\b{l}(\b{x})$, then
$$ \left[\matrix{y_1\cr y_2 \cr \vdots \cr y_m}\right]\,=\,
\left[\matrix{l_{1,1}&l_{1,2}&\ldots&l_{1,n}\cr
l_{2,1}&l_{2,2}&\ldots&l_{2,n}\cr
\vdots&\vdots&\ddots&\vdots\cr
l_{m,1}&l_{m,2}&\ldots&l_{m,n}}\right]
\left[\matrix{x_1\cr x_2\cr \vdots \cr x_n}\right]$$
according to $(1)$ and the defintion of matrix-vector vector multiplication;
indeed, the definition of matrix-vector multiplication is motivated by $(1)$.
\state{Proposition.} Suppose $\b{l}:{\bf R}^n\rightarrow{\bf R}^m$.  Then
$\b{l}$ is linear if and only if each component $l_i, \ i=1,\ldots,m$ of 
$\b{l}$ is linear.
\proof You do it as an exercise.  This is one of those situations where, once
you understand what it says, it's obvious and where, if you don't understand
what it says, no amount of explanation will help until you do! \endproof
\end


A {\bf covector} $\omega$ is a real valued function on ${\bf R}^n$, so
$$ \omega:{\bf R}^n\rightarrow{\bf R},$$
having the following two properties:
\vs
(1) $\omega(c\b{x})\,=\,c\omega(\b{x})$ for any scalar $c$ and and vector 
$\b{x}$ and 
\vs
(2) $\omega(\b{x}+\b{y})\,=\,\omega(\b{x})+\omega(\b{y})$ for any vectors
$\b{x}$ and $\b{y}$.
\vs
Let $c$ be a scalar and let $\omega$ be a covector.  Set
$$ (c\omega)(\b{x})\,=\,c(\omega(\b{x}))\qquad\hbox{for $\b{x}\in{\bf R}^n$}$$
and check that $c\omega$ is a covector.  Suppose $\omega_1$ and $\omega_2$
are covectors. Set
$$ (\omega_1+\omega_2)(\b{x})\,=\,\omega_1(\b{x})+\omega_2(\b{x})
\qquad\hbox{for $\b{x}\in{\bf R}^n$}$$
and check that $\omega_1+\omega_2$ is a covector.  Then check that covectors
behave the same way as vectors with respect to these operations.  For each 
$i=1,\ldots,n$ we let
$$\b{e}^i$$
be the covector whose value on the vector $\b{x}$ is the $i$-th component of
$\b{x}$.  Verify that
$$ \b{x}\,=\,\b{e}^1(\b{x})\b{e}_1+\cdots\b{e}^n(\b{x})\b{e}_n
\,=\,\sum_{i=1}^n\b{e}^i(\b{x})\b{e}_i$$
for any vector $\b{x}$ and verify that
$$ \omega\,=\,\omega(\b{e}_1)\b{e}^1+\cdots\omega(\b{e}_n)\b{e}^n
\,=\,\sum_{i=1}^n\omega(\b{e}_i)\b{e}^i$$
for any covector $\omega$.

\end