\magnification 1800
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein
\vsize 9.5truein
\input ../../../PAPERS/GOLD/psfig

\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}

\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\a{{\bf a}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{\vskip -5pt}
\def\v{{\bf v}}
\def\w{{\bf w}}

%%\input amsmath
%%\usepackage{amssymb}
%%\def\R{\mathbb{R}}

{\bf 1.1}  Vectors:

Let ${\bf v} = \left(\matrix{1 \cr 2}\right)$.

\vskip -1.1in
\rightline{\psfig{figure=graph.eps,width=2truein}
\hskip 0.3in \psfig{figure=graph.eps,width=2truein}}

If $\v$ = velocity in m/sec of an object, then the object is moving east 
at a rate of 1 m/sec and 
north at a rate of 2m/sec


Speed of the object =  

\vfill
A vector can be described by its Euclidean coordinates OR by its length and direction.

Let ${\bf w} = \left(\matrix{3 \cr -1}\right)$.  Then $\v + \w =  \left(\matrix{1 \cr 2}\right) + 
\left(\matrix{3 \cr -1}\right) = $

\centerline{\psfig{figure=graph.eps,width=2truein}
\hskip 0.3in \psfig{figure=graph.eps,width=2truein}
\hskip 0.3in \psfig{figure=graph.eps,width=2truein}}

$\v - \w =  \left(\matrix{1 \cr 2}\right) - \left(\matrix{3 \cr -1}\right) $ = 

\centerline{\psfig{figure=graph.eps,width=2truein}
\hskip 0.3in \psfig{figure=graph.eps,width=2truein}
\hskip 0.3in \psfig{figure=graph.eps,width=2truein}}

\eject

{\bf 2.1}  Let $f:  X \rightarrow Y$ where $X \subset \R^n$, $Y \subset \R^m$

Graph of $f = \{ (\x, f(\x)) ~|~ \x \in X  \} \subset \R^n \times \R^m$

Domain of $f = X$, \hfil  
Codomain of $f = Y$ 

Range of $f$ = Image of $f = f(X)$

\rightline{$ = \{y \in Y ~|~ $ there exists $x \in X$ such that $f(x) = y\}$.}

\vskip 5pt
\hrule
\vskip -5pt

$f$ is a function if for all $x$ in domain of $f$, $f(x)$ has a unique value.

\centerline{I.e, for all $x, y \in X$, if $x = y$, then $f(x) = f(y)$}
\centerline{and for all $x \in X$, $f(x)$ is defined.}
\vfill
\vskip 5pt
\hrule
\vskip -5pt

$f$ is 1:1 if $f(x) = f(y)$ implies $x = y$.

\hskip 20pt   $f$ gives a one-to-one correspondence between $X$ and $f(X)$.

\hskip 20pt   Given $b \in Y$, $f(x) = b$ has at most one solution 

\hskip 40pt  Side-note: $f(x) = b$ has exactly one solution if $b \in f(X)$.
. 
\hskip 40pt  Side-note: $f(x) = b$ has no solution if $b \not\in f(X)$.
\vfill

\vskip 5pt
\hrule
\vskip -5pt

$f$ is onto if $f(X) = Y$ (i.e., image of $f$ = codomain of $f$).

\hskip 20pt  Given $b \in Y$, $f(x) = b$ has at least one solution. 
\eject
%%\vskip 5pt
%%\hrule
%%\vskip -5pt
\parskip 12pt
Ex 1:  $f:\R^n \rightarrow \R$, $f(\x) = ||\x|| = \sqrt{x_1^2 + x_2^2 + ... + x_n^2}$

\hskip 20pt  Domain = \hskip 40pt  Codomain = 
\hskip 40pt   Image = 

\hskip 20pt Is $f$ 1:1?  

\hskip 30pt Proof:

\vskip 10pt

\hskip 20pt   Is $f$ onto?

\hskip 30pt Proof:

\vskip 10pt

\hskip 30pt Alternate Proof:

\vskip 10pt

\vfil


Ex 2:  $g(x, y) = (x^2y, x^4 - y, x^6)$

\hskip 20pt  Domain = \hskip 40pt  Codomain = 
\hskip 40pt   

\hskip 20pt Is $g$ 1:1?  

\hskip 30pt Proof:

\vskip 10pt

\hskip 20pt   Is $g$  onto?

\hskip 30pt Proof:


\eject

\parskip 15pt


Ex 3:  $h(\x) = \left(\matrix{1 & 2 & 3 \cr 4 & 5 & 6}\right)\left(\matrix{x \cr y \cr z }\right)$

\hskip 20pt  I.e, $h(\x) = (x + 2y + 3z, 4x +5y + 6z)$.

\hskip 20pt  Domain = 
\hskip 40pt  Codomain = 
\hskip 40pt  Image =

\hskip 20pt  Is $h$ onto?
\hskip 80pt  Is $h$ 1:1?

How many solutions does $h(\x) = {\bf b}$ have?

I.e., how many solutions does $\left(\matrix{1 & 2 & 3 \cr 4 & 5 & 6}\right)\left(\matrix{x \cr y \cr z }\right)
 = \left(\matrix{b_1 \cr b_2 }\right)$ have?

I.e, how many solutions does the following system of equations have:
\u\u
\hskip 60pt  $x + 2y + 3z = b_1$, 
\u
\hskip 60pt  $4x +5y + 6z = b_2$.


Does  $\left(\matrix{1 \cr 4 }\right) x + \left(\matrix{2 \cr 5 
}\right)y 
+ \left(\matrix{3 \cr 6 }\right)z$ span all of $\R^2$?


Is $\{ \left(\matrix{1 \cr 4 }\right), \left(\matrix{2 \cr 5 
}\right),  \left(\matrix{3 \cr 6 }\right) \}$ linearly independent?


\eject

Definitions:  

\vskip -5pt

If the codomain of $f$ is $\R$ (i.e., $f:  X \rightarrow \R$), we say that $f$ is {\it real-valued} or {\it scalar valued}.

 Suppose $f:  X \subset \R^2 \rightarrow \R$ and $c$ is a constant scalar.  

The {\it level curve at height $c$ of $f$}  is the curve in $\R^2$ defined by $f(x, y) = c$.  That is,

\rightline{the level curve at height $c$ of $f = \{(x, y) \in \R^2 ~|~  f(x, y) = c\}$.}

The {\it contour curve at height $c$ of $f$}  is the curve in $\R^3$ defined by the two equations, $z = f(x, y), z = c$.  That is,

\centerline{the contour curve at height $c$ of $f = $}

\rightline{ $= \{(x, y, z) \in \R^2 ~|~  z = f(x, y) = c\}$}

\rightline{$= \{(x, y, f(x, y)) \in \R^3 ~|~  f(x, y) = c\}$.}


Recall the graph of 
$f = \{ (x, y, z) ~|~  z = f(x, y) \}$

\rightline{$= \{ (x, y, f(x, y)) ~|~  (x, y) \in X \}$$\subset \R^2 \times \R$}


The {\it section of the graph of $f$ by the plane $x = c$}  is the set of points in $\R^3$ defined by the two equations, $z = f(x, y), x = c$.  That is,

\centerline{the section by $x = c$ is $ \{(x, y, z) \in \R^2 ~|~  z = f(x, y), x = c\}$}

\rightline{$= \{(c, y, f(c, y)) \in \R^3 ~|~  (c, y) \in X \}$.}




The section by $y = c$ is $ \{(x, y, z) \in \R^2 ~|~  z = f(x, y), y = c\}$

\rightline{$= \{(x, c, f(x, c)) \in \R^3 ~|~  (x,c) \in X \}$.}








 



\end