10
20
30
40
50
60
A man drops a ball from
the top of a building.
After ½ second, the ball has fallen 4 feet.
10
1/2
20
30
40
50
60
4
After 1 second, the ball has fallen 16 feet.
10
10
1/2
20
20
30
30
40
40
50
50
60
60
1
4
16
After 3/2 second, the
ball has fallen 36 feet.
10
10
10
1/2
20
20
20
30
30
30
40
40
40
1
4
16
3/2 36
50
50
50
60
60
60
After 2 seconds, the
ball has fallen 64 feet.
10
10
10
10
1/2
20
20
20
20
30
30
30
30
40
40
40
40
1
4
16
3/2 36
50
50
50
50
60
60
60
60
2
64
Motion is described as a
set of ordered pairs.
10
20
30
{ ( 1/2 , 4 ), ( 1 ,16 ), ( 3/2 , 36 ), ( 2 , 64 ) }
1/2
1
4
16
40
3/2 36
50
60
2
64
Motion is described as a
set of ordered pairs.
10
20
30
{ ( 1/2 , 4 ), ( 1 ,16 ), ( 3/2 , 36 ), ( 2 , 64 ) }
Sometimes there is a pattern, and we
can write an equation:
1/2
1
4
16
40
d
=
16 t2
3/2 36
50
60
t is time in seconds
d is distance in feet
2
64
More generally, a function is defined as a
set of ordered pairs.
10
20
30
{ ( 1/2 , 4 ), ( 1 ,16 ), ( 3/2 , 36 ), ( 2 , 64 ) }
When we write an equation for a function,
the solutions (ordered pairs) define the
function.
1/2
1
4
16
40
d
=
16 t2
3/2 36
50
60
t is time in seconds
d is distance in feet
2
64
A function is defined as a
set of ordered pairs.
10
20
30
40
{ ( 1/2 , 4 ), ( 1 ,16 ), ( 3/2 , 36 ), ( 2 , 64 ) }
The DOMAIN of the function =
{ 1/2 , 1 , 3/2 , 2 }
The RANGE of the function =
{ 4 ,16 , 36 , 64 }
1/2
1
4
16
3/2 36
50
60
2
64
10
20
30
The DOMAIN
of the function =
1/2
1
3/2
2
The RANGE
of the function =
4
16
36
64
The function is a mapping that relates every
Domain element t to a unique corresponding
Range element, denoted f(t) and called the
image of t
1/2
1
4
16
40
3/2 36
50
60
2
64
10
20
30
The DOMAIN
of the function =
1/2
1
3/2
2
The RANGE
of the function =
4
16
36
64
The function is a mapping that relates every
Domain element t to a unique corresponding
Range element, denoted f(t) and called the
image of t
1/2
1
4
16
40
50
60
4 is the image of ½
16 is the image of 1
36 is the image of 3/2
64 is the image of 2
4=f(½)
16 = f ( 1 )
36 = f ( 3/2 )
64 = f ( 2 )
3/2 36
2
64
10
20
30
40
50
60
The DOMAIN
of the function =
1/2
1
3/2
2
The RANGE
of the function =
4
16
36
64
The function (of high school algebra fame)
relates a set of real numbers to
another set of real numbers.
1/2
1
4
16
Next we will examine a mapping that links
a set of vectors to another set of vectors. In doing 3/2 36
so, we use much of the same terminology that we
used in the study of functions. A function is a type
of mapping.
2 64
A farmer plans to purchase a herd of cows.
He considers 2 breeds:
Purple cows and Brown cows
Each day a purple cow
will eat 1 bale of hay and
will produce 2 bottles of milk
Each day a brown cow
will eat 2 bales of hay and
will produce 3 bottles of milk
purple
brown
 50 
 
 70 
 190 


 310 
A herd comprised of 50 purple and 70 brown cows
will consume 190 bales of hay and produce 310 bottles of milk.
 50 
 
 70 
 190 


 310 
purple
brown
 50 
 
 70 
100 


 30 
 190 


 310 
 160 


 290 
A herd comprised of 100 purple and 30 brown cows
will consume 160 bales of hay and produce 290 bottles of milk.
100 


 30 
 160 


 290 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
A herd comprised of 80 purple and 150 brown cows
will consume 380 bales of hay and produce 610 bottles of milk.
 80 


150 
 380 


 610 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
The DOMAIN of the mapping:
These vectors describe the
composition of the herd,
and this determines
 190 


 310 
 160 


 290 
 380 


 610 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
The DOMAIN of the mapping:
These vectors describe the
composition of the herd,
and this determines
 190 


 310 
 160 


 290 
 380 


 610 
The RANGE of the mapping:
These vectors describe the
daily food intake and
milk yield.
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
Just as it is sometimes possible to find an equation to define a function,
it is sometimes possible to produce a matrix to define a vector space
mapping.
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
Just as it is sometimes possible to find an equation to define a function,
it is sometimes possible to produce a matrix to define a vector space
mapping.
# bales of hay = 1 (# purple cows) + 2 (# brown cows)
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
Just as it is sometimes possible to find an equation to define a function,
it is sometimes possible to produce a matrix to define a vector space
mapping.
# bales of hay = 1 (# purple cows) + 2 (# brown cows)
# bottles of milk = 2 (# purple cows) + 3 (# brown cows)
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
Just as it is sometimes possible to find an equation to define a function,
it is sometimes possible to produce a matrix to define a vector space
mapping.
# bales of hay = 1 (# purple cows) + 2 (# brown cows)
# bottles of milk = 2 (# purple cows) + 3 (# brown cows)
 # bales   1 2  # purple 

  


 # bottles   2 3  # brown 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
eg:
 190 


 310 
 160 


 290 
 380 


 610 
 # bales   1 2 100 

  


 # bottles   2 3  30 
 # bales   1 2  # purple 

  


 # bottles   2 3  # brown 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
eg:
 190 


 310 
 160 


 290 
 380 


 610 
 160   1 2 100 

  


 290   2 3  30 
100 
,
to find the vector related to 
30


100   1 2 
by

multiply 
30
2
3

 

 # bales   1 2  # purple 

  


 # bottles   2 3  # brown 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
For every domain element v ( a vector in R2 whose entries are the
numbers of each breed of cow) there is a unique corresponding
range element Av ( a vector in R2 whose entries are the numbers of
bales consumed and bottles produced.)
A
 # bales   1 2  # purple 

  


 # bottles   2 3  # brown 
purple
brown
 50 
 
 70 
100 


 30 
 80 


150 
 190 


 310 
 160 


 290 
 380 


 610 
eg:
100   1 2 100   160 
100 
  

  
  the image of 

A
 30   2 3  30   290 
 30 
A
 # bales   1 2  # purple 

  


 # bottles   2 3  # brown 