1.8 Introduction to Linear Transformations
Since the matrix A acts on the vector x, we
can look at A as a type of function with
vectors as its inputs.
The difference between the matrix equation
Ax = b and the vector equation
x1v1  x2 v 2    x p v p  b is only notation.
But, a matrix equation Ax = b can arise in
linear algebra and its applications in a way
that is not directly connected to linear
combinations of vectors.
We think of A as an object that acts by
multiplication on the vector x to produce the
vector b.
 2  4
A  3  6 u  2 v  2
 3 ,
1 
Ex:
,


 
1  2
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Then,
 2  4
 8 
3  6 2   12

  3 



1  2
  4 
and
 2  4
0 
 3  6   2   0 

 1   
1  2   0 .
Notation and Terminology:
Definition: A matrix transformation
(transformation) T from Rn to Rm is a rule that
assigns to each vector x in Rn a vector T(x) in
Rm .
►Rn is the domain of T.
►Rm is the codomain of T.
►T(x) in Rm is the image of x under the
transformation T.
►The set of all images T(x) is the range of T.
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Example: Let
1 0 
A  2 1 x  2
1 
,
 
0 1
We define a transformation
T: R2 →to R3 by T(x) = Ax
Then
1 0 
 2
 2  


T x   Ax  2 1    5
1

0 1
1
Or
 1  0   2 
2 2  11  5 
 
0 1 1 
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Example: Let
 2
3 
 1 2
 3
A
,
u


 

5
10

15


1 ,
 3
 2 
c 
b

0 
 10 ,
Then, define a transformation
T: R3 →to R2 by T(x) = Ax
►Find an x in R3 whose image under T is b.
Solve T(x) = b for x.
Or solve Ax = b for x.
 x1 
3    2 
 1 2
 5 10  15  x2    10

x  

 3
The augmented matrix is and the reduced row
echelon form are below.
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2  1  2 3 2
 1 2
 5 10  15  10 ~ 0 0 0 0


 
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The solution is
x1  2 x2  3x3  2
x2 is free
x3 is free
To get a specific x, choose values for x2 and
x3.
Since zero is easy to work with, let x2 = x3 = 0.
 2
x  0
0 .
Is there more than one x in R3 whose image
under T is b?
Is there an x in R3 whose image under T is c?
This is another way of asking is Ax = c
consistent?
The augmented matrix is and the reduced row
echelon form are below.
1
 5

2
10
3 3 1
~

 15 0 0
2
0
3 0

0 1
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Since there is no solution, we say c is not in
the range of T.
Transformations have many applications
including computer graphics. The basic
transformation in R2 is:
1 0
A

0
1


T: R2 →to R2 defined by T(x) = Ax
1 0  x1 
1 
0 
Ax  
 x1    x2  



 0 1   x2 
0 
1 
 x1 
Then   
So,
 x2 
1 0
A
 does not change the vector it
0
1


multiplies. We call it the 2x2 identity matrix.
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Consider:
0.5 0 
A

0
0
.
5


This transformation T: R2 →to R2 defined by
T(x) = Ax
is an example of a contraction. It can be used
to move a point x toward 0.
Properties of Linear Transformations
If A is an m x n matrix then the transformation
T(x) = Ax has the following properties
T(u + v) = A(u + v)
= A(u) + A(v)
= T(u) + T(v)
and
T(cu) = A(cu)
= cA(u) = cT(u)
for all u, v in Rn and all scalars c.
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Definition: A transformation T is linear if:
T(u + v) = T(u) + T(v)  u, v in the domain of
T
and
T(cu) = cT(u)
 u, v in the domain of T and all scalars c.
►Every matrix transformation is a linear
transformation.
Results: if T is a linear transformation, then:
T(0) = 0
And
T(cu + dv) = cT(u) + dT(v)
Proof:
T(0) = T(0u)
= 0T(u)
=0
And
T(cu + dv) = T(cu) + T(dv)
= cT(u) + dT(v)
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Example: Let
0 
1 
0 y1  0 y 2  1
1 
 
 
e1    e 2   
1
2 ,
1 ,
0  ,
Suppose T: R2 →to R3 is a linear
transformation which maps e1 into y1 and e2
into y2.
 x1 
 3
and  


Find the images of 2
 x2  under T.
ˆ ˆ
Note: What we used to call i , j are now
1 
0 
e1   
e2   
0 and
1
We are given: T(e1) = y1 and T(e2) = y2.
3
To find the image of  2  under T, write
3
2
 
as a linear combination of e1 and e2.
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3 1 0
2  30  21
     
3e1  2e 2
Now, we can apply T.
 3 
T      T 3e1  2e 2 
 2 
 3T e1   2T e 2 
 3y 1  2 y 2
 3  0 




 0    2 
6 2
3
 2
8 
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For the general case:
 x1 
write  x2  as a linear combo of e1 and e2.
 x1 
1
0
 x   x1 0  x2 1
 
 
 2
x1e1  x2e 2
  x1  
T      T  x1e1  x2e 2 
  x2  
 x1T e1   x2T e 2   x1y1  x2 y 2
 x1   0   x1 
 0   x    x 
2
   2 

2 x1   x2  2 x1  x2 
Example: A non-linear T:
Define: T: R2 →to R3 s.t.
T x1, x2 , x3    x1  x3 ,2  5x2 
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We could also write
 x1 
   x1  x3 
T  x2   

 x   2  5 x2 
 3
To show T is not linear, we need to provide a
counterexample where T fails to meet one of
the requirements of linearity:
 T(0) = 0
 T(u + v) = T(u) + T(v)
 T(cu) = cT(u)
Take T(0) = 0
 0
   0  0  0 0
T  0  
  
 0   2  0  2 0
 
So, T is not linear.
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For the third property, let c = –1 and
1
u  1
1
T(cu)=
 1 
  1 
  
 
T   11   T   1 
 1 
  1 
  
 
  1  1   2 

 
 2  5   3
But
 1 
 1 
  
 
T   11   1T  1 
 1 
 1 
  
 
 1  1    2
 1
 
2  5  7
Only one of these is necessary.
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