Transcript Document

Chap. 6
Linear Transformations
Linear Algebra
Ming-Feng Yeh
Department of Electrical Engineering
Lunghwa University of Science and Technology
6.1 Introduction to
Linear Transformation

Learn about functions that map a vector space V
into a vector space W --- T: V  W
V: domain of T
range
v
T: V  W
w
image of v
W: codomain of T
Ming-Feng Yeh
Chapter 6
6-2
Section 6-1
Map



If v is in V and w is in W s.t. T(v) = w, then
w is called the image of v under T.
The set of all images of vectors in V is called the
range of T.
The set of all v in V s.t. T(v) = w is called the
preimage of w.
Ming-Feng Yeh
Chapter 6
6-3
Section 6-1
Ex. 1: A function from
2
R
into
2
R
For any vector v = (v1, v2) in R2, and
let T: R2 R2 be defined by
T(v1, v2) = (v1  v2, v1 + 2v2)
 Find the image of v = (1, 2)
T(1, 2) =(1 2, 1+2·2) = (3, 3)
 Find the preimage of w = (1, 11)
T(v1, v2) = (v1  v2, v1 + 2v2) = (1, 11)
 v1  v2 = 1; v1 + 2v2 = 11
 v1 = 3; v2 = 4
Ming-Feng Yeh
Chapter 6
6-4
Section 6-1
Linear Transformation


Let V and W be vector spaces. The function
T: V  W is called a linear transformation of V
into W if the following two properties are true for
all u and v in V and for any scalar c.
1. T(u + v) = T(u) + T(v)
2. T(cu) = cT(u)
A linear transformation is said to be operation
reserving (the operations of addition and scalar
multiplication).
Ming-Feng Yeh
Chapter 6
6-5
Section 6-1
Ex. 2: Verifying a linear
2
2
transformation from R into R
Show that the function T(v1, v2) = (v1  v2, v1 + 2v2) is a linear
transformation from R2 into R2.
Let v = (v1, v2) and u = (u1, u2)
 Vector addition: v + u = (v1 + u1, v2 + u2)
T(v + u) = T(v1 + u1, v2 + u2)
= ( (v1 + u1)  (v2 + u2), (v1 + u1) + 2(v2 + u2) )
= (v1  v2 , v1 + 2v2) + (u1  u2, u1 +2u2)
= T(v) + T(u)
 Scalar multiplication: cv = c(v1, v2) = (cv1, cv2)
T(cv) = (cv1  cv2, cv1+ 2cv2) = c(v1  v2, v1+ 2v2) = cT(v)
Therefore T is a linear transformation.
Ming-Feng Yeh
Chapter 6
6-6
Section 6-1
Ex. 3: Not linear transformation



f(x) = sin(x)
In general, sin(x1 + x2)  sin(x1) + sin(x2)
f(x) = x2
2
2
2
In general, ( x1  x2 )  x1  x2
f(x) = x + 1
f(x1 + x2) = x1 + x2 + 1
f(x1) + f(x2) = (x1 + 1) + (x2 + 1) = x1 + x2 + 2
Thus, f(x1 + x2)  f(x1) + f(x2)
Ming-Feng Yeh
Chapter 6
6-7
Section 6-1
Linear Operation &
Zero / Identity Transformation



A linear transformation T: V  V from a vector
space into itself is called a linear operator.
Zero transformation (T: V  W):
T(v) = 0, for all v
Identity transformation (T: V  V):
T(v) = v, for all v
Ming-Feng Yeh
Chapter 6
6-8
Section 6-1
Thm 6.1: Linear transformations
Let T be a linear transformation from V into W,
where u and v are in V. Then the following
properties are true.
1. T(0) = 0
2. T(v) = T(v)
3. T(u  v) = T(u)  T(v)
4. If v = c1v1 + c2v2 + … + cnvn, then
T(v) = c1T(v1) + c2T(v2) + … + cnT(vn)
Ming-Feng Yeh
Chapter 6
6-9
Section 6-1
Proof of Theorem 6.1
1. Note that 0v = 0. Then it follows that
T(0) = T(0v) = 0T(v) = 0
2. Follow from v = (1)v, which implies that
T(v) = T[(1)v] = (1)T(v) = T(v)
3. Follow from u  v = u + (v), which implies that
T(u  v) = T[u + (1)v] = T(u) + (1)T(v)
= T(u)  T(v)
4. Left to you
Ming-Feng Yeh
Chapter 6
6-10
Section 6-1
Remark of Theorem 6.1


A linear transformation T: V  W is determined
completely by its action on a basis of V.
If {v1, v2, …, vn} is a basis for the vector space V
and if T(v1), T(v2), …, T(vn) are given, then T(v) is
determined for any v in V.
Ming-Feng Yeh
Chapter 6
6-11
Section 6-1
Ex 4: Linear transformations
and bases
Let T: R3  R3 be a linear transformation s.t.
T(1, 0, 0) = (2, 1, 4); T(0, 1, 0) = (1, 5, 2);
T(0, 0, 1) = (0, 3, 1). Find T(2, 3, 2).
 (2, 3, 2) = 2(1, 0, 0) + 3(0, 1, 0) 2(0, 0, 1)
T(2, 3, 2) = 2T(1, 0, 0) + 3T(0, 1, 0) 2T(0, 0, 1)
= 2(2, 1, 4) + 3(1, 5, 2) 2(0, 3, 1)
= (7, 7, 0)
Ming-Feng Yeh
Chapter 6
6-12
Section 6-1
Ex 5: Linear transformation
defined by a matrix
The function T: R2  R3 is defined as follows
0
 3
 v1 


T ( v)  Av   2
1  
v2 

 1  2
 Find T(v), where v = (2, 1)
0
 3
6
 2  


T ( v)  Av   2
1    3
 1

 1  2
0
Therefore, T(2, 1) = (6, 3, 0)
Ming-Feng Yeh
Chapter 6
6-13
Section 6-1
Example 5 (cont.)
Show that T is a linear transformation from R2 to
R3.
1. For any u and v in R2, we have
T(u + v) = A(u + v) = Au +Av = T(u) +T(v)
2. For any u in R2 and any scalar c, we have
T(cu) = A(cu) = c(Au) = cT(u)
Therefore, T is a linear transformation from R2 to R3.

Ming-Feng Yeh
Chapter 6
6-14
Section 6-1
Thm 6.2: Linear transformation
given by a matrix
Let A be an m  n matrix. The function T defined by
T(v) = Av
is a linear transformation from Rn into Rm.
In order to conform to matrix multiplication with an
m  n matrix, the vectors in Rn are represented by
m  1 matrices and the vectors in Rm are
represented by n  1 matrices.
Ming-Feng Yeh
Chapter 6
6-15
Section 6-1
Remark of Theorem 6.2



The m  n matrix zero matrix corresponds to the zero
transformation from Rn into Rm.
The n  n matrix identity matrix In corresponds to the
identity transformation from Rn into Rn.
An m  n matrix A defines a linear transformation from
Rn into Rm.
 a11 a12  a1n   v1   a11v1  a12v2  a1n vn 
a
 v   a v  a v   a v 
a

a
22
2n   2 
22 2
2n n 
Av   21
  21 1
 

      


  

am1 am 2  amn  vn  am1v1  am 2v2  amnvn 
Rm
Rn
Ming-Feng Yeh
Chapter 6
6-16
Section 6-1
Ex 7: Rotation in the plane
Show that the linear transformation T: R2  R2 given by the
cos  sin  
matrix A  
has the property that it rotates

 sin  cos 
every vector in counterclockwise about the origin through
the angle . Sol: Let v  ( x, y )  (r cos , r sin  )
cos  sin   r cos  y
T ( v )  Av  
  r sin  
sin

cos

T ( x, y )



 r cos cos   r sin  sin  


r
sin

cos


r
cos

sin



r cos(   )


r
sin(



)


Ming-Feng Yeh
Chapter 6
( x, y )


x
6-17
Section 6-1
Ex 8: A projection in
3
R
The linear transformation T: R3  R3 given by
z
1 0 0
A  0 1 0
( x, y , z )
0 0 0
is called a projection in R3.
If v = (x, y, z) is a vector in R3,
then T(v) = (x, y, 0). In other words,
x
T ( x, y , z )
T maps every vector in R3 to its
 ( x , y ,0 )
orthogonal projection in the xy - plane.
Ming-Feng Yeh
Chapter 6
6-18
y
Section 6-1
Ex 9: Linear transformation
from Mm,n to Mn,m
Let T: Mm,n  Mn,m be the function that maps m  n
matrix A to its transpose. That is, T ( A)  AT
Show that T is a linear transformation.
pf: Let A and B be m  n matrix.
 T ( A  B)  ( A  B)T  AT  BT  T ( A)  T ( B)
and T (cA)  (cA)T  c( AT )  cT ( A)
 T is a linear tra nsformatio n form M m , n into M n , m
Ming-Feng Yeh
Chapter 6
6-19
6.2 The Kernel and Range of
a Linear Transformation



Definition of Kernel of a Linear Transformation
Let T:V  W be a linear transformation. Then
the set of all vectors v in V that satisfy T(v) = 0 is
called the kernel of T and is denoted by ker(T).
The kernel of the zero transformation T: V  W consists
of all of V because T(v) = 0 for every v in V. That is,
ker(T) = V.
The kernel of the identity transformation T: V  V
consists of the single element 0. That is, ker(T) = {0}.
Ming-Feng Yeh
Chapter 6
6-20
Section 6-2
Ex 3: Finding the kernel
Find the kernel of the projection T: R3  R3 given by
T(x, y, z) = (x, y, 0).
z
Sol: This linear transformation
projects the vector (x, y, z)
(0,0, z )
in R3 to the vector (x, y, 0)
in xy-plane. Therefore,
(0,0,0)
ker(T) = { (0, 0, z) : zR}
x
Ming-Feng Yeh
Chapter 6
y
6-21
Section 6-2
Ex 4: Finding the kernel
Find the kernel of T: R2  R3 given by
T(x1, x2) = (x1  2x2, 0, x1).
Sol: The kernel of T is the set of all x = (x1, x2) in R2
s.t. T(x1, x2) = (x1  2x2, 0, x1) = (0, 0, 0).
Therefore, (x1, x2) = (0, 0).
 ker(T) = { (0, 0) } = { 0 }
Ming-Feng Yeh
Chapter 6
6-22
Section 6-2
Ex 5: Finding the kernel
Find the kernel of T: R3  R2 defined by T(x) = Ax, where
 1  1  2
A


1
2
3


Sol: The kernel of T is the set of all x = (x1, x2, x3) in R3 s.t.
T(x1, x2, x3) = (0, 0). That is,
 x1 
 x1   t 
 1  1  2    0     
x2       x2    t , t  R
 1 2


3
0


 x3 
 x3   t 
Therefore, ker(T) = {t(1, 1,1): tR} = span{(1, 1,1) }
Ming-Feng Yeh
Chapter 6
6-23
Section 6-2
Thm 6.3: Kernel is a subspace
The kernel of a linear transformation T: V  W is a
subspace of the domain V.
pf: 1. ker(T) is a nonempty subset of V.
2. Let u and v be vectors in ker(T). Then
T(u + v) = T(u) + T(v) = 0 + 0 = 0 (vector addition)
Thus, u + v is in the kernel
3. If c is any scalar, then T(cu) = cT(u) = c0 = 0
(scalar multiplication), Thus, cu is in the kernel.
 The kernel of T sometimes called the nullspace of T.
Ming-Feng Yeh
Chapter 6
6-24
Section 6-2
Ex 6: Finding a basis for kernel
Let T: R5  R4 be defined by T(x) = Ax, where x is in R5
 1
 2
and A  
 1

 0
2
0 1  1
1
3 1 0
.
0 2 0
1

0
0 2
8
Find a basis for ker(T) as a subspace of R5.
Ming-Feng Yeh
Chapter 6
6-25
Section 6-2
Example 6 (cont.)
 1
 2

 1

 0
2
0
1
3
0 2
0
0
1
1
0
2
 x1  0  x1 
 2  1
 1      
 1  2
x
x
0
2
2
   
0      
 x3   0   x3   s  1  t  0
1      
   
0   4
  x4  0  x4 

8
 x5  0  x5 
 0  1
Thus one basis for the kernel T is given by
B = { (2, 1, 1, 0, 0), (1, 2, 0, 4, 1) }
Ming-Feng Yeh
Chapter 6
6-26
Section 6-2
Solution Space

A basis for the kernel of a linear transformation
T(x) = Ax was found by solving the homogeneous
system given by Ax = 0.

It is the same produce used to find the solution
space of Ax = 0.
Ming-Feng Yeh
Chapter 6
6-27
Section 6-2
The Range of a Linear Transform
Thm 6.4: Range is a subspace
The range of a linear transformation T: V  W is a subspace
of the domain W.
 range(T) = { T(v): v is in V }
 ker(T) is a subspace of V.
pf: 1. range(T) is a nonempty because T(0) = 0.
2. Let T(u) and T(v) be vectors in range(T). Because u and v
are in V, it follows that u + v is also in V. Hence the sum
T(u) + T(v) = T(u + v) is in the range of T. (vector addition)
3. Let T(u) be a vector in the range of T and let c be a scalar.
Because u is in V, it follows that cu is also in V. Hence,
cT(u) = T(cu) is in the range of T. (scalar multiplication)
Ming-Feng Yeh
Chapter 6
6-28
Section 6-2
Figure 6.6
Kernel
ker(T) is a subspace of V
Codomain
V
Domain
T: V  W
Range
Ming-Feng Yeh
0
W
range(T) is a subspace of W
Chapter 6
6-29
Section 6-2
Column Space

To find a basis for the range of a linear transformation
defined by T(x) = Ax, observe that the range consists of all
vectors b such that the system Ax = b is consistent.
 a11 a12
a
a22
21

Ax 
 


am1 am 2

 a1n   x1 
 a11 
 a1n   b1 
a 
a  b 
 a2 n   x2 
 x1  21     xn  2 n    2   b
  
   
    
 
 
   
 amn   xn 
am1 
amn  bm 
b is in the range of T if and only if b is a linear
combination of the column vectors of A.
Ming-Feng Yeh
Chapter 6
6-30
Section 6-2
Corollary of Theorems 6.3 & 6.4
Let T: Rn  Rm be the linear transformation given by
T(x) = Ax.
[Theorem 6.3] The kernel of T is equal to the
solution space of Ax = 0.
[Theorem 6.4] The column space of A is equal to
the range of T.
Ming-Feng Yeh
Chapter 6
6-31
Section 6-2
Ex 7: Finding a basis for range
Let T: R5  R4 be the linear transform given in Example 6.
Find a basis for the range of T.
Sol: The row echelon of A:
0 1  1  1 0 2
 1 2
 2 1
 0 1  1
3
1
0

A
 1 0  2 0
1 0 0 0

 
0 2
8 0 0 0
 0 0
One basis for the range of T is
B = { (1, 2, 1, 0), (2, 1, 0, 0), (1, 1, 0, 2) }
Ming-Feng Yeh
Chapter 6
0  1
0  2
1
4

0
0
6-32
Section 6-2
Rank and Nullity
Let T: V  W be a linear transformation.
 The dimension of the kernel of T is called the
nullity of T and is denoted by nullity(T).
 The dimension of the range of T is called the
rank of T and is denoted by rank(T).
Ming-Feng Yeh
Chapter 6
6-33
Section 6-2
Thm 6.5: Sum of rank and nullity

Let T: V  W be a linear transformation from an
n-dimension vector space V into a vector space W.
Then the sum of the dimensions of the range and
the kernel is equal to the dimension of the domain.
That is,
rank(T) + nullity(T) = n
or
dim(range) + dim(kernel) = dim(domain)
Ming-Feng Yeh
Chapter 6
6-34
Section 6-2
Proof of Theorem 6.5
The linear transformation from an n-dimension vector
space into an m-dimension vector space can be represented
by a matrix, i.e., T(x) = Ax where A is an m  n matrix.
Assume that the matrix A has a rank of r. Then,
rank(T) = dim(range of T) = dim(column space) = rank(A) = r
From Thm 4.7, we have
nullity(T) = dim(kernel of T) = dim(solution space) = n  r
Thus,
rank(T) + nullity(T) = n + (n  r) = n
Ming-Feng Yeh
Chapter 6
6-35
Section 6-2
Ex 8: Finding the rank & nullity
Find the rank and nullity of T: R3  R3 defined by
1 0  2
0 1 1 
A

the matrix


0 0 0 
Sol: Because rank(A) = 2, the rank of T is 2.
The nullity is dim(domain) – rank = 3 – 2 = 1.
Ming-Feng Yeh
Chapter 6
6-36
Section 6-2
Ex 8: Finding the rank & nullity
Let T: R5  R7 be a linear transformation
 Find the dimension of the kernel of T if the
dimension of the range is 2.
dim(kernel) = n – dim(range) = 5 – 2 = 3
 Find the rank of T if the nullity of T is 4
rank(T) = n – nullity(T) = 5 – 4 = 1
 Find the rank of T if ker(T) = {0}
rank(T) = n – nullity(T) = 5 – 0 = 5
Ming-Feng Yeh
Chapter 6
6-37
Section 6-2
One-to-One & Onto Linear Transformation
One-to-One Mapping

A linear transformation T :VW is said to be
one-to-one if and only if for all u and v in V,
T(u) = T(v) implies that u = v.
V
V
W
W
T
T
Not one-to-one
One-to-one
Ming-Feng Yeh
Chapter 6
6-38
Section 6-2
Thm 6.6: One-to-one Linear
transformation
Let T :VW be a linear transformation. Then T is one-to-one
if and only if ker(T) = {0}.
pf: 」Suppose T is one-to-one. Then T(v) = 0 can have
only one solution: v = 0. In this case, ker(T) = {0}.
」Suppose ker(T) = {0} and T(u) = T(v). Because T is a
linear transformation, it follows that
T(u – v) = T(u) – T(v) = 0
This implies that u – v lies in the kernel of T and
must therefore equal 0. Hence u – v = 0 and u = v,
and we can conclude that T is one-to-one.
Ming-Feng Yeh
Chapter 6
6-39
Section 6-2
Example 10


The linear transformation T: Mm,n  Mn,m given
by T ( A)  AT is one-to-one because its kernel
consists of only the m  n zero matrix.
The zero transformation T: R3  R3 is not oneto-one because its kernel is all of R3.
Ming-Feng Yeh
Chapter 6
6-40
Section 6-2
Onto Linear Transformation




A linear transformation T :VW is said to be onto
if every element in W has a preiamge in V.
T is onto W when W is equal to the range of T.
[Thm 6.7] Let T :VW be a linear transformation,
where W is finite dimensional. Then T is onto if
and only if the rank of T is equal to the dimension
of W, i.e., rank(T) = dim(W).
One-to-one: ker(T) = {0} or nullity(T) = 0
Ming-Feng Yeh
Chapter 6
6-41
Section 6-2
Thm 6.8: One-to-one and onto
linear transformation

Let T :VW be a linear transformation with
vector spaces V and W both of dimension n.
Then T is one-to-one if and only if it is onto.
pf: 」If T is one-to-one, then ker(T) = {0} and
dim(ker(T)) = 0. In this case,
dim(range of T) = n – dim(ker(T)) = n = dim(W).
By Theorem 6.7, T is onto.
」If T is onto, then dim(range of T) = dim(W) = n.
Which by Theorem 6.5 implies that dim(ker(T)) = 0
By Theorem 6.6, T is onto-to-one.
Ming-Feng Yeh
Chapter 6
6-42
Section 6-2
Example 11
The linear transformation T:RnRm is given by T(x) = Ax.
Find the nullity and rank of T and determine whether T is
one-to-one, onto, or either.
1 2 0
1 2
1 2 0
1 2 0 




0 1 1
(a) A  0 1 1, (b) A  0 1, (c) A  
,
(
d
)
A




0
1

1


0 0 1
0 0
0 0 0
T:RnRm
dim(domain)
rank(T)
nullity(T)
one-to-one
onto
(a) T:R3 R3
3
3
0
Yes
Yes
(b) T:R2 R3
2
2
0
Yes
No
(c) T:R3 R2
3
2
1
No
Yes
(d) T:R3 R3
3
2
1
No
No
Ming-Feng Yeh
Chapter 6
6-43
Section 6-2
Isomorphisms of Vector Spaces
Isomorphism
Def: A linear transformation T :VW that is one-toone and onto is called isomorphism. Moreover, if
V and W are vector spaces such that there exists an
isomorphism from V to W, then V and W are said
to be isomorphic to each other.
Theorem 6.9: Isomorphism Spaces & Dimension
Two finite-dimensional vector spaces V and W are
isomorphic if and only if they are of the same
dimension.
Ming-Feng Yeh
Chapter 6
6-44
Section 6-2
Ex 12: Isomorphic Vector Spaces
The following vector spaces are isomorphic to each
other.
 R4 = 4-space
 M4,1 = space of all 4  1 matrices
 M2,2 = space of all 2  2 matrices
 P3 = space of all polynomials of degree 3 or less
 V = {(x1, x2, x3, x4, 0): xi is a real number}
(subspace of R5)
Ming-Feng Yeh
Chapter 6
6-45
6.3 Matrices for
Linear Transformation



Which one is better?
T ( x1 , x2 , x3 )  (2 x1  x2  x3 ,  x1  3x2  2 x3 , 3x2  4 x3 )
 2 1  1  x1  Simpler to write. Simpler
T (x)  Ax   1 3  2  x2  to read, and more adapted
 0 3 4   x3  for computer use.
The key to representing a linear transformation T:VW
by a matrix is to determine how it acts on a basis of V.
Once you know the image of every vector in the basis, you
can use the properties of linear transformations to
determine T(v) for any v in V.
Ming-Feng Yeh
Chapter 6
6-46
Section 6-3
Thm 6.10: Standard matrix for
a linear transformation
Let T: RnRm be a linear transformation such that
1  a11 
0  a12 
0  a1n 
0   a 
1  a 
0   a 
T (e1 )  T (   )   21 , T (e 2 )  T (   )   22 , , T (e n )  T (   )   2 n 
   
   
   
   
   
   
a
a
0  m1 
0   m 2 
1 amn 
Then the m  n matrix whose n columns corresponds to T (ei ),
 a11 a12  an1 
is such that T(v) = Av for every
a

a

a
v in Rn. A is called the standard
22
n2 
A   21
 
   
matrix for T.


am1 am 2  amn 
Ming-Feng Yeh
Chapter 6
6-47
Section 6-3
Proof of Theorem 6.10
Let v  v1 v2  vn   v1e1  v2e2    vnen  Rn .
Because T is a linear transformation, we have
T ( v )  T (v1e1  v2e 2    vne n )
 T (v1e1 )  T (v2e 2 )    T (vne n )
 v1T (e1 )  v2T (e 2 )    vnT (e n )
On the other hand,
 a11 a12  an1   v1 
 a11 
 a12 
 a1n 
a
 v 
a 
a 
a 
a

a
22
n2   2 
Av   21
 v1  21   v2  22     vn  2 n 
 
  
  
  
   

 
 
 
 
a
a

a
v
a
a
m2
mn   n 
 m1
 m1 
 m2 
amn 
 v1T (e1 )  v2T (e 2 )    vnT (e n )  T ( v )
T
Ming-Feng Yeh
Chapter 6
6-48
Section 6-3
Example 1
Find the standard matrix for the linear transformation
T: R3R2 defined by T(x, y, z) = ( x – 2y, 2x + y)
 1 
Sol:


1 


T (e1 )  T (1, 0, 0)  (1, 2)  T (e1 )  T  0    
 0    2 
  
 0  
      2
T (e 2 )  T (0, 1, 0)  (2, 1)  T (e 2 )  T  1    
 0   1
  
 0  
    0 
T (e3 )  T (0, 0, 1)  (0, 0)  T (e3 )  T  0    
 1    0 
 
Ming-Feng Yeh
Chapter 6
6-49
Section 6-3
Example 1 (cont.)
 1  2 0
A  T (e1 )T (e 2 )T (e3 )  

2
1
0


Note that
 x
 x
 1  2 0    x  2 y 


A y   
y  



2
1
0
2
x

y
 z 

 z  
 
which is equivalent to T(x, y, z) = ( x – 2y, 2x + y).
Ming-Feng Yeh
Chapter 6
6-50
Section 6-3
Example 2
The linear transformation T: R2R2 is given by projecting
each point in R2 onto to the x-axis. Find the standard
y
matrix for T.
Sol: This linear transformation is given by
( x, y )
T(x, y) = (x, 0).
Therefore, the standard matrix for T is
1 0
A  T (1, 0)T (0, 1)  

0
0


(x, 0)
Ming-Feng Yeh
Chapter 6
6-51
x
Section 6-3
Composition of Linear Transformation
Composition of linear
transformation



The composition T, of T1: RnRm with T2: RmRp is
defined by T(v) = T2( T1(v) ) = T2  T1
where v is a vector in Rn.
The domain of T is defined to the domain of T1.
The composition is not defined unless the range of T1 lies
within the domain of T2.
T2
T1
R
n
v
Rm
w
Rp u
T
Ming-Feng Yeh
Chapter 6
6-52
Section 6-3
Theorem 6.11
Let T1: RnRm and T2: RmRp be linear
transformation with standard matrix A1 and A2.
The composition T: RnRp, defined by
T(v) = T2( T1(v) ), is linear transformation.
Moreover, the standard matrix of A for T is given
by the matrix product A = A2A1.
Ming-Feng Yeh
Chapter 6
6-53
Section 6-3
Proof of Theorem 6.11
1. Let u and v be vectors in Rn and let c be any scalar.
Because T1 and T2 are linear transformation,
T(u + v) = T2(T1(u + v)) = T2(T1(u) + T1(v))
= T2(T1(u)) + T2(T1(v)) = T(u) + T(v).
T(cv) = T2(T1(cv)) = T2(cT1(v)) = cT2(T1(v)) = cT(v).
Thus, T is a linear transformation.
2. T(v) = T2(T1(v)) = T2(A1v) = A2(A1v) = A2A1v

In general, the composition T2  T1 is not the same as
T1  T2 .
Ming-Feng Yeh
Chapter 6
6-54
Section 6-3
Example 3
Let T1 and T2 be linear transformation R3 from R3 such that
T1 ( x, y, z)  (2x  y, 0, x  z) and T2 ( x, y, z)  ( x  y, z, y)
Find the standard matrices for the compositions T  T2  T1
and T   T1  T2 .
Sol: The standard matrices for T1 and T2 are
 2 1 0
 1  1 0
A1  0 0 0, A2  0 0 1
1 0 1
0
1 0
 2 1 0
2  2
T  A  A1 A2  1 0 1, T   A  A2 A1  0
0
0 0 0
 1
0
Ming-Feng Yeh
Chapter 6
6-55
1
0
0
Section 6-3
Inverse Linear Transformation

One benefit of matrix representation is that it can
represent the inverse of a linear transformation.
[Definition] If T1:RnRn and T2:RnRn are linear
transformations such that T2(T1(v)) = v and
T1(T2(v)) = v, then T2 is called the inverse of T1
and T1 is said to be invertible.
Ming-Feng Yeh
Chapter 6
6-56
Section 6-3
Theorem 6.12

Let T:RnRn be linear transformation with
standard matrix A. Then the following conditions
are equivalent.
1. T is invertible.
2. T is an isomorphism.
3. A is invertible.
And, if T is invertible with standard matrix A, then
1
1
T
A
the standard matrix for
is
.
Ming-Feng Yeh
Chapter 6
6-57
Section 6-3
Example 4
The linear transformation T: R3R3 is defined by
T ( x, y , z )  ( 2 x  3 y  z , 3 x  3 y  z , 2 x  4 y  z )
Show that T is invertible, and find its inverse.
2 3 1
Sol:
A  3 3 1  det( A)  0
1
0
2 4 1
 1
A is invertible. Its inverse is A1   1
0
1
 6  2  3
1
Therefore T is invertible and its standard matrix is A .
Ming-Feng Yeh
Chapter 6
6-58
Section 6-3
Example 4 (cont.)
Using the standard matrix for the inverse, we can find the rule
for T 1 by computing the image of an arbitrary vector
v  ( x, y, z ).
1
0  x    x  y 
 1
T 1 ( v)   1
0
1  y     x  z 
 6  2  3  z  6 x  2 y  3 y 
T 1 ( x, y, z )  ( x  y,  x  z, 6 x  2 y  3z ).
Ming-Feng Yeh
Chapter 6
6-59
Section 6-3
Nonstandard Bases and General Vector Spaces
Nonstandard Bases





Finding a matrix for a linear transformation T:VW,
where B and B  are ordered bases for V and W, respectively.
The coordinate matrix of v relative to B is [v]B.
To represent the linear transformation T, A must be
multiplied by a coordinate matrix relative to B.
The result of the multiplication will be a coordinate matrix
relative to B .
[T ( v)]B  A[ v]B . A is called the matrix of T relative to the
bases B and B .
Ming-Feng Yeh
Chapter 6
6-60
Section 6-3
Transformation Matrix
Let V and W be finite-dimensional vector spaces with bases
B and B , respectively, where B  {v1 , v 2 , ..., v n }
If T:VW is a linear transformation such that
 a11 
 a12 
 a1n 
a 
a 
a 
[T ( v1 )]B    21 , [T ( v 2 )]B    22 , ,[T ( v n )]B    2 n ,
  
  
  
 
 
 
a
a
m 2 columns
 n

amn  to
then the mmn1 matrix whose
correspond
 a11 a12
a
a22
21

A
 


am1 am 2
Ming-Feng Yeh
 an1 
 an 2  is s.t.
,
[T (every
v)]B  vAin
[ v]V.
for
B

 

 amn 
Chapter 6
[T ( v)]B ,
6-61
Section 6-3
Example 5
Let T: R2R2 be a linear transformation defined by
T ( x, y )  ( x  y, 2 x  y ). Find the matrix of T relative to the
bases B  {(1, 2), (1, 1)} and B  {(1, 0), (0, 1)}
v1
v2
w1 w2
Sol:
T ( v1 )  T (1, 2)  (3, 0)  3w1  0w 2
T ( v 2 )  T (1, 1)  (0,  3)  0w1  3w 2
Therefore the coordinate matrices of T(v1) and T(v2) relative
to B  are [T ( v1 )]B   3 0T , [T ( v 2 )]B   0  3T .
3 0 
The matrix for T relative to B and B  is A  

0

3


Ming-Feng Yeh
Chapter 6
6-62
Section 6-3
Example 6
For the linear transformation T: R2R2 given in Example 5,
use the matrix A to find T(v), where v = (2, 1).
Sol: B  {(1, 2), (1, 1)}
1
v  (2, 1)  1(1, 2)  1(1, 1)  [ v]B   
 1
0  1 3
3
[T ( v)]B   A[ v]B  
 



3  3  1 3
 B  {(1, 0), (0, 1)}  T ( v)  3(1, 0)  3(0, 1)  (3, 3).
Ex 5 : T ( x, y )  ( x  y, 2 x  y )  T (2, 1)  (3, 3)
Ming-Feng Yeh
Chapter 6
6-63
6.4 Transition Matrix and
Similarity



The matrix of a linear transformation T:VV
depends on the basis of V.
The matrix of T relative to a basis B is different
from the matrix of T relative to another basis B 
Is is possible to find a basis B such that the matrix
of T relative to B is diagonal?
Ming-Feng Yeh
Chapter 6
6-64
Section 6-4
Transition Matrices
A linear transformation is defined by T:VV
Matrix of T relative to B: A
T ( v)B  Av B
Matrix of T relative to B  : A
T ( v)B  Av B
Transition matrix from B  to B: P
1
1 [ v ]B  P[ v ]B  ; [ v ]B   P [ v ]B
Transition matrix from B to B : P

[v]B
A
P 1
P
[v]B
Ming-Feng Yeh
[T ( v)]B
A
[T ( v)]B
Chapter 6
T ( v)B  AvB 

1
T ( v)B  P APvB 
 A  P 1 AP
6-65
Section 6-4
Example 1
Find the matrix A for T: R2R2, T ( x, y )  (2 x  2 y,  x  3 y )
relative to the basis B  {(1, 0), (1, 1)}.
 2  2
Sol: The standard matrix for T is A  


1
3


The transition matrix from B  to the standard basis
1 1
 1  1
1
B  {(1, 0), (0, 1)} is P  
P 


0
1
0
1




Therefore the matrix for T relative to B  is
 3  2
A  P AP  


1
2


1
Ming-Feng Yeh
Chapter 6
6-66
Section 6-4
Example 2
Let B  {( 3, 2), (4,  2)} and B  {( 1, 2), (2,  2)}.
be bases for
R2,
 2 7 
and let A  
 be the matrix for

3
7


T: R2R2 relative to B. Find A.
Sol: In Example 5 in Section 4.7,
3  2 1   1 2
P
,P 


2

1

2
3




The matrix of T relative to B  is given by
 2 1
1
A  P AP  


1
3


Ming-Feng Yeh
Chapter 6
6-67
Section 6-4
Example 3
For the linear transformation T: R2R2 given in Example 2,
find [v]B , [T ( v)]B , and [T ( v)]B for the vector v whose
T
coordinate matrix [ v]B    3  1
Sol:
3
[ v ]B  P[ v ]B   
2
 2
[T ( v )]B  A[ v ]B  
 3
 2  3  7 
 



 1    1   5 
7   7   21




7    5   14 
  1 2  21  7 
[T ( v )]B   P [T ( v )]B  
 



 2 3   14   0 
 A[ v ]B 
1
Ming-Feng Yeh
Chapter 6
6-68
Section 6-4
Similar Matrices
Similar Matrices
[Definition] For square matrices A and A of order n,
A is said to be similar to A if there exists an invertible matrix
1
P such that A  P AP.
[Theorem 6.13] Let A, B, and C be square matrices of order n,
Then the following properties are true.
1. A is similar to A.
2. If A is similar to B, then B is similar to A.
3. If A is similar to B and B is similar to C,
then A is similar to C.
Ming-Feng Yeh
Chapter 6
6-69
Section 6-4
Example 5
1 3 0 
Suppose A  3 1 0  is the matrix for T: R3R3 relative


0 0  2
the standard basis. Find the matrix for T relative to the basis
B  {(1, 1, 0), (1,  1, 0), (0, 0, 1)}
Sol: The transition matrix from B  to the standard matrix is
 12 12
1 1 0

P  1  1 0  P 1   12  12
0 0
0 0 1
Ming-Feng Yeh
0
0
4 0

0  A  P 1 AP  0  2 0 
0 0  2
1
Chapter 6
6-70
6.5 Applications of Linear
Transformation
The geometry of linear transformations in the plane
Reflection in the y-axis
T ( x, y )  (  x , y )
Reflection in the x-axis
T ( x, y )  ( x ,  y )
  1 0  x    x 
 0 1  y    y 

   
y
( x, y )
(  x, y )
1 0   x   x 
0  1  y    y 

   
y
( x, y )
x
x
Ming-Feng Yeh
Chapter 6
( x,  y )
6-71
Section 6-5
Reflection in the Plane
Reflection in the line y = x
y
T ( x, y )  ( y , x )
0 1   x   y 
1 0  y    x 

   
( y, x)
( x, y )
x
yx
Ming-Feng Yeh
Chapter 6
6-72
Section 6-5
Expansions & Contractions
in the Plane -- Horizontal
T ( x, y )  (kx, y )
y
( kx, y )
k 0  x  kx
 0 1  y    y 

   
y
( x, y )
( x, y )
x
x
Contraction: 0 < k <1
Ming-Feng Yeh
( kx, y )
Expansion: k >1
Chapter 6
6-73
Section 6-5
Expansions & Contractions
in the Plane -- Vertical
T ( x, y )  ( x, ky)
y
1 0   x   x 
0 k   y   ky

   
y
( x, y )
( kx, y )
( kx, y )
( x, y )
x
Contraction: 0 < k <1
Ming-Feng Yeh
x
Expansion: k >1
Chapter 6
6-74
Section 6-5
Shears in the Plane
Horizontal shear:
T ( x, y )  ( x  ky, y )
Vertical shear:
T ( x, y )  ( x, y  kx)
1 k   x   x  ky
0 1   y    y 

  

y
( x, y )
( x  ky, y )
 1 0  x   x 
k 1  y   kx  y 

  

y
( x, kx  y )
( x, y )
x
x
Ming-Feng Yeh
Chapter 6
6-75