Transcript MM_Ch06x

Chapter 6
Linear Transformations
6.1 Introduction to Linear Transformations
6.2 The Kernel and Range of a Linear Transformation
6.3 Matrices for Linear Transformations
6.4 Transition Matrices and Similarity
6.5 Applications of Linear Transformations
6.1
6.1 Introduction to Linear Transformations

A function T that maps a vector space V into a vector space W:
mapping
T : V 
W,
V ,W : vector spaces
V: the domain (定義域) of T

W: the codomain (對應域) of T
Image of v under T (在T映射下v的像):
If v is a vector in V and w is a vector in W such that
T ( v)  w,
then w is called the image of v under T
(For each v, there is only one w)

The range of T (T的值域):
The set of all images of vectors in V (see the figure on the
next slide)
6.2


The preimage of w (w的反像):
The set of all v in V such that T(v)=w
(For each w, v may not be unique)
The graphical representations of the domain, codomain, and range
※ For example, V is R3, W is R3, and T
is the orthogonal projection of any
vector (x, y, z) onto the xy-plane, i.e.
T(x, y, z) = (x, y, 0)
(we will use the above example many
times to explain abstract notions)
※ Then the domain is R3, the codomain
is R3, and the range is xy-plane (a
subspace of the codomian R3)
※ (2, 1, 0) is the image of (2, 1, 3)
※ The preimage of (2, 1, 0) is (2, 1, s),
6.3
where s is any real number

Ex 1: A function from R2 into R2
T : R2  R2
v  (v1 , v2 )  R 2
T (v1 , v2 )  (v1  v2 , v1  2v2 )
(a) Find the image of v=(-1,2) (b) Find the preimage of w=(-1,11)
Sol:
(a) v  (1, 2)
 T ( v)  T (1, 2)  (1  2,  1  2(2))  (3, 3)
(b) T ( v)  w  (1, 11)
T (v1 , v2 )  (v1  v2 , v1  2v2 )  (1, 11)
 v1  v2  1
v1  2v2  11
 v1  3, v2  4 Thus {(3, 4)} is the preimage of w=(-1, 11)
6.4

Linear Transformation (線性轉換):
V , W： vector spaces
T : V  W： A linear tra nsformatio n of V into W if the
following two properties are true
(1) T (u  v)  T (u)  T ( v), u, v V
(2) T (cu)  cT (u),
c  R
6.5

Notes:
(1) A linear transformation is said to be operation preserving
(because the same result occurs whether the operations of addition
and scalar multiplication are performed before or after the linear
transformation is applied)
T (u  v)  T (u)  T ( v)
Addition
in V
Addition
in W
T (cu)  cT (u)
Scalar
multiplication
in V
Scalar
multiplication
in W
(2) A linear transformation T : V  V from a vector space into
itself is called a linear operator (線性運算子)
6.6

Ex 2: Verifying a linear transformation T from R2 into R2
T (v1 , v2 )  (v1  v2 , v1  2v2 )
Pf:
u  (u1 , u2 ), v  (v1 , v2 ) : vector in R 2 , c : any real number
(1) Vector addition :
u  v  (u1 , u2 )  (v1 , v2 )  (u1  v1 , u2  v2 )
T (u  v)  T (u1  v1 , u2  v2 )
 ((u1  v1 )  (u2  v2 ), (u1  v1 )  2(u2  v2 ))
 ((u1  u2 )  (v1  v2 ), (u1  2u2 )  (v1  2v2 ))
 (u1  u2 , u1  2u2 )  (v1  v2 , v1  2v2 )
 T (u)  T ( v)
6.7
(2) Scalar multiplica tion
cu  c(u1 , u2 )  (cu1 , cu2 )
T (cu)  T (cu1 , cu2 )  (cu1  cu2 , cu1  2cu2 )
 c(u1  u 2 , u1  2u 2 )
 cT (u)
Therefore, T is a linear transformation
6.8

Ex 3: Functions that are not linear transformations
(a) f ( x)  sin x
sin( x1  x2 )  sin( x1 )  sin( x2 )
(f(x) = sin x is not a linear




sin( 2  3 )  sin( 2 )  sin( 3 ) transformation)
(b) f ( x)  x 2
( x1  x2 ) 2  x12  x22
(1  2) 2  12  22
(f(x) = x2 is not a linear transformation)
(c) f ( x)  x  1
f ( x1  x2 )  x1  x2  1
f ( x1 )  f ( x2 )  ( x1  1)  ( x2  1)  x1  x2  2
f ( x1  x2 )  f ( x1 )  f ( x2 )
In fact, f (cx)  cf ( x)
(f(x) = x+1 is not a linear transformation,
although it is a linear function)
6.9

Notes: Two uses of the term “linear”.
(1) f ( x)  x  1 is called a linear function because its graph
is a line
(2) f ( x)  x  1 is not a linear transformation from a vector
space R into R because it preserves neither vector
addition nor scalar multiplication
6.10

Zero transformation (零轉換):
T :V W

Identity transformation (相等轉換):
T :V V

T ( v)  0, v V
T ( v)  v, v  V
Theorem 6.1: Properties of linear transformations
T : V  W , u, v  V
(1) T (0)  0 (T(cv) = cT(v) for c=0)
(2) T ( v)  T ( v) (T(cv) = cT(v) for c=-1)
(3) T (u  v)  T (u)  T ( v) (T(u+(-v))=T(u)+T(-v) and property (2))
(4) If v  c1v1  c2 v2    cn vn ,
then T ( v)  T (c1v1  c2v2    cn vn )
 c1T (v1 )  c2T (v2 )    cnT (vn )
(Iteratively using T(u+v)=T(u)+T(v) and T(cv) = cT(v))
6.11

Ex 4: Linear transformations and bases
Let T : R 3  R 3 be a linear transformation such that
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)
Sol:
(2,3,2)  2(1,0,0)  3(0,1,0)  2(0,0,1)
 According to the fourth property on the previous slide that 


 T (c1v1  c2v2   cnvn )  c1T (v1 )  c2T (v2 )   cnT (vn ) 
T (2,3,2)  2T (1,0,0)  3T (0,1,0)  2T (0,0,1)
 2(2,1,4)  3(1,5,2)  2T (0,3,1)
 (7,7,0)
6.12

Ex 5: A linear transformation defined by a matrix
0
3
v1 

2
3


The function T : R  R is defined as T ( v)  Av  2
1  

 v2 
  1  2
(a) Find T ( v), where v  (2, 1)
(b) Show that T is a linear transformation form R 2 into R3
Sol:
2
3
R
vector
R
vector
(a) v  (2, 1)
0
3
6
2
T ( v)  Av   2
1     3

  1  
  1  2
0
T (2,1)  (6,3,0)
(b) T (u  v)  A(u  v)  Au  Av  T (u)  T ( v)
T (cu)  A(cu)  c( Au)  cT (u)
(vector addition)
(scalar
multiplication)6.13

Theorem 6.2: The linear transformation defined 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

Note:
R n vector
 a11
 a21
Av  
 
am1
R m vector
a12  a1n   v1   a11v1  a12v2    a1n vn 
a22  a2 n  v2   a21v1  a22v2    a2 n vn 
   


    


am 2  amn  vn  am1v1  am 2 v2    amnvn 
T ( v )  Av
T : Rn 
 R m
※ If T(v) can represented by Av, then T is a linear transformation
※ If the size of A is m×n, then the domain of T is Rn and the
codomain of T is Rm
6.14

Ex 7: Rotation in the plane
Show that the L.T. T : R 2  R 2 given by the matrix
cos 
A
 sin 
 sin  
cos  
has the property that it rotates every vector in R2
counterclockwise about the origin through the angle 
Sol:
(Polar coordinates: for every point on the xyv  ( x, y )  (r cos  , r sin  ) plane, it can be represented by a set of (r, α))
r：the length of v (  x 2  y 2 )
：the angle from the positive
x-axis counterclockwise to
the vector v
T(v
)
v
6.15
cos   sin    x  cos 
T ( v)  Av  




 sin  cos    y   sin 
r cos  cos   r sin  sin  

r sin  cos   r cos  sin  
r cos(   )

 r sin(    ) 
 sin   r cos  
cos    r sin  
according to the addition
formula of trigonometric
identities (三角函數合角公式)
r：remain the same, that means the length of T(v) equals the
length of v
 +：the angle from the positive x-axis counterclockwise to
the vector T(v)
Thus, T(v) is the vector that results from rotating the vector v
counterclockwise through the angle 
6.16

Ex 8: A projection in R3
The linear transformation T : R 3  R 3 is given by
1 0 0
A  0 1 0


0
0
0


is called a projection in R3
1 0 0   x   x 
If v is ( x, y, z ), Av  0 1 0   y    y 
0 0 0  z   0 
※ In other words, T maps every vector in R3
to its orthogonal projection in the xyplane, as shown in the right figure
6.17

Ex 9: The transpose function is a linear transformation from
Mmn into Mn m
T ( A)  AT
(T : M mn  M nm )
Show that T is a linear transformation
Sol:
A, B  M mn
T ( A  B)  ( A  B)T  AT  BT  T ( A)  T ( B)
T (cA)  (cA)T  cAT  cT ( A)
Therefore, T (the transpose function) is a linear transformation
from Mmn into Mnm
6.18
Keywords in Section 6.1:

function: 函數

domain: 定義域

codomain: 對應域

image of v under T: 在T映射下v的像

range of T: T的值域

preimage of w: w的反像

linear transformation: 線性轉換

linear operator: 線性運算子

zero transformation: 零轉換

identity transformation: 相等轉換
6.19
6.2 The Kernel and Range of a Linear Transformation

Kernel of a linear transformation T (線性轉換T的核空間):
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)
ker(T )  {v | T ( v)  0, v V }
※ For example, V is R3, W is R3, and T is the
orthogonal projection of any vector (x, y, z)
onto the xy-plane, i.e. T(x, y, z) = (x, y, 0)
※ Then the kernel of T is the set consisting of
(0, 0, s), where s is a real number, i.e.
ker(T )  {(0, 0, s) | s is a real number}
6.20

Ex 1: Finding the kernel of a linear transformation
T ( A)  AT (T : M 32  M 23 )
Sol:
0 0
ker(T )  0 0


0
0



Ex 2: The kernel of the zero and identity transformations
(a) If T(v) = 0 (the zero transformation T : V  W ), then
ker(T )  V
(b) If T(v) = v (the identity transformation T : V  V ), then
ker(T )  {0}
6.21

Ex 5: Finding the kernel of a linear transformation
 x1 
 1 1 2  
3
2
T (x)  Ax  
x
(
T
:
R

R
)
2


 1 2 3   x 
 3
ker(T )  ?
Sol:
ker(T )  {( x1 , x2 , x3 ) | T ( x1 , x2 , x3 )  (0, 0), and ( x1, x2 , x3 )  R 3}
T ( x1 , x2 , x3 )  (0,0)
 x1 
 1  1  2   0
x2   
 1 2

3    0
 x3 
6.22
 1 1 2 0 G.-J. E. 1 0 1 0 
 1 2 3 0  0 1 1 0 




 x1   t   1 
  x2    t   t  1
     
 x3   t   1 
 ker(T )  {t (1,1,1) | t is a real number }
 span{( 1,1,1)}
6.23

Theorem 6.3: The kernel is a subspace of V
The kernel of a linear transformation T : V  W is a
subspace of the domain V
Pf:
 T (0)  0 (by Theorem 6.1)  ker(T ) is a nonempty subset of V
Let u and v be vectors in the kernel of T . Then
T (u  v)  T (u)  T ( v)  0  0  0
T is a linear transformation
T (cu)  cT (u)  c0  0
(u  ker(T ), v  ker(T )  u  v  ker(T ))
(u  ker(T )  cu  ker(T ))
Thus, ker(T ) is a subspace of V (according to Theorem 4.5
that a nonempty subset of V is a subspace of V if it is closed
under vector addition and scalar multiplication)
6.24

Ex 6: Finding a basis for the kernel
Let T : R 5  R 4 be defined by T (x)  Ax, where x is in R 5 and
1
2
A
 1

0
1  1
1 0 
0 2 0 1 

0 0 2 8
2
1
0
3
Find a basis for ker(T) as a subspace of R5
Sol:
To find ker(T) means to find all x satisfying T(x) = Ax = 0.
Thus we need to form the augmented matrix  A 0 first
6.25
A
1
2

 1

0
0 
2 0
1 3
0 2
0 0
1 1
1 0
0 1
2 8
0
1
0  G.-J. E. 0

0
0


0
0
 x1   2s  t 
 2  1 
 x   s  2t 
1 2
 2 

   
x   x3    s   s  1   t  0 
  

   
 x4   4t 
 0   4
 x5   t 
 0   1 
0 2
1 1
0 0
0 0
s
0 1
0 2
1 4
0 0
0
0 
0

0
t
B  (2, 1, 1, 0, 0), (1, 2, 0,  4, 1): one basis for the kernel of T
6.26

Corollary to Theorem 6.3:
Let T : R n  R m be the linear fransforma tion given by T (x)  Ax.
Then the kernel of T is equal to the solution space of Ax  0
T (x)  Ax (a linear transformation T : R n  R m )
 ker(T )  NS ( A)  x | Ax  0, x  R n  (subspace of R n )
※ The kernel of T equals the nullspace of A (which is defined
in Theorem 4.16 on p.239) and these two are both subspaces
of Rn )
※ So, the kernel of T is sometimes called the nullspace of T
6.27

Range of a linear transformation T (線性轉換T的值域):
Let T : V  W be a linear tra nsformatio n. Then the set of all
vectors w in W that are images of any vector s in V is called the
range of T and is denoted by range (T )
range (T )  {T ( v) | v V }
※ For the orthogonal projection of
any vector (x, y, z) onto the xyplane, i.e. T(x, y, z) = (x, y, 0)
※ The domain is V=R3, the codomain
is W=R3, and the range is xy-plane
(a subspace of the codomian R3)
※ Since T(0, 0, s) = (0, 0, 0) = 0, the
kernel of T is the set consisting of
(0, 0, s), where s is a real number
6.28

Theorem 6.4: The range of T is a subspace of W
The range of a linear tra nsformatio n T : V  W is a subspace of W
Pf:
T (0)  0 (Theorem 6.1)
 range (T ) is a nonempty subset of W
Since T (u) and T ( v) are vectors in range( T ), and we have
because u  v V
T (u)  T ( v)  T (u  v)  range (T )
T is a linear transformation
cT (u)  T (cu)  range (T )
because cu V
 Range of T is closed under vector addition 


 because T (u), T ( v), T (u  v)  range(T ) 
 Range of T is closed under scalar multi


 plication because T (u) and T (cu)  range(T ) 
Thus, range(T ) is a subspace of W (according to Theorem 4.5
that a nonempty subset of W is a subspace of W if it is closed
under vector addition and scalar multiplication)
6.29

Notes:
T : V  W is a linear transformation
(1) ker(T ) is subspace of V (Theorem 6.3)
(2) range (T ) is subspace of W (Theorem 6.4)
6.30

Corollary to Theorem 6.4:
Let T : R n  R m be the linear tra nsformatio n given by T (x)  Ax.
The range of T is equal to the column space of A, i.e. range (T )  CS ( A)
(1) According to the definition of the range of T(x) = Ax, we know that the
range of T consists of all vectors b satisfying Ax=b, which is equivalent to
find all vectors b such that the system Ax=b is consistent
(2) Ax=b can be rewritten as
 a11 
 a12 
a 
a 
Ax  x1  21   x2  22  
 
 
 
 
 am1 
 am 2 
 a1n 
a 
 xn  2 n   b
 
 
 amn 
Therefore, the system Ax=b is consistent iff we can find (x1, x2,…, xn)
such that b is a linear combination of the column vectors of A, i.e. b  CS ( A)
Thus, we can conclude that the range consists of all vectors b, which is a linear
combination of the column vectors of A or said b  CS ( A) . So, the column space
6.31
of the matrix A is the same as the range of T, i.e. range(T) = CS(A)

Use our example to illustrate the corollary to Theorem 6.4:
※ For the orthogonal projection of any vector (x, y, z) onto the xyplane, i.e. T(x, y, z) = (x, y, 0)
※ According to the above analysis, we already knew that the range
of T is the xy-plane, i.e. range(T)={(x, y, 0)| x and y are real
numbers}
※ T can be defined by a matrix A as follows
1 0 0 
A  0 1 0 , such that
0 0 0
1 0 0   x   x 
0 1 0   y    y 

   
0 0 0   z   0 
※ The column space of A is as follows, which is just the xy-plane
1 
0
0  x1 
CS ( A)  x1 0  x2 1   x3 0   x2  , where x1 , x2  R
0
0
0  0 
6.32

Ex 7: Finding a basis for the range of a linear transformation
Let T : R5  R 4 be defined by T (x)  Ax, where x is R 5 and
1
2
A
 1

0
1 1
1 3 1 0 
0 2 0 1 

0 0 2 8
2
0
Find a basis for the range of T
Sol:
Since range(T) = CS(A), finding a basis for the range of T is
equivalent to fining a basis for the column space of A
6.33
1
2
A
 1

0
2 0
1 3
0 2
0 0
1 1
1
1 0  G.-J. E. 0

0
0 1


2 8
0
c1 c2 c3 c4 c5
0 2
1 1
0 0
0 0
0 1
0 2 
B
1 4

0 0
w1 w2 w3 w4 w5
 w1 , w2 , and w4 are indepdnent, so w1 , w2 , w4  can
form a basis for CS ( B)
Row operations will not affect the dependency among columns
 c1 , c2 , and c4 are indepdnent, and thus c1 , c2 , c4  is
a basis for CS ( A)
That is, (1, 2,  1, 0), (2, 1, 0, 0), (1, 1, 0, 2) is a basis
for the range of T
6.34

Rank of a linear transformation T:V→W (線性轉換T的秩):
rank(T )  the dimension of the range of T  dim(range(T ))
According to the corollary to Thm. 6.4, range(T) = CS(A), so dim(range(T)) = dim(CS(A))

Nullity of a linear transformation T:V→W (線性轉換T的核次數):
nullity(T )  the dimension of the kernel of T  dim(ker(T ))
According to the corollary to Thm. 6.3, ker(T) = NS(A), so dim(ker(T)) = dim(NS(A))

Note:
If T : R n  R m is a linear transformation given by T (x)  Ax, then
rank(T )  dim(range(T ))  dim(CS ( A))  rank( A)
nullity(T )  dim(ker(T ))  dim( NS ( A))  nullity( A)
※ The dimension of the row (or column) space of a matrix A is called the rank of A
※ The dimension of the nullspace of A ( NS ( A)  {x | Ax  0}) is called the nullity
6.35
of A

Theorem 6.5: Sum of rank and nullity
Let T: V →W be a linear transformation from an n-dimensional
vector space V (i.e. the dim(domain of T) is n) into a vector
space W. Then
rank(T )  nullity(T )  n
(i.e. dim(range of T )  dim(kernel of T )  dim(domain of T ))
※ You can image that the dim(domain of T)
should equals the dim(range of T)
originally
※ But some dimensions of the domain of T
is absorbed by the zero vector in W
※ So the dim(range of T) is smaller than
the dim(domain of T) by the number of
how many dimensions of the domain of
T are absorbed by the zero vector,
which is exactly the dim(kernel of T) 6.36
Pf:
Let T be represente d by an m  n matrix A, and assume rank( A)  r
(1) rank(T )  dim(range of T )  dim(column space of A)  rank( A)  r
(2) nullity(T )  dim(kernel of T )  dim(null space of A)  n  r
 rank (T )  nullity (T )  r  (n  r )  n
according to Thm. 4.17 where
rank(A) + nullity(A) = n
※ Here we only consider that T is represented by an m×n matrix A. In
the next section, we will prove that any linear transformation from an
n-dimensional space to an m-dimensional space can be represented
by m×n matrix
6.37

Ex 8: Finding the rank and nullity of a linear transformation
Find the rank and nullity of the linear tra nsformatio n T : R 3  R 3
define by
1 0  2
A  0 1 1 
0 0 0 
Sol:
rank (T )  rank ( A)  2
nullity (T )  dim( domain of T )  rank (T )  3  2  1
※ The rank is determined by the number of leading 1’s, and the
nullity by the number of free variables (columns without leading
1’s)
6.38

Ex 9: Finding the rank and nullity of a linear transformation
Let T : R 5  R 7 be a linear transformation
(a) Find the dimension of the kernel of T if the dimension
of the range of T is 2
(b) Find the rank of T if the nullity of T is 4
(c) Find the rank of T if ker(T )  {0}
Sol:
(a) dim(domain of T )  n  5
dim(kernel of T )  n  dim(range of T )  5  2  3
(b) rank(T )  n  nullity(T )  5  4  1
(c) rank(T )  n  nullity(T )  5  0  5
6.39

One-to-one (一對一):
A function T : V  W is called one-to-one if the preimage of
every w in the range consists of a single vector. This is equivalent
to saying that T is one-to-one iff for all u and v in V , T (u)  T ( v)
implies that u  v
one-to-one
not one-to-one
6.40

Theorem 6.6: One-to-one linear transformation
Let T : V  W be a linear transformation. Then
T is one-to-one iff ker(T )  {0}
Pf:
() Suppose T is one-to-one
Then T ( v)  0 can have only one solution : v  0
i.e. ker(T )  {0}
Due to the fact
that T(0) = 0 in
Thm. 6.1
() Suppose ker(T )={0} and T (u)=T (v)
T (u  v)  T (u)  T ( v)  0
T is a linear transformation, see Property 3 in Thm. 6.1
 u  v  ker(T )  u  v  0  u  v
 T is one-to-one (because T (u)  T ( v) implies that u  v)
6.41

Ex 10: One-to-one and not one-to-one linear transformation
(a) The linear transformation T : M mn  M nm given by T ( A)  AT
is one-to-one
because its kernel consists of only the m×n zero matrix
(b) The zero transformation T : R3  R3 is not one-to-one
because its kernel is all of R3
6.42

Onto (映成):
A function T : V  W is said to be onto if every element
in W has a preimage in V
(T is onto W when W is equal to the range of T)

Theorem 6.7: Onto linear transformations
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
rank (T )  dim( range of T )  dim( W )
The definition of
the rank of a linear
transformation
The definition
of onto linear
transformations
6.43

Theorem 6.8: One-to-one and onto linear transformations
Let T : V  W be a linear transformation with vector space 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
Thm. 6.5
dim( range (T ))  n  dim(ker( T ))  n  dim( W )
Consequently, T is onto
According to the definition of dimension
(on p.227) that if a vector space V
consists of the zero vector alone, the
dimension of V is defined as zero
() If T is onto, then dim( range of T )  dim( W )  n
Thm. 6.5
dim(ker( T ))  n  dim( range of T )  n  n  0  ker(T )  {0}
Therefore, T is one-to-one
6.44

Ex 11:
The linear transformation T : R n  R m is given by T (x)  Ax. Find the nullity
and rank of T and determine whether T is one-to-one, onto, or neither
1 2 0 
(a) A  0 1 1 
0 0 1 
1 2 
(b) A  0 1 
0 0 
Sol:
dim(Rn)
=
=n
1 2 0 
(c) A  

0 1 1
= dim(range
of T)
= # of
leading 1’s
T:Rn→Rm
dim(domain
of T) (1)
(a) T:R3→R3
3
3
(b) T:R2→R3
2
(c) T:R3→R2
(d) T:R3→R3
= (1) – (2) =
dim(ker(T))
rank(T)
nullity(T)
(2)
1 2 0 
(d) A  0 1 1 
0 0 0
If nullity(T)
=
dim(ker(T))
=0
If rank(T) =
dim(Rm) =
m
1-1
onto
0
Yes
Yes
2
0
Yes
No
3
2
1
No
Yes
3
2
1
No
No
6.45

Isomorphism (同構):
A linear tra nsformatio n T : V  W that is one to one and onto
is called an isomorphis m. Moreover, if V and W are vector spaces
such that there exists an isomorphis m from V to W , then V and W
are said to be isomorphic to each other
Theorem 6.9: Isomorphic spaces (同構空間) and dimension
Two finite-dimensional vector space V and W are isomorphic
if and only if they are of the same dimension
Pf:
() Assume that V is isomorphic to W , where V has dimension n

 There exists a L.T. T : V  W that is one to one and onto
T is one-to-one
dim(V) = n
 dim(ker( T ))  0
 dim( range of T )  dim( domain of T )  dim(ker( T ))  n  0  6.46
n
T is onto
 dim( range of T )  dim(W )  n
Thus dim(V )  dim(W )  n
() Assume that V and W both have dimension n
Let B   v1 , v 2 ,
B '  w1 , w 2 ,
,v n  be a basis of V and
,w n  be a basis of W
Then an arbitrary vector in V can be represente d as
v  c1 v1  c2 v 2    cn v n
and you can define a L.T. T : V  W as follows
w  T ( v)  c1w1  c2 w 2   cn w n (by defining T (v i )  w i )
6.47
Since B ' is a basis for V, {w1, w2,…wn} is linearly
independent, and the only solution for w=0 is c1=c2=…=cn=0
So with w=0, the corresponding v is 0, i.e., ker(T) = {0}
 T is one - to - one
By Theorem 6.5, we can derive that dim(range of T) =
dim(domain of T) – dim(ker(T)) = n –0 = n = dim(W)
 T is onto
Since this linear transformation is both one-to-one and onto,
then V and W are isomorphic
6.48

Note
Theorem 6.9 tells us that every vector space with dimension
n is isomorphic to Rn

Ex 12: (Isomorphic vector spaces)
The following vector spaces are isomorphic to each other
(a) R4  4 - space
(b) M 41  space of all 4 1 matrices
(c) M 22  space of all 2  2 matrices
(d) P3 ( x)  space of all polynomials of degree 3 or less
(e) V  {( x1 , x2 , x3 , x4 , 0), xi are real numbers}(a subspace of R 5 )
6.49
Keywords in Section 6.2:

kernel of a linear transformation T: 線性轉換T的核空間

range of a linear transformation T: 線性轉換T的值域

rank of a linear transformation T: 線性轉換T的秩

nullity of a linear transformation T: 線性轉換T的核次數

one-to-one: 一對一

onto: 映成

Isomorphism (one-to-one and onto): 同構

isomorphic space: 同構空間
6.50
6.3 Matrices for Linear Transformations

Two representations of the linear transformation T:R3→R3 :
(1) T ( x1 , x2 , x3 )  (2 x1  x2  x3 ,  x1  3x2  2 x3 ,3x2  4 x3 )
 2 1 1  x1 
(2) T (x)  Ax   1 3 2   x2 
 0 3 4   x3 

Three reasons for matrix representation of a linear transformation:

It is simpler to write

It is simpler to read

It is more easily adapted for computer use
6.51

Theorem 6.10: Standard matrix for a linear transformation
Let T : R n  R m be a linear trtansformation such that
 a11 
 a12 
a 
a 
T (e1 )   21  , T (e2 )   22  ,
 
 
 
 
a
 m1 
 am 2 
where {e1 , e2 ,
 a1n 
a 
, T (en )   2 n  ,
 
 
 amn 
, en } is a standard basis for R n . Then the m  n
matrix whose n columns correspond to T (ei ),
A  T (e1 ) T (e2 )
 a11
a
T (en )    21


 am1
a12
a22
am 2
a1n 
a2 n 
,


amn 
is such that T ( v)  Av for every v in R n , A is called the
standard matrix for T (T的標準矩陣)
6.52
Pf:
 v1 
1 
0
v 
0
1 
v   2   v1    v2   
 
 
 
 
 
 
v
0
 
0
 n
0
0
 vn    v1e1  v2e2 
 
 
1 
T is a linear transformation  T ( v)  T (v1e1  v2e2 
If A  T (e1 ) T (e2 )
 a11
 a21
Av  
 
am1
a12
a22

am 2
 vnen
 vne n )
 T (v1e1 )  T (v2e2 ) 
 T (vne n )
 v1T (e1 )  v2T (e2 ) 
 vnT (en )
T (en ) , then
 a1n   v1   a11v1  a12v2    a1n vn 
 a2 n  v2   a21v1  a22v2    a2 n vn 
   

     


 amn  vn  am1v1  am 2 v2    amnvn 
6.53
 a11 
 a12 
 a1n 
a 
a 
a 
 v1  21   v2  22    vn  2 n 
 
 
 
 
 
 
 am1 
 am 2 
 amn 
 v1T (e1 )  v2T (e2 )   vnT (e n )
Therefore, T ( v)  Av for each v in R n

Note
Theorem 6.10 tells us that once we know the image of every
vector in the standard basis (that is T(ei)), you can use the
properties of linear transformations to determine T(v) for any
v in V
6.54

Ex 1: Finding the standard matrix of a linear transformation
Find the standard matrix for the L.T. T : R3  R 2 defined by
T ( x, y , z )  ( x  2 y , 2 x  y )
Sol:
Vector Notation
T (e1 )  T (1, 0, 0)  (1, 2)
T (e2 )  T (0, 1, 0)  (2, 1)
T (e3 )  T (0, 0, 1)  (0, 0)
Matrix Notation
1 
1 


T (e1 )  T ( 0)   
2

0
0 
 2


T (e2 )  T ( 1 )   
1

0
0 
0 


T (e3 )  T ( 0 )   
0

1 
6.55
A  T (e1 ) T (e2 ) T (e3 ) 
1 2 0 


2
1
0



Check:
 x
 x
1  2 0    x  2 y 
A y   
y 

  2 1 0   2 x  y 
z
z
i.e., T ( x, y, z )  ( x  2 y, 2 x  y )

Note: a more direct way to construct the standard matrix
1  2 0  1x  2 y  0 z
A
2 1 0  2 x  1y  0 z
※ The first (second) row actually
represents the linear transformation
function to generate the first (second)
component of the target vector
6.56

Ex 2: Finding the standard matrix of a linear transformation
The linear transformation T : R 2  R 2 is given by projecting
each point in R 2 onto the x - axis. Find the standard matrix for T
Sol:
T ( x, y )  ( x, 0)

1 0 
A  T (e1 ) T (e2 )  T (1, 0) T (0, 1)   

0
0


Notes:
(1) The standard matrix for the zero transformation from Rn into Rm
is the mn zero matrix
(2) The standard matrix for the identity transformation from Rn into
Rn is the nn identity matrix In
6.57

Composition of T1:Rn→Rm with T2:Rm→Rp :
T ( v)  T2 (T1 ( v)), v  R n
This composition is denoted by T  T2 T1

Theorem 6.11: Composition of linear transformations (線性轉換
的合成)
Let T1 : R n  R m and T2 : R m  R p be linear transformations
with standard matrices A1 and A2 ,then
(1) The composition T : R n  R p , defined by T ( v)  T2 (T1 ( v)),
is still a linear transformation
(2) The standard matrix A for T is given by the matrix product
A  A2 A1
6.58
Pf:
(1) (T is a linear transformation)
Let u and v be vectors in R n and let c be any scalar. Then
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)
(2) ( A2 A1 is the standard matrix for T )
T ( v)  T2 (T1 ( v))  T2 ( A1v)  A2 A1v  ( A2 A1 ) v

Note:
T1  T2  T2  T1
6.59

Ex 3: The standard matrix of a composition
Let T1 and T2 be linear transformations from R 3 into R 3 s.t.
T1 ( x, y, z)  (2x  y, 0, x  z)
T2 ( x, y, z)  ( x  y, z, y)
Find the standard matrices for the compositions T  T2 T1
and T '  T1 T2
Sol:
2
A1  0
1
1
A2  0
0
1 0
0 0 (standard matrix for T1 )
0 1
 1 0
0 1 (standard matrix for T2 )
1 0
6.60
The standard matrix for T  T2  T1
1  1 0 2 1 0 2 1 0
A  A2 A1  0 0 1 0 0 0  1 0 1


 

0
1
0
1
0
1
0
0
0


 

The standard matrix for T '  T1  T2
2 1 0 1  1 0 2  2 1
A'  A1 A2  0 0 0 0 0 1  0 0 0
1 0 1 0 1 0 1 0 0
6.61

Inverse linear transformation (反線性轉換):
If T1 : R n  R n and T2 : R n  R n are L.T. s.t. for every v in R n
T2 (T1 ( v))  v and T1 (T2 ( v))  v
Then T2 is called the inverse of T1 and T1 is said to be invertible

Note:
If the transformation T is invertible, then the inverse is
unique and denoted by T–1
6.62

Theorem 6.12: Existence of an inverse transformation
Let T : R n  R n be a linear transformation with standard matrix A,
Then the following condition are equivalent
※ For (2)  (1), you can imagine that
since T is one-to-one and onto, for every
w in the codomain of T, there is only one
(2) T is an isomorphism
preimage v, which implies that T-1(w) =v
can be a L.T. and well-defined, so we can
(3) A is invertible
infer that T is invertible
※ On the contrary, if there are many
preimages for each w, it is impossible to
find a L.T to represent T-1 (because for a
L.T, there is always one input and one
Note:
output), so T cannot be invertible
(1) T is invertible

If T is invertible with standard matrix A, then the standard
matrix for T–1 is A–1
6.63

Ex 4: Finding the inverse of a linear transformation
The linear transformation T : R3  R3 is defined by
T ( x1 , x2 , x3 )  (2 x1  3x2  x3 , 3x1  3x2  x3 , 2 x1  4 x2  x3 )
Show that T is invertible, and find its inverse
Sol:
The standard matrix for T
2 3 1
A  3 3 1
2 4 1
 2 x1  3x2  x3
 3 x1  3x2  x3
 2 x1  4 x2  x3
 2 3 1 1 0 0
 A I 3    3 3 1 0 1 0


2
4
1
0
0
1


6.64
1 0 0 1 1 0 
G.-J. E.

 0 1 0 1 0 1    I
0 0 1 6 2 3
A1 
Therefore T is invertible and the standard matrix for T 1 is A1
0
 1 1
A1   1 0
1


 6  2  3
0   x1    x1  x2 
 1 1
T 1 ( v)  A1v   1 0
1   x2     x1  x3 

  

6

2

3
x
6
x

2
x

3
x
2
3

 3   1
In other word s,
T 1 ( x1 , x2 , x3 )  ( x1  x2 ,  x1  x3 , 6 x1  2 x2  3x3 )
※ Check T-1(T(2, 3, 4)) = T-1(17, 19, 20) = (2, 3, 4)
6.65

The matrix of T relative to the bases B and B‘:
T :V  W
B  {v1 , v 2 ,
, vn}
(a linear transformation)
(a nonstandard basis for V )
The coordinate matrix of any v relative to B is denoted by [v]B


 if v can be represeted as c v  c v 
1 1
2 2



 c1  
c  
 cn v n , then [v]B   2  
 
  
cn  
A matrix A can represent T if the result of A multiplied by a
coordinate matrix of v relative to B is a coordinate matrix of v
relative to B’, where B’ is a basis for W. That is,
T (v)B '  A[ v]B ,
where A is called the matrix of T relative to the bases B and B’ (T對
應於基底B到B'的矩陣)
6.66

Transformation matrix for nonstandard bases (the
generalization of Theorem 6.10, in which standard bases are
considered) :
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 s.t.
T ( v1 )B '
 a11 
a 
  21  ,
 
 
 am1 
T ( v 2 )B '
 a12 
a 
  22  ,
 
 
 am 2 
,
T ( v n )B '
 a1n 
a 
  2n 
 
 
 amn 
then the m n matrix who se n columns correspond to T (vi )B '
6.67
A  [T ( v1 )]B [T ( v 2 )]B
 a11
a
[T ( v n )]B    21


 am1
a12
a22
am 2
a1n 
a2 n 


amn 
is such that T ( v)B '  A[ v]B for every v in V
※The above result state that the coordinate of T(v) relative to the basis B’
equals the multiplication of A defined above and the coordinate of v
relative to the basis B.
※ Comparing to the result in Thm. 6.10 (T(v) = Av), it can infer that the
linear transformation and the basis change can be achieved in one step
through multiplying the matrix A defined above (see the figure on 6.74
for illustration)
6.68

Ex 5: Finding a matrix relative to nonstandard bases
Let T : R2  R 2 be a linear transformation defined by
T ( x1 , x2 )  ( x1  x2 , 2x1  x2 )
Find the matrix of T relative to the basis B  {(1, 2), ( 1, 1)}
and B '  {(1, 0), (0, 1)}
Sol:
T (1, 2)  (3, 0)  3(1, 0)  0(0, 1)
T (1, 1)  (0,  3)  0(1, 0)  3(0, 1)
T (1, 2)B '  03, T (1, 1)B '  03
 
 
the matrix for T relative to B and B'
3 0 
A  T (1, 2)B ' T (1, 2)B '   
0  3
6.69

Ex 6:
For the L.T. T : R 2  R 2 given in Example 5, use the matrix A
to find T ( v), where v  (2, 1)
Sol:
v  (2, 1)  1(1, 2)  1(1, 1)
B  {(1, 2), (1, 1)}
1
 v B   
 1
3 0   1  3
 T ( v)B '  Av B  
 



0  3  1 3
 T ( v)  3(1, 0)  3(0, 1)  (3, 3)
B' {(1, 0), (0, 1)}

Check:
T (2, 1)  (2  1, 2(2)  1)  (3, 3)
6.70

Notes:
(1) In the special case where V  W (i.e., T : V  V ) and B  B ',
the matrix A is called the matrix of T relative to the basis B
(T 對應於基底B的矩陣)
(2) If T : V  V is the identity transformation
B  {v1 , v 2 , , v n }: a basis for V
 the matrix of T relative to the basis B
A  T ( v1 ) B T ( v 2 ) B
1 0
0 1
T ( v n )B   

0 0
0
0 
 In


1
6.71
Keywords in Section 6.3:

standard matrix for T: T 的標準矩陣

composition of linear transformations: 線性轉換的合成

inverse linear transformation: 反線性轉換


matrix of T relative to the bases B and B' : T對應於基底B到
B'的矩陣
matrix of T relative to the basis B: T對應於基底B的矩陣
6.72
6.4 Transition Matrices and Similarity
T :V  V
B  {v1 , v 2 ,
, vn}
B '  {w1 , w 2 ,
( a linear transformation)
( a basis of V )
, w n } (a basis of V )
A  T ( v1 )B T ( v 2 ) B
T ( v n )B 
A '  T (w1 ) B ' T (w 2 ) B '
( matrix of T relative to B)
(T相對於B的矩陣)
T (w n )B ' 
(matrix of T relative to B ')
(T相對於B’的矩陣)
T ( v)B  A vB , and T ( v)B '  A  vB '
P   w1 B
P 1   v1 B '
 w 2 B
 v 2 B '
 w n B 
( transition matrix from B ' to B )
 v n B ' 
(從B'到B的轉移矩陣)
( transition matrix from B to B ')
(從B到B’的轉移矩陣)
from the definition of the transition matrix on p.254 in the
text book or on Slide 4.108 and 4.109 in the lecture notes
 vB  P  vB ' , and  vB '  P1  vB
6.73

Two ways to get from v B ' to T ( v)B ' :
(1) (direct) : A '[ v]B '  [T ( v)]B '
(2) (indirect) : P 1 AP[ v]B '  [T ( v)]B '  A '  P 1 AP
indirect
direct
6.74

Ex 1: (Finding a matrix for a linear transformation)
Find the matrix A' for T： R 2  R 2
T ( x1, x2 )  (2x1  2x2 ,  x1  3x2 )
reletive to the basis B'  {(1, 0), (1, 1)}
Sol:
(1) A '  T (1, 0) B '
T (1, 1)B ' 
3
T (1, 0)  (2,  1)  3(1, 0)  1(1, 1)  T (1, 0)B '   
 1
  2
T (1, 1)  (0, 2)  2(1, 0)  2(1, 1)  T (1, 1)B '   
2
 3  2
 A'  T (1, 0)B ' T (1, 1)B '   
 1 2 
6.75
(2) standard matrix for T (matrix of T relative to B  {(1, 0), (0, 1)})
 2  2
A  T (1, 0) T (0, 1)  
 1 3 
transition matrix from B' to B
1 1
P  (1, 0)B (1, 1)B   

0
1


transition matrix from B to B '
※ Solve a(1, 0) + b(0, 1) = (1, 0) 
(a, b) = (1, 0)
※ Solve c(1, 0) + d(0, 1) = (1, 1) 
(c, d) = (1, 1)
1 1
P   (1, 0) B '  (0, 1) B '   

0
1


matrix of T relative B'
1
※ Solve a(1, 0) + b(1, 1) = (1, 0) 
(a, b) = (1, 0)
※ Solve c(1, 0) + d(1, 1) = (0, 1) 
(c, d) = (-1, 1)
1  1  2  2 1 1  3  2
A'  P AP  







0
1

1
3
0
1

1
2



 

1
6.76

Ex 2: (Finding a matrix for a linear transformation)
Let B  {(3, 2), (4,  2)} and B '  {( 1, 2), (2,  2)} be basis for
 2
R , and let A  
 3
Find the matrix of T
2
7
2
2
be
the
matrix
for
T
:
R

R
relative to B.

7
relative to B '
Sol:
Because the specific function is unknown, it is difficult to apply
the direct method to derive A’, so we resort to the indirect
method where A’ = P-1AP
 3  2
transition matrix from B' to B : P  (1, 2)B (2,  2)B   

2

1


  1 2
1
transition matrix from B to B': P  (3, 2)B ' (4,  2)B '   


2
3


matrix of T relative to B':
  1 2  2 7 3  2  2 1
1
A'  P AP  







 2 3   3 7 2  1  1 3
6.77

Ex 3: (Finding a matrix for a linear transformation)
For the linear transformation T : R 2  R 2 given in Ex.2, find  v B ,
T ( v)B , and T ( v)B ' , for the vector v whose coordinate matrix is
 v B '
Sol:
 3
 
 1
3  2  3  7
vB  PvB '  2  1   1    5

   
 2 7  7  21
T ( v)B  AvB    3 7   5   14

  

  1 2  21  7 
1
T ( v)B '  P T ( v)B   2 3  14   0 


  
or T ( v)B '  A' vB '
 2 1  3  7

 



 1 3   1  0 
6.78

Similar matrix (相似矩陣):
For square matrices A and A’ of order n, A’ is said to be
similar to A if there exist an invertible matrix P s.t. A’=P-1AP

Theorem 6.13: (Properties of similar matrices)
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
Pf for (1) and (2) (the proof of (3) is left in Exercise 23):
(1) A  I n AI n (the transition matrix from A to A is the I n )
(2) A  P 1BP  PAP 1  P( P 1BP) P 1  PAP 1  B
 Q1 AQ  B (by defining Q  P 1 ), thus B is similar to A
6.79

Ex 4: (Similar matrices)
 2 2
 3 2
(a) A  
and A '  
are similar


 1 3 
 1 2 
1 1
1
because A'  P AP, where P  

0
1


 2 7 
 2 1
(b) A  
and A '  
are similar


 3 7 
 1 3
 3  2
because A'  P AP, where P  

2

1


1
6.80

Ex 5: (A comparison of two matrices for a linear transformation)
1 3 0 
Suppose A   3 1 0  is the matrix for T : R 3  R 3 relative
0 0 2 
to the standard basis. Find the matrix for T relative to the basis
B '  {(1, 1, 0), (1,  1, 0), (0, 0, 1)}
Sol:
The transitio n matrix from B' to the standard matrix
P  (1, 1, 0)B
(1,  1, 0)B
 12 12
 P 1   12  12

0 0
0
0

1
1 1 0
(0, 0, 1)B   1  1 0
0 0 1
6.81
matrix of T relative to B ' :
 12 12 0  1 3 0  1 1 0 
A '  P 1 AP   12  12 0   3 1 0  1 1 0 
0 0 1  0 0 2  0 0 1 
4 0 0 
(A’ is a diagonal matrix, which is


  0 2 0  simple and with some
 0 0 2  computational advantages)
※ You have seen that the standard matrix for a linear
transformation T:V →V depends on the basis used for V.
What choice of basis will make the standard matrix for T as
simple as possible? This case shows that it is not always
the standard basis.
6.82

Notes: Diagonal matrices have many computational advantages
over nondiagonal ones (it will be used in the next chapter)
 d1k 0

k
0
d
k
2
(1) D  


 0 0
 d11 0

1
 0 d2
1
(3) D  

0 0

 d1 0
0 d
2
for D  


0 0
0

0

k
d n 
0

0
 , di  0

1 
dn 
0
0 


dn 
(2) DT  D
6.83
Keywords in Section 6.4:

matrix of T relative to B: T 相對於B的矩陣

matrix of T relative to B' : T 相對於B'的矩陣

transition matrix from B' to B : 從B'到B的轉移矩陣

transition matrix from B to B' : 從B到B'的轉移矩陣

similar matrix: 相似矩陣
6.84
6.5 Applications of Linear Transformation

The geometry of linear transformation in the plane (p.407-p.110)
Reflection in x-axis, y-axis, and y=x
Horizontal and vertical expansion and contraction
Horizontal and vertical shear
Computer graphics (to produce any desired angle of view of a 3D figure)




6.85