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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
Elementary Linear Algebra
R. Larsen et al. (6 Edition)
投影片設計編製者
淡江大學 電機系 翁慶昌 教授
6.1 Introduction to Linear Transformations

Function T that maps a vector space V into a vector space W:
T : V mapping
W ,
V ,W : vector space
V: the domain of T
W: the codomain of T
Elementary Linear Algebra: Section 6.1, pp.361-362
1/78

Image of v under T:
If v is in V and w is in W such that
T ( v)  w
Then w is called the image of v under T .


the range of T:
The set of all images of vectors in V.
the preimage of w:
The set of all v in V such that T(v)=w.
Elementary Linear Algebra: Section 6.1, p.361
2/78

Ex 1: (A function from R2 into R2 )
T : R 2  R 2 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).
Elementary Linear Algebra: Section 6.1, p.362
3/78

Linear Transformation (L.T.):
V ,W： vector space
T : V  W： V to W linear tra nsformatio n
(1) T (u  v)  T (u)  T ( v), u, v V
(2) T (cu)  cT (u),
c  R
Elementary Linear Algebra: Section 6.1, p.362
4/78

Notes:
(1) A linear transformation is said to be operation preserving.
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.
Elementary Linear Algebra: Section 6.1, p.363
5/78

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)
Elementary Linear Algebra: Section 6.1, p.363
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(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.
Elementary Linear Algebra: Section 6.1, p.363
7/78

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
sin( 2  3 )  sin( 2 )  sin( 3 )
linear tra nsformatio n
(b) f ( x)  x 2
( x1  x2 ) 2  x12  x22
(1  2) 2  12  22
 f ( x)  x 2 is not linear
tra nsformatio n
(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 )  f ( x)  x  1 is not
linear tra nsformatio n 8/78
Elementary Linear Algebra: Section 6.1, p.363

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.
Elementary Linear Algebra: Section 6.1, p.364
9/78

Zero transformation:
T :V W

Identity transformation:
T :V V

T ( v)  0, v V
T ( v)  v, v  V
Thm 6.1: (Properties of linear transformations)
T : V  W , u, v  V
(1) T (0)  0
(2) T ( v)  T ( v)
(3) T (u  v)  T (u)  T ( v)
(4) If v  c1v1  c2 v2    cn vn
Then T ( v)  T (c1v1  c2 v2    cn vn )
 c1T (v1 )  c2T (v2 )    cnT (vn )
Elementary Linear Algebra: Section 6.1, p.365
10/78

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)
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)
(T is a L.T.)
 (7,7,0)
Elementary Linear Algebra: Section 6.1, p.365
11/78

Ex 5: (A linear transformation defined by a matrix)
0
3
v 
The function T : R 2  R 3 is defined as T ( v)  Av   2
1  1 

 v2 
  1  2
(a) Find T ( v) , where v  (2,1)
(b) Show that T is a linear transformatio n form R 2 into R 3
Sol: (a) v  (2,1)
R 2 vector R 3 vector
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)
Elementary Linear Algebra: Section 6.1, p.366
(vector addition)
(scalar multiplication)
12/78

Thm 6.2: (The 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.

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
Elementary Linear Algebra: Section 6.1, p.367
13/78

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:
v  ( x, y )  (r cos  , r sin  )
(polar coordinates)
r： the length of v
：the angle from the positive
x-axis counterclockwise to
the vector v
Elementary Linear Algebra: Section 6.1, p.368
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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  
r：the length of T(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 .
Elementary Linear Algebra: Section 6.1, p.368
15/78

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.
Elementary Linear Algebra: Section 6.1, p.369
16/78

Ex 9: (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 is a linear transformation from Mmn into Mn m.
Elementary Linear Algebra: Section 6.1, p.369
17/78
Keywords in Section 6.1:

function: 函數

domain: 論域

codomain: 對應論域

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
18/78
6.2 The Kernel and Range of a Linear Transformation

Kernel of a linear transformation T:
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 }

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  


Elementary Linear Algebra: Section 6.2, p.375
19/78

Ex 2: (The kernel of the zero and identity transformations)
(a) T(v)=0 (the zero transformation T : V  W )
ker(T )  V
(b) T(v)=v (the identity transformation T : V  V )
ker(T )  {0}

Ex 3: (Finding the kernel of a linear transformation)
T ( x, y, z )  ( x, y,0)
(T : R3  R3 )
ker(T )  ?
Sol:
ker(T )  {( 0,0, z ) | z is a real number }
Elementary Linear Algebra: Section 6.2, p.375
20/78

Ex 5: (Finding the kernel of a linear transformation)
 x1 
 1  1  2  
T (x)  Ax  
x2

3  
 1 2
 x3 
ker(T )  ?
(T : R 3  R 2 )
Sol:
ker(T )  {( x1 , x2 , x3 ) | T ( x1 , x2 , x3 )  (0,0), x  ( x1 , x2 , x3 )  R 3}
T ( x1 , x2 , x3 )  (0,0)
 x1 
 1  1  2   0
x2   
 1 2

3    0
 x3 
Elementary Linear Algebra: Section 6.2, p.376
21/78
 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)}
Elementary Linear Algebra: Section 6.2, p.377
22/78

Thm 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.
 T (0)  0 (Theorem 6.1)
 ker(T ) is a nonempty subset of V
Pf:
Let u and v be vectors in the kernel of T . then
T (u  v)  T (u)  T ( v)  0  0  0  u  v  ker(T )
T (cu)  cT (u)  c0  0
Thus, ker(T ) is a subspace of V .

 cu  ker(T )
Note:
The kernel of T is sometimes called the nullspace of T.
Elementary Linear Algebra: Section 6.2, p.377
23/78

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.
Elementary Linear Algebra: Section 6.2, p.377
24/78
Sol:
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
0 2
1 1
0 0
0 0
s
0 1
0 2
1 4
0 0
t
0
0

0
0
 x1   2s  t 
  2  1 
 x2   s  2t 
1 2
 
x   x3    s   s  1   t  0 
 x4    4t 
 0    4
 x5   t 
 0   1 
B  (2, 1, 1, 0, 0), (1, 2, 0,  4, 1): one basis for the kernel of T
Elementary Linear Algebra: Section 6.2, p.378
25/78

Corollary to Thm 6.3:
Let T : R n  R m be the L.T given by T (x)  Ax
Then the kernel of T is equal to the solution space of Ax  0
T (x)  Ax (a linear transformati on T : R n  R m )
 Ker (T )  NS ( A)  x | Ax  0, x  R m  (subspace of R m )

Range of a linear transformation T:
Let T : V  W be a L.T.
Then the set of all vectors w in W that are images of vector
in V is called the range of T and is denoted by range(T )
range(T )  {T ( v) | v  V }
Elementary Linear Algebra: Section 6.2, p.378
26/78

Thm 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 (Thm.6. 1)
 range(T ) is a nonempty subset of W
Let T (u) and T ( v) be vector in the range of T
T (u  v)  T (u)  T ( v)  range(T ) (u V , v V  u  v V )
T (cu)  cT (u)  range(T )
(u V  cu V )
Therefore, range(T ) is W subspace .
Elementary Linear Algebra: Section 6.2, p.379
27/78

Notes:
T : V  W is a L.T.
(1) Ker (T ) is subspace of V
(2)range(T ) is subspace of W

Corollary to Thm 6.4:
Let T : R n  R m be the L.T. given by T (x)  Ax
Then the range of T is equal to the column space of A
 range(T )  CS ( A)
Elementary Linear Algebra: Section 6.2, p.379
28/78

Ex 7: (Finding a basis for the range of a linear transformation)
Let T : R 5  R 4 be defined by T (x)  Ax, where x is 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 the range of T.
Elementary Linear Algebra: Section 6.2, p.379
29/78
Sol:
1
2
A
 1
 0
c1
2 0
1 3
0 2
0 0
c2 c3
1  1
1
1 0  G . J . E 0
 
0 1
0
2 8 
0
c4 c5
w1
0  1
0 2
B
1 4
0 0 
w3 w4 w5
0 2
1 1
0 0
0 0
w2
 w1 , w2 , w4 is a basis for CS ( B)
c , c , c is a basis for CS ( A)
1
2
4
 (1, 2, 1, 0), (2, 1, 0, 0), (1, 1, 0, 2)is a basis for the range of T
Elementary Linear Algebra: Section 6.2, pp.379-380
30/78

Rank of a linear transformation T:V→W:
rank (T )  the dimension of the range of T

Nullity of a linear transformation T:V→W:
nullity (T )  the dimension of the kernel of T

Note:
Let T : R n  R m be the L.T. given by T (x)  Ax, then
rank (T )  rank ( A)
nullity (T )  nullity ( A)
Elementary Linear Algebra: Section 6.2, p.380
31/78
Thm 6.5: (Sum of rank and nullity)
Let T : V  W be a L.T. form an n - dimensiona l vector space V
into a vector space W . then
rank (T )  nullity (T )  n
dim( range of T )  dim( kernel of T )  dim( domain of T )
Pf:

Let T is represente d by an m  n matrix A
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( solution space of A)
 nr
 rank (T )  nullity (T )  r  (n  r )  n
Elementary Linear Algebra: Section 6.2, p.380
32/78

Ex 8: (Finding the rank and nullity of a linear transformation)
Find the rank and nullity of the L.T. 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
Elementary Linear Algebra: Section 6.2, p.381
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
Ex 9: (Finding the rank and nullity of a linear transformation)
Let T : R 5  R 7 be a linear transformatio n.
(a ) Find the dimension of the kernel of T if the dimension
of the range 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 )  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
Elementary Linear Algebra: Section 6.2, p.381
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
One-to-one:
A function T : V  W is called one - to - one if the preimage of
every w in the range consists of a single vector.
T is one - to - one iff for all u and v inV, T (u)  T ( v)
implies that u  v.
one-to-one
Elementary Linear Algebra: Section 6.2, p.382
not one-to-one
35/78

Onto:
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.)
Elementary Linear Algebra: Section 6.2, p.382
36/78

Thm 6.6: (One-to-one linear transformation)
Let T : V  W be a L.T.
Then T is 1 - 1 iff Ker (T )  {0}
Pf:
Suppose T is 1 - 1
Then T (v)  0 can have only one solution : v  0
i.e. Ker (T )  {0}
Suppose Ker (T )  {0} and T (u )  T (v)
T (u  v)  T (u )  T (v)  0
T is a L.T.
 u  v  Ker (T )  u  v  0
 T is 1-1
Elementary Linear Algebra: Section 6.2, p.382
37/78

Ex 10: (One-to-one and not one-to-one linear transformation)
(a ) The L.T. 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 R 3 .
Elementary Linear Algebra: Section 6.2, p.382
38/78

Thm 6.7: (Onto linear transformation)
Let T : V  W be a L.T., where W is finite dimensiona l.
Then T is onto iff the rank of T is equal to the dimension of W .

Thm 6.8: (One-to-one and onto linear transformation)
Let T : V  W be a L.T. with vect or 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
dim( range(T ))  n  dim( Ker (T ))  n  dim( W )
Consequent ly, T is onto.
If T is onto, then dim( range of T )  dim(W )  n
dim( Ker (T ))  n  dim( range of T )  n  n  0
Therefore, T is one - to - one.
Elementary Linear Algebra: Section 6.2, p.383
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
Ex 11:
The L.T. 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 0 
(c ) A  
0 1  1
Sol:
1 2
(b) A  0 1


0 0
1 2
(d ) A  0 1

0 0
0
1

0
T:Rn→Rm
dim(domain of T)
rank(T)
nullity(T)
1-1
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
Elementary Linear Algebra: Section 6.2, p.383
40/78


Isomorphism:
A linear transformatio 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.
Thm 6.9: (Isomorphic spaces 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( Ker (T ))  0
 dim( range of T )  dim( domain of T )  dim( Ker (T ))  n  0  n
Elementary Linear Algebra: Section 6.2, p.384
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 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 v1 , v2 , , vn  be a basis of V, and
let w1 , w2 , , wn  be a basis of W .
Then an arbitrary vector in V can be represente d as
v  c1v1  c2 v2    cn vn
and you can define a L.T. T : V  W as follows.
T ( v )  c1w1  c2 w2    cn wn
It can be shown that this L.T. is both 1-1 and onto.
Thus V and W are isomorphic.
Elementary Linear Algebra: Section 6.2, p.384
42/78

Ex 12: (Isomorphic vector spaces)
The following vector spaces are isomorphic to each other.
(a) R 4  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 polynomial s of degree 3 or less
(e)V  {( x1 , x2 , x3 , x4 , 0), xi is a real number }(subspace of R 5 )
Elementary Linear Algebra: Section 6.2, p.385
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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: 同構的空間
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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.
Elementary Linear Algebra: Section 6.3, p.387
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
Thm 6.10: (Standard matrix for a linear transformation)
Let T : R n  R m be a linear trt ansformati on such that
 a11 
 a12 
 a1n 
 a21 
 a22 
 a2 n 
T (e1 )   , T (e2 )   ,  , T (en )   ,
  
  
  
am1 
am 2 
amn 
Then the m n matrix who se n columns
 a11
 a21
A  T (e1 ) T (e2 )  T (en )  
 
am1
correspond to T (ei )
a12  a1n 
a22  a2 n 

   
am 2  amn 
is such that T ( v )  Av for every v in R n .
A is called the standard matrix for T .
Elementary Linear Algebra: Section 6.3, p.388
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Pf:
 v1 
 v2 
v     v1e1  v2 e2    vn en

vn 
T is a L.T.  T (v )  T (v1e1  v2 e2    vn en )
 T (v1e1 )  T (v2 e2 )    T (vn en )
 v1T (e1 )  v2T (e2 )    vnT (en )
 a11
 a21
Av  
 
am1
a12
a22

am 2
 a1n   v1   a11v1  a12v2    a1n vn 
 a2 n  v2   a21v1  a22v2    a2 n vn 
   

     


 amn  vn  am1v1  am 2 v2    amnvn 
Elementary Linear Algebra: Section 6.3, p.388
47/78
 a11 
 a12 
 a1n 
 a21 
 a22 
 a2 n 
 v1    v2      vn  
  
  
  
am1 
am 2 
amn 
 v1T (e1 )  v2T (e2 )    vnT (en )
Therefore, T ( v)  Av for each v in R n
Elementary Linear Algebra: Section 6.3, p.389
48/78

Ex 1: (Finding the standard matrix of a linear transformation)
Find the standard matrix for the L.T. T : R3  R 2 define 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)
Elementary Linear Algebra: Section 6.3, p.389
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
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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:
1  2 0  1x  2 y  0 z
A
2 1 0  2 x  1y  0 z
Elementary Linear Algebra: Section 6.3, p.389
50/78

Ex 2: (Finding the standard matrix of a linear transformation)
The linear tra nsformatio n 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 zero transformation from Rn into Rn
is the nn identity matrix In
Elementary Linear Algebra: Section 6.3, p.390
51/78

Composition of T1:Rn→Rm with T2:Rm→Rp :
T ( v)  T2 (T1 ( v)), v  R n
T  T2  T1 , domain of T  domain of T1

Thm 6.11: (Composition of linear transformations)
Let T1 : R n  R m and T2 : R m  R p be L.T.
with standard matrices A1 and A2 , then
(1)The compositio n T : R n  R p , defined by T ( v)  T2 (T1 ( v)), is a L.T.
(2) The standard matrix A for T is given by the matrix product A  A2 A1
Elementary Linear Algebra: Section 6.3, p.391
52/78
Pf:
(1)( T is a L.T.)
Let u and v be vectors in R n and let c be any scalar the n
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
Elementary Linear Algebra: Section 6.3, p.391
53/78

Ex 3: (The standard matrix of a composition)
Let T1 and T2 be L.T. from R3 into R3 s.t.
Sol:
T1 ( x, y, z)  (2x  y, 0, x  z)
T2 ( x, y, z)  ( x  y, z, y)
Find the standard matrices for the compositio ns
T  T2  T1 and T '  T1  T2 ,
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
Elementary Linear Algebra: Section 6.3, p.392
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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
Elementary Linear Algebra: Section 6.3, p.392
55/78

Inverse linear transformation:
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 .
Elementary Linear Algebra: Section 6.3, p.392
56/78

Thm 6.12: (Existence of an inverse transformation)
Let T : R n  R n be a L.T. with standard matrix A,
Then the following condition are equivalent .
(1) T is invertible.
(2) T is an isomorphism.
(3) A is invertible.

Note:
If T is invertible with standard matrix A, then the standard
matrix for T–1 is A–1 .
Elementary Linear Algebra: Section 6.3, p.393
57/78

Ex 4: (Finding the inverse of a linear transformation)
The L.T. 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
 3x1  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


Elementary Linear Algebra: Section 6.3, p.393
58/78
0
1 0 0  1 1
. J .E
G
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 )
Elementary Linear Algebra: Section 6.3, p.394
59/78

the matrix of T relative to the bases B and B':
T :V  W
B  {v1 , v2 ,, vn }
(a L.T. )
(a basis for V )
B'  {w1 , w2 ,, wm } (a basis for W )
Thus, the matrix of T relative to the bases B and B' is
A  T (v1 )B ' , T (v2 )B ' ,, T (vn )B '  M mn
Elementary Linear Algebra: Section 6.3, p.394
60/78

Transformation matrix for nonstandard bases:
Let V and W be finite - dimensiona l vector spaces with basis B and B' ,
respective ly, where B  {v1 , v2 ,, vn }
If T : V  W is a L.T. s.t.
T (v1 )B '
 a11 
 a12 
 a1n 
 a21 
 a22 
 a2 n 
  , T (v2 )B '   ,, T (vn )B '   
  
  
  
am1 
am 2 
amn 
then the m n matrix who se n columns correspond to T (vi )B '
Elementary Linear Algebra: Section 6.3, p.395
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 a11
 a21
A  T (e1 ) T (e2 )  T (en )  
 
am1
a12
a22

am 2
 a1n 
 a2 n 

  
 amn 
is such that T ( v)B '  A[ v]B for every v in V .
Elementary Linear Algebra: Section 6.3, p.395
62/78

Ex 5: (Finding a matrix relative to nonstandard bases)
Let T： R 2  R 2 be a L.T. 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
Elementary Linear Algebra: Section 6.3, p.395
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
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)
Elementary Linear Algebra: Section 6.3, p.395
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
Notes:
(1)In the special case where V  W and B  B' ,
the matrix A is alled the matrix of T relative to the basis B
(2)T : V  V : the identity t ransformat ion
B  {v1 , v2 , , vn } : a basis for V
 the matrix of T relative to the basis B
1 0  0
0 1  0 
  In
A  T (v1 )B , T (v2 )B , , T (vn )B   
   


0 0  1 
Elementary Linear Algebra, Section 6.3, p.396
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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
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6.4 Transition Matrices and Similarity
T :V  V
B  {v1 , v2 ,, vn }
( a L.T. )
( a basis of V )
B'  {w1 , w2 ,, wn } (a basis of V )
A  T (v1 )B , T (v2 )B ,, T (vn )B 
A'  T (w1 )B ' , T ( w2 )B ' ,, T (wn )B ' 
P  w1 B , w2 B ,, wn B 
P 1  v1 B ' , v2 B ' ,, vn B ' 
( matrix of T relative to B)
(matrix of T relative to B' )
( transitio n matrix from B' to B )
( transitio n matrix from B to B' )
vB  PvB ' , vB '  P 1vB
T ( v)B  AvB
T ( v)B '  A' vB '
Elementary Linear Algebra: Section 6.4, p.399
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
Two ways to get from v B ' to T ( v)B ':
(1)(direct )
A'[ v]B '  [T ( v)] B '
indirect
(2)(indirect)
P 1 AP[ v]B '  [T ( v)]B '
1
 A'  P AP
Elementary Linear Algebra: Section 6.4, pp.399-400
direct
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
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:
(I) 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 
Elementary Linear Algebra: Section 6.4, p.400
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(II) 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'
1  1
P 

0
1


matrix of T relative B'
1
1  1  2  2 1 1  3  2
A'  P AP  







0
1

1
3
0
1

1
2



 

1
Elementary Linear Algebra: Section 6.4, p.400
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
Ex 2: (Finding a matrix for a linear transformation)
Let B  {( 3, 2), (4,  2)} and B'  {( 1, 2), (2,  2)} be basis for R 2 ,
 2 7 
2
2
and let A  
be
the
matrix
for
T
:
R

R
relative to B.

  3 7
Find the matrix of T relative to B'.
Sol:
transition matrix from B' to B : P  (1, 2)B
 3  2
(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
A'  P AP  







 2 3   3 7 2  1  1 3
1
Elementary Linear Algebra: Section 6.4, p.401
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
Ex 3: (Finding a matrix for a linear transformation)
For the linear tra nsformatio n 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
 3
vB '   
  1
Sol:
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 
Elementary Linear Algebra: Section 6.4, p.401
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

Similar matrix:
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 1 AP
Thm 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:
(1) A  I n AI n
(2) A  P 1BP  PAP1  P( P 1BP ) P 1
PAP1  B  Q 1 AQ  B (Q  P 1 )
Elementary Linear Algebra: Section 6.4, p.402
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
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
Elementary Linear Algebra: Section 6.4, p.403
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
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
Elementary Linear Algebra: Section 6.4, p.403
1 1 0
(0, 0, 1)B   1  1 0
0 0 1
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matrix of T relative to B':
 12 12
A'  P 1 AP   12  12
0 0
0
4 0
 0  2 0 
0 0  2
Elementary Linear Algebra: Section 6.4, p.403
0 1 3 0  1 1 0
0 3 1 0  1  1 0
1 0 0  2 0 0 1
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
Notes: Computational advantages of diagonal matrices:
d1k
0
k
(1) D  

 0
0
d 2k

0
 0
 0

  
 d nk 
0




d1
0
D

 0
0
d2

0
 0
 0

 
 d n 
(2) DT  D
 d11
0
(3) D 1  

0

1
d2

0
Elementary Linear Algebra: Section 6.4, p.404
0
0
, d i  0

1 
dn 
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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
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