Linear Transformations

Report
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 : vectorspace
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: vect orspace
T : V  W: V toW linear transformation
(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 ) : vectorin R2 , c : any real number
(1)Vectoraddition :
u  v  (u1 , u2 )  (v1 , v2 )  (u1  v1 , u2  v2 )
T (u  v )  T (u1  v1 , u 2  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
6/78
(2) Scalar multiplication
cu  c(u1 , u2 )  (cu1 , cu 2 )
T (cu)  T (cu1 , cu 2 )  (cu1  cu 2 , cu1  2cu 2 )
 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 )
(b) f ( x)  x 2
( x1  x2 )2  x12  x22
(1  2)  1  2
2
2
2
linear tra nsformatio n
 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
Elementary Linear Algebra: Section 6.1, p.363
linear tra nsformatio n 8/78

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  c2v2    cn vn
T henT ( v)  T (c1v1  c2v2    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 : R3  R3 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)
 (7,7,0)
Elementary Linear Algebra: Section 6.1, p.365
(T is a L.T.)
11/78

Ex 5: (A linear transformation defined by a matrix)
0
3
 v1 
2
3


1  
The function T : R  R is defined as T ( v)  Av  2

 v 2 
  1  2
(a) Find T ( v) , where v  (2,1)
(b) Show that T is a linear transformation form R 2 into R3
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  a12 v2    a1n vn 
a22  a2 n  v2   a21v1  a22 v2    a2 n vn 
   


    


am 2  amn  vn  am1v1  am 2 v2    amn vn 
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
14/78
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 : M3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 )  R3}
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 (T heorem6.1)
 ker(T ) is a nonemptysubset 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
T (cu)  cT (u)  c0  0
Thus, ker(T ) is a subspace of V .

 u  v  ker(T )
 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 3 1 0 
0 2 0 1 

0 0 2 8
2
0
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 thekernelof T
Elementary Linear Algebra: Section 6.2, p.378
25/78

Corollary to Thm 6.3:
Let T : R n  R m be theL.T given by T (x)  Ax
T hen thekernelof T is equal to thesolutionspace of Ax  0
T (x)  Ax (a linear tr ansformati 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 .
T hen theset of all vectorsw in W thatare images of vector
in V is called therange of T and is denotedby 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)
T herange of a linear transformation T : V  W is a subspace of W .
Pf:
T (0)  0 (T hm.6.1)
 range(T ) is a nonemptysubset of W
Let T (u) andT ( v) be vectorin therange 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 )
T herefore,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 theL.T .given by T (x)  Ax
T hen therange of T is equal to thecolumnspace 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 3 1 0 
0 2 0 1 

0 0 2 8
2
0
Find a basis for the range of T.
Elementary Linear Algebra: Section 6.2, p.379
29/78
Sol:
1
2
A
 1
 0
2 0
1 3
0 2
0 0
1  1
1
1 0  G . J . E 0
 
0 1
0
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 , 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 therangeof T
Elementary Linear Algebra: Section 6.2, pp.379-380
30/78

Rank of a linear transformation T:V→W:
rank(T )  thedimensionof therange of T

Nullity of a linear transformation T:V→W:
nullity(T )  thedimensionof thekernelof T

Note:
Let T : R n  R m be theL.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 .forman n - dimensional vectorspaceV
intoa vectorspaceW . then
rank(T )  nullity(T )  n
dim(range of T )  dim(kernelof T )  dim(domain of T )
Pf:

Let T is represented 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(kernelof T )  dim(solutionspace 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 t herank and nullit yof t heL.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
33/78

Ex 9: (Finding the rank and nullity of a linear transformation)
Let T : R 5  R 7 be a linear tra nsformatio 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 functionT : V  W is called one- to - oneif thepreimageof
every win therange consistsof a single vector.
T is one- to - oneiff for all u and v inV, T (u)  T ( v)
implies thatu  v.
one-to-one
Elementary Linear Algebra: Section 6.2, p.382
not one-to-one
35/78

Onto:
A functionT : V  W is said to be ontoif everyelement
in w has a preimagein 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 .
T henT is 1 - 1 iff Ker (T )  {0}
Pf:
SupposeT is 1 -1
T henT (v)  0 can haveonlyonesolution: v  0
i.e. Ker(T )  {0}
Suppose Ker(T )  {0} andT (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 : R 3  R 3 is not one - to - one.
Because its kernel is all of R3 .
Elementary Linear Algebra: Section 6.2, p.382
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
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
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

Isomorphism:
A linear transformation 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
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
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)  spaceof all polynomial
s of degree 3 or less
(e)V  {( x1, x2 , x3 , x4 , 0), xi is a real number}(subspace of R5 )
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 )  (2x1  x2  x3 , x1  3x2  2x3 ,3x2  4x3 )
 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 trtansformation such that
 a11 
 a12 
 a1n 
 a21 
 a22 
 a2 n 
T (e1 )   , T (e2 )   ,  , T (en )   ,
  
  
  
am1 
am 2 
amn 
Then them n matrix whose n columnscorrespondtoT (ei )
 a11 a12  a1n 
 a21 a22  a2 n 
A  T (e1 ) T (e2 )  T (en )  







am1 am 2  amn 
is such thatT ( v)  Av for every v in R n .
A is called thestandardmatrixfor 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  a12 v2    a1n vn 
 a2 n  v2   a21v1  a22 v2    a2 n vn 
   

     


 amn  vn  am1v1  am 2 v2    amn vn 
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 )
T herefore,T ( v)  Av for each v in Rn
Elementary Linear Algebra: Section 6.3, p.389
48/78

Ex 1: (Finding the standard matrix of a linear transformation)
Find thestandardmatrixfor theL.T .T : R3  R2 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
49/78
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)
T helinear transformation T : R 2  R 2 is given by projecting
each pointin R 2 ontothe x - axis. Find thestandardmatrixfor 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  Rn
T  T2  T1 , domainof T  domainof T1

Thm 6.11: (Composition of linear transformations)
Let T1 : R n  R m and T2 : R m  R p be L.T .
with standardmatricesA1 and A2 , then
(1)Thecomposition T : Rn  R p , definedby T ( v)  T2 (T1 ( v)), is a L.T.
(2) T hestandardmatrixA forT is given by thematrixproduct A  A2 A1
Elementary Linear Algebra: Section 6.3, p.391
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Pf:
(1)( T is a L.T .)
Let u and v be vectorsin 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 thestandardmatrixfor 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 andT2 be L.T.fromR3 into R3 s.t.
Sol:
T1 ( x, y, z)  (2 x  y, 0, x  z)
T2 ( x, y, z)  ( x  y, z, y)
Find thestandardmatricesfor thecompositions
T  T2  T1 and T '  T1  T2 ,
2
A1  0
1
1
A2  0
0
1 0
0 0 (st andardmat rixfor T1 )
0 1
 1 0
0 1 (st andardmat rixfor T2 )
1 0
Elementary Linear Algebra: Section 6.3, p.392
54/78
T hestandardmatrixfor 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


 

T hestandardmatrixforT '  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 : Rn  Rn andT2 : Rn  Rn are L.T .s.t.for everyv inRn
T2 (T1 ( v))  v and T1 (T2 ( v))  v
T henT2 is called theinverseof T1 andT1 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 standardmatrixA,
T hen thefollowingconditionare 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)
TheL.T.T: R3  R3 is definedby
T ( x1, x2 , x3 )  (2x1  3x2  x3 , 3x1  3x2  x3 , 2x1  4x2  x3 )
Show that T is invertible, and find its inverse.
Sol:
T hestandardmatrixfor 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

T hereforeT is invertibleand thestandardmatrixfor 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1 v   1 0
1   x2     x1  x3 

  

6

2

3
x
6
x

2
x

3
x
2
3

 3   1
In other words,
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 forV )
B'  {w1 , w2 ,, wm } (a basis forW )
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 andW be finite- dimensional vectorspaces with basis B and B' ,
respectively,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 them n matrix whose n columnscorrespondto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 everyv in V .
Elementary Linear Algebra: Section 6.3, p.395
62/78

Ex 5: (Finding a matrix relative to nonstandard bases)
Let T: R2  R2 be a L.T .definedby
T ( x1 , x2 )  ( x1  x2 , 2 x1  x2 )
Find thematrixof T relativeto thebasis
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 theL.T .T: R 2  R 2 given in Example5, use thematrix A
to findT ( 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 thespecialcase whereV  W and B  B' ,
thematrix A is alled thematrixof T relativeto thebasis B
(2)T : V  V : theidentity ransformat
t
ion
B  {v1 , v2 , , vn } : a basis forV
 thematrixof T relativeto thebasis 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
65/78
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' 
( matrixof T relativeto B)
(matrixof T relativeto B' )
( transition matrixfromB' to B )
( transition matrixfromB 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
67/78

Two ways to get from v B ' to T ( v)B ':
(1)(direct)
indirect
A'[ v]B '  [T ( v)]B '
(2)(indirect)
P 1 AP[ v]B '  [T ( v)]B '
1
 A'  P AP
Elementary Linear Algebra: Section 6.4, pp.399-400
direct
68/78

Ex 1: (Finding a matrix for a linear transformation)
Find thematrixA' for T: R 2  R 2
T ( x1 , x2 )  (2x1  2 x2 ,  x1  3x2 )
reletiveto thebasis 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
69/78
(II)standardmatrixfor T ( matrixof T relativeto B  {(1, 0), (0, 1)})
 2  2
A  T (1, 0) T (0, 1)  
 1 3 
transitionmatrixfrom B' to B
1 1
P  (1, 0)B (1, 1)B   

0
1


transitionmatrixfrom B to B'
1  1
P 

0
1


matrixof T relativeB'
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
70/78

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
relativeto B.

  3 7
Find thematrixof T relativeto B'.
Sol:
3  2
transitionmatrixfrom B' to B : P  (1, 2)B (2,  2)B   

2

1


  1 2
1
transitionmatrixfrom 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
71/78

Ex 3: (Finding a matrix for a linear transformation)
For thelinear transformation T : R 2  R 2 given in Ex.2,find v B、
T ( v)B and T ( v)B ' , for thevectorv whose coordinatematrixis
 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
72/78


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 AIn
(2) A  P 1 BP  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
73/78

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
74/78

Ex 5: (A comparison of two matrices for a linear transformation)
1 3 0 
Suppose A  3 1 0  is thematrixfor T : R 3  R 3 relative
0 0  2
to thestandardbasis. Find thematrixfor T relativeto thebasis
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
1
 12
2
1
1
 P   2  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
75/78
matrixof T relativeto 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
76/78

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 
77/78
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
78/78

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