Linear transformations and matrices

Chapter 6
Linear Transformations
6.1 Introduction to Linear Transformations
6.2 The Kernel and Range of a Linear Transformation
6.3 Isomorphisms
6.3 Matrices for Linear Transformations
6.4 Transition Matrices and Similarity
6.5 Applications of Linear Transformations
6.1

6.2
6.1 Introduction to Linear Transformations
◼ A function T that maps a vector space V into a vector space W:
mapping
: , , : vector spacesT V W V W⎯⎯⎯→
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)

6.3
※ 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),
where s is any real number
◼ 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

6.4
◼ Ex 1: A function from R2 into R2
22
: RRT →
)2,(),( 212121 vvvvvvT +−=
2
21 ),( Rvv =v
(a) Find the image of v=(-1,2) (b) Find the preimage of w=(-1,11)
Sol:
(a) ( 1, 2)
( ) ( 1, 2) ( 1 2, 1 2(2)) ( 3, 3)T T
= −
 = − = − − − + = −
v
v
(b) ( ) ( 1, 11)T = = −v w
)11,1()2,(),( 212121 −=+−= vvvvvvT
112
1
21
21
=+
−=−
vv
vv
4,3 21 == vv Thus {(3, 4)} is the preimage of w=(-1, 11)

6.5
◼ Linear Transformation (線性轉換):
truearepropertiestwofollowing
theifintoofnnsformatiolinear traA：:
spacesvector：,
WVWVT
WV
→
VTTT +=+ vuvuvu ,),()()((1)
RccTcT = ),()()2( uu

6.6
◼ Notes:
(1) A linear transformation is said to be operation preserving
)()()( vuvu TTT +=+
Addition
in V
Addition
in W
)()( uu cTcT =
Scalar
multiplication
in V
Scalar
multiplication
in W
(2) A linear transformation from a vector space into
itself is called a linear operator (線性運算子)
VVT →:
(because the same result occurs whether the operations of addition
and scalar multiplication are performed before or after the linear
transformation is applied)

6.7
◼ Ex 2: Verifying a linear transformation T from R2 into R2
Pf:
)2,(),( 212121 vvvvvvT +−=
numberrealany:,invector:),(),,( 2
2121 cRvvuu == vu
1 2 1 2 1 1 2 2
(1) Vector addition :
( , ) ( , ) ( , )u u v v u v u v+ = + = + +u v
)()(
)2,()2,(
))2()2(),()((
))(2)(),()((
),()(
21212121
21212121
22112211
2211
vu
vu
TT
vvvvuuuu
vvuuvvuu
vuvuvuvu
vuvuTT
+=
+−++−=
+++−+−=
++++−+=
++=+

6.8
),(),(
tionmultiplicaScalar)2(
2121 cucuuucc ==u
)(
)2,(
)2,(),()(
2121
212121
u
u
cT
uuuuc
cucucucucucuTcT
=
+−=
+−==
Therefore, T is a linear transformation

6.9
◼ Ex 3: Functions that are not linear transformations
(a) ( ) sinf x x=
2
(b) ( )f x x=
(c) ( ) 1f x x= +
)sin()sin()sin( 2121 xxxx ++
)sin()sin()sin( 3232

++
2
2
2
1
2
21 )( xxxx ++
222
21)21( ++
1)( 2121 ++=+ xxxxf
2)1()1()()( 212121 ++=+++=+ xxxxxfxf
)()()( 2121 xfxfxxf ++
(f(x) = sin x is not a linear transformation)
(f(x) = x2 is not a linear transformation)
(f(x) = x+1 is not a linear transformation,
although it is a linear function)
In fact, ( ) ( )f cx cf x

6.10
◼ Notes: Two uses of the term “linear”.
(1) is called a linear function because its graph
is a line
1)( += xxf
(2) is not a linear transformation from a vector
space R into R because it preserves neither vector
addition nor scalar multiplication
1)( += xxf

6.11
◼ Zero transformation (零轉換):
VWVT → vu,,:
( ) ,T V=  v 0 v
◼ Identity transformation (相等轉換):
VVT →: VT = vvv ,)(
◼ Theorem 6.1: Properties of linear transformations
WVT →:
00 =)((1)T
)()((2) vv TT −=−
)()()((3) vuvu TTT −=−
)()()(
)()(then
,If(4)
2211
2211
2211
nn
nn
nn
vTcvTcvTc
vcvcvcTT
vcvcvc
+++=
+++=
+++=



v
v
(T(cv) = cT(v) for c=-1)
(T(u+(-v))=T(u)+T(-v) and property (2))
(T(cv) = cT(v) for c=0)
(Iteratively using T(u+v)=T(u)+T(v) and T(cv) = cT(v))

6.12
◼ Ex 4: Linear transformations and bases
Let be a linear transformation such that33
: RRT →
)4,1,2()0,0,1( −=T
)2,5,1()0,1,0( −=T
)1,3,0()1,0,0( =T
Sol:
)1,0,0(2)0,1,0(3)0,0,1(2)2,3,2( −+=−
)0,7,7(
)1,3,0(2)2,5,1(3)4,1,2(2
)1,0,0(2)0,1,0(3)0,0,1(2)2,3,2(
=
−−+−=
−+=−
T
TTTT
Find T(2, 3, -2)
1 1 2 2 1 1 2 2
According to the fourth property on the previous slide that
( ) ( ) ( ) ( )n n n nT c v c v c v c T v c T v c T v
 
 
+ + + = + + + 

6.13
◼ Ex 5: A linear transformation defined by a matrix
The function is defined as32
: RRT → 











−−
==
2
1
21
12
03
)(
v
v
AT vv
2 3
(a) Find ( ), where (2, 1)
(b) Show that is a linear transformation form into
T
T R R
= −v v
Sol:
(a) (2, 1)= −v








=



−







−−
==
0
3
6
1
2
21
12
03
)( vv AT
)0,3,6()1,2( =−T
vector2
R vector3
R
(b) ( ) ( ) ( ) ( )T A A A T T+ = + = + = +u v u v u v u v
)()()()( uuuu cTAccAcT ===
(vector addition)
(scalar multiplication)

6.14
◼ Theorem 6.2: The linear transformation defined by a matrix
Let A be an mn matrix. The function T defined by
vv AT =)(
is a linear transformation from Rn into Rm
◼ Note:












+++
+++
+++
=
























=
nmnmm
nn
nn
nmnmm
n
n
vavava
vavava
vavava
v
v
v
aaa
aaa
aaa
A









2211
2222121
1212111
2
1
21
22221
11211
v
vv AT =)( mn
RRT ⎯→⎯:
vectorn
R vectorm
R
※ 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

6.15
Show that the L.T. given by the matrix
has the property that it rotates every vector in R2
counterclockwise about the origin through the angle 
◼ Ex 7: Rotation in the plane
22
: RRT →



 −
=


cossin
sincos
A
Sol:
( , ) ( cos , sin )x y r r = =v
(Polar coordinates: for every point on the xy-
plane, it can be represented by a set of (r, α))
r：the length of v ( )
：the angle from the positive
x-axis counterclockwise to
the vector v
2 2
x y= +
v
T(v)

6.16




+
+
=




+
−
=







 −
=






 −
==
)sin(
)cos(
sincoscossin
sinsincoscos
sin
cos
cossin
sincos
cossin
sincos
)(










r
r
rr
rr
r
r
y
x
AT vv
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 
according to the addition
formula of trigonometric
identities (三角函數合角公式)

6.17
is called a projection in R3
◼ Ex 8: A projection in R3
The linear transformation is given by33
: RRT →








=
000
010
001
A
1 0 0
If is ( , , ), 0 1 0
0 0 0 0
x x
x y z A y y
z
     
     = =     
          
v v
※ In other words, T maps every vector in R3
to its orthogonal projection in the xy-plane,
as shown in the right figure

6.18
Show that T is a linear transformation
◼ Ex 9: The transpose function is a linear transformation from
Mmn into Mn m
):()( mnnm
T
MMTAAT  →=
Sol:
nmMBA ,
)()()()( BTATBABABAT TTT
+=+=+=+
)()()( AcTcAcAcAT TT
===
Therefore, T (the transpose function) is a linear transformation
from Mmn into Mnm

6.19
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

6.20
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 be a linear transformation. Then the set of
all vectors v in V that satisfy is called the kernel
of T and is denoted by ker(T)
WVT →:
0v =)(T
},)(|{)ker( VTT == v0vv
※ 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( ) {(0,0, ) | is a real number}T s s=

6.21
◼ Ex 2: The kernel of the zero and identity transformations
(a) If T(v) = 0 (the zero transformation ), thenWVT →:
VT =)ker(
(b) If T(v) = v (the identity transformation ), thenVVT →:
}{)ker( 0=T
◼ Ex 1: Finding the kernel of a linear transformation
):()( 3223  →= MMTAAT T
Sol:








=
00
00
00
)ker(T

6.22
◼ Ex 5: Finding the kernel of a linear transformation
1
3 2
2
3
1 1 2
( ) ( : )
1 2 3
x
T A x T R R
x
 
− −   = = →   −    
x x
?)ker( =T
Sol:
3
1 2 3 1 2 3 1 2 3ker( ) {( , , ) | ( , , ) (0,0), and ( , , ) }T x x x T x x x x x x R= = 
)0,0(),,( 321 =xxxT




=












−
−−
0
0
321
211
3
2
1
x
x
x

6.23








−=








−=









1
1
1
3
2
1
t
t
t
t
x
x
x
)}1,1,1span{(
}numberrealais|)1,1,1({)ker(
−=
−= ttT
G.-J. E.1 1 2 0 1 0 1 0
1 2 3 0 0 1 1 0
− − −   
⎯⎯⎯→   −   

6.24
◼ Theorem 6.3: The kernel is a subspace of V
The kernel of a linear transformation is a
subspace of the domain V
WVT →:
)16.Theoremby()( 00 =T
Pf:
VT ofsubsetnonemptyais)ker(
Then.ofkernelin thevectorsbeandLet Tvu
000vuvu =+=+=+ )()()( TTT
00uu === ccTcT )()(
Thus, ker( ) is a subspace of (according to Theorem 4.5
that a nonempty subset of is a subspace of if it is closed
under vector addition and scalar multiplication)
T V
V V
T is a linear transformation
( ker( ), ker( ) ker( ))T T T   + u v u v
( ker( ) ker( ))T c T  u u

6.25
◼ Ex 6: Finding a basis for the kernel












−−
−
=
=→
82000
10201
01312
11021
andiniswhere,)(bydefinedbe:Let 545
A
RATRRT xxx
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 first A 0

6.26
 
G.-J. E.
1 2 0 1 1 0 1 0 2 0 1 0
2 1 3 1 0 0 0 1 1 0 2 0
1 0 2 0 1 0 0 0 0 1 4 0
0 0 0 2 8 0 0 0 0 0 0 0
A =
− −   
   − −   ⎯⎯⎯→
   − −
   
   
0
s t
1
2
3
4
5
2 2 1
2 1 2
1 0
4 0 4
0 1
x s t
x s t
x s ts
x t
x t
− + −       
       +       
       = = = +
       
− −       
             
x
  TB ofkernelfor thebasisone:)1,4,0,2,1(),0,0,1,1,2( −−=

6.27
◼ Corollary to Theorem 6.3:
0x
xx
=
=→
AT
ATRRT mn
ofspacesolutionthetoequalisofkernelThen the
.)(bygiventionfransformalinearthebe:Let
 
( ) (a linear transformation : )
ker( ) ( ) | 0, (subspace of )
n m
n n
T A T R R
T NS A A R R
= →
 = = =  
x x
x x x
※ 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

6.28
◼ Range of a linear transformation T (線性轉換T的值域):
)(rangebydenotedisandofrange
thecalledisinsany vectorofimagesarethatinvectors
allofsetThen then.nsformatiolinear traabe:Let
TT
VW
WVT
w
→
}|)({)(range VTT = vv
※ For the orthogonal projection of
any vector (x, y, z) onto the xy-
plane, 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

6.29
WWVT ofsubspaceais:nnsformatiolinear traaofrangeThe →
◼ Theorem 6.4: The range of T is a subspace of W
Pf:
( ) (Theorem 6.1)T =0 0
WT ofsubsetnonemptyais)(range
haveweand),range(invectorsare)(and)(Since TTT vu
)(range)()()( TTTT +=+ vuvu
)(range)()( TcTcT = uu
Thus, range( ) is a subspace of (according to Theorem 4.5
that a nonempty subset of is a subspace of if it is closed
under vector addition and scalar multiplication)
T W
W W
T is a linear transformation
Range of is closed under vector addition
because ( ), ( ), ( ) range( )
T
T T T T
 
 
+  u v u v
Range of is closed under scalar multi-
plication because ( ) and ( ) range( )
T
T T c T
 
 
 u u
because V+ u v
because c Vu

6.30
◼ Notes:
ofsubspaceis)ker()1( VT
: is a linear transformationT V W→
ofsubspaceis)(range)2( WT
(Theorem 6.3)
(Theorem 6.4)

6.31
◼ Corollary to Theorem 6.4:
)()(rangei.e.,ofspacecolumnthetoequalisofrangeThe
.)(bygivennnsformatiolinear trathebe:Let
ACSTAT
ATRRT mn
=
=→ xx
(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
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.
11 12 1
21 22 2
1 2
1 2
n
n
n
m m mn
a a a
a a a
A x x x
a a a
     
     
     = + + + =
     
     
     
x b
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 . So, the column space
of the matrix A is the same as the range of T, i.e. range(T) = CS(A)
( )CS Ab
( )CS Ab

6.32
◼ Use our example to illustrate the corollary to Theorem 6.4:
※ For the orthogonal projection of any vector (x, y, z) onto the xy-
plane, 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
※ The column space of A is as follows, which is just the xy-plane
1 0 0 1 0 0
0 1 0 , such that 0 1 0
0 0 0 0 0 0 0
x x
A y y
z
       
       = =       
              
1
1 2 3 2 1 2
1 0 0
( ) 0 1 0 , where ,
0 0 0 0
x
CS A x x x x x x R
       
       = + + =        
              

6.33
◼ Ex 7: Finding a basis for the range of a linear transformation
5 4 5
Let : be defined by ( ) ,where is and
1 2 0 1 1
2 1 3 1 0
1 0 2 0 1
0 0 0 2 8
T R R T A R
A
→ =
− 
 
 =
 − −
 
 
x x x
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

6.34
G.-J. E.
1 2 0 1 1 1 0 2 0 1
2 1 3 1 0 0 1 1 0 2
1 0 2 0 1 0 0 0 1 4
0 0 0 2 8 0 0 0 0 0
A B
− −   
   −   = ⎯⎯⎯→ =
   − −
   
   
54321 ccccc 54321 wwwww
 
 
1 2 4 1 2 4
1 2 4 1 2 4
, , and are indepdnent, so , , can
form a basis for ( )
Row operations will not affect the dependency among columns
, , and are indepdnent, and thus , , is
a b
w w w w w w
CS B
c c c c c c


 
asis for ( )
That is, (1, 2, 1, 0), (2, 1, 0, 0), (1, 1, 0, 2) is a basis
for the range of
CS A
T
−

6.35
◼ Rank of a linear transformation T:V→W (線性轉換T的秩):
rank( ) the dimension of the range of dim(range( ))T T T= =
◼ Nullity of a linear transformation T:V→W (線性轉換T的核次數):
nullity( ) the dimension of the kernel of dim(ker( ))T T T= =
◼ Note:
If : is a linear transformation given by ( ) , then
dim(range( )) dim( ( ))rank( ) rank( )
null dim(kerity( ) nullity( )) dim( ( ) () )
n m
T R R T A
T CS A
T NST A
T A
A
=
=
→ =
= =
= =
x x
※ 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 ( ) is called the nullity
of A
( ) { | 0}NS A A= =x x
According to the corollary to Thm. 6.3, ker(T) = NS(A), so dim(ker(T)) = dim(NS(A))
According to the corollary to Thm. 6.4, range(T) = CS(A), so dim(range(T)) = dim(CS(A))

6.36
◼ Theorem 6.5: Sum of rank and nullity
(i.e. dim(range of ) dim(kernel of
rank( )
)
n
dim(domain o
ullity
f ))
)(
T
T
T T
T n
+ =
+ =
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
※ 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)

6.37
Pf:
rAAnmT = )rank(assumeand,matrixanbydrepresentebeLet
(1) rank( ) dim(range of ) dim(column space of ) rank( )T T A A r= = = =
nrnrTT =−+=+ )()(nullity)(rank
(2) nullity( ) dim(kernel of ) dim(null space of )T T A n r= = = −
※ 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
according to Thm. 4.17 where
rank(A) + nullity(A) = n

6.38
◼ Ex 8: Finding the rank and nullity of a linear transformation









 −
=
→
000
110
201
bydefine
:nnsformatiolinear tratheofnullityandranktheFind 33
A
RRT
Sol:
123)(rank)ofdomaindim()(nullity
2)(rank)(rank
=−=−=
==
TTT
AT
※ 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)

6.39
◼ Ex 9: Finding the rank and nullity of a linear transformation
5 7
Let : be a linear transformation
(a) Find the dimension of the kernel of if the dimension
of the range of is 2
(b) Find the rank of if the nullity of is 4
(c) Find the rank of if ke
T R R
T
T
T T
T
→
r( ) { }T = 0
Sol:
(a) dim(domain of ) 5
dim(kernel of ) dim(range of ) 5 2 3
T n
T n T
= =
= − = − =
(b) rank( ) nullity( ) 5 4 1T n T= − = − =
(c) rank( ) nullity( ) 5 0 5T n T= − = − =

6.40
A function : is called one-to-one if the preimage of
every in the range consists of a single vector. This is equivalent
to saying that is one-to-one iff for all and in , ( ) ( )
implies
T V W
T V T T
→
=
w
u v u v
that =u v
◼ One-to-one (一對一):
one-to-one not one-to-one

6.41
◼ Theorem 6.6: One-to-one linear transformation
Let : be a linear transformation.Then
is one-to-one iff ker( ) { }
T V W
T T
→
= 0
Pf:
( )Suppose is one-to-oneT
Then ( ) can have only one solution :T = =v 0 v 0
}{)ker(i.e. 0=T
( )Suppose ker( )={ } and ( )= ( )T T T 0 u v
0vuvu =−=− )()()( TTT
ker( )T −   − =  =u v u v 0 u v
is one-to-one (because ( ) ( ) implies that )T T T = =u v u v
T is a linear transformation, see Property 3 in Thm. 6.1
Due to the fact
that T(0) = 0 in
Thm. 6.1

6.42
◼ Ex 10: One-to-one and not one-to-one linear transformation
(a) The linear transformation : given by ( )
is one-to-one
T
m n n mT M M T A A → =
3 3
(b) The zero transformation : is not one-to-oneT R R→
because its kernel consists of only the m×n zero matrix
because its kernel is all of R3

6.43
inpreimageahasin
elementeveryifontobetosaidis:functionA
VW
WVT →
◼ Onto (映成):
(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
)dim()ofrangedim()(rank WTT ==
The definition of
the rank of a linear
transformation
The definition
of onto linear
transformations

6.44
◼ Theorem 6.8: One-to-one and onto linear transformations
Let : be a linear transformation with vector space and
both of dimension .Then is one-to-one if and only if it is onto
T V W V W
n T
→
Pf:
( ) If is one-to-one, then ker( ) { }and dim(ker( )) 0T T T = =0
)dim())dim(ker())(rangedim(
6.5Thm.
WnTnT ==−=
Consequently, is ontoT
}{)ker(0)ofrangedim())dim(ker(
6.5Thm.
0==−=−= TnnTnT
Therefore, is one-to-oneT
nWTT == )dim()ofrangedim(thenonto,isIf)(
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

6.45
◼ Ex 11:
The linear transformation : is given by ( ) .Find the nullity
and rank of and determine whether is one-to-one, onto, or neither
n m
T R R T A
T T
→ =x x
1 2 0
(a) 0 1 1
0 0 1
A
 
 =  
  
1 2
(b) 0 1
0 0
A
 
 =  
  
1 2 0
(c)
0 1 1
A
 
=  − 
1 2 0
(d) 0 1 1
0 0 0
A
 
 =  
  
Sol:
T:Rn→Rm dim(domain
of T) (1)
rank(T)
(2)
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
= dim(range
of T)
= # of
leading 1’s
= (1) – (2) =
dim(ker(T))
If nullity(T)
= dim(ker(T))
= 0
If rank(T) =
dim(Rm) =
m
= dim(Rn)
= n

6.46
◼ Isomorphism (同構):
othereachtoisomorphicbetosaidare
andthen,tofrommisomorphisanexiststheresuch that
spacesvectorareandifMoreover,m.isomorphisancalledis
ontoandonetooneisthat:nnsformatiolinear traA
WVWV
WV
WVT →
◼ Theorem 6.9: Isomorphic spaces (同構空間) and dimension
Pf:
nVWV dimensionhaswhere,toisomorphicisthatAssume)(
ontoandonetooneisthat:L.T.aexistsThere WVT →
is one-to-oneT
nnTTT
T
=−=−=
=
0))dim(ker()ofdomaindim()ofrangedim(
0))dim(ker(
Two finite-dimensional vector space V and W are isomorphic
if and only if they are of the same dimension
dim(V) = n

6.47
nWV dimensionhavebothandthatAssume)(
is ontoT
nWT == )dim()ofrangedim(
nWV == )dim()dim(Thus
 
 
1 2
1 2
Let , , , be a basis of and
' , , , be a basis of
n
n
B V
B W
=
=
v v v
w w w
nnccc
V
vvvv +++= 2211
asdrepresentebecaninvectorarbitraryanThen
1 1 2 2
and you can define a L.T. : as follows
( ) (by defining ( ) )n n i i
T V W
T c c c T
→
= = + + + =w v w w w v w

6.48
Since 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}
By Theorem 6.5, we can derive that dim(range of T) =
dim(domain of T) – dim(ker(T)) = n –0 = n = dim(W)
Since this linear transformation is both one-to-one and onto,
then V and W are isomorphic
'B
one-to-oneisT
ontoisT

6.49
◼ Ex 12: (Isomorphic vector spaces)
4
(a) 4-spaceR =
4 1(b) space of all 4 1 matricesM  = 
2 2(c) space of all 2 2 matricesM  = 
3(d) ( ) space of all polynomials of degree 3 or lessP x =
5
1 2 3 4(e) {( , , , , 0), are real numbers}(a subspace of )iV x x x x x R=
The following vector spaces are isomorphic to each other
◼ Note
Theorem 6.9 tells us that every vector space with dimension
n is isomorphic to Rn

6.50
◼ 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

Isomorphisms
◼ V, W T:V->W one-to-one and onto (invertible). Then T is an
isomorphism. V,W are isomorphic.
◼ Isomorphic relation is an equivalence relation: V~V, V~W <-
> W~V, V~W, W~U -> V~W.

◼ Theorem 10: Every n-dim vector space over F is isomorphic
to Fn. (noncanonical)
◼ Proof: V n-dimensional
◼ Let B={a1,…,an} be a basis.
◼ Define T:V -> Fn by
◼ One-to-one
◼ Onto

◼ Example: isomorphisms
There will be advantages in looking this way!

6.54
6.3 Matrices for Linear Transformations
1 2 3 1 2 3 1 2 3 2 3(1) ( , , ) (2 , 3 2 ,3 4 )T x x x x x x x x x x x= + − − + − +
◼ Three reasons for matrix representation of a linear transformation:
1
2
3
2 1 1
(2) ( ) 1 3 2
0 3 4
x
T A x
x
−   
   = = − −   
      
x x
◼ It is simpler to write
◼ It is simpler to read
◼ It is more easily adapted for computer use
◼ Two representations of the linear transformation T:R3→R3 :

6.55
◼ Theorem 6.10: Standard matrix for a linear transformation
Let : be a linear trtansformation such thatn m
T R R→
11 12 1
21 22 2
1 2
1 2
( ) , ( ) , , ( ) ,
n
n
n
m m mn
a a a
a a a
T T T
a a a
     
     
     = = =
     
     
     
e e e
1 2 nwhere{ , , , }is a standard basis for .Then the
matrix whose columns correspond to ( ),
n
i
R m n
n T
e e e
e
is such that ( ) for every in , is called the
standard matrix for ( )
n
T A R A
T T
=v v v
的標準矩陣
 
11 12 1
21 22 2
1 2
1 2
( ) ( ) ( ) ,
n
n
n
m m mn
a a a
a a a
A T T T
a a a
 
 
 = =
 
 
 
e e e

6.56
Pf:
1
2
1 2 1 1 2 2
1 0 0
0 1 0
0 0 1
n n n
n
v
v
v v v v v v
v
       
       
       = = + + + = + + +
       
       
      
v e e e
1 1 2 2
1 1 2 2
1 1
is a linear transformation ( ) ( )
( ) ( ) ( )
( )
n n
n n
T T T v v v
T v T v T v
v T
 = + + +
= + + +
= +
v e e e
e e e
e 2 2( ) ( )n nv T v T+ +e e












+++
+++
+++
=
























=
nmnmm
nn
nn
nmnmm
n
n
vavava
vavava
vavava
v
v
v
aaa
aaa
aaa
A









2211
2222121
1212111
2
1
21
22221
11211
v
 1 2If ( ) ( ) ( ) , thennA T T T= e e e

6.57
11 12 1
21 22 2
1 2
1 2
1 1 2 2( ) ( ) ( )
n
n
n
m m mn
n n
a a a
a a a
v v v
a a a
v T v T v T
     
     
     = + + +
     
     
     
= + + +e e e
n
RAT ineachfor)(Therefore, vvv =
◼ 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

6.58
◼ Ex 1: Finding the standard matrix of a linear transformation
3 2
Find the standard matrix for the L.T. : defined byT R R→
)2,2(),,( yxyxzyxT +−=
Sol:
1( ) (1, 0, 0) (1, 2)T T= =e
2( ) (0, 1, 0) ( 2, 1)T T= = −e
3( ) (0, 0, 1) (0, 0)T T= =e
1
1
1
( ) ( 0 )
2
0
T T
 
  = =   
   
e
2
0
2
( ) ( 1 )
1
0
T T
 
−  = =   
   
e
3
0
0
( ) ( 0 )
0
1
T T
 
  = =   
   
e
Vector Notation Matrix Notation

6.59
 1 2 3( ) ( ) ( )
1 2 0
2 1 0
A T T T=
− 
=  
 
e e e
◼ Note: a more direct way to construct the standard matrix
zyx
zyx
A
012
021
012
021
++
+−





 −
=




+
−
=











 −
=








yx
yx
z
y
x
z
y
x
A
2
2
012
021
i.e., ( , , ) ( 2 ,2 )T x y z x y x y= − +
◼ Check:
※ The first (second) row actually
represents the linear transformation
function to generate the first (second)
component of the target vector

6.60
◼ Ex 2: Finding the standard matrix of a linear transformation
2 2
2
The linear transformation : is given by projecting
each point in onto the x -axis.Find the standard matrix for
T R R
R T
→
Sol:
)0,(),( xyxT =
   1 2
1 0
( ) ( ) (1, 0) (0, 1)
0 0
A T T T T
 
= = =  
 
e e
◼ 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

6.61
◼ Composition of T1:Rn→Rm with T2:Rm→Rp :
n
RTTT = vvv )),(()( 12
2 1This composition is denoted by T T T=
◼ Theorem 6.11: Composition of linear transformations (線性轉換
的合成)
1 2
1 2
Let : and : be linear transformations
with standard matrices and ,then
n m m p
T R R T R R
A A
→ →
2 1(1) The composition : , defined by ( ) ( ( )),
is still a linear transformation
n p
T R R T T T→ =v v
2 1
(2) The standard matrix for is given by the matrix productA T
A A A=

6.62
Pf:
(1) ( is a linear transformation)
Let and be vectors in and let be any scalar.Thenn
T
R cu v
2 1(2) ( is the standard matrix for )A A T
)()())(())((
))()(())(()(
1212
11212
vuvu
vuvuvu
TTTTTT
TTTTTT
+=+=
+=+=+
)())(())(())(()( 121212 vvvvv cTTcTcTTcTTcT ====
vvvvv )()())(()( 12121212 AAAAATTTT ====
◼ Note:
1221 TTTT  

6.63
◼ Ex 3: The standard matrix of a composition
3 3
1 2Let and be linear transformations from into s.t.T T R R
),0,2(),,(1 zxyxzyxT ++=
),z,(),,(2 yyxzyxT −=
2 1
1 2
Find the standard matrices for the compositions
and '
T T T
T T T
=
=
Sol:
)formatrixstandard(
101
000
012
11 TA










=
)formatrixstandard(
010
100
011
22 TA









 −
=

6.64
12formatrixstandardThe TTT =
21'formatrixstandardThe TTT =








=















 −
==
000
101
012
101
000
012
010
100
011
12 AAA









 −
=









 −










==
001
000
122
010
100
011
101
000
012
' 21AAA

6.65
◼ Inverse linear transformation (反線性轉換):
1 2If : and : are L.T.s.t.for every inn n n n n
T R R T R R R→ → v
))((and))(( 2112 vvvv == TTTT
invertiblebetosaidisandofinversethecalledisThen 112 TTT
◼ Note:
If the transformation T is invertible, then the inverse is
unique and denoted by T–1

6.66
◼ Theorem 6.12: Existence of an inverse transformation
Let : be a linear transformation with standard matrix ,
Then the following condition are equivalent
n n
T R R A→
◼ Note:
If T is invertible with standard matrix A, then the standard
matrix for T–1 is A–1
(1) T is invertible
(2) T is an isomorphism
(3) A is invertible
※ 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
preimage v, which implies that T-1(w) =v
can be a L.T. and well-defined, so we can
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
output), so T cannot be invertible

6.67
◼ Ex 4: Finding the inverse of a linear transformation
3 3
The linear transformation : is defined byT R R→
)42,33,32(),,( 321321321321 xxxxxxxxxxxxT ++++++=
Sol:
142
133
132
formatrixstandardThe










=A
T
321
321
321
42
33
32
xxx
xxx
xxx
++
++
++
 








=
100142
010133
001132
3IA
Show that T is invertible, and find its inverse

6.68
G.-J. E. 1
1 0 0 1 1 0
0 1 0 1 0 1
0 0 1 6 2 3
I A−
− 
   ⎯⎯⎯→ − =   
 − − 
11
isformatrixstandardtheandinvertibleisTherefore −−
ATT








−−
−
−
=−
326
101
011
1
A








−−
+−
+−
=
















−−
−
−
== −−
321
31
21
3
2
1
11
326326
101
011
)(
xxx
xx
xx
x
x
x
AT vv
)326,,(),,(
s,other wordIn
3213121321
1
xxxxxxxxxxT −−+−+−=−
※ Check T-1(T(2, 3, 4)) = T-1(17, 19, 20) = (2, 3, 4)

6.69
◼ The matrix of T relative to the bases B and B‘:
1 2
: (a linear transformation)
{ , , , } (a nonstandard basis for )
The coordinate matrix of any relative to is denoted by [ ]
n
B
T V W
B V
B
→
= v v v
v v
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,
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)
  '
( ) [ ] ,BB
T A=v v
1
2
1 1 2 2if can be represeted as , then [ ]n n B
n
c
c
c c c
c
  
  
  + + + =
  
   
  
v v v v v

6.70
◼ Transformation matrix for nonstandard bases (the
generalization of Theorem 6.10, in which standard bases are
considered) :
     
11 12 1
21 22 2
1 2' ' '
1 2
( ) , ( ) , , ( )
n
n
nB B B
m m mn
a a a
a a a
T T T
a a a
     
     
     = = =
     
     
     
v v v
1 2
Let and be finite-dimensional vector spaces with bases and ',
respectively,where { , , , }n
V W B B
B = v v v
If : is a linear transformation s.t.T V W→
  '
)(tocorrespondcolumnssematrix whothen the BivTnnm

6.71
  '
is such that ( ) [ ] for every inBB
T A V=v v v
 
11 12 1
21 22 2
1 2
1 2
[ ( )] [ ( )] [ ( )]
n
n
B B n B
m m mn
a a a
a a a
A T T T
a a a
  
 
 
 = =
 
 
 
v v 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)

6.72
◼ Ex 5: Finding a matrix relative to nonstandard bases
2 2
Let : be a linear transformation defined byT R R→
)2,(),( 212121 xxxxxxT −+=
Find the matrix of relative to the basis {(1, 2), ( 1, 1)}
and ' {(1, 0), (0, 1)}
T B
B
= −
=
Sol:
)1,0(3)0,1(0)3,0()1,1(
)1,0(0)0,1(3)0,3()2,1(
−=−=−
+==
T
T
    



−
=−



=
3
0
)1,1(,
0
3
)2,1( '' BB TT
'andtorelativeformatrixthe BBT
     



−
==
30
03
)2,1()2,1( '' BB TTA

6.73
◼ Ex 6:
2 2
For the L.T. : given in Example 5, use the matrix
to find ( ),where (2, 1)
T R R A
T
→
=v v
Sol:
)1,1(1)2,1(1)1,2( −−==v
  



−
=
1
1
Bv
    



=



−



−
==
3
3
1
1
30
03
)( ' BB AT vv
)3,3()1,0(3)0,1(3)( =+= vT )}1,0(),0,1{('=B
)}1,1(),2,1{( −=B
)3,3()12(2),12()1,2( =−+=T
◼ Check:

6.74
◼ Notes:
(1) In the special case where (i.e., : ) and ',
the matrix is called the matrix of relative to the basis
( )
V W T V V B B
A T B
T B
= → =
對應於基底的矩陣
     
1 2
1 2
(2) If : is the identity transformation
{ , , , }: a basis for
the matrix of relative to the basis
1 0 0
0 1 0
( ) ( ) ( )
0 0 1
n
n nB B B
T V V
B V
T B
A T T T I
→
=

 
 
  = = =   
 
 
v v v
v v v

6.75
◼ 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

6.76
6.4 Transition Matrices and Similarity
1 2
1 2
: ( a linear transformation)
{ , , , } ( a basis of )
' { , , , } (a basis of )
n
n
T V V
B V
B V
→
=
=
v v v
w w w
     1 2( ) ( ) ( ) ( matrix of relative to )nB B B
A T T T T B =  v v v
     1 2' ' '
' ( ) ( ) ( ) (matrix of relative to ')nB B B
A T T T T B =  w w w
     1 2 ( transition matrix from ' to )nB B B
P B B =  w w w
     1
1 2' ' '
( transition matrix from to ')nB B B
P B B−
 =  v v v
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
       1
' '
, andB B B B
P P−
 = =v v v v
       ' '
( ) , and ( )B B B B
T A T A = =v v v v
(T相對於B的矩陣)
(T相對於B’的矩陣)
(從B'到B的轉移矩陣)
(從B到B’的轉移矩陣)

6.77
◼ Two ways to get from to :
' '(1) (direct) : '[ ] [ ( )]B BA T=v v
  'Bv   ')( BT v
1 1
' '(2) (indirect): [ ] [ ( )] 'B BP AP T A P AP− −
=  =v v
direct
indirect

6.78
◼ Ex 1: (Finding a matrix for a linear transformation)
Sol:
  



−
=−=−=
1
3
)0,1()1,1(1)0,1(3)1,2()0,1( 'BTT
22
：formatrixtheFind RRTA' →
)3,22(),( 212121 xxxxxxT +−−=
)}1,1(),0,1{('basisthetoreletive =B
  


−
=+−==
2
2
)1,1()1,1(2)0,1(2)2,0()1,1( 'BTT
     



−
−
==
21
23
)1,1()0,1(' '' BB TTA
   ' '
(1) ' (1, 0) (1, 1)B B
A T T =  

6.79
(2) standard matrix for (matrix of relative to {(1, 0), (0, 1)})T T B =
  



−
−
==
31
22
)1,0()0,1( TTA
     





==
10
11
)1,1()0,1(
to'frommatrixtransition
BBP
BB
   1
' '
transition matrix from to '
1 1
(1, 0) (0, 1)
0 1B B
B B
P− − 
 = =   
 






−
−
=











−
−





 −
== −
21
23
10
11
31
22
10
11
'
'relativeofmatrix
1
APPA
BT
※ 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)
※ 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)

6.80
Sol:
2 2 2
Let {( 3, 2), (4, 2)} and ' {( 1, 2), (2, 2)}be basis for
2 7
, and let be the matrix for : relative to .
3 7
Find the matrix of relative to '
B B
R A T R R B
T B
= − − = − −
− 
= → − 
     





−
−
=−−=
12
23
)2,2()2,1(:to'frommatrixtransition BBPBB
     





−
−
=−−=−
32
21
)2,4()2,3(:'tofrommatrixtransition ''
1
BBPBB






−
=





−
−






−
−






−
−
== −
31
12
12
23
73
72
32
21
'
:'torelativeofmatrix
1
APPA
BT
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

6.81
Sol:
 
   
 
2 2
'
'
For the linear transformation : given in Ex.2, find ,
( ) , and ( ) , for the vector whose coordinate matrix is
3
1
B
B B
B
T R R
T T
→
− 
=  − 
v
v v v
v
    



−
−
=



−
−




−
−
==
5
7
1
3
12
23
'BB P vv
    



−
−
=



−
−




−
−
==
14
21
5
7
73
72
)( BB AT vv
    


−
=



−
−




−
−
== −
0
7
14
21
32
21
)()( 1
' BB TPT vv
    




−
=





−
−






−
==
0
7
1
3
31
12
')(or '' BB AT vv

6.82
◼ 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) (the transition matrix from to is the )n n nA I AI A A I=
1 1 1 1 1
1 1
(2) ( )
(by defining ), thus is similar to
A P BP PAP P P BP P PAP B
Q AQ B Q P B A
− − − − −
− −
=  =  =
 = =

6.83
◼ Ex 4: (Similar matrices)
2 2 3 2
(a) and ' are similar
1 3 1 2
A A
− −   
= =   − −   






== −
10
11
where,'because 1
PAPPA
2 7 2 1
(b) and ' are similar
3 7 1 3
A A
−   
= =   − −   






−
−
== −
12
23
where,'because 1
PAPPA

6.84
◼ Ex 5: (A comparison of two matrices for a linear transformation)
3 3
1 3 0
Suppose 3 1 0 is the matrix for : relative
0 0 2
to the standard basis.Find the matrix for relative to the basis
' {(1, 1, 0), (1, 1, 0), (0, 0, 1)}
A T R R
T
B
 
 = → 
 − 
= −
Sol:
      










−=−=
100
011
011
)1,0,0()0,1,1()0,1,1(
matrixstandardthetofrommatrixntransitioThe
BBBP
B'








−= −
100
0
0
2
1
2
1
2
1
2
1
1
P

6.85
1 1
2 2
1 1 1
2 2
matrix of relative to ':
0 1 3 0 1 1 0
' 0 3 1 0 1 1 0
0 0 1 0 0 2 0 0 1
4 0 0
0 2 0
0 0 2
T B
A P AP−
     
     = = − −     
     −    
 
 = − 
 − 
※ 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.
(A’is a diagonal matrix, which is
simple and with some
computational advantages)

6.86
◼ Notes: Diagonal matrices have many computational advantages
over nondiagonal ones (it will be used in the next chapter)
1
2
0 0
0 0
for
0 0 n
d
d
D
d
 
 
 =
 
 
 
1
2
0 0
0 0
(1)
0 0
k
k
k
k
n
d
d
D
d
 
 
 =
 
 
  
(2) T
D D=
1
2
1
1
1
1
0 0
0 0
(3) , 0
0 0 n
d
d
i
d
D d−
 
 
 
=  
 
 
 

6.87
◼ 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

Linear transformations and matrices

More Related Content

What's hot (20)

Similar to Linear transformations and matrices (20)

More from EasyStudy3 (20)

Recently uploaded (20)

Linear transformations and matrices