linear tranformation- vc&la
TRANSCRIPT
Linear TransformationBy Kaushal Patel
2
Linear Transformation
Zero transformation:
VWVT vu, ,:
VT vv ,0)( Identity transformation:
VVT : VT vvv ,)( 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
3
The Kernel and Range of a Linear Transformation
› Kernel of a linear transformation T:Let be a linear transformationWVT :
Then the set of all vectors v in V that satisfy is called the kernel of T and is denoted by ker(T).
0)( vT
} ,0)(|{)ker( VTT vvv Ex 1: (Finding the kernel of a linear transformation)
):( )( 3223 MMTAAT T
Sol:
00
00
00
)ker(T
4
The kernel is a subspace of V
The kernel of a linear transformation is a
subspace of the domain V.
)16. Theorem( 0)0( TPf:VT ofsubset nonempty a is )ker(
then. of kernel in the vectorsbe and Let Tvu
000)()()( vuvu TTT
00)()( ccTcT uu )ker(Tc u
)ker(T vu
. of subspace a is )ker(Thus, VT
Note:
The kernel of T is sometimes called the nullspace of T.
WVT :
5
Ex 6: (Finding a basis for the kernel)
82000
10201
01312
11021
and Rin is where,)(by defined be :Let 545
A
ATRRT xxx
Find a basis for ker(T) as a subspace of R5.
6
Sol:
000000041000020110010201
082000010201001312011021
0
.. EJG
A
s t
14
021
00112
4
22
5
4
3
2
1
ts
tt
ststs
xxxxx
x
TB of kernel for the basis one:)1 ,4 ,0 ,2 ,1(),0 ,0 ,1 ,1 ,2(
7
Range of a linear transformation T:
)(by denoted is and T of range thecalled is Vin
vector of images arein W that w vectorsall ofset Then the
L.T. a be :Let
Trange
WVT
}|)({)( VTTrange vv
8
. :Tn nsformatiolinear tra a of range The WWV fo ecapsbus a si
The range of T is a subspace of W
Pf:)1Thm.6.( 0)0( T
WTrange ofsubset nonempty a is )(
TTT of range in the vector be )( and )(Let vu
)()()()( TrangeTTT vuvu
)()()( TrangecTcT uu
),( VVV vuvu
)( VcV uu
.subspace is )( Therefore, WTrange
9
Notes:
of subspace is )()1( VTKer
L.T. a is : WVT
of subspace is )()2( WTrange
10
Ex 7: (Finding a basis for the range of a linear transformation)
82000
10201
01312
11021
and is where,)(by defined be :Let 545
A
RATRRT xxx
Find a basis for the range of T.
11
Sol:
BA EJG
0000 04100 02011 01020 1
82000102010131211021
..
54321 ccccc54321 wwwww
)(for basis a is , ,
)(for basis a is , ,
421
421
ACSccc
BCSwww
T of range for the basis a is )2 ,0 ,1 ,1( ),0 ,0 ,1 ,2( ),0 ,1 ,2 ,1(
12
Rank of a linear transformation T:V→W:
TTrank of range theofdimension the)( Nullity of a linear transformation T:V→W:
TTnullity of kernel theofdimension the)(
Note:
)()(
)()(
then,)(by given L.T. thebe :Let
AnullityTnullity
ArankTrank
ATRRT mn
xx
Rank and Nullity of Linear Transformation
13
then. space vector a into
space vector ldimensiona-nan form L.T. a be :Let
W
VWVT
Sum of rank and nullity
Pf:
AmatrixnmT an by drepresente is Let
) ofdomain dim() of kerneldim() of rangedim(
)()(
TTT
nTnullityTrank
rArank )( Assume
rArank
ATTrank
)(
) of spacecolumn dim() of rangedim()((1)
nrnrTnullityTrank )()()(
rn
ATTnullity
) of spacesolution dim() of kerneldim()()2(
14
Ex 8: (Finding the rank and nullity of a linear transformation)
000
110
201
by define : L.T. theofnullity andrank theFind 33
A
RRT
Sol:
123)() ofdomain dim()(
2)()(
TrankTTnullity
ArankTrank
15
Ex 9: (Finding the rank and nullity of a linear transformation)
}0{)( if ofrank theFind )(
4 is ofnullity theif ofrank theFind )(
2 is range theof
dimension theif of kernel theofdimension theFind )(
n.nsformatiolinear tra a be :Let 75
TKerTc
TTb
Ta
RRT
Sol:
325) of rangedim() of kerneldim(
5) ofdomain dim()(
TnT
Ta
145)()()( TnullitynTrankb
505)()()( TnullitynTrankc
16
One-to-one Transformation
vector.single a of consists range in the every w
of preimage theif one-to-one called is :function A WVT
. that implies
)()( inV, vandu allfor iff one-to-one is
vu
vu
TTT
one-to-one not one-to-one
17
Onto Transformation
in preimage a has in
element every if onto be tosaid is :function A
V
WVT
w
(T is onto W when W is equal to the range of T.)
18
One-to-one linear transformation
}0{)( iff 1-1 is TThen
L.T. a be :Let
TKer
WVT
Pf:1-1 is Suppose T
0 :solution oneonly havecan 0)(Then vvT
}0{)( i.e. TKer
)()( and }0{)( Suppose vTuTTKer
0)()()( vTuTvuT
L.T. a is T0)( vuTKervu
1-1 is T
19
Ex 10: (One-to-one and not one-to-one linear transformation)
one.-to-one is
)(by given : L.T. The )( Tmnnm AATMMTa
matrix. zero only the of consists kernel its Because nm
one.-to-onenot is :ation transformzero The )( 33 RRTb . of all is kernel its Because 3R
20
Thm 6.7: (Onto linear transformation)
. ofdimension the toequal is ofrank theiff onto is Then
l.dimensiona finite is whereL.T., a be :Let
WTT
WWVT
Thm 6.8: (One-to-one and onto linear transformation)
onto. isit ifonly and if one-to-one is Then .dimension
ofboth and spaceor with vectL.T. a be :Let
Tn
WVWVT
Pf:0))(dim( and }0{)( then one,-to-one is If TKerTKerT
)dim())(dim())(dim( WnTKernTrange onto. is ly,Consequent T
0) of rangedim())(dim( nnTnTKerone.-to-one is Therefore,T
nWTT )dim() of rangedim( then onto, is If
21
Ex 11:
neither.or onto, one,-to-one is whether determine and of
rank andnullity theFind ,)(by given is : L.T. The
TT
ATRRT mn xx
100110021
)( Aa
001021
)( Ab
110021
)( Ac
000110021
)( Ad
Sol:
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
22
Isomorphism:
other.each toisomorphic be tosaid are
and then , to from misomorphisan exists theresuch that
spaces vector are and if Moreover, m.isomorphisan called is
onto and one toone is that :n nsformatiolinear traA
WVWV
WV
WVT
Isomorphic spaces and dimension
Pf:.dimension has where, toisomorphic is that Assume nVWV
onto. and one toone is that : L.T. a exists There WVT one-to-one is T
nnTKerTT
TKer
0))(dim() ofdomain dim() of rangedim(
0))(dim(
Two finite-dimensional vector space V and W are isomorphic if
and only if they are of the same dimension.
Bijective Transformation
23
.dimension haveboth and that Assume nWV
onto. is T
nWT )dim() of rangedim(
nWV )dim()dim( Thus
. of basis a be , , ,let
and V, of basis a be , , ,Let
21
21
Wwww
vvv
n
n
nnvcvcvc
V
2211
as drepresente becan in vector arbitrary an Then
v
nnwcwcwcT
WVT
2211)(
follows. as : L.T. a definecan you and
v
It can be shown that this L.T. is both 1-1 and onto.Thus V and W are isomorphic.
24
Inverse linear Transformation
in every for s.t. L.T. are : and : If 21nnnnn RRRTRRT v
))(( and ))(( 2112 vvvv TTTT
invertible be tosaid is and of inverse thecalled is Then 112 TTT
Note:
If the transformation T is invertible, then the inverse is
unique and denoted by T–1 .
25
Existence of an inverse transformation
.equivalent arecondition following Then the
,matrix standard with L.T. a be :Let ARRT nn
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.
26
Finding the inverse of a linear transformation
by defined is ： L.T. The 33 RRT
)42 ,33 ,32(),,( 321321321321 xxxxxxxxxxxxT
Sol:
142
133
132
for matrix standard The
A
T
321
321
321
42
33
32
xxx
xxx
xxx
100142010133001132
3IA
Show that T is invertible, and find its inverse.
27
1..
326100101010011001
AIEJG
11 is for matrix standard theand invertible is Therefore ATT
326101011
1A
321
31
21
3
2
111
326326101011
)(xxx
xxxx
xxx
AT vv
)326 , ,(),,(
s,other wordIn
32131213211 xxxxxxxxxxT
28
.for matrix standard theFind axis.- x theonto in point each
projectingby given is :n nsformatiolinear tra The2
22
TR
RRT
Sol:
)0 ,(),( xyxT
0001
)1 ,0()0 ,1()()( 21 TTeTeTA
Notes: (1) The standard matrix for the zero transformation from Rn into Rm
is the mn zero matrix.
(2) The standard matrix for the zero transformation from Rn into Rn
is the nn identity matrix In
Finding the matrix of a linear transformation
29
Composition of T1:Rn→Rm with T2:Rm→Rp :nRTTT vvv )),(()( 12
112 ofdomain ofdomain , TTTTT
Composition of linear transformations
then, and matrices standardwith
L.T. be : and :Let
21
21
AA
RRTRRT pmmn
L.T. a is )),(()(by defined ,:n compositio The(1) 12 vv TTTRRT pn
12product matrix by thegiven is for matrix standard The )2( AAATA
30
Pf:
nscalar theany be clet and in vectorsbe and Let
L.T.) a is ((1)nR
T
vu
)for matrix standard theis )(2( 12 TAA
)()())(())((
))()(())(()(
1212
11212
vuvu
vuvuvu
TTTTTT
TTTTTT
)())(())(())(()( 121212 vvvvv cTTcTcTTcTTcT
vvvvv )()())(()( 12121212 AAAAATTTT
Note:
1221 TTTT
31
The standard matrix of a compositions.t. into from L.T. be and Let 33
21 RRTT
) ,0 ,2(),,(1 zxyxzyxT ) ,z ,(),,(2 yyxzyxT
,' and
nscompositio for the matrices standard theFind
2112 TTTTTT Sol:
)for matrix standard(
101
000
012
11 TA
)for matrix standard(
010
100
011
22 TA
32
12for matrix standard The TTT
21'for matrix standard The TTT
000101012
101000012
010100011
12 AAA
001
000
122
010
100
011
101
000
012
' 21AAA
33
Matrices for Linear Transformations
)43,23,2(),,()1( 32321321321 xxxxxxxxxxxT
Three reasons for matrix representation of a linear transformation:
3
2
1
430231112
)()2(xxx
AT xx
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 :
34
Definition 1: A nonzero vector x is an eigenvector (or characteristic vector) of a square matrix A if there exists a scalar λ such that Ax = λx. Then λ is an eigenvalue (or characteristic value) of A.
Note: The zero vector can not be an eigenvector even though A0 = λ0. But λ = 0 can be an eigenvalue.
Example:
Show x 2
1
is an eigenvector for A
2 4
3 6
Solution : Ax 2 4
3 6
2
1
0
0
But for 0, x 02
1
0
0
Thus,x is an eigenvector of A,and 0 is an eigenvalue .
Definitions
35
An n×n matrix A multiplied by n×1 vector x results in another n×1 vector y=Ax. Thus A can be considered as a transformation matrix.
In general, a matrix acts on a vector by changing both its magnitude and its direction. However, a matrix may act on certain vectors by changing only their magnitude, and leaving their direction unchanged (or possibly reversing it). These vectors are the eigenvectors of the matrix.
A matrix acts on an eigenvector by multiplying its magnitude by a factor, which is positive if its direction is unchanged and negative if its direction is reversed. This factor is the eigenvalue associated with that eigenvector.
Geometric interpretation of Eigenvalues and Eigenvectors
36
Let x be an eigenvector of the matrix A. Then there must exist an eigenvalue λ such that Ax = λx or, equivalently,
Ax - λx = 0 or
(A – λI)x = 0
If we define a new matrix B = A – λI, then
Bx = 0
If B has an inverse then x = B-10 = 0. But an eigenvector cannot be zero.
Thus, it follows that x will be an eigenvector of A if and only if B does not have an inverse, or equivalently det(B)=0, or
det(A – λI) = 0
This is called the characteristic equation of A. Its roots determine the eigenvalues of A.
Eigenvalues
37
Eigenvalues: examples
Example 1: Find the eigenvalues of
two eigenvalues: 1, 2
Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2 =…= λk. If that happens, the eigenvalue is said to be of multiplicity k.
51
122A
)2)(1(23
12)5)(2(51
122
2
AI
38
Eigenvectors
Example 1 (cont.):
00
41
41
123)1(:1 AI
0,1
4
,404
2
11
2121
ttx
x
txtxxx
x
00
31
31
124)2(:2 AI
0,1
3
2
12
ss
x
xx
To each distinct eigenvalue of a matrix A there will correspond at least one eigenvector which can be found by solving the appropriate set of homogenous equations. If λi is an eigenvalue then the corresponding eigenvector xi is the
solution of (A – λiI)xi = 0
39
Example 2 (cont.): Find the eigenvectors of
Recall that λ = 2 is an eigenvector of multiplicity 3.
Solve the homogeneous linear system represented by
Let . The eigenvectors of = 2 are of the form
and t not both zero.
0
0
0
000
000
010
)2(
3
2
1
x
x
x
AI x
txsx 31 ,
,
1
0
0
0
0
1
0
3
2
1
ts
t
s
x
x
x
x
200
020
012
A
40
Definition: The trace of a matrix A, designated by tr(A), is the sum of the elements on the main diagonal.
Property 1: The sum of the eigenvalues of a matrix equals the trace of the matrix.
Property 2: A matrix is singular if and only if it has a zero eigenvalue.
Property 3: The eigenvalues of an upper (or lower) triangular matrix are the elements on the main diagonal.
Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ is an eigenvalue of matrix A-1.
Properties of Eigenvalues and Eigenvectors
41
Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of kA where k is any arbitrary scalar.
Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of Ak for any positive integer k.
Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT.
Property 9: The product of the eigenvalues (counting multiplicity) of a matrix equals the determinant of the matrix.
42
Theorem: Eigenvectors corresponding to distinct (that is, different) eigenvalues are linearly independent.
Theorem: If λ is an eigenvalue of multiplicity k of an n n matrix A then the number of linearly independent eigenvectors of A associated with λ is given by m = n - r(A- λI). Furthermore, 1 ≤ m ≤ k.
Example 2 (cont.): The eigenvectors of = 2 are of the form
s and t not both zero.
= 2 has two linearly independent eigenvectors
,
1
0
0
0
0
1
0
3
2
1
ts
t
s
x
x
x
x
Linearly independent eigenvectors
43
Diagonalization
Diagonalizable matrix:
A square matrix A is called diagonalizable if there exists an
invertible matrix P such that P-1AP is a diagonal matrix.
(P diagonalizes A) Notes:
(1) If there exists an invertible matrix P such that ,
then two square matrices A and B are called similar.
(2) The eigenvalue problem is related closely to the
diagonalization problem.
APPB 1
44
A diagonalizable matrix
200013031
A
Sol: Characteristic equation:
0)2)(4(200
013031
I 2
A
2 ,2 ,4 :sEigenvalue 321
Ex.5) p.403 (See
0
1
1
:rEigenvecto 4)1( 1
p
45
1
0
0
,
0
1
1
:igenvectorE 2)2( 32
pp
200
020
004
100
011
011
][ 1321 APPpppP
400
020
002
010
101
101
][ (2)
200
040
002
100
011
011
][ (1)
1132
1312
APPpppP
APPpppP
Notes:
46
Condition for Diagonalization
An nn matrix A is diagonalizable if and only if
it has n linearly independent eigenvectors. Pf:
ablediagonaliz is )( A
),,,( and ][Let
diagonal is s.t. invertiblean exists there
2121
1
nn diagDpppP
APPDP
][
00
0000
][
2211
2
1
21
nn
n
n
ppp
pppPD
47
][][2121 nn
ApApAppppAAP
) of rseigenvecto are of tor column vec the..(
,,2 ,1 ,
APpei
nipAp
PDAP
i
iii
t.independenlinearly are ,,, invertible is 21 npppP
rs.eigenvectot independenlinearly has nA
n
npppnA
,, seigenvalue ingcorrespondh wit
,, rseigenvectot independenlinearly has )(
21
21
nipAp iii ,,2 ,1 , i.e.
][Let 21 n
pppP
48
PDppp
ppp
ApApAp
pppAAP
n
n
nn
n
n
00
00
00
][
][
][
][
2
1
21
2211
21
21
ablediagonaliz is
invertible is tindependenlinearly are ,,,1
11
A
DAPP
Pppp n
Note: If n linearly independent vectors do not exist,
then an nn matrix A is not diagonalizable.
49
A matrix that is not diagonalizable
10
21
able.diagonaliznot ismatrix following that theShow
A
Sol: Characteristic equation:
0)1(10
21I 2
A
1 :Eigenvalue 1
0
1 :rEigenvecto
00
10~
00
20I 1pAIA
A does not have two (n=2) linearly independent eigenvectors,
so A is not diagonalizable.
50
Steps for diagonalizing an nn square matrix:
n ,,, 21
Step 2: Let ][ 21 npppP
Step 1: Find n linearly independent eigenvectors
for A with corresponding eigenvalues
nppp ,,, 21
Step 3:
nipApDAPP iii
n
,,2 ,1 , where ,
00
00
00
2
1
1
Note:
The order of the eigenvalues used to form P will determine
the order in which the eigenvalues appear on the main diagonal of D.
51
Diagonalizing a matrix
diagonal. is such that matrix a Find
113
131
111
1APPP
A
Sol: Characteristic equation:
0)3)(2)(2(113
131111
I
A
3 ,2 ,2 :sEigenvalue 321
52
21
000010101
~313111
111I1 A
1
0
1
:rEigenvecto
1
0
1
0 1
3
2
1
pt
t
t
x
x
x
22
0001001
~113151
113I 4
141
2 A
4
1
1
:rEigenvecto
4
1
1
241
41
41
3
2
1
pt
t
t
t
x
x
x
53
33
000110
101~
413101
112I3 A
1
1
1
:rEigenvecto
1
1
1
3
3
2
1
pt
t
t
t
x
x
x
300
020
002
141
110
111
][Let
1
321
APP
pppP
54
Notes: k is a positive integer
kn
k
k
k
n d
dd
D
d
dd
D
00
0000
00
0000
)1( 2
1
2
1
1
1
1
1111
111
1
1
)()()(
)())((
)(
)2(
PPDA
PAP
APAAP
APPPPPAPPAP
APPAPPAPP
APPD
APPD
kk
k
kk
55
Sufficient conditions for Diagonalization
If an nn matrix A has n distinct eigenvalues, then the
corresponding eigenvectors are linearly independent and
A is diagonalizable.
56
Determining whether a matrix is diagonalizable
300
100
121
A
Sol: Because A is a triangular matrix,
its eigenvalues are the main diagonal entries.
3 ,0 ,1 321
These three values are distinct, so A is diagonalizable. (Thm.7.6)
57
Finding a diagonalizing matrix for a linear transformation
diagonal. is torelative
for matrix thesuch that for basis a Find
)33()(
bygiven n nsformatiolinear tra thebe Let
3
321321321321
33
B
TRB
xxx, xx, xxxx,x,xxT
RT:R
Sol:
113
131
111
)()()(
bygiven is for matrix standard The
321 eTeTeTA
T
From Ex. 5, there are three distinct eigenvalues
so A is diagonalizable. (Thm. 7.6)3,2 ,2 321
58
300
020
002
][ ][ ][
][ ][ ][
])([ )]([ ])([
332211
321
321
BBB
BBB
BBB
ppp
ApApAp
pTpTpTD
The matrix for T relative to this basis is
)}1 ,1 ,1(),4 ,1 ,1(),1 ,0 ,1{(},,{ 321 pppB
Thus, the three linearly independent eigenvectors found in Ex. 5
can be used to form the basis B. That is
)1 ,1 ,1(),4 ,1 ,1(),1 ,0 ,1( 321 ppp