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Ex 1: (Verifying eigenvalues and eigenvectors)
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7 - 1
5.1 Eigenvalues and Eigenvectors 5.2 Diagonalization 5.3 Symmetric Matrices and Orthogonal Diagonalizatio
n
Chapter 5 Eigenvalues and EigenvectorsChapter 5 Eigenvalues and Eigenvectors
7 - 2
5.1 Eigenvalues and Eigenvectors Eigenvalue problem:
If A is an nn matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x?
Eigenvalue and eigenvector:
A : an nn matrix : a scalarx: a nonzero vector in Rn
xAx
Eigenvalue
Eigenvector
Geometrical Interpretation
7 - 3
Ex 1: (Verifying eigenvalues and eigenvectors)
10
02A
01
1x
11 20120
201
1002 xAx
Eigenvalue
22 )1(1011
010
1002 xAx
Eigenvalue
Eigenvector
Eigenvector
10
2x
7 - 4
Thm 5.1: (The eigenspace of A corresponding to )
If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of .
Pf:x1 and x2 are eigenvectors corresponding to
) , ..( 2211 xAxxAxei
) toingcorrespondr eigenvectoan is ..( )()( )1(
21
21212121
λxxeixxxxAxAxxxA
) toingcorrespondr eigenvectoan is ..( )()()()( )2(
1
1111
cxeicxxcAxccxA
7 - 5
Ex 3: (An example of eigenspaces in the plane)
Find the eigenvalues and corresponding eigenspaces of
1001
A
yx
yx
A 1001
v
If ) ,(v yx
0
100
1001 xxx
For a vector on the x-axis Eigenvalue 11
Sol:
7 - 6
For a vector on the y-axis
yyy0
100
1001
Eigenvalue 12
Geometrically, multiplying a vector (x, y) in R2 by the matrix A corresponds to a reflection in the y-axis.
The eigenspace corresponding to is the x-axis.
The eigenspace corresponding to is the y-axis.
11
12
7 - 7
Thm 5.2: (Finding eigenvalues and eigenvectors of a matrix AMnn )
0)Idet( A(1) An eigenvalue of A is a scalar such that .
(2) The eigenvectors of A corresponding to are the nonzero
solutions of .
Characteristic polynomial of AMnn:
11 1 0det( I ) ( I ) n n
nA A c c c L
Characteristic equation of A:
0)Idet( A
0)I( xA
Let A is an nn matrix.
If has nonzero solutions iff . 0)I( xA 0)Idet( A0)I( xAxAx
Note:(homogeneous system)
7 - 8
Ex 4: (Finding eigenvalues and eigenvectors)
51
122A
Sol: Characteristic equation:
0)2)(1(23
51122
)I(
2
A
Eigenvalue: 2 ,1 21
2 ,1
7 - 9
2)2( 2
0 ,133
00
31124
)I(
2
1
2
12
tttt
xx
xx
xA
1)1( 1
0 ,144
00
41123
)I(
2
1
2
11
tttt
xx
xx
xA
7 - 10
200020012
A
Sol: Characteristic equation:
0)2(200
020012
I 3
A
Eigenvalue: 2
Ex 5: (Finding eigenvalues and eigenvectors)
Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue?
7 - 11
The eigenspace of A corresponding to :2
000
000000010
)I(3
2
1
xxx
xA
0, ,100
001
03
2
1
tsts
t
s
xxx
2 toingcorrespondA of eigenspace the:,100
001
Rtsts
Thus, the dimension of its eigenspace is 2.
7 - 12
Notes:
(1) If an eigenvalue 1 occurs as a multiple root (k times) for
the characteristic polynominal, then 1 has multiplicity k.
(2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.
7 - 13
Ex 6 : Find the eigenvalues of the matrix A and find a basis
for each of the corresponding eigenspaces.
30010201105100001
A
Sol: Characteristic equation:
0)3)(2()1(
30010201
105100001
I
2
A
Eigenvalue: 3 ,2 ,1 321
7 - 14
1)1( 1
0000
20010101
105000000
)I(
4
3
2
1
1
xxxx
xA
0, ,
1202
0010
2
2
4
3
2
1
tsts
tt
st
xxxx
1202
,
0010
is a basis for the eigenspace of A corresponding to 1
7 - 15
2)2( 2
0000
10010001
105100001
)I(
4
3
2
1
2
xxxx
xA
0 ,
0150
0
50
4
3
2
1
tttt
xxxx
0150
is a basis for the eigenspace of A corresponding to 2
7 - 16
3)3( 3
0000
00010101
105200002
)I(
4
3
2
1
3
xxxx
xA
0 ,
105
0
050
4
3
2
1
tt
t
t
xxxx
105
0
is a basis for the eigenspace of A corresponding to 3
7 - 17
Thm 5.3: (Eigenvalues for triangular matrices)If A is an nn triangular matrix, then its eigenvalues are the entries on its main diagonal.
Ex 7: (Finding eigenvalues for diagonal and triangular matrices)
335011002
)( Aa
3000004000000000002000001
)( Ab
Sol:)3)(1)(2(
335011002
I )(
Aa
3 ,1 ,2 321
3 ,4 ,0 ,2 ,1 )( 54321 b
7 - 18
Eigenvalues and eigenvectors of linear transformations:
. of eigenspace thecalled is vector)zero (with the of rseigenvecto all setof theand
, toingcorrespond ofr eigenvectoan called is vector The.)(such that vector nonzero a is thereif :n nsformatiolinear tra a of eigenvaluean called is number A
TTVVT
xxxx
7 - 19
Ex 8: (Finding eigenvalues and eigenspaces)
.200
013031
seigenspace ingcorrespond and seigenvalue theFind
A
Sol:
200013031
AI )4()2( 2
2 ,4 seigenvalue 21
2for Basis )}1 ,0 ,0(),0 ,1 ,1{(4for Basis )}0 ,1 ,1{(follows. as are seigenvalue twofor these seigenspace The
22
11
BB
7 - 20
Notes:
)}1 ,0 ,0(),0 ,1 ,1(),0 ,1 ,1{('
diagonal. is,basis the torelative ofmatrix the,Then 8. Ex.in found rseigenvectot independen
linear threeof up made of basis thebelet and 8, Ex.in A is
matrix standard n whosensformatiolinear tra thebe Let (1)3
33
B
B'TA'
RB'
RT:R
200020004
'A
A of rsEigenvecto A of sEigenvalue
. of seigenvalue theare matrix theof entries diagonalmain The (2) AA'
7 - 21
Keywords in Section 5.1:
eigenvalue problem: 特徵值問題 eigenvalue: 特徵值 eigenvector: 特徵向量 characteristic polynomial: 特徵多項式 characteristic equation: 特徵方程式 eigenspace: 特徵空間 multiplicity: 重數
7 - 22
5.2 Diagonalization Diagonalization problem:
For a square matrix A, does there exist an invertible matrix P such that P-1AP is diagonal?
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
7 - 23
Thm 5.4: (Similar matrices have the same eigenvalues)
If A and B are similar nn matrices, then they have the same eigenvalues.
Pf:
APPBBA 1similar are and
A
APPAPPPAP
PAPAPPPPAPPB
I
III
)I(III111
1111
Thus A and B have the same eigenvalues.
7 - 24
Ex 1: (A diagonalizable matrix)
200013031
A
Sol: Characteristic equation:
0)2)(4(200
013031
I 2
A
2 ,2 ,4s:eigenvalue The 321
reigenvecto the4)1(
011
1p
7 - 25
r eigenvecto the2)2(
100
,01
132 pp
200020004
such that
, 100011011
][
1
321
APP
pppP
200040002
100011011
][
1
312
APP
pppP Note: If
7 - 26
Thm 5.5: (Condition for diagonalization)
An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.
Pf:ablediagonaliz is )( A
there exists an invertible such that is diagonalLet and
1
1 2 1 2[ ] ( , , , )n n
P D P APP p p p D diag
L L
1
21 2
1 1 2 2
0 00 0
[ ]
0 0[ ]
n
n
n n
PD p p p
p p p
LLL M M O ML
L
7 - 27
1 2[ ]nAP Ap Ap ApAP PD
Q L
are eigenvectors of , 1, 2, ,
( . . the column vector of )i i i
i
Ap p i ni e p P A
K
is invertible are linearly independent.1 2 , , , nP p p pQ Lrs.eigenvectot independenlinearly has nA
has linearly independent eigenvectors 1 2
1 2
( ) , , with corresponding eigenvalues , ,
n
n
A n p p p
LL
, 1, 2, , i i iAp p i n K
Let 1 2[ ]nP p p p L
7 - 28
1 2 1 2
1 1 2 2
1
21 2
[ ] [ ][ ]
0 00 0
[ ]
0 0
n n
n n
n
n
AP A p p p Ap Ap App p p
p p p PD
L LL
LLL M M O ML
are linearly independent is invertible
is diagonalizable
1 1
1
, , , np p p P
P AP DA
Q L
7 - 29
Ex 4: (A matrix that is not diagonalizable)
1021
able.diagonaliznot ismatrix following that theShow
A
Sol: Characteristic equation:
0)1(1021I 2
A
1:eigenvalue The 1
1
0 2 0 1 1I ~ eigenvector
0 0 0 0 0A I A p
A does not have two linearly independent eigenvectors, so A is not diagonalizable.
7 - 30
Steps for diagonalizing an nn square matrix:
Step 2: Let1 2[ ]nP p p p L
Step 1: Find n linearly independent eigenvectors
for A with corresponding eigenvalues.1 2, , np p pL
Step 3: 1
21
0 00 0
0 0 n
P AP D
LL
M M O ML
where, , 1, 2, , i i iAp p i n K
7 - 31
Ex 5: (Diagonalizing a matrix)
diagonal. is such that matrix a Find
113131111
1APPP
A
Sol: Characteristic equation:
0)3)(2)(2(113
131111
I
A
3 ,2 ,2s:eigenvalue The 321
7 - 32
21
000010101
~313111
111I1 A
101
r eigenvecto 0 1
3
2
1
pt
t
xxx
22
0001001
~113151
113I 4
141
2 A
41
1r eigenvecto 24
141
3
2
1
pt
tt
xxx
7 - 33
33
000110
101~
413101
112I3 A
111
r eigenvecto 3
3
2
1
pttt
xxx
300020002
s.t.
, 141110111
][
1
321
APP
pppP
7 - 34
Notes:
1 1
2 2
0 0 0 00 0 0 0
(1)
0 0 0 0
k
kk
kn n
d dd d
D D
d d
L LL L
M M O M M M O ML L
1
11
)2(
PPDA
PAPDAPPDkk
kk
7 - 35
Thm 5.6: (Sufficient conditions for diagonalization)
If an nn matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and A is diagonalizable.
Ex 7: (Determining whether a matrix is diagonalizable)
300100121
A
Sol: Because A is a triangular matrix, its eigenvalues are
3 ,0 ,1 321
These three values are distinct, so A is diagonalizable.
7 - 36
Ex 8: (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
BTRB
xxx, xx, xxxx,x,xxTRT:R
Sol:
113131111
bygiven is for matrix standard The
A
T
From Ex. 5 you know that A is diagonalizable. Thus, the three linearly independent eigenvectors found in Ex. 5 can be used to form the basis B. That is
7 - 37
)}1 ,1 ,1(),4 ,1 ,1(),1 ,0 ,1{( B
300020002
][ ][ ][][ ][ ][
321
321
BBB
BBB
AvAvAvvTvTvTD
The matrix for T relative to this basis is
7 - 38
Keywords in Section 5.2:
diagonalization problem: 對角化問題 diagonalization: 對角化 diagonalizable matrix: 可對角化矩陣
7 - 39
5.3 Symmetric Matrices and Orthogonal Diagonalization
Symmetric matrix:A square matrix A is symmetric if it is equal to its transpose:
TAA
Ex 1: (Symmetric matrices and nonsymetric matrices)
502031210
A
13
34B
501041123
C
(symmetric)
(symmetric)
(nonsymmetric)
7 - 40
Thm 5.7: (Eigenvalues of symmetric matrices)
If A is an nn symmetric matrix, then the following properties are true.
(1) A is diagonalizable.
(2) All eigenvalues of A are real.
(3) If is an eigenvalue of A with multiplicity k, then has k linearly independent eigenvectors. That is, the eigenspace of has dimension k.
7 - 41
Ex 2: Prove that a symmetric matrix is diagonalizable.
bccaA
Pf: Characteristic equation:
0)( 22
cabbabccaAI
022
222
22222
4)(
42
442)(4)(
cba
cbaba
cabbabacabba
As a quadratic in , this polynomial has a discriminant of
7 - 42
04)( (1) 22 cba
0 , cba
diagonal. ofmatrix a is 0
0
aa
A
04)( )2( 22 cba
The characteristic polynomial of A has two distinct real roots, which implies that A has two distinct real eigenvalues. Thus, A is diagonalizable.
7 - 43
A square matrix P is called orthogonal if it is invertible and
Orthogonal matrix:
TPP 1
Thm 5.9: (Properties of symmetric matrices)
Let A be an nn symmetric matrix. If 1 and 2 are distinct eig
envalues of A, then their corresponding eigenvectors x1 and x2 a
re orthogonal.
Thm 5.8: (Properties of orthogonal matrices)
An nn matrix P is orthogonal if and only if its column vectors form an orthogonal set.
7 - 44
Ex 5: (An orthogonal matrix)
535
534
532
51
52
32
32
31
0P
Sol: If P is a orthogonal matrix, then I 1 TT PPPP
I100010001
00
535
32
534
51
32
532
52
31
535
534
532
51
52
32
32
31
TPP
535
32
3
5345
132
2
53252
31
1 0 , ,Let ppp
7 - 45
1
0 roducesp
321
323121
ppp
pppppp
set. lorthonormaan is } , ,{ 321 ppp
7 - 46
Thm 5.10: (Fundamental theorem of symmetric matrices)
Let A be an nn matrix. Then A is orthogonally diagonalizable and has real eigenvalue if and only if A is symmetric.
Orthogonal diagonalization of a symmetric matrix:Let A be an nn symmetric matrix.(1) Find all eigenvalues of A and determine the multiplicity of each.(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector. (3) For each eigenvalue of multiplicity k2, find a set of k linearly indepen
dent eigenvectors. If this set is not orthonormal, apply Gram-Schmidt orthonormalization process.
(4) The composite of steps 2 and 3 produces an orthonormal set of n eigenvectors. Use these eigenvectors to form the columns of P. The matrix will be diagonal.DAPPAPP T 1
7 - 47
Ex 7: (Determining whether a matrix is orthogonally diagonalizable)
111101111
1A
081812125
2A
102
0233A
2000
4A
Orthogonally diagonalizable
Symmetricmatrix
7 - 48
Ex 9: (Orthogonal diagonalization)
142412222
A. esdiagonaliz that matrix orthogonalan Find
A
P
Sol:0)6()3( )1( 2 AI
)2 ofty multiplici a has( 3 ,6 21
) , ,( )2 ,2 ,1( ,6 )2( 32
32
31
1
1111
vvuv
)1 ,0 ,2( ),0 ,1 ,2( ,3 )3( 322 vv
Linear Independent
7 - 49
Gram-Schmidt Process:
)1 , ,( ),0 ,1 ,2( 54
52
222
233322
wwwwvvwvw
) , ,( ),0 , ,(53
553
453
2
3
335
15
2
2
22
wwu
wwu
535
32
534
51
32
532
52
31
0P
300030006
1APP
7 - 50
Keywords in Section 5.3:
symmetric matrix: 對稱矩陣 orthogonal matrix: 正交矩陣 orthonormal set: 單範正交集 orthogonal diagonalization: 正交對角化