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Free video Lecture
By:video.edhole.com
Chapter 7
Eigenvalues and Eigenvectors
7.1 Eigenvalues and Eigenvectors
7.2 Diagonalization
7.3 Symmetric Matrices and Orthogonal Diagonalization
Elementary Linear Algebra 投影片設計編製者R. Larsen et al. (6 Edition) 淡江大學 電機系 翁慶昌 教授
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7.1 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 scalar
x: a nonzero vector in Rn
xAx
Eigenvalue
Eigenvector
Geometrical Interpretation
Elementary Linear Algebra: Section 7.1, pp.421-422
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Ex 1: (Verifying eigenvalues and eigenvectors)
10
02A
01
1x
11 201
202
01
1002
xAx
Eigenvalue
22 )1(10
11
010
1002
xAx
Eigenvalue
Eigenvector
Eigenvector
10
2x
Elementary Linear Algebra: Section 7.1, p.423
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Thm 7.1: (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
λxxei
xxxxAxAxxxA
) toingcorrespondr eigenvectoan is ..(
)()()()( )2(
1
1111
cxei
cxxcAxccxA
Elementary Linear Algebra: Section 7.1, p.424
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Ex 3: (An example of eigenspaces in the plane)
Find the eigenvalues and corresponding eigenspaces of
10
01A
y
x
y
xA
10
01v
If ) ,(v yx
0
100
10
01 xxx
For a vector on the x-axis Eigenvalue 11
Sol:
Elementary Linear Algebra: Section 7.1, p.425
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For a vector on the y-axis
yyy
01
00
10
01
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
Elementary Linear Algebra: Section 7.1, p.425
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Thm 7.2: (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:
011
1)I()Idet( cccAA nn
n
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)
Elementary Linear Algebra: Section 7.1, p.426
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Ex 4: (Finding eigenvalues and eigenvectors)
51
122A
Sol: Characteristic equation:
0)2)(1(23
51
122)Idet(
2
A
2 ,1 :sEigenvalue 21
2 ,1
Elementary Linear Algebra: Section 7.1, p.426
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2)2( 2
0 ,1
33
00
31~
31
124
0
0
31
124)I(
2
1
2
12
ttt
t
x
x
x
xxA
1)1( 1
0 ,1
44
00
41~
41
123
0
0
41
123)I(
2
1
2
11
ttt
t
x
x
x
xxA
Elementary Linear Algebra: Section 7.1, p.426
:Check xAxi
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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?
Elementary Linear Algebra: Section 7.1, p.428
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The eigenspace of A corresponding to :2
0
0
0
000
000
010
)I(
3
2
1
x
x
x
xA
0, ,
1
0
0
0
0
1
0
000
000
010
~
000
000
010
3
2
1
tsts
t
s
x
x
x
2 toingcorrespondA of eigenspace the:,
1
0
0
0
0
1
Rtsts
Thus, the dimension of its eigenspace is 2.
Elementary Linear Algebra: Section 7.1, p.428
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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.
Elementary Linear Algebra: Section 7.1, p.428
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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
3 ,2 ,1 :sEigenvalue 321 Elementary Linear Algebra: Section 7.1, p.429
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1)1( 1
0
0
0
0
2001
0101
10500
0000
)I(
4
3
2
1
1
x
x
x
x
xA
0, ,
1
2
0
2
0
0
1
0
2
2
0000
0000
2100
2001
~
2001
0101
10500
0000
4
3
2
1
tsts
t
t
s
t
x
x
x
x
1
2
0
2
,
0
0
1
0
is a basis for the eigenspace of A corresponding to 1
Elementary Linear Algebra: Section 7.1, p.429
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2)2( 2
0
0
0
0
1001
0001
10510
0001
)I(
4
3
2
1
2
x
x
x
x
xA
0 ,
0
1
5
0
0
5
0
0000
1000
0510
0001
~
1001
0001
10510
0001
4
3
2
1
tt
t
t
x
x
x
x
0
1
5
0
is a basis for the eigenspace of A corresponding to 2
Elementary Linear Algebra: Section 7.1, p.429
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3)3( 3
0
0
0
0
0001
0101
10520
0002
)I(
4
3
2
1
3
x
x
x
x
xA
0 ,
1
0
5
0
0
5
0
0000
0100
5010
0001
~
0001
0101
10520
0002
4
3
2
1
tt
t
t
x
x
x
x
1
0
5
0
is a basis for the eigenspace of A corresponding to 3
Elementary Linear Algebra: Section 7.1, p.429
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Thm 7.3: (Eigenvalues of 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)
335
011
002
)( Aa
3000004000000000002000001
)( Ab
Sol:
)3)(1)(2(
335
011
002
I )(
Aa
3 ,1 ,2 321
3 ,4 ,0 ,2 ,1 )( 54321 bElementary Linear Algebra: Section 7.1, p.430
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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
T
TVVT
x
xxx
Elementary Linear Algebra: Section 7.1, p.431
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Ex 8: (Finding eigenvalues and eigenspaces)
.
200
013
031
seigenspace ingcorrespond and seigenvalue theFind
A
Sol:
)4()2(
200
013
0312
AI
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
B
B
Elementary Linear Algebra: Section 7.1, p.431
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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'
Elementary Linear Algebra: Section 7.1, p.431
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Keywords in Section 7.1:
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
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7.2 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
Elementary Linear Algebra: Section 7.2, p.435
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Thm 7.4: (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(III
111
1111
A and B have the same characteristic polynomial. Thus A and B have the same eigenvalues.
Elementary Linear Algebra: Section 7.2, p.436
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Ex 1: (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
Elementary Linear Algebra: Section 7.2, p.436
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Ex.5) p.403 (See
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:
Elementary Linear Algebra: Section 7.2, p.436
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Thm 7.5: (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
Elementary Linear Algebra: Section 7.2, p.437
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][][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
Elementary Linear Algebra: Section 7.2, p.438
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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
Elementary Linear Algebra: Section 7.2, p.438
Note: If n linearly independent vectors do not exist,
then an nn matrix A is not diagonalizable.video.edhole.com
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Ex 4: (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.Elementary Linear Algebra: Section 7.2, p.439
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Steps for diagonalizing an nn square matrix:
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
Elementary Linear Algebra: Section 7.2, p.439
n ,,, 21
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.video.edhole.com
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Ex 5: (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
Elementary Linear Algebra: Section 7.2, p.440
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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
Elementary Linear Algebra: Section 7.2, p.440
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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
Elementary Linear Algebra: Section 7.2, p.440
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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
Elementary Linear Algebra: Section 7.2, p.445
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Thm 7.6: (Sufficient conditions for diagonalization)
If an nn matrix A has n distinct eigenvalues, then the
corresponding eigenvectors are linearly independent and
A is diagonalizable.
Elementary Linear Algebra: Section 7.2, p.442
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Ex 7: (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)
Elementary Linear Algebra: Section 7.2, p.443
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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
B
TRB
xxx, xx, xxxx,x,xxT
RT:R
Sol:
113
131
111
)()()(
bygiven is for matrix standard The
321 eTeTeTA
T
Elementary Linear Algebra: Section 7.2, p.443
From Ex. 5, there are three distinct eigenvalues
so A is diagonalizable. (Thm. 7.6)
3,2 ,2 321
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300
020
002
][ ][ ][
][ ][ ][
])([ )]([ ])([
332211
321
321
BBB
BBB
BBB
ppp
ApApAp
pTpTpTD
The matrix for T relative to this basis is
Elementary Linear Algebra: Section 7.2, p.443
)}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
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Keywords in Section 7.2:
diagonalization problem: 對角化問題 diagonalization: 對角化 diagonalizable matrix: 可對角化矩陣
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7.3 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
1334
B
501041123
C
(symmetric)
(symmetric)
(nonsymmetric)
Elementary Linear Algebra: Section 7.3, p.447
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Thm 7.7: (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.
Elementary Linear Algebra: Section 7.3, p.447
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Ex 2: Prove that a symmetric matrix is diagonalizable.
bcca
A
Pf: Characteristic equation:
0)( 22
cabbabc
caAI
022
222
22222
4)(
42
442)(4)(
cba
cbaba
cabbabacabba
As a quadratic in , this polynomial has a discriminant of
Elementary Linear Algebra: Section 7.3, p.448
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04)( (1) 22 cba
0 , cba
diagonal. ofmatrix a is 0
0
a
aA
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.
Elementary Linear Algebra: Section 7.3, p.448
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A square matrix P is called orthogonal if it is invertible and
Orthogonal matrix:
TPP 1
Elementary Linear Algebra: Section 7.3, p.449
. 01
10 because orthogonal is
01
10 (a) 1
TPPP
Ex 4: (Orthogonal matrices)
.
5
30
5
40105
40
5
3
because orthogonal is
5
30
5
40105
40
5
3
(b) 1
TPPP
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Thm 7.8: (Properties of orthogonal matrices)
An nn matrix P is orthogonal if and only if
its column vectors form an orthogonal set.
Elementary Linear Algebra: Section 7.3, p.450
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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
0
0
535
32
534
51
32
532
52
31
535
534
532
51
52
32
32
31
TPP
535
32
3
534
51
32
2
532
52
31
1 0 , ,Let ppp
Elementary Linear Algebra: Section 7.3, p.451
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1
0
roducesp
321
323121
ppp
pppppp
set. lorthonormaan is } , ,{ 321 ppp
Elementary Linear Algebra: Section 7.3, p.451
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Thm 7.9: (Properties of symmetric matrices)
Let A be an nn symmetric matrix. If 1 and 2 are distinct
eigenvalues of A, then their corresponding eigenvectors x1
and x2 are orthogonal.
Elementary Linear Algebra: Section 7.3, p.452
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.orthogonal are seigenvaluedistinct toingcorrespond
01
10 of rseigenvecto any two that Show
A
Ex 6: (Eigenvectors of a symmetric matrix)
Sol: Characteristic function
0)4)(2(8631
13I 2
A
4 ,2 :sEigenvalue21
0 ,1
1 x
00
11~
11
11I2 (1)
111
ssA
0 ,1
1 x
00
11~
11
11I4 (2)
222
ttA
.orthogonal are xand x 0xx2121
stst
t
t
s
s
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Thm 7.10: (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 independent 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
Elementary Linear Algebra: Section 7.3, p.453
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Ex 7: (Determining whether a matrix is orthogonally diagonalizable)
111101111
1A
081812125
2A
102023
3A
2000
4A
Orthogonally diagonalizable
Symmetricmatrix
Elementary Linear Algebra: Section 7.3, p.454
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Ex 9: (Orthogonal diagonalization)
142
412
222
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
v
vuv
)1 ,0 ,2( ),0 ,1 ,2( ,3 )3( 322 vv
Linear IndependentElementary Linear Algebra: Section 7.3, p.455
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Gram-Schmidt Process:
)1 , ,( ),0 ,1 ,2( 54
52
222
233322
www
wvvwvw
) , ,( ),0 , ,(53
553
453
2
3
335
15
2
2
22
w
wu
w
wu
535
32
534
51
32
532
52
31
321
0
] [ (4) pppP
300
030
006
1 APPAPP T
Elementary Linear Algebra: Section 7.3, p.456
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Keywords in Section 7.3:
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
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