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Chapter 6 Eigenvalues
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Outlines
System of linear ODE (Omit) Diagonalization Hermitian matrices Ouadratic form Positive definite matrices
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Motivations To simplify a linear dynamics such
that it is as simple as possible. To realize a linear system characteristics
e.g.,
the behavior of system dynamics.
x Ax
zecharacteripofroot
bapeqch
byyay
)(
0)(..
0
2
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§6-1 Eignvalues & Eignvectors
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Example: In a town, each year 30% married women get divorced 20% single women get married In the 1st year, 8000 married women 2000
single women. Total population remains constant
, where
represent the numbers of married & single
women after i years, respectively.
,
,
Let ,m ii
s i
ww
w
, ,&m i s iw w
0 1
8000 0.7 0.2 &
2000 0.3 0.8i iw w w
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If
Question: Why does converge?
Why does it converge to the same limit vector even when the initial condition is different?
1 0
1 12 12
0.7 0.2 8000,
0.3 0.8 2000
6000 4000, , & , 12
4000 6000
i i i
n
w Aw w w
w W w w n
0 14
14
10000 4000 (steady-state vector)
0 6000
, 14i
w w
w w i
iw
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Ans: Choose a basis
Given an initial for some
for example,
Question: How does one know choosing such a basis?
1 2 1 1 2 2
2 1 1, &
3 1 2x x Ax x Ax x
1
1)4000(
3
22000
2000
8000
0 0 1 1 2 2w w c x c x 1 2&c c
0 1 1 2 2 1 1 n n nA w c A x c A x c x as n
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Def: Let . A scalar is said to be an
eigenvalue or characteristic value of A if
such that
The vector is said to be an eigenvector
or characteristic vector belonging to .
is called an eigen pair of A.
Question: Given A, How to compute eigenvalues
& eigenvectors?
Ax x
nnFA
0x
x
( , )x
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is an eigen pair of
Note that, is a polynomial, called
characteristic polynomial of A, of degree n in
Thus, by FTA, A has exactly n eigenvalues including
multiplicities. is a eigenvector associated
with eigenvalue while is eigenspace of A.
, 0
( ) 0, 0
( ) is singular
det( ) 0
Ax x x
A I x x
A I
A I
nnFA
.( ) det( )p A I
( ) \ 0x N A I
)( IAN
( , )x
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Example:
To find the eigenspace of 2:( i.e., )
2
4 2Let
1 1
( ) det( )
4 2 det
1 1
5 6 ( 2)( 3)
2 & 3 are eignvalues of
A
p A I
A
1
2
2 2 0( 2 ) 0
1 1 0
1( 2 )
1
xA I x
x
N A I span
)2( IAN
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To find the eigenspace of 3(i.e., )
Let
1
2
1 2 0( 3 ) 0
1 2 0
2( 3 )
1
xA I x
x
N A I span
)3( IAN
1
1 2
1 1
2 0
0 3
p
p Ap
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Let , then
( ) is an eignvalue of ;
( ) ( ) 0 has a nontrivil solution;
( ) ( ) 0
( ) is singular;
( ) det( ) 0
( ) ( ) < ;
( ) ( ) 1
i A
ii A I x
iii N A I
iv A I
v A I
vi rank A I n
vii Nullity A I
nnFA
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Let .
If is an eigenvalue of A with eigenvector
Then
This means that is also an eigen-pair of
A.
Ax x
Ax Ax Ax x x
nnA
.x
C
( , )x
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Let .
where are eigenvalues of A.
(i) Let
(ii) Compare with the coefficient of
, we have
nnA
1( ) det( ) ( )
n
ii
p A I
n 1
n
iiA
1
det0
1 1( 1)n n )(
11
Atracean
iii
n
ii
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Theorem 6.1.1: Let A & B be n×n matrices, if B is similar to A, then and consequently A & B have the same eigenvalues.
Pf: Let for some nonsingular matrix S.
1
1
1
( ) det( )
det( )
det ( )
det( )det( )det( )
det( ) ( )
B
A
P B I
S AS I
S A I S
S A I S
A I P
)()( BA PP
ASSB 1
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§6-3 Diagonalization
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Diagonalization
Goal: Given find a nonsingular matrix
S, such that is a diagonal
matrix.
Question1: Are all matrices diagonalizable?
Question2: What kinds of A are diagonalizable?
Question3: How to find S if A is diagonalizable?
nnFA
DASS 1
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NOT all matrices are diagonalizable
e.g., Let
If A is diagonalizable,
nonsingular matrix S,
00
10A
2
11
0
0
d
dASS
1 2
1 2
1 2
1 1
2
( ) 0
det( ) 0
0
0
0
trace A d d
A d d
d d
dA S S O
d
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To answer Q2,
Suppose that A is diagonalizable.
nonsingular matrix S,
Let
This gives a condition for diagonalizability and a way to
find S.
1 nS S S
AS SD
)( 11
ndddiagDASS
1 nAS AS
1 1 n nd S d S
, , 1, , , are eigenpairs of A. i id S i n
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Theorem6.3.1: If are distinct eigenvalues of an
matrix A with corresponding eigenvectors ,
then are linearly independent.
Pf: Suppose that are linearly dependent
not all zero,
Suppose that
.... golw
2 1
1 12
( ) 0
( ) 0
kk
i i ii i
k
ii
A I c x
c x
1, , k nn
1
0k
i ii
c x
1, , kx x
1, , kx x
1, , kx x
1 kc c
1 0c
1
1
1
contradiction!
0 & are distinct.
0 ( )
are linear independent.
!i
k
x
c
x x
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Theorem 6.3.2: Let is diagonalizable
A has n linearly independent eigenvectors.
Note : Similarity transformation
Change of coordinate
diagonalization
( ) ( )
V V
E E
L
A
B
1A SBS
( ) ( )
V V
F F
LSI
nnFA
y Px
1I S
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Remarks: Let , and
(i) is an eigenpair of A for
(ii) The diagonalizing matrix S is not unique because
Its columns can be reordered or multiplied by
an nonzero scalar
(iii) If A has n distinct eigenvalues , A is diagonalizable.
If the eigenvalues are not distinct , then may or may
not diagonalizable depending on whether A has n
linearly independent eigenvectors or not.
(iv)
1SDSA )( 1 ndiagD 1 nS S S
i iS
ni ,,1
1 11 , .k k k k
nA SD S S diag S k
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Example: Let
For
For
Let
112
202
213
A
det 0 0, 1, 1A I
1
1
1
)0(,0 SpanIAN
1
2
0
,
0
2
1
)(,1 SpanIAN
1
1 1 0 0 0 0
1 2 2 0 1 0
1 0 1 0 0 1
S D S AS
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Def: If an matrix A has fewer than n linearly independent eigenvectors,we say that A is defective.
e.g.
(i) is defective
(ii) is defective
1 1 0
0 1 1
0 0 1
A
n n
00
10A
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Example 4: Let A & B both have the same eigenvalues Nullity (A-2I)=1 The eigenspace associated with has only one dimension. A is NOT diagonalizable However, Nullity (B-2I)=2 B is diagonalizable
263
041
002
&
201
040
002
BA
4,2,2
2
001
020
000
)2(
rankIArank
1
063
021
000
)2(
rankIBrank
2
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Question:
Are the following matrices diagonalizable ?
00
00
01
)(
00
10
01
)( BiiAi
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The Exponential of a Matrix
Motivation:The general solution of
is
The unique solution of
is
Question: What is and how to compute ?
.
x Ax
( ) Atx t e c
0 0( )
x Ax
x t x
0( )0( ) A t tx t e x
Ate Ate
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Note that
Define
0 !i
ia
i
ae
......!2!
2
0
AAI
i
Ae
i
iA
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Suppose that A is diagonalizable with
1SDSA
1 ,k kA SD S k
1
1
0 0
1
1
( )! !
0
0
n
i i i iAt
i i
t
t
Dt
A t D te S S
i i
e
S S
e
Se S
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Example 6: Compute
Sol: The eigenvalues of A are
with eigenvectors
0,1
1 2 3 1 0 ;
1 1 0 0A XDX X D
1 2
2 3
1 1x and x
2 6, for
1 3Ate A
1 10
0 1
3 2 6 6
1 3 2
A D ee Xe X X X
e e
e e
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§6-4 Hermitian Matrices
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Hermitian matrices
Let , then A can be written as
, where
e.g. , 2 3
4 7 2
2 3 1 1
4 7 1 2
i iA
i i
i
m nA C iCBA , m nB C R
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Let , then
e.g. ,
ii
ii
ii
iiH
273
42
274
32
m nA C TH AA
HHH
HHH
HH
ACAC
BABA
AA
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Def:
(a) A is said to be Hermitian if
(b) A is said to be skew-Hermitian if
(c) A is said to be unitary if
( i.e. its column vectors form an orthonormal set in )
HAA
AAH
H HA A I AA nC
Let n nA F
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Theorem6.4.1: Let , then
(i)
(ii) eigenvectors belonging to distinct eigenvalues
are orthogonal.
Pf : (i) Let be an eigenpair of A,
(ii) Let and be two eigenpairs of A,
HAA RA
, x
2 2 2
( )
H H
H H H H H
Ax x x Ax x x
x Ax x A x x Ax
x x x
1 1, x 2 2, x
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Theorem6.4.1 Pf :
(ii) Let and be two eigenpairs of A,1 1, x 2 2, x
1 1 2 1 1 2 1 2 1 2
1 2 2 1 2
1 2
1 2 1 2 1 2
=( ) ( )
, 0
H HH H H
H H
H
x x x x Ax x x A x
x Ax x x
x x x x x x
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Theorem: Let and
then
Pf : (i) Let be an eigenpair of A,
HA A be an eignvalue of ,A is purely imaginary.
( , )x 2
2 2 2
2
( ) =
is pure-imaginary.
H H
H H H H H
HH
H
x Ax x x x
x Ax x A x x A x
x x x
x Ax
x
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Theorem6.4.3 (Schur’s Theorem): Let , then unitary matrix , is upper triangular.
Pf : The proof is by mathematical induction on n. (i) The result is obvious if n=1;
(ii) Assume the hypothesis holds for k×k matrices;
(iii) let A be a (k+1)×(k+1) matrix.
nnCA AUU HU
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Proof of Schur’s Theorem
Let be an eigenpair of A with
Using the Gram-Schmidt process, construct an orthonormal
basis of
Let
1
0
0
H
k k
W AWM
1 1,w 1 1.w
1 2 1{ , , , }kw w w 1.k
1 2 1 ,kW w w w
1 1 1 1 1 is unitary and H HW W Aw W w e
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Proof of Schur’s Theorem By the induction hypothesis (ii)
1 1 1 1 1 unitary , , where is upper triangular.k k HV V MV T T
1
1 0 0
0Define ,
0
VV
1
1 1
1
0
is unitary and
0
H H H
T
V V W AWV TV MV
Let is unitary and .HU WV U U AU T
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Theorem6.4.4: (Spectral Theorem)
If , then unitary matrix U that diagonalizes A .
Pf : By Theorem 6.4.3 ,
unitary matrix ,
where T is upper triangular .
T is a diagonal matrix.
AAH
TAUUU H
TAUU
UAUAUUTH
HHHHH
AAH
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Cor.6.4.5: Let A be real symmetric matrix . Then (i) (ii) an orthogonal matrix U, is a diagonal matrix.proof :
RA
TU AU
is real symmetric
is Hermitian, and its eignvalues must be real
and eignvectors may be chosen to be real.
By Th 6.4.4 ,
the diagonalizing matrix U must be orthogonal.
A
A
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Example 4:
Find an orthogonal matrix U that diagonalizes A.
Sol : (i)
(ii)
021
232
120
ALet
1 2 1( 5 ) , ,
6 6 6
( ) 1 , 0 , 1 , 2 , 1 , 0
T
T T
N A I Span
N A I Span
2( ) det( ) ( 1) (5 )
1, 1, 5
p A I
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Example 4:
Sol :
(iii) By Gram-Schmidt process,
and 5, 1, 1 .TU AU diag
1 1 1 1 1( ) , 0, , , ,
2 2 3 3 3
TT
N A I span
3
1 1 1
6 2 32 1
The columns of 0 form an orthonormal eigenbasis of 6 3
1 1 1
6 2 3
U
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Question:In addition to Hermitian matrices ,
is there any other matrices possessing
orthonormal eigenbasis?
Note : If A has orthonormal eigenbasis
where U is unitary &diagonal.HA UDU
H H H H
H H H H
H H H H
A A UD U UDU
UD DU UDD U
UDU UD U AA
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Def: A is said to be normal if
Remark : Hermitian, Skew- Hermitian and Unitary
matrices are all normal.
HH AAAA
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Theorem6.4.6: A is normal A possesses an orthonormal eigenbasisPf :
If A has an orthonormal eigenbasis, then ,where U is unitary & diagonal.
""
HA UDU
H H H H
H H H H
H H H H
A A UD U UDU
UD DU UDD U
UDU UD U AA
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proof of Theorem6.4.6: By Th.6.4.3, unitary U,
Compare the diagonal elements of
0,
is diagonal.
ijT i j
T
"" is upper triangular.
T is also normal.
H
H H H H
T U AU
A A AA T T TT
22
11 11
2 22
2 21 2
2 2
1
n
jj
n
i ji j
n
in nni
t t
t t
t t
has an orthonormal eigenbasis (WHY?)A
&H HT T TT