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Positive Semidefinite matrix
nHA
nCzAzz 0*
A is a positive semidefinite matrix
(also called nonnegative definite matrix)
Positive definite matrix
nHA
nCzAzz 0*
A is a positive definite matrix
Negative semidefinite matrix
nHA
nCzAzz 0*
A is a negative semidefinite matrix
Negative definite matrix
nHA
nCzAzz 0*
A is a negative definite matrix
Positive semidefinite matrix
nT RxAxx 0
A is a positive semidefinite matrix
A is real symmetric matrix
Positive definite matrix
nT RxAxx 0
A is a positive definite matrix
A is real symmetric matrix
Question
nT RxAxx 0nCzAzz 0*
Yes
Is It true that
)(RMALetn
?
Proof of Question
)(
)()()()(
))((
)()(
)(
,,
,
0)(
)(
***
*
yAxAxyiAyyAxx
AxiyyAxiAyyAxx
yAxixAyiyAyxAx
yAyyAixxAiyxAx
yiAxAiyx
iyxAiyx
zAzAzz
iyxz
Ryxwhereiyxz
writecanweCzanyFor
RxAxxthatAssume
clearisIt
TTTT
TTTT
TTTTTTTTTTTT
TTTTTTTT
TTTT
TTT
T
TT
n
n
nT
?
Proof of Question
)(
))((
)()(*
AxyyAxiAyyAxx
AxiyyAxiAyyAxx
AyyyAixAxiyAxx
iAyxAiyx
iyxAiyxAzz
TTTT
TTTT
TTTT
TT
TT
?
Fact 1.1.6
The eigenvalues of a Hermitian (resp. positive semidefinite , positive definit
e) matrix are all real (resp. nonnegative, positive)
Proof of Fact 1.1.6
edeifnegativeisAif
esemideifnegativeisAif
edeifpositiveisAif
esemideifpositiveisAif
andRz
Azz
numberrealaisAzz
AzzzAzAzzSince
zwherez
Azz
zzzAzz
CzzAz
thenreigenvectongcorresponithebez
andAofeigenvalueanbeLet
HALet
n
n
int0
int0
int0
int0
,
.
,)(
0,
0,
,
.
2
*
*
*****
2
2
*
2**
Exercise n
nCzRAzzHA *
)(CMHn
nCzHzz 0*
From this exercise we can redefinite:
H is a positive semidefinite
注意 )(RMA
n
nT RxAxx 0
A is symmetric
注意 之反例
2
2
1
1
2
21
2
1
210
01
10R
01
10But is not
symmetric
Proof of Exercise
n
n
n
n
n
n
n
HAHence
AA
AA
CzzAAz
CzAzzzAz
CzAzzAzzthen
CzRAzzthatAssume
numberrealaisAzz
thenAzzzAzAzz
CzanyFor
HAthatAssume
0
0
0)(
)(
)(
.
,)(
.)(
*
*
**
***
***
*
*
*****
Remark
Let A be an nxn real matrix. If λ is a real eigenvalue of A, then there must exist a corresponding real eigenvector.
However, if λ is a nonreal eigenvalue of A, then it cannot have a real eigenvector.
Explain of Remark p.1
A, λ : real Az= λz, 0≠z (A- λI)z=0 By Gauss method, we obtain that z is a real vector.
Explain of Remark p.2
A: real, λ is non-real Az= λz, 0≠z z is real, which is impossible
Elementary symmetric function
nnS
21211),,,(
21211
212),,,(
iinii
nS
kiii
nkiiink
S
21211
21),,,(
kth elementary symmetric function
KxK Principal Minor
nxnijaALet
niiianyFork
211
kikiikiiki
kiiiiii
kiiiiii
aaa
aaa
aaa
21
22212
12111
det
kxk principal minor of A
Lemma p.1
nMALet
AofvectorcolumnithbeaLeti
niiianyFork
211
kj
kj
j iiijife
iiijifabLet
,,,
,,,
21
21
vectordardsithbeeLeti
tan
Lemma p.2
nbbbThen
21det
kiiibyindexedcolumnsand
rowswithorprincipalthe
,,,
min
21
Explain Lemma
4442
3432
2422
4442
3432
2422
1412
0
1
0
det
00
10
00
01
det
aa
aa
aa
aa
aa
aa
aa
4442
2422detaa
aa
The Sum of KxK Principal Minors
Aoforsprincipalkxkall
ofsumthebeAELetk
min
)(
Theorem
nMALet
AofpolynomialsticcharacterithebexcLetA)(
in
n
n
ii
in
AtSttcThen
),,,()1()(
211
AofseigenvaluethebeLetn
,,,21
inn
ii
in tAEt
)()1(1
Proof of Theorem p.1
inn
ini
in
in
ijjj
n
i nijjj
n
nA
tSt
tt
ttttc
121
211 211
21
),,,()1(
)())((
)())(()()1(
Proof of Theorem p.2
LemmapreviousbytAEt
jjjkifte
jjjkifabwhere
bbbt
ateateate
AtItc
eeeILet
Aofvectorcolumnithisawhere
aaaALet
inn
ii
in
ik
ik
k
n
n
i nijjj
n
nn
A
n
i
n
,)()1(
,,,
,,,
det
det
)det()(
,)2(
1
21
21
211 211
2211
21
21
Rank P.1
rankA:=the maximun number of linear independent column vectors =the dimension of the column space = the maximun number of linear independent row vectors =the dimension of the row space
result
result
Rank P.2
rankA:=the number of nonzero rows in a row-echelon (or the reduced row echlon form of A)
Rank P.3
rankA:=the size of its largest nonvanishing minor
(not necessary a principal minor)
=the order of its largest nonsigular
submatrix.
See next page
Rank P.4
00
10A
1x1 minorNot principal
minor
rankA=1
Theorem
Let A be an nxn sigular matrix.Let s be the algebraic multiple of eigenvalue 0 of A.Then A has at least one nonsingular(nonzero)principal submatrix(minor) oforder n-s.
Proof of Theorem p.1
snorderoforprincipal
nonzerooneleastathasA
AE
tAEtAEt
formtheofistc
smultipleoftcofzeroaisSince
tAEttc
sn
s
sn
snnn
A
A
n
i
in
i
in
A
min
0)(
)()1()(
)(
,)(0
)()1()(
1
1
1
Geometric multiple
Let A be a square matrix and λ be aneigenvalue of A, then the geometric multiple of λ=dimN(λI-A)
the eigenspace of A corresponding to λ
Diagonalizable
matrixdiagonalaisAPP
tsPgularnon
ifablediagonalizisA
1
.sin
Exercise
A and have the same characteristic polynomial and moreov
er the geometric multiple and algebraic multiple are similarily invariants.
APP 1
Proof of Exercise p.1
)(
)det(
det)det(det
))(det(
)det(
)det()()1(
1
1
11
1
1
xc
AxI
PAxIP
PAxIP
APPxPP
APPxIxc
A
APP
Proof of Exercise p.2
(2)Since A and have the same
characteristic polynomial, they have
the same eigenvalues and the algebraic
multiple of each eigenvalue is the same.
APP 1
Proof of Exercise p.3
)(dim)(dim
)(dim)(dim
)(dim)(dim
,,,,
)(
)(
)(
,,2,1
)(,,,
)(dim
)3(
1
1
1
21
1
1
21
1
1
APPEAEHence
AEAPPEhaveweSimilarly
APPEAE
ntindenpendelinearllyisPXPXPXSince
AEXP
PXXPA
XXAPP
riFor
APPEforbasisabeXXXLet
APPErLet
APPandAofeigenvalueanbeLet
r
i
ii
ii
r
Explain: geom.mult=alge.mult in diagonal matrix
2lg3
25
))1,0,0,1,0((5
))1,0,0,1,0((dim
))3,2,2,3,2()2,2,2,2,2((dim
))3,2,2,3,2(2(dim
2
32lg
)3,2,2,3,2(2
ofmultipleebraicaThe
diagrank
diagN
diagdiagN
diagIN
ofmultiplegeometricThe
ofmultipleebraicaThe
diagofeigenvalueanis
Fact
For a diagonalizable(square) matrix,the algebraic multiple and the geometri
c multiple of each of its eigenvalues areequal.
Corollary
Let A be a diagonalizable(square) matrix
and if r is the rank of A, then A has at least one nonsingular principalSubmatrix of order r.
Proof of Corollary p.1
rsnorderofsubmatrix
principalnonsigularoneleastathasA
TheorempreviousBy
sn
ofmultipleebraican
ofmultiplegeometricn
ANnrankAr
Aofeigenvalueofmultiple
ebraicathebesLet
0lg
0
)(dim
0
lg
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