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