Positive Semidefinite matrix

transcript

Positive Semidefinite matrix nHA

nCzAzz 0*

A is a positive semidefinite matrix(also called nonnegative definite matrix)

Positive definite matrix

nCzAzz 0*

A is a positive definite matrix

Negative semidefinite matrix nHA

nCzAzz 0*

A is a negative semidefinite matrix

Negative definite matrix

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*

Is It true that)(RMALet n

Proof of Question

)()()()(

))(()()(

yAxAxyiAyyAxxAxiyyAxiAyyAxx

yAxixAyiyAyxAxyAyyAixxAiyxAx

yiAxAiyxiyxAiyx

zAzAzziyxz

RyxwhereiyxzwritecanweCzanyForRxAxxthatAssume

clearisIt

TTTTTTTTTTTT

TTTTTTTT

Proof of Question

))(()()(*

AxyyAxiAyyAxxAxiyyAxiAyyAxxAyyyAixAxiyAxx

iAyxAiyxiyxAiyxAzz

Fact 1.1.6 The eigenvalues of a Hermitian (resp. positive semidefinite , positive definite) matrix are all real (resp. nonnegative, positive)

Proof of Fact 1.1.6

edeifnegativeisAifesemideifnegativeisAif

edeifpositiveisAifesemideifpositiveisAif

andRzAzz

numberrealaisAzzAzzzAzAzzSince

zwherezAzz

zzzAzz

CzzAzthenreigenvectongcorresponithebez

andAofeigenvalueanbeLetHALet

int0int0

Exercise n

n CzRAzzHA *

)(CMH n

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

21 00110

10But is not

symmetric

Proof of Exercise

HAHenceAAAA

CzzAAzCzAzzzAzCzAzzAzzthen

CzRAzzthatAssumenumberrealaisAzz

thenAzzzAzAzz

CzanyForHAthatAssume

RemarkLet 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.1A, 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.2A: real, λ is non-real

Az= λz, 0≠z z is real, which is impossible

Elementary symmetric function

nnS 21211 ),,,(

212 ),,,( iinii

nkiiinkS

21121 ),,,(

kth elementary symmetric function

KxK Principal Minor

nxnijaALet niiianyFor k 211

kikiikiiki

kiiiiii

aaaaaa

kxk principal minor of A

Lemma p.1nMALet

AofvectorcolumnithbeaLet i

niiianyFor k 211

j iiijifeiiijifa

bLet,,,,,,

vectordardsithbeeLet i tan

Lemma p.2 nbbbThen 21det

kiiibyindexedcolumnsand

rowswithorprincipalthe

Explain Lemma

00100001

detaaaaaa

aaaaaaaa

2422detaaaa

The Sum of KxK Principal Minors

AoforsprincipalkxkallofsumthebeAELet k

Theorem nMALet

AofpolynomialsticcharacterithebexcLet A )(

A tSttcThen

),,,()1()( 211

AofseigenvaluethebeLet n ,,, 21

in tAEt

)()1(1

Proof of Theorem p.1

i nijjj

211 211

),,,()1(

)())((

)())(()()1(

LemmapreviousbytAEt

jjjkiftejjjkifa

bwhere

ateateateAtItc

eeeILetAofvectorcolumnithisawhere

aaaALet

i nijjj

,)()1(

,,,,,,

det)det()(

211 211

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

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

1x1 minorNot principal

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.

snorderoforprincipalnonzerooneleastathasA

AEtAEtAEt

formtheofistcsmultipleoftcofzeroaisSince

tAEttc

0)()()1()(

)(,)(0

)()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

Exercise A and have the same characteristic polynomial and moreover the geometric multiple and algebraic multiple are similarily invariants.

Proof of Exercise p.1

)()det(

det)det(det))(det()det(

)det()()1(

PAxIPPAxIPAPPxPP

APPxIxc

Proof of Exercise p.2 (2)Since A and have the same characteristic polynomial, they have the same eigenvalues and the algebraicmultiple of each eigenvalue is the same.

Proof of Exercise p.3

)(dim)(dim)(dim)(dim

)(dim)(dim,,,,

)(,,2,1

)(,,,)(dim

APPEAEHenceAEAPPEhaveweSimilarly

APPEAEntindenpendelinearllyisPXPXPXSince

AEXPPXXPAXXAPP

riForAPPEforbasisabeXXXLet

APPErLetAPPandAofeigenvalueanbeLet

Explain: geom.mult=alge.mult in diagonal matrix

2lg325

))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(dim2

32lg)3,2,2,3,2(2

ofmultipleebraicaThe

diagrankdiagN

diagdiagNdiagIN

ofmultiplegeometricTheofmultipleebraicaThediagofeigenvalueanis

Fact For a diagonalizable(square) matrix,the algebraic multiple and the geometric multiple of each of its eigenvalues areequal.

Corollary Let A be a diagonalizable(square) matrixand if r is the rank of A, then A has at least one nonsingular principalSubmatrix of order r.

Proof of Corollary p.1

rsnorderofsubmatrixprincipalnonsigularoneleastathasA

TheorempreviousBysn

ofmultipleebraicanofmultiplegeometricn

ANnrankArAofeigenvalueofmultiple

ebraicathebesLet

)(dim0

Positive Semidefinite matrix

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