Post on 04-Jan-2016
Chapter 2
Nonnegative Matrices
2-1
Introduction
Entrywise nonnegative
,0nmA
(entrywise ) nonnengative means
jiaij ,0 different from positive semidefinite
Strictly positive
0nmA
strictly positive means
jiaij ,0
different from positive definite
Remark
000 AA
e.g.
01
10
nonzero, nonnegative but not positive
semipositive≡nonzero, nonnegative
BA
0 BA
jiba ijij ,
00,0 ABBA
CBCABCACCBA ,0,
000,0 AAxandxA
000,0 AABandBA
Remark
000 BorAAB
e.g.
00
10BA
2-2
Perron’s Theorem
spectral radius
)(CMA n
spectral radius
)(max)( AA
譜半徑
Example
30
01 iA
3)( A
ijnm aA
ijnmaA
BAAB
Proven in next page
BAABHence
BA
ba
ba
ABAB
ij
n
kkjik
n
kkjik
ijij
1
1
Collatz Wielandt
0A
collatz weilandt
XAXtsXA ..000)(
Lemma 2.2.2 (1)
)()( AA
0A
Proven in next page
)(
0
A
AzA
zA
Az
zz
Lemma 2.2.2 (2)
)(A
0A
Proven in next page ( 證明很重要 )
is closed and bounded above
.)(
)(
,
01,lim
1,..00
,
,lim)(
:)(
)(
)(0
closedisAHence
Awthen
wXAX
klettingbyobtainwe
XwAXSince
XandXthenXXLet
XesubsequencconvergentahasX
XXwAXtsX
Nkfor
thenwwandAwIf
AofCloseness
A
A
kkk
k
k
iii
ik
ii
kkkkk
kkNkk
.)(
)(
.max
00
)(
,111
:)(
aboveboundedisAHence
Aforboundupperanis
xe
xe
xe
xAew
eAethen
AofsumcolumnimunthebeLet
wxe
xAe
xewwxeAxe
xsomeforwxAx
AwanyFor
ReLet
aboveboundedisA
TT
TT
TT
TT
TTT
nT
Remark
andAAA )()(max)(0
0A
Proven in next page
uAAutsu )(..0
)(00
0
0,00
)(
.0)()(
)(
0)(
0
,
..00
),(max
max
max
max
max
max
max
max
max
max
max
max
max
Aandu
Auu
AuuandASince
uAu
A
smallsuffiAuAuA
AuAuA
uAuA
ABut
vesemipositiisuAu
thenuAuSuppose
uAuthatshowTo
uAutsu
thenALet
generalized eigenvector
NkAsomefor ),(
u is called generalized eigenvector ofA if
0)( uIA k
Remark
AJAPP 1
Proven in next page
the columns of P are the
generalized eigenvectors of A.
11
23323
12212
111
21
21
1
)(
)(
)(
0)(
*0
01
01
kkkkk
nk
nk
A
A
PPIAPPAP
PPIAPPAP
PPIAPPAP
PIAPAP
PPPP
PPPPA
PJAP
JAPP
tocorrklenghofchainJordancalled
PPP
PIAPIAPIAP
and
PIA
PIA
PIA
PIA
kk
kk
kkk
kk
.
)(,,)(,)(,
0)(
0)(
0)(
0)(
121
12
33
22
1
Remark
eigenvaluegleisA sin)(
The geometric multiple of λ =1 and
there is no generalized eigenvector
other than eigenvector corr. to λ
Remark
.sin)( eigenvaluegleisA
0A
Proven in next page
1)(
)()()(0
..
,,0
)(,)(..,
.
1)(1
AofmultiplegeometricHence
vuAvuABut
positivenotbutvesemipositiisvu
tsRchoosecanwethen
tindependenlineararevuandu
vAAvuAAutsRvu
notSuppose
AofmultiplegeometricthatshowoT
n
.sin)(
)(max)(,
)()(
.0))((
)0()(
)(
0
,.sin.
)(
)())((..
,
)(.
2
eigenvaluegleisATherefore
AAthatbecauseimpossibleiswhich
AA
smallsuffallforyAAy
uyAAy
uyAAy
ythatassumemaywe
smallsuffrgchooByRranyfor
ruybyyreplacemaywethatNote
uyAAy
uAAuwhereuyIAAtsRy
thennotSuppose
Atorrecorseigenvectothanother
reigenvectodgeneralizenohasAthatshowTo
n
Remark
)()( AandA
0A
Proven in next page
)()( AandA
)(
argarg
,,1
)(.
)(
,)(
)(,)(
)(
)(
..0
1
1111
A
othereachofmultiplearezandz
zz
zazazaza
nkfor
Atocorresp
AofreigenvectoaniszandzAAz
zAAzzzA
thenAIf
AthenAifthatshowTo
A
A
zAAzz
zAztsCz
n
nknknknk
n
Remark
If A>0, then A has no nonnegative
eigenvector other than
(multiple of) u , where u>0 and
uAAu )( Proven in next page( 證明很特
別 )
uofmultiplethanother
reigenvectoenonnegativnohasAHence
impossibleiswhich
XvceA
XvA
XvAAXvXv
vAAvthen
vAvAtsvLet
Athen
uAAuanduwhere
uofmultipleanotisXandXAX
tsXthatSuppose
T
T
TTT
TT
T
0sin,)(
0))((
)(
)(
)(..0
)(
)(0
..00
Theorem 2.2.1 p.1
(Perron’s Thm)thenAIf ,0
0)( A
)()( AA (b)
(c)
(a)
uAAutsu )(..0
eigenvaluesimpleaisA)(
)(),()( AAA
(f)
(g)
(e)
1,)(,)(
,)(
lim
vuandvAvAuAAu
whereuvA
A
TT
Tm
m
(d)
A has no nonnegative eigenvector
other than (multiples of) u.
Norm on a vector space
Vxx 0
(i)
(iii)
(ii) scalarxx
is a norm on V
= hold iff x=0
Vyxyxyx ,
d
we introduce a metric
is a metric space
yxyxd ),(
with
on V, by
dV ,
kasxxk
kasxxd k 0,
Convergent matrix sequence
LA Nkk
NkMA nk
can be interpreted in
where
ijk
ijk
lanjiji
)(lim,1,
one of the following equivalent way:
(i)
ijkijk lLaA ,)(
kasLAk 0
is in any fixed norm of
where
nM
The topology of
(ii)
nM is independent of
(the maximum norm)
to be
ijnjiaA
,1max
we obtain (i)
In (ii), take
Bounded matrix sequence
NkkA
,2,1,1..0 )( knjiMatsM kij
(ii)
(i)
,2,1..0 kMAtsM k
is bounded means
Fact 2.2.4
kk
kk
kkk
BABA
limlimlim
)lim)(lim(lim kk
kk
kkk
BABA
(ii)
(i)
kp
k
k
k
k
k
k
p
k
A
A
A
A
A
A
lim
lim
lim
lim
2
1
2
1
(iii)
Apply of Fact 2.2.4 (ii)
PgularnonsomeforJAPP A sin1
LAkk
lim
PAPPAP kk
kk
)lim(lim 11
and P is nonsigularIf
then
convergent problem of A is corresponding to convergent problem of
AJ
Theorem 2.2.3
)(1 A 1)( A
nMA
NkkA
Let
(i) The sequence
converges to the zero matrix iff
11
1)( A
1)( A
NkkA
(ii) converges iff
or
and 1 is the only eigenvalue
with modulus 1 and the corresp.
Jordan blocks are all
)(A
1)( A
1)( A
NkkA
(iii) is bounded iff
either
and
if then
or
1 1)( Av
Lemma 2.2.5
1)(lim
km
kJ
1
1
0)(lim
km
kJ (i) If
(ii) If
then
and m=1, then
1 convergesk
1
1
Nkk
mJ )(
(iii) If
the sequence
and m=1, then
is bounded
Note:
In this case, the seqence does not
converge if
explain in next page
θ
2m
1
1
Nkk
mJ )(
(iv) If
then the sequence
or
is unbounded
and
)0(
0
1
0
010
1
01
)(
)(
1
00miifNN
i
kN
i
k
NwhereNI
J
i
iim
i
ikik
i
ik
k
k
km
0)(lim
0!
)1()1(
01
2
2
12
)(
1
2
1
21
121
km
k
ikik
k
k
kk
k
kk
kkk
mkkkk
km
J
kasi
ikkki
k
and
k
kk
kk
m
kkk
J
unboundedisJ
kaskk
mandCase
unboundedisJCase
k
kk
kk
m
kkk
J
iv
Nkk
m
k
Nkk
m
k
kk
k
kk
kkk
mkkkk
km
)(
21:2
)(1:1
2
2
12
)(
)(
1
1
2
1
21
121
Exercise 2.2.7
)(),( AA
nMA
)(A
)0)(( A
eigenvalue and
is non-nipotent
for every
Suppose that )(A is a simple
Exercise 2.2.7k
k AA
)(lim
what can you tell about the vector
Prove that
x and y?
exists and is of the form *xy
2-3
Nonnegative Matrices
Lemma 2.2.2
)(A0A
is closed , bounded above and
If , then
)()(: AA
Lemma 2.3.1
0AIf , then
)()(max AA
XXA
AX
A
A
XXA
A
XAAX
tsXthen
AAtsSuppose
AAshowtoremainsIt
AAhavealreadyWe
0
2
0
0
0
00
)()(
)(
)(
..00
)()(..0
0)()(
)()(
00,
0
0)(
lim
3.2.2
1)(
)(
)(
,2,1)(
,
0
00
0
Xasoncontraditiaiswhich
X
A
A
TheoremBy
A
A
A
ABut
kXXA
A
yInductivel
k
k
k
Lemma 2.3.2
BA0If , then
)()( BA
)()(
)()(max)(
)()(
)(
)(..00
)()(
BA
BBA
BA
BXXA
BXAXso
BABut
AXXAtsX
AA
Fact
)()()( 2121 AAAA
)(),(max)( 2121 AAAA
Corollary 2.3.3
0, AMA n
)()( AB
, and B is a principal submatrix of A
If
then
In particular )(max1
Aaniii
)()(
)()ˆ()(
,00
0ˆ
**
*
AB
ABB
thenAB
BLet
BAthatassummay
generalityofloseWithout
Exercise 2.3.4
BAMBA n 0,,
)()( BA then
If
Hint: There is some α>1 such that
BAA 0
)()(
),()()()(0
)()()(
,2.3.2
0
,1,1,0;min
,10
0)(:2
)()(0
,0)(:1
0)(,
0,0
BAhaveweso
BAAA
BAA
LemmaBy
BAA
thennjiaa
bLet
njibaSince
AthatAssumeCase
BA
thenAthatAssumeCase
BhaveweThmPerronby
BBASince
ijij
ij
ijij
Theorem 2.3.5 (Perron-Frobenius Thm)
XAAX )(
0A
..00 tsX
)()( AA , thenIf
and
)(#)()(lim
)()()()(
0
,1
,2,1
321
321
existsAA
AAAA
AAAAAlso
keachforAthen
jiak
AandMALet
kFor
kk
k
ijijknk
)(),#(#)(#
)#(#)(
)(
,
)(1,00
,
1,,1,11)(
..0
,
AandBy
Ahence
A
xAxkLetting
xAxAandxexthen
xx
sayesubsequencconvergentahasIt
sequenceboundedaisx
ewherexeandxAxA
tsx
ThmPerronbykeachFor
kkkk
k
iiiiT
i
Nkk
Tk
Tkkkk
k
Ri (A)
n
jija
1
)(ARi = i th row sum of A
Cj (A)
n
iija
1
)(AC j = j th column sum of A
Corollary 2.3.6
)(max)()(min11
ARAAR ini
ini
)(max)()(min11
ACAAC jnj
jnj
0 AandMA n
Then
Let
and
)(max)()(min
)(
,0
)(
)(
)(..00
Re
,111
),(max,)(min
)(max)()(min)1(
11
11
11
ARAAR
RAr
havewesoezBut
eRzezAerz
eRzAezerz
zAAzthen
zAzAtszisthere
ThmFrobeniusPerronBy
Aere
thenRe
andARRARrLet
ARAARthatshowTo
ini
ini
T
TTT
TTT
TT
T
nT
ini
ini
ini
ini
)(max)()(min
)()(),()(
)(max)()(min
)(max)()(min
),2(
)(max)()(min)2(
11
11
11
11
ACAAC
AAtctcSince
ACAAC
ARAAR
haveweBy
ACAACthatshowTo
jnj
jnj
TAA
jnj
Tj
nj
Tj
nj
TTj
nj
jnj
jnj
T
Matrix norm
nM
)()()( BNANABN
is called a matrix norm if N( - ) is a
A norm N( - ) on
norm on
, and N( - ) is submultiplicative i.e.
nM
Matrix norm Induced by Vector norm
nC
nM
AXAx 1max
be a (vector) norm on
Let
Define on by
matrix norm induced by the vector norm
Proposition of matrix norm induced by vector norm
BABA
BAAB
BAx
ABxAB
xBABxABxAABx
xAAx
x
AxAxA
BA
BxAx
BxAx
xBABA
x
xx
xx
x
x
0
01
11
1
1
max
)(
maxmax
maxmax
max
)(max
Remark 2.3.7
nMA
nM
)(AA
is a marix norm on If
then
)(
..0)(
AA
A
AXAXXA
XAX
tsXchooseandALet
not Euclidean
matrix normcorrect proof in
next page( 很重要 )
)(
..0)(
AAHence
A
BBABBAthen
BXXX
AXAXXA
XXXAAB
MXXXBLet
XAX
tsXchooseandALet
n
Special norm:l∞,lp
llp p
ini
n
l
1
2
1
max
pn
i
p
i
n
pl
1
1
2
1
Special Matrix norm
nM
nCl
be the matrix norm on
Let
induced by the norm of
Corollary
0, AMA n
)(A
If the row sums of A are constant
Let
then A row sum of A
)(),#(#)(#
)#(#)(
max
1
1
1
:
:
)(#)(
)(
1
111
1
21
ArandBy
Arso
AAxrHence
eandrAeBut
raXaxa
nisomeforxaAX
XwithCxxxXFor
pf
ArClaim
Arr
Ar
reAethen
AofsumrowcommontheberLet
x
n
jij
n
jij
n
jjij
n
jjij
nn
Exercise 2.3.8 p.1
1
A
n
iij
njaA
111max
max absolute column sum of A
Exercise 2.3.8 p.2
A
n
jij
niaA11
max
max absolute row sum of A
n
iij
njx
n
iij
nj
n
iijjj
n
iij
n
iij
nj
n
iij
nj
n
iij
nj
n
jj
n
iij
nj
j
n
j
n
iij
njj
n
j
n
iij
n
i
n
jjij
n
i
n
jjij
n
ii
nTn
n
iij
nj
aAxAHence
aaAAethen
njsomeforaaLet
a
xaxa
xaxa
xaxaAxAx
xwithRxxxxanyFor
aAthatshowTo
11111
11111
0111
11
111111
1 111 1
1 11 111
121
111
maxmax
max
1max
max
maxmax
max
1
max)1(
1
000
0
n
jij
nix
n
jij
ni
n
jjii
n
jji
n
jiji
nTniii
n
jji
n
jij
ni
n
jij
ni
n
jij
ni
n
jjij
ni
n
jjij
nii
ni
nTn
n
iij
ni
aAxAHence
aAy
thenaAyand
iaaAyandythen
RaaayLet
aatsniLet
axa
xaxaAxAx
xwithRxxxxanyFor
aAthatshowTo
111
11
1
11
21
1110
1111
11111
21
11
maxmax
max
,
1
)sgn()sgn()sgn(
max..
maxmax
maxmaxmax
1
max)2(
00
0
000
0
Exercise 2.3.9
)(A
0, AMA n
Prove that if A has a positive eigenvector, then the corresponding eigenvalue is
Let
[Hint: Apply the Perron-Frobenius Thm
to AT ]
)(#)(
0,000
)(#)(
)(
)(
)(
)(..00
..)(
,0
byA
uvvanduSince
uvAuv
uvAuv
uvAAuv
vAAv
vAvAtsv
ThmFrobeniusPerronBy
uAutsA
thenAofeigenvalueanbeuLet
T
TT
TT
TT
TT
T
Remark 2.3.10
1)( AA
AA)(
0, AMA n
If A has equal row sums, then
Let
If A has equal column sums, then
AAHence
AA
CorollaryBy
ArrA
Ar
ExercisebyArandreAethen
sumrowcommontheberLet
solutioneAlternativ
ArAExerciseBy
AofreigenvectopositiveiseSince
ExercisebyArandreAethen
sumrowcommontheberLet
AAthatshowTo
)(
)(
,6.3.2
)(
)(
8.3.2,
.
:
.)(9.3.2
,
8.3.2,
.
)()1(
1
1
1
1
1
1
1
)(
)(,6.3.2
)(
)(
8.3.2
.
:
.)()(9.3.2
,
8.3.2
.
)()1(
AAHence
AACorollaryBy
ArrA
Ar
ExercisebyArandreeAthen
sumrowcommontheberLet
solutioneAlternativ
ArAAExerciseBy
AofreigenvectopositiveiseSince
ExercisebyArandreeAthen
sumcolumncommontheberLet
AAthatshowTo
T
T
T
a row stochastic matrix
0, AMA n
with row sums all equal to 1,then
A is called a row stochastic matrix.
If
1)( AandeAe
a column stochastic matrix
0, AMA n
with column sums all equal to 1,then
A is called a column stochastic matrix.
If
1)( AandeAe TT
Exercise 2.3.11
)(min1
ARini
)(A
AB
))(
,,)(
,)(
(21 ARARAR
diagn
[ Hint: Let
Deduce Corollary 2.3.6 from Remark
2.3.9 and Lemma 2.3.2
To show that inequality
consider B=DA, where D is the diagonal
matrix
show that
))(
,,)(
,)(
(
0:2
)(0:1
0),(min
)()(min)1(
21
1
1
ARARARdiagDwhere
DABConsider
Case
ACase
thenARLet
AARthatshowTo
n
ini
ini
)(),#(#)(#
)#(#)(,9.3.2
,,,2,1)(
)(#)()(
0
,,,2,11)(
0
)(
)(
)(
22
11
Aandby
BExerciseby
eeBandniBRBut
AB
AB
niAR
Since
AAR
AAR
AAR
Bthen
i
i
nn
0)(
0)(0
,0)(0
0)()(
),,,(
0),(max
)(max)()1(
21
1
1
ARif
ARifc
andARif
ARifARd
withddddiagDwhere
CDABConsider
thenARLet
ARAthatshowTo
i
ii
i
iii
n
ini
ini
)(max)(
)(),#(#)(#
)#(#)(,9.3.2
,,,2,1)(
)(#)()(
0
0)(1)(
100)(
1ARAHence
Aandby
BExerciseby
eeBandniBRBut
AB
ABthen
ARifAR
andnjaARSince
ini
i
ii
iji
Diagonally Similar p.1
ADDB 1
00,0 BDA
00,0 BDA
nMBA , are diagonal similar
In particular
if there is nonsingular matrix D s.t.
Diagonally Similar p.2
00,0 BDA
00,0 BDA
preserves the class of nonnegative
(as well as , positive) matrices.
In particular
nonnegative diagonal similarity
Corollary 2.3.12
nCx
n
i i
ijj
nj
n
i i
ijj
nj x
axA
x
ax
1111max)(min
n
jjij
ini
n
jjij
inixa
xAxa
x 1111
1max)(
1min
0, AMA nThen for any positive vector
and
we have
n
j i
ijj
nj
n
j i
ijj
nj
n
jjij
ini
n
jjij
ini
n
jjij
ini
n
jjij
ini
jiji
n
nTn
x
axA
x
ax
haveweSimilarlly
xax
Axax
xax
ADDxax
CorollaryBy
xax
ADDofentryji
thenxxxdiagD
letandRxxxxLet
1111
1111
11
1
11
1
21
21
maxmin
,
1max
1min
1max
1min
6.3.2
1),(
,),,,(
0),,,(
Exercise 2.3.13 p.1
nTn Rxxxx ,,, 21
For any semipositive vector
Wielandt numbers of A with respect to x are defined and denoted respectively by:
the upper and the lower Collatz-
0, AMA nLet
Exercise 2.3.13 p.2
xAxxRA :0inf)(
wxAxwxrA :0sup)(
(we adopt the convention that inf ψ=∞)
Prove that for any semipositive x, we have
Exercise 2.3.13
0:)(
max)( ii
iA x
x
AxxR
0:min)( ii
iA x
x
Axxr
0;max)(
max
0..
0..
,,2,1
0
ii
iA
i
i
x
ii
i
iii
ii
xx
AxxRHence
x
Ax
xtsix
Ax
xtsixAx
nixAx
xAx
i
0;min)(
min
0..
0..
,,2,1
0
ii
iA
i
i
x
ii
i
iii
ii
xx
AxxrHence
x
Ax
xtsix
Ax
xtsixAx
nixAx
xAx
i
Exercise 2.3.14 p.1
)(A
Axxts ..0
0, AMA n
(i) Prove that if
Let
for some positive vector x then
)(
min)(
,12.3.2
min
0sin,,,1
,,1
1
1
AHence
x
AxA
haveweCorollaryBy
x
Ax
xcenix
Ax
nixAx
Axx
i
i
ni
i
i
ni
i
i
ii
Exercise 2.3.14 p.2
)(A
xAxts ..0
0, AMA n
(ii) Prove that if
Let
for some positive vector x then
)(
max)(
,12.3.2
max
0sin,,,1
,,1
1
1
AHence
x
AxA
haveweCorollaryBy
x
Ax
xcenix
Ax
nixAx
xAx
i
i
ni
i
i
ni
i
i
ii
Exercise 2.3.14 p.3
)(A
0, AMA n
(iii) Use parts (i) and (ii) to deduce that
Let
if A has a positive eigenvector thenthe corresponding eigenvalue is
)(
)()(
),()(
0,00
)(
,
AHence
AandA
haveweiiandiBy
uAuanduAu
uandASince
AsomeforuAu
thenAofreigenvectopositiveabeuLet
Exercise 2.3.15 p.1
2221
1211
aa
aa
2221
1211
aa
aa
are diagonally similar.
Show that the matrices
and
.
10
01
10
01
2221
1211
2221
1211
2221
1211
2221
1211
similardiagonallyare
aa
aaand
aa
aathen
aa
aa
aa
aa
Exercise 2.3.15 p.2
333231
232221
131211
aaa
aaa
aaa
are diagonally similar ?
Are the matrices
and
333231
232221
131211
aaa
aaa
aaa
.
1,,
1,1,1
..),,(
sin
.
323121
31
231
121
1
332321
31311
3
3231
2221211
2
3131
12121
111
3
2
1
333231
232221
131211
13
12
11
333231
232221
131211
321
similardiagonally
notarematricestwothisHence
impossibleiswhich
ddanddddd
ddanddddd
adaddad
dadadad
daddada
d
d
d
aaa
aaa
aaa
d
d
d
aaa
aaa
aaa
tsddddiagD
matrixdiagonalgularnonisthereThen
matricesdiagonalarematricestwotheseSuppose