Stability of reducing subspaces

We cha ac e ize he s abili y o educing subspaces o a ec angula
ma ix pencil o complex ma ices λB −A, excep o he special case in
which he pencil has no eigen alues and only has one ow and one column
minimal indices and bo h a e di e en om ze o.
Keywo ds: ec angula ma ix pencils, K onecke canonical o m, educing
subspaces, de la ing subspaces, s abili y, pe u ba ion.
MSC AMS: 15A22, 15A60, 47A55, 47A15, 93B10
1 In oduc ion
Gi en wo ma ices A, B ∈Cm×n, we call ma ix pencil he i s o de ma ix
polynomial λB −A. Fo simplici y, we will deno e he se o ma ix pencils
o he o m λB −A, wi h A, B ∈Cm×n, by P[λ]m×n. We de ine he no mal
ank o a pencil λB −A∈ P[λ]m×n, and we deno e i by n ank(λB −A), o
be he g ea es o de o he mino s o λB −A ha a e di e en om he ze o
polynomial. I m=nand n ank(λB −A) = n, he pencil λB −A∈ P[λ]m×n
is said o be egula . O he wise, he pencil is said o be singula .
No e by C(λ) he ield o a ional ac ions in λ. I we conside λB −Aas a
linea map om he ec o space C(λ)nin o C(λ)m, bo h o e he ield C(λ),
we ha e
n ank(λB −A) = dimC(λ)Im(λB −A),
(see [4]). We de ine he nulli y o λB −Aby ν(λB −A) := dimC(λ)Ke (λB −A).
F om
n= dimC(λ)Ke (λB −A) + dimC(λ)Im(λB −A),
ν(λB −A) = n−n ank(λB −A).
∗Wo k suppo ed by he Spanish Minis y o Educa ion and Science P ojec MTM2010–
19356–C02–01, and he Basque Go e nmen P ojec s GIC10/169–IT–361–10 and GIC13/IT–
710–13.
†Depa men o Ma hema ical Enginee ing and Compu e Science, The Public Uni e si y
o Na a e, Campus de A osad´ıa, 31006 Pamplona, Spain. [email p o ec ed]
‡Depa men o Applied Ma hema ics and S a is ics, Uni e si y o he Basque Coun y
UPV/EHU, Facul y o Pha macy, 7 Paseo de la Uni e sidad, 01006 Vi o ia-Gas eiz, Spain,
[email p o ec ed], [email p o ec ed]
1
This is he accep ed manusc ip o he a icle ha appea ed in inal o m in Linea Algeb a and i s
Applica ions 470 : 252-299 (2015), which has been published in inal o m a h ps://doi.o g/10.1016/
j.laa.2014.10.026. © 2014 Else ie unde CC BY-NC-ND license (h p://c ea i ecommons.o g/licenses/
by-nc-nd/4.0/)
S abili y o educing subspaces∗
Go ka A men ia,
† Juan-Miguel G acia,
‡ F ancisco-En ique Velasco‡
Oc obe 4, 2014
Dedica ed o P o esso Leiba Rodman on he occasion o his 65 h bi hday
Abs ac
As usual we iden i y a ma ix M∈Cm×nwi h he linea map x7→ Mx om
Cn≡Cn×1in o Cm≡Cm×1. Le Nbe a subspace o Cn, we de ine M(N)
as he subspace o Cm o med by all ma ix p oduc s Mx wi h x∈ N. Van
Doo en p o ed ha
dim(A(N) + B(N)) ≥dim N − ν(λB −A),(1)
whe e A(N) + B(N) is he sum o hese subspaces o Cm. See [17, Equa ion
(2.16) on page 63 and in he line ollowing (2.25 a) and (2.25 b) on page 65]. In
he case he equali y holds in (1), he subspace Nis called a educing subspace
o he pencil (see [17]) o , equi alen ly, ha Nis a (λB −A)- educing subspace.
Obse e ha i he pencil is egula hen ν(λB −A) = 0. So, in his case N
is a educing subspace i and only i dim(A(N) + B(N)) = dim N. These
educing subspaces a e also called de la ing subspaces o egula pencils (see
[16]). In o de o simpli y he e on, we will deno e by (λB −A)(N) he subspace
A(N) + B(N).
We use he ope a o no m induced by he Euclidean no ms on Cmand Cn,
also called he spec al no m,
kMk:= max
x∈Cn
kxk2=1
kMxk2.
The gap be ween subspaces Mand N(in Cn) is de ined as
θ(M,N) := kPM−PNk
whe e PMand PNa e he o hogonal p ojec o s on Mand N, espec i ely.
Le λB −A∈ P[λ]m×na ma ix pencil. A educing subspace No Cn o
(λB −A) is said o be s able i o e e y ε > 0 he e exis s a δ > 0 such ha
e e y ma ix pencil λB0−A0∈ P[λ]m×n ha sa is ies
kA0−Ak+kB0−Bk< δ
has a (λB0−A0)- educing subspace N0 o which he inequali y θ(N0,N)< ε
holds. In he same way, we will say ha he subspace No Cnis Lipschi z
s able i he e exis K, ε > 0 such ha e e y ma ix pencil λB0−A0∈ P[λ]m×n
ha sa is ies kA0−Ak+kB0−Bk< ε has a (λB0−A0)- educing subspace N0
o which he inequali y
θ(N0,N)≤K(kA0−Ak+kB0−Bk)
holds. To simpli y, we will o en say ha a subspace Nis (λB −A)-s able o
mean ha Nis a (λB −A)- educing subspace and is s able o pe u ba ions in
he ma ices Band A. Fo a clea mo i a ion see Chap e s 13 and 15 o [5].
A p e ious pape on his opic was published by he second and hi d au-
ho s [7]. A cha ac e iza ion o he s abili y o Lipschi z s abili y o de la ing
subspaces o a egula ma ix pencil was al eady gi en he e. In he cu en
pape , we add ess his s abili y p oblem o he case o educing subspaces o
singula pencils. Be o e s a ing he main esul o his pape , ecall some p op-
e ies o he pencils o ma ices.
Two ma ix pencils λB −A, λD −C∈ P[λ]m×na e said o be s ic ly
equi alen i he e exis in e ible ma ices P∈Cm×m,Q∈Cn×nsuch ha
2
λD −C=P(λB −A)Q. Rema k ha wo s ic ly equi alen pencils ha e he
same no mal ank. Hence a subspace Nis (λB −A)- educing i and only i he
subspace Q−1(N) is (λD −C)- educing. Se C:= C∪ {∞}. We will use he
no a ion ∞B−A:= B. We will say ha he elemen α∈Cis an eigen alue o
he pencil λB −Ai
ank(αB −A)<n ank(λB −A).
An eigen alue αo λB −Ais ini e i α∈C, and i is in ini e i α=∞. We
call spec um o he pencil λB −Aand we deno e i by Λ(λB −A), he se o
i s eigen alues. I is a subse o C.
The well-known K onecke canonical o m o he s ic equi alence o ma ix
pencils is gi en in he ollowing esul .
Lemma 1 (K onecke canonical o m [4]).Gi en a ma ix pencil λB −A∈
P[λ]m×n, he e always exis in e ible ma ices P∈Cm×m,Q∈Cn×n, such
ha P(λB −A)Qhas he o m


λBc−AcO O
O λB −A O
O O λB −A 
,(2)
whe e
λBc−Ac= (O(n1− 0)×( 0− 1),diag(L 1, L 2, . . . , L 1)) ∈ P[λ](n1− 0)×n1; (3)
L i:= λ(I i, O)−(O, I i)∈ P[λ] i×( i+1); and he pencil
λB −A := λN −I O
O λI −J∈ P[λ]n2×n2,(4)
wi h a nilpo en ma ix N, is egula ; and, las ly,
λB −A := O(s0−s1)×(m3−s0)
diag(LT
`1, LT
`2, . . . , LT
`s1)∈ P[λ]m3×(m3−s0).(5)
Rema k 2. The sequences
( 1, 2, . . . , 1,
0− 1
z }| {
0,...,0),(`1, `2, . . . , `s1,
s0−s1
z }| {
0,...,0)
a e he column minimal indices and he ow minimal indices, espec i ely. The
elemen a y di iso s o he ma ices Nand J, a e called in ini e elemen a y
di iso s and ini e elemen a y di iso s, espec i ely, o he ma ix pencil. Recall
ha Λ(λB −A) deno es he spec um o he pencil λB −A, hen Λ(λB −A) =
Λ(λB −A ).
Thus, a comple e sys em o in a ian s o he s ic equi alence o wo
ma ix pencils is o med by he ini e sequences o ow and column minimal
indices and he sys em o elemen a y di iso s ( ini e and in ini e). Fo some
pa icula pencils some o hese in a ian can be absen . S ic ly speaking i is
no he same he se o ow minimal indices and he sequence o ow minimal
indices. Bu om now on we will loosely speak and — o example— we will
say ha a pencil has wo ow minimal indices o mean ha he sequence o ow
minimal indices has wo e ms, which migh be equals.
3
Rema k 3. Since 0=ν(λB −A), whe e 0is he numbe o column minimal
indices (see [4]), om (1) we deduce ha o each subspace No Cn,
dim(A(N) + B(N)) ≥dim N − 0.
Le Xbe a basis ma ix o N. As dim(A(N) + B(N)) = ank(AX, BX) we
ha e
ank(AX, BX)≥ ank(X)− 0.(6)
As a consequence, he subspace Nis (λB −A)- educing i and only i
ank(AX, BX) = ank(X)− 0.(7)
Recall ha a ma ix pencil λB −A∈ P[λ]m×nis said o be igh egula i
n ank(λB −A) = n, o equi alen ly ν(λB −A) = 0, o equi alen ly, i has no
column minimal index. In an analogous way, we will say ha a pencil λB −Ais
le egula i λBT−ATis igh egula , ha is, i λB −Ahas no ow minimal
index.
Ano he p e ious wo k on he opic o he s abili y o educing subspaces was
he one by Demmel in [2]. He s udied he s abili y o some educing subspaces
o singula ma ix pencils, bu unde addi ional condi ions. We will explain i
b ie ly. Le λB −Abe a singula pencil and le Nbe a educing subspace o
his pencil. Then, acco ding o ou no a ions, he e is no loss o gene ali y i we
suppose ha λB −Aand Nha e he o m
λB −A=



λBc−AcO O O
O λB1
−A1
λB2
−A2
O
O O λB3
−A3
O
O O O λB −A



,
N=*



In1O
O In2
O O
O O



+,
whe e he (m1×n1)-pencil λBc−Aconly has column minimal indices, λB −A
only has ow minimal indices and he pencil
λB1
−A1
λB2
−A2
O λB3
−A3

is egula , whe e λB1
−A1
is a pencil o size n2×n2. In [2] i is supposed ha
Λ(λB1
−A1
)∩Λ(λB3
−A3
) = ∅.
Unde hese hypo heses, in Theo em 6, page 26 o [2], some esul s a e gi en
on he s abili y o he subspace N, bu assuming also ha he pe u bed pencils
ha e educing subspaces o he same dimension as N.
The e is an ample li e a u e on he use o educing subspaces o ma ix pen-
cils as a ool o ac o izing a ional ma ices and o sol ing Ricca i equa ions.
One can see many e e ences in he book by Ionescu, Oa ˇa and Weiss [10]. See
also [13].
Wi h heses no a ions, he main esul o he pape is he ollowing.
4
Theo em 4. Le λB −A∈ P[λ]m×nbe a singula ma ix pencil. The ollowing
asse ions a e ue:
(1) I he pencil has ow minimal indices, column minimal indices and eigen al-
ues, hen no educing subspace is s able.
(2) I he pencil has no ow minimal index, hen he unique s able and Lipschi z
s able educing subspace is Cn.
(3) I he pencil has no column minimal index, hen he unique s able and Lip-
schi z s able educing subspace is {0}.
(4) I he pencil has column minimal indices and, a leas , wo ow minimal
indices, hen no educing subspace is s able.
(5) I he pencil only has one ow minimal index which is equal o ze o, and
has no eigen alues, hen he only s able and Lipschi z s able educing subspace
is Cn.
(6) I he pencil only has one ow minimal index which is di e en han ze o, has
no eigen alues and has a leas wo column minimal indices, hen no educing
subspace is s able.
(7) I he pencil only has one ow minimal index which is di e en han ze o, and
has one column minimal index, which is equal o ze o, and has no eigen alues,
hen he unique s able and Lipschi z s able educing subspace is Ke A.
The o ganiza ion o his pape is he ollowing. In Sec ion 2 algeb aic p op-
e ies o he educing subspaces o pencils o linea maps a e es ablished. In
Sec ion 3, hese p ope ies a e ansla ed in o e ms o ma ix pencils. In Sec-
ion 4 he p oblem o he s abili y o educing subspaces is add essed by means
o con e ging sequences o ma ix pencils and basis ma ices o subspaces. In
Sec ions 5 o 9 he p oo o Theo em 4 (Main Theo em) is de eloped. In Sec ion
5 Asse ions (1), (2) and (3) o he Theo em a e p o ed. In Sec ions 6, 7, 8 and
9 Asse ions (4), (5), (6) and (7), espec i ely, a e p o ed.
2 P ope ies o he educing subspaces o linea
map pencils
In his sec ion we gi e a cha ac e iza ion o he educing subspaces o pencils
o linea maps. I s p oo will be made in he ollowing sec ion, ansla ing hese
esul s o he ma ix pencils.
Fi s , ema k ha he concep s o no mal ank and educing subspace can
be ex ended o he case o a pai o linea maps. Le Uand Vbe ec o spaces
o e Cand le A,B:U → V be linea maps. The no mal ank o he pencil o
linea maps λB−Ais de ined by
n ank(λB−A) := max
z∈C ank(zB−A),
whe e zB−A:U → V is a linea map o each z∈C. A pencil o linea maps
λB−Ais said o be egula i dim U= dim Vand he linea map zB−A:U → V
is in e ible o e e y z∈C, excep o a mos a ini e numbe o complex
numbe s. O he wise, we will say ha he pencil is singula . Fo each x∈ U we
de ine
(λB−A)(x) := B(x) + A(x).
5

F om his de ini ion i is deduced ha o e e y subspace No U, (λB−
A)(N) = B(N)+A(N). The e o e, he subspace No Uis said o be (λB−A)-
educing i
dim(λB−A)(N) = dim N − min
z∈CdimCKe (zB−A).
To w i e he s a emen s o he main heo ems in his sec ion, we need some
p e ious de ini ions and no a ions. Le Ukdeno e he Ca esian p oduc U ×
· · · × U,k− imes. Gi en a pai o linea maps A,B:U → V and α∈C, o
k= 1,2,..., conside he linea maps
Tk
λB−A:Uk→ Vk+1,Pk,α
λB−A,Pk,∞
λB−A:Uk→ Vk
de ined o x= (x1, x2, . . . , xk)∈ Ukby means o
Tk
λB−A(x) := (B(x1),−A(x1) + B(x2),...,−A(xk−1) + B(xk),−A(xk)) ,(8)
Pk,α
λB−A(x) := ((αB−A)(x1),B(x1)+(αB−A)(x2),...,B(xk−1)+(αB−A)(xk)) ,(9)
Pk,∞
λB−A(x) := (B(x1),−A(x1) + B(x2),...,−A(xk−1) + B(xk)) .(10)
Gi en x= (x1, x2, . . . , xk)∈ Uk, o i= 1,2, . . . , k we de ine he p ojec ions
πk
i(x) = xi. Now, o e e y α∈¯
C:= C∪ {∞} and k= 1,2, . . ., we de ine he
subspaces:
Sk
λB−A:=
k
X
i=1
πk
iKe (Tk
λB−A),(11)
Sk,α
λB−A:=
k
X
i=1
πk
iKe (Pk,α
λB−A),(12)
Dk
λB−A:= Sk
λB−A+X
α∈Λ(λB−A)
Sk,α
λB−A.(13)
Wi h hese no a ions we ob ain he i s esul in his sec ion.
Theo em 5. Gi en wo linea maps A,B:U → V, hen (a) The subspaces
Sn
λB−Aand Dn
λB−Aa e (λB−A)- educing.
(b) Fo e e y (λB−A)- educing subspace Nwe ha e Sn
λB−A⊂ N ⊂ Dn
λB−A.
Rema k 6. Theo em 5 will be p o en by means o ma ix pencils in Theo-
ems 12 and 20 in Sec ion 3.
P oposi ion 7. To p o e Theo em 5 he e is no loss o gene ali y i , ins ead
o he linea map pencil λB−A, we conside he linea map pencil λD−C=
P◦(λB−A)◦Q(wi h P,Qin e ible ans o ma ions o Vand U, espec i ely).
P oo . Conside he linea maps Q1:Uk→ Ukand P1:Vk+1 → Vk+1 de ined
by
Q1(x1, . . . , xk) := (Q(x1),...,Q(xk))
and
P1(y1, . . . , yk+1) := (P(y1),...,P(yk+1)).
6
F om (8) we immedia ely deduce
Tk
λD−C=P1◦Tk
λB−A◦Q1.
The e o e,
Tk
λB−A(x)=0⇔P1◦Tk
λB−A◦Q1◦Q−1
1(x)=0⇔Tk
λD−C◦Q−1
1(x) = 0.
Tha is, Ke (Tk
λD−C) = Q−1
1(Ke (Tk
λB−A)). , om (11) we in e ha
Sk
λD−C=Q−1(Sk
λB−A).(14)
Using he same a gumen s om (9), (10) and (12), we ob ain Sk,α
λD−C=Q−1(Sk,α
λB−A),
and subs i uing (14) in (13), we ha e
Dk
λD−C=Q−1(Dk
λB−A).(15)
As Nis a (λB−A)- educing subspace i and only i Q−1(N) is (λD−C)-
educing, om (14) and (15) we conclude ha Sn
λB−Aand Dn
λB−Aa e (λB−A)-
educing i and only i Sn
λD−Cand Dn
λD−Ca e (λD−C)- educing. Mo eo e i
is clea ha Sn
λB−A⊂ N ⊂ Dn
λB−Ai and only i Sn
λD−C⊂Q−1(N)⊂ Dn
λD−C.
2
Fo he second esul we need some no a ions. Le Kbe a di ec complemen
o Sn
λB−Ain Dn
λB−Aand le πK:Dn
λB−A→ K be he p ojec ion o e Kalong
Sn
λB−A. Tha is, Im πK=Kand Ke πK=Sn
λB−A. Deno e
HλB−A:= (λB−A)(Dn
λB−A),MλB−A:= (λB−A)(Sn
λB−A).(16)
Now, le Lbe a di ec complemen o MλB−Ain HλB−Aand le πL:HλB−A→
Lbe he p ojec ion o e Lalong MλB−A. Wi h hese no a ions, we ha e he
ollowing esul .
Theo em 8. Le A,B:U → V be linea maps. Then a subspace No Uis
(λB−A)- educing i and only i he subspace πK(N)o Kis de la ing o he
egula pencil
πL◦(λB−A)|K:K → L.
Rema k 9. Theo em 8 will be p o ed in Theo em 20 in Sec ion 3.
P oposi ion 10. The conclusions o Theo em 8do no depend on he choice
o he subspaces K,L.
P oo . Le K1be ano he di ec complemen o Sn
λB−Ain DλB−Aand le
πK1:DλB−A→ K1be he p ojec ion o e K1along Sn
λB−A. In he same
manne , le L1be ano he di ec complemen o MλB−Ain HλB−Aand le
πL1:HλB−A→ L1be he p ojec ion o e L1along MλB−A. Then, (see [15,
Rema k 2, p. 402]), he e exis in e ible linea maps
Q:K → K1,P:L→L1
such ha
∀x∈ K, x −Q(x)∈ Sn
λB−A,∀y∈ L, y −P(y)∈ MλB−A,(17)
7
and mo eo e ,
πK1=Q◦πK, πL1=P◦πL.(18)
See i s ha he pencils πL◦(λB−A)|Kand πL1◦(λB−A)|K1a e s ic ly
equi alen . So, one is egula i and only i he o he is. Obse e ha Qa e P
a e in e ible, i su ices o see ha P◦πL◦(λB−A)|K=πL1◦(λB−A)|K1◦Q.
Gi en ha πL1=P◦πLby (18), i is su icien o p o e ha
πL◦(λB−A)|K=πL◦(λB−A)|K1◦Q.(19)
Le x∈ K. Then, as Q(x)∈ K1, o p o e (19) i su ices o see ha (πL◦
(λB−A))(x−Q(x)) = 0. Bu gi en ha , by (17) , x−Q(x)∈ Sn
λB−A, om
he no a ions o (16) we see ha (λB−A)(x−Q(x)) ∈ MλB−A. The e o e
(πL◦(λB−A))(x−Q(x)) = 0, which p o es (19).
Now see ha
(πL1◦(λB−A))(πK1(N)) = (P◦πL◦(λB−A))(πK(N)).(20)
As by (18), we ha e
πK1(N) = Q(πK(N)) = πK(N)+(Q−I)(πK(N)),
and πL1=P◦πL, we deduce ha
(πL1◦(λB−A))(πK1(N)) = (P◦πL◦(λB−A))(πK(N))
+ (P◦πL◦(λB−A))((Q−I)πK(N)).
(21)
Now, as (17) implies (Q−I)(πK(N)) ⊂(Q−I)(K)⊂ Sn
λB−A, om he no a ions
o (16) we ob ain
(πL◦(λB−A))((Q−I)πK(N)) ⊂(πL◦(λB−A))(Sn
λB−A) = πL(M(B,A)) = {0}.
This las exp ession oge he wi h (21) yields (20).
Finally, as πK1(N) = Q(πK(N)) wi h Qin e ible, we see ha dim(πK1(N)) =
dim(πK(N)). This ac oge he wi h (20) implies
dim ((πL1◦(λB−A))(πK1(N))) = dim(πK1(N))
m
dim ((P◦πL◦(λB−A))(πK(N))) = dim(πK(N)).
Consequen ly, πK(N) is πL◦(λB−A)|K-de la ing i and only i πK1(N) is
πL1◦(λB−A)|K1-de la ing. 2
P oposi ion 11. In he conclusions o Theo em 8 he e is no loss o gene ali y
i we conside he s ic ly equi alen pencil λD−C=P◦(λB−A)◦Q, wi h
in e ible ans o ma ions Pand Qo Vand U, espec i ely.
P oo . Obse e ha om (14), (15) and (16) we ob ain
HλD−C=P(HλB−A),MλD−C=P(MλB−A).(22)
The e o e, as Dn
λB−A=Sn
λB−A⊕ K and HλB−A=MλB−A⊕ L, om (14),
(15) and (22) we deduce ha
8
Dn
λD−C=Sn
λD−C⊕Q−1(K),HλD−C=MλD−C⊕P(L).(23)
To p o e he P oposi ion, see i s ha
πQ−1(K)(Q−1(N)) = (Q−1◦πK)(N).(24)
Le xbe a ec o o N. Then, by 5(b), we see x=y+zwi h y∈ Sn
λB−A
and z∈ K. Hence πK(x) = πK(z) = z. On he o he hand, as Q−1(x) =
Q−1(y) + Q−1(z), wi h Q−1(y)∈ Sn
λD−Cand Q−1(z)∈Q−1(K), we in e ha
πQ−1(K)(Q−1(x)) = πQ−1(K)(Q−1(z)) = Q−1(z) = Q−1(πK(x)),
which p o es (24).
Now conside he in e ible linea maps
Q−1
1:= Q−1|K:K → Q−1(K),P1:= P|L:L → P(L).
Wi h he no a ions o (16), see ha
πP(L)◦PHλB−A=P1◦πLHλB−A.(25)
Le x=y+zbe a ec o o HλB−Awi h y∈ MλB−Aand z∈ L. Then, as
πL(y) = 0 and πL(z) = z, we ha e
P1(πL(x)) = P1(πL(y)) + P1(πL(z)) = P1(z) = P(z).(26)
On he o he hand, as y∈ MλB−A, by (22), we see ha P(y)∈ MλD−C.
The e o e πP(L)(P(y)) = 0. Mo eo e , since z∈ L, i ollows P(z)∈P(L).
Hence πP(L)(P(z)) = P(z). Thus πP(L)(P(x)) = P(z). This equali y oge he
wi h (26) p o es (25).
Now see ha
πP(L)◦(λD−C)Q−1(K)◦Q−1
1=P1◦πL◦(λB−A)|K.(27)
Le x∈ K. Then, as Q−1
1(x) = Q−1(x)∈Q−1(K), we ha e
(πP(L)◦(λD−C))(Q−1
1(x)) =
(πP(L)◦P◦P−1◦(λD−C)◦Q−1)(x) =
(πP(L)◦P◦(λB−A))(x).
The e o e, using (25), we conclude ha
(πP(L)◦(λD−C))(Q−1
1(x)) = (P1◦πL◦(λB−A))(x).
Hence, he pencils πP(L)◦(λD−C)Q−1Kand πL◦(λB−A)|Ka e s ic ly
equi alen . The e o e one is egula i and only i he o he is oo. Finally,
om (27) we ha e
(πP(L)◦(λD−C))(πQ−1(K)(Q−1(N)) = (P1◦πL◦(λB−A))(πK(N)),
gi en ha P1is in e ible we deduce ha he subspaces
(πP(L)◦(λD−C))(πQ−1(K)(Q−1(N))
and (πL◦(λB−A))(πK(N)) ha e he same dimension. Mo eo e , om (24)
we deduce ha πQ−1(K)(Q−1(N)) and πK(N) ha e he same dimension, we
conclude ha πQ−1(K)(Q−1(N)) is πP(L)◦(λD−C)Q−1(K)-de la ing i and
only i πK(N) is πL◦(λB−A)|K-de la ing. 2
9
hence, by Lemma 22 he e exis in e ible ma ices Pand Qsuch ha Bc=
AcQ,B =PA and Λ(P)∩Λ(Q) = ∅. As a consequence,
n1− 0= ank AcAcQ
A Y31 PA Y31= ank AcO
A Y31 PA Y31 −A Y31Q.
The e o e, since ank(Ac) = n1− 0, we ha e PA Y31 −A Y31Q=O. Now, as
Λ(P)∩Λ(Q) = ∅, we see ha A Y31 =O, and hence Y31 =O.
To conclude he p oo i su ices o p o e ha we can choose Y21 =O.
Pa i ioning Y21 acco ding o (43),
Y21 =



O
V
O
W



,
as Y31 and Y32 a e ze o ma ices, om (40), (41) and (43) we deduce ha
ank 





AcO O BcO O
O Iq1O N3V N1O
V O O N2V O O
J3W O J1O O Iq2
J2W O O W O O






=n1+q1+q2− 0.
The e o e, as by Lemma 22 we ha e Bc=AcQ o some ma ix Q, hen
n1− 0= ank 

AcAcQ
V N2V
J2W W 
= ank 

AcO
V N2V−V Q
J2W W −J2WQ
.
Hence, as ank(Ac) = n1− 0we ha e N2V−V Q =Oand W−J2WQ =O.
O equi alen ly WQ−1−J2W=O. Choosing Qin such a way ha Λ(J2)∩
(Λ(Q)∪Λ(Q−1)) = ∅, we deduce ha Vand Wa e null ma ices and he e o e
Y21 =O. The con e se is immedia e. 2
4 P ope ies o he s abili y
In his sec ion we gi e some auxilia y esul s abou he s abili y o educing
subspaces. Fi s , obse e ha om (7), i Xis a basis ma ix o he subspace
N, hen Nis (λB −A)- educing i and only i ank(AX, BX) = ank(X)− 0.
This de ini ion enables us o make a e o mula ion o he concep o s abili y
and Lipschi z s abili y o a educing subspace in e ms o limi s o sequences
o ma ices. To do so, we will use he ollowing esul on he con e gence o a
sequence o subspaces ha one deduces s aigh o wa dly om ([1], Sec ion 1.5,
p. 29–31), ([5], Theo em 13.5.1) and ([3], Theo em I-2-6).
P oposi ion 23. Le Nbe a p-dimensional subspace o Cnand le {Nq}∞
q=1 be
a sequence o subspaces o Cn ha con e ges o Nin he gap me ic. Then, o
each X∈Cn×p, basis ma ix o N, he e exis a sequence o ma ices {Xq}∞
q=1
con e ging o X, wo posi i e cons an s K1,K2, and a posi i e in ege q0, such
ha o q≥q0,Xqis a basis ma ix o Nq, and
K1kXq−Xk ≤ θ(Nq,N)≤K2kXq−Xk.
16

F om P oposi ion 23, we can e o mula e he concep o s able and Lipschi z
s able subspace in e ms o he con e gence o sequences o ma ices. The esul
is he ollowing.
P oposi ion 24. Le λB −A∈ P[λ]m×nbe a ma ix pencil and le Nbe a
(λB −A)- educing subspace such ha dim N=p. Then Nis (λB −A)-s able
i and only i o e e y basis ma ix X∈Cn×po N, and o e e y sequence o
ma ix pencils λBq−Aq→λB −A, he e exis a sequence o ma ices Xq→X
and a posi i e in ege q0, such ha o q≥q0:Xqis a ma ix o ank pand he
subspace hXqiis (λBq−Aq)- educing.
Mo eo e , Nis (λB−A)-Lipschi z s able i and only i he e exis a cons an
K > 0and a posi i e in ege q0such ha o q≥q0
kXq−Xk ≤ K(kAq−Ak+kBq−Bk).
In addi ion, i X=Ip
0, hen o q≥q0we can choose Xq=Ip
Yq, whe e
Yq→0.
We will see some esul s ha will simpli y he s a emen s o Theo em 4 and
some p oo s. The i s is he ollowing, which can be p o ed om P oposi ion 24
and using he echniques employed in he p oo o P oposi ion 3.3 o [18].
P oposi ion 25. Le λB −A∈ P[λ]m×nbe a ma ix pencil and le λD −C∈
P[λ]m×nbe a pencil s ic ly equi alen o λB −A; ha is o say, λD −C=
P(λB −A)Qwi h P∈Cm×mand Q∈Cn×nin e ible ma ices. Le Nbe a
(λB −A)- educing subspace. Then, Nis (λB −A)-s able (o Lipschi z s able)
i and only i Q−1Nis (λB −A)-s able (o Lipschi z s able).
Rema k 26. As a consequence o his P oposi ion, when s udying he s abili y
(o Lipschi z s abili y) o a educing subspace, no gene ali y is los i we conside
ano he s ic ly equi alen pencil and he co esponding ans o med subspace.
To p o e he ollowing esul we need some p e ious no a ions. Gi en a
ma ix pencil λB −A∈ P[λ]m×n, we deno e by CS(λB −A) he se o all
sequences o ma ix pencils ha con e ge o λB −A. Le ˜
CS(λB −A) be a
subse o CS(λB −A). We will say ha a se G ⊂ ˜
CS(λB −A) is a Lipschi z
gene a o subse o ˜
CS(λB −A) i o e e y sequence
{(Aq, Bq)}∞
q=1 ∈˜
CS(λB −A),
he e exis sequences
{(λBq−Aq)}∞
q=1 ∈ G and {(Pq, Qq)}∞
q=1 con e ging o (Im, In),
and he e exis a posi i e in ege numbe q0and a cons an K > 0, ha depends
on he p eceding sequences, such ha o q≥q0,
λBq−Aq=Pq(λBq−Aq)Qq,
max{kPq−Imk,kQq−Imk} ≤ K(kAq−Ak+kBq−Bk).
Wi h he p eceding no a ion we ha e he ollowing p oposi ion, whose demon-
s a ion is simila o ha one o P oposi ion 3.5 o [8] using P oposi ion 25.
17
P oposi ion 27. Le λB −A∈ P[λ]m×nbe a ma ix pencil, and le Nbe a
(λB −A)− educing subspace and Xa basis ma ix o N. Le ˜
CS(λB −A)be a
subse o CS(λB −A)and Ga Lipschi z gene a o subse o ˜
CS(λB −A). Then
he asse ions below a e equi alen .
(i) Fo e e y sequence {(λBq−Aq)}∞
q=1 ∈˜
CS(λB −A), he e exis a sequence
o ma ices Xq→X, a cons an K1>0and a posi i e in ege q1, such
ha o q≥q1, he subspace hXqiis (λBq−Aq)− educing, and
kXq−Xk ≤ K1(kAq−Ak+kBq−Bk).
(ii) Fo e e y sequence {(λBq−Aq)}∞
q=1 ∈ G, he e exis a sequence o ma ices
Xq→X, a cons an K2>0and a posi i e in ege q2, such ha o q≥q2,
he subspace Xqis (λBq−Aq)− educing, and
kXq−Xk ≤ K2(kAq−Ak+kBq−Bk).
In addi ion, i G1is a Lipschi z gene a o subse o G hen G1is a Lipschi z
gene a o subse o ˜
CS(λB −A).
Rema k 28. In he abo e esul s, he exis ence o a posi i e in ege q0is
equi ed in such a way ha he esul s a e ue o q≥q0. To simpli y, wi hou
loss o gene ali y, we will assume he ea e ha q0= 1.
5 P oo o Theo em 4. Asse ions (1), (2) and
(3).
To p o e Asse ions (1), (2) and (3) o Theo em 4, we need some lemmas. Le
λB −Abe a pencil in he o m (2) and le Nbe a subspace (λB −A)- educing,
which, by Theo em 20 can be pu in he o m (37).
Lemma 29. Wi h he p e ious no a ions, i he subspace Nis (λB −A)-s able,
hen hXiis (λB −A )-s able subspace.
P oo . Conside an a bi a y sequence o ma ix pencils λBq
−Aq
con e ging
o λB −A as q→ ∞. F om now on we will summa ize his wi h he no a ion
Sq→L o mean ha Sqis a sequence o ma hema ical objec s con e ging o
he limi Lwhen q→ ∞. Then
λBq−Aq:= 

λBc−Ac0 0
0λBq
−Aq
0
0 0 λB −A 
→λB −A.
Now, as Nis a subspace (λB −A)-s able, he e exis s a sequence o subspaces
Nq→ N such ha Nqis a (λBq−Aq)- educing subspace o e e y q. By he
o m o λBq−Aq, om Theo em 20 we know ha he e exis s a sequence o
ma ices Xq→Xwhe e
Nq=*

In1O
O Xq
O O 
+,
hXqibeing a (λBq
−Aq
)-de la ing subspace. Hence he subspace hXqiis (λB −
A )-s able. 2
18
Rema k 30. Obse e ha by [5, Theo em 14.3.1, p. 429] and [7], i hXiis
(λB −A )-s able, hen hXiis isola ed; ha is, he e exis s a neighbou hood
o he subspace hXisuch ha he unique (λB −A )-de la ing subspace ha is
in his neighbou hood is hXii sel .
Lemma 31. Conside a pencil λB −A∈ P[λ]m×nand a (λB −A)- educing
subspace N, bo h in he o m
λB −A=
n1n2
m1λD −C O
m2O λF −E,N=X
O,(44)
wi h X∈Cn1×pa ma ix o ank p, he pencil λD −Cis le egula and
he pencil λF −Eonly has ow minimal indices. Then he subspace hXiis
(λD −C)- educing. Mo eo e , i he subspace Nis (λB −A)-s able, hen hXi
is (λD −C)-s able.
P oo . As λD −Cis le egula and λF −Eonly has ow minimal indices,
i ollows ha ν(λB −A) = ν(λD −C). Hence, as Nis (λB −A)- educing we
ha e dim(A(N)+B(N)) = p−ν(λD−C). The e o e, om (44) we deduce ha
ank(CX, DX) = ank(X)−ν(λD −C), ha is, hXiis (λD −C)- educing.
Conside now an a bi a y sequence λDq−Cq→λD−C. Then he sequence
λBq−Aq=λDq−CqO
O λF −E→λB −A, (45)
and mo eo e , as λF −Eonly has ow minimal indices,
ν(λBq−Aq) = ν(λDq−Cq).(46)
Now hen, as Nis (λB−A)-s able, he e exis s a sequence o subspaces Nq→ N
such ha Nqis (λBq−Aq)- educing o e e y q. Due o he o m o λBq−Aq,
gi en in (45), om Theo em 20 we see ha he e exis s a sequence o ma ices
Xq→Xsuch ha o e e y q
Nq=Xq
O.(47)
The e o e, because Nqis a (λBq−Aq)- educing subspace, om (46) i ollows
ha dim(Aq(Nq) + Bq(Nq)) = p−ν(λDq−Cq). Tha is, om (45), (46)
and (47), we in e ha
ank(Xq)−ν(λDq−Cq) = p−ν(λDq−Cq) = ank CqXqDqXq
O O = ank(CqXq, DqXq).
Hence hXqiis (λDq−Cq)- educing o e e y q. Las ly, as Xq→Xwe ha e hXi
is (λD −C)-s able. 2
Lemma 32. Conside a pencil λB −A∈ P[λ]m×nand a (λB −A)- educing
subspace N, bo h in he o m
λB −A=
n1n2
m1λD −C O
m2O λF −E,N=In1O
O X,(48)
19
wi h X∈Cn2×pa ma ix o ank p, he pencil λD −Cis le egula and he
pencil λF −Eis igh egula . Then he subspace hXiis (λF −E)- educing.
Mo eo e , i he subspace Nis (λB −A)-s able we ha e hXiis (λF −E)-s able.
P oo . Obse e i s ha ν(λB −A) = ν(λD −C) = n1−m1. Thus, as Nis
(λB −A)- educing, i ollows ha dim(A(N) + B(N)) = p+m1, and hence,
om (48), we see ha
ank C O D O
O EX O FX=p+m1.
The e o e, as ank(C, D) = m1and ν(λF −E) = 0, we ha e ank(EX, FX) =
p= ank(X)−ν(λF −E), ha is, hXiis (λF −E)- educing.
Conside now an a bi a y sequence λFq−Eq→λF −E. Then he sequence
λBq−Aq=λD −C O
O λFq−Eq→λB −A, (49)
and mo eo e , as λFq−Eqis igh egula ,
ν(λBq−Aq) = ν(λD −C) = n1−m1.(50)
Now, gi en ha Nis (λB −A)-s able, om (48) we deduce ha he e exis wo
sequences o ma ices, Xq→Xand Yq→O, such ha o e e y q he subspace
Nq=In1O
YqXq.(51)
is (λBq−Aq)- educing. The e o e, om (49) and (51) we in e ha
p+m1= ank C O D O
EqYqEqXqFqYqFqXq≥ ank(C, D)+ ank(EqXq, FqXq).
Now, as ank(C, D) = m1and ν(λFq−Eq) = 0 we ha e
ank(EqXq, FqXq) = p= ank(Xq)−ν(λFq−Eq),
he o e hXqiis (λFq−Eq)- educing o e e y q. Finally, since Xq→X, i
ollows ha hXiis (λF −E)-s able. 2
Lemma 33. Suppose ha he ma ix pencil λB −A∈ P[λ]m×nhas no ow
minimal indices. Then he unique (λB −A)-s able subspace is Cn.
P oo . Le Nbe a (λB −A)-s able subspace. Gi en ha he pencil has no
ow minimal indices, om Rema k 26, (2) and Theo em 20 we can assume ha
λB −Aand Na e in he o m
λB −A=
n1n2
m1λBc−AcO
n2O λB −A ,N=*
n1p
n1In1O
n2O X +,(52)
whe e λBc−Acis a pencil wi h only column minimal indices, λB −A is
a egula pencil, Xis a ma ix o ank pand hXiis a (λB −A )-de la ing
20
subspace. Thus by [7] we can assume ha
λB −A =



λN1−Ip1λN3O O
O λN2−I 1O O
O O λIp2−J1−J3
O O O λI 2−J2



, X =



Ip1O
O O
O Ip2
O O



,
(53)
wi h N1,N2and N3nilpo en ma ices and p1+p2=p.
Conside now wo a bi a y sequences o ma ices Eq→O∈Cq2×n1and
Fq→O∈Cq1×n1. As he sequence
λBq−Aq=





λBc−AcO O O O
O λN1−Ip1λN3O O
λFqO λN2−I 1O O
O O O λIp2−J1−J3
EqO O O λI 2−J2






(54)
con e ges o λB −Aand he subspace Nis (λB −A)-s able, i ollows ha
he e exis sequences o ma ices Yq, Zq, Uq, Vq, Wq, Hqo adequa e sizes, all o
hem con e ging o O, such ha o e e y q he subspace
Nq=*





In1O O
O Ip1O
YqUqWq
O O Ip2
ZqVqHq





+(55)
is (λBq−Aq)- educing, ha is dim(Aq(Nq)+Bq(Nq)) = dim(Nq)−ν(λBq−Aq).
Bu as he pencil λB −Ahas no ow minimal indices, we ha e ν(λBq−Aq) =
ν(λB −A) = n1−m1. The e o e dim(Aq(Nq)+Bq(Nq)) = n1+p+(n1−m1) =
m1+p. Thus om (54) and (55) we conclude ha
ank 





n1p1p1n1p2p2
m1AcO O BcO O
p1O Ip1O N3YqN1+N3UqN3Wq
1YqUqWqFq+N2YqN2UqN2Wq
p2J3ZqJ3VqJ1+J3HqO O Ip2
2Eq+J2ZqJ2VqJ2HqZqVqHq






=m1+p.
(56)
Since ank(Ac) = ank(Bc) = m1, om (56) we see ha
ank 



Ip1O N1+N3UqN3Wq
UqWqN2UqN2Wq
J3VqJ1+J3HqO Ip2
J2VqJ2HqVqHq



=p.
Consequen ly, om (53), he subspace Mqgene a ed by he columns o he
ma ix
Xq=



Ip1O
UqWq
O Ip2
VqHq




21

is (λB −A )-s able. Hence, by Rema k 30 we in e ha hXqi=hXi o e e y
q, and he e o e he ma ices o he sequences Uq, Vq, Wq, Hqa e all ze o. So,
om (56) we deduce ha
ank 

n1n1
m1AcBc
1YqFq+N2Yq
2Eq+J2ZqZq
=m1.(57)
A his poin , no e ha o p o e he lemma i su ices o e i y ha 1= 2=
0. Fo he sake o con adic ion, assume i s ha 1>0, hen as ank(Ac) =
ank(Bc) = m1, om (57) we deduce ha
ank AcBc
YqFq+N2Yq=m1.(58)
Now, by Lemma 22 he e exis s a ma ix Q∈Cn1×n1wi h 0 ∈Λ(Q) such
ha Bc=AcQ. Then om (58) we immedia ely ob ain
m1= ank(Ac) = ank AcAcQ
YqFq+N2Yq= ank AcO
YqFq+N2Yq−YqQ,
and hence Fq+N2Yq−YqQ=O. In conclusion, i i we e ue ha 1>0,
we would ha e p o ed ha o e e y sequence o ma ices Fq→O he e exi s a
sequence o ma ices Yq→Osuch ha o each qi sa is ies Fq+N2Yq−YqQ=
O, wi h Λ(N2)∩Λ(Q)6=∅, which is impossible. The e o e, 1= 0.
I i we e ue ha 2>0, as by Lemma 22 he e exis s a ma ix Q∈Cn1×n1
wi h Λ(J2)∩Λ(Q)6=∅, so ha Ac=BcQ, applying he p e ious easoning we
would lead o a con adic ion. Thus 2= 0. 2
Lemma 34. Gi en a ma ix pencil λB −A∈ P[λ]m×n, suppose ha i has no
column minimal indices. Then he unique (λB −A)-s able subspace is {0}.
P oo . Le Nbe a (λB −A)-s able subspace. We can now p oceed analogously
o he p oo o he p e ious lemma. So, we can assume ha λB −Aand Na e
in he o m
λB −A=
n2n3
n2λB −A O
m3O λB −A ,N=*
p
n2X
n3O+,(59)
whe e λB −A is a egula pencil and hXiis a (λB −A )-de la ing subspace,
bo h in he o m (53). Mo eo e , λB −A is a pencil ha only has ow minimal
indices.
Now conside wo a bi a y sequences o ma ices Eq→O∈Cm3×p2and
Fq→O∈Cm3×p1. As he sequence
λBq−Aq=





λN1−Ip1λN3O O O
O λN2−I 1O O O
O O λIp2−J1−J3O
O O O λI 2−J2O
λFqO−EqO λB −A






,
(60)
22
con e ges o λB−Aand he subspace Nis (λB−A)-s able, he e exis sequences
o ma ices Yq, Zq, Uq, Vq, Wq, Hq, o adequa e sizes, ha con e ge o O, such
ha o e e y q he subspace
Nq=*





Ip1O
UqWq
O Ip2
VqHq
YqZq





+(61)
is (λBq−Aq)- educing; ha is dim(Aq(Nq)+Bq(Nq)) = dim(Nq)−ν(λBq−Aq).
Bu as he pencil λB −Ahas no column minimal indices, hen ν(λBq−Aq) =
ν(λB −A) = 0. The e o e dim(AqNq+BqNq) = p. Thus om (60) and (61)
we see ha
ank 





p1p2p1p2
p1Ip1O N1+N3UqN3Wq
1UqWqN2UqN2Wq
p2J3VqJ1+J3HqO Ip2
2J2VqJ2HqVqHq
m3A YqEq+A ZqFq+B YqB Zq






=p. (62)
Now, as he sequences Yq, Zq, Uq, Vq, Wq, Hqcon e ge o O, om (62)
ank 



Ip1O N1+N3UqN3Wq
UqWqN2UqN2Wq
J3VqJ1+J3HqO Ip2
J2VqJ2HqVqHq



=p;
ha implies o (53) ha he subspace Mqgene a ed by he ma ix
Xq=



Ip1O
UqWq
O Ip2
VqHq




is (λB −A )-de la ing. Hence, like in he p e ious lemma, all he e ms o he
sequences o ma ices Uq, Vq, Wq, Hqa e O. Thus om (62) we in e ha
ank 

p1p2p1p2
p1Ip1O N1O
p2O J1O Ip2
m3A YqEq+A ZqFq+B YqB Zq
=p;
ha is
Fq+B Yq−A YqN1=O, Eq+A Zq−B ZqJ1=O. (63)
Now, by Lemma 22 he e exis ma ices P, Q ∈Cm3×m3wi h 0 ∈Λ(P) and
Λ(J2)∩Λ(Q)6=∅such ha B =PA and A =QB . Then we immedia ely
see om (63) ha o e e y pai o sequences o ma ices Eq, Fq→O he e a e
sequences Yq, Zq→O ha sa is y
Fq+PA Yq−A YqN1=O, Eq+QB Zq−B ZqJ1=O,
23
o e e y q. This con adic s o he choice o Pand Q. Consequen ly p1=p2= 0
and he e o e N={0}.2
We a e now eady o p o e Asse ions (1) (2) and (3) o Theo em 4.
P oo Theo em 4: Asse ions (1), (2) and (3).
Fi s , no e ha Asse ions (2) and (3) ollow s aigh o wa d om Lem-
mas 33 and 34, espec i ely.
Second, o p o e Asse ion (1), by Rema k 28, we can assume ha λB −Ais
in he o m gi en in (2), wi h n1, n2, m3nonze o . Le Nbe a (λB−A)- educing
subspace. Then, by Theo em 20, we can assume ha
N=*

n1p
n1In1O
n2O X
m3O O 
+,
whe e ank(X) = p. Suppose ha Nis (λB −A)-s able. Hence, as he pencil
diag(λBc−Ac, λB −A ) is le egula and he pencil λB −A only has ow
minimal indices, applying Lemma 31, we deduce ha he subspace
In1O
O X
is diag(λBc−Ac, λB −A )-s able. The e o e, by Lemma 33 we ha e X=In2.
On he o he hand, applying Lemma 32 o he pencils λBc−Ac(le egula )
and diag(λB −A , λB −A ) ( igh egula ), since Nis (λB −A)- educing,
we see ha he subspace gene a ed by he columns o he ma ix In2
Ois
diag(λB −A , λB −A )-s able, which con adic s Lemma 34. 2
6 P oo o Theo em 4: Asse ion (4).
In his sec ion we p o e Asse ion (4) o Theo em 4. The e o e in all he sec ion
we will assume ha he pencil λB −Ahas ow minimal indices, a leas wo
column minimal indices, and no eigen alues. The ollowing esul will allow us
o simpli y he p oo s.
Lemma 35. Le λB −A∈ P[λ]m×nbe a pencil wi hou eigen alues and le N
be a (λB −A)− educing subspace, which a e gi en by
λB −A=
n1n2
m1λD −C0
n20λF −E,N=*
p1p2
n1X0
n20Y+,
whe e X, Y a e ma ices o ull column ank. Le M:= hXi. Then i Nis
(λB −A)−s able i ollows ha Mis (λD −C)−s able.
P oo . No e ha as he pencil λF −Ehas no eigen alues, om (31) and (33)
we see ha Dk
λB−A=Sk
λB−A. Hence by Theo em 20 we in e ha
hYi=Dn
λF −E=Sn
λF −E.(64)
24
Conside now an a bi a y sequence λDq−Cq→λD −C. Then as he
sequence
λBq−Aq=λDq−Cq0
0λF −E
con e ges o λB −Aand Nis (λB −A)−s able, he e exis sequences Xq→X,
Yq→Y,Zq→0, Vq→0, such ha o e e y q he subspace
Nq:= XqZq
VqYq (65)
is (λBq−Aq)− educing. The e o e, by Theo em 20, Nq⊂ Dn
λBq−Aq. Now
applying Lemma 14 and (64), we ob ain
Nq⊂ Dn
λBq−Aq=Dn
λDq−Cq
O⊕O
Dn
λF −E⊂In1
O⊕O
Y=In1O
O Y .
Fo his eason om (65) he e exis ma ices o adequa e sizes Qi,i= 1,2,3,4,
such ha XqZq
VqYq=In1O
O Y Q1Q2
Q3Q4.
Obse e ha Yq=Y Q4. Mo eo e , as Yq→Yand Yis o ull column ank, we
deduce ha Q4is in e ible. Hence, as Vq=Y Q3, i ollows ha Vq=YqQ−1
4Q3.
Thus, in (65), i we sub ac o he i s column he second one mul iplied by
Q−1
4Q3we ob ain
Nq=Xq−ZqQ−1
4Q3Zq
O Y Q4=Xq−ZqQ−1
4Q3ZqQ−1
4
O Y .(66)
In he same way, om Theo em 20, Lemma 14 and (64) we ob ain
Nq⊃ Sn
λBq−Aq=Sn
λDq−Cq
O⊕O
Sn
λF −E=Sn
λDq−Cq
O⊕O
Y⊃O
Y.
F om (66) we deduce ha he e exis ma ices o adequa e sizes P1, P2such ha
O
Y=Xq−ZqQ−1
4Q3ZqQ−1
4
O Y P1
P2.
Hence P2=Ip2and ZqQ−1
4=−(Xq−ZqQ−1
4Q3)P1. The e o e deno ing ˜
Xq:=
Xq−ZqQ−1
4Q3→X, om (66) we see ha
Nq=˜
XqO
O Y .
Finally, as Nqis (λB −A)− educing
ank Cq˜
XqO Dq˜
XqO
O EY O F Y =p1+p2−ν(λDq−Cq)−ν(λF −E).(67)
Bu gi en ha hYiis a (λF−E)− educing subspace, i ollows ha ank(EY, FY ) =
p2−ν(λF −E). Thus om (67) we conclude ha ank(Cq˜
Xq, Dq˜
Xq) = p1−
ν(λDq−Cq); ha is he subspace D˜
XqEis (λDq−Cq)− educing. Consequen ly
Mis (λD −C)−s able. 2
25
o p o e he lemma we can assume ha (G, H) = (G, H). Deno e α:=
(a1, a2,· · · , ah) wi h ai∈C1× i. Then, because n ank(λD −C) = m, we ha e
n ank 






λI 1−G1−H10 0 · · · 0 0
0 0 I 2−G2−H2· · · 0 0
.
.
..
.
..
.
..
.
.....
.
..
.
.
0 0 0 0 . . . I h−Gh−Hh
a10a20. . . ah0







=m.
(73)
Now, as
(λI i−Gi,−Hi) = 




λ−1 0 . . . 0 0
0λ−1. . . 0 0
.
.
..
.
..
.
.....
.
..
.
.
0 0 0 . . . λ −1





,
i we deno e by ai:= (bi1, bi2, . . . , bi i), making ans o ma ions by columns in
he ma ix o (73), we deduce ha
n ank 






0−I 10 0 · · · 0 0
000−I 2· · · 0 0
.
.
..
.
..
.
..
.
.....
.
..
.
.
0000. . . 0−I h
p1(λ)? p2(λ)? . . . ph(λ)?







= 1+ 2+· · ·+ h=m,
wi h pi(λ) = bi1+bi2λ+· · · +bi iλ i−1. The e o e pi(λ) = 0, o equi alen ly
α= 0. 2
Conside now a sequence λBq−Aq→λB −A. F om (72), we can assume
ha o e e y q,
λBq−Aq=λIq
mβq
αqδq−EqFq
ηqθq.
Now, as m= n ank(λBq−Aq)≥ ank(Aq)≥ ank(Eq, Fq)≥ ank(E, F) = m,
i is immedia e o see ha he e exis s a sequence o ma ices
Pq=Im0
ξq1→Im+1,
such ha o e e y q
PqAq=EqFq
0 0 .
Hence, by P oposi ion 27, i su ices o conside sequences o he o m Pq(λBq−
Aq) o s udy he s abili y o he subspace Cn, ha is,
λBq−Aq=λIq
mβq
αqδq−EqFq
0 0 →λB −A. (74)
On he o he hand, as he sequence o ma ices
Qq=(Iq
m)−1−(Iq
m)−1βq
0 1 →Im+1,
32

by P oposi ion 27 i is su icien o conside sequences o he o m (λBq−Aq)Qq;
ha is, om (74),
λBq−Aq=λIm0
αqδq−EqFq
0 0 →λB −A.
Bu , since m= n ank(λBq−Aq)≥ ank(Bq)≥m, we ha e δq= 0, o e e y q.
,
λBq−Aq=λIm0
αq0−EqFq
0 0 .
Now, since (Eq, Fq) is con ollable and n ank(λB −A) = m, by Lemma 37 i
ollows ha αq= 0 o e e y q. Hence, i su ices o conside sequences o he
o m
λBq−Aq=λIm0
0 0−EqFq
0 0 .
Finally, as ank(AqIn, BqIn) = m=n−ν(λBq−Aq), hen he subspace Cnis
(λBq−Aq)- educing and, , Cnis (λB −A)-s able.
8 P oo o Asse ion (6) o Theo em 4
In his sec ion we will p o e ha i he pencil λB −A∈ P[λ]m×nhas only one
ow minimal index which is di e en han ze o, a leas wo column minimal
indices and no eigen alues, hen i has no any s able educing subspace. Fi s
no e ha om Lemma 35, in an analogous way as in Rema k 36, we can assume
ha λB −Aonly has wo column minimal indices. Hence we conside h ee
subcases: (a) wo column minimal indices which a e equal o ze o; (b) one
column minimal index which is equal o ze o and ano he column minimal index
which is di e en han ze o; (c) bo h column minimal indices which a e di e en
han ze o.
8.1 Two column minimal indices which a e equal o ze o
Deno e D:= Ik
0, C := 0
Ik∈C(k+1)×k, in his case we can assume ha he
pencil λB −Aand he unique educing subspace Na e o he o m
λB −A=λ(0, D)−(0, C)∈ P[λ](k+1)×(2+k),N=I2
0,
espec i ely.
Conside he sequences o ma ices
aq=




1/q 0
0 0
.
.
..
.
.
0 0





, bq=




0 0
.
.
..
.
.
0 0
0 1/q





∈C(k+1)×2,
he sequence λBq−Aq=λ(bq, D)−(aq, C) con e ges o λB −Aand mo eo e
ν(λBq−Aq) = 1. Now, i Nis (λB −A)-s able, by P oposi ion 24 he e exis s
a sequence o ma ices Xq→0∈Ck×2such ha o e e y q he subspace
33
Nq=DI2
XqEis (λBq−Aq)- educing; ha is, dim(Aq(Nq) + Bq(Nq)) = 1.
The e o e, i we de ine Xq:= (xq
ij)1≤i≤k,j=1,2so ha Nqcan be (λBq−Aq)-
educing, i mus be sa is ied
ank 






1/q 0xq
11 xq
12
xq
11 xq
12 xq
21 xq
22
.
.
..
.
..
.
..
.
.
xq
k−1,1xq
k−1,2xq
k1xq
k2
xq
k1xq
k20 1/q







= 1.
Hence xq
ij = 0. Thus, we conclude ha ank 1/q 0
0 1/q = 1, which is a con a-
dic ion. In conclusion, Nis no (λB −A)-s able.
8.2 One column minimal index which is equal o ze o and
ano he column minimal index which is di e en han
ze o
De ine λD −C:= λ[In,0] −[0, In] and λF −E:= λIm
0−0
Im, in his case
i ollows ha he pencil λB −Aand i s unique educing subspace Nha e he
o m
λB −A=0λD −C0
0 0 λF −E,N=*

1 0
0In+1
0 0 
+,
espec i ely.
Conside he sequences
aq=


1/q
.
.
.
0


∈C(m+1)×1, bq=


0· · · 0 0
.
.
....0 0
0· · · 0 1/q


∈C(m+1)×(n+1).
Then
λBq−Aq=λ0D0
0bqF−0C0
aq0E→λB −A,
and mo eo e ν(λBq−Aq) = 1. Hence, i Nis (λB −A)-s able, he e exis
sequences o ma ices Xq→0∈Cm×1and Yq→0∈Cm×(n+1) such ha he
subspace
Nq:= *

1 0
0In+1
XqYq
+,
is (λBq−Aq)- educing; ha is, dim(Aq(Nq) + Bq(Nq)) = n+ 1. Thus
ank 0C0D
aq+EXqEYqF Xqbq+FYq=n+ 1.(75)
De ine Xq:= (xq
1, xq
2, . . . , xq
m)T. Then as ank C=n, om (75) we ha e
1≥ ank(aq+EXq) = ank 1/q xq
1· · · xq
m−1xq
m
xq
1xq
2· · · xq
m0T
.
34
The e o e xq
i= 0 and Xq= 0. Now, deno e by Yq
n+1 := (yq
1, yq
2, . . . , yq
m)T he
las column o Yq. As ank D=n, om (75) we see ha
1≥ ank(aq, bq+FY q
n+1) = ank 1/q 0· · · 0 0
yq
1yq
2· · · yq
m1/qT
,
which is a con adic ion. Thus, N=Cn+1 is no (λB −A)-s able.
8.3 Two column minimal indices which a e di e en han
ze o.
De ine λD −C:= λ[In,0] −[0, In], λF −E:= λ[Ip,0] −[0, Ip] and λH −G:=
λIm
0−0
Im, in his case we in e ha he pencil λB −Aand i s unique
educing subspace Nha e he o m
λB −A=

λD −C0 0
0λF −E0
0 0 λH −G
,N=*

In+1 0
0Ip+1
0 0 
+,
espec i ely.
Conside he sequences
aq=




1/q 0· · · 0
0 0 · · · 0
.
.
..
.
....0
0 0 · · · 0





∈C(m+1)×(n+1), bq=


0· · · 0 0
.
.
....0 0
0· · · 0 1/q


∈C(m+1)×(p+1).
Then
λBq−Aq=λ

D0 0
0F0
0bqH
−

C0 0
0E0
aq0G
→λB −A,
and mo eo e , ν(λBq−Aq) = 1. Hence, i Nis (λB −A)-s able, he e exis
sequences o ma ices Xq→0∈Cm×(n+1) and Yq→0∈Cm×(p+1) such ha
he subspace
Nq:= *

In+1 0
0Ip+1
XqYq
+
is (λBq−Aq)- educing; ha is, dim(Aq(Nq) + Bq(Nq)) = n+p+ 1. The e o e
ank 

C0D0
0E0F
aq+GXqGYqHXqbq+HYq
=n+p+ 1.
Deno e by Xq= (xij) and Yq= (ykl), om he p e ious equali y i ollows ha
ank









0In0 0 In0 0 0
0 0 0 Ip0 0 Ip0
1/q 0···0 0 0 ···0x11 ···x1nx1,n+1 y11 ···y1py1,p+1
x11 x12 ···x1,n+1 y11 y12 ···y1,p+1 x21 ···x2nx2,n+1 y21 ···y2py2,p+1
.
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
.
xm1xm2···xm,n+1 ym1ym2···ym,p+1 0···0 0 0 ···0 1/q









=n+p+1.
35
Obse e now ha , choosing he subma ix o med by he n+p+ 2 i s
columns, we deduce immedia ely ha yi1= 0 o i= 1,2, . . . , m. In he same
way, wi h he n+p+ 2 las columns we see ha xi,n+1 = 0 o i= 1,2, . . . , m.
Hence wi h he en ies 1 co esponding o he places (n, n + 1) and (n+ 1,2n+
p+4) we can educe he p e ious ma ix o one on he same o m, bu educing
he sizes om n o n−1 and om p o p−1, and whose ank is n+p−1.
Following his p ocess we each he case whe e a leas one column minimal
index is equal o ze o, which is al eady sol ed in Subsec ions 8.1 and 8.2.
9 P oo o Asse ion (7) o Theo em 4
In his sec ion we will analyze he case o a ma ix pencil wi h only one ow
minimal index which is di e en han ze o, and one column minimal index which
is equal o ze o. P e iously we will in oduce some auxilia y esul s. We begin
by s a ing some bounds abou he maximum modulus o a oo o a polynomial,
ha can be seen in [14], Sec ion 8, pp. 243–247.
Lemma 38. Le (z) = a0+a1z+· · ·+an−1zn−1+znbe a polynomial o deg ee
nwi h coe icien s in Cdis inc om he polynomial zn. Deno e
α:= max
0≤k≤n−1 n
k−1
|ak|!1/(n−k)
.
Assume ha znis a oo o maximum modulus o (z). Then
(21/n −1)α < |zn| ≤ (21/n −1)−1α.
In o de o p o e Lemma 41, we need he ollowing wo lemmas. The i s
one, Lemma 39, is deduced immedia ely om (34).
Lemma 39. Conside he ma ix pencil λB −A∈ P[λ]m×nand he ma ix
Tk
λB−Ade ined in (28). Then
(i) I ν(Tk
λB−A)=0, hen λB −Ahas no any column minimal indices
≤k−1.
(ii) I λB −A=λ(Ik,0) −(0, Ik)∈ P[λ]k×(k+1), hen
ν(Tp
λB−A) = (0i p≤k
1i p≥k+ 1.
The second one, Lemma 40, can be seen in [5], Theo em 13.5.1, p. 406.
Lemma 40. Le F∈Cp×qand le X∈Cq× be a basis ma ix o Ke F.
Conside a sequence Fq→Fsuch ha , o e e y q,ν(Fq) = ν(F). Then he e
exis a sequence Xq→Xand a posi i e cons an K1such ha , o e e y q,Xq
is a basis ma ix o Ke Fqand
kXq−Xk ≤ K1kFq−Fk.
36
Based on hese esul s we will p o e he ollowing lemma.
Lemma 41. Conside he pencil λB −A=λ(Ik,0) −(0, Ik)∈ P[λ]k×(k+1).
Then o each sequence λBq−Aq→λB−A he e exis wo sequences o ma ices
Pq→Ikand Qk→Ik+1 such ha , o e e y q, we ha e P−1
q(λBq−Aq)Qq=
λB −A; mo eo e , he e exis s a cons an K > 0 ha sa is ies
max{kPq−Ikk,kQq−Ik+1k} ≤ K(kAq−Ak+kBq−Bk).
P oo . No e i s ha by Lemma 39, ν(Tk+1
λB−A) = 1. Deno e by {e1, e1, . . . , ek+1}
he ec o s o he canonical basis o Ck+1, i is clea ha
Ke (Tk+1
λB−A) = *




ek+1
ek
.
.
.
e1




+.(76)
Now conside a sequence λBq−Aq→λB −A. Since n ank(λBq−Aq) = k, i
ollows ha he pencil λBq−Aqhas a leas a column minimal index. Mo eo e ,
o e e y pwe ha e ν(Tp
λBq−Aq)≤ν(Tp
λB−A), by Lemma 39, i ollows ha he
pencil λBq−Aqhas no any column minimal indices < k. Tha is, i has
one column minimal index which is equal o k; hence ν(Tk+1
λBq−Aq) = ν(Tk+1
λB−A).
Since Tk+1
λBq−Aq→Tk+1
λB−Aand o e e y q he ma ices Tk+1
λBq−Aqha e he same
nulli y, i ollows om (76) and Lemma 40 ha he e exis s a basis ma ix o
he subspace Ke (Tk+1
λBq−Aq)





xq
k+1
xq
k
.
.
.
xq
1





con e ging o 




ek+1
ek
.
.
.
e1





such ha kxq
i−eik ≤ K1(kAq−Ak+kBq−Bk). Now le
Pq:= (Aqxq
2,· · · , Aqxq
k+1), Qq:= (xq
1,· · · , xq
k+1).
I is ob ious ha P−1
q(λBq−Aq)Qq=λB−Aand kQq−Ik+1k ≤ K1(kAq−Ak+
kBq−Bk). I su ices o demons a e ha kPq−Ikk ≤ K2(kAq−Ak+kBq−Bk)
o conclude he p oo o he lemma.
In ac , deno ing by ( 1, 2, . . . , k) he canonical basis o Ck, i ollows ha
kPq−Ikk ≤
k+1
X
i=2
kAqxq
i− i−1k=
k+1
X
i=2
kAqxq
i−Aeik.
Now
kAqxq
i−Aeik ≤ kAqxq
i−Aqeik+kAqei−Aeik≤kAqkkxq
i−eik+kAq−Akkeik
≤(kAq−Ak+kAk)kxq
i−eik+kAq−Ak ≤ K2(kAq−Ak+kBq−Bk),
2
Wi h hese p e ious esul s we a e eady o p o e Asse ion (7) o Theo em 4.
P oo o Asse ion (7) o Theo em 4
37

De ine D:= In
0and C:= 0
In, bo h ma ices o C(n+1)×n. Hence,
λB −A=λ(0, D)−(0, C)∈ P[λ](n+1)×(n+1).
The unique educing subspace o λB −Ais N=he1i, wi h e1 he i s canonical
ec o o Cn+1. No e ha ν(λB −A) = 1. Now conside a sequence (λBq−
Aq)→(λB −A). Then, by Lemma 40 and by P oposi ion 27, when s udying
he Lipschi z s abili y o he subspace N, no gene ali y is los i we only conside
sequences o he o m λBq−Aq=λ(εq, D)−(δq, C). Ope a ing wi h he columns
o D, by P oposi ion 27, we can assume ha
λBq−Aq=λ0In
aq0−bq0
HqIn,(77)
wi h Hq= (cq
1, cq
2, . . . , cq
n)T∈Cn.
No e ha making ow ope a ions i is immedia e o see ha
de (λBq−Aq) = aqλn+1 −
n
X
i=1
cq
iλi−bq.
The e o e, ν(λBq−Aq) = 1 i and only i aq=bq=cq
i= 0, which is equi alen o
λBq−Aq=λB−A. Fo his case, i is clea ha Nq=Nis a educing subspace
o λBq−Aq. Thus, om he e on, we will assume ha ν(λBq−Aq) = 0. In
o de o p o e ha Nis Lipschi z s able, i su ices o ind sequences o complex
numbe s xq
i,i= 1,2, . . . , n, such ha o e e y q, he subspace
Nq:= *






1
xq
1
xq
2
.
.
.
xq
n






+
is (λBq−Aq)− educing; ha is, since ν(λBq−Aq) = 0, i ollows ha dim Aq(Nq)+
Bq(Nq) = 1 holds. O , which is he same, om (77)
ank









bqxq
1
cq
1+xq
1xq
2
cq
2+xq
2xq
3
.
.
..
.
.
cq
n−1+xq
n−1xq
n
cq
n+xq
naq









= 1,(78)
and, mo eo e , ha he e exis s a cons an K > 0 such ha ,
|xq
i| ≤ K(kBq−Bk+kAq−Ak), i = 1,2, . . . , n. (79)
No e i s ha i aq= 0, i su ices o ake xq
i= 0 o each i. On he o he
hand, i bq= 0, i is su icien o choose xq
i=−cq
i o each i. Hence, we will
assume ha aqbq6= 0. In o de o (78) o hold, since aq6= 0, we sea ch o he
xq
iin such a way ha he i s column is p opo ional o he second one. No e
38
ha he p opo ionali y ac o is bq
1/xq
1. Now, doing ope a ions in (78), by a
induc ion p ocess i is p o ed ha
xq
k=
(xq
1)k+
k−1
X
i=1
cq
ibi−1
q(xq
1)k−i
bk−1
q
, k = 2,3, . . . , n, (80)
and o xq
1we ha e
(xq
1)n+1 +
n
X
i=1
cq
ibi−1
q(xq
1)n−i+1 −aqbn
q= 0.(81)
Conside he polynomial
q(z) := zn+1 +
n
X
i=1
cq
ibi−1
qzn−i+1 −aqbn
q.
We ind a bound o he maximum modulus o i s oo s. De ine Bk
n+1 :=
n+1
k−1/(n−k+1), by Lemma 38,
α= max{B0
n+1 |aqbn
q|1/(n+1), B1
n+1 |cq
nbn−1
q|1/n,...,
. . . , |Bk
n+1 |cq
n−k+1bn−k
q|1/(n−k+1), . . . , Bn
n+1 |cq
1|}.(82)
A e ha , we choose xq
1as one o he oo s o (z) ha ha e maximum modulus.
By Lemma 38 and (82) i is clea ha xq
1sa is ies (79).
Le k∈ {2,3, . . . , n}. Then, combining (80) and (81) we in e ha
xq
k=−
n
X
i=k
cq
ibi−1
q(xq
1)n−i+1 −aqbn
q
bk−1
q(xq
1)n+1−k=−
n
X
i=k
cq
ibi−k
q
(xq
1)i−k+aqbn−k+1
q
(xq
1)n−k+1 .(83)
In o de o conclude his case, i su ices o see ha each summand o (83) is
bounded by K(kBq−Bk+kAq−Ak), o a posi i e cons an K.
Fi s , by Lemma 38 and (82) i ollows ha he e exis s a posi i e cons an
Lsuch ha |xq
1|−1≤L|aqbn
q|−1/(n+1). The e o e,

aqbn−k+1
q
(xq
1)n−k+1 
≤L
aqbn−k+1
q
(aqbn
q)(n−k+1)/(n+1) 
=Lak/(n+1)
qb(n−k+1)/(n+1)
q≤K(kBq−Bk+kAq−Ak).
Second, ollowing Lemma 38 and (82) again, we see ha he e exis s a posi i e
cons an Lisuch ha |xq
1|−1≤Li|cq
ibi−1
q|−1/i. Thus,

cq
ibi−k
q
(xq
1)i−k
≤Li
cq
ibi−k
q
(cq
ibi−1
q)(i−k)/i 
=Li(cq
i)k/ib(i−k)/i
q≤K(kBq−Bk+kAq−Ak).
2
Acknowledgemen
The au ho s hank he e e ee o he de ailed help in imp o ing he w i ing o
his a icle.
39
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