On eigenvalues of generalized shift linear vector isomorphisms
Fatemeh Ayatollah Zadeh Shirazi∗
Faculty of Mathematics, Statistics and Computer Science, College of Science, University of Tehran, Tehran, Iran. E-mail: [email protected]
Elham Soleimani
Faculty of Mathematics, Statistics and Computer Science, College of Science, University of Tehran, Tehran, Iran. E-mail: [email protected]
Abstract Our main aim is to compute eigenvalues of generalized shift isomorphismσϕ:VΓ→
VΓwithσ
ϕ((xα)α∈Γ) = (xϕ(α))α∈Γ((xα)α∈Γ∈VΓ) whereV is a vector space (over fieldF), Γ is a nonempty arbitrary set andϕ: Γ→Γ is an arbitrary bijection.
Keywords. Eigenvalue, Generalized shift, Isomorphism.
2010 Mathematics Subject Classification. 15A18.
1. Introduction
Let’s mention that one–sided shift {1, . . . , k}N→ {1, . . . , k}N (x1,x2,x3,···)7→(x2,x3,x4,···)
and two–sided shift
{1, . . . , k}Z→ {1, . . . , k}Z (xn)n∈Z7→(xn+1)n∈Z
are two well-known operators in dynamical systems,
er-godic theory [5], etc. For arbitrary setX with at least two elements, nonempty set Γ and ϕ : Γ → Γ generalized shift σϕ : XΓ→XΓ
(xα)α∈Γ7→(xϕ(α))α∈Γ
has been introduced for
the first time in [2] as a generalization of one–sided and two–sided shifts. It’s evident that ifX has group structure, thenσϕ:XΓ →XΓ is group homomorphism (see e.g.,
[1]) and for topological spaceX,σϕ:XΓ→XΓ is continuous, whereXΓ is equipped
with product topology, see e.g. [3].
Convention. In the following text, supposeV(6= 0) is a linear vector space over field
F, Γ is an arbitrary set with at least two elements andϕ: Γ→Γ is an arbitrary map. Generalized shift σϕ : VΓ → VΓ with σϕ((xα)α∈Γ) = (xϕ(α))α∈Γ is a linear vector
space homomorphism since for (xα)α∈Γ,(yα)α∈Γ∈VΓ andr∈F, letzα=xα+ryα,
now we have:
σϕ((xα)α∈Γ+r(yα)α∈Γ) = σϕ((zα)α∈Γ) = (zϕ(α))α∈Γ
= (xϕ(α)+ryϕ(α))α∈Γ= (xϕ(α))α∈Γ+r(yϕ(α))α∈Γ
= σϕ((xα)α∈Γ) +rσϕ((yα)α∈Γ).
∗Corresponding author.
Forα, β∈Γ, letα∼ϕβ or brieflyα∼βif there existn, m >0 withϕn(α) =ϕm(β).
Clearly∼ϕis an equivalence relation on Γ. Forα∈Γ by
α ∼ϕ
we mean the equivalence
class of αin ∼ϕ. In other words
α ∼ϕ
=S
{ϕn(α) :n ∈Z}. In particular one may
considerϕ| α
∼ϕ :
α ∼ϕ
→ α
∼ϕ
.
2. Eigenvalues of generalized shifts
In this section we compute eigenvalues of homomorphismσϕ:VΓ →VΓ.
Lemma 2.1 (Decomposition Lemma). For r∈F,r is an eigenvalue of σϕ :VΓ→
VΓ if and only if there exists α∈Γ such that r is an eigenvalue ofσ
ϕ| α
∼ϕ
:V∼αϕ →
V∼αϕ. I.e.,
Eigen(σϕ, VΓ) =
[
α∈Γ
Eigen(σϕ| α
∼ϕ
, V∼αϕ),
where for vector spaceW and homomorphismh:W →W, Eigen(h, W)denotes the collection of all eigenvalues ofh:W →W.
Proof. For α∈Γ, supposer∈Eigen(σϕ| α
∼ϕ
, V α
∼ϕ). Choose (x λ)λ∈ α
∼ϕ ∈V α
∼ϕ \ {0}
withϕ| α
∼ϕ((xλ)λ∈ α
∼ϕ) =r(xλ)λ∈ α
∼ϕ. Forλ∈Γ\
α ∼ϕ
letxλ:= 0. For allλ∈Γ\
α ∼ϕ
we haveϕ(λ)∈Γ\ α ∼ϕ
which leads to
xλ=xϕ(λ)=rxλ, (λ∈Γ\
α ∼ϕ
). (*)
Using (*) and (xϕ(λ))λ∈ α
∼ϕ = (rxλ)λ∈ α
∼ϕ we haveσϕ((xλ)λ∈Γ) =r(xλ)λ∈Γ. Moreover (xλ)λ∈Γ6= 0 since (xλ)λ∈ α
∼ϕ 6= 0. Hence ris an eigenvalue ofσϕ:V
Γ→VΓ.
Conversely, if r ∈ F an eigenvalue of σϕ : VΓ → VΓ, there exist (xλ)λ∈Γ ∈ VΓ\ {0} such that σϕ((xλ)λ∈Γ) = r(xλ)λ∈Γ. Choose α ∈ Γ such that xα 6= 0. Using
σϕ| α
∼ϕ
((xλ)λ∈ α
∼ϕ) = r(xλ)λ∈ α
∼ϕ, we have r ∈ Eigen(σϕ|∼αϕ
, V ∼αϕ) which completes
the proof.
Remark 2.2. We have the following statements [2]:
•the mappingσϕ:VΓ→VΓ is injective if and only ifϕ: Γ→Γ is surjective;
•the mappingσϕ:VΓ→VΓ is bijective if and only ifϕ: Γ→Γ is injective.
Corollary 2.3. By Remark 2.2,0 is an eigenvalue of σϕ :VΓ →VΓ if and only if
Definition 2.4. Forα, β∈Γ let:
m(α, β) =
min{n >0 :ϕn(α) =β}, ∃n >0(ϕn(α) =β),
0, otherwise.
It’s evident that for distinctα, β ∈Γ we have m(α, β) +m(β, α) >0 if and only if
β∈ α ∼ϕ
.
Definition 2.5. Forα∈Γ we say:
•αis a periodic point ofϕif there existsn≥1 withϕn(α) =α,
•αis a quasi–periodic point ofϕif there exist n > m≥1 withϕn(α) =ϕm(α). By W(Γ, ϕ) we mean the collection of all non–quasi–periodic points of ϕ : Γ → Γ. ConsequentlyQ(Γ, ϕ) denotes the set of all quasi–periodic points of ϕ: Γ→Γ and byP(Γ, ϕ) we mean the collection of all periodic points ofϕ: Γ→Γ.
Remark 2.6. Forα∈Γ we have:
• α∈Q(Γ, ϕ) if and only if α
∼ϕ
⊆Q(Γ, ϕ).
• α∈W(Γ, ϕ) if and only if α
∼ϕ
⊆W(Γ, ϕ).
In particular byW(Γ, ϕ)∩Q(Γ, ϕ) =∅andW(Γ, ϕ)∪Q(Γ, ϕ) = Γ, hence if∼ϕ= Γ×Γ,
then Γ =W(Γ, ϕ) or Γ =Q(Γ, ϕ).
Proof. Consider α, β ∈Γ with β ∈ α ∼ϕ
, so there existn, m >1 such that ϕn(α) =
ϕm(β). Now we have the following cases:
•Case 1: α∈Q(Γ, ϕ). In this case there existsp > q≥1 withϕp(α) =ϕq(α), thus:
ϕp+m(β)ϕ
m(β)=ϕn(α)
= ϕp+n(α)ϕ
p(α)=ϕq(α)
= ϕp+n(α)ϕ
n(α)=ϕm(β)
= ϕp+n(β)
andβ∈Q(Γ, ϕ).
• Case 2: α∈W(Γ, ϕ). If β /∈W(Γ, ϕ) = Γ\Q(Γ, ϕ), thenβ ∈Q(Γ, ϕ) and using Case 1 we haveα∈Q(Γ, ϕ) which leads to contradictionα∈Q(Γ, ϕ)∩W(Γ, ϕ) =∅.
Henceβ∈W(Γ, ϕ).
Lemma 2.7. If ϕ : Γ → Γ is bijective and ∼ϕ= Γ×Γ with W(Γ, ϕ) 6= ∅, then
F\ {0}= Eigen(σϕ, VΓ).
Proof. Suppose ϕ : Γ → Γ is bijective and ∼ϕ= Γ×Γ with W(Γ, ϕ) 6= ∅. By
Corollary2.3we have 0∈/Eigen(σϕ, VΓ) and Eigen(σϕ, VΓ)⊆F\ {0}.
Chooseθ ∈ W(Γ, ϕ) 6=∅, x∈ V \ {0} and r ∈F\ {0}. Since ∼ϕ= Γ×Γ we have
Γ = θ
∼ϕ
, moreover by Remark 2.6 we have θ
∼ϕ
⊆ W(Γ, ϕ). As a matter of fact
Γ ={ϕn(θ) :n∈
Z}. Let:
xα=
rm(θ,α)x, m(θ, α)>0, r−m(α,θ)x, m(α, θ)>0,
and in brief words:
xα=rnx (ϕn(θ) =α, n∈Z).
Forα∈Γ there exists unique n∈Zwithϕn(θ) =α, thusϕn+1(θ) =ϕ(α) and
xϕ(α)=xϕn+1(θ)=rn+1x=r(rnx) =rxϕn(θ)=rxα.
Hence σϕ((xα)α∈Γ) = (xϕ(α))α∈Γ = (rxα)α∈Γ = r(xα)α∈Γ. By (xα)α∈Γ 6= 0 and σϕ((xα)α∈Γ) =r(xα)α∈Γ we haver∈Eigen(σϕ, VΓ), which completes the proof.
Lemma 2.8.Ifϕ: Γ→Γis bijective withW(Γ, ϕ)6=∅, thenF\{0}= Eigen(σϕ, VΓ).
Proof. By Corollary 2.3 we have Eigen(σϕ, VΓ) ⊆ F \ {0}. Choose θ ∈ W(Γ, ϕ),
then ϕ|θ
∼ : θ
∼ →
θ
∼ is bijective and θ ∈ W(ϕ|∼θ), thus by Lemma 2.8 we have F\ {0}= Eigen(σϕ|θ
∼
, V ∼θ). Using Lemma2.1we haveF\ {0}= Eigen(σ
ϕ|θ
∼
, V∼θ)⊆
Eigen(σϕ, VΓ) which completes the proof.
Lemma 2.9. Ifϕ: Γ→Γis bijective and∼ϕ= Γ×ΓwithW(Γ, ϕ) =∅, then (where
forr∈F\ {0} byo(r)we mean order ofr in commutative multiplying groupF\ {0}
and for finite setA,|A|denotes cardinality ofA): 1. P(Γ, ϕ) = Γis finite,
2. for allα ∈Γ we have {ϕn(α) : 0≤n < |Γ|}= Γ(= {ϕn(α) :n ≥0}) and
ϕ|Γ|(α) =α,
3. Eigen(σϕ, VΓ) ={r∈F\ {0}:o(r)is finite ando(r)divides|Γ|},
Proof. 1. Since W(Γ, ϕ) =∅, we have Q(Γ, ϕ) = Γ and for α ∈Γ = Q(Γ, ϕ) there existn > m≥1 withϕn(α) =ϕm(α), thusϕn−m(α) =αand α∈P(Γ, ϕ).
2. Use (1).
3. Supposer∈Eigen(σϕ, VΓ), by Corollary2.3we haver6= 0. There exists nonzero
(xα)α∈Γ ∈ VΓ with σϕ((xα)α∈Γ = r(xα)α∈Γ. By (2) we have ϕ|Γ|(α) = α for all α∈Γ, thus
σϕ|Γ|((xα)α∈Γ) = (xϕ|Γ|α)α∈Γ = (xα)α∈Γ.
On the other handσϕ|Γ|((xα)α∈Γ) =r|Γ|(xα)α∈Γ= (r|Γ|xα)α∈Γ. Hence: ∀α∈Γ (r|Γ|xα=xα).
Chooseβ∈Γ withxβ6= 0. Byr|Γ|xβ=xβ we have r|Γ|= 1 ando(r) divides|Γ|.
Conversely, forr∈F\{0}ifo(r) divides|Γ|, then choose fixx∈V\{0}andθ∈Γ. For
α∈Γ let xα=rm(θ,α)x. Usingσϕ((xα)α∈Γ) =r(xα)α∈Γ we haver∈Eigen(σϕ, VΓ)
which completes the proof.
Lemma 2.10. Forϕ: Γ→Γ ifP(Γ, ϕ) = Γ, then:
• ϕ: Γ→Γ is bijective,
• Eigen(σϕ, VΓ) ={r∈F\ {0}:o(r)is finite and there exists α∈Γ such that
o(r)divides |α ∼|}.
Theorem 2.11(Main Theorem). For isomorphismσϕ:VΓ →VΓ (soϕ: Γ→Γ is
bijective) we have:
Eigen(σϕ, VΓ) =
F\ {0}, W(ϕ,Γ)6=∅,
{r∈F\ {0}:∃θ∈P(ϕ,Γ)(o(r)|m(θ, θ))}, W(ϕ,Γ) =∅.
In particularEigen(σϕ, VΓ)does not depend onV.
Proof. Note that for periodic pointαofϕwe havem(α, α) =|{ϕn(α) :n≥0}|. Also
by bijection ofϕwe haveQ(ϕ,Γ) =P(ϕ,Γ). Use Lemmas2.8 and2.10to complete
the proof.
Corollary 2.12. For isomorphism σϕ : VΓ → VΓ and r, s ∈ F \ {0} with r ∈
Eigen(σϕ, VΓ)if o(s)|o(r), thens∈Eigen(σϕ, VΓ).
Proof. Ifr∈Eigen(σϕ, VΓ), then by Theorem2.11there existsα∈Γ witho(r)|m(α, α)
thuso(s)|m(α, α) too, which leads tos∈Eigen(σϕ, VΓ).
Acknowledgment
The primary form of this paper has been prersented in “4th International Conference of National Sciences (ICNS2019)–Mathematics & Computer” (University of Kurdis-tan, Sanandaj, Iran, 18–19 April 2019) [4].
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