# Primal Topologies on Finite-Dimensional Vector Spaces Induced by Matrices

Authors: Luis Mejías; Jorge Vielma; Ángel Guale; Ebner Pineda
Journal: International Journal of Mathematics and Mathematical Sciences
 (2023)

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## Abstract

Given an matrixA, considered as a linear map A:ℝn⟶ℝn, then A induces a topological space structure on ℝn which differs quite a lot from the usual one (induced by the Euclidean metric). This new topological structure on ℝn has very interesting properties with a nice special geometric flavor, and it is a particular case of the so called “primal space,” In particular, some algebraic information can be shown in a topological fashion and the other way around. If X is a non-empty set and f:X⟶X is a map, there exists a topology τf induced on X by f, defined by τf=U⊂X:f−1U⊂U. The pair X,τf is called the primal space induced by f. In this paper, we investigate some characteristics of primal space structure induced on the vector space ℝn by matrices; in particular, we describe geometrical properties of the respective spaces for the case.


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## Body

## 1. Introduction

In this paper we present some elementary properties of a special topological space structure induced by matrices on the vector spaceℝn. Such a structure is a particular case of a more general construction: If X is a non-empty set and f:X⟶X is a map, the collection τf=U⊂X:f−1U⊂U is a topology for X, which is named the primal topology induced by f, see Echi [1] and Echi and Turki [2]. The topological space X,τf is an example of the so called Alexandroff space, that is, a space where intersections of arbitrary collections of open sets is an open set. These spaces have, in general terms, a complicated structure, because in most of the cases they do not satisfy the axiom T1. For general properties of functional Alexandroff spaces see Kukiela [3] and, Ayatollah Zadeh Shirazi and Golestani [4].Examples of primal topologies with a description of their main properties are presented in Section2 of this paper. Some of such spaces are part of active investigations. In particular, Example 5 shows how Vielma et al. [5] used a primal to topology on the set of the positive integers to obtain a topological version of the Collatz conjecture.In our investigation we work with the primal topology generated by takingX=ℝn and f a matrix (identified with a linear map on ℝn). Thus, we obtain the matrix primal space ℝn,τA and explore how its properties are derived from those of A, and the other way around.In Section3 we show some basic results which require the knowledge of elementary concepts from linear algebra and topology. In most of the cases those propositions have a nice geometric flavor, even though matrix primal spaces are quite different from Euclidean spaces. In Section 4 we consider the problem for the case of the vector space ℝ2 and we conclude investigating properties of continuous maps on matrix primal spaces. A few of the propositions presented in this paper are suitable for generalizations which deserve to be investigated as, for example, in the context of vector spaces of infinite dimensions.


## 2. Primal Spaces

For an arbitrary topological space it is not true that arbitrary intersections of open sets is an open set, so those with this special property have been singled out. A topological spaceX,τ is said to be an Alexandroff space if any arbitrary intersection of open sets is an open set. General properties of Alexandroff spaces are presented in Kukiela [3]. There are some special examples of such structures, in particular the so, called “functional Alexandroff spaces,” which are determined in the following way: let X be a non-empty set and f:X⟶X a map, we consider the topology τf on X such that the closed sets are those E⊂X such that fE⊂E. Some of the most important properties of functional Alexandroff spaces are found in Ayatollah Zadeh Shirazi and Golestani [4]. Another approach to these spaces by focusing on the open sets, was introduced by Echi [1] based on the fact that for any set X≠∅ and any map f:X⟶X, the collection τf=U⊂X:f−1U⊂U is a topology for X.The topologyτf is called the primal topology induced by f and X,τf is said to be a primal space. Sometimes we use the notation Xf as an abbreviation for X,τf, which is very convenient when we deal with more than one primal topology on the same set. Obviously the map f:Xf⟶Xf is continuous.Next, we describe the most elementary open and closed sets of primal spaces. Let us denote byℕ0=ℕ∪0 the set of non-negative integers and by fr the r-fold composite f∘⋯∘f, for r>0 and f0 the identity map. Given p∈X, the orbit of p, is defined by(1)Ωp=y∈X:y=frpforr∈ℕ0.ThusΩp is the smallest closed set containing p, that is clp, the closure of p.If there exists a numberr>0 such that frp=p we say that Ωp is a periodic orbit and p itself is referred to as a periodic point. If Ωp contains just one element, then p is said to be a fixed point.On the other hand, we define(2)Γp=x∈X:p=frxforr∈ℕ0,which is the smallest open set containing p. We use Ωfp and Γfp whenever we want to emphasize that these are subsets of the space Xf.Some general properties of primal spaces have been developed over the last few years, for example, Echi and Turki [2] characterize compactness with the following result.Theorem 1.
LetX,τf a primal space. Then the following statements are equivalent:(1)
The spaceX,τf is compact,(2)
The collection of periodic points is finite and for eachx∈X, frx is periodic, for some r.
In particular, this theorem implies that compact primal spaces have finitely many connected components each of them containing a periodic point.
In general, primal spaces are somehow elusive, especially in relation with separation axioms. As a matter of fact As a matter of, most of the cases the space is not evenT1 (Example 1). In order to illustrate this issue we introduce a few examples, the first two being the extreme cases.Example 1.
IfX≠∅ and q is an element of X, consider the map f:X⟶X, such that fx=q for all x∈X (constant). Then for all x≠q we have that x∈τf. Furthermore, the space X,τf satisfies the axiom T0 but it is not a Hausdorff space.Example 2.
For anyX≠∅ the primal topology induced by the identity map on X is the discrete topology. In this case, each p∈X is a fixed point. This is the only case in which a primal space is T1, because any set x is closed if and only if Ωx=x, and this is true if and only if the map that induces the topology is the identity.Example 3.
Lets:ℕ⟶ℕ be the map defined by sx=x+1. This space is suitable for a nice graphic representation:(3)1⟶2⟶3⟶⋯⟶p⟶p+1⟶⋯,where p⟶p+1 means sp=p+1.
Note that for allp∈ℕ we have Γp⊂Γp+1 and the space ℕ,τs is connected, but it is not compact.Example 4.
Letℤn denote the integers modulo n with its usual multiplication. For any k∈ℤn, define mk:ℤn⟶ℤn by mkx=kx. Then the space ℤn,τmk is compact because ℤn is finite.
This space is suitable for a nice graphic representation with specific values ofn and k. For example, for the space ℤ9,τm3, where Γ0=ℤ9:(4)12↓↓4⟶3⟶0←6←5↑↑78.
We have a very different situation with the spaceℤ9,τm2, which has three connected components, namely Γ0, Γ1, and Γ3:(5)1⟶2⟶40↑↓3⇆65←7←8.
It is easy to verify thatℤ9,τm2 and ℤ9,τm5 are homeomorphic.The next example is associated to a very famous problem known as “the Collatz conjecture” which was set in the first half of the 20th century and still remains unsolved.Example 5.
LetC:ℕ⟶ℕ the map defined by(6)Cx=x2,ifxis even,3x+1,ifxis odd.
The Collatz conjecture states that for allp∈ℕ there exists an r∈ℕ such that Crp=1. This problem may be stated in terms of primal spaces as follows: For all p∈ℕ we have that 1∈Ωp. The next diagram shows part of Γ1 which is a connected component of ℕ,τC:(7)⋯⟶6⟶3↓⋯⟶20⟶10⟶5↓⋯⟶64⟶32⟶16⟶8⟶4⟶2↖↓1.
Separation properties of the spaceℕ,τC were determined by Vielma and Guale [6] by using some ideas introduced of Di Maio [7]. Furthermore, it is not known if the space ℕ,τC is connected; neither is known if it is compact. These facts set a very important problem because Vielma and Guale [6] used the result of Echi and Turki cited above (Theorem 1), to prove that the Collatz conjecture is equivalent to connectedness of the primal space ℕ,τC. Some other properties of this space related to the conjecture are shown in [5].


## 3. Matrix Primal Spaces

In this section we explore the construction of primal topologies on the vector spaceℝn induced by linear maps. First, we establish some conventions of notation. Let ℬn=e1,e2,…,en be the standard basis of ℝn, that is, ei=0,…,0,1,0,…,0, where 1 appears at the i-th position. If T:ℝn⟶ℝn is a linear map, we denote by μT the matrix associated to T with respect to the standard basis in ℝn. As usual, we identify T with A=μT, in particular if no confusion arises we use both τT and τA to denote the primal topology induced by T and the respective primal space ℝn,τT=ℝn,τA is denoted by ℝA2. Obviously, we have that 0 is always a fixed point. The next result provides information about Γ0 from invertibility of T.Theorem 2.
IfT:ℝn⟶ℝn is a non-invertible linear transformation, then Γ0 is a nontrivial invariant linear subspace of ℝTn.Proof.
Ifx,y∈Γ0, there exist r,s∈ℕ such that Trx=Tsy=0. Then(8)Tr+sx+y=Tr+sx+Tr+sy=0,and x+y∈Γ0. On the other hand, if a∈ℝ and x∈Γ0, there exists k∈ℕ such that Tkx=0. Thus(9)Tkax=aTkx=0,therefore ax∈Γ0.
Ifx∈Γ0 and Tkx=0 for a number k∈ℕ, then Tk−1Tx=0. Thus Tx∈Γ0, so Γ0 is closed and, by definition, this implies that TΓ0⊂Γ0. In other words words Γ0 is invariant under T.Now, in the following two propositions we prove that invertibility ofT is equivalent to some topological information about both kernel and image of T.Theorem 3.
LetT:ℝn⟶ℝn be a linear map. Then the following propositions hold:(1)
IfT is invertible then KerT is open in ℝTn.(2)
IfKerT is open in ℝTn and KerT⊂ImT, then T is invertible.Proof.
Proposition (1) is obvious becauseKerT=0=Γ0∈τT.
To prove the second suppose that there existsx∈ℝn, x≠0 such thatx∈KerT. By assumption, KerT⊂ImT, so there exists y such that Ty=x. Theny∈Γx⊂Γ0=KerT, therefore Ty=0, which is a contradiction.Note that the hypothesis of openness ofKerT in the second part of Theorem [3] is not a sufficient condition. Let us consider, for example, T:ℝ2⟶ℝ2 given by Tx1,x2=x1,0. Then KerT=0,x2:x2∈ℝ=Γ0,0 (is open), but KerT∩ImT=0,0.Theorem 4.
Ann×n matrix A is invertible if and only if ImA is open in ℝA2.Proof.
Proof IfA is invertible, then ImA=ℝn∈τA.
Now suppose thatImA∈τA. Let cj be any column vector of A. Then Γcj⊂ImA. But Aej=cj, so ej∈Γcj. Thus ℝn⊂ImA⊂ℝn, that is ImA=ℝn, and therefore A is invertible.


## 4. Matrix Primal Topologies for the Plane

In this section we introduce a pretty long list of examples in order to illustrate in detail some specific facts about the primal spaceℝA2, for various matrices. Sometimes we make comparative references to the Euclidean topology ϵ for ℝ2 (the one induced by the usual metric) and denote by ℝε2 the space ℝ2,ϵ. Let us begin with two non-invertible matrices.Example 6.
Let℘1 be the orthogonal projection of ℝ2 on the x1-axis:(10)℘1=1000.
Note that the collection of singlegonsx1,x2 with x2≠0 and the vertical lines Γk,0, k∈ℝ form a basis ℬ for τ℘1. On the other hand, the collection ℬ′=0,x2,x1,x2:x1,x2∈ℝ is a basis for τ℘1′, the complement topology of τ℘1, that is the one whose elements are the τ℘1-closed sets.
Something similar happens with the orthogonal projection℘2 on the x2-axis. Even in the much more general case, of the orthogonal projection ℘k on the xk-axis in ℝn, 1≤k≤n, given by the matrix with entries.(11)aij=1,ifi=j=k,0,otherwise.Now, let us consider a very simple invertible matrix.Example 7.
Letρi be the reflection through the xi-axis:(12)ρi=−1i+100−1i.
Elements of the primal topologyτρi are subsets of ℝ2 which are symmetric with respect to the xi-axis, because ρi=ρi−1, and therefore, all the open sets are also closed.The next two examples illustrate the orbits for some general2×2 diagonal matrices and describe them as sets in the Euclidean space ℝɛ2.Example 8.
For a positive numberλ, let δλ be the matrix:(13)δλ=λ00λ.
We have three different cases depending onλ.Case 1. Ifλ=1, then δ1 is the identity and every x∈ℝ2 is a fixed point.Case 2. Ifλ>1, there are not periodic orbits other than Ω0. For each x∈ℝ2, x≠0 we have that the sequence δλnx is divergent in the space ℝɛ2.Case 3. Ifλ<1, there are not periodic orbits other than Ω0. For each x∈ℝ2, x≠0 we have that the sequence δλnx converges to 0 in the space ℝɛ2.Example 9.
Given two positive numbersλ1 and λ2, consider the matrix:(14)A=λ100λ2.
The description of the spaceℝA2 requires to consider a few cases.Case 1. Ifλ1=1 and λ2<1, the only periodic orbits Ωx are those of points x with x2=0 (in fact, fixed points). If x2≠0 then the sequence Anx converges to x1,0, in the sense of the standard space ℝɛ2.Case 2. Ifλ1=1 and λ2>1, we also have that the only periodic orbits Ωx are those of points x with x2=0. If x2≠0 then the sequence Anx diverges in the space ℝɛ2.Case 3. Ifλ1,λ2>1, the only periodic orbit is Ω0. In any other case, the sequence Anx diverges in ℝϵ2.Case 4. Ifλ1,λ2<1, we also have that the only periodic orbit is Ω0, but the sequence Anx converges to 0 in the space ℝɛ2.Case 5. Ifλ1>1 and λ2<1, the only periodic orbit is Ω0. In the space ℝɛn, for x1=0 the sequence Anx converges to 0, and it diverges if x1≠0.
Of course, we may consider a diagonal matrix as the one in Example9 with no necessarily positive eigenvalues.Example 10.
Given the matrix(15)A=λ100λ2,
It turns out that some families of sets which are either open or closed are symmetric with respect to the origin or the axes, particularly the circleSr and the square Kr, r>0:(16)Sr=x1,x2∈ℝ2:x12+x22≤r2.and(17)Kr=x1,x2∈ℝ2:x1+x2≤r.Case 1. Ifλ1,λ2≥1, we have that Ax≥x, for all x∈ℝ2. then Sr is open in ℝA2 for any r>0. On the other hand, if x=x1,x2∈A−1Kr there is a point y=y1,y2∈Kr such that Ax=y, that is(18)y1,y2=Ax1,x2=λ1x1,λ2x2.Then(19)x1+x2≤λ1x1+λ2x2=y1+y2≤r,andA−1K⊂Kr, so Kr is open in ℝA2.Case 2. Ifλ1,λ2≤1, A is a contraction, Ax≤x, for all x∈ℝ2, and therefore Sr is closed in ℝA2 for any r>0.
In this case we have thatKr is also closed in ℝA2 In fact, if x=x1,x2∈Kr, we have that Ax1,x2=λ1x1,λ2x2∈K and(20)λ1x1+λ2x2≤x1+x2≤r,so AK⊂Kr. Similarly, we may proof that for all r1,r2∈ℝ, the diamond shaped set(21)Kr1,r2=x1,x2∈ℝ2:r2x1+r1x2≤r1,r2,is closed.
Let us note that all these cases may be deduced also from the fact that for anya,b>0 the rectangle R=−a,a×−b,b is closed. for if x1,x2∈R we have we have(22)x1λ1≤x1≤aandx2λ2≤x2≤b,
ThusAR⊂R.Example 11.
Leta be a positive number and let A be the upper triangular matrix(23)A=1a01.
For any pointx=x1,x2 we have(24)Γx=x1−nax2,x2:n∈ℤ,n≥0.
Then the only periodic orbits inℝA2 are those of points on the x1-axis.
Similarly, ifb is a positive number and B is the matrix(25)B=10b1.
then for anyx=x1,x2 we have Γx=x1,−nbx1+x2:n∈ℤ,n≥0, and therefore, the only periodic orbits in ℝB2 are those of points on the x2-axis.Example 12.
Forθ∈ℝ, we consider the rotation Aθ given by(26)Aθ=cosθ−sinθsinθcosθ.
Recall that the collection of such matrices is an abelian subgroup of the general linear groupGL2, because for any α and β we have(27)Aα⋅Aβ=Aα+β.
We denote by eitherℝ2,τθ or ℝθ2 the space ℝ2 with the primal topology τθ, and we call it a primal rotation space. If θ=rπ, with r∈ℚ, r≠0, for each x∈ℝ2 the orbit Ωx is finite, and it is both open and closed for Ωx=Γx.
Some more details about the spaceℝθ2 may be given if we have specific information about r. Let us consider, for instance, that p∈ℤ,q∈ℕ, p≠0, and θ=2pπ/q. then U∈τθ if and only if AθU=U. Moreover for all n∈ℕ we have(28)Ωxin the spaceℝθ2=Ωxin the spaceℝ2npπ/q2,and(29)Γxin the spaceℝθ2=Γxin the spaceℝ2npπ/q2,
Therefore the spacesℝθ2 and ℝ2pnπ/q2 are homeomorphic. Let us note that for all x(30)Ωx=Γx=x,Ax,A2x,…,Aq−1x,
and therefore for allx∈ℝθ2, x≠0, Ωx is the set of vertices of a regular polygon with q vertices and center in the origin, with a vertex in x.


## 5. Continuity

In this section we study some facts about continuity of functions defined on matrix primal spaces which are deduce from some general propositions. We begin with an example which shows that even in very simple cases things may not be as one might expect.Example 13.
Letι:ℝ2⟶ℝ2 be the map defined by ιx=x for all x, but seen as a function ι:ℝπ/22⟶ℝπ2. Then ι is not continuous, for instance, Γπe1=e1,−e1∉τπ/2, because nontrivial open sets in ℝπ/22 contain at least 4 elements.Before addressing some general problems, we deal with a few particular results. Thus the, following lemmas offer a hint about hypothesis on the matricesA and B in order to deduce the continuity of some linear maps f:ℝAn⟶ℝBn.Lemma 1.
LetX be a set, f,g:X⟶X two maps. Let τf and τg be the respective primal topologies induced by f and g. If h:Xf⟶Xg is a function such that h∘f=g∘h, then for all x,y∈X if y∈Ωfx we have that hy∈Ωghx.Proof.
Suppose thatx∈X and y∈Ωfx. then there exists a k∈ℕ0 such that fkx=y. Then(31)hy=h∘fkx=h∘f∘fk−1x=g∘h∘fk−1x=g∘h∘f∘fk−2x=g2∘h∘fk−3x=gk∘hx.
Thusy∈Ωghx.Lemma 2.
LetXf,Xg be primal spaces and h:Xf⟶Xg a homeomorphism. Then hΩfx=Ωghx.Proof.
This is a direct consequence of the fact that para any setA⊂X, since h is a homemorphism, we have that hclA=clhA.Corollary 1.
LetXf,Xg be primal spaces and let be h:Xf⟶Xg a homeomorphism. Then we have that for all x,y∈Xx∈Ωfy if and only if hx∈Ωghy.Now we return to the problem of continuity of functions in the context of matrix primal spaces.Theorem 5.
LetA,B two square matrices. If A and B are similar matrices then the primal spaces ℝAn and ℝBn are homeomorphic. Furthermore, if A=P−1BP then P:ℝAn⟶ℝBn is a homeomorphism.Proof.
By hypothesis,P:ℝAn⟶ℝBn is invertible. If V∈τB, then(32)A−1P−1V=PA−1V=BP−1V=P−1B−1V⊂P−1V,therefore P−1V∈τA, so P is continuous.
LetF be a closed set in ℝAn, that is AF⊂F. Now we have(33)BPF=BPF=PAF=PAF⊂PF.
ThenP is a closed map, which means that P−1 is continuous and therefore P is a homeomorphism. □Theorem 6.
IfA and B are two n×n matrices which commute with each other, then the maps A:ℝBn⟶ℝBn and B:ℝAn⟶ℝAn are both continuous.Proof.
Of course, it is enough to prove thatA:ℝBn⟶ℝBn is continuous.
LetV∈τB and x∈B−1A−1V, then ABx∈V and B−1ABx∈B−1V. But A and B commute, thus Ax∈B−1V⊂V, and therefore x∈A−1V, which means that A−1V∈τB.Let us note that the hypothesis of commutativity of the matricesA and B in Theorem 6 cannot be dropped. In fact, if we consider A and B as in Example 11 with a=b=1, that is(34)A=1101andB=1011,then AB≠BA and ΓB0,1=0,1∈τB, but A−1ΓB0,1=−1,1∉τB because 1,2∈ΓB−1,1.Moreover, invertibility of the matricesA and B in Theorem 6 is not a necessary condition: the orthogonal projections (Example 6) ℘j:ℝPi2⟶ℝPi2, with i,j∈1,2, i≠j are continuous and ℘1℘2=℘2℘1.Lemma 3.
LetA and B be two invertible n×n matrices. If τB⊂τAB−1, then τB⊂τA.Proof.
Suppose thatV∈τB. Then V∈τAB−1, which means that(35)AB−1−1V⊂V,BA−1V⊂V,A−1V⊂B−1V⊂V,because V∈τB, in other words, V∈τA.Now we obtain the following result, which somehow fixes the problem set in Example13 related to the function ι:ℝAn⟶ℝBn defined by ιx=x.Corollary 2.
LetA and B be two invertible n×n matrices. If τB⊂τAB−1 then the map ι:ℝAn⟶ℝBn is continuous.Theorem 7.
LetA and B be two invertible n×n matrices which commute with each other. If τA⊂τBA−1 and τB⊂τAB−1, then the spaces ℝAn and ℝBn are homeomorphic.Proof.
Let us consider the compositeφ=A°ι:ℝAn⟶ℝBn:(36)ℝAn⟶ιℝBn⟶AℝBn.
Obviously,φ is bijective. By Corollary 2 we have that ι is continuous and, by Theorem 6, we have that A is continuous, therefore φ is continuous.
On the other hand, we haveτA⊂τBA−1 then Theorem [6] implies that A−1 is continuous, and so is φ−1.



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