# Ordered Structures and Projections

Authors: M. Yazi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We associate a covering relation to the usual order relation defined in the set of all idempotent endomorphisms (projections) of a finite-dimensional
vector space. A characterization is given of it. This characterization makes
this order an order verifying the Jordan-Dedekind chain condition. We give
also a property for certain finite families of this order. More precisely, the
family of parts intervening in the linear representation of diagonalizable
endomorphism, that is, the orthogonal families forming a decomposition of
the identity endomorphism.


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## Body

## 1. Introduction

In this paper,
we consider a poset𝒫, there is a set of all idempotent endomorphisms (projections) of a
finite-dimensional vector space E endowed with a reflexive, symmetric, and
transitive binary relation (denoted ≤) ([1, 2]). Given two elements p and q of 𝒫,  p≤q⇔p∘q=q∘p=p is equivalent to Imp⊆Imq and kerq⊆kerp. We say that q covers p (denoted p≺q) if p≤q,p≠q,
and the interval ]p,q[ is empty. So an element p of 𝒫 is an atom (resp., a coatom) if p coverers 0𝒫 (resp., covered by idE) where 0𝒫≡ zero endomorphism (resp., idE≡ identity endomorphism)
is the least element (resp., the greatest
element)
of 𝒫. When the bounds of two elements p and q exist we denote their meet (resp., their join)
by p∧q (resp., p∨q) ([3, 4]). We remark that the usual order relation defined in 𝒫 between two elements p and q is expressed by relations of inclusion between
their kernels and their ranges, then between elements of Es: set of subspaces of E which is well known that when this later is
endowed with the relation of set inclusion, the couple (Es,⊆) is a geometric lattice ([5–7]) where the covering relation
is defined for two subspaces F and G by F≺G⇔F⊂G and dimG=dimF+1.The extension
of the notion of covering constitutes my principal motivation in this paper.
Indeed, the main result of this paper, given in Section2, is a
characterization of covering relation defined by means of the one from (Es,⊆) ([6, 8]). We use this result to show that 𝒫 is a graded poset with the rank function R defined by R(p)=dimImp, for all p∈𝒫; and all maximal chains between the same endpoints have the same finite
length ([3, 4]). As a final point, when two
elements p and q of 𝒫 satisfy p∘q=q∘p we observe, as we will show in Section 3, that
the poset  𝒫 possesses some properties, among other things
is the covering property ([5, 6]).


## 2. Covering Relation

Proposition 2.1.
IfE is of finite dimension, the ordered set 𝒫 verifies the ascending chain condition.Proof.
Let(pi)i∈ℕ be an increasing
sequence of elements from 𝒫. Then p1≤p2≤⋯≤pi≤pi+1≤⋯ is equivalent to Imp1⊆Imp2⊆⋯⊆Impi⊆Impi+1⊆⋯ and ⋯⊆kerpi+1⊆kerpi⊆⋯⊆kerp2⊆kerp1. Since E is of finite dimension, then (Es,⊆) verifies the ascending chain condition, that
is, there exists i0∈ℕ such that Impi0=Impi0+1=⋯ for all i≥i0. Hence, for all i≥i0, the elements kerpi are two by two isomorph. Since they are two by
two comparable, they are equal. Hence, pi=pi0 for all i≥i0.We will prove later that this proposition is a
consequence of Proposition4.1.Before stating a theorem, we remark that for two
elementsp and q of 𝒫 such that p<q, one has this following equivalence: Imp≺Imq⇔kerq≺kerp. From this, we have the following theorem.
Theorem 2.2.
Ifp and q are two elements of such that p<q, the two following properties are equivalent:(i)
]Imp,Imq[=∅;(ii)
]p,q[=∅.Proof.
(i)⇒(ii) Assume that q does not cover p, that is, there exists in 𝒫 an element r such that p<r<q, or equivalently, Imp⊂Imr⊂Imq and kerq⊂kerr⊂kerp. Hence, we have a contradiction.
Conversely,
assume thatImq does not cover Imp, it follows from the precedent remark that kerp does not cover kerq.
The chains of(Es,⊆) of respective endpoints (Imp,Imq) and (kerq,kerp) are then at least of length two. There exists
then at least  a couple of elements (Es,⊆) such that(2.1)Imp⊂F⊂Imq,ker⁡q⊂G⊂ker⁡pwith dimG=dimkerp−1 and dimF=dimImp+1.
Amongts these
couples, we prove that it exists one which is the direct sum ofE, that is F⊕G=E. From the relation kerq⊂kerp, it follows kerp=kerq⊕(Imq∩kerp). Then E=Imp⊕kerq⊕(Imq∩kerp).
Setr1=dimImp,r2=dimkerp,s1=dimImq,s2=dimkerq with s1+s2=r1+r2=dimE, and (g1,g2,…,gs2) a basis of kerq,(gs2+1,…,gr2) a basis of Imq∩kerp and (f1,…,fr1) a basis of Imp. If we consider the union of the precedent
bases, we obtain obviously a base of E. Thus, the subspaces F and G defined as follows: F=[f1,…,fr1,gr2] subspace generated by (f1,…,fr1,gr2),G=[g1,…,gs2,gs2+1,…,gr2−1] subspace generated by (g1,…,gr2−1) form, in fact, a couple of Es for which E is a direct sum, that is, E=F⊕G; moreover, verifying the relations Imp⊂F⊂Imq and kerq⊂G⊂kerp. Thus, the existence of the projection on F along G.
Hence, we have a contradiction.Corollary 2.3.
The elements of𝒫 for which the image is a line are atoms. Dually, the elements of 𝒫 for which the image is a hyperplane are coatoms.


## 3. Covering Property

Several ordered sets having the property of lattice
satisfy the covering property (e.g.,(Es,⊆)). In the following result, we prove that one
also finds it in 𝒫 when two arbitrary elements commute.Proposition 3.1.
Letr and p be two elements of 𝒫. If r is an atom such that p∧r=r∧p=0, then the covering property is verified, that
is, p≺p∨r.
Dually, ifr is a coatom, then p∧r≺p.Proof.
It is
well known thatp∨r and p∧r exist, and we have the
following:(3.1)p∨r=p+r−p∘r,withIm(p∨r)=Imp+Imr,ker⁡(p∨r)=ker⁡p∩Imr,p∧r=p∘r,withIm(p∧r)=Imp∩Imr,ker⁡(p∧r)=ker⁡p+ker⁡r.
If r is an atom, the subspaces Imr (resp., kerr) represent in (Es,⊆) an atom (line) (resp., a coatom is a
hyperplane in (Es,⊆)). So, the assumption p∧r=0 implies(3.2)ker⁡p+ker⁡r=E,(3.3)Imp∩Imr={0}Hence, in (Es,⊆) the equality (3.2) implies Imp≺Imp+Imr or equivalently Imp≺Im(p∨r). Furthermore, the equality (3.3) implies kerp∩kerr≺kerp or equivalently ker(p∨r)≺kerp. Then from Theorem 2.2
p≺p∨r.
Using a similar
reasoning as above we obtainp∧r≺p.Proposition 3.2.
Letp,q,s be three elements  of 𝒫 such that p∘q=q∘p. If s≺p and s≺q, s=p∧q. Dually, if p≺s and q≺s, s=p∨q.Proof.
From Theorem2.2, s≺p and s≺q imply Ims≺Imp,kerp≺kers and Ims≺Imq, kerq≺kers. So, in (Es,⊆) we have Ims=Imp∩Imq and kers=kerp+kerq. We have then s=p∧q. In a similar reasoning we can prove that s=p∨q if p≺s and q≺s.


## 4. Rank Function

Theorem2.2 allows us to define in 𝒫 a rank function ℛ defined by ℛ(p)=dimImp,p∈𝒫. We can easily prove that(4.1)ℛ(0p)=0=dim⁡0Pp≠q,p≤q⁢implyℛ(p)<ℛ(q),p,q∈P(strict⁢isotonicity)p≺q⁢impliesℛ(q)=ℛ(p)+1.Proposition 4.1.
𝒫 is of finite length and of length equal to the dimension of E.Proof.
Let a
maximal chain of𝒫: 0𝒫≺p1≺p2≺⋯≺pi≺pi+1≺⋯≺idE. By applying ℛ to this sequence, we get 0<ℛ(p1)=1<2=ℛ(p2)<⋯<dimE, which proves that the length of 𝒫 is equal to the dimension of E.𝒫 being of finite length, it verifies the ascending and descending chain
condition.Corollary 4.2.
The poset𝒫 with the rank function ℛ is of finite length, it verifies then the
Jordan-chain condition, that is, all maximal chain between the same endpoints
have the same finite length.Corollary 4.3.
Ifp and q are two elements of 𝒫 such that p∘q=q∘p,ℛ(p∨q)=ℛ(p)+ℛ(q).Proof.
This
follows from a property of the dimension of an endomorphism of a
finite-dimensional vector space.It is well known that whenp,q,
and r are three elements commuting two by two, each
one of them commutes with the supremum and infimum of the two others. We
propose to generalize this result in the theorem below.Theorem 4.4.
Ifp1,p2,…,pm is a family of m elements of 𝒫 commuting two by two, that is, pi∘pj=pj∘pii≠j,1≤i,j≤n ,
then(4.2)∨i=1mpi=∑i=1mpi−∑i≠jpi∘pj+∑i≠j≠kpi∘pj∘pk−⋯+(−1)m−1p1∘p2∘⋯∘pm.In particular, if this family is
orthogonal, then ∨i=1mpi=∑i=1mpi.Proof.
For
proving the equality we use induction onm≥2. Since by assumption the elements pi commuting two by two, we have for   m=2,p1∨p2=p1+p2−p1∘p2.
For m=3,(4.3)p1∨p2∨p3=(p1∨p2)∨p3=(p1∨p2)+p3−(p1∨p2)∘p3=p1+p2−p1∘p2+p3−p1∘p3−p2∘p3+p1∘p2∘p3=∑i=13pi−∑1≤i≠j≤3pi∘pj+p1∘p2∘p3.Assume that the equality is
verified until the order m−1,(4.4)∨i=1m−1pi=∑i=1m−1pi−∑1≤i≠j≤m−1pi∘pj+∑1≤i≠j≠k≤m−1pi∘pj∘pk−⋯+(−1)m−2p1∘p2∘⋯∘pm−1.Then(4.5)∨i=1mpi=(∨i=1m−1pi)∨pm=∨i=1m−1pi+pm−(∨i=1m−1pi)∘pm=∑i=1m−1pi−∑1≤i≠j≤m−1pi∘pj+∑1≤i≠j≠k≤m−1pi∘pj∘pk−⋯+(−1)m−2p1∘p2∘⋯∘pm−1+pm−(∑i=1m−1pi)∘pm+∑1≤i≠j≤m−1(pi∘pj)∘pm−∑1≤i≠j≠k≤m−1(pi∘pj∘pk)∘pm+⋯+(−1)m−1p1∘p2∘⋯∘pm−1∘pm.=∑i=1m−1pi−∑1≤i≠j≤m−1pi∘pj−(∑i=1m−1pi)∘pm+∑1≤i≠j≠k≤m−1pi∘pj∘pk+∑1≤i≠j≤m−1(pi∘pj)∘pm−∑1≤i≠j≠k≤m−1(pi∘pj∘pk)∘pm+⋯+(−1)m−1p1∘p2∘⋯∘pm−1∘pm.It follows from the
combinatorial relation Cms=Cm−1s+Cm−1s−1 for all s≤m that (4.6)∑1≤i≠j≠k≤m−1pi∘pj∘pk+(∑1≤i≠j≤m−1pi∘pj)∘pm=∑1≤i≠j≠k≤mpi∘pj∘pk. Hence we have
the desired equality.It is clear that if the familyp1,p2,…,pm is orthogonal (pi∘pj=pj∘pi=0𝒫,i≠j), the equality proved before becomes ∨i=1mpi=∑i=1mpi. It is well known in Linear algebra that if (p1,…,pm) is an orthogonal family verifying in addition ∨i=1mpi=idE,
then this family characterizes diagonalizable endomorphisms.


## 5. Conclusion

The consequence of the preceding theorem can
constitute a characterization of a diagonalizable endomorphism of a finite
dimensional vector space by means of the ordered set of the idempotent
endomorphism in the same space. It is thus natural to ask the following
question. Can one in the general case for a given orderϑ of finite length satisfying in addition the
condition of Jordan-Dedekind identify a finite family x1,…,xp such that ∨xi=1ϑ and xi∧xj=0ϑ for i≠j, where 1ϑ and 0ϑ are the bounds of ϑ.
If the answer is affirmative, can one in this case establish a junction between
the Linear algebra and the ordered structures so that the identification of a
diagonalizable endomorphism is concrete.



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