Volume 2008, Article ID 783041,6pages doi:10.1155/2008/783041
Research Article
Ordered Structures and Projections
M. Yazi
Probability and Statistics Department, Faculty of Mathematics, University of Sciences and Technology USTHB, 16111 Algiers, Algeria
Correspondence should be addressed to M. Yazi,mohyazi@gmail.com
Received 28 July 2007; Accepted 4 March 2008
Recommended by Pentti Haukkanen
We associate a covering relation to the usual order relation defined in the set of all idempotent endomorphismsprojectionsof 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.
Copyrightq2008 M. Yazi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we consider a posetP, there is a set of all idempotent endomorphisms projec-tionsof a finite-dimensional vector spaceEendowed with a reflexive, symmetric, and tran-sitive binary relationdenoted≤ 1,2. Given two elementspandq ofP,p ≤q ⇔ p◦q
q◦ppis equivalent to Imp⊆Imqand kerq⊆kerp.We say thatqcoverspdenotedp≺q ifp≤q, p /q, and the intervalp, qis empty. So an elementpofPis an atomresp., a coatom ifpcoverers 0Presp., covered by idEwhere 0P ≡zero endomorphismresp., idE ≡identity
endomorphismis the least elementresp., the greatest elementofP. When the bounds of two elementspandqexist we denote their meetresp., their joinbyp∧qresp.,p∨q 3,4. We remark that the usual order relation defined inPbetween two elementspandqis expressed by relations of inclusion between their kernels and their ranges, then between elements ofEs:
set of subspaces ofEwhich is well known that when this later is endowed with the relation of set inclusion, the coupleEs,⊆is a geometric lattice5–7where the covering relation is defined for two subspacesFandGbyF≺G⇔F⊂Gand dimGdimF1.
is a graded poset with the rank functionR defined by Rp dim Imp,for all p ∈ P; and all maximal chains between the same endpoints have the same finite length3,4. As a fi-nal point, when two elementspandq ofPsatisfyp◦q q◦pwe observe, as we will show inSection 3, that the posetPpossesses some properties, among other things is the covering property5,6.
2. Covering relation
Proposition 2.1. IfEis of finite dimension, the ordered setPverifies the ascending chain condition.
Proof. Letpii∈Nbe an increasing sequence of elements fromP. Thenp1≤p2≤ · · · ≤pi≤pi1≤ · · · is equivalent to Imp1 ⊆ Imp2 ⊆ · · · ⊆ Impi ⊆ Impi1 ⊆ · · · and· · · ⊆ kerpi1 ⊆ kerpi ⊆ · · · ⊆ kerp2 ⊆ kerp1.SinceEis of finite dimension, thenEs,⊆verifies the ascending chain
condition, that is, there existsi0 ∈Nsuch that Impi0 Impi01 · · · for alli ≥ i0.Hence, for
alli≥i0,the elements kerpiare two by two isomorph. Since they are two by two comparable, they are equal. Hence,pipi0for alli≥i0.
We will prove later that this proposition is a consequence ofProposition 4.1.
Before stating a theorem, we remark that for two elementspandqofPsuch thatp < q, one has this following equivalence: Imp ≺ Imq ⇔ kerq ≺ kerp. From this, we have the following theorem.
Theorem 2.2. Ifpandqare two elements of such thatp < q,the two following properties are equivalent:
i Imp,Imq∅;
ii p, q∅.
Proof. i⇒iiAssume thatqdoes not coverp,that is, there exists inPan elementrsuch that
p < r < q,or equivalently, Imp ⊂ Imr ⊂ Imq and kerq ⊂ kerr ⊂ kerp.Hence, we have a contradiction.
Conversely, assume that Imqdoes not cover Imp,it follows from the precedent remark that kerpdoes not cover kerq.
The chains ofEs,⊆of respective endpointsImp,Imqandkerq,kerpare then at least of length two. There exists then at least a couple of elementsEs,⊆such that
Imp⊂F⊂Imq,
kerq⊂G⊂kerp 2.1
with dimGdim kerp−1 and dimFdim Imp1.
Amongts these couples, we prove that it exists one which is the direct sum of E,that isF⊕G E.From the relation kerq ⊂ kerp,it follows kerp kerq⊕Imq∩kerp.Then
EImp⊕kerq⊕Imq∩kerp.
Setr1 dim Imp,r2 dim kerp,s1 dim Imq,s2 dim kerqwiths1s2 r1r2 dimE,andg1, g2, . . . , gs2a basis of kerq,gs21, . . . , gr2a basis of Imq∩kerpandf1, . . . , fr1
a basis of Imp.If we consider the union of the precedent bases, we obtain obviously a base ofE.Thus, the subspacesFandGdefined as follows:F f1, . . . , fr1, gr2subspace generated
in fact, a couple ofEsfor whichEis a direct sum, that is,E F⊕G; moreover, verifying the relations Imp⊂F⊂Imqand kerq⊂G⊂kerp.Thus, the existence of the projection onFalong
G. Hence, we have a contradiction.
Corollary 2.3. The elements ofPfor which the image is a line are atoms. Dually, the elements ofPfor which the image is a hyperplane are coatoms.
3. Covering property
Several ordered sets having the property of lattice satisfy the covering propertye.g.,Es,⊆.
In the following result, we prove that one also finds it inPwhen two arbitrary elements com-mute.
Proposition 3.1. Letrandpbe two elements ofP. Ifris an atom such thatp∧rr∧p0,then the covering property is verified, that is,p≺p∨r.
Dually, ifris a coatom, thenp∧r≺p.
Proof. It is well known thatp∨randp∧rexist, and we have the following:
p∨rpr−p◦r, with Imp∨r ImpImr, kerp∨r kerp∩Imr,
p∧rp◦r, with Imp∧r Imp∩Imr, kerp∧r kerpkerr. 3.1
Ifr is an atom, the subspaces Imr resp., kerrrepresent inEs,⊆an atom line resp., a coatom is a hyperplane inEs,⊆. So, the assumptionp∧r0 implies
kerpkerrE, 3.2
Imp∩Imr{0} 3.3
Hence, inEs,⊆the equality3.2implies Imp≺ImpImror equivalently Imp≺Imp∨r.
Furthermore, the equality3.3implies kerp∩kerr≺kerpor equivalently kerp∨r≺kerp. Then fromTheorem 2.2p≺p∨r.
Using a similar reasoning as above we obtainp∧r≺p.
Proposition 3.2. Letp, q, sbe three elements ofPsuch thatp◦qq◦p.Ifs≺pands≺q,sp∧q. Dually, ifp≺sandq≺s,sp∨q.
Proof. FromTheorem 2.2,s ≺ pands ≺ qimply Ims ≺ Imp, kerp ≺ kersand Ims ≺ Imq, kerq ≺kers.So, inEs,⊆we have ImsImp∩Imqand kerskerpkerq.We have then
sp∧q.In a similar reasoning we can prove thatsp∨qifp≺sandq≺s.
4. Rank function
Theorem 2.2allows us to define inPa rank functionRdefined byRp dim Imp,p∈ P.We can easily prove that
R0p 0dim 0P
p /q, p≤q implyRp<Rq, p, q∈ Pstrict isotonicity
p≺q impliesRq Rp 1.
Proposition 4.1. Pis of finite length and of length equal to the dimension ofE.
Proof. Let a maximal chain ofP: 0P≺p1≺p2≺ · · · ≺pi≺pi1≺ · · · ≺idE.By applyingRto this sequence, we get 0<Rp1 1<2Rp2<· · ·<dimE,which proves that the length ofPis equal to the dimension ofE.
Pbeing of finite length, it verifies the ascending and descending chain condition.
Corollary 4.2. The posetPwith the rank functionRis 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. Ifpandqare two elements ofPsuch thatp◦qq◦p,Rp∨q Rp Rq.
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, andrare 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, . . . , pmis a family ofmelements ofPcommuting two by two, that is,pi◦pj
pj◦pii /j,1≤i, j≤n, then
m
i1
pi m
i1
pi−
i/j
pi◦pj
i/j/k
pi◦pj◦pk− · · · −1m−1p1◦p2◦ · · · ◦pm. 4.2
In particular, if this family is orthogonal, then∨mi1pimi1pi.
Proof. For proving the equality we use induction onm≥2.Since by assumption the elements
picommuting two by two, we have form2, p1∨p2p1p2−p1◦p2. Form3,
p1∨p2∨p3
p1∨p2
∨p3
p1∨p2
p3−
p1∨p2
◦p3
p1p2−p1◦p2p3−p1◦p3−p2◦p3p1◦p2◦p3
3
i1
pi− 1≤i/j≤3
pi◦pjp1◦p2◦p3.
4.3
Assume that the equality is verified until the orderm−1,
m−1
i1
pim−1
i1
pi− 1≤i/j≤m−1
pi◦pj
1≤i/j/k≤m−1
pi◦pj◦pk
− · · · −1m−2p1◦p2◦ · · · ◦pm−1.
Then m i1 pi m−1 i1 pi ∨pm
m−1
i1
pipm− m−1
i1
pi
◦pm
m−1
i1
pi−
1≤i/j≤m−1
pi◦pj
1≤i/j/k≤m−1
pi◦pj◦pk− · · · −1m−2p1◦p2◦ · · · ◦pm−1
pm− m−1
i1
pi
◦pm
1≤i/j≤m−1
pi◦pj◦pm− 1≤i/j/k≤m−1
pi◦pj◦pk◦pm
· · · −1m−1p
1◦p2◦ · · · ◦pm−1◦pm
m−1
i1
pi− 1≤i/j≤m−1
pi◦pj− m−1 i1 pi ◦pm
1≤i/j/k≤m−1
pi◦pj◦pk
1≤i/j≤m−1
pi◦pj◦pm−
1≤i/j/k≤m−1
pi◦pj◦pk◦pm· · · −1m−1p1◦p2◦ · · · ◦pm−1◦pm.
4.5
It follows from the combinatorial relationCsmCsm−1Csm−1−1for alls≤mthat
1≤i/j/k≤m−1
pi◦pj◦pk
1≤i/j≤m−1
pi◦pj
◦pm
1≤i/j/k≤m
pi◦pj◦pk. 4.6
Hence we have the desired equality.
It is clear that if the familyp1, p2, . . . , pmis orthogonalpi◦pj pj◦pi 0P, i /j, the equality proved before becomes ∨mi1pi mi1pi.It is well known in Linear algebra that if
p1, . . . , pmis an orthogonal family verifying in addition∨mi1piidE, then this family
charac-terizes 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 idem-potent endomorphism in the same space. It is thus natural to ask the following question. Can one in the general case for a given orderϑof finite length satisfying in addition the condition of Jordan-Dedekind identify a finite familyx1, . . . , xpsuch that∨xi1ϑandxi∧xj0ϑfori /j,
where 1ϑand 0ϑare the bounds ofϑ. If the answer is affirmative, can one in this case establish
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