Linear Projectors in the max-plus Algebra G. Cohen S. Gaubert J.-P. Quadrat Centre Automatique et Systemes` INRIA Ecole´ des Mines de Paris Rocquencourt Fontainebleau, France Le Chesnay, France [email protected] Stephane.Gaubert Jean-Pierre.Quadrat @inria.fr
Abstract In this paper, we consider the linear projection problem. Consider three semimodules , , and two linear maps In general semimodules, we say that the image of a linear B, C: U X Y operator B and the kernel of a linear operator C are direct B C factors if every equivalence class modulo C crosses the . (2) image of B at a unique point. For linear maps represented U → X → Y by matrices over certain idempotent semifields such as the We say that im B and ker C are direct factors if for all x (max, )-semiring, we give necessary and sufficient con- , there is a unique im B such that Cx C . When∈ ditions+ for an image and a kernel to be direct factors. We X ∈ C = it is the case: 1. the map 5B : , x z, which is characterize the semimodules that admit a direct factor (or C 2 C XC → X 7→C C linear, satisfies (5B ) 5B , 5B B B, C5B C (5B equivalently, the semimodules that are the image of a lin- is the projector onto im=B, parallel to= ker C); 2.= we have ear projector): their matrices have a g-inverse. We give the isomorphism / ker C im B; in particular, if and simple effective tests for all these properties, in terms of are free finitelyX generated' semimodules, then theU linear matrix residuation. mapX B, which can be identified with a matrix, yields a parametrization of the ‘abstract’ object / ker C. In [5], a first answer to the projectionX problem was 1 Introduction given, in a nonlinear setting. If , , are complete lattices, and if B, C are (possiblyUnonlinearX Y ) residuated Classical linear control theory is built on a firm algebraic maps (see 2.2 below), it was shown that im B and ker C ground: vector spaces, and modules. There is some ev- are direct factors§ iff, setting 5 B (C B)] C, (where idence that the construction of a ‘geometric approach’ F ] denotes the residuated map= of a map F), we have of (max, )-linear discrete event systems, in the spirit + C 5 C and 5 B B. Even in the simplest case of Wonham [14], requires the analogue of module the- of =-linear operators= over free finitely generated - ory, for semimodules over idempotent semirings, such as Rmax Rmax semimodules (that is, when B and C are matrices with the ‘(max, ) semiring’ Rmax (R , max, ). ] + = ∪ { ∞} + entries in max), the operator 5 B (C B) C is a By comparison with modules, the theory of semimodules R complicated object (a min-max function= in the sense of over idempotent semirings is an essentially fresh subject, Olsder and Gunawardena — see e.g. [10]), and the test in which even the most basic questions are yet unsolved. C 5 C, 5 B B, is computationally difficult (an Clearly, the image of a linear map F : example= of non trivial= direct factors was given in [5], for should be defined as usual: im F F(x) Xx → Y. ( )2, ( )3, the proof that im B But what is the kernel of F? Some= authors{ [7,| 12]∈ X de-} Rmax Rmax andU = kerYC=are direct factorsX = involved a tedious computa- fine ker F x F(x) ε , where ε is the zero tion of equivalence classes, together with a geometrical element of =.{ This∈ notionX | is essentially= } non pertinent for argument). -linearY maps, since ker F is in general trivial, even Rmax In this paper, we give a much simpler test for matrices: for ‘strongly’ non injective maps. im B and ker C are direct factors iff there exist two matri- Consider now the following alternative definition ces L, K such that B LCB and C CBK (we denote = = 2 with the same symbol B the matrix B and the linear map ker F (x, x 0) F(x) F(x 0) . (1) C = ∈ X = x Bx). Then, 5B LC BK. The existence of 7→ = = Clearly, ker F is a semimodule congruence, and we can the matrices K, L can be checked very simply (in polyno- define the quotient semimodule / ker F. Now, triv- mial time) using residuation of matrices (and not of linear ially, the canonical isomorphism theoremX holds: im F maps). / ker F. Thus, in general semimodules, (1) seems' to As a by-product, we solve the following problem, Xbe the appropriate definition. For control applications, it which was left open in [5]: given a matrix B, does there is indeed appropriate, since F typically represents an ob- exist a projector onto im B?; or, equivalently does there servation map, by which we wish to quotient some state exist a matrix C such that im B and ker C are direct fac- space. tors? The answer is positive iff B admits a g-inverse, that is, iff B BXB, for some matrix X. The existence of a For the sake of symmetry, we will complete an idem- g-inverse= can also be checked (simply) in polynomial time potent semifield with a maximal element (for “top”), using matrix residuation. which satisfies aD , a ,> and a The proofs are critically based on a linear extension a , a >=>( ) ∀ε ∈. WeD∪ denote {>} >= theorem, which states that a linear form F on a finitely > this dioid,=> and∀ we∈ willD∪{>} call it\{ the} top completionD =ofD∪{>}. n D generated subsemimodule of (Rmax) can be represented In , the product also distributes with respect to : by a row vector G: F(x) Gx. This result was D ∧ = a(c d) ac ad , proved by Kim [8, Lemma 1.3.2] for matrices with en- a, b, c , ∧ = ∧ (4) ∀ ∈ D (c d)a ca da tries in the Boolean semiring. Cao, Kim and Roush [4, ∧ = ∧ Th. 4.7.4] proved a variant of this result for the semiring (this property does not hold in general dioids). ([0, 1], max, ). As in the case of [8, 4], the proof con- sists in proving that the maximal linear subextension is an extension. This seems to require very strong properties on 2.2 Residuation the dioid (lattice distributivity, invertibility of product). Definition 1. We say that a dioid is residuated if Note that certain results pertaining to kernels rely upon D a linear extension theorem whereas this theorem is not re- 1. for all a and b in , x ax b admits a maxi- D { ∈ D | } quired to prove dual results on images mal element denoted a b; \ In 2, we introduce the algebraic notions used in the 2. x xa b admits a maximal element de- paper.§ In 3, we prove the linear extension theorem, and noted{ ∈bD/a;| } derive factorization§ theorems for linear maps. In 4, we § characterize direct factors. In 5, we relate the existence 3. ( , ) is a lattice. of projectors to the existence of§ g-inverses. D Then, the maximal element of x axc b ex- ists and can be denoted a b/c which{ ∈ canD be| read indiffer-} 2 Algebraic Preliminaries ently as (a b)/c or a (b/\c). The top\ completion\ of an idempotent semifield is S 1 S residuated, with a x a x if a is invertible, ε x , We briefly and informally recall the few algebraic results and x ε if \x = , (similar\ formulæ=> needed here. More details can be found in [1] for dioids for />\). = 6= > >\>=> and ordered sets, and in [7] for semirings and semimod- Consider the following linear equations in X: ules. A seminal reference in residuation theory is [3]. See also [6]. AX B , (5a) = A semiring is a set equipped with two laws , , XC D , (5b) such that: ( , ) is a commutativeS monoid (the zero is de- = noted ε); ( S, ) is a (possibly noncommutative) monoid AXC F , (5c) S = (the unit is denoted e); is right and left distributive over where X, A, ... are (possibly rectangular) matrices with ; and the zero is absorbing. A semiring in which non entries in a residuated dioid . zero elements have an inverse is a semifield. A semiring We extend the and / notationD to matrices: is idempotent if a , a a a. Idempotent semir- \ S ∀ ∈ S = ings are also called dioids. In this paper, we will mostly A B def X AX B , (6a) consider dioids such as Rmax, which is also a semifield. \ = { | } D/C def _ X XC D , (6b) = { | } 2.1 Order properties of dioids def A F/C _ X AXC F . (6c) \ = { | } A dioid (or more generally, an idempotent additive _ monoid) is equipped with the natural order relation: Explicitly, we have the following formulæ, which relate the residuation of matrices to the residuation of scalars: a b a b b . (3) ⇐⇒ = (A B)ij Aki Bkj , (7a) \ = \ Then, a b coincides with the upper bound a b for the k ∨ ^ natural order . Note that ε is the bottom element of : (D/C)ij Dil/Cjl , (7b) = x ,ε x. D l ∀ ∈ D ^ Moreover, if is a semifield, ( , ) is a lattice. In- (A F/C)ij Aki Fkl/C jl . (7c) D D 1 1 1 \ = \ deed, for all non zero a, b, a b (a b ) kl ^ (a 1 b 1) 1;ifaor b is zero, ∧a b= ε. This∨ shows that= To decide whether the matrix equations (5) have a so- ( , ) is a lattice. We say that the∧ idempotent= semifield lution, it suffices to check that the maximal subsolution isD distributive if the lattice ( , ) is distributive [1]. D D satisfies the equality. Proposition 2. Take five matrices A, B, C, D, F as 3 Linear Extension Theorem and above, with entries in a residuated dioid. Then: Factorization of Linear Maps X, AX B A ( A B ) B , (8a) ∃ = ⇐⇒ \ = X, XC D ( D / C ) C D , (8b) Let us begin with an elementary and apparently innocent ∃ = ⇐⇒ = X, AXC F A ( A F / C ) C F . (8c) property. ∃ = ⇐⇒ \ = Proposition 3. Let denote an arbitrary semiring. Con- sider a free -semimoduleS , two -semimodules , , 2.3 Semimodules and two linearS maps F : F S, G : .G TheH F → H G → H In this section, denotes an arbitrary semiring. A right following assertions are equivalent: S -semimodule,oraright semimodule over , is a com- 1. im F im G; mutativeS monoid ( , ), together with anS external law E , (u, s) u.s, which satisfies, for all u,v , 2. there exists a linear map H : such that F E S → E 7→ ∈E s,t , u.(st) (u.s).t, (u v).s u.s v.s, G H. F → G = u.(s ∈ tS) u.s u=.t, ε.s ε, u.ε ε, u.e= u. Left -semimodules= are= defined dually.= For= simplicity, Proof. Clearly, (2) implies (1). Conversely, taking a basis we willS simply speak of semimodule when the underlying of , ui i I , for all i I , there exists hi im G such F { } ∈ ∈ ∈ that F(ui ) G(hi ). Since is free, setting H(ui ) hi , semiring and the side (right vs. left) are clear from the = F = context. S for all i I , we define a linear map H : , which satisfies ∈F G H. F → G In a semimodule over a dioid, addition is idempotent. = Indeed, a a a.e a.e a.(e e) a.e a. = = = = The dual of Prop. 3 requires some conditions on the A map F from a (right -) semimodule to a (right semiring, as shown by the following counterexample. -) semimodule is linearS if it is additiveE ( u,v S F ∀ ∈ , F(u v) F(u) F(v)) and right-homogeneous Example 4. Let , , denote three free semimodules, E = ( u , s , F(u.s) F(u).s). The set of linear and consider twoF mapsG HF : , G : . The ∀ ∈ E ∈ S = maps is denoted Hom ( , ). inclusion ker G ker F needH not→ implyF theH existence→ G of a E → F E F A generating family of a semimodule is a family linear map H : such that F H G. Consider E G → F = ui i I of elements of such that each element v the semiring Nmax (N , max, ) equipped with { } ∈ E ∈ E = ∪ { ∞} + writes as a finite linear combination v i I ui .si , with the laws max, , Nmax, = ∈ = =+ F=G=H= si (‘finite’ means that i I si ε is finite, even G(x) x 2, F(x) x 1. We have ker G ker F ∈ S { ∈ | 6= L} = + = + if I is infinite). A generating family ui i I is a ba- (in fact, ker G ker F (x, x) x Nmax ) but there { } ∈ = = { | ∈ } sis if i I ui .si i I ui .ti , with i I si ε and exists no linear map H such that F H G. Indeed, any ∈ = ∈ { ∈ | 6= } = i I ti ε finite, implies si ti , for all i I .A linear map H : Nmax Nmax writes H(x) x a where { ∈ L| 6= } L = ∈ → = + semimodule is finitely generated (f.g., for short) if it has a a H(0) Nmax. We obtain F(0) 1 2 a:a finite generating family. A semimodule is free if it has a contradiction.= ∈ = = + basis. The term free for arises from the following universal We will derive the dual of Prop. 3 from the following E semimodule version of the Hahn-Banach theorem. property: given an arbitrary family gi i I of elements of a semimodule , there is a unique linear{ } ∈ map F : F E → F Theorem 5 (Linear Extension). Let be a distributive such that F(ui ) gi , i I . S = ∀ ∈ idempotent semifield. Let , denote two free f.g. - All semimodules with a basis of n elements are isomor- semimodules, and let F Gbe a f.g. subsemimodule.S phic to n, equipped with the laws: u,v n,s , H G S ∀ ∈ S ∈ S For all F Hom ( , ), there exists G Hom ( , ) (u v)i ui vi, (u.s)i ui.s. A linear map ∈ H F ∈ G F = = such that x , G(x) F(x). F : p n writes F(x) Ax, where A is a n p ∀ ∈ H = S → S = matrix with entries in . Proof. It suffices to prove the result when n and S We will use the following notation, for matrices: . Since is f.g., we have G =imSH, for someF = SH n pH. Clearly, G we can assumeH that= H has no T transpose: (A )ij Aji , zero row.∈ InS this case, for all 1 i n, kernel: ker A = (x, y) ( n)2 Ax Ay , = ∈ S n | = image: im A Ax x . L(i) j 1 j p, Hij ε ∅ . = { | ∈ S } = 6= 6= That is, matrix A is identified with the linear map x Let H j denote the j-th column of H, and let F(H) denote 7→ Ax. We will use this convention systematically in the se- the row-vector whose j-th entry is F(H j ). Wehaveto 1 n quel. prove the existence of a row-vector G such that ∈ S F(H) GH . (9) = Using (8b), this is equivalent to 4 Direct Factors in Semimodules
F(H) F(H)/H H . (10) and Linear Projectors = Using (7b), we get: Definition 7. Let X be a subset in a semimodule and E be an equivalence relation in . We say that x inX has 1 a projection on X parallel to E Xif there exists in XXsuch F(H)/H H F(H)l Hkl Hkj . j = k l L(k) ! that E x. We say that X crosses E if there exists such ∈ M ^ a projection for all x in . We say that X is transverse Using the distributivity of product with respect to E X ∧ to if the projection of any x is unique whenever it (see (4)), we obtain: exists. Finally, we say that X and E are direct factors if existence and uniqueness of the projection is ensured for 1 F(H)/H H F(H)l Hkl Hkj . (11) all x in . j = k l L(k) X M ∈^ In the previous definition, consider X im B for B = ∈ Let 8 denote the set of maps ϕ : 1,... ,n kL(k), Hom ( , ) and E ker C for C Hom ( , ), { }→∪ such that ϕ(k) L(k), for all k. Since the lattice ( , ) whereU ,X and are= semimodules over∈ a semiringX Y . is distributive, we∈ have: S U X Y C S If im B and ker C are direct factors, 5B denotes the cor- responding projector. It is straightforward to check that 1 C F(H)/H H F(H)ϕ(k)Hk ϕ(k) Hkj 5 Hom ( , ). Also, from the very definition, it j B = ϕ 8 k ∈ X X ^∈ M comes that 1 F(Hϕ(k))Hk ϕ(k) Hkj C C B 5 B C C5 . (13) = ϕ 8 k B B ^∈ M = ; = 1 However, the existence of a linear projector 5 (that is, F Hϕ(k) Hk ϕ(k) Hkj 2 such that 5 5) satisfying (13) is not a sufficient con- = ϕ 8 k ! = ^∈ M dition for im B and ker C to be direct factors. Indeed, for any B and C, the identity over satisfies (13). (the last equality is by linearity of F on im H). To show X that (F(H)/H)H F(H) (the other inequality is triv- Theorem 8 (Existence). Let denote an arbitrary ial), it remains to check that for all ϕ and j, semiring. Let B Hom ( , S) and C Hom ( , ) ∈ U X ∈ X Y 1 where , , are free f.g. -semimodules. The follow- H ϕ(k)Hk ϕ(k) Hkj H j . (12) U X Y S ing statements are equivalent: k M 1. there exists K Hom ( , ) such that If Hij ε, then the inequality (12) is plain. If Hij ε, ∈ X U = 1 6= we choose k i, and obtain Hiϕ(i)Hi ϕ(i) Hij Hij, which C CBK (14) shows that (12)= holds and the proof is complete.= = ; 2. im C im CB; Corollary 6. Let be an idempotent distributive semi- = field. Let , , Sdenote three free f.g. -semimodules, 3. im B crosses ker C. and considerF G twoH maps F : , G S: . The Moreover, if is a residuated dioid, a practical test for following assertions are equivalent:H → F H → G checking that theS above conditions hold true is by trying 1. ker G ker F; the equality 2. there exists a linear map H : such that F C CB (CB) C . (15) = \ H G. G → F = Proof. Proof. Clearly, 2 implies 1. Conversely, assume that 1 2 The assumption implies that im C im CB ker G ker F. Then, there exists a map K ⇒ ∈ but the converse inclusion is trivial. Hence equality holds Hom (im G, ) such that K (G(x)) F(x), for all x true. . Indeed, forF any y G(x) im=G, define K (y) ∈ H = ∈ = 2 3 From the assumption, it follows that for all x , F(x). Since ker G ker F, the value K (y) is indepen- ⇒ ∈ X dent of the choice of x such that y F(x). Clearly, there exists u such that Cx CBu. The projection = of x we are looking∈ U for is Bu=. the map K is linear. By Theorem 5, K admits a linear = extension H Hom ( , ). For all x ,wehave 3 1 By assumption, for all x , there exists u H G(x) K∈(G(x)) G FF(x), hence H ∈G H F. such⇒ that Cx CBu. That is to∈ X say, im C im CB∈ U. = = = From Prop. 3, it= follows that there exists K Hom ( , ) such that C CBK. ∈ X U = The practical test follows from statement 1, together However, while the pair of leftmost equations have been with (8a). proved to be a test that im B and ker C are direct factors, there is no evidence at this moment that the other pairs of Theorem 9 (Uniqueness). The mappings B and C are as equations can play the same role. in the previous theorem. But now is an idempotent dis- tributive semifield. The followingS statements are equiva- Remark 12 (Duality). If is an idempotent distributive S lent: semifield, by transposition, it is straightforward to check that im B crosses ker C if, and only if, im C T is transverse 1. im B is transverse to ker C; to ker BT . Likewise, im B is transverse to ker C if, and T T 2. ker B ker CB; only if, im C crosses ker B . Finally, im B and ker C = are direct factors if, and only if, im C T and ker BT are 3. there exists L Hom ( , ) such that C T BT T ∈ Y X so. In this case, 5B 5C T (in general, (M/N) T T = = B LCB . (16) N M ). = \ A practical test for checking that they hold true is by trying 5 Direct Factors and g-Inverses the equality
B B/(CB) CB . (17) = In this section, we answer the question of when a semi- Proof. module im B admits a direct factor ker C. Unlike in the case of usual linear spaces, this question cannot receive a 1 2 By assumption, if there exist u and u0 such that positive answer for any linear operator B. An explicit test ⇒ CBu CBu0, since x Bu has a unique projection on is given to characterize homomorphisms such that their = = im B parallel to ker C, it should be that Bu Bu0. This images admit a direct factor. = means that ker CB ker B. But the converse inclusion is Definition 13. Let , denote two semimodules over an trivial, hence equality holds true. arbitrary semiring U. LetX B Hom ( , ). S ∈ U X 2 3IfkerB ker CB, from Cor. 6, there exists L 1. An element F Hom ( , ) such that BFB B is Hom⇒ ( , ) such= that B LCB. ∈ ∈ X U = Y X = called a g-inverse of B; 3 1 For some x, suppose there exist two projections ⇒ 2. when B admits a g-inverse, it is called regular; Bu and Bu0 on im B parallel to ker C. Then CBu CBu, hence LCBu LCBu , thus Bu Bu and the= 0 = 0 = 0 3. a g-inverse F which satisfies FBF F is called a projection is unique. reflexive g-inverse; = The practical test follows from statement 1, together 4. when is a dioid1, an element F Hom ( , ) such with (8b). that BFBS B is called a g-subinverse∈ of B.X U Corollary 10. If im B and ker C are direct factors, then Theorem 14. Let , denote free f.g. semimodules over U X 5C LC BK B/(CB) C B (CB) C . (18) a residuated dioid . Then: B = = = = \ D Proof. If B and C are direct factors, then (14) and (16) 1. Any B Hom ( , ) admits a maximal g-subinverse. ∈ g U X g 5 5 It is denoted B . In matrix terms: B B B/B. both hold true. Consider BK. From (16), = \ LCBK, and then from (14), 5= LC. Thus, 5 BK = 2. When B is regular, Bg is the greatest g-inverse and LC 52. In addition, this shows= that, for any= x, 5=x Br, defined by Bg BBg, is the greatest reflexive g-inverse belongs= to im B and, moreover, (14) shows that C5x of B. Cx, which means that the projection on im B is parallel= to C ker C. Thus this projector 5 is indeed 5B . The last two Proof. expressions in (18) result from the choice of maximal L and K previously mentioned. 1. This is an immediate consequence of (8c). g r Remark 11. Gathering the results in (15) on the one 2. If B is a g-inverse, then B is the maximal reflexive r hand, of (13) and (18) on the other hand, one has that g-inverse: indeed, B is a reflexive g-inverse since r g g g C CB (CB) C C B/(CB) C . (19a) BB B (BB B)B B BB B B , = \ = = = = BrBBr Bg(BBgB)BgBBg Bg(BBgB)Bg Similarly, with (17) on the one hand, (13) and (18) again = = BgBBg Br . on the other hand, one obtains = = 1In this case, the monoid (Hom ( , ), ) is naturally ordered by B B/(CB) CB B (CB) C B . (19b) F G iff F G G. U X = = \ = It is maximal since, if F is another reflexive g-inverse, Observe that the image of any homomorphism being in- then F Bg because Bg is the maximal g-inverse. It variant by translation along the first diagonal, it is enough follows that to represent im B by its projection on any plane orthogo- nal to that diagonal (see figure). F FBF BgBBg Br . = = Finally, if B is regular, then BBrB B and Br BBr References Br. = = [1] F. Baccelli, G. Cohen, G.J. Olsder, and J.P. Quadrat. Theorem 15. Let be a distributive semifield, , be Synchronization and Linearity. Wiley, 1992. free f.g. -semimodules,S and B Hom ( , ). ThisU XB is regular iffS im B admits a direct∈ factor kerU CX, where C [2] R.B. Bapat and T.E.S. Raghavan. Nonnegative Ma- Hom ( , ) and isafreef.g. -semimodule. One can∈ trices and Applications. Encyclopedia of Mathemat- take forXCYany reflexiveY g-inverseS of B. ics and its Applications 64, Cambridge University Press, 1997. Proof. Let B be any reflexive g-inverse of B. Then the statements of Theorems 8 and 9 are satisfied with , [3] T.S. Blyth and M.F. Janowitz. Residuation Theory. C B , L B, K B , and ker B is the directY factor= U Pergamon press, 1972. = = = we are looking for. [4] Z.Q. Cao, K.H. Kim, and F.W. Roush. Incline alge- Conversely, if im B admits a direct factor ker C,by bra and applications. Ellis Horwood, 1984. (16), B LCB, and then by (18), B BKB, which shows that= B is regular. = [5] G. Cohen, S. Gaubert, and J.P Quadrat. Kernels, images and projections in dioids. Proceedings of Remark 16. When B is put in the form WODES’96, Edinburgh, August 1996. IEE. D ε [6] R.A. Cuninghame-Green. Minimax Algebra. Num- P Q, εε ber 166 in Lecture notes in Economics and Mathe- matical Systems. Springer, 1979. where P and Q are permutation matrices and D has no zero rows and columns, it is easy to check that Dg — and a [7] J.S. Golan. The theory of semirings with applica- fortiori Dr Dg —haveno entries and that a particular tions in mathematics and theoretical computer sci- reflexive g-inverse of B is > ence, volume 54. Longman Sci & Tech., 1992.
r [8] K.H. Kim. Boolean Matrix Theory and Applications. 1 D ε 1 B0 Q P . Marcel Dekker, New York, 1982. = εε [9] V. Kolokoltsov and V Maslov. The general form of Remark 17. The maximal reflexive g-inverse does not al- endomorphism in the space of continuous functions ways coincides with the maximal g-inverse. For example, with values in a numerical semiring with idempotent take Rmax, and consider S = addition. Dokl, Akad. Nauk SSSR, 295(2):283–287, 010 g 1 20 r 1 20 1987. Engl. transl., Soviet. Math. Dokl 36 (1988), B 021 ,B 2 2 2 B 2 2 2 . = 000 = 1 2 1 6= = 2 2 1 no. 1, 55–59. Example 18. The following matrix with entries in Rmax, [10] J. Gunawardena. Min-max functions. Discrete Event
0 n 0 Dynamic Systems, 4:377–406, 1994. B 00n , = n00 [11] V. Maslov and S. Samborski˘ı, editors. Idempotent is regular whenever n 0 and not regular otherwise: analysis, volume 13 of Adv. in Sov. Math. AMS, RI, 1992. g 2n 2n n nnn B n 2n 2n if n 0 and nnn if n < 0 . = 2n n 2n nnn [12] M.A. Shubin. Algebraic remarks on idempotent semirings and the kernel theorem in spaces of bounded functions. V. Maslov and S. Samborski˘ı, editors, Idempotent analysis, volume 13 of Adv. in Sov. Math. AMS, RI, 1992. [13] E. Wagneur. Moduloids and pseudomodules. 1. di- mension theory. Discrete Math., 98:57–73, 1991. [14] W.M. Wonham. Linear multivariable control: a ge- ometric approach. Springer, 1985. regular case nonregular case