Reduction of linear systems based on Serres theorem Alban Quadrat - PowerPoint PPT Presentation

Reduction of linear systems based on Serre’s theorem Alban Quadrat INRIA Sophia Antipolis, APICS Project, 2004 route des lucioles, BP 93, 06902 Sophia Antipolis cedex, France. Alban.Quadrat@sophia.inria.fr Work in collaboration with: Mohamed S. Boudellioua , Department of Mathematics and Statistics, Sultan Qaboos University, PO Box 36, Al-Khodh, 123, Muscat, Oman. boudell@squ.edu.om AACA 2009, Hagenberg (13-17/07/09) Alban Quadrat

Smith forms and reduction problem • When is a polynomial matrix equivalent to its Smith form diag ( γ 1 , . . . , γ q ) , where γ i = α i /α i − 1 , α i = gcd of the i × i -minors of R ( α 0 = 1)? • Theorem: (Boudellioua, 05) Let D = R [ x 1 , . . . , x n ], R ∈ D p × p be a full row rank matrix. Then, the assertions are equivalent 1 There exist U ∈ GL p ( D ) and V ∈ GL p ( D ) satisfying: � � 0 I p − 1 U R V = . 0 det R 2 There exists Λ ∈ D p admitting a left-inverse over D such that − Λ) ∈ D p × ( p +1) admits a right-inverse over D . P = ( R Alban Quadrat

Outline of the lecture • The purpose of this lecture is to: 1 Explain the relations between the previous result and a Serre’s theorem (S´ eminaire Dubreuil-Pisot 60-61) based on the concept of Baer extensions. 2 Simplify and generalize this result to a general full row rank matrix R ∈ D q × p over an Ore algebra D . 3 Constructively solve the problem for important cases. • Implementation in the forthcoming package Serre . • For the reduction problem, the next results are more efficient than the general ones obtained in Cluzeau-Q., LAA 08, based on idempotents of the endomorphism ring end D ( M ), where: M = D 1 × p / ( D 1 × q R ) ( OreMorphisms ) . Alban Quadrat

Generalization of a Serre’s result • Theorem: Let R ∈ D q × p be a full row rank matrix, Λ ∈ D q , − Λ) and the two left D -modules M = D 1 × p / ( D 1 × q R ) P = ( R and E = D 1 × ( p +1) / ( D 1 × q P ) defining an extension of D by M : β α 0 − → D − → E − → M − → 0 . We have the equivalent assertions: 1 E is stably free of rank p + 1 − q : E ⊕ D 1 × q ∼ = D 1 × ( p +1) . 2 P = ( R − Λ) admits a right-inverse over D . 3 ext 1 D ( E , D ) = D q / ( P D p +1 ) = 0. 4 ext 1 D ( M , D ) = D q / ( R D p ) is the cyclic right D -module generated by ρ (Λ), where ρ denotes the projection: ρ : D q − → ext 1 D ( M , D ) = D q / ( R D p ) . The previous equivalences only depend on the residue class ρ (Λ). Alban Quadrat

Main result • Theorem: Let R ∈ D q × p be a full row rank matrix and Λ ∈ D q satisfying that there exists U ∈ GL p +1 ( D ) such that: ( R − Λ) U = ( I q 0) . Let us denote by � � S 1 Q 1 U = ∈ GL p +1 ( D ) , S 2 Q 2 where: S 1 ∈ D p × q , S 2 ∈ D 1 × q , Q 1 ∈ D p × ( p +1 − q ) , Q 2 ∈ D 1 × ( p +1 − q ) . Then, we have: M = D 1 × p / ( D 1 × q R ) ∼ = L = D 1 × ( p +1 − q ) / ( D Q 2 ) The converse result also holds. These results only depend on: ρ (Λ) ∈ ext 1 D ( M , D ) = D q / ( R D p ) . Alban Quadrat

Corollaries • Corollary: We have the following isomorphism: ψ : M = D 1 × p / ( D 1 × q R ) L = D 1 × ( p +1 − q ) / ( D Q 2 ) − → π ( λ ) �− → κ ( λ Q 1 ) . Its inverse ψ − 1 : L − → M is defined by ψ − 1 ( κ ( µ )) = π ( µ T 1 ): � � R − Λ U − 1 = T 1 ∈ D ( p +1 − q ) × p , T 2 ∈ D ( p +1 − q ) . , T 1 T 2 • Corollary: Let F be a left D -module and the linear systems: ker F ( R . ) = { η ∈ F p | R η = 0 } , � ker F ( Q 2 . ) = { ζ ∈ F p +1 − q | Q 2 ζ = 0 } . Then, we have the isomorphism ker F ( R . ) ∼ = ker F ( Q 2 . ) and: ker F ( R . ) = Q 1 ker F ( Q 2 . ) , ker F ( Q 2 . ) = T 1 ker F ( R . ) . Alban Quadrat

Ring conditions • Proposition: Let R ∈ D q × p be a full row rank matrix and Λ ∈ D q such that P = ( R − Λ) ∈ D q × ( p +1) admits a right-inverse over D . Moreover, if D is either a 1 principal left ideal domain, 2 commutative polynomial ring with coefficients in a field, 3 Weyl algebra A n ( k ) or B n ( k ), where k is a field of characteristic 0, and p − q ≥ 1, then there exists U ∈ GL p +1 ( D ) satisfying that P U = ( I q 0). • The matrix U can be obtained by means of: 1 a Jacobson form ( Jacobson ), 2 the Quillen-Suslin theorem ( QuillenSuslin ), 3 Stafford’s theorem ( Stafford ). Alban Quadrat

Example: Wind tunnel model • The wind tunnel model (Manitius, IEEE TAC 84):  x 1 ( t ) + a x 1 ( t ) − k a x 2 ( t − h ) = 0 , ˙   x 2 ( t ) − x 3 ( t ) = 0 , ˙ x 3 ( t ) + ω 2 x 2 ( t ) + 2 ζ ω x 3 ( t ) − ω 2 u ( t ) = 0 .  ˙  • Let us consider D = Q ( a , k , ω, ζ )[ ∂, δ ], the system matrix   ∂ + a − k a δ 0 0  ∈ D 3 × 4 , R = 0 ∂ − 1 0    ω 2 − ω 2 0 ∂ + 2 ζ ω and the finitely presented D -module M = D 1 × 4 / ( D 1 × 3 R ). • The D -module ext 1 D ( M , D ) = D 3 / ( R D 4 ) is a Q ( a , k , ω, ζ )- 0) T ) is a basis. vector space of dimension 1 and ρ ((1 0 Alban Quadrat

Example: Wind tunnel model 0) T and P = ( R • Let us consider Λ = (1 0 − Λ). • The matrix P admits the following right-inverse S : 0 0 0   0 0 0     ∈ D 5 × 3 .  0 − 1 0  S =     − ∂ +2 ζ ω − 1 0   ω 2 ω 2   − 1 0 0 • According to Quillen-Suslin theorem, E = D 1 × 5 / ( D 1 × 3 P ) is free D -module of rank 2. Alban Quadrat

Example: Wind tunnel model • Computing a basis of E , we obtain that U ∈ GL 5 ( D ),  0 0 0 − 1 0  ω 2 0 0 0 0     ω 2 ∂   0 − 1 0 0 U = ,     ∂ 2 + 2 ζ ω ∂ + ω 2 − ∂ +2 ζ ω − 1 0 0   ω 2 ω 2   − ω 2 k a δ − 1 0 0 − ( ∂ + a ) satisfies that P U = ( I 3 0) ( OreModules , QuillenSuslin ). • The wind tunnel model is equivalent to the sole equation: ( ∂ + a ) ζ 1 + ω 2 k a δ ζ 2 = 0 ζ 1 ( t ) + a ζ 1 ( t ) + ω 2 k a ζ 2 ( t − h ) = 0 . ˙ ⇔ Alban Quadrat

Algorithmic issue 1 Consider an ansatz Λ ∈ D q of a given order. 2 Compute a Gr¨ obner basis of ext 1 D ( M , D ) = D q / ( R D p ). 3 Compute the normal form Λ ∈ D q of ρ (Λ). 4 Compute the obstructions to freeness of the left D -module E = D 1 × ( p +1) / ( D 1 × q ( R − Λ)) ( π -polynomials). 5 Solve the systems in the arbitrary coefficients obtained by making the obstructions vanish. 6 If a solution Λ ⋆ exists, then compute U ∈ GL p +1 ( D ) satisfying 0) and return Q 2 ∈ D 1 × ( p +1 − q ) . that ( R − Λ) U = ( I q • Remark: If ext 1 D ( M , D ) = D q / ( R D p ) is 0-dimensional, then we take Λ to be a generic combination of a basis of ext 1 D ( M , D ). Alban Quadrat

Example: Transmission line • Let us consider a general transmission line: ∂ V ∂ x + L ∂ I  ∂ t + R ′ I = 0 ,   C ∂ V ∂ t + G V + ∂ I  ∂ x = 0 .  • Let D = Q ( L , R ′ , C , G )[ ∂ t , ∂ x ] and M = D 1 × 2 / ( D 1 × 2 R ), where: � � L ∂ t + R ′ ∂ x ∈ D 2 × 2 . R = C ∂ t + G ∂ x β ) T , P = ( R − Λ) ∈ A 2 × 3 . • We consider A = D [ α, β ], Λ = ( α • If we denote by N = A 1 × 2 / ( A 1 × 3 P T ), then we have: ext 1 ext 2 A ( N , A ) = 0 , A ( N , A ) = A / ( L 1 , L 2 ) , L 1 = ( C α 2 − L β 2 ) ∂ t + G α 2 − R ′ β 2 , � L 2 = ( C α 2 − L β 2 ) ∂ x + ( L G − R ′ C ) α β. Alban Quadrat

Example: Transmission line • We consider β = C � = 0, α 2 = L C � = 0 and R ′ C − L G � = 0. • Over B = D [ α ] / ( α 2 − L C ), we have ext 2 B ( B ⊗ D N , B ) = 0, i.e., E = B 1 × 3 / ( B 1 × 2 P ) is a projective B -module, and thus, is free. • Then, we have:  − α L  1 − C α S =  ,   R ′ C − L G  ( α ∂ x + L C ∂ t + R ′ C ) − ( α ∂ x + C L ∂ t + L G ) α C Q 1 = α ∂ x − L C ∂ t − R ′ C C ∂ x − α C ∂ t − α G , t − ( L C + R ′ C ) ∂ t − R ′ G . Q 2 = ∂ 2 x − L C ∂ 2 • The transmission line is equivalent to the sole equation: t − ( L C + R ′ C ) ∂ t − R ′ G ) Z ( t , x ) = 0 . ( ∂ 2 x − L C ∂ 2 Alban Quadrat

Torsion-free degree • Theorem: ext 1 D ( M , D ) is 0-dimensional iff the torsion-free degree of M is n − 1 (the last but one step before projectiveness). 1 n = 2, M is torsion-free, 2 n = 3, M is reflexive, . . . Then, we can constructively check whether or not M (ker F ( R . )) can be generated by 1 relation (1 equation)! • If M = D 1 × p / ( D 1 × q R ) is free of rank p − q , i.e., there exists V ∈ GL p ( D ) satisfying that R V = ( I q 0), then we have: � � 0 V ( R 0) = ( I q 0) , 0 1 = D 1 × ( p − q ) (0 equation!) . ⇒ M ∼ = D 1 × ( p +1 − q ) / ( D (0 . . . 1)) ∼ Alban Quadrat

Example: String with an interior mass • Model of a string with an interior mass (Fliess et al, COCV 98):  φ 1 ( t ) + ψ 1 ( t ) − φ 2 ( t ) − ψ 2 ( t ) = 0 ,   φ 1 ( t ) + ˙ ˙  ψ 1 ( t ) + η 1 φ 1 ( t ) − η 1 ψ 1 ( t ) − η 2 φ 2 ( t ) + η 2 ψ 2 ( t ) = 0 ,  φ 1 ( t − 2 h 1 ) + ψ 1 ( t ) − u ( t − h 1 ) = 0 ,     φ 2 ( t ) + ψ 2 ( t − 2 h 2 ) − v ( t − h 2 ) = 0 . • Let us consider D = Q ( η 1 , η 2 )[ ∂, σ 1 , σ 2 ], the system matrix 1 1 − 1 − 1 0 0   ∂ + η 1 ∂ − η 1 − η 2 η 2 0 0    ∈ D 4 × 6 , R =   σ 2  1 0 0 − σ 1 0  1  σ 2 0 0 1 0 − σ 2 2 and the finitely presented D -module M = D 1 × 6 / ( D 1 × 4 R ). Alban Quadrat