A constructive version of Warfield's Theorem

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A constructive version of Warfield’s Theorem

Cyrille Chenavier

To cite this version:

Cyrille Chenavier. A constructive version of Warfield’s Theorem. 2019. �hal-02120656v5�

A constructive version of Warfield’s Theorem

Cyrille Chenavier^∗

∗Inria Lille - Nord Europe, Villeneuve d’Ascq, France (e-mail:

[email protected]).

Abstract:Within the algebraic analysis approach to linear system theory, a multidimensional linear system can be studied by means of its associated finitely presented left module. Deep connections exist between module isomorphisms and equivalent matrices. In the present paper, we introduce a constructive proof of a result due to Warfield which controls the size of equivalent matrices involved in the study of isomorphic modules. We illustrate our constructive proof with an example coming from differential equations with constant coefficients.

Keywords:Linear functional systems, module isomorphisms, matrix equivalence 1. INTRODUCTION

A linear multidimensional system, such as a linear system of differential equations or partial derivative equations, maybe described by a matrix of functional operators.

Indeed, a system ofqequations withpunknown functions η₁,· · ·, η_p over a ring of operators D is written

ker_F(R.) :={η∈ F^p|Rη= 0}, (1) where R∈D^q×p andF is the functional space where we are looking for the solutions. The system (1) can be studied by mean of algebraic tools, using the finitely presented left moduleM :=D^1×p/(D^1×qR) withpgenerators submitted to theqrelations specified by the lines ofR. Indeed, from the properties of free and quotient modules, the abelian group kerF(R.) is isomorphic to homD(M,F), that is the abelian group of left D-linear maps from M to F, see Malgrange (1963). Hence, systemic properties of (1) can be studied by mean of modules properties, which can be computed using effective homological algebra and Groebner bases theory, see Chyzak et al. (2005).

The abelian group hom_D(M,F) only depend on M and F, in particular it does not depend on the matrixR. That means that two matrices R and R⁰ defining isomorphic modules M 'M⁰ have the same algebraic properties. A particular example for the isomorphism M ' M⁰ is the case whereR andR⁰ areequivalent, that isR=Y R⁰X⁻¹ for invertible matrices X and Y. A result due to Fitting (1936) asserts a weak converse implication: if R and R⁰ define isomorphic modules, then they can be enlarged by 0 and identity blocs leading to equivalent matricesLandL⁰. An effective version of this result was obtained in Cluzeau and Quadrat (2011).

The purpose of the present paper is to introduce an effective version of a result of Warfield (1978), which asserts that the number of 0 and the size of identity blocs inLandL⁰maybe reduced, while keeping equivalent matrices. This result is based on an algebraic invariant of D, called the stable rank, see McConnell and Robson (2001). For a general presentation of the problem, we refer to Nicolas (2014).

The paper is organised as follows. In Section 2, we recall the notions of isomorphic left D-modules and equivalent matrices, as well as the effective version of Fitting’s result and the statement of Warfield’s result. In Section 3, we present the constructive version of Warfield’s result for removing 0 and identity blocs. In Section 4, we illustrate this constructive version with a differential equation with constant coefficients. Section 5 contains proofs of formulas used in Section 3.

2. ISOMORPHISMS AND EQUIVALENT MATRICES In this section, we recall the characterization of morphisms between finitely presented left D-modules, as well as results of Fitting and Warfield which rely isomorphic left D-modules to matrix conjugation.

2.1 Effective version of Fitting’s Theorem

Consider two left D-modulesM and M⁰ with finite pre- sentations:

D^1×q D^1×p M 0,

D^1×q0 D^1×p0 M⁰ 0,

.R π

.R⁰ π⁰

namely, exact sequences (see Rotman (2009)), where R ∈ D^q×p, (.R)(µ) = µR, for every µ ∈D^1×q and π is the natural projection onM =D^1×p/(D^1×qR) (similarly forR⁰ andπ⁰).

From Rotman (2009), there exists f ∈ hom_D(M, M⁰) if and only if there exist matricesP ∈D^p×p0 andQ∈D^q×q0 such thatRP =QR⁰ and

∀λ∈D^1×p, f(π(λ)) =π⁰(λπ).

Hence, the following diagram is exact and commutative:

D^1×q D^1×p M 0

D^1×q0 D^1×p0 M⁰ 0

.R

.Q

π

.P f

.R⁰ π⁰

We letn:=q+p⁰+p+q⁰ andm:=p+p⁰. The twon×m matrices

L:=





R 0 0 id_p0

0 0 0 0



, L⁰:=





0 0 0 0 idp 0 0 R⁰



 (2) induce finite presentations: M ' D^1×m/(D^1×nL) and M⁰ 'D^1×m/(D^1×nL⁰). In Cluzeau and Quadrat (2011), an effective version of a result due to Fitting (1936) is given. If f is an isomorphism, then L and L⁰ are equivalent: there exist 6 matricesR2∈D^r×q,R⁰₂∈D^r0×q0, Z₂ ∈ D^p×q, Z₂ ∈ D^q×r, Z₂⁰ ∈ D^q0×r0, Z ∈ D^p×q and Z⁰ ∈ D^p0×q0 and two pairs of invertible matrices (X, X⁰) and (Y, Y⁰) of size mandn

X:=

_id

p P

−P⁰ id_p0−P⁰P

, Y :=





id_q 0 R Q

0 id_p0 −P⁰ Z⁰

−Z P 0 P Z⁰−ZQ

−Q⁰ −R⁰ 0 Z⁰₂R⁰₂



^,

X⁰:=

_id

p−P P⁰ −P P⁰ id_p0

, Y⁰:=





Z2R2 0 −R −Q P⁰Z−Z⁰Q⁰ 0 P⁰ −Z⁰ Z −P id_p 0 Q⁰ R⁰ 0 id_q0



^,

(3)

such that L⁰ = Y⁰LX. In other words, the following diagram is exact and commutative

D^1×n D^1×m M 0

D^1×n D^1×m M⁰ 0

.L .Y

π⊕0

.X f

.L⁰ .Y⁰

0⊕π⁰

.X f⁰

2.2 Warfield’s Theorem

A result due to Warfield (1978) asserts that that the size of 0 an id blocs in (2) can be reduced, whereas the new matrices are still equivalent. This result is based on the notion ofstable rank. The definition of the latter requires to introduce various notions that we present now.

A column vector u := (u1· · ·uk)^T ∈ D^k×1 is called unimodular if there exists a line v ∈ D^1×k such that vu = 1. Moreover, u is said to be stable if there exist d₁,· · · , d_k−1∈D such that (u₁+d₁u_k· · ·u_k−1+d_k−1u_k) is unimodular. An integerris said to be in the stable rank ofD if whenever k > r, every columnu∈D^k×1 is stable.

The stable rank sr(D) of D is the smallest integer in the stable rank ofD.

Assume that the two matrices (2) are equivalent, then Warfield’s Theorem asserts that if there exist two integers randssuch that







s≤min(p+q⁰, q+p⁰)

sr(D)≤max(p+q⁰−s, q+p⁰−s) r≤min(p, p⁰)

sr(D)≤max(p−r, p⁰−r)

then the following (n−r−s)×(m−r) matrices are equivalent

L:=

R 0 0 id_p0−r

0 0

!

, L⁰:=

0 0

idp−r 0 0 R⁰

!

,

and induce finite presentations ofM andM⁰, respectively.

In the next section, we introduce a procedure which computes two pairs of invertible matrices (X, X⁰) and (Y , Y⁰) such thatL⁰ =Y⁰LX.

3. EFFECTIVE WARFIELD’S THEOREM Throughout this Section, we fix some notations. LetMand M⁰ be two leftD-modules, isomorphic withf :M →^∼ M⁰, with inverse f⁰, finitely presented by matrices R ∈D^q×p andR⁰∈D^q0×p0, respectively, and letL,L⁰,X,Y,X⁰and Y⁰ be the matrices defined in (2) and (3).

Given a nonzero integerk, we letk:=k−1. For 1≤i≤k, thei-th vector of the canonical basis ofD^1×kis writtene^k_i. Moreover, thei-th component ofv∈D^1×k orv∈D^k×1is writtenvi. Finally, for a matrixA∈D^k×k0, the coefficient at position (i, j), the i-th row and the j-th column are writtenAij,Ai. andA.j, respectively.

3.1 Reduction of the zero bloc

In this section, we present the procedure for removing s lines of 0 inLandL⁰, where sis such that

s≤min(p+q⁰, q+p⁰),

sr(D)≤max(p+q⁰−s, q+p⁰−s). (4)

Without lost of generalities, we supposeq+p⁰≤p+q⁰. We letn:= (q+p⁰+p+q⁰)−s, so thatn= (q+p⁰+p+q⁰)−s, and we defineL_s, L⁰_s∈D^n×mas follows:

Ls:=

R 0 0 id_p0

0 0

!

, L⁰_s:=

0 0 idp 0 0 R⁰

!

,

that is,LsandL⁰_shave respectivelyp+q⁰−sandq+p⁰−s lines of zeros.

Our objective is to constructn-square matricesYsandY_s⁰, inverse to each other and such that the following diagram is exat and commutative:

D^1×n D^1×m M 0

D^1×n D^1×m M⁰ 0

.L_s

.Ys

π⊕0

.X f

.L⁰_s .Y_s⁰

0⊕π⁰

.X⁰ f⁰ (5)

For that, we let Y₀ := Y, Y₀⁰ := Y⁰ and we assume by induction thatYs and Y_s⁰ have been constructed and are such that (5) is exact and commutative. We decomposeY_s andY_s⁰ as follows:

Y_s= Y1

Y2

Y3

!

, Y_s⁰= Y₁⁰ Y₂⁰ Y₃⁰

,

where Y1, Y2 and Y3 (respectively, Y₁⁰, Y₂⁰ and Y₃⁰) have q +p⁰, p+ q⁰ −s and 1 lines (respectively, columns), respectively. Finally, we let

k:=q+p⁰−s.

Proposition 1. There exist c∈D and d,u∈D^{1×(p+q0−s)} such that

c(Y₁⁰)k. d

(Y1).k

(Y2).k+u^T(Y3)k

= 1. (6)

Proof. Getting the coefficient at position (k, k) in the relation Y_s⁰Ys = idn, we get the following relation:

(Y₁⁰)_k.(Y₁)_.k+ (Y₂⁰)_k.(Y₂)_.k+ (Y₃⁰)_k(Y₃)_k = 1. Hence, the left D-module N := D/D

(Y₁⁰)k.(Y1).k

is generated by [(Y₂)_ik]_N, 1 ≤i ≤ p+q⁰−s, and [(Y₃)_k]. From (4), we have sr(D)≤p+q⁰−s, and from (McConnell and Robson, 2001, Lemma 11.4.6), sr(D) is in the stable range of N.

Hence, there existsu:= (u1,· · · , up+q⁰−s) such that N is generated by [(Y2)ik+ui(Y3)k], 1≤i≤p+q⁰−s. Thus, there existc∈Dandd:= (d₁,· · · , d_p+q⁰_−s) such that (6) holds.

With the notations of Proposition 1, we introduce the lines

`˜∈D<sup>1×n</sup> and`∈D^1×n defined as follows:

`˜:= c(Y₁⁰)k. d

, `:= c(Y₁⁰)k. d 0

,

as well as the matrices U, U⁰ ∈ D^n×n and F ∈ D^n×n defined as follows:

U:=

id_q+p0 0 0 0 idp+q⁰−s u^T

0 0 1

!

, U⁰:=

id_q+p0 0 0 0 idp+q⁰−s −u^T

0 0 1

!

,

F:=

Y1

Y2+u^TY3

.

We point out that U and U⁰ are inverses to each other, and that F = (idn 0)U Y_s. Moreover, from ` = ˜`(id_n 0) and (6), we get

1 = ˜`F(e<sup>n</sup><sub>k</sub>)<sup>T</sup> =`U Y_s(eⁿ_k)^T. (7) Finally, we consider ψ, ψ⁰ ∈D^n×n, ι, ι⁰ ∈D^n×n defined as follows

ψ:=

_id

n−F(eⁿ_k)^T`˜

`˜

, ι:= id_n−F(eⁿ_k)^T` F(e˜ ⁿ_k)^T

,

ψ⁰:= id_n−(eⁿ_k)^T`U Y_s

_{idk−1 0}

0 0

0 id_p+q0

, ι⁰:=

_id

k 0 0 0 0 id_p+q0

.

Proposition 2. We have the following relations:

(1) ιψ= idn, (2) ι⁰ψ⁰= id_n,

(3) ker(.ψ) =D`, (4) ker(.ψ<sup>0</sup>) =D`U Y_s.

Proof. By computing matrix products, ιψ is equal to (idn−F(eⁿ_k)^T`)˜<sup>2</sup>+F(e<sup>n</sup><sub>k</sub>)<sup>T</sup>`. Moreover, from (7),˜ F(eⁿ_k)^T`˜

is a projector, so that idn−F(eⁿ_k)^T`˜is also a projector, whence Point 1.

We have ι⁰(eⁿ_k)^T = 0, from which we deduce Point 2 by computing the matrix product.

Let us show Point 3. Considering the isomorphismD^1×n' D^1×n ⊕D, we have ψ = ψ1ψ2, where ψ1 ∈ D^n×n and ψ₂∈D^n×n are defined as follows:

ψ1:=

id_n−F(eⁿ_k)^T`˜0

0 1

, ψ2:=

id_n

`˜

.

From (7), im(.ψ₁) is included in ker(.F(eⁿ_k)^T)⊕Dand the restriction of .ψ2 to the latter is injective: for (u, x) ∈

ker(.F(eⁿ_k)^T)⊕D

∩ ker(.ψ2), we have u + x`˜ = 0, so that x`F(e˜ ⁿ_k)^T = 0, which, by (7), gives x = 0 and u = 0. Hence, we have ker(.ψ) = ker(.ψ1), that is ker

id_n−.F(eⁿ_k)^T`˜

⊕ 0. We conclude by showing ker

idn−.F(eⁿ_k)^T`˜

= D`: the right to left inclusion is˜ due to (7), and the other is due to x= xF(e<sup>n</sup><sub>k</sub>)<sup>T</sup>`, for˜ everyx∈ker

idn−.F(eⁿ_k)^T`˜ .

Let us show Point 4. From (7), we haveD`U Y_s⊆ker(.ψ⁰).

The converse inclusion is due to the relationx=xk`U Ys, for everyx∈ker(.ψ<sup>0</sup>). Indeed, the firstk−1 and the last p+q<sup>0</sup>columns ofxandx<sub>k</sub>`U Y_sare equal sincex∈ker(.ψ⁰) implies

x

idk−1 0

0 0

0 id_p+q0

!

=x_k`U Y_s

idk−1 0

0 0

0 id_p+q0

!

.

Moreover, thek-th column ofxk`U Y<sub>s</sub>isxk`U Y_s(eⁿ_k)^T, that isxk from (7).

Theorem 3. With the previous notations, we let Ys:=ιU Ysψ⁰, Y_s⁰:=ι⁰Y_s⁰U⁰ψ.

The following diagram is exact and commutative:

D^1×n D^1×m M 0

D^1×n D^1×m M⁰ 0

.L_s .Ys

π⊕0

.X f

.L⁰_s .Y_s⁰

0⊕π⁰

.X⁰ f⁰

In particular, we have

L⁰_s=Y_s⁰L_sX.

Proof. We only have to show that the diagram is com- mutative.

First, we show thatYs and Y_s⁰ are inverse to each other.

From Proposition 2, the lines of the following diagram are exact

D D^1×n D^1×n 0

D D^1×n D^1×n 0

.`

id_D

ψ

.U Y_s .Ys

.`U Y_s ψ⁰ .Y⁰

sU⁻ .Y_s⁰ (8)

Moreover, it is also commutative. Indeed,ψYs is equal to ψιU Y_sψ⁰ and from 1 of Proposition 2, we have im(.ψι− idn)⊆ker(.ψ). By commutativity of the left rectangle and

by exactness of the lines of (8), we have (ψι−idn)U Y_sψ⁰= 0, so that ψY_s = U Y_sψ⁰. In the same manner, we show that Y_s⁰U⁰ψ = ψ⁰Y_s⁰. By commutativity and exactness of (8) and from the equations U Y_sY_s⁰U⁰ =Y_s⁰U⁰U Y_s = id_n, we getYsY_s⁰ =Y_s⁰Ys= idn.

In Section 5.1 we prove the following relations:

ιL_s=Ls, U L_s=L_s, Y_sL⁰_s=L_sX, ψ⁰L⁰_s=L⁰_s, ι⁰L⁰_s=L⁰_s, Y_s⁰L_s=L⁰_sX⁰, U⁰L_s=L_s, ψL_s=L_s.

(9)

Hence, YsL⁰_s = LsX and Y_s⁰Ls = L⁰_sX⁰ follow from commutativity of

D^1×n D^1×n D^1×n D^1×n D^1×n

D^1×m D^1×m D^1×m D^1×m D^1×m

Ls ι

L_s U ψ

L_s Y_s

U⁰

L⁰ s

ψ⁰ Y⁰

s ι⁰ L⁰_s

idm id_m X id_m

X⁰

3.2 Reduction of the identity bloc

In this section, we assume that s zero lines have already been removed from L and L⁰, where s is as in (4). For simplicity, we writeK,K⁰,Z andZ⁰instead ofL_s,L⁰_s,Y_s andY_s⁰, respectively. Hence, we have

K= R 0

0 idp⁰

0 0

!

, K⁰= 0 0 idp 0 0 R⁰

!

,

whereKandK⁰ have respectivelyp+q⁰−sandq+p⁰−s lines of zeros, and Z, Z⁰ are (q+p⁰ +p+q⁰ −s)-square matrices such as in (5).

We present the procedure for removingr lines of identity blocs ofK andK⁰, whereris such that

r≤min(p, p⁰)

sr(D)≤max(p−r, p⁰−r). (10)

Without lost of generality, we assume thatp≤p⁰. We let n1:= (q+p⁰+p+q⁰−s)−randn2:=p+p⁰−r, so that n₁= (q+p⁰+p+q⁰−s)−r,n₂=p+p⁰−r, and we define K_r, K_r⁰ ∈D^n1×n2 as follows:

Kr:=

R 0 0 id_p0−r

0 0

!

, K_r⁰ :=

0 0

idp−r 0 0 R⁰

!

.

Our objective is to construct two pairs of square matrices (Z_r, Z_r⁰), (X_r, X_r⁰), of sizen₁ and n₂, respectively, invert- ible to each other and such that the following diagram is exact and commutative:

D^1×n1 D^1×n2 M 0

D^1×n1 D^1×n2 M⁰ 0

.Kr

.Z_r

π⊕0

.X_r f

.K⁰_r .Z⁰_r

0⊕π⁰

.X_r⁰ f⁰ (11)

We let Z0 := Z, Z₀⁰ := Z⁰, X0 = X, X₀⁰ := X⁰ and we assume by induction that Z_r, Z_r⁰, X_r and X_r⁰ have been constructed and are such that (11) is exact and commutative. We decompose these matrices as follows:

X_r=

X1 X2

X3 X4

X5 X6

!

, Z_r=





Z1 Z2 Z3

Z4 Z5 Z6

Z7 Z8 Z9

Z10 Z11 Z12



^,

X_r⁰ =

_X⁰

1 X₂⁰ X₃⁰ X₄⁰ X₅⁰ X₆⁰

, Z_r⁰ =

Z₁⁰ Z₂⁰ Z₃⁰ Z⁰₄ Z₅⁰ Z₆⁰ Z₇⁰ Z⁰₈ Z₉⁰ Z₁₀⁰ Z⁰₁₁ Z₁₂⁰

!

,

where the column and line separations are the following:

• the column (respectively, line) blocs of X_r from left to right (respectively, top to bottom) havep−rand p⁰ columns (respectively,p,p⁰−rand 1 lines),

• the column (respectively, line) blocs of X_r⁰ from left to right (respectively, top to bottom) have p, p⁰−r and 1 columns (respectively,p−randp⁰ lines),

• the column (respectively, line) blocs ofZrfrom left to right (respectively, top to bottom) haveq+p⁰−s,p−r andq⁰ colums (respectively,q,p⁰−r, 1 andp+q⁰−s lines),

• the column (respectively, line) blocs of Z_r⁰ from left to right (respectively, top to bottom) haveq,p⁰−r, 1 andp+q⁰−scolumns (respectively,q+p⁰−s,p−r andq⁰ lines).

Finally, we let

k1:=q+p⁰−s+p−r, k2:=p−r.

Lemma 4. We have

X3=Z5, X5=Z8 (12)

and

(Z₅⁰)k₂.(Z2).k₂+ (Z⁰₈)k₂.(Z11).k₂= (X₁⁰)k₂.(X1).k₂. (13)

Proof. From KrXr =ZrK_r⁰ andK_r⁰X_r⁰ =Z_r⁰Kr, we get respectively





RX1 RX2

X3 X4

X5 X6

0 0



=





Z2 Z3R⁰ Z5 Z6R⁰ Z8 Z9R⁰ Z11 Z12R⁰



,

0 0 0

X₁⁰ X₂⁰ X₃⁰ R⁰X₄⁰ R⁰X₅⁰ R⁰X₆⁰

!

=

Z₁⁰R Z₂⁰ Z⁰₃ Z₅⁰R Z₆⁰ Z⁰₇ Z₉⁰R Z₁₀⁰ Z₁₁⁰

!

.

(14)

Hence, (12) holds. Moreover, from Z_r⁰Zr = idn₁ and X_r⁰Xr = idn₂, the coefficients at positions (k1, k1) and (k2, k2) of Z_r⁰Z_r and X_r⁰X_r, respectively, are equal to 1.

Hence, we get

(Z₅⁰)k₂.(Z2).k₂+ (Z₆⁰)k₂.(Z5).k₂+ (Z₇⁰)k₂.(Z8).k₂+ (Z⁰₈)k₂.(Z11).k₂

=(X₁⁰)k₂.(X1).k₂+ (X⁰₂)k₂.(X3).k₂+ (X₃⁰)k₂.(X5).k₂.

From (14), we also have X₂⁰ = Z₆⁰ andX₃⁰ =Z₇⁰, so that (13) holds.

Proposition 5. There exist c ∈ D and d,u ∈ D^1×(p0−r) such that

c(Z₅⁰)k₂. d c(Z₈⁰)k₂.

^(Z2).k2

(Z5)_.k2+u^T(Z8)_k2 (Z11).k₂

!

(15)

and

c(X₁⁰)k₂. d

(X1).k₂

(X3).k₂+u^T(X5)k₂

= 1. (16)

Proof. By computing the matrix products and from Lemma 4, the left-hand sides of (15) and (16) are equal.

Moreover, we show (15) as we did for Proposition 1 by taking the coefficient at position (k₁, k₁) of Z_r⁰Z_r, and considering the leftD-module

N:=D/D

(Z₅⁰)_k2.(Z2)_.k2+ (Z₈⁰)_k2.(Z11)_.k2

.

With the notations of Proposition 5, we introduce the lines

`˜1 ∈ D<sup>1×n1</sup>, `1 ∈ D^1×n1, ˜`2 ∈ D<sup>1×n2</sup> and `2 ∈ D^1×n2 defined as follows:

`˜1:= c(Z₅⁰)k2. d c(Z₈⁰)k2.

, `1:= c(Z₅⁰)k2. d 0 c(Z₈⁰)k2.

,

`˜2:= c(X₁⁰)k₂. d

, `2:= c(X₁⁰)k₂. d 0

,

as well as the matrices U₁, U₁⁰ ∈ D^n1×n1, F₁ ∈ D^n1×n1, U₂, U₂⁰ ∈D^n2×n2 andF₂∈D^n2×n2:

U1:=







idq 0 0 0

0 idp⁰−r u^T 0

0 0 1 0

0 0 0 id_p+q0−s





^{, U2:=}

idp 0 0 0 id_p0−r u^T

0 0 1

!

,

U₁⁰ :







idq 0 0 0

0 id_p0−r −u^T 0

0 0 1 0

0 0 0 id_p+q0−s





^{, U}

0 2:=

idp 0 0 0 idp⁰−r −u^T

0 0 1

!

,

F1:=

Z1 Z2 Z3

Z4+u^TZ7 Z5+u^TZ8 Z6+u^TZ9

Z10 Z11 Z12

!

,

F2:=

X1 X2

X3+u^TX5 X4+u^TX6

.

We point out that Ui and U_i⁰, i ∈ {1,2}, are inverses to each other and

F1=

idq 0 0 0

0 idp⁰−r 0 0 0 0 0 id_p+q0−s

!

U1Zr, F2=

_id

p 0 0

0 id_p0−r 0

U2Xr.

Adapting the arguments for proving (7), we show

1 = ˜`1F1(e<sup>n</sup><sub>k1</sub>)<sup>T</sup> =`1U1Zr(eⁿ_k1)^T

1 = ˜`2F2(e<sup>n</sup><sub>k2</sub>)<sup>T</sup> =`2U2X_r(eⁿ_k2)^T.

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We decompose id_n1−F₁ eⁿ_k

1

^T`˜₁∈D^n1×n1 as follows

id_n1−F1 eⁿ_k

1

^T_˜

`1= ϕ1 ϕ2

=

_ϕ

3

ϕ4

,

where ϕ1 and ϕ2 (respectively, ϕ3 and ϕ4) have q + p⁰ −r and p+q⁰ −s columns (respectively, lines). Fi-

nally, we consider ψ1, ψ₁⁰ ∈ D^n1×n1, ι1, ι⁰₁ ∈ D^n1×n1, ψ2, ψ⁰₂∈D^n2×n2, andι2, ι⁰₂∈D^n2×n2 defined as follows:

ψ₁:=

ϕ3

˜`1

ϕ4

!

, ψ₁⁰:= id_n1−(eⁿ_k¹

1)^T`₁U₁Z_r

id_k

1 0

0 0

0 id_q0

!

,

ψ₂:=

id_n

2−F2(eⁿ_k²

2)^T`˜2

`˜₂

, ψ₂⁰:= idn2−(eⁿ_k²

2)^T`2U2X_r

id_k

2 0

0 0

0 id_p0

!

,

ι1:= ϕ₁ F₁(eⁿ_k¹

1)^T ϕ₂

, ι2:= id_n

2−F₂(eⁿ_k²

2)^T˜`₂ F₂(eⁿ_k²

2)^T

,

ι⁰₁:=

_id

k1 0 0 0 0 id_q0

, ι⁰₂:=

_id

k2 0 0 0 0 id_p0

.

By adapting the arguments of the proof of Proposition 2, we get:

Proposition 6. We have the following relations (1) ι1ψ1= idn₁,ι2ψ2= idn₂,

(2) ι⁰₁ψ₁⁰ = id_n1,ι⁰₂ψ₂⁰ = id_n2, (3) ker(.ψ1) =D`1, ker(.ψ2) =D`2,

(4) ker(.ψ₁⁰) =D`1U1Z<sub>r</sub>, ker(.ψ<sup>0</sup><sub>2</sub>) =D`2U2X_r.

Theorem 7. With the previous notations, we let Zr:=ι1U1Zrψ⁰₁, Xr:=ι2U2Xrψ₂⁰, Z_r⁰ :=ι⁰₁Z_r⁰U₁⁰ψ1, X_r⁰ :=ι⁰₂X_r⁰U₂⁰ψ2. The following diagram is exact and commutative:

D^1×n1 D^1×n2 M 0

D^1×n1 D^1×n2 M⁰ 0

.K_r .Zr

π⊕0

.Xr f

.K_r⁰ .Z_r⁰

0⊕π⁰ .X_r⁰ f⁻

In particular, we have

K_r⁰ =Z_r⁰KrXr.

Proof. We only have to show that the diagram is com- mutative.

We show that Z_r⁰ and Z_r⁰ (respectively, X_r and X_r⁰) are inverse to each over in the same manner that we did in the proof of Theorem 3.

In Section 5.1 we prove the following relations:

ι1K_r=Krι2, ι⁰₁K_r⁰ =K_r⁰ι⁰₂, U1K_r=K_rU2, Z⁰_rK_r=K_r⁰X_r⁰, Z_rK⁰

r=K_rX_r, U₁⁰K_r=K_rU₂⁰, ψ₁⁰K_r⁰ =K_r⁰ψ⁰₂, ψ1Kr=Krψ2.

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Hence, we deduceZ_rK_r⁰ =K_rX_randZ_r⁰K_r=K_r⁰X_r⁰ from commutativity of:

D^1×n1 D^1×n1 D^1×n1 D^1×n1 D^1×n

D^1×n2 D^1×n2 D^1×n2 D^1×n2 D^1×n2

K_r ι₁

K_r U₁ ψ₁

K_r Z_r

U₁⁰

K⁰ r

ψ⁰₁

Z_r⁰

K_r⁰ ι⁰₁

ι2 U2

ψ2

X_r

U₂⁰

ψ₂⁰

X⁰

r ι⁰₂