Column reduction of polynomial matrices

Transcription

1 Column reduction of polynomial matrices WHL Neven Afd Informatica NLR POBox AD Emmeloord The Netherlands fax C Praagman Department of Econometrics University of Groningen POBox AV Groningen The Netherlands cpraagman@ecorugnl fax: July Abstract A few years ago Beelen developed an algorithm to determine a minimal basis for the kernel of a polynomial matrix (see [1 3] In this paper we use a modified version of this algorithm to find a column reduced polynomial matrix unimodularly equivalent to a given polynomial matrix 1 Introduction For us the problem considered in this paper finding a column reduced polynomial matrix unimodularly equivalent to a given one has its roots in linear systems theory For instance in the book of Kailath [7] one can find several examples of the importance of column or row reduced polynomial matrices Our direct interest stems from the behavioral approach to systems theory see Willems [ ] Assume that we are interested in the behavior of a set of variables w : T R q and from physical or economical considerations we can derive a number a linear differential or difference equations with constant coefficients that these variables have to satisfy: P (τw(t =0 where P R g q [s] and τ = d/dt or the shift: τw(t=w(t+1 For obvious reasons we like to minimize the number and the order of the equations Clearly the set of solutions does not change if we premultiply the equation by an invertible operator It turns out (see [13] that the invertible operators are unimodular polynomial matrices and minimality is reached if we find a unimodular U such that UP is row reduced Transposing leads to the problem we consider here Some preliminary results preceding this paper have been reported on several occasions see for the original idea Beelen van den Hurk Praagman [2] and for successive improvements: Neven [8] Praagman [9 10] all correspondence should be sent to the second author 1

2 2 Preliminaries Let us start with defining the notions mentioned in the introduction: Definition 1 Let P R m n [s] Then d(p the degree of P is defined as the maximum of the degrees of its entries and d j (P the j th column degree of P as the maximum of the degrees in the j th column δ(p is the array of integers obtained by arranging the column degrees of P in non-decreasing order Definition 2 Let U R m m [s] Then U is unimodular if det(u R\{0} Let P (s = diag(s d1(p s dn(p then P P is a proper rational matrix Definition 3 Let P R m n [s] Then the leading column coefficient matrix of P Γ(P is defined as: Γ(P :=P P ( If P =( 0 P TT a permutation matrix and Γ(P has full column rank then P is called column reduced With a little abuse of terminology we will call a matrix Q a basis for the module M ifthe columns of Q form a basis of M: Definition 4 Let M be a submodule of R n [s] Then Q R n r [s] is called a basis of M if rank Q = r andm=im Q If moreover Q is column reduced then Q is called a minimal basis of M Note that if Q(s has full column rank for all s C thenmis a direct summand of R n [s] so in that case Q is a minimal polynomial basis in the sense of Forney [4] or Beelen [1] The theorem (Wolovich [16] Kailath [7] stating that every polynomial matrix is unimodularly equivalent to a column reduced one was the starting point of the investigations leading to this paper Since we need a slightly stronger formulation of the theorem than usually is proven we give a proof here: Theorem 1 Let P R m n [s] then there exists a U R n n [s] unimodular such that R := PU is column reduced Furthermore δ(r δ(p totally Proof By induction on δ(p lexicographically If P is column reduced there is nothing to prove So suppose that in Γ(P there is a linear dependence between its nonzero columns: k a kγ k (P = 0 Let δ j be the largest column degree involved ie such that a j 0 then replacing the jth column of P by k a ks δj(p δ k(p P k yields a P unimodularly equivalent to P for which δ(p <δ(p both totally and lexicographically which proves the theorem The above proof is constructive but unfortunately it has awkward numerical properties as was pointed out in Van Dooren [11] The idea on which this paper is based is the following: calculate a minimal basis for the module ker(p I m :={v R n+m [s] (P Iv=0} see also [2] The first observation is that if (U t R t t is such a basis then U is unimodular Lemma 1 Let (U t R t t be a basis for ker(p I m thenuis unimodular Proof If U is not unimodular then there exists a λ C and a v C n such that U(λv =0 and hence such that R(λv = P (λu(λv = 0 Define w C n+m [s] byw=(u t R t t vthen w(λ = 0 hence w(s =(s λx(s But then x ker(p I m so x =(U t R t t yforsome y C n [s] implying that (s λy(s =v a contradiction Of course R = PU but although (U t R t t is minimal and hence column reduced this does not necessarily hold for R But if R is not column reduced then for b>d(u(u t s b R t t cannot be a minimal basis for ker(s b P I m Calculating a minimal basis for ker(s b P I m yield a pair (U b R b such that s b PU b = s b R b where again we may hope that R b is column reduced 2

3 The first part of this paper is devoted to the proof that indeed for b large enough R b is column reduced and to the investigation into the nature of this b In the second part we describe an algorithm to calculate ker(s b P I m in a numerically reliable way and we estimate the computational effort involved Since the effort increases quickly with the growth of b we develop in the third part an iterative algorithm that consists of calculating ker(s b P I m forb=1 until R b is column reduced as was suggested already in [2] 3 Column Reduction 31 Invariants of polynomial matrices In this section we introduce the concepts of left and right minimal indices and of elementary exponents which will play a role in the next section: Definition 5 Let P R m n [s] Its right minimal indices κ := (κ 1 κ q are defined by κ = δ(q where Q is a minimal basis for ker(p Itsleft minimal indices are the right minimal indices of P t Clearly q equals n r(p with r(p := rank(p Next we define the notion of elementary exponent closely related to elementary divisors Therefore we introduce the homogeneous polynomial associated to P : Let P R m n [s]: then P h R m n [s t] is defined by P (s =P d s d + +P 0 P h (s t =P d s d +P d 1 s d 1 t+ +P 0 t d Let i be the greatest common divisor of the i i minors of P h and define 0 =1 Then i divides i+1 and let i / i 1 =: c i Π(as bt li(b/a where the product is taken over all pairs (1band(01 and 1/0 is denoted by Definition 6 The factors (as bt li(b/a withl i (b/a 0are called the elementary divisors of P and the integers l i the elementary exponents of P Remark It is well known (see Gantmacher [5] that there exist unimodular matrices U and V such that UPV (s=diag( i(s 1/ i 1(s 1 in particular this implies that P and CP have the same finite elementary divisors (ie those for which a 0 for any unimodular C Of course the same holds if one reverses the role of s and t: There exist unimodular S T such that S(tP h (1tT(t= diag( i(1t/ i 1(1t Definition 7 Let P R m n [s] then the structural indices of P are its left and right minimal indices and its elementary exponents Following Beelen [1] we define for each matrix polynomial P R m n [s] of degree d>0its linearization L P R md n+(m 1d [s] by: I 0 0 sp d si I sp d 1 L P (s:= 0 0 si sp 1 + P 0 Remark The concept of linearization of a polynomial matrix is widely used Not always the same definition is used see for another implementation for instance the book of Kailath [7] But basically all linearizations amount to the same kind of construction There is a close relationship between the structural indices of a polynomial matrix and those of its linearization: 3

4 Theorem 2 Let P R m n [s] andletl P be its linearization Then a The right minimal indices of P and L P are equal; b The elementary divisors of P and L P are equal; c The left minimal indices of L P exceed those of P by d(p 1 Proof Premultiplying L P by the unimodular matrix C(s defined by I 0 0 si I C(s := 0 s d 1 I I yields that Note that v := (v 1 v d I 0 0 sp d 0 I s 2 P d +sp d 1 C(sL P (s = 0 0 P(s ker(l P { vd ker(p v i := (s i P d + +sp d i+1 v d i<d Since Pv d = 0 it follows that v i = (P d i + + s i d P 0 v d so d(v = d(v d proving that κ(l P =κ(p Note that CL P and L P have the same finite elementary divisors and that there exist a unimodular D such that CL P D = diag{iip} This implies immediately that the finite elementary divisors of P and L P are the same By symmetry of s and t the same holds for the infinite elementary divisors Let V t be a minimal polynomial basis of P t then clearly (0 0V t is a minimal polynomial basis for (CL P t and hence it follows that C t (0 0V t =(s d 1 V V t is a minimal polynomial basis for (L P t which yields the third statement As an immediate consequence of theorem 2 we find: Theorem 3 Let P R m n [s] be a polynomial matrix Then the sum of its structural indices equals r(p d(p Proof It can be deduced immediately from the well known Kronecker normal form for matrix pencils ([5] that the theorem holds for polynomial matrices of degree 1 Then r(l P =r(cl P = m(d(p 1 + r(p hence its number of left minimal indices (and that of P ism r(pfrom theorem 2 we conclude that the sum the structural indices of P equals the sum of the structural indices of L P minus (m r(p (d(p 1 hence equals Corollary 1 κ i r d and l j r d md m + r md + m + rd r = rd 4

5 32 The associated polynomial matrices For each b 1 we associate to P R m n [s] a matrix polynomial defined by: P b (s :=(s b P(s I Note that P b has no left minimal indices and that all its elementary divisors have the form t ω Denote its right minimal indices by ε(b =(ε 1 (bε n (b and its elementary exponents by ω(b =(ω 1 (bω m (b Lemma 2 Let P R m n [s] have rank r letκ 1 κ 2 κ n r be its right minimal indices and let l i = l i ( Then the structural indices of the associated matrices P b satisfy for all b and i : a if κ i b then ε i (b =κ i b if i>n r then b ε i (b c for all iε i (b b+d d for all iε i (b ε i (b+1 ε i (b+1 e if i r then ω i (b =min(l i b+d and if i>rthen ω i (b =b+d Proof a Note first that any nullvector ( of P extends to a nullvector of P b by adding a number of zeros U So κ i ε i (b Let V b = be a minimal basis for ker(p b Clearly s b divides R b so Rb if any column in V b hasdegreelessthanbit has zeros in R b and therefore is made up of a nullvector of P Hence ε i (b κ i if ε i (b <bif κ i = b then b ε i (b κ i b b This inequality follows from the reasoning above I c Since s b is a basis for ker(p P b U d The fact that s 1 is a basis for ker(p R b+1 implies the first inequality the second b+1 U is an immediate consequence of the fact that is a basis for ker(p sr b+1 b e Note that P h b (s t =(sb P h (s t t b+d I All nonzero order i minors of P h b are of the form t k an order i k minor of P h So i (P b = gcd(t k(b+d s (i kb i k (Pk =1i = gcd(t i(b+d t (i 1(b+d+l1 q 1 (st (i 2(b+d+l2+l1 q 2 (s t i r(b+d+lr+l1 q r (s = t (i g(b+d+lg+l1 gcd(q 1 (sq r (s where g r is such that l g b + d and if g<rthen l g+1 >b+d From this it is clear that for i g ω i (b =l i and for i>gω i (b=b+d proving e Let us analyze the situation: the sum of the structural indices of P b equals (b + dm by theorem 3 and those of P b+1 sum up to (b +1+dm Each index is a nondecreasing function of b and increases at most by 1 in view of lemma 2 These observations give a unique description of the behavior of the structural indices for largeb which enables us to prove the following generalization of theorem 1 5

6 Theorem 4 Let P R m n [s] have rank r and degree d andletv b =(U t R t t be a minimal polynomial basis for ker P b Then U b is unimodular Moreover if b max(κ n r +1l r d then R b is column reduced and if R b is column reduced then b l r d Proof The unimodularity of U b follows as in [2] Since ij ε i(b +ω j (b=(b+dmit is clear that for b max(κ n r +1l r d the structural indices ε n r+1 ε n ω r+1 ω m have to increase with increasing b Since s b divides R b we have in the first place that R b =(0 and further by the observation made in the preceding sentence that diag(is h I V b is a minimal basis for ker P b+h Ifhis large enough this implies that s h R b and hence R b is column reduced If on the other hand R b is column reduced then R b contains n r zero columns implying that all the κ i occur in ε(b and hence ω r (b =l r b+d implying the final statement Remark The bound given here depends on κ n r and l r which are in general unknown As a direct consequence of this theorem and the corollary of theorem 3 we find that if b exceeds r d(p then R b is column reduced If P has full column rank we find this yields b>n d(p a worse bound than was found in [2] But it is not hard to see that max(κ jl i d(p never exceeds the bound given there since the κ only occur if P does not have full column rank If r<nour bound will be much better in general 4 Calculation of a minimal basis Let Q R m n [s] and assume that we want to calculate a minimal basis for ker Q The procedure described in [1] reads as follows: i Linearize Q to L Q and find orthogonal matrices U and V such that UL Q V is in a generalized Schur form: upper triangular staircase form with constant right invertible matrices along the block diagonal ii Find a minimal basis for the kernel of this matrix iii Calculate a minimal basis for ker Q starting from the minimal basis found in the preceding step Since in our case the polynomial matrix P b has some special features this procedure works extremely well if we bring some minor modifications in the algorithm kerpol described in [1] 41 Linearization In the first place we introduce in a slightly different linearization of P b P(s =P d s d +P d 1 s d 1 ++P 0 Define : Let P be given by H b (s = A b s E b = sp d I 0 0 sp d 1 si I 0 sp 0 si I si I L((b + dm n 6

7 where L(m n :={H R m m+n [s] degr H =1H(0 = (0 I} As in the proof of theorem 2 we see that sp d I 0 0 C b (sh b (s = s 2 P d +sp d 1 0 I 0 s b+d P 0 I I 0 0 and if V is a basis of ker(h b then V tb := V is a basis for ker(s 0 0 I b P I Andasintheorem2ifV is minimal then V tb is also minimal So the problem reduces to finding a minimal basis for ker H b 42 Minimal basis of the associated pencil A minimal basis of ker H b can be found by constructing an orthogonal matrix U such that UH b Ũ Ũ = diag(iu t is in an upper staircase form in which the constant part equals E := (0 I as in H b Crucial in that respect is the following theorem on the reduction to a staircase form similar to theorems in [1 11] Since our pencil has a special form the theorem gives a slightly stronger statement and therefore we give a complete proof Theorem 5 Let H = sa E L(m n Then there exists an orthogonal matrix U suchthat: sa 11 sa 12 I sa 13 sa 1l+2 UHŨ = 0 sa 22 sa 23 I 0 0 sa ll sa ll+1 I sa ll sn I with A jj R mj mj 1 [s] right invertible and N R m l+1 m l+1 Moreover m j 1 m j =#{i N ε i =j 1} for j =1l Here m 0 := n Before we prove this theorem we state a lemma which will be needed in the theorem but which on the other hand needs this theorem to be proven The proof of the theorem and the lemma will be proven by a simultaneous induction step Note that H has full row rank Lemma 3 Let ( H L(m n and let V R (m+n n [s] be a minimal polynomial basis for ker H V0 Then V (0 = with V 0 0 R n n Since V is a minimal polynomial basis this implies immediately: Corollary 2 V 0 is invertible Proof (of the theorem and the lemma By induction on m + n If m + n = morn = 0 then we can take l =0andN=AandthenkerH=0so there is nothing left to prove Now let n>0 Partition A in the following way: A =(A 1 A 2 with A 2 square and A 1 R m n [s] then there exists an orthogonal U 1 R m m such that U 1 A 1 = invertible It is easy to see that U 1 HŨ1 = ( A11 0 sa11 sa 12 E 12 0 sa 22 E 22 with A 11 R m1 n right with sa 22 E 22 L(m m1m1 m<n+m so the induction hypothesis on A 22 yields the first statement of the theorem Let K 11 be a left invertible real matrix such that Im K 11 =kera 11 let 7

8 A 11 be a right inverse of A 11 and let V 2 be a minimal basis for sa 22 E 22 where we have used the analogous partitioning of E then ( K11 A V := 11 (E 12 sa 12 V 2 0 sv 2 is a basis for UHŨ Let us show that it is minimal Clearly it is column reduced since its leading column matrix equals K11 Γ(V := 0 Γ(V 2 For all nonzero λ V (λ has full column rank since both K 11 and V 2 (λ have ToseethatV(0 has full column rank suppose that (K 11 A 11 E w1 12V 2 (0 =0 w 2 Premultiplying this equation by A 11 yields that w 2 = 0 since E 12 V 2 (0 is invertible by the induction hypothesis applied on the lemma and its corollary But then K 11 w 1 =0and therefore w 1 =0for K 11 is left invertible The second statement of the theorem is now obvious: the dimension of the space of nullvectors of H equals the number of columns of K 11 and that is exactly m 1 m 0 and the induction hypothesis yields the rest The statement of the lemma is true for V and for any other basis of ker UHŨ since the property is invariant under column manipulations But because of the structure of Ũ the statement also holds for bases of ker H for these have the form ŨV(sQ and ŨV(0 = V (0! Clearly the numbers m i do not depend on the particular choice of U Denote these invariants by µ 0 (Hµ l (H In the proof of this theorem we already showed how a basis for ker H is constructed: Let K ii beabasisforker(a ii and A ii a right inverse for A ii Define V ii (s :=s i K ii then i V ji (s :=A jj ( A jk V ki (s+s 1 V j+1i (s k=j+1 V 11 V 12 V 1l 0 V 22 V := 0 Vll 0 0 is a minimal basis for HŨ 43 The kernel of the original matrix polynomial In the special case that H = H b ( V1 I 0 0 := V I I 0 0 U t V is a minimal basis for (s b P I (see [2] Note that V 1 =(V 11 V 22 V 1l and that V 2 = s b PV 1 8

9 44 Numerical properties Since our method to calculate a generalized Schur form is essentially the same as the method proposed in [1] we can conclude that this step of the algorithm is numerically stable But if we consider the complete algorithm then numerical stability is already hard to define For how do we define a small disturbance of a polynomial matrix? If we define two polynomial matrices to be close if the degrees of all entries are the same and the coefficients are close then there is no chance that we can prove that the algorithm is stable for if P is not column reduced and P = RU then a lot of the higher order terms in the product RU have to vanish exactly a property that will not be satisfied for arbitrary small perturbations of U and R Probably a better idea is to return to the original problem we posed in the introduction: we describe a phenomenon by a set of equations expressed in P What we do request of R is that it describes almost the same phenomenon For this we need to set up a topology on solution sets of difference or differential operators which would go beyond the scope of this paper An intermediate approach is the definition in Van Dooren and DeWilde [12] where distance between polynomial matrices is defined in terms of their linearizations But also in this set-up our problem is an ill-posed one In Geurts and Praagman [6] a Fortran implementation of the algorithm is described In section 6 we describe some examples that have been calculated using this implementation A fair number of examples suggested that the algorithm behaves very well but we also encountered examples in which the algorithm did not behave very well A full description of the implementation and both favorable and unfavorable examples are included in [6] In the algorithm described in Beelen van den Hurk and Praagman [2] a different linearization is used This also leads to different answers but to our present knowledge there is no difference in the quality of the answers The research that we report in this paper is by no means finished We still have to find out whether a different linearization can lead to better answers More basically we have to find a good definition of the condition of this problem in terms of the coefficients of P This will give us the right tools to investigate the nature of the problems that arise in the above mentioned examples Finally we like to implement a number of basically different algorithms to make a comparison For instance our problem is closely related to coprime factorization of rational matrices (see [3] or to the problem of finding minimal solutions to rational matrix equations (see [7] Therefore algorithms for these problems possibly could be adapted to our problem 5 An iterative algorithm In principle the problem we posed is solved taking b large enough and calculating the kernel of P b to yield a column reduced R unimodularly equivalent to P Unfortunately this procedure has a severe drawback: The effort that is needed to calculate ker P b is proportional to (m(b + d 3 so if we take the lowest upperbound for b that we can derive from P directly we get (mnd 3 Therefore an alternative idea was suggested in [2] 51 The idea The idea suggested in [2] was to start with b =0and to increase b by one if the calculation of a minimal basis for H b did not lead to a column reduced R Practical evidence confirms the idea that in most cases a small b already produces a column reduced R But in [2] also an example was given of a matrix polynomial for which b had to be at least (n 2d Comparing the CPU times in the examples did not lead to a clear decision (based on a number of examples! about the best way to proceed in general In this paper we improve on this idea by using information already obtained in the previous step for the computation of ker H b+1 9

10 52 The iterative step Assume that we have terminated the algorithm at step b and we proceed with a = b+h Instepb we have determined a U b andaĥbsuch that Ĥb = U b H b Ũ b where Ĥb has in its leading columns a generalized Schur form structure Clearly Hb 0 H a = sn J with and 0 0 I N = 0 Rhm ((b+dm+n 0 0 I 0 0 si I J = 0 Rhm hm [s] si I so if we define U a = diag(u b I then Ĥb 0 Ĥ a := U a H a Ũ a = snũb J If U b =(U ij U ij R mi m then NŨ b =( 0 U t b+d1 U t b+db+d 0 0 so at first sight it seems that only the first block column of Ĥ a preserves the desired structure But fortunately it turns out that we can select U b in such a way that U b+d1 = U b+d2 = = U b+db =0 Theorem 6 Let H L(k + hm n m k have the following structure: H 0 H = sn J with H L(k n N R hm (n+k and J R hm hm [s] with the block structure as above Then there exists an orthogonal matrix U such that UHŨ has a generalized Schur form which displays the following structure: with U = U ij U ij R mi nj m i = µ i (H i=1l m l+1 = hm + k µ i (H n 1 = k n 2 = =n h+1 = m U ij = 0 if j>i+k h 10

11 Proof By induction on hm + k For h = k = 0 there is nothing to prove so assume that hm + k>0 Let sh 1 be the matrix containing the first n columns of H and let H 1 = QR be a QR-decomposition of H 1 : Q orthogonal and R upper staircase: R =(R t 000 t with R R m1 n of full rank Define U = diag(q t I let N = U HŨ = ( N1 0 R 0 0 H H 0 0 sn 1 Q J sn 2 J 2 and J = J1 0 then sn 2 J 2 Define H H = 1 0 sn J 2 with H1 H H := L(k+m m sn 1 Q J 1 m 1 2 and N =(0N 2 R (h 1 (k+m Since (h 1m+m+k m 1 <hm+kthe induction hypothesis yields that there exists an orthogonal U with the following structure: U =(U ij U ij R m i n j with m i = µ i (H fori=1l 1 m l = hm + k m 1 µ i (H n 1 = k + m m 1 n 2 = =n h =m Uij = 0 if j>i Let V be the orthogonal matrix diag(iu and U = VU Divide V into blocks as follows: V 11 V 1h+1 I 0 0 = 0 U11 U 1h R(k+hm (k+hm V l+11 V l+1h+1 0 Ul1 Ulh with V 11 R m1 m1 etc and V ij = Ui 1j 1 j 2 then V ij R mi nj with m i = µ i 1 (H i 2 n 1 = k n j = m j 2 V ij = 0 if j>i From U = VU it follows immediately that U ij = V ij if j>1 U 1j = V 1j Q 11

12 so we see that U has the requested structure if we can prove that m i = µ i (H Since UHŨ has a generalized Schur form this follows immediately from theorem 5 Remark Note that the proof gives a constructive way to find U In the sequel we will use this several times As a consequence we find that if we have a generalized Schur form for H b and we search one for H a then we can divide the orthogonal U that we constructed into blocks as above and then ( U 0 0 I H a I U t I = sa 11 sa 12 I sa 1b sa 22 sa 1h 0 sab+1b sub+1l+1 t suhl+1 t I 0 0 si I 0 si I and hence we only have to work on the matrix in the box in the lower right corner with sa b+1b+1 as first diagonal block having size ((a + dm m 1 m b ((a + dm m 1 m b 1 instead of on H a L((a + dm n Remark Note that in the algorithm from [2] the information cannot be carried over to the next step since there another linearization is used destroying the structure of U Due to this feature the b th step there is of order (b + d 3 m 3 while in our algorithm each step is of order d 3 m 3 Sincebcanincreaseupto rd this means that in the worst case the algorithm in [2] is of order r 4 d 4 m 3 while the algorithm presented here is of order rd 4 m 3 Note that the one step procedure ie starting with b = rd given in [2] is of order r 3 d 3 m 3 Of course the one step procedure has much lower order in our set up since we thoroughly exploit the zero structure of the initial matrix A quick calculation yields that the one step procedure is of order rd 4 m 3 too This means that the iterative algorithm is even in the worst case of the same order as the one step procedure unlike the situation in [2] As said before at each step we gain at least a factor 4 since in the first place we can use Householder transformations instead of Givens transformations and further we do not have to find the postmultiplication separately but we can use Ũ 6 Examples The algorithm as described in section 5 has been implemented in Fortran For a description of the subroutine see Geurts and Praagman [6] The following example has been calculated in double precision on a VAX with machine precision in the order of 29D 39 Example 1 Let the polynomial matrix P be given by: ( s P (s = 4 +6s 3 +13s 2 +12s+4 s 3 4s 2 5s 2 0 s+2 In Kailath [7] page 386 we can find (if we correct a small typo thatpu 0 = R 0 with 1 0 U 0 (s = s+2 1 ( R 0 (s= 0 (s 3 +4s 2 +5s+2 s 2 +4s+4 s+2 12

13 Clearly R 0 is column reduced and U 0 unimodular This example was also treated in [2] The program with a prescribed tolerance of 10D 12 below which matrix entries are considered to be zero yields the following solution: ( R(s = ( U(s = s s s s (s 3 +4s 2 +5s (s 2 +4s s s s then PU R =0up to the tolerance and U is unimodular: ( s U(s = s This solution was found without iterations ie for b =1and equals the solution found in [2] Example 2 In this example we took P (s = 1 s s s as in example 63 from [2] After several iterations the program found for b =5: 1 s 2 s s 6 U(s= 0 1 s s s R(s =diag{ } Our algorithm leads to success for the same b as in [2] and yields a solution in which the special structure of P is reflected References [1] ThGJ Beelen New algorithms for computing the Kronecker structure of a pencil with applications to systems and control theory PhD thesis Eindhoven University of Technology 1987 [2] ThGJ Beelen GJHH van den Hurk and C Praagman A new method for computing a column reduced polynomial matrix Systems and Control Letters 10: [3] ThGJ Beelen and GW Veltkamp Numerical computation of a coprime factorization of a transfer function matrix Systems and Control Letters 9: [4] GD Forney Minimal bases of rational vector spaces with applications to multivariable linear systems SIAM Journal of Control and Optimization 13: [5] FR Gantmacher Theory of matrices III Chelsea New York 1959 [6] AJ Geurts and C Praagman A Fortran 77 package for column reduction of polynomial matrices Transactions on Mathematical Software 1997 To appear 13

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