Generic degree structure of the minimal polynomial nullspace basis: a block Toeplitz matrix approach

Generic degree structure of the minimal polynomial nullspace basis: a block Toeplitz matrix approach Bhaskar Ramasubramanian, Swanand R Khare and Madhu N Belur Abstract This paper formulates the problem of determining the degrees of all polynomial entries in a minimal polynomial basis of a polynomial matrix by using only the degree structure of the given matrix Using a block Toeplitz matrix structure corresponding to the given polynomial matrix, it is shown that the degrees of the elements in its minimal polynomial basis, in the generic case, depend only on the degree structure of the given matrix, and not explicitly on the polynomial coefficients The genericity assumption ensures that numerical issues in determination of the nullspace do not arise like for a specific case Instead of the generic case, when dealing with a specific polynomial matrix, this method gives an upper bound on the degree structure of its minimal polynomial basis The Toeplitz structure effectively reduces the problem of the computation of a right annihilator of a polynomial matrix to a relatively simpler, equivalent problem of computing an annihilator of a constant matrix Index Terms minimal polynomial basis, right annihilator, genericity, degree structure I INTRODUCTION The problem of computating a minimal polynomial basis of a given polynomial matrix is of importance in systems theory (see for instance, [Kai8], [Var91], [Kuc79]) In this paper, an approach to determine the degrees of polynomial entries in each polynomial vector in a minimal polynomial basis of a given polynomial matrix is discussed The work done in this field so far has involved prior knowledge of the entries of the polynomial matrix, after which algorithms have been developed Most notable among these have been [AH9], [AVV5], [DD83], [BV87], [BD88] In [DD83], [BV87], [BD88], the matrix pencil approach is used, which involves converting the given polynomial matrix into an equivalent matrix pencil The problem of computing a nullspace basis is solved for this matrix pencil and the results are transformed to the polynomial matrix case In [AH9], [AVV5], using the relation between the nullspace of a given polynomial matrix and the kernels of associated Toeplitz matrices, an algorithm to compute a minimal polynomial basis for the nullspace of the given polynomial matrix is proposed Here, the LQ factorization (see [GL7]) of these Toeplitz matrices is used to obtain kernels of these matrices and in turn, a minimal polynomial Bhaskar Ramasubramanian is in the Department of Electrical and Computer Engineering, University of Maryland, College Park, MD, USA Swanand R Khare is in the Faculty of Engineering, University of Alberta, Canada Madhu N Belur is in the Department of Electrical Engineering, Indian Institute of Technology Bombay, India Fax: +9157377 Email address: belur@eeiitbacin (corresponding author) basis The primary concern in these references is numerically robust algorithms and it is acknowledged that numerically reliable/stable rank-revealing algorithms are central to these methods In this paper, we do not pursue numerical aspects of the degree structure determination Instead, we restrict ourselves to the so-called generic case and we propose an algorithm for determining the degree structure working with only the degrees of the entries in the original polynomial matrix The next section gives a brief summary of the topics that would be used in the paper after which the problem is formulated The following sections discuss the construction of the block Toeplitz matrix and an algorithm for determining the degree structure of the minimal polynomial basis of a given polynomial matrix (see [KPB1b]), which is then modified for the generic case The paper ends with a few examples which verify the results proposed II PRELIMINARIES AND PROBLEM FORMULATION This section provides a brief introduction to the wellknown notions of a polynomial matrix, genericity of a property (Definition 4) and annihilators of polynomial matrices The problem to be solved is stated at the end of the section A Polynomial matrices The commutative ring of polynomials in the indeterminate s with coefficients from the field of real numbers is denoted by R[s] Then, R [s] denotes the set of matrices with m rows and n columns, with each entry being an element from R[s] In this paper, m < n shall be assumed, unless stated otherwise The set {q 1 (s),, q k (s)} R n [s] is a polynomially independent set if for polynomials α 1 (s),, α k (s), the polynomial combination i α i(s)q i (s) = implies that the polynomials α i (s) =, identically for all i = 1,, k The rank, also called the normal rank, of a polynomial matrix R(s) R [s], is defined as the number of polynomially independent rows in R(s) B Minimal polynomial basis Definition 1: The degree of a polynomial vector v(s) R n [s], denoted by d v, is the maximum amongst degrees of its polynomial components The degree of a polynomial matrix R(s) R [s] is also defined in a similar way Given a polynomial matrix R(s) R [s] of degree d we can write R(s) as a matrix polynomial as R(s) = R + R 1 s + R s + + R d s d

The (right) null space of R(s) is defined as {q(s) R n [s] R(s)q(s) = } (1) For a given polynomial matrix R(s) R [s], with rank m, any set of n m polynomially independent vectors satisfying Equation (1) forms a basis for the nullspace Definition : Minimal Polynomial Basis Let R R [s] be a given polynomial matrix of degree d and rank m Let q 1, q,, q n m be polynomially independent vectors in the nullspace of R arranged in non-decreasing order of their degrees a 1, a,, a n m respectively Then the set {q 1, q,, q n m } is said to be a minimal polynomial basis if for any other set of n m polynomially independent vectors from the nullspace with degrees b 1 b b n m, it turns out that a i b i for i = 1,,, n m The minimal polynomial basis is unique as far as the degrees of the polynomial vectors are concerned The degrees of the vectors in a minimal polynomial basis are called the right minimal indices or Forney indices of R(s) (see [For75], [MMMM6]) C Genericity of parameters Genericity of parameters is a key assumption made in this paper Definition 3: An algebraic variety is the set of solutions E q R n, to a system of polynomial equations If we consider an algebraic variety S R n, the zero equation in the variables will render the variety trivial The fact that a non-trivial algebraic variety in R n (or C n ) is a set of measure zero is used to define genericity of a property Definition 4: ([Wil97, Page 344]) Consider a property P in terms of variables p 1, p,, p n R Property P is said to be satisfied generically if the set of values p 1, p,, p n that do not satisfy P form a non-trivial algebraic variety in R n Example 5: A simple illustration of the property of genericity is the fact that any two nonzero polynomials are generically coprime Another instance is when one chooses all the entries in a square matrix generically from R, the matrix is generically non-singular Left primeness of a wide polynomial matrix is also a generic property III PROBLEM FORMULATION AND SUMMARY OF MAIN RESULTS Given R R [s], define the matrix D Z such that an entry in D is defined as the degree of the corresponding entry in R Notice that all entries in D are nonnegative The only exception is when an entry in R is the zero polynomial, in which case the corresponding entry in D is assigned the value We now define the following sets: Z + = {z Z z } Z + = Z + { } Thus, given R R [s], one can construct a unique D Z + Definition 31: The matrix D constructed from the polynomial matrix R as described above is called the degree structure of the polynomial matrix R We now pose the following question: Problem 3: Let R R [s] be a given polynomial matrix with degree structure D Z + Let Q R n (n m) [s] be such that the columns of Q form a minimal n (n m) polynomial basis for the nullspace of R Let K Z + denote the degree structure of Q Then, can we determine K from D? This problem is the same as computing right minimal indices for a given polynomial matrices and is addressed in the papers [AVV5], [AH9], [KPB1a] Given R R [s], it is clear that it has a unique degree structure D Z +, associated to it It is also clear that given a degree structure D Z +, there exist many R s with appropriate dimension such that their associated degree structures is D Two given polynomial matrices R 1, R R [s] are said to be related, denoted R 1 R, if their degree structures D is same It can be easily checked that this is an equivalence relation The equivalence class for R 1 R [s] is defined as: D(R 1 ) = {R R [s] R 1 R} We now state another interesting problem about inferring the degree structure for minimal polynomial basis for any matrix in a given equivalence class D(R) Problem 33: Let R R [s] be a given polynomial matrix with degree structure D Z + Let R 1, R D(R) and Q 1, Q R n (n m) [s] be such that the columns of Q 1 and Q form minimal polynomial bases for R 1 and R respectively Let K 1 and K denote the degree structures of Q 1 and Q respectively Then, is K 1 = K generically? Problem 33 is settled in positive in the following theorem Theorem 34: Given degree structures K 1, K of minimal polynomial bases corresponding to the same degree structure D, K 1 = K In the following section, we describe the construction of block Toeplitz matrices from a given polynomial matrix, and the relation of kernels of these Toeplitz matrices with the nullspace of the given polynomial matrix IV CONSTRUCTION OF TOEPLITZ MATRICES A polynomial matrix R R [s] with degree d can be written as a matrix polynomial as: R = R + R 1 s + + R d s d where R i R for i =, 1,, d We construct a sequence of real structured matrices from the given polynomial matrix as follows: R R 1 A A =,A 1 =, A = A R d A A A, ()

where s in the above equation are zero matrices of size m n and A i R (d+i+1)m (i+1)n for i The kernels of these structured matrices are related to the nullspace of the given polynomial matrix Consider a polynomial vector v of degree i in the nullspace of R Writing this polynomial vector as a vector polynomial as v = v + v 1 s + v i s i, we have Rv = d i R j s j v j s j = (3) j= j= Comparing the coefficients on both sides in Equation (3), we have A i v c = where v c = v v 1 R(i+1)n Thus, any vector in the kernel of A i is a degree i vector in the nullspace of R, and vice-versa Further, given a vector v with degree i in the nullspace of R, the vectors sv, s v, also belong to the nullspace of R This fact is reflected in the following observation- for any i, if v c R (i+1)n is in the kernel of A i,[ then ] from[ the ] structure of these vc matrices, it is clear that and belong to the kernel v c of A i+1 These constant vectors are related to v and sv in the nullspace of R with degrees i and i+1 respectively Thus, the problem of computing the nullspace of a polynomial matrix is equivalent to computing the kernels of constant matrices (Refer [AH9], [KPB1a] for a detailed analysis) We now elaborate on the computation of a minimal polynomial nullspace basis for the case of polynomial matrices of size 1 3 Let R R 1 3 [s] be a given polynomial matrix with degree d The structured matrices A i as in Equation () are constructed For each i, the matrix A i R (d+i+1) 3(i+1) The dimensions of matrices A i for different i are listed in the following table We now compute a stage i for which v i i size of A i (d + 1) 3 1 (d + ) 6 (d + 3) 9 i (d + i + 1) 3(i + 1) the matrix A i becomes a wide matrix In such a case, the following relation is satisfied 3(i + 1) d + i + 1 i d For instance, when d = 3, from Equation (4), the matrix A i is wide for i 1 In fact, for i = 1, A i is of size 5 6 and A is of the size 6 9 Thus if the matrix A is full (4) rank, then at stage 1, the dimension of the nullspace of A 1 is 1 This vector corresponds to a vector of degree 1 in the nullspace of the polynomial matrix R Further, the nullspace dimension of A is 3 Then from above discussion, it is clear that there is a degree polynomial vector in a minimal polynomial nullspace basis of R In general, at stage i + 1 such that i satisfies condition in Equation (4), one obtains both the vectors in a minimal polynomial nullspace basis of R Therefore, i + 1 is an upper bound for the highest right minimal index For more general cases and the algorithm to compute a minimal polynomial nullspace basis, the reader is referred to [AH9], [KPB1a] These facts are illustrated in the following examples Example 41: Let R = [ s + 1 s 4 s 3 + s 1 ] be the given polynomial matrix Construct matrix A as in Equation () as 1 4 1 A = 1 1 1 1 R4 3 Note that A is full column rank and A 1 R 5 6 has 1-dimensional nullspace spanned by [4 1 1 1 ] T The corresponding degree 1 polynomial vector in a minimal 4 1 nullspace basis is given by 1 + 1 s Further, A R 6 9 has a 3-dimensional nullspace spanned by the vectors v 1, v, v 3 given by v 1 = [4 1 1 1 ] T v = [ 4 1 1 1 ] T v 3 = [4 1 131 33 3] T Clearly, the first two vectors v 1 and v correspond to the degree 1 vector in the minimal polynomial basis The vector v 3 corresponds to a degree vector in the minimal polynomial nullspace basis of R V DEGREE STRUCTURE OF A MINIMAL POLYNOMIAL NULLSPACE BASIS FOR A MATRIX OF SIZE 1 3 In this section, the block Toeplitz approach is used to find the degree structure of the minimal polynomial basis when 1 3 D Z + This leads to a closed form solution for the degree structure of the minimal polynomial basis for the D of given dimensions, which is presented at the end of the section Let R R 1 3 [s] be a given polynomial matrix with degree structure D = [ a b c ] Without loss of generality, we assume that a, b, c are in non-decreasing order Further, it is assumed that none of a, b or c corresponds to the zero polynomial Let Q R 3 [s] be such that the columns of Q constitute a minimal polynomial basis for the nullspace of R Let K be the degree of structure of Q Clearly, the matrix K is a 3 matrix Then we want to infer the degree structure K from the degree structure D In the following subsections,

we discuss cases when c is even and odd separately While investigating these cases, it is assumed that b c The case when b < c is dealt with separately A Case 1: when c is even The matrix A is constructed as outlined in Equation () It is of the following form: deg a b c 1 a A = a + 1 b b + 1 c The structure of A has certain features, listed below: (a) A has dimensions (c + 1) 3 (b) The first (a + 1) rows of A has all elements nonzero (c) The next (b a) rows correspond to terms greater than degree a, and will have the first column, while the remaining two columns will be nonzero (d) The last (c b) rows will have the first two columns, and only the last column will be nonzero Every subsequent A i will be a matrix of dimension (c + i + 1) 3(i + 1) The construction is continued till a stage i when A i is a wide matrix At this juncture, c + i + 1 < 3(i + 1) In fact, c being even ensures the following: 3(i + 1) (c + i + 1) = (5) i + c = (6) i = c It is already known that at this stage, the degree of the minimal polynomial basis vector is i Theorem 51: Let R R 1 3 [s] be a given polynomial matrix with the degree structure D = [a b c] where a b c and c is even Let Q R 3 [s] with the degree structure K be such that the columns of Q constitute a minimal polynomial nullspace basis for R Then, c K = c c (8) b c b c Proof: We construct the matrix A i as in Equation () for i satisfying Equation (7) As mentioned earlier, A i is a matrix of size (c + i + 1) 3(i + 1) with structure as shown c (7) below: A i = A A A Consider the last row of A i The only nonzero element in this row is at the (c + i + 1)(3(i + 1)) th position This means that the annihilator of the constant matrix A i has to necessarily have a zero at the corresponding position(s) This effectively eliminates the last column of A i as it will get multiplied to zero Now, consider the second last row of the original A i After the last column has been eliminated, the only nonzero element in this row is at the (c+i)(3i) th position The annihilator will have a zero at the corresponding position This effectively eliminates column number 3i Continuing in a similar fashion, the modified A i will continue to have only one nonzero element in a row for (c b) steps, thus effectively eliminating (c b) columns at the indices 3(i+1), 3i, 3(i 1) and so on The annihilator in this case has dimensions 3(i+1) The rows of this matrix can be partitioned into (i+1) blocks, each of dimension 3, corresponding to the coefficients of various degree terms in the nullspace polynomial, ie the first block corresponds to the degree zero terms, the second block corresponds to the coefficients of s, and so on The positions corresponding to the columns eliminated will have a zero, while the others will be nonzero (c b) columns have been eliminated from A i Noticing carefully, it is seen that the columns eliminated all correspond to those that will be getting multiplied to the highest degree term c, in the specified degree structure D Thus the degree of the term(s) in K getting multiplied to c will be equal to i (c b), while the other terms will have degree i It has already been shown in Equation (7) that i = c, thus completing the proof B Case : when c is odd We construct the matrix A as in Equation () The construction of A i s is continued till a stage when the A i is a wide matrix At this stage: (9) c + i + 1 < 3(i + 1) (1) In fact, c being odd leads to the following At the i th stage, 3(i + 1) (c + i + 1) = 1 (11) i + c = 1 (1) i = c 1 (13) At this stage, however, only one vector in the kernel has been got To get the other vector, A i+1 is constructed At

this stage, 3(i + 1) (c + i + 1) = 3 (14) i + c = 3 (15) i = c + 1 (16) Theorem 5: The degree structure of the minimal polynomial basis is given by c 1 c 1 c+1 c+1 K = (17) c 1 c+1 (c b) (c b) Proof: The proof is similar to that of Theorem 51 C Case 3: when b c Let c = b + k, where k is a positive integer When c is even, i = c c 1 (Equation (7)), and when c is odd, i = (Equation (13)) In each case, i (c b) < This means that the minimal polynomial basis will have one of its terms as the zero polynomial, and the corresponding entry in the degree structure of the minimal polynomial will be Once the presence of the zero polynomial in the minimal polynomial basis is confirmed, the problem reduces to that of finding the degree structure of the minimal polynomial basis of a 1 matrix The examples in the following section illustrate the various cases outlined in V-A, V-B and V-C D Examples Example 53: Given D = [ 1 ], to find K =, such that DK = 1) A = ) A 1 =, ie A 1 = A A 3) A 1 is a wide matrix Thus, we stop the iteration here We now have (D + D 1 s + D s )K =, where D i corresponds to the coefficients of the degree i terms in the given polynomial matrix 4) We need a constant matrix P such that A 1 P = Observing the last row of A 1, we find that for this row to be annihilated, the corresponding element(s) in K must be zero This effectively eliminates the last column of A 1, as it will get multiplied to zero This is shown below: * * * * * * * * * * * * The resulting 3 5 matrix got after eliminating the last row and last column of A 1 will have a 5 kernel with non-zero elements 5) Thus P =, thus yielding the degrees of the 1 1 elements in the right kernel of D as K = 1 1 Example 54: Consider D = [ 4 ] 1) In this case, we get A =, a 7 9 matrix ) From the structure of the last row of A, last term of P will be zero After removing the last row and last column of A, and observing the resulting 6 8 matrix, it is seen that the 8 th row has only one non-zero element in the 6 th column, thus forcing the 6 th element in P also to be zero The residual 5 7 matrix will have a right annihilator with all non-zero elements This is shown below: * * * * * * * * * * * * * * * * * * * * * * * * * * * 3) Thus P =, yielding K = Example 55: Consider D = [ 1 3 ]

1) A 1 =, a 5 6 matrix The element in P corresponding to the last column of A 1 will be zero ) We get P 1 = K 1 = 1 1 * * * * * * * * * * * * * * * * * *, hence the first column of K is However, from the given dimensions of D, we need two vectors in the right kernel A 1 however gives us only one vector Thus we proceed and construct A 3) A =, a 6 9 matrix There are three vectors in the kernel of A Two of them are however, exactly P 1, appropriately padded with zeros After elimination of the last row of A as shown below, * * * * * * * * * * * * * * * * * * * * * * * * * * * we get the third vector as P = second column, K = 1, giving the VI SATURATION Given a degree structure D, one can get the degree structure of its minimal polynomial basis, K as described above One can use the same algorithm to determine the degree structure of the minimal left indices of K and the corresponding degree structure, say D 1 While it is seen that the highest value in the row vectors (ie the degree of the polynomial vectors) in D and D 1 remain the same, D and D 1 are not necessarily identical We introduce the notion of saturation in this context Definition 61: Let D be a given degree structure and K be the corresponding degree structure for its minimal polynomial nullspace basis Then, the degree structure D is said to be saturated if any degree structure D 1 such that K is the corresponding degree structure for the minimal nullspace basis satisfies 1 D D 1 Remark 6: If the degree structure of the minimal polynomial basis of the polynomial matrix corresponding to D is given by K, then the degree structure of the minimal polynomial basis of the polynomial matrix corresponding to K T is given by D T (D K D K ) In this case, an unsaturated D would imply that there exists some degree of freedom to change one or more coefficients from to a nonzero value, That is, the degree of certain entries can be raised up to a certain limit while maintaining the same degree structure of the minimal polynomial basis The case when the given D is a 1 3 matrix is discussed in some detail below The following theorem [KPB1b] is a useful result: (n 1) n Theorem 63: Given D Z +, the degree structure of a polynomial matrix R R (n 1) n [s] The degree structure of its minimal polynomial basis is: K = [ deg(det(r 1 )) deg(det(r )) deg(det(r n )) ] T (18) where R i is the minor obtained from R by removing the i th column Proof: The reader is referred to [KPB1b] for the proof in greater detail Only Case 1 discussed there is relevant to this paper as genericity ensures coprimeness of the elements in R and Q As mentioned earlier, saturation can be interpreted as the degree of freedom offered to replace zeros by nonzeros in the degree structure of D while maintaining that of K Proposition 64: When D = [ a b c ] 1 3 Z +, and c b, the saturated D is given by D sat = [ b b c ] Proof: From V-A and V-B, the degree structure of K does not depend on a Further, the degrees in the first two rows of K are the same Now, using 63, it can be seen that the degree structure of D sat is as mentioned An alternate method of proving the same result is by constructing the block Toeplitz matrices for K, and finding the degree structure of its minimal polynomial basis 1 The relation here is a component-wise relation

Remark 65: It is to be noted that the idea of saturation is based on the premise that the degree structures of the minimal polynomial bases of D and D sat are identical The above fact can be used to check the veracity of 64 by replacing as many by in the degree structure, D, at the same time ensuring that no change is effected in K In fact, if the degree structure is a 1 3 matrix, we can conclude it is saturated if the first two columns have the same structure of s and s A Illustrations of saturation Example 66: For D = [ 1 ], D sat = [ 1 1 ] We have not been able to devise a method of determining the degree structure of D sat for D of higher dimensions by 1 4 mere inspection Two cases for D Z + are presented below: Example 67: For D = [ 1 3 ] [ ] [, ] D sat = [ 3 ] However for D = 1 4, Dsat = 1 1 4 Forming the block Toeplitz matrices corresponding to the given D s will confirm the correctness of the D sat matrices as indicated above The presence of the zero polynomial in the minimal polynomial basis could be a possible influence on the degree structure of D sat VII CONCLUDING REMARKS It has been shown that the degree structure of the minimal polynomial basis of a given polynomial matrix depends only on the degree structure of the given polynomial matrix and not on the coefficients While in the generic case, the degree structure got by the use of the block Toeplitz algorithm discussed is exact, the same degree structure serves as an upper bound for non-generic (ie specific) cases with the same degree structure The upper bound on the degree structure is particularly helpful in the sense that it limits the dimension of the space in which the solution (the minimal polynomial basis) is sought The case when the specified degree structure was a 1 3 matrix was studied in detail and a closed form solution was given for the degree structure of its minimal polynomial basis The case of saturation of a degree structure was also examined for this case, and this can be subsequently extended to other cases The key assumption of genericity of parameters ensured that at all stages, the matrices in consideration had full rank This meant that the results did not have to be modified for cases when the matrices were not full rank or lost rank at some point(s) in the solution space The methods adopted in this work were dependent only on the zero-nonzero nature of the coefficients and did not involve extensive numerical computations at any stage Absence of numerical investigation thus makes our results just generic, instead of the specific case: nevertheless, degree structures obtained using our approach is valid for a dense set of polynomial matrices with the given polynomial degree structure ACKNOWLEDGEMENTS We thank Dr C Praagman for making available the Scilab codes (see [Pra11]) to compute the minimal polynomial basis of a given polynomial matrix REFERENCES [AH9] J C Zúñiga Anaya and D Henrion An improved Toeplitz algorithm for polynomial matrix 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