- Email: [email protected]
Expressing Polynomials as the Permanent of low rank Square Matrices
Electronic Notes in Discrete Mathematics 36 (2010) 73–80 www.elsevier.com/locate/endm
Expressing Polynomials as the Permanent of low rank Square Matrices Mumtaz Ahmad 1 LORIA INRIA Nancy Grand Est Nancy, France
Abstract It is known that the problem of computing the permanent of a given matrix is #P hard. However, Alexander I. Barvinok has proven that if we fix the rank of the matrix then its permanent can be computed in strongly polynomial time. Barvinok’s algorithm [1] computes the permanent of square matrices of fixed rank by constructing polynomials. We study the problem of expressing polynomials as the permanent of low rank square matrices and vice versa. We prove that the permanent of a square matrix with rank 1 is a monomial and the permanent of a square matrix (with integer entries) that has not full rank, is a polynomial with even coefficients. We also prove that, for a polynomial f ∈ k[x], there exist a square matrix of rank 2, whose permanent is the polynomial f . Our results contribute in computing the permanent of a square matrix efficiently. Keywords: Permanent, Polynomials, Algorithm complexity, Combinatorial optimization.
1
Email: [email protected]
1571-0653/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2010.05.010
74
1
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
Introduction
Permanents made their first appearance in 1812, in the famous memoirs of Binet [3] and Cauchy [4]. Since then, permanents have been studied by researchers because of its applications. For example, in graph theory, the permanent of the adjacency matrix of a bipartite graph is exactly the number of perfect matching in the graph. The permanent have very important applications in physics and chemistry where statistical methods are used to study phenomena that is the outcome of the combined action of a very large number of items. These problems are enumeration problems involving lattices and sometimes can be solved using permanents of corresponding incidence matrices [7,8]. The methods for computing permanents have become more attractive to researchers [5,6]. The permanent of a matrix is defined in a way similar to the determinant but without the alternating signs. The permanent have not so many properties like the determinant of matrices. Definition 1.1 Let A = (aij ) be a real n × n square matrix. The expression per(A) =
n
ai,g(i)
g∈Sn i=1
is called the permanent of A. (see [7]). For example, if ⎛ ⎞ ab A=⎝ ⎠ cd then per(A) = ad + bc. Leslie Valiant [2] proved that the problem of computing the permanent of a matrix is #P hard and remains #P complete even if the matrix is restricted to have entries that are all 0, 1. Since then, a lot of efforts are being made by researchers and complexity theorists to study special and easy cases. Alexander I. Barvinok, has proven that if we fix the rank of the matrix then its permanent can be computed in strongly polynomial time. Barvinok’s algorithm computes the permanent of a square matrix (with fixed rank) by constructing polynomials. The technique used, can be interpreted as a Gaussian elimination method for computing the permanent. We study the problem of expressing polynomials as the permanent of low rank square matrices and vice versa. We prove that the permanent of a square matrix with rank 1 is a monomial. We prove that the permanent of square matrices (with integer
75
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
entries) that does not have full rank is a polynomial with even coefficients. We also prove that, for a polynomial f ∈ k[x], where k[x] is a ring of polynomials, there exist a square matrix of rank 2, whose permanent is the polynomial f , the entries of such matrix belong to the set k ∪ [x]. Next, in section 2, we present the permanent of square matrices of rank 1 as monomials, while, in section 3, we discuss the case for square matrices of rank > 1.
2
Polynomials as the permanent of square matrices with rank 1
Theorem 2.1 The permanent of a square matrix of size n with rank 1 is a monomial. Proof. Consider a square matrix A of size n with rank 1, we can write matrix A as given below. ⎛ ⎞ a ⎜ 1 ⎜ ⎜ a2 ⎜ ⎜ A = ⎜ a3 ⎜ ⎜ .. ⎜. ⎝ an
λ2 a1 λ3 a1 · · · λn a1 λ2 a2 λ3 a2 · · · λ2 a3 λ3 a3 · · · .. .. . ··· .
λ2 an λ3 an · · ·
⎟ ⎟ λ n a2 ⎟ ⎟ ⎟ λ n a3 ⎟ ⎟ .. ⎟ . ⎟ ⎠ λn an
The rank of the matrix A is equal 1, since each column of matrix A is a multiple of the first column. To find the permanent of matrix A, we proceed as follows: ⎞ ⎛ a a · · · a1 ⎟ ⎜ 1 1 ⎟ ⎜ ⎜ a2 a2 · · · a2 ⎟ ⎟ per(A) = λ2 · λ3 · λ4 · · · λn · per ⎜ ⎜ .. .. .⎟ ⎜ . . · · · .. ⎟ ⎠ ⎝ an an · · · an ⎛ 1 ⎜ ⎜ ⎜1 per(A) = (λ2 · · · λn ) · (a1 · a2 · · · an )per ⎜ ⎜ .. ⎜. ⎝ 1
⎞
1 ··· 1
⎟ ⎟ 1 · · · 1⎟ ⎟ .. ⎟ .. . · · · .⎟ ⎠ 1 ··· 1
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M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
per(A) = (λ2 · · · λn ) · (a1 · a2 · · · an )(n!) per(A) = n!(a1 )(λ2 · a2 )(λ3 · a3 ) · · · (λn · an ) Hence, the permanent of a square matrix A with rank 1 is a monomial.
3
2
Polynomials as the permanent of square matrices with rank > 1
Theorem 3.1 If a square matrix A does not have full rank and its constant entries are only integers, then per(A) is a polynomial with even coefficients. Proof. Suppose that A is a square matrix of rank k and size greater than k. If the size of a matrix A is greater than its rank, then the matrix A will not have full rank and if the matrix has not full rank, then its determinant will be zero. ⇒ det(A) = 0 Now, observe the definitions of the permanent and determinant per(A) =
n
ai,π(i) and det(A) =
π∈Sn i=1
sgn(π)
π∈Sn
n
ai,π(i)
i=1
In determinant, we use sign ±, i.e, sgn ∈ {−1, +1} but not for the permanent. Working with modulo 2, we have, 1 + 1 ≡ 0 ⇒ 1 ≡ −1 Hence Therefore,
per(A) ≡ det(A)(mod 2) per(A) ≡ 0(mod 2)
⇒ an xn + an−1 xn−1 + · · · + a2 x2 + a1 x1 + a0 ≡ 0(mod 2) It is only possible if the coefficients of the polynomials will be even.
2
Theorem 3.2 Let k be an algebraically closed field and k[x] be the ring of polynomials over k. Then for any f ∈ k[x], there exists a matrix M of rank 2, the entries of M being in the set k ∪ {x}, such that f = per(M ).
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
77
Proof. Consider the following four square matrices, each of size n. ⎞ ⎛ ⎛ ⎞ a a . . . an x x ... x ⎟ ⎜ 1 2 ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ a1 a2 . . . a n ⎟ ⎜x x . . . x⎟ ⎟ ⎜ ⎜ ⎟ X = ⎜. . . .⎟ A = ⎜ . . . .. ⎟ . . . . . . . . .⎟ ⎜. . ⎜. . . .⎟ ⎠ ⎝ ⎝ ⎠ x x ... x a 1 a2 . . . a n ⎞ ⎞ ⎛ ⎛ c c ... c 1 1 ... 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎜c c . . . c⎟ ⎜1 1 . . . 1⎟ ⎟ ⎜ ⎜ B = ⎜ . . . . ⎟ C = ⎜. . . .⎟ ⎟ ⎜ .. .. . . .. ⎟ ⎜ .. .. . . .. ⎟ ⎠ ⎠ ⎝ ⎝ c c ... c 1 1 ... 1 such that a1 , . . . , an , and c ∈ k. The square matrices X, A, B and C can be grouped into a single matrix M of size 2n and rank 2 as given below. ⎞ ⎛ . X .. A ⎟ ⎜ ⎜ . ⎟ M = ⎜. . . .. . . .⎟ ⎠ ⎝ .. B . C The permanent of matrix M will be polynomial in variable x and of degree n. Suppose that f (x) = f0 + f1 x + f2 x2 + f3 x3 + . . . + fn xn and
per(M ) = p0 + p1 x + p2 x2 + p3 x3 + . . . + pn xn (1) m Observe that a term x , where 0 ≤ m < n, can be obtained when a permutation takes exactly m entries from matrix X, i,e. from exactly m rows and m columns of X. Since in the permutation we have to take exactly one entry from each row and column, so in such a permutation, there will be entries from n − m
n rows of A and n − m columns of B. n − m rows
n of A can be chosen in m ways, and n − m columns of B can be chosen in m ways; also there has to be chosen n − m rows of B and n − m columns of A. Suppose a permutation takes the following columns of A, i1 , i2 , i3 . . . in−m
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M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
If these columns of A are kept fixing, then out of m rows and m columns of X, m! permutations are possible. Similarly, m! permutations are possible out of the m entries of the matrix C. In matrix A, (n − m)! permutations are possible and same in matrix B, for any choice of rows and columns of B. Hence the coefficient of xm can be determined as under. 3 n ((m!)2 )((n − m)!)2 cm ai1 ai2 · · · ain−m m Now, if the columns of matrix A are varied, the coefficient of xm in per(M ) can be obtained as Hence 3 n pm = (m!)2 {(n − m)!}2 cm ai1 ai2 · · · ain−m (2) m 0
2
n−m
Also,
pn = (n!)2 cn (3) per(M ) will be equal to f ⇐⇒ pi = fi , 0 ≤ i ≤ n. So we have to satisfy p i = fi Now pn = fn =⇒ (n!)2 cn = fn
1 fn n =⇒ c = 2 (n!) And, pm = fm So from (2) ai1 ai2 · · · ain−m = (−1)n−m f´m , 0 ≤ m < n (4) 0
where f´m = (−1)n−m n 3 m
fm (m!)2 {n − m}2 cm
,0 ≤ m < n
Let f´ be the polynomial f´(x) = f´0 + f´1 (x) + f´2 x2 + . . . + f´n−1 xn−1 + xn
(5)
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
79
Since f´ ∈ k[x], and k is an algebraically closed field, so f´ can be split into linear factors over k. Let f´(x) = (x − a1 )(x − a2 ) . . . (x − an ) where ai ∈ k, 1 ≤ i ≤ n from (5) and (6), it follows that f´m = (−1)n−m
ai1 ai2 · · · ain−m , 0 ≤ m < n
(6)
(7)
0
From (7), it is clear that, if rank of the matrix A equals 1 , then eq(4) is satisfied. Hence f = per(M ). 2 Next, we prove by an example, that the polynomial x2 +c can be expressed as the permanent of a square matrix with rank 2 and of any size ≥ 2. Example 3.3 Consider a matrix A of size 2 × 2 and rank 2. ⎛ ⎞ c x A = ⎝ ⎠ , per(A) = c + x2 x1 We extend the size of the matrix A up to n, keeping rank of the matrix equals 2, as follows. ⎞ ⎛ c x 0 ⎟ ⎜ ⎟ ⎜ A3 = ⎜x 1 1/2⎟ ⎠ ⎝ x 1 1/2
, per(A3 ) = c + x2 3×3
Similarly, matrix An of size n and rank 2, whose permanent is a polynomial x2 + c, can be written as ⎞ ⎛ c x 0 0 ··· 0 ⎟ ⎜ ⎟ ⎜ ⎜x 1 1/2 1/3 · · · 1/(n − 1)⎟ ⎟ ⎜ ⎟ ⎜ An = ⎜x 1 1/2 1/3 · · · 1/(n − 1)⎟ , size = n × n ⎟ ⎜ ⎟ ⎜ .. .. .. .. .. .. ⎟ ⎜. . . . . . ⎠ ⎝ x 1 1/2 1/3 · · · 1/(n − 1) per(An ) = c(1 ·
1 1 1 1 1 1 · ··· )((n − 1)!) + x · x(1 · · · · · )((n − 1)!) 2 3 n−1 2 3 n−1
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1 1 )((n − 1)!) + x2 ( )((n − 1)!) = x2 + c (n − 1)! (n − 1)! Hence, x2 + c = per(A) = per(A3 ) = per(A4 ) = per(A5 ), . . . , = per(An ). per(An ) = c(
4
Conclusions and Future work
In this paper, we focus on expressing polynomials as the permanent of low rank square matrices, particularly, we deal with matrices of rank 1 and 2. For the ring of polynomials, we express polynomials as the permanent of square matrices with rank 2 but still a lot of work is to be done in this area of research. For example, how the polynomials can be expressed as the permanent of low rank square matrices with entries (x or 1). With such restriction, we are still unable to express polynomial x3 + 1 as the permanent of a square matrix with rank 2 and size 3 × 3.
References [1] Barvinok, A. I., Two algorithmic results for the traveling salesman problem, Math. Oper. Res. 21 (1996), 65–84. [2] Valiant, L., The complexity of computing the permanent, Theoretical Computer Science. 8 (1979), 189–201. [3] Binet, J. P. M., M´emoire sur un syst`eme de formules analytiques, et leur application ` a des consid´erations g´eometriques, J. Ec. Polyt. 9 (1812), 280–302. [4] Cauchy, A. L., M´emoire sur les fonctions qui ne peuvent obtenir que deux valeurs ´egales et de signes contraires par suite des transpositions op´er´ees entre les variables qu’elles renferment, J. Ec. polyt. 10 (1812), 29–112. [5] Brualdi, R. A., and H. J. Ryser, Combinatorial Matrix Theory, Cambridge University Press (1991). [6] Karmarkar, N., R.M. Karp, R. Lipton, L. Lovasz and M. Luby, A Monte Carlo algorithm for estimating the permanent, SIAM Journal on Computing. 22 (1993), 284–293. [7] Minc, H., “Permanents”, Addison-Wesley, Cambridge, MA, (1978). [8] Forbert, H., and D. Mark, Calculation of the permanent of a sparse positive matrix, Computer Physics Communications. 150 (2003), 267–273.
Expressing Polynomials as the Permanent of low rank Square Matrices Mumtaz Ahmad 1 LORIA INRIA Nancy Grand Est Nancy, France
Abstract It is known that the problem of computing the permanent of a given matrix is #P hard. However, Alexander I. Barvinok has proven that if we fix the rank of the matrix then its permanent can be computed in strongly polynomial time. Barvinok’s algorithm [1] computes the permanent of square matrices of fixed rank by constructing polynomials. We study the problem of expressing polynomials as the permanent of low rank square matrices and vice versa. We prove that the permanent of a square matrix with rank 1 is a monomial and the permanent of a square matrix (with integer entries) that has not full rank, is a polynomial with even coefficients. We also prove that, for a polynomial f ∈ k[x], there exist a square matrix of rank 2, whose permanent is the polynomial f . Our results contribute in computing the permanent of a square matrix efficiently. Keywords: Permanent, Polynomials, Algorithm complexity, Combinatorial optimization.
1
Email: [email protected]
1571-0653/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2010.05.010
74
1
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
Introduction
Permanents made their first appearance in 1812, in the famous memoirs of Binet [3] and Cauchy [4]. Since then, permanents have been studied by researchers because of its applications. For example, in graph theory, the permanent of the adjacency matrix of a bipartite graph is exactly the number of perfect matching in the graph. The permanent have very important applications in physics and chemistry where statistical methods are used to study phenomena that is the outcome of the combined action of a very large number of items. These problems are enumeration problems involving lattices and sometimes can be solved using permanents of corresponding incidence matrices [7,8]. The methods for computing permanents have become more attractive to researchers [5,6]. The permanent of a matrix is defined in a way similar to the determinant but without the alternating signs. The permanent have not so many properties like the determinant of matrices. Definition 1.1 Let A = (aij ) be a real n × n square matrix. The expression per(A) =
n
ai,g(i)
g∈Sn i=1
is called the permanent of A. (see [7]). For example, if ⎛ ⎞ ab A=⎝ ⎠ cd then per(A) = ad + bc. Leslie Valiant [2] proved that the problem of computing the permanent of a matrix is #P hard and remains #P complete even if the matrix is restricted to have entries that are all 0, 1. Since then, a lot of efforts are being made by researchers and complexity theorists to study special and easy cases. Alexander I. Barvinok, has proven that if we fix the rank of the matrix then its permanent can be computed in strongly polynomial time. Barvinok’s algorithm computes the permanent of a square matrix (with fixed rank) by constructing polynomials. The technique used, can be interpreted as a Gaussian elimination method for computing the permanent. We study the problem of expressing polynomials as the permanent of low rank square matrices and vice versa. We prove that the permanent of a square matrix with rank 1 is a monomial. We prove that the permanent of square matrices (with integer
75
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
entries) that does not have full rank is a polynomial with even coefficients. We also prove that, for a polynomial f ∈ k[x], where k[x] is a ring of polynomials, there exist a square matrix of rank 2, whose permanent is the polynomial f , the entries of such matrix belong to the set k ∪ [x]. Next, in section 2, we present the permanent of square matrices of rank 1 as monomials, while, in section 3, we discuss the case for square matrices of rank > 1.
2
Polynomials as the permanent of square matrices with rank 1
Theorem 2.1 The permanent of a square matrix of size n with rank 1 is a monomial. Proof. Consider a square matrix A of size n with rank 1, we can write matrix A as given below. ⎛ ⎞ a ⎜ 1 ⎜ ⎜ a2 ⎜ ⎜ A = ⎜ a3 ⎜ ⎜ .. ⎜. ⎝ an
λ2 a1 λ3 a1 · · · λn a1 λ2 a2 λ3 a2 · · · λ2 a3 λ3 a3 · · · .. .. . ··· .
λ2 an λ3 an · · ·
⎟ ⎟ λ n a2 ⎟ ⎟ ⎟ λ n a3 ⎟ ⎟ .. ⎟ . ⎟ ⎠ λn an
The rank of the matrix A is equal 1, since each column of matrix A is a multiple of the first column. To find the permanent of matrix A, we proceed as follows: ⎞ ⎛ a a · · · a1 ⎟ ⎜ 1 1 ⎟ ⎜ ⎜ a2 a2 · · · a2 ⎟ ⎟ per(A) = λ2 · λ3 · λ4 · · · λn · per ⎜ ⎜ .. .. .⎟ ⎜ . . · · · .. ⎟ ⎠ ⎝ an an · · · an ⎛ 1 ⎜ ⎜ ⎜1 per(A) = (λ2 · · · λn ) · (a1 · a2 · · · an )per ⎜ ⎜ .. ⎜. ⎝ 1
⎞
1 ··· 1
⎟ ⎟ 1 · · · 1⎟ ⎟ .. ⎟ .. . · · · .⎟ ⎠ 1 ··· 1
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M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
per(A) = (λ2 · · · λn ) · (a1 · a2 · · · an )(n!) per(A) = n!(a1 )(λ2 · a2 )(λ3 · a3 ) · · · (λn · an ) Hence, the permanent of a square matrix A with rank 1 is a monomial.
3
2
Polynomials as the permanent of square matrices with rank > 1
Theorem 3.1 If a square matrix A does not have full rank and its constant entries are only integers, then per(A) is a polynomial with even coefficients. Proof. Suppose that A is a square matrix of rank k and size greater than k. If the size of a matrix A is greater than its rank, then the matrix A will not have full rank and if the matrix has not full rank, then its determinant will be zero. ⇒ det(A) = 0 Now, observe the definitions of the permanent and determinant per(A) =
n
ai,π(i) and det(A) =
π∈Sn i=1
sgn(π)
π∈Sn
n
ai,π(i)
i=1
In determinant, we use sign ±, i.e, sgn ∈ {−1, +1} but not for the permanent. Working with modulo 2, we have, 1 + 1 ≡ 0 ⇒ 1 ≡ −1 Hence Therefore,
per(A) ≡ det(A)(mod 2) per(A) ≡ 0(mod 2)
⇒ an xn + an−1 xn−1 + · · · + a2 x2 + a1 x1 + a0 ≡ 0(mod 2) It is only possible if the coefficients of the polynomials will be even.
2
Theorem 3.2 Let k be an algebraically closed field and k[x] be the ring of polynomials over k. Then for any f ∈ k[x], there exists a matrix M of rank 2, the entries of M being in the set k ∪ {x}, such that f = per(M ).
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
77
Proof. Consider the following four square matrices, each of size n. ⎞ ⎛ ⎛ ⎞ a a . . . an x x ... x ⎟ ⎜ 1 2 ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎜ a1 a2 . . . a n ⎟ ⎜x x . . . x⎟ ⎟ ⎜ ⎜ ⎟ X = ⎜. . . .⎟ A = ⎜ . . . .. ⎟ . . . . . . . . .⎟ ⎜. . ⎜. . . .⎟ ⎠ ⎝ ⎝ ⎠ x x ... x a 1 a2 . . . a n ⎞ ⎞ ⎛ ⎛ c c ... c 1 1 ... 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎜ ⎜c c . . . c⎟ ⎜1 1 . . . 1⎟ ⎟ ⎜ ⎜ B = ⎜ . . . . ⎟ C = ⎜. . . .⎟ ⎟ ⎜ .. .. . . .. ⎟ ⎜ .. .. . . .. ⎟ ⎠ ⎠ ⎝ ⎝ c c ... c 1 1 ... 1 such that a1 , . . . , an , and c ∈ k. The square matrices X, A, B and C can be grouped into a single matrix M of size 2n and rank 2 as given below. ⎞ ⎛ . X .. A ⎟ ⎜ ⎜ . ⎟ M = ⎜. . . .. . . .⎟ ⎠ ⎝ .. B . C The permanent of matrix M will be polynomial in variable x and of degree n. Suppose that f (x) = f0 + f1 x + f2 x2 + f3 x3 + . . . + fn xn and
per(M ) = p0 + p1 x + p2 x2 + p3 x3 + . . . + pn xn (1) m Observe that a term x , where 0 ≤ m < n, can be obtained when a permutation takes exactly m entries from matrix X, i,e. from exactly m rows and m columns of X. Since in the permutation we have to take exactly one entry from each row and column, so in such a permutation, there will be entries from n − m
n rows of A and n − m columns of B. n − m rows
n of A can be chosen in m ways, and n − m columns of B can be chosen in m ways; also there has to be chosen n − m rows of B and n − m columns of A. Suppose a permutation takes the following columns of A, i1 , i2 , i3 . . . in−m
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M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
If these columns of A are kept fixing, then out of m rows and m columns of X, m! permutations are possible. Similarly, m! permutations are possible out of the m entries of the matrix C. In matrix A, (n − m)! permutations are possible and same in matrix B, for any choice of rows and columns of B. Hence the coefficient of xm can be determined as under. 3 n ((m!)2 )((n − m)!)2 cm ai1 ai2 · · · ain−m m Now, if the columns of matrix A are varied, the coefficient of xm in per(M ) can be obtained as Hence 3 n pm = (m!)2 {(n − m)!}2 cm ai1 ai2 · · · ain−m (2) m 0
2
n−m
Also,
pn = (n!)2 cn (3) per(M ) will be equal to f ⇐⇒ pi = fi , 0 ≤ i ≤ n. So we have to satisfy p i = fi Now pn = fn =⇒ (n!)2 cn = fn
1 fn n =⇒ c = 2 (n!) And, pm = fm So from (2) ai1 ai2 · · · ain−m = (−1)n−m f´m , 0 ≤ m < n (4) 0
where f´m = (−1)n−m n 3 m
fm (m!)2 {n − m}2 cm
,0 ≤ m < n
Let f´ be the polynomial f´(x) = f´0 + f´1 (x) + f´2 x2 + . . . + f´n−1 xn−1 + xn
(5)
M. Ahmad / Electronic Notes in Discrete Mathematics 36 (2010) 73–80
79
Since f´ ∈ k[x], and k is an algebraically closed field, so f´ can be split into linear factors over k. Let f´(x) = (x − a1 )(x − a2 ) . . . (x − an ) where ai ∈ k, 1 ≤ i ≤ n from (5) and (6), it follows that f´m = (−1)n−m
ai1 ai2 · · · ain−m , 0 ≤ m < n
(6)
(7)
0
From (7), it is clear that, if rank of the matrix A equals 1 , then eq(4) is satisfied. Hence f = per(M ). 2 Next, we prove by an example, that the polynomial x2 +c can be expressed as the permanent of a square matrix with rank 2 and of any size ≥ 2. Example 3.3 Consider a matrix A of size 2 × 2 and rank 2. ⎛ ⎞ c x A = ⎝ ⎠ , per(A) = c + x2 x1 We extend the size of the matrix A up to n, keeping rank of the matrix equals 2, as follows. ⎞ ⎛ c x 0 ⎟ ⎜ ⎟ ⎜ A3 = ⎜x 1 1/2⎟ ⎠ ⎝ x 1 1/2
, per(A3 ) = c + x2 3×3
Similarly, matrix An of size n and rank 2, whose permanent is a polynomial x2 + c, can be written as ⎞ ⎛ c x 0 0 ··· 0 ⎟ ⎜ ⎟ ⎜ ⎜x 1 1/2 1/3 · · · 1/(n − 1)⎟ ⎟ ⎜ ⎟ ⎜ An = ⎜x 1 1/2 1/3 · · · 1/(n − 1)⎟ , size = n × n ⎟ ⎜ ⎟ ⎜ .. .. .. .. .. .. ⎟ ⎜. . . . . . ⎠ ⎝ x 1 1/2 1/3 · · · 1/(n − 1) per(An ) = c(1 ·
1 1 1 1 1 1 · ··· )((n − 1)!) + x · x(1 · · · · · )((n − 1)!) 2 3 n−1 2 3 n−1
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1 1 )((n − 1)!) + x2 ( )((n − 1)!) = x2 + c (n − 1)! (n − 1)! Hence, x2 + c = per(A) = per(A3 ) = per(A4 ) = per(A5 ), . . . , = per(An ). per(An ) = c(
4
Conclusions and Future work
In this paper, we focus on expressing polynomials as the permanent of low rank square matrices, particularly, we deal with matrices of rank 1 and 2. For the ring of polynomials, we express polynomials as the permanent of square matrices with rank 2 but still a lot of work is to be done in this area of research. For example, how the polynomials can be expressed as the permanent of low rank square matrices with entries (x or 1). With such restriction, we are still unable to express polynomial x3 + 1 as the permanent of a square matrix with rank 2 and size 3 × 3.
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