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Extremal sparsity of the companion matrix of a polynomial < Chao Ma, Xingzhi Zhan ∗ Department of Mathematics, East China Normal University, Shanghai 200241, China

ARTICLE INFO

ABSTRACT

Article history: Received 16 July 2012 Accepted 15 August 2012 Available online 20 September 2012

Let C be the companion matrix of a monic polynomial p over a field F . We prove that if A is a matrix whose entries are rational functions of the coefficients of p over F and whose characteristic polynomial is p, then A has at least as many nonzero entries as C . © 2012 Elsevier Inc. All rights reserved.

Submitted by R.A. Brualdi AMS classification: 15A18 12E05 05C50 05C05 Keywords: Polynomial Companion matrix Sparsity Spanning branching Transcendence degree

The companion matrix of a monic polynomial

p(x) = xn + a1 xn−1 + · · · + an−1 x + an over a field is defined to be ⎡ ⎤ 0 0 · · · 0 −an ⎢ ⎥ ⎢ ⎥ ⎢ 1 0 · · · 0 −an−1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 1 · · · 0 − a C (p) = ⎢ n−2 ⎥ ⎥. ⎢. . . .. ⎥ ⎢ . . . .. ⎥ . . ⎢. . . ⎥ ⎣ ⎦ 0 0 · · · 1 −a1

< This research was supported by the NSFC Grant 10971070. ∗ Corresponding author. E-mail addresses: [email protected] (C. Ma), [email protected] (X. Zhan). 0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.08.017

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C. Ma, X. Zhan / Linear Algebra and its Applications 438 (2013) 621–625

It is well known [5, p. 147] that the characteristic polynomial of C (p) is p(x). Because of this relation, companion matrices can be used to study properties of polynomials. For example, matrix tools are used in [7] to locate the roots of a complex polynomial via its companion matrix. The companion matrix C (p) is very sparse, i.e., it has many zero entries. If we regard the coefficients a1 , . . . , an of p(x) as distinct indeterminates, then C (p) has 2n − 1 nonzero entries. We will show that the companion matrix is the sparsest in a sense to be described below. Let F be a field and x1 , . . . , xn be distinct indeterminates. We denote by F [x1 , . . . , xn ] the ring of polynomials in x1 , . . . , xn over F , and by F (x1 , . . . , xn ) the field of rational functions in x1 , . . . , xn over F :

f F (x1 , . . . , xn ) = f , g ∈ F [x1 , . . . , xn ], g = 0 . g Denote by Mn (E ) the set of n × n matrices whose entries are elements of a given field E . Our main result is the following theorem. Theorem 1. Let F be a field, a1 , . . . , an be distinct indeterminates, and A characteristic polynomial of A is xn

+ a1 xn−1 + · · · + an−1 x + an ,

∈ Mn (F (a1 , . . . , an )). If the (1)

then A has at least 2n − 1 nonzero entries. To prove Theorem 1 we need several lemmas. Lemma 2. The polynomial in (1) is irreducible over F (a1 , . . . , an ). Proof. Let p(x) = xn + a1 xn−1 + · · · + an−1 x + an . If n = 1, p(x) is obviously irreducible. Now suppose n ≥ 2, and we first show that p(x) has no factors of degree 1. Otherwise p(x) has a root f /g where f , g ∈ F [a1 , . . . , an ] with g = 0. Then f n /g n + a1 f n−1 /g n−1 + · · · + an−1 f /g + an = 0. Hence fn

+ a1 f n−1 g + · · · + an−1 fg n−1 + an g n = 0.

(2)

Note that f = 0. For a nonzero polynomial h ∈ F [a1 , . . . , an ] we use degan h to denote the degree of an in h. Let degan f = d and degan g = e, and let u be the polynomial on the left-hand side of (2). If d ≤ e then degan u

= degan (an g n ) = ne + 1 ≥ 1,

contradicting (2); if d degan u

> e then

= degan (f n ) = nd ≥ 2,

contradicting (2). Thus, p(x) has no factors of degree 1. It follows that the lemma holds also for n = 2 or n = 3. We use induction on n. Next suppose n ≥ 4 and assume that the lemma holds for the degree n − 1. To the contrary, suppose that p(x) is reducible: ⎛ ⎞⎛ ⎞ k n −k

bi xk−i ⎠ ⎝ cj xn−k−j ⎠ , (3) xn + a1 xn−1 + · · · + an−1 x + an = ⎝ i=0

j=0

where bi , cj ∈ F (a1 , . . . , an ) with b0 = c0 = 1. Since we have proved that p(x) has no factors of degree q 1, 2 ≤ k ≤ n − 2. Given h ∈ F [a1 , . . . , an ], we may write h as h = h0 + h1 an + h2 a2n + · · · + hq an m with each hi ∈ F [a1 , . . . , an−1 ]. Thus, if h = 0, h = an w for some nonnegative integer m and

C. Ma, X. Zhan / Linear Algebra and its Applications 438 (2013) 621–625

w ∈ F [a1 , . . . , an ] with w(a1 , . . . , an−1 , 0) written as v = v1 /arn where r is an integer, v1

=

bi

fi s

ani

, cj =

gj tj

an

623

= 0. Hence every nonzero v ∈ F (a1 , . . . , an ) can be ∈ F (a1 , . . . , an ) and v1 (a1 , . . . , an−1 , 0) = 0. Now let

,

where fi , gj ∈ F (a1 , . . . , an ), si and tj are integers such that if bi = 0 then fi (a1 , . . . , an−1 , 0) = 0 and if bi = 0 then fi = 0 and si = −1, and if cj = 0 then gj (a1 , . . . , an−1 , 0) = 0 and if cj = 0 then gj = 0 and tj = −1. Since b0 = c0 = 1, we set f0 = g0 = 1 and s0 = t0 = 0. We will show that there are no positive integers among s0 , . . . , sk , t0 , . . . , tn−k . To the contrary we first suppose that there is at least one positive integer among s0 , . . . , sk . Let i0 be the largest subscript such that si0 = max{s0 , . . . , sk }. Then si0 ≥ 1, bi0 = 0 and fi0 (a1 , . . . , an−1 , 0) = 0. We distinguish two cases. Case 1. There is at least one nonnegative integer among t1 , . . . , tn−k . Let j0 be the largest subscript such that tj0 = max{t0 , . . . , tn−k }. Then tj0 ≥ 0, cj0 = 0 and gj0 (a1 , . . . , an−1 , 0) = 0. Comparing the coefficients of xn−i0 −j0 on both sides of (3) we have ai0 +j0

=

bi cj

i+j=i0 +j0

=

fi gj

i+j=i0 +j0

si +tj

an

.

(4)

+ tj0 ≥ si + tj for all i, j with i + j = i0 + j0 and equality holds if and only if i = i0 and si +tj = j0 . Multiplying both sides of (4) by an 0 0 and then setting an = 0 we obtain

Note that si0 j

0

= fi0 (a1 , . . . , an−1 , 0)gj0 (a1 , . . . , an−1 , 0),

a contradiction. Case 2. t1 , . . . , tn−k are all negative integers. Comparing the coefficients of xn−i0 on both sides of (3) we have ai0

=

i+j=i0

bi cj

=

i+j=i0

fi gj si +tj

an

.

(5)

> si + tj for all i = 0, 1, . . . , k and j = 1, . . . , n − k. Multiplying both sides of (5) by and then setting an = 0 we obtain 0 = fi0 (a1 , . . . , an−1 , 0), a contradiction. Thus we have proved that s0 , . . . , sk are all non-positive integers. In the same way we can prove that t0 , . . . , tn−k are all non-positive integers. Consequently, all the bi (a1 , . . . , an−1 , 0) and cj (a1 , . . . , an−1 , 0) are well defined and they are elements of F (a1 , . . . , an−1 ). Denote b˜ i = bi (a1 , . . . , an−1 , 0) and c˜j = cj (a1 , . . . , an−1 , 0). Setting an = 0 in (3) we have Note that si0

si an 0

⎛

xn

+ a1 xn−1 + · · · + an−1 x = ⎝

k

i=0

⎞⎛

b˜ i xk−i ⎠ ⎝

n −k j=0

⎞

c˜j xn−k−j ⎠ .

(6)

Considering the constant term we get b˜ k c˜n−k = 0. Hence b˜ k = 0 or c˜n−k = 0. In either case (6) shows that xn−1 + a1 xn−2 + · · · + an−2 x + an−1 is reducible over F (a1 , . . . , an−1 ), which contradicts the induction hypothesis. This proves that p(x) is irreducible.

We will use a little graph theory [1,9]. A branching is an oriented tree having a root of in-degree 0 and all other vertices of in-degree 1. A spanning branching of a digraph is a branching that includes all vertices of the digraph. If (a, b) is an arc of a digraph, then a is called an in-neighbor of b and b is called an out-neighbor of a. The following lemma is well known ([1, p. 108] or [9, p. 90]) and easy to prove. Lemma 3. In a strongly connected digraph, every vertex is the root of a spanning branching.

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C. Ma, X. Zhan / Linear Algebra and its Applications 438 (2013) 621–625

We denote by D(A) the digraph of a matrix A = (aij ) of order n. The vertices of D(A) are 1, 2, . . . , n and (s, t ) is an arc if and only if ast = 0. A(i, j) will mean the entry of the matrix A in row i and column j. Lemma 4. Let E be a field. If the digraph of a matrix A ∈ Mn (E ) has a spanning branching whose arcs are (i1 , j1 ), . . . , (in−1 , jn−1 ), then there exists a nonsingular diagonal matrix G ∈ Mn (E ) such that GAG−1 (ik , jk ) = 1 for k = 1, . . . , n − 1. Proof. Let B be the spanning branching of D(A). We renumber the vertices of B as follows. The root of B is 1. If 1 has t out-neighbors, number them as 2, 3, . . . , t + 1 in any order. Then number the outneighbors of 2, 3, . . . , t +1 successively. Continuing in this way we will finally renumber all the vertices of D(A), since B is a spanning branching. Thus we obtain a new digraph D with a spanning branching B . The arcs of B are (r1 , 2), (r2 , 3), . . . , (rn−1 , n) which satisfy the key condition 1 ≤ rk ≤ k for each k = 1, 2, . . . , n − 1. In particular, r1 = 1. Moreover, there is a permutation matrix P such that D = D(PAP T ) where P T denotes the transpose of P . Let PAP T = (aij ). Then ark ,k+1 = 0 for k = 1, . . . , n − 1. We define n nonzero elements d1 , . . . , dn successively by setting d1 = 1, and dk+1 = drk ark ,k+1 for k = 1, . . . , n − 1. This is well defined since 1 ≤ rk ≤ k. Let H = diag(d1 , d2 , . . . , dn ). We have HPAP T H −1 (rk , k + 1) = 1 for k = 1, . . . , n − 1. Let G = P T HP . Then G is a nonsingular diagonal matrix and GAG−1 (ik , jk ) = 1 for k = 1, . . . , n − 1. Results similar to Lemma 4 are known ([3, p. 259], [8, p. 3]). We will use a little algebra [6]. Let F ⊆ E be a field extension. We denote by trd(E /F ) the transcendence degree of E over F . Let a1 , . . . , an be distinct indeterminates. Since {a1 , . . . , an } is a transcendence basis of F (a1 , . . . , an ) over F [6, p. 317], we have trd(F (a1 , . . . , an )/F ) = n. Given e1 , . . . , ek ∈ E , we denote by F (e1 , . . . , ek ) the subfield of E defined by

f (e1 , . . . , ek ) F (e1 , . . . , ek ) = f , g ∈ F [x1 , . . . , xk ], g (e1 , . . . , ek ) = 0 . g (e1 , . . . , ek ) It is easy to show that trd(F (e1 , . . . , ek )/F )

≤ k.

Proof of Theorem 1. Recall that a square matrix R is said to be reducible if there exists a permutation matrix P such that PRP T is of the form ⎡ ⎤ R1 0 ⎣ ⎦, R2 R3 where R1 and R3 are square matrices of order at least 1, i.e., they do appear, and 0 is a zero matrix. A square matrix that is not reducible is called irreducible. Obviously, the characteristic polynomial of a reducible matrix is reducible. Now by Lemma 2, the matrix A is irreducible. Since the digraph of an irreducible matrix is strongly connected [4, p. 55], D(A) is strongly connected. By Lemma 3, D(A) has a spanning branching. Then by Lemma 4, there exists a nonsingular diagonal matrix G ∈ Mn (F (a1 , . . . , an )) such that A = GAG−1 has at least n − 1 entries equal to 1. Suppose that A has precisely m nonzero entries. Note that A and A have the same characteristic polynomial in (1) and have the same number m of nonzero entries. Let the nonzero entries of A be e1 , e2 , . . . , em−n+1 , 1, . . . , 1. Of course, every ej ∈ F (a1 , . . . , an ) and hence F (e1 , . . . , em−n+1 ) ⊆ F (a1 , . . . , an ). On the other hand, since each of the coefficients a1 , . . . , an of the polynomial in (1) is the value of a polynomial over F in the entries of A , we have F (a1 , . . . , an ) ⊆ F (e1 , . . . , em−n+1 ). Hence F (e1 , . . . , em−n+1 ) = F (a1 , . . . , an ). Finally from n

= trd(F (a1 , . . . , an )/F ) = trd(F (e1 , . . . , em−n+1 )/F ) ≤ m − n + 1 ≥ 2n − 1.

we obtain m

In the proof of Theorem 1 we have used a method in [2,3].

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Acknowledgment The authors are grateful to Dr. Lingfei Jin for helpful discussions. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory, GTM 244, Springer, 2008. [2] T. Britz, The inverse of a non-singular free matrix, Linear Algebra Appl. 338 (2001) 245–249. [3] T. Britz, J.J. McDonald, D.D. Olesky, P. van den Driessche, Minimal spectrally arbitrary sign patterns, SIAM J. Matrix Anal. Appl. 26 (1) (2004) 257–271. [4] R.A. Brualdi, H.J. Ryser, Combinatorial Matrix Theory, Cambridge University Press, 1991. [5] R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, 1985. [6] T.W. Hungerford, Algebra, GTM 73, Springer, 1980. [7] F. Kittaneh, Singular values of companion matrices and bounds on zeros of polynomials, SIAM J. Matrix Anal. Appl. 16 (1) (1995) 333–340. [8] B.L. Shader, Notes on the 2n-conjecture, in: AIM Workshop, 2006. Available from: