Determinants and polynomial root structure

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1 International Journal of Mathematical Education in Science and Technology, Vol 36, No 5, 2005, Determinants and polynomial root structure L G DE PILLIS Department of Mathematics, Harvey Mudd College, Claremont, CA 91711, USA (Received 3 December 2003) A little known property of determinants is developed in a manner accessible to beginning undergraduates in linear algebra Using the language of matrix theory, a classical result by Sylvester that describes when two polynomials have a common root is recaptured Among results concerning the structure of polynomial roots, polynomials with pairs of roots that are either multiplicative inverses or additive inverses are completely characterized 1 Introduction Just when you think you have seen all possible applications to motivate students in the study of determinants (matrix invertibility, volumes in R 3, solutions of systems) there comes to light a classical but little known link between determinants and the relationships of polynomial roots It was Sylvester who developed the mathematical object called a resultant, although in a form which is slightly different from that seen in modern notation The resultant is the determinant of a special matrix, which is constructed very simply from the coefficient vectors of two polynomials, and can be used to determine when these two polynomials have a common root Even though the proof may seem straightforward, the result is surprising In light of this, the concept of the resultant, which is almost never seen by the undergraduate (or even graduate) student, should be afforded greater visibility The purpose of this paper is to render this surprising and amusing result more accessible to the beginning student in linear algebra Additionally, among results concerning the structure of polynomial roots, we completely characterize polynomials with pairs of roots that are either multiplicative inverses or additive inverses The properties and applications of the Sylvester resultant continue to be of current interest to researchers [1 14] In this paper, we pose and solve (cf Theorem 1) the following resultant-related problem Consider the coefficients {a i } defined by the binomial polynomial ð1 þ tþ 2Nþ1 ¼ 1 þ a 1 t þ a 2 t 2 þþa 2N t 2N þ a 2Nþ1 t 2Nþ1 ð1:1þ depillis@hmcedu International Journal of Mathematical Education in Science and Technology ISSN X print/issn online # 2005 Taylor & Francis Group Ltd DOI: /

2 470 L G de Pillis Use the odd coefficients fa 1, a 3,, a 2Nþ1 g and the even coefficients f1, a 2,, a 2N g, respectively, to generate uniquely all columns of the 2N 2N matrix a a 2 a 3 1 a 1 a 2 a 3 a 2N 2 a 2N 1 1 a 1 ð1:2þ a 2N a 2Nþ1 a 2N 2 a 2N 1 a 2 a a 2N a 2Nþ a 2N 2 a 2N a 2N a 2Nþ1 Question Is this matrix invertible? The problem can be framed as a special case of Sylvester s result We will completely answer the question in the broadest setting In fact, we will focus on the most general class of matrices which preserves the zero pattern of (12) but allows all scalars in the defining first two columns to be arbitrary We note that in the context of abstract ring theory, Ratliff and Rush [11, 12] address related questions about the determinants of such classes of structured matrices However, the context is abstract and less accessible to the undergraduate, and the term resultant is not specifically used The general step-down matrix Given the pair of vectors x, y, 2 R Nþ1 where x ¼fx 0 x 1 x 2 x i x N 1 x N g T y ¼fy 0 y 1 y 2 y i y N 1 y N g T ð1:3þ then x, y uniquely determine the N pairs of columns in R 2N which define the 2N 2N step-down matrix 2 x 0 y x 1 y 1 x 0 y 0 x 1 y 1 x N 1 y N 1 x 0 y 0 Mðx, yþ ¼ x N y N x N 1 y N 1 x 1 y x N y N ð1:4þ x N 1 y N x N y N This type of matrix is known as a resultant For the purposes of this paper, however, we will replace the term resultant by the more descriptive term step-down matrix Classic concepts related to the step-down matrix are known,

3 Determinants and polynomial root structure 471 and discussions can be found in various references, for example, [15, p ]; [7, p ]; and [6, p 60 62] This list is by no means exhaustive However, the form of the presentation in most classic works may not be readily accessible to the undergraduate In this paper, we develop concepts related to the step-down matrix in a brief, self-contained and tractable manner that can be appreciated by the student Clearly, if matrix (12) (or (14)) is to be invertible at all, it is necessary that the first two columns be linearly independent in R 2N But linear independence of the defining columns is not sufficient Nonetheless, we shall see (Theorem 1) that for the general step-down matrix, a complete and straight-forward characterization of invertibility exists Additionally, the theory we develop allows us to characterize in terms of determinants, those polynomials with pairs of roots that are either multiplicative inverses ðr,1=rþ or additive inverses ðþr, rþ (cf Theorem 2, Theorem 3) 2 Preliminaries A more convenient sawtooth configuration will result if we re-order the columns of Mðx; yþ of (4) (we thereby alter the determinant by only a 1 factor) The desired matrix is 2 3 x 0 y 0 x 0 y 0 x N 1 x 0 y N 1 y 0 Mðx; yþ ¼ x N x N 1 y N y 6 N 1 7 ð2:5þ 6 4 x N y N x N 1 y N x N y N The first N columns of Mðx; yþ are linearly independent, as are the last N columns Let us denote the defining column vectors ~X and ~Y 2 R 2N by ~X ¼½x 0, x 1,, x N, 0,,0 fflfflfflffl{zfflfflfflffl} Nþ1 N 1 ~Y ¼½y 0, y 1,, y fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} N,0,,0 fflfflfflffl{zfflfflfflffl} Nþ1 N 1 If we define the 2N 2N shift matrix S by S ¼ Š T Š T ð2:6þ ð2:7þ ð2:8þ

4 472 L G de Pillis then sawtooth matrix (25) can be denoted Mðx; yþ ¼½~X, S ~X, S 2 ~X,, S N 1 ~X, ~Y, S ~Y, S fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2 ~Y,, S N 1 ~Y Š fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} N columns N columns ð2:9þ With these conventions, we set down Lemma 1 Step-down matrix Mðx; yþ of (29) has zero determinant if and only if there exist sequences f i g, f j g, i, j ¼ 0, 1,, N 1 such that X N 1 is i ~X X N 1 js j ~Y ¼ 0 2 R 2N ð2:10þ Moreover, the scalar sequences are non-trivial in the sense that some i0 6¼ 0 and some j0 6¼ 0, 0 4 i 0, j 0 4 N 1 Proof To say that detðmðx; yþþ ¼ 0 is to say that some of the columns of (29) are linearly dependent Write such a linear combination using 0 is for the first N columns of M and using 0 js for the last N columns This gives us equation (210) But not all i ¼ 0, or else the last N columns of Mðx; yþ in (29) are linearly dependent a property that cannot hold for a triangular sawtooth set of vectors This contradiction establishes that some i0 6¼ 0 Similarly, not all j ¼ 0 lest we produce the contradiction that the first N columns of Mðx; yþ in (29) are linearly dependent This ends the proof Notation We use the symbol ~e i to denote the canonical vector with zeros everywhere except for a 1 in the ith place That is, Then clearly, ~e i ¼½0,,0, {z} 1,0,,0Š T 2 R 2N ð2:11þ ith place ~e kþ1 ¼ S k ~e 1, k ¼ 0, 1,, ð2n 1Þ ð2:12þ With this notation in hand, we come to Lemma 2 The vectors ~X and ~Y of (26) and (27) can be written as ~X ¼ X N x k¼0 ks k ~e 1, ~Y ¼ X N y l¼0 ls l ~e 1 : ð2:13þ fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Q x ðsþ Q y ðsþ Proof ~X ¼½x 0, x 1,, x N,0,,0Š T from (26) ¼ XN k¼0 x k ~e kþ1 from (211) ð2:14þ ¼ XN k¼0 ðx k S k Þ~e 1 from (212)

5 Determinants and polynomial root structure 473 Similarly, ~Y ¼½y 0, y 1,, y N,0,,0Š T from (27) ¼ XN l¼0 y l ~e lþ1 from (211) ð2:15þ ¼ XN l¼0 ðy l S l Þ~e 1 from (212) which establishes (213) The proof is done Combining Lemmas 1 and 2, we arrive at the following Lemma 3 Matrix Mðx; yþ of (29) has zero determinant if and only if there exist non-trivial scalar sequences f i g, f j g, i, j ¼ 0, 1,, N 1 such that X N 1 X is i N k¼0 x ks k X N 1 X js j N y l¼0 ls l ¼ 0 ð2:16þ fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Q x ðsþ Q y ðsþ Proof We first replace ~X and ~Y in (210) by (213) to obtain the following restatement of Lemma 1 Matrix Mðx; yþ of (29) has zero determinant if and only if there exist non trivial scalar sequences f i g and f j g, such that X N 1 X is i N k¼0 x ks k ~e 1 X N 1 X js j N y l¼0 ls l ~e 1 ¼ 0 2 R 2N fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Q x ðsþ Q y ðsþ Now multiply out the matrix polynomials Q x ðsþ and Q y ðsþ to obtain 2N 1 X 2N 1 X a i S i {z} ~e 1 ~e iþ1 2N 1 X 2N 1 a i ~e iþ1 X b i S i ~e 1 {z} ~e iþ1 ¼ 0 2 R 2N which implies b i ~e iþ1 ¼ 0 2 R 2N since S i ~e 1 ¼ ~e iþ1 ð2:17þ Here, a i and b i are the convolutions of the i with the x k and the j with the y l respectively The vectors f~e 1, ~e 2,, ~e 2N g above form a basis for R 2N This guarantees that, in (217), a i ¼ b i for all i ¼ 0, 1,, ð2n 1Þ ð2:18þ Since coefficients fa i g, and fb i g are a result of multiplying out the matrix products Q x ðsþ and Q y ðsþ, we conclude from (218) that Q x ðsþ ¼ Q y ðsþ This establishes equality (216) and the lemma is proved Now for some notation and final observations Notation The symbols C½SŠ and C½zŠ denote the polynomial rings C½SŠ ¼span½I 2N, S, S 2,, S 2N 1 Š ð2:19þ

6 474 L G de Pillis and C½zŠ ¼span½1, z, z 2,, z 2N 1 Š ð2:20þ where C is the field of complex numbers, S is the 2N 2N shift matrix (28), and z is an indeterminate As usual, deg(p), the degree of a polynomial P 2 C½SŠ (resp deg(p), the degree of p 2 C½zŠ) is that integer denoting the highest non-zero power used in the unique expansion of P (resp of p) The following lemma formally establishes the isomorphism between C½SŠ and C½zŠ Lemma 4 Vector spaces, C½SŠ and C½zŠ, of dimension 2N, have respective bases fs i g and fz i g, i ¼ 0, 1, 2,, ð2n 1Þ One vector space isomorphism :C½SŠ7! C½zŠ, is defined by linear extension of the one one map ðs i Þ¼z i, i ¼ 0, 1, 2, ð2n 1Þ ð2:21þ Under this isomorphism, the multiplicative ring structure of C½SŠ is preserved in the sense that ðp 1 P 2 Þ¼ðP 1 ÞðP 2 Þ2C½zŠ ð2:22þ for all ðp 1 Þ, ðp 2 Þ, ðp 1 P 2 Þ2C½SŠ, that is, whenever degðp 1 P 2 Þ¼degðP 1 Þþ degðp 2 Þ 4 ð2n 1Þ: Proof From the definition (221), we see that (222) holds for the special case P 1 ¼ S i 1 and P 2 ¼ S i 2, where 0 4 i 1, i 2, 4 ð2n 1Þ and i 1 þ i 2 4 ð2n 1Þ Linear extension establishes (222) for general P 1 and P 2 where degðp 1 P 2 Þ¼ degðp 1 ÞþdegðP 2 Þ 4 ð2n 1Þ: 3 Main result Theorem 1 Given vectors x, y 2 R Nþ1 of (13), generate the 2N 2N step-down matrix Mðx; yþ of (14) Then detðmðx; yþþ ¼ 0 ð3:23þ if and only if the polynomials p x ðtþ ¼x 0 þ x 1 t þ x 2 t 2 þþx N t N and ð3:24þ p y ðtþ ¼y 0 þ y 1 t þ y 2 t 2 þþy N t N have at least one root in common Proof We have from (216) of Lemma 3, that detðmðx; yþþ ¼ 0 if and only if for non-trivial scalar sequences f i g and f j g, Q x ðsþ ¼Q y ðsþ That is, X N 1 is i X N fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} x k¼0 ks k fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} Q x ðsþ iff X N 1 iz i X N fflfflfflfflfflffl{zfflfflfflfflfflffl} x k¼0 kz k fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} p ðzþ q x ðzþ ¼ X N 1 js j fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} ¼ X N 1 jz j fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} p ðzþ X N l¼0 y ls l fflfflfflfflfflffl{zfflfflfflfflfflffl} Q y ðsþ X N l¼0 y lz l fflfflfflfflffl{zfflfflfflfflffl} q y ðzþ from Lemma 3 from ð2:22þ

7 Determinants and polynomial root structure 475 Factoring of polynomials p ðzþ, q x ðzþ, p ðzþ, q y ðzþ above, gives us the equality A 1 ðz r 1 Þðz r N 1 Þ A fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2 ðz s 1 Þðz s N Þ fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} p ðzþ q x ðzþ¼x 0 þx 1 zþþx N z N ¼ A 3 ðz ~r 1 Þðz ~r N 1 Þ A fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 4 ðz ~s 1 Þðz ~s N Þ fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} p ðzþ q y ðzþ¼y 0 þy 1 zþþy N z N ð3:25þ for scalars A 1, A 2, A 3, A 4 Factored polynomials (325) are equal if and only if A 1 A 2 ¼ A 3 A 4 and (most importantly) the roots agree That is, fr 1, r 2, ; r N 1, s 1, s 2,, s fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} N g¼f~r 1, ~r 2, ; ~r N 1, ~s 1, ~s 2,, ~s N g fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} q x ðzþ roots q y ðzþ roots ð3:26þ A simple counting argument establishes that the roots {s i } and f~s j g have an element in common In fact, whenever we match the N-element set f~s 1, ~s 2,, ~s N g from the RHS of (326), with any N elements from the LHS, at least one ~s j0 must be chosen This says that that polynomials (324) have a common root (some s i0 ¼ ~s j0 ) if and only if detðmðx; yþþ ¼ 0 The theorem is proved 4 Applications 41 Lacunary polynomials We characterize polynomials having certain roots which lie on a circle centered at the origin in the complex plane Definition Choose integers m and k so that 1 4 m < k ð4:27þ Polynomial P(x) is a lacunary polynomial when it can be written in the following form: PðxÞ ¼ XN a jk x jk þ a jkþm x jkþm ð4:28þ Note, for example, that if one chooses m ¼ 1 and k ¼ 2, then P(x) is a polynomial of degree at most 2N þ 1, containing successively increasing powers of x Generate from the coefficients of P(x) two new polynomials, by splitting the coefficient sequence into two distinct sequences and constructing standard polynomials with successively increasing powers, namely: P 1 ðxþ ¼ XN a jk x j P 2 ðxþ ¼ XN a jkþm x j ð4:29þ ð4:30þ

8 476 L G de Pillis Define the coefficient vectors of P 1 (x) and P 2 (x) by a 1 ¼fa 0 a k a 2k a Nk g T, ð4:31þ a 2 ¼fa m a kþm a 2kþm a Nkþm g T : ð4:32þ We now have Theorem 2 Given coefficient vectors a 1 of (431) and a 2 of (432), generate step-- down matrix Mða 1, a 2 Þ according to (14) Then detðmða 1, a 2 ÞÞ ¼ 0 ð4:33þ if and only if P(x) in (428) has the k roots rk 1 i2j e k, j ¼ 0, 1,, k 1 ð4:34þ where p i ¼ ffiffiffiffiffiffi <2 a 2 R > 0 The roots in (434) lie on a circle of radius rk 1 centred at the origin Proof By Theorem 1, detðmða 1, a 2 ÞÞ ¼ 0 if and only if P 1 (x) and P 2 (x) share a common root P 1 (x) and P 2 (x) share root r precisely when for some degree N 1 polynomials ~P 1 ðxþ and ~P 2 ðxþ, P 1 ðxþ ¼ðx rþ ~P 1 ðxþ P 2 ðxþ ¼ðx rþ ~P 2 ðxþ by (429) and (430) Also, by (428) we may write PðxÞ ¼P 1 ðx k Þþx m P 2 ðx k Þ Substitution of (435) and (436) into (437) yields PðxÞ ¼ðx k rþ ~P 1 ðx k Þþx m ðx k rþ ~P 2 ðx k Þ ¼ðx k rþð ~P 1 ðx k Þþx m ~P 2 ðx k ÞÞ ð4:35þ ð4:36þ ð4:37þ ð4:38þ ð4:39þ By the Fundamental Theorem of Algebra, the form of P(x) in (439) reveals that P(x) has the k roots rk 1 i2j e k j ¼ 0, 1,, k 1 In the other direction, if P(x) has roots of the form rk 1 i2j e k j ¼ 0, 1,, k 1 then (439) and (438) hold for some polynomials ~P 1 and ~P 2 By (428), (429), and (430), equation (437) holds, which in turn implies (435) and (436) hold From these it is clear that P 1 (x) and P 2 (x) have root r in common By Theorem 1, detðmða 1, a 2 ÞÞ ¼ 0 The theorem is proved 42 Polynomials with additive inverse root pairs Corollary to Theorem 2 Given a polynomial PðxÞ ¼a 0 þ a 1 x þ a 2 x 2 þa 2Nþ1 x 2Nþ1

9 let a even and a odd represent the coefficient vectors for the even and odd portions of P(x), that is a even ¼fa 0 a 2 a 2N g T ð4:40þ a odd ¼fa 1 a 3 a 2Nþ1 g T ð4:41þ Let Mða even, a odd Þ be the step-down matrix generated by a even and a odd Then detðmða even, a odd ÞÞ ¼ 0 ð4:42þ if and only if P(x) contains a root pair of additive inverses ðþ, Þ Proof P(x) can be written in the form (428) with m ¼ 1andk¼2 That is, PðxÞ ¼a 0 þ a 1 x þ a 2 x 2 þþa 2Nþ1 x 2Nþ1 ¼ XN a 2j x 2j þ a 2jþ1 x 2jþ1 Clearly, for m ¼ 1 and k ¼ 2, a 1 ¼ a even by (431) and (440) Similarly, a 2 ¼ a odd by (432) and (441) By Theorem p 2, detðmða even, a odd ÞÞ ¼ 0 if and only if P(x) has the ðk ¼ 2Þ roots ðr 1 ¼ ffiffi r e 0 p ¼, r 2 ¼ ffiffi r e i ¼ Þ Clearly, r 1 þ r 2 ¼ 0 The corollary is proved In the next section, we characterize polynomials with root pairs that are multiplicative, as opposed to additive, inverses 43 Polynomials with multiplicative inverse root pairs We characterize polynomials having roots which are multiplicative inverses Suppose P(x) is a degree N polynomial We wish to determine whether P(x) has a root pair of multiplicative inverses ðr,1=rþ for some r 6¼ 0 Theorem 3 Suppose P(x) is a degree N polynomial with non-zero roots Let a ¼fa 0 a 1 a N g T ð4:43þ be the vector of coefficients of P(x) Create vector a rev ¼fa N a N 1 a 0 g T ð4:44þ to be the reverse image of a Let Mða, a rev Þ be the step-down matrix generated by a and a rev Then P(x) has a root pair of multiplicative inverses ðr,1=rþ (for some r 6¼ 0) if and only if Proof so that Determinants and polynomial root structure 477 detðmða, a rev ÞÞ ¼ 0 P(x) is degree N and has only non-zero roots Define polynomial Q(x) QðxÞ ¼x N Pð1=xÞ

10 478 L G de Pillis Clearly, if the coefficient vector for P(x) is given by a in (443), then the coefficient vector for Q(x) is given by (444) Why: PðxÞ ¼a 0 þ a 1 x þ a 2 x 2 þþa N x N ð4:45þ QðxÞ ¼x N Pð1=xÞ ð4:46þ ¼ x N ða 0 þ a 1 =x þ a 2 =x 2 þþa N =x N Þ ð4:47þ ¼ a N þ a N 1 x þ a N 2 x 2 þþa 0 x N ð4:48þ By the definition of Q(x), for each root r i of P(x), ði ¼ 1, 2,, NÞ, 1/r i is a root of Q(x) By Theorem 1, the step-down matrix Mða, a rev Þ has zero determinant if and only if P(x) and Q(x) have a root in common P(x) andq(x) share a root when for some i and some j, r i ðof PðxÞÞ ¼ 1=r j ðof QðxÞÞ By the construction of Q(x), the equality holds precisely when r j is a root of P(x) Therefore, P(x) has root pair ðr j, r i Þ, or equivalently, ðr j,1=r j Þ The theorem is proved 44 Palindromic polynomials Definition Suppose P(x) is a degree N polynomial with non-zero roots Let the coefficient vector of P(x) be given by a (443), and the reverse image of a given by a rev (444) Polynomial P(x) ispalindromic when a ¼ a rev The following well known result is now immediate due to Theorem 3 Corollary to Theorem 3 Suppose polynomial P(x) of degree N is palindromic Then P(x) has at least one pair of roots which are multiplicative inverses, that is, of the form ðr,1=rþ Proof Create the step-down matrix Mða, a rev Þ Because a ¼ a rev, it is immediate that det ðmþ ¼0 Apply Theorem 3, which implies all palindromic polynomials contain at least one root pair in which the roots are multiplicative inverses 5 Examples The following examples illustrate that even in the case of N ¼ 2 and N ¼ 3, the classical determinant has novel and interesting properties One of our examples (example 3) supplies the complete solution of the special problem posed at the beginning of this paper (12) Recall that a defining pair of vectors in R Nþ1 induce a step-down matrix (14) of dimension 2N 2N 51 Example matrices Our theorem provides an alternate view of the determinant of 2 2 matrices Accordingly, consider the case when N ¼ 1 Choose ½x 0, x 1 Š T, ½y 0, y 1 Š T 2 R 2 and construct 2 2 matrix M ¼ x 0 y 0 x 1 y 1

11 Determinants and polynomial root structure 479 Now the roots of q x ðtþ ¼x 0 þ x 1 t and q y ðtþ ¼y 0 þ y 1 t are tabulated as follows: Roots of q x ðtþ Roots of q y ðtþ ð1þ x 1, y 1 6¼ 0 ðx 0 =x 1 Þ ðy 0 =y 1 Þ ð2þ x 1 ¼ y 1 ¼ 0 empty empty ð3þ x 1 ¼ 0, y 1 6¼ 0 empty ðy 0 =y 1 Þ ð4þ x 1 6¼ 0, y 1 ¼ 0 ðx 0 =x 1 Þ empty Consider the shared roots indicated in case ð1þ above We recover the familiar condition that detðmþ ¼0 if and only if ðx 0 =x 1 Þ¼ðy 0 =y 1 Þ or, in its more familiar form, x 0 y 1 ¼ x 1 y 0 Case (2) occurs when a row of M is zero According to Theorem 1, detðmþ ¼0 since it is vacuously true that q x (t) and q y (t) have shared roots Polynomials q x (t) and q y (t) have no common root in cases (3) and (4), indicating the determinant will be non-zero 52 Example matrices Consider the case N ¼ 2 Choose the linearly independent vectors ½1, x 1,0Š T, ½0, y 1,1Š T 2 R 3, from which matrix x M ¼ 1 y x 1 y is constructed Since detðmþ ¼y 1 x 1 1 it is clear that detðmþ ¼0 if and only if x 1 y 1 ¼ 1 ð5:49þ Through Theorem 1, we arrive at the same conclusion regarding the invertibility of M The roots of q x ðtþ ¼1 þ x 1 t and q y ðtþ ¼y 1 t þ t 2 are tabulated as follows: Roots of q x ðtþ Roots of q y ðtþ ð1þ x 1, y 1 6¼ 0 ð1=x 1 Þ 0, y 1 ð2þ x 1 ¼ y 1 ¼ 0 empty 0, 0 ð3þ x 1 ¼ 0, y 1 6¼ 0 empty 0, y 1 ð4þ x 1 6¼ 0, y 1 ¼ 0 ð1=x 1 Þ 0, 0 Only in case (1) could q x (t) and q y (t) potentially share a common root, and in that case this happens precisely when ð1=x 1 Þ¼ y 1 ð5:50þ or equivalently x 1 y 1 ¼ 1 ð5:51þ This is exactly the condition in equation (549) which guarantees that detðmþ ¼0

12 480 L G de Pillis 53 Example 3 Binomial coefficients Let us return to the original question posed in the introduction of this paper, which for completeness we re-state here: Consider the coefficients {a i } defined by the binomial polynomial PðtÞ ¼ð1þtÞ 2Nþ1 ¼ 1 þ a 1 t þ a 2 t 2 þþa 2N t 2N þ a 2Nþ1 t 2Nþ1 Here, we see that a i ¼ 2N þ 1, i ¼ 0, 1,,2N þ 1 i Use the even coefficient vector x ¼f1, a 2,, a 2N g, and the odd coefficient vector y ¼fa 1, a 3,, a 2Nþ1 g respectively, to uniquely generate all columns of the 2N 2N matrix a a 2 a 3 1 a 1 a 2 a 3 a 2N 2 a 2N 1 1 a 1 Mðx, yþ ¼ ð5:52þ a 2N a 2Nþ1 a 2N 2 a 2N 1 a 2 a a 2N a 2Nþ a 2N 2 a 2N a 2N a 2Nþ1 Question Is this matrix invertible? With the theory developed in this paper, this question is now easily answered in the affirmative By the Corollary to Theorem 2, detðmþ ¼0 if and only if P(t) contains a root pair of additive inverses Clearly, since the only roots of PðtÞ ¼ðt þ 1Þ 2Nþ1 are r ¼ 1, P(t) contains no such root pair It follows immediately that matrix Mðx, yþ will be invertible for any N 5 1 Acknowledgments The author heartily thanks Professor John de Pillis of UC Riverside for initially raising the question about the invertibility of step-down matrices, and for invaluable conversations during the writing of this paper References [1] Akritas, A, 1991, Sylvester s form of the resultant and the matrix-triangularization subresultant PRS method, in Computer Aided Proofs in Analysis, vol 28, IMA Vol Math Appl (Springer), pp 5 11 [2] Akritas, AG, 1993, Sylvester s forgotten form of the resultant, Fibonacci Quarterly, 31,

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