Determinants and polynomial root structure

International Journal of Mathematical Education in Science and Technology, Vol. 36, No. 5, 2005, 469–481

Determinants and polynomial root structure L. G. DE PILLIS Department of Mathematics, Harvey Mudd College, Claremont, CA 91711, U.S.A (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 R3 , 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 {ai} defined by the binomial polynomial ð1 þ tÞ2Nþ1 ¼ 1 þ a1 t þ a2 t2 þ þ a2N t2N þ a2Nþ1 t2Nþ1

E-mail: [email protected] International Journal of Mathematical Education in Science and Technology ISSN 0020–739X print/ISSN 1464–5211 online # 2005 Taylor & Francis Group Ltd http://www.tandf.co.uk/journals DOI: 10.1080/00207390412331303531

ð1:1Þ

470

L. G. de Pillis

Use the odd coefficients fa1 , a3 , . . . , a2Nþ1 g and the even coefficients f1, a2 , . . . , a2N g, respectively, to generate uniquely all columns of the 2N 2N matrix 3 2 1 a1 0 0 0 0 6 .. 7 .. 6 a . 7 a3 1 a1 . 7 6 2 6 . .. .. 7 .. .. 7 6 . 6 . . a2 . 7 . . a3 7 6 7 6 .. .. .. 6 a2N2 a2N1 . 1 a1 7 . . 7 6 ð1:2Þ 7 6 .. 7 6 . a2 a2Nþ1 a2N2 a2N1 a3 7 6 a2N 7 6 6 .. 7 .. .. 6 0 . 7 . . 0 a2N a2Nþ1 7 6 7 6 . .. .. .. .. 7 6 . 4 . . a2N2 a2N1 5 . . . 0

0

0

0

a2N

a2Nþ1

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 (1.2) 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 RNþ1 where x ¼ fx0

x1

x2

. . . xi

. . . xN1

xN gT

y ¼ fy0

y1

y2

...

...

y N gT

then x, y uniquely determine 2N 2N step-down matrix 2 x0 6 6 x 6 1 6 6 .. 6 . 6 6 6 6 xN1 Mðx, yÞ ¼ 6 6 6 x 6 N 6 6 6 0 6 6 . 6 . 4 . 0

yi

yN1

ð1:3Þ

the N pairs of columns in R2N which define the y0

0

0

0 .. . .. .

y1 .. .

x0

y0

yN1

x1 .. .

y1 .. .

yN

xN1

yN1

0 .. .

xN .. .

yN .. .

.. . .. . .. . .. . .. .

xN1

0

0

0

xN

x0 x1 .. .

0 .. . .. .

3

7 7 7 7 7 7 7 7 7 y0 7 7 7 y1 7 7 7 .. 7 . 7 7 7 7 yN1 5

ð1:4Þ

yN

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,

Determinants and polynomial root structure

471

and discussions can be found in various references, for example, [15, p. 102–107]; [7, p. 397–444]; 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 (1.2) (or (1.4)) is to be invertible at all, it is necessary that the first two columns be linearly independent in R2N . 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 3 2 x0 y0 7 6 . .. 7 6 .. x0 . y0 7 6 7 6 .. .. .. .. 7 6 7 6 xN1 . x . y . . y 0 N1 0 7 6 7 6 . . .. .. 7 ð2:5Þ Mðx; yÞ ¼ 6 yN xN1 yN1 7 6 xN 7 6 6 .. .. 7 .. .. 6 . yN . 7 xN . . 7 6 7 6 .. .. 7 6 . xN1 . yN1 5 4 xN yN 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 R2N by X~ ¼ ½x0 , x1 , . . . , xN , 0, . . . , 0T |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflffl{zfflfflfflffl}

ð2:6Þ

Y~ ¼ ½y0 , y1 , . . . , yN , 0, . . . , 0T |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflffl{zfflfflfflffl}

ð2:7Þ

Nþ1

Nþ1

N1

N1

If we define the 2N 2N shift matrix S by 2 3 0 0 61 0 7 6 7 6 7 6 7 1 0 S¼6 7 .. .. 6 7 4 5 . . 0 1 0

ð2:8Þ

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L. G. de Pillis

then sawtooth matrix (2.5) can be denoted Mðx; yÞ ¼ ½X~ , S X~ , S2X~ , . . . , S N1 X~ , Y~ , S Y~ , S2Y~ , . . . , SN1Y~ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} N columns

ð2:9Þ

N columns

With these conventions, we set down Lemma 1. Step-down matrix Mðx; yÞ of (2.9) has zero determinant if and only if there exist sequences fi g, fj g, i, j ¼ 0, 1, . . . , N 1 such that XN1 XN1 i j ~ X Y~ ¼ 0 2 R2N S S ð2:10Þ i j i¼0 j¼0 |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} P ðSÞ

P ðSÞ

Moreover, the scalar sequences are non-trivial in the sense that some i0 6¼ 0 and some j0 6¼ 0, 0 4 i0 , j0 4 N 1. Proof. To say that detðMðx; yÞÞ ¼ 0 is to say that some of the columns of (2.9) are linearly dependent. Write such a linear combination using 0i s for the first N columns of M and using 0j s for the last N columns. This gives us equation (2.10). But not all i ¼ 0, or else the last N columns of Mðx; yÞ in (2.9) 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 (2.9) are linearly dependent. This ends the proof. œ 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, 1 , 0, . . . , 0T 2 R2N e~i ¼ ½0, . . . , 0, |{z}

ð2:11Þ

ith place

Then clearly, e~kþ1 ¼ Sk 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 (2.6) and (2.7) can be written as XN XN X~ ¼ x S k e~1 , Y~ ¼ y S l e~1 : k¼0 k l¼0 l |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Qx ðSÞ

ð2:13Þ

Qy ðSÞ

Proof. X~ ¼ ½x0 , x1 , . . . , xN , 0, . . . , 0T ¼

N X xk e~kþ1

from (2.11)

k¼0

¼

N X ðxk S k Þ~e1 k¼0

from (2.12)

from (2.6)

ð2:14Þ


473

Similarly, Y~ ¼ ½ y0 , y1 , . . . , yN , 0, . . . , 0T ¼

N X yl e~lþ1

from (2.7)

from (2.11)

ð2:15Þ

l¼0

¼

N X ðyl S l Þ~ e1

from (2.12)

l¼0

œ

which establishes (2.13). The proof is done. Combining Lemmas 1 and 2, we arrive at the following

Lemma 3. Matrix Mðx; yÞ of (2.9) has zero determinant if and only if there exist non-trivial scalar sequences fi g, fj g, i, j ¼ 0, 1, . . . , N 1 such that XN1 XN XN1 XN i k j l S x S S y S ¼0 ð2:16Þ i k j l i¼0 k¼0 j¼0 l¼0 |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} P ðSÞ

Qx ðSÞ

P ðSÞ

Qy ðSÞ

Proof. We first replace X~ and Y~ in (2.10) by (2.13) to obtain the following restatement of Lemma 1. Matrix Mðx; yÞ of (2.9) has zero determinant if and only if there exist non–trivial scalar sequences fi g and fj g, such that XN1 XN XN1 XN i k j l ~ S x S S y S e e~1 ¼ 0 2 R2N i 1 j i¼0 k¼0 k j¼0 l¼0 l |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} P ðSÞ

Qx ðSÞ

P ðSÞ

Qy ðSÞ

Now multiply out the matrix polynomials P ðSÞQx ðSÞ and P ðSÞQy ðSÞ to obtain 2N1 X i¼0

ai |{z} Si e~1

2N1 X i¼0

2N1 X

e~iþ1

ai e~iþ1

i¼0

bi |{z} S i e~1 ¼ 0 2 R2N which implies

2N1 X

e~iþ1

ð2:17Þ

bi e~iþ1 ¼ 0 2 R2N since Si e~1 ¼ e~iþ1

i¼0

Here, ai and bi are the convolutions of the i with the xk and the j with the yl respectively. The vectors f~ e1 , e~2 , . . . , e~2N g above form a basis for R2N . This guarantees that, in (2.17), ai ¼ bi for all i ¼ 0, 1, . . . , ð2N 1Þ

ð2:18Þ

Since coefficients fai g, and fbi g are a result of multiplying out the matrix products P ðSÞQx ðSÞ and P ðSÞQy ðSÞ, we conclude from (2.18) that P ðSÞQx ðSÞ ¼ P ðSÞQy ðSÞ. This establishes equality (2.16) and the lemma is proved. œ Now for some notation and final observations. Notation.

The symbols C½S and C½z denote the polynomial rings C½S ¼ span½I2N , S, S 2 , . . . , S 2N1

ð2:19Þ

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L. G. de Pillis

and C½z ¼ span½1, z, z2 , . . . , z2N1

ð2:20Þ

where C is the field of complex numbers, S is the 2N 2N shift matrix (2.8), and z is an indeterminate. As usual, deg(P), the degree of a polynomial P 2 C½S (resp. deg(p), the degree of p 2 C½z) 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 C½S and C½z. Lemma 4. Vector spaces, C½S and C½z, of dimension 2N, have respective bases fSi g and fzi g, i ¼ 0, 1, 2, . . . , ð2N 1Þ. One vector space isomorphism : C½S 7 ! C½z, is defined by linear extension of the one–one map ðSi Þ ¼ zi , i ¼ 0, 1, 2 . . . , ð2N 1Þ

ð2:21Þ

Under this isomorphism, the multiplicative ring structure of C½S is preserved in the sense that ðP1 P2 Þ ¼ ðP1 Þ ðP2 Þ 2 C½z

ð2:22Þ

for all ðP1 Þ, ðP2 Þ, ðP1 P2 Þ 2 C½S, that is, whenever degðP1 P2 Þ ¼ degðP1 Þ þ degðP2 Þ 4 ð2N 1Þ: Proof. From the definition (2.21), we see that (2.22) holds for the special case P1 ¼ S i1 and P2 ¼ Si2 , where 0 4 i1 , i2 , 4 ð2N 1Þ and i1 þ i2 4 ð2N 1Þ. Linear extension establishes (2.22) for general P1 and P2 where degðP1 P2 Þ ¼ degðP1 Þ þ degðP2 Þ 4 ð2N 1Þ: œ

3. Main result Theorem 1. Given vectors x, y 2 RNþ1 of (1.3), generate the 2N 2N step-down matrix Mðx; yÞ of (1.4). Then detðMðx; yÞÞ ¼ 0

ð3:23Þ

if and only if the polynomials px ðtÞ ¼ x0 þ x1 t þ x2 t2 þ þ xN tN and py ðtÞ ¼ y0 þ y1 t þ y2 t2 þ þ yN tN

ð3:24Þ

have at least one root in common. Proof. We have from (2.16) of Lemma 3, that detðMðx; yÞÞ ¼ 0 if and only if for non-trivial scalar sequences fi g and fj g, P ðSÞ Qx ðSÞ ¼ P ðSÞ Qy ðSÞ. That is, XN XN1 XN XN1 Si x Sk ¼ Sj y Sl from Lemma 3 i¼0 i k¼0 k j¼0 j l¼0 l |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl} P ðSÞ

iff

Qx ðSÞ

P ðSÞ

Qy ðSÞ

XN1

XN XN1 XN i zi x zk ¼ zj yl z l k¼0 k j¼0 j |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflffll¼0 {zfflfflfflfflfflffl} i¼0

p ðzÞ

qx ðzÞ

p ðzÞ

qy ðzÞ

from ð2:22Þ

475


Factoring of polynomials p ðzÞ, qx ðzÞ, p ðzÞ, qy ðzÞ above, gives us the equality A1 ðz r1 Þ ðz rN1 Þ A2 ðz s1 Þ ðz sN Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} qx ðzÞ¼x0 þx1 zþþxN zN

p ðzÞ

ð3:25Þ ¼ A3 ðz r~1 Þ ðz r~N1 Þ A4 ðz s~1 Þ ðz s~N Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} qy ðzÞ¼y0 þy1 zþþyN zN

p ðzÞ

for scalars A1 , A2 , A3 , A4 . Factored polynomials (3.25) are equal if and only if A1 A2 ¼ A3 A4 and (most importantly) the roots agree. That is, fr1 , r2 , . . . ; rN1 , s1 , s2 , . . . , sN g ¼ f~r1 , r~2 , . . . ; r~N1 , s~1 , s~2 , . . . , s~N g |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} qx ðzÞ roots

ð3:26Þ

qy ðzÞ roots

A simple counting argument establishes that the roots {si} and f~sj g have an element in common. In fact, whenever we match the N-element set f~s1 , s~2 , . . . , s~N g from the RHS of (3.26), with any N elements from the LHS, at least one s~j0 must be chosen. This says that that polynomials (3.24) have a common root (some si0 ¼ s~j0 ) if and only if detðMðx; yÞÞ ¼ 0. The theorem is proved. œ

4. Applications 4.1. 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 14m < k

ð4:27Þ

Polynomial P(x) is a lacunary polynomial when it can be written in the following form: PðxÞ ¼

N X

ajk xjk þ ajkþm xjkþm

ð4:28Þ

j¼0

œ 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: P1 ðxÞ ¼

N X

ajk xj

ð4:29Þ

ajkþm xj

ð4:30Þ

j¼0

P2 ðxÞ ¼

N X j¼0

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L. G. de Pillis

Define the coefficient vectors of P1(x) and P2(x) by a1 ¼ fa0 ak a2k . . . aNk gT ,

ð4:31Þ T

a2 ¼ fam akþm a2kþm . . . aNkþm g :

ð4:32Þ

We now have Theorem 2. Given coefficient vectors a1 of (4.31) and a2 of (4.32), generate step-down matrix Mða1 , a2 Þ according to (1.4). Then detðMða1 , a2 ÞÞ ¼ 0

ð4:33Þ

if and only if P(x) in (4.28) has the k roots 1 i2j

rk e k , j ¼ 0, 1, . . . , k 1

ð4:34Þ

where i¼

pffiffiffiffiffiffiffi 1

0 4 < 2

a2R>0

1 k

The roots in (4.34) lie on a circle of radius r centred at the origin. Proof. By Theorem 1, detðMða1 , a2 ÞÞ ¼ 0 if and only if P1(x) and P2(x) share a common root. P1(x) and P2(x) share root r precisely when for some degree N 1 polynomials P~ 1 ðxÞ and P~ 2 ðxÞ, P1 ðxÞ ¼ ðx rÞP~ 1 ðxÞ P2 ðxÞ ¼ ðx rÞP~ 2 ðxÞ

ð4:35Þ ð4:36Þ

by (4.29) and (4.30). Also, by (4.28) we may write PðxÞ ¼ P1 ðxk Þ þ xm P2 ðxk Þ

ð4:37Þ

Substitution of (4.35) and (4.36) into (4.37) yields PðxÞ ¼ ðxk rÞP~ 1 ðxk Þ þ xm ðxk rÞP~ 2 ðxk Þ ¼ ðxk rÞðP~ 1 ðxk Þ þ xm P~ 2 ðxk ÞÞ

ð4:38Þ ð4:39Þ

By the Fundamental Theorem of Algebra, the form of P(x) in (4.39) reveals that P(x) has the k roots 1 i2j

rk e k

j ¼ 0, 1, . . . , k 1

In the other direction, if P(x) has roots of the form 1 i2j

rk e k

j ¼ 0, 1, . . . , k 1

then (4.39) and (4.38) hold for some polynomials P~ 1 and P~ 2 . By (4.28), (4.29), and (4.30), equation (4.37) holds, which in turn implies (4.35) and (4.36) hold. From these it is clear that P1(x) and P2(x) have root r in common. By Theorem 1, detðMða1 , a2 ÞÞ ¼ 0. The theorem is proved. œ 4.2. Polynomials with additive inverse root pairs Corollary to Theorem 2.

Given a polynomial

PðxÞ ¼ a0 þ a1 x þ a2 x2 þ a2Nþ1 x2Nþ1

477


let aeven and aodd represent the coefficient vectors for the even and odd portions of P(x), that is aeven ¼ fa0 a2 . . . a2N gT

ð4:40Þ

aodd ¼ fa1 a3 . . . a2Nþ1 gT

ð4:41Þ

Let Mðaeven , aodd Þ be the step-down matrix generated by aeven and aodd . Then detðMðaeven , aodd 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 (4.28) with m ¼ 1 and k ¼ 2. That is, PðxÞ ¼ a0 þ a1 x þ a2 x2 þ þ a2Nþ1 x2Nþ1 ¼

N X

a2j x2j þ a2jþ1 x2jþ1

j¼0

Clearly, for m ¼ 1 and k ¼ 2, a1 ¼ aeven by (4.31) and (4.40). Similarly, a2 ¼ aodd by (4.32) and (4.41). By Theorem pffiffi 2, detðMðapeven ffiffi , aodd ÞÞ ¼ 0 if and only if P(x) has the ðk ¼ 2Þ roots ðr1 ¼ r e0 ¼ , r2 ¼ r ei ¼ Þ. Clearly, r1 þ r2 ¼ 0. The corollary is proved. œ In the next section, we characterize polynomials with root pairs that are multiplicative, as opposed to additive, inverses. 4.3. 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 ¼ fa0 a1 . . . aN gT

ð4:43Þ

be the vector of coefficients of P(x). Create vector arev ¼ faN aN1 . . . a0 gT

ð4:44Þ

to be the reverse image of a. Let Mða, arev Þ be the step-down matrix generated by a and arev . Then P(x) has a root pair of multiplicative inverses ðr, 1=rÞ (for some r 6¼ 0) if and only if detðMða, arev ÞÞ ¼ 0 Proof. P(x) is degree N and has only non-zero roots. Define polynomial Q(x) so that QðxÞ ¼ xN Pð1=xÞ

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L. G. de Pillis

Clearly, if the coefficient vector for P(x) is given by a in (4.43), then the coefficient vector for Q(x) is given by (4.44). Why: PðxÞ ¼ a0 þ a1 x þ a2 x2 þ þ aN xN

ð4:45Þ

QðxÞ ¼ xN Pð1=xÞ N

ð4:46Þ 2

N

¼ x ða0 þ a1 =x þ a2 =x þ þ aN =x Þ 2

N

¼ aN þ aN1 x þ aN2 x þ þ a0 x

ð4:47Þ ð4:48Þ

By the definition of Q(x), for each root ri of P(x), ði ¼ 1, 2, . . . , NÞ, 1/ri is a root of Q(x). By Theorem 1, the step-down matrix Mða, arev Þ has zero determinant if and only if P(x) and Q(x) have a root in common. P(x) and Q(x) share a root when for some i and some j, ri ðof PðxÞÞ ¼ 1=rj ðof QðxÞÞ By the construction of Q(x), the equality holds precisely when rj is a root of P(x). Therefore, P(x) has root pair ðrj , ri Þ, or equivalently, ðrj , 1=rj Þ. The theorem is proved. œ 4.4. 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 (4.43), and the reverse image of a given by arev (4.44). Polynomial P(x) is palindromic when a ¼ arev . 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, arev Þ. Because a ¼ arev , it is immediate that det ðMÞ ¼ 0. Apply Theorem 3, which implies all palindromic polynomials contain at least one root pair in which the roots are multiplicative inverses. œ

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 (1.2). Recall that a defining pair of vectors in RNþ1 induce a step-down matrix (1.4) of dimension 2N 2N. 5.1. Example 1 – 2 3 2 matrices Our theorem provides an alternate view of the determinant of 2 2 matrices. Accordingly, consider the case when N ¼ 1. Choose ½x0 , x1 T , ½y0 , y1 T 2 R2 and construct 2 2 matrix x0 y0 M¼ x1 y1

479


Now the roots of qx ðtÞ ¼ x0 þ x1 t and qy ðtÞ ¼ y0 þ y1 t are tabulated as follows:

ð1Þ ð2Þ ð3Þ ð4Þ

x1 , y1 6¼ 0 x1 ¼ y1 ¼ 0 x1 ¼ 0, y1 6¼ 0 x1 6¼ 0, y1 ¼ 0

Roots of qx ðtÞ

Roots of qy ðtÞ

ðx0 =x1 Þ empty empty ðx0 =x1 Þ

ðy0 =y1 Þ empty ðy0 =y1 Þ empty

Consider the shared roots indicated in case ð1Þ above. We recover the familiar condition that detðMÞ ¼ 0 if and only if ðx0 =x1 Þ ¼ ðy0 =y1 Þ or, in its more familiar form, x0 y1 ¼ x1 y0 . Case (2) occurs when a row of M is zero. According to Theorem 1, detðMÞ ¼ 0 since it is vacuously true that qx(t) and qy(t) have shared roots. Polynomials qx(t) and qy(t) have no common root in cases (3) and (4), indicating the determinant will be non-zero. 5.2. Example 2 – 3 3 matrices Consider the case N ¼ 2. Choose the linearly independent vectors ½1, x1 , 0T , ½0, y1 , 1T 2 R3 , from which matrix 2 3 1 0 0 0 6 x1 y 1 1 0 7 7 M¼6 4 0 1 x1 y1 5 0 0 0 1 is constructed. Since detðMÞ ¼ y1 x1 1 it is clear that detðMÞ ¼ 0 if and only if x1 y 1 ¼ 1

ð5:49Þ

Through Theorem 1, we arrive at the same conclusion regarding the invertibility of M. The roots of qx ðtÞ ¼ 1 þ x1 t and qy ðtÞ ¼ y1 t þ t2 are tabulated as follows:

ð1Þ ð2Þ ð3Þ ð4Þ

Roots of qx ðtÞ

Roots of qy ðtÞ

ð1=x1 Þ empty empty ð1=x1 Þ

0, y1 0, 0 0, y1 0, 0

x1 , y1 6¼ 0 x1 ¼ y1 ¼ 0 x1 ¼ 0, y1 6¼ 0 x1 6¼ 0, y1 ¼ 0

Only in case (1) could qx(t) and qy(t) potentially share a common root, and in that case this happens precisely when ð1=x1 Þ ¼ y1

ð5:50Þ

x1 y1 ¼ 1

ð5:51Þ

or equivalently

This is exactly the condition in equation (5.49) which guarantees that detðMÞ ¼ 0.

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L. G. de Pillis

5.3. 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 {ai} defined by the binomial polynomial PðtÞ ¼ ð1 þ tÞ2Nþ1 ¼ 1 þ a1 t þ a2 t2 þ þ a2N t2N þ a2Nþ1 t2Nþ1 Here, we see that ai ¼

2N þ 1 , i ¼ 0, 1, . . . , 2N þ 1 i

Use the even coefficient vector x ¼ f1, a2 , . . . , a2N g, and the odd coefficient vector y ¼ fa1 , a3 , . . . , a2Nþ1 g respectively, to uniquely generate all columns of the 2N 2N matrix 3 2 1 a1 0 0 0 0 6 .. 7 .. 6 a . 7 a3 1 a1 . 7 6 2 6 . .. .. 7 .. .. 7 6 . 6 . . a2 . 7 . . a3 7 6 7 6 .. .. .. 6 a2N2 a2N1 . 1 a1 7 . . 7 6 ð5:52Þ Mðx, yÞ ¼ 6 7 .. 7 6 . a2 a2Nþ1 a2N2 a2N1 a3 7 6 a2N 7 6 6 .. 7 .. .. 6 0 . 7 . . 0 a2N a2Nþ1 7 6 7 6 . .. .. .. .. 7 6 . 5 4 . . a . . . a 2N2

0

0

0

0

a2N

2N1

a2Nþ1

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 ¼ 0 if and only if P(t) contains a root pair of additive inverses. Clearly, since the only roots of PðtÞ ¼ ðt þ 1Þ2Nþ1 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 U.C. Riverside for initially raising the question about the invertibility of step-down matrices, and for invaluable conversations during the writing of this paper.

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