# A Novel Necessary and Sufficient Condition for the Positivity of a Binary Quartic Form

Authors: Yang Guo
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, by considering the common points of two conics instead of the roots of the binary quartic form, we propose a novel necessary and sufficient condition for the positivity of a binary quartic form using the theory of the pencil of conics. First, we show the degenerate members of the pencil of conics according to the distinct natures of the common points of two base conics. Then, the inequalities about the parameters of the degenerate members are obtained according to the properties of the degenerate conics. Last, from the inequalities, we derive a novel criterion for determining the positivity of a binary quartic form without the discriminant.


---

## Body

## 1. Introduction

Inℝx,y, let the polynomial(1)fx,y=x4+a3x3y+a2x2y2+a1xy3+a0y4,be a monic binary quartic form on real number field ℝ. We say that fx,y is positive definite (positive semidefinite) if for any real numbers x,y (not both 0), we have fx,y>0fx,y≥0.The problem of determining the global positivity or nonnegativity of forms has many applications. For instance, in the study of two-dimensional nonlinear systems and other mathematical and physical fields, we often need to deal with the positive definite or negative definite problems of a binary quartic form [1, 2]. So far, about the positive definite problem of a binary quartic form, there has existed some literatures. For example, Chin [3] used the canonical forms of a real quartic equation to propose a necessary and sufficient condition for no real roots. Gadenz and Li [4] published a systematic method for determining the positive definiteness of the binary quartic form. Their method adopted the testing of the permanence and variations of sign in the sequence of principal minors of the corresponding Hankel matrix. Jury and Mansour [5] used Ferrari’s solution to obtain condition for positivity for a quartic equation. In the same year, Fuller [6] also gave similar criteria in terms of bigradient determinants. Ku [7] gave a criterion for the positive definiteness of the general quartic form from the known solution. However, his result was not complete. Wang and Qi [8] gave a complete and improved result.In the above literatures, one of the most common results obtained by different methods can be summarized as follows. Let(2)Vx,y=c0x4+4c1x3y+6c2x2y2+4c3xy3+c4y4,be a binary quartic form on real number field ℝ. In terms of the coefficients of (2), the following quantities can be defined [7]:(3)G=c02c3−3c0c1c2+2c13;H=c0c2−c12;I=c0c4−4c1c3+3c22;J=detc0c1c2c1c2c3c2c3c4;Δ=I3−27J2.Supposec0>0. Then, the quartic form Vx,y is positive definite if and only if one of the following three conditions holds [3, 5, 6, 8]:(1)
Δ=0,G=0,12H2−c02I=0,H>0(2)
Δ>0,H≥0(3)
Δ>0,H<0,12H2−c02I<0Compared with the criterion in [4], this criterion based on the discriminant of a quartic polynomial is superior in that the conditions contained in this criterion are stated in the form of a set of inequalities which are explicit functions of the coefficients of the quartic form. Besides for this common criterion, Hasan and Hasan [9] presented an alternative criterion. They made use of some of the results of the quadratic programming theory to show that testing for positive semidefiniteness of the quartic form (1) was reduced to a test whether there is a real number λ such that the parametric matrix(4)Mλ=1a32−λ2a32a2+λa12−λ2a12a0,is positive semidefinite. Essentially, Hasan et al.’s method needs to solve the inequality equations coming from the constraints that the parametric matrix Mλ is positive semidefinite (positive definite).Now, the natural question to ask is whether there exists a concrete real numberλ obtained by the coefficients of the form (1) such that the form (1) is positive semidefinite (positive definite) if and only if the matrix Mλ is a positive semidefinite (positive definite) matrix. If yes, the criterion in [9] will be simplified. In this paper, we will see that the answer to this question is yes. We use the relationship between the common points of two conics and the degenerate members of the pencil of conics to address this question and further derive a novel criterion for determining the positivity of a binary quartic form without the discriminant. We show that given a monic binary quartic form (1), the following quantities can be defined:(5)b0=14−a12+a1a2a3−a0a32;b1=144a0−a22−a1a3;b2=a22;λ0=4b2+23b1+4b223.Then, the form (1) is positive semidefinite (positive definite) if and only if the matrix(6)1a32a2−λ02a32λ0a12a2−λ02a12a0,is a positive semidefinite (positive definite) matrix (Theorem 1).More concretely, the form (1) is positive semidefinite if and only if(7)λ0≥a324,−14λ03+b2λ02+b1λ0+b0≥0.In particular, the form (1) is positive definite if and only if (Theorem 2)(8)λ0>a324,−14λ03+b2λ02+b1λ0+b0>0.


## 2. Notations and Preliminaries

Throughout this paper, letA1 and A2 denote respectively two real symmetric matrices as follows:(9)A1=1a32a22a320a12a22a12a0,A2=00−12010−1200,and let the matrix Mλ be(10)Mλ=A1+λA2=1a32a2−λ2a32λa12a2−λ2a12a0,where λ is a variable real parameter. Since every 3×3 real symmetric matrix denotes a conic in the complex projective plane, Mλ is a pencil of conics determined by the base conics A1, A2.Obviously, using the matrixMλ, the binary quartic form (1) can be represented as follows:(11)fx,y=x2,xy,y2Mλx2xyy2.On the other hand, for arbitraryx, y not all zero, x2,xy,y2T is the projective coordinate of the point of the conic A2, so we can use the relative position of the conics A1, A2 to address the determination of positivity of the binary quartic form (1). We assert that the binary quartic form (1) is positive semidefinite if and only if there is no real simple (or transversal) intersection point between the conic A1 and the conic A2. In particular, the form (1) is positive definite if and only if there is no real common point between A1 and A2.It is well known that the conicsA1, A2 have four common points (which need not all be distinct). Further, the relative position of the conics A1, A2 has nine cases according to the nature (generic or nongeneric, real or complex) of their common points [10, 11]. They are presented in Table 1.Table 1
Nine cases of the intersection between the conicA1 and the conic A2.
CaseNature of the common point1Four real simple points2Two pairs of complex conjugate simple points3Two real simple points + a pair of complex conjugate simple points4Two real simple points + a real double point (simple-contact)5A pair of complex conjugate simple points + a real double point (simple-contact)6Two real double points (double contact)7A pair of complex conjugate double points (double contact)8A real simple point + a triple point (three-point contact)9A quadruple point (four-point contact)Now, we discuss the degenerate members of the pencilMλ since they are closely related to the relative position of the conics A1, A2. The pencil Mλ contains three degenerate members (which need not all be distinct) [12]. Their parameters are the roots of the cubic equation detMλ=0. The degenerate conics are the line-pairs through four common points of A1 and A2. So, according to the nature of the common point in Table 1, the degenerate member in the pencil Mλ may concretely consist of one of as follows:(1)
A real line-pair (two real lines)(2)
A complex line-pair (two complex lines)(3)
A complex conjugate line-pair (two complex conjugate lines)(4)
A real repeated lineThe middle column of Table2 gives the concrete type of every degenerate member in nine cases. The different type of degenerate member Mλi can provide important information about its parameter λi. More formally, we have the following lemmas.Table 2
Degenerate conics in the pencilMλ and properties of three roots of detMλ=0.
CaseaDegenerate conicProperty ofλ1,λ2,λ31Three real line-pairsλ1<λ2<λ3≤a32/42One real line-pair + two complex conjugate line-pairsλ1≤a32/4<λ2<λ3orλ1<a32/4≤λ2<λ33One real line-pair + two complex line-pairsλ1≤a32/4,λ2=λ¯3b4Two real line-pairsλ1=λ2<λ3≤a32/4orλ1<λ2=λ3<a32/45One real line-pair + one complex conjugate line-pairλ1≤a32/4<λ2=λ36One real repeated line + one real line-pairλ1<a32/4=λ2=λ37One real repeated line + one complex conjugate line-pairλ1=λ2=a32/4<λ38One real line-pairλ1=λ2=λ3<a32/49One real repeated lineλ1=λ2=λ3=a32/4aCorresponding to Table 1, bλ2,λ3 are complex conjugate.Lemma 1.
If the matrixMλi is a degenerate conic consisting of a real line-pair, then the rank of Mλi is 2 and λi≤a32/4.Lemma 2.
If the matrixMλi is a degenerate conic consisting of a complex conjugate line-pair, then the rank of Mλi is 2 and λi≥a32/4.Lemma 3.
The matrixMλi is a degenerate conic consisting of a real repeated line if and only if λi=a32/4 is a root of multiplicity ≥2 of detMλ=0.The first two lemmas can be proved by using Sylvester’s criterion ([13], Theorem 7.2.5) and the facts that the matrix Mλi is a degenerate conic consisting of a real line-pair if and only if Mλi is indefinite with the rank 2 and the matrix Mλi is a degenerate conic consisting of a complex conjugate line-pair if and only if  Mλi is positive semidefinite with the rank 2. Lemma 3 can be proved by using direct computation and the fact that the matrix Mλi is a degenerate conic consisting of a real repeated line if and only if Mλi is positive semidefinite with the rank 1.From the above three lemmas, it is easy to obtain the relation betweena32/4 and the parameters λ1, λ2, λ3 of the three degenerate members of the pencil  Mλ in every case that is given in the middle column of Table 2. The results are presented in the last column of Table 2. It is worth noting that this is a one-to-one correspondence between the middle column of Table 2 and the last column of Table 2.We denote bygλ the determinant of the parameter matrix Mλ, namely,(12)gλ=detMλ=detA1+λA2=−14λ3+b2λ2+b1λ+b0,where b0=−a12+a1a2a3−a0a32/4, b1=4a0−a22−a1a3/4, and b2=a2/2. Using the resultant of gλ and its formal derivative g′λ, we may obtain the discriminant of gλ as follows [14]:(13)Dgλ=163b1+4b223−27b0+36b1b2+32b232432.SinceDgλ<0 if and only if gλ=0 has a pair of complex conjugate roots, we assert that 3b1+4b22≥0 for the other eight cases, except for Case 3 in Table 2.Now let(14)λ0=4b2+23b1+4b223.Obviously, when3b1+4b22≥0, λ0 is the largest of the two real roots (which need not be distinct) of g′λ=0; otherwise, λ0 is a complex number.Further, according tog−∞>0, g+∞<0 and the inequalities about three roots λ1, λ2, λ3 in Table 2, we may get the properties of λ0 and gλ0 in the nine cases. For example, for λ1≤a32/4, λ2=λ¯3 in Case 3, i.e., Dgλ<0, then one of the following two configurations happens:(1)
If3b1+4b22<0, then λ0 is a nonreal number(2)
If3b1+4b22≥0, then λ0<a32/4 or λ0≥a32/4, gλ0<0For the other eight cases in Table2, λ0 must be real and furthermore the relation between λ0 and a32/4 and the relation between gλ0 and 0 can be obtained easily. The results are presented in the last column of Table 3.Table 3
Properties ofλ0 and gλ0.
CaseaProperty ofλ1, λ2, λ3Property ofλ0 and gλ01λ1<λ2<λ3≤a32/4λ0<a32/4, gλ0>02λ1≤a32/4<λ2<λ3 or λ1<a32/4≤λ2<λ3λ0>a32/4, gλ0>03λ1≤a32/4,λ2=λ¯3λ0 is nonreal or λ0<a32/4 or λ0≥a32/4, gλ0<04λ1=λ2<λ3≤a32/4 or λ1<λ2=λ3<a32/4λ0<a32/4, gλ0≥05λ1≤a32/4<λ2=λ3λ0>a32/4, gλ0=06λ1<a32/4=λ2=λ3λ0=a32/4, gλ0=07λ1=λ2=a32/4<λ3λ0>a32/4, gλ0>08λ1=λ2=λ3<a32/4λ0<a32/4, gλ0=09λ1=λ2=λ3=a32/4λ0=a32/4, gλ0=0aCorresponding to Tables 1 and 2.


## 3. The Main Results

From the foregoing discussion, we know that the binary quartic form (1) is positive semidefinite if and only if there is no real simple intersection point between the conic A1 and the conic A2, i.e., Case 2, Case 5, Case 6, Case 7, and Case 9 in Table 1. In particular, the form (1) is positive definite if and only if there is no real common point between the conic A1 and the conic A2, i.e., Case 2 and Case 7. Consequently, we can use the properties of λ0 and gλ0 in Table 3 to obtain a simplified necessary and sufficient condition compared with [9] for determining the positivity of the form (1).Theorem 1.
Given a monic binary quartic form,(15)fx,y=x4+a3x3y+a2x2y2+a1xy3+a0y4.Let(16)b1=144a0−a22−a1a3,b2=a22,λ0=4b2+23b1+4b223.Then, the form (15) is positive semidefinite (positive definite) if and only if the 3×3 real symmetric matrix(17)Mλ0=1a32a2−λ02a32λ0a12a2−λ02a12a0,is a positive semidefinite (positive definite) matrix.Proof.
One direction of the proof is obvious since ifλ0 is a real number satisfying the matrix Mλ0 which is positive semidefinite (positive definite), then according to the properties of positive semidefinite matrix (positive definite matrix), we have λ0≥a32/4 and gλ0=detMλ0≥0 (λ0>a32/4 and gλ0>0). Further, from Table 3, we know that one of Case 2, Case 5, Case 6, Case 7, and Case 9 occurs (one of Case 2 and Case 7 occurs). That means that the form (15) is positive semidefinite (positive definite).
To prove the converse, we just need to show that ifλ0≥a32/4 and gλ0≥0, then the matrix Mλ0 is positive semidefinite. On one hand, if λ0>a32/4 and gλ0≥0, then according to Sylvester’s criterion, we have Mλ0 which is positive semidefinite. On the other hand, when λ0=a32/4 and gλ0≥0, since ga32/4 can be represented as follows:(18)ga324=det1a324a2−a328a32a324a124a2−a328a12a0=−12568a1−4a2a3+a332,we have ga32/4=0. So, we may get a1=4a2a3−a33/8. Further, according to λ0=4b2+23b1+4b22/3=a32/4, we can also get a0=4a2−a322/64. That means that Mλ0 is a positive semidefinite matrix with the rank 1.
Thus, if the form (15) is positive semidefinite, then there is no real simple intersection point between two base conics, i.e., Case 2, Case 5, Case 6, Case 7, and Case 9 in Table 1. Further, from Table 3, we have λ0≥a32/4 and gλ0≥0, namely, the matrix Mλ0 is positive semidefinite. In particular, if the form (15) is positive definite, then there is no real common point between two base conics, i.e., Case 2 and Case 7. Further, from Table 3, we have λ0>a32/4 and gλ0>0. Consequently, according to Sylvester’s criterion, the matrix Mλ0 is positive definite. This completes the proof of the theorem.
From the proof of Theorem1, we can obtain directly a novel criterion for determining the positivity of a monic binary quartic form.Theorem 2.
Given a monic binary quartic form,(19)fx,y=x4+a3x3y+a2x2y2+a1xy3+a0y4.Let(20)b0=14−a12+a1a2a3−a0a32,b1=144a0−a22−a1a3,b2=a22,λ0=4b2+23b1+4b223.Then, the form (19) is positive semidefinite if and only if(21)λ0≥a324,−14λ03+b2λ02+b1λ0+b0≥0.In particular, the form (19) is positive definite if and only if(22)λ0>a324,−14λ03+b2λ02+b1λ0+b0>0.It is well known that the negativity of−x4+a3x3y+a2x2y2+a1xy3+a0y4 is equivalent to the positivity of x4−a3x3y−a2x2y2−a1xy3−a0y4. Consequently, according to the obtained conclusions, we may obtain directly the following corollaries about the negativity of a binary quartic form.Corollary 1.
Given a binary quartic form,(23)fx,y=−x4+a3x3y+a2x2y2+a1xy3+a0y4.Let(24)b1=−144a0+a22+a1a3,b2=−a22,λ0=4b2+23b1+4b223.Then, the form (23) is negative semidefinite (negative definite) if and only if the matrix(25)−1a32a2+λ02a32−λ0a12a2+λ02a12a0,is a negative semidefinite (negative definite) matrix.Corollary 2.
Given a binary quartic form,(26)fx,y=−x4+a3x3y+a2x2y2+a1xy3+a0y4.Let(27)b0=14−a12−a1a2a3+a0a32,b1=−144a0+a22+a1a3,b2=−a22,λ0=4b2+23b1+4b223.Then, the form (26) is negative semidefinite if and only if(28)λ0≥a324,−14λ03+b2λ02+b1λ0+b0≥0.In particular, the form (26) is negative definite if and only if(29)λ0>a324,−14λ03+b2λ02+b1λ0+b0>0.


## 4. Examples

The following examples illustrate the proposed method.Example 1 (see [8]).
Consider the binary quartic form(30)fx,y=x4+xy3+y4.
According to Theorem2, we have b0=−1/4, b1=1, b2=0, and further λ0=2/3. Since λ0>a32/4=0 and gλ0=163−9/36>0, we assert that the form must be positive definite (one can check that fx,y=0 has two pairs of complex conjugate roots in 1-dimensional projective space).Example 2.
Consider the binary quartic form,(31)fx,y=x4−8x3y+26x2y2−40xy3+25y4.
According to Theorem2, we have b0=1280, b1=−224, b2=13, and further λ0=56/3. Since λ0>a32/4=16 and gλ0=64/27>0, we assert that the form must be positive definite (one can check that fx,y=0 has two complex conjugate double roots).Example 3 (see [9]).
Consider the binary quartic form,(32)fx,y=x4+x3y+xy3+y4.
According to Theorem2, we have b0=−1/2, b1=3/4, b2=0, and further λ0=1. Since λ0>a32/4=1/4 and gλ0=0, we assert that the form is positive semidefinite (one can check that fx,y=0 has a real double root and a pair of complex conjugate roots).Example 4.
Consider the binary quartic form,(33)fx,y=x4+4x3y+2x2y2−4xy3+y4.
According to Theorem2, we have b0=−16, b1=4, b2=1, and further λ0=4. Since λ0=a32/4=4 and gλ0=0, we assert that the form is positive semidefinite (one can check that fx,y=0 has two real double roots).Example 5.
Consider the binary quartic form,(34)fx,y=x4+4x3y+6x2y2+4xy3+y4.
According to Theorem2, we have b0=16, b1=−12, b2=3, and further λ0=4. Since λ0=a32/4=4 and gλ0=0, we assert that the form is positive semidefinite (one can check that fx,y=0 has a quadruple root).Example 6.
Consider the binary quartic form,(35)fx,y=−x4+6x3y−13x2y2+24xy3−36y4.
According to Corollary2, we have b0=0, b1=−169/4, b2=13/2, and further λ0=13. Since λ0>a32/4=9 and gλ0=0, we assert that the form is negative semidefinite (one can check that fx,y=0 has a real double root and a pair of complex conjugate roots).


## 5. Conclusion

In this paper, we obtained a simplified necessary and sufficient condition compared with [9] for the positivity of a binary quartic form. We considered the common points of two conics instead of roots of the given quartic form and further a novel criterion for determining the positivity was derived by using the theory of the pencil of conics. Our method required neither to use the discriminant of a quartic polynomial like the literatures [3, 5, 6, 8] nor to solve a set of inequality equations like the literature [9]. The examples illustrated the correctness of our criterion.



---

*Source: 2339746-2021-11-30.xml*