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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# On the Yang-Baxter-like matrix equation for rank-two matrices

Duanmei Zhou
/ Guoliang Chen
/ Jiu Ding
• Department of Mathematics, East China Normal University, Shanghai 200241, China
• Department of Mathematics, The University of Southern Mississippi, Hattiesburg, MS 39406-5045, USA
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Published Online: 2017-03-29 | DOI: https://doi.org/10.1515/math-2017-0026

## Abstract

Let A = PQT, where P and Q are two n × 2 complex matrices of full column rank such that QTP is singular. We solve the quadratic matrix equation AXA = XAX. Together with a previous paper devoted to the case that QTP is nonsingular, we have completely solved the matrix equation with any given matrix A of rank-two.

Keywords: Rank-two matrix; Matrix equation; Jordan form

MSC 2010: 15A18; 15A24; 65F15

## 1 Introduction

Recently [1], the authors have found all the solutions of the Yang-Baxter-like matrix equation $AXA=XAX,$(1)

where the given n × n complex matrix A = PQT, with two n × 2 matrices P and Q, satisfies the assumption that the 2 × 2 matrix QTP is nonsingular. In the above situation, the eigenvalue 0 of A is semi-simple with multiplicity n − 2. This leaves the case unsolved that QTP is singular, which makes it much more challenging to solve the corresponding matrix equation. The purpose of this paper is to find all the solutions of (1) for the remaining case that the algebraic multiplicity of the eigenvalue 0 of A is more than n − 2.

The equation (1) has a similar format to the classical Yang-Baxter equation [2]. Yang [3] in 1967 first considered a one dimensional quantum mechanical many body problem with a combination of delta functions as the potential and found a factorization of the scattering matrix, and the Yang-Baxter equation was obtained as a consistence property for the factorization. Then Baxter in 1972 solved an eight-vertex model in statistical mechanics, resulting in a similar matrix equation, which, together with that from [3], was first called the Yang-Baxter equation by Russian researchers at the end of the 1970s. Since then the Yang-Baxter equation has been extensively investigated by mathematicians and physicists in knot theory, braid group theory, and quantum group theory in the past thirty years; see, e.g., [2, 4-8] and the references therein. In the past several years, the quadratic matrix equation (1) has been studied with linear algebra techniques; see, for example, the references [9-14].

Solving (1) is a tough job in general since we need to solve a large system of n2 quadratic equations with n2 variables if we multiply out its both sides. By restricting the task to only finding the solutions that commute with A, several solution results have been obtained in [10] for matrices A of special Jordan forms, and a more general result was proved in [15] for the class of diagonalizable matrices. However, no general result has been found so far for non-commuting solutions for arbitrary matrices. Thus, it is our hope to find all the solutions of (1) for general matrices A. When A is a rank one matrix, all the solutions of (1) have been found in [13]. The case of A being of rank-two turns out to be much more tedious to analyze, and some special cases have been solved in our previous paper [1]. In the current paper we continue to study the rank two case to find all the solutions for the remaining Jordan form structure of A.

Matrices A of rank at most two in the equation (1) have appeared in the classic Yang-Baxter equation (see, e.g., the references in [2, 5]). For example, the two classes of 4 × 4 matrices $a00c0bd00eb0f00aandt110t1200t110t12t210t2200t210t22$(2)

with certain values of the given parameters have been studied in [16-18] for completely integrable systems and inverse scattering problems. Each matrix in the second class is actually the tensor product TI2 of the 2 × 2 matrix T = [tij] with the 2 × 2 identity matrix I2, which constitutes a basic operation in the context of the classical Yang-Baxter equation. Our complete solutions to the quadratic matrix equation with a given rank-two matrix are expected to find applications in such physical applications. We shall pick up one matrix from (2) to apply our result in Section 5.

It is well known [10] that solving the Yang-Baxter-like matrix equation for any given matrix A is equivalent to solving the same equation with A replaced by a matrix that is similar to A, and all the solutions of the first equation are similar to those of the second one with the same similarity matrix. Thus, solving (1) for the given A can be reduced to solving the same equation with the Jordan form of A. Our approach in this paper will follow the above principle. That is, since any matrix is similar to its simplest possible canonical form J, the Jordan form of the matrix, which is called a Jordan matrix here for simplicity, is to be found. For the purpose of the present paper that will supplement the work of [1], we shall solve the following simpler Yang-Baxter-like matrix equation $JYJ=YJY$(3)

with J satisfying the conditions:

1. J is a Jordan matrix of rank-2.

2. The eigenvalue 0 of J has algebraic multiplicity at least n − 1.

If we can find the solutions of (3) for such Jordan matrices, then the solutions of (1) are available immediately for any A that is similar to J.

It turns out that the Jordan matrices J that satisfy the conditions (i) and (ii) above are $J=diag(0,Λ)$(4)

such that Λ is one of the following three matrices $Λ1≡010001000,Λ2≡01000000λ,λ≠0;Λ3≡0100000000010000$(5)

and the diagonal block 0 in (4) is either (n − 3) × (n − 3) or (n − 4) × (n − 4) accordingly. With respect to each of the three cases for J, there exists a corresponding nonsingular similarity matrix W = [w1, …, wn] connecting A to J, so that $A=PQT=WJW−1.$(6)

More specifically, in the case that Λ = Λ1, the columns w1,…, wn − 2 of W are eigenvectors of A and the columns wn − 1 and wn of W are generalized eigenvectors of A with degrees 2 and 3 respectively, namely $Awn−1≠0,A2wn−1=0,A2wn≠0,A3wn=0,$

all associated with eigenvalue 0. When Λ = Λ2, the first n − 2 columns of W are eigenvectors of A associated with eigenvalue 0, the column wn − 1 is a generalized eigenvector of degree 2 with respect to eigenvalue 0, and the column wn is an eigenvector of A associated with eigenvalue λ ≠ 0. If Λ = Λ3, then the columns w1, …, wn − 3, and wn − 1 of W are eigenvectors of A, and the columns wn − 2 and wn of W are generalized eigenvectors of degrees 2, all with respect to eigenvalue 0.

In the next three sections we look for the solutions of the Yang-Baxter-like matrix equation (1) when the corresponding Jordan matrix J of A is given by (4) with Λ = Λ1, Λ2, Λ3, respectively; such cases will be referred to as type I, type II, and type III for the given matrix A. We present two examples of our solution result in Section 5 and conclude with Section 6.

## 2 Solutions when A is of type I

In this and subsequent sections we let A = PQT in (1) with P = [p1, p2] and Q = [q1, q2] of rank-2 such that det QTP = 0. Let J be the Jordan form of A given by (4), where the diagonal zero sub-matrix 0 is either (n − 3) × (n − 3) and the sub-matrix Λ = Λ1 or Λ2 defined by (5), or the zero sub-matrix 0 is (n − 4) × (n − 4) and Λ = Λ3 in (5). As pointed out in Section 1, it is enough to find all the solutions of the equation (3) with J the Jordan form of A, so we just focus on solving (3).

Let Y be partitioned the same way as J into the 2 × 2 block matrix $Y=[MUVTK],$(7)

where M is (n − 3) × (n − 3) or (n − 4) × (n − 4), U = [u1, u2, u3] or [u1, u2, u3, u4], V = [v1, v2, v3] or [v1, v2, v3, v4], and K is 3 × 3 or 4 × 4, depending on the size of Λ. Then (3) with J partitioned by (4) is $000TΛMUVTK000TΛ=MUVTK000TΛMUVTK,$

which is equivalent to the system $UΛVT=0,UΛK=0,KΛVT=0,ΛKΛ=KΛK.$(8)

In the current section we assume that J = diag (0, Λ1), and the other two cases that J = diag (0, Λ2) and J = diag (0, Λ3) will be investigated in Sections 3 and 4, respectively. So we solve (8) with Λ = Λ1. Because of the special zero structure of Λ1, the two unknown vectors u3 and v1 actually do not appear at all in the above system, so they always appear as free variables in the solutions. In addition, the first equation is independent of K and is in fact $u1v2T+u2v3T=0.$

The last equation of (8), $ΛKΛ=KΛK,$(9)

is itself a Yang-Baxter-like matrix equation of small size when Λ = Λj with j = 1, 2, 3, and finding all of its solutions is the first step for solving (8).

#### Lemma 2.1

The solutions K of the equation (9) with Λ = Λ1 are $K1=xyz00−cyx00c,x≠0;K2=00z00g00c,c≠0;K3=0yz00g000.$

#### Proof

Write $K=xyzefgabc.$

Then, the matrix equation Λ1KΛ1 = KΛ1K becomes $0ef0ab000=ex+ayfx+bygx+cye2+afef+bfeg+cfae+abaf+b2ag+bc.$

Note that there is no z in the above equation, so all the solutions have z as an arbitrary parameter. The two equations ef + bf = a and ae + ab = 0 imply a = 0, so b = 0 from 0 = af + b2 = b2 and e = 0 since 0 = e2 + af = e2. Thus we have the remaining three equations $fx=0,gx+cy=f,cf=0$

from which f = 0. Hence gx + cy = 0. If x ≠ 0, then g = −cy/x, which gives the first matrix K1 in the lemma. When x = 0, we have cy = 0. So c ≠ 0 or c = 0, resulting in the other two matrices K2 and K3, respectively. □

By Lemma 2.1, all the solutions of the last equation of (8) are K1, K2, and K3. Substituting such matrices into the first three equations of (8) and solving them respectively, we obtain the following result.

#### Theorem 2.2

Suppose A = PQT is such that Λ = Λ1 in its Jordan form (4). Then all the solutions of (1) are X = WYW−1, where W is as given by (6) and Y is partitioned as (7) in which M is an arbitrary (n − 3) × (n − 3) matrix such that Y = $Mu1yxu1u3v1Txyz−yxv3T00−yxcv3T00c,Mu1u2u3v1Txyz0T0000T000,x≠0,u2≠yxu1,$(10) $M00u3v1T00zv2T00gv3T00c,Mu1−gcu1u3v1T00zgcv3T00gv3T00c,c≠0,u1≠0,$(11) $M00u3v1T00zv2T00gv3T000,M0u2u3v1T0yzv2T00g0T000,$(12) $Mu1u2u3v1T00z−u1Hu2∥u1∥2v3T000v3T000,Mu1u2u3v1T0yz0T0000T000,y≠0,u1≠0.$(13)

Here uH is the conjugate transpose of u.

#### Proof

We just solve the first three equations of (8) with K = K1, K2, and K3 in succession. When K = K1, those equations of (8) are $(u2−yxu1)v3T=0,(u2−yxu1)c=0,v2=−yxv3$

after simplification. From the first equation, u2yu1/x = 0 or v3 = 0. If the former is satisfied, then u2 = yu1/x and the second equation is satisfied. This gives the first solution matrix of (10). In the case that v3 = 0, either u2 = yu1/x or c = 0 from the second equation. The former case still leads to the first matrix of (10), while the latter gives the second matrix of (10).

Suppose K = K2. Then K2Λ1VT = 0, so (8) is reduced to $u1v2T+u2v3T=0,gu1+cu2=0.$(14)

Thus z is a free variable in the solutions. Since c ≠ 0, (14) is equivalent to u2 = −gu1/c and u1(v2gv3/c)T = 0. Letting u1 = 0 gives the first matrix of (11), the second matrix of (11) is the consequence of u1 ≠ 0, and v2gv3/c = 0.

If K = K3, then (8) becomes $u1v2T+u2v3T=0,gu1=0,yv3T=0.$(15)

Letting u1 = 0, u2 = 0, and y = 0 gives the first matrix of (12), and the choice of u1 = 0 and v3 = 0 produces the second matrix of (12). If u1 ≠ 0, then the first equation of (15) implies that ${v}_{2}=-\left({u}_{1}^{H}{u}_{2}\right){v}_{3}/\parallel {u}_{1}{\parallel }^{2},$ and the other two equations give that either g = 0 and y = 0, from which we get the first matrix of (13), or g = 0 and y ≠ 0, which gives the second one. □

## 3 Solutions when A is of type II

We now consider the second case that the Jordan form of the matrix A is J = diag (0, Λ2), so we solve (8) with Λ = Λ2. Clearly, the structure of Λ makes u2 and v1 free vectors in all the solutions of (8), and the first equation of (8) is now $u1v2T+λu3v3T=0.$

#### Lemma 3.1

The solutions K of the equation (9) with Λ = Λ2 are $K4≡0y00f00b0;K5≡0yz0f0000,z≠0;K6≡xyz0−λbzx00b0,x≠0;K7≡λyz00g000,g≠0;K8≡0y00f000λ;K9≡xy000000λ,x≠0;K10≡0y00λ0ab0,a≠0.$

#### Proof

The equation Λ2KΛ2 = KΛ2K now becomes $0eλg0000λaλ2c=ex+λazfx+λbzgx+λcze2+λagef+λbgeg+λcgae+λacaf+λbcag+λc2.$(16)

Since the unknown y does not appear in the above, it is a free variable in all the solutions. To solve for all the other unknowns, let a = 0 first. Then e = 0 from e2 + λag = 0. Because of the equation λ2c = ag + λc2, there are two possibilities that c = 0 or c = λ. Assume first that c = 0. Since a = 0 and e = 0, the system (16) is reduced to $fx=−λbz,bg=0,g(x−λ)=0.$

If g = 0, then the above is just fx = −λbz. When x = 0, we get K4 and K5, and x ≠ 0 implies K6. In the case that g ≠ 0, we obtain K7. The other possibility of c = λ implies K8 and K9.

Now let a ≠ 0. Then e = −λ c via equating entries (3, 1) of the both sides of (16). Since ag = −e2/λ from comparing the (2, 1) entries, it follows from equating entries (3, 3) in (16) that $λ2c=ag+λc2=−(−λc)2/λ+λc2=0.$

So c = 0 and then e = 0. By equating (3, 2) entries, we have f = λ. Also z = 0 and g = 0 from comparing entries (1, 1) and (2, 1) of (16). Finally x = 0 via comparing the entries (1, 2) of (16), thus arriving at the corresponding solution matrix K10. □

Lemma 3.1 gives all the solutions K4, …, K10 of the last equation of (8) with Λ = Λ2. Substituting them for K in the system and solving the resulting three equations in succession, we are lead to the next theorem.

#### Theorem 3.2

Suppose A = PQT is such that its Jordan form is given by (4) with Λ = Λ2. Then all the solutions of (1) are X = WYW−1, where W is given by (6) and Y is partitioned as (7) in which M is an arbitrary (n − 3) × (n − 3) matrix such that Y= $M0u20v1T0y0v2T0f0v3T0b0,Mu1u20v1T0y00T000v3T0b0,Mu1u2u3v1T0y0v2T0f0−u3Hu1λ∥u3∥2v2T0−fu3Hu1λ∥u3∥20,$(17) with u3 ≠ 0,

$M0u2u3v1T0yzv2T0f00T000,Mu1u2u3v1T0yz0T0000T000,z≠0,u1≠0,$(18) $Mu1u2u3v1Txyz0T0000T000,Mu1u2zxu1v1Txyz−λzxv3T0−λbzx0v3T0b0,M0u2u3v1Tλyz0T00g0T000$(19)

with x ≠ 0 and g ≠ 0, $M0u20v1T0y0v2T0f00T00λ,Mu1u20v1T0y00T0000T00λ,u1≠0,$(20) $Mu1u20v1Txy00T0000T00λ,M0u20v1T0y00T0λ0v3Tab0,x≠0,a≠0.$(21)

#### Proof

When K = K4, the system (8) is reduced to $u1v2T+λu3v3T=0,fu1+λbu3=0,$

which does not contain y so it will appear in all the solutions. Letting u3 = 0. Then v3 and b can be arbitrary, and the above system becomes ${u}_{1}{v}_{2}^{T}=0$ and fu1 = 0. If u1 = 0, then v2 and f are arbitrary, giving the first solution matrix of (17). If v2 = 0, then u1 is arbitrary and f = 0, which results in the second matrix of (17). Now let u3 ≠ 0. Then from the above system, $v3T=−u3Hu1λ∥u3∥2v2T,b=−fu3Hu1λ∥u3∥2.$

This gives the third matrix of (17).

Next suppose K = K5. Then (8) is simply ${u}_{1}{v}_{2}^{T}=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f\phantom{\rule{thinmathspace}{0ex}}{u}_{1}=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{v}_{3}=0$ since λ ≠ 0 and z ≠ 0. All the solutions are those in (18). Assume K = K6. Then (8) can be written as $u1v2T+λu3v3T=0,bu3−bzxu1=0,xv2+λzv3=0.$

Since x ≠ 0, we can solve v2 out from the last equation and substitute it into the first one in the above system, getting $(u3−Zxu1)v3T=0,(u3−Zxu1)b=0,v2=−λzxv3.$

Letting b = 0 and v3 = 0 above gives the first matrix of (19), and if u3 = zu1/x, then b and v3 are arbitrary, giving the second one of (19).

For K = K7, the system is simplified to ${u}_{1}{v}_{2}^{T}+\lambda {u}_{3}{v}_{3}^{T}=0,{u}_{1}=0,{v}_{2}+z{v}_{3}^{T}=0,{v}_{3}=0$ since g ≠ 0, whose solutions are given by the third matrix of (19). When K = K8, the system (8) is actually u3 = 0, v3 = 0, ${u}_{1}{v}_{2}^{T}=0,\phantom{\rule{thinmathspace}{0ex}}f\phantom{\rule{thinmathspace}{0ex}}{u}_{1}=0$ since λ ≠ 0. The solutions are the two matrices of (20). With K = K9, the system becomes ${v}_{2}=0,{v}_{3}=0,{u}_{1}{v}_{2}^{T}+\lambda {u}_{3}{v}_{3}^{T}=0,{u}_{3}=0$ since λ ≠ 0 and x ≠ 0, so the first matrix of (21) is obtained. Finally, as K = K10, we have ${u}_{1}=0,{v}_{2}=0,{u}_{1}{v}_{2}^{T}+\lambda {u}_{3}{v}_{3}^{T}=0,{u}_{3}=0$ since a ≠ 0, thus obtaining the second matrix of (21). □

## 4 Solutions when A is of type III

Unlike the previous two cases that involve only 3 × 3 matrices Λ1 and Λ2, the third case that we shall study involves the 4 × 4 matrix Λ3. The zero structure of Λ3 ensures that u2, u4, v1,v3 are not present in the equations, so they are free vectors in the solutions. The following lemma gives all the solutions of the last equation of (8).

#### Lemma 4.1

The solutions K of the equation (9) with Λ = Λ3 are $K11≡0a120a140a220a240a320a340a420a44;K12≡0a120a140a220a240a32a33a340000,a33≠0;K13≡0a12a13a140a220a240a32a33a340000,a13≠0;K14≡0a12a13a140000a31a32a33a340000,a13a31≠0;K15≡0a120a140−a33a42a310−a33a44a31a31a32a33a340a420a44,a31≠0;K16≡a11a12a13a140−a13a42a110−a13a44a11a31a32a13a31a11a340a420a44,a11≠0;K17≡a11a12a13a140000a31a32a33a340000,a11≠0,a11a13a31a33≠0.$

#### Proof

Denote $K=a11a12a13a14a21a22a23a24a31a32a33a34a41a42a43a44.$

Then the equation K Λ3 K = Λ3KΛ3 is just $a11a21+a13a41a11a22+a13a42a11a23+a13a43a11a24+a13a44a21a21+a23a41a21a22+a23a42a21a23+a23a43a21a24+a23a44a31a21+a33a41a31a22+a33a42a31a23+a33a43a31a24+a33a44a41a21+a43a41a41a22+a43a42a41a23+a43a43a41a24+a43a44=0a210a2300000a410a430000.$(22)

Comparing the (2,1),(4,3),(2,3),(4,1) entries above gives the four equations $a212=−a23a41,a432=−a23a41,a23(a21+a43)=0,a41(a21+a43)=0.$(23)

We discuss two cases separately. First assume a41 = 0. Then a21 = a43 = 0 from (23). If a23 ≠ 0, then a11 = a31 = a44 = 0 from equating entries (1,3),(3,3),(2,4) of both sides in (22), from which a23 = a11 a24 + a13a44 = 0, a contradiction. Thus a23 = 0 and (22) is reduced to $a11a22+a13a42=0,a31a22+a33a42=0,a11a24+a13a44=0,a31a24+a33a44=0$(24)

with a21 = a23 = a41 = a43 = 0.

Now we assume a41≠ 0. Then the last equation of (23) implies that a21 = −a43. On the other hand, the entries (4,2) and (1,1) equations of (22) gives that a22 = −a43a42/a41 and a13 = −a11a21/a41. Hence $a21=a11a22+a13a42=−a11a43a42a41−a11a21a42a41=a11a21a42a41−a11a21a42a41=0,$

and so a43 = 0. On the other hand, the entries (3,1) and (4,2) equalities give a33 = a22 = 0, so a41 = a31a22 + a33a42 = 0 from the entry (3,2) equality of (22), contradictory to the assumption that a41≠ 0.

Therefore, (24) is the only equation for us to solve. Let a11 = 0. Then (24) becomes a13a42 = 0,a31a22 + a42 = 0,a13a44 = 0, a31a24 + a33a44 = 0. If a13 = 0, then the above system is a31a22 + a33a42 = 0, a31a24 + a33a44 = 0. The first case is a31 = 0. Then a33a42 = 0 and a33a44 = 0. If a33 = 0, then we have solution matrix K11; otherwise K12 occurs. For the second case a31 ≠ 0, there hold a22 = −a33a42/a31 and a24 = −a33a44/a31, giving K15. If a13 ≠ 0, then a42 = 0 and a44 = 0, so a31a22 = 0, a31a24 = 0. The case that a31 = 0 implies K13 and when a31 ≠ 0, we obtain K14.

Now assume that a11 ≠ 0. Then a22 = −a13a42/a11 and a24= −a13a44/a11, so a31a22 + a33a42 = 0 and a31a24 + a33a44 = 0, or equivalently, $a42a33−a13a31a11=0,a44a33−a13a31a11=0.$

In the case that a33 = a13a31/a11, the numbers a42 and a44 are arbitrary and the matrix K16 appears, and otherwise a42 = a44 = 0, so a22 = a24 = 0, resulting in K17. This completes the proof. □

Lemma 4.1 gives all the solutions K11, …, K17 of the last equation of (8) with Λ = Λ3, which can be used to prove the next result.

#### Theorem 4.2

Suppose Λ = Λ3 in the Jordan form (4) of A = PQT. Then all the solutions of (1) are X = WYW−1, where Y is partitioned as (7) in which M is an arbitrary (n − 4) × (n − 4) matrix such that Y = $M0u20u4v1T0a120a14v2T0a220a24v3T0a320a34v4T0a420a44,M0u2u3u4v1T0a120a14v2T0a220a24v3T0a320a340T0000,u3≠0,$(25) $Mu1u2u3u4v1T0a120a14−u1Hu3∥u1∥2v4T0−u1Hu3∥u1∥2a420−u1Hu3∥u1∥2a44v3T0a320a34v4T0a420a44,u1≠0,$(26) $M0u2u3u4v1T0a120a14v2T0a220a24v3T0a32a33a340T0000,Mu1u2u3u4v1T0a120a140T0000v3T0a32a33a340T0000,u1≠0,a33≠0,$(27) $M0u2u3u4v1T0a12a13a14v2T0a220a24v3T0a32a33a340T0000,Mu1u2u3u4v1T0a12a13a140T0000v3T0a32a33a340T0000,u1≠0,a13≠0,$(28) $Mu1u2u3u4v1T0a12a13a140T0000v3Ta31a32a33a340T0000,Mu1u2a33a31u1u4v1T0a120a14−a33a31v4T0−a33a42a310−a33a44a31v3Ta31a32a33a34v4T0a420a44$(29) with a13 ≠ 0; a31 ≠ 0, $Mu1u2u3u4v1T0a120a140T0000v3Ta31a32a33a340T0000,Mu1u2u3u4v1Ta11a12a13a140T0000v3Ta31a32a13a31a11a340T0000$(30)

with a31 ≠ 0, u3a33u1/a31 in the left matrix, and a11 ≠ 0, u3a13u1/a11 in the second one, $Mu1u2a13a11u1u4v1Ta11a12a13a14−a13a11v4T0−a13a42a110−a13a44a11v3Ta31a32a13a31a11a34v4T0a420a44,Mu1u2u3u4v1Ta11a12a13a140T0000v3Ta31a32a33a340T0000$(31)

with a11 ≠ 0,a33a13a31/a11.

#### Proof

Clearly the system (8) does not involve u2, u4, v1,v3, so they are free vectors in all the solutions. The first equation of (8) is ${u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}=0.$ Now we solve the first three equations of (8) with K = K11, …, K17 separately.

When K = K11, the system (8) is reduced to $u1v2T+u3v4T=0,a22U1+a42u3=0,a24u1+a44u3=0.$

Assume that u1 = 0. Then u3v4T = 0, a42u3 = 0, a44u3 = 0. If we let u3 = 0, then v4, a42, a44 can be arbitrary, getting the first matrix of (25). Otherwise v4 = 0, a42 = a44 = 0, which results in the second matrix in (25). Now assume that u1 ≠ 0. Then ${v}_{2}=-\left({u}_{1}^{H}{u}_{3}\right){v}_{4}/\parallel {u}_{1}{\parallel }^{2},{a}_{22}=-\left({u}_{1}^{H}{u}_{3}\right){a}_{42}/\parallel {u}_{1}{\parallel }^{2},{a}_{24}=-\left({u}_{1}^{H}{u}_{3}\right){a}_{44}/\parallel {u}_{1}{\parallel }^{2}.$ This gives (26).

For K = K12, (8) becomes $u1v2T+u3v4T=0,a22U1=0,a24U1=0,a33v4T=0.$

Since a33≠ 0, we have v4 = 0, so the above is simplified to ${u}_{1}{v}_{2}^{T}=0,{a}_{22}{u}_{1}=0,{a}_{24}{u}_{1}=0.$ Letting u1 = 0 gives the first matrix of (27), and otherwise we have v2 = 0, a22 = a24 = 0, leading to the second matrix in (27).

If K = K13, then (8) is simplified to $v4=0,u1v2T=0,a22U1=0,a24u1=0.$

u1 = 0 produces the left matrix of (28) and otherwise, v2 = 0, a22 = a24 = 0, so the second matrix in (28).

With K = K14, we have $u1v2T+u3v4T=0,a13v4T=0,a31v2T+a33v4T=0.$

Since a13 ≠ 0 and a31 ≠ 0, we have v2 = v4 = 0, and the first matrix of (29) is obtained.

The choice of K = K15 gives the system $u1v2T+u3v4T=0,a42u3−a33a31u1=0,a44u3−a33a31u1=0,a31v2T+a33v4T=0.$

Since a31 ≠ 0, from the last equation, v2 = −a33v4/a31. Substituting into the first equation, we obtain (u3a33u1/a31) ${v}_{4}^{T}=0.$ So if u3a33u1/a31 = 0, then v4, a42, a44 are arbitrary and the second matrix of (29) is true. Otherwise, v4 = 0,a42 = a44 = 0, which gives the first matrix of (30).

In the case K = K16, (8) can be written as $u1v2T+u3v4T=0,a42u3−a13a11u1=0,a44u3−a13a11u1=0,a11v2T+a13v4T=0,a31v2T+a13a11v4T=0.$

By the forth equation, v2 = −a13v4/a11. Substituting into the first one gives $\left({u}_{3}-{a}_{13}{u}_{1}/{a}_{11}\right){v}_{4}^{T}=0.$ Hence, if U3a13u1/a11, then v4 = 0, a42 = a44 = 0 and we get the right matrix of (30). Otherwise v4, a42 a44 are arbitrary, so the left matrix of (31).

The last case is K = K17. Then $u1v2T+u3v4T=0,a11v2T+a13v4T=0,a31v2T+a33v4T=0,$

the solutions of which are given by the last matrix of (31). □

## 5 Examples

We give two examples to illustrate our results. The first one is artificial for the use of Theorem 3.1. Let p1 = [0, 1, 1, 0]T, p2 = [0, 0, 0, 1]T, q1 = [1, −1, 1, 0]T, and q2 = [−1, 1, −1, 1]T, so that $A=p1q1T+p2q2T=00001−1101−110−11−11=WJW−1=100011000110001100000010000000011000−11001−110−11−11.$

By Theorem 3.1 and multiplying WYW−1 out, all the solutions of (1) are $X=m+u2−u2u20m+v1+u2+y−u2−yu2+y0v1+v2+y+f−y−fy+f0v2+v3+f+b−f−bf+b0,X=m−u1+u2u1−u2u20m+v1−u1+u2+yu1−u2−yu2+y0v1+y−yy0v3+b−bb0,X=m−u1+u2−u3u1−u2+u3u2−u3u3m+v1−u1+u2+y−u3u1−u2−y+u3u2+y−u3u3v1+v2+y+f−y−fy+f0(v2+f)(1−u1/u3)f(u1/u3−1)f(1−u1/u3)0,u3≠0,X=m+u2−u3u3−u2u2−u3u3m+v1+u2+y−u3−zu3+z−u2−yu2+y−u3−zu3+zv1+v2+y+f−zz−y−fy+f−zzv2+f−ff0,z≠0,X=m−u1+u2−u3u1−u2+u3u2−u3u3m+v1−u1+u2+y−u3−zu1−u2−y+u3+zu2+y−u3−zu3+zv1+y−zz−yy−zz0000,u1≠0,z≠0,X=m−u1+u2−u3u1−u2+u3u2−u3u3m+v1−u1−x+u2+y−u3−zu1+x−u2−y+u3+zu2+y−u3−zu3+zv1−x+y−zx+z−yy−zz0000,x≠0,X=m−u1+u2−zxu1u1−u2+zxu1u2−zxu1zxu1m+v1−u1−x+u2+y−zxu1−zu1+x−u2−y+zxu1+zu2+y−zxu1−zzxu1+zv1−zxv3−x+y+zxb−zx+z−y+zxby−z−zxbzv3(1−zx)−zxbzxb−zxb0,x≠0,X=m+u2−u3u3−u2u2−u3u3m+v1−1+u2+y−u3−z1−u2−y+u3+zu2+y−u3−zu3+zv1−1+y−z−g1+z−y+gy−z−gz+g−gg−gg,g≠0,X=m+u2−u2u20m+v1+u2+y−u2−yu2+y0v1+v2+y+f−y−fy+f0v2+f−11−ff−11,X=m−u1+u2u1−u2u20m+v1−u1+u2+yu1−u2−yu2+y0v1+y−yy0−11−11,u1≠0,X=m−u1+u2u1−u2u20m+v1−u1−x+u2+yu1+x−u2−yu2+y0v1−x+yx−yy0−11−11,x≠0,X=m+u2−u2u20m+v1+u2+y−u2−yu2+y0v1+y+1−y−1y+10v3−a+b+1a−b−1b+10,a≠0.$

The matrix in our second example has appeared in the application of the classical Yang-Baxter equation to the inverse scattering theory [18]. Let p1 = [1, 0, 0, 0]T, p2 = [0, 1, 0, 0]T, q1 = [0, 0, 1, 0]T, and q2 = [0, 0, 0, 1]T, so $A=p1q1T+p2q2T=0010000100000000=WJW−1=100000100100000101000000000100001000001001000001.$

Note that A is a member of the second matrix class in (2) as the tensor product J0I2 of the 2 × 2 Jordan block J0 with eigenvalue 0 and I2. Since J = Λ3 in this example, all the solutions Y of (3) are exactly K11, …, K17 as given by Lemma 4.1. Multiplying X = WYW−1 = WKkW out with k = 11, …, 17, we find all the solutions of (1): $X=00a12a1400a32a3400a22a2400a32a34;00a12a140a33a32a3400a22a240000,a33≠0;X=0a13a12a140a33a32a3400a22a240000,a13≠0;0a13a12a14a31a33a32a3400000000,a13a31≠0;X=00a12a14a31a33a32a3400−a33a42a31−a33a44a3100a42a44,a31≠0;X=a11a13a12a14a31a13a31a11a32a3400−a13a42a11−a13a44a1100a42a44,a11≠0;X=a11a13a12a14a31a33a32a3400000000,a11≠0,a11a13a31a33≠0.$

## 6 Conclusions

As continuation of our earlier paper in which all the solutions of the Yang-Baxter-like matrix equation (1) were found when A = PQT, with P and Q being n × 2 and of rank-two such that QTP is nonsingular, we have found in this paper all the solutions of (1) when QTP is singular. For each of the resulting three kinds of Jordan forms of A we solved the corresponding simplified matrix equation, thus obtaining all the structures of the solution matrices.

The various structures of the solution matrices reflect the complicated structure of the algebraic varieties as solutions of polynomial systems of multi-variable. Although we were not able to use algebraic geometry to find all the solutions of (1) in the general case, we have been successful in finding all the solutions when the rank of A is at most 2. Our future work will be devoted to the exploration of solving the Yang-Baxter-like matrix equation with a given matrix A of rank-k.

## Acknowledgement

This work is supported by the National Natural Science Foundation of China (Nos. 11501126, 11471122, 11661007, 11661008), the Youth Natural Science Foundation of Jiangxi Province (Nos. 20151BAB211011, 20161BAB211025), the Science Foundation of Jiangxi Provincial Department of Education (Nos. GJJ150979, GJJ150982), the 111 Project, and the Foundation for Supporting Local Colleges and Universities of China— Applied Mathematics Innovative Team Building.

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Accepted: 2017-02-07

Published Online: 2017-03-29

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 340–353, ISSN (Online) 2391-5455,

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