# All Solutions of the Yang–Baxter-Like Matrix Equation for Nilpotent Matrices of Index Two

Authors: Duanmei Zhou; Jiawen Ding
Journal: Complexity
 (2020)

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## Abstract

LetA be a nilpotent matrix of index two, and consider the Yang–Baxter-like matrix equation AXA=XAX. We first obtain a system of matrix equations of smaller sizes to find all the solutions of the original matrix equation. When A is a nilpotent matrix with rank 1 and rank 2, we get all solutions of the Yang–Baxter-like matrix equation.


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## Body

## 1. Introduction

We are interested in finding all solutions of the quadratic matrix equation:(1)AXA=XAX,where the given A and the unknown X are n×n square complex matrices with n≥2. The above equation (1) is called the Yang–Baxter-like matrix equation because it is similar to the classical parameter-free Yang–Baxter equation [1, 2]. The Yang–Baxter equation was first introduced by Yang in 1967 and then by Baxter in 1972 in the study of statistical mechanics. The Yang–Baxter equation has been extensively researched by mathematicians and physicists in knot theory, braid group theory, and so on [3–6].Obviously, the Yang–Baxter matrix equation has two trivial solutionsX=0 and X=A. However, we are interested in finding nontrivial solutions. Finding all solutions of equation (1) is a hard work for a general matrix A. Indeed, we can reformulate (1) into a system of polynomial equations, so it is equivalent to solving a system of n2 quadratic polynomial equations in n2 variables. To find all solutions is not an easy task even if for a 3×3 matrix [7]. Most solutions obtained so far are commuting ones for particular choices of matrices A. See, for example, [8] for diagonalizable matrix and [9, 10] for nilpotent matrix. In [11], infinitely many solutions of (1) were obtained for any semisimple eigenvalues of the given matrix. A family of commuting solutions of (1) were constructed for those eigenvalues of A that are non-semisimple in [12, 13]. Some researchers have also proposed some numerical methods for finding commuting solutions. For instance, in [14], when A is a nonsingular matrix such that its inverse is a stochastic matrix, Ding and Rhee found nontrivial solutions of (1) via Brouwer’s fixed point theorem. In [15], when A is a diagonalisable matrix, the authors proposed numerical methods to calculate solutions of (1) by applying the mean ergodic theorem. When A is a low rank matrix, all solutions of (1) have been found in [16–18] for the noncommuting case. In [19], the authors have obtained explicit solutions when A is an idempotent matrix. However, for a general matrix A, it is difficult to characterize and determine all the solutions of (1), even if for nilpotent matrix.The purpose of this paper is to find all the solutions of equation (1) under the assumption that A is a nilpotent matrix of index 2. We first give a system of matrix equations of smaller sizes to find all solutions of the original matrix equation in Section 2. In the next two sections, we study all solutions for AXA=AXA when A is rank one and rank two, respectively. Finally, we present two examples of our solution result in Section 5 and conclude with Section 6.


## 2. All Solutions for the Nilpotent Matrix of Index 2

LetA≠0 be a nilpotent matrix with index 2. Clearly, matrix A has 0 as its only eigenvalue and the minimal polynomial of A is mt=t2. So, we can write the Jordan canonical form J of A as(2)J=0J0⋱J0,in which the 2×2 Jordan block(3)J0=0100appears k times with k the rank of A, and 0 denotes the n−2k×n−2k zero matrix. So, there exists a nonsingular matrix W such that(4)W−1AW=J≡diag0,J0,…,J0.Lemma 1.
Let two matricesA and X satisfy AXA=XAX. Then, for any nonsingular matrix U, the matrices B=U−1AU and Y=U−1XU satisfy(5)BYB=YBY.Conversely, ifY satisfies the above equations for a given B=U−1AU, then X=UYU−1 satisfies AXA=XAX.According to Lemma1, we know that solving (1) can be reduced to solving the simplified matrix equation:(6)YJY=JYJ.So, solutionX to (1) can be expressed as X=WYW−1, where W satisfies A=WJW−1. Thus, in our analysis below, we find all solutions of (6). Let Y be partitioned the same way as J into the k+1×k+1 block matrix:(7)Y=MQ1⋯QkP1Y11⋯Y1k⋮⋮⋱⋮PkYk1⋯Ykk,where M is n−2k×n−2k, Qi=q2i−1,q2i∈ℂn−2k×2 and Pi=p2i−1,p2iT∈ℂ2×n−2k for i=1,…,k, and(8)Yij=bijcijsijtij,i,j=1,2,…,k.Then, the equationJYJ=YJY becomes(9)00⋯00J0Y11J0⋯J0Y1kJ0⋮⋮⋱⋮0J0Yk1J0⋮J0YkkJ0=∑i=1kQiJ0Pi∑i=1kQiJ0Yi1⋯∑i=1kQiJ0Yik∑i=1kY1iJ0Pi∑i=1kY1iJ0Yi1⋯∑i=1kY1iJ0Yik⋮⋮⋱⋮∑i=1kYkiJ0Pi∑i=1kYkiJ0Yi1⋯∑i=1kYkiJ0Yik,which is equivalent to the system(10)∑i=1kQiJ0Pi=0,∑i=1kYjiJ0P=0,j=1,2,…,k,i∑i=1kQiJ0Yij=0,j=1,2,…,k,∑i=1kYliJ0Yij=J0YljJ0,l,j=1,2,…,k.Note that the above system does not containM. This means that M is arbitrary. Because finding all solutions of (10) is very difficult, we focus on finding all solutions of (10) when k=1 and k=2.


## 3. All Solutions for the Nilpotent Matrix of Index 2 with Rank 1

Whenk=1, this means that the rank of A is 1. Then,(11)J=000J0,where 0 is the n−2×n−2 zero matrix. Let Y be partitioned the same way as J into the 2×2 block matrix:(12)Y=MQPZ,where M∈ℂn−2×n−2, Q=q1q2∈ℂn−2×2, P=p1p2T∈ℂ2×n−2, and Z∈ℂ2×2. According to (4), we obtain(13)QJ0P=0,QJ0Z=0,ZJ0P=0,ZJ0Z=J0ZJ0.Obviously, the last equation of (13) is also a Yang–Baxter matrix equation. Next, we solve the last equation in (13).Lemma 2.
The solutionsZ of the equation ZJ0Z=J0ZJ0 are(14)Z1=0z20z4,z4≠0,Z2=0z200,Z3=z1z200,z1≠0.Proof.
Let(15)Z=z1z2z3z4.
Then, the equationZJ0Z=J0ZJ0 becomes(16)z1z2z3z40100z1z2z3z4=0100z1z2z3z40100.
That is,(17)z1z3z1z4z32z3z4=0z300.
This leads toz3=0 and z4z1=0, and z2 is arbitrary.
By Lemma2, all solutions of the last equation of (13) are Z1, Z2, and Z3. Substituting such matrices into the first three equations of (13), we obtain the following result.Theorem 1.
Suppose thatA is a nilpotent matrix with rank 1. Then, all solutions of (1) are X=WYW−1, where Y is partitioned as (12) in which M is an arbitrary n−2×n−2 matrix such that Y=(18)M0q2p1T0z2p2T0z4,z4≠0,(19)M0q2p1T0z2p2T00,Mq1q2p1T0z2000,(20)Mq1q2p1Tz1z2000,z1≠0.Proof.
WhenZ=Z1, then ZJ0=0. So, we just need to solve the first two equations in (13). The second equation (13) is QJ0Z=0. That is,(21)0q1z4=0.
Sincez4≠0, we have q1=0. Thus, QJ0=0. This gives (18).
LetZ=Z2. Then, ZJ0=0 and J0Z=0. So, the second and the third equations of (13) are satisfied. Next, we only have to solve the first equation of (13), i.e.,(22)QJ0P=q1q20100p1Tp2T=q1p2T=0,from which q1=0, p2 is arbitrary or p2=0, q1 is arbitrary. This gives (19).
IfZ=Z3, then J0Z=0. So, the second equation of (13) is satisfied. From the first equation ZJ0P=0, we obtain(23)z1p2T0=0,which leads to z1p2T=0. Since z1≠0, we obtain p2=0. Thus, J0P=0. This gives (20).


## 4. All Solutions for the Nilpotent Matrix of Index 2 with Rank 2

Whenk=2, this means(24)J=0000J0000J0,where 0 is the n−4×n−4 zero matrix. Let Λ=J000J0. Then,(25)J=000Λ.LetY be partitioned the same way as J into the 2×2 block matrix:(26)Y=MUVTK,where M is n−4×n−4, U=u1,u2,u3,u4∈ℂn−4×4, V=v1,v2,v3,v4,∈ℂn−4×4, and K is 4×4. Then, JYJ=YJY is equivalent to the system:(27)UΛVT=0,UΛK=0,KΛVT=0,ΛKΛ=KΛK.All solutions of this system were obtained in [18]. We have the following results.Lemma 3 (see Lemma 4.1 in [18]).
The solutionsK of the equation ΛKΛ=KΛK are(28)K1=0a120a140a220a240a320a340a420a44,K2=0a120a140a220a240a32a33a340000,a33≠0,K3=0a12a13a140a220a240a32a33a340000,a13≠0,K4=0a12a13a140000a31a32a33a340000,a13a31≠0,K5=0a120a140−a33a42a310−a33a44a31a31a32a33a340a420a44,a31≠0,K6=a11a12a13a140−a13a42a110−a13a44a11a31a32a13a31a11a340a420a44,a11≠0,K7=a11a12a13a140000a31a32a33a340000,a11≠0,a11a13a31a33≠0.Theorem 2 (see Theorem 4.2 in [18]).
SupposeA is a nilpotent matrix with rank 2, and is partitioned as (4). Then, all solutions of (1) are X=WYW−1, where Y is partitioned as (10) in which M is an arbitrary n−4×n−4 matrix such that Y=(29)M0u20u4v1T0a120a14v2T0a220a24v3T0a320a34v4T0a420a44,M0u2u3u4v1T0a120a14v2T0a220a24v3T0a320a340T0000,u3≠0,Mu1u2u3u4v1T0a120a14−u1Hu3u12v4T0−u1Hu3u12a420−u1Hu3u12a44v3T0a320a34v4T0a420a44,u1≠0,M0u2u3u4v1T0a120a14v2T0a220a24v3T0a32a33a340T0000,Mu1u2u3u4v1T0a120a140T0000v3T0a32a33a340T0000,u1≠0,a33≠0,M0u2u3u4v1T0a12a13a14v2T0a220a24v3T0a32a33a340T0000,Mu1u2u3u4v1T0a12a13a140T0000v3T0a32a33a340T0000,u1≠0,a13≠0,Mu1u2u3u4v1T0a12a13a140T0000v3Ta31a32a33a340T0000,Mu1u2a33a31u1u4v1T0a120a14−a33a31v4T0−a33a42a310−a33a44a31v3Ta31a32a33a34v4T0a420a44,a13≠,a31≠0,Mu1u2u3u4v1T0a120a140T0000v3Ta31a32a33a340T0000,Mu1u2u3u4v1Ta11a12a13a140T0000v3Ta31a32a13a31a11a340T0000,with a13≠0, u3≠a33u1/a31 in the left matrix, and a11≠0, u3≠a13u1/a11 in the second one,(30)Mu1u2a13a11u1u4v1Ta11a12a13a14−a13a11v4T0−a13a42a110−a13a44a11v3Ta31a32a13a31a11a34v4T0a420a44,Mu1u2u3u4v1Ta11a12a13a140T0000v3Ta31a32a33a340T0000,a11≠0,a33≠a13a31a11.


## 5. Examples

We give two examples to illustrate our results.Example 1.
Let(31)A=101101−10−1.
Then, there exists(32)W=11001−1−1−11,J=000001000,such that A=WJW−1. By Theorem 1, we obtain all solutions of (1) as(33)X=q2+z2−m−p1−m−p1+q2+z2z2−z4−p1+p2−p1+p2+z2−z4−q2−z2+z4m+p1−p2m+p1−p2−q2−z2+z4,z4≠0,X=q2+z2−m−p1−m−p1+q2+z2z2−p1+p2−p1+p2+z2−q2−z2m+p1−p2m+p1−p2−q2−z2,X=q1+q2+z2q1−p1−mq1−p1−m+q2+z2z2−p1z2−p1−q1−q2−z2m+p1−q1m+p1−q1−q2−z2,X=q1+q2+z1+z2q1−p1−m+z1q1−p1−m+q2+z1+z2z1+z2z1−p1z1−p1+z2−q1−q2−z1−z2m+p1−q1−z1m+p1−q1−q2−z1−z2,z1≠0.
The second example is a matrix with rank 2.Example 2.
Let(34)A=0000000010000010000000000=WJW−1=100000100000010001000000100000001000000000001000001000001000000100010000001.
All solutionsY of (11) are K1,…,K7 as given by Lemma 3. Multiplying X=WYW−1=WKkW−1 out with k=1,…,7, we find all solutions of (1) are(35)X=m00u2u4v100a12a14v300a32a34v200a22a24v400a42a44,m0u3u2u4v100a12a14v300a32a34v200a22a2400000,u3≠0,X=mu1u2u3u4v100a12a14v300a32a34−u1Hu3‖u1‖2v4T00−u1Hu3‖u1‖2a42−u1Hu3‖u1‖2a44v4T00a42a44,u1≠0,X=m0u3u2u4v100a12a14v30a33a32a34v200a22a2400000,mu1u2u3u4v100a12a14v30a33a32a340000000000,u1≠0,a33≠0,X=m0u3u2u4v10a13a12a14v30a33a32a34v200a22a2400000,mu1u3u2u4v10a13a12a14v30a33a32a340000000000,u1≠0,a13≠0,X=mu1u3u2u4v10a13a12a14v3a31a33a32a340000000000,mu1a33a31u1u2u4v100a12a14v3a31a33a32a34−a33a31v4T00−a33a42a31−a33a44a31v3a31a32a33a34v400a42a44,a13≠,a31≠0,X=mu1u3u2u4v100a12a14v3a31a33a32a340000000000,mu1u3u2u4v1a11a13a12a14v3Ta31a13a31a11a32a340000000000,with a13≠0, u3≠a33u1/a31 in the left matrix, and a11≠0u3≠a13u1/a11 in the second one,(36)X=mu1a13a11u1u2u4v1a11a13a12a14v3a31a13a31a11a32a34−a13a11v4T00−a13a42a11−a13a44a11v400a42a44,mu1u3u2u4v1a11a13a12a14v3a31a33a32a340000000000,a11≠0,a33≠a13a31a11.


## 6. Conclusions

In this paper, we obtain all solutions for the Yang–Baxter-like (1) when A is a nilpotent matrix of index 2. We first obtain a system of matrix equations of smaller sizes to find all the solutions of the original matrix equation. For a special case, that is, k=1 and k=2, we derive all solutions of the Yang–Baxter-like matrix equation in detail. However, for an arbitrary nilpotent matrix, finding all solutions of (1) is a hard work. We hope to solve this problem in the future.



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*Source: 2585602-2020-06-18.xml*