allsolutionsoftheyang–baxter...

7
Research Article All Solutions of the Yang–Baxter-Like Matrix Equation for Nilpotent Matrices of Index Two Duanmei Zhou 1,2 andJiawenDing 1 1 College of Mathematics and Computer Science, Gannan Normal University, Ganzhou 341000, China 2 Department of Mathematics, Shanghai University, Shanghai 200444, China Correspondence should be addressed to Duanmei Zhou; [email protected] Received 14 January 2020; Accepted 2 June 2020; Published 18 June 2020 Academic Editor: Sigurdur F. Hafstein Copyright © 2020 Duanmei Zhou and Jiawen Ding. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let A 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. 1.Introduction We are interested in finding all solutions of the quadratic matrix equation: AXA XAX, (1) where the given A and the unknown X are n × n square complex matrices with n 2. e 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]. e Yang–Baxter equation was first introduced by Yang in 1967 and then by Baxter in 1972 in the study of statistical me- chanics. e 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 solutions X 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 n 2 quadratic polynomial equations in n 2 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. e 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 isrankoneandranktwo, respectively. Finally, we present two examples of our so- lution result in Section 5 and conclude with Section 6. Hindawi Complexity Volume 2020, Article ID 2585602, 7 pages https://doi.org/10.1155/2020/2585602

Upload: others

Post on 16-Jul-2020

20 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

Research ArticleAll Solutions of the YangndashBaxter-Like Matrix Equation forNilpotent Matrices of Index Two

Duanmei Zhou 12 and Jiawen Ding1

1College of Mathematics and Computer Science Gannan Normal University Ganzhou 341000 China2Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Duanmei Zhou gzzdm2008163com

Received 14 January 2020 Accepted 2 June 2020 Published 18 June 2020

Academic Editor Sigurdur F Hafstein

Copyright copy 2020 Duanmei Zhou and Jiawen Ding -is is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in anymedium provided the original work isproperly cited

Let A be a nilpotent matrix of index two and consider the YangndashBaxter-like matrix equation AXA XAX We first obtaina system of matrix equations of smaller sizes to find all the solutions of the original matrix equation When A is a nilpotent matrixwith rank 1 and rank 2 we get all solutions of the YangndashBaxter-like matrix equation

1 Introduction

We are interested in finding all solutions of the quadraticmatrix equation

AXA XAX (1)

where the given A and the unknown X are n times n squarecomplex matrices with nge 2-e above equation (1) is calledthe YangndashBaxter-like matrix equation because it is similar tothe classical parameter-free YangndashBaxter equation [1 2]-eYangndashBaxter equation was first introduced by Yang in 1967and then by Baxter in 1972 in the study of statistical me-chanics -e YangndashBaxter equation has been extensivelyresearched by mathematicians and physicists in knot theorybraid group theory and so on [3ndash6]

Obviously the YangndashBaxter matrix equation has twotrivial solutions X 0 andX A However we are interestedin finding nontrivial solutions Finding all solutions ofequation (1) is a hard work for a general matrix A Indeed wecan reformulate (1) into a system of polynomial equations soit is equivalent to solving a system of n2 quadratic polynomialequations in n2 variables To find all solutions is not an easytask even if for a 3 times 3 matrix [7] Most solutions obtained sofar are commuting ones for particular choices of matrices ASee for example [8] for diagonalizable matrix and [9 10] for

nilpotent matrix In [11] infinitely many solutions of (1) wereobtained for any semisimple eigenvalues of the given matrixA family of commuting solutions of (1) were constructed forthose eigenvalues of A that are non-semisimple in [12 13]Some researchers have also proposed some numericalmethods for finding commuting solutions For instance in[14] when A is a nonsingular matrix such that its inverse isa stochastic matrix Ding and Rhee found nontrivial solutionsof (1) via Brouwerrsquos fixed point theorem In [15] when A isa diagonalisable matrix the authors proposed numericalmethods to calculate solutions of (1) by applying the meanergodic theoremWhenA is a low rankmatrix all solutions of(1) have been found in [16ndash18] for the noncommuting case In[19] the authors have obtained explicit solutions when A is anidempotent matrix However for a general matrix A it isdifficult to characterize and determine all the solutions of (1)even if for nilpotent matrix

-e purpose of this paper is to find all the solutions ofequation (1) under the assumption that A is a nilpotentmatrix of index 2 We first give a system of matrix equationsof smaller sizes to find all solutions of the original matrixequation in Section 2 In the next two sections we study allsolutions forAXA AXA when A is rank one and rank tworespectively Finally we present two examples of our so-lution result in Section 5 and conclude with Section 6

HindawiComplexityVolume 2020 Article ID 2585602 7 pageshttpsdoiorg10115520202585602

2 All Solutions for the Nilpotent Matrix ofIndex 2

Let Ane 0 be a nilpotent matrix with index 2 Clearly matrix A

has 0 as its only eigenvalue and the minimal polynomial of A ism(t) t2 So we can write the Jordan canonical form J ofA as

J

0

J0

J0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(2)

in which the 2 times 2 Jordan block

J0 0 1

0 01113890 1113891 (3)

appears k times with k the rank of A and 0 denotes the(n minus 2k) times (n minus 2k) zero matrix So there exists a non-singular matrix W such that

Wminus1

AW J equiv diag 0 J0 J0( 1113857 (4)

Lemma 1 Let two matrices A and X satisfy AXA XAXen for any nonsingular matrixU the matrices B Uminus1 AU

and Y Uminus1XU satisfy

BYB YBY (5)

Conversely if Y satisfies the above equations for a givenB Uminus1 AU then X UYUminus1 satisfies AXA XAX

According to Lemma 1 we know that solving (1) can bereduced to solving the simplified matrix equation

YJY JYJ (6)

So solution X to (1) can be expressed as X WYWminus1where W satisfies A WJWminus1 -us in our analysis belowwe find all solutions of (6) Let Y be partitioned the same wayas J into the (k + 1) times (k + 1) block matrix

Y

M Q1 middot middot middot Qk

P1 Y11 middot middot middot Y1k

⋮ ⋮ ⋱ ⋮

Pk Yk1 middot middot middot Ykk

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7)

where M is (n minus 2k) times (n minus 2k) Qi [q2iminus1 q2i] isin C(nminus2k)times2

and Pi [p2iminus1 p2i]T isin C2times(nminus2k) for i 1 k and

Yij bij cij

sij tij

⎡⎣ ⎤⎦ i j 1 2 k (8)

-en the equation JYJ YJY becomes

0 0 middot middot middot 0

0 J0Y11J0 middot middot middot J0Y1kJ0

⋮ ⋮ ⋱ ⋮

0 J0Yk1J0 ⋮ J0YkkJ0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1113936k

i1QiJ0Pi 1113936

k

i1QiJ0Yi1 middot middot middot 1113936

k

i1QiJ0Yik

1113936k

i1Y1iJ0Pi 1113936

k

i1Y1iJ0Yi1 middot middot middot 1113936

k

i1Y1iJ0Yik

⋮ ⋮ ⋱ ⋮

1113936k

i1YkiJ0Pi 1113936

k

i1YkiJ0Yi1 middot middot middot 1113936

k

i1YkiJ0Yik

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(9)

which is equivalent to the system

1113936k

i1QiJ0Pi 0

1113936k

i1YjiJ0P 0 j 1 2 ki

1113936k

i1QiJ0Yij 0 j 1 2 k

1113936k

i1YliJ0Yij J0YljJ0 l j 1 2 k

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(10)

Note that the above system does not contain M -ismeans 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 AllSolutions for theNilpotentMatrixof Index2 with Rank 1

When k 1 this means that the rank of A is 1 -en

J 0 0

0 J01113890 1113891 (11)

where 0 is the (n minus 2) times (n minus 2) zero matrix Let Y be par-titioned the same way as J into the 2 times 2 block matrix

Y M Q

P Z1113890 1113891 (12)

where M isin C(nminus 2)times(nminus 2) Q q1 q21113858 1113859 isin C(nminus 2)times2 P

p1 p21113858 1113859T isin C2times(nminus 2) and Z isin C2times2 According to (4) we

obtainQJ0P 0

QJ0Z 0

ZJ0P 0

ZJ0Z J0ZJ0

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(13)

Obviously the last equation of (13) is also a YangndashBaxtermatrix equation Next we solve the last equation in (13)

Lemma 2 e solutions Z of the equation ZJ0Z J0ZJ0 are

2 Complexity

Z(1)

0 z2

0 z41113890 1113891 z4 ne 0

Z(2)

0 z2

0 01113890 1113891

Z(3)

z1 z2

0 01113890 1113891 z1 ne 0

(14)

Proof Let

Z z1 z2

z3 z41113890 1113891 (15)

-en the equation ZJ0Z J0ZJ0 becomesz1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

(16)

-at isz1z3 z1z4

z23 z3z4

1113890 1113891 0 z3

0 01113890 1113891 (17)

-is leads to z3 0 and z4z1 0 and z2 is arbitraryBy Lemma 2 all solutions of the last equation of (13) are

Z(1) Z(2) and Z(3) Substituting such matrices into the firstthree equations of (13) we obtain the following result

Theorem 1 Suppose that A is a nilpotent matrix with rank 1en all solutions of (1) are X WYWminus1 where Y is par-titioned as (12) in which M is an arbitrary (n minus 2) times (n minus 2)

matrix such that Y

M 0 q2

pT1 0 z2

pT2 0 z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0 (18)

M 0 q2

pT1 0 z2

pT2 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M q1 q2

pT1 0 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(19)

M q1 q2

pT1 z1 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0 (20)

Proof When Z Z(1) then ZJ0 0 So we just need tosolve the first two equations in (13) -e second equation(13) is QJ0Z 0 -at is

0 q1z41113858 1113859 0 (21)

Since z4 ne 0 we have q1 0 -us QJ0 0 -is gives(18)

Let Z Z(2) -en ZJ0 0 and J0Z 0 So the secondand the third equations of (13) are satisfied Next we onlyhave to solve the first equation of (13) ie

QJ0P q1 q21113858 11138590 1

0 01113890 1113891

pT1

pT2

⎡⎣ ⎤⎦ q1pT2 0 (22)

from which q1 0 p2 is arbitrary or p2 0 q1 is arbitrary-is gives (19)

If Z Z(3) then J0Z 0 So the second equation of (13)is satisfied From the first equation ZJ0P 0 we obtain

z1pT2

0⎡⎣ ⎤⎦ 0 (23)

which leads to z1pT2 0 Since z1 ne 0 we obtain p2 0

-us J0P 0 -is gives (20)

4 AllSolutions for theNilpotentMatrixof Index2 with Rank 2

When k 2 this means

J

0 0 0

0 J0 0

0 0 J0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (24)

where 0 is the (n minus 4) times (n minus 4) zero matrix Let

Λ J0 00 J0

1113890 1113891 -en

J 0 0

0 Λ1113890 1113891 (25)

Let Y be partitioned the same way as J into the 2 times 2block matrix

Y M U

VT K1113890 1113891 (26)

where M is (n minus 4) times (n minus 4) U u1 u2 u3 u41113858 1113859 isinC(nminus 4)times4 V v1 v2 v3 v41113858 1113859 isin C(nminus 4)times4 and K is 4 times 4-en JYJ YJY is equivalent to the system

UΛVT 0

UΛK 0

KΛVT 0

ΛKΛ KΛK

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(27)

All solutions of this system were obtained in [18] Wehave the following results

Lemma 3 (see Lemma 41 in [18]) e solutions K of theequation ΛKΛ KΛK are

Complexity 3

K1

0 a12 0 a14

0 a22 0 a24

0 a32 0 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

K2

0 a12 0 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a33 ne 0

K3

0 a12 a13 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne 0

K4

0 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13a31 ne 0

K5

0 a12 0 a14

0 minusa33a42

a310 minus

a33a44

a31

a31 a32 a33 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a31 ne 0

K6

a11 a12 a13 a14

0 minusa13a42

a110 minus

a13a44

a11

a31 a32a13a31

a11a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

K7

a11 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a11 a13

a31 a33

11138681113868111386811138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868111386811138681113868ne 0

(28)

Theorem 2 (see Theorem 42 in [18]) Suppose A is a nil-potent matrix with rank 2 and is partitioned as (4) en allsolutions of (1) are X WYWminus1 where Y is partitioned as

(10) in which M is an arbitrary (n minus 4)times (n minus 4) matrix suchthat Y

M 0 u2 0 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M 0 u2 u3 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

minusuH1 u3

u1

2v

T4 0 minus

uH1 u3

u1

2a42 0 minus

uH1 u3

u1

2a44

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

M 0 u2 u3 u4

vT1 0 a12 0 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

M 0 u2 u3 u4

vT1 0 a12 a13 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2a33

a31u1 u4

vT1 0 a12 0 a14

minusa33

a31v

T4 0 minus

a33a42

a310 minus

a33a44

a31

vT3 a31 a32 a33 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32

a13a31

a11a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(29)

4 Complexity

with a13 ne 0 u3 ne a33u1a31 in the left matrix and a11 ne 0u3 ne a13u1a11 in the second one

M u1 u2a13

a11u1 u4

vT1 a11 a12 a13 a14

minusa13

a11v

T4 0 minus

a13a42

a110 minus

a13a44

a11

vT3 a31 a32

a13a31

a11a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(30)

5 Examples

We give two examples to illustrate our results

Example 1 Let

A

1 0 1

1 0 1

minus1 0 minus1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (31)

-en there exists

W

1 1 0

0 1 minus1

minus1 minus1 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

J

0 0 0

0 0 1

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

such that A WJWminus1 By-eorem 1 we obtain all solutionsof (1) as

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minus z4 minusp1 + p2 minusp1 + p2 + z2 minus z4

minusq2 minus z2 + z4 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2 + z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minusp1 + p2 minusp1 + p2 + z2

minusq2 minus z2 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z2 q1 minus p1 minus m q1 minus p1 minus m + q2 + z2

z2 minusp1 z2 minus p1

minusq1 minus q2 minus z2 m + p1 minus q1 m + p1 minus q1 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z1 + z2 q1 minus p1 minus m + z1 q1 minus p1 minus m + q2 + z1 + z2

z1 + z2 z1 minus p1 z1 minus p1 + z2

minusq1 minus q2 minus z1 minus z2 m + p1 minus q1 minus z1 m + p1 minus q1 minus q2 minus z1 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0

(33)

-e second example is a matrix with rank 2

Example 2 Let

A

0 0 0 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

WJWminus1

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

All solutionsY of (11) areK1 K7 as given by Lemma 3Multiplying X WYWminus1 WKkWminus1 out with k 1 7we find all solutions of (1) are

Complexity 5

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 2: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

2 All Solutions for the Nilpotent Matrix ofIndex 2

Let Ane 0 be a nilpotent matrix with index 2 Clearly matrix A

has 0 as its only eigenvalue and the minimal polynomial of A ism(t) t2 So we can write the Jordan canonical form J ofA as

J

0

J0

J0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(2)

in which the 2 times 2 Jordan block

J0 0 1

0 01113890 1113891 (3)

appears k times with k the rank of A and 0 denotes the(n minus 2k) times (n minus 2k) zero matrix So there exists a non-singular matrix W such that

Wminus1

AW J equiv diag 0 J0 J0( 1113857 (4)

Lemma 1 Let two matrices A and X satisfy AXA XAXen for any nonsingular matrixU the matrices B Uminus1 AU

and Y Uminus1XU satisfy

BYB YBY (5)

Conversely if Y satisfies the above equations for a givenB Uminus1 AU then X UYUminus1 satisfies AXA XAX

According to Lemma 1 we know that solving (1) can bereduced to solving the simplified matrix equation

YJY JYJ (6)

So solution X to (1) can be expressed as X WYWminus1where W satisfies A WJWminus1 -us in our analysis belowwe find all solutions of (6) Let Y be partitioned the same wayas J into the (k + 1) times (k + 1) block matrix

Y

M Q1 middot middot middot Qk

P1 Y11 middot middot middot Y1k

⋮ ⋮ ⋱ ⋮

Pk Yk1 middot middot middot Ykk

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(7)

where M is (n minus 2k) times (n minus 2k) Qi [q2iminus1 q2i] isin C(nminus2k)times2

and Pi [p2iminus1 p2i]T isin C2times(nminus2k) for i 1 k and

Yij bij cij

sij tij

⎡⎣ ⎤⎦ i j 1 2 k (8)

-en the equation JYJ YJY becomes

0 0 middot middot middot 0

0 J0Y11J0 middot middot middot J0Y1kJ0

⋮ ⋮ ⋱ ⋮

0 J0Yk1J0 ⋮ J0YkkJ0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1113936k

i1QiJ0Pi 1113936

k

i1QiJ0Yi1 middot middot middot 1113936

k

i1QiJ0Yik

1113936k

i1Y1iJ0Pi 1113936

k

i1Y1iJ0Yi1 middot middot middot 1113936

k

i1Y1iJ0Yik

⋮ ⋮ ⋱ ⋮

1113936k

i1YkiJ0Pi 1113936

k

i1YkiJ0Yi1 middot middot middot 1113936

k

i1YkiJ0Yik

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(9)

which is equivalent to the system

1113936k

i1QiJ0Pi 0

1113936k

i1YjiJ0P 0 j 1 2 ki

1113936k

i1QiJ0Yij 0 j 1 2 k

1113936k

i1YliJ0Yij J0YljJ0 l j 1 2 k

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(10)

Note that the above system does not contain M -ismeans 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 AllSolutions for theNilpotentMatrixof Index2 with Rank 1

When k 1 this means that the rank of A is 1 -en

J 0 0

0 J01113890 1113891 (11)

where 0 is the (n minus 2) times (n minus 2) zero matrix Let Y be par-titioned the same way as J into the 2 times 2 block matrix

Y M Q

P Z1113890 1113891 (12)

where M isin C(nminus 2)times(nminus 2) Q q1 q21113858 1113859 isin C(nminus 2)times2 P

p1 p21113858 1113859T isin C2times(nminus 2) and Z isin C2times2 According to (4) we

obtainQJ0P 0

QJ0Z 0

ZJ0P 0

ZJ0Z J0ZJ0

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(13)

Obviously the last equation of (13) is also a YangndashBaxtermatrix equation Next we solve the last equation in (13)

Lemma 2 e solutions Z of the equation ZJ0Z J0ZJ0 are

2 Complexity

Z(1)

0 z2

0 z41113890 1113891 z4 ne 0

Z(2)

0 z2

0 01113890 1113891

Z(3)

z1 z2

0 01113890 1113891 z1 ne 0

(14)

Proof Let

Z z1 z2

z3 z41113890 1113891 (15)

-en the equation ZJ0Z J0ZJ0 becomesz1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

(16)

-at isz1z3 z1z4

z23 z3z4

1113890 1113891 0 z3

0 01113890 1113891 (17)

-is leads to z3 0 and z4z1 0 and z2 is arbitraryBy Lemma 2 all solutions of the last equation of (13) are

Z(1) Z(2) and Z(3) Substituting such matrices into the firstthree equations of (13) we obtain the following result

Theorem 1 Suppose that A is a nilpotent matrix with rank 1en all solutions of (1) are X WYWminus1 where Y is par-titioned as (12) in which M is an arbitrary (n minus 2) times (n minus 2)

matrix such that Y

M 0 q2

pT1 0 z2

pT2 0 z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0 (18)

M 0 q2

pT1 0 z2

pT2 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M q1 q2

pT1 0 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(19)

M q1 q2

pT1 z1 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0 (20)

Proof When Z Z(1) then ZJ0 0 So we just need tosolve the first two equations in (13) -e second equation(13) is QJ0Z 0 -at is

0 q1z41113858 1113859 0 (21)

Since z4 ne 0 we have q1 0 -us QJ0 0 -is gives(18)

Let Z Z(2) -en ZJ0 0 and J0Z 0 So the secondand the third equations of (13) are satisfied Next we onlyhave to solve the first equation of (13) ie

QJ0P q1 q21113858 11138590 1

0 01113890 1113891

pT1

pT2

⎡⎣ ⎤⎦ q1pT2 0 (22)

from which q1 0 p2 is arbitrary or p2 0 q1 is arbitrary-is gives (19)

If Z Z(3) then J0Z 0 So the second equation of (13)is satisfied From the first equation ZJ0P 0 we obtain

z1pT2

0⎡⎣ ⎤⎦ 0 (23)

which leads to z1pT2 0 Since z1 ne 0 we obtain p2 0

-us J0P 0 -is gives (20)

4 AllSolutions for theNilpotentMatrixof Index2 with Rank 2

When k 2 this means

J

0 0 0

0 J0 0

0 0 J0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (24)

where 0 is the (n minus 4) times (n minus 4) zero matrix Let

Λ J0 00 J0

1113890 1113891 -en

J 0 0

0 Λ1113890 1113891 (25)

Let Y be partitioned the same way as J into the 2 times 2block matrix

Y M U

VT K1113890 1113891 (26)

where M is (n minus 4) times (n minus 4) U u1 u2 u3 u41113858 1113859 isinC(nminus 4)times4 V v1 v2 v3 v41113858 1113859 isin C(nminus 4)times4 and K is 4 times 4-en JYJ YJY is equivalent to the system

UΛVT 0

UΛK 0

KΛVT 0

ΛKΛ KΛK

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(27)

All solutions of this system were obtained in [18] Wehave the following results

Lemma 3 (see Lemma 41 in [18]) e solutions K of theequation ΛKΛ KΛK are

Complexity 3

K1

0 a12 0 a14

0 a22 0 a24

0 a32 0 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

K2

0 a12 0 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a33 ne 0

K3

0 a12 a13 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne 0

K4

0 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13a31 ne 0

K5

0 a12 0 a14

0 minusa33a42

a310 minus

a33a44

a31

a31 a32 a33 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a31 ne 0

K6

a11 a12 a13 a14

0 minusa13a42

a110 minus

a13a44

a11

a31 a32a13a31

a11a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

K7

a11 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a11 a13

a31 a33

11138681113868111386811138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868111386811138681113868ne 0

(28)

Theorem 2 (see Theorem 42 in [18]) Suppose A is a nil-potent matrix with rank 2 and is partitioned as (4) en allsolutions of (1) are X WYWminus1 where Y is partitioned as

(10) in which M is an arbitrary (n minus 4)times (n minus 4) matrix suchthat Y

M 0 u2 0 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M 0 u2 u3 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

minusuH1 u3

u1

2v

T4 0 minus

uH1 u3

u1

2a42 0 minus

uH1 u3

u1

2a44

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

M 0 u2 u3 u4

vT1 0 a12 0 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

M 0 u2 u3 u4

vT1 0 a12 a13 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2a33

a31u1 u4

vT1 0 a12 0 a14

minusa33

a31v

T4 0 minus

a33a42

a310 minus

a33a44

a31

vT3 a31 a32 a33 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32

a13a31

a11a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(29)

4 Complexity

with a13 ne 0 u3 ne a33u1a31 in the left matrix and a11 ne 0u3 ne a13u1a11 in the second one

M u1 u2a13

a11u1 u4

vT1 a11 a12 a13 a14

minusa13

a11v

T4 0 minus

a13a42

a110 minus

a13a44

a11

vT3 a31 a32

a13a31

a11a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(30)

5 Examples

We give two examples to illustrate our results

Example 1 Let

A

1 0 1

1 0 1

minus1 0 minus1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (31)

-en there exists

W

1 1 0

0 1 minus1

minus1 minus1 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

J

0 0 0

0 0 1

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

such that A WJWminus1 By-eorem 1 we obtain all solutionsof (1) as

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minus z4 minusp1 + p2 minusp1 + p2 + z2 minus z4

minusq2 minus z2 + z4 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2 + z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minusp1 + p2 minusp1 + p2 + z2

minusq2 minus z2 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z2 q1 minus p1 minus m q1 minus p1 minus m + q2 + z2

z2 minusp1 z2 minus p1

minusq1 minus q2 minus z2 m + p1 minus q1 m + p1 minus q1 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z1 + z2 q1 minus p1 minus m + z1 q1 minus p1 minus m + q2 + z1 + z2

z1 + z2 z1 minus p1 z1 minus p1 + z2

minusq1 minus q2 minus z1 minus z2 m + p1 minus q1 minus z1 m + p1 minus q1 minus q2 minus z1 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0

(33)

-e second example is a matrix with rank 2

Example 2 Let

A

0 0 0 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

WJWminus1

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

All solutionsY of (11) areK1 K7 as given by Lemma 3Multiplying X WYWminus1 WKkWminus1 out with k 1 7we find all solutions of (1) are

Complexity 5

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 3: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

Z(1)

0 z2

0 z41113890 1113891 z4 ne 0

Z(2)

0 z2

0 01113890 1113891

Z(3)

z1 z2

0 01113890 1113891 z1 ne 0

(14)

Proof Let

Z z1 z2

z3 z41113890 1113891 (15)

-en the equation ZJ0Z J0ZJ0 becomesz1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

z1 z2

z3 z41113890 1113891

0 1

0 01113890 1113891

(16)

-at isz1z3 z1z4

z23 z3z4

1113890 1113891 0 z3

0 01113890 1113891 (17)

-is leads to z3 0 and z4z1 0 and z2 is arbitraryBy Lemma 2 all solutions of the last equation of (13) are

Z(1) Z(2) and Z(3) Substituting such matrices into the firstthree equations of (13) we obtain the following result

Theorem 1 Suppose that A is a nilpotent matrix with rank 1en all solutions of (1) are X WYWminus1 where Y is par-titioned as (12) in which M is an arbitrary (n minus 2) times (n minus 2)

matrix such that Y

M 0 q2

pT1 0 z2

pT2 0 z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0 (18)

M 0 q2

pT1 0 z2

pT2 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M q1 q2

pT1 0 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(19)

M q1 q2

pT1 z1 z2

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0 (20)

Proof When Z Z(1) then ZJ0 0 So we just need tosolve the first two equations in (13) -e second equation(13) is QJ0Z 0 -at is

0 q1z41113858 1113859 0 (21)

Since z4 ne 0 we have q1 0 -us QJ0 0 -is gives(18)

Let Z Z(2) -en ZJ0 0 and J0Z 0 So the secondand the third equations of (13) are satisfied Next we onlyhave to solve the first equation of (13) ie

QJ0P q1 q21113858 11138590 1

0 01113890 1113891

pT1

pT2

⎡⎣ ⎤⎦ q1pT2 0 (22)

from which q1 0 p2 is arbitrary or p2 0 q1 is arbitrary-is gives (19)

If Z Z(3) then J0Z 0 So the second equation of (13)is satisfied From the first equation ZJ0P 0 we obtain

z1pT2

0⎡⎣ ⎤⎦ 0 (23)

which leads to z1pT2 0 Since z1 ne 0 we obtain p2 0

-us J0P 0 -is gives (20)

4 AllSolutions for theNilpotentMatrixof Index2 with Rank 2

When k 2 this means

J

0 0 0

0 J0 0

0 0 J0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (24)

where 0 is the (n minus 4) times (n minus 4) zero matrix Let

Λ J0 00 J0

1113890 1113891 -en

J 0 0

0 Λ1113890 1113891 (25)

Let Y be partitioned the same way as J into the 2 times 2block matrix

Y M U

VT K1113890 1113891 (26)

where M is (n minus 4) times (n minus 4) U u1 u2 u3 u41113858 1113859 isinC(nminus 4)times4 V v1 v2 v3 v41113858 1113859 isin C(nminus 4)times4 and K is 4 times 4-en JYJ YJY is equivalent to the system

UΛVT 0

UΛK 0

KΛVT 0

ΛKΛ KΛK

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(27)

All solutions of this system were obtained in [18] Wehave the following results

Lemma 3 (see Lemma 41 in [18]) e solutions K of theequation ΛKΛ KΛK are

Complexity 3

K1

0 a12 0 a14

0 a22 0 a24

0 a32 0 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

K2

0 a12 0 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a33 ne 0

K3

0 a12 a13 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne 0

K4

0 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13a31 ne 0

K5

0 a12 0 a14

0 minusa33a42

a310 minus

a33a44

a31

a31 a32 a33 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a31 ne 0

K6

a11 a12 a13 a14

0 minusa13a42

a110 minus

a13a44

a11

a31 a32a13a31

a11a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

K7

a11 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a11 a13

a31 a33

11138681113868111386811138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868111386811138681113868ne 0

(28)

Theorem 2 (see Theorem 42 in [18]) Suppose A is a nil-potent matrix with rank 2 and is partitioned as (4) en allsolutions of (1) are X WYWminus1 where Y is partitioned as

(10) in which M is an arbitrary (n minus 4)times (n minus 4) matrix suchthat Y

M 0 u2 0 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M 0 u2 u3 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

minusuH1 u3

u1

2v

T4 0 minus

uH1 u3

u1

2a42 0 minus

uH1 u3

u1

2a44

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

M 0 u2 u3 u4

vT1 0 a12 0 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

M 0 u2 u3 u4

vT1 0 a12 a13 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2a33

a31u1 u4

vT1 0 a12 0 a14

minusa33

a31v

T4 0 minus

a33a42

a310 minus

a33a44

a31

vT3 a31 a32 a33 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32

a13a31

a11a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(29)

4 Complexity

with a13 ne 0 u3 ne a33u1a31 in the left matrix and a11 ne 0u3 ne a13u1a11 in the second one

M u1 u2a13

a11u1 u4

vT1 a11 a12 a13 a14

minusa13

a11v

T4 0 minus

a13a42

a110 minus

a13a44

a11

vT3 a31 a32

a13a31

a11a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(30)

5 Examples

We give two examples to illustrate our results

Example 1 Let

A

1 0 1

1 0 1

minus1 0 minus1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (31)

-en there exists

W

1 1 0

0 1 minus1

minus1 minus1 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

J

0 0 0

0 0 1

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

such that A WJWminus1 By-eorem 1 we obtain all solutionsof (1) as

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minus z4 minusp1 + p2 minusp1 + p2 + z2 minus z4

minusq2 minus z2 + z4 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2 + z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minusp1 + p2 minusp1 + p2 + z2

minusq2 minus z2 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z2 q1 minus p1 minus m q1 minus p1 minus m + q2 + z2

z2 minusp1 z2 minus p1

minusq1 minus q2 minus z2 m + p1 minus q1 m + p1 minus q1 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z1 + z2 q1 minus p1 minus m + z1 q1 minus p1 minus m + q2 + z1 + z2

z1 + z2 z1 minus p1 z1 minus p1 + z2

minusq1 minus q2 minus z1 minus z2 m + p1 minus q1 minus z1 m + p1 minus q1 minus q2 minus z1 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0

(33)

-e second example is a matrix with rank 2

Example 2 Let

A

0 0 0 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

WJWminus1

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

All solutionsY of (11) areK1 K7 as given by Lemma 3Multiplying X WYWminus1 WKkWminus1 out with k 1 7we find all solutions of (1) are

Complexity 5

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 4: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

K1

0 a12 0 a14

0 a22 0 a24

0 a32 0 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

K2

0 a12 0 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a33 ne 0

K3

0 a12 a13 a14

0 a22 0 a24

0 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne 0

K4

0 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13a31 ne 0

K5

0 a12 0 a14

0 minusa33a42

a310 minus

a33a44

a31

a31 a32 a33 a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a31 ne 0

K6

a11 a12 a13 a14

0 minusa13a42

a110 minus

a13a44

a11

a31 a32a13a31

a11a34

0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

K7

a11 a12 a13 a14

0 0 0 0

a31 a32 a33 a34

0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a11 a13

a31 a33

11138681113868111386811138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868111386811138681113868ne 0

(28)

Theorem 2 (see Theorem 42 in [18]) Suppose A is a nil-potent matrix with rank 2 and is partitioned as (4) en allsolutions of (1) are X WYWminus1 where Y is partitioned as

(10) in which M is an arbitrary (n minus 4)times (n minus 4) matrix suchthat Y

M 0 u2 0 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M 0 u2 u3 u4

vT1 0 a12 0 a14

v2T 0 a22 0 a24

vT3 0 a32 0 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

minusuH1 u3

u1

2v

T4 0 minus

uH1 u3

u1

2a42 0 minus

uH1 u3

u1

2a44

vT3 0 a32 0 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

M 0 u2 u3 u4

vT1 0 a12 0 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

M 0 u2 u3 u4

vT1 0 a12 a13 a14

vT2 0 a22 0 a24

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 0 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

M u1 u2 u3 u4

vT1 0 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2a33

a31u1 u4

vT1 0 a12 0 a14

minusa33

a31v

T4 0 minus

a33a42

a310 minus

a33a44

a31

vT3 a31 a32 a33 a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

M u1 u2 u3 u4

vT1 0 a12 0 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32

a13a31

a11a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(29)

4 Complexity

with a13 ne 0 u3 ne a33u1a31 in the left matrix and a11 ne 0u3 ne a13u1a11 in the second one

M u1 u2a13

a11u1 u4

vT1 a11 a12 a13 a14

minusa13

a11v

T4 0 minus

a13a42

a110 minus

a13a44

a11

vT3 a31 a32

a13a31

a11a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(30)

5 Examples

We give two examples to illustrate our results

Example 1 Let

A

1 0 1

1 0 1

minus1 0 minus1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (31)

-en there exists

W

1 1 0

0 1 minus1

minus1 minus1 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

J

0 0 0

0 0 1

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

such that A WJWminus1 By-eorem 1 we obtain all solutionsof (1) as

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minus z4 minusp1 + p2 minusp1 + p2 + z2 minus z4

minusq2 minus z2 + z4 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2 + z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minusp1 + p2 minusp1 + p2 + z2

minusq2 minus z2 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z2 q1 minus p1 minus m q1 minus p1 minus m + q2 + z2

z2 minusp1 z2 minus p1

minusq1 minus q2 minus z2 m + p1 minus q1 m + p1 minus q1 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z1 + z2 q1 minus p1 minus m + z1 q1 minus p1 minus m + q2 + z1 + z2

z1 + z2 z1 minus p1 z1 minus p1 + z2

minusq1 minus q2 minus z1 minus z2 m + p1 minus q1 minus z1 m + p1 minus q1 minus q2 minus z1 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0

(33)

-e second example is a matrix with rank 2

Example 2 Let

A

0 0 0 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

WJWminus1

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

All solutionsY of (11) areK1 K7 as given by Lemma 3Multiplying X WYWminus1 WKkWminus1 out with k 1 7we find all solutions of (1) are

Complexity 5

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 5: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

with a13 ne 0 u3 ne a33u1a31 in the left matrix and a11 ne 0u3 ne a13u1a11 in the second one

M u1 u2a13

a11u1 u4

vT1 a11 a12 a13 a14

minusa13

a11v

T4 0 minus

a13a42

a110 minus

a13a44

a11

vT3 a31 a32

a13a31

a11a34

vT4 0 a42 0 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

M u1 u2 u3 u4

vT1 a11 a12 a13 a14

0T 0 0 0 0

vT3 a31 a32 a33 a34

0T 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(30)

5 Examples

We give two examples to illustrate our results

Example 1 Let

A

1 0 1

1 0 1

minus1 0 minus1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ (31)

-en there exists

W

1 1 0

0 1 minus1

minus1 minus1 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

J

0 0 0

0 0 1

0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

such that A WJWminus1 By-eorem 1 we obtain all solutionsof (1) as

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minus z4 minusp1 + p2 minusp1 + p2 + z2 minus z4

minusq2 minus z2 + z4 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2 + z4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z4 ne 0

X

q2 + z2 minusm minus p1 minusm minus p1 + q2 + z2

z2 minusp1 + p2 minusp1 + p2 + z2

minusq2 minus z2 m + p1 minus p2 m + p1 minus p2 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z2 q1 minus p1 minus m q1 minus p1 minus m + q2 + z2

z2 minusp1 z2 minus p1

minusq1 minus q2 minus z2 m + p1 minus q1 m + p1 minus q1 minus q2 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

X

q1 + q2 + z1 + z2 q1 minus p1 minus m + z1 q1 minus p1 minus m + q2 + z1 + z2

z1 + z2 z1 minus p1 z1 minus p1 + z2

minusq1 minus q2 minus z1 minus z2 m + p1 minus q1 minus z1 m + p1 minus q1 minus q2 minus z1 minus z2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦ z1 ne 0

(33)

-e second example is a matrix with rank 2

Example 2 Let

A

0 0 0 0 0

0 0 0 1 0

0 0 0 0 1

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

WJWminus1

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

0 0 0 0 0

0 0 1 0 0

0 0 0 0 0

0 0 0 0 1

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

1 0 0 0 0

0 1 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 0 0 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

All solutionsY of (11) areK1 K7 as given by Lemma 3Multiplying X WYWminus1 WKkWminus1 out with k 1 7we find all solutions of (1) are

Complexity 5

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 6: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

X

m 0 0 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u3 ne 0

X

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 0 a32 a34

minusuH1 u3

u12v

T4 0 0 minus

uH1 u3

u12a42 minus

uH1 u3

u12a44

vT4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0

X

m 0 u3 u2 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u2 u3 u4

v1 0 0 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a33 ne 0

X

m 0 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

v2 0 0 a22 a24

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 0 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

u1 ne 0 a13 ne 0

X

m u1 u3 u2 u4

v1 0 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1a33

a31u1 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

minusa33

a31v

T4 0 0 minus

a33a42

a31minus

a33a44

a31

v3 a31 a32 a33 a34

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a13 ne a31 ne 0

X

m u1 u3 u2 u4

v1 0 0 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

vT3 a31

a13a31

a11a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(35)

with a13 ne 0 u3 ne a33u1a31 in the left matrix anda11 ne 0 u3 ne a13u1a11 in the second one

X

m u1a13

a11u1 u2 u4

v1 a11 a13 a12 a14

v3 a31a13a31

a11a32 a34

minusa13

a11v

T4 0 0 minus

a13a42

a11minus

a13a44

a11

v4 0 0 a42 a44

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m u1 u3 u2 u4

v1 a11 a13 a12 a14

v3 a31 a33 a32 a34

0 0 0 0 0

0 0 0 0 0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

a11 ne 0

a33 nea13a31

a11

(36)

6 Conclusions

In this paper we obtain all solutions for the YangndashBaxter-like (1) when A is a nilpotent matrix of index 2 We firstobtain a system of matrix equations of smaller sizes to findall the solutions of the original matrix equation For a specialcase that is k 1 and k 2 we derive all solutions of theYangndashBaxter-like matrix equation in detail However for anarbitrary nilpotent matrix finding all solutions of (1) isa hard work We hope to solve this problem in the future

Conflicts of Interest

-e authors declare that there are no conflicts of interestregarding the publication of this paper

Acknowledgments

-is work was supported by the National Natural ScienceFoundation of China (no 11861008) China PostdoctoralScience Foundation (no 2018M641974) Natural ScienceFoundation of Jiangxi Province (no 20192BAB201008)China Scholarship Council (no 201909865004) Researchfund of Gannan Normal University (Nos YJG-2018-11 and18zb04) and Key Disciplines Coordinate Innovation Pro-jects of Gannan Normal University

6 Complexity

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7

Page 7: AllSolutionsoftheYang–Baxter …downloads.hindawi.com/journals/complexity/2020/2585602.pdfwithrank1andrank2,wegetallsolutionsoftheYang–Baxter-likematrixequation. 1.Introduction

References

[1] R J Baxter ldquoPartition function of the Eight-Vertex latticemodelrdquo Annals of Physics vol 70 no 1 pp 193ndash228 1972

[2] C N Yang ldquoSome exact results for the many-body problem inone dimension with repulsive delta-function interactionrdquoPhysical Review Letters vol 19 no 23 pp 1312ndash1315 1967

[3] L D Faddeev ldquoHistory and perspectives of quantum groupsrdquoMilan Journal of Mathematics vol 74 no 1 pp 279ndash2942006

[4] F Felix Nonlinear Equations Quantum Groups and Dualityeorems A Primer on the Yang-Baxter Equation VDMVerlag Saarbrucken Germany 2009

[5] M Jimbo Introduction to the Yang-Baxter Equation BraidGroup Knot eory and Statistical Physics II pp 153ndash176World Scientic Singapore 1994

[6] C Yang and M Ge Braid Group Knot eory and StatisticalMechanics World Scientific Singapore 1989

[7] A Cibotarica J Ding J Kolibal and N Rhee ldquoSolutions ofthe Yang-Baxter matrix equation for an idempotentrdquo Nu-merical Algebra Control and Optimization Journal vol 3no 2 pp 235ndash245 2013

[8] Q Dong and J Ding ldquoComplete commuting solutions of theYang-Baxter-like matrix equation for diagonalizable matri-cesrdquo Computers amp Mathematics with Applications vol 72no 1 pp 194ndash201 2016

[9] Q Dong J Ding and Q Huang ldquoCommuting solutions ofa quadratic matrix equation for nilpotent matricesrdquo AlgebraColloquium vol 25 no 1 pp 31ndash44 2018

[10] D Zhou and J Ding ldquoSolving the Yang-Baxter-like matrixequation for nilpotent matrices of index threerdquo InternationalJournal of Computer Mathematics vol 95 no 2 pp 303ndash3152018

[11] J Ding and C Zhang ldquoOn the structure of the spectral so-lutions of the Yang-Baxter matrix equationrdquo Applied Math-ematics Letters vol 35 pp 86ndash89 2014

[12] Q Dong ldquoProjection-based commuting solutions of theYang-Baxter matrix equation for non-semisimple eigen-valuesrdquo Applied Mathematics Letters vol 64 pp 231ndash2342017

[13] D Zhou G Chen G Yu and J Zhong ldquoOn the projection-based commuting solutions of the Yang-Baxter matrixequationrdquo Applied Mathematics Letters vol 79 pp 155ndash1612018

[14] J Ding and N H Rhee ldquoA nontrivial solution to a stochasticmatrix equationrdquo East Asian Journal on Applied Mathematicsvol 2 no 4 pp 277ndash284 2012

[15] J Ding and N H Rhee ldquoComputing solutions of the Yang-Baxter-like matrix equation for diagonalisable matricesrdquo EastAsian Journal on AppliedMathematics vol 5 no 1 pp 75ndash842015

[16] H Tian ldquoAll solutions of the Yang-Baxter-like matrixequation for rank-onematricesrdquoAppliedMathematics Lettersvol 51 pp 55ndash59 2016

[17] D Zhou G Chen and J Ding ldquoSolving the Yang-Baxter-likematrix equation for rank-two matricesrdquo Journal of Compu-tational and Applied Mathematics vol 313 pp 142ndash151 2017

[18] D Zhou G Chen and J Ding ldquoOn the Yang-Baxter-likematrix equation for rank-two matricesrdquo Open Mathematicsvol 15 no 1 pp 340ndash353 2017

[19] S I A Mansour J Ding and Q Huang ldquoExplicit solutions ofthe Yang-Baxter-like matrix equation for an idempotentmatrixrdquo Applied Mathematics Letters vol 63 pp 71ndash76 2017

Complexity 7