Evolution algebras whose evolution operator is a homomorphism

Received: 31 August 2021 Accepted: 17 September 2021 DOI: 10.1002/cmm4.1200 SPECIAL ISSUE PAPER Evolution algebras whose evolution operator is a homomorphism Desamparados Fernández-Ternero Víctor Manuel Gómez-Sousa Juan Núñez-Valdés Department of Geometry and Topology, Faculty of Mathematics, University of Seville, Seville, Spain Correspondence Víctor Manuel Gómez-Sousa, Department of Geometry and Topology, Faculty of Mathematics, University of Seville, Seville, Spain. Email: victor[email protected] Abstract This article deals with the evolution operator of evolution algebras. We give a theorem that allows to characterize these algebras when this operator is a homomorphism of algebras of rank n−2 and this result in turn allows us to extend the classification of this type of algebras, given in a previous result by ourselves in 2021, up to the case of dimension 4. For this purpose, we analyze and make use of an algorithm for the degenerate case. A computational study of the procedure is also made. KEYWORDS algorithm, classification, evolution algebras, evolution operator, homomorphism 1INTRODUCTION On the one hand, evolution algebras, which were firstly introduced by Tian in his Ph.D. thesis1in 2004, have many connections with different mathematical theories, which makes them especially interesting for finding links between concepts from different branches of Mathematics, such as graph theory, group theory, dynamical systems, stochastic processes and mathematical physics.2-6 On the other hand, they have also many applications to different scientific disciplines, such as physics, engineering and, above all, biology.7-10 Both aspects explain the growing interest of researchers in them today. However, only the classification of these algebras up to dimension 3 has been obtained.11-13 This article is related to this topic, being the main objective to classify a special family of evolution algebras in dimension 4. All the information of these algebras is collected in the so-called structure matrix, from which the main operator, that reveals the dynamics of the algebra, is defined. In this article we will focus on the study of this operator, specifically in the case where it is a homomorphism of algebras. This fact has been previously analyzed in Reference 14, hence our purpose is to obtain new results in this subject and develop the methods outlined in that paper for higher-dimensional evolution algebras. The structure of this article is as follows. In Section 2 we briefly introduce the main definitions and properties of evolution algebras. Section 3 begins with the review of the main results obtained in Reference 14, to later present a new result that allows to extend the classification of these algebras up to dimension 4. For this purpose, we analyze and make use of an algorithm for the degenerate case, doing a computational study of the procedure. At the end of this section, we explicitly classify these algebras in dimension 4. Finally, the article ends with some conclusions in Section 4. This is an open access article under the terms of the Creative Commons Attribution-NonCommercial-NoDerivs License, which permits use and distributioninanymedium, provided the original work is properly cited, the use is non-commercial and no modifications or adaptations are made. © 2021 The Authors. Computational and Mathematical Methods published by John Wiley & Sons Ltd. Comp and Math Methods. 2021;3:e1200. wileyonlinelibrary.com/journal/cmm4 1of14 https://doi.org/10.1002/cmm4.1200 2of14 FERNÁNDEZ-TERNERO et al. 2PRELIMINARIES ON EVOLUTION ALGEBRAS Let E≡(E,+,⋅)be an algebra over a field K.ItissaidthatEis an evolution algebra if there exists a basis ={ei∶i∈Λ} of E,whereΛis an index set, such that ei⋅ej=0, if i≠j. The basis is called a natural basis. Since is a basis, the product ej⋅ej=e2 jcan be written as ∑i∈Λ aijei,withaij ∈K, where only a finite quantity of aij, called structure constants, are nonzero for each j∈Λfixed. So, the product on Eis determined by the structure matrix A =(aij). Tian defined in Reference 15 the evolution operator associated with as the endomorphism L∶E→Ewhich maps each generator into its square, that is, L(ej)=e2 j=∑i∈Λ aijei, for all j∈Λ. The matrix representation of the evolution operator with respect to the basis is the structure matrix A. Throughout this article we will differentiate between two types of algebras, the nondegenerate ones, when e2 j≠0, for all j∈Λ, and the degenerate ones, when there exists jsuch that e2 j=0. In general, evolution algebras are neither associative nor power-associative, but they are commutative and hence flexible. 3EVOLUTION ALGEBRAS WHOSE EVOLUTION OPERATOR IS A HOMOMORPHISM OF ALGEBRAS An endomorphism g∶E→Eis called a homomorphism of algebras if g(x⋅y)=g(x)⋅g(y), for all x,y∈E. In this section, we study those evolution algebras over Csuch that L∈Der(E), that is, the evolution operator is a derivation. This condition, as shown in Reference 14, is equivalent to A(cj⊙(cj− 1))=0,∀j,(1) A(ci⊙cj)=0,∀i≠j,(2) where cjis the jth column of Aand ⊙denotes the Hadamard product, that is, the element-by-element multiplication. A study is carried out in Reference 14 to characterize this type of algebras, up to ordering of the basis elements, reaching the following results in the nondegenerate case. Proposition 1. Let E be an evolution algebra of dimension n such that L ∈End(E).Letc kbe a nonnull column that can be written as a linear combination of others columns. Then, Ack=Ac(2) k=0. In particular, if E is nondegenerate and A has rank n −k, then there are at least k +1of these columns. Theorem 1. Let E be an evolution algebra with structure matrix A =(aij)such that L is an automorphism. Then, L ∈End(E) if and only if A is obtained by permuting the columns of the identity matrix. Theorem 2. Let n ≥2. Let E be a nondegenerate evolution algebra, with natural basis ={ei∶i=1,…,n}and structure matrix A =(aij)with rank n −1. Then, L ∈End(E)if and only if there exists a rearrangement of the basis such that the structure matrix has the form (A1A2 OP ), where •A1=(a−a a−a),forsomea≠0. •A2is a 2×(n−2)matrix whose two rows are equal. •Oisthe(n−2)×2zero matrix. •Pisa(n−2)×(n−2)matrix obtained by permuting the columns of the identity matrix of size n −2. Theorem 3. Let E be a nondegenerate evolution algebra of dimension n. Suppose the structure matrix A has range 1. Then, L∈End(E)if and only if  𝜆=(1,𝜆 2,…,𝜆 n)tis orthogonal to c1and c(2) 1, where cj=𝜆jc1, for all j ≠1. If Eis degenerate, the following algorithm to obtain all the structure matrices Aof these algebras in dimension nis given. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License FERNÁNDEZ-TERNERO et al. 3of14 Step 1 Obtain all matrices A′of dimension n−1, both degenerate and nondegenerate. Step 2 For each A′obtained, find the vector space orthogonal to {c′ j⊙(c′ j−1),c′ i⊙c′ j;i≠j}. Then, the searched matrices are those of the form A=(A′0 vt0),withvbelonging to this orthogonal space. Additionally, the complete classification for dimensions less than four is given in Reference 14. 3.1 Characterization for rank n−2 We give next a characterization for rank n−2, which will be used later to obtain the complete classification of these algebras in dimension 4. Theorem 4. Let n ≥4. Let E be a nondegenerate evolution algebra with natural basis ={ei∶i=1,…,n}and structure matrix A =(aij)of rank n −2. Then, L ∈End(E)if and only if there exists a rearrangement of the elements of the basis  such that one of the following cases is obtained 1. The structure matrix has the form (A1A2 OP ), where •Oisthe(n−r)×r zero matrix. •Pisa(n−r)×(n−r)matrix obtained by permuting the columns of the identity matrix of size n −r. •A1and A2are matrices of dimension r ×randr×(n−r), respectively, and match one of these descriptions (a) r =3and there exist a,b≠0,witha≠−b, such that A1=⎛⎜⎜⎜⎝ ab−a−b ab−a−b ab−a−b⎞⎟⎟⎟⎠ . A2has all its rows the same. (b) r =3and there exist a,b≠0, such that A1=⎛⎜⎜⎜⎝ a−ab a−ab 000 ⎞⎟⎟⎟⎠ . A2has the first two rows the same. In the third row there is at most a 1. (c) r =3and there exist a,b,c≠0and different from each other, such that ⎛⎜⎜⎜⎝ aa𝜆a𝜇 bb𝜆b𝜇 cc𝜆c𝜇⎞⎟⎟⎟⎠ , with 𝜆=− a(c−a) b(c−b)and 𝜇=− a(b−a) c(b−c). Column j of A2is of the form (𝛼j+𝛽ja,𝛼 j+𝛽jb,𝛼 j+𝛽jc)t, where there is at most a nonnull 𝛼jand equal to 1. (d) r =4and there exist a,b,c,d, with ad −bc ≠0, such that A1=⎛⎜⎜⎜⎜⎜⎝ a−ac−c a−ac−c b−bd−d b−bd−d ⎞⎟⎟⎟⎟⎟⎠ . A2has its first two rows the same and its last two rows the same. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 4of14 FERNÁNDEZ-TERNERO et al. 2. Same assumptions as (1b), with a 1 in the third row of A2and canceling in P the column that corresponds to this element. 3. Same assumptions as (1c), with an 𝛼j=1and canceling in P the column that corresponds to this element. Proof. Let V=null(A)and Ω={i∶∃v=(v1,…,vn)t∈V,vi≠0}.Ifi∈Ω, then there exists v=(v1,…,vn)t∈Cnsuch that Av =0andvi≠0. From Av =0, we get that ∑n i=1vici=0. Since vi≠0, Proposition (1) implies ci∈V.Thatis,ifi∈Ω, then ci∈V. Let us see that |Ω|≤4. By contradiction, suppose |Ω|>4. Then, there are more than four columns that belong to V. Since the dimension of Vis 2, the rank of the matrix formed by these columns is less than or equal to 2. This implies that the rank of A is less than 2 +(n−4)=n−2, which is a contradiction. In addition, |Ω|≠1, since otherwise any vector of Vwould have all its coordinates null, except one, from which it follows that Vwould be contained in a one-dimensional space, which is not possible since its dimension is 2. Similarly, |Ω|≠2, since if Ω={i,j},thenVis the space generated by eiand ej.FromAej=0, it would be deduced that cj=0, which is not possible. Therefore, |Ω|is 3 or 4. We analyze these two cases •If |Ω|=3, we can assume by a rearrangement of the basis that Ω={1,2,3}.Then,c1,c2,c3,c(2) 1,c(2) 2,c(2) 3∈V,and therefore these columns can only have their first three entries nonnull. We split the matrix Aas follows A=(A1A2 OP ), with A1of dimension 3 ×3. Clearly Ois the zero matrix. Note that if v=(v1,v2,v3,0,…,0)t∈V, then the product Av =0 is equivalent to the product A1(v1,v2,v3)t=0. Since the only nonnull coordinates of the vectors in Vare the first three ones, we decompose Rnas Rn=R3×Rn−3, and we identify the subspace Vwith its projection in R3. Let j≥4. From the condition (1), all the coordinates of cj⊙(cj− 1), except the first three ones, are null. That is, aij is 1 or 0, for all i≥4. Considering the constraints of the conditions (1) and (2) to A1, we deduce that this matrix must be the structure matrix of a nondegenerate evolution algebra of dimension 3 whose evolution operator is a homomorphism of algebras. Moreover, the rank of this matrix must be less than or equal to 2, since c1,c2,c3∈Vand dim(V)=2. We know all these matrices (up to rearrangement of the basis vectors) so we can distinguish the following cases –IfA1is of the form ⎛⎜⎜⎜⎝ a−a𝜆 a−a𝜆 a−a1⎞⎟⎟⎟⎠ , for some a≠0. In this case c3∉V, which is a contradiction. –IfA1is of the form ⎛⎜⎜⎜⎜⎝ ab−a−b ab−a−b ab−a−b ⎞⎟⎟⎟⎟⎠ , for some a,b≠0, with a≠−b. Let j≥4. Since c1⊙cj∈V,then(a1j,a2j,a3j)∈V. From this, along with cj⊙(cj− 1)∈V, it follows that (a2 1j,a2 2j,a2 3j)∈V.Thatis {a(a1j−a3j)+b(a2j−a3j)=0, a(a2 1j−a2 3j)+b(a2 2j−a2 3j)=0. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License FERNÁNDEZ-TERNERO et al. 5of14 Let us see that the only solution to this system is a1j=a2j=a3j, so all the rows in A2are equal. The second equation is equivalent to a(a1j−a3j)(a1j+a3j)+b(a2j−a3j)(a2j+a3j)=0. From the first equation, we obtain that b(a2j−a3j)=−a(a1j−a3j), so the previous equation remains a(a1j−a3j)(a1j−a2j)=0, from which it follows that a1j=a2j=a3j. Let us see the structure of the submatrix P. From the condition (1), it follows that all the entries in this matrix are zeros and ones. From the condition (2), it follows that there are not more than a 1 per row, since otherwise the product of the corresponding columns would have one of its last n−3 coordinates equal to 1, which cannot be true because Ω={1,2,3}. In addition, any column cj,withj≥4, must have a 1, since otherwise that column would be a multiple of c1and the rank of Awould be less than n−2. Thus, Pis a matrix obtained by permuting the columns of the identity matrix of size n−4. –IfA1is of the form ⎛⎜⎜⎜⎜⎝ a−ab a−ab 000 ⎞⎟⎟⎟⎟⎠ , for some a,b≠0. Let j≥4. Since c1⊙cj∈V,then(a1j,a2j,0)∈V,fromwhichwegeta1j=a2j.Sincecj⊙(cj− 1)∈V,then u∶= ⎛⎜⎜⎜⎜⎝ 0 0 a3j(a3j−1) ⎞⎟⎟⎟⎟⎠ =⎛⎜⎜⎜⎜⎝ a1j(a1j−1) a1j(a1j−1) a3j(a3j−1) ⎞⎟⎟⎟⎟⎠ −⎛⎜⎜⎜⎜⎝ a1j(a1j−1) a1j(a1j−1) 0 ⎞⎟⎟⎟⎟⎠ ∈V. From Au =0, we get that a3j(a3j−1)c3=0, so a3j∈{0,1}. Therefore, the first and second rows of A2are the same, and in the third one a 1, at most, can appear, since otherwise the condition (2) would not be fulfilled. Only in the case in which this 1 exists (say, in position (3,j0)), it could happen that the matrix Pis not a permutation by columns of the identity matrix, since the 1 located in the j0th column could be removed from P, leaving the corresponding column null. –IfA1is of the form ⎛⎜⎜⎜⎜⎝ aa𝜆a𝜇 bb𝜆b𝜇 cc𝜆c𝜇 ⎞⎟⎟⎟⎟⎠ , for some b,c≠0, with b≠c,𝜆=− a(c−a) b(c−b)and 𝜇=− a(b−a) c(b−c).Since𝜆and 𝜇cannot be null, also b≠a≠c. Then, {c1,c(2) 1}is a basis of V, since a vector and its square are linearly independent if and only if the vector has at least two different nonnull coordinates. In addition, we have that a≠0 by the definition of Ω. Let j≥4. Since c1⊙cj∈V, then there exist 𝛼jand 𝛽jsuch that ⎛⎜⎜⎜⎝ aa1j ba2j ca3j⎞⎟⎟⎟⎠ =𝛼j⎛⎜⎜⎜⎝ a b c⎞⎟⎟⎟⎠ +𝛽j⎛⎜⎜⎜⎝ a2 b2 c2⎞⎟⎟⎟⎠ . 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 6of14 FERNÁNDEZ-TERNERO et al. Therefore ⎧ ⎪ ⎨ ⎪ ⎩ a1j=𝛼j+𝛽ja, a2j=𝛼j+𝛽jb, a3j=𝛼j+𝛽jc. Since cj⊙(cj−1) ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ∈V =(𝛼j+𝛽jc1)⊙(𝛼j+𝛽jc1− 1) =𝛽2 jc(2) 1 ⏟⏟⏟ ∈V +𝛽j(2𝛼j−1)c1 ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ∈V +𝛼j (2)−𝛼j, then 𝛼j (2)−𝛼j∈V.Since |||||||| a2a1 b2b1 c2c1|||||||| =(a−b)(a−c)(b−c)≠0, then there are no nonzero constant vectors in V,so𝛼j∈{0,1}. Since ci⊙cj ⏟⏟⏟ ∈V =(𝛼i+𝛽ic1)⊙(𝛼j+𝛽jc1) =𝛽i𝛽jc(2) 1 ⏟⏟⏟ ∈V +(𝛼i𝛽j+𝛼j𝛽i)c1 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ∈V +𝛼i⊙𝛼j, then 𝛼i⊙𝛼j∈V, so there can only exist, at most, one nonnull 𝛼j. Only if there is a 𝛼j0=1, it could happen that the matrix Pis not a permutation by columns of the identity matrix, since the 1 located in the j0th column could be removed from P, leaving this column null. •If |Ω|=4, we can assume by a rearrangement of the basis that Ω={1,2,3,4}.Then,cj,c(2) j∈V, for j=1,2,3,4. Analogously to the previous case, we split the matrix Aas A=(A1A2 OP ), with A1of dimension 4 ×4andOthe zero matrix, and we identify the subspace Vwith its projection in R4. Rows of A1provide implicit equations of V, so the rank of A1must be 2. Suppose, by reordering the basis, that the first two rows of A1are linearly independent. We distinguish two cases – All the nonnull entries of cjare equal, for every j∈{1,2,3,4}.Letusseti0∈{3,4}. Suppose that the row i0is not null. Since it is a linear combination of the first two rows and the nonnull elements in a column are equal, the only possibility is that row i0is equal to row 1 or row 2. Let us rearrange the first four elements of the basis in such a way that c1is the column with the highest number of zeros, regardless of the fact that maybe the first two rows are not linearly independent. * Iftherearenozerosinc1, then there are no zeros in A1and all rows are equal, which is not possible due to the rank of A1is 2. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License FERNÁNDEZ-TERNERO et al. 7of14 * If there is a 0 in c1at position i0,fromAc1=0 we obtain that c1+c2+c3+c4−ci0=0, that is to say, there are three columns that add up to 0. If there is a column without zeros, we obtain that c1+c2+c3+c4=0, so ci0=0, which cannot be possible. Therefore there is a zero in all columns. Since c1⊙cj∈V, if these zeros are in a position i1≠i0, we obtain that c1+c2+c3+c4−ci0−ci1=0, from which we get ci1=0. Since this is not possible, all zeros are in the same row, and the other three rows of A1would be the same, which is a contradiction due to the rank of A. * If there are three zeros in c1,fromAc1=0 we get that there is a null column. * There cannot be four zeros in c1because c1≠0. Therefore there are two zeros in c1. A similar reasoning to the one done above shows us that, if there are zeros in one column, there must be two of them and in the same or opposite positions as in c1. Reordering the elements 2, 3, 4 of the basis (this leaves the first column with the same number of zeros), we can assume that c1=(a,a,0,0)tor c1=(0,0,a,a)t,witha≠0. Since c1∈V,itisnecessarythatc2=−c1. In these cases A1can be ⎛⎜⎜⎜⎜⎜⎝ a−ac−c a−ac−c 00c−c 00c−c ⎞⎟⎟⎟⎟⎟⎠ ,⎛⎜⎜⎜⎜⎜⎝ a−a00 a−a00 00c−c 00c−c ⎞⎟⎟⎟⎟⎟⎠ , ⎛⎜⎜⎜⎜⎜⎝ 00c−c 00c−c a−ac−c a−ac−c ⎞⎟⎟⎟⎟⎟⎠ or ⎛⎜⎜⎜⎜⎜⎝ 00c−c 00c−c a−a00 a−a00 ⎞⎟⎟⎟⎟⎟⎠ . Recall that the rows of these matrices determine the implicit equations of V. From condition (2) we get that the first two rows of A2are equal, and the last two as well. The matrix Pis the expected one. –Thereexistsj∈{1,2,3,4}such that cjhas at least two distinct nonnull values. By reordering the basis, we assume that this column is the first one, c1=(a,b,c,d)t, and that a≠b.Then,{c1,c(2) 1}is a basis of V.Alsoa,b,c,d≠0, by the definition of Ω.Let𝜆j,𝜇 jbe such that cj=𝜆jc1+𝜇jc(2) 1.Letj0be such that 𝜇j0≠0 (there must exist at least one or the rank of A1would be 1). Then c1⊙cj0 ⏟⏟⏟ ∈V =𝜆j0c(2) 1 ⏟⏟⏟ ∈V +𝜇j0c(3) 1, that is to say, c(3) 1∈V.Then 0=|||||||| a3a2a b3b2b c3c2c|||||||| =abc(a−b) ⏟⏞⏞⏞⏟⏞⏞⏞⏟ ≠0 (a−c)(b−c), 0=|||||||| a3a2a b3b2b d3d2d|||||||| =abd(a−b) ⏟⏞⏞⏞⏟⏞⏞⏞⏟ ≠0 (a−d)(b−d). Hence c,d∈{a,b}. Let us see the different possibilities *Ifc=d=a,thenc1=(a,b,a,a), and another basis of Vis {(1,0,1,1)t,(0,1,0,0)t}.SinceA(0,1,0,0)t=0, then c2=0, which is a contradiction. *Ifc=d=b,thenc1=(a,b,b,b), and another basis of Vis {(1,0,0,0)t,(0,1,1,1)t}.SinceA(1,0,0,0)t=0, then c1=0, which is a contradiction. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License 8of14 FERNÁNDEZ-TERNERO et al. TABLE 1 Computing time Input Computing time (s) Input Computing time (s) n=2 0.063 n=80 26.703 n=60.219 n=100 44.469 n=10 0.406 n=120 69.309 n=20 1.375 n=140 101.391 n=30 3.078 n=160 143.172 n=40 5.531 n=180 195.469 n=60 13.656 n=200 258.422 Therefore {c,d}={a,b}. By a rearrangement of the basis we can assume that c1=(a,a,b,b).Them {(1,1,0,0)t,(0,0,1,1)t}is a basis of V, from which it follows that c2=−c1and c4=−c3. Hence A1is of the form ⎛⎜⎜⎜⎜⎜⎝ a−ac−c a−ac−c b−bd−d b−bd−d ⎞⎟⎟⎟⎟⎟⎠ , with ad −bc ≠0 so that the rank of A1is 2. The first two rows of A2are equal, and the last two as well. The matrix Pis the expected one. ▪ 3.2 The algorithm for the degenerate case In this subsection we implement the algorithm for the degenerate case in Python and we make a computational study of its complexity. The code to execute the second step of the algorithm for a given matrix is as follows from sympy import Matrix,matrix_multiply_elementwise,ones def algorithm(A): n=A.shape[0] M1 =Matrix([[matrix_multiply_elementwise(A.col(i),A.col(i)-ones(n,1)).T] for iin range(n)]) M2 =Matrix([[matrix_multiply_elementwise(A.col(i),A.col(j)).T] for iin range(n) for jin range(i+1,n)]) M=Matrix([[M1],[M2]]) return M.nullspace() To do the complexity study, we have considered the family of evolution algebras whose structure matrix is a block matrix with blocks of the form (a−a a−a). It is easy to check that all these matrices correspond to evolution algebras whose evolution operator is a homomorphism of algebras. The algorithm has been implemented with an Intel Core i5-1135G7 processor (frequency of 2.4 GHz) and 16 GB of RAM. Table 1 shows the computing time used to return the output of the whole procedure according to the dimension nof the algebra. 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License FERNÁNDEZ-TERNERO et al. 9of14 TABLE 2 Real computing time and expected time by the cubic fit in higher dimensions Input Computing time (s) Expected time (s) Relative error n=220 334.894 333.528 0.0041 n=240 424.325 422.120 0.0052 n=260 529.316 525.170 0.0078 n=280 649.094 643.775 0.0082 n=300 785.872 779.031 0.0087 FIGURE 1 Least square cubic polynomial fit to the experimental data and test data for predictions Let us compute the complexity of the algorithm taking into account the number of operations carried out in the worst case. For this purpose, we use the big Onotation: Given two functions f,g∶R→R,wesaythatf(x)=O(g(x)) if there exist M∈R+and x0∈Rsuch that |f(x)|≤Mg(x), for all x≥x0. In our algorithm, the first loop is executed ntimes, and the second loop is executed n(n−1) 2=O(n2)times. Functions matrix−multiply−elementwise,col,Thas complexity O(n), so the complexity of the loops are O(n2)and O(n3), respectively. If function nullspace runs in O(n3), then the complexity of the algorithm is O(n3). To check if this assumption is correct, we calculate the least square cubic polynomial fit to our experimental data in Table 1, and we make predictions for larger dimensions (n=220,240,260,280,300). As can be seen in Table 2 and Figure 1, our data fits this model, so our algorithm has complexity O(n3). 3.3 Four-dimensional evolution algebras In this subsection we give the complete classification of these algebras in dimension 4, up to reordering of the basis elements. 3.3.1 Nondegenerate evolution algebras ByTheorem1,thoseofmaximumrankare ⎛⎜⎜⎜⎜⎜⎝ 1000 0100 0010 0001 ⎞⎟⎟⎟⎟⎟⎠ ,⎛⎜⎜⎜⎜⎜⎝ 1000 0100 0001 0010 ⎞⎟⎟⎟⎟⎟⎠ ,⎛⎜⎜⎜⎜⎜⎝ 1000 0001 0100 0010 ⎞⎟⎟⎟⎟⎟⎠ ,⎛⎜⎜⎜⎜⎜⎝ 0100 1000 0001 0010 ⎞⎟⎟⎟⎟⎟⎠ ,and ⎛⎜⎜⎜⎜⎜⎝ 0001 1000 0100 0010 ⎞⎟⎟⎟⎟⎟⎠ . 25777408, 2021, 6, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/cmm4.1200 by Universidad De Sevilla, Wiley Online Library on [14/04/2023]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License