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## Discussion on Scientific Approach

In the appendix, we discuss the scientific approach of this survey. First of
all, we would like to clarify our motivation — mHealth sensing, where we
include a brief discussion on the comparisons between mHealth sensing and
sensing techniques in general medical settings. Later, we review the
scientific procedures and criteria that we select publications for review.
Finally, we review the scientific way that we built the two taxonomy systems.

###  -A Research Definition: mHealth Sensing versus Medical Sensing

In this paper, we give a comprehensive survey on _mHealth Sensing_ techniques,
where the topic (e.g., mHealth Sensing) is close but significantly from “
_Medical Sensing_ ”. Here we discuss the major differences between the two
types of sensing techniques in a structured manner, including “target
populations” who need the two sensing techniques, “deployment contexts” that
the two sensing techniques are adopted, “medical goals” that the two sensing
techniques aim to meet, and “methodologies” that the two sensing techniques
propose to collect data.

As shown in Table IV, the _mHealth Sensing_ and _Medical Sensing_ techniques
vary significantly. Specifically, _mHealth Sensing_ techniques are majorly
designed for voluntary/active users in some daily/commercial scenarios, for
health-related behaviors monitoring and intervention, by leveraging
wearable/mobile devices. The overall goals of mHealth Sensing are improving
health status and well-beings by studying the innovative applications of
mobile sensing techniques to collect behavioral and environmental data. In
contrast, the _Medical Sensing_ takes care of patients in medicare context
such as hospital or clinical scenarios, where it aims at providing the medical
diagnoses and treatments through professional devices, where the overall goals
of Medical Sensing is to design new measurement and instrument techniques for
medical purposes. Thus, the mHealth Sensing mainly discussed in this paper
essentially differ from Medical Sensing.

TABLE IV: mHealth Sensing versus Medical Sensing | mHealth Sensing | Medical Sensing
---|---|---
Populations | Voluntary or active users | Patients in medical care
Contexts | Daily or commercial | Hospital or clinical
Coverage | Monitoring and intervention | Diagnosis and treatment
Goals | Health issues for well-being | Medical issues for clinics
Methods | Wearable/mobile devices | Professional devices

###  -B Research Methods: Selection Procedure and Criteria for Related Works

In tis survey, we refer more than 300 technical papers and review them from
applications and taxonomies perspectives. Generally, we collect these
publications in scientific way as follows.

  * •

First, we cover several notable publications from our previous works in
_mHealth Sensing_ , mobile systems design, and crowd sensing and data
analytics. Specifically, we include our works on mobile systems design to
understand the dynamics and personalization of health and well-being [284, 7,
52, 82, 83, 241, 27, 199, 268, 267]. We also introduce our works in smart
healthcare, elderly care and context-aware computing for health [103, 100,
102, 46, 284, 235, 307, 236, 237, 286], and the fundamentals of mobile crowd
sensing and crow data analysis for health [282, 57, 234, 284, 283, 235, 7, 4,
82, 52, 236, 126, 132, 131]. The inclusion of our previous works further
confirms the professionality of our group to carry out this survey.

  * •

Besides, we also cover some notable works from pioneering research groups in
mHealth sensing domains. Specifically, we cover works from Campbell’s lab that
focus on developing mobile sensing technology capable of accessing mental
health [1, 10, 84, 209, 246, 310, 295, 311, 242, 201], Murphy’s works on
decision making and interventions in mHealth [43, 300, 18, 142, 301, 247, 143,
277], Choudhury’s works on the mHealth sensing systems for people’s context,
activities and social networks [9, 10, 11, 146, 88, 62, 84, 136, 246, 316, 49,
138, 317, 311, 141, 139], as well as Mascolo’s works on mobile and wearable
systems design to human behaviour understanding from mHealth perspectives [14,
332, 114, 329, 5, 115]. Of-course, we also include and discuss works from
other important groups.

  * •

Later, publications are selected from leading technical conferences and
journals in the related areas, such as IEEE Internet of Things Journal [235,
224, 195, 281, 45, 153], IEEE Communications Magazine [1, 67, 46, 52, 237],
IEEE Pervasive Computing [331, 9], IEEE Sensors [334, 249, 211, 279], IEEE
Transactions on Mobile Computing [57, 284, 236], Proceedings of the ACM
Transactions on Interactive, Mobile, Wearable and Ubiquitous Technologies
(aka., ACM UbiComp) [56, 88, 282, 296, 7, 84, 4, 272, 246, 43, 83, 310, 286,
87, 360, 220], as well some medical journals such as JAMA [20, 35, 218, 127],
Journal of Medical Internet Research [78, 54, 98, 80, 82, 137, 266, 107, 274,
208], New England Journal of Medicine [251, 70], the Lancet [97, 128, 337],
Nature and Nature Communication [79, 240, 121, 118, 139], Frontiers in
neuroscience [335, 81]. Of-course, we also include and discuss works from
other important journals and conferences.

  * •

Finally, we also cover the publications contributed by the related projects
funded by National Science Foundation (NSF), National Institute of Health
(NIH) and other funding agency. Actually, we pay special attentions to the
works from Mobile Sensor Data to Knowledge (MD2K)
[361]111https://md2k.org/research-agenda/pubs.html which was established by
the National Institutes of Health Big Data to Knowledge Initiative, and also
the Trustworthy Health & Wellness (THaW)222https://thaw.org/,
https://www.zotero.org/groups/2647330/thaw/library that aims at “Making mobile
health effective and secure”
333https://www.nsf.gov/discoveries/disc_summ.jsp?cntn_id=137188.

Based on the above procedures, we collect, review and discuss related works in
mHealth sensing areas.

###  -C Scientific Methods in Taxonomy Systems

In this paper, we followed a “Descriptive and Mapping Reviews” 444Chapter 9
Methods for Literature Reviews, Handbook of eHealth Evaluation: An Evidence-
based Approach [Internet], https://www.ncbi.nlm.nih.gov/books/NBK481583/
pattern to organize the survey, where we extracted a body of knowledge from
existing research works on mHealth sensing. The body of knowledge included a
list of health issues related to mHealth sensing and the publications, two
taxonomy systems summarizing and classifying the research topics in mHealth
sensing area.

Actually, we first reviewed the existing mHealth sensing apps from the
perspectives of health issues, where a particular attention to the significant
health issues that have been widely studied in mHealth sensing has been paid.
We summarized and generated the list of health issues that we took care of
from the collected publications (see also in Appendix B). We also discussed
the important health issues, such as addictive behaviors, that we might ignore
in this survey.

For the two taxonomy systems, we followed simple “Narrative Reviews” 555Samuel
J. Stratton, MD, MPH, Literature Reviews: Methods and Applications,
https://www.cambridge.org/core/services/aop-cambridge-
core/content/view/70581E0B68B491693E8360DE39E0D6E4/S1049023X19004588a.pdf/literature_reviews_methods_and_applications.pdf
strategies on classifying literature works: (1) what technical problems they
aimed to solve in the paper (objectives), and (2) how they solved the problem
(methodologies). With these two strategies in mind, we develop the taxonomy
systems on “mHealth Sensing Objectives” and “D&I Issues of mHealth Sensing”
that classify the existing works according to their sensing problems and
solutions respectively. Of-course the interplays between objectives and D&I
issues are also discussed in an ad-hoc manner. After-all, we demonstrate our
visions in the area.


# Set of evolution operators of an evolution algebra

D. Fernández-Ternero and V.M. Gómez-Sousa and J. Núñez-Valdés  Dpto. de
Geometría y Topología, Universidad de Sevilla, C/ Tarfia S/N, Sevilla, 41012,
Spain.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

An automorphism defined on an evolution algebra can provide both a finite
number and an infinite number of evolution operators on it. This question is
dealt with in the paper, as well as others more related to the evolution
operators of evolution algebras. After defining the concept of the set of
evolution operators of an evolution algebra, differences between trivial and
non-trivial sets of evolution operators are also covered. Some properties of
these concepts are studied and several examples of the above issues are shown.

###### Key words and phrases:

Evolution algebra, evolution operator, automorphism

###### 2010 Mathematics Subject Classification:

47C05, 17D92, 17A36, 17A20

The first and the second authors were partially supported by MICINN Spain
Research Project PID2020-118753GB-I00. All authors were partially suported by
Junta de Andalucía Research Group FQM–326.

##  1\. Introduction

Evolution algebras were firstly introduced in the Ph.D. Thesis of Jianjun Paul
Tian in 2004 ([8]). These algebras are non-associative and commutative, but
unlike many other non-associative algebras such as Lie, Malcev or Jordan
algebras, evolution algebras are not described by a series of identities. An
evolution algebra $E\equiv(E,+,\cdot)$ over a field $\mathbb{K}$ is a
$\mathbb{K}$-algebra such that there exists a natural basis, that is, a basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$, such that $e_{i}\cdot e_{j}=0$, for all
$i\neq j$. The scalars $a_{ij}$ such that $e_{j}^{2}=e_{j}\cdot
e_{j}=\sum_{i=1}^{n}a_{ij}e_{i}$ will be called the structure constants, and
the matrix $A=(a_{ij})$ is said to be the structure matrix. Furthermore, the
evolution algebra is called non-degenerate if $e_{j}^{2}\neq 0$, for all $j$.
Otherwise, the algebra is called degenerate. Therefore, all the dynamic
information of an evolution algebra is collected in this matrix, from which
the operator that is the object of study in this paper is defined.

Fixed a natural basis, the linear map $L:E\rightarrow E$ which maps each
generator into its square, $L(e_{j})=e_{j}^{2}=\sum_{i=1}^{n}a_{ij}e_{i}$, is
called the evolution operator associated with $\mathcal{B}$. It is obvious
that $M_{\mathcal{B}}(L)=A$, that is, the matrix representation of this linear
transformation is the structure matrix.

Topics related to the evolution operator of evolution algebras are scant in
the literature to date, being [5, 6, 4, 7, 9] some of the several papers that
deal with this subject. In this work we delve into different aspects related
to this operator in order to solve this shortage and provide new properties
that make easier the study of this field.

Since the evolution operator depends on the chosen natural basis, it is not
unique. Therefore, the following questions naturally arise

  * •

Do evolution operators share any properties?

  * •

When do two evolution operators represent the same linear map?

  * •

When writing all evolution operators in the same basis, how are they related?

  * •

Is there any way to relate evolution operators and automorphisms?

The aim of this paper is to answer these questions in a general way. To do so,
in Section 2 we briefly introduce the main definitions and properties that
will be used, to later study, in Section 3, the connection between
automorphisms and evolution operators. Finally, in Section 4 we study the
structure of the set of all evolution operators in order to find connections
between these linear maps.

##  2\. Preliminaries

Below we define some concepts that will be used repeatedly throughout the
paper.

######  Definition 1.

Let $A=(a_{ij})$ and $B=(b_{ij})$ be two $n\times m$ matrices. We define the
following matrix operations

  * •

The Hadamard product $A\odot B=(a_{ij}b_{ij})$.

  * •

The Hadamard power $A^{(n)}=\underbrace{A\odot\dots\odot A}_{n\text{ times}}$.

  * •

$A*B=(c_{ij}^{k})$ a $n\times\frac{m(m-1)}{2}$ matrix whose columns are
indexed by pairs $(i,j)$ with $i<j$, considering the lexicographical order.
The column $(i,j)$ is $A_{*i}\odot B_{*j}$, that is, the Hadamard product of
the $i$th column of $A$ and the $j$th column of $B$.

The following result, whose proof can be found at [3], allows us to relate the
product of the algebra to the Hadamard product and the evolution operator.

######  Proposition 1.

Let $E$ be an evolution algebra with natural basis $\mathcal{B}$ and structure
matrix $A$. Then, the following diagram is commutative

${E\times
E}$${E}$${E}$$\scriptstyle{\odot}$$\scriptstyle{\cdot}$$\scriptstyle{L}$

where $\odot:E\times E\rightarrow E$ is defined as

$x\odot
y=\left(\sum_{j=1}^{n}x_{j}e_{j}\right)\odot\left(\sum_{j=1}^{n}y_{j}e_{j}\right)=\sum_{j=1}^{n}x_{j}y_{j}\,e_{j}.$

Equivalently, in matrix form, we get

$\left(x\cdot y\right)_{\mathcal{B}}=A\left(x_{\mathcal{B}}\odot
y_{\mathcal{B}}\right).$

Recall that, given two algebras $A$ and $A^{\prime}$, a linear map
$f:A\rightarrow A^{\prime}$ is said to be an homomorphism of algebras
(homomorphism for short) if $f(x\cdot y)=f(x)\cdot f(y)$, for every $x,y\in
A$. A homomorphism from an algebra to itself is an endomorphism and an
automorphism is a biyective endomorphism.

The following characterization of endomorphisms is well know. We take
advantage to give a shorter alternative proof to the one shown in [8].

######  Proposition 2.

Let $E$ be an evolution algebra with natural basis $\mathcal{B}$ and structure
matrix $A$. Then, a linear map $g:E\rightarrow E$ is an homomorphism if and
only if $A(G*G)=0$ and $AG^{(2)}=GA$, where $G=M_{\mathcal{B}}(g)$.

###### Proof.

A linear map $g$ is an homomorphism if and only if $g(e_{i})\cdot
g(e_{j})=g(e_{i}\cdot e_{j})=g(0)=0$ and $g(e_{j})^{2}=g(e_{j}^{2})$. Taking
coordinates with respect to $\mathcal{B}$ and making use of Proposition 1, the
first equality is

$0=\left(g(e_{i})\cdot
g(e_{j})\right)_{\mathcal{B}}=A\left(g(e_{i})_{\mathcal{B}}\odot
g(e_{j})_{\mathcal{B}}\right)=A\left(G_{*i}\odot G_{*j}\right),$

so it is equivalent to $A(G*G)=0$. Since

$\left(g(e_{j})^{2}\right)_{\mathcal{B}}=Ag(e_{j})_{\mathcal{B}}^{(2)}=AG_{*j}^{(2)},$
$\left(g(e_{j}^{2})\right)_{\mathcal{B}}=G\left(e_{j}^{2}\right)_{\mathcal{B}}=GA_{*j},$

the second equality is equivalent to $AG^{(2)}=GA$. ∎

We end these preliminaries with the following results given in [2].

######  Theorem 3.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A=(a_{ij})$. Let
$\mathcal{B}^{\prime}=\\{\eta_{j}=\sum_{i=1}^{n}g_{ij}e_{i}:j=1,\dots,n\\}$ be
another basis of $E$, with $supp(\eta_{j})=\\{i:g_{ij}\neq
0\\}=\\{i_{1},\dots,i_{r}\\}$.

  * •

If $\eta_{j}^{2}=0$, then $e_{i_{1}}^{2}=\dots=e_{i_{r}}^{2}=0$.

  * •

If $\eta_{j}^{2}\neq 0$, then
$rank(\\{e_{i_{1}}^{2},\dots,e_{i_{r}}^{2}\\})=1$.

######  Definition 2.

Let $E$ be an evolution algebra with natural basis $\mathcal{B}$.

  * •

We say that $E$ has a unique natural basis if any other natural basis of $E$
is equal to $B$ up to rearrangement and product by nonzero scalars.

  * •

We say that $E$ has Property (2LI) if for any two different vectors
$e_{i},e_{j}$ of $\mathcal{B}$, $\\{e_{i}^{2},e_{j}^{2}\\}$ is linearly
independent.

######  Corollary 4.

Let $E$ be a non-degenerate evolution algebra. Then the following assertions
are equivalent

  1. (1)

$E$ has a unique natural basis.

  2. (2)

$E$ has Property (2LI).

Degenerate evolution algebras do not satisfy any of these assertions.

##  3\. Automorphisms and evolution operators

######  Proposition 5.

Let $g:E\rightarrow E$ be an automorphism of an evolution algebra $E$ with
natural basis $\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix
$A=(a_{ij})$. Then, $\mathcal{B^{\prime}}=\\{g(e_{i}):i=1,\dots,n\\}$ is a
natural basis of $E$ and the evolution operator $L^{\prime}$ associated with
this basis fulfills that

$M_{\mathcal{B^{\prime}}}(L^{\prime})=A=M_{\mathcal{B}}(L).$

Furthermore, $M_{\mathcal{B}}(L^{\prime})=GAG^{-1}$ and these two operators
are the same linear map if and only if $AG=GA$, where $G=M_{\mathcal{B}}(g)$.

###### Proof.

If $i\neq j$, then $g(e_{i})\cdot g(e_{j})=g(e_{i}\cdot e_{j})=g(0)=0$, so
$\mathcal{B^{\prime}}$ is a natural basis.
Secondly, the evolution operator $L^{\prime}$ maps each element of
$\mathcal{B^{\prime}}$ onto its square

$g(e_{i})^{2}=g(e_{i}^{2})=g\left(\sum_{k=1}^{n}a_{ki}e_{k}\right)=\sum_{k=1}^{n}a_{ki}g(e_{k}),$

that is to say,
$[g(e_{i})^{2}]_{\mathcal{B^{\prime}}}=(a_{1i},\dots,a_{ni})^{t}$, which is
the $i$-th column of $A$. Hence it follows that
$M_{\mathcal{B^{\prime}}}(L^{\prime})=A$.
Finally, by the definition of $\mathcal{B^{\prime}}$, we have that
$G=M(\mathcal{B^{\prime}},\mathcal{B})$, so
$M_{\mathcal{B}}(L^{\prime})=M(\mathcal{B^{\prime}},\mathcal{B})M_{\mathcal{B^{\prime}}}(L^{\prime})M(\mathcal{B},\mathcal{B^{\prime}})=GAG^{-1}$.
This operator is the same as $L$ if and only if $A=GAG^{-1}$, or what is the
same, $AG=GA$. ∎

######  Remark 3.

On the same assumptions as above, $G^{n}AG^{-n}$ is the matrix of an evolution
operator with respect to the basis $\mathcal{B}$, for all $n\in\mathbb{Z}$,
since $g^{n}$ is also an automorphism.

We show below two examples in which $G^{n}AG^{-n}$ provides a finite number of
evolution operators and an infinite number of them, respectively.

######  Example 4.

Let $E$ be the evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2},e_{3}\\}$ and structure matrix

$A=\begin{pmatrix}a&a&b\\\ (\sqrt{2}-1)a&(\sqrt{2}-1)a&(\sqrt{2}-1)b\\\
c&c&d\end{pmatrix}\\!\\!,$

for some $a,b,c,d\in\mathbb{C}$. Let us consider the invertible linear map

$G=\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\\
\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\\ 0&0&1\end{pmatrix}\\!\\!.$

which satisfies that $G^{-1}=G$. This map is an automorphism, since

$A(G*G)=A\begin{pmatrix}\frac{1}{2}&0&0\\\ -\frac{1}{2}&0&0\\\
0&0&0\end{pmatrix}=0\textup{ and }AG^{(2)}=A=GA.$

Then,
$\mathcal{B^{\prime}}=\left\\{\frac{1}{\sqrt{2}}(e_{1}+e_{2}),\frac{1}{\sqrt{2}}(e_{1}-e_{2}),e_{3}\right\\}$
is a natural basis whose evolution operator satisfies
$M_{\mathcal{B^{\prime}}}(L^{\prime})=A$. The expression of this operator with
respect to the basis $\mathcal{B}$ is

$M_{\mathcal{B}}(L^{\prime})=GAG^{-1}=GAG=AG=\begin{pmatrix}\sqrt{2}a&0&b\\\
(2-\sqrt{2})a&0&(\sqrt{2}-1)b\\\ \sqrt{2}c&0&d\end{pmatrix}\\!\\!,$

which is different from $A$ if and only if $a$ or $c$ are nonzero. With this
assumption, $L$ and $L^{\prime}$ are not the same linear map, and there are
two different evolution operators. Note that, since $G^{2}=G$, the matrix
$G^{n}AG^{-n}$ is $A$ or $AG$.

The following example shows that there can be infinite evolution operators,
different from each other, but with the same matrix with respect to its
natural basis.

######  Example 5.

Let $E$ be the evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2},e_{3}\\}$ and structure matrix

$A=\begin{pmatrix}1&1&1\\\ -\frac{i}{2}&-\frac{i}{2}&-\frac{i}{2}\\\
-\frac{i}{2}&-\frac{i}{2}&-\frac{i}{2}\end{pmatrix}\\!\\!,$

Let us consider the invertible linear map

$G=\begin{pmatrix}\frac{5}{3}&-\frac{4i}{3}&0\\\ 0&0&1\\\
-\frac{4i}{3}&-\frac{5}{3}&0\end{pmatrix}\\!\\!,$

which satisfies that $G^{-1}=G^{t}$. This map is an automorphism, since

$A(G*G)=A\begin{pmatrix}-\frac{20i}{9}&0&0\\\ 0&0&0\\\
\frac{20i}{9}&0&0\end{pmatrix}=0\textup{ and }AG^{(2)}=A=GA.$

Then,
$\mathcal{B^{\prime}}=\left\\{\frac{1}{3}(5e_{1}-4ie_{3}),-\frac{1}{3}(4ie_{1}+5e_{3}),e_{2}\right\\}$
is a natural basis whose evolution operator satisfies
$M_{\mathcal{B^{\prime}}}(L^{\prime})=A$. The expression of this operator with
respect to the basis $\mathcal{B}$ is

$M_{\mathcal{B}}(L^{\prime})=GAG^{-1}=GAG^{t}=AG^{t}=\begin{pmatrix}\frac{5-4i}{3}&1&-\frac{5+4i}{3}\\\
-\frac{4+5i}{6}&-\frac{i}{2}&\frac{-4+5i}{6}\\\
-\frac{4+5i}{6}&-\frac{i}{2}&\frac{-4+5i}{6}\end{pmatrix}\\!\\!,$

which is different from $A$, so $L$ and $L^{\prime}$ are not the same linear
map. Let us see that the matrix $G^{n}AG^{-n}$, for $n\geq 0$, provides
infinite evolution operators, all of them different from each other.
Since $GA=A$, then $G^{n}AG^{-n}=AG^{-n}=A(G^{t})^{n}$. By contradiction,
suppose there exist $m>n\geq 0$, such that $A(G^{t})^{m}=A(G^{t})^{n}$, or
what is the same

(1) $\displaystyle A((G^{t})^{k}-I)=0,$

where $k=m-n$. The eigenvalues of $G^{t}$ are $1$, $\lambda$ and
$\frac{1}{\lambda}$, where

$\lambda=\frac{1}{3}+\frac{2\sqrt{2}}{3}\,i.$

Since $G^{t}$ has three different eigenvalues, it is diagonalizable, so there
exists $P$ such that $G^{t}=PDP^{-1}$, with
$D=diag(1,\lambda,\frac{1}{\lambda})$. Then, (1) is equivalent to
$AP(D^{k}-I)P^{-1}=0$, and thus, $AP(D^{k}-I)=0$. Let us denote the $i$-th
column of $P$ as $p_{i}$. Then we have

$0=A\begin{pmatrix}\vrule&\vrule&\vrule\\\ p_{1}&p_{2}&p_{3}\\\
\vrule&\vrule&\vrule\end{pmatrix}\begin{pmatrix}0&0&0\\\ 0&\lambda^{k}-1&0\\\
0&0&\frac{1}{\lambda^{k}}-1\end{pmatrix}=$
$=A\begin{pmatrix}0&\vrule&\vrule\\\
0&(\lambda^{k}-1)p_{2}&(\frac{1}{\lambda^{k}}-1)p_{3}\\\
0&\vrule&\vrule\end{pmatrix}\\!\\!.$

We distinguish two cases

  * •

If $\lambda^{k}\neq 1$, then $p_{2},p_{3}\in null(A)$. Therefore,
$0=AP(D-I)P^{-1}=A(G^{t}-I)$, which is a contradiction due to $AG^{t}\neq A$.

  * •

If $\lambda^{k}=1$, since the argument of $\lambda$ is
$\alpha=\arccos{\left(\frac{1}{3}\right)}$, then $k\alpha=2\pi l$, with
$l\in\mathbb{Z}$. This is equivalent to $\frac{\alpha}{\pi}=\frac{2l}{k}$,
which is a contradiction because $\frac{\alpha}{\pi}$ is not a rational
number.

In the previous example we have made use of the following proposition, which
was given by Aigner and Ziegler in [1].

######  Proposition 6.

For every odd integer $n\geq 3$, the number
$\frac{1}{\pi}\arccos{\left(\frac{1}{\sqrt{n}}\right)}$ is irrational.

As a consequence of Proposition 5 we get the following result.

######  Corollary 7.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ satisfying Property (2LI). Then
$Aut(E)\subseteq S_{n}\rtimes(\mathbb{K}^{\times})^{n}$.

###### Proof.

Let $g$ be an automorphism. Then,
$\mathcal{B}^{\prime}=\\{g(e_{i}):i=1,\dots,n\\}$ is a natural basis by
Proposition 5. Since $E$ has Property (2LI), then $E$ has a unique natural
basis by Corollary 4. That is, $\mathcal{B}^{\prime}$ is obtained from
$\mathcal{B}$ by a rearrangement and product by nonzero scalars. ∎

######  Proposition 8.

Let $E$ be a real evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A=(a_{ij})$. Let
suppose there exists $\pi\in S_{n}$ such that $a_{i\pi(i)}\neq 0$, for all
$i=1,\dots,n$. Then, $Aut(E)\cap(\mathbb{R}^{\times})^{n}=\\{id\\}$.

###### Proof.

A linear map $g$ satisfies $g\in Aut(E)\cap(\mathbb{R}^{\times})^{n}$ if and
only if $G=M_{\mathcal{B}}(g)=Diag(\lambda)$, for some
$\lambda=(\lambda_{i})_{i=1}^{n}$, $\lambda_{i}\neq 0$, $A(G*G)=0$ and
$AG^{(2)}=GA$. The identity $A(G*G)=0$ holds trivially and $AG^{(2)}=GA$ is
equivalent to $a_{ij}\lambda_{j}^{2}=\lambda_{i}a_{ij}$, that is $a_{ij}=0$ or
$\lambda_{i}=\lambda_{j}^{2}$.

Le us write $\pi$ as a product of disjoint cycles, $\pi=c_{1}\circ\dots\circ
c_{r}$. For all $i$ there is a unique $j$ such that $i\in supp(c_{j})$. Let
$l_{j}$ be the length of the cycle $c_{j}$. Then,

$\lambda_{i}=\lambda_{c_{j}(i)}^{2}=\dots=\lambda_{c_{j}^{l_{j}}(i)}^{2^{l_{j}}}=\lambda_{i}^{2^{l_{j}}}.$

The only real non-zero solution of the equation
$\lambda_{i}=\lambda_{i}^{2^{l_{j}}}$ is $\lambda_{i}=1$, so $g=id$. ∎

######  Proposition 9.

Let $E$ be a real evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i\in\Lambda=\\{1,\dots,n\\}\\}$ and structure matrix
$A=(a_{ij})$ without null rows. Then,
$Aut(E)\cap(\mathbb{R}^{\times})^{n}=\\{id\\}$.

###### Proof.

We proceed by induction in $n$. For $n=1$, it is trivial. We assume that the
statement holds for dimension $n-1$ in order to prove that it holds for
dimension $n$.

If $A=Diag(\lambda)P$, with $\lambda=(\lambda_{i})_{i\in\Lambda}$,
$\lambda_{i}\neq 0$ and $P$ being a permutation matrix, then the result
follows from Proposition 8.

In another case, as in the proof of Proposition 8, we have $g\in
Aut(E)\cap(\mathbb{R}^{\times})^{n}$ if and only if
$G=M_{\mathcal{B}}(g)=Diag(\lambda)$ and either $a_{ij}=0$ or
$\lambda_{i}=\lambda_{j}^{2}$, for all $i,j\in\Lambda$. Since $A\neq
Diag(\lambda)P$, there exists $k\in\Lambda$ such that the matrix
$A^{\prime}=(a_{ij})_{i,j\in\Lambda\setminus\\{k\\}}$ has no null rows. By
induction hypothesis, $\lambda_{i}=1$, for all $i\in\Lambda\setminus\\{k\\}$.
Since row $k$ is not null, there exists $l$ such that $a_{kl}\neq 0$ and then
$\lambda_{k}=\lambda_{l}^{2}$. If $k\neq l$, then
$\lambda_{k}=\lambda_{l}^{2}=1^{2}=1$. If $k=l$, then
$\lambda_{k}=\lambda_{k}^{2}$ and the only non-zero solution of this equation
is $\lambda_{k}=1$. Therefore, $\lambda_{i}=1$, for all $i\in\Lambda$, so
$g=id$. ∎

######  Example 6.

Let $E$ be the complex evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2}\\}$ and structure matrix

$A=\begin{pmatrix}0&a\\\ b&0\end{pmatrix}\\!\\!,$

with $a,b\neq 0$. Let us consider the linear map $g:E\rightarrow E$ with
$G=M_{\mathcal{B}}(g)=Diag(\lambda_{1},\lambda_{2})$ and
$\lambda_{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$,
$\lambda_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$. Then, $g\in
Aut(E)\cap(\mathbb{C}^{\times})^{n}$.

######  Example 7.

Let $E$ be the real evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2}\\}$ and structure matrix

$A=\begin{pmatrix}a&b\\\ 0&0\end{pmatrix}\\!\\!,$

with $a,b\neq 0$. Let us consider the linear map $g:E\rightarrow E$ with
$G=M_{\mathcal{B}}(g)=Diag(1,-1)$. Then, $g\in
Aut(E)\cap(\mathbb{R}^{\times})^{n}$.

######  Proposition 10.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A=(a_{ij})$. Then,
$S_{n}\subseteq Aut(E)$ if and only if $A=\alpha J+\beta I$, where $J$ is the
all-ones matrix and $\alpha,\beta\in\mathbb{K}$.

###### Proof.

Let $P$ be a permutation matrix. This permutation is an automorphism if and
only if $A(P*P)=0$ and $AP^{(2)}=PA$. The first equality is trivially
satisfied and the second is equivalent to $AP=PA$. The only matrices that
commute with any permutation matrix are those of the form $\alpha J+\beta I$.
∎

##  4\. Set of evolution operators

######  Proposition 11.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A=(a_{ij})$. Let
$\mathcal{B^{\prime}}=\\{\eta_{i}:i=1,\dots,n\\}$ be another basis of $E$ and
$G=M_{\mathcal{B}}(g)$, where $g$ is the linear map $g(e_{i})=\eta_{i}$. Then,
$\mathcal{B^{\prime}}$ is a natural basis if and only if $A(G*G)=0$. In this
case, the evolution operator $L^{\prime}$ associated with this basis fulfills
that

$M_{\mathcal{B^{\prime}}}(L^{\prime})=G^{-1}AG^{(2)},M_{\mathcal{B}}(L^{\prime})=AG^{(2)}G^{-1}.$

Furthermore, these two operators are the same linear map if and only if
$AG=AG^{(2)}$.

###### Proof.

The basis $\mathcal{B^{\prime}}$ is a natural basis if and only if
$\eta_{i}\cdot\eta_{j}=0$, for all $i\neq j$. By the definition of $g$, this
is the same as $g(e_{i})\cdot g(e_{j})=0$. Taking coordinates with respect to
$\mathcal{B}$,

$0=\left(g(e_{i})\cdot
g(e_{j})\right)_{\mathcal{B}}=A\left(g(e_{i})_{\mathcal{B}}\odot
g(e_{j})_{\mathcal{B}}\right)=A\left(G_{*i}\odot G_{*j}\right),$

so this is equivalent to $A(G*G)=0$. Since for all $j=1,\dots,n$

$\left(\eta_{j}^{2}\right)_{\mathcal{B^{\prime}}}=M(\mathcal{B},\mathcal{B^{\prime}})\left(\eta_{j}^{2}\right)_{\mathcal{B}}=G^{-1}A\left(\eta_{j}\right)_{\mathcal{B}}^{(2)}=G^{-1}AG_{*j}^{(2)},$

then $M_{\mathcal{B^{\prime}}}(L^{\prime})=G^{-1}AG^{(2)}$. Finally,

$M_{\mathcal{B}}(L^{\prime})=M(\mathcal{B^{\prime}},\mathcal{B})M_{\mathcal{B^{\prime}}}(L^{\prime})M(\mathcal{B},\mathcal{B^{\prime}})=GG^{-1}AG^{(2)}G^{-1}=AG^{(2)}G^{-1}.$

This operator is the same as $L$ if and only if $A=AG^{(2)}G^{-1}$, or what is
the same, $AG=AG^{(2)}$. ∎

######  Definition 8.

Given an evolution algebra $E$ with natural basis $\mathcal{B}$ and structure
matrix $A$, we define the set of all evolution operators with respect to its
own natural basis as

$\mathcal{L}=\left\\{G^{-1}AG^{(2)}:A(G*G)=0,rank(G)=n\right\\}$

and with respect to the natural basis $\mathcal{B}$ as

$\mathcal{L}_{\mathcal{B}}=\left\\{AG^{(2)}G^{-1}:A(G*G)=0,rank(G)=n\right\\}.$

######  Proposition 12.

All evolution operators of an evolution algebra $E$ have the same rank.

###### Proof.

Let $\mathcal{B},\mathcal{B}^{\prime}$ be two natural basis and $L,L^{\prime}$
the evolution operators associated with these basis, respectively. Let
$A:=M_{\mathcal{B}}(L)$ and
$A^{\prime}:=M_{\mathcal{B}^{\prime}}(L^{\prime})=G^{-1}AG^{(2)}$. Then,

$rank(A^{\prime})=rank(G^{-1}AG^{(2)})=rank(AG^{(2)})\leq rank(A).$

Exchanging the roles of $A$ and $A^{\prime}$ it follows that
$rank(A^{\prime})\leq rank(A)$. Therefore $rank(A)=rank(A^{\prime})$. ∎

######  Proposition 13.

$ADiag(\lambda)\in\mathcal{L}_{\mathcal{B}}$ for all
$\lambda=(\lambda_{i})_{i=1}^{n}\in\mathbb{C}^{n}$ with $\lambda_{i}\neq 0$,
for all $i=1,\dots,n$.

###### Proof.

Let $G=Diag(\lambda)$. Then, $G*G$ is the zero matrix and $A(G*G)=0$, so
$AG^{(2)}G^{-1}\in\mathcal{L}_{\mathcal{B}}$. That is to say,

$\displaystyle AG^{(2)}G^{-1}$
$\displaystyle=ADiag\left(\lambda\right)^{(2)}Diag\left(\lambda\right)^{-1}=$
$\displaystyle=ADiag\left(\lambda^{(2)}\right)Diag\left(\lambda^{(-1)}\right)=$
$\displaystyle=ADiag\left(\lambda^{(2)}\lambda^{(-1)}\right)=ADiag(\lambda)\in\mathcal{L}_{\mathcal{B}}.$

∎

######  Definition 9.

$\mathcal{L}_{\mathcal{B}}$ is said to be trivial if

$\mathcal{L}_{\mathcal{B}}=\left\\{ADiag(\lambda):\lambda=(\lambda_{i})_{i=1}^{n}\in\mathbb{C}^{n},\lambda_{i}\neq
0\right\\},$

and semitrivial if

$\mathcal{L}_{\mathcal{B}}\subseteq\left\\{ADiag(\lambda):\lambda=(\lambda_{i})_{i=1}^{n}\in\mathbb{C}^{n}\right\\}.$

######  Proposition 14.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A=(a_{ij})$
satisfying Property (2LI). Then, $\mathcal{L}_{\mathcal{B}}$ is trivial.

###### Proof.

If $E$ has Property (2LI) then $E$ has a unique natural basis by Corollary 4,
that is $G=Diag(\lambda)P$, with $\lambda_{i}\neq 0$ for all $i=1,\dots,n$ and
$P$ a permutation matrix. Then,

$\displaystyle AG^{(2)}G^{-1}$
$\displaystyle=A(Diag(\lambda)P)^{(2)}(Diag(\lambda)P)^{-1}=$
$\displaystyle=ADiag(\lambda)^{(2)}PP^{-1}Diag(\lambda)^{-1}=$
$\displaystyle=ADiag(\lambda^{(2)})Diag(\lambda^{(-1)})=ADiag(\lambda).$

∎

The following examples show that if $E$ does not satisfy (2LI) then
$L_{\mathcal{B}}$ may or may not be trivial.

######  Example 10.

Let $E$ be the evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2}\\}$ and structure matrix

$A=\begin{pmatrix}a&0\\\ b&0\end{pmatrix}\\!\\!,$

with $a\neq 0$ or $b\neq 0$. Then, $null(A)=\\{x_{1}=0\\}$ and
$G=\begin{pmatrix}g_{1}&g_{2}\\\ g_{3}&g_{4}\end{pmatrix}$ satisfies
$A(G*G)=0$ if and only if $g_{1}g_{2}=0$. It is easy to see that

  * •

If $g_{1}=0$, then

$\displaystyle AG^{(2)}G^{-1}$ $\displaystyle=\begin{pmatrix}a&0\\\
b&0\end{pmatrix}\begin{pmatrix}0&g_{2}^{2}\\\
g_{3}^{2}&g_{4}^{2}\end{pmatrix}\frac{1}{det(G)}\begin{pmatrix}g_{4}&-g_{2}\\\
-g_{3}&0\end{pmatrix}=$
$\displaystyle=\frac{1}{-g_{2}g_{3}}\begin{pmatrix}-ag_{2}^{2}g_{3}&0\\\
-bg_{2}^{2}g_{3}&0\end{pmatrix}=\begin{pmatrix}ag_{2}&0\\\
bg_{2}&0\end{pmatrix}\\!\\!.$

  * •

If $g_{2}=0$, then

$\displaystyle AG^{(2)}G^{-1}$ $\displaystyle=\begin{pmatrix}a&0\\\
b&0\end{pmatrix}\begin{pmatrix}g_{1}^{2}&0\\\
g_{3}^{2}&g_{4}^{2}\end{pmatrix}\frac{1}{det(G)}\begin{pmatrix}g_{4}&0\\\
-g_{3}&g_{1}\end{pmatrix}=$
$\displaystyle=\frac{1}{g_{1}g_{4}}\begin{pmatrix}ag_{1}^{2}g_{4}&0\\\
bg_{1}^{2}g_{4}&0\end{pmatrix}=\begin{pmatrix}ag_{4}&0\\\
bg_{4}&0\end{pmatrix}\\!\\!.$

Therefore $\mathcal{L}_{\mathcal{B}}$ is trivial.

######  Example 11.

Let $E$ be the evolution algebra with natural basis
$\mathcal{B}=\\{e_{1},e_{2}\\}$ and structure matrix

$A=\begin{pmatrix}a&a\\\ b&b\end{pmatrix}\\!\\!,$

with $a\neq 0$ or $b\neq 0$. Then, $G=\begin{pmatrix}1&-1\\\ 1&1\end{pmatrix}$
satisfies $A(G*G)=0$ and

$AG^{(2)}G^{-1}=\begin{pmatrix}a&a\\\ b&b\end{pmatrix}\begin{pmatrix}1&1\\\
1&1\end{pmatrix}\frac{1}{det(G)}\begin{pmatrix}1&1\\\
-1&1\end{pmatrix}=\begin{pmatrix}0&2a\\\ 0&2b\end{pmatrix}\\!\\!.$

Note that this matrix is not of the form $ADiag(\lambda_{1},\lambda_{2})$ with
$\lambda_{i}\neq 0$, so $\mathcal{L}_{\mathcal{B}}$ is not trivial.
Nevertheless, $AG^{(2)}G^{-1}=ADiag(0,2)$, so $\mathcal{L}_{\mathcal{B}}$
could be semitrivial.

######  Proposition 15.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$. Then, $\mathcal{L}_{\mathcal{B}}$ is
semitrivial.

###### Proof.

Let $\mathcal{B}^{\prime}=\\{\eta_{i}:i=1,\dots,n\\}$ be another natural basis
and let us write $e_{j}=\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\eta_{i_{k}}$,
with $g_{i_{k}j}^{\prime}\neq 0$. We distinguish two cases

  * •

If $e_{j}^{2}=0$, then we deduce that $\eta_{i_{k}}^{2}=0$ by Theorem 3.
Therefore

$\displaystyle L^{\prime}(e_{j})$
$\displaystyle=L^{\prime}\left(\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\eta_{i_{k}}\right)=\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}L^{\prime}(\eta_{i_{k}})=\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\eta_{i_{k}}^{2}=0,$
$\displaystyle L(e_{j})$ $\displaystyle=e_{j}^{2}=0.$

Hence $L^{\prime}(e_{j})=L(e_{j})$.

  * •

If $e_{j}^{2}\neq 0$, then we deduce that $\eta_{i_{k}}^{2}=\alpha_{i_{k}}v$
for some $v\in E$ by Theorem 3. Therefore

$\displaystyle L^{\prime}(e_{j})$
$\displaystyle=\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\eta_{i_{k}}^{2}=\left(\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\alpha_{i_{k}}\right)v,$
$\displaystyle L(e_{j})$
$\displaystyle=e_{j}^{2}=\left(\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\eta_{i_{k}}\right)^{2}=\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime
2}\eta_{i_{k}}^{2}=\left(\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime
2}\alpha_{i_{k}}\right)v\neq 0.$

Hence $L^{\prime}(e_{j})=\lambda_{j}L(e_{j})$ with
$\displaystyle\lambda_{j}=\frac{\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\alpha_{i_{k}}}{\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime
2}\alpha_{i_{k}}}$.

Therefore, there exists $\lambda_{j}$ such that
$L^{\prime}(e_{j})=\lambda_{j}L(e_{j})$, for all $j=1,\dots,n$. From this we
deduce that $L_{\mathcal{B}}$ is semitrivial, since
$M_{\mathcal{B}}(L^{\prime})=M_{\mathcal{B}}(L)Diag(\lambda)=ADiag(\lambda)$,
with $\lambda=(\lambda_{j})_{j=1}^{n}$. ∎

######  Remark 12.

In the previous proof, we can not claim that $E$ is trivial, since it is
necessary that $\lambda_{j}\neq 0$, that is,
$\sum_{k=1}^{r_{j}}g_{i_{k}j}^{\prime}\alpha_{i_{k}}\neq 0$, for all
$j=1,\dots,n$ with $e_{j}^{2}\neq 0$.

######  Proposition 16.

Let $E$ be an evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$. Let suppose there exist $r,s$ and
$\alpha\neq 0$ such that $e_{r}^{2}=\alpha e_{s}^{2}$ and $e_{s}^{2}\neq 0$.
Then, $\mathcal{L}_{\mathcal{B}}$ is not trivial.

###### Proof.

Let $\beta$ be such that $\beta^{2}=\alpha$ and consider the basis
$\mathcal{B}^{\prime}=\\{\eta_{i}:i=1,\dots,n\\}$ with

$\eta_{i}=\begin{cases}e_{i}&\text{ if }i\neq r,s,\\\ e_{r}+\beta e_{s}&\text{
if }i=r,\\\ e_{r}-\beta e_{s}&\text{ if }i=s.\end{cases}$

Then, $\mathcal{B}^{\prime}$ is a natural basis due to

$\eta_{r}\cdot\eta_{s}=(e_{r}+\beta e_{s})\cdot(e_{r}-\beta
e_{s})=e_{r}^{2}-\beta^{2}e_{s}^{2}=e_{r}^{2}-\alpha e_{s}^{2}=0.$

Furthermore, since $\eta_{r}^{2}=\eta_{s}^{2}=(1+\alpha)e_{s}^{2}$, we have
that

$L^{\prime}(e_{s})=L^{\prime}\left(\frac{1}{2\beta}(\eta_{r}-\eta_{s})\right)=\frac{1}{2\beta}(\eta_{r}^{2}-\eta_{s}^{2})=0.$

Since $e_{s}^{2}\neq 0$, then $\mathcal{L}_{\mathcal{B}}$ is not trivial. ∎

######  Corollary 17.

Let $E$ be a non-degenerate evolution algebra with natural basis
$\mathcal{B}=\\{e_{i}:i=1,\dots,n\\}$ and structure matrix $A$ not satisfying
Property (2LI). Then, $\mathcal{L}_{\mathcal{B}}$ is not trivial.

######  Corollary 18.

Let $E$ be a non-degenerate evolution algebra with natural basis
$\mathcal{B}$. Then the following assertions are equivalent

  1. (1)

$E$ has a unique natural basis.

  2. (2)

$E$ has Property (2LI).

  3. (3)

$\mathcal{L}_{\mathcal{B}}$ is trivial.

As a consequence, the fact of $\mathcal{L}_{\mathcal{B}}$ being trivial is an
intrinsic property of the algebra, that is, it does not depend on the natural
basis $\mathcal{B}$.

##  5\. Conclusions

In this work we have related evolution operators with homomorphisms, in
addition to introducing new concepts such as the set of evolution operators of
an evolution algebra, studying whether it is trivial or not. To the best of
our knowledge, there are no other papers in the literature studying the
relations between the different evolution operators, and our work can serve as
a starting point for research on this topic, of which some questions remain
open. For example, it can be studied if the fact of
$\mathcal{L}_{\mathcal{B}}$ being trivial or semitrivial depends on the
natural basis $\mathcal{B}$ in the degenerate case.

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