# Classification in chains of three-dimensional real evolution algebras

B. A. Narkuziyev, U.A. Rozikov  B.A. Narkuziyev, V.I.Romanovskiy Institute of
Mathematics of Uzbek Academy of Sciences, Tashkent, Uzbekistan.
<EMAIL_ADDRESS>U.Rozikova,b,c a V.I.Romanovskiy Institute of Mathematics
of Uzbek Academy of Sciences; b AKFA University, 1st Deadlock 10, Kukcha
Darvoza, 100095, Tashkent, Uzbekistan; c Faculty of Mathematics, National
University of Uzbekistan<EMAIL_ADDRESS>
###### Abstract.

A chain of evolution algebras (CEA) is an uncountable family (depending on
time) of evolution algebras on the field of real numbers. The matrix of
structural constants of a CEA satisfies Kolmogorov-Chapman equation. In this
paper, we consider three CEAs of three-dimensional real evolution algebras.
These CEAs depend on several (non-zero) functions defined on the set of time.
For each chain we give full classification (up to isomorphism) of the algebras
depending on the time-parameter. We find concrete functions ensuring that the
corresponding CEA contains all possible three-dimensional evolution algebras.

###### Key words and phrases:

Evolution algebra, chain of evolution algebra, Kolmogorov-Chapman equation,
isomorphism of algebras.

###### 2010 Mathematics Subject Classification:

17D92; 17D99;

##  1\. Introduction

An evolution algebra (EA) (see [22]) is an abstract system, it gives an
insight for the study of non-Mendelian genetics. In the relation between EAs
and Markov processes, the Chapman-Kolmogorov equation gives the fundamental
relationship between the probability transitions (kernels). There are many
recent papers devoted to the theory of evolution algebras (see for example
[1], [2], [12], [13], [17], [19] and the references therein).

In [3] a notion of a chain of evolution algebras (CEA) is introduced. Later,
in [8] and [9] (see also [19]) the notion of CEA was generalized and a concept
of flow of arbitrary finite-dimensional algebras is introduced. It is known
that each EA is determined by a quadratic matrix of structural constants. A
CEA is a (uncountable) family of EAs depending on the two-dimensional time
$(s,t)$, $0\leq s\leq t$.

The matrices of structural constants (depending on $(s,t)$) of a CEA satisfy
the Chapman-Kolmogorov equation. In other words, a CEA is a continuous-time
dynamical system which in a fixed time is an EA.

In [18] a wide class of finite-dimensional CEAs is constructed. In [3]-[6],
[8]-[11], [14]-[16], [18]-[21] several new classes of CEAs and flows of
algebras are given. These investigations are used to develop the theory of
Markov processes of cubic matrices [4], [11]. The recent book [19] contains a
systematic presentation of these algebraic and probabilistic approaches to
study population dynamics.

In this paper we consider three CEAs constructed in [6].

Before formulation of our main problem, let us give basic notations. Following
[3] we consider a family $\\{E^{[s,t]}:s,t\in\mathbb{R},0\leq s\leq t\\}$ of
$n$-dimensional evolution algebras over the field $\mathbb{R}$, with the basis
$e_{1},e_{2},\dots,e_{n}$ and the multiplication table

$e_{i}e_{i}=\sum_{j=1}^{n}a_{i,j}^{[s,t]}e_{j},\,\,\,\,\,\,i=1,\dots,n;\,\,\,\,\,e_{i}e_{j}=0,\,\,\,i\neq
j.$

Here parameters $s,t$ are considered as time.

Denote by $\mathcal{M}^{[s,t]}=\left(a_{i,j}^{[s,t]}\right)_{i,j=1,\dots n}$
the matrix of structural constants of $E^{[s,t]}$.

######  Definition 1.

A family $\\{E^{[s,t]}:s,t\in\mathbb{R},0\leq s\leq t\\}$ of $n$-dimensional
evolution algebras over the field $\mathbb{R}$, is called a chain of evolution
algebras (CEA) if the matrix $\mathcal{M}^{[s,t]}$ of structural constants
satisfies the Chapman-Kolmogorov equation

$\mathcal{M}^{[s,t]}=\mathcal{M}^{[s,\tau]}\mathcal{M}^{[\tau,t]},\,\,\,for\,\,\,any\,\,\,s<\tau<t.$
(1.1)

In this paper we study the following known CEAs (constructed in [6]):

$E_{i}^{[s,t]}$, which correspond to the $\mathcal{M}_{i}^{[s,t]},\,\,i=1,2,3$
defined as:

$\mathcal{M}_{1}^{[s,t]}=\frac{h(t)}{2}\left(\begin{array}[]{cccc}\frac{1}{h(s)}+f(s)&\
\ \ \frac{1}{h(s)}+f(s)&\ \ \ \frac{1}{h(s)}+f(s)\\\\[5.69054pt]
\frac{1}{h(s)}-g(s)&\ \ \ \frac{1}{h(s)}-g(s)&\ \ \
\frac{1}{h(s)}-g(s)\\\\[5.69054pt] g(s)-f(s)&\ \ \ g(s)-f(s)&\ \ \
g(s)-f(s)\end{array}\right)$

where $h$, $g$ and $f$ are arbitrary functions with $h(s)\neq 0$ ;

$\mathcal{M}_{2}^{[s,t]}=\frac{1}{2}\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cccccc}1+\psi(s)&1+\psi(s)&1+\psi(s)\\\\[5.69054pt]
1-\varphi(s)&1-\varphi(s)&1-\varphi(s)\\\\[5.69054pt]
\varphi(s)-\psi(s)&\varphi(s)-\psi(s)&\varphi(s)-\psi(s)\end{array}\right),\,\,\,\,\,\mbox{if}\,\,\,\,\,s\leq
t<a,\\\\[5.69054pt] \left(\begin{array}[]{cccccc}0&0&0\\\ 0&0&0\\\
0&0&0\end{array}\right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mbox{if}\,\,\,\,\,t\geq
a,\end{array}\right.$

where $a>0$ and $\varphi$, $\psi$ are arbitrary functions.

$\mathcal{M}_{3}^{[s,t]}=\theta(s)\left(\begin{array}[]{cccccc}\eta(t)&\vartheta(t)&\kappa(t)\\\\[2.84526pt]
\varphi_{1}(s)\eta(t)&\varphi_{1}(s)\vartheta(t)&\varphi_{1}(s)\kappa(t)\\\\[5.69054pt]
\varphi_{2}(s)\eta(t)&\varphi_{2}(s)\vartheta(t)&\varphi_{2}(s)\kappa(t)\end{array}\right)$

where

$\theta(s)=\frac{1}{\eta(s)+\varphi_{1}(s)\vartheta(s)+\varphi_{2}(s)\kappa(s)}$

and $\eta$, $\vartheta$, $\kappa$, $\varphi_{1}$, $\varphi_{2}$ are arbitrary
functions with $\eta(s)+\varphi_{1}(s)\vartheta(s)+\varphi_{2}(s)\kappa(s)\neq
0$.

In [7] (see also [1] for the complex field case) three-dimensional real
evolution algebras with $dim(E^{2})=1$ are classified and twelve pairwise non-
isomorphic evolution algebras are described. They are given by the following
matrices of structural constants:

$E_{1}:\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\ -1&-1&0\\\ \ 0&\
0&0\end{array}\right)$, $E_{2}:\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\
-1&-1&0\\\ \ 1&\ 1&0\end{array}\right)$, $E_{3}:\left(\begin{array}[]{cccccc}\
1&\ 1&0\\\ -1&-1&0\\\ -1&-1&0\end{array}\right)$,
$E_{4}:\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
0&0&0\end{array}\right)$,
$E_{5}:\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
1&0&0\end{array}\right)$, $E_{6}:\left(\begin{array}[]{cccccc}\ 1&0&0\\\ \
0&0&0\\\ -1&0&0\end{array}\right)$,
$E_{7}:\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
1&0&0\end{array}\right)$, $E_{8}:\left(\begin{array}[]{cccccc}\ 1&0&0\\\ \
1&0&0\\\ -1&0&0\end{array}\right)$,

$E_{9}:\left(\begin{array}[]{cccccc}\ 1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right)$, $E_{10}:\left(\begin{array}[]{cccccc}0&0&0\\\
0&0&0\\\ 1&0&0\end{array}\right)$,
$E_{11}:\left(\begin{array}[]{cccccc}0&0&0\\\ 1&0&0\\\
1&0&0\end{array}\right)$, $E_{12}:\left(\begin{array}[]{cccccc}\ 0&0&0\\\ \
1&0&0\\\ -1&0&0\end{array}\right).$

The main problem of this paper: For each CEA $E_{i}^{[s,t]}$, $i=1,2,3$
(listed above), to classify all EAs involved in the CEA. That is for each
algebra $E_{j}$, $j=1,2,\dots,12$ to describe the set of two-dimensional times
$(s,t)$ for which $E_{i}^{[s,t]}$ is isomorphic to the algebra $E_{j}$.

##  2\. On classification of algebras in chains $E_{i}^{[s,t]},\ i=1,2,3$.

To give classification of algebras in three-dimensional real chains of
evolution algebras we shall prove some lemmas, which are important to prove
the main theorems of this section.

###  2.1. The case $E_{1}^{[s,t]}$.

Any structural constants matrix of $E_{1}^{[s,t]}$ has the form as in the
following lemma.

######  Lemma 1.

The real evolution algebra corresponding to the matrix

$\mathcal{M}=\left(\begin{array}[]{cccccc}\lambda&\lambda&\lambda\\\
\mu&\mu&\mu\\\ \gamma&\gamma&\gamma\end{array}\right)$ with conditions
$\lambda+\mu+\gamma\neq 0$ is isomorphic to one of the following algebras:

  * (a)

$E_{4}$ if one of the following conditions is hold:

  * 1)

$\lambda\neq 0,\ \mu=0,\ \gamma=0$;

  * 2)

$\mu\neq 0,\ \lambda=0,\ \gamma=0$;

  * 3)

$\gamma\neq 0,\ \lambda=0,\ \mu=0$;

  * (b)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\lambda=0,\ \mu\gamma>0$;

  * 2)

$\mu=0,\ \lambda\gamma>0$;

  * 3)

$\gamma=0,\ \lambda\mu>0$;

  * (c)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\lambda=0,\ \mu\gamma<0$;

  * 2)

$\mu=0,\ \lambda\gamma<0$;

  * 3)

$\gamma=0,\ \lambda\mu<0$;

  * (d)

$E_{7}$ if the following conditions is hold:

  * $\lambda\mu\gamma\neq 0,\ \lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)>0,\ \gamma(\lambda+\mu)>0;$

  * (e)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$\lambda\mu\gamma\neq 0,\ \lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)>0,\
\gamma(\lambda+\mu)<0;$

  * 2)

$\lambda\mu\gamma\neq 0,\ \lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)<0,\
\gamma(\lambda+\mu)>0;$

  * 3)

$\lambda\mu\gamma\neq 0,\ \lambda+\mu=0;$

  * (f)

$E_{9}$ if the following conditions is hold:

  * $\lambda\mu\gamma\neq 0,\ \lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)<0,\ \gamma(\lambda+\mu)<0.$

###### Proof.

Let $E_{\mathcal{M}}$ be the evolution algebra with basis
$\\{e_{1},e_{2},e_{3}\\}$. Then, the multiplication table in $E_{\mathcal{M}}$
is

$\begin{array}[]{c}e_{1}e_{1}=\lambda e_{1}+\lambda e_{2}+\lambda e_{3},\
e_{2}e_{2}=\mu e_{1}+\mu e_{2}+\mu e_{3},\\\\[2.84526pt] e_{3}e_{3}=\gamma
e_{1}+\gamma e_{2}+\gamma e_{3},\
e_{1}e_{2}=e_{1}e_{3}=e_{2}e_{3}=0.\end{array}$ (2.1)

Let us consider $E_{i},\ i=1,...,12$ listed in previous section. First we
consider $E_{1}$, the multiplication table in $E_{1}$ is

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime}+e_{2}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=-e_{1}^{\prime}-e_{2}^{\prime},\
e_{3}^{\prime}e_{3}^{\prime}=0,\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$ (2.2)

Assume that $E_{\mathcal{M}}$ is isomorphic to the algebra $E_{1}$. Then there
exists a change of basis as follows

$\begin{array}[]{c}e_{1}^{\prime}=x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3},\\\
e_{2}^{\prime}=y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3},\\\
e_{3}^{\prime}=z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3}\end{array}$ (2.3)

where determinant of the change is non-zero, i.e.,

$x_{1}y_{2}z_{3}+x_{3}y_{1}z_{2}+x_{2}y_{3}z_{1}\neq
x_{3}y_{2}z_{1}+x_{2}y_{1}z_{3}+x_{1}y_{3}z_{2}.$ (2.4)

From the equalities (2.1) and (2.3) we obtain the following equations:

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=(x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3})(x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3})=(\lambda
x_{1}^{2}+\mu x_{2}^{2}+\gamma x_{3}^{2})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{2}^{\prime}e_{2}^{\prime}=(y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3})(y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3})=(\lambda
y_{1}^{2}+\mu y_{2}^{2}+\gamma y_{3}^{2})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{3}^{\prime}e_{3}^{\prime}=(z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3})(z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3})=(\lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma z_{3}^{2})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=(x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3})(y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3})=(\lambda
x_{1}y_{1}+\mu x_{2}y_{2}+\gamma
x_{3}y_{3})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{1}^{\prime}e_{3}^{\prime}=(x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3})(z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3})=(\lambda
x_{1}z_{1}+\mu x_{2}z_{2}+\gamma
x_{3}z_{3})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{2}^{\prime}e_{3}^{\prime}=(y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3})(z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3})=(\lambda
y_{1}z_{1}+\mu y_{2}z_{2}+\gamma
y_{3}z_{3})(e_{1}+e_{2}+e_{3}),\\\\[2.84526pt]
e_{1}^{\prime}+e_{2}^{\prime}=(x_{1}+y_{1})e_{1}+(x_{2}+y_{2})e_{2}+(x_{3}+y_{3})e_{3},\\\\[2.84526pt]
-e_{1}^{\prime}-e_{2}^{\prime}=(-x_{1}-y_{1})e_{1}+(-x_{2}-y_{2})e_{2}+(-x_{3}-y_{3})e_{3}.\end{array}$

Consequently, being $\\{{e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}}\\}$ is
the basis of evolution algebra and from the equalities (2.2) we have

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}+\mu x_{2}^{2}+\gamma
x_{3}^{2}=x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}\\\ \lambda y_{1}^{2}+\mu
y_{2}^{2}+\gamma y_{3}^{2}=-x_{1}-y_{1}=-x_{2}-y_{2}=-x_{3}-y_{3}\\\ \lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma z_{3}^{2}=0\\\ \lambda x_{1}y_{1}+\mu
x_{2}y_{2}+\gamma x_{3}y_{3}=0\\\ \lambda x_{1}z_{1}+\mu x_{2}z_{2}+\gamma
x_{3}z_{3}=0\\\ \lambda y_{1}z_{1}+\mu y_{2}z_{2}+\gamma y_{3}z_{3}=0\
.\end{array}\right.$ (2.5)

If we add the right and left sides of the first and second equations of the
system (2.5), respectively, we obtain the following equation

$\lambda(x_{1}^{2}+y_{1}^{2})+\mu(x_{2}^{2}+y_{2}^{2})+\gamma(x_{3}^{2}+y_{3}^{2})=0.$

From this equation we get

$\begin{array}[]{c}\lambda(x_{1}+y_{1})^{2}-2\lambda
x_{1}y_{1}+\mu(x_{2}+y_{2})^{2}-2\mu
x_{2}y_{2}+\gamma(x_{3}+y_{3})^{2}-2\gamma x_{3}y_{3}=0\\\
\lambda(x_{1}+y_{1})^{2}+\mu(x_{2}+y_{2})^{2}+\gamma(x_{3}+y_{3})^{2}-2(\lambda
x_{1}y_{1}+\mu x_{2}y_{2}+\gamma x_{3}y_{3})=0.\end{array}$

Denote $x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}=k$. If we use the fourth equation
of the system (2.5), then this equation will be

$\lambda k^{2}+\mu k^{2}+\gamma k^{2}=0,\ k^{2}(\lambda+\mu+\gamma)=0,\
k^{2}(\lambda+\mu+\gamma)=0.$

By the condition of the lemma $\lambda+\mu+\gamma\neq 0$, then $k=0$. It means
that

$x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}=0,\ x_{1}=-y_{1},\ x_{2}=-y_{2},\
x_{3}=-y_{3}.$

It is a contradiction to (2.4), hence $E_{\mathcal{M}}$ is not isomorphic to
the algebra $E_{1}$.

Now, let us consider the algebras $E_{2}$ and $E_{3}$. For these algebras
systems of equations similar to (2.5) are as follows, respectively

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}+\mu x_{2}^{2}+\gamma
x_{3}^{2}=x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}\\\ \lambda y_{1}^{2}+\mu
y_{2}^{2}+\gamma y_{3}^{2}=-x_{1}-y_{1}=-x_{2}-y_{2}=-x_{3}-y_{3}\\\ \lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma
z_{3}^{2}=x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}\\\ \lambda x_{1}y_{1}+\mu
x_{2}y_{2}+\gamma x_{3}y_{3}=0\\\ \lambda x_{1}z_{1}+\mu x_{2}z_{2}+\gamma
x_{3}z_{3}=0\\\ \lambda y_{1}z_{1}+\mu y_{2}z_{2}+\gamma
y_{3}z_{3}=0\end{array}\right.$

and

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}+\mu x_{2}^{2}+\gamma
x_{3}^{2}=x_{1}+y_{1}=x_{2}+y_{2}=x_{3}+y_{3}\\\ \lambda y_{1}^{2}+\mu
y_{2}^{2}+\gamma y_{3}^{2}=-x_{1}-y_{1}=-x_{2}-y_{2}=-x_{3}-y_{3}\\\ \lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma
z_{3}^{2}=-x_{1}-y_{1}=-x_{2}-y_{2}=-x_{3}-y_{3}\\\ \lambda x_{1}y_{1}+\mu
x_{2}y_{2}+\gamma x_{3}y_{3}=0\\\ \lambda x_{1}z_{1}+\mu x_{2}z_{2}+\gamma
x_{3}z_{3}=0\\\ \lambda y_{1}z_{1}+\mu y_{2}z_{2}+\gamma y_{3}z_{3}=0\
.\end{array}\right.$

Similar to the above, it is not difficult to show that $E_{\mathcal{M}}$ is
not isomorphic to the algebras $E_{2}$ and $E_{3}$.

Let us consider the next cases.

Case 1. One of $\lambda,\ \mu,\ \gamma$ is not equal to zero.

Case 1.1. Let $\lambda\neq 0,\ \mu=0,\ \gamma=0$. We consider the algebra
$E_{4}$. The multiplication table in $E_{4}$ is

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=0,\
e_{3}^{\prime}e_{3}^{\prime}=0,\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

In this case the system of equations similar to (2.5) is as follows

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}=x_{1}=x_{2}=x_{3}\\\ \lambda
y_{1}^{2}=0\\\ \lambda z_{1}^{2}=0\\\ \lambda x_{1}y_{1}=0\\\ \lambda
x_{1}z_{1}=0\\\ \lambda y_{1}z_{1}=0\end{array}\right.$

If we solve the first equation of this system then there will be two
solutions:

1) If $x_{1}=0$ then $x_{2}=x_{3}=0$. It is contradiction to (2.4).

2) If $x_{1}=\frac{1}{\lambda}$ then $x_{2}=x_{3}=\frac{1}{\lambda}$ and
$y_{1}=z_{1}=0$. In this case the coefficients of the change of basis (2.3)
are as follows $x_{1}=x_{2}=x_{3}=\frac{1}{\lambda},$ $y_{1}=z_{1}=0$ and
$y_{2},\ y_{3},\ z_{2},\ z_{3}$ are arbitrary real numbers such that satisfy
the condition $y_{2}z_{3}\neq y_{3}z_{2}.$

Without loss of generality we can take $y_{2}=1,\ y_{3}=0,\ z_{2}=0,z_{3}=1$.
So by the change of basis
$e_{1}^{\prime}=\frac{1}{\lambda}e_{1}+\frac{1}{\lambda}e_{2}+\frac{1}{\lambda}e_{3}$,
$e_{2}^{\prime}=e_{2},\ e_{3}^{\prime}=e_{3}$ we can see that the algebras
$E_{\mathcal{M}}$ and $E_{4}$ are isomorphic.

Case 1.2. If $\lambda=0,\ \gamma=0,\ \mu\neq 0$ then we take the following
basis change $\\{e_{2},\ e_{1},\ e_{3}\\}$ and we have case similar to the
Case 1.1.

Case 1.3. If $\lambda=0,\ \mu=0,\ \gamma\neq 0$ then we take the following
basis change $\\{e_{3},\ e_{2},\ e_{1}\\}$ and we have case similar to the
Case 1.1.

Case 2. Two of $\lambda,\ \mu,\ \gamma$ are not equal to zero.

Case 2.1. Let $\lambda=0,\ \mu\gamma>0$. We consider the algebra $E_{5}$. The
multiplication table in $E_{5}$ is

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=0,\
e_{3}^{\prime}e_{3}^{\prime}=e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

In this case the system of equations similar to (2.5) is as follows

$\left\\{\begin{array}[]{ll}\mu x_{2}^{2}+\gamma
x_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \mu y_{2}^{2}+\gamma y_{3}^{2}=0\\\ \mu
z_{2}^{2}+\gamma z_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \mu x_{2}y_{2}+\gamma
x_{3}y_{3}=0\\\ \mu x_{2}z_{2}+\gamma x_{3}z_{3}=0\\\ \mu y_{2}z_{2}+\gamma
y_{3}z_{3}=0\ .\end{array}\right.$ (2.6)

From the first equation of this system we have the following solutions:

1) $x_{1}=x_{2}=x_{3}=0$. It is contradiction to (2.4).

2) $x_{1}=x_{2}=x_{3}=\frac{1}{\mu+\gamma}$. In this case from the remaining
equations of (2.6) we obtain the following system

$\left\\{\begin{array}[]{ll}\mu y_{2}^{2}+\gamma y_{3}^{2}=0\\\ \mu
z_{2}^{2}+\gamma z_{3}^{2}=\frac{1}{\mu+\gamma}\\\ \mu y_{2}+\gamma y_{3}=0\\\
\mu z_{2}+\gamma z_{3}=0\\\ \mu y_{2}z_{2}+\gamma
y_{3}z_{3}=0.\end{array}\right.$

Solution of this system is $y_{2}=0,\ y_{3}=0,\
z_{2}=\pm\sqrt{\frac{\gamma}{\mu}}\frac{1}{\mu+\gamma},\
z_{3}=\mp\sqrt{\frac{\mu}{\gamma}}\frac{1}{\mu+\gamma}$.

We may take the coefficients of the change of basis (2.3) as follows

$\begin{array}[]{ll}x_{1}=x_{2}=x_{3}=\frac{1}{\mu+\gamma},\ y_{1}=\alpha,\
y_{2}=0,\ y_{3}=0,\\\\[5.69054pt] z_{1}=\beta,\
z_{2}=\pm\sqrt{\frac{\gamma}{\mu}}\frac{1}{\mu+\gamma},\
z_{3}=\mp\sqrt{\frac{\mu}{\gamma}}\frac{1}{\mu+\gamma},\end{array}$

where $\alpha\neq 0,\ \alpha,\beta\in\mathbb{R}.$ So, we may take the change
of basis

$e_{1}^{\prime}=\frac{1}{\mu+\gamma}e_{1}+\frac{1}{\mu+\gamma}e_{2}+\frac{1}{\mu+\gamma}e_{3},\
e_{2}^{\prime}=\alpha e_{1},\ e_{3}^{\prime}=\beta
e_{1}+\sqrt{\frac{\gamma}{\mu}}\frac{1}{\mu+\gamma}e_{2}-\sqrt{\frac{\mu}{\gamma}}\frac{1}{\mu+\gamma}e_{3}$

and the determinant of the matrix of this change is equal to
$D=\frac{\alpha}{(\mu+\gamma)\sqrt{\mu\gamma}}\neq 0$. By this change of basis
we can see that $E_{\mathcal{M}}$ and $E_{5}$ are isomorphic.

Case 2.2. Let $\lambda=0,\ \mu\gamma<0$. By the condition of the lemma
$\mu+\gamma\neq 0$. In these case we may take the change of basis

$e_{1}^{\prime}=\frac{1}{\mu+\gamma}e_{1}+\frac{1}{\mu+\gamma}e_{2}+\frac{1}{\mu+\gamma}e_{3},\
e_{2}^{\prime}=\alpha e_{1},\ e_{3}^{\prime}=\beta
e_{1}+\sqrt{-\frac{\gamma}{\mu}}\frac{1}{\mu+\gamma}e_{2}-\sqrt{-\frac{\mu}{\gamma}}\frac{1}{\mu+\gamma}e_{3}$

and the determinant of the matrix of this change is equal to
$D=\frac{\alpha}{(\mu+\gamma)^{2}}(\sqrt{-\frac{\gamma}{\mu}}+\sqrt{-\frac{\mu}{\gamma}})\neq
0$. By this change of basis we can see that $E_{\mathcal{M}}$ is isomorphic to
$E_{6}$.

For the cases

  * (i)

$\mu=0,\ \lambda\gamma>0$ and $\mu=0,\ \lambda\gamma<0$;

  * (ii)

$\gamma=0,\ \lambda\mu>0$ and $\gamma=0,\ \lambda\mu<0$,

respectively, if we take the changes of basis as follows $\\{e_{2},\ e_{1},\
e_{3}\\}$ , $\\{e_{3},\ e_{2},\ e_{1}\\}$ then we will have cases similar to
the Case 2.1 and Case 2.2.

Case 3. None of $\lambda,\ \mu,\ \gamma$ is zero, i.e $\lambda\mu\gamma\neq
0$.

Case 3.1. Let

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)>0,\ \gamma(\lambda+\mu)>0.$ (2.7)

We consider the algebra $E_{7}$. The multiplication table in $E_{7}$ is

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=e_{1}^{\prime},\
e_{3}^{\prime}e_{3}^{\prime}=e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

In this case the system of equations similar to (2.5) is as follows

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}+\mu x_{2}^{2}+\gamma
x_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \lambda y_{1}^{2}+\mu y_{2}^{2}+\gamma
y_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \lambda z_{1}^{2}+\mu z_{2}^{2}+\gamma
z_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \lambda x_{1}y_{1}+\mu x_{2}y_{2}+\gamma
x_{3}y_{3}=0\\\ \lambda x_{1}z_{1}+\mu x_{2}z_{2}+\gamma x_{3}z_{3}=0\\\
\lambda y_{1}z_{1}+\mu y_{2}z_{2}+\gamma y_{3}z_{3}=0\ .\end{array}\right.$
(2.8)

If we solve the first equation of this system then:

1) $x_{1}=x_{2}=x_{3}=0$. It is contradiction to (2.4).

2) $x_{1}=x_{2}=x_{3}=\frac{1}{\lambda+\mu+\gamma}$. In this case from the
remaining equations of (2.8) we obtain the following system

$\left\\{\begin{array}[]{ll}\lambda y_{1}^{2}+\mu y_{2}^{2}+\gamma
y_{3}^{2}=\frac{1}{\lambda+\mu+\gamma}\\\ \lambda z_{1}^{2}+\mu
z_{2}^{2}+\gamma z_{3}^{2}=\frac{1}{\lambda+\mu+\gamma}\\\ \lambda y_{1}+\mu
y_{2}+\gamma y_{3}=0\\\ \lambda z_{1}+\mu z_{2}+\gamma z_{3}=0\\\ \lambda
y_{1}z_{1}+\mu y_{2}z_{2}+\gamma y_{3}z_{3}=0\ .\end{array}\right.$ (2.9)

Without loss of generality we solve the system (2.9) for the case $y_{3}=0$.
Then we have

$\left\\{\begin{array}[]{ll}\lambda y_{1}^{2}+\mu
y_{2}^{2}=\frac{1}{\lambda+\mu+\gamma}\\\ \lambda y_{1}+\mu y_{2}=0\\\ \lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma z_{3}^{2}=\frac{1}{\lambda+\mu+\gamma}\\\
\lambda z_{1}+\mu z_{2}+\gamma z_{3}=0\\\ \lambda y_{1}z_{1}+\mu y_{2}z_{2}=0\
.\end{array}\right.$ (2.10)

From the second equation of the system (2.10) we find
$y_{1}=-\frac{\mu}{\lambda}y_{2}$ and put it in the first equation of this
system then we have the next equation

$\frac{\mu^{2}}{\lambda}y_{2}^{2}+\mu y_{2}^{2}=\frac{1}{\lambda+\mu+\gamma}.$

The solution of this equation is
$y_{2}=\pm\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}$. Note
that, if the condition (2.7) is hold then $\lambda\mu>0$ . Thus
$y_{1}=\mp\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}.$

So, we may take
$y_{1}=-\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}$ and
$y_{2}=\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}$. From the
last equation of (2.10) we have

$-\lambda\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}z_{1}+\mu\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}z_{2}=0.$

It means that $z_{1}=z_{2}$. From this equality and from the third and fourth
equations of the system (2.10) we have the next system of equations

$\left\\{\begin{array}[]{ll}\lambda z_{1}^{2}+\mu z_{1}^{2}+\gamma
z_{3}^{2}=\frac{1}{\lambda+\mu+\gamma}\\\ \lambda z_{1}+\mu z_{1}+\gamma
z_{3}=0.\end{array}\right.$

Solution of this system is
$z_{1}=\pm\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma},\
z_{3}=\mp\sqrt{\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}.$

Thus, one of the solution of the system (2.8) is

$\begin{array}[]{ccc}x_{1}=x_{2}=x_{3}=\frac{1}{\lambda+\mu+\gamma},\
y_{1}=-\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}},\
y_{2}=\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}},\
y_{3}=0,\\\\[2.84526pt]
z_{1}=z_{2}=\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma},\
z_{3}=-\sqrt{\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}.\end{array}$

Consequently, if we take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=-\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}-\sqrt{\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3}\end{array}$
(2.11)

then the algebra $E_{\mathcal{M}}$ will be isomorphic to $E_{7}$ . Indeed,
according to the (2.1)

$\begin{array}[]{ccc}e_{1}^{\prime}e_{1}^{\prime}=\frac{1}{({\lambda+\mu+\gamma})^{2}}e_{1}^{2}+\frac{1}{({\lambda+\mu+\gamma})^{2}}e_{2}^{2}+\frac{1}{({\lambda+\mu+\gamma})^{2}}e_{3}^{2}=\frac{e_{1}+e_{2}+e_{3}}{\lambda+\mu+\gamma}=e_{1}^{\prime},\\\\[2.84526pt]
e_{2}^{\prime}e_{2}^{\prime}=\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}e_{1}^{2}+\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}e_{2}^{2}=\frac{\mu(e_{1}+e_{2}+e_{3})}{(\lambda+\mu)(\lambda+\mu+\gamma)}+\frac{\lambda(e_{1}+e_{2}+e_{3})}{(\lambda+\mu)(\lambda+\mu+\gamma)}=e_{1}^{\prime},\\\\[2.84526pt]
e_{3}^{\prime}e_{3}^{\prime}=\frac{\gamma}{\lambda+\mu}\frac{e_{1}^{2}}{(\lambda+\mu+\gamma)^{2}}+\frac{\gamma}{\lambda+\mu}\frac{e_{2}^{2}}{(\lambda+\mu+\gamma)^{2}}+\frac{\lambda+\mu}{\gamma}\frac{e_{3}^{2}}{(\lambda+\mu+\gamma)^{2}}=\frac{\lambda\gamma}{(\lambda+\mu)(\lambda+\mu+\gamma)^{2}}(e_{1}+e_{2}+e_{3})+\\\\[2.84526pt]
+\frac{\mu\gamma}{(\lambda+\mu)(\lambda+\mu+\gamma)^{2}}(e_{1}+e_{2}+e_{3})+\frac{\lambda+\mu}{(\lambda+\mu+\gamma)^{2}}(e_{1}+e_{2}+e_{3})=\frac{e_{1}+e_{2}+e_{3}}{\lambda+\mu+\gamma}=e_{1}^{\prime}\end{array}$

and $e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.$
The determinant of the matrix of the change (2.11) is $D=\frac{\pm
1}{(\lambda+\mu+\gamma)\sqrt{(\lambda+\mu+\gamma)\lambda\mu\gamma}}\neq 0.$

Case 3.2. If

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)>0,\ \gamma(\lambda+\mu)<0,\
\lambda\mu>0$

then similarly to the above, we may take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=-\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}+\sqrt{-\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3}.\end{array}$
(2.12)

The determinant of the matrix of the change (2.12) is not equal to zero and
for this basis it is not difficult to check that

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=e_{1}^{\prime},\
e_{3}^{\prime}e_{3}^{\prime}=-e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

It means that in this case the algebra $E_{\mathcal{M}}$ is isomorphic to
$E_{8}$ .

Case 3.3. If

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)>0,\ \gamma(\lambda+\mu)<0,\
\lambda\mu<0.$

then we take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=\sqrt{\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}+\sqrt{-\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3}.\end{array}$

By this change of basis we can see that the algebra $E_{\mathcal{M}}$ is
isomorphic to $E_{8}$ again.

Case 3.4. Let

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)<0,\ \gamma(\lambda+\mu)>0.$
(2.13)

Note that, if the condition (2.13) is hold then $\lambda\mu<0$. Then similarly
to the above, in this case the algebra $E_{\mathcal{M}}$ is isomorphic to
$E_{8}$ . For this we take the change of basis:

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}-\sqrt{\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{-\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{-\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2}.\end{array}$
(2.14)

The determinant of the matrix of the change (2.14) is not equal to zero and
for this basis it is easy to check that

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=e_{1}^{\prime},\
e_{3}^{\prime}e_{3}^{\prime}=-e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

Case 3.5. If

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)<0,\ \gamma(\lambda+\mu)<0,\
\lambda\mu>0$

then we take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=-\sqrt{-\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{-\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}+\sqrt{-\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3}.\end{array}$
(2.15)

The determinant of the matrix of the change (2.15) is not equal to zero and
for this basis it is not difficult to check that

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=e_{1}^{\prime},\
e_{2}^{\prime}e_{2}^{\prime}=-e_{1}^{\prime},\
e_{3}^{\prime}e_{3}^{\prime}=-e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

It means that in this case the algebra $E_{\mathcal{M}}$ is isomorphic to
$E_{9}$.

Case 3.6. If

$\lambda\mu(\lambda+\mu)(\lambda+\mu+\gamma)<0,\ \gamma(\lambda+\mu)<0,\
\lambda\mu<0$

then we take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\lambda+\mu+\gamma}e_{1}+\frac{1}{\lambda+\mu+\gamma}e_{2}+\frac{1}{\lambda+\mu+\gamma}e_{3},\\\\[2.84526pt]
e_{2}^{\prime}=\sqrt{-\frac{\mu}{\lambda(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{1}+\sqrt{-\frac{\lambda}{\mu(\lambda+\mu)(\lambda+\mu+\gamma)}}e_{2},\\\\[2.84526pt]
e_{3}^{\prime}=\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{1}+\sqrt{-\frac{\gamma}{\lambda+\mu}}\frac{1}{\lambda+\mu+\gamma}e_{2}+\sqrt{-\frac{\lambda+\mu}{\gamma}}\frac{1}{\lambda+\mu+\gamma}e_{3}.\end{array}$

By this change of basis we can see that the algebra $E_{\mathcal{M}}$ is
isomorphic to $E_{9}$ again.

Case 3.7. Let $\lambda+\mu=0$.

Case 3.7.1. If $\lambda(\lambda+\gamma)>0$ then we take the following change
of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\gamma}e_{1}+\frac{1}{\gamma}e_{2}+\frac{1}{\gamma}e_{3},\
e_{2}^{\prime}=\frac{1}{\sqrt{\lambda(\lambda+\gamma)}}e_{1}-\frac{\lambda}{\gamma}\frac{1}{\sqrt{\lambda(\lambda+\gamma)}}e_{3},\\\\[2.84526pt]
e_{3}^{\prime}=\frac{1}{\gamma}\sqrt{\frac{\lambda}{\lambda+\gamma}}e_{1}+\frac{\lambda+\gamma}{\lambda\gamma}\sqrt{\frac{\lambda}{\lambda+\gamma}}e_{2}+\frac{1}{\gamma}\sqrt{\frac{\lambda}{\lambda+\gamma}}e_{3}.\end{array}$

By this change of basis we can see that the algebra $E_{\mathcal{M}}$ is
isomorphic to $E_{8}$.

Case 3.7.2. If $\lambda(\lambda+\gamma)<0$ then we take the following change
of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{1}{\gamma}e_{1}+\frac{1}{\gamma}e_{2}+\frac{1}{\gamma}e_{3},\\\\[5.69054pt]
e_{2}^{\prime}=\frac{1}{\gamma}\sqrt{\frac{-\lambda}{\lambda+\gamma}}e_{1}+\frac{\lambda+\gamma}{\lambda\gamma}\sqrt{\frac{-\lambda}{\lambda+\gamma}}e_{2}+\frac{1}{\gamma}\sqrt{\frac{-\lambda}{\lambda+\gamma}}e_{3},\\\\[5.69054pt]
e_{3}^{\prime}=\frac{1}{\sqrt{-\lambda(\lambda+\gamma)}}e_{1}-\frac{\lambda}{\gamma}\frac{1}{\sqrt{-\lambda(\lambda+\gamma)}}e_{3}.\end{array}$

By this change of basis we can see that the algebra $E_{\mathcal{M}}$ is
isomorphic to $E_{8}$ again.

Case 3.7.3. If $\lambda+\gamma=0$ then we take the following change of basis

$\begin{array}[]{ccc}e_{1}^{\prime}=\frac{-1}{\lambda}e_{1}+\frac{-1}{\lambda}e_{2}+\frac{-1}{\lambda}e_{3},\
e_{2}^{\prime}=\frac{1}{\sqrt{2}\lambda}e_{2}-\frac{1}{\sqrt{2}\lambda}e_{3},\\\\[2.84526pt]
e_{3}^{\prime}=\frac{\sqrt{2}}{\lambda}e_{1}+\frac{1}{\sqrt{2}\lambda}e_{2}+\frac{1}{\sqrt{2}\lambda}e_{3}.\end{array}$

In this case again the algebra $E_{\mathcal{M}}$ is isomorphic to the algebra
$E_{8}$.

Now, we consider the algebra $E_{10}$. The multiplication table in $E_{10}$ is

$\begin{array}[]{c}e_{1}^{\prime}e_{1}^{\prime}=0,\
e_{2}^{\prime}e_{2}^{\prime}=0,\
e_{3}^{\prime}e_{3}^{\prime}=e_{1}^{\prime},\\\\[2.84526pt]
e_{1}^{\prime}e_{2}^{\prime}=0,\ e_{1}^{\prime}e_{3}^{\prime}=0,\
e_{2}^{\prime}e_{3}^{\prime}=0.\end{array}$

Assume that $E_{\mathcal{M}}$ is isomorphic to the algebra $E_{10}$. Then
there exists a change of basis as (2.3). In this case we have the following
system of equations similar to (2.5)

$\left\\{\begin{array}[]{ll}\lambda x_{1}^{2}+\mu x_{2}^{2}+\gamma
x_{3}^{2}=0\\\ \lambda y_{1}^{2}+\mu y_{2}^{2}+\gamma y_{3}^{2}=0\\\ \lambda
z_{1}^{2}+\mu z_{2}^{2}+\gamma z_{3}^{2}=x_{1}=x_{2}=x_{3}\\\ \lambda
x_{1}y_{1}+\mu x_{2}y_{2}+\gamma x_{3}y_{3}=0\\\ \lambda x_{1}z_{1}+\mu
x_{2}z_{2}+\gamma x_{3}z_{3}=0\\\ \lambda y_{1}z_{1}+\mu y_{2}z_{2}+\gamma
y_{3}z_{3}=0\ .\end{array}\right.$ (2.16)

If we use $x_{1}=x_{2}=x_{3}$ and the first equation of (2.16) then we have
the next equation

$x_{1}^{2}(\lambda+\mu+\gamma)=0.$

We know that by the condition of the lemma $\lambda+\mu+\gamma\neq 0$. Hence
$x_{1}=0$. It means that $x_{1}=x_{2}=x_{3}=0$. It is contradiction to (2.4),
hence our assumption is not true.

In the same way, we can see that the algebra $E_{\mathcal{M}}$ is not
isomorphic to the algebras $E_{11}$ and $E_{12}$. ∎

Now, using this lemma we prove the following theorem which gives
classification of the CEA: $E_{1}^{[s,t]}$, which correspond to the
$\mathcal{M}_{1}^{[s,t]}$.

######  Theorem 1.

For given values $(s,t)$ of time the algebra $E_{1}^{[s,t]}$ is isomorphic to

  * (a)

$E_{4}$ if one of the following conditions holds:

  * 1)

$g(s)=f(s)=\frac{1}{h(s)}$;

  * 2)

$g(s)=f(s)=-\frac{1}{h(s)}$;

  * 3)

$g(s)=-f(s)=\frac{1}{h(s)}$;

  * (b)

$E_{5}$ if one of the following conditions holds:

  * 1)

$f(s)=-\frac{1}{h(s)},\ f^{2}(s)>g^{2}(s)$;

  * 2)

$g(s)=\frac{1}{h(s)},\ g^{2}(s)>f^{2}(s)$;

  * 3)

$g(s)=f(s),\ \frac{1}{h^{2}(s)}>f^{2}(s)$;

  * (c)

$E_{6}$ if one of the following conditions holds:

  * 1)

$f(s)=-\frac{1}{h(s)},\ f^{2}(s)<g^{2}(s)$;

  * 2)

$g(s)=\frac{1}{h(s)},\ g^{2}(s)<f^{2}(s)$;

  * 3)

$g(s)=f(s),\ \frac{1}{h^{2}(s)}<f^{2}(s)$;

  * (d)

$E_{7}$ if the following conditions hold:

  * $(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\ (g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))>0\\\ \frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))>0$;

  * (e)

$E_{8}$ if one of the following conditions holds:

  * 1)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
(g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))<0\\\
\frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))>0$;

  * 2)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
(g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))>0\\\
\frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))<0$;

  * 3)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
\frac{2}{h(s)}+f(s)-g(s)=0;$

  * (f)

$E_{9}$ if the following conditions hold:

  * $(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\ (g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))<0\\\ \frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))<0$.

###### Proof.

We mention that in $\mathcal{M}_{1}^{[s,t]}$ the parameter functions $h,\ g$
and $f$ are arbitrary with $h(s)\neq 0$.

Let us denote

$\lambda(s,t)=\frac{1}{2}h(t)(\frac{1}{h(s)}+f(s)),\
\mu(s,t)=\frac{1}{2}h(t)(\frac{1}{h(s)}-g(s)),\
\gamma(s,t)=\frac{1}{2}h(t)(g(s)-f(s)).$

Note that $\lambda(s,t)+\mu(s,t)+\gamma(s,t)=\frac{h(t)}{h(s)}\neq 0$.

a) If $g(s)=f(s)=\frac{1}{h(s)}$ then $\mu(s,t)=\gamma(s,t)=0$ and
$\lambda(s,t)=\frac{h(t)}{h(s)}\neq 0$. By Lemma 1, it is easy to see that
$E_{1}^{[s,t]}$ is isomorphic to $E_{4}$.

The cases $g(s)=f(s)=-\frac{1}{h(s)}$ and $g(s)=-f(s)=\frac{1}{h(s)}$ are
similarly.

b) Let us $f(s)=-\frac{1}{h(s)},\ f^{2}(s)>g^{2}(s)$. It means that
$\lambda(s,t)=0$ and $\mu(s,t)\gamma(s,t)>0$. Indeed

$\lambda(s,t)=\frac{1}{2}h(t)(\frac{1}{h(s)}+f(s))=\frac{1}{2}h(t)(\frac{1}{h(s)}-\frac{1}{h(s)})=0$

and

$\begin{array}[]{cc}\mu(s,t)\gamma(s,t)=\frac{1}{4}h^{2}(t)(\frac{1}{h(s)}-g(s))(g(s)-f(s))=\\\\[5.69054pt]
=\frac{1}{4}h^{2}(t)(-f(s)-g(s))(g(s)-f(s))=\frac{1}{4}h^{2}(t)(f^{2}(s)-g^{2}(s))>0\
.\end{array}$

In this case by Lemma 1, $E_{1}^{[s,t]}$ is isomorphic to $E_{5}$.

If the conditions $g(s)=\frac{1}{h(s)},\ g^{2}(s)>f^{2}(s)$ and $g(s)=f(s),\
\frac{1}{h^{2}(s)}>f^{2}(s)$ are hold then $\mu(s,t)=0,\
\lambda(s,t)\gamma(s,t)>0$ and $\gamma(s,t)=0,\ \lambda(s,t)\mu(s,t)>0$
respectively. In these cases by Lemma 1, $E_{1}^{[s,t]}$ is isomorphic to
$E_{5}$ again.

c) Similar to the previous case, if $f(s)=-\frac{1}{h(s)},\ f^{2}(s)<g^{2}(s)$
then $\lambda(s,t)=0$ and $\mu(s,t)\gamma(s,t)<0$. If the conditions
$g(s)=\frac{1}{h(s)},\ g^{2}(s)<f^{2}(s)$ and $g(s)=f(s),\
\frac{1}{h^{2}(s)}<f^{2}(s)$ are hold then $\mu(s,t)=0,\
\lambda(s,t)\gamma(s,t)<0$ and $\gamma(s,t)=0,\ \lambda(s,t)\mu(s,t)<0$
respectively. In these cases $E_{1}^{[s,t]}$ is isomorphic to $E_{6}$.

d) If

$\begin{array}[]{cc}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq
0,\ (g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))>0\\\\[5.69054pt]
\frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))>0\end{array}$

then

$\lambda(s,t)\mu(s,t)\gamma(s,t)=h^{3}(t)(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq
0$

and

$\gamma(s,t)(\lambda(s,t)+\mu(s,t))=\frac{1}{4}h^{2}(t)(g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))>0,$
$\begin{array}[]{cc}\lambda(s,t)\mu(s,t)(\lambda(s,t)+\mu(s,t))(\lambda(s,t)+\mu(s,t)+\gamma(s,t))=\\\\[5.69054pt]
=\frac{1}{8}\frac{h^{4}(t)}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))>0.\end{array}$

In this case by Lemma 1, $E_{1}^{[s,t]}$ is isomorphic to $E_{7}$.

e) In the same way it is not difficult to see that if one of the following
conditions is hold

  * (i)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
(g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))<0\\\
\frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))>0$;

  * (ii)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
(g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))>0\\\
\frac{2}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))<0$;

  * (iii)

$(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq 0,\
\frac{2}{h(s)}+f(s)-g(s)=0$

then $E_{1}^{[s,t]}$ is isomorphic to $E_{8}$.

f) If

$\begin{array}[]{cc}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(g(s)-f(s))\neq
0,\ (g(s)-f(s))(\frac{2}{h(s)}+f(s)-g(s))<0\\\\[5.69054pt]
\frac{1}{h(s)}(\frac{1}{h(s)}+f(s))(\frac{1}{h(s)}-g(s))(\frac{2}{h(s)}+f(s)-g(s))<0\end{array}$

then $E_{1}^{[s,t]}$ is isomorphic to $E_{9}$. ∎

######  Remark 1.

Note that by Lemma 1, $E_{1}^{[s,t]}$ is not isomorphic to the algebras
$E_{1},$ $E_{2},$ $E_{3},$ $E_{10},$ $E_{11},$ $E_{12}$ for any $0\leq s\leq
t$.

To illustrate the essence of Theorem 1, consider the following example.

######  Example 1.

Let $g(s)=4s-16,\ f(s)=4s^{2}-24s+32,\ \frac{1}{h(s)}=s+1$ then we obtain:

  * (a)

$E_{4}$ when $s=3$;

  * (b)

$E_{5}$ when $s=4$;

  * (c)

$E_{6}$ when $s=\frac{11}{4}$ and $s=\frac{17}{3}$;

  * (d)

$E_{7}$ when $s\in(3;4)$;

  * (e)

$E_{8}$ when $s\in[0;\frac{11}{4})\cup(4;\frac{17}{3})$;

  * (f)

$E_{9}$ when $s\in(\frac{11}{4};3)\cup(\frac{17}{3};\infty)$;

where $0\leq s\leq t$ (see Fig. 1).

Figure 1. The partition of the set $\\{(s,t):0\leq s\leq t\\}$ corresponding
to classification of EAs in the CEA $E_{1}^{[s,t]}$.

###  2.2. The case $E_{2}^{[s,t]}$.

In this case by Lemma 1 we get

######  Lemma 2.

The real evolution algebra corresponding to the matrix

$\left(\begin{array}[]{cccccc}1+a&1+a&1+a\\\ 1-b&1-b&1-b\\\
b-a&b-a&b-a\end{array}\right)$ is isomorphic to one of the following algebras:

  * (a)

$E_{4}$ if one of the following conditions is hold:

  * 1)

$a=b=1$;

  * 2)

$a=b=-1$;

  * 3)

$a=-1,\ b=1;$

  * (b)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$a=-1,\ |b|<1$;

  * 2)

$b=1,\ |a|<1$;

  * 3)

$a=b,\ |a|<1$;

  * (c)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$a=-1,\ |b|>1$;

  * 2)

$b=1,\ |a|>1$;

  * 3)

$a=b,\ |a|>1$;

  * (d)

$E_{7}$ if the following condition is hold:

  * $(1+a)(1-b)(b-a)\neq 0,\ (1+a)(1-b)(2+a-b)>0,\ (b-a)(2+a-b)>0$;

  * (e)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$(1+a)(1-b)(b-a)\neq 0,\ (1+a)(1-b)(2+a-b)>0,\ (b-a)(2+a-b)<0$;

  * 2)

$(1+a)(1-b)(b-a)\neq 0,\ (1+a)(1-b)(2+a-b)<0,\ (b-a)(2+a-b)>0$;

  * 3)

$(1+a)(1-b)(b-a)\neq 0,\ 2+a-b=0$;

  * (f)

$E_{9}$ if the following conditions is hold:

  * $(1+a)(1-b)(b-a)\neq 0,\ (1+a)(1-b)(2+a-b)<0,\ (b-a)(2+a-b)<0$.

Now, by this lemma we obtain the following theorem which gives classification
of the CEA: $E_{2}^{[s,t]}$. Let’s denote by $E_{0}$ the trivial evolution
algebra (i.e. with zero multiplication).

######  Theorem 2.

For given values $(s,t)\in\\{(s,t):0\leq s\leq t<a\\}$ of time the algebra
$E_{2}^{[s,t]}$ is isomorphic to

  * (a)

$E_{4}$ if one of the following conditions is hold:

  * 1)

$\psi(s)=\varphi(s)=1$;

  * 2)

$\psi(s)=\varphi(s)=-1$;

  * 3)

$\psi(s)=-1,\ \varphi(s)=1;$

  * (b)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\psi(s)=-1,\ |\varphi(s)|<1$;

  * 2)

$\varphi(s)=1,\ |\psi(s)|<1$;

  * 3)

$\psi(s)=\varphi(s),\ |\psi(s)|<1$;

  * (c)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\psi(s)=-1,\ |\varphi(s)|>1$;

  * 2)

$\varphi(s)=1,\ |\psi(s)|>1$;

  * 3)

$\psi(s)=\varphi(s),\ |\psi(s)|>1$;

  * (d)

$E_{7}$ if the following condition is hold:

  * $(1+\psi(s))(1-\varphi(s))(\varphi(s)-\psi(s))\neq 0,\\\ (1+\psi(s))(1-\varphi(s))(2+\psi(s)-\varphi(s))>0,\ (\varphi(s)-\psi(s))(2+\psi(s)-\varphi(s))>0$;

  * (e)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$(1+\psi(s))(1-\varphi(s))(\varphi(s)-\psi(s))\neq 0,\\\
(1+\psi(s))(1-\varphi(s))(2+\psi(s)-\varphi(s))>0,\
(\varphi(s)-\psi(s))(2+\psi(s)-\varphi(s))<0$;

  * 2)

$(1+\psi(s))(1-\varphi(s))(\varphi(s)-\psi(s))\neq 0,\\\
(1+\psi(s))(1-\varphi(s))(2+\psi(s)-\varphi(s))<0,\
(\varphi(s)-\psi(s))(2+\psi(s)-\varphi(s))>0$;

  * 3)

$(1+\psi(s))(1-\varphi(s))(\varphi(s)-\psi(s))\neq 0,\
2+\psi(s)-\varphi(s)=0$;

  * (f)

$E_{9}$ if the following condition is hold:

  * 1)

$(1+\psi(s))(1-\varphi(s))(\varphi(s)-\psi(s))\neq 0,\\\
(1+\psi(s))(1-\varphi(s))(2+\psi(s)-\varphi(s))<0,\
(\varphi(s)-\psi(s))(2+\psi(s)-\varphi(s))<0$.

For all $(s,t)\in\\{(s,t):t\geq a\\}$ the algebra $E_{2}^{[s,t]}$ is
isomorphic to $E_{0}$.

###### Proof.

The proof of the theorem is similar to that of Theorem 1. ∎

######  Remark 2.

By Lemma 2, $E_{2}^{[s,t]}$ is not isomorphic to the algebras $E_{1},$
$E_{2},$ $E_{3},$ $E_{10},$ $E_{11},$ $E_{12}$ for any $0\leq s\leq t$.

To illustrate the essence of Theorem 2, consider the following example.

######  Example 2.

Let $\varphi(s)=s^{2}-8s+13,\ \psi(s)=s^{2}-5$ then for given

$(s,t)\in\\{(s,t):0\leq s\leq t<a\\}$ we obtain:

  * (a)

$E_{4}$ when $s=2$;

  * (b)

$E_{5}$ when $s=\frac{9}{4}$;

  * (c)

$E_{6}$ when $s=6$;

  * (d)

$E_{7}$ when $s\in(2;\frac{9}{4})$;

  * (e)

$E_{8}$ when $s\in(\frac{9}{4};6)$;

  * (f)

$E_{9}$ when $s\in[0;2)\cup(6;\infty)$;

and for all $(s,t)\in\\{(s,t):t\geq a\\}$ we obtain $E_{0}$ (see Fig. 2).

Figure 2. The partition of time set corresponding to the classification of the
CEA $E_{2}^{[s,t]}$

Since the two different chains we considered above contain EAs isomorphic to
the same algebras. But they do not include all twelve algebras $E_{i}$
$i=1,2,\dots,12$. Below we show that $E_{3}^{[s,t]}$ is reach enough, one can
choose its parameter functions such that the chain includes all the algebras.

###  2.3. The case $E_{3}^{[s,t]}$.

######  Lemma 3.

For the real evolution algebra corresponding to the matrix

$\left(\begin{array}[]{cccccc}\alpha&\beta&\gamma\\\
\lambda\alpha&\lambda\beta&\lambda\gamma\\\
\mu\alpha&\mu\beta&\mu\gamma\end{array}\right)$ we have
I. If $\alpha\neq 0,\ \alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}=0$ then it is
isomorphic to one of the following algebras:

  * (a)

$E_{1}$ if one of the following conditions is hold:

  * 1)

$\mu=0,\ \lambda\beta\gamma\neq 0$;

  * 2)

$\lambda=0,\ \beta\gamma\mu\neq 0$;

  * 3)

$\gamma=\mu=0,\ \lambda\beta\neq 0$;

  * 4)

$\lambda=\beta=0,\ \gamma\mu\neq 0$;

  * (b)

$E_{2}$ if one of the following conditions is hold:

  * 1)

$\gamma=0,\ \lambda\beta\mu\neq 0,\ \mu>0$;

  * 2)

$\beta=0,\ \lambda\gamma\mu\neq 0,\ \lambda>0$;

  * 3)

$\lambda\beta\gamma\mu\neq 0$;

  * (c)

$E_{3}$ if one of the following conditions is hold:

  * 1)

$\gamma=0,\ \lambda\beta\mu\neq 0,\ \mu<0$;

  * 2)

$\beta=0,\ \lambda\gamma\mu\neq 0,\ \lambda<0$.

II. If $\alpha\neq 0,\ \alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}\neq 0$ then
it is isomorphic to one of the following algebras:

  * (d)

$E_{4}$ if the following condition is hold:

  * $\lambda=0,\ \mu=0$;

  * (e)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\lambda=0,\ \mu>0$;

  * 2)

$\mu=0,\ \lambda>0$;

  * (f)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\lambda=0,\ \mu<0$;

  * 2)

$\mu=0,\ \lambda<0$;

  * (g)

$E_{7}$ if the following condition is hold:

  * $\lambda>0,\ \mu>0$;

  * (h)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$\lambda>0,\ \mu<0,\ \alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$;

  * 2)

$\lambda<0,\ \mu>0,\ \alpha^{2}+\lambda\beta^{2}<0,\
\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$;

  * 3)

$\lambda<0,\ \mu>0,\ \alpha^{2}+\lambda\beta^{2}>0$;

  * 4)

$\lambda<0,\ \mu<0,\ \alpha^{2}+\lambda\beta^{2}>0,\
\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$;

  * 5)

$\lambda<0,\ \mu<0,\ \alpha^{2}+\lambda\beta^{2}<0$;

  * 6)

$\lambda\neq 0,\ \mu\neq 0,\ \alpha^{2}+\lambda\beta^{2}=0$;

  * (i)

$E_{9}$ if one of the following conditions is hold:

  * 1)

$\lambda>0,\ \mu<0,\ \alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$;

  * 2)

$\lambda<0,\ \mu>0,\ \alpha^{2}+\lambda\beta^{2}<0,\
\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$;

  * 3)

$\lambda<0,\ \mu<0,\ \alpha^{2}+\lambda\beta^{2}>0,\
\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$.

III. If $\alpha=0,\ \beta\neq 0$ then it is isomorphic to one of the following
algebras:

  * (j)

$E_{2}$ if the following condition is hold:

  * $\mu\neq 0,\ \gamma\neq 0,\ \lambda>0,\ \lambda\beta^{2}+\mu\gamma^{2}=0$;

  * (k)

$E_{3}$ if the following condition is hold:

  * $\mu\neq 0,\ \gamma\neq 0,\ \lambda<0,\ \lambda\beta^{2}+\mu\gamma^{2}=0$;

  * (l)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\mu=0,\ \lambda>0$;

  * 2)

$\lambda=0,\ \gamma\neq 0,\ \mu>0$;

  * (m)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\mu=0,\ \lambda<0$;

  * 2)

$\lambda=0,\ \gamma\neq 0,\ \mu<0$;

  * (n)

$E_{7}$ if the following condition is hold:

  * $\lambda>0,\ \mu>0$;

  * (o)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$\lambda<0,\ \mu<0$;

  * 2)

$\lambda\beta^{2}+\mu\gamma^{2}>0,\ \lambda\mu<0$;

  * (p)

$E_{9}$ if the following condition is hold:

  * $\lambda\beta^{2}+\mu\gamma^{2}<0,\ \lambda\mu<0$;

  * (q)

$E_{10}$ if the following condition is hold:

  * $\lambda=0,\ \mu=0$;

  * (r)

$E_{11}$ if the following condition is hold:

  * $\lambda=0,\ \gamma=0,\ \mu>0$;

  * (s)

$E_{12}$ if the following condition is hold:

  * $\lambda=0,\ \gamma=0,\ \mu<0$;

###### Proof.

Let

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&\gamma\\\
\lambda\alpha&\lambda\beta&\lambda\gamma\\\
\mu\alpha&\mu\beta&\mu\gamma\end{array}\right)$

be the matrix of structural constants of the evolution algebra $E_{M_{B}}$
relative to a basis $B=\\{e_{1},e_{2},e_{3}\\}$. Then, the multiplication
table in $E_{M_{B}}$ is

$\begin{array}[]{c}e_{1}e_{1}=\alpha e_{1}+\beta e_{2}+\gamma e_{3},\
e_{2}e_{2}=\lambda\alpha e_{1}+\lambda\beta e_{2}+\lambda\gamma
e_{3},\\\\[2.84526pt] e_{3}e_{3}=\mu\alpha e_{1}+\mu\beta e_{2}+\mu\gamma
e_{3},\ e_{1}e_{2}=e_{1}e_{3}=e_{2}e_{3}=0.\end{array}$ (2.17)

We consider all possible cases.

Case 1. Suppose that $\alpha\neq 0$.

Case 1.1. Assume that $\alpha^{2}+\beta^{2}\lambda+\gamma^{2}\mu=0$. In this
case may be one of the following cases:

  * 1)

$\mu=0,\ \lambda\beta\gamma\neq 0;$

  * 2)

$\lambda=0,\ \beta\gamma\mu\neq 0;$

  * 3)

$\gamma=0,\ \lambda\beta\mu\neq 0;$

  * 4)

$\beta=0,\ \lambda\gamma\mu\neq 0;$

  * 5)

$\gamma=\mu=0,\ \lambda\beta\neq 0;$

  * 6)

$\lambda=\beta=0,\ \gamma\mu\neq 0;$

  * 7)

$\lambda\beta\gamma\mu\neq 0.$

Case 1.1.1. If $\mu=0,\ \lambda\beta\gamma\neq 0$ then
$\alpha^{2}+\beta^{2}\lambda=0$. This implies that
$\lambda=-\frac{\alpha^{2}}{\beta^{2}}$. In this case the matrix $M_{B}$ will
be

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&\gamma\\\
-\frac{\alpha^{3}}{\beta^{2}}&-\frac{\alpha^{2}}{\beta}&-\frac{\gamma\alpha^{2}}{\beta^{2}}\\\
0&0&0\end{array}\right).$

If we take the change of basis $B^{\prime}=\\{\frac{1}{\alpha}e_{1},\
\frac{\beta}{\alpha^{2}}e_{2},\ \frac{\gamma}{\alpha^{2}}e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&\ 1\\\ -1&-1&-1\\\ \ 0&\
0&\ 0\end{array}\right).$

Let take another change of basis $B^{{}^{\prime\prime}}=\\{-e_{2},\
-e_{1}-e_{3},\ e_{3}\\}$ and we find a structure matrix with more zeros. Then

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\
-1&-1&0\\\ \ 0&\ 0&0\end{array}\right).$

Consequently, in this case $E_{M_{B}}$ is isomorphic to $E_{1}$.

Case 1.1.2. Let $\lambda=0,\ \beta\gamma\mu\neq 0$ then
$\alpha^{2}+\gamma^{2}\mu=0$. From this equality we find
$\mu=-\frac{\alpha^{2}}{\gamma^{2}}$. In this case the structure matrix is:

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&\gamma\\\ 0&0&0\\\
-\frac{\alpha^{3}}{\gamma^{2}}&-\frac{\alpha^{2}\beta}{\gamma^{2}}&-\frac{\alpha^{2}}{\gamma}\end{array}\right).$

If we take the change of basis $B^{\prime}=\\{\frac{1}{\alpha}e_{1},\
\frac{\beta}{\alpha^{2}}e_{2},\ \frac{\gamma}{\alpha^{2}}e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}1&1&1\\\ 0&0&0\\\
-1&-1&-1\end{array}\right).$

Let $B^{{}^{\prime\prime}}=\\{-e_{3},\ -e_{1}-e_{2},\ e_{2}\\}$, then by this
change of basis we have

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\
-1&-1&0\\\ \ 0&\ 0&0\end{array}\right).$

So, in this case also $E_{M_{B}}$ is isomorphic to $E_{1}$.

Case 1.1.3. Let $\gamma=0,\ \lambda\beta\mu\neq 0$. Then
$\alpha^{2}+\beta^{2}\lambda=0$ and $\lambda=-\frac{\alpha^{2}}{\beta^{2}}$.
In this case the structure matrix is:

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&0\\\
-\frac{\alpha^{3}}{\beta^{2}}&-\frac{\alpha^{2}}{\beta}&0\\\
\mu\alpha&\mu\beta&0\end{array}\right).$

Assume $\mu>0$. If we take the change of basis
$B^{\prime}=\\{\frac{1}{\alpha}e_{1},\ \frac{\beta}{\alpha^{2}}e_{2},\
\frac{1}{\alpha\sqrt{\mu}}e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\ -1&-1&0\\\ \ 1&\
1&0\end{array}\right).$

Thus in this case $E_{M_{B}}$ is isomorphic to $E_{2}$.

Assume $\mu<0$. If we take $B^{\prime}=\\{\frac{1}{\alpha}e_{1},\
\frac{\beta}{\alpha^{2}}e_{2},\ \frac{1}{\alpha\sqrt{-\mu}}e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\ -1&-1&0\\\
-1&-1&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{3}$.

Case 1.1.4. Let $\beta=0,\ \lambda\gamma\mu\neq 0$. Then
$\alpha^{2}+\gamma^{2}\mu=0$ and $\mu=-\frac{\alpha^{2}}{\gamma^{2}}$. We have

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&0&\gamma\\\
\lambda\alpha&0&\lambda\gamma\\\
-\frac{\alpha^{3}}{\gamma^{2}}&0&-\frac{\alpha^{2}}{\gamma}\end{array}\right).$

If we take the basis $B^{\prime}=\\{e_{1},\ e_{3},\ e_{2}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\alpha&\gamma&0\\\
-\frac{\alpha^{3}}{\gamma^{2}}&-\frac{\alpha^{2}}{\gamma}&0\\\
\lambda\alpha&\lambda\gamma&0\\\ \end{array}\right).$

This case is similar to the case 1.1.3. $E_{M_{B}}$ is isomorphic to $E_{2}$
when $\lambda>0$, isomorphic to $E_{3}$ when $\lambda<0$.

Case 1.1.5. Let $\gamma=\mu=0,\ \lambda\beta\neq 0$. Then
$\alpha^{2}+\beta^{2}\lambda=0$ and $\lambda=-\frac{\alpha^{2}}{\beta^{2}}$.
In this case the structure matrix is:

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&0\\\
-\frac{\alpha^{3}}{\beta^{2}}&-\frac{\alpha^{2}}{\beta}&0\\\
0&0&0\end{array}\right).$

If we take the basis $B^{\prime}=\\{\frac{1}{\alpha}e_{1},\
\frac{\beta}{\alpha^{2}}e_{2},\ e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\ -1&-1&0\\\ \ 0&\
0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{1}$.

Case 1.1.6. Let $\lambda=\beta=0,\ \gamma\mu\neq 0$. Taking the change of
basis $B^{\prime}=\\{e_{1},\ e_{3},\ e_{2}\\}$ then we are in the case 1.1.5.

Case 1.1.7. Let $\lambda\beta\gamma\mu\neq 0.$ If we take the basis
$B^{\prime}=\\{\frac{1}{\alpha}e_{1},\ \frac{\beta}{\alpha^{2}}e_{2},\
\frac{\gamma}{\alpha^{2}}e_{3}\\}$ then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&1\\\\[2.84526pt]
\frac{\lambda\beta^{2}}{\alpha^{2}}&\frac{\lambda\beta^{2}}{\alpha^{2}}&\frac{\lambda\beta^{2}}{\alpha^{2}}\\\\[2.84526pt]
\frac{\mu\gamma^{2}}{\alpha^{2}}&\frac{\mu\gamma^{2}}{\alpha^{2}}&\frac{\mu\gamma^{2}}{\alpha^{2}}\end{array}\right).$

From the equality $\alpha^{2}+\beta^{2}\lambda+\gamma^{2}\mu=0$ we have
$\frac{\lambda\beta^{2}}{\alpha^{2}}=-1-\frac{\mu\gamma^{2}}{\alpha^{2}}$.
Hence

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&1\\\\[2.84526pt]
-1-\frac{\mu\gamma^{2}}{\alpha^{2}}&-1-\frac{\mu\gamma^{2}}{\alpha^{2}}&-1-\frac{\mu\gamma^{2}}{\alpha^{2}}\\\\[2.84526pt]
\frac{\mu\gamma^{2}}{\alpha^{2}}&\frac{\mu\gamma^{2}}{\alpha^{2}}&\frac{\mu\gamma^{2}}{\alpha^{2}}\end{array}\right).$

Now, we take the change of basis $B^{{}^{\prime\prime}}=\\{e_{1}^{\prime},\
e_{2}^{\prime},\ e_{3}^{\prime}\\}$ such that which the matrix is following

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{\alpha^{2}(K+1)}{2(\alpha^{2}+\mu\gamma^{2})}&\frac{\alpha^{2}(K-1)}{2(\alpha^{2}+\mu\gamma^{2})}&\frac{\alpha^{2}(K+1)}{2(\alpha^{2}+\mu\gamma^{2})}\\\\[5.69054pt]
\frac{\alpha^{2}(K-1)}{2(\alpha^{2}+\mu\gamma^{2})}&\frac{\alpha^{2}(K+1)}{2(\alpha^{2}+\mu\gamma^{2})}&\frac{\alpha^{2}(K-1)}{2(\alpha^{2}+\mu\gamma^{2})}\\\\[5.69054pt]
-\frac{\mu\gamma^{2}}{\alpha^{2}}&\ 0&1\end{array}\right),$

where
$K=\frac{\mu\gamma^{2}}{\alpha^{2}}(1+\frac{\mu\gamma^{2}}{\alpha^{2}})^{2}$
and note that
$|P_{B^{{}^{\prime\prime}}B^{\prime}}|=\frac{\mu\gamma^{2}(\alpha^{2}+\mu\gamma^{2})}{\alpha^{4}}\neq
0.$ Then we obtain

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&\ 1&0\\\
-1&-1&0\\\ \ 1&\ 1&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{2}$.

Case 1.2. Assume that $\alpha^{2}+\beta^{2}\lambda+\gamma^{2}\mu\neq 0$. We
consider the next cases.

Case 1.2.1. Suppose $\lambda=0,\ \mu=0$. In this case the structure matrix is:

$M_{B}=\left(\begin{array}[]{cccccc}\alpha&\beta&\gamma\\\ 0&0&0\\\
0&0&0\end{array}\right).$

Consider the change of basis
$B^{\prime}=\\{\frac{1}{\alpha}e_{1}+\frac{\beta}{\alpha^{2}}e_{2}+\frac{\gamma}{\alpha^{2}}e_{3};e_{2}-e_{3};4e_{2}+e_{3}\\}$.
Then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
0&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{4}$.

Case 1.2.2. Suppose $\lambda=0,\ \mu\neq 0$. Then
$\alpha^{2}+\gamma^{2}\mu\neq 0$.
For
$B^{\prime}=\\{\frac{1}{\alpha}e_{1}+\frac{\beta}{\alpha^{2}}e_{2}+\frac{\gamma}{\alpha^{2}}e_{3};e_{2};-\frac{\gamma\mu}{\alpha}e_{1}+e_{2}+e_{3}\\}$
the structure matrix is

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}1+\frac{\mu\gamma^{2}}{\alpha^{2}}&0&0\\\
0&0&0\\\ \mu(\alpha^{2}+\mu\gamma^{2})&0&0\end{array}\right).$

Case 1.2.2.1. Assume that $\mu>0$. Consider
$B^{{}^{\prime\prime}}=\\{\frac{\alpha^{2}}{\alpha^{2}+\mu\gamma^{2}}e_{1};\
e_{2};\ \frac{\alpha}{\sqrt{\mu}(\alpha^{2}+\mu\gamma^{2})}e_{3}\\}$. Then
$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to $E_{5}$.

Case 1.2.2.2. Assume that $\mu<0$. Consider
$B^{{}^{\prime\prime}}=\\{\frac{\alpha^{2}}{\alpha^{2}+\mu\gamma^{2}}e_{1};\
e_{2};\ \frac{\alpha}{\sqrt{-\mu}(\alpha^{2}+\mu\gamma^{2})}e_{3}\\}$. Then
$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
-1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to $E_{6}$.

Case 1.2.3. Assume that $\lambda\neq 0$ and $\mu=0$. If we take the basis
$B^{\prime}=\\{e_{1},\ e_{3},\ e_{2}\\}$ then we are in the same conditions as
in Case 1.2.2. Namely, $E_{M_{B}}$ is isomorphic to $E_{5}$ and $E_{6}$
respectively $\lambda>0$ and $\lambda<0$.

Case 1.2.4. Assume that $\lambda\neq 0$ , $\mu\neq 0$ and
$\alpha^{2}+\lambda\beta^{2}\neq 0$. If $B^{\prime}$ is the change of basis
such that

$P_{B^{\prime}B}=\left(\begin{array}[]{cccccc}\frac{1}{\alpha}&\frac{\beta}{\alpha^{2}}&\frac{\gamma}{\alpha^{2}}\\\\[2.84526pt]
-\frac{\lambda\beta}{\alpha}&1&0\\\\[2.84526pt]
\frac{-\mu\gamma}{\alpha^{2}+\lambda\beta^{2}}&\frac{-\beta\mu\gamma}{\alpha(\alpha^{2}+\lambda\beta^{2})}&\frac{1}{\alpha}\end{array}\right)$

then we obtain

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{1}{\alpha^{2}}(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})&0&0\\\\[2.84526pt]
\lambda(\alpha^{2}+\lambda\beta^{2})&0&0\\\\[2.84526pt]
\frac{\mu(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}{\alpha^{2}+\lambda\beta^{2}}&0&0\end{array}\right).$

Note that
$|P_{B^{\prime}B}|=\frac{1}{\alpha^{4}}(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})\neq
0$.

Case 1.2.4.1. If $\lambda>0$ and $\mu>0$ then consider the change of basis
$B^{{}^{\prime\prime}}=\\{e_{1}^{{}^{\prime\prime}},\
e_{2}^{{}^{\prime\prime}},\ e_{3}^{{}^{\prime\prime}}\\}$

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{\alpha^{2}}{\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}}&0&0\\\\[5.69054pt]
0&\frac{\alpha}{\sqrt{\lambda(\alpha^{2}+\lambda\beta^{2})(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}}&0\\\\[5.69054pt]
0&\
0&\frac{\alpha\sqrt{\alpha^{2}+\lambda\beta^{2}}}{\sqrt{\mu}(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}\end{array}\right)$

and the structure matrix is

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{7}$.

Case 1.2.4.2. If $\lambda>0$ and $\mu<0$. Assume that
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$. Consider the change of basis
$B^{{}^{\prime\prime}}=\\{e_{1}^{{}^{\prime\prime}},\
e_{2}^{{}^{\prime\prime}},\ e_{3}^{{}^{\prime\prime}}\\}$

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{\alpha^{2}}{\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}}&0&0\\\\[5.69054pt]
0&\frac{\alpha}{\sqrt{\lambda(\alpha^{2}+\lambda\beta^{2})(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}}&0\\\\[5.69054pt]
0&\
0&\frac{\alpha\sqrt{\alpha^{2}+\lambda\beta^{2}}}{\sqrt{-\mu}(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}\end{array}\right)$

and the structure matrix is

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{8}$.

Case 1.2.4.3. If $\lambda>0$ and $\mu<0$. Assume that
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$. Consider the change of basis
$B^{{}^{\prime\prime}}=\\{e_{1}^{{}^{\prime\prime}},\
e_{2}^{{}^{\prime\prime}},\ e_{3}^{{}^{\prime\prime}}\\}$

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{\alpha^{2}}{\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}}&0&0\\\\[5.69054pt]
0&\frac{\alpha}{\sqrt{-\lambda(\alpha^{2}+\lambda\beta^{2})(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}}&0\\\\[5.69054pt]
0&\
0&\frac{\alpha\sqrt{\alpha^{2}+\lambda\beta^{2}}}{\sqrt{-\mu}(\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2})}\end{array}\right)$

and the structure matrix is

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{9}$.

Case 1.2.4.4. If $\lambda<0$ and $\mu>0$. This case is similar to the Case
1.2.4.2 and Case 1.2.4.3. There are only differences in conditions.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
-1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}<0$ and
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}<0$ and
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}>0$ and if we take
the change of basis $\\{e_{1},\ e_{3},\ e_{2}\\}$ then we have
$\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\ -1&0&0\end{array}\right)$.

Case 1.2.4.5. If $\lambda<0$ and $\mu<0$. In this case the same as above,
There are differences in conditions.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
-1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}>0$ and
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}<0$.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}>0$ and
$\alpha^{2}+\lambda\beta^{2}+\mu\gamma^{2}>0$.
The structure matrix is $\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
1&0&0\end{array}\right)$ when $\alpha^{2}+\lambda\beta^{2}<0$ and if we take
the change of basis $\\{e_{1},\ e_{3},\ e_{2}\\}$ then we have
$\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\ -1&0&0\end{array}\right)$.

Case 1.2.5. Suppose that $\lambda\neq 0,\ \mu\neq 0$ and
$\alpha^{2}+\lambda\beta^{2}=0$. Then $\lambda\beta\mu\gamma\neq 0$ and so
$\lambda=-\frac{\alpha^{2}}{\beta^{2}}$. If we take the change of basis
$B^{\prime}=\\{e_{1}^{\prime},\ e_{2}^{\prime},\ e_{3}^{\prime}\\}$ such that

$P_{B^{\prime}B}=\left(\begin{array}[]{cccccc}\frac{1}{\alpha}&\frac{\beta}{\alpha^{2}}&\frac{\gamma}{\alpha^{2}}\\\\[2.84526pt]
-\frac{\mu}{2\alpha}&\frac{\beta\mu}{2\alpha^{2}}&\frac{1}{\gamma}\\\\[2.84526pt]
\frac{2\alpha^{2}-\mu\gamma^{2}}{2\alpha^{3}}&\frac{\beta(2\alpha^{2}+\mu\gamma^{2})}{2\alpha^{4}}&\frac{\gamma}{\alpha^{2}}\end{array}\right)$

then we obtain

$\begin{array}[]{cccccc}e_{1}^{\prime}e_{1}^{\prime}=\frac{1}{\alpha^{2}}e_{1}^{2}+\frac{\beta^{2}}{\alpha^{4}}e_{2}^{2}+\frac{\gamma^{2}}{\alpha^{4}}e_{3}^{2}=(\frac{1}{\alpha^{2}}+\lambda\frac{\beta^{2}}{\alpha^{4}}+\mu\frac{\gamma^{2}}{\alpha^{4}})(\alpha
e_{1}+\beta e_{2}+\gamma
e_{3})=\frac{\mu\gamma^{2}}{\alpha^{2}}e_{1}^{\prime},\\\\[5.69054pt]
e_{2}^{\prime}e_{2}^{\prime}=\frac{\mu^{2}}{4\alpha^{2}}e_{1}^{2}+\frac{\beta^{2}\mu^{2}}{4\alpha^{4}}e_{2}^{2}+\frac{1}{\gamma^{2}}e_{3}^{2}=(\frac{\mu^{2}}{4\alpha^{2}}+\lambda\frac{\beta^{2}\mu^{2}}{4\alpha^{4}}+\frac{\mu}{\gamma^{2}})(\alpha
e_{1}+\beta e_{2}+\gamma
e_{3})=\frac{\mu\alpha^{2}}{\gamma^{2}}e_{1}^{\prime},\\\\[5.69054pt]
e_{3}^{\prime}e_{3}^{\prime}=\frac{(2\alpha^{2}-\mu\gamma^{2})^{2}}{4\alpha^{6}}e_{1}^{2}+\frac{\beta^{2}(2\alpha^{2}+\mu\gamma^{2})^{2}}{4\alpha^{8}}e_{2}^{2}+\frac{\gamma^{2}}{\alpha^{4}}e_{3}^{2}=(\frac{(2\alpha^{2}-\mu\gamma^{2})^{2}}{4\alpha^{6}}+\lambda\frac{\beta^{2}(2\alpha^{2}+\mu\gamma^{2})^{2}}{4\alpha^{8}}+\\\\[5.69054pt]
+\mu\frac{\gamma^{2}}{\alpha^{4}})(\alpha e_{1}+\beta e_{2}+\gamma
e_{3})=-\frac{\mu\gamma^{2}}{\alpha^{4}}(\alpha e_{1}+\beta e_{2}+\gamma
e_{3})=-\frac{\mu\gamma^{2}}{\alpha^{2}}e_{1}^{\prime}.\\\ \end{array}$

Moreover

$\begin{array}[]{cccccc}e_{1}^{\prime}e_{2}^{\prime}=-\frac{\mu}{2\alpha^{2}}e_{1}^{2}+\frac{\beta^{2}\mu}{2\alpha^{4}}e_{2}^{2}+\frac{1}{\alpha^{2}}e_{3}^{2}=e_{1}^{2}(-\frac{\mu}{2\alpha^{2}}+\lambda\frac{\beta^{2}\mu}{2\alpha^{4}}+\frac{\mu}{\alpha^{2}})=e_{1}^{2}(\lambda\frac{\beta^{2}\mu}{2\alpha^{4}}+\frac{\mu}{2\alpha^{2}})=0\\\\[5.69054pt]
e_{1}^{\prime}e_{3}^{\prime}=\frac{2\alpha^{2}-\mu\gamma^{2}}{2\alpha^{4}}e_{1}^{2}+\frac{\beta^{2}(2\alpha^{2}+\mu\gamma^{2})}{2\alpha^{6}}e_{2}^{2}+\frac{\gamma^{2}}{\alpha^{4}}e_{3}^{2}=\frac{2\alpha^{2}(\alpha^{2}+\lambda\beta^{2})+\mu\gamma^{2}(\alpha^{2}+\lambda\beta^{2})}{2\alpha^{6}}e_{1}^{2}=0\\\\[5.69054pt]
e_{2}^{\prime}e_{3}^{\prime}=\frac{-\mu(2\alpha^{2}-\mu\gamma^{2})}{4\alpha^{4}}e_{1}^{2}+\frac{\mu\beta^{2}(2\alpha^{2}+\mu\gamma^{2})}{4\alpha^{6}}e_{2}^{2}+\frac{1}{\alpha^{2}}e_{3}^{2}=e_{1}^{2}(\frac{-2\mu\alpha^{2}+\mu^{2}\gamma^{2}}{4\alpha^{4}}+\frac{2\alpha^{2}\mu\lambda\beta^{2}+\mu^{2}\gamma^{2}\lambda\beta^{2}}{4\alpha^{6}}+\\\\[5.69054pt]
+\frac{\mu}{\alpha^{2}})=e_{1}^{2}(\frac{-2\mu\alpha^{4}+2\mu\alpha^{2}\lambda\beta^{2}}{4\alpha^{6}}+\frac{\mu}{\alpha^{2}})=\frac{-2\mu\alpha^{4}+2\mu\alpha^{2}\lambda\beta^{2}+4\mu\alpha^{4}}{4\alpha^{6}}e_{1}^{2}=\frac{2\mu\alpha^{2}(\alpha^{2}+\lambda\beta^{2})}{4\alpha^{6}}=0,\end{array}$

note that $|P_{B^{\prime}B}|=-\frac{\beta\mu\gamma}{\alpha^{5}}\neq 0$. So, we
have

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{\mu\gamma^{2}}{\alpha^{2}}&0&0\\\\[2.84526pt]
\frac{\mu}{\gamma^{2}}\alpha^{2}&0&0\\\\[2.84526pt]
-\frac{\mu\gamma^{2}}{\alpha^{2}}&0&0\end{array}\right).$

Considering the change of basis
$B^{{}^{\prime\prime}}=\\{\frac{\alpha^{2}}{\mu\gamma^{2}}e_{1},\
\frac{1}{\mu}e_{2},\ \frac{-\alpha^{2}}{\mu\gamma^{2}}e_{3}\\}$ and we obtain

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&0&0\\\ \ 1&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{8}$.

Case 2. Suppose that $\alpha=0$. In this case the structure matrix of the
evolution algebra is

$M_{B}=\left(\begin{array}[]{cccccc}0&\beta&\gamma\\\
0&\lambda\beta&\lambda\gamma\\\ 0&\mu\beta&\mu\gamma\end{array}\right).$

Necessarily that $\beta^{2}+\gamma^{2}\neq 0$. Without loss of generality we
assume $\beta\neq 0$.

Case 2.1. Assume $\lambda\neq 0$. Consider the change of basis
$B^{\prime}=\\{e_{2},\ e_{3},\ e_{1}\\}$. Then

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\lambda\beta&\lambda\gamma&0\\\
\mu\beta&\mu\gamma&0\\\ \beta&\gamma&0\end{array}\right).$

Case 2.1.1. Suppose $\mu=0$, $\gamma\neq 0$ and $\lambda>0$. Consider the
change of basis
$B^{{}^{\prime\prime}}=\\{e_{1}^{{}^{\prime\prime}},\
e_{2}^{{}^{\prime\prime}},\ e_{3}^{{}^{\prime\prime}}\\}$ such that

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{1}{\lambda\beta}&\frac{\gamma}{\lambda\beta^{2}}&0\\\
0&1&0\\\ 0&0&\frac{\sqrt{\lambda}}{\lambda\beta}\end{array}\right).$

We have

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{5}$.

If $\lambda<0$ then we take as
$B^{{}^{\prime\prime}}=\\{e_{1}^{{}^{\prime\prime}},\
e_{2}^{{}^{\prime\prime}},\ e_{3}^{{}^{\prime\prime}}\\}$ the following

$P_{B^{{}^{\prime\prime}}B^{\prime}}=\left(\begin{array}[]{cccccc}\frac{1}{\lambda\beta}&\frac{\gamma}{\lambda\beta^{2}}&0\\\
0&1&0\\\ 0&0&\frac{\sqrt{-\lambda}}{\lambda\beta}\end{array}\right).$

We have

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 0&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{6}$.

Case 2.1.2. The cases $\mu=0,\ \gamma=0,\lambda>0$ and $\mu=0,\
\gamma=0,\lambda<0$ are similar to the previous case. In this case $E_{M_{B}}$
is isomorphic to $E_{5}$ and $E_{6}$ respectively.

Case 2.1.3. Assume $\mu\neq 0,\ \gamma=0$. In this case if $\lambda>0,\ \mu>0$
then for $B^{{}^{\prime\prime}}=\\{\frac{1}{\lambda\beta}e_{1};\
\frac{1}{\sqrt{\mu\lambda}\beta}e_{2};\ \frac{1}{\sqrt{\lambda}\beta}e_{3}\\}$
we have

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{7}$.

Similarly, if $\lambda<0,\ \mu<0$ and $\lambda>0,\ \mu<0$ then the structure
matrix will be

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&0&0\\\ \ 1&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{8}$.

If $\lambda<0,\ \mu>0$ then the structure matrix will be

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right).$

In this case $E_{M_{B}}$ is isomorphic to $E_{9}$.

Case 2.1.4. Assume $\mu\neq 0$, $\gamma\neq 0$. Now, consider the change of
basis $B^{{}^{\prime\prime}}=\\{\frac{1}{\lambda\beta}e_{1};\ e_{2};\
e_{3}\\}$. Then

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&\frac{\gamma}{\lambda\beta^{2}}&0\\\
\mu\lambda\beta^{2}&\mu\gamma&0\\\
\lambda\beta^{2}&\gamma&0\end{array}\right).$

Case 2.1.4.1. Suppose that $\lambda\beta^{2}+\mu\gamma^{2}=0$. And so
$\mu=-\frac{\lambda\beta^{2}}{\gamma^{2}}$. It follows that

$M_{B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&\frac{\gamma}{\lambda\beta^{2}}&0\\\
-\frac{\lambda^{2}\beta^{4}}{\gamma^{2}}&-\frac{\lambda\beta^{2}}{\gamma}&0\\\
\lambda\beta^{2}&\gamma&0\end{array}\right).$

Let’s $\lambda>0$, if we take $B^{{}^{\prime\prime\prime}}=\\{e_{1};\
\frac{\gamma}{\lambda\beta^{2}}e_{2},\ \frac{1}{\sqrt{\lambda}\beta}\\}$ then
$M_{B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&1&0\\\
-1&-1&0\\\ \ 1&1&0\end{array}\right).$
In this case $E_{M_{B}}$ is isomorphic to $E_{2}$.

Let’s $\lambda<0$, in this case we may take
$B^{{}^{\prime\prime\prime}}=\\{e_{1};\ \frac{\gamma}{\lambda\beta^{2}}e_{2},\
\frac{1}{\sqrt{\lambda}\beta}\\}$, then we have
$M_{B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}\ 1&1&0\\\
-1&-1&0\\\ -1&-1&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic
to $E_{3}$.

Case 2.1.4.2. Suppose that $\lambda\beta^{2}+\mu\gamma^{2}\neq 0$ and
$\left\\{\begin{array}[]{ll}\lambda\mu>0\\\
\lambda\beta^{2}+\mu\gamma^{2}>0\end{array}\right.$. If we take
$B^{{}^{\prime\prime\prime}}$ is the change of basis such that

$P_{B^{{}^{\prime\prime\prime}}B^{{}^{\prime\prime}}}=\left(\begin{array}[]{cccccc}1&\frac{\gamma}{\lambda\beta^{2}}&0\\\
-\mu\gamma&1&0\\\ 0&0&1\end{array}\right)$

then we obtain that

$M_{B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}\frac{\lambda\beta^{2}+\mu\gamma^{2}}{\lambda\beta^{2}}&0&0\\\
\mu(\lambda\beta^{2}+\mu\gamma^{2})&0&0\\\
\lambda\beta^{2}&0&0\end{array}\right).$

Now, consider the change of basis $B^{*}=\\{e_{1}^{*},\ e_{2}^{*},\
e_{3}^{*}\\}$ such that

$P_{B^{*}B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}\frac{\lambda\beta^{2}}{\lambda\beta^{2}+\mu\gamma^{2}}&0&0\\\
0&\sqrt{\frac{\lambda}{\mu}}\frac{\beta}{\lambda\beta^{2}+\mu\gamma^{2}}&0\\\
0&0&\frac{1}{\sqrt{\lambda\beta^{2}+\mu\gamma^{2}}}\end{array}\right)$

and the structure matrix is $M_{B^{*}}=\left(\begin{array}[]{cccccc}1&0&0\\\
1&0&0\\\ 1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to
$E_{7}$.

Similarly, it is not difficult to show that if
$\left\\{\begin{array}[]{ll}\lambda\mu<0\\\
\lambda\beta^{2}+\mu\gamma^{2}>0\end{array}\right.$ and
$\left\\{\begin{array}[]{ll}\lambda\mu>0\\\
\lambda\beta^{2}+\mu\gamma^{2}<0\end{array}\right.$ then the structure matrix
is $M_{B^{*}}=\left(\begin{array}[]{cccccc}1&0&0\\\ 1&0&0\\\
-1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to $E_{8}$.

If $\left\\{\begin{array}[]{ll}\lambda\mu<0\\\
\lambda\beta^{2}+\mu\gamma^{2}<0\end{array}\right.$ then the structure matrix
is $M_{B^{*}}=\left(\begin{array}[]{cccccc}1&0&0\\\ -1&0&0\\\
-1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to $E_{9}$.

Case 2.2. Now, we consider the case $\lambda=0$.

Case 2.2.1. Assume that $\mu\gamma\neq 0$. Consider the change of basis
$B^{\prime}\\{e_{3},\ e_{2},\ e_{1}\\}$ and we obtain

$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}\mu\gamma&\mu\beta&0\\\ 0&0&0\\\
\gamma&\beta&0\end{array}\right).$

If we take $B^{{}^{\prime\prime}}=\\{\mu\gamma e_{1}+\mu\beta e_{2};\ e_{2},\
\mu e_{3}\\}$ then

$M_{B^{*}}=\left(\begin{array}[]{cccccc}\mu^{2}\gamma^{2}&0&0\\\ 0&0&0\\\
\mu&0&0\end{array}\right).$

Let $\mu>0$, consider the change of basis
$B^{{}^{\prime\prime\prime}}=\\{\frac{1}{\mu^{2}\gamma^{2}}e_{1};\ e_{2};\
\frac{1}{\sqrt{\mu}\mu\gamma}e_{3}\\}$. Then
$M_{B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\
0&0&0\\\ 1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to
$E_{5}$.

Similarly, if $\mu<0$ then we have
$M_{B^{{}^{\prime\prime\prime}}}=\left(\begin{array}[]{cccccc}1&0&0\\\
0&0&0\\\ -1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to
$E_{6}$.

Case 2.2.2. Suppose that $\mu\gamma=0$.

Case 2.2.2.1. Assume that $\mu=0$ and $\gamma$ is an arbitrary real number.
Then the structure matrix is
$M_{B}=\left(\begin{array}[]{cccccc}0&\beta&\gamma\\\ 0&0&0\\\
0&0&0\end{array}\right).$ Let’s take the change of basis $B^{\prime}=\\{\beta
e_{2}+\gamma e_{3};\ \frac{1}{\beta}e_{3};\ e_{1}\\}$ then
$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}0&0&0\\\ 0&0&0\\\
1&0&0\end{array}\right).$ In this case $E_{M_{B}}$ is isomorphic to $E_{10}$.

Case 2.2.2.2. Assume that $\mu>0$. It follows that $\gamma=0$. For
$B^{\prime}=\\{\beta e_{2};\ e_{1};\ \frac{1}{\sqrt{\mu}}e_{3}\\}$
we have $M_{B^{\prime}}=\left(\begin{array}[]{cccccc}0&0&0\\\ 1&0&0\\\
1&0&0\end{array}\right).$
It means that in this case $E_{M_{B}}$ is isomorphic to $E_{11}$.

Case 2.2.2.3. Assume that $\mu<0$. It follows that $\gamma=0$. For
$B^{\prime}=\\{\beta e_{2};\ e_{1};\ \frac{1}{\sqrt{-\mu}}e_{3}\\}$ we have
$M_{B^{\prime}}=\left(\begin{array}[]{cccccc}0&0&0\\\ 1&0&0\\\
-1&0&0\end{array}\right).$
It means that in this case $E_{M_{B}}$ is isomorphic to $E_{12}$. ∎

Using Lemma 3 we get the following theorem.

######  Theorem 3.

For given values
$(s,t)\in\\{(s,t):\eta(s)+\varphi_{1}(s)\vartheta(s)+\varphi_{2}(s)\kappa(s)\neq
0,$ $\eta(t)\neq 0,$
$\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)=0\\}$
of time, $E_{3}^{[s,t]}$ is isomorphic to

  * (a)

$E_{1}$ if one of the following conditions is hold:

  * 1)

$\varphi_{2}(s)=0,\ \varphi_{1}(s)\vartheta(t)\kappa(t)\neq 0$;

  * 2)

$\varphi_{1}(s)=0,\ \varphi_{2}(s)\vartheta(t)\kappa(t)\neq 0$;

  * 3)

$\kappa(t)=\varphi_{2}(s)=0,\ \varphi_{1}(s)\vartheta(t)\neq 0$;

  * 4)

$\varphi_{1}(s)=\vartheta(t)=0,\ \kappa(t)\varphi_{2}(s)\neq 0$;

  * (b)

$E_{2}$ if one of the following conditions is hold:

  * 1)

$\kappa(t)=0,\ \varphi_{1}(s)\varphi_{2}(s)\vartheta(t)\neq 0,\
\varphi_{2}(s)>0$;

  * 2)

$\vartheta(t)=0,\ \varphi_{1}(s)\varphi_{2}(s)\kappa(t)\neq 0,\
\varphi_{1}(s)>0$;

  * 3)

$\varphi_{1}(s)\varphi_{2}(s)\vartheta(t)\kappa(t)\neq 0$;

  * (c)

$E_{3}$ if one of the following conditions is hold:

  * 1)

$\kappa(t)=0,\ \varphi_{1}(s)\varphi_{2}(s)\vartheta(t)\neq 0,\
\varphi_{2}(s)<0$;

  * 2)

$\vartheta(t)=0,\ \varphi_{1}(s)\varphi_{2}(s)\kappa(t)\neq 0,\
\varphi_{1}(s)<0$;

For given values
$(s,t)\in\\{(s,t):\eta(s)+\varphi_{1}(s)\vartheta(s)+\varphi_{2}(s)\kappa(s)\neq
0,$ $\eta(t)\neq 0,$
$\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)\neq
0\\}$ of time, $E_{3}^{[s,t]}$ is isomorphic to

  * (d)

$E_{4}$ if the following condition is hold:

  * $\varphi_{1}(s)=0,\ \varphi_{2}(s)=0$;

  * (e)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\varphi_{1}(s)=0,\ \varphi_{2}(s)>0$;

  * 2)

$\varphi_{2}(s)=0,\ \varphi_{1}(s)>0$;

  * (f)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\varphi_{1}(s)=0,\ \varphi_{2}(s)<0$;

  * 2)

$\varphi_{2}(s)=0,\ \varphi_{1}(s)<0$;

  * (g)

$E_{7}$ if the following condition is hold:

  * $\varphi_{1}(s)>0,\ \varphi_{2}(s)>0$;

  * (h)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$\varphi_{1}(s)>0,\ \varphi_{2}(s)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)>0$;

  * 2)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)>0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)>0$;

  * 3)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)>0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)>0$;

  * 4)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)>0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)<0$;

  * 5)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)<0$;

  * 6)

$\varphi_{1}(s)\neq 0,\ \varphi_{2}(s)\neq 0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)=0$;

  * (i)

$E_{9}$ if one of the following conditions is hold:

  * 1)

$\varphi_{1}(s)>0,\ \varphi_{2}(s)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)<0$;

  * 2)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)>0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)<0$;

  * 3)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)<0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)>0,\
\eta^{2}(t)+\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)>0$.

For given values
$(s,t)\in\\{(s,t):\eta(s)+\varphi_{1}(s)\vartheta(s)+\varphi_{2}(s)\kappa(s)\neq
0,\ \eta(t)=0,\ \vartheta(t)\neq 0\\}$ of time , $E_{3}^{[s,t]}$ is isomorphic
to

  * (j)

$E_{2}$ if the following condition is hold:

  * $\varphi_{2}(s)\neq 0,\ \kappa(t)\neq 0,\ \varphi_{1}(s)>0,\ \varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)=0$;

  * (k)

$E_{3}$ if the following condition is hold:

  * $\varphi_{2}(s)\neq 0,\ \kappa(t)\neq 0,\ \varphi_{1}(s)<0,\ \varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)=0$;

  * (l)

$E_{5}$ if one of the following conditions is hold:

  * 1)

$\varphi_{2}(s)=0,\ \varphi_{1}(s)>0$;

  * 2)

$\varphi_{1}(s)=0,\kappa(t)\neq 0,\ \varphi_{2}(s)>0$;

  * (m)

$E_{6}$ if one of the following conditions is hold:

  * 1)

$\varphi_{2}(s)=0,\ \varphi_{1}(s)<0$;

  * 2)

$\varphi_{1}(s)=0,\ \kappa(t)\neq 0,\ \varphi_{2}(s)<0$;

  * (n)

$E_{7}$ if the following condition is hold:

  * $\varphi_{1}(s)>0,\ \varphi_{2}(s)>0$;

  * (o)

$E_{8}$ if one of the following conditions is hold:

  * 1)

$\varphi_{1}(s)<0,\ \varphi_{2}(s)<0$;

  * 2)

$\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)>0,\
\varphi_{1}(s)\varphi_{2}(s)<0$;

  * (p)

$E_{9}$ if the following condition is hold:

  * $\varphi_{1}(s)\vartheta^{2}(t)+\varphi_{2}(s)\kappa^{2}(t)<0,\ \varphi_{1}(s)\varphi_{2}(s)<0$;

  * (q)

$E_{10}$ if the following condition is hold:

  * $\varphi_{1}(s)=0,\ \varphi_{2}(s)=0$;

  * (r)

$E_{11}$ if the following condition is hold:

  * $\varphi_{1}(s)=0,\ \kappa(t)=0,\ \varphi_{2}(s)>0$;

  * (s)

$E_{12}$ if the following condition is hold:

  * $\varphi_{1}(s)=0,\ \kappa(t)=0,\ \varphi_{2}(s)<0$;

###### Proof.

The proof follows from Lemma 3. ∎

To illustrate the essence of Theorem 3, consider the following example.

######  Example 3.

Let $\eta(t)=\left\\{\begin{array}[]{ll}t+1,\ 0\leq t<6\\\\[2.84526pt] 0,\ \ \
\ \ \ t\geq 6\end{array}\right.$,
$\vartheta(t)=\left\\{\begin{array}[]{ll}\frac{1}{\sqrt{2}}(t+1),\ 0\leq
t<2\\\\[5.69054pt] \sqrt{(t-1)^{2}+4},\ 2\leq t<3\\\\[2.84526pt] 0,\ \ \ \ \ \
3\leq t<6\\\\[2.84526pt] t-2,\ \ t\geq 6\end{array}\right.$,

$\kappa(t)=\left\\{\begin{array}[]{ll}0,\ \ \ \ \ \ 0\leq t<1\\\\[2.84526pt]
t-3,\ 1\leq t<6\\\\[2.84526pt] 0,\ \ \ \ \ \ t\geq 6\end{array}\right.$,
$\varphi_{1}(s)=\left\\{\begin{array}[]{ll}-2,\ \ \ \ 0\leq s<3\\\\[2.84526pt]
0,\ \ \ \ \ \ 3\leq s\leq 5\\\\[2.84526pt] s-5,7\ ,\ s>5\end{array}\right.$,

$\varphi_{2}(s)=\left\\{\begin{array}[]{ll}-1,\ \ \ \ 0\leq s<1\\\\[2.84526pt]
0,\ \ \ \ \ \ 1\leq s<2\\\\[2.84526pt] 1`,\ \ \ \ \ \ 2\leq s<3\\\\[2.84526pt]
s-4,\ s\geq 3\end{array}\right.$.

Figure 3. The partition corresponding to the classification of CEA
$E_{3}^{[s,t]}$

We consider the next cases (see Fig. 3):

Case 1. If $0\leq s<1$ then we have

  * (a)

$E_{3}$ when $0\leq t<1$;

  * (b)

$E_{8}$ when $1\leq t<3$;

  * (c)

$E_{9}$ when $3\leq t<6$;

  * (d)

$E_{8}$ when $t\geq 6$;

Case 2. If $1\leq s<2$ then we have

  * (a)

$E_{1}$ when $1\leq t<2$;

  * (b)

$E_{6}$ when $t\geq 2$;

Case 3. If $2\leq s<3$ then we have

  * (a)

$E_{2}$ when $2\leq t<3$;

  * (b)

$E_{8}$ when $3\leq t<6$;

  * (c)

$E_{9}$ when $t\geq 6$;

Case 4. If $3\leq s<4$ then we have

  * (a)

$E_{6}$ when $3\leq t<6$;

  * (b)

$E_{12}$ when $t\geq 6$;

Case 5. If $s=4$ then we have

  * (a)

$E_{4}$ when $4\leq t<6$;

  * (b)

$E_{10}$ when $t\geq 6$;

Case 6. If $4<s\leq 5$ then we have

  * (a)

$E_{5}$ when $4<t<6$;

  * (b)

$E_{11}$ when $t\geq 6$;

Case 7. If $5<s<5,7$ then we have

  * (a)

$E_{8}$ when $5<t<6$;

  * (b)

$E_{9}$ when $t\geq 6$;

Case 8. If $s=5,7$ then we have

  * (a)

$E_{5}$ when $5,7\leq t<6$;

  * (b)

$E_{11}$ when $t\geq 6$;

Case 9. If $s>5,7$ then we have $E_{7}$ when $t>5,7$.

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