In this section, we establish a DT to a CNLSE. The CNLSE can be used to describe the evolution of localised waves in a two-mode non-linear fibre, two-component Bose–Einstein condensate, and other coupled non-linear systems [32], [33]. The CNLSE is considered as follows:

$$\{\begin{array}{c}\hfill i\frac{\partial {q}_{1}}{\partial t}+\frac{{\partial}^{2}{q}_{1}}{\partial {x}^{2}}+2\left({q}_{1}^{*}{q}_{1}+{q}_{2}^{*}{q}_{2}\right){q}_{1}=0,\hfill \\ \hfill i\frac{\partial {q}_{2}}{\partial t}+\frac{{\partial}^{2}{q}_{2}}{\partial {x}^{2}}+2\left({q}_{1}^{*}{q}_{1}+{q}_{2}^{*}{q}_{2}\right){q}_{2}=0,\hfill \end{array}$$(1)

which governs the simultaneous propagation of two orthogonal components of an electric field in optical fibre [34] and is also important in describing the effects of averaged random birefringence on an orthogonally polarised pulse in fibres [35].

Consider the isospectral problem of (1); the Lax pairs of the system is in the form of

$${\phi}_{x}=U\phi =\left(\begin{array}{ccc}\hfill -2i\lambda \hfill & \hfill {q}_{1}\hfill & \hfill {q}_{2}\hfill \\ \hfill -q_{1}{}^{*}\hfill & \hfill i\lambda \hfill & \hfill 0\hfill \\ \hfill -q_{2}{}^{*}\hfill & \hfill 0\hfill & \hfill i\lambda \hfill \end{array}\right)\phi ,$$(2)

$${\phi}_{t}=V\phi =\left(\begin{array}{ccc}\hfill -2i{\lambda}^{2}+\frac{i}{3}\left({q}_{1}^{*}{q}_{1}+{q}_{2}^{*}{q}_{2}\right)\hfill & \hfill {q}_{1}\lambda +\frac{i}{3}{q}_{1x}\hfill & \hfill {q}_{2}\lambda +\frac{i}{3}{q}_{2x}\hfill \\ \hfill -{q}_{1}^{*}\lambda +\frac{i}{3}{q}_{1x}^{*}\hfill & \hfill i{\lambda}^{2}-\frac{i}{3}{q}_{1}^{*}{q}_{1}\hfill & \hfill -\frac{i}{3}{q}_{1}^{*}{q}_{2}\hfill \\ \hfill -{q}_{2}^{*}\lambda +\frac{i}{3}{q}_{2x}^{*}\hfill & \hfill -\frac{i}{3}{q}_{1}{q}_{2}^{*}\hfill & \hfill i{\lambda}^{2}-\frac{i}{3}{q}_{2}^{*}{q}_{2}\hfill \end{array}\right)\phi .$$(3)

Here, *q*_{1}(*x*, *t*), *q*_{2}(*x*, *t*) are potentials, *λ* is a spectral parameter, and *φ*=(*φ*_{1}, *φ*_{2}, *φ*_{3})^{T} is a column vector solution of (2) and (3) associated with an eigenvalue *λ*.

The aim of this section is to construct DT for the CNLSEs (2) and (3), which is satisfied with the 3×3 matrix transformation on *φ*, $\tilde{U},$ and $\tilde{V}.$ Now, we recommend a gauge transformation of the CNLSEs (2) and (3):

$$\stackrel{~}{{\phi}_{n}}=T{\phi}_{n},T=\left(\begin{array}{ccc}\hfill {T}_{11}\hfill & \hfill {T}_{12}\hfill & \hfill {T}_{13}\hfill \\ \hfill {T}_{21}\hfill & \hfill {T}_{22}\hfill & \hfill {T}_{23}\hfill \\ \hfill {T}_{31}\hfill & \hfill {T}_{32}\hfill & \hfill {T}_{33}\hfill \end{array}\right),$$(4)

$${\phi}_{x}=\stackrel{~}{U}\phi ,\stackrel{~}{U}=\left({T}_{x}+TU\right){T}^{-1},$$(5)

$${\phi}_{t}=\stackrel{~}{V}\phi ,\stackrel{~}{V}=\left({T}_{t}+TV\right){T}^{-1}.$$(6)

If the fact that $\tilde{U},$ $\tilde{V}$ and *U*, *V* have the same types, the system (4) is called DT of CNLSE.

Let *ψ*=(*ψ*_{1}, *ψ*_{2}, *ψ*_{3})^{T}, *ϕ*=(*ϕ*_{1}, *ϕ*_{2}, *ϕ*_{3})^{T}, and *X*=(*X*_{1}, *X*_{2}, *X*_{3})^{T} are three basic solutions of the CNLSEs (2) and (3); thus, we give the following linear algebraic system:

$$\{\begin{array}{c}{\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}\left({A}_{11}^{\left(i\right)}+{A}_{12}^{\left(i\right)}{M}_{j}^{\left(1\right)}+{A}_{13}^{\left(i\right)}{M}_{j}^{\left(2\right)}\right){\lambda}_{j}^{i}=-{\lambda}_{j}^{N},\hfill \\ {\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}\left({A}_{21}^{\left(i\right)}+{A}_{22}^{\left(i\right)}{M}_{j}^{\left(1\right)}+{A}_{23}^{\left(i\right)}{M}_{j}^{\left(2\right)}\right){\lambda}_{j}^{i}=-{M}_{j}^{\left(1\right)}{\lambda}_{j}^{N},\hfill \\ {\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}\left({A}_{31}^{\left(i\right)}+{A}_{32}^{\left(i\right)}{M}_{j}^{\left(1\right)}+{A}_{33}^{\left(i\right)}{M}_{j}^{\left(2\right)}\right){\lambda}_{j}^{i}=-{M}_{j}^{\left(2\right)}{\lambda}_{j}^{N},\hfill \end{array}$$(7)

with

$$\begin{array}{cc}& {M}_{j}^{\left(1\right)}=\frac{{\psi}_{2}+{\nu}_{j}^{\left(1\right)}{\varphi}_{2}+{\nu}_{j}^{\left(2\right)}{X}_{2}}{{\psi}_{1}+{\nu}_{j}^{\left(1\right)}{\varphi}_{1}+{\nu}_{j}^{\left(2\right)}{X}_{1}},\hfill \\ & {M}_{j}^{\left(2\right)}=\frac{{\psi}_{3}+{\nu}_{j}^{\left(1\right)}{\varphi}_{3}+{\nu}_{j}^{\left(2\right)}{X}_{3}}{{\psi}_{1}+{\nu}_{j}^{\left(1\right)}{\varphi}_{1}+{\nu}_{j}^{\left(2\right)}{X}_{1}},0\le j\le 3N,\hfill \end{array}$$(8)

where *λ*_{j} and ${\nu}_{j}^{\mathrm{(}k\mathrm{)}}\mathrm{(}i\ne k,\text{\hspace{0.17em}}{\lambda}_{i}\ne {\lambda}_{j},\text{\hspace{0.17em}}{\nu}_{i}^{\mathrm{(}k\mathrm{)}}\ne {\nu}_{j}^{\mathrm{(}k\mathrm{)}},\text{\hspace{0.17em}}k\ne \mathrm{1,2}\mathrm{)}$ should choose appropriate parameters, so that the determinants of coefficients for (7) are non-zero.

Defining a 3×3 matrix *T*, and the *T* is of the following form:

$$\{\begin{array}{c}\hfill {T}_{11}={\lambda}^{N}+{\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{11}^{\left(i\right)}{\lambda}^{i},{T}_{12}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{12}^{\left(i\right)}{\lambda}^{i},{T}_{13}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{13}^{\left(i\right)}{\lambda}^{i},\hfill \\ \hfill {T}_{21}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{21}^{\left(i\right)}{\lambda}^{i},{T}_{22}={\lambda}^{N}+{\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{22}^{\left(i\right)}{\lambda}^{i},{T}_{23}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{23}^{\left(i\right)}{\lambda}^{i},\hfill \\ \hfill {T}_{31}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{31}^{\left(i\right)}{\lambda}^{i},{T}_{32}={\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{32}^{\left(i\right)}{\lambda}^{i},{T}_{33}={\lambda}^{N}+{\sum}_{i=\mathrm{\hspace{0.33em}0}}^{N-\mathrm{\hspace{0.33em}1}}{A}_{33}^{\left(i\right)}{\lambda}^{i},\hfill \end{array}$$(9)

where *N* is a natural number and ${A}_{mn}^{i}\mathrm{(}m\mathrm{,}\text{\hspace{0.17em}}n=\mathrm{1,}\text{\hspace{0.17em}}\mathrm{2,}\text{\hspace{0.17em}}3.m\ge 0\mathrm{)}\mathrm{}$ are the functions of *n* and *t*. Through calculations, we find

$$\mathrm{det}T={\prod}_{j=\mathrm{\hspace{0.33em}1}}^{3N}\left(\lambda -{\lambda}_{j}\right)$$(10)

which proves that *λ*_{j} (*j*=1≤*j*≤3*N*) are 3*N* roots of det*T*. Based on these conditions, we will prove that $\tilde{U}$ and $\tilde{V}$ have the same forms with *U* and *V*, respectively.

**Proposition 1.** The matrix $\tilde{U}$ defined by (5) has the same type as *U*; that is

$$\tilde{U}=\mathrm{(}\begin{array}{lll}-2i\lambda \hfill & \tilde{{q}_{1}}\hfill & \tilde{{q}_{2}}\hfill \\ -\tilde{{q}_{1}{}^{\ast}}\hfill & i\lambda \hfill & 0\hfill \\ -\tilde{{q}_{1}{}^{\ast}}\hfill & 0\hfill & i\lambda \hfill \end{array}\mathrm{)}\mathrm{,}$$(11)

in which the transformation formulae between old and new potentials are shown as

$$\{\begin{array}{c}\hfill \stackrel{~}{{q}_{1}}={q}_{1}+3i{A}_{12},\hfill \\ \hfill \stackrel{~}{{q}_{2}}={q}_{2}+3i{A}_{13},\hfill \\ \hfill \stackrel{~}{q_{1}{}^{*}}=q_{1}{}^{*}+3i{A}_{21},\hfill \\ \hfill \stackrel{~}{q_{2}{}^{*}}=q_{2}{}^{*}+3i{A}_{31}.\hfill \end{array}$$(12)

The transformation (33) is used to get a DT of the spectral problem (5).

**Proof.** By assuming ${T}^{-1}=\frac{{T}^{\ast}}{\text{det}T}$ and

$$\left({T}_{x}+TU\right){T}^{*}=\left(\begin{array}{ccc}\hfill {B}_{11}\left(\lambda \right)\hfill & \hfill {B}_{12}\left(\lambda \right)\hfill & \hfill {B}_{13}\left(\lambda \right)\hfill \\ \hfill {B}_{21}\left(\lambda \right)\hfill & \hfill {B}_{22}\left(\lambda \right)\hfill & \hfill {B}_{23}\left(\lambda \right)\hfill \\ \hfill {B}_{31}\left(\lambda \right)\hfill & \hfill {B}_{32}\left(\lambda \right)\hfill & \hfill {B}_{33}\left(\lambda \right)\hfill \end{array}\right).$$(13)

It is easy to verify that *B*_{sl} (1≤*s*, l≤3) are 3*N*-order or 3*N*+1-order polynomials in *λ*. Hence, by (2), (7), and (8), we see that

$$\{\begin{array}{c}\hfill {M}_{jx}^{\left(1\right)}=-{q}_{1}^{*}+3i{\lambda}_{j}{M}_{j}^{\left(1\right)}-{q}_{1}{M}_{j}^{\left(1\right)}{M}_{j}^{\left(1\right)}-{q}_{2}{M}_{j}^{\left(1\right)}{M}_{j}^{\left(2\right)},\hfill \\ \hfill {M}_{jx}^{\left(2\right)}=-{q}_{2}^{*}+3i{\lambda}_{j}{M}_{j}^{\left(2\right)}-{q}_{2}{M}_{j}^{\left(2\right)}{M}_{j}^{\left(2\right)}-{q}_{1}{M}_{j}^{\left(1\right)}{M}_{j}^{\left(2\right)}.\hfill \end{array}$$(14)

Through an accurate calculation, *λ*_{j} (1≤*j*≤3) is the root of *B*_{sl} (1≤*s*, l≤3). Equation (13) has the following structure:

$$\mathrm{(}{T}_{x}+TU\mathrm{)}{T}^{\ast}=\mathrm{(}\text{det}T\mathrm{)}C\mathrm{(}\lambda \mathrm{)},$$(15)

where

$$C\left(\lambda \right)=\left(\begin{array}{ccc}\hfill {C}_{11}^{\left(1\right)}\left(\lambda \right)+{C}_{11}^{\left(0\right)}\hfill & \hfill {C}_{12}^{\left(0\right)}\hfill & \hfill {C}_{13}^{\left(0\right)}\hfill \\ \hfill {C}_{21}^{\left(0\right)}\hfill & \hfill {C}_{22}^{\left(1\right)}\left(\lambda \right)+{C}_{22}^{\left(0\right)}\hfill & \hfill {C}_{23}^{\left(0\right)}\hfill \\ \hfill {C}_{31}^{\left(0\right)}\hfill & \hfill {C}_{32}^{\left(0\right)}\hfill & \hfill {C}_{33}^{\left(1\right)}\left(\lambda \right)+{C}_{33}^{\left(0\right)}\hfill \end{array}\right),$$(16)

and ${C}_{mn}^{\mathrm{(}k\mathrm{)}}$ (*m*, *n*=1, 2, *k*=0, 1) satisfy the functions without *λ*. So, the following (15) is obtained:

$$\left({T}_{x}+TU\right)=C\left(\lambda \right)T.$$(17)

Through comparing the coefficients of *λ* in (17), we see

$$\{\begin{array}{c}\hfill {C}_{11}^{\left(1\right)}=-2i\lambda ,{C}_{11}^{\left(0\right)}=0,{C}_{12}^{\left(0\right)}={q}_{1}+3i{A}_{12}=\stackrel{~}{{q}_{1}},{C}_{13}^{\left(0\right)}={q}_{2}+3i{A}_{13}=\stackrel{~}{{q}_{2}},\hfill \\ \hfill {C}_{21}^{\left(0\right)}=-q_{1}{}^{*}-3i{A}_{21}=-\stackrel{~}{q_{1}{}^{*}},{C}_{22}^{\left(1\right)}=i,{C}_{22}^{\left(0\right)}=0,{C}_{23}^{\left(0\right)}=0,\hfill \\ \hfill {C}_{31}^{\left(0\right)}=-q_{2}{}^{*}-3i{A}_{31}=-\stackrel{~}{q_{2}{}^{*}},{C}_{32}^{\left(0\right)}=0,{C}_{33}^{\left(1\right)}=i,{C}_{33}^{\left(0\right)}=0.\hfill \end{array}$$(18)

In the following section, we assume that the new matrix $\tilde{U}$ has the same type with *U*, which means that they have the same structures only ${\text{q}}_{1},{\text{q}}_{2},{\text{q}}_{1}^{*},{\text{q}}_{2}^{*},$ of *U* transformed into ${\tilde{q}}_{1}\mathrm{,}{\tilde{q}}_{2}\mathrm{,}{\widehat{q}}_{1}^{\ast}\mathrm{,}\text{\hspace{0.17em}}{\tilde{q}}_{2}^{\ast}$ of $\tilde{U}.$ After careful calculation, we compare the ranks of *λ*^{N} and get the objective equations as follows:

$$\{\begin{array}{l}\tilde{{q}_{1}}={q}_{1}+3i{A}_{12}\mathrm{,}\hfill \\ \tilde{{q}_{2}}={q}_{2}+3i{A}_{13}\mathrm{,}\hfill \\ \tilde{{q}_{1}{}^{\ast}}\mathrm{=}{q}_{1}{}^{\ast}+3i{A}_{21}\mathrm{,}\hfill \\ \tilde{{q}_{2}{}^{\ast}}={q}_{2}{}^{\ast}+3i{A}_{31}\mathrm{.}\hfill \end{array}$$(19)

From (11) and (33), we know that $\tilde{U}=C\mathrm{(}\lambda \mathrm{}\mathrm{)}\mathrm{.}$ The proof is completed.

**Proposition 2.** Under the transformation (19), the matrix defined $\tilde{V}$ by (6) has the same form as *V*, that is

$$\stackrel{~}{V}=\left(\begin{array}{ccc}\hfill -2i{\lambda}^{2}+\frac{i}{3}(\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{1}}+\stackrel{~}{{q}_{2}^{*}}\stackrel{~}{{q}_{2})}\hfill & \hfill \stackrel{~}{{q}_{1}}\lambda +\frac{i}{3}\stackrel{~}{{q}_{1x}}\hfill & \hfill \stackrel{~}{{q}_{2}}\lambda +\frac{i}{3}\stackrel{~}{{q}_{2x}}\hfill \\ \hfill -\stackrel{~}{{q}_{1}^{*}}\lambda +\frac{i}{3}\stackrel{~}{{q}_{1x}^{*}}\hfill & \hfill i{\lambda}^{2}-\frac{i}{3}\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{1}}\hfill & \hfill -\frac{i}{3}\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{2}}\hfill \\ \hfill -\stackrel{~}{{q}_{2}^{*}}\lambda +\frac{i}{3}\stackrel{~}{{q}_{2x}^{*}}\hfill & \hfill -\frac{i}{3}\stackrel{~}{{q}_{1}}\stackrel{~}{{q}_{2}^{*}}\hfill & \hfill i{\lambda}^{2}-\frac{i}{3}\stackrel{~}{{q}_{2}^{*}}\stackrel{~}{{q}_{2}}\hfill \\ \hfill \hfill \end{array}\right).$$(20)

**Proof.** We assume the new matrix $\tilde{V}$ also have the same form with *V*. If we obtain the similar relations between ${q}_{1}\mathrm{,}\text{\hspace{0.17em}}{q}_{2}\mathrm{,}\text{\hspace{0.17em}}{q}_{1}^{\ast}\mathrm{,}\text{\hspace{0.17em}}{q}_{2}^{\ast}$ of *V* and ${\tilde{q}}_{1}\mathrm{,}\text{\hspace{0.17em}}{\tilde{q}}_{2}\mathrm{,}\text{\hspace{0.17em}}{\widehat{q}}_{1}^{\ast}\mathrm{,}\text{\hspace{0.17em}}{\tilde{q}}_{2}^{\ast}$ of $\tilde{V}$ like (33), we can prove that the gauge transformation under *T* turns the Lax pairs *U*, *V* into new Lax pairs $\tilde{U}\mathrm{,}\text{\hspace{0.17em}}\tilde{V}$ with the same types.

By assuming ${T}^{-1}=\frac{{T}^{\ast}}{\text{det}T}$ and

$$\left({T}_{t}+TV\right){T}^{*}=\left(\begin{array}{ccc}\hfill {E}_{11}\left(\lambda \right)\hfill & \hfill {E}_{12}\left(\lambda \right)\hfill & \hfill {E}_{13}\left(\lambda \right)\hfill \\ \hfill {E}_{21}\left(\lambda \right)\hfill & \hfill {E}_{22}\left(\lambda \right)\hfill & \hfill {E}_{23}\left(\lambda \right)\hfill \\ \hfill {E}_{31}\left(\lambda \right)\hfill & \hfill {E}_{32}\left(\lambda \right)\hfill & \hfill {E}_{33}\left(\lambda \right)\hfill \end{array}\right).$$(21)

It is easy to verify that *E*_{sl} (1≤*s*, *l*≤3) are 3*N*+1-order or 3*N*+2-order polynomials in *λ*. Based on (3), (7), and (8), we see that

$$\{\begin{array}{c}{M}_{jt}^{\left(1\right)}=-{q}_{1}^{*}i\lambda +\frac{1}{3}{q}_{1x}^{*}\left(3{\lambda}^{2}-\frac{1}{3}\left({q}_{1}^{2}+{\sum}_{i=\mathrm{\hspace{0.33em}1}}^{2}{q}_{i}^{2}\right)\right){M}_{j}^{\left(1\right)}\hfill \\ -\left({q}_{1}i\lambda -\frac{1}{3}{q}_{1x}\right){M}_{j}^{\left(1\right)}{M}_{j}^{\left(1\right)}+\frac{1}{3}{q}_{1}^{*}{q}_{2}{M}_{j}^{\left(2\right)}-\left({q}_{2}i\lambda -\frac{1}{3}{q}_{2x}\right){M}_{j}^{\left(1\right)}{M}_{j}^{\left(2\right)},\hfill \\ {M}_{jt}^{\left(2\right)}=-{q}_{2}^{*}i\lambda +\frac{1}{3}{q}_{2x}^{*}\left(3{\lambda}^{2}-\frac{1}{3}\left({q}_{2}^{2}+{\sum}_{i=\mathrm{\hspace{0.33em}1}}^{2}{q}_{i}^{2}\right)\right){M}_{j}^{\left(2\right)}\hfill \\ -\left({q}_{2}i\lambda -\frac{1}{3}{q}_{2x}\right){M}_{j}^{\left(2\right)}{M}_{j}^{\left(2\right)}+\frac{1}{3}{q}_{2}^{*}{q}_{2}{M}_{j}^{\left(1\right)}-\left({q}_{1}i\lambda -\frac{1}{3}{q}_{1x}\right){M}_{j}^{\left(1\right)}{M}_{j}^{\left(2\right)}.\hfill \end{array}$$(22)

Through an accurate calculation, *λ*_{j} (*j*=1≤*j*≤3) is the root of *E*_{sl} (*s*, *l*=1≤*j*≤3). Thus, (21) has the following structure:

$$\left({T}_{t}+TV\right){T}^{*}=\left(\mathrm{det}T\right)F\left(\lambda \right),$$(23)

where

$$F\left(\lambda \right)=\left(\begin{array}{ccc}\hfill {F}_{11}^{\left(2\right)}{\lambda}^{2}+{F}_{11}^{\left(1\right)}\lambda +{F}_{11}^{\left(0\right)}\hfill & \hfill {F}_{12}^{\left(1\right)}\lambda +{F}_{12}^{\left(0\right)}\hfill & \hfill {F}_{13}^{\left(1\right)}\lambda +{F}_{13}^{\left(0\right)}\hfill \\ \hfill {F}_{21}^{\left(1\right)}\lambda +{F}_{21}^{\left(0\right)}\hfill & \hfill {F}_{22}^{\left(2\right)}{\lambda}^{2}+{F}_{22}^{\left(1\right)}\lambda +{F}_{22}^{\left(0\right)}\hfill & \hfill {F}_{23}^{\left(0\right)}\hfill \\ \hfill {F}_{31}^{\left(1\right)}\lambda +{F}_{31}^{\left(0\right)}\hfill & \hfill {F}_{32}^{\left(0\right)}\hfill & \hfill {F}_{33}^{\left(2\right)}{\lambda}^{2}+{F}_{33}^{\left(1\right)}\lambda +{F}_{33}^{\left(0\right)}\hfill \end{array}\right),$$(24)

and ${F}_{mn}^{\mathrm{(}k\mathrm{)}}$ (*m*, *n*=1, 2, *k*=0, 1) satisfy the functions without *λ*. So, the following (23) is obtained:

$$\left({T}_{t}+TV\right)=F\left(\lambda \right)T.$$(25)

Through comparing the coefficients of *λ* in (25), we get the objective equations as follows:

$$\{\begin{array}{cc}{F}_{11}^{\left(2\right)}\hfill & =-2i,{F}_{11}^{\left(1\right)}=0,\hfill \\ {F}_{11}^{\left(0\right)}\hfill & =\frac{i}{3}\left({q}_{1}^{*}{q}_{1}+{q}_{2}^{*}{q}_{2}\right)-{A}_{12}{q}_{1}^{*}-{A}_{13}{q}_{2}^{*}-{A}_{21}\stackrel{~}{{q}_{1}}-{A}_{31}\stackrel{~}{{q}_{2}}\hfill \\ & =\frac{i}{3}(\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{1}}+\stackrel{~}{{q}_{2}^{*}}\stackrel{~}{{q}_{2})},\hfill \\ {F}_{12}^{\left(1\right)}\hfill & ={q}_{1}+3i{A}_{12}=\stackrel{~}{{q}_{1}},{F}_{12}^{\left(0\right)}=\frac{i}{3}{q}_{1x}+{A}_{11}{q}_{1}-{A}_{22}\stackrel{~}{{q}_{1}}-{A}_{32}\stackrel{~}{{q}_{2}}\hfill \\ & =\frac{i}{3}\stackrel{~}{{q}_{1x}},\hfill \\ {F}_{13}^{\left(1\right)}\hfill & ={q}_{2}+3i{A}_{13}=\stackrel{~}{{q}_{2}},{F}_{13}^{\left(0\right)}=\frac{i}{3}{q}_{2x}+{A}_{11}{q}_{2}-{A}_{23}\stackrel{~}{{q}_{1}}-{A}_{33}\stackrel{~}{{q}_{2}}=\frac{i}{3}\stackrel{~}{{q}_{2x}},\hfill \\ {F}_{21}^{\left(1\right)}\hfill & =-q_{1}{}^{*}-3i{A}_{21}=-\stackrel{~}{{q}_{1}^{*}},{F}_{21}^{\left(0\right)}=\frac{i}{3}{q}_{1x}^{*}-{A}_{22}{q}_{1}^{*}-{A}_{23}{q}_{2}^{*}+{A}_{11}\stackrel{~}{{q}_{1}^{*}}\hfill \\ & =\frac{i}{3}\stackrel{~}{{q}_{1x}^{*}},\hfill \\ {F}_{22}^{\left(2\right)}\hfill & =i,{F}_{22}^{\left(1\right)}=0,{F}_{22}^{\left(0\right)}={A}_{21}{q}_{1}-\frac{i}{3}{q}_{1}^{*}{q}_{1}+{A}_{12}\stackrel{~}{{q}_{1}^{*}}=-\frac{i}{3}\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{1}},\hfill \\ {F}_{23}^{\left(0\right)}\hfill & ={A}_{21}{q}_{2}-\frac{i}{3}{q}_{1}^{*}{q}_{2}+{A}_{13}\stackrel{~}{{q}_{1}^{*}}=-\frac{i}{3}\stackrel{~}{{q}_{1}^{*}}\stackrel{~}{{q}_{2}},\hfill \\ {F}_{31}^{\left(1\right)}\hfill & =-q_{2}{}^{*}-3i{A}_{13}=-\stackrel{~}{q_{2}{}^{*}},{F}_{31}^{\left(0\right)}=-{A}_{32}{q}_{1}^{*}+\frac{i}{3}{q}_{2x}^{*}-{A}_{33}{q}_{2}^{*}+{A}_{11}\stackrel{~}{{q}_{2}^{*}}\hfill \\ & =\frac{i}{3}\stackrel{~}{{q}_{2x}^{*}},\hfill \\ {F}_{32}^{\left(0\right)}\hfill & ={A}_{31}{q}_{1}-\frac{i}{3}{q}_{2}^{*}{q}_{1}+{A}_{12}\stackrel{~}{{q}_{2}^{*}}=-\frac{i}{3}\stackrel{~}{{q}_{2}^{*}}\stackrel{~}{{q}_{1}},\hfill \\ {F}_{33}^{\left(2\right)}\hfill & =i,{F}_{33}^{\left(1\right)}=0,{F}_{33}^{\left(0\right)}={A}_{31}{q}_{2}-\frac{i}{3}{q}_{2}^{*}{q}_{2}+{A}_{13}\stackrel{~}{{q}_{2}^{*}}=-\frac{i}{3}\stackrel{~}{{q}_{2}^{*}}\stackrel{~}{{q}_{2}}.\hfill \end{array}$$(26)

In the following section, we assume the new matrix $\tilde{V}$ has the same type with *V*, which means that they have the same structures only ${q}_{1}\mathrm{,}\text{\hspace{0.17em}}{q}_{2}\mathrm{,}\text{\hspace{0.17em}}{q}_{1}^{\ast}\mathrm{,}\text{\hspace{0.17em}}{q}_{2}^{\ast}$ of *V* transformed into ${\tilde{q}}_{1}\mathrm{,}\text{\hspace{0.17em}}{\tilde{q}}_{2}\mathrm{,}\text{\hspace{0.17em}}{\widehat{q}}_{1}^{\ast}\mathrm{,}\text{\hspace{0.17em}}{\tilde{q}}_{2}^{\ast}$ of $\tilde{V}.$ From (33) and (20), we know that $\tilde{V}=F\mathrm{(}\lambda \mathrm{}\mathrm{)}\mathrm{.}$ The proof is completed.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.