text:
\begin{document}

\title{Entwining Yang-Baxter maps and integrable lattices}

\author{Theodoros E. Kouloukas and Vassilios G. Papageorgiou}

\maketitle
\begin{center}
\emph{Department of Mathematics, University of Patras \\ GR-265 00 Patras, \
Greece}
\end{center}

\bigskip

\begin{abstract}
Yang--Baxter (YB) map systems (or set-theoretic analoga of entwining YB structures) are presented.
They admit zero curvature representations with spectral parameter
depended Lax triples $L_1, \ L_2, \ L_3$ derived from symplectic leaves of $2 \times 2$ binomial matrices
equipped with the Sklyanin bracket. A unique factorization condition of the Lax triple
implies a 3-dimensional compatibility property of these maps. In case  $L_1=L_2=L_3$ this property yields the set-theoretic
quantum Yang-Baxter equation, i.e. the YB map property. By considering periodic `staircase' initial value problems on quadrilateral
lattices, these maps give rise to multidimensional integrable mappings which preserve the spectrum of the corresponding
monodromy matrix.
\end{abstract}

\section{Introduction}
The connection between set theoretical solutions of the quantum Yang-Baxter equations
and integrable mappings has been recently investigated by many authors.
First Veselov in \cite{ves2,ves3} proved that for such
a solution, admitting a Lax matrix, there is a family of
commuting transfer maps which preserve the spectrum of the
corresponding monodromy matrix. He also proposed the shorter term `Yang Baxter map'
for a set-theoretic solution of the quantum Yang-Baxter equation. In the present article we
present system of maps that admit a Lax triple of matrices
$L_1,\ L_2$, $L_3$. They satisfy 3-dimensional compatibility conditions that constitute
set-theoretic analoga of entwining quantum Yang-Baxter equations. The latter term was
introduced in \cite{Nich} for systems of quantum YB equations. Such entwining structures
for the quantum Yang-Baxter equation appeared already in \cite{Frei-Mail, cnp, vlad} and their study continued in
\cite{hlav, hlav2}.

The basic definitions and notations about Lax pairs and integrable mappings are introduced in section
\ref{secLax}.The transfer map and the monodromy matrix are
defined for a periodic `staircase' initial value problem on a lattice for each pair $L_i, ~ L_j$ and
integrals of motion are obtained from the spectrum of the monodromy matrix, which are in involution with respect to the Sklyanin bracket.
Section \ref{sec3-d} deals with the the 3-d compatibility condition (entwining Yang-Baxter equation)
of maps that admit Lax pairs giving rise to Lax triples.
A construction of Lax triples and their corresponding set theoretical solutions of the entwining YB equation is presented in section \ref{secbin}, while in the next section the whole theory is applied to a concrete example.
We end  in section \ref{persp} by giving some conclusions and perspectives for future work on this subject.

\section{Lax pairs and integrable mappings} \label{secLax}
Let $S$ be the map
\begin{equation} \label{mapphi}
S:((x,\alpha),(y,\beta))\mapsto((u,\alpha),(v,\beta))=
(u(x,\alpha, y,\beta),v(x,\alpha, y,\beta))
\end{equation}
where $x, \ y $ belong to a set $\mathcal{X}$, from our consideration the set $\mathcal{X}$ has the structure
of an algebraic variety,  and the parameters $\alpha, \beta \in
\mathbb{C}^m$. We usually keep the parameters separately and denote
$S((x,\alpha),(y,\beta))$ by $S_{\alpha,\beta}(x,y)$.
We can represent $S_{\alpha,\beta}(x,y)$ as a map assigned to the
edges of an elementary quadrilateral as in Fig.\ref{fig:map}.
\\
\begin{figure}[h]
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\move (7 1) \linewd 0.05 \arrowheadtype t:F \avec(1 7) \lpatt( )
\move (0 0) \linewd 0.1 \lvec(8 0)  \lvec (8 8)  \lvec (0 8)  \lvec (0 0) \lpatt( )

\htext (2.8 -1.9){$(x;\alpha)$} \htext (8.71 3.5){$(y;\beta)$}
\htext (2.8 8.5){$(u;\alpha)$} \htext (-3.7 3.5){$(v;\beta)$}
\htext (4.3 3.8){$S_{\alpha,\beta}$}
}
\caption{A map assigned to the edges of a quadrilateral}
\label{fig:map}
\end{figure}
\begin{definition}
The ordered pair of square matrices $(L(x,\alpha,\zeta) ,  M(x, \alpha, \zeta))$ that depends on a point $x \in \mathcal{X}$, on a parameter
$\alpha \in \mathbb{C}^m$ and on a spectral parameter $\zeta \in \mathbb{C}$ is called a {\em Lax pair}
for the parametric map $S_{\alpha,\beta}$, if
\begin{equation} \label{laxmat}
L(u,\alpha,\zeta)M(v,\beta,\zeta)=M(y,\beta,\zeta)L(x,\alpha,\zeta),
\end{equation}
for any $\zeta\in \mathbb{C}$. Furthermore if equation
is equivalent to $(u,\ v)=S_{\alpha,\beta}(x,y)$ then
we will call $(L(x,\alpha,\zeta) ,  M(x, \alpha, \zeta))$ a {\em strong Lax pair}.
\end{definition}
We usually omit the spectral parameter $\zeta$ and denote the
matrices $L(x,\alpha,\zeta)$ and $M(x, \alpha, \zeta)$ by
$L(x;\alpha)$ and $M(x;\alpha)$ respectively.
\begin{unnumrem}
It is instructive to think of a Lax pair as functions $L,M:\mathcal{X} \times \mathbb{C}^m
\rightarrow Mat({k \times k})$.
In this consideration if $L=M$ the definition coincides with the definition of the Lax matrix
which is given in \cite{ves4}. The aim of this work is to study the case where $L \neq M$.
\end{unnumrem}

Let $S_{\alpha,\beta}$ be a map that admits the Lax pair
$L(x;\alpha) , \ M(x; \alpha)$. Next we consider the standard
periodic `staircase' initial value problem, as in \cite{pnc}, for
integrable lattices difference equations.  Initial values
$x_1,...x_n$ and $y_1,...,y_n$ are assigned to the edges of a
`staircase' on a quadrilateral lattice as in Fig. \ref{fig:mapping},
with periodic boundary conditions $x_{n+1}=x_1, \ y_{n+1}=y_1$. The
edge with the $x_i$ value carries the parameter $\alpha_i$, while
the one with the $y_i$ value the parameter $\beta_i$, for
$i=1,...,n$. Now, having in mind the representation of
$S_{\alpha_i,\beta_i}$ to the edges of a quadrilateral as in Fig.
\ref{fig:map}, we can compute the values of the next level of the
lattice according to Fig. \ref{fig:mapping}. By $(x_i',y_i')$ we
denote the values $S_{\alpha_i,\beta_i}(x_i,y_i)$ for $i=1,...,n$,
while $k$-primed variables $x_i^{(k)}$ and $y_i^{(k)}$ denote the
corresponding values:
$$(x_i^{(k)},y_{j}^{(k)})=S_{\alpha_i,\beta_{j}}(x_i^{(k-1)},y_{j}^{(k-1)})$$
with $j \equiv i+k-1 (mod n)$.

\bigskip

\begin{figure}[h]
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\htext (0.15 0.55){$y_1'$} \htext (1.5 0.55){$x_1'$}\htext (2.2
0.55){$y_2'$} \htext (3.5 0.55){$x_2'$} \htext (4.2 0.55){$y_3'$}
\htext (5.5 0.55){$x_3'$} \htext (7.6 0.55){$y_n'$} \htext (9
0.55){$x_n'$}\htext (9.65 0.55){$y_1'$} \htext (11 0.55){$x_1'$}

\htext (1.1 1.54){$y_2''$}\htext (2.5 1.55){$x_1''$} \htext (3.2
1.55){$y_3''$} \htext (4.5 1.6){$x_2''$}  \htext (8.5
1.55){$y_1''$}\htext (10.1 1.55){$x_n''$}

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 \move(0.935 -0.57) \avec(0.935 0.57)
\htext (1,02 -0.13) {\small{$S_{\alpha_1,\beta_1} $}} \move(2.935 -0.57) \avec(2.935 0.57) \htext (3.02
-0.13) {\small{$S_{\alpha_2,\beta_2}$}} \move(2 0.57) \avec(2 1.57)

}
 \caption{n-period mapping} \label{fig:mapping}
\end{figure}

\bigskip

For any $n$-periodic `staircase' initial value problem we define the
{\em `transfer' map}:
\begin{equation} \label{transfer}
T_n:(x_1,...,x_n,y_1,...,y_n)\mapsto (x'_1,...,x'_{n},y'_{2},...,y'_n,y'_1)
\end{equation}
and the {\em k-`transfer' map}:
\begin{equation} \label{ktransfer}
T^{k}_n:(x_1,...,x_n,y_1,...,y_n)\mapsto (x^{(k)}_1,...,x^{(k)}_{n},
\underbrace{y^{(k)}_{d+1},...,y^{(k)}_n}_{n-d},\underbrace{y^{(k)}_1,...,y^{(k)}_d}_{d}),
\end{equation}
with $d \equiv k (modn)$ and $T^{1}_n=T_n$.

We observe that
$$T_n^n(x_1,...,x_n,y_1,...,y_n)=(x_1^{(n)},...,x_n^{(n)},y_1^{(n)},...,y_n^{(n)}).$$
We also define the {\em monodromy matrix}
$M_n(x_1,...,x_n,y_1,...,y_n)= \overset{\curvearrowleft
}{\prod\limits_{i=1}^{n}}M(y_{i};\beta_i)L(x_{i};\alpha_i ),$ where
${\curvearrowleft }$ indicates that the elements
$M(y_{i};\beta)L(x_{i};\alpha )$ in the product are arranged from
right to left.

So, for example, for the 1-periodic initial value problem on the
lattice the transfer map will be
$T_1(x,y)=(x',y')=S_{\alpha,\beta}(x,y),$ with corresponding
monodromy matrix $M_1(x,y)=M(y;\beta)L(x;\alpha )$, while for the
2-periodic case, $$T_2(x_1,x_2,y_1,y_2)=(x'_1,x'_2,y'_2,y'_1),$$ where
$x'_i$ and $y'_i$ are given by:
$(x'_i,y'_i)=S_{\alpha_i,\beta_i}(x_i,y_i)$, for $i=1,2$. The
monodromy matrix in this case is
$M_2(x_1,x_2,y_1,y_2)=M(y_2;\beta_2)L(x_2;\alpha_2
)M(y_1;\beta_1)L(x_1;\alpha_1 )$.

The equivalent proposition to the one in \cite{ves2} for the
$n$-periodic `staircase' initial value problem holds.
\begin{proposition} \label{transfer}
The transfer map preserves the spectrum of the monodromy matrix.
\end{proposition}

\begin{proof} \label{integ}
The definition of the transfer map and the monodromy matrix implies:
$$M_n(T_n(x_1,...,x_n,y_1,...,y_n))M(y'_1;\beta_1)=M(y'_1;\beta_1)
\overset{\curvearrowleft
}{\prod\limits_{i=1}^{n}}L(x'_{i};\alpha_i)M(y'_{i};\beta_i ).$$
Since $L(x;\alpha)$, $M(x;\alpha)$ is a Lax pair for the map and
$(x_i',y_i')=S_{\alpha_i,\beta_i}(x_i,y_i)$ we have that
$L(x'_{i};\alpha_i)M(y'_{i};\beta_i
)=M(y_{i};\beta_i)L(x_{i};\alpha_i )$, so
\begin{eqnarray*}
M_n(T_n(x_1,...,x_n,y_1,...,y_n))M(y'_1;\beta_1)&=& M(y'_1;\beta_1)
\overset{\curvearrowleft
}{\prod\limits_{i=1}^{n}}M(y_{i};\beta_i)L(x_{i};\alpha_i ) \\ &=&
M(y'_1;\beta_1)M_n(x_1,...,x_n,y_1,...,y_n),
\end{eqnarray*}
or $
M_n(T_n(x_1,...,x_n,y_1,...,y_n))=M(y'_1;\beta_1)M_n(x_1,...,x_n,y_1,...,y_n)M^{-1}(y'_1;\beta_1)$.
\end{proof}
Similarly proposition \ref{transfer} holds also for any $k$-transfer map. So
the spectrum of the monodromy matrix gives integrals of the transfer map. If we derive
$N$ functionally independent integrals,
the transfer map is integrable provided that
a symplectic
structure $\omega$ exists in a 2N-dimensional phase space such that:
i) the transfer map is symplectic with respect to $\omega$ and ii) the integrals are
in involution. In the case of polynomial Lax matrices that we are dealing bellow,
the Sklyanin bracket (\ref{sklyanin})
give rise to a symplectic structure after reduction to the symplectic leaves \cite{skly3}.

\section{Entwining Yang--Baxter equation and Lax triples} \label{sec3-d}

Now we consider three parametric maps of the form (\ref{mapphi}),
$S_{\alpha,\beta}, \ R_{\alpha,\beta}, \ T_{\alpha,\beta}:
\mathcal{X} \times \mathcal{X} \rightarrow \mathcal{X} \times \mathcal{X}$.
In correspondence with \cite{Nich} we call the equation
\begin{equation} \label{eYBeq}
T_{\beta,\gamma}^{23}\circ R_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12}\circ R_{\alpha,\gamma}^{13}\circ T_{\beta,\gamma}^{23},
\end{equation}
\textit{the entwining quantum Yang-Baxter equation} or just the entwining YB equation.
Here by $S^{ij}$ (respectively $T^{ij}$ and $R^{ij}$)
for $i,j=1,2,3$, we denote the map that acts as $S$ (resp. $T$ and $R$)
on the $i$ and
$j$ factor of $\mathcal{X} \times \mathcal{X} \times \mathcal{X}$
and identically on the others.

A \textit{Lax triple} (resp. \textit{strong Lax triple})
of three maps $(S_{\alpha,\beta}, R_{\alpha,\beta},  T_{\alpha,\beta})$
is a triple of matrices $$(L_1(x;\alpha),  L_2(x;\alpha), L_3(x;\alpha)),$$
such that the pairs $(L_1(x;\alpha),  L_2(x;\alpha))$, $(L_1(x;\alpha),  L_3(x;\alpha))$,
$(L_2(x;\alpha),  L_3(x;\alpha))$ are Lax pairs (resp. strong Lax pairs) of the maps
$S_{\alpha,\beta}, \ R_{\alpha,\beta}, \ T_{\alpha,\beta}$ respectively.

\begin{proposition} \label{3product}
Let
$(S_{\alpha,\beta},  R_{\alpha,\beta},  T_{\alpha,\beta})$ be
three maps on $\mathcal{X} \times \mathcal{X}$ that admit the Lax triple
$(L_1(x;\alpha),  L_2(x;\alpha), L_3(x;\alpha)).$
If the equation
\begin{equation}
L_1( \hat{x}; \alpha )L_2( \hat{y}; \beta )L_3(\hat{z}; \gamma )=
L_1(x; \alpha )L_2(y; \beta )L_3(z; \gamma ) \label{xyz}
\end{equation}
implies that $\hat{x}=x, \ \hat{y}=y$ and $\hat{z}=z$, then
$S_{\alpha,\beta}, \ R_{\alpha,\beta}, \ T_{\alpha,\beta}$
satisfy the entwining YB equation (\ref{eYBeq}).
\end{proposition}

\begin{proof}
Let
\begin{eqnarray*}
S_{\alpha,\beta}^{12}(x,y,z)&=&(x',y',z), \\
R_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}(x,y,z)&=&(x'',y',z'), \\
T_{\beta,\gamma}^{23}\circ R_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}(x,y,z)&=&
(x'',y'',z'')
\end{eqnarray*}
We can represent these maps
as chains of maps at the down, back, left faces of a cube as
in $(i)$ of Fig.\ref{fig:3d}. All the parallel edges to the $x$ (resp.
$y,z$) axis carry the parameter $\alpha$ (resp. $\beta$, $\gamma$).
From the corresponding Lax pairs we have
$L_2(y;\beta)L_1(x;\alpha)=L_1(x';\alpha)L_2(y';\beta)$, so
$L_3(z;\gamma)L_2(y;\beta)L_1(x;\alpha)=(L_3(z;\gamma)L_1(x';\alpha))L_2(y';\beta)
=L_1(x'';\alpha)(L_3(z';\gamma)L_2(y';\beta))
=L_1(x'';\alpha)L_2(y''\beta)L_3(z'';\gamma)$.

We also assume that
\begin{eqnarray*}
T_{\beta,\gamma}^{23}(x,y,z)&=&(x,\tilde{y},\tilde{z}), \\
R_{\alpha,\gamma}^{13}\circ T_{\beta,\gamma}^{23}(x,y,z)&=&(\tilde{x},\tilde{y},\tilde{\tilde{z}}), \\
S_{\alpha,\beta}^{12}\circ R_{\alpha,\gamma}^{13}\circ T_{\beta,\gamma}^{23}(x,y,z)&=&
(\tilde{\tilde{x}},\tilde{\tilde{y}},\tilde{\tilde{z}})
\end{eqnarray*}
These maps are represented at right, front and upper faces of the cube as in
$(ii)$ of Fig.\ref{fig:3d}.
Similarly from the Lax pairs we get
$L_3(z;\gamma)L_2(y;\beta)L_1(x;\alpha)=
L_1(\tilde{\tilde{x}};\alpha)L_2(\tilde{\tilde{y}};\beta)L_3(\tilde{\tilde{z}};\gamma)$.
So finally we have that
$$L_1(x'';\alpha)L_2(y''\beta)L_3(z'';\gamma)=
L_1(\tilde{\tilde{x}};\alpha)L_2(\tilde{\tilde{y}};\beta)L_3(\tilde{\tilde{z}};\gamma)$$
which implies (from the assumptions of the proposition) that $x''= \tilde{\tilde{x}},  \
y''=\tilde{\tilde{y}}, \ z''=\tilde{\tilde{z}}$. i.e.
$T_{\beta,\gamma}^{23}\circ R_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12}\circ R_{\alpha,\gamma}^{13}\circ T_{\beta,\gamma}^{23}$.
\end{proof}

\begin{figure}[h]
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\htext (-0.05 -0.65) {$x$}
\htext (0.8 -0.35) {$y$}
\htext (1.05 0.45) {$z$}
\htext (0.35 -0.05) {$\tilde{z}$}
\htext (0.75 0.6) {$\tilde{y}$}
\htext (-0.05 0.35) {$\tilde{x}$}
\htext (-0.6 -0.05) {$\tilde{\tilde{z}}$}
\htext (-0.4 0.7) {$\tilde{\tilde{y}}$}
\htext (0.4 1.05) {$\tilde{\tilde{x}}$}

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\htext (-2.45 0.45) {$z'$}
\htext (-2.91 0.75) {$y''$}
\htext (-3.18 -0.05) {$z''$}

\htext (0.6 0.05) {$T^{23}_{\beta,\gamma}$}
\htext (-0.2 -0.2) {$R^{13}_{\alpha,\gamma}$}
\htext (0.0 0.6) {$S_{\alpha,\beta}^{12}$}

\htext (-2.5 -0.4) {$S_{\alpha,\beta}^{12}$}
\htext (-2.2 0.3) {$R^{13}_{\alpha,\gamma}$}
\htext (-2.9 0.05) {$T^{23}_{\beta,\gamma}$}

\htext (-3.2 -0.78)  {$(i)$}
\htext (-0.75 -0.78) {$(ii)$}

}
\caption{Cubic representation of the 3-d compatibility condition}
\label{fig:3d}
\end{figure}

\newpage

\begin{corollary} \label{corYB}
If $(L_1(x;\alpha),  L_2(x;\alpha), L_3(x;\alpha))$ is a strong Lax triple that satisfies
Prop. \ref{3product} and
$L_1(x;\alpha)= L_2(x; \alpha)=L_3(x; \alpha)$, then
$S_{\alpha,\beta}:\mathcal{X} \times \mathcal{X} \rightarrow \mathcal{X} \times \mathcal{X}$
is a parametric Yang-Baxter map.
\end{corollary}

\begin{proof}
Since the Lax pairs $(L_1(x;\alpha),  L_2(x;\alpha))$, $(L_1(x;\alpha),  L_3(x;\alpha))$,
$(L_2(x;\alpha),  L_3(x;\alpha))$ of $S_{\alpha,\beta}, \ R_{\alpha,\beta}$ and $T_{\alpha,\beta}$
are a strong, then for $L_1=L_2=L_3$,
$S_{\alpha,\beta}= R_{\alpha,\beta}=T_{\alpha,\beta}$. So equation
(\ref{eYBeq}) becomes the Yang-Baxter equation
\begin{equation} \label{YB}
S_{\beta,\gamma}^{23}\circ S_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12}\circ S_{\alpha,\gamma}^{13}\circ S_{\beta,\gamma}^{23}.
\end{equation}
\end{proof}
In this case the matrix $L_1(x;\alpha)=L_2 (x; \alpha)=L_3(x; \alpha)$ is called a (strong)
Lax matrix of the Yang-Baxter map
$S_{\alpha,\beta}$ \cite{ves4}.

\section{Binomial Lax pairs and triples} \label{secbin}
A matrix re-factorization procedure provides a way of constructing strong Lax pairs and
consequently Lax triples.
Following the lines of \cite{kp1,ves1} we consider the set $\mathcal{L}^2$ of first degree
$2 \times 2$ polynomial
matrices $X-\zeta A$, and we denote by $p_{X}^A$ the polynomial
$$p_{X}^A(\zeta):=\det(X-\zeta A)=
f_{2}(X;A) \zeta^{2}-f_{1}(X;A) \zeta + f_{0}(X;A).$$
For $X=[x_{ij}]$ and $A=[\alpha_{ij}]$,
$$f_{0}(X;A)= \det X, \
f_{1}(X;A)=a_{22}x_{11}-a_{21}x_{12}-a_{12}x_{21}+a_{11}x_{22}, \
f_{2}(X;A)= \det A.$$
We also consider the matrix functions functions $\Pi_1,\ \Pi_2,$ with
\begin{eqnarray} \label{p1p2}
\Pi _{1}(X,Y)&=& f_{2}(X;A)(YA+BX)-f_{1}(X;A)AB, \\
\Pi _{2}(X,Y)&=& f_{2}(X;A)YX-f_{0}(X;A)AB .
\end{eqnarray}

\begin{proposition} \label{UV}
Let $A, \ B$ be invertible $2 \times 2$ matrices, such that $AB=BA$ and
$X, Y \in Mat( 2\times 2)$ with $\det \Pi _{1}(X,Y) \neq 0$ . Then
\begin{equation}
(U-\zeta A)(V-\zeta B)= (Y- \zeta B)(X-\zeta A), \label{fact}
\end{equation}
and
$p_{U}^{A}(\zeta)=p_{X}^{A}(\zeta)$
(equivalently $p_{V}^{B}(\zeta)=p_{Y}^{B}(\zeta)$),
iff
\begin{eqnarray}  \label{U}
U=U(X,Y) &:=& \Pi _{2}(X,Y) \Pi _{1}(X,Y)^{-1}A, \\
V=V(X,Y) &:=& A^{-1}(YA+BX-U(X,Y)B). \label{V}
\end{eqnarray}
\end{proposition}
The proof is given in Appendix A. For $A=B$, the proof has appeared in \cite{kp1}.
The fact that $p_{U}^{A}(\zeta)=p_{X}^{A}(\zeta)$ and
$p_{V}^{B}(\zeta)=p_{Y}^{B}(\zeta)$, or equivalently
$$f_{i}(U;A)=f_{i}(X;A), \ f_{i}(V;B)=f_{i}(Y;B), \ \text{for} \ i=0,1,2, $$
is crucial and leads to the construction of Lax pairs and symplectic solutions of the entwining YB equation.

A Poisson structure on $\mathcal{L}^2$ is defined by the Sklyanin bracket \cite{skly2}:
\begin{equation}
\{L(\zeta ) \ \overset{\otimes }{,} \ L(\eta)\}=[\frac{r}{\zeta
-\eta},L(\zeta )\otimes L(\eta)],  \label{sklyanin}
\end{equation}
where $L(\zeta)=X-\zeta A$ and $r$ denotes the permutation matrix,
 $r(x\otimes y) = y\otimes x$.
There are six Casimir functions on $\mathcal{L}^2$ which are
the elements $a_{ij}$ of the matrix $A$ and
the coefficients of the polynomial $p_{X}^A(\zeta)$,
$f_{0}(X;A), \ f_{1}(X;A)$.
For any constant matrix $A$ we denote by $\mathcal{L}^2_{A}$ the level set
$\mathcal{L}^2_{A}=\{X-\zeta A \ / \ X \in Mat(2\times2) \}$.

By a direct computation we can prove that the map $(X,Y)\mapsto (U,V)$, for $U$, $V$,
(\ref{U}) and (\ref{V}) respectively, is a Poisson
map $\mathcal{L}^2\times \mathcal{L}^2
\rightarrow  \mathcal{L}^2\times \mathcal{L}^2$ (we extend the Sklyanin bracket to the cartesian
product  $\mathcal{L}^2 \times \mathcal{L}^2$ in the natural way).

Let $A_1,A_2,A_3$ be three invertible matrices that
commute with each other.
We restrict to a level set on $\mathcal{L}^2_{A_1}$, of the Casimir functions $f_0$ and $f_1$,
by solving the system $f_{0}(X;{A_1})=\alpha_{0}$, $f_{1}(X;{A_1})=\alpha_{1}$,
with respect to two elements of $X$. We denote the two remaining elements of $X$
by $x_1$ and $x_2$.
In this way we define the matrix $L_1'(x_{1},x_{2};\bar{\alpha})$, with
$\bar{\alpha}=(\alpha_0,\alpha_1)$, such that
\begin{equation*}
f_{0}(L_1'(x_{1},x_{2};\bar{\alpha});A_1) = \alpha_{0} \ \text{and} \
f_{1}(L_1'(x_{1},x_{2};\bar{\alpha});A_1) = \alpha_{1}
\end{equation*}
and the two dimensional symplectic leaves of
$\mathcal{L}^2_{A_1}$
$$\Sigma_{A_1}(\bar{\alpha})=\{L'(x_{1},x_{2};\bar{\alpha}) -\zeta A_1 \
/ \ x_1,x_2 \in I\subset \mathbb{C} \},$$
with respect to the reduced Sklyanin structure (\ref{sklyanin}).
In a similar way we define the matrices $L_2'(x_{1},x_{2};\bar{\alpha})$ and
$L_3'(x_{1},x_{2};\bar{\alpha})$
from the restriction on the level sets of the Casimir functions on
$\mathcal{L}^2_{A_2}$ and $\mathcal{L}^2_{A_3}$ respectively i.e.
\begin{eqnarray*}
f_{0}(L_2'(x_{1},x_{2};\bar{\alpha});A_2) = \alpha_{0},  \
f_{1}(L_2'(x_{1},x_{2};\bar{\alpha});A_2) = \alpha_{1},  \\
f_{0}(L_3'(x_{1},x_{2};\bar{\alpha});A_3) = \alpha_{0},  \
f_{1}(L_3'(x_{1},x_{2};\bar{\alpha});A_3) = \alpha_{1} \
\end{eqnarray*}
and the corresponding symplectic leaves $\Sigma_{A_2}(\bar{\alpha})$ and $\Sigma_{A_3}(\bar{\alpha})$.
\begin{theorem} \label{lpun}
The equations
\begin{equation} \label{uniq}
(L_i'(u^{ij}_{1},u^{ij}_{2};\bar{\alpha})-\zeta A_i) (L_j'(v^{ij}_{1},v^{ij}_{2};\bar{\beta})-\zeta A_j)
=
(L_j'(y_{1},y_{2};\bar{\beta})-\zeta A_j) (L_i'(x_{1},x_{2};\bar{\alpha})-\zeta A_i),
\end{equation}
for $i,j=1,2,3$, are uniquely solvable with respect to $u^{ij}_1,\ u^{ij}_2, \ v^{ij}_1$ and $v^{ij}_2$.
The parametric maps $(S_{\bar{\alpha},\bar{\beta}},R_{\bar{\alpha},\bar{\beta}},T_{\bar{\alpha},\bar{\beta}})$,
with
\begin{eqnarray*}
S_{\bar{\alpha},\bar{\beta}}:((x_1,x_2),(y_1,y_2)) &\mapsto & ((u^{12}_1,u^{12}_2),(v^{12}_1,v^{12}_2)), \\
R_{\bar{\alpha},\bar{\beta}}:((x_1,x_2),(y_1,y_2)) &\mapsto & ((u^{13}_1,u^{13}_2),(v^{13}_1,v^{13}_2)), \\
T_{\bar{\alpha},\bar{\beta}}:((x_1,x_2),(y_1,y_2)) &\mapsto & ((u^{23}_1,u^{23}_2),(v^{23}_1,v^{23}_2))
\end{eqnarray*}
are symplectic and admit the strong Lax triple
$(L_1(x_1,x_2,\bar{\alpha}),L_2(x_1,x_2,\bar{\alpha}),L_3(x_1,x_2,\bar{\alpha}))$,
where $$L_i(x_1,x_2,\bar{\alpha})= L_i'(x_1,x_2,\bar{\alpha})-\zeta A_i, \ \text{for} \ i=1,2,3.$$
Moreover they satisfy the entwining YB equation
$T_{\beta,\gamma}^{23} R_{\alpha,\gamma}^{13} S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12} R_{\alpha,\gamma}^{13} T_{\beta,\gamma}^{23}$.
\end{theorem}
The proof of this theorem is given in appendix B.

As it is remarked in \cite{kp1}, there are cases where limits of quadrirational YB maps give rise to new
degenerate maps. A similar procedure can be applied here as well. For example we can
consider commuting invertible matrices $A_i=A_i(\varepsilon)$, depending on a
parameter $\varepsilon$, such that $\lim_{\varepsilon \rightarrow 0}detA_i(\varepsilon)=0$,
for some $i \in \{1,2,3\}$, and construct the corresponding maps
$(S_{\bar{\alpha},\bar{\beta}}(\varepsilon),R_{\bar{\alpha},\bar{\beta}}(\varepsilon),
T_{\bar{\alpha},\bar{\beta}}(\varepsilon))$
that we described. A new solution of the entwining YB equation can be derived by taking
the limit (if it exists) of $(S_{\bar{\alpha},\bar{\beta}}(\varepsilon),R_{\bar{\alpha},\bar{\beta}}(\varepsilon),
T_{\bar{\alpha},\bar{\beta}}(\varepsilon))$ for $\varepsilon \rightarrow 0$.

\section{Integrable rational maps on $\mathbb{C}^2 \times \mathbb{C}^2$} \label{secent}
We give here an example of two solutions of the entwining YB equation from a pair of
commuting matrices.

We consider $X-\zeta A_i \in \mathcal{L}^2_{A_i}$, for $i=1,2,$
with
\begin{equation*}
X=
\left(
\begin{array}{ll}
 x_1 & x_2 \\
 x_3 & x_4
\end{array}
\right) , \
A_1=
\left(
\begin{array}{ll}
 1 & 0 \\
 0 & 1
\end{array}
\right)
\ \text{and}  \
A_2=
\left(
\begin{array}{ll}
 1 & 0 \\
 0 & \varepsilon
\end{array}
\right) , \ \varepsilon \neq 0.
\end{equation*}
First we will construct symplectic rational maps on symplectic leaves of
$\mathcal{L}^2_{A_1} \times \mathcal{L}^2_{A_1}$,
$\mathcal{L}^2_{A_2} \times \mathcal{L}^2_{A_2}$,
$\mathcal{L}^2_{A_1} \times \mathcal{L}^2_{A_2}$ and $\mathcal{L}^2_{A_2} \times \mathcal{L}^2_{A_1}$
respectively and their corresponding Lax pairs. Next by considering two cases
$A_3=A_1$ and $A_3=A_2$ we derive two solutions of the entwining YB equation. All these
maps are integrable.

\subsection{ A Yang-Baxter map on $\Sigma_{A_1}(\alpha,1) \times \Sigma_{A_1}(\beta,1)$}
The Casimir functions on $\mathcal{L}^2_{A_1}$ are
$$f_0(X;A_1)=x_1x_4-x_2x_3, \ f_1(X;A_1)=x_1+x_4.$$
We set $f_0(X;A_1)=\alpha$, $f_1(X;A_1)=1$ and solve with
respect to $x_3$ and $x_4$ to get the matrix
\begin{equation}
\label{L1}
L_1'(x_1,x_2;\alpha)=
\left(
\begin{array}{cc}
 {x_1} & {x_2} \\
 -\frac{\alpha+({x_1}-1) {x_1}}{{x_2}} & 1-{x_1}
\end{array}
\right)
\end{equation}
and the Lax matrix
\begin{equation} \label{laxL1}
L_1(x_1,x_2;\alpha)=L'(x_1,x_2;\alpha)-\zeta A_1.
\end{equation}
As expected from theorem \ref{lpun} and corollary \ref{corYB}, the equation
$$L_1(u'_1,u'_2;\alpha)L_1(v'_1,v'_2;\beta)=L_1(y_1,y_2;\beta)L_1(x_1,x_2;\alpha)$$
is uniquely solvable with respect to $u'_1,u'_2,v'_1,v'_2$ and implies the parametric
Yang-Baxter map
\begin{equation} \label{YB1}
R_{\alpha,\beta}((x_1,x_2),(y_1,y_2))=((u'_1,u'_2),(v'_1,v'_2)),
\end{equation}
where
\begin{eqnarray*}
u'_1 &=&  y_{1}+\frac{x_{2}y_{2}}{N}(\alpha-\beta)  (x_{1}+y_{1}-1)  ,\\
u'_2 &=& \frac{y_{2}}{N} \left(\alpha (x_{2}+y_{2})^2+(x_{2} (y_{1}-1)-x_{1} y_{2}) (x_{2} y_{1}-x_{1}
   y_{2}+y_{2})\right), \\
v'_1 &=& x_{1}-\frac{x_{2}y_{2}}{N}(\alpha-\beta)  (x_{1}+y_{1}-1) , \\
v'_2 &=& \frac{x_{2}}{N} \left(\beta (x_{2}+y_{2})^2+(x_{2} (y_{1}-1)-x_{1} y_{2}) (x_{2} y_{1}-x_{1}
   y_{2}+y_{2})\right), \\
   N &=& \alpha {y_2}^2 + \beta {x_2}^2+(\alpha+\beta -1) {y_2} {x_2}+(x_2 y_1 - x_1 y_2)
   (x_2 y_1 - x_1 y_2+y_2-x_2).
\end{eqnarray*}
We can verify that $u'_1=U'_{11}$, $u'_2=U'_{12}, \ v'_1=V'_{11}$ and $v'_2=V'_{12}$, where $U'_{ij}$ and $V'_{ij}$
are the $ij$ elements of the matrices
$$U':=U(L'(x_{1},x_{2};\alpha),L'(y_1,y_2,\beta)), \  V':=V(L'(x_{1},x_{2};\alpha),L'(y_1,y_2,\beta))$$
defined by (\ref{U}) and (\ref{V}) respectively.

\subsection{ A Yang-Baxter map on $\Sigma_{A_2}(\alpha,1) \times \Sigma_{A_2}(\beta,1)$}
The construction of the following Yang-Baxter map appears in \cite{kp1}.
The Casimir functions on $\mathcal{L}^2_{A_2}$ are
$$f_0(X;A_2)=x_1 x_4-x_2 x_3, \ f_1(X;A_2)=\varepsilon x_1+x_4.$$
We set here as well $f_0(X;A_2)=\alpha$, $f_1(X;A_2)=1$ and solve
with respect to $x_1,\ x_4$. By performing the change of
variables: $x_2 \mapsto x_1$ and $x_3 \mapsto x_2$ we derive
the matrix
\begin{equation} \label{Me}
M_{\varepsilon}'(x_1,x_2;\alpha)=\left(
\begin{array}{cc}
 \frac{1-(1-4 \varepsilon (\alpha+{x_1} {x_2}))^{1/2}}{2 \varepsilon} & {x_1} \\
 {x_2} & \frac{1}{2} (1-4 \varepsilon (\alpha+{x_1} {x_2}))^{1/2}+\frac{1}{2}
 \end{array} \right)
\end{equation}
and the corresponding Lax matrix $M_{\varepsilon}(x_1,x_2;\alpha)=M_{\varepsilon}'(x_1,x_2;\alpha)-\zeta A_2$.
In this way we obtain the non-degenerate Yang-Baxter map
\begin{equation} \label{YB2}
R_{\alpha,\beta}^{\varepsilon}((x_1,x_2),(y_1,y_2))=((\mathbf{u}_1,\mathbf{u}_2),(\mathbf{v}_1,\mathbf{v}_2)),
\end{equation}
with  $\mathbf{u}_1=\mathbf{U}_{12}$, $\mathbf{u}_2=\mathbf{U}_{21},
\ \mathbf{v}_1=\mathbf{V}_{12}$, $\mathbf{v}_2=\mathbf{V}_{21}$ the corresponding elements of
$\mathbf{U}=U(M_{\varepsilon}'(x_1,x_2;\alpha),M_{\varepsilon}'(y_1,y_2;\beta))$,
$\mathbf{V}=V(M_{\varepsilon}'(x_1,x_2;\alpha),M_{\varepsilon}'(y_1,y_2;\beta))$. As in \cite{kp1} we
take the limit of (\ref{YB2}), for $\varepsilon \rightarrow 0$, in order to derive the degenerate Yang-Baxter map
\begin{equation} \label{YB3}
\bar{R}_{\alpha,\beta}((x_1,x_2),(y_1,y_2))=
\lim_{\varepsilon \rightarrow 0}R_{\alpha,\beta}^{\varepsilon}((x_1,x_2),(y_1,y_2))
=((\bar{u}_1,\bar{u}_2),(\bar{v}_1,\bar{v}_2))
\end{equation}
with
\begin{eqnarray*}
\bar{u}_{1} =  y_{1}-\frac{(\alpha-\beta)x_{1}}
{1+x_{1}y_{2}}, \
\bar{u}_{2}= y_{2}, \
\bar{v}_{1} = x_{1}, \
\bar{v}_{2} = x_{2}+\frac{(\alpha-\beta)y_{2}}{1+x_{1}y_{2}},
\end{eqnarray*}
and strong Lax matrix
\begin{equation} \label{LaxL2}
L_2(x_1,x_2;\alpha)=\lim_{\varepsilon \rightarrow 0} M_{\varepsilon}(x_1,x_2;\alpha)=
\left(
\begin{array}{cc}
 {x_1} x_2+\alpha-\zeta & {x_1} \\
 x_2 & 1
\end{array}
\right).
\end{equation}

\subsection{The maps on $\Sigma_{A_1}(\alpha,1) \times \Sigma_{A_2}(\beta,1)$
and on $\Sigma_{A_2}(\alpha,1) \times \Sigma_{A_1}(\beta,1)$}
The equation
$$L_1(u_1,u_2;\alpha)L_2(v_1,v_2;\beta)=L_2(y_1,y_2;\beta)L_1(x_1,x_2;\alpha)$$
with $L_1, \ L_2$ (\ref{laxL1}) and (\ref{LaxL2}) respectively, yields the unique solution
\begin{eqnarray*}
u_1 &=& {y_2} {y_1}-\frac{{x_1} {y_1}}{{x_2}}+\frac{(\alpha+\beta ({x_2}
   {y_2}-{x_1}))(x_2+y_1)}{x_2( {x_2} {y_2}-{x_1}-\beta +1)}, \\
u_2 &=& -\frac{\alpha ({x_2}+{y_1})^2+(\beta {x_2}+{y_1} {y_2} {x_2}-{x_1} {y_1}+{y_1}) ({x_2} (\beta+{y_1}
   {y_2}-1)-{x_1} {y_1})}{{x_2} ({x_2} {y_2}-{x_1}-\beta+1)}, \\
v_1 &=& x_2+y_1, \\
 v_2 &=& \frac{{x_1}}{{x_2}}-\frac{\alpha+\beta ({x_2} {y_2}-{x_1})}{{x_2} ({x_2} {y_2}-{x_1}-\beta +1)}.
\end{eqnarray*}
This solution can be obtained by the limit for $\varepsilon \rightarrow 0$ of the $U_{11}$, $U_{12}$, $V_{12}$
and $V_{21}$ elements of the matrices
$$U:=U(L_1'(x_{1},x_{2};\alpha),M'_{\varepsilon}(y_1,y_2,\beta)), \
V:=V(L_1'(x_{1},x_{2};\alpha),M'_{\varepsilon}(y_1,y_2,\beta)),$$
defined by (\ref{U}) and (\ref{V}) with $L_1', \ M'_{\varepsilon}$ (\ref{L1}) and (\ref{Me}) respectively.
So the map
\begin{equation} \label{exmap}
S_{\alpha,\beta}:((x_1,x_2),(y_1,y_2)) \mapsto ((u_1,u_2),(v_1,v_2))
\end{equation}
admits the
strong Lax pair $(L_1(x_1,x_2;\alpha),L_2(x_1,x_2;\alpha))$.

Furthermore the twisted equation
$$L_2(\tilde{u}_1,\tilde{u}_2;\alpha)L_1(\tilde{v}_1,\tilde{v}_2;\beta)=L_1(y_1,y_2;\beta)L_2(x_1,x_2;\alpha)$$
implies
\begin{eqnarray*}
\tilde{u}_1 &=& \frac{(\alpha {x_1}-{y_1} {x_1}-{y_2}) {y_2}}{\beta {x_1}+({y_1}-1) {y_1} {x_1}+(\alpha+{y_1}-1) {y_2}},\\
\tilde{u}_2 &=& {x_2}-\frac{\beta+({y_1}-1) {y_1}}{{y_2}}, \\
\tilde{v}_1 &=& \frac{\beta (\alpha {x_1}-{y_2})+({y_1} {x_1}-{x_1}+{y_2}) (\alpha {y_1}+{x_2} {y_2})+{x_1} {x_2} \left({x_1}
   \left({y_1}^2-{y_1}+\beta\right)+{y_1} {y_2}\right)}{\beta {x_1}+({y_1}-1) {y_1} {x_1}+(\alpha+{y_1}-1) {y_2}}, \\
\tilde{v}_2 &=& {x_1}-\frac{(\alpha {x_1}-{y_1} {x_1}-{y_2}) {y_2}}{\beta {x_1}+({y_1}-1) {y_1} {x_1}+(\alpha+{y_1}-1) {y_2}}
\end{eqnarray*}
and the map $T_{\alpha,\beta}$,
\begin{equation}\label{extw}
{T}_{\alpha,\beta}:((x_1,x_2),(y_1,y_2))
\mapsto ((\tilde{u}_1,\tilde{u}_2),(\tilde{v}_1,\tilde{v}_2)),
\end{equation}
with strong Lax pair $(L_2(x_1,x_2;\alpha),L_1(x_1,x_2;\alpha))$.

\subsection{Solutions of the entwining YB equation and integrability}
We consider two cases. First we choose a third matrix $A_3=A_1$ and the corresponding
Lax matrix $L_3(x_1,x_2;\alpha)=L_1(x_1,x_2;\alpha)$. Theorem \ref{lpun} implies that
the maps $(S_{\alpha,\beta},R_{\alpha,\beta},T_{\alpha,\beta})$ (\ref{exmap},  \ref{YB1} and \ref{extw}
respectively) admits the strong Lax triple
$(L_1(x_1,x_2;\alpha),L_2(x_1,x_2;\alpha),L_1(x_1,x_2;\alpha))$ and satisfy the entwining Yang-Baxter equation
$$T_{\beta,\gamma}^{23}\circ R_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12}\circ R_{\alpha,\gamma}^{13} \circ T_{\beta,\gamma}^{23}.$$

Now if choose $A_3=A_2, \ L_3(x_1,x_2;\alpha)=L_2(x_1,x_2;\alpha)$, we derive the strong Lax triple
$(L_1(x_1,x_2;\alpha),L_2(x_1,x_2;\alpha),L_2(x_1,x_2;\alpha))$ of the maps
$(S_{\alpha,\beta},S_{\alpha,\beta},\bar{R}_{\alpha,\beta})$, from (\ref{exmap}) and (\ref{YB3}), that satisfy the
entwining YB equation
$$\bar{R}_{\beta,\gamma}^{23}\circ S_{\alpha,\gamma}^{13}\circ S_{\alpha,\beta}^{12}=
S_{\alpha,\beta}^{12}\circ S_{\alpha,\gamma}^{13} \circ \bar{R}_{\beta,\gamma}^{23}.$$

The map $S_{\alpha,\beta}$ (\ref{exmap}) is symplectic with respect to the reduced Sklyanin bracket
on $\Sigma_{A_1}(\alpha,1) \times \Sigma_{A_2}(\beta,1)$:
\begin{equation} \label{pstr}
\{ x_1,x_2 \}=-x_2, \  \{y_1,y_2 \}=1, \ \{x_i,y_j \}=0, \ i,j=1,2 \ \ (x_2 \neq 0).
\end{equation}
For the 1-periodic `staircase' initial value problem
the trace of the corresponding
monodromy matrix, $M_1(\mathbf{x},\mathbf{y})=L_2(y_1,y_2;\beta)L_1(x_1,x_2;\alpha)$,
gives the two first integrals:
\begin{eqnarray*}
J^s_1(x_1,x_2,y_1,y_2) &=& x_2 y_2+x_1( y_1 y_2+\beta-1)-
\frac{y_1}{x_2}(x_1^2-x_1+\alpha), \\
J^s_2(x_1,x_2,y_1,y_2) &=& x_1+y_1 y_2,
\end{eqnarray*}
which are in involution with respect to the Poisson bracket (\ref{pstr}).

Similarly
$\bar{R}_{\alpha,\beta}$, $T_{\alpha,\beta}$
are integrable and $R_{\alpha,\beta}$ is an
involution i.e. $R_{\alpha,\beta}\circ R_{\alpha,\beta}=Id$.
The reduced Sklyanin bracket on the corresponding symplectic leaves is
given by the brackets of the coordinates
\begin{eqnarray}
&&\{ x_1,x_2 \}=-x_2, \  \{y_1,y_2 \}=-y_2, \ ~ \{x_i,y_j \}=0, \ \text{on} \
 \Sigma_{A_1}(\alpha,1) \times \Sigma_{A_1}(\beta,1), \label{brR}\\
&& \{ x_1,x_2 \}=1, \ \ \ \ ~  \{y_1,y_2 \}=1, \ \ ~ ~ \ \  \{x_i,y_j \}=0, \ \text{on} \
 \Sigma_{A_2}(\alpha,1) \times \Sigma_{A_2}(\beta,1), \label{brRR}\\
&&\{ x_1,x_2 \}=1, \ \ \ \ ~ \{y_1,y_2 \}=-y_2,  \ ~ \{x_i,y_j \}=0, \ \text{on} \
 \Sigma_{A_2}(\alpha,1) \times \Sigma_{A_1}(\beta,1) \label{brT}.
\end{eqnarray}
The corresponding integrals
of the YB map $\bar{R}_{\alpha,\beta}$ are
$$
J^{\bar{R}}_1 = \alpha  y_1 y_2 + b x_1 x_2+(x_2 y_1+1) (x_1 y_2+1),  \
J^{\bar{R}}_2 = x_1 x_2+y_1 y_2,
$$
and of the map $T_{\alpha,\beta}$ are
$$J^T_1 = x_2 y_2+y_1( x_1 x_2 + \alpha -1) -
\frac{x_1}{y_2}(y_1^2-  y_1+ \beta),  \
J^T_2 = y_1+x_1 x_2.$$
The integrals of each pair are in involution with respect to the
Poisson brackets (\ref{brRR}) and (\ref{brT}) respectively.

\section{Perspectives} \label{persp}
We presented the construction of entwining Yang-Baxter maps following the procedure of using Lax matrices
as symplectic leaves of the Sklyanin bracket of polymonial matrices. These maps give rise to, in principle, asymmetric
discrete integrable systems on the 2-dimensional integer lattice. We believe an extensive study as well as their relation to scalar asymmetric integrable systems on quad-graphs deserves more attention.

\section{Acknowledgments}
The first author acknowledges partial support from the State Scholarships Foundation of
Greece.

\section{Appendix A}
The proof of proposition \ref{UV}.

\begin{proof}
From equation (\ref{fact}) we derive the system :
\begin{equation} \label{syst}
UV=YX, \ \ UB+AV=YA+BX
\end{equation}
which implies
\begin{equation} \label{tetr}
(UA^{-1})^2 AB=UA^{-1}(YA+BX)-YX.
\end{equation}
Since $p_{U}^{A}(\zeta)=p_{X}^{A}(\zeta)$, then $f_{i}(U;A)=f_{i}(X;A)$
so from the Cayley-Hamilton theorem we have that:
\begin{equation} \label{cal}
f_{2}(X;A)(UA^{-1})^{2}-f_{1}(X;A)U A^{-1} + f_{0}(X;A)I=0.
\end{equation}
By solving (\ref{tetr}) and(\ref{cal}) with respect to $U$ we get that:
$$U=(f_{2}(X;A)YX-f_{0}(X;A)AB)
(f_{2}(X;A)(YA+BX)-f_{1}(X;A)AB))^{-1}A$$
and from (\ref{syst}) that $V = A^{-1}(YA+BX-U(X,Y)B)$.

From the other hand let us assume that $U$ and $V$ are given by (\ref{U}) and  (\ref{V}).
By performing some calculations we have that
\begin{eqnarray*}
\Pi _{2}(X,Y) &=& \Pi _{2}(X,Y)+\Pi _{1}(X,Y)A^{-1}X-\Pi _{1}(X,Y)A^{-1}X  \\
&=& \Pi _{1}(X,Y)A^{-1}X - BA(f_{2}(X;A)
(A^{-1}X)^{2}-f_{1}(X;A) A^{-1}X + f_{0}(X;A)I) \\
&=& \Pi _{1}(X,Y)A^{-1}X-BAp_{X}^{A}(A^{-1}X)
\end{eqnarray*}
Cayley-Hamilton theorem states that $p_{X}^{A}(A^{-1}X)=0$ so
$\Pi _{2}(X,Y)=\Pi _{1}(X,Y)A^{-1}X$
and from (\ref{U}) we get $UA^{-1}=\Pi_1A^{-1}X \Pi_1^{-1}$ which means that
$f_{i}(U;A)=f_{i}(X;A)$ or $p_{U}^{A}(\zeta)=p_{X}^{A}(\zeta)$ and
$p_{V}^{B}(\zeta)=p_{Y}^{B}(\zeta)$.
Also from (\ref{U}) and (\ref{V}) we derive that $UB+AV=YA+BX$ and
$$UA^{-1}(f_{2}(X;A)(UB+AV)-f_{1}(X;A)AB)=
f_{2}(X;A)YX-f_{0}(X;A)AB$$
or
$$(f_{2}(X;A)(UA^{-1})^2-f_{1}(X;A)UA^{-1}+f_{0}(X;A)I)AB=
f_2(X;A)(YX-UV).$$
Since $f_i(X;A)=f_i(U;A)$
from Cayley-Hamilton theorem we get $UV=YX$.
\end{proof}

\section{Appendix B}
The proof of theorem \ref{lpun}.
\begin{proof}
The proof of the existence and uniqueness of the solutions of (\ref{uniq})
is a consequence of the
construction of the matrices $L_i'(x_1,x_2,\bar{\alpha})$ in addition
with Prop. \ref{UV},
for $X=L_i'(x_1,x_2,\bar{\alpha})$, $Y =L_j'(x_1,x_2,\bar{\alpha})$,
$A=A_i$ and $B=A_j$.
The elements $u^{ij}_1, \ u^{ij}_2, \ v^{ij}_1$ and $v^{ij}_2$
can be determined by the corresponding elements of
the matrices
$$U(L'_i(x_{1},x_{2};\bar{\alpha}),L'_j(y_1,y_2,\bar{\beta})), \
V(L'_i(x_{1},x_{2};\bar{\alpha}),L'_j(y_1,y_2,\bar{\beta})),$$
of (\ref{U}) and (\ref{V}). Since the map of Prop. \ref{UV}, $(X,Y) \mapsto (U,V)$, is Poisson,
the reduced map
$S_{\bar{\alpha},\bar{\beta}}:  \Sigma_{A_1}(\bar{\alpha}) \times \Sigma_{A_2}(\bar{\beta})
\rightarrow  \Sigma_{A_1}(\bar{\alpha}) \times \Sigma_{A_2}(\bar{\beta})$
is symplectic (resp. the maps $R_{\bar{\alpha},\bar{\beta}}$ and
$T_{\bar{\alpha},\bar{\beta}}$).

In order to prove the entwining YB property it suffices to show that the equation
\begin{eqnarray} \label{pren}
(L_1'(x'_{1},x'_{2};\bar{\alpha})-\zeta A_1) (L'_2(y'_{1},y'_{2};\bar{\beta})-\zeta A_2) (L'_3(z'_{1},z'_{2};\bar{\gamma})-\zeta A_3)
\nonumber \\
=
(L'(x_{1},x_{2};\bar{\alpha})-\zeta A_1) (L'_2(y_{1},y_{2};\bar{\beta})-\zeta A_2) (L'_3(z_{1},z_{2};\bar{\gamma})-\zeta A_3)
\end{eqnarray}
implies that $x_i'=x_i, \ y_i'=y_i, \ z_i'=z_i$, for $i=1,2$, then the proof follows
from Prop. \ref{3product}.
We set
\begin{eqnarray*}
X=L'_1(x_{1},x_{2};\bar{\alpha}), \ Y=L'_2(y_{1},y_{2};\bar{\beta}), \ Z=L'_3(z_{1},z_{2};\bar{\gamma}), \ \\
X'=L'_1(x'_{1},x'_{2};\bar{\alpha}), \ Y'=L'_2(y'_{1},y'_{2};\bar{\beta}), \ Z'=L'_3(z_{1},z_{2};\bar{\gamma}),
\end{eqnarray*}
and
$XYZ=K$, $XYA_3+XA_2Z+A_1YZ=L$, $XA_2A_3+A_1 YA_3+ A_1 A_2 Z=M$.
Substituting to (\ref{pren}) we come up to the system :
$$X^{\prime }Y^{\prime }Z^{\prime }=K, \
X^{\prime }Y^{\prime }A_3+X^{\prime }A_2Z^{\prime }+A_1Y^{\prime }Z^{\prime }=L, \
X^{\prime }A_2A_3+A_1 Y^{\prime }A_3+ A_1 A_2 Z^{\prime }=M$$
which implies that
\begin{equation} \label{tritis}
(X'A_1^{-1})^{3}A_1A_2A_3 -(X'A_1^{-1})^{2}M+X'A_1^{-1}L=K.
\end{equation}

Since $$\det(X'-\zeta A_1)=\det(X-\zeta A_1)= \alpha_2\zeta^{2}- \alpha_1 \zeta + \alpha_0,$$
with $\alpha_2=f_{2}(X;A_1), \ \alpha_1= f_{1}(X;A_1), \  \alpha_0= f_{0}(X;A_1)$,
from Cayley-Hamilton theorem we have that
\begin{equation*}
p_{X'}^{A_1}(X'A_1^{-1})=\alpha_2(X'A_1^{-1})^{2}-\alpha_1(X'A_1^{-1})+\alpha_0 I=0.
\end{equation*}
By evaluating the powers of $X'A_1^{-1}$ from the last equation, equation (\ref{tritis}) becomes

$$X'A_1^{-1}[\alpha_{2}^{2}L-\alpha_{2}\alpha_{1}M+(\alpha_{1}^{2}-
\alpha_{2}\alpha_{0})A_1 A_2 A_3]=
\alpha_{2}^{2}K-\alpha_{2}\alpha_{0}M+\alpha_{1}\alpha_{0}A_1 A_2 A_3$$
or
\begin{eqnarray} \label{res}
 X'A_1^{-1}[\alpha_{2}^{2}L-\alpha_{2}\alpha_{1}M+
(\alpha_{1}^{2}-\alpha_{2}\alpha_{0})A_1 A_2 A_3] \ \ \ ~~ ~\nonumber \\ =
XA_1^{-1}[\alpha_{2}^{2}L-\alpha_{2}\alpha_{1}M+(
\alpha_{1}^{2}-\alpha_{2}\alpha_{0}) A_1 A_2 A_3]
+ Q
\end{eqnarray}
where
\begin{equation*}
Q = \alpha_{2}^{2}K-\alpha_{2}\alpha_{0}M+\alpha_{1}\alpha_{0}A_1 A_2 A_3 -
XA_1^{-1}[\alpha_{2}^{2}L-\alpha_{2}\alpha_{1}M+
(\alpha_{1}^{2}-\alpha_{2}\alpha_{0})A_1 A_2 A_3 ].
\end{equation*}
If we replace again $K, L, M$ by $XYZ$, $XYA_3+XA_2Z+A_1YZ$,
$XA_2A_3+A_1 YA_3+ A_1 A_2 Z$ respectively, we can factorize $Q$ as follows
\begin{eqnarray*}
Q &=& (\alpha_{2}(XA_1^{-1})^{2}-\alpha_{1}XA_1^{-1}+\alpha_{0}I)
(\alpha_{1}A_1 A_2 A_3-\alpha_{2}(A_1 Y A_3+A_1 A_2 Z)) \\
&=& p_{X}^{A_1}(XA_1^{-1})(\alpha_{1}A_1 A_2 A_3 -\alpha_{2}(A_1 Y A_3+A_1 A_2 Z))
\end{eqnarray*}
and since $p_{X}^{A_1}(XA_1^{-1})=0$, $Q=0$.
So from (\ref{res})
we conclude that
$X'=X$. In a similar way we can prove that $Z'=Z$
which means that also $Y'=Y$.
\end{proof}


text:
\begin{document}
\maketitle

\vspace{-20pt}
\begin{center}
{\footnotesize Instituto de Matem\'aticas, Pontificia Universidad Cat\'olica de Valpara\'iso, Blanco Viel 596, Cerro Bar\'on, Valpara\'iso, Chile. \\
E-mails: gamaliel.cerda.m@mail.pucv.cl
}\end{center}

\vspace{5pt}

\begin{abstract}
In this paper, we give quadratic approximation of generalized Tribonacci sequence $\{V_{n}\}_{n\geq0}$ defined by Eq. (\ref{eq:7}) and use this result to give the matrix form of the $n$-th power of a companion matrix of $\{V_{n}\}_{n\geq0}$. Then we re-prove the cubic identity or Cassini-type formula for $\{V_{n}\}_{n\geq0}$ and the Binet's formula of the generalized Tribonacci quaternions.
\end{abstract}

\medskip
\noindent
\subjclass{\footnotesize {\bf Mathematical subject classification:}
Primary: 11B39; Secondary: 15A24, 40C05.}

\medskip
\noindent
\keywords{\footnotesize {\bf Key words:} Binet's formula, companion matrix, generalized Tribonacci sequence, Narayana number, Padovan number, quadratic approximation, Tribonacci number.}
\medskip

\vspace{5pt}

\section{Introduction}\label{sec:1}
\setcounter{equation}{0}

Let $Q=\left(
\begin{array}{ccc}
1& 1& 1 \\
1& 0 & 0 \\
0 & 1& 0
\end{array}
\right)$ be a companion matrix of the Tribonacci sequence $\{T_{n}\}_{n\geq0}$ defined by the third-order linear recurrence relation
\begin{equation}\label{eq:1}
T_{0}=T_{1}=0,\ T_{2}=1,\ T_{n}=T_{n-1}+T_{n-2}+T_{n-3}\ (n\geq3).
\end{equation}
Then, by an inductive argument (\cite{Spi}, \cite{Wa}), the $n$-th power $Q^{n}$ has the matrix form
\begin{equation}\label{eq:2}
Q^{n}=\left(
\begin{array}{ccc}
T_{n+2}& T_{n+1}+T_{n}& T_{n+1} \\
T_{n+1}&T_{n}+T_{n-1}& T_{n} \\
T_{n} & T_{n-1}+T_{n-2}& T_{n-1}
\end{array}
\right)\ (n\geq2).
\end{equation}
This property provides an alternate proof of the Cassini-type formula for $\{T_{n}\}_{n\geq0}$:
\begin{equation}\label{eq:3}
T_{n}^{3}+T_{n-1}^{2}T_{n+2}+T_{n-2}T_{n+1}^{2}-2T_{n-1}T_{n}T_{n+1}-T_{n-2}T_{n}T_{n+2}=1.
\end{equation}

Now, let's think of the other access method in order to give the matrix form Eq. (\ref{eq:2}) of $Q^{n}$. This method give the motivation of our research. That is, our research is based on the following observation: It is well known [5] that the usual Tribonacci numbers can be expressed using Binet's formula
\begin{equation}\label{eq:4}
T_{n}=\frac{\alpha^{n}}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{\omega_{1}^{n}}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}+\frac{\omega_{2}^{n}}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})}.
\end{equation}
where $\alpha$, $\omega_{1}$ and $\omega_{2}$ are the roots of the cubic equation $x^{3}-x^{2}-x-1=0$. Furthermore, $\alpha=\frac{1}{3}+A_{T}+B_{T}$, $\omega_{1}=\frac{1}{3}+\epsilon A_{T}+\epsilon^{2} B_{T}$ and $\omega_{2}=\frac{1}{3}+\epsilon^{2}A_{T}+\epsilon B_{T}$, where $$A_{T}=\sqrt[3]{\frac{19}{27}+\sqrt{\frac{11}{27}}},\ B_{T}=\sqrt[3]{\frac{19}{27}-\sqrt{\frac{11}{27}}},$$ and $\epsilon=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$.

From the Binet's formula Eq. (\ref{eq:4}), using the classic identities $\alpha+\omega_{1}+\omega_{2}=1$, $\alpha\omega_{1}+\alpha\omega_{2}+\omega_{1}\omega_{2}=-1$, we have for any integer $n\geq2$:
\begin{align*}
\alpha T_{n}&+(1+\omega_{1}\omega_{2})T_{n-1}+T_{n-2}\\
&=\frac{\alpha^{n-2}(\alpha^{3}+(1+\omega_{1}\omega_{2})\alpha+1)}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{\omega_{1}^{n-2}(\alpha\omega_{1}^{2}+(1+\omega_{1}\omega_{2})\omega_{1}+1)}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}\\
&\ \ +\frac{\omega_{2}^{n-2}(\alpha\omega_{2}^{2}+(1+\omega_{1}\omega_{2})\omega_{2}+1)}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})}\\
&=\alpha^{n-1}.
\end{align*}
Then, we obtain
\begin{equation}\label{eq:5}
\alpha T_{n}+(1+\omega_{1}\omega_{2})T_{n-1}+T_{n-2}=\alpha^{n-1}\ (n\geq2).
\end{equation}
Multipling Eq. (\ref{eq:5}) by $\alpha$, using $\alpha\omega_{1}\omega_{2}=1$, and if we change $\alpha$, $\omega_{1}$ and $\omega_{2}$ role above process, we obtain the quadratic approximation of $\{T_{n}\}_{n\geq0}$
\begin{equation}\label{eq:6}
\textrm{Quadratic app. of $\{T_{n}\}$}: \left\{
\begin{array}{c }
\alpha^{n}=T_{n}\alpha^{2}+(T_{n-1}+T_{n-2})\alpha+T_{n-1},\\
\omega_{1}^{n}=T_{n}\omega_{1}^{2}+(T_{n-1}+T_{n-2})\omega_{1}+T_{n-1},\\
\omega_{2}^{n}=T_{n}\omega_{2}^{2}+(T_{n-1}+T_{n-2})\omega_{2}+T_{n-1},
\end{array}
\right.
\end{equation}
where $\alpha$, $\omega_{1}$ and $\omega_{2}$ are the roots of the cubic equation $x^{3}-x^{2}-x-1=0$.

In Eq. (\ref{eq:6}), if we change $\alpha$, $\omega_{1}$ and $\omega_{2}$ into the companion matrix $Q$ and change $T_{n-1}$ into the matrix $T_{n-1}I$, where $I$ is the $3\times 3$ identity matrix, then we obtain the matrix form Eq. (\ref{eq:2}) of $Q^{n}$
$$Q^{n}=T_{n}Q^{2}+(T_{n-1}+T_{n-2})Q+T_{n-1}I\left(=\left(
\begin{array}{ccc}
T_{n+2}& T_{n+1}+T_{n}& T_{n+1} \\
T_{n+1}&T_{n}+T_{n-1}& T_{n} \\
T_{n} & T_{n-1}+T_{n-2}& T_{n-1}
\end{array}
\right)\right).$$

The Tribonacci sequence has been generalized in many ways, for example, by changing the recurrence relation while preserving the initial terms, by altering the initial terms but maintaining the recurrence relation, by combining of these two techniques, and so on (for more details see \cite{Ge,Pe,Sha,Wa}).

In this paper, we consider one type of generalized Tribonacci sequences. In fact, the sequence $\{V_{n}\}_{n\geq0}$ defined by Shannon and Horadam \cite{Sha} depending on six positive integer parameters $V_{0}$, $V_{1}$, $V_{2}$, $r$, $s$ and $t$ used in the third-order linear recurrence relation:
\begin{equation}\label{eq:7}
V_{n}=rV_{n-1}+sV_{n-2}+tV_{n-3}\ \ (n\geq3).
\end{equation}

In this paper, as mentioned above, we provide quadratic approximation of $\{V_{n}\}_{n\geq0}$ and use this result to give the matrix form of the $n$-th power of a companion matrix of $\{V_{n}\}_{n\geq0}$. Then, we re-prove the Cassini-type formula for the sequence $\{V_{n}\}_{n\geq0}$ and Binet's formula of the generalized Tribonacci quaternions.

\section{Quadratic approximation of the generalized Tribonacci sequences $\{V_{n}\}$}
\setcounter{equation}{0}

We consider the generalized Tribonacci sequence $\{V_{n}(V_{0},V_{1},V_{2};r,s,t)\}$, or briefly $\{V_{n}\}$, defined as in (\ref{eq:7}), where $V_{0}$, $V_{1}$, $V_{2}$ are arbitrary integers and $r$, $s$, $t$, are real numbers. This sequence has been studied by Shannon and Horadam \cite{Sha}, Yalavigi \cite{Ya} and Pethe \cite{Pe}. If we set $r=s=t=1$ and $V_{0}=V_{1}=0$, $V_{2}=1$, then $\{V_{n}\}$ is the well-known Tribonacci sequence which has been considered extensively (see, for example, \cite{Fe}).

As the elements of this Tribonacci-type number sequence provide third order iterative relation, its characteristic equation is $x^{3}-rx^{2}-sx-t=0$, whose roots are $\alpha=\alpha(r,s,t)=\frac{r}{3}+A_{V}+B_{V}$, $\omega_{1}=\frac{r}{3}+\epsilon A_{V}+\epsilon^{2} B_{V}$ and $\omega_{2}=\frac{r}{3}+\epsilon^{2}A_{V}+\epsilon B_{V}$, where $$A_{V}=\sqrt[3]{\frac{r^{3}}{27}+\frac{rs}{6}+\frac{t}{2}+\sqrt{\Delta}},\ B_{V}=\sqrt[3]{\frac{r^{3}}{27}+\frac{rs}{6}+\frac{t}{2}-\sqrt{\Delta}},$$ with $\Delta=\Delta(r,s,t)=\frac{r^{3}t}{27}-\frac{r^{2}s^{2}}{108}+\frac{rst}{6}-\frac{s^{3}}{27}+\frac{t^{2}}{4}$ and $\epsilon=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$.

In this paper, $\Delta(r,s,t)>0$, then the cubic equation $x^{3}-rx^{2}-sx-t=0$ has one real and two nonreal solutions, the latter being conjugate complex. Thus, the Binet formula for the generalized Tribonacci numbers can be expressed as:
\begin{equation}\label{eq:8}
V_{n}=\frac{P\alpha^{n}}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{Q\omega_{1}^{n}}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}+\frac{R\omega_{2}^{n}}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})},
\end{equation}
where $P=V_{2}-(\omega_{1}+\omega_{2})V_{1}+\omega_{1}\omega_{2}V_{0}$, $Q=V_{2}-(\alpha+\omega_{2})V_{1}+\alpha\omega_{2}V_{0}$ and $R=V_{2}-(\alpha+\omega_{1})V_{1}+\alpha\omega_{1}V_{0}$.

In fact, the generalized Tribonacci sequence is the generalization of the well-known sequences like Tribonacci, Padovan, Narayana and third-order Jacobsthal. For example, $\{V_{n}(0,0,1;1,1,1)\}_{n\geq0}$, $\{V_{n}(0,1,0;0,1,1)\}_{n\geq0}$, are Tribonacci and Padovan sequences, respectively. The Binet formula for the generalized Tribonacci sequence is expressed as follows:
\begin{lemma}
The Binet formula for the generalized Tribonacci sequence $\{V_{n}\}_{n\geq0}$ is:
\begin{equation}\label{eq:9}
V_{n}=V_{2}U_{n}+(sV_{1}+tV_{0})U_{n-1}+tV_{1}U_{n-2},
\end{equation}
and
\begin{equation}\label{eq:10}
U_{n}=\frac{\alpha^{n}}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{\omega_{1}^{n}}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}+\frac{\omega_{2}^{n}}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})},
\end{equation}
where $\alpha$, $\omega_{1}$ and $\omega_{2}$ are the roots of the cubic equation $x^{3}-rx^{2}-sx-t=0$.
\end{lemma}
\begin{proof}
The validity of this formula can be confirmed using the recurrence relation. Furthermore, $\{U_{n}\}_{n\geq0}=\{V_{n}(0,0,1;r,s,t)\}_{n\geq0}$.
\end{proof}

In \cite{Sha}, using an inductive argument, authors give the matrix form of the $n$-th power of a companion matrix $M=\left(
\begin{array}{ccc}
r& s& t \\
1& 0 & 0 \\
0 & 1& 0
\end{array}
\right)$ of $\{V_{n}\}_{n\geq0}$
\begin{equation}\label{eq:11}
\left(
\begin{array}{c}
V_{n+2} \\
V_{n+1} \\
V_{n}
\end{array}
\right)=\left(
\begin{array}{ccc}
r& s& t \\
1& 0 & 0 \\
0 & 1& 0
\end{array}
\right)^{n}\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)
\end{equation}
and
\begin{equation}\label{eq:12}
\left(
\begin{array}{ccc}
r& s& t \\
1& 0 & 0 \\
0 & 1& 0
\end{array}
\right)^{n}=\left(
\begin{array}{ccc}
U_{n+2}& sU_{n+1}+tU_{n}& tU_{n+1} \\
U_{n+1}&sU_{n}+tU_{n-1}& tU_{n} \\
U_{n} & sU_{n-1}+tU_{n-2}& tU_{n-1}
\end{array}
\right)\ (n\geq2),
\end{equation}
where $U_{n}$ is defined by Eq. (\ref{eq:10}).

And then give the Cassini-type identity for $\{U_{n}\}$ by taking determinant both sides of the matrix form Eq. (\ref{eq:12})
\begin{equation}\label{eq:13}
U_{n}^{3}+U_{n-1}^{2}U_{n+2}+U_{n-2}U_{n+1}^{2}-2U_{n-1}U_{n}U_{n+1}-U_{n-2}U_{n}U_{n+2}=t^{n-2},
\end{equation}
for $n\geq2$.

More generally,
\begin{equation}\label{eq:14}
\left(
\begin{array}{ccc}
V_{n+4}& V_{n+3}+V_{n+2}& V_{n+3} \\
V_{n+3}& V_{n+2}+V_{n+1}& V_{n+2} \\
V_{n+2} & V_{n+1}+V_{n}& V_{n+1}
\end{array}
\right) =M^{n}\left(
\begin{array}{ccc}
V_{4}& V_{3}+V_{2}& V_{3} \\
V_{3}& V_{2}+V_{1}& V_{2} \\
V_{2} & V_{1}+V_{0}& V_{1}\end{array}
\right),
\end{equation}
for $n\geq0$. Or equivalently, we can write the Cassini-type identity for $\{V_{n}\}$ by taking determinant both sides of the matrix form Eq. (\ref{eq:14})
\begin{equation}\label{eq:15}
V_{n+2}^{3}+V_{n+1}^{2}V_{n+4}+V_{n}V_{n+3}^{2}-V_{n+2}(2V_{n+1}V_{n+3}+V_{n}V_{n+4})=t^{n}g(0),
\end{equation}
where $g(0)=V_{2}^{3}+V_{1}^{2}V_{4}+V_{0}V_{3}^{2}-V_{2}(2V_{1}V_{3}+V_{0}V_{4})$.

Now, we give the quadratic approximation of $\{V_{n}\}$ and then use this result to obtain the matrix form Eq. (\ref{eq:14}).
\begin{theorem}\label{teoP}
Let $\{V_{n}\}_{n\geq0}$, $\alpha$, $\omega_{1}$ and $\omega_{2}$ be as above. Then, we have for all integer $n\geq 0$
\begin{equation}\label{eq:16}
\textrm{Quadratic app. of $\{V_{n}\}$}: \left\{
\begin{array}{c }
P\alpha^{n+2}=\alpha^{2} V_{n+2}+\alpha(sV_{n+1}+tV_{n})+ tV_{n+1},\\
Q\omega_{1}^{n+2}=\omega_{1}^{2} V_{n+2}+\omega_{1}(sV_{n+1}+tV_{n})+ tV_{n+1},\\
R\omega_{2}^{n+2}=\omega_{2}^{2} V_{n+2}+\omega_{2}(sV_{n+1}+tV_{n})+ tV_{n+1},
\end{array}
\right.
\end{equation}
where $P=V_{2}-(\omega_{1}+\omega_{2})V_{1}+\omega_{1}\omega_{2}V_{0}$, $Q=V_{2}-(\alpha+\omega_{2})V_{1}+\alpha\omega_{2}V_{0}$ and $R=V_{2}-(\alpha+\omega_{1})V_{1}+\alpha\omega_{1}V_{0}$.
\end{theorem}
\begin{proof}
Using the Binet's formula Eq. (\ref{eq:8}), we have
\begin{align*}
\alpha V_{n+2}&+(s+\omega_{1}\omega_{2})V_{n+1}+tV_{n}\\
&=\frac{P\alpha^{n}(\alpha^{3}+(s+\omega_{1}\omega_{2})\alpha+t)}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{Q\omega_{1}^{n}(\alpha\omega_{1}^{2}+(s+\omega_{1}\omega_{2})\omega_{1}+t)}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}\\
&\ \ +\frac{R\omega_{2}^{n}(\alpha\omega_{2}^{2}+(s+\omega_{1}\omega_{2})\omega_{2}+t)}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})}\\
&=(V_{2}-(\omega_{1}+\omega_{2})V_{1}+\omega_{1}\omega_{2}V_{0})\alpha^{n+1},
\end{align*}
the latter given that $\alpha\omega_{1}^{2}+(s+\omega_{1}\omega_{2})\omega_{1}+t=0$ and $\alpha\omega_{2}^{2}+(s+\omega_{1}\omega_{2})\omega_{2}+t=0$. Then, we get
\begin{equation}\label{eq:17}
\alpha V_{n+2}+(s+\omega_{1}\omega_{2})V_{n+1}+tV_{n}=(V_{2}-(\omega_{1}+\omega_{2})V_{1}+\omega_{1}\omega_{2}V_{0})\alpha^{n+1}.
\end{equation}
Multiplying Eq. (\ref{eq:17}) by $\alpha$ and using $\alpha\omega_{1}\omega_{2}=t$, we have
\begin{align*}
P\alpha^{n+2}&=\alpha^{2}V_{n+2}+\alpha(s+\omega_{1}\omega_{2})V_{n+1}+\alpha tV_{n}\\
&=\alpha^{2} V_{n+2}+\alpha(sV_{n+1}+tV_{n})+ tV_{n+1},
\end{align*}
where $P=V_{2}-(\omega_{1}+\omega_{2})V_{1}+\omega_{1}\omega_{2}V_{0}$. If we change $\alpha$, $\omega_{1}$ and $\omega_{2}$ role above process, we obtain the desired result Eq. (\ref{eq:16}).
\end{proof}

We can re-prove equations Eq. (\ref{eq:14}) and Eq. (\ref{eq:15}) by using the above quadratic approximation of $\{V_{n}\}$ in Eq. (\ref{eq:16}).
\begin{corollary}
Let $M=\left(
\begin{array}{ccc}
r& s& t \\
1& 0 & 0 \\
0 & 1& 0
\end{array}
\right)$ be a companion matrix of $\{V_{n}\}$. Then the matrix form of the $n$-th power $M^{n}$ is given by Eq. (\ref{eq:14}) and the Cassini-type formula for $\{V_{n}\}$ is given by Eq. (\ref{eq:15}).
\end{corollary}
\begin{proof}
In Eq. (\ref{eq:16}), if we change $\alpha$, $\omega_{1}$ and $\omega_{2}$ into the matrix $M$ and change $tV_{n+1}$ into the matrix $tV_{n+1}I$, then we have
\begin{equation}\label{eq:18}
M^{n}(V_{2}M^{2}+(sV_{1}+tV_{0})M+tV_{1}I)=V_{n+2}M^{2}+(sV_{n+1}+tV_{n})M+tV_{n+1}I.
\end{equation}
In fact, Eq. (\ref{eq:18}) holds for the following reason: Since $$M\left(
\begin{array}{c}
V_{n+1} \\
V_{n} \\
V_{n-1}
\end{array}
\right)=\left(
\begin{array}{c}
V_{n+2} \\
V_{n+1} \\
V_{n}
\end{array}
\right)\ \textrm{and}\ M^{n}\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)=\left(
\begin{array}{c}
V_{n+2} \\
V_{n+1} \\
V_{n}
\end{array}
\right),$$ we have
\begin{align*}
M^{n}(V_{2}M^{2}&+(sV_{1}+tV_{0})M+tV_{1}I)\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)\\
&=V_{2}M^{n+2}\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)+(sV_{1}+tV_{0})M^{n+1}\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)+tV_{1}M^{n}\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)\\
&=V_{2}\left(
\begin{array}{c}
V_{n+4} \\
V_{n+3} \\
V_{n+2}
\end{array}
\right)+(sV_{1}+tV_{0})\left(
\begin{array}{c}
V_{n+3} \\
V_{n+2} \\
V_{n+1}
\end{array}
\right)+tV_{1}\left(
\begin{array}{c}
V_{n+2} \\
V_{n+1} \\
V_{n}
\end{array}
\right)\\
&=\left(
\begin{array}{c}
V_{2}V_{n+4}+(sV_{1}+tV_{0})V_{n+3}+tV_{1}V_{n+2} \\
V_{2}V_{n+3}+(sV_{1}+tV_{0})V_{n+2}+tV_{1}V_{n+1} \\
V_{2}V_{n+2}+(sV_{1}+tV_{0})V_{n+1}+tV_{1}V_{n}
\end{array}
\right).
\end{align*}
Using $V_{n+3}=rV_{n+2}+sV_{n+1}+tV_{n}$ and $V_{n+4}=(r^{2}+s)V_{n+2}+(rs+t)V_{n+1}+rtV_{n}$, we have
\begin{align*}
M^{n}(V_{2}M^{2}&+(sV_{1}+tV_{0})M+tV_{1}I)\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right)\\
&=\left(
\begin{array}{c}
V_{4}V_{n+2}+(sV_{n+1}+tV_{n})V_{3}+tV_{n+1}V_{2} \\
V_{3}V_{n+2}+(sV_{n+1}+tV_{n})V_{2}+tV_{n+1}V_{1} \\
V_{2}V_{n+2}+(sV_{n+1}+tV_{n})V_{1}+tV_{n+1}V_{0}
\end{array}
\right)\\
&=(V_{n+2}M^{2}+(sV_{n+1}+tV_{n})M+tV_{n+1}I)\left(
\begin{array}{c}
V_{2} \\
V_{1} \\
V_{0}
\end{array}
\right).
\end{align*}
Thus from Eq. (\ref{eq:18}) we have
\begin{align*}
M^{n}(V_{2}M^{2}+(sV_{1}+&tV_{0})M+tV_{1}I)=M^{n}\left(
\begin{array}{ccc}
V_{4}&sV_{3}+tV_{2}&tV_{3} \\
V_{3}&sV_{2}+tV_{1}&tV_{2}\\
V_{2}&sV_{1}+tV_{0}&tV_{1}
\end{array}
\right)\\
&=M^{n}\left(
\begin{array}{ccc}
V_{4}&V_{3}+V_{2}&V_{3} \\
V_{3}&V_{2}+V_{1}&V_{2}\\
V_{2}&V_{1}+V_{0}&V_{1}
\end{array}
\right)\left(
\begin{array}{ccc}
1&0&0 \\
0&t&0\\
0&s-t&t
\end{array}
\right)
\end{align*}
and
\begin{align*}
V_{n+2}M^{2}+(sV_{n+1}&+tV_{n})M+tV_{n+1}I\\
&=\left(
\begin{array}{ccc}
V_{n+4}&sV_{n+3}+tV_{n+2}&tV_{n+3} \\
V_{n+3}&sV_{n+2}+tV_{n+1}&tV_{n+2}\\
V_{n+2}&sV_{n+1}+tV_{n}&tV_{n+1}
\end{array}
\right)\\
&=\left(
\begin{array}{ccc}
V_{n+4}&V_{n+3}+V_{n+2}&V_{n+3} \\
V_{n+3}&V_{n+2}+V_{n+1}&V_{n+2}\\
V_{n+2}&V_{n+1}+V_{n}&V_{n+1}
\end{array}
\right)\left(
\begin{array}{ccc}
1&0&0 \\
0&t&0\\
0&s-t&t
\end{array}
\right).
\end{align*}
Since the matrix $\left(
\begin{array}{ccc}
1&0&0 \\
0&t&0\\
0&s-t&t
\end{array}
\right)$ is invertible, we obtain the desired result Eq. (\ref{eq:14}) and by taking determinant both sides of the matrix form Eq. (\ref{eq:14}) we obtain the desired result Eq. (\ref{eq:15}). The proof is completed.
\end{proof}

\begin{remark}
For some positive integer $k$, if $r=k$, $s=0$ and $t=1$, then $\{U_{n}\}$ is the $k$-Narayana sequence $\{b_{k,n}\}$ (for more details see \cite{Ra}).

It is well known that the usual $k$-Narayana numbers can be expressed using Binet's formula
\begin{equation}\label{eq:Nara}
\begin{aligned}
b_{k,n}&=\frac{\alpha_{k}^{n}}{(\alpha_{k}-\omega_{k,1})(\alpha_{k}-\omega_{k,2})}-\frac{\omega_{k,1}^{n}}{(\alpha_{k}-\omega_{k,1})(\omega_{k,1}-\omega_{k,2})}\\
&\ \ +\frac{\omega_{k,2}^{n}}{(\alpha_{k}-\omega_{k,2})(\omega_{k,1}-\omega_{k,2})}.
\end{aligned}
\end{equation}
where $\alpha_{k}$, $\omega_{k,1}$ and $\omega_{k,2}$ are the roots of the cubic equation $x^{3}-kx^{2}-1=0$. Furthermore, $\alpha_{k}=\frac{k}{3}+A_{b}+B_{b}$, $\omega_{k,1}=\frac{k}{3}+\epsilon A_{b}+\epsilon^{2} B_{b}$ and $\omega_{k,2}=\frac{k}{3}+\epsilon^{2}A_{b}+\epsilon B_{b}$, where $$A_{b}=\sqrt[3]{\frac{k^{3}}{27}+\frac{1}{2}+\sqrt{\frac{k^{3}}{27}+\frac{1}{4}}},\ B_{b}=\sqrt[3]{\frac{k^{3}}{27}+\frac{1}{2}-\sqrt{\frac{k^{3}}{27}+\frac{1}{4}}},$$ and $\epsilon=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$.

From the Binet's formula Eq. (\ref{eq:Nara}), using the identities $\alpha_{k}+\omega_{k,1}+\omega_{k,2}=k$, $\alpha_{k}\omega_{k,1}+\alpha_{k}\omega_{k,2}+\omega_{k,1}\omega_{k,2}=0$, we have for any integer $n\geq2$:

\begin{align*}
\alpha_{k} b_{k,n+2}&+(\omega_{k,1}\omega_{k,2})b_{k,n+1}+b_{k,n}\\
&=\frac{\alpha_{k}^{n}(\alpha_{k}^{3}+(\omega_{k,1}\omega_{k,2})\alpha_{k}+1)}{(\alpha_{k}-\omega_{k,1})(\alpha_{k}-\omega_{k,2})}-\frac{\omega_{k,1}^{n}(\alpha_{k}\omega_{k,1}^{2}+(\omega_{k,1}\omega_{k,2})\omega_{k,1}+1)}{(\alpha_{k}-\omega_{k,1})(\omega_{k,1}-\omega_{k,2})}\\
&\ \ +\frac{\omega_{k,2}^{n}(\alpha_{k}\omega_{k,2}^{2}+(\omega_{k,1}\omega_{k,2})\omega_{k,2}+1)}{(\alpha_{k}-\omega_{k,2})(\omega_{k,1}-\omega_{k,2})}\\
&=\alpha_{k}^{n+1},
\end{align*}
the latter given that $\alpha_{k}\omega_{k,1}^{2}+(\omega_{k,1}\omega_{k,2})\omega_{k,1}+1=0$ and $\alpha_{k}\omega_{k,2}^{2}+(\omega_{k,1}\omega_{k,2})\omega_{k,2}+1=0$. Then, we get
\begin{equation}\label{eq:new}
\alpha_{k} b_{k,n+2}+(\omega_{k,1}\omega_{k,2})b_{k,n+1}+b_{k,n}=\alpha_{k}^{n+1}.
\end{equation}

Multipling Eq. (\ref{eq:new}) by $\alpha_{k}$, using $\alpha_{k}\omega_{k,1}\omega_{k,2}=1$, and if we change $\alpha_{k}$, $\omega_{k,1}$ and $\omega_{k,2}$ role above process, we obtain the quadratic approximation of $\{b_{k,n}\}_{n\geq0}$
\begin{equation}\label{eq:6}
\textrm{Quadratic app. of $\{b_{k,n}\}$}: \left\{
\begin{array}{c }
\alpha_{k}^{n}=b_{k,n}\alpha_{k}^{2}+b_{k,n-2}\alpha+b_{k,n-1},\\
\omega_{k,1}^{n}=b_{k,n}\omega_{k,1}^{2}+b_{k,n-2}\omega_{k,1}+b_{k,n-1},\\
\omega_{k,2}^{n}=b_{k,n}\omega_{k,2}^{2}+b_{k,n-2}\omega_{k,2}+b_{k,n-1},
\end{array}
\right.
\end{equation}
where $\alpha_{k}$, $\omega_{k,1}$ and $\omega_{k,2}$ are the roots of the cubic equation $x^{3}-x^{2}-x-1=0$.

In Eq. (\ref{eq:6}), if we change $\alpha$, $\omega_{1}$ and $\omega_{2}$ into the companion matrix $Q$ and change $T_{n-1}$ into the matrix $T_{n-1}I$, where $I$ is the $3\times 3$ identity matrix, then we obtain the matrix form Eq. (\ref{eq:2}) of $Q^{n}$
$$Q^{n}=b_{k,n}Q^{2}+b_{k,n-2}Q+b_{k,n-1}I\left(=\left(
\begin{array}{ccc}
b_{k,n+2}& b_{k,n}& b_{k,n+1} \\
b_{k,n+1}& b_{k,n-1}& b_{k,n} \\
b_{k,n} & b_{k,n-2}& b_{k,n-1}
\end{array}
\right)\right),$$
where $Q=\left(
\begin{array}{ccc}
k& 0& 1 \\
1& 0& 0\\
0 & 1& 0
\end{array}
\right)$.
\end{remark}

The next corollary gives an alternative proof of the Binet's formula for the generalized Tribonacci quaternions (see \cite[Theorem 2.1]{Ce1}).
\begin{corollary}
For any integer $n\geq 0$, the $n$-th generalized Tribonacci quaternion is
\begin{equation}\label{binQ}
Q_{V,n}=\frac{P\underline{\alpha}\alpha^{n}}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{Q\underline{\omega_{1}}\omega_{1}^{n}}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}+\frac{R\underline{\omega_{2}}\omega_{2}^{n}}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})},
\end{equation}
where $P$, $Q$ and $R$ as in Eq. (\ref{eq:8}), $\underline{\alpha}=1+\alpha \textbf{i}+\alpha^{2} \textbf{j}+\alpha^{3} \textbf{k}$ and $\underline{\omega_{1,2}}=1+\omega_{1,2} \textbf{i}+\omega_{1,2}^{2} \textbf{j}+\omega_{1,2}^{3} \textbf{k}$. If $V_{0}=V_{1}=0$ and $V_{2}=1$, we get the classic Tribonacci quaternion.
\end{corollary}
\begin{proof}
For the Eq. (\ref{eq:16}), we have
\begin{align*}
&\alpha^{2} Q_{V,n+2}+\alpha \left( sQ_{V,n+1}+tQ_{V,n}\right) +tQ_{V,n+1}\\
&=\alpha^{2} \left(V_{n+2}+V_{n+3}\textbf{i}+V_{n+4} \textbf{j}+V_{n+5}\textbf{k}\right)\\
&\ \ +\alpha \left(sV_{n+1}+tV_{n} +(sV_{n+2}+tV_{n+1})\textbf{i}+(sV_{n+3}+tV_{n+2})\textbf{j}+(sV_{n+4}+tV_{n+3})\textbf{k}\right)\\
&\ \ +t \left(V_{n+1}+V_{n+2}\textbf{i}+V_{n+3} \textbf{j}+V_{n+4}\textbf{k}\right)\\
&=\alpha^{2} V_{n+2}+\alpha \left( sV_{n+1}+tV_{n}\right) +tV_{n+1}+\left(\alpha^{2} V_{n+3}+\alpha \left( sV_{n+2}+tV_{n+1}\right) +tV_{n+2}\right)\textbf{i}\\
&\ \ + \left(\alpha^{2} V_{n+4}+\alpha \left( sV_{n+3}+tV_{n+2}\right) +tV_{n+3}\right)\textbf{j}\\
&\ \ +\left(\alpha^{2} V_{n+5}+\alpha \left( sV_{n+4}+tV_{n+3}\right) +tV_{n+4}\right)\textbf{k}.
\end{align*}

From the identity $P\alpha^{n+2}=\alpha^{2} V_{n+2}+\alpha(sV_{n+1}+tV_{n})+ tV_{n+1}$ in Theorem \ref{teoP}, we obtain
\begin{equation}\label{ge1}
\alpha^{2} Q_{V,n+2}+\alpha \left( sQ_{V,n+1}+tQ_{V,n}\right) +tQ_{V,n+1}=P\underline{\alpha}\alpha^{n+2}.
\end{equation}
Similarly, we have
\begin{equation}\label{ge2}
\omega_{1}^{2} Q_{V,n+2}+\omega_{1} \left( sQ_{V,n+1}+tQ_{V,n}\right) +tQ_{V,n+1}=Q\underline{\omega_{1}}\omega_{1}^{n+2},
\end{equation}
\begin{equation}\label{ge3}
\omega_{2}^{2} Q_{V,n+2}+\omega_{2} \left( sQ_{V,n+1}+tQ_{V,n}\right) +tQ_{V,n+1}=R\underline{\omega_{2}}\omega_{2}^{n+2}.
\end{equation}
Subtracting Eq. (\ref{ge2}) from Eq. (\ref{ge1}) gives
\begin{equation}\label{d1}
(\alpha+\omega_{1}) Q_{V,n+2}+\left( sQ_{V,n+1}+tQ_{V,n}\right) =\frac{P\underline{\alpha}\alpha^{n+2}-Q\underline{\omega_{1}}\omega_{1}^{n+2}}{\alpha-\omega_{1}}.
\end{equation}
Similarly, subtracting Eq. (\ref{ge3}) from Eq. (\ref{ge1}) gives
\begin{equation}\label{d2}
(\alpha+\omega_{2}) Q_{V,n+2}+\left( sQ_{V,n+1}+tQ_{V,n}\right) =\frac{P\underline{\alpha}\alpha^{n+2}-R\underline{\omega_{2}}\omega_{2}^{n+2}}{\alpha-\omega_{2}}.
\end{equation}
Finally, subtracting Eq. (\ref{d2}) from Eq. (\ref{d1}), we obtain
\begin{align*}
Q_{V,n+2}&=\frac{1}{\omega_{1}-\omega_{2}}\left( \frac{P\underline{\alpha}\alpha^{n+2}-Q\underline{\omega_{1}}\omega_{1}^{n+2}}{\alpha-\omega_{1}}-\frac{P\underline{\alpha}\alpha^{n+2}-R\underline{\omega_{2}}\omega_{2}^{n+2}}{\alpha-\omega_{2}}\right)\\
&=\frac{P\underline{\alpha}\alpha^{n+2}}{(\alpha-\omega_{1})(\alpha-\omega_{2})}-\frac{Q\underline{\omega_{1}}\omega_{1}^{n+2}}{(\alpha-\omega_{1})(\omega_{1}-\omega_{2})}+\frac{R\underline{\omega_{2}}\omega_{2}^{n+2}}{(\alpha-\omega_{2})(\omega_{1}-\omega_{2})}.
\end{align*}
 So, the corollary is proved.
\end{proof}

config: arxiv


TITLE: To prove or give counterexample that $R \subset Dom(R) \times Ran(R)$
QUESTION [0 upvotes]: To prove or give counterexample that $R \subset Dom(R) \times Ran(R)$
Let (a,b) \in R. So $a \in A$ and $b \in B$ where $R :A \rightarrow B$. Since Dom(R) is asubset of A. Let assume that $a \not \in Dom(R) $. So (a,b) doesnot belong to $ Dom(R) \times Ran(R)$
Is this correct ? Thanks for help

REPLY [3 votes]: If $(a,b)\in R$ then $a\in Dom(R)$ by the definition of "domain" and $b \in Ran(R)$ by the definition of "range", so $(a,b)\in Dom(R) \times Ran(R)$, so $R \subseteq Dom(R) \times Ran(R)$. QED

REPLY [1 votes]: I can see that you're trying to use a proof by contradiction, but it isn't very clear what you're trying to contradict. It's easier to use a direct proof, and it would look something like this:

Proof. Let $R:A\rightarrow B$ be a mapping from $A$ to $B$ where $R=\{a\in A, b\in B: R(a)=b\}$.
Let $(a,b)\in R$. Then by definition, $R(a)=b$, so $a$ is in the domain of $R$ and $b$ is in the range of $R$. So $(a,b)\in$ Dom$(R)\times$Dom$(R)$. Therefore $R\subseteq$ Dom$(R)\times$Dom$(R)$.


TITLE: Show that $(A ∩ B) ▵ C = (A ▵ C) ▵ (A \setminus B)$
QUESTION [2 upvotes]: I want to show the following equality (using logical connectives, not venn diagrams)
Show that:
$$(A ∩ B) ▵ C = (A ▵ C) ▵ (A \setminus B)$$
$A ▵ B$ is defined as: $(A ∪ B) \setminus (A ∩ B)$
My attempt:
If I expand the symmetric difference symbol $▵$ in the original equality I get:
$$((A ∩ B) ∪ C) \setminus ((A ∩ B) ∩ C) =$$
$$(((A ∪ C) \setminus (A ∩ C)) ∪ (A \setminus B)) \setminus (((A ∪ C) \setminus (A ∩ C)) ∩ (A \setminus B))$$
So now I want to write the second statement in the equality with logical connectives and reduce it to the first one:
$$((A \land B) \lor C) \land \lnot ((A \land B) \land C) =$$
$$(((A \lor C) \land \lnot (A \land C)) \lor (A \land \lnot B)) \land \lnot (((A \lor C) \land \lnot (A \land C)) \land (A \land \lnot B)) =$$
But I am stuck here. I don't really know where to begin...

REPLY [3 votes]: When things are finite and you are in doubt, resort to brute-force (though, this isn't actually good advice...)


TITLE: Confusion of proof on nowhere dense sets
QUESTION [3 upvotes]: I am reading the book `Measure and Category' by Oxtoby. In the first chapter, they focus solely on the real line. I am confused by the following proof:

I understand that, since $ A_1, A_2 $ are nowhere dense, given an interval $ I $, both sets are not dense on $ I $.
Therefore, there are subintervals $ I_1 ,I_2 $ that lie on the complement of $ A_1, A_2$.
The proof then says `Hence $ I_2 \subset I -(A_1 \cup A_2) $. I am struggling to see how that follows from the previous sentence. Surely, $  I -(A_1 \cup A_2) $ need not contain $  I -A_1 $?

REPLY [2 votes]: To complement the answers here, I'd like to add an alternative proof to the statement in question following one of the comments. We do not require the full generality of Baire's theorem, but only the fact that $\mathbb{R}$ is separable.  One equivalence to saying that a  $A$ is nowhere dense is that its closure has empty interior: $\operatorname{int}(\overline{A})=\emptyset$. Suppose $A$ and $B$ are nowhere dense and $U$ is any open set $U\subset \overline{A\cup B}=\overline{A}\cup\overline{B}$. Then
$\Big(\overline{A}\Big)^c \cap \Big(\overline{B}\Big)^c\subset U^c$. The sets $\Big(\overline{A}\Big)^c$ and $\Big(\overline{B}\Big)^c$ are both open and dense in $\mathbb{R}$. It follows that $\Big(\overline{A}\Big)^c \cap \Big(\overline{B}\Big)^c$ is open and dense in $\mathbb{R}$. Since $U^c$ is closed and contains a dense set, it follows that $U^c=\mathbb{R}$. Consequently $U=\emptyset$. This shows that $\overline{A\cup B}$ has empty interior, that is $A\cup B$ is also nowhere dense.


TITLE: Convergence of infinite product of cyclotomic polynomials
QUESTION [1 upvotes]: From dabbling with p-adics, I remember mentions of $\infty $ being the 'infinite prime'. So hence something like: $\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{d| p_\infty \\ d \neq 1}} \Phi_{d}(x) = \Phi_{p_\infty}(x) = \sum_{n=0}^{\infty -1} x^n$ would perhaps make some abstract sense in some formal power series something-or-other. However, if we instead claim that $\substack{n | \infty \\ n \in \mathbb{N}}$ we get  $\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{n=2}}^\infty \Phi_{n}(x)$, which would be appealing (though to my knowledge, simply untrue) if we work with p-adics etc. For reals, with for instance $x = \frac{1}{100} \in \mathbb{R}$, the RHS seems to hover around the 'expected asymptote' value of $\frac{100}{99}$ with increasing variance as $n$ increases. For the 2-adics, setting $x=2$ and taking the logarithm gives an infinite sum of infinite sums, with each $\log(\Phi_{n}(x))$ producing a sum starting with a 2-adic valuation given by the first non-zero power of $x$ in that polynomial and thence growing, and there are infinitely many of them with $x^1$ (e.g. all primes), so the sums would not converge (unless there could be cancellation in some space? dunno).
So after all these incoherent babble, I was wondering what is known about the convergence of the product of all cyclotomic polynomials with or without $\Phi_1(x)$ -- and whether the first or second representation of $\frac{x^\infty -1}{x - 1}$ make any sense.
Thanks!

REPLY [1 votes]: The infinite products $\prod_{n=1}^\infty \Phi_n(x)$ and $\prod_{n=2}^\infty \Phi_n(x)$ diverge for all $x \ne 0$.  In order to converge, you would certainly need $\Phi_n(x) \to 1$ as $n \to \infty$, but if $n$ is prime $\Phi_n(x) = 1 + x + x^2 + \ldots + x^{n-1} = 1 + x + O(x^2)$.
Note that for an infinite product "converge" means "converge to a nonzero limit".


TITLE: Linear Regression: Correlation between predictors and residuals
QUESTION [0 upvotes]: I am reading Chapter 3 from Elements of Statistical Learning. In the explanation for Forward Stagewise Regression and Least Angle Regression, the authors explain that reducing the correlation between the predictors and the residuals amounts to moving in the direction of the standard linear regression. I also read online that in the standard linear regression (minimizing least squares), the predictors are uncorrelated with the residuals. I am not able to prove this. I have tried the following so far:
$r_i = y_i - x_i^T\hat{\beta}$
$x_i^Tr_i = x_i^Ty_i - x_i^T\hat{\beta}$
The above expression does not evaluate to $0$ for the general case. Ideally, I think I should compute $E[rX]$. I don't know how to get an empirical formula for it, since r is a vector and X is a matrix. Could someone please help me out? I am really confused. Also, it would be helpful if someone could point me to a resource explaining these correlations in more detail. I know that they should approximately have a Gaussian distribution since they are an estimate of the noise.

REPLY [0 votes]: The fact that
$r^TX=0$ (as proved in @d.k.o.'s answer) means that the residual vector $r$ is orthogonal to every column in the design matrix $X$. This alone does not imply that the sample correlation (between $r$ and each column in $X$) is zero -- for that you need the residuals to sum to zero, which isn't always true. For example, in simple linear regression with no intercept, the residuals don't necessarily sum to zero.
A sufficient condition for the residuals to sum to zero is that a vector of all ones exists in the column space of $X$. This is the case for example, if there is an intercept term in your model.
To prove this last assertion: If a vector $\bf 1$ of all ones exists in the column space of $X$, then ${\bf 1} = Xv$ for some vector $v$. Then
$$\sum_i r_i=r^T{\bf 1}=(y-Hy)^TXv=y^T\underbrace{(I-H)X}_0v=0,$$
where we abbreviate $H:=X(X^TX)^{-1}X^T$ and use the facts $H^T=H$ and $HX=X$.


TITLE: Can a function have uncountably many jump discontinuities?
QUESTION [2 upvotes]: I understand the proof that shows that if $f$ is monotone on an interval, then it has at most countably many jump discontinuities. I feel like a similar idea also allows us to show that any function can have at most countably many jump discontinuities. Here is my attempt at a proof.
If a jump continuity of $f$ exists at $x$, this means that $f(x^-)$ and $f(x^+)$ exist, but $f(x^-)\neq f(x^+)$. The existence of left and right hand limits means that for any $\epsilon > 0$, we can find a $\delta > 0$ such that if $y\in (x,x+\delta)$, then $|f(x^+) - f(y)| < \epsilon $ and if $y\in (x-\delta, x)$, then $|f(x^-)-f(y)| < \epsilon$. Thus, the only discontinuity (jump or otherwise) in the interval $(x-\delta, x+\delta)$ is at $x$.
I believe that this means that the jump discontinuities only exist within disjoint open intervals, of which there are at most countably many. If there does appear to be an issue at the boundary of the interval, for example, in a case where $x<y$ are both jump discontinuities and $x+\delta_x = y-\delta_y$, then we can make $\delta_x$ or $\delta_y$ a bit smaller to prevent this overlap.
Is this proof correct, or am I missing something? This statement seems a bit strong, especially since I've only heard a similar statement for monotone functions on an interval. If it's not true, can someone give a counterexample?

REPLY [0 votes]: Your assertion that the only discontinuity in $(x-\delta, x+\delta)$ is at $x$ is incorrect. Consider for example the function $f$ such that $f(x)=x$ on $\mathbb{Q}$ and $f(x)=1-x$ on $\mathbb{R} \backslash \mathbb{Q}$. It is discontinuous everywhere except $x=1/2$.
I'm not sure whether your assertion is correct, though.


TITLE: Proving equivalence relation for 7 | (3a + 4b)
QUESTION [1 upvotes]: I know this might be quite trivial, but I just can't seem to figure out how to prove $$R = \{(a,b) \in \mathbb{Z} \times \mathbb{Z} : 3a + 4b \text{ is divisible by } 7\}$$ is a symmetric relation, i.e., if $aRb$, then $bRa$, where $\mathbb{Z}$ is the set of integers

REPLY [1 votes]: Hint:
$(3a+4b)+(4a+3b)=7(a+b)$.

REPLY [1 votes]: Hint: $3a+4b=7c\implies3b+4a=7(a+b-c)$

REPLY [0 votes]: HINT
Assume $aRb$. Then $7 \mid (3a+4b)$. What can you say about $7\mid(4a+3b)$ instead?


TITLE: (Another) Multivariable delta-epsilon proof
QUESTION [0 upvotes]: Q: Find the limit if it exists
 $$ \lim \limits_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2} $$
What I have done so far:

Setting y = 0 & x = 0 I find that along these axes it looks like the limit will approach $0$ therefore:
$$ \lim \limits_{(x,y) \to (0,0)} \frac{x^3 + y^3}{x^2 + y^2} = 0$$

$$\mid \frac{x^3 + y^3}{x^2 + y^2} - 0 \mid < \epsilon  \quad \quad \quad 0 < \sqrt{x^2 + y^2} < \delta $$
Now since $ x^2 + y^2 \geq 0 $ we cans simplify to:
$$ \mid x^3 + y^3 \mid \leq \sqrt{ (x^3 + y^3)^2 } = \quad .... $$
I do not know how to proceed; how do I modify the $\sqrt{ (x^3 + y^3)^2 } $ into a $\sqrt{x^2 + y^2} $ ?

REPLY [1 votes]: For both $x,y\ne 0$, then
\begin{align*}
\dfrac{|x|^{3}+|y|^{3}}{x^{2}+y^{2}}&=\dfrac{|x|^{3}}{x^{2}+y^{2}}+\dfrac{|y|^{3}}{x^{2}+y^{2}}\\
&\leq\dfrac{|x|^{3}}{x^{2}}+\dfrac{|y|^{3}}{y^{2}}\\
&=|x|+|y|\\
&\leq\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+y^{2}}\\
&=2\sqrt{x^{2}+y^{2}},
\end{align*}
the above inequality still holds for either $x=0$ or $y=0$, so we conclude that
\begin{align*}
\dfrac{|x|^{3}+|y|^{3}}{x^{2}+y^{2}}\leq 2\sqrt{x^{2}+y^{2}},~~~~(x,y)\ne(0,0).
\end{align*}


TITLE: How to solve this equation: $\frac{\sqrt[5]{x^3 \sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$
QUESTION [3 upvotes]: Please, help me to solve this equation:
$$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$$
I tried to shorten fraction, but I get very weird numbers like
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$$ and I'm stuck there :(

REPLY [3 votes]: If $x\gt 0$, then:

$\sqrt[3]{x^{-2}} = x^{-2/3}$.
$x\sqrt[3]{x^{-2}} = xx^{-2/3} = x^{1/3}$.
$\sqrt{x\sqrt[3]{x^{-2}}} = x^{1/6}$.
$x^3\sqrt{x\sqrt[3]{x^{-2}}} = x^{19/6}$.
$\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}} = x^{19/30}$.
$x\sqrt[3]{x} = xx^{1/3} = x^{4/3}$.
$\sqrt[4]{x\sqrt[3]{x}} = x^{1/3}$.

So the quotient is equal to $x^{19/30}/x^{1/3} = x^{(19/30)-(1/3)} = x^{9/30}$. Your equation is then equivalent to
$$x^{9/30} = 3$$
which can be solved by raising both sides to the $30/9$.
If $x\lt 0$, then you need to throw in a few absolute values, since for example, $\sqrt{x\sqrt[3]{x^{-2}}} = |x|^{1/6}$, instead of $x^{1/6}$.