# Linear maps preserving $\ell_{p}$-norm parallel vectors

Chi-Kwong Li, Ming-Cheng Tsai, Ya-Shu Wang and Ngai-Ching Wong  Department of
Mathematics, The College of William & Mary, Williamsburg, VA 13187, USA.
<EMAIL_ADDRESS>General Education Center, National Taipei University of
Technology, Taipei 10608, Taiwan<EMAIL_ADDRESS>Department of
Applied Mathematics, National Chung Hsing University, Taichung 40227, Taiwan.
<EMAIL_ADDRESS>Department of Applied Mathematics, National Sun Yat-sen
University, Kaohsiung, 80424, Taiwan; Department of Healthcare Administration
and Medical Information, Kaohsiung Medical University, 80708 Kaohsiung,
Taiwan<EMAIL_ADDRESS>
###### Abstract.

Two vectors $\mathbf{x},\mathbf{y}$ in a normed vector space are parallel if
there is a scalar $\mu$ with $|\mu|=1$ such that
$\|\mathbf{x}+\mu\mathbf{y}\|=\|\mathbf{x}\|+\|\mathbf{y}\|$; they form a
triangle equality attaining (TEA) pair if
$\|\mathbf{x}+\mathbf{y}\|=\|\mathbf{x}\|+\|\mathbf{y}\|$. In this paper, we
characterize linear maps on ${\mathbb{F}}^{n}={\mathbb{R}}^{n}$ or
${\mathbb{C}}^{n}$, equipped with the $\ell_{p}$-norm for $p\in[1,\infty]$,
preserving parallel pairs or preserving TEA pairs. Indeed, any linear map will
preserve parallel pairs and TEA pairs when $1<p<\infty$. For the
$\ell_{1}$-norm, TEA preservers form a semigroup of matrices in which each row
has at most one nonzero entries; adding rank one matrices to this semigroup
will be the semigroup of parallel preserves. For the $\ell_{\infty}$-norm, a
nonzero TEA preserver, or a parallel preserver of rank greater than one, is
always a multiple of an $\ell_{\infty}$-norm isometry, except when
${\mathbb{F}}^{n}={\mathbb{R}}^{2}$. We also have a characterization for the
exceptional case. The results are extended to linear maps of the infinite
dimensional spaces $\ell_{1}(\Lambda)$, $c_{0}(\Lambda)$ and
$\ell_{\infty}(\Lambda)$.

###### Key words and phrases:

norm parallelism, parallel pair preservers, triangle equality attaining pair
preservers

###### 1991 Mathematics Subject Classification:

15A86, 15A60.

##  1\. Introduction

In a normed vector space $({\bf V},\|\cdot\|)$ over the real field
${\mathbb{F}}={\mathbb{R}}$ or the complex field ${\mathbb{F}}={\mathbb{C}}$,
two vectors $\mathbf{x},\mathbf{y}\in{\bf V}$ form a _parallel_ pair if

$\|\mathbf{x}+\mu\mathbf{y}\|=\|\mathbf{x}\|+\|\mathbf{y}\|\quad\hbox{ for
some }\mu\in{\mathbb{F}}\hbox{ with }|\mu|=1;$ (1.1)

the vectors $\mathbf{x},\mathbf{y}\in{\bf V}$ form a _triangle equality
attaining_ (_TEA_) pairs if

$\|\mathbf{x}+\mathbf{y}\|=\|\mathbf{x}\|+\|\mathbf{y}\|.$ (1.2)

We are interested in those linear maps
$T:({\mathbb{F}}^{n},\|\cdot\|)\rightarrow({\mathbb{F}}^{n},\|\cdot\|)$
preserving parallel pairs, namely,

$(T\mathbf{x},T\mathbf{y})\hbox{ is a parallel
pair}\quad\hbox{whenever}\quad(\mathbf{x},\mathbf{y})\hbox{ is a parallel
pair,}$ (1.3)

or preserving TEA pairs, namely,

$(T\mathbf{x},T\mathbf{y})\hbox{ is a TEA
pair}\quad\hbox{whenever}\quad(\mathbf{x},\mathbf{y})\hbox{ is a TEA pair.}$
(1.4)

Recall that a norm $\|\cdot\|$ on ${\bf V}$ is strictly convex if for any two
nonzero vectors $\mathbf{x},\mathbf{y}\in{\bf V}$ satisfying
$\|\mathbf{x}+\mathbf{y}\|=\|\mathbf{x}\|+\|\mathbf{y}\|$ we have
$\mathbf{x}=t\mathbf{y}$ for some $t>0$. It is easy to see that if the norm
$\|\cdot\|$ is strictly convex, then any linear map $T$ will satisfy (1.3) and
(1.4). For example, the $\ell_{p}$-norms on ${\mathbb{F}}^{n}$ defined by
$\|\mathbf{x}\|_{p}=(\sum_{j=1}^{n}|x_{j}|^{p})^{1/p}$ when $p\geq 1$, and
$\|\mathbf{x}\|_{\infty}=\max\\{|x_{j}|:1\leq j\leq n\\}$, for
$\mathbf{x}=(x_{1},\dots,x_{n})^{{\mathrm{t}}}$, are strict convex except for
$p=1,\infty$.

In this paper, we characterize linear maps on ${\mathbb{F}}^{n}$ preserving
parallel pairs or preserving TEA pairs with respect to the $\ell_{1}$-norm and
$\ell_{\infty}$-norm, respectively. In the following, let
$\\{\mathbf{e}_{1},\ldots,\mathbf{e}_{n}\\}$ denote the standard basis for
${\mathbb{F}}^{n}$, and ${\mathbf{M}}_{n}$ be the algebra of $n\times n$
matrices with entries from ${\mathbb{F}}$. We identify linear maps from
${\mathbb{F}}^{n}$ to ${\mathbb{F}}^{n}$ with matrices in ${\mathbf{M}}_{n}$.
For $\mathbf{u}\in{\mathbb{F}}^{n}$ and $A\in{\mathbf{M}}_{n}$, we let
$\mathbf{u}^{\mathrm{t}}$ and $A^{\mathrm{t}}$ denote their transposes. Below
are our findings. The proofs are given in Section 2.

######  Theorem 1.1.

Let
$T:({\mathbb{F}}^{n},\|\cdot\|_{1})\rightarrow({\mathbb{F}}^{n},\|\cdot\|_{1})$
be a linear map.

  * (a)

$T$ preserves TEA pairs if and only if each row of $T$ has at most one nonzero
entry.

  * (b)

$T$ preserves parallel pairs if and only if each row of $T$ has at most one
nonzero entry, or $T={\mathbf{v}}{\mathbf{u}}^{\mathrm{t}}$ for some column
vectors ${\mathbf{u}},{\mathbf{v}}\in{\mathbb{F}}^{n}$.

By the above result, for the $\ell_{1}$-norm, TEA preservers form a semigroup
of matrices such that each row has at most one nonzero entries; adding rank
one matrices to this semigroup will be the semigroup of parallel preserves.

Recall that a matrix in ${\mathbf{M}}_{n}$ is a _monomial_ matrix if each row
and each column of it has exactly one nonzero entry, and a monomial matrix is
a _generalized permutation matrix_ if all its nonzero entries have modulus
one. By Theorem 1.1, an invertible linear map $T$ preserves TEA or parallel
pairs of $({\mathbb{F}}^{n},\|\cdot\|_{1})$ if and only if it is a monomial
matrix. For $({\mathbb{F}}^{n},\|\cdot\|_{\infty})$, we have the following.

######  Theorem 1.2.

Let
$T:({\mathbb{F}}^{n},\|\cdot\|_{\infty})\rightarrow({\mathbb{F}}^{n},\|\cdot\|_{\infty})$
be a linear map.

  1. (a)

$T$ preserves parallel pairs if and only if there is $\gamma\geq 0$ and a
generalized permutation matrix $Q$ such that one of the following forms holds:

    1. (a.1)

$T$ has the form $\mathbf{x}\mapsto\gamma Q\mathbf{x}$,

    2. (a.2)

$n=2$ and $T$ has the form $\mathbf{x}\mapsto\gamma CQ\mathbf{x}$, where
$C=\begin{pmatrix}1&\beta\\\ \overline{\beta}&1\end{pmatrix}$ for some scalar
$\beta$ with $|\beta|<1$.

    3. (a.3)

$T=\mathbf{v}{\mathbf{u}}^{\mathrm{t}}$ for some nonzero column vectors
${\mathbf{u}},{\mathbf{v}}\in{\mathbb{F}}^{n}$.

  2. (b)

$T$ preserves TEA pairs if and only if one of the following holds.

    1. (b.1)

$T$ has the form in (a.1).

    2. (b.2)

${\mathbb{F}}^{n}={\mathbb{R}}^{2}$ and $T$ has the form in (a.2).

    3. (b.3)

${\mathbb{F}}^{n}={\mathbb{R}}^{2}$ and $T$ has the form in (a.3), where
$\mathbf{u}=(u_{1},u_{2})^{{\mathrm{t}}}$ with $|u_{1}|=|u_{2}|$.

######  Corollary 1.3.

Let $n\geq 3$. The following conditions are equivalent to each other for a
nonzero linear map
$T:({\mathbb{F}}^{n},\|\cdot\|_{\infty})\rightarrow({\mathbb{F}}^{n},\|\cdot\|_{\infty})$.

  * (a)

$T$ preserves TEA pairs.

  * (b)

$T$ preserves parallel pairs and its range space has dimension larger than
one.

  * (c)

There is $\gamma>0$ and a generalized permutation matrix
$Q\in{\mathbf{M}}_{n}$ such that $T$ has the form $\mathbf{x}\mapsto\gamma
Q\mathbf{x}$.

It is known that an isometry $T$ for the normed space
$({\mathbb{F}}^{n},\|\cdot\|_{p})$, with $p\in[1,\infty]$ and $p\neq 2$, is a
generalized permutation matrix; see, e.g., [4]. By Corollary 1.3, for $({\bf
V},\|\cdot\|)=({\mathbb{F}}^{n},\|\cdot\|_{\infty})$ with $n\geq 3$, a
bijective linear map satisfies (1.3) if and only if it satisfies (1.4); such a
map is a scalar multiple of an $\ell_{\infty}$-isometry. On the other hand, by
the remark after Theorem 1.1, for $({\mathbb{F}}^{n},\|\cdot\|_{1})$ with
$n\geq 2$, a bijective linear map $T$ satisfying (1.3) (or, equivalently,
(1.4)) if and only if $T$ is a monomial matrix. Note, however, that by Theorem
1.1, the range space of linear TEA/parallel pair preservers on
$({\mathbb{F}}^{n},\|\cdot\|_{1})$ can have arbitrary dimension in contrast to
the $\ell_{\infty}$–norm case.

More generally, it would be interesting to determine a norm $\|\cdot\|$ on
${\mathbb{F}}^{n}$ such that the set of bijective linear maps
$T:{\mathbb{F}}^{n}\to{\mathbb{F}}^{n}$ satisfying (1.3) or (1.4) consisting
of scalar multiples of isometries. For more results concerning the study of
parallel pairs and their preservers; see [2, 8, 9, 10, 12, 11] and [4, 6].

In Section 3, we provide similar characterizations of parallel/TEA pair linear
preservers of the infinite dimensional Banach “sequence spaces”
$\ell_{1}(\Lambda)$, $c_{0}(\Lambda)$ and $\ell_{\infty}(\Lambda)$, where the
index set $\Lambda$ can be uncountably infinite. The case for the
$\ell_{1}$–norm is similar, while for the $\ell_{\infty}$–norm case, we need
to assume the linear map $T$ of $c_{0}(\Lambda)$ or $\ell_{\infty}(\Lambda)$
preserves parallel/TEA pairs in both directions; namely,

$(T\mathbf{x},T\mathbf{y})\hbox{ is a parallel/TEA pair}\quad\hbox{if and only
if}\quad(\mathbf{x},\mathbf{y})\hbox{ is a parallel/TEA pair.}$

Under this stronger assumption, we see that $\ell_{\infty}$–norm parallel/TEA
linear preservers of $c_{0}(\Lambda)$ and $\ell_{\infty}(\Lambda)$ are scalar
multiples of isometries, in line with the finite dimensional case.

##  2\. Preservers of parallel/TEA vectors

We begin with the following observation.

######  Lemma 2.1.

Let
$\mathbf{x}=(x_{1},\dots,x_{n})^{{\mathrm{t}}},\mathbf{y}=(y_{1},\dots,y_{n})^{{\mathrm{t}}}\in{\mathbb{F}}^{n}$.

  1. (a)

$\mathbf{x},\mathbf{y}$ are parallel (resp. TEA) with respect to the
$\ell_{1}$–norm if and only if there is a unimodular scalar $\mu$ (resp.
$\mu=1$) such that $\mu\overline{x_{k}}y_{k}\geq 0$ for all $k=1,\ldots,n$.

  2. (b)

$\mathbf{x},\mathbf{y}$ are parallel (resp. TEA) with respect to the
$\ell_{\infty}$–norm if and only if $\|\mathbf{x}\|_{\infty}=|x_{k}|$ and
$\|\mathbf{y}\|_{\infty}=|y_{k}|$ (resp. such that $\overline{x_{k}}y_{k}\geq
0$) for some $k$ between $1$ and $n$.

As direct consequences of Lemma 2.1, a linear map $T$ of $\mathbb{F}^{n}$
preserves $\ell_{1}$–norm (resp. $\ell_{\infty}$–norm) TEA pairs if, and only
if, $PTQ$ does for any monomial matrices (resp. generalized permutation
matrices) $P,Q$.

###### Proof of Theorem 1.1.

(a) Suppose each row of the $n\times n$ matrix $T$ has at most one nonzero
entries. Then there are monomial matrices $P,Q$ such that
$PTQ=\begin{bmatrix}\mathbf{T}_{1}&\mathbf{T}_{2}&\cdots&\mathbf{T}_{n}\end{bmatrix}$
in which the column vectors
$\mathbf{T}_{1},\mathbf{T}_{2},\ldots,\mathbf{T}_{k}$ are nonzero and
satisfying that

$\displaystyle\mathbf{T}_{1}=\mathbf{e}_{1}+\cdots+\mathbf{e}_{n_{1}},\quad\mathbf{T}_{2}=\mathbf{e}_{n_{1}+1}+\cdots+\mathbf{e}_{n_{2}},\quad\dots,\quad\mathbf{T}_{k}=\mathbf{e}_{n_{k-1}+1}+\cdots+\mathbf{e}_{n_{k}},$
and $\displaystyle\mathbf{T}_{k+1}=\cdots=\mathbf{T}_{n}={0},$

where $k\leq n$ and $1\leq n_{1}<n_{2}<\cdots<n_{k}\leq n$. Note that $T$
preserves $\ell_{1}$–norm TEA pairs exactly when $PTQ$ does. We may replace
$T$ by $PTQ$, and assume that
$T=\begin{bmatrix}\mathbf{T}_{1}&\mathbf{T}_{2}&\cdots&\mathbf{T}_{n}\end{bmatrix}$.
Let $\mathbf{x}=(x_{1},\dots,x_{n})^{\mathrm{t}}$ and
$\mathbf{y}=(y_{1},\dots,y_{n})^{\mathrm{t}}\in{\mathbb{F}}^{n}$ form a TEA
pair; or equivalently, $\bar{x}_{j}y_{j}\geq 0$ for $j=1,\dots,n$. Then

$\displaystyle T\mathbf{x}$
$\displaystyle=x_{1}T_{1}+\cdots+x_{k}T_{k}=(\underbrace{x_{1},\ldots,x_{1}}_{n_{1}},\,\underbrace{x_{2},\ldots,x_{2}}_{n_{2}-n_{1}},\,\ldots,\underbrace{x_{k},\ldots,x_{k}}_{n_{k}-n_{k-1}},\,0,\ldots,0)^{\mathrm{t}},$
$\displaystyle T\mathbf{y}$
$\displaystyle=y_{1}T_{1}+\cdots+y_{k}T_{k}=(\underbrace{y_{1},\ldots,y_{1}}_{n_{1}},\
\underbrace{y_{2},\ldots,y_{2}}_{n_{2}-n_{1}},\
\ldots,\underbrace{y_{k},\ldots,y_{k}}_{n_{k}-n_{k-1}},\
0,\ldots,0)^{\mathrm{t}}$

clearly form a TEA pair.

Conversely, suppose $T$ preserves TEA pairs. Assume the contrary that $T$ has
a row with two nonzero entries. We may replace $T$ by $PTQ$ for suitable
monomial matrices $P$ and $Q$ and assume that $T$ has the $(1,1)$th and the
$(1,2)$th entries equal to $1$. Since
$\mathbf{x}=(2,-1,0,\dots,0)^{\mathrm{t}}$ and
$\mathbf{y}=(1,-2,0,\dots,0)^{\mathrm{t}}$ form a TEA pair, so do
$T\mathbf{x}$ and $T\mathbf{y}$. But the first entries of $T\mathbf{x}$ and
$T\mathbf{y}$ are $1$ and $-1$, respectively. So, $T\mathbf{x}$ and
$T\mathbf{y}$ do not form a TEA pair, a desired contradiction.

(b) If each row of $T$ has at most one nonzero entry, then $T$ will preserve
TEA pairs. Hence, $T$ will also preserve parallel pairs. On the other hand, if
$T=\mathbf{v}\mathbf{u}^{\mathrm{t}}$ for some
$\mathbf{u},\mathbf{v}\in{\mathbb{F}}^{n}$, then
$T\mathbf{x}=(\mathbf{u}^{\mathrm{t}}\mathbf{x})\mathbf{v}$ and
$T\mathbf{y}=(\mathbf{u}^{\mathrm{t}}\mathbf{y})\mathbf{v}$ are both scalar
multiples of $\mathbf{v}$, and thus always parallel, for any
$\mathbf{x},\mathbf{y}\in{\mathbb{F}}^{n}$.

Conversely, let $T$ be a linear parallel pair preserver with rank larger than
$1$. We will show that every row of $T$ has at most one nonzero entry. Suppose
on contrary that $T$ has a row with more than one nonzero entries. We may
replace $T$ by $PTQ$ for some suitable monomial matrices $P$ and $Q$ and
assume that the first row of $T$ has the maximum number of nonzero entries
among all the rows. Moreover, we may also assume that all these nonzero
entries in the first row are $1$ and lie in the $(1,1)$th, $(1,2)$th, …,
$(1,k)$th positions.

Since $T$ has rank at least two, there is a row, say, the second row, which is
not equal to a multiple of the first row. We consider two cases.

Case 1. The first $k$ entries of the second row are not all equal. We may
replace $T$ by $TQ$ for a permutation matrix of the form $Q=Q_{1}\oplus
I_{n-k}$ such that the $(2,1)$ entry is nonzero and different from the $(2,2)$
entry. Further replace $T$ by $PT$ for an invertible diagonal matrix $P$ and
assume that the leading $2\times 2$ matrix of $T$ equals
$\begin{pmatrix}1&1\cr 1&a\cr\end{pmatrix}$ for some $a\neq 1$.

Let $\mathbf{x}=(m,\bar{a},0,\dots,0)^{\mathrm{t}}$ and
$\mathbf{y}=(1,m\bar{a},0,\dots,0)^{\mathrm{t}}$ with $m>0$. Then $\mathbf{x}$
and $\mathbf{y}$ are parallel, and so are the vectors $T\mathbf{x}$ and
$T\mathbf{y}$. The first two entries of $T\mathbf{x}$ are $m+\bar{a}$ and
$m+|a|^{2}$, and the first two entries of $T\mathbf{y}$ are $1+m\bar{a}$ and
$1+m|a|^{2}$. The second entries of $T\mathbf{x}$ and $T\mathbf{y}$ are always
positive. It forces
$\overline{(m+\bar{a})}(1+m\bar{a})=m(1+|a|^{2})+a+m^{2}\bar{a}\geq 0$ for all
$m>0$. Consequently, $a\geq 0$.

Furthermore, if $m>0$, then $\mathbf{x}=(m,-1,0,\dots,0)^{\mathrm{t}}$ and
$\mathbf{y}=(1,-m,0,\dots,0)^{\mathrm{t}}$ are parallel, and so are
$T\mathbf{x}$ and $T\mathbf{y}$. The first two entries of $T\mathbf{x}$ are
$m-1$ and $m-a$, and the first two entries of $T\mathbf{y}$ are $1-m$ and
$1-am$. It follows that $(m-a)(1-am)\leq 0$ for all $m>0$ with $m\neq 1$.
Since $a\neq 1$ and $a\geq 0$, we see that $m=(1+a)/2\neq 1$ and
$(m-a)(1-am)=\frac{1}{4}(1-a)(2-a-a^{2})=\frac{1}{4}(1-a)^{2}(2+a)>0$, which
is a contradiction.

Case 2. The first $k$ entries of the (nonzero) second row of $T$ are the same
scalar $\gamma$. If $\gamma\neq 0$ then all other entries of the second row of
$T$ are zeros due to the assumption that the first row of $T$ has maximal
number of nonzero entries among all rows of $T$. But then the second row is
$\gamma$ times the first row, a contradiction. Hence, $\gamma=0$. Suppose the
$({2},j)$th entry of $T$ equals $a\neq 0$ for some $j>k$. We may replace $T$
by $TQ$ for a suitable permutation matrix $Q$ and assume that the leading
$2\times 3$ matrix of $T$ is $\begin{pmatrix}1&1&0\cr 0&0&a\cr\end{pmatrix}$.
Let $\mathbf{x}=(2,-1,1,0,\dots,0)^{\mathrm{t}}$ and
$\mathbf{y}=(1,-2,1,0,\dots,0)^{\mathrm{t}}$. Then $\mathbf{x}$ and
$\mathbf{y}$ are parallel and so are $T\mathbf{x}$ and $T\mathbf{y}$. Now,
$T\mathbf{x}$ has the first two entries equal to $1$ and $a$, whereas
$T\mathbf{y}$ has the first two entries equal to $-1$ and $a$. Thus,
$T\mathbf{x}$ and $T\mathbf{y}$ cannot be parallel, which is a contradiction.

We need to establish two more lemmas to prove Theorem 1.2.

######  Lemma 2.2.

If ${\bf V}$ is a subspace of ${\mathbb{F}}^{n}$ such that any two elements in
${\bf V}$ are parallel with respect to $\|\cdot\|_{\infty}$, then ${\rm
dim}\,{\bf V}\leq 1$.

###### Proof.

We are going to show that any nonzero $\mathbf{u},\mathbf{v}\in{\bf V}$ are
linearly dependent. To this end, we may replace $(\mathbf{u},\mathbf{v})$ by
$(\alpha\mathbf{u},\beta\mathbf{v})$ for some nonzero scalars $\alpha,\beta$,
and assume that $\mathbf{u},\mathbf{v}$ are unit vectors with
$\|\mathbf{u}+\mathbf{v}\|_{\infty}=2$. Since
$\|\mathbf{u}+\mathbf{v}\|_{\infty}=2$, we may further replace
$\mathbf{u},\mathbf{v}$ by $Q\mathbf{u},Q\mathbf{v}$ for a suitable
generalized permutation matrix $Q$ and assume that
$\mathbf{u}=(1,\dots,1,u_{k+1},\dots,u_{n})^{{\mathrm{t}}}$ and
$\mathbf{v}=(1,\dots,1,v_{k+1},\dots,v_{n})^{{\mathrm{t}}}$ with $|v_{j}|<1$
for $j=k+1,\dots,n$, and $k\geq 1$.

By assumption, $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$ are
parallel with respect to the $\ell_{\infty}$–norm. There exists a unimodular
scalar $\alpha$ such that

$2+\|\mathbf{u}-\mathbf{v}\|_{\infty}=\|\mathbf{u}+\mathbf{v}\|_{\infty}+\|\mathbf{u}-\mathbf{v}\|_{\infty}=\|(\mathbf{u}+\mathbf{v})+\alpha(\mathbf{u}-\mathbf{v})\|_{\infty}=|(u_{i}+v_{i})+\alpha(u_{i}-v_{i})|$

for some $i$. If $k+1\leq i\leq n$, then
$|(u_{i}+v_{i})+\alpha(u_{i}-v_{i})|<2+\|\mathbf{u}-\mathbf{v}\|_{\infty}$, a
contradiction. Consequently, $1\leq i\leq k$. Then
$|(u_{i}+v_{i})+\alpha(u_{i}-v_{i})|=2$, and thus $\mathbf{u}=\mathbf{v}$.

######  Lemma 2.3.

Let
$T:({\mathbb{F}}^{n},\|\cdot\|_{\infty})\rightarrow({\mathbb{F}}^{n},\|\cdot\|_{\infty})$
be a nonzero linear map. Then $T$ is invertible if one of the following holds.

  * (a)

${\mathbb{F}}^{n}\neq{\mathbb{R}}^{2}$ and $T$ preserves TEA pairs,

  * (b)

$T$ preserves parallel pairs with range space of dimension larger than one.

###### Proof.

Recall that $\mathbf{e}_{j}$ denotes the coordinate vector with the $j$th
coordinate $1$ and all others $0$ for $j=1,\ldots,n$, and let
$\mathbf{e}=\sum_{j=1}^{n}\mathbf{e}_{j}$ be the constant one vector in
${\mathbb{F}}^{n}$. If $J$ is a subset of $N=\\{1,\ldots,n\\}$, let
$\mathbf{e}_{J}=\sum_{j\in J}\mathbf{e}_{j}$. In particular,
$\mathbf{e}=\mathbf{e}_{N}$.

(a) Suppose $T\mathbf{v}=0$ for some unit vector
$\mathbf{v}=\sum_{j=1}^{n}v_{j}\mathbf{e}_{j}\in{\mathbb{F}}^{n}$. We claim
that $T$ is a zero map. Replacing $T$ by $TR$ for a suitable general
permutation matrix $R$, we can assume that $v_{1}=1\geq v_{2}\geq\cdots\geq
v_{n}\geq 0$.

Suppose first that $v_{1}=\cdots=v_{k}=1>v_{k+1}\geq\cdots\geq v_{n}\geq 0$
for some $k<n$. Since $\xi\mathbf{v}+\mathbf{e}_{n}$ and
$\xi\mathbf{v}-\mathbf{e}_{n}$ form a TEA pair for large $\xi>0$, so do
$T(\xi\mathbf{v}+\mathbf{e}_{n})=T(\mathbf{e}_{n})$ and
$T(\xi\mathbf{v}-\mathbf{e}_{n})=-T(\mathbf{e}_{n})$. It follows
$0=\|T(\mathbf{e}_{n})+T(-\mathbf{e}_{n})\|_{\infty}=\|T(\mathbf{e}_{n})\|_{\infty}+\|T(-\mathbf{e}_{n})\|_{\infty}$,
and thus $T(\mathbf{e}_{n})=0$. With $\mathbf{e}_{n}$ taking the role of
$\mathbf{v}$, we see that $T\mathbf{e}_{j}=0$ for all $j=1,\ldots,n-1$, and
thus $T=0$.

Suppose next that $\mathbf{v}=\mathbf{e}$ and $n\geq 3$. For any subset $J,K$
of $N$ such that $J\cup K\neq N$, we have $\mathbf{e}-\mathbf{e}_{J}$ and
$\mathbf{e}-\mathbf{e}_{K}$ form a TEA pair, and so do
$T(\mathbf{e}-\mathbf{e}_{J})=-T(\mathbf{e}_{J})$ and
$T(\mathbf{e}-\mathbf{e}_{K})=-T(\mathbf{e}_{K})$. Hence,

$\|T(\mathbf{e}_{J})+T(\mathbf{e}_{K})\|_{\infty}=\|T(\mathbf{e}_{J})\|_{\infty}+\|T(\mathbf{e}_{K})\|_{\infty}.$

An inductive argument with the fact $T(\mathbf{e})=0$ gives

$\|T(\mathbf{e}_{1})\|_{\infty}=\|T(\mathbf{e}_{2})+\cdots+T(\mathbf{e}_{n-1})\|_{\infty}=\|T(\mathbf{e}_{2})\|_{\infty}+\cdots+\|T(\mathbf{e}_{n-1})\|_{\infty}.$

We also have similar equalities for all other
$\|T(\mathbf{e}_{j})\|_{\infty}$. Summing up these $n\geq 3$ equalities, we
have

$\sum_{j=1}^{n}\|T(\mathbf{e}_{j})\|_{\infty}=(n-1)\sum_{j=1}^{n}\|T(\mathbf{e}_{j})\|_{\infty},$

and thus all $\|T(\mathbf{e}_{j})\|_{\infty}=0$. This also forces $T=0$.

Finally, suppose $n=2$, ${\mathbb{F}}={\mathbb{C}}$ and
$\mathbf{v}=\mathbf{e}=\mathbf{e}_{1}+\mathbf{e}_{2}$. Since the vectors
$\mathbf{e}-\frac{1+\sqrt{3}i}{2}\mathbf{e}_{1}$ and
$\mathbf{e}-\frac{1-\sqrt{3}i}{2}\mathbf{e}_{1}$ attain the triangle equality,
so do
$T(\mathbf{e}-\frac{1+\sqrt{3}i}{2}\mathbf{e}_{1})=-\frac{1+\sqrt{3}i}{2}T(\mathbf{e}_{1})$
and
$T(\mathbf{e}-\frac{1-\sqrt{3}i}{2}\mathbf{e}_{1})=-\frac{1-\sqrt{3}i}{2}T(\mathbf{e}_{1})$,
which implies $T(\mathbf{e}_{1})=0$. Consequently, $T(\mathbf{e}_{2})=0$, and
thus $T=0$ again.

In conclusion, a nonzero linear map $T$ preserving TEA pairs is invertible
unless ${\mathbb{F}}^{n}={\mathbb{R}}^{2}$.

(b) If $n=2$ and the range space of $T$ has dimension larger than one, then
$T$ is invertible. Suppose $n>2$ and $T$ is not invertible. Let $\mathbf{v}$
be a nonzero vector such that $T\mathbf{v}=0$. We may replace $T$ by the map
$\mathbf{x}\mapsto T(\alpha Q\mathbf{x})$ for some $\alpha>0$ and generalized
permutation matrix $Q$, and assume that
$\mathbf{v}=(v_{1},\dots,v_{n})^{\mathrm{t}}$ with
$v_{1}=\cdots=v_{k}=1>v_{k+1}\geq\cdots\geq v_{n}\geq 0$ where $1\leq k\leq
n$. We are verifying that the range space of $T$ has dimension at most one.

Case 1. Suppose $k=1$. Then for any
$\mathbf{x}=(0,x_{2},\dots,x_{n})^{\mathrm{t}},\mathbf{y}=(0,y_{2},\dots,y_{n})^{\mathrm{t}}$
in the linear span $E$ of $\mathbf{e}_{2},\ldots,\mathbf{e}_{n}$, the vectors
$r\mathbf{v}+\mathbf{x},r\mathbf{v}+\mathbf{y}$ are parallel for sufficiently
large $r>0$. Consequently, $T(r\mathbf{v}+\mathbf{x})=T(\mathbf{x})$ and
$T(r\mathbf{v}+\mathbf{y})=T(\mathbf{y})$ are also parallel. By Lemma 2.2, the
space $T(E)$ has dimension at most one. Since ${\mathbb{F}}^{n}$ is spanned by
$\mathbf{v}$ and $E$ and $T\mathbf{v}=0$, we conclude that
$T({\mathbb{F}}^{n})$ has dimension at most one.

Case 2. Suppose $\mathbf{v}=\mathbf{e}_{1}+\mathbf{e}_{2}$. In view of Case 1,
we may assume $T(\mathbf{e}_{1})=-T(\mathbf{e}_{2})\neq 0$. We claim that
$T(\mathbf{e}_{1})$ and $T(\mathbf{u})$ are linearly dependent for any norm
one vector $\mathbf{u}=(0,0,u_{3},\ldots,u_{n})^{\mathrm{t}}$. Consequently,
being the span of $T(\mathbf{e}_{1}),T(\mathbf{e}_{2})$, and all such
$T(\mathbf{u})$, the range space $T({\mathbb{F}}^{n})$ has dimension at most
one.

Consider $\mathbf{x}=\alpha\mathbf{e}_{1}+\beta\mathbf{u}$ and
$\mathbf{y}=\gamma\mathbf{e}_{1}+\delta\mathbf{u}$ for any scalars
$\alpha,\beta,\gamma$ and $\delta$. If $\mathbf{x},\mathbf{y}$ are parallel,
so are $T(\mathbf{x}),T(\mathbf{y})$. If $\mathbf{x}$ is not parallel with
$\mathbf{y}$ then we can assume that
$\mathbf{x}=\mathbf{e}_{1}+\beta\mathbf{u}$ and
$\mathbf{y}=\gamma\mathbf{e}_{1}+\mathbf{u}$ with $|\beta|,|\gamma|<1$. If
$\gamma\neq 0$ then $\mathbf{x}$ is parallel with
$s\gamma\mathbf{v}+\mathbf{y}$ for $s\geq\frac{1}{|\gamma|}-1$. We see that
$T(\mathbf{x})$ is parallel with
$T(s\gamma\mathbf{v}+\mathbf{y})=T(\mathbf{y})$. In case when $\gamma=0$, we
see that $T(\mathbf{x})$ is parallel with
$T(s\epsilon\mathbf{v}+\epsilon\mathbf{e}_{1}+\mathbf{u})=T(\epsilon\mathbf{e}_{1}+\mathbf{u})=\epsilon
T(\mathbf{e}_{1})+T(\mathbf{y})$ whenever $0<\epsilon<1$ and
$s\geq\frac{1}{\epsilon}-1$. In other words,

$\|T(\mathbf{x})+\mu_{\epsilon}(\epsilon
T(\mathbf{e}_{1})+T(\mathbf{y}))\|_{\infty}=\|T(\mathbf{x})\|_{\infty}+\|\epsilon
T(\mathbf{e}_{1})+T(\mathbf{y})\|_{\infty}$

for some unimodular scalar $\mu_{\epsilon}$. Choosing a sequence
$\epsilon_{n}\to 0^{+}$ with $\mu_{\epsilon_{n}}$ converging to some
unimodular $\mu$, we see that $T(\mathbf{x})$ and $T(\mathbf{y})$ are
parallel. Therefore, in any case $T(\mathbf{x})$ is parallel with
$T(\mathbf{y})$. Hence $T(\mathbf{e}_{1})$ and $T(\mathbf{u})$ are linearly
dependent by Lemma 2.2, as claimed.

Case 3. Suppose that
$\mathbf{v}=\mathbf{e}_{1}+\mathbf{e}_{2}+a_{3}\mathbf{e}_{3}+\cdots+a_{n}\mathbf{e}_{n}$
with $1\geq a_{3}\geq a_{4}\geq\cdots\geq a_{m}\geq 0$. In view of Case 2, we
can assume that $a_{3}>0$.

Consider $\mathbf{x}=\alpha\mathbf{e}_{1}+\beta\mathbf{e}_{2}$ and
$\mathbf{y}=\gamma\mathbf{e}_{1}+\delta\mathbf{e}_{2}$. We claim that
$T(\mathbf{x})$ and $T(\mathbf{y})$ are parallel. If $\mathbf{x},\mathbf{y}$
are parallel, then it is the case. Otherwise, we can assume that
$\mathbf{x}=\mathbf{e}_{1}+\mu\mathbf{e}_{2}$ and
$\mathbf{y}=\nu\mathbf{e}_{1}+\mathbf{e}_{2}$ with $|\mu|,|\nu|<1$. If
$|1-\mu|\geq|1+\mu|a_{3}$, then $\mathbf{x}-(1+\mu)\mathbf{v}/2$ and
$\mathbf{y}$ are parallel, and so are
$T(\mathbf{x})=T(\mathbf{x}-(1+\mu)\mathbf{v}/2)$ and $T(\mathbf{y})$. If
$|1-\nu|\geq|1+\nu|a_{3}$, then $\mathbf{x}$ and
$\mathbf{y}-(1+\nu)\mathbf{v}/2$ are parallel, and so are $T(\mathbf{x})$ and
$T(\mathbf{y})=T(\mathbf{y}-(1+\nu)\mathbf{v}/2)$. If $|1-\mu|<|1+\mu|a_{3}$
and $|1-\nu|<|1+\nu|a_{3}$ then $\mathbf{x}-(1+\mu)\mathbf{v}/2$ and
$\mathbf{y}-(1+\nu)\mathbf{v}/2$ are parallel, then so are
$T(\mathbf{x})=T(\mathbf{x}-(1+\mu)\mathbf{v})$ and
$T(\mathbf{y})=T(\mathbf{y}-(1+\nu)\mathbf{v})$. We see that $T(\mathbf{x})$
and $T(\mathbf{y})$ are parallel in all cases. Hence $T(\mathbf{e}_{1})$ and
$T(\mathbf{e}_{2})$ are linearly dependent by Lemma 2.2. Consequently, there
is a nontrivial linear combination
$\mathbf{v}^{\prime}=\xi\mathbf{e}_{1}+\eta\mathbf{e}_{2}$ belongs to the
kernel of $T$. We can then reduce the situation to either Case 1 (if
$|\xi|>|\eta|$) or Case 2 (if $|\xi|=|\eta|$).

We thus conclude that a linear parallel pair preserver is invertible if its
range space has dimension larger than one.

Consider the linear map $T:{\mathbb{R}}^{2}\to{\mathbb{R}}^{2}$ defined by
$(\mathbf{x},\mathbf{y})^{\mathrm{t}}\mapsto(\mathbf{x}-\mathbf{y},0)^{\mathrm{t}}$.
It is easy to see that $T({\mathbb{R}}^{2})$ has dimension one and $T$
preserves TEA/parallel pairs for the $\ell_{\infty}$–norm. This example says
that Lemma 2.3 does not hold in the missing cases.

######  Lemma 2.4.

Let $A=(a_{rs})\in{\mathbf{M}}_{n}$. Suppose either $a_{jj}>|a_{jk}|$ whenever
$j\neq k$, or $a_{jj}>|a_{kj}|$ whenever $j\neq k$. The following conditions
are equivalent.

  * (a)

For any $\mathbf{x}=(x_{1},\dots,x_{n})^{{\mathrm{t}}}\in{\mathbb{F}}^{n}$
with
$\mathbf{y}=A^{{\mathrm{t}}}\mathbf{x}=(y_{1},\dots,y_{n})^{{\mathrm{t}}}$, we
have

$\displaystyle\|\mathbf{x}\|_{\infty}=|x_{r}|>|x_{s}|\ \text{whenever $s\neq
r$}\quad\implies\quad\|\mathbf{y}\|_{\infty}=|y_{r}|\geq|y_{s}|\
\text{whenever $s\neq r$.}$ (2.1)

  * (b)

Either $n=2$ and $A=A^{*}$ with $a_{11}=a_{22}>|a_{12}|$, or $A=a_{11}I_{n}$.

###### Proof.

The implication (b) $\Rightarrow$ (a) is clear if $A=a_{11}I_{n}$. Suppose
$n=2$ and $A=A^{*}$ with $a_{11}=a_{22}>|a_{12}|$. Observe that

$\displaystyle\begin{pmatrix}y_{1}\\\
y_{2}\end{pmatrix}=A^{{\mathrm{t}}}\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix}=\begin{pmatrix}a_{11}&\overline{a_{12}}\\\
a_{12}&a_{11}\end{pmatrix}\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix}=\begin{pmatrix}a_{11}x_{1}+\overline{a_{12}}x_{2}\\\
{a_{12}}x_{1}+a_{11}x_{2}\end{pmatrix},$

and

$\displaystyle|y_{1}|\geq|y_{2}|$ $\displaystyle\Longleftrightarrow\quad$
$\displaystyle|a_{11}x_{1}+\overline{a_{12}}x_{2}|\geq|{a_{12}}x_{1}+a_{11}x_{2}|$
$\displaystyle\Longleftrightarrow\quad$ $\displaystyle
a_{11}^{2}|x_{1}|^{2}+|a_{12}|^{2}|x_{2}|^{2}+2a_{11}\operatorname{Re}\,({a_{12}}x_{1}\bar{x}_{2})\geq|a_{12}|^{2}|x_{1}|^{2}+a_{11}^{2}|x_{2}|^{2}+2a_{11}\operatorname{Re}\,({a_{12}}x_{1}\bar{x}_{2})$
$\displaystyle\Longleftrightarrow\quad$
$\displaystyle(a_{11}^{2}-|a_{12}|^{2})|x_{1}|^{2}\geq(a_{11}^{2}-|a_{12}|^{2})|x_{2}|^{2}$
$\displaystyle\Longleftrightarrow\quad$ $\displaystyle|x_{1}|\geq|x_{2}|.$

Consequently, (b) $\Rightarrow$ (a) also holds in this case.

We are going to prove (a) $\Rightarrow$ (b). We may replace $A$ by
$P^{{\mathrm{t}}}AP$ with a suitable permutation matrix $P$ and assume the
first row $\mathbf{v}_{1}$ of $A^{{\mathrm{t}}}$ has the maximal
$\ell_{1}$–norm. Then further replace $A$ by $DAD^{*}$ with a suitable
diagonal $D\in{\mathbf{M}}_{n}$ with $D^{*}D=I_{n}$ and assume that the first
row of $A^{{\mathrm{t}}}$ has nonnegative entries.

By the continuity and an induction argument, for
$\mathbf{x}=(x_{1},\dots,x_{n})^{{\mathrm{t}}}$,
$\mathbf{y}=A^{{\mathrm{t}}}\mathbf{x}=(y_{1},\dots,y_{n})^{{\mathrm{t}}}$ and
$k=1,\ldots,m$, we have

$\displaystyle|x_{1}|=\cdots=|x_{k}|>|x_{r}|\ \text{for all
$r>k$}\quad\implies\quad|y_{1}|=\cdots=|y_{k}|\geq|y_{r}|\ \text{for all
$r>k$.}$ (2.2)

Let $\mathbf{x}=(1,\dots,1)^{{\mathrm{t}}}$ and
${\mathbf{y}}=A^{{\mathrm{t}}}\mathbf{x}$. If
$\mathbf{v}_{1},\dots,\mathbf{v}_{n}$ are the rows of $A^{{\mathrm{t}}}$, then

$\displaystyle\|\mathbf{v}_{1}\|_{1}$
$\displaystyle=a_{11}+\cdots+a_{n1}=|y_{1}|=|y_{j}|=|a_{1j}+\cdots+a_{nj}|$
$\displaystyle\leq|a_{1j}|+\cdots+|a_{nj}|=\|\mathbf{v}_{j}\|_{1}\leq\|\mathbf{v}_{1}\|_{1}\quad\text{for
$j\geq 1$.}$

Since $a_{jj}>0$, we see that $a_{ij}\geq 0$ for all $i\neq j$, and all row
sums of $A$ are equal, say, to $s>0$.

Taking $\mathbf{x}=(1,1,0,\ldots,0)^{{\mathrm{t}}}$, with (2.2) we have
$a_{11}+a_{21}=a_{12}+a_{22}$. Similarly, taking
$\mathbf{x}=(1,-1,0,\ldots,0)^{{\mathrm{t}}}$, we have
$|a_{11}-a_{21}|=|a_{12}-a_{22}|$. It follows from either the assumption
$a_{11}>a_{12}$ and $a_{22}>a_{21}$, or the assumption $a_{11}>a_{21}$ and
$a_{22}>a_{12}$ that $a_{11}-a_{21}=a_{22}-a_{12}$. Consequently,
$a_{11}=a_{22}$ and $a_{12}=a_{21}$. The assertion follows when $n=2$.

Suppose $n\geq 3$. Apply the same argument to other pairs $(i,j)$ with $i\neq
j$ instead of $(1,2)$, we see that

$a_{11}=\cdots=a_{nn}\quad\text{and}\quad a_{jk}=a_{kj}\quad\text{whenever
$j\neq k$.}$

For a fixed $j=1,\dots,n$, we take
$\mathbf{x}=(1,\ldots,1,\underbrace{-1}_{j\text{th}},1,\ldots,1)^{{\mathrm{t}}}$
and ${\mathbf{y}}=A^{{\mathrm{t}}}\mathbf{x}$. For distinct indices $j,k,l$,
we have $|y_{l}|=s-2a_{j{l}}=|y_{k}|=s-2a_{jk}$. It follows
$a_{j{l}}=a_{jk}=a_{kj}$, and thus $a_{jk}=a_{12}$ are all equal for $j\neq
k$. Consider $\mathbf{u}=(1,-1,\ldots,-1)^{{\mathrm{t}}}$ and
${\mathbf{v}}=A^{{\mathrm{t}}}\mathbf{u}=(v_{1},v_{2},\ldots,v_{n})^{{\mathrm{t}}}$.
Then $|v_{1}|=|v_{2}|$ implies either

$a_{11}-(n-1)a_{12}=a_{11}+(n-3)a_{12}\quad\text{or}\quad(n-1)a_{12}-a_{11}=a_{11}+(n-3)a_{12}.$

Since $n\geq 3$, either $a_{12}=0$ or $a_{11}=a_{12}$. But $a_{11}>a_{12}$.
This implies that $A=a_{11}I_{n}$.

We are now ready to present the

###### Proof of Theorem 1.2.

(a) It is clear that $T$ preserves parallel pairs if $T$ has the form in (a.1)
or (a.3). Suppose $n=2$ and $T$ has the form in (a.2). Then, $T$ preserves
parallel pairs in $({\mathbb{F}}^{2},\|\cdot\|_{\infty})$ by the implication
from (b) to (a) in Lemma 2.4; indeed, with a continuity argument the condition
(2.1) implies that

$\|\mathbf{x}\|_{\infty}=|x_{r}|\geq|x_{s}|\ \text{whenever $s\neq
r$}\quad\implies\quad\|\mathbf{y}\|_{\infty}=|y_{r}|\geq|y_{s}|\
\text{whenever $s\neq r$.}$

Conversely, suppose $T$ preserves parallel pairs. If $T$ is not invertible
then it follows from Lemma 2.3 that $T$ is either the zero map or has the form
in (a.3). Suppose from now on $T$ is invertible.

If $T^{-1}\mathbf{e}_{j}={\mathbf{x}}_{j}$, then ${\mathbf{x}}_{i}$ and
${\mathbf{x}}_{j}$ cannot be parallel for any $i\neq j$. Otherwise,
$T{\mathbf{x}}_{i}=\mathbf{e}_{i}$ and $T{\mathbf{x}}_{j}=\mathbf{e}_{j}$ were
parallel. Thus, the vectors ${\mathbf{x}}_{i}$ and ${\mathbf{x}}_{j}$ cannot
attain the $\ell_{\infty}$-norm at a same coordinate. So, there is a
permutation $\sigma$ on $\\{1,\ldots,n\\}$ such that each ${\mathbf{x}}_{j}$
attains its norm at its $\sigma(j)$th coordinate but no other. Consequently,
there is a generalized permutation matrix $Q\in{\mathbf{M}}_{n}$ such that the
map $L$ defined by ${z}\mapsto QT^{-1}{z}$ will send $\mathbf{e}_{j}$ to a
vector ${\mathbf{y}}_{j}=(a_{j1},\dots,a_{jn})^{{\mathrm{t}}}$ such that
$a_{jj}>|a_{ji}|$ for all $i\neq j$. Clearly, $L^{-1}$ defined by
${\mathbf{y}}\mapsto T(Q^{-1}{\mathbf{y}})$ preserves parallel pairs.

Let $A=(a_{ij})\in{\mathbf{M}}_{n}$ so that
$L({\mathbf{x}})=A^{{\mathrm{t}}}{\mathbf{x}}$. If
${\mathbf{x}}_{j}=(x_{j1},\ldots,x_{jn})^{{\mathrm{t}}}$ satisfies
$|{x}_{jj}|>|{x}_{ji}|$ for all $i\neq j$, then we claim that
$L({\mathbf{x}}_{j})=A^{{\mathrm{t}}}{\mathbf{x}}_{j}$ is parallel to
${\mathbf{y}}_{j}$ and thus $\mathbf{e}_{j}$, but not any other
${\mathbf{y}}_{i}$. Otherwise, the fact $L({\mathbf{x}}_{j})$ is parallel to
${\mathbf{y}}_{i}$ for some $i\neq j$ would imply that ${\mathbf{x}}_{j}$ is
parallel to $L^{-1}({\mathbf{y}}_{i})=\mathbf{e}_{i}$, which is impossible.
Thus, the matrix $A=(a_{ij})$ satisfies the hypothesis (a) of Lemma 2.4. If
$n\geq 3$, we see that $A=\gamma I_{n}$. with $\gamma=a_{11}>0$. If $n=2$, we
see that $A$ is Hermitian with $a_{11}=a_{22}$. So, $T$ has the form in (a.2).

(b) It is clear that if $T$ assumes the form in (b.1) then it preserves TEA
pairs. It is also the case if $T$ assumes the form in (b.3) by direct
verification. Suppose ${\mathbb{F}}^{n}={\mathbb{R}}^{2}$ and $T=\gamma CQ$ as
in (a.2). Clearly, $T$ satisfies (1.4) if and only if $C$ satisfies (1.4).
There is $S={\rm diag}\,(1,\pm 1)$ such that $SCS$ has the form
$\hat{C}=\begin{pmatrix}1&\beta\cr\beta&1\cr\end{pmatrix}$ with
$0\leq\beta<1$. The map ${\mathbf{x}}\mapsto C{\mathbf{x}}$ satisfies (1.4) if
and only if the map ${\mathbf{x}}\mapsto\hat{C}{\mathbf{x}}$ does. Now, if the
nonzero vectors $\mathbf{x},\mathbf{y}$ in ${\mathbb{R}}^{2}$ satisfy that
$\|{\mathbf{x}}+{\mathbf{y}}\|_{\infty}=\|{\mathbf{x}}\|_{\infty}+\|{\mathbf{y}}\|_{\infty}$,
then we may replace $({\mathbf{x}},{\mathbf{y}})$ by
$(\xi{\mathbf{x}}/\|\mathbf{x}\|_{\infty},\xi{\mathbf{y}}/\|\mathbf{y}\|_{\infty})$
with $\xi\in\\{-1,1\\}$ and assume that
${\mathbf{x}}=(1,x_{2})^{{\mathrm{t}}},{\mathbf{y}}=(1,y_{2})^{{\mathrm{t}}}$
with $|x_{2}|,|y_{2}|\leq 1$, or
${\mathbf{x}}=(x_{1},1)^{{\mathrm{t}}},{\mathbf{y}}=(y_{1},1)^{{\mathrm{t}}}$
with $|x_{1}|,|y_{1}|\leq 1$. One can check that
$\|\hat{C}{\mathbf{x}}+\hat{C}{\mathbf{y}}\|_{\infty}=\|\hat{C}{\mathbf{x}}\|_{\infty}+\|\hat{C}{\mathbf{y}}\|_{\infty}$.
Thus, $T$ preserves TEA as well.

Conversely, suppose the map $T$ is nonzero and satisfies (1.4). Then $T$ will
preserve parallel pairs. Thus, it will be of the form (a.1), (a.2), or (a.3).
We will show that (a.2) is impossible in the complex case unless it reduces to
the form in (a.1), and there are additional restrictions for ${\mathbf{u}}$ if
(a.3) holds.

Suppose ${\mathbb{F}}^{n}={\mathbb{C}}^{2}$ and $T=\gamma CQ$ has the form in
(a.2) in which $C=\begin{pmatrix}1&\beta\\\ \overline{\beta}&1\end{pmatrix}$
for some complex scalar $\beta$ with $|\beta|<1$. In this case, the map
$\mathbf{x}\mapsto Cx$ also preserves TEA pairs. Consider
${\mathbf{x}}=(1,0)^{{\mathrm{t}}}$, ${\mathbf{y}}=(1,1)^{{\mathrm{t}}}$ and
${z}=(1,i)^{{\mathrm{t}}}$, and their images
$Cx=(1,\overline{\beta})^{\mathrm{t}}$,
$Cy=(1+\beta,\overline{\beta}+1)^{\mathrm{t}}$ and
$Cz=(1+i\beta,\overline{\beta}+i)^{\mathrm{t}}$. Note that
$\mathbf{x},\mathbf{y}$ and $\mathbf{x},\mathbf{z}$ are both TEA pairs, while
$Cx,Cy$ and $Cx,Cz$ form TEA pairs exactly when the first coordinates of $Cy$
and $Cz$ assume positive values. This forces $\beta=0$, and thus $T=\gamma Q$
reduces to the form in (a.1).

Suppose $T=\mathbf{v}{\mathbf{u}}^{\mathrm{t}}$ assumes the form in (a.3) for
some vectors $\mathbf{u}=(u_{1},u_{2})^{\mathrm{t}}$ and $\mathbf{v}$ in
${\mathbb{F}}^{n}$. In this case, $T$ is not invertible. By Lemma 2.3,
${\mathbb{F}}^{n}={\mathbb{R}}^{2}$. Since ${y}_{1}=(1,1)^{{\mathrm{t}}}$ and
${\mathbf{y}}_{2}=(1,-1)^{{\mathrm{t}}}$ form a TEA pair, so are
$T\mathbf{y}_{1}=(u_{1}+u_{2})\mathbf{v}$ and
$T\mathbf{y}_{2}=(u_{1}-u_{2})\mathbf{v}$. This forces $u_{1}+u_{2}$ and
$u_{1}-u_{2}$ have the same sign. Similarly, ${\mathbf{y}}_{1}$ and
$-{\mathbf{y}}_{2}$ also form a TEA pair, and thus $u_{1}+u_{2}$ and
$-u_{1}+u_{2}$ also have the same sign. If $u_{1}+u_{2}\neq 0$ then
$u_{2}=u_{1}$. In any case, we have $|u_{1}|=|u_{2}|$ as asserted.

##  3\. The infinite dimensional cases

Let the underlying field ${\mathbb{F}}$ be either ${\mathbb{R}}$ or
${\mathbb{C}}$, and let $\Lambda$ be a finite or an infinite index set. When
$1\leq p<\infty$, let $\ell_{p}(\Lambda)$ be the (real or complex) Banach
space of $p$-summable families $\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$
(of real or complex numbers) with $\ell_{p}$–norm

$\|\mathbf{x}\|_{p}=\left(\sum_{\lambda\in\Lambda}|x_{\lambda}|^{p}\right)^{{1}/{p}}<+\infty.$

Note that the above sum is finite only if there are at most countably many
coordinates $x_{\lambda}\neq 0$. Let $\ell_{\infty}(\Lambda)$ be the Banach
space of uniformly bounded family
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ with the $\ell_{\infty}$–norm

$\|\mathbf{x}\|_{\infty}=\sup_{\lambda\in\Lambda}|x_{\lambda}|<+\infty.$

We are also interested in the Banach subspace $c_{0}(\Lambda)$ of
$\ell_{\infty}(\Lambda)$ consisting of essentially null families
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ for which for any $\epsilon>0$
there are at most finitely many coordinates $x_{\lambda}$ with
$|x_{\lambda}|\geq\epsilon$. It is plain that the vector spaces satisfying

$\ell_{1}(\Lambda)\subseteq\ell_{p}(\Lambda)\subseteq
c_{0}(\Lambda)\subseteq\ell_{\infty}(\Lambda),\quad\text{whenever
$1<p<\infty$}.$

When $\Lambda$ is a finite set, all above spaces coincide; otherwise, all
inclusions are proper. We write $\ell_{p}$ and $c_{0}$ for
$\ell_{p}(\mathbb{N})$ and $c_{0}(\mathbb{N})$ as usual.

As in the finite dimensional case, the $\ell_{p}$–norm is strictly convex when
$1<p<\infty$. Two nonzero $\mathbf{x},\mathbf{y}$ in $\ell_{p}(\Lambda)$ are
parallel (resp. TEA), exactly when there is a scalar $t$ (resp. $t>0$) such
that $\mathbf{x}=t\mathbf{y}$. Therefore, any linear map of
$\ell_{p}(\Lambda)$ preserves parallel pairs and TEA pairs when $1<p<\infty$.

We study the cases when $p=1$ and $p=\infty$, and $\Lambda$ is an _infinite_
index set below. As in Section 2, we start with the following observations.
Note that $\ell_{\infty}(\Lambda)$ is isometrically isomorphic to the Banach
space $C(\beta\Lambda)$ of continuous functions on the Stone-Cech
compactification $\beta\Lambda$ of $\Lambda$, which consists of all
ultrafilters of the discrete space $\Lambda$.

######  Lemma 3.1.

  1. (a)

$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda},\mathbf{y}=(y_{\lambda})_{\lambda\in\Lambda}$
in $\ell_{1}(\Lambda)$ are parallel (resp. TEA) with respect to the
$\ell_{1}$–norm if and only if there is a unimodular scalar $\mu$ (resp.
$\mu=1$) such that $\mu\overline{x_{\lambda}}y_{\lambda}\geq 0$ for all
$\lambda\in\Lambda$.

  2. (b)

$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda},\mathbf{y}=(y_{\lambda})_{\lambda\in\Lambda}$
in $c_{0}(\Lambda)$ are parallel (resp. TEA) with respect to the
$\ell_{\infty}$–norm if and only if there is an index $\lambda$ in $\Lambda$
such that
$|\overline{x_{\lambda}}y_{\lambda}|=\|\mathbf{x}\|_{\infty}\|\mathbf{y}\|_{\infty}$
(resp.
$\overline{x_{\lambda}}y_{\lambda}=\|\mathbf{x}\|_{\infty}\|\mathbf{y}\|_{\infty}$).

  3. (c)

$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda},\mathbf{y}=(y_{\lambda})_{\lambda\in\Lambda}$
in $\ell_{\infty}(\Lambda)$ are parallel (resp. TEA) with respect to the
$\ell_{\infty}$–norm if and only if there is an ultrafilter $\mathfrak{U}$ on
$\Lambda$ such that
$\lim_{\mathfrak{U}}|\overline{x_{\lambda}}y_{\lambda}|=\|\mathbf{x}\|_{\infty}\|\mathbf{y}\|_{\infty}$
(resp.
$\lim_{\mathfrak{U}}\overline{x_{\lambda}}y_{\lambda}=\|\mathbf{x}\|_{\infty}\|\mathbf{y}\|_{\infty}$).

For each $\alpha\in\Lambda$, let
$\mathbf{e}_{\alpha}=(e_{\alpha,\lambda})_{\lambda\in\Lambda}$ be the
$\alpha$–coordinate vector with coordinates $e_{\alpha,\lambda}=1$ when
$\alpha=\lambda$ and $0$ elsewhere. We can identify any bounded linear map $T$
of $\ell_{1}(\Lambda)$ or $c_{0}(\Lambda)$ as the infinite “matrix”
$(t_{\alpha\beta})$ with
$t_{\alpha\beta}=\mathbf{e}_{\alpha}^{\mathrm{t}}(T\mathbf{e}_{\beta})$, where
$\mathbf{e}_{\alpha}^{\mathrm{t}}$ denotes the linear functional
$(x_{\lambda})_{\lambda\in\Lambda}\mapsto x_{\alpha}$ for any
$\alpha\in\Lambda$. However, there are unbounded linear maps such that their
representation “matrices” are zero. For example, consider any unbounded linear
functional $f$ of $c_{0}$ vanishing on the subspace of all finite sequences,
that is, $f(\mathbf{e}_{n})=0$ for all $n=1,2,\ldots$. Then the unbounded
linear map $\mathbf{x}\mapsto f(\mathbf{x})\mathbf{e}_{1}$ has zero “matrix”.

On the other hand, the Banach dual space of $\ell_{1}(\Lambda)$ is
$\ell_{\infty}(\Lambda)$ when one identify
$\mathbf{u}=(u_{\lambda})_{\lambda\in\Lambda}\in\ell_{\infty}(\Lambda)$ with
the bounded linear functional
$\mathbf{u}^{\mathrm{t}}=\sum_{\lambda\in\Lambda}u_{\lambda}\mathbf{e}_{\lambda}^{\mathrm{t}}$
(converging in the weak* topology
$\sigma(\ell_{\infty}(\Lambda),\ell_{1}(\Lambda)$). In this case, a nonzero
bounded linear operator $S$ of $\ell_{\infty}(\Lambda)$ can have zero
representation “matrix”. For example, let $g$ be any nonzero bounded linear
functional of $\ell_{\infty}$ vanishing on the essential null sequence space
$c_{0}$, and $S\mathbf{x}=g(\mathbf{x})\mathbf{e}_{1}$. However, a
$\sigma(\ell_{\infty}(\Lambda),\ell_{1}(\Lambda))$–$\sigma(\ell_{\infty}(\Lambda),\ell_{1}(\Lambda))$
continuous linear map is determined by its representation “matrix”.

Using the terminology in the finite dimensional case, we call a “matrix”
$U=(u_{\alpha\beta})$ a “ _monomial matrix_ ” if for each $\alpha\in\Lambda$
there is exactly one $\beta\in\Lambda$ such that $u_{\alpha\beta}\neq 0$. A
“monomial matrix” $U$ is a “ _generalized permutation matrix_ ” if all its
nonzero entries $|u_{\alpha\beta}|=1$, and it is a “ _diagonal unitary matrix_
” if $|u_{\alpha\alpha}|=1$ for each $\alpha\in\Lambda$. We also assume that
the linear map $U$ is bounded or
$\sigma(\ell_{\infty}(\Lambda),\ell_{1}(\Lambda))$–$\sigma(\ell_{\infty}(\Lambda),\ell_{1}(\Lambda))$
continuous, depending on the context, so that the representation “ _monomial
matrix_ ” $(u_{\alpha\beta})$ determines $U$. It is clear that a linear map
$T$ of $\ell_{1}(\Lambda)$ (resp. $c_{0}(\Lambda)$ or
$\ell_{\infty}(\Lambda)$) preserves parallel pairs or TEA pairs if, and only
if, $\gamma UTV$ does whenever $\gamma>0$, and $U,V$ are some invertible
“monomial matrices” (resp. “generalized permutation matrices”).

######  Theorem 3.2.

Let $T=(t_{\alpha\beta}):\ell_{1}(\Lambda)\to\ell_{1}(\Lambda)$ be a bounded
linear map.

  * (a)

$T$ preserves TEA pairs if and only if for each $\alpha\in\Lambda$ there is at
most one $\beta\in\Lambda$ such that $t_{\alpha\beta}\neq 0$.

  * (b)

$T$ preserves parallel pairs if and only if $T$ preserves TEA, or
$T={\mathbf{v}}{\mathbf{u}}^{\mathrm{t}}$ for some
$\mathbf{u}\in\ell_{\infty}(\Lambda)$ and $\mathbf{v}\in\ell_{1}(\Lambda)$.

###### Proof.

(a) Suppose for each $\alpha\in\Lambda$ there is at most one $\beta\in\Lambda$
such that $t_{\alpha\beta}\neq 0$. Write such $\beta=\alpha^{\prime}$ in this
case. Let
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda},\mathbf{y}=(y_{\lambda})_{\lambda\in\Lambda}$
be a TEA pair in $\ell_{1}(\Lambda)$. By Lemma 3.1,
$\overline{x_{\lambda}}y_{\lambda}\geq 0$ for all $\lambda\in\Lambda$. For
each $\lambda\in\Lambda$, the $\lambda$-coordinate of $T\mathbf{x}$ and
$T\mathbf{y}$ are $t_{\lambda\lambda^{\prime}}x_{\lambda^{\prime}}$ and
$t_{\lambda\lambda^{\prime}}y_{\lambda^{\prime}}$, respectively. Since
$\overline{t_{\lambda\lambda^{\prime}}x_{\lambda^{\prime}}}t_{\lambda\lambda^{\prime}}y_{\lambda^{\prime}}=\overline{x_{\lambda^{\prime}}}y_{\lambda^{\prime}}|t_{\lambda\lambda^{\prime}}|^{2}\geq
0$ for all $\lambda\in\Lambda$, we see that $T\mathbf{x},T\mathbf{y}$ form a
TEA pair. The converse follows from exactly the same arguments for the finite
dimensional case given in the proof of Theorem 1.1(a).

(b) It suffices to verify the necessity for the case when $T$ does not
preserve TEA. Suppose $T$ preservers parallel pairs, and the
${\alpha_{1}}$-row $(t_{{\alpha_{1}}\beta})_{\beta\in\Lambda}$ of its matrix
representation has more than one nonzero entries. Suppose the range of $T$ has
dimension at least two, and thus there is another ${\alpha_{2}}$–row
$(t_{{\alpha_{2}}\beta})_{\beta\in\Lambda}$ of $T$ linearly independent from
the ${\alpha_{1}}$-row. We are going to derive a contradiction.

Since the ${\alpha_{1}}$–row and the ${\alpha_{2}}$–row of $T$ are linearly
independent, there are distinct indices $\beta_{1},\beta_{2}$ such that the
$2\times 2$ matrix

$\begin{pmatrix}t_{{\alpha_{1}}\beta_{1}}&t_{{\alpha_{1}}\beta_{2}}\\\
t_{{\alpha_{2}}\beta_{1}}&t_{{\alpha_{2}}\beta_{2}}\end{pmatrix}$

is invertible. If both $t_{{\alpha_{1}}\beta_{1}},t_{{\alpha_{2}}\beta_{1}}$,
or both $t_{{\alpha_{1}}\beta_{2}},t_{{\alpha_{2}}\beta_{2}}$, are nonzero,
then by replacing $T$ with $PTQ$ for some suitable invertible “monomial
matrix” $P,Q$, we can assume that the above matrix assumes the form

$\begin{pmatrix}1&1\\\ 1&a\end{pmatrix}$

for some scalar $a\neq 1$. Then the argument in Case 1 of the proof of Theorem
1.1(b) derives a desired contradiction. If
$t_{{\alpha_{1}}\beta_{2}}=t_{{\alpha_{2}}\beta_{1}}=0$, say, then we search
for an other index $\beta_{3}$ with $t_{{\alpha_{1}}\beta_{3}}\neq 0$, and
such $\beta_{3}$ exists by assumption. If $t_{{\alpha_{2}}\beta_{3}}\neq 0$,
then it comes back to the first case above, and we are done. Suppose
$t_{{\alpha_{2}}\beta_{3}}=0$. By replacing $T$ with $P^{\prime}TQ^{\prime}$
for some suitable invertible “monomial matrices” $P^{\prime},Q^{\prime}$, we
may assume that $T$ has a “submatrix” of the form

$\begin{pmatrix}1&1&0\\\ 0&0&a\end{pmatrix}$

with $a\neq 0$ for the indices ${\alpha_{1}},{\alpha_{2}}$, and
$\beta_{1},\beta_{2},\beta_{3}$. Then the argument in Case 2 of the proof of
Theorem 1.1(b) provides us a contradiction, as well.

######  Lemma 3.3.

Let $\Lambda$ be an infinite index set.

  1. (a)

If ${\bf V}$ is a subspace of $\ell_{\infty}(\Lambda)$ such that any two
elements in ${\bf V}$ are parallel with respect to $\|\cdot\|_{\infty}$, then
${\rm dim}\,{\bf V}\leq 1$.

  2. (b)

Let $T:\ell_{\infty}(\Lambda)\to\ell_{\infty}(\Lambda)$ or
$T:c_{0}(\Lambda)\to c_{0}(\Lambda)$ be a nonzero linear map. Then $T$ is
injective if one of the following holds.

    1. (i)

$T$ preserves TEA pairs.

    2. (ii)

$T$ preserves parallel pairs with range space of dimension larger than one.

###### Proof.

(a) It follows from the proof of Lemma 2.2.

(b) We discuss only the case $T$ is a linear map of $\ell_{\infty}(\Lambda)$,
since the other case is similar.

(i) Let $T$ be a linear TEA preserver of $\ell_{\infty}(\Lambda)$ such that
$T(\mathbf{v})=0$ for some $\mathbf{v}=(v_{\lambda})_{\lambda\in\Lambda}$ with
$\|\mathbf{v}\|_{\infty}=\sup_{\lambda\in\Lambda}|v_{\lambda}|=1$. We will
show that $T=0$.

Suppose
$\Lambda^{\prime}=\\{\lambda\in\Lambda:|v_{\lambda}|<1\\}\neq\emptyset$. For
any $\lambda^{\prime}\in\Lambda^{\prime}$, since
$\xi\mathbf{v}+\mathbf{e}_{\lambda^{\prime}}$ and
$\xi\mathbf{v}-\mathbf{e}_{\lambda^{\prime}}$ form a TEA pair for large
$\xi>0$, so do
$T(\xi\mathbf{v}+\mathbf{e}_{\lambda^{\prime}})=T(\mathbf{e}_{\lambda^{\prime}})$
and
$T(\xi\mathbf{v}-\mathbf{e}_{\lambda^{\prime}})=-T(\mathbf{e}_{\lambda^{\prime}})$.
It forces $T(\mathbf{e}_{\lambda^{\prime}})=0$. For any
$\mathbf{u}=\sum_{\lambda\neq\lambda^{\prime}}v_{\lambda}\mathbf{e}_{\lambda}$
with zero $\lambda^{\prime}$–coordinate, we have
$\xi\mathbf{e}_{\lambda^{\prime}}+\mathbf{u}$ and
$\xi\mathbf{e}_{\lambda^{\prime}}-\mathbf{u}$ form a TEA pair for large
$\xi>0$, and so do
$T(\xi\mathbf{e}_{\lambda^{\prime}}+\mathbf{u})=T(\mathbf{u})$ and
$T(\xi\mathbf{e}_{\lambda^{\prime}}-\mathbf{u})=-T(\mathbf{u})$. It follows
$T(\mathbf{u})=0$. In general, for any
$\mathbf{x}=x_{\lambda^{\prime}}\mathbf{e}_{\lambda^{\prime}}+\mathbf{u}$ such
that $\mathbf{u}=\mathbf{x}-x_{\lambda^{\prime}}\mathbf{e}_{\lambda^{\prime}}$
has zero ${\lambda^{\prime}}$–coordinate,

$T(\mathbf{x})=x_{\lambda^{\prime}}T(\mathbf{e}_{\lambda^{\prime}})+T(\mathbf{u})=0.$

Hence $T=0$. Therefore, we may assume $\Lambda^{\prime}=\emptyset$.

Replacing $T$ with $TQ$ for some suitable “generalized permutation matrix”
$Q$, we may further assume that $\mathbf{v}=\mathbf{e}_{\Lambda}$, that is all
coordinates of $\mathbf{v}$ is $1$. For any distinct
$\lambda_{1},\lambda_{2}\in\Lambda$, since
$\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{2}}$ and
$\mathbf{e}_{\lambda_{1}}+\mathbf{e}_{\lambda_{2}}$ form a TEA pair, so do
$T(\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{2}})=-T(\mathbf{e}_{\lambda_{2}})$
and
$T(\mathbf{e}_{\lambda_{1}}+\mathbf{e}_{\lambda_{2}})=-T(\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{1}}-\mathbf{e}_{\lambda_{2}})$,
since $T(\mathbf{e}_{\Lambda})=0$. Therefore,

$\|T(\mathbf{e}_{\lambda_{1}})\|_{\infty}=\|T(\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{1}})\|_{\infty}=\|T(\mathbf{e}_{\lambda_{2}})\|_{\infty}+\|T(\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{1}}-\mathbf{e}_{\lambda_{2}})\|_{\infty}$

for any distinct $\lambda_{1},\lambda_{2}\in\Lambda$. Exchanging the roles of
$\lambda_{1}$ and $\lambda_{2}$, we see that

$T(\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{1}}-\mathbf{e}_{\lambda_{2}})=0.$

Replacing $\mathbf{v}=\mathbf{e}_{\Lambda}$ with
$\mathbf{v}^{\prime}=\mathbf{e}_{\Lambda}-\mathbf{e}_{\lambda_{1}}-\mathbf{e}_{\lambda_{2}}\neq
0$ (since $\Lambda$ has more than two elements), and arguing as above, we see
that $\Lambda^{\prime}=\\{\lambda_{1},\lambda_{2}\\}\neq\emptyset$, and then
$T=0$.

(ii) Suppose $T$ preserves parallel pairs. Assume $T$ is not injective, and
$\mathbf{v}=(v_{\lambda})_{\lambda\in\Lambda}$ is a norm one element such that
$T\mathbf{v}=0$. Let
$\Lambda^{\prime}=\\{\lambda\in\Lambda:|v_{\lambda}|<1\\}$. For any
$\mathbf{x},\mathbf{y}\in\ell_{\infty}(\Lambda)$ such that their
$\lambda$–coordinates
$\mathbf{e}_{\lambda}^{\mathrm{t}}\mathbf{x}=\mathbf{e}_{\lambda}^{\mathrm{t}}\mathbf{y}=0$
for all $\lambda$ outside $\Lambda^{\prime}$, we see that
$\xi\mathbf{v}+\mathbf{x}$ and $\xi\mathbf{v}+\mathbf{y}$ are parallel for
large $\xi>0$, and so are $T(\xi\mathbf{v}+\mathbf{x})=T(\mathbf{x})$ and
$T(\xi\mathbf{v}+\mathbf{y})=T(\mathbf{y})$. It follows from part (a) that the
space $\\{T(\mathbf{x}):\mathbf{e}_{\lambda}^{\mathrm{t}}\mathbf{x}=0\
\text{for all $\lambda$ outside $\Lambda^{\prime}$}\\}$ has dimension at most
one.

If $\Lambda=\Lambda^{\prime}$ then we are done. If
$\Lambda^{\prime\prime}=\Lambda\setminus\Lambda^{\prime}$ is nonempty, then by
replacing $T$ with $TQ$ for some “generalized permutation matrix” $Q$, we may
assume that $v_{\lambda}=1$ for all $\lambda\in\Lambda^{\prime\prime}$. Then
the proof of Lemma 2.3(b) shows that the range of $T$ has dimension at most
one.

We note that unlike the finite dimensional case, an injective TEA/parallel
pair linear preserver of $c_{0}(\Lambda)$ or $\ell_{\infty}(\Lambda)$ can be
non-surjective. For an example, consider the isometric right shift operator
$L$ of $\ell_{\infty}$ or $c_{0}$ by sending $\mathbf{e}_{n}$ to
$\mathbf{e}_{n+1}$ for $n=1,2,\ldots$. However, the following result
demonstrates that TEA/parallel pair linear preservers of $c_{0}(\Lambda)$ or
$\ell_{\infty}(\Lambda)$ are automatically bounded, and indeed scalar
multiples of injective isometries.

######  Theorem 3.4.

Let $\Lambda$ be an infinite index set. Let $T:c_{0}(\Lambda)\rightarrow
c_{0}(\Lambda)$ be a nonzero linear map. The following conditions are
equivalent to each other.

  1. (a)

$(T(\mathbf{u}),T(\mathbf{v}))$ is a TEA pair if and only if
$(\mathbf{u},\mathbf{v})$ is a TEA pair, for any $\mathbf{u},\mathbf{v}\in
c_{0}(\Lambda)$.

  2. (b)

$(T(\mathbf{u}),T(\mathbf{v}))$ is a parallel pair if and only if
$(\mathbf{u},\mathbf{v})$ is a parallel pair, for any
$\mathbf{u},\mathbf{v}\in c_{0}(\Lambda)$.

  3. (c)

$T$ is a scalar multiple of a (not necessarily surjective) linear isometry.

In this case, there is $\gamma>0$, a subset $\Lambda_{1}$ of $\Lambda$, a
family $\\{\mu_{\lambda}:\lambda\in\Lambda_{1}\\}$ of unimodular scalars, and
a surjective map $\tau:\Lambda_{1}\to\Lambda$ such that for any
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}\in c_{0}(\Lambda)$, the image
$\mathbf{y}=T(\mathbf{x})=(y_{\lambda})_{\lambda\in\Lambda}$ has coordinates

$\displaystyle y_{\beta}\ $
$\displaystyle=\gamma\mu_{\beta}x_{\tau(\beta)}\quad\text{for all
$\beta\in\Lambda_{1}$},$ (3.1) and $\displaystyle|y_{\beta^{\prime}}|\ $
$\displaystyle\leq\gamma\quad\text{when
$\beta^{\prime}\in\Lambda\setminus\Lambda_{1}$.}$ (3.2)

###### Proof.

The implications (c) $\implies$ (a) $\implies$ (b) are plain. We are verifying
(b) $\implies$ (c). Note that $T$ is injective. Indeed, if $T(\mathbf{x})=0$
for some nonzero $\mathbf{x}\in c_{0}(\Lambda)$, then the fact $T(\mathbf{x})$
and $T(\mathbf{e}_{\lambda})$ are parallel would imply that $\mathbf{x}$ and
$\mathbf{e}_{\lambda}$ are parallel for every $\lambda\in\Lambda$. But this
contradicts to the fact that $\Lambda$ is infinite and $\mathbf{x}$ is
essentially null. In particular, for any nonzero
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}\in c_{0}(\Lambda)$, its peak set

$\operatorname{Pk}(\mathbf{x})=\\{\lambda\in\Lambda:|x_{\lambda}|=\|\mathbf{x}\|_{\infty}\\}.$

is a nonempty proper subset of $\Lambda$.

For any $\alpha,\beta\in\Lambda$, we claim that

$\operatorname{Pk}(T(\mathbf{e}_{\alpha}))\cap\operatorname{Pk}(T(\mathbf{e}_{\beta}))=\emptyset\quad\text{whenever
$\alpha\neq\beta$.}$

In fact, if
$\lambda\in\operatorname{Pk}(T(\mathbf{e}_{\alpha}))\cap\operatorname{Pk}(T(\mathbf{e}_{\beta}))$
then both $T(\mathbf{e}_{\alpha})$ and $T(\mathbf{e}_{\beta})$ attain their
norms at the $\lambda$-coordinate, and thus they are parallel. This forces
$\mathbf{e}_{\alpha}$ and $\mathbf{e}_{\beta}$ are parallel, a contradiction.

Consider the disjoint union

$\Lambda_{1}=\bigcup_{\lambda\in\Lambda}\operatorname{Pk}(T(\mathbf{e}_{\lambda})).$

We define a surjective map $\tau:\Lambda_{1}\to\Lambda$ such that

$\tau(\beta)=\alpha\quad\text{if and only
if}\quad\beta\in\operatorname{Pk}(T(\mathbf{e}_{\alpha})).$

There is a unimodular scalar $\mu_{\lambda}$ such that the norm attaining
$\lambda$–coordinate of
$\overline{\mu_{\lambda}}T(\mathbf{e}_{\tau(\lambda)})$ is positive for every
$\lambda$ in $\Lambda_{1}$. Replacing $T$ with $QT$ for a suitable “diagonal
unitary matrix” $Q$, we can assume that all $\mu_{\lambda}=1$.

Let $\alpha_{1}=\tau(\beta_{1}),\alpha_{2}=\tau(\beta_{2})$ and
$\alpha_{3}=\tau(\beta_{3})$ be three distinct indices for
$\beta_{1},\beta_{2}$ and $\beta_{3}$ in $\Lambda_{1}$. Consider the linear
map $L:{\rm
span}\,(\mathbf{e}_{\alpha_{1}},\mathbf{e}_{\alpha_{2}},\mathbf{e}_{\alpha_{3}})\to{\rm
span}\,(\mathbf{e}_{\beta_{1}},\mathbf{e}_{\beta_{2}},\mathbf{e}_{\beta_{3}})$
defined by taking only the $\beta_{1}$–, $\beta_{2}$– and
$\beta_{3}$–coordinates of $T(\mathbf{x})$ when $\mathbf{x}\in{\rm
span}\,(\mathbf{e}_{\alpha_{1}},\mathbf{e}_{\alpha_{2}},\mathbf{e}_{\alpha_{3}})$.
We can identify $L$ with a $3\times 3$ matrix $A$ satisfying the assumption in
Lemma 2.4, from which we have a positive $\gamma$ such that
$L(\mathbf{e}_{\alpha_{j}})=\gamma\mathbf{e}_{\beta_{j}}$ for $j=1,2,3$.

The above argument shows that for any $\alpha$ in $\Lambda$, the
$\beta$-coordinate of $T(\mathbf{e}_{\alpha})$ is a fixed nonzero scalar
$\gamma$ whenever $\tau(\beta)=\alpha$, and all the other
$\beta^{\prime}$-coordinates with $\beta^{\prime}\in\Lambda_{1}$ are zero. In
other words,

$\displaystyle
T(\mathbf{e}_{\alpha})=\gamma\sum_{\tau(\beta)=\alpha}\mathbf{e}_{\beta}+\mathbf{t}_{\alpha}\quad\text{for
every $\alpha\in\Lambda$,}$ (3.3)

where $\mathbf{t}_{\alpha}\in c_{0}(\Lambda\setminus\Lambda_{1})$, that is,
all $\lambda$-coordinate of $\mathbf{t}_{\alpha}$ with $\lambda\in\Lambda_{1}$
are zero. Since the peak set
$\operatorname{Pk}(T(\mathbf{e}_{\alpha}))\subseteq\Lambda_{1}$, we have
$\|\mathbf{t}_{\alpha}\|_{\infty}<\gamma$. Note that the above sum must be
finite, as $T(\mathbf{e}_{\alpha})$ is essentially null. Replacing $T$ by
$T/\gamma$, we can assume that $\gamma=1$, and thus all
$\|T(\mathbf{e}_{\alpha})\|_{\infty}=1$.

We claim that $\|T(\mathbf{x})\|_{\infty}\leq 1$ whenever
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ in $c_{0}(\Lambda)$ has norm
one. To see this, we first assume that $x_{\alpha_{1}}=1$ and
$x_{\alpha_{2}}=x_{\alpha_{3}}=0$ with $\alpha_{1}=\tau(\beta_{1})$,
$\alpha_{2}=\tau(\beta_{2})$, $\alpha_{3}=\tau(\beta_{3})$ for distinct
indices $\alpha_{1},\alpha_{2},\alpha_{3}\in\Lambda$ and
$\beta_{1},\beta_{2},\beta_{3}\in\Lambda_{1}$. Since $\mathbf{x}$ and
$\mathbf{e}_{\alpha_{1}}$ are parallel, so are $T(\mathbf{x})$ and
$T(\mathbf{e}_{\alpha_{1}})=\mathbf{e}_{\beta_{1}}$. In particular,
$\|T(\mathbf{x})\|_{\infty}$ is attained at the $\beta_{1}$-coordinate of
$T(\mathbf{x})$. On the other hand, $\mathbf{x}$ is not parallel with
$\mathbf{e}_{\alpha_{2}},\mathbf{e}_{\alpha_{3}}$, and thus $T(\mathbf{x})$ is
not parallel with
$T(\mathbf{e}_{\alpha_{2}})=\mathbf{e}_{\beta_{2}},T(\mathbf{e}_{\alpha_{3}})=\mathbf{e}_{\beta_{3}}$.
With $\mathbf{x}$ playing the role of $\mathbf{e}_{\alpha_{1}}$, and
$T(\mathbf{x})$ playing the role of $\mathbf{e}_{\beta_{1}}$, the above
argument shows that the $\beta_{1}$-coordinate of $T(\mathbf{x})$ is
$\|T(\mathbf{x})\|_{\infty}=1$. In general, for any norm one
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ in $c_{0}(\Lambda)$, we may
choose distinct indices $\alpha_{1}=\tau(\beta_{1})$,
$\alpha_{2}=\tau(\beta_{2})$, $\alpha_{3}=\tau(\beta_{3})$, and assume its
$\alpha_{1}$-coordinate equals $1$. Then

$\displaystyle\Big{|}\|T(\mathbf{x})\|_{\infty}-\|T(\mathbf{x}-x_{\alpha_{2}}\mathbf{e}_{\alpha_{2}}-x_{\alpha_{3}}\mathbf{e}_{\alpha_{3}})\|_{\infty}\Big{|}\leq\|T(x_{\alpha_{2}}\mathbf{e}_{\alpha_{2}}+x_{\alpha_{3}}\mathbf{e}_{\alpha_{3}})\|_{\infty}$

implies

$\displaystyle\Bigg{|}\|T(\mathbf{x})\|_{\infty}-1\Bigg{|}$
$\displaystyle\leq\left\|x_{\alpha_{2}}\left(\sum_{\tau(\beta_{2})=\alpha_{2}}\mathbf{e}_{\beta_{2}}+\mathbf{t}_{\alpha_{2}}\right)+x_{\alpha_{3}}\left(\sum_{\tau(\beta_{3})=\alpha_{3}}\mathbf{e}_{\beta_{3}}+\mathbf{t}_{\alpha_{3}}\right)\right\|_{\infty}$
$\displaystyle\leq|x_{\alpha_{2}}|+|x_{\alpha_{3}}|.$

Since $\mathbf{x}$ is essentially null, we can choose as small as possible
$|x_{\alpha_{2}}|,|x_{\alpha_{3}}|$. It follows that
$\|T(\mathbf{x})\|_{\infty}=1$, as claimed. In particular, $T$ is an isometry.

Going back to the original bounded linear map $T$, the formula (3.3) becomes

$T(\mathbf{e}_{\alpha})=\gamma\sum_{\tau(\beta)=\alpha}\mu_{\beta}\mathbf{e}_{\beta}+\mathbf{t}_{\alpha},$

where $\mathbf{t}_{\alpha}\in c_{0}(\Lambda\setminus\Lambda_{1})$ with
$\|\mathbf{t}_{\alpha}\|_{\infty}<\gamma$ for every $\alpha\in\Lambda$. For
any
$\mathbf{x}=(x_{\alpha})_{\alpha\in\Lambda}=\sum_{\alpha\in\Lambda}x_{\alpha}\mathbf{e}_{\alpha}$
in $c_{0}(\Lambda)$, the boundedness of $T$ ensures that

$T(\mathbf{x})=\sum_{\alpha\in\Lambda}x_{\alpha}T(\mathbf{e}_{\alpha})=\sum_{\alpha\in\Lambda}x_{\alpha}\left(\gamma\sum_{\tau(\beta)=\alpha}\mu_{\beta}\mathbf{e}_{\beta}+\mathbf{t}_{\alpha}\right)=\sum_{\alpha\in\Lambda}\left(\big{(}\sum_{\tau(\beta)=\alpha}\gamma\mu_{\beta}x_{\tau(\beta)}\mathbf{e}_{\beta}\big{)}+x_{\alpha}\mathbf{t}_{\alpha}\right).$

It follows (3.1).

To verify (3.2), we may assume that there is
$\beta^{\prime}\in\Lambda\setminus\Lambda_{1}$ and a norm one
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ in $c_{0}(\Lambda)$ such that
$\mathbf{y}=T(\mathbf{x})$ has $\beta^{\prime}$–coordinate with
$|y_{\beta^{\prime}}|>\gamma$. Since the $\beta$–coordinates of $\mathbf{y}$
are bounded by $\gamma$ for all $\beta\in\Lambda_{1}$ due to (3.1), we see
that $\mathbf{y}=T(\mathbf{x})$ and $T(\mathbf{e}_{\alpha})$ have disjoint
peak sets, and thus they are not parallel to each other for any $\alpha$ in
$\Lambda=\tau(\Lambda_{1})$. However, if $\mathbf{x}$ attains its norm at the
${\alpha_{1}}$–coordinate, then $\mathbf{x}$ is parallel with
$\mathbf{e}_{\alpha_{1}}$, and thus $T(\mathbf{x})$ is parallel with
$T(\mathbf{e}_{\alpha_{1}})$, a contradiction. We thus establish (3.2). It is
now clear that $T/\gamma$ is an into isometry.

Finally, we note that a (scalar multiple of an) into isometry of
$c_{0}(\Lambda)$ assumes the stated forms (3.1) and (3.2), due to
Holsztynski’s Theorem (see, e.g., [3]).

The case for linear parallel/TEA pair preservers of $\ell_{\infty}(\Lambda)$
seems to be more complicated, as an element
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}$ in $\ell_{\infty}(\Lambda)$
might have empty peak set, that is, all its coordinates
$|x_{\lambda}|<\|x\|_{\infty}$. This makes the argument in the proof of
Theorem 3.4 cannot be transported directly. However,
$\ell_{\infty}(\Lambda)\cong C(\beta\Lambda)$, and we can apply the following
result for abelian C*-algebras.

######  Theorem 3.5 ([7]).

Let $X,Y$ be compact Hausdorff spaces containing of at least three points. Let
$T:C(X)\to C(Y)$ be a bijective linear map such that both $T,T^{-1}$ preserve
parallel pairs. Then there is a homeomorphism $\sigma:Y\to X$ and a constant
unimodulus function $h\in C(Y)$ such that

$Tf(y)=h(y)f(\sigma(y))\quad\text{for any $f\in C(X)$ and $y\in Y$.}$

Below is an infinite dimensional analog of Corollary 1.3.

######  Corollary 3.6.

Let $\Lambda$ be an infinite index set. Let
$T:\ell_{\infty}(\Lambda)\rightarrow\ell_{\infty}(\Lambda)$ be a bijective
linear map. The following conditions are equivalent to each other.

  1. (a)

Both $T$ and $T^{-1}$ send TEA pairs to TEA pairs.

  2. (b)

Both $T$ and $T^{-1}$ send parallel pairs to parallel pairs.

  3. (c)

$T$ is a scalar multiple of a surjective linear isometry.

In this case, there is $\gamma>0$, a family
$\\{\mu_{\lambda}:\lambda\in\Lambda\\}$ of unimodular scalars, and a bijective
map $\tau:\Lambda\to\Lambda$ such that for any
$\mathbf{x}=(x_{\lambda})_{\lambda\in\Lambda}\in\ell_{\infty}(\Lambda)$, the
image $\mathbf{y}=T(\mathbf{x})=(y_{\lambda})_{\lambda\in\Lambda}$ has
coordinates

$\displaystyle y_{\lambda}=\gamma\mu_{\lambda}x_{\tau(\lambda)}\quad\text{for
all $\lambda\in\Lambda$}.$ (3.4)

In other words, $T$ is a scalar multiple of a “generalized permutation
matrix”.

###### Proof.

While the implications (c) $\implies$ (a) $\implies$ (b) are plain, Theorem
3.5 establishes (b) $\implies$ (c) when we identify $\ell_{\infty}(\Lambda)$
with $C(\beta\Lambda)$. We note that $\Lambda$ consists of all isolated points
of $\beta\Lambda$. Thus the homeomorphism $\sigma$ induces a bijective map
$\tau$ from $\Lambda$ onto itself to implement (3.4).

######  Problem 3.7.

Can one replace the two direction preserver conditions by that the linear map
$T$ sends parallel/TEA pairs to parallel/TEA pairs in Theorem 3.4 and
Corollary 3.6, as in the finite dimensional case?

## Acknowledgment

Li is an affiliate member of the Institute for Quantum Computing, University
of Waterloo; his research was partially supported by the Simons Foundation
Grant 851334. M.-C. Tsai, Y.-S. Wang and N.-C. Wong are supported by Taiwan
NSTC grants 112-2115-M-027-002, 113-2115-M-005-008-MY2 and
112-2115-M-110-006-MY2, respectively.

## References

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# Quasideterminant Solutions of Noncommutative Equation of Langmuir
Oscillations

Irfan Mahmood  and  Hira Sohail  Centre for High Energy Physics, University of
the Punjab, 54590, Lahore, Pakistan<EMAIL_ADDRESS>
###### Abstract.

In this article we present nonncommutative analogue of equation of Langmuir
oscillations and its Darbboux solutions in additive form with their $N$-th
generalization in terms of quasideterminants. We also derive the
noncommutative ricatti equation from its linear system that yields the
Bäcklund transformation for the commutative version of equation of Langmuir
oscillations with classical limit. The last section, involves the derivation
of reduction of that Bäcklund transformation to the Darboux solutions under
commutative limit.

Keywords: Noncommutative equation of Langmuir oscillations , Darboux
transformation, Quasideterminats, Riccati equation, Bäcklund transformation

##  1\. Introduction

The differential difference equation

$u_{nt}=u_{n}(u_{n-1}-u_{n+1})$ (1)

appears in analysis of spectrum structure associated to the Langmuir
oscillations in plasma which completely integrable in classical framework as
it possesses Lax representation [1]. The non-abelian analogue of equation (1)

$u_{nt}=u_{n-1}u_{n}-u_{n}u_{n+1}$ (2)

has been computed [2] from the compatibility of following linear system

$u_{n}\psi_{n+1}=\lambda\psi_{n}-\psi_{n-1}$ (3)

$\psi_{(n)t}=-u_{n}u_{n+1}\psi_{n+2}.$ (4)

with its connection to discrete nonlinear Schr$\ddot{o}$dinger equation [3,
4]. Moreover Darboux transformation for that non-abelian version has been
presented in [2] in multiplicative form as

$u_{n}\left[1\right]=\varphi_{n-2}\varphi^{-1}_{n}u_{n}(\varphi_{n-1}\varphi^{-1}_{n+1})^{-1}.$
(5)

that generates all solutions with non-zero seed solution, means $u_{n}\neq 0$.
Initialy, That Darboux method was developed by [5] to find transformations on
potential of the Schr$\ddot{o}$dinger equation which satisfies Korteweg-de
Vries equation [6] in framework of Lax formalism. Later on few remarkable
results on DT were analysed by [7] to reveal its importance in theory of
integrable system and the more efficient implementations on various nonlinear
physical systems were developed by V. Matveev [8] to construct the exact
solutions of these systems.. The successful implementations of these
transformations have been shown in the analysis of various mathematical
features of graphene [9] and has also fruitful applications in cavity quantum
electrodynamics [10, 11] for the dynamical analysis of the propagation of
associated disturbance Moreover these transformations significantly extended
to construct the determinantal solutions of noncommutative integrable systems
such as in case of noncommutative Painlevé second equation [12] with its
associated nocommutative Toda equation [13]
In this article, we construct the Darboux transformation (DT) for the
noncommutative analogue of equation (1) in additive form as

$u_{n}\left[1\right]=u_{n}-\varphi^{\prime}_{n}\varphi^{-1}_{n-1}.$ (6)

which generates all possible solutions in zero background as seed solution
$u_{n}=0$, that is initial trivial solution of equation (2) and $\varphi_{n}$,
$\varphi_{n}$ are the particular solutions at particular values of $\lambda$.
Further the NC DT (6) is generalized to $N$-th form fin terms of
quasideterminat with exact solution in commutative case. The end section
encloses the derivation of Ricatti equation associated to noncommutative
analogue of equation (1) that yields its Backlund transformation which are
reducible to NC DT (6) .

##  2\.  Noncommutative Darboux Solutions of Equations of Langmuir
Oscillations

For the noncommutative extension of equation (2), we consider $u_{i}$ and
independent variable $t$ are purely noncommuting objects such as
$[u_{i},t]\neq 0$ and the time derivation is defined as
$\partial_{t}f^{-1}(t)=-f^{-1}\partial_{t}ff^{-1}$, further the fields and
their derivatives are also noncommuting elements. From the compatibility
condition of linear systems (3) and (4) in nocommutative frame work, we obtain

$u_{nt}=u_{n-1}u_{n}-u_{n}u_{n+1}$ (7)

the NC verion of equation (2) and the Darboux solution of above equation can
be derived through its associated linear systems with Darboux transformation
[15] on arbitrary function $\psi_{n}$ defining in NC framework as

$\psi_{n}\left[1\right]=\psi_{n}-\varphi_{n}\varphi^{-1}_{n+1}\psi_{n+1}$ (8)

Now under above transformation (8) the linear system (3) can be written in
following form

$u_{n}\left[1\right]\psi_{n+1}\left[1\right]=\lambda\psi_{n}\left[1\right]-\psi_{n-1}\left[1\right].$
(9)

Now after substituting the values for transformed eigenfuctions from (8) into
above transformed expression and then with the help of system (4) the
resulting expression yields Darboux transformation on $u_{n}$ as below

$u_{n}\left[1\right]=u_{n}-\lambda\varphi_{n}\varphi^{-1}_{n-1}.$ (10)

we can also express above result as follow, taking
$\lambda\varphi_{n}=\varphi^{\prime}_{n}$

$u_{n}\left[1\right]=u_{n}-\varphi^{\prime}_{n}\varphi^{-1}_{n-1}.$ (11)

The above transformation involves new solution $u_{n}\left[1\right]$ , old
solution $u_{n}$ of equation (2) also called the seed solution and the
particular solutions of linear systems (3) and (4). Here the comparison of
Darboux solution (11) and with result on Darboux solution obtained in [2]
shows a difference, the transformations (11) are additive holds for all seed
solution even for the trivial solution $u_{n}=0$ of equation (7) in NC frame
as well as in non-abelian case and also in classical framework under
commutative limit.

The one fold Darboux transformation (8) with its second iteration can be
expressed in form of quasideterminat as below with setting
$\psi_{n}=\psi_{0}$,,$\psi_{n+1}=\psi^{\prime}_{0}$ and $\varphi_{n}=\psi_{1}$
, $\varphi_{n+1}=\psi^{\prime}_{1}$ and defining $\lambda\psi=\psi^{\prime}$ ,
then we can present one fold Darboux transformation in terms of
quasideterminant as

$\psi_{n}\left[1\right]=\begin{vmatrix}\psi_{0}&\psi_{1}\\\
\lambda_{0}\psi_{0}&{\boxed{\lambda_{1}\psi_{1}}}\end{vmatrix}$ (12)

and the two fold NC Darboux transformation can be evaluated as

$\psi_{n}\left[2\right]=\begin{vmatrix}\psi_{0}&\psi_{1}&\psi_{2}\\\
\lambda_{0}\psi_{0}&\lambda_{1}\psi_{1}&\lambda_{2}\psi_{2}\\\
\lambda^{2}_{0}\psi_{0}&\lambda^{2}_{1}\psi_{1}&{\boxed{\lambda^{2}_{2}\psi_{2}}}\end{vmatrix}$
(13)

further can be generalized to $N$-th form as below

$\psi_{n}\left[N\right]=\begin{vmatrix}\psi_{0}&\psi_{1}&\cdots&\psi_{N-1}&\psi_{N}\\\
\lambda_{0}\psi_{0}&\lambda_{1}\psi_{1}&\cdots&\lambda_{N-1}\psi_{N-1}&\lambda_{N}\psi_{N}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
\lambda^{N-1}_{0}\psi_{0}&\lambda^{N-1}_{1}\psi_{1}&\cdots&\lambda^{N-1}_{N-1}\psi_{N-1}&\lambda^{N-1}_{N}\psi_{N}\\\
\lambda^{N}_{0}\psi_{0}&\lambda^{N}_{1}\psi_{1}&\cdots&\lambda^{N}_{N-1}\psi_{N-1}&{\boxed{\lambda^{N}_{N}\psi_{N}}}\end{vmatrix}$
(14)

Now in the similar way the one fold Darboux solution (6) can be generalized to
$N$-th form in terms of quasideterminants as

$u_{n}\left[N+1\right]=u_{n}\left[N\right]-\psi^{\prime}_{n}\left[N\right]\psi_{n-1}\left[N\right]^{-1}.$
(15)

Here for $N=0$, we have $u_{n}\left[0\right]=u_{n}$, initial solution and
$\psi_{n}\left[0\right]=\varphi_{n}$ ,
$\psi_{n-1}\left[0\right]=\varphi_{n-1}$ are the particular solutions.
Further, we may construct the $N$ fold expression for
$\psi_{n-1}\left[N\right]$ with the help of (8) with the replacement of $n$ by
$n-1$ and setting $\psi_{n-1}=\lambda_{0}\psi_{0}$

$\psi_{n-1}\left[N\right]=\begin{vmatrix}\psi_{N}&\psi_{N-1}&\cdots&\psi_{1}&\psi_{0}\\\
\lambda_{N}\psi_{N}&\lambda_{N-1}\psi_{N-1}&\cdots&\lambda_{1}\psi_{N-1}&\lambda_{0}\psi_{0}\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\
\lambda^{N-1}_{N}\psi_{N}&\lambda^{N-1}_{N-1}\psi_{N-1}&\cdots&\lambda^{N-1}_{1}\psi_{1}&\lambda^{N-1}_{0}\psi_{0}\\\
\lambda^{N}_{N}\psi_{N}&\lambda^{N}_{N-1}\psi_{N-1}&\cdots&\lambda^{N}_{1}\psi_{1}&{\boxed{\lambda^{N}_{0}\psi_{0}}}\end{vmatrix}$
(16)

here we can assume $\psi_{n-1}\left[0\right]=\varphi_{n-1}$ as the initial
untransformed solution, that is particular solution of the linear system.

##  3\.  NC Riccati euation

To construct the NC riccati equation, let us start with setting

$R_{n}=\psi_{n}\psi^{-1}_{n-1}$ (17)

and taking time derivation of above expression (17), we obtain

$R^{\prime}_{n}=u_{n-1}R_{n}-\lambda u_{n}+\lambda^{2}R_{n}-\lambda
R^{2}_{n}-R_{n}u_{n}$ (18)

by using the values from linear systems (2) and (3), where $R^{\prime}=R_{t}$.
Now substituting the value for $(u_{n}-\lambda R_{n})$ from system (3),
finally we get NC riccati equation associated to equation (7) as below

$R^{\prime}_{n}=u_{n-1}R_{n}-R_{n}u_{n}+\lambda R_{n-1}-\lambda R^{2}_{n}$
(19)

In next section, we show the reduction of above NC riccati equation (19) into
Bäcklund transformation under the commutative limt, further can be simplified
to the Darboux transformtion for the classical analogue of equation (7) .

##  4\.  NC Ricatti equation with commutative limit

This can be shown that for the classical version (1) the Darboux solution (6)
will take the following form

$u_{n}\left[1\right]=u_{n}-\frac{\varphi^{\prime}_{n}}{\varphi_{n-1}}.$ (20)

where $u_{n}$ and $\varphi_{n}$ are scalars with same linear systems (3) and
(4). Now the NC ricatti equation under the commutative limit becomes

$R^{\prime}_{n}=(u_{n-1}-u_{n})R_{n}+\lambda R_{n-1}-\lambda R^{2}_{n}.$ (21)

The above ricatti equation (21) under charge partity time reversal (CPT)
symmetry transformation [16] becomes

$-R^{\prime}_{n}=(u_{n-1}\left[1\right]-u_{n}\left[1\right])R_{n}+\lambda
R_{n-1}-\lambda R^{2}_{n}.$ (22)

where $u_{n}\left[1\right]$ is new solution and $R^{\prime}_{n}$ is replaced
by $-R^{\prime}_{n}$ . Now subtracting (21) from (22) , we get

$u_{n-1}\left[1\right]-u_{n-1}=(u_{n}\left[1\right]-u_{n})+(\lambda\left[1\right]-\lambda)\frac{R_{n-1}}{R_{n}}-(\lambda\left[1\right]-\lambda)R_{n}.$
(23)

that equation may be regarded as Bäcklund transformation for equation (1) with
Darboux solution (20). Now we assume that
$\lambda\left[1\right]-\lambda=\epsilon$ a very small difference, then above
expression can be written as

$\frac{u_{n-1}\left[1\right]-u_{n}\left[1\right]}{\epsilon}=-(\frac{u_{n-1}-u_{n}}{\epsilon})+\frac{R_{n-1}}{R_{n}}-R_{n}$
(24)

and under the limiting case $\epsilon 0$, then finally we get

$\frac{du_{n}\left[1\right]}{dt}=\frac{du_{n}}{dt}+\frac{R_{n-1}}{R_{n}}-R_{n}.$
(25)

The above expression becomes equivalent to the Darboux solution (20),
$u_{n}\left[1\right]=u_{n}-\frac{\varphi^{\prime}_{n}}{\varphi_{n-1}}$, with
condition
$\frac{R_{n-1}}{R_{n}}-R_{n}=\frac{d}{dt}\frac{\varphi^{\prime}_{n}}{\varphi_{n-1}}$
and taking constant of integration as zero. Here it has been shown that the
commutative version of ricatti equation yields the Darboux solution through
the Bäcklund transformation in classical framework.

##  5\. Conclusion:

In this paper, the noncommutative analogue equations of Langmuir oscillations
has been presented with its Darboux transformation in additive structure as
most of the integrable posses in noncommutative as well as in classical
frameworks. Further $N$-fold darboux solutions have presented in term of
quasideterminants applicable for zero background seed solution. Moreover, the
associated NC ricatti equation is presented which reduced to the Darboux
expression through the Bäcklund transformation in classical framework. Further
motivation involves its connection to Discrete noncommutative NLS equation as
possesses in classical case and also to investigate its solutions in terms of
quantum determinants for its matrix version.

Acknowledgments:I am very thankful to the Punjab University 54590 on providing
me facilities. My special thanks to Science and technology commission of
Shanghai, China recruited me to work on project, No $.20590742900$, as Belt
and Road Yung Scientist fromm 2020 to 2023. This work is completed as one of
the elements of that project.

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