# Extremal problems for the eccentricity matrices of complements of trees

Iswar Mahato Department of Mathematics, Indian Institute of Technology
Kharagpur, Kharagpur 721302, India. Email<EMAIL_ADDRESS><EMAIL_ADDRESS>M. Rajesh Kannan Department of Mathematics, Indian
Institute of Technology Hyderabad, Hyderabad 502285, India. Email:
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

The eccentricity matrix of a connected graph $G$, denoted by $\mathcal{E}(G)$,
is obtained from the distance matrix of $G$ by keeping the largest nonzero
entries in each row and each column, and leaving zeros in the remaining ones.
The $\mathcal{E}$-eigenvalues of $G$ are the eigenvalues of $\mathcal{E}(G)$,
in which the largest one is the $\mathcal{E}$-spectral radius of $G$. The
$\mathcal{E}$-energy of $G$ is the sum of the absolute values of all
$\mathcal{E}$-eigenvalues of $G$. In this article, we study some of the
extremal problems for eccentricity matrices of complements of trees and
characterize the extremal graphs. First, we determine the unique tree whose
complement has minimum (respectively, maximum) $\mathcal{E}$-spectral radius
among the complements of trees. Then, we prove that the
$\mathcal{E}$-eigenvalues of the complement of a tree are symmetric about the
origin. As a consequence of these results, we characterize the trees whose
complement has minimum (respectively, maximum) least $\mathcal{E}$-eigenvalues
among the complements of trees. Finally, we discuss the extremal problems for
the second largest $\mathcal{E}$-eigenvalue and the $\mathcal{E}$-energy of
complements of trees and characterize the extremal graphs. As an application,
we obtain a Nordhaus-Gaddum type lower bounds for the second largest
$\mathcal{E}$-eigenvalue and $\mathcal{E}$-energy of a tree and its
complement.

AMS Subject Classification (2010): 05C50, 05C35.

Keywords. Complements of trees, Eccentricity matrix, $\mathcal{E}$-spectral
radius, Second largest $\mathcal{E}$-eigenvalue, Least
$\mathcal{E}$-eigenvalue, $\mathcal{E}$-energy.

##  1 Introduction

Throughout the paper, we consider finite, simple, and connected graphs. Let
$G=(V(G),E(G))$ be a graph with the vertex set
$V(G)=\\{v_{1},v_{2},\ldots,v_{n}\\}$ and the edge set
$E(G)=\\{e_{1},e_{2},\ldots,e_{m}\\}$. The number of vertices in $G$ is the
order of $G$. If two vertices $u$ and $v$ in $G$ are adjacent, then we write
$u\sim v$, otherwise $u\nsim v$. The adjacency matrix $A(G)$ of $G$ is the
$n\times n$ matrix with its rows and columns indexed by the vertices of $G$,
and the entries are defined as

$A(G)_{uv}=\begin{cases}\text{$1$}&\quad\text{if $u\sim v$,}\\\
\text{0}&\quad\text{otherwise.}\end{cases}$

Let $\lambda_{1}(G)\geq\lambda_{2}(G)\geq\ldots\geq\lambda_{n}(G)$ be the
eigenvalues of the adjacency matrix $A(G)$ of $G$. The largest eigenvalue of
$A(G)$ is the spectral radius of $G$. The energy (or the $A$-energy) of $G$ is
defined as
$E_{A}(G)=\sum_{i=1}^{n}\mathinner{\\!\left\lvert\lambda_{i}(G)\right\rvert}$.
The distance between the vertices $u$ and $v$ in $G$, denoted by $d_{G}(u,v)$,
is the length of a shortest path between them in $G$, and define
$d_{G}(u,u)=0$ for all $u\in V(G)$. The distance matrix $D(G)$ of $G$ is the
$n\times n$ matrix whose rows and columns are indexed by the vertices of $G$
and the $(u,v)$-th entry is equal to $d_{G}(u,v)$. Let $N_{G}(v)$ denote the
collection of vertices that are adjacent to the vertex $v$ in $G$, and
$N_{G}(v)$ is called the neighborhood of $v$ in $G$. The eccentricity
$e_{G}(v)$ of a vertex $v\in V(G)$ is the maximum distance from $v$ to all
other vertices of $G$. The maximum eccentricity of all vertices of $G$ is the
diameter of $G$, which is denoted by $\operatorname{diam}(G)$. The diametrical
path is a path whose length is equal to the diameter of $G$.

The eccentricity matrix of a graph $G$ on $n$ vertices, denoted by
$\mathcal{E}(G)$, is the $n\times n$ matrix whose rows and columns are indexed
by the vertices of $G$, and the entries are defined as

$\mathcal{E}(G)_{uv}=\begin{cases}\text{$d_{G}(u,v)$}&\quad\text{if
$d_{G}(u,v)=\min\\{e_{G}(u),e_{G}(v)\\}$,}\\\
\text{0}&\quad\text{otherwise.}\end{cases}$

The eccentricity matrix $\mathcal{E}(G)$ of a graph $G$ is a real symmetric
matrix, and hence all of its eigenvalues are real. The eigenvalues of
$\mathcal{E}(G)$ are the $\mathcal{E}$-eigenvalues of $G$, in which the
largest one is the $\mathcal{E}$-spectral radius of $G$. The set of all
$\mathcal{E}$-eigenvalues of $G$ is the $\mathcal{E}$-spectrum of $G$. If
$\xi_{1}>\xi_{2}>\ldots>\xi_{k}$ are the distinct $\mathcal{E}$-eigenvalues of
$G$, then we write the $\mathcal{E}$-spectrum of $G$ as

$\operatorname{Spec}_{\varepsilon}(G)=\left\\{{\begin{array}[]{cccc}\xi_{1}&\xi_{2}&\ldots&\xi_{k}\\\
m_{1}&m_{2}&\ldots&m_{k}\\\ \end{array}}\right\\},$

where $m_{i}$ is the multiplicity of $\xi_{i}$ for $i=1,2,\ldots,k$.

Spectral extremal problems are one of the interesting problems in spectral
graph theory. Recently, the extremal problems for eccentricity matrices of
graphs have gained significant importance and attracted the attention of
researchers. In [23], Wei et al. considered the extremal problem for
$\mathcal{E}$-spectral radius of trees and determined the trees with minimum
$\mathcal{E}$-spectral radius among all trees on $n$ vertices. Also, they
characterized the trees with minimum $\mathcal{E}$-spectral radius among the
trees with a given diameter. In [14], the authors studied the minimal problem
for $\mathcal{E}$-spectral radius of graphs with a given diameter and
characterized the extreme graphs. Moreover, they identified the unique
bipartite graph with minimum $\mathcal{E}$-spectral radius. Wang et al. [19]
characterized the graphs with minimum and second minimum
$\mathcal{E}$-spectral radius as well as the graphs with maximum least and
second least $\mathcal{E}$-eigenvalues. Recently, He and Lu [4] considered the
maximal problem for the $\mathcal{E}$-spectral radius of trees with the fixed
odd diameter and determined the extremal trees. Wei et al. [25] characterized
the trees with second minimum $\mathcal{E}$-spectral radius and identified the
trees with the small matching number having the minimum $\mathcal{E}$-spectral
radius. Very Recently, Wei and Li [24] studied the relationship between the
majorization and $\mathcal{E}$-spectral radius of complete multipartite graphs
and determined the extremal complete multipartite graphs with minimum and
maximum $\mathcal{E}$-spectral radius. Mahato and Kannan [16] considered the
extremal problem for the second largest $\mathcal{E}$-eigenvalue of trees and
determined the unique tree with minimum second largest
$\mathcal{E}$-eigenvalue among all trees on $n$ vertices other than the star.
For more advances on the eccentricity matrices of graphs, we refer to [7, 8,
13, 15, 17, 18, 21, 22].

The eccentricity energy (or the $\mathcal{E}$-energy) of a graph $G$ is
defined [20] as

$E_{\mathcal{E}}(G)=\sum_{i=1}^{n}\mathinner{\\!\left\lvert\xi_{i}\right\rvert},$

where $\xi_{1},\xi_{2},\ldots,\xi_{n}$ are the $\mathcal{E}$-eigenvalues of
$G$. Recently, many researchers focused on the eccentricity energy of graphs.
In [20], Wang et al. studied the $\mathcal{E}$-energy of graphs and obtained
some bounds for the $\mathcal{E}$-energy of graphs and determined the
corresponding extremal graphs. Lei et al. [8] obtained an upper bound for the
$\mathcal{E}$-energy of graphs and characterized the extremal graphs. Very
recently, Mahato and Kannan [16] studied the minimization problem for the
$\mathcal{E}$-energy of trees and characterized the trees with minimum
$\mathcal{E}$-energy among all trees on $n$ vertices. For more details about
the $\mathcal{E}$-energy of graphs, we refer to [14, 15, 18].

Motivated by the above-mentioned works, in this article, we study the extremal
problems for eccentricity matrices of complements of trees and characterize
the extremal graphs for the $\mathcal{E}$-spectral radius, second largest
$\mathcal{E}$-eigenvalue, least $\mathcal{E}$-eigenvalue and
$\mathcal{E}$-energy among the complements of trees. The extremal problems for
the complements of trees with respect to the other graph matrices have been
studied in [2, 9, 10, 11]. For a tree $T$, let $T^{c}$ be the complement of
$T$. Throughout the paper, we always assume that $T^{c}$ is connected; hence,
$T$ is not isomorphic to the star graph. Let $\mathcal{T}_{n}^{c}$ denote the
complements of all trees on $n$ vertices. Note that if $T$ is a tree with
$\operatorname{diam}(T)>3$, then $\mathcal{E}(T^{c})=2A(T)$. If
$\operatorname{diam}(T)=3$, then $\mathcal{E}(T^{c})\geq 2A(T)$ entrywise.
Since $A(T)$ is irreducible, therefore $\mathcal{E}(T^{c})$ is also
irreducible.

The article is organized as follows: In section $2$, we collect needed
notations and some preliminary results. In section $3$, we characterize the
extremal graphs with the minimum and maximum $\mathcal{E}$-spectral radius
among the complements of trees. As a consequence, we determine the unique
graphs with the minimum and maximum least $\mathcal{E}$-eigenvalues among the
complements of trees. We discuss the extremal problems for the second largest
$\mathcal{E}$-eigenvalue and the $\mathcal{E}$-energy of complements of trees
in section $4$ and section $5$, respectively.

##  2 Preliminaries

In this section, we introduce some notations and collect some preliminary
results, which will be used in the subsequent sections. First, we define the
quotient matrix and equitable partition.

######  Definition 2.1 (Equitable partition [1]).

Let $A$ be a real symmetric matrix whose rows and columns are indexed by
$X=\\{1,2,\ldots,n\\}$. Let $\Pi=\\{X_{1},X_{2},\ldots,X_{m}\\}$ be a
partition of $X$. The characteristic matrix $C$ is the $n\times m$ matrix
whose $j$-th column is the characteristic vector of $X_{j}$
$(j=1,2,\ldots,m)$. Let $A$ be partitioned according to $\Pi$ as

$A=\left[{\begin{array}[]{cccc}A_{11}&A_{12}&\ldots&A_{1m}\\\
A_{21}&A_{22}&\ldots&A_{2m}\\\ \vdots&\ldots&\ddots&\vdots\\\
A_{m1}&A_{m2}&\ldots&A_{mm}\\\ \end{array}}\right],$

where $A_{ij}$ denotes the submatrix (block) of $A$ formed by rows in $X_{i}$
and the columns in $X_{j}$. If $q_{ij}$ denotes the average row sum of
$A_{ij}$, then the matrix $Q=(q_{ij})$ is called the quotient matrix of $A$.
If the row sum of each block $A_{ij}$ is a constant, then the partition $\Pi$
is called _equitable partition_.

In the following theorem, we state a well-known result about the spectrum of a
quotient matrix corresponding to an equitable partition.

######  Theorem 2.1 ([1]).

Let $Q$ be a quotient matrix of any square matrix $A$ corresponding to an
equitable partition. Then the spectrum of $A$ contains the spectrum of $Q$.

Figure 1: The tree $T_{n,3}^{a,b}$, where $a+b=n-4$. Figure 2: The tree
$D_{n,d}^{a,b}$, where $a+b=n-d-1$ and $d\geq 4$.

Let $K_{n}$, $P_{n}$, and $K_{1,n-1}$ denote the complete graph, the path, and
the star on $n$ vertices, respectively. For $d=3$, let $T_{n,d}^{a,b}$ be the
tree obtained from $P_{4}=v_{0}v_{1}v_{2}v_{3}$ by attaching $a$ pendant
vertices to $v_{1}$ and $b$ pendent vertices to $v_{2}$, where $a+b=n-4$ and
$b\geq a\geq 0$. For $d\geq 4$, let $D_{n,d}^{a,b}$ be the tree obtained from
$P_{d+1}=v_{0}v_{1}v_{2}\ldots v_{d}$ by attaching $a$ pendant vertices to
$v_{1}$ and $b$ pendent vertices to $v_{d-1}$, where $a+b=n-d-1$ and $b\geq
a\geq 0$. The trees $T_{n,3}^{a,b}$ and $D_{n,d}^{a,b}$ are depicted in Figure
1 and Figure 2, respectively. For a real number $x$, let $\lfloor x\rfloor$
denote the greatest integer less than or equal to $x$, and $\lceil x\rceil$
denote the least integer greater than or equal to $x$. In the following lemma,
we compute the $\mathcal{E}$-spectrum of the complement of the tree
$T_{n,3}^{a,b}$, where $a+b=n-4$ and $b\geq a\geq 0$.

######  Lemma 2.1.

For $b\geq a\geq 0$ with $a+b=n-4$, the $\mathcal{E}$-spectrum of
$(T_{n,3}^{a,b})^{c}$ is given by

$\operatorname{Spec}_{\mathcal{E}}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=\left\\{{\begin{array}[]{ccc}-\sqrt{\frac{4n+1\pm\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}&0&\sqrt{\frac{4n+1\pm\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}\\\
1&n-4&1\\\ \end{array}}\right\\}.$

###### Proof.

Let $T_{n,3}^{a,b}$ be the tree obtained from $P_{4}=v_{0}v_{1}v_{2}v_{3}$ by
attaching $a$ pendant vertices $u_{1},\ldots,u_{a}$ to $v_{1}$ and $b$ pendent
vertices $w_{1},\ldots,w_{b}$ to $v_{2}$, where $a+b=n-4$ and $b\geq a\geq 0$.
Then the eccentricity matrix of $(T_{n,3}^{a,b})^{c}$ is given by

$\mathcal{E}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=\blockarray{ccccccccccc}&u_{1}\ldots
u_{a}v_{0}v_{1}v_{2}v_{3}w_{1}\ldots w_{b}\\\
\block{c(cccccccccc)}u_{1}0\ldots 002000\ldots 0\\\
\vdots\vdots\ddots\vdots\vdots\vdots\vdots\vdots\vdots\ddots\vdots\\\
u_{a}0\ldots 002000\ldots 0\\\ v_{0}0\ldots 002000\ldots 0\\\ v_{1}2\ldots
220300\ldots 0\\\ v_{2}0\ldots 003022\ldots 2\\\ v_{3}0\ldots 000200\ldots
0\\\ w_{1}0\ldots 000200\ldots 0\\\
\vdots\vdots\ddots\vdots\vdots\vdots\vdots\vdots\vdots\ddots\vdots\\\
w_{b}0\ldots 000200\ldots 0\\\ ~{}.$

It is easy to see that the rank of
$\mathcal{E}\big{(}(T_{n,3}^{a,b})^{c}\big{)}$ is $4$. Therefore, $0$ is an
eigenvalue of $\mathcal{E}\big{(}(T_{n,3}^{a,b})^{c}\big{)}$ with multiplicity
$n-4$.

If $U=\\{u_{1},\ldots,u_{a},v_{0}\\}$ and $W=\\{v_{3},w_{1},\ldots,w_{b}\\}$,
then $\Pi_{1}=U\cup\\{v_{1}\\}\cup\\{v_{2}\\}\cup W$ is an equitable partition
of $\mathcal{E}\big{(}(T_{n,3}^{a,b})^{c}\big{)}$ with the quotient matrix

$Q_{1}=\left({\begin{array}[]{cccc}0&2&0&0\\\ 2(a+1)&0&3&0\\\ 0&3&0&2(b+1)\\\
0&0&2&0\\\ \end{array}}\right).$

By a direct calculation, the eigenvalues of $Q_{1}$ are

$\displaystyle\pm\sqrt{\frac{4a+4b+17+\sqrt{(4a+4b+17)^{2}-64(a+1)(b+1)}}{2}}\quad\text{and}$
$\displaystyle\pm\sqrt{\frac{4a+4b+17-\sqrt{(4a+4b+17)^{2}-64(a+1)(b+1)}}{2}}.$

Now, the proof follows from Theorem 2.1 by substituting $a+b=n-4$. ∎

In the following theorem, we find the adjacency energy of the path graph on
$n$ vertices. We include proof of this result for the sake of completeness.

######  Theorem 2.2.

The energy of a path $P_{n}$ on $n$ vertices is given by

$\displaystyle
E_{A}(P_{n})=\begin{cases}\text{$2\Big{(}\cot\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$2\Big{(}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is even.}\end{cases}$

###### Proof.

We know that the eigenvalues of $P_{n}$ are $2\cos{\frac{\pi k}{n+1}}$,
$k=1,2,\ldots,n$. By using the formula
$\cos{x}+\cos{2x}+\ldots+\cos{nx}=\sin\big{(}\frac{nx}{2}\big{)}\csc\big{(}\frac{x}{2}\big{)}\cos\big{(}\frac{(n+1)x}{2}\big{)}$(for
a proof of this identity, we refer to [6]), we have

$\displaystyle E_{A}(P_{n})=$
$\displaystyle~{}\begin{cases}\text{$4\Big{(}\cos\big{(}{\frac{\pi}{n+1}}\big{)}+\cos\big{(}{\frac{2\pi}{n+1}}\big{)}+\ldots+\cos\big{(}{\frac{(n-1)\pi}{2(n+1)}}\big{)}\Big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$4\Big{(}\cos\big{(}{\frac{\pi}{n+1}}\big{)}+\cos\big{(}{\frac{2\pi}{n+1}}\big{)}+\ldots+\cos\big{(}{\frac{n\pi}{2(n+1)}}\big{)}\Big{)}$}&\quad\text{if
$n$ is even;}\end{cases}$ $\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$4\sin\big{(}{\frac{(n-1)\pi}{4(n+1)}}\big{)}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}\cos\big{(}{\frac{\pi}{4}}\big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$4\sin\big{(}{\frac{n\pi}{4(n+1)}}\big{)}\csc\big{(}{\frac{(n+2)\pi}{4(n+1)}}\big{)}\cos\big{(}{\frac{\pi}{4}}\big{)}$}&\quad\text{if
$n$ is even;}\end{cases}$ $\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$2\Big{(}\cos\big{(}{\frac{\pi}{2(n+1)}}\big{)}-\sin\big{(}{\frac{\pi}{2(n+1)}}\big{)}\Big{)}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$2\Big{(}\sin\big{(}\frac{\pi}{2}\big{)}-\sin\big{(}{\frac{\pi}{2(n+1)}}\big{)}\Big{)}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}$}&\quad\text{if
$n$ is even;}\end{cases}$ $\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$2\Big{(}\cot\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$2\Big{(}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is even.}\end{cases}$

∎

Now, we collect some results related to the spectral radius, second-largest
eigenvalue, and energy for the adjacency matrices of trees.

######  Theorem 2.3 ([12]).

Let $T$ be a tree on $n$ vertices. Then $\lambda_{1}(T)\geq
2\cos{\frac{\pi}{n+1}}$ with equality if and only if $T\cong P_{n}$.

######  Theorem 2.4 ([5, Theorem 2]).

Let $T$ be a tree on $n\geq 4$ vertices and $T\ncong K_{1,n-1}$. Then
$\lambda_{1}(T)\leq\sqrt{\frac{n-1+\sqrt{n^{2}-6n+13}}{2}}$ with equality if
and only if $T\cong T_{n,3}^{0,n-4}$.

######  Theorem 2.5 ([26, Theorem 2]).

Let $T$ be a tree of order $n\geq 4$ and $T\ncong K_{1,n-1},T_{n,3}^{0,n-4}$.
Then $\lambda_{2}(T)\geq 1$.

######  Theorem 2.6 ([5, Theorem 10]).

Let $T$ be a tree with $n\geq 4$ vertices.

  1. 1.

If $T\ncong T_{n,5}^{s-1,s-1}$, then $\lambda_{2}(T)\leq\sqrt{\frac{n-3}{2}}$.
The equality holds if and only if $n=2s+3$ and $T\cong
T_{n,4}^{s-1,s-1},T_{n,5}^{s-2,s-1},T_{n,6}^{s-2,s-2}$.

  2. 2.

If $T\cong T_{n,5}^{s-1,s-1}$, then $\lambda_{2}(T)=x>\sqrt{\frac{n-3}{2}}$,
where $x$ is the positive root of the equation $x^{3}+x^{2}-(s+1)-s=0$.

######  Lemma 2.2 ([3, Proposition 4]).

Let $T$ be a tree on $n\geq 2$ vertices such that $T$ is not isomorphic to
$K_{1,n-1},T_{n,3}^{0,n-4},T_{n,3}^{1,n-3}$ and $T_{n,4}^{0,n-5}$. Then

$E_{A}(T)>E_{A}(T_{n,4}^{0,n-5})>E_{A}(T_{n,3}^{1,n-3})>E_{A}(T_{n,3}^{0,n-4})>E_{A}(K_{1,n-1}).$

######  Lemma 2.3 ([3, Proposition 3]).

Let $T$ be a tree on $n\geq 2$ vertices such that $T\ncong K_{1,n-1},P_{n}$.
Then

$E_{A}(K_{1,n-1})<E_{A}(T)<E_{A}(P_{n}).$

Next, we state a result about the minimum second largest
$\mathcal{E}$-eigenvalue of trees.

######  Theorem 2.7 ([16]).

Let $T$ be a tree on $n\geq 4$ vertices other than the star. Then

$\xi_{2}(T)\geq\sqrt{\frac{13n-35-\sqrt{169n^{2}-974n+1417}}{2}}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

The following theorem is about the characterization of trees with minimum
$\mathcal{E}$-energy.

######  Theorem 2.8 ([16]).

Let $T$ be a tree on $n\geq 5$ vertices. Then

$E_{\mathcal{E}}(T)\geq 2\sqrt{13n-35+8\sqrt{n-3}}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

##  3 Extremal problems for $\mathcal{E}$-spectral radius and least
$\mathcal{E}$-eigenvalue

In this section, we characterize the graphs with the minimum and maximum
$\mathcal{E}$-spectral radius among the complements of all trees on $n$
vertices. As a consequence, we determine the graphs for which the least
$\mathcal{E}$-eigenvalues attain the minimum and maximum value among
$\mathcal{T}_{n}^{c}$, where $\mathcal{T}_{n}^{c}$ denote the complements of
all trees on $n$ vertices. First, we give an ordering of the complements of
trees with diameter $3$ according to their $\mathcal{E}$-spectral radius.

######  Theorem 3.1.

Let $T$ be a tree with diameter $3$, that is, $T\cong T_{n,3}^{a,b}$ with
$a+b=n-4$, $b\geq a\geq 0$. Then the complements of $T_{n,3}^{a,b}$ can be
ordered with respect to their $\mathcal{E}$-spectral radius as follows:

$\xi_{1}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}>\xi_{1}\big{(}(T_{n,3}^{1,n-3})^{c}\big{)}>\ldots>\xi_{1}\Big{(}\big{(}T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}\big{)}^{c}\Big{)}.$

###### Proof.

By Lemma 2.1 , we have

$\displaystyle\xi_{1}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}$
$\displaystyle=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64(n-3+a(n-4-a))}}{2}}\qquad\text{(since,
$a+b=n-4$)}.$

Now, consider the function $f(x)=4n+1+\sqrt{(4n+1)^{2}-64(n-3+x(n-4-x))}$,
where $0\leq x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,

$f^{\prime}(x)=\frac{-32(n-4-2x)}{\sqrt{(4n+1)^{2}-64(n-3+x(n-4-x))}}\leq 0.$

Hence, $f(x)$ is an decreasing function of $x$, where $0\leq
x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,
$\xi_{1}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=\sqrt{\frac{f(a)}{2}}$ is an
decreasing function for $0\leq a\leq\lfloor\frac{n-4}{2}\rfloor$. This
completes the proof. ∎

As a consequence of the above result, we get the following corollaries.

######  Corollary 3.1.

Let $T$ be a tree on $n\geq 4$ vertices with $\operatorname{diam}(T)=3$. Then

$\xi_{1}(T^{c})\leq\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64(n-3)}}{2}}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

The proof follows from Lemma 2.1 and Theorem 3.1. ∎

######  Corollary 3.2.

If $T$ is a tree with diameter $3$, then

$\displaystyle\xi_{1}(T^{c})\geq\begin{cases}\text{$\sqrt{\frac{4n+1+\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1+\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

The equality holds if and only if $T\cong
T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$.

###### Proof.

If $T\cong T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$,
by Lemma 2.1 it follows that

$\displaystyle\xi_{1}(T^{c})=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64\big{(}\lceil\frac{n-2}{2}\rceil\lfloor\frac{n-2}{2}\rfloor\big{)}}}{2}}$
$\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$\sqrt{\frac{4n+1+\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1+\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

If $T\ncong T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$,
then, by Theorem 3.1, it follows that
$\xi_{1}(T^{c})>\xi_{1}\Big{(}\big{(}T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}\big{)}^{c}\Big{)}$.
This completes the proof. ∎

In the following theorem, we characterize the trees whose complement has the
minimum $\mathcal{E}$-spectral radius among the complements of all trees on
$n$ vertices. Note that if $T$ is a tree with $\operatorname{diam}(T)>3$, then
$\mathcal{E}(T^{c})=2A(T)$. If $\operatorname{diam}(T)=3$, then
$\mathcal{E}(T^{c})\geq 2A(T)$ entrywise.

######  Theorem 3.2.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{1}(T^{c})\geq\xi_{1}(P_{n}^{c})$

with equality if and only if $T\cong P_{n}$.

###### Proof.

Since $P_{4}$ is the only tree on $4$ vertices with connected complement, we
assume that $n\geq 5$. For $n\geq 5$, we have
$\xi_{1}(P_{n}^{c})=2\lambda_{1}(P_{n})=4\cos{\frac{\pi}{n+1}}<4$.

If $T$ is a tree with diameter $3$, by Corollary 3.2 it follows that

$\xi_{1}(T^{c})\geq\begin{cases}\text{$\sqrt{\frac{4n+1+\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1+\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

It is easy to check that
$\sqrt{\frac{4n+1+\sqrt{72n-47}}{2}}>\sqrt{\frac{4n+1+\sqrt{72n-63}}{2}}>4$.
Therefore, $\xi_{1}(T^{c})\geq\xi_{1}(P_{n}^{c})$.

If $T$ is a tree with $\operatorname{diam}(T)\geq 4$, then the proof follows
from Theorem 3.2. ∎

Now, we determine the unique tree whose complement has maximum
$\mathcal{E}$-spectral radius in $\mathcal{T}_{n}^{c}$.

######  Theorem 3.3.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{1}(T^{c})\leq\xi_{1}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

If $T$ is a tree with diameter $3$, then the proof follows from Corollary 3.1.
If $T$ is a tree with $\operatorname{diam}(T)\geq 4$, then
$\xi_{1}(T^{c})=2\lambda_{1}(T)$ and the proof follows from Theorem 2.4. ∎

Now, we consider the extremal problem for the least $\mathcal{E}$-eigenvalue
of complements of trees and characterize the extremal graphs. First, we show
that the $\mathcal{E}$-eigenvalues of the complements of trees are symmetric
about the origin.

######  Theorem 3.4.

Let $T$ be a tree of order $n\geq 4$. Then the eigenvalues of
$\mathcal{E}(T^{c})$ are symmetric about the origin, that is, if $\lambda$ is
an eigenvalue of $\mathcal{E}(T^{c})$ with multiplicity $k$, then $-\lambda$
is also an eigenvalue of $\mathcal{E}(T^{c})$ with multiplicity $k$.

###### Proof.

If $T$ is a tree with diameter $3$, then the proof follows from Lemma 2.1. Let
$T$ be a tree with $\operatorname{diam}(T)\geq 4$. Then
$\mathcal{E}(T^{c})=2A(T)$. Since every tree is a bipartite graph, the
eigenvalues of $A(T)$ are symmetric about the origin, and hence the
$\mathcal{E}$-eigenvalues of $T^{c}$ are also symmetric about the origin. ∎

Let $\xi_{n}(T^{c})$ denote the least $\mathcal{E}$-eigenvalue of
$\mathcal{E}(T^{c})$. In the following theorems, we characterize the trees
whose complements have minimum and maximum least $\mathcal{E}$-eigenvalue in
$\mathcal{T}_{n}^{c}$, respectively.

######  Theorem 3.5.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{n}(T^{c})\geq\xi_{n}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

The proof follows from Theorem 3.3 and Theorem 3.4. ∎

######  Theorem 3.6.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{n}(T^{c})\leq\xi_{n}(P_{n}^{c})$

with equality if and only if $T\cong P_{n}$.

###### Proof.

The proof follows from Theorem 3.2 and Theorem 3.4. ∎

##  4 Extremal problems for the second largest $\mathcal{E}$-eigenvalue

In this section, we study the extremal problems for the second largest
$\mathcal{E}$-eigenvalue of complements of trees and characterize the extremal
graphs. First, we give an ordering of the complements of trees with diameter
$3$ according to their second-largest $\mathcal{E}$-eigenvalues.

######  Theorem 4.1.

Let $T$ be a tree with diameter $3$, that is, $T\cong T_{n,3}^{a,b}$ with
$a+b=n-4$, $b\geq a\geq 0$. Then the complements of the trees $T_{n,3}^{a,b}$
can be ordered with respect to their second largest $\mathcal{E}$-eigenvalues
as follows:

$\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}<\xi_{2}\big{(}(T_{n,3}^{1,n-3})^{c}\big{)}<\ldots<\xi_{2}\Big{(}\big{(}T_{n,3}^{\lfloor\frac{n-2}{2}\rfloor,\lceil\frac{n-2}{2}\rceil}\big{)}^{c}\Big{)}.$

###### Proof.

By Lemma 2.1, it follows that

$\displaystyle\xi_{2}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}$
$\displaystyle=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64(n-3+a(n-4-a))}}{2}}\qquad\text{(since,
$a+b=n-4$)}.$

Now, consider the function $f(x)=4n+1-\sqrt{(4n+1)^{2}-64(n-3+x(n-4-x))}$,
where $0\leq x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,

$f^{\prime}(x)=\frac{32(n-4-2x)}{\sqrt{(4n+1)^{2}-64(n-3+x(n-4-x))}}\geq 0.$

Hence, $f(x)$ is an increasing function of $x$ for $0\leq
x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,
$\xi_{2}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=\sqrt{\frac{f(a)}{2}}$ is an
increasing function for $0\leq a\leq\lfloor\frac{n-4}{2}\rfloor$. This
completes the proof. ∎

As a consequence of the above result, we get the following corollaries.

######  Corollary 4.1.

If $T$ is a tree with diameter $3$, then

$\xi_{2}(T^{c})\geq 2\sqrt{4n+1-\sqrt{(4n+1)^{2}-64(n-3)}}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

The proof follows from Lemma 2.1 and Theorem 4.1. ∎

######  Corollary 4.2.

If $T$ is a tree with diameter $3$, then

$\displaystyle\xi_{2}(T^{c})\leq\begin{cases}\text{$\sqrt{\frac{4n+1-\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1-\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

The equality holds if and only if $T\cong
T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$.

###### Proof.

If $T\cong T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$,
by Lemma 2.1 it follows that

$\displaystyle\xi_{2}(T^{c})=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64\big{(}\lceil\frac{n-2}{2}\rceil\lfloor\frac{n-2}{2}\rfloor\big{)}}}{2}}$
$\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$\sqrt{\frac{4n+1-\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1-\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

If $T\ncong T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$,
by Theorem 4.1 it follows that
$\xi_{2}(T^{c})<\xi_{2}\Big{(}\big{(}T_{n,3}^{\lfloor\frac{n-2}{2}\rfloor,\lceil\frac{n-2}{2}\rceil}\big{)}^{c}\Big{)}$.
This completes the proof. ∎

In the following theorem, we determine the unique tree whose complement has
minimum second largest $\mathcal{E}$-eigenvalue in $\mathcal{T}_{n}^{c}$.

######  Theorem 4.2.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{2}(T^{c})\geq\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

If $T$ is a tree on $n\geq 4$ vertices with diameter $3$, then the proof
follows from Corollary 4.1.

Let $T$ be a tree with $\operatorname{diam}(T)\geq 4$. Then
$\mathcal{E}(T^{c})=2A(T)$, and hence by Theorem 2.5, we have
$\xi_{2}(T^{c})\geq 2$. Note that $\sqrt{(4n-7)^{2}+144}>4n-7$ and hence
$(4n+1-\sqrt{(4n-7)^{2}+144})<8$. Therefore,
$\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}=\sqrt{\frac{4n+1-\sqrt{(4n-7)^{2}+144}}{2}}<2$.
Thus, $\xi_{2}(T^{c})>\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$. This
completes the proof. ∎

Next, we characterize the maximal graphs for the second largest
$\mathcal{E}$-eigenvalue of complements of trees.

######  Theorem 4.3.

Let $T$ be a tree with $n\geq 4$ vertices.

  1. 1.

If $T\ncong T_{n,5}^{s-1,s-1}$, then $\xi_{2}(T)\leq\sqrt{2(n-3)}$. The
equality holds if and only if $n=2s+3$ and $T\cong
T_{n,4}^{s-1,s-1},T_{n,5}^{s-2,s-1},T_{n,6}^{s-2,s-2}$.

  2. 2.

If $T\cong T_{n,5}^{s-1,s-1}$, then $\xi_{2}(T)=2x>\sqrt{2(n-3)}$, where $x$
is the positive root of the equation $x^{3}+x^{2}-(s+1)-s=0$.

###### Proof.

If $T$ is a tree on $n\geq 4$ vertices with diameter $3$, by Corollary 4.2 it
follows that

$\displaystyle\xi_{2}(T^{c})\leq\begin{cases}\text{$\sqrt{\frac{4n+1-\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1-\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

It is easy to check that
$\sqrt{\frac{4n+1-\sqrt{72n-47}}{2}}<\sqrt{\frac{4n+1-\sqrt{72n-63}}{2}}<\sqrt{2(n-3)}$
for $n\geq 4$. Therefore, $\xi_{2}(T^{c})<\sqrt{2(n-3)}$.

Let $T$ be a tree with $\operatorname{diam}(T)\geq 4$. Then
$\xi_{2}(T^{c})=2\lambda_{2}(A(T))$ and the proof follows from Theorem 2.6. ∎

Now, we give a Nordhaus-Gaddum type lower bound for the second largest
$\mathcal{E}$-eigenvalue of a tree and its complement, which directly follows
from Theorem 2.7 and Theorem 4.2.

######  Theorem 4.4.

Let $T$ be a tree of order $n\geq 4$. Then

$\xi_{2}(T)+\xi_{2}(T^{c})\geq\Bigg{(}\sqrt{\frac{13n-35-\sqrt{169n^{2}-974n+1417}}{2}}+\sqrt{\frac{4n+1-\sqrt{16n^{2}-56n+193}}{2}}\Bigg{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

##  5 Extremal problems for $\mathcal{E}$-energy

In this section, we characterize the complements of trees with the minimum and
maximum $\mathcal{E}$-energy among the complements of all trees on $n$
vertices, respectively. To begin with, in the following lemma, we give an
ordering of the complements of trees with diameter $3$ according to their
$\mathcal{E}$-energy.

######  Theorem 5.1.

Let $T$ be a tree with diameter $3$, that is, $T\cong T_{n,3}^{a,b}$ with
$a+b=n-4$, $b\geq a\geq 0$. Then the complements of $T_{n,3}^{a,b}$ can be
ordered with respect to their $\mathcal{E}$-energy as follows:

$E_{\mathcal{E}}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}<E_{\mathcal{E}}\big{(}(T_{n,3}^{1,n-3})^{c}\big{)}<\ldots<E_{\mathcal{E}}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}.$

###### Proof.

By Lemma 2.1, it follows that
$E_{\mathcal{E}}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=2\Big{(}\xi_{1}\big{(}(T_{n,3}^{a,b})^{c}\big{)}+\xi_{2}\big{(}(T_{n,3}^{a,b})^{c}\big{)}\Big{)}$,
where

$\displaystyle\xi_{1}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}\quad\text{and}$
$\displaystyle\xi_{2}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64(a+1)(b+1)}}{2}}.$

Therefore,

$\displaystyle E_{\mathcal{E}}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=$
$\displaystyle~{}2\sqrt{4n+1+8\sqrt{(a+1)(b+1)}}$ $\displaystyle=$
$\displaystyle~{}2\sqrt{4n+1+8\sqrt{n-3+a(n-4-a)}}\quad(\text{since,~{}
$a+b=n-4$}).$

Now, consider the function $f(x)=4n+1+8\sqrt{n-3+x(n-4-x)}$, where $0\leq
x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,

$f^{\prime}(x)=\frac{8(n-4-2x)}{4n+1+8\sqrt{n-3+x(n-4-x)}}\geq 0.$

Hence, $f(x)$ is an increasing function of $x$, where $0\leq
x\leq\lfloor\frac{n-4}{2}\rfloor$. Therefore,
$E_{\mathcal{E}}\big{(}(T_{n,3}^{a,b})^{c}\big{)}=2\sqrt{f(a)}$ is an
increasing function for $0\leq a\leq\lfloor\frac{n-4}{2}\rfloor$. This
completes the proof. ∎

As a consequence of the above result, we get the following corollaries.

######  Corollary 5.1.

If $T$ is a tree with diameter $3$, then

$E_{\mathcal{E}}(T^{c})\geq 2\sqrt{4n+1+8\sqrt{n-3}}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

By Lemma 2.1, it follows that

$\displaystyle\xi_{1}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64(n-3)}}{2}},\quad\text{and}$
$\displaystyle\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64(n-3)}}{2}}.$

Thus,

$\displaystyle E_{\mathcal{E}}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}=$
$\displaystyle~{}2\Big{(}\xi_{1}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}+\xi_{2}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}\Big{)}=2\sqrt{4n+1+8\sqrt{n-3}}.$

Now, the proof follows from Theorem 5.1. ∎

######  Corollary 5.2.

If $T$ is a tree with diameter $3$, then

$\displaystyle
E_{\mathcal{E}}(T^{c})\leq\begin{cases}\text{$2\sqrt{8n-7}$}&\quad\text{if $n$
is even,}\\\ \text{$2\sqrt{4n+1+4\sqrt{n^{2}-4n+3}}$}&\quad\text{if $n$ is
odd.}\end{cases}$

The equality holds if and only if $T\cong
T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil}$.

###### Proof.

It follows from Lemma 2.1 that

$\displaystyle\xi_{1}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1+\sqrt{(4n+1)^{2}-64\big{(}\lceil\frac{n-2}{2}\rceil\lfloor\frac{n-2}{2}\rfloor\big{)}}}{2}}$
$\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$\sqrt{\frac{4n+1+\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1+\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

and

$\displaystyle\xi_{2}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}=$
$\displaystyle~{}\sqrt{\frac{4n+1-\sqrt{(4n+1)^{2}-64\big{(}\lceil\frac{n-2}{2}\rceil\lfloor\frac{n-2}{2}\rfloor\big{)}}}{2}}$
$\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$\sqrt{\frac{4n+1-\sqrt{72n-63}}{2}}$}&\quad\text{if
$n$ is even,}\\\ \text{$\sqrt{\frac{4n+1-\sqrt{72n-47}}{2}}$}&\quad\text{if
$n$ is odd.}\end{cases}$

Therefore,

$\displaystyle
E_{\mathcal{E}}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}=$
$\displaystyle~{}2\Big{(}\xi_{1}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}+\xi_{2}\big{(}(T_{n,3}^{\lfloor\frac{n-4}{2}\rfloor,\lceil\frac{n-4}{2}\rceil})^{c}\big{)}\Big{)}$
$\displaystyle=$
$\displaystyle~{}\begin{cases}\text{$2\sqrt{8n-7}$}&\quad\text{if $n$ is
even,}\\\ \text{$2\sqrt{4n+1+4\sqrt{n^{2}-4n+3}}$}&\quad\text{if $n$ is
odd.}\end{cases}$

Now, the proof follows from Theorem 5.1. ∎

Next, we find the $\mathcal{E}$-energy of $(T_{n,4}^{0,n-5})^{c}$ by computing
the $\mathcal{E}$-spectrum of $(T_{n,4}^{0,n-5})^{c}$. This will be used in
the proof of Theorem 5.2.

Figure 3: The tree $T_{n,4}^{0,n-5}$.

######  Lemma 5.1.

For $n\geq 5$, the $\mathcal{E}$-energy of
$E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}$ is given by

$E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=2\sqrt{4(n-1)+8\sqrt{2n-7}}.$

###### Proof.

The eccentricity matrix of $(T_{n,4}^{0,n-5})^{c}$ is given by

$\mathcal{E}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=\blockarray{ccccccccc}&v_{0}v_{1}v_{2}v_{3}v_{4}w_{1}\ldots
w_{n-5}\\\ \block{c(cccccccc)}v_{0}020000\ldots 0\\\ v_{1}202000\ldots 0\\\
v_{2}020200\ldots 0\\\ v_{3}002022\ldots 2\\\ v_{4}000200\ldots 0\\\
w_{1}000200\ldots 0\\\
\vdots\vdots\vdots\vdots\vdots\vdots\vdots\ddots\vdots\\\ w_{n-5}000200\ldots
0\\\ ~{}.$

It is easy to see that the rank of
$\mathcal{E}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}$ is $4$, and hence $0$ is an
eigenvalue of $\mathcal{E}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}$ with
multiplicity $n-4$.

If $W=\\{v_{4},w_{1},\ldots,w_{n-5}\\}$, then
$\Pi_{2}=\\{v_{0}\\}\cup\\{v_{1}\\}\cup\\{v_{2}\\}\cup\\{v_{3}\\}\cup U$ is an
equitable partition of $\mathcal{E}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}$ with
the quotient matrix

$Q_{2}=\left({\begin{array}[]{ccccc}0&2&0&0&0\\\ 2&0&2&0&0\\\ 0&2&0&2&0\\\
0&0&2&0&2(n-4)\\\ 0&0&0&2&0\\\ \end{array}}\right).$

Now, the eigenvalues of $Q_{2}$ are

$\displaystyle-\sqrt{2n-2\pm 2\sqrt{n^{2}-10n+29}},\quad
0,\quad\text{and}\quad\sqrt{2n-2\pm 2\sqrt{n^{2}-10n+29}}.$

Therefore, by Theorem 2.1, we have

$\operatorname{Spec}_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=\left\\{{\begin{array}[]{ccc}-\sqrt{2n-2\pm
2\sqrt{n^{2}-10n+29}}&0&\sqrt{2n-2\pm 2\sqrt{n^{2}-10n+29}}\\\ 1&n-4&1\\\
\end{array}}\right\\}.$

Hence,
$E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=2\Big{(}\sqrt{2n-2+2\sqrt{n^{2}-10n+29}}+\sqrt{2n-2-2\sqrt{n^{2}-10n+29}}\Big{)}$.
If $\alpha=\sqrt{2n-2+2\sqrt{n^{2}-10n+29}}$ and
$\beta=\sqrt{2n-2-2\sqrt{n^{2}-10n+29}}$, then $\alpha^{2}+\beta^{2}=4n-4$ and
$2\alpha\beta=8\sqrt{2n-7}$. Therefore,

$\displaystyle E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=$
$\displaystyle~{}2(\alpha+\beta)$ $\displaystyle=$
$\displaystyle~{}2\Big{(}\sqrt{\alpha^{2}+\beta^{2}+2\alpha\beta}\Big{)}$
$\displaystyle=$ $\displaystyle~{}2\bigg{(}\sqrt{4n-4+8\sqrt{2n-7}}\bigg{)}.$

∎

######  Theorem 5.2.

Let $T$ be a tree of order $n\geq 4$. Then

$E_{\mathcal{E}}(T^{c})\geq
E_{\mathcal{E}}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

###### Proof.

For $n=4$, $T_{n,3}^{0,0}$ is the only tree with a connected complement. For
$n=5$ and $6$, it is easy to see that
$E_{\mathcal{E}}(T^{c})>E_{\mathcal{E}}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$
(see Appendix). So, let us assume that $T$ is a tree on $n\geq 7$ vertices.

If $T$ is a tree with $\operatorname{diam}(T)=3$, then the proof follows from
Corollary 5.1. If $T$ is a tree with $\operatorname{diam}(T)\geq 4$, then, by
Lemma 2.2, it follows that

$E_{\mathcal{E}}(T^{c})=2E_{A}(T)\geq
2E_{A}(T_{n,4}^{0,n-5})=E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}.$

Again, by Lemma 5.1, we have
$E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}=2\sqrt{4(n-1)+8\sqrt{2n-7}}$.
Note that $4(n-1)+8\sqrt{2n-7}>4n+1+8\sqrt{n-3}$ for $n\geq 7$ and hence
$E_{\mathcal{E}}\big{(}(T_{n,4}^{0,n-5})^{c}\big{)}>E_{\mathcal{E}}\big{(}(T_{n,3}^{0,n-4})^{c}\big{)}$
for $n\geq 7$. This completes the proof. ∎

In the following theorem, we characterize the complements of trees with
maximum $\mathcal{E}$-energy.

######  Theorem 5.3.

Let $T$ be a tree of order $n\geq 4$. Then

$E_{\mathcal{E}}(T^{c})\leq E_{\mathcal{E}}(P_{n}^{c})$

with equality if and only if $T\cong P_{n}$.

###### Proof.

For $n=4$, $P_{4}$ is the only tree with a connected complement. For $n=5$ and
$6$, it is easy to see that $E_{\mathcal{E}}(T^{c})\leq
E_{\mathcal{E}}(P_{n}^{c})$ with equality if and only if $T\cong P_{n}$ (see,
Appendix). So, let us assume that $T$ is a tree on $n\geq 7$ vertices. For
$n\geq 7$, it follows from Theorem 2.2 that

$\displaystyle E_{\mathcal{E}}(P_{n}^{c})=2E_{A}(P_{n})=$
$\displaystyle~{}\begin{cases}\text{$4\Big{(}\cot\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is odd,}\\\
\text{$4\Big{(}\csc\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}$}&\quad\text{if
$n$ is even;}\end{cases}$ $\displaystyle\geq$
$\displaystyle~{}\Big{(}\cot\big{(}{\frac{\pi}{2(n+1)}}\big{)}-1\Big{)}.$

We know that $\cot{x}>\big{(}\frac{1}{x}+\frac{1}{x-\pi}\big{)}$ for
$0<x<\frac{\pi}{2}$. Since $0<\frac{\pi}{2(n+1)}<\frac{\pi}{2}$, therefore

$E_{\mathcal{E}}(P_{n}^{c})\geq
4\bigg{(}\cot\Big{(}\frac{\pi}{2(n+1)}\Big{)}-1\bigg{)}>4\Big{(}\frac{4n(n+1)}{(2n+1)\pi}-1\Big{)}.$
(1)

Let $T$ be a tree with $\operatorname{diam}(T)=3$. Therefore, by Corollary 5.2
it follows that

$\displaystyle E_{\mathcal{E}}(T^{c})\leq$
$\displaystyle~{}\begin{cases}\text{$2\sqrt{8n-7}$}&\quad\text{if $n$ is
even,}\\\ \text{$2\sqrt{4n+1+4\sqrt{n^{2}-4n+3}}$}&\quad\text{if $n$ is
odd;}\end{cases}$ $\displaystyle\leq$ $\displaystyle~{}2\sqrt{8n-7}.$

By using (1) one can check that

$\displaystyle
E_{\mathcal{E}}(P_{n}^{c})>4\Big{(}\frac{4n(n+1)}{(2n+1)\pi}-1\Big{)}>2\sqrt{8n-7}\geq
E_{\mathcal{E}}(T^{c})\quad\text{for $n\geq 7$}.$

Thus, for any tree $T$ with diameter $3$,
$E_{\mathcal{E}}(P_{n}^{c})>E_{\mathcal{E}}(T^{c})$.

Let $T$ be a tree with $\operatorname{diam}(T)\geq 4$. Therefore,
$\mathcal{E}(T^{c})=2A(T)$ and hence $E_{\mathcal{E}}(T^{c})=2E_{A}(T)$. Now,
by Lemma 2.3 it follows that

$E_{\mathcal{E}}(T^{c})=2E_{A}(T)\leq
2E_{A}(P_{n})=E_{\mathcal{E}}(P_{n}^{c})$

and the equality holds if and only if $T\cong P_{n}$. This completes the
proof. ∎

Finally, we obtain a Nordhaus-Gaddum type lower bound for the
$\mathcal{E}$-energy of a tree and its complement, which directly follows from
Theorem 2.8 and Theorem 5.2.

######  Theorem 5.4.

Let $T$ be a tree of order $n\geq 4$. Then

$E_{\mathcal{E}}(T)+E_{\mathcal{E}}(T^{c})\geq
2\bigg{(}\sqrt{13n-35+8\sqrt{n-3}}+\sqrt{4n+1+8\sqrt{n-3}}\bigg{)}$

with equality if and only if $T\cong T_{n,3}^{0,n-4}$.

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## Appendix

Figure 4: List of trees on $5$ and $6$ vertices with connected complements. Trees($T$) | $E_{\mathcal{E}}(T^{c})$ | Trees($T$) | $E_{\mathcal{E}}(T^{c})$
---|---|---|---
$T_{1}$ | $\approx 10.4528$ | $T_{5}$ | $\approx 13.798$
$T_{2}$ | $\approx 10.9284$ | $T_{6}$ | $\approx 12.3108$
$T_{3}$ | $\approx 11.6372$ | $T_{7}$ | $\approx 13.9756$
$T_{4}$ | $=12$ |  |
Table 1: $\mathcal{E}$-energy of complements of the trees $T_{1}$-$T_{7}$.