Sequence of Bounds for Spectral Radius and Energy of Digraph

Abstract

The graph spectra analyze the structure of the graph using eigenspectra. The spectral graph theory deals with the investigation of graphs in terms of the eigenspectrum. In this paper, the sequence of lower bounds for the spectral radius of digraph $\mathcal{D}$ having at least one doubly adjacent vertex in terms of indegree is proposed. Particularly, it is exhibited that $\rho (\mathcal{D})\ge {\alpha }^j=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^m{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^m{({\chi }_j^{(p)})}^2}}}$, such that equality is attained iff $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}+$ {$\mathcal{DE}\notin$ Cycle}, where each component of associated graph is a k-regular or $(k_1,k_2)$ semiregular bipartite. By utilizing the sequence of lower bounds of the spectral radius of $\mathcal{D}$, the sequence of upper bounds of energy of $\mathcal{D}$, where the sequence decreases when $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j$ and increases when $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}>{\alpha }^j$, are also proposed. All of the obtained inequalities are elaborated using examples. We also discuss the monotonicity of these sequences.

1. Introduction

Spectral graph theory plays a crucial role in numerous domains by providing innovative methods and frameworks for different methodologies such as algebra, geometry, computational mathematics, decision theory, optimization, social systems, computer science, and topology. A graph is an ordered pair $\mathfrak{G}=(\mathcal{V},\mathcal{E})$, where $\mathcal{V}$ is referred to as a vertex set and $\mathcal{E}$ referred to as edge set. Graphs with directed edges are directed graphs or digraphs and are expressed as $\mathcal{D}=({\mathcal{V}}_{\mathcal{D}},{\mathcal{E}}_{\mathcal{D}})$, $|{\mathcal{V}}_{\mathcal{D}}|,|{\mathcal{E}}_{\mathcal{D}}|$ are referred as order and size of $\mathcal{D}$, which play a crucial role in various fields including computational science, regulatory theory, game theory, systems biology, etc., due to their wide range of applications and valuable insights. They can be used to model direct relationships in networks like social networks, data flow, transportation systems, communication networks, supply chains, etc. They can help to provide insides into the structure to optimize and analyze these systems. There are various algorithms that use the concept of digraphs, such as Tarjan’s algorithm for computing strongly connected components of a digraph, the Dijkstra algorithm for computing the shortest path, and the Ford–Fulkerson algorithm for computing the maximum flow, etc. Many terminologies and concepts have been proposed and extended for digraphs [1,2,3,4,5,6,7,8]. A digraph $\mathcal{D}=({\mathcal{V}}_{\mathcal{D}},{\mathcal{E}}_{\mathcal{D}})$ is simple if it has no loops or multiple diedges, and is connected if there is a dipath between every two vertices. Throughout this article, we investigate only finite, simple, and unweighted digraphs.

If diedge $({\vartheta }_1,{\vartheta }_2)\in {\mathcal{E}}_{\mathcal{D}}$, or $({\vartheta }_2,{\vartheta }_1)\in {\mathcal{E}}_{\mathcal{D}}$, then we say ${\vartheta }_1$ and ${\vartheta }_2$ are adjacent. For ${\vartheta }_p\in {\mathcal{V}}_{\mathcal{D}}$, the set of doubly adjacent vertices is given as

$$\overleftrightarrow{{\aleph }_p}=\{{\vartheta }_{\ell }\in {\mathcal{V}}_{\mathcal{D}}:({\vartheta }_p,{\vartheta }_{\ell }),({\vartheta }_{\ell },{\vartheta }_p)\in {\mathcal{E}}_{\mathcal{D}}\}.$$

The bijection between graph and symmetric digraph is denoted as $\mathfrak{G}⇝\overleftrightarrow{\mathfrak{G}}$. This indicates $\mathcal{V}(\overleftrightarrow{\mathfrak{G}})=\mathcal{V}(\mathfrak{G})$, and ${\vartheta }_p{\vartheta }_{\ell }\in \mathcal{E}(\mathfrak{G})$ associates with symmetric diedges ${\vartheta }_p{\vartheta }_{\ell }$ and ${\vartheta }_{\ell }{\vartheta }_p$. Due to this bijection, a simple and connected graph is considered to be a symmetric digraph. It is to be mentioned that, as the adjacency matrix of a simple connected graph is symmetric, so all of its eigenvalues must be real.

For a digraph $\mathcal{D},{\mathcal{V}}_{\mathcal{D}}=\{{\vartheta }_1,{\vartheta }_2,\dots {\vartheta }_{\mathfrak{U}}\}$, we can express the adjacency matrix $\mathfrak{A}(\mathfrak{D})=\left[a_{\mathfrak{p}\mathfrak{q}}\right]$ as

$$a_{\mathfrak{p}\mathfrak{q}}=\left\{\begin{array}{cc}1,\hfill & \mathrm{as}{\vartheta }_{\mathfrak{p}}{\vartheta }_{\mathfrak{q}}\in \Omega ,\hfill \\ 0,\hfill & \mathrm{as}{\vartheta }_{\mathfrak{p}}{\vartheta }_{\mathfrak{q}}\notin \Omega \hfill \end{array}\right.$$

A digraph $\mathcal{D}$ is referred to as bipartite if ${\mathcal{V}}_{\mathcal{D}}={\mathcal{V}}_{1\mathcal{D}}\cup {\mathcal{V}}_{2\mathcal{D}}$, such that if $({\vartheta }_p,{\vartheta }_{\ell })\in {\mathcal{E}}_{\mathcal{D}}$, then ${\vartheta }_p\in {\mathcal{V}}_{1\mathcal{D}}$ and ${\vartheta }_{\ell }\in {\mathcal{V}}_{2\mathcal{D}}$.

The degree of ${\vartheta }_p\in {\mathcal{V}}_{\mathcal{D}}$ is expressed as

$$d_p=d_p^{+}+d_p^{-},$$

where $d_p^{-}=d^{-}({\vartheta }_p)$ is indegree and $d_p^{+}=d^{+}({\vartheta }_p)$ is outdegree. A bipartite digraph is $(f,g)$ semiregular bipartite digraph, or simply a bipartite semiregular digraph, if

$$d_p=\left\{\begin{array}{cc}f,\hfill & \mathrm{if}{\vartheta }_p\in {\mathcal{V}}_{1\mathcal{D}}\hfill \\ g,\hfill & \mathrm{if}{\vartheta }_p\in {\mathcal{V}}_{2\mathcal{D}}.\hfill \end{array}\right.$$

The $(d_1^{-},d_2^{-},d_3^{-},\dots d_{\mathfrak{U}}^{-})$ and $(d_1^{+},d_2^{+},d_3^{+},\dots d_{\mathfrak{U}}^{+})$ are the sequences of indegrees and outdegrees of ${\vartheta }_1,{\vartheta }_2,\dots {\vartheta }_{\mathfrak{U}}$, respectively. The energy of a digraph is computed as:

$$\mathfrak{E}(\mathcal{D})=\sum _{p=1}^{\mathfrak{U}}\left|\mathcal{R}e({\mathbf{t}}_p)\right|.$$

The largest absolute value of the eigenvalue of a digraph is referred to as spectral radius, computed as $\rho (\mathcal{D})=max|{\mathbf{t}}_p|$, such that ${\mathbf{t}}_p$ refers to an eigenvalue. Gudiño and Rada [6] proposed the lower bound of digraph $\rho (\mathcal{D})\ge {\displaystyle \frac{{\mathfrak{c}}_2}{\mathfrak{U}}}$, where regular graphs satisfied the equality. Furthermore, in [9], a lower bound of $\rho (\mathcal{D})$ in terms of ${\mathfrak{c}}_2$ is estimated as $\rho (\mathcal{D})\ge \sqrt{{\displaystyle \frac{{\Sigma }_{\mathfrak{l}=1}^{\mathfrak{U}}{({\mathfrak{c}}_2^{(\mathfrak{l})})}^2}{\mathfrak{U}}}}$, such that regular graphs or semiregular bipartite graphs attained the equality. Recently, in [10], a sequence of bounds of the $\mathfrak{E}(\mathcal{D})$ was proposed with the help of the already-established lower bounds of $\rho (\mathcal{D})$.

Gutman, in [11], described the energy of a graph as

$$\mathfrak{E}(\mathfrak{G})=\sum _{p=1}^{\mathfrak{U}}\left|{\mathcal{H}}_p\right|,$$

(1)

where ${\mathcal{H}}_1,{\mathcal{H}}_2,\dots ,{\mathcal{H}}_{\mathfrak{U}}$ are eigenvalues. More information about $\mathfrak{E}(\mathcal{D})$ can be found in [11,12,13,14], and related to $\rho (\mathcal{D})$ in [15]. The energy of digraph $\mathcal{D}$ is described in [13] as

$$\mathfrak{E}(\mathcal{D})=\sum _{p=1}^{\mathfrak{U}}\left|\mathcal{R}e({\mathbf{z}}_p)\right|,$$

(2)

where $\mathcal{R}e({\mathbf{z}}_p)$ refers to real part of eigenvalue ${\mathbf{z}}_p$.

McClelland inequality is illustrated by Rada [11] as

$$\mathfrak{E}(\mathcal{D})\le \sqrt{{\displaystyle \frac{\mathfrak{U}(\mathbf{e}+{\mathfrak{c}}_2)}{2}}},$$

(3)

where equality is achieved iff $\mathcal{D}$ can be expressed as the direct sum of ${\displaystyle \frac{\mathfrak{U}}{2}}$ copies of $\overleftrightarrow{{\mathcal{K}}_2}$.

In [6], the energy and spectral radius are related as

$$\mathfrak{E}(\mathcal{D})\le \rho +\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{\rho }^2)}.$$

(4)

In [6], for ${\displaystyle \frac{{\mathfrak{c}}_2}{\mathfrak{U}}}\le \rho$, Equation (4) becomes

$$\mathfrak{E}(\mathcal{D})\le {\displaystyle \frac{{\mathfrak{c}}_2}{\mathfrak{U}}}+\sqrt{(\mathfrak{U}-1)(e-{({\displaystyle \frac{{\mathfrak{c}}_2}{\mathfrak{U}}})}^2)}$$

(5)

Here, equality is achieved in case $\mathcal{D}=\varnothing$ or $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$, with $\mathfrak{G}$ being ${\displaystyle \frac{\mathfrak{U}}{2}}{K}_2$ or $K_{\mathfrak{U}}$.

In [9], the upper bound of $\mathfrak{E}$ is given as

$$\mathfrak{E}(\mathcal{D})\le \sqrt{{\displaystyle \frac{1}{\mathfrak{U}}}\sum _{p=1}^{\mathfrak{U}}{({\mathfrak{c}}_2^{(p)})}^2}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{\displaystyle \frac{1}{\mathfrak{U}}}\sum _{p=1}^{\mathfrak{U}}{({\mathfrak{c}}_2^{(p)})}^2)},$$

(6)

Equality in Equation (6) is satisfied iff $D=\overleftrightarrow{G}$ with $\mathfrak{G}$ is ${\displaystyle \frac{\mathfrak{U}}{2}}{K}_2$ or $K_{\mathfrak{U}}$. In [10], the bounds in terms of sequence for $\mathfrak{E}(\mathcal{D})$ are estimated:

$$\mathfrak{E}(\mathcal{D})\le {\gamma }^{(y)}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\gamma }^{(y)})}^2)}$$

(7)

Inspired and motivated by the work already performed in [9,16,17], we seek to derive a more refined sequence of bounds for $\rho (D)$ and $\mathfrak{E}(\mathcal{D})$.

2. Spectral Radius of Digraph

The adjacency spectral radius is pivotal in examining the flow dynamics of digraphs, providing important insights into the structure to determine the most efficient paths for transmitting information. Through the study of spectral radius, a more optimized system for communication transportation and other network-driven activities can be designed. We can say that spectral radius can be a powerful tool for improving the reliability and efficiency of digraph-based systems.

We can construct a sequence for ${\vartheta }_p\in {\mathcal{V}}_{\mathcal{D}}$ as ${\chi }_1^{(p)}=d_p^{-}$, and ${\chi }_{j+1}^{(p)}={\sum }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_p}}{\chi }_j^{(\ell )}$, for all $j\ge$ 1.

Remark 1. 

The geometric symmetrization of $\mathfrak{A}=\left[a_{p\ell }\right]$ is $\mathfrak{S}(\mathfrak{A})=\left[s_{p\ell }\right]$, such that

$$s_{p\ell }=\sqrt{a_{p\ell }{a}_{\ell p}},$$

where $p,\ell \in \{1,2,\dots ,\mathfrak{U}\}$.

It is easy to see that the following are satisfied for a digraph $\mathfrak{D}$ with adjacency matrix $\mathfrak{A}$:

1. 

${\chi }_{j+1}^{(\ell )}={\Sigma }_{p=1}^{\mathfrak{U}}{\chi }_j^{(p)}{s}_{p\ell }$, for each vertex ${\vartheta }_{\ell }\in {\mathcal{V}}_{\mathcal{D}}$ and $j\ge 1$.

2. 

${\sum }_{p=1}^{\mathfrak{U}}({\chi }_{j+2}^{(p)}{\chi }_j^{(p)})={\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2$.

3. 

$\rho (\mathfrak{A})\ge \rho (\mathfrak{S}(\mathfrak{A}))=\sqrt{\rho (\mathfrak{S}{(\mathfrak{A})}^2)}$ [18].

Remark 2. 

For a digraph $\mathcal{D}$, we can write

$${\alpha }^{(1)}=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_1^{(p)})}^2}}}, {\alpha }^{(2)}=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_3^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}}}, \cdots , {\alpha }^{(j)}=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}.$$

(8)

Lemma 1. 

Let ${\left\{{\alpha }^{(j)}\right\}}_{j\ge 1}$ be the sequence as expressed in Equation (8). Then, $\forall j\ge 1$, we have

$${\alpha }^{(1)}\le {\alpha }^{(2)}\le {\alpha }^j\le {\alpha }^{(j+1)}\le \cdots$$

Proof. 

It is easy to see that

$$\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}\le \sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+2}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}}}$$

(9)

for $j=1,2,3,\cdots$

By leveraging the Cauchy–Schwarz inequality, we can rewrite Equation (9) as

$$\sum _{p=1}^{\mathfrak{U}}({\chi }_j^{(p)})({\chi }_{j+2}^{(p)})\le \sqrt{\sum _{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}\sqrt{\sum _{p=1}^{\mathfrak{U}}{({\chi }_{j+2}^{(p)})}^2}.$$

(10)

From Equation (10) and Remark 1, we have

$$\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+2}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}}}=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+2}^{(p)})}^2{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}$$

$$\ge \sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+2}^{(p)}{\chi }_j^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}$$

$$=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}.$$

Hence, ${\alpha }^{(j)}\le {\alpha }^{(j+1)}$.

This indicates that ${\alpha }^{(j)}$ is increasing. □

Lemma 2. 

For ${\left\{{\alpha }^{(j)}\right\}}_{j\ge 1}$ in Equation (8), $li{m}_{k\to \infty }=\rho (\mathfrak{S}(\mathfrak{A}))$.

Proof. 

The result follows by adopting the same line of reasoning as used in proving Proposition 4 of [19]. □

Theorem 1. 

For a digraph $\mathcal{D}$ having at least one doubly adjacent vertex, and the sequence ${\left\{{\alpha }^{(j)}\right\}}_{j\ge 1}$ as expressed in Equation (8), we have

$$\rho (\mathcal{D})\ge {\alpha }^{(j)}=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}},$$

(11)

$\forall j\ge$1. Here, the necessary and sufficient conditions for equality to behold in Equation (11) is $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}+$ {$\mathcal{DE}\notin$ Cycle}, such that every $\mathcal{CC}s$ of $\mathfrak{G}$ is one of the following:

(i). 

k-regular;

(ii). 

$(k_1,k_2)$ semiregular bipartite.

This satisfies $k^2=k_1{k}_2={({\alpha }^j)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}$.

Proof. 

Applying Rayleigh–Ritz theorem gives us

$$\sqrt{\rho (\mathfrak{S}{(\mathfrak{A})}^2)}=\sqrt{ma{x}_{x\ne 0}{\displaystyle \frac{x^T\mathfrak{S}{(\mathfrak{A})}^{2}x}{x^{T}x}}}$$

$$\ge \sqrt{{\displaystyle \frac{c^T\mathfrak{S}{(\mathfrak{A})}^{2}c}{c^{T}c}}}$$

$$=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}$$

where

$$c={\chi }_j^p={({\chi }_j^{(1)},{\chi }_j^{(2)},\dots ,{\chi }_j^{(\mathfrak{U})})}^T,\forall j\ge 1$$

Hence, the result. Moreover, for equality to be satisfied in Equation (11), we must have

$$\rho {(\mathfrak{S}(\mathfrak{A}))}^2={\displaystyle \frac{c^T\mathfrak{S}{(\mathfrak{A})}^{2}c}{c^{T}c}}.$$

Here, for $\rho (\mathfrak{S}{(\mathfrak{A})}^2)$, c is positive eigenvector of $\mathfrak{S}{(\mathfrak{A})}^2$ corresponding to $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$, which implies that the multiplicity of $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$ is either one or two. This leads to three cases.

Case 1. Assume that $\mathcal{D}$ is strongly connected. This shows that $\mathfrak{A}$ is irreducible. Since $\mathfrak{A}>\mathfrak{S}(\mathfrak{A})$, and using Corollary 2.1.5 from [20], we can reduce that to $\rho (\mathfrak{A})>\rho (\mathfrak{S}(\mathfrak{A}))$, which provides a contradiction. Hence, for equality, $\mathfrak{A}=\mathfrak{S}(\mathfrak{A})$. This implies that $\mathfrak{A}$ becomes a symmetric matrix. Consequently, all eigenvalues are real. This further shows $\mathcal{D}$ is a symmetric digraph, and can be represented as $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$, such that $\mathfrak{G}$ is a connected, simple, and unweighted. If $\rho (\mathfrak{S}{(\mathfrak{A})}^2)$ has the one multiplicity, then c is a positive eigenvector of $\mathfrak{S}{(\mathfrak{A})}^2$, corresponding to $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$. This implies $\rho (\mathfrak{S}{(\mathfrak{A})}^2)={\displaystyle \frac{{({\chi }_{j+1}^{(p)})}^2}{{({\chi }_j^{(p)})}^2}}$ for each ${\vartheta }_p\in {\mathcal{V}}_{\mathcal{D}}$. Hence, $\mathfrak{G}$ is k-regular, satisfying $k^2={({\alpha }^j)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}$. If the multiplicity of $\rho (\mathfrak{S}{(\mathfrak{A})}^2)$ is two, then $-\rho (\mathfrak{S}(\mathfrak{A}))$ must be an eigenvalue of $\mathfrak{S}(\mathfrak{A})$. Hence, $\mathfrak{G}$ is bipartite ([21], Theorem 3.5). Now, for the sake of simplicity, we consider that

$$\mathfrak{S}(\mathfrak{A})=\left(\begin{array}{cc}0& \mathcal{B}\\ {\mathcal{B}}^T& 0\end{array}\right),$$

where $\mathcal{B}$ is a matrix of order ${\mathfrak{U}}_1\times {\mathfrak{U}}_2$. Since c is a positive eigenvector of $\mathfrak{S}{(\mathfrak{A})}^2$ corresponding to $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$, such that $\mathfrak{S}{(\mathfrak{A})}^{2}c=\rho {(\mathfrak{S}(\mathfrak{A}))}^{2}c$, we obtain

$$\mathcal{B}{\mathcal{B}}^T{c}_{{\mathfrak{U}}_1}=\rho {(\mathfrak{S}(\mathfrak{A}))}^2{c}_{{\mathfrak{U}}_1},\mathcal{B}{\mathcal{B}}^T{c}_{{\mathfrak{U}}_2}=\rho {(\mathfrak{S}(\mathfrak{A}))}^2{c}_{{\mathfrak{U}}_2},$$

(12)

where $c=(c_{{\mathfrak{U}}_1}^T,c_{{\mathfrak{U}}_2}^T)$. Since $\mathcal{B}{\mathcal{B}}^T$ and ${\mathcal{B}}^T\mathcal{B}$, both have the same nonzero eigenvalues, so $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$ is the spectral radius of both $\mathcal{B}{\mathcal{B}}^T$ and ${\mathcal{B}}^T\mathcal{B}$ with multiplicity one. By using the Equation (12), we obtain

$${\mathcal{B}}^T\mathcal{B}{\mathcal{B}}^T{c}_{{\mathfrak{U}}_1}=\rho {(\mathfrak{S}(\mathfrak{A}))}^2{\mathcal{B}}^T{c}_{{\mathfrak{U}}_1},\mathcal{B}{\mathcal{B}}^T\mathcal{B}{c}_{{\mathfrak{U}}_2}=\rho {(\mathfrak{S}(\mathfrak{A}))}^2\mathcal{B}{c}_{{\mathfrak{U}}_2}.$$

(13)

Hence, ${\mathcal{B}}^T{c}_{{\mathfrak{U}}_1}$ and $\mathcal{B}{c}_{{\mathfrak{U}}_2}$ are eigenvectors of ${\mathcal{B}}^T\mathcal{B}$ and $\mathcal{B}{\mathcal{B}}^T$, respectively, corresponding to $\rho {(\mathfrak{S}(\mathfrak{A}))}^2$. Hence, $\mathcal{B}{c}_{{\mathfrak{U}}_2}=k_1{c}_{{\mathfrak{U}}_1}$ and ${\mathcal{B}}^T{c}_{{\mathfrak{U}}_1}=k_2{c}_{{\mathfrak{U}}_2}$, where $k_1,k_2$ are two positive constants. This implies that $\mathfrak{G}$ is $(k_1,k_2)$ semiregular bipartite, satisfying $k_1{k}_2={({\alpha }^j)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}.$

Case 2. Suppose $\mathcal{D}$ is expressed as direct sum of $\mathfrak{b}$$\mathcal{SCC}s$${\mathcal{D}}_1,{\mathcal{D}}_2,\dots ,{\mathcal{D}}_{\mathfrak{b}}$ with adjacency matrices ${\mathfrak{A}}_1,{\mathfrak{A}}_2,{\mathfrak{A}}_3,{\mathfrak{A}}_4,\dots ,{\mathfrak{A}}_{\mathfrak{b}}$ of order ${\mathfrak{U}}_j$ satisfying ${\sum }_{j=1}^{\mathfrak{b}}{\mathfrak{U}}_j=\mathfrak{U}$. Thus, we can write

$${\mathfrak{A}}^2=\left(\begin{array}{ccccc}{\mathfrak{A}}_1^2& & & & \\ & {\mathfrak{A}}_2^2& & & \\ & & .& & \\ & & & .& \\ & & & & {\mathfrak{A}}_{\mathfrak{b}}^2\end{array}\right),$$

where all other elements are 0.

Since we assume that the equality in Equation (11) is to be attained, we can write

$$\sqrt{\rho (\mathfrak{S}{(\mathfrak{A})}^2)}=\sqrt{ma{x}_{x\ne 0}{\displaystyle \frac{x^T\mathfrak{S}{(\mathfrak{A})}^{2}x}{x^{T}x}}}=\sqrt{{\displaystyle \frac{c^T\mathfrak{S}{(\mathfrak{A})}^{2}c}{c^{T}c}}}$$

$$=\sqrt{\sum _{j=1}^{\mathfrak{b}}{\displaystyle \frac{c_{{\mathfrak{U}}_j}^{T}S{(A_j)}^2{c}_{{\mathfrak{U}}_j}}{c^{T}c}}{\displaystyle \frac{c_{{\mathfrak{U}}_j}^T{c}_{{\mathfrak{U}}_j}}{c_{{\mathfrak{U}}_j}^T{c}_{{\mathfrak{U}}_j}}}}$$

$$\le \sqrt{\sum _{j=1}^{\mathfrak{b}}{\displaystyle \frac{\rho (\mathfrak{S}{({\mathfrak{A}}_j)}^2)c_{{\mathfrak{U}}_j}^T{c}_{{\mathfrak{U}}_j}}{c^{T}c}}}\le \sqrt{ma{x}_j\rho (\mathfrak{S}{({\mathfrak{A}}_j)}^2)}$$

$$=ma{x}_j\sqrt{\rho (\mathfrak{S}{({\mathfrak{A}}_j)}^2)}=\sqrt{\rho (\mathfrak{S}{(\mathfrak{A})}^2)}$$

$$=\sqrt{\rho (\mathfrak{A}{(\mathcal{D})}^2)}.$$

Then, ∀ j = 1, 2, …, $\mathfrak{b}$.

$$\rho (\mathfrak{A}(\mathcal{D}))=\sqrt{\rho (\mathfrak{A}{(\mathcal{D})}^2)}$$

$$=\sqrt{\rho ({\mathfrak{A}}_j^2)}=\sqrt{\rho (\mathfrak{S}({\mathfrak{A}}_j^2))}$$

$$=\sqrt{\sum _{j=1}^{\mathfrak{b}}{\displaystyle \frac{c_{{\mathfrak{U}}_j}^T\mathfrak{S}{({\mathfrak{A}}_j)}^2{c}_{{\mathfrak{U}}_j}}{{\mathfrak{U}}_j}}}$$

By applying Case 1, the required result is attained.

Case 3. Lastly, assume that digraph $\stackrel{‘}{\mathcal{D}}$ attained from $\mathcal{D}$ by discarding those $\mathcal{DE}$ which do not lie on a cycle. Hence, $\mathfrak{S}(\mathfrak{A}(\mathcal{D}))=\mathfrak{S}(\mathfrak{A}(\stackrel{‘}{\mathcal{D}}))$. It can be easily seen that cycle arrangement $\mathcal{D}$ and $\stackrel{‘}{\mathcal{D}}$ coincide. Furthermore, since $\stackrel{‘}{\mathcal{D}}$ is the direct sum of $\mathcal{SCC}s$, by applying case 2, the eigenvalues of $\mathcal{D}$ and $\stackrel{‘}{\mathcal{D}}$ coincide.

Hence, $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$+ {$\mathcal{DE}\notin$ cycle}.

The proof of converse part is straightforward.  □

Corollary 1. 

For the sequences $({\chi }_1^{(1)},{\chi }_1^{(2)},{\chi }_1^{(3)},\dots ,{\chi }_1^{(\mathfrak{U})})$ and $({\chi }_2^{(1)},{\chi }_2^{(2)},{\chi }_2^{(3)},\dots ,{\chi }_2^{(\mathfrak{U})})$, corresponding to a digraph $\mathcal{D}$ with at least one doubly adjacent vertex, we have

$$\rho (\mathcal{D})\ge \sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_1^{(p)})}^2}}}={\alpha }^1.$$

(14)

Here, the necessary and sufficient conditions for equality in Equation (14) is $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$+{$\mathcal{DE}\notin$ Cycle}, so that every $\mathcal{CC}$ of $\mathfrak{G}$ is either k-regular graph or $(k_1,k_2)$ semi-regular bipartite graph, such that $k^2=k_1{k}_2={({\alpha }^1)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{(2)}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{(1)}^{(p)})}^2}}$.

Corollary 2. 

For the sequences $({\chi }_3^{(1)},{\chi }_3^{(2)},{\chi }_3^{(3)},\dots ,{\chi }_3^{(\mathfrak{U})})$ and $({\chi }_2^{(1)},{\chi }_2^{(2)},{\chi }_2^{(3)},\dots ,{\chi }_2^{(\mathfrak{U})})$, corresponding to a digraph $\mathcal{D}$ having at least one doubly adjacent vertex, we have

$$\rho (\mathcal{D})\ge \sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_3^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}}}={\alpha }^2$$

(15)

Here, the necessary and sufficient conditions for equality to behold in Equation (15) is $\mathcal{D}=\overleftrightarrow{G}$+{$\mathcal{DE}\notin$ cycle}, so that every $\mathcal{CC}s$ of $\mathfrak{G}$ is either a k-regular graph or a $(k_1,k_2)$-semi-regular bipartite graph, such that $k^2=k_1{k}_2={({\alpha }^2)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_3^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}}$.

Remark 3. 

It is easy to see that the sequence of bounds corresponding to $\rho (D)$ is increasing,

$${\alpha }^1\le {\alpha }^2\le {\alpha }^3\le \dots \le {\alpha }^j\dots \le \rho (D)$$

(16)

Theorem 1 can be illustrated by the following examples.

Example 1. 

Take a digraph $(2,3)$ semi-regular bipartite digraph (${\mathcal{D}}_1$) having

$${\mathcal{V}}_{{\mathcal{D}}_1}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_3,{\vartheta }_4,{\vartheta }_5\},$$

with partition as ${\mathcal{V}}_{{\mathcal{D}}_1}={\mathcal{V}}_{1,{\mathcal{D}}_1}\cup {\mathcal{V}}_{2,{\mathcal{D}}_1}$, where ${\mathcal{V}}_{1,{\mathcal{D}}_1}=\{{\vartheta }_1,{\vartheta }_2\}$ is set of all those vertices having three out-degrees, and ${\mathcal{V}}_{2,{\mathcal{D}}_1}=\{{\vartheta }_3,{\vartheta }_4,{\vartheta }_5\}$ is a set of all vertices having two out-degrees, as exhibited in Figure 1.

Figure 1. (2, 3) semi-regular bipartite digraph (${\mathcal{D}}_1)$.

For each vertex, $\overleftrightarrow{{\aleph }_p}$ is computed as:

$$\overleftrightarrow{{\aleph }_1}=\{{\vartheta }_3,{\vartheta }_4,{\vartheta }_5\},\overleftrightarrow{{\aleph }_2}=\{{\vartheta }_3,{\vartheta }_4,{\vartheta }_5\},\overleftrightarrow{{\aleph }_3}=\{{\vartheta }_1,{\vartheta }_2\},\overleftrightarrow{{\aleph }_4}=\{{\vartheta }_1,{\vartheta }_2\},and\overleftrightarrow{{\aleph }_5}=\{{\vartheta }_1,{\vartheta }_2\}.$$

Also,

$${\chi }_1^{(1)}=d_1^{-}=3, {\chi }_1^{(2)}=d_2^{-}=3, {\chi }_1^{(3)}=d_3^{-}=2, {\chi }_1^{(4)}=d_4^{-}=2, {\chi }_1^{(5)}=d_5^{-}=2,$$

or ${(3,3,2,2,2)}^T$.

$${\chi }_2^{(1)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_1^{(\ell )}={\chi }_1^{(3)}+{\chi }_1^{(4)}+{\chi }_1^{(5)}=6,$$

$${\chi }_2^{(2)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_1^{(j)}={\chi }_1^{(3)}+{\chi }_1^{(4)}+{\chi }_1^{(5)}=6,$$

$${\chi }_2^{(3)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}=6,$$

$${\chi }_2^{(4)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}=6,$$

$${\chi }_2^{(5)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_5}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}=6,$$

or ${(6,6,6,6,6)}^T$. Similarly, we have

$${\chi }_3^{(1)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_2^{(\ell )}={\chi }_2^{(3)}+{\chi }_2^{(4)}+{\chi }_2^{(5)}=18,$$

$${\chi }_3^{(2)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_2^{(3)}+{\chi }_2^{(4)}+{\chi }_2^{(5)}=18,$$

$${\chi }_3^{(3)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}=12,$$

$${\chi }_3^{(4)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}=12,$$

$${\chi }_3^{(5)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_5}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}=12,$$

or ${(18,18,12,12,12)}^T$.

$${\chi }_4^{(1)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_3^{(\ell )}={\chi }_3^{(3)}+{\chi }_3^{(4)}+{\chi }_3^{(5)}=36,$$

$${\chi }_4^{(2)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_3^{(\ell )}={\chi }_3^{(3)}+{\chi }_3^{(4)}+{\chi }_3^{(5)}=36,$$

$${\chi }_4^{(3)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_3^{(\ell )}={\chi }_3^{(1)}+{\chi }_3^{(2)}=36,$$

$${\chi }_4^{(4)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_3^{(\ell )}={\chi }_3^{(1)}+{\chi }_3^{(2)}=36,$$

$${\chi }_4^{(5)}={\Sigma }_{v_{\ell }\in \overleftrightarrow{{\aleph }_5}}{\chi }_3^{(\ell )}={\chi }_3^{(1)}+{\chi }_3^{(2)}=36,$$

or ${(36,36,36,36,36)}^T$.

Hence, for j = 1,

$$\rho (D_1)\ge {\alpha }^1=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_1^{(i)})}^2}}}=2.449489743,$$

and for k = 2,

$$\rho (D_1)\ge {\alpha }^2=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_3^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_2^{(i)})}^2}}}=2.449489743,$$

Hence, we conclude

$${\alpha }^1={\alpha }^2={\alpha }^3={\alpha }^4={\alpha }^5,\dots \dots \dots$$

On the other hand, by direct computation, $\sigma ({\mathcal{D}}_1)=\{0,0,0,\sqrt{6},-\sqrt{6}\}$. Here, $\rho ({\mathcal{D}}_1)=\sqrt{6}=2.449489743$. This shows that our result is compatible with equality to be held in Theorem 1.

Hence, ${\mathcal{D}}_1={\overleftrightarrow{K}}_2{,}_3+\{\varnothing \}$.

Example 2. 

Take a digraph ${\mathcal{D}}_2=\overleftrightarrow{{\mathcal{K}}_3}+\mathbf{e}$, where $\overleftrightarrow{{\mathcal{K}}_3}$ is a symmetric digraph corresponding to ${\mathcal{K}}_3$, and $\mathbf{e}={\vartheta }_3{\vartheta }_4$ is a $\mathcal{DE}\notin$ cycle with ${\mathcal{V}}_{{\mathcal{D}}_2}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_3,{\vartheta }_4\}$, as exhibited in Figure 2.

Figure 2. ${\mathcal{D}}_2=\overleftrightarrow{{\mathcal{K}}_3}+\mathbf{e}$.

For each vertex, $\overleftrightarrow{{\aleph }_p}$ is computed as

$$\overleftrightarrow{{\aleph }_1}=\{{\vartheta }_2,{\vartheta }_3\},\overleftrightarrow{{\aleph }_2}=\{{\vartheta }_1,{\vartheta }_3\},\overleftrightarrow{{\aleph }_3}=\{{\vartheta }_1,{\vartheta }_2\}\mathrm{and}\overleftrightarrow{{\aleph }_4}=\varnothing .$$

Also,

$${\chi }_1^{(1)}=d_1^{-}=2,{\chi }_1^{(2)}=d_2^{-}=2,{\chi }_1^{(3)}=d_3^{-}=2,{\chi }_1^{(4)}=d_4^{-}=1$$

$${\chi }_2^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_1^{(\ell )}={\chi }_1^{(2)}+{\chi }_1^{(3)}=4,$$

$${\chi }_2^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(3)}=4,$$

$${\chi }_2^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}=4,$$

$${\chi }_2^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_1^{(\ell )}=0.$$

Similarly, we have

$${\chi }_3^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_2^{(\ell )}={\chi }_2^{(2)}+{\chi }_2^{(3)}=8=2^3,$$

$${\chi }_3^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(3)}=8=2^3,$$

$${\chi }_3^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}=8=2^3,$$

$${\chi }_3^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}=0.$$

Furthermore, we now have

$${\chi }_j^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(2)}+{\chi }_{j-1}^{(3)}=2^j,$$

$${\chi }_j^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(1)}+{\chi }_{j-1}^{(3)}=2^j,$$

$${\chi }_j^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(1)}+{\chi }_{j-1}^{(2)}=2^j,$$

$${\chi }_j^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_{j-1}^{(\ell )}=0.$$

This implies

$$\rho (D_2)\ge {\alpha }^j=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}=\sqrt{{\displaystyle \frac{{(2^{j+1})}^2}{{(2^j)}^2}}}=2.$$

Also, $\sigma ({\mathcal{D}}_2)=\{2,-1,-1,0\}$. Hence, $\rho ({\mathcal{D}}_2)=2$ which shows that the equality in Theorem 1 holds.

3. Energy of Digraphs

Digraph energy is important for improving the reliability and efficiency of networks and directed networks. By studying and analyzing the digraph energy, the strength of vertex connectivity can be evaluated, critical pathways can be identified, and optimization of networks can be improved in terms of improving transmission information or increasing resilience to failures. In this section, we drive a sequence of bounds for $\mathfrak{E}(\mathcal{D})$ by leveraging a sequence of bounds of $\rho (\mathcal{D})$, as computed in the previous section.

The following lemma is fundamental in constructing the upper bounds for $\mathfrak{E}(\mathcal{D})$.

Lemma 3 

([11]). Suppose $\mathcal{D}$ is a digraph with $|{\mathcal{V}}_{\mathcal{D}}|=\mathfrak{U}$, $|{\mathcal{E}}_{\mathcal{D}}|=e$. So, we have

1. 

${\sum }_{\mathfrak{p}=1}^{\mathfrak{U}}{(\mathcal{R}e({\mathbf{t}}_{\mathfrak{p}}))}^2-{\sum }_{\mathfrak{p}=1}^{\mathfrak{U}}{(Im({\mathbf{t}}_{\mathfrak{p}}))}^2={\mathfrak{c}}_2$;

2. 

${\sum }_{\mathfrak{p}=1}^{\mathfrak{U}}{(\mathcal{R}e({\mathbf{t}}_{\mathfrak{p}}))}^2+{\sum }_{\mathfrak{p}=1}^{\mathfrak{U}}{(Im({\mathbf{t}}_{\mathfrak{p}}))}^2\le \mathbf{e}$.

Here, ${\mathbf{t}}_p$ is an eigenvalue.

Theorem 2. 

Consider the sequence ${\left\{{\alpha }^j\right\}}_{j\ge 1}$, as expressed in Equation (8), corresponding to a digraph $\mathcal{D}$ having at least one doubly adjacent vertex. Then,

$$\mathfrak{E}(\mathcal{D})\le {\alpha }^{(j)}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)},$$

(17)

$\forall j\ge 1$. The necessary and sufficient condition for equality in Equation (17) to be attained is $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$+{$\mathcal{DE}\notin$ Cycle}, such that $\mathcal{CC}$ of $\mathfrak{G}$ is a complete $K_{\mathfrak{U}}$, or ${\displaystyle \frac{\mathfrak{U}}{2}}{K}_2$, or a k-regular complete with at least two nontrivial eigenvalues, whose absolute value is $\sqrt{{\displaystyle \frac{\mathbf{e}-{({\alpha }^j)}^2}{\mathfrak{U}-1}}}$.

Proof. 

Since $\mathfrak{A}$ is adjacency matrix of $\mathcal{D}$, $\mathfrak{A}\ge 0$, and $\rho =\rho (\mathcal{D})$ is highest eigenvalue of $\mathcal{D}$, then $\mathcal{R}e({\mathbf{t}}_1)\ge \mathcal{R}e({\mathbf{t}}_2)\ge \dots \mathcal{R}e({\mathbf{t}}_{\mathfrak{U}})$ are the real values of eigenvalues ${\mathbf{t}}_1,{\mathbf{t}}_2,{\mathbf{t}}_3,\dots {\mathbf{t}}_{\mathfrak{U}}$. By Cauchy–Schwarz inequality, we obtain

$$(\sum _{\mathfrak{p}=2}^{\mathfrak{U}}|\mathcal{R}e({\mathbf{t}}_p){|)}^2\le (\mathfrak{U}-1)\sum _{p=2}^{\mathfrak{U}}{\left|\mathcal{R}e({\mathbf{t}}_p)\right|}^2$$

(18)

Applying Lemma 3 (2) gives us

$$\sum _{p=2}^{\mathfrak{U}}\mathcal{R}e{({\mathbf{t}}_p)}^2\le \mathbf{e}-{\rho }^2,$$

(19)

such that $\mathbf{e}$ are $\mathcal{DE}$. Using Equations (18) and (19), we have

$$\sum _{p=2}^{\mathfrak{U}}\left|\mathcal{R}e({\mathbf{t}}_p)\right|\le \sqrt{(\mathfrak{U}-1)\sum _{p=2}^{\mathfrak{U}}(\mathcal{R}e{({\mathbf{t}}_p)}^2)}\le \sqrt{(\mathfrak{U}-1)(\mathbf{e}-{\rho }^2)}.$$

So,

$$\mathfrak{E}(\mathcal{D})\le \rho +\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{\rho }^2)}.$$

(20)

Consider a correspondence such that $\varphi (H)=H+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-H^2)}$, $H\in [0,\sqrt{\mathbf{e}}]$. It is clear that $\varphi (H)$ is strictly increasing in $[0,\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}]$, and decreasing in $[\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}},\sqrt{\mathbf{e}}]$. Here arise two cases.

Case 1. Assume that $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j$. By Theorem 1 and Equation (19), $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j\le \rho \le \sqrt{\mathbf{e}}$. Since $\varphi$ is strictly decreasing and using Equation (20), we have

$$\mathfrak{E}(\mathcal{D})\le \varphi (\rho )\le \varphi ({\alpha }^j)={\alpha }^j+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)},$$

(21)

This implies that inequality in Equation (17) is attained as

$$\mathfrak{E}(\mathcal{D})\le {\alpha }^j+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)}.$$

Assume the equality in Equation (17) satisfies that from Equation (21); we have $\varphi (\rho )=\varphi ({\alpha }^j)$, and this implies $\rho ={\alpha }^j$.

Hence, from Theorem 1, $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$+{$\mathcal{DE}\notin$ Cycle}, such that every $\mathcal{CC}s$ of $\mathfrak{G}$ is k-regular or a $(k_1,k_2)$ semi-regular bipartite, so that $k^2=k_1{k}_2={({\alpha }^j)}^2={\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(i)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}.$

Hence, $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$ is a symmetric digraph. For a symmetric digraph $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}$, we have $\mathbf{e}=$ number of $\mathcal{DE}$=${\sum }_{p=1}^{\mathfrak{U}}{d}_p^{-}$=sum of all in-degrees of each vertex. Hence,

$$\mathfrak{E}(\mathcal{D})\le {\alpha }^{(j)}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)}$$

$$\le {\alpha }^{(j)}+\sqrt{(\mathfrak{U}-1)(\sum _{p=1}^{\mathfrak{U}}{d}_p^{-}-{({\alpha }^j)}^2)}.$$

If $\mathcal{D}$ is completely connected, or $\mathcal{D}={\displaystyle \frac{\mathfrak{U}}{2}}{K}_2$, then equality is attained directly.

Case 2. Finally, suppose that $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}>{\alpha }^j$. Since sequence ${\left\{{\alpha }^j\right\}}_{j\ge 1}$ is increasing, we have

$$0\le {\alpha }^{(1)}\le {\alpha }^{(j)}<\sqrt{{\displaystyle \frac{e}{\mathfrak{U}}}}.$$

As $\varphi (H)$ is increasing in $[0,\sqrt{{\displaystyle \frac{e}{\mathfrak{U}}}}]$, therefore

$$\varphi ({\alpha }^{(0)})\le \varphi ({\alpha }^{(1)})\le \varphi ({\alpha }^j).$$

Moreover, for attaining the equality in Equation (17), we must have

$$\mathfrak{E}(\mathcal{D})=\varphi ({\alpha }^{(1)})=\varphi ({\alpha }^j).$$

The result follows by adopting the same line of reasoning using the same methodology as used in Theorem 11 in [10] and Theorem 3.1 in [17]. It follows satisfied equality. The proof of converse part is straightforward. □

Theorem 2 can be illustrated by the following examples.

Example 3. 

Take a digraph ${\mathcal{D}}_3=\overleftrightarrow{{\mathcal{K}}_4}$, where $\overleftrightarrow{{\mathcal{K}}_4}$ is a symmetric digraph corresponding to ${\mathcal{K}}_4$ with ${\mathcal{V}}_{{\mathcal{D}}_3}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_3,{\vartheta }_4\}$ with $\mathbf{e}=12$ and $\mathfrak{U}=4$, as exhibited in Figure 3.

Figure 3. ${\mathcal{D}}_3=\overleftrightarrow{{\mathcal{K}}_4}$.

For each vertex, $\overleftrightarrow{{\aleph }_p}$ is computed as:

$$\overleftrightarrow{{\aleph }_1}=\{{\vartheta }_2,{\vartheta }_3,{\vartheta }_4\},\overleftrightarrow{{\aleph }_2}=\{{\vartheta }_1,{\vartheta }_3,{\vartheta }_4\},\overleftrightarrow{{\aleph }_3}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_4\},and\overleftrightarrow{{\aleph }_4}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_3\}$$

Also,

$${\chi }_1^{(1)}=d_1^{-}=3,{\chi }_1^{(2)}=d_2^{-}=3,{\chi }_1^{(3)}=d_3^{-}=3,{\chi }_1^{(4)}=d_4^{-}=3.$$

or ${(3,3,3,3)}^T$.

$${\chi }_2^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_1^{(\ell )}={\chi }_1^{(2)}+{\chi }_1^{(3)}+{\chi }_1^{(4)}=9=3^2,$$

$${\chi }_2^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(3)}+{\chi }_1^{(4)}=9=3^2,$$

$${\chi }_2^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}+{\chi }_1^{(4)}=9=3^2,$$

$${\chi }_2^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}+{\chi }_1^{(3)}=9=3^2.$$

or ${(3^2,3^2,3^2,3^2)}^T$.

Similarly, we have

$${\chi }_3^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_2^{(\ell )}={\chi }_2^{(2)}+{\chi }_2^{(3)}+{\chi }_2^{(4)}=27=3^3,$$

$${\chi }_3^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(3)}+{\chi }_2^{(4)}=27=3^3,$$

$${\chi }_3^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}+{\chi }_2^{(4)}=27=3^3,$$

$${\chi }_3^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_1^{(\ell )}={\chi }_2^{(1)}+{\chi }_2^{(2)}+{\chi }_2^{(3)}=27=3^3,$$

or ${(3^3,3^3,3^3,3^3)}^T$.

Furthermore, we now have

$${\chi }_j^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(2)}+{\chi }_{j-1}^{(3)}+{\chi }_{j-1}^{(4)}=3^j,$$

$${\chi }_j^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(1)}+{\chi }_{j-1}^{(3)}+{\chi }_{j-1}^{(4)}=3^j,$$

$${\chi }_j^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(1)}+{\chi }_{j-1}^{(2)}+{\chi }_{j-1}^{(4)}=3^j,$$

$${\chi }_j^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_{j-1}^{(\ell )}={\chi }_{j-1}^{(1)}+{\chi }_{j-1}^{(2)}+{\chi }_{j-1}^{(3)}=3^j,$$

or ${(3^j,3^j,3^j,3^j)}^T$.

So,

$${\alpha }^j=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^3}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^3}}}=\sqrt{{\displaystyle \frac{{(3^{j+1})}^3}{{(3^j)}^3}}}=3.$$

Also, $\sigma ({\mathcal{D}}_3)=\{3,-1,-1,-1\}$. Hence, $\mathfrak{E}({\mathcal{D}}_3)=6$. By using the above data, we can compute

$${\alpha }^{(j)}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)}=3+\sqrt{(4-1)(12-3^2)}=6.$$

Hence,

$$\mathfrak{E}(\mathcal{D})={\alpha }^{(j)}+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)},$$

(22)

$\forall j\ge 1$, which shows that the equality in Theorem 2 holds. Hence, the necessary and sufficient conditions for equality in Equation (17) to be attained are ${\mathcal{D}}_3=\overleftrightarrow{\mathfrak{G}}$+{$\mathcal{DE}\notin$ Cycle}, such that $\mathcal{CC}$ of $\mathfrak{G}$ is a complete $K_{\mathfrak{4}}$, or a three-regular complete with at least two nontrivial eigenvalues whose absolute value is $\sqrt{{\displaystyle \frac{\mathbf{e}-{({\alpha }^j)}^2}{\mathfrak{U}-1}}}=1$.

Corollary 3. 

Consider the sequence ${\delta }_j={\alpha }^j+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)}$, for $j=1,2,\dots$ of upper bounds for $\mathfrak{E}(\mathcal{D})$ corresponding to a digraph $\mathcal{D}$ having at least one doubly adjacent vertex. Then, ${\delta }_j$ is decreasing for $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j$, and increasing for $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}>{\alpha }^j$.

Proof. 

From Lemma 2 and Remark 3,

$${\alpha }^{(1)}\le {\alpha }^{(2)}\le {\alpha }^{(3)}\le \dots \le {\alpha }^k\dots \le \rho (D).$$

Consider ${\delta }_j={\alpha }^j+\sqrt{(\mathfrak{U}-1)(\mathbf{e}-{({\alpha }^j)}^2)}$, for $j=1,2,\dots$.

i:

If $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j$, then case 1 of Theorems 1 and 2, ${\left\{{\delta }_j\right\}}_{j\ge 1}$, is decreasing as

$$\mathfrak{E}(\mathcal{D})\le {\delta }_j\le {\delta }_{j-1}\le {\delta }_{j-2}\le \dots \le {\delta }_1.$$

ii:

If $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}>{\alpha }^j$, then case 2 of Theorems 1 and 2, ${\left\{{\delta }_j\right\}}_{j\ge 1}$, is increasing as

$$\mathfrak{E}(\mathcal{D})\le {\delta }_1\le {\delta }_2\le \dots \le {\delta }_{j-1}\le {\delta }_j.$$

□

The following example is the illustration of Theorem 2 and Corollary 3.

Take a digraph ${\mathcal{D}}_4$ having at least one doubly adjacent vertex with ${\mathcal{V}}_{{\mathcal{D}}_4}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_3,{\vartheta }_4\}$, as exhibited in Figure 4.

Figure 4. ${\mathcal{D}}_4$.

For each vertex, $\overleftrightarrow{{\aleph }_p}$ is computed as

$$\overleftrightarrow{{\aleph }_1}=\{{\vartheta }_2,{\vartheta }_3\},\overleftrightarrow{{\aleph }_2}=\{{\vartheta }_1,{\vartheta }_3,{\vartheta }_4\},\overleftrightarrow{{\aleph }_3}=\{{\vartheta }_1,{\vartheta }_2,{\vartheta }_4\},and\overleftrightarrow{{\aleph }_4}=\{{\vartheta }_2,{\vartheta }_3\}.$$

Also,

$${\chi }_1^{(1)}=d_1^{-}=2,{\chi }_1^{(2)}=d_2^{-}=3,{\chi }_1^{(3)}=d_3^{-}=3,{\chi }_1^{(4)}=d_4^{-}=3,$$

or ${(2,3,3,3)}^T$

$${\chi }_2^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_1^{(\ell )}={\chi }_1^{(2)}+{\chi }_1^{(3)}=6,$$

$${\chi }_2^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_1^{(\ell )}={\chi }_1^{(3)}+{\chi }_1^{(4)}=8,$$

$${\chi }_2^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_1^{(\ell )}={\chi }_1^{(1)}+{\chi }_1^{(2)}+{\chi }_1^{(4)}=8,$$

$${\chi }_2^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_1^{(\ell )}={\chi }_1^{(2)}+{\chi }_1^{(3)}=6,$$

or ${(6,8,8,6)}^T$.

$${\chi }_3^{(1)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_1}}{\chi }_2^{(\ell )}={\chi }_2^{(2)}+{\chi }_2^{(3)}=16,$$

$${\chi }_3^{(2)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_2}}{\chi }_2^{(\ell )}={\chi }_3^{(3)}+{\chi }_3^{(4)}=20,$$

$${\chi }_3^{(3)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_3}}{\chi }_2^{(\ell )}={\chi }_3^{(1)}+{\chi }_3^{(2)}+{\chi }_3^{(4)}=20,$$

$${\chi }_3^{(4)}={\Sigma }_{{\vartheta }_{\ell }\in \overleftrightarrow{{\aleph }_4}}{\chi }_2^{(\ell )}={\chi }_3^{(2)}+{\chi }_3^{(3)}=16,$$

or ${(16,20,20,16)}^T$.

Also, $\sigma ({\mathcal{D}}_4)=\{1+\sqrt{3},-1,-1,1-\sqrt{3}\}$, so

$$\rho ({\mathcal{D}}_4)=1+\sqrt{3}\approx 2.7320508075,$$

and

$$\mathfrak{E}({\mathcal{D}}_4)=5.464100808.$$

The above calculation implies that, for $j=$1,2,3,…,

$$\begin{array}{c}{\alpha }^{(1)}=2.54000254\le {\alpha }^{(2)}=2.561249695\le {\alpha }^{(3)}=2.561440144\le {\alpha }^{(4)}=2.56150939\\ \le \dots \le \rho ({\mathcal{D}}_4)=1+\sqrt{3}\approx 2.7320508075.\end{array}$$

Hence, this shows that the sequence of bounds is increasing for $\rho ({\mathcal{D}}_4)$. And, by using Theorem 3.0.2, the upper bounds for the $\mathfrak{E}({\mathcal{D}}_4)=5.46410\le 5.98899={\delta }_j\le \dots \le \dots 6.21062={\delta }_4\le 6.21069={\delta }_3\le 6.21090={\delta }_2\le 6.23393={\delta }_1$. This sequence of upper bounds for $\mathfrak{E}({\mathcal{D}}_4)$ decreases as we proved, or we can directly compute that this sequence decreases by using case 1 of Corollary 3.

4. Conclusions

The spectral radius $\rho (\mathcal{D})$ is an essential tool in spectral graph theory, for exploring valuable insights into the structural and dynamic characteristics of the graph. It assists as the connectivity measure, reflecting how well the vertices are linked and their accessibility within the graph. This study focuses on deriving the sequence of lower bounds of $\rho (\mathcal{D})$ of $\mathcal{D}$ having at least one doubly adjacent vertex in terms of in-degrees of their corresponding vertices, and it is particularly exhibited that $\rho (\mathcal{D})\ge {\alpha }^j=\sqrt{{\displaystyle \frac{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_{j+1}^{(p)})}^2}{{\sum }_{p=1}^{\mathfrak{U}}{({\chi }_j^{(p)})}^2}}}$, such that equality is attained iff $\mathcal{D}=\overleftrightarrow{\mathfrak{G}}+$ {$\mathcal{DE}\notin$ Cycle}, where each component of the associated graph is a k-regular or $(k_1,k_2)$ semiregular bipartite, providing a deeper understanding of its behavior in graphs with particular structural characteristics. Using these bounds, we constructed a sequence of upper bounds for $\mathfrak{E}(\mathcal{D})$, where the sequence decreases when $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}\le {\alpha }^j$ and increases when $\sqrt{{\displaystyle \frac{\mathbf{e}}{\mathfrak{U}}}}>{\alpha }^j$. All obtained results have been proved by examples.