Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X  

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Some Properties of The Spectrum of The Power Digraph 𝚪(𝒏, 𝒌) 
 

Sanjay Kumar Thakur 𝟏∗, Gautam Chandra Ray 𝟐, Pinkimani Goswami 𝟑 

1Department of Mathematics, Science College, Kokrajhar, India, e-mail: sanj26sc@yahoo.com 

2Department of Mathematics, CIT, Kokrajhar, India, e-mail: gc.roy@cit.ac.in 

3Department of Mathematics, University of Science and Technology, Baridua,  

e-mail: pinkimanigoswami@yahoo.com  ∗corresponding author   

 

Article History: 

Received: 23-05-2024 

Revised: 10-07-2024 

Accepted: 22-07-2024 

Abstract: 

For every positive integer 𝑛 and 𝑘 , a power digraph modulo 𝑛, denoted by Γ(𝑛, 𝑘) is 

constructed with the vertex set ℤ𝑛 = {0,1,2,⋯ , 𝑛 − 1}, and a directed edge from a vertex 

𝑥 to a vertex 𝑦 exists if and only if 𝑥𝑘 ≡ 𝑦(𝑚𝑜𝑑 𝑛), where 𝑥, 𝑦 ∈ ℤ𝑛. In this work, we 

define the out-adjacency (𝐴Γ
+) and the in-adjacency (𝐴Γ

−) matrices of the digraph Γ(𝑛, 𝑘) 

and some results on 𝐴Γ
+ and 𝐴Γ

− are discussed. It is proved that the matrices 𝐴Γ
+ and 𝐴Γ

− are 

singular if 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝. Some spectral properties of Γ(𝑛, 𝑘) are also 

presented. Moreover, it is proved that the algebraic multiplicity of 1 as an eigenvalue of 

𝐴Γ
+ is the number of components of the digraph Γ(𝑛, 𝑘).   

Keywords: Digraph, Adjacency matrices of power digraph (mod 𝑛), eigenvalues.   

AMS Subject classification: 11A07, 05C50 

1. Introduction 

In recent years, exploring the interconnections between Graph theory, Group theory, and Number 

theory has emerged as an attractive and effective study area, for example, [3, 4, 6, 8, 10, 11, 13, 14, 

17, 18, 20]. In this article, for each positive integers 𝑛 and 𝑘, we consider a power digraph modulo 𝑛 

denoted by Γ(𝑛, 𝑘) whose vertex set is ℤ𝑛 = {0,1,2,⋯ , 𝑛 − 1} and the ordered pair (𝑥, 𝑦) is a directed 

arc (or directed edge) of Γ(𝑛, 𝑘) from 𝑥 to 𝑦 iff 𝑥𝑘 ≡ 𝑦(𝑚𝑜𝑑 𝑛), where 𝑥, 𝑦 ∈ ℤ𝑛. In [3, 6, 9, 12, 14, 

17, 18, 21] some properties of the power digraph Γ(𝑛, 𝑘) were studied.  

The adjacency matrix is a commonly used matrix representation for graphs, and numerous researchers 

have investigated the connection between the eigenvalues of the adjacency matrix and the graph’s 

structures in the past, for example, [1, 2, 7]. In the case of a multidigraph 𝐺 with 𝑛 vertices, the 

adjacency matrix of 𝐺 defined in [1] as the 𝑛 × 𝑛 matrix 𝐴(𝐺) = [𝑎𝑖𝑗], where 𝑎𝑖𝑗 represents the 

number of directed edges that start at the vertex 𝑖 and ends at the vertex 𝑗. It is important to note that 

based on this definition, the adjacency matrix of a multidigraph is not symmetric in general. So, it may 

have complex eigenvalues. Furthermore, a graph is completely determined by its adjacency 

eigenvalues and corresponding eigenvectors. This is evident from the fact that a graph 𝐺 can be 

uniquely determined by 𝐴(𝐺). In the case of an undirected simple graph 𝐺, 𝐴(𝐺) is symmetric. 

It is important to mention that the study of adjacency matrices of Γ(𝑛, 𝑘), the power digraph modulo 

𝑛 is still open. In this paper, we aim to define the adjacency matrices of the digraph Γ(𝑛, 𝑘) and try to 

explore some properties associated with them.  



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We organize the rest of the paper as follows: In Section 2, we provide some definitions and results 

from Graph Theory and Matrix Theory. In Section 3, we define the out-adjacency (𝐴Γ
+) and the in-

adjacency (𝐴Γ
−) matrices of the digraph Γ(𝑛, 𝑘) and some results on 𝐴Γ

+ and 𝐴Γ
− are discussed. It is 

proved that the matrices 𝐴Γ
+ and 𝐴Γ

− are singular if 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝. In Section 4, 

some spectral properties of Γ(𝑛, 𝑘) are presented. It is also proved that the algebraic multiplicity of 1 

as an eigenvalue of 𝐴Γ
+ is the number of components of the digraph Γ(𝑛, 𝑘). 

2. Preliminaries 

For each positive integers 𝑛 and 𝑘, we consider a power digraph modulo 𝑛 denoted by Γ(𝑛, 𝑘) (in 

short, directed graph Γ(𝑛, 𝑘) or digraph Γ(𝑛, 𝑘)) whose vertex set is ℤ𝑛 and any two vertices 𝑥, 𝑦 ∈ ℤ𝑛 

are connected by a directed arc from 𝑥 to 𝑦 if and only if 𝑥𝑘 ≡ 𝑦(𝑚𝑜𝑑 𝑛). 

We denote the vertex set of the digraph Γ(𝑛, 𝑘) by 𝑉(Γ(𝑛, 𝑘)) or by 𝑉(Γ) (= ℤ𝑛) and the arc set by 

𝐴(Γ(𝑛, 𝑘)) or by 𝐴(Γ). The distinct vertices  𝑣1, 𝑣2, 𝑣3, … , 𝑣𝑡 in 𝑉(Γ) will form a cycle of length 𝑡 if  

𝑣1
𝑘 ≡ 𝑣2(𝑚𝑜𝑑 𝑛) 

𝑣2
𝑘 ≡ 𝑣3(𝑚𝑜𝑑 𝑛) 

𝑣3
𝑘 ≡ 𝑣4(𝑚𝑜𝑑 𝑛) 

⋮ 

𝑣𝑡
𝑘 ≡ 𝑣1(𝑚𝑜𝑑 𝑛) 

We call a cycle of length 𝑡 as a t- cycle and a cycle of length 1 is named as a fixed point (or a self-

loop). A vertex is isolated if it is not connected to any other vertex in Γ(𝑛, 𝑘). Some researchers have 

developed theorems to find the number of fixed points of the digraph Γ(𝑛, 𝑘), denoted by 𝐿(𝑛) for 

some values of 𝑘 see [5, 15, 16, 19, 20]. From these theorems, it is clear that 0 is always a fixed point 

of Γ(𝑛, 𝑘) and so the number of fixed points, 𝐿(𝑛) > 0. 

The in-degree of a vertex 𝑣 ∈   𝑉(Γ), denoted by 𝑑Γ
−(𝑣) is the number of directed arcs incident into 

the vertex 𝑣 and the out-degree of a vertex 𝑣, denoted by 𝑑Γ
+(𝑣) is the number of directed arcs incident 

out of the vertex 𝑣. Since the residue of a number modulo 𝑛 is unique, so 𝑑Γ
+(𝑣) = 1 and 𝑑Γ

−(𝑣) ≥ 0 

for each vertex 𝑣 ∈   𝑉(Γ). Also, for an isolated fixed point 𝑣 ∈   𝑉(Γ) , 𝑑Γ
+(𝑣) = 𝑑Γ

−(𝑣) = 1. The total 

degree (or simply degree) of a vertex 𝑣 ∈   𝑉(Γ), denoted by 𝑑Γ(𝑣) is the sum of out-degree and in-

degree of 𝑣 i.e. 𝑑Γ(𝑣) = 𝑑Γ
+(𝑣) + 𝑑Γ

−(𝑣).  

A component of a digraph is a subdigraph which is a maximal connected subgraph of the associated 

nondirected graph.  

As the out-degree of each vertex of the digraph Γ(𝑛, 𝑘) is equal to 1, the number of components of 

Γ(𝑛, 𝑘) equals the number of all cycles. The cycles may or may not be isolated. 

We call a digraph regular if the in-degree of each vertex is equal to 1. Every component of such a 

digraph is a cycle. A digraph is semi-regular if there exists a positive integer 𝑑 such that each vertex 

either has in-degree 0 or 𝑑. 



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For 𝑛 > 1, let us divide the digraph Γ(𝑛, 𝑘) into two subdigraphs Γ1(𝑛, 𝑘) and Γ2(𝑛, 𝑘), where Γ1(𝑛, 𝑘) 

is the subdigraph induced on the set of the vertices 𝑣 ∈ ℤ𝑛 such that gcd(𝑣, 𝑛) = 1 and Γ2(𝑛, 𝑘) is the 

subdigraph induced on the set of the vertices 𝑣 ∈ ℤ𝑛 such that gcd(𝑣, 𝑛) ≠ 1. Clearly, the vertex set 

of Γ1(𝑛, 𝑘) is the unit group ℤ𝑛
∗  with order 𝜙(𝑛), where 𝜙(𝑛) denotes the Euler’s totient function. 

Also, 1 and (𝑛 − 1) are vertices of Γ1(𝑛, 𝑘) and 0 is always a vertex of Γ2(𝑛, 𝑘). One can easily observe 

that Γ1(𝑛, 𝑘) ∪ Γ2(𝑛, 𝑘) = Γ(𝑛, 𝑘) and Γ1(𝑛, 𝑘) ∩ Γ2(𝑛, 𝑘) = 𝜙.  

From definition of Γ(𝑛, 𝑘), it is clear that |𝐴(Γ)| = 𝑛. Since the number of arcs in a directed graph 

equals the number of their tails (or their heads), we have the following theorem. 

Theorem 2.1. [22] (Handshaking theorem) In the digraph Γ(𝑛, 𝑘),  

 ∑𝑣∈  𝑉(Γ) 𝑑Γ
+(𝑣) = ∑𝑣∈  𝑉(Γ) 𝑑Γ

−(𝑣) = |𝐴(Γ)|  

 A directed walk in a digraph D is an alternating sequence 𝑣1, 𝑒1, 𝑣2, 𝑒2, 𝑣3, … , 𝑒𝑛−1, 𝑣𝑛 of vertices and 

arcs in which each arc 𝑒𝑖 is 𝑣𝑖𝑣𝑖+1. A directed path is a walk in which all vertices are distinct. If there 

is a directed path from a vertex 𝑢 to a vertex 𝑣, then 𝑣 is said to be reachable from 𝑢. 

In a digraph D, a semi-walk is an alternating sequence 𝑣1, 𝑒1, 𝑣2, 𝑒2, 𝑣3, … , 𝑒𝑛−1, 𝑣𝑛 of vertices and arcs 

in which each arc 𝑒𝑖 may be either 𝑣𝑖𝑣𝑖+1 or 𝑣𝑖+1𝑣𝑖. A semi-path is a semi-walk in which all vertices 

are distinct. 

A digraph is strongly connected (or strong) if every two vertices are mutually reachable. A digraph is 

unilaterally connected (or unilateral) if for any two vertices at least one is reachable from the other. A 

digraph is weakly connected (or weak) if every two vertices are joined by a semi-path. 

Every strongly connected (or strong) digraph is a unilateral digraph and every unilateral digraph is 

weak. But the converse statements are not true. 

A digraph is disconnected if it is not even weak. 

 Note 2.1. From the definition of the digraph Γ(𝑛, 𝑘), it is clear that Γ(𝑛, 𝑘) is a disconnected graph, 

and the components of Γ(𝑛, 𝑘) are weakly connected. 

A tree is a connected acyclic graph. A tree in which one vertex has been designated as the root is a 

rooted tree. The edges of a rooted tree can be assigned a natural orientation, either away from or 

towards the root, in which case the structure becomes a directed rooted tree. When a directed rooted 

tree has an orientation away from the root, it is called an arborescence or out-tree and when it has an 

orientation towards the root, it is called an anti-arborescence or in-tree. A vertex in a rooted tree is 

called a leaf if 𝑑Γ
−(𝑣) = 0. 

A block diagonal matrix is a square matrix of the form  

 B = [

𝐴11 0 0 ⋯ 0
0 𝐴22 0 ⋯ 0
⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴𝑚𝑚

]  

Where 𝐴11, 𝐴22, ⋯ , 𝐴𝑚𝑚 are square matrices lying along the diagonal and all other entries of the  

matrix is 0 (zero matrices). Determinant of B is given by  



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 det(𝐵) = det(𝐴11) × det(𝐴22) × ⋯× det(𝐴𝑚𝑚).  

3. Adjacency matrices of the digraph 𝚪(𝒏, 𝒌) 

In this section, we try to define adjacency matrices of the digraph Γ(𝑛, 𝑘) and try to study some 

properties associated with them. 

Definition 3.1.  We define the out-adjacency matrix of the digraph Γ(𝑛, 𝑘) as an 𝑛 × 𝑛 matrix [𝑎𝑖𝑗] 

such that  

𝑎𝑖𝑗 = { 
1, if (vi, vj) ∈ A(Γ)

0, otherwise
 

 We denote this matrix by 𝐴+(Γ(𝑛, 𝑘)) or by 𝐴Γ
+.  

 Definition 3.2.  We define the in-adjacency matrix of the digraph Γ(𝑛, 𝑘) as an 𝑛 × 𝑛 matrix [𝑎𝑖𝑗] 

such that  

𝑎𝑖𝑗 = {
 1, if (vj, vi) ∈ A(Γ)

 0, otherwise
 

 We denote this matrix by 𝐴−(Γ(𝑛, 𝑘)) or by 𝐴Γ
−.  

           From definition of Γ(𝑛, 𝑘) it is clear that Γ(𝑛, 𝑘) is a disconnected graph, so the out-adjacency 

matrix 𝐴Γ
+ can also be defined as a block diagonal matrix [𝐴𝑖𝑗]𝑚×𝑚 i.e. 𝐴Γ

+ = [𝐴𝑖𝑗]𝑚×𝑚, such that  

𝐴𝑖𝑗 = {
 [𝑎𝑢𝑣]𝑞×𝑞 , for  i = j;  q ≤ m ≤ n

 0, for  i ≠ j.
 

 where,  

𝑎𝑢𝑣 = {
 1, if there is a directed arc from uth vertex to vth vertex.
 0, otherwise.

 

Similarly, the in-adjacency matrix 𝐴Γ
− can be defined as a block diagonal matrix [𝐴𝑖𝑗]𝑚×𝑚 i.e.  

𝐴Γ
− = [𝐴𝑖𝑗]𝑚×𝑚, such that  

𝐴𝑖𝑗 = {
 [𝑎𝑢𝑣]𝑞×𝑞 , for  i = j;  q ≤ m ≤ n

 0, for  i ≠ j.
 

 where,  

𝑎𝑢𝑣 = {
 1, if there is a directed arc from vth vertex to uth vertex.
 0, otherwise.

 

We have the following observations about 𝐴Γ
+ and 𝐴Γ

− of a digraph Γ(𝑛, 𝑘): 

i. Each non-zero element on the main diagonal of 𝐴Γ
+ and 𝐴Γ

− represents a loop at the corresponding 

vertex. 

ii. The number of non-zero entries of either 𝐴Γ
+ or 𝐴Γ

− equals the number of directed arcs in Γ(𝑛, 𝑘).  



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iii. Permutations of any rows together with a permutation of the corresponding columns do not alter 

the power digraph Γ(𝑛, 𝑘); indicating that the permutation simply rearranges the vertices. 

iv. The out-adjacency matrix 𝐴Γ
+ (or in-adjacency matrix 𝐴Γ

−) is not unique (follows from iii.).  

v. The out-adjacency (or in-adjacency) matrix 𝐴Γ
+ (or 𝐴Γ

−) of the digraph Γ(𝑛, 𝑘) can be written as a 

block-diagonal matrix with diagonal elements as the out-adjacency (or in-adjacency) matrices of the 

component digraphs of the digraph Γ(𝑛, 𝑘).  

Example 3.1.  Let us consider the digraph Γ(6, 2).   

 

Figure  1: Digraph Γ(6, 2) with components Γ1, Γ2, Γ3, Γ4. 

Here, 

𝐴Γ
+ = 

0 1 2 3 4 5

0 1 0 0 0 0 0

1 0 1 0 0 0 0

2 0 0 0 0 1 0

3 0 0 0 1 0 0

4 0 0 0 0 1 0

5 0 1 0 0 0 0

 
 
 
 
 
 
 
 
 

,   𝐴Γ
− = 

0 1 2 3 4 5

0 1 0 0 0 0 0

1 0 1 0 0 0 1

2 0 0 0 0 0 0

3 0 0 0 1 0 0

4 0 0 1 0 1 0

5 0 0 0 0 0 0

 
 
 
 
 
 
 
 
 

, and (𝐴Γ
+)𝑡 = 

0 1 2 3 4 5

0 1 0 0 0 0 0

1 0 1 0 0 0 1

2 0 0 0 0 0 0

3 0 0 0 1 0 0

4 0 0 1 0 1 0

5 0 0 0 0 0 0

 
 
 
 
 
 
 
 
 

 

Let us apply the elementary operations 𝑅2 ↔ 𝑅6 and then 𝐶2 ↔ 𝐶6; 𝑅3 ↔ 𝑅6 and then 𝐶3 ↔ 𝐶6 and 

finally 𝑅5 ↔ 𝑅6 and then 𝐶5 ↔ 𝐶6 to the matrix 𝐴Γ
+, we get the following matrix  

                                      

0 5 1 3 2 4

0 1 0 0 0 0 0

5 0 0 1 0 0 0

1 0 0 1 0 0 0

3 0 0 0 1 0 0

2 0 0 0 0 0 1

4 0 0 0 0 0 1

 
 
 
 
 
 
 
 
 

, which gives us the digraph Γ(6, 2). 

Hence, permuting rows together with the corresponding columns, the matrix 𝐴Γ
+ can be written as  



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𝐴Γ
+ =

0 5 1 3 2 4

0 1 0 0 0 0 0

5 0 0 1 0 0 0

1 0 0 1 0 0 0

3 0 0 0 1 0 0

2 0 0 0 0 0 1

4 0 0 0 0 0 1

 
 
 
 
 
 
 
 
 

=

[
 
 
 
 
𝐴Γ1
+ 0 0 0

0 𝐴Γ2
+ 0 0

0 0 𝐴Γ3
+ 0

0 0 0 𝐴Γ4
+
]
 
 
 
 

 

Where, 𝐴Γ1
+  = [1],     𝐴Γ2

+  = [
0 1
0 1

],     𝐴Γ3
+  = [1] ,      and 𝐴Γ4

+  = [
0 1
0 1

] are out-adjacency matrices of 

the component digraphs Γ1, Γ2, Γ3 , and Γ4 respectively. Thus, the out-adjacency matrix 𝐴Γ
+ of the 

digraph Γ(6, 2) can be written as a block-diagonal matrix with diagonal elements as the out-adjacency 

matrices of the component digraphs of the digraph Γ(6, 2). 

Similarly, the in-adjacency matrix 𝐴Γ
− of the digraph Γ(6, 2) can be written as a block-diagonal matrix 

with diagonal elements as the in-adjacency matrices of the component digraphs of the digraph 

Γ(6, 2) 𝑖. 𝑒. 

𝐴Γ
− = 

0 5 1 3 2 4

0 1 0 0 0 0 0

5 0 0 0 0 0 0

1 0 1 1 0 0 0

3 0 0 0 1 0 0

2 0 0 0 0 0 0

4 0 0 0 0 1 1

 
 
 
 
 
 
 
 
 

=

[
 
 
 
 
𝐴Γ1
− 0 0 0

0 𝐴Γ2
− 0 0

0 0 𝐴Γ3
− 0

0 0 0 𝐴Γ4
−
]
 
 
 
 

 

Where, 𝐴Γ1
−  = [1],     𝐴Γ2

−  = [
0 0
1 1

],     𝐴Γ3
−  = [1],     and 𝐴Γ4

−  = [
0 0
1 1

] are in-adjacency matrices of the 

component digraphs Γ1, Γ2, Γ3, and Γ4 respectively. 

Result 3.1.   (𝐴Γ
+)𝑡 = 𝐴Γ

− and (𝐴Γ
−)𝑡 = 𝐴Γ

+.  

Proof. Clearly, the matrices 𝐴Γ
+,  𝐴Γ

−, and (𝐴Γ
+)𝑡 are of the same order 𝑛 × 𝑛. 

Also, the (𝑖, 𝑗)𝑡ℎ element of (𝐴Γ
+)𝑡 = the (𝑗, 𝑖)𝑡ℎ element of 𝐴Γ

+ 

                                                  = the (𝑖, 𝑗)𝑡ℎ element of 𝐴Γ
−.       [By definition of 𝐴Γ

−] 

 Hence, (𝐴Γ
+)𝑡 = 𝐴Γ

−. 

Similarly, it can be shown that (𝐴Γ
−)𝑡 = 𝐴Γ

+.                                                                   

Result 3.2.   Let 𝐴Γ be the adjacency matrix of the underlying graph 𝐺 of the digraph Γ(𝑛, 𝑘),  

then 𝐴Γ = 𝐴Γ
+ + 𝐴Γ

−.  

Proof. Let 𝐴Γ
+ = [𝑎𝑖𝑗]𝑛×𝑛,  

 



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where,  

 𝑎𝑖𝑗 = {
 1, if there is a directed arc from the vertex vi to the vertex vj.

 0, otherwise.
 

 and,  𝐴Γ
− = [𝑏𝑖𝑗]𝑛×𝑛,  

where,  

 𝑏𝑖𝑗 = {
 1, if there is a directed arc from the  vertex vj to the vertex vi.

 0, otherwise.
 

 Also, let 𝐴Γ = [𝑐𝑖𝑗]𝑛×𝑛,  

where,  

 𝑐𝑖𝑗 = {

 2, if there is a loop at the vertex vi.
 1, if there is an edge between the vertices vi and vj.

 0, otherwise.

 

 Clearly, the matrices 𝐴Γ and 𝐴Γ
+ + 𝐴Γ

− are of the same order 𝑛 × 𝑛.  

We have, 

          𝐴Γ
+ + 𝐴Γ

− = [𝑎𝑖𝑗]𝑛×𝑛 + [𝑏𝑖𝑗]𝑛×𝑛 

 = [𝑎𝑖𝑗 + 𝑏𝑖𝑗]𝑛×𝑛 

 = [𝑑𝑖𝑗]𝑛×𝑛, where, 𝑑𝑖𝑗 = {

2, if there is a loop at the vertex vi.
1, if there is an edge between the vertices vi and vj.

0, otherwise.

 

 = [𝑐𝑖𝑗]𝑛×𝑛 

 = 𝐴Γ 

Hence, 𝐴Γ = 𝐴Γ
+ + 𝐴Γ

−.                                                                                                                      

Remark 3.1.   

       i.  𝐴Γ is symmetric i.e. (𝐴Γ)
𝑡 = 𝐴Γ  

       ii. 𝐴Γ = 𝐴Γ
+ + (𝐴Γ

+)𝑡  

       iii.  𝐴Γ = 𝐴Γ
− + (𝐴Γ

−)𝑡  

Result 3.3.    The sum of entries in the 𝑖𝑡ℎ row of 𝐴Γ
+ is 1.  

Proof. Let 𝐴Γ
+ = [𝑎𝑖𝑗] be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘) and let 𝑅𝑖 = [𝑎𝑖1, 𝑎𝑖2, ⋯ , 𝑎𝑖𝑛] 

be the 𝑖𝑡ℎ row of 𝐴Γ
+ corresponding to the vertex 𝑣𝑖 ∈ 𝑉(Γ). As the residue of a number modulo 𝑛 is 

unique, the number of directed arcs leaving the vertex 𝑣𝑖 is exactly one. It contributes thereby 1 exactly 

in one of the entries of 𝑅𝑖 and 0 in the remaining entries of 𝑅𝑖. Thus, ∑𝑛𝑗=1 𝑎𝑖𝑗 = 1.  

 



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Corollary 3.1.   The sum of entries in the 𝑖𝑡ℎ row of 𝐴Γ
+ is 𝑑Γ

+(𝑣𝑖), where 𝑑Γ
+(𝑣𝑖) is the out-degree of 

the 𝑖𝑡ℎ vertex 𝑣𝑖 .  

Proof. As the out-degree of each vertex 𝑣𝑖 ∈ Γ(𝑛, 𝑘) is 1, so we have  

 𝑑Γ
+(𝑣𝑖) = 1,    ∀ 𝑣𝑖 ∈ 𝑉(Γ). 

 ⇒ 𝑑Γ
+(𝑣𝑖) = 1 = ∑

𝑛
𝑗=1 𝑎𝑖𝑗     [By Result 3.3] 

            𝑖. 𝑒.  ∑𝑛𝑗=1 𝑎𝑖𝑗 = 𝑑Γ
+(𝑣𝑖).         

 Result 3.4.  The sum of entries in the 𝑖𝑡ℎ row of 𝐴Γ
− is 𝑑Γ

−(𝑣𝑖), where 𝑑Γ
−(𝑣𝑖) is the in-degree of the 

𝑖𝑡ℎ vertex 𝑣𝑖 .  

Proof. Let 𝐴Γ
− = [𝑎𝑖𝑗] be an in-adjacency matrix of the digraph Γ(𝑛, 𝑘) and let 𝑅𝑖 = [𝑎𝑖1, 𝑎𝑖2, ⋯ , 𝑎𝑖𝑛] 

be the 𝑖𝑡ℎ row of the matrix 𝐴Γ
− corresponding to the vertex 𝑣𝑖 ∈ 𝑉(Γ). We now consider the sum 

∑𝑛𝑗=1 𝑎𝑖𝑗. Clearly, 1 is added to this sum ∑𝑛𝑗=1 𝑎𝑖𝑗 exactly once for each directed arc coming to the 

vertex 𝑣𝑖 and thereby using the definition of the in-degree of a vertex the result follows immediately 

i.e. ∑𝑛𝑗=1 𝑎𝑖𝑗 = 𝑖𝑛𝑑𝑒𝑔(𝑣𝑖) = 𝑑Γ
−(𝑣𝑖).        

Result 3.5.  The sum of entries in the 𝑗𝑡ℎ column of 𝐴Γ
+ is 𝑑Γ

−(𝑣𝑗), where 𝑑Γ
−(𝑣𝑗) is the in-degree of 

the 𝑗𝑡ℎ vertex 𝑣𝑗 .  

 Proof. Let 𝐴Γ
+ = [𝑎𝑖𝑗] be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘) and let 𝐶𝑗 = [

𝑎1𝑗
𝑎2𝑗
⋮
𝑎𝑛𝑗

] be the 

𝑗𝑡ℎcolumn of 𝐴Γ
+ corresponding to the vertex 𝑣𝑗 ∈ 𝑉(Γ). We now consider the sum ∑𝑛𝑖=1 𝑎𝑖𝑗 . Clearly, 

1 is added to this sum ∑𝑛𝑖=1 𝑎𝑖𝑗 exactly once for each directed arc coming to the vertex 𝑣𝑗  and thereby 

using the definition of the in-degree of a vertex the result follows immediately i.e.                              

∑𝑛𝑖=1 𝑎𝑖𝑗 = 𝑖𝑛𝑑𝑒𝑔(𝑣𝑗) = 𝑑Γ
−(𝑣𝑗).  

Result 3.6.  The sum of entries in the 𝑗𝑡ℎ column of 𝐴Γ
− is 1.  

Proof. Let 𝐴Γ
− = [𝑎𝑖𝑗] be an in-adjacency matrix of the digraph Γ(𝑛, 𝑘) and let 𝐶𝑗 = [

𝑎1𝑗
𝑎2𝑗
⋮
𝑎𝑛𝑗

] be the 𝑗𝑡ℎ 

column of 𝐴Γ
− corresponding to the vertex 𝑣𝑗 ∈ 𝑉(Γ). As the residue of a number modulo 𝑛 is unique, 

the number of directed arcs leaving the vertex 𝑣𝑗  is exactly one. It contributes thereby 1 exactly in one 

of the entries of 𝐶𝑗 and 0 in the remaining entries of 𝐶𝑗. Thus, ∑𝑛𝑖=1 𝑎𝑖𝑗 = 1.      

Corollary 3.2.   The sum of entries in the 𝑗𝑡ℎ column of 𝐴Γ
− is 𝑑Γ

−(𝑣𝑗), where 𝑑Γ
−(𝑣𝑗) is the in-degree 

of the 𝑗𝑡ℎ vertex 𝑣𝑗 .  

Result 3.7.  The sum of all entries in the matrix 𝐴Γ
+ is ∑𝑛𝑖=1 𝑑Γ

+(𝑣𝑖).  



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Proof. Let 𝐴Γ
+ = [𝑎𝑖𝑗]𝑛×𝑛 be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘). Suppose 𝑅1, 𝑅2, ⋯ , 𝑅𝑛 

be the 𝑛-rows of the matrix 𝐴Γ
+. By Corollary 3.1., the sum of entries in the 𝑖𝑡ℎ row (i.e. 𝑅𝑖 ) is 𝑑Γ

+(𝑣𝑖), 

for all 𝑖 = 1,2,⋯ , 𝑛 and consequently, the sum of entries in all these rows is 𝑑Γ
+(𝑣1) + 𝑑Γ

+(𝑣2) + ⋯+

𝑑Γ
+(𝑣𝑛) = ∑

𝑛
𝑖=1 𝑑Γ

+(𝑣𝑖) i.e. ∑𝑛𝑖=1 ∑
𝑛
𝑗=1 𝑎𝑖𝑗 = ∑

𝑛
𝑖=1 𝑑Γ

+(𝑣𝑖).  

Result 3.8.  The sum of all entries in the matrix 𝐴Γ
+ is ∑𝑛𝑖=1 𝑑Γ

−(𝑣𝑖).  

Proof. The result can be easily established using Result 3.4.     

Remark 3.2.  If 𝐴Γ
+ = [𝑎𝑖𝑗]𝑛×𝑛 be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘) then 

    i.  ∑𝑛𝑖=1 ∑
𝑛
𝑗=1 𝑎𝑖𝑗 = ∑

𝑛
𝑖=1 𝑑Γ

+(𝑣𝑖) = ∑
𝑛
𝑖=1 𝑑Γ

−(𝑣𝑖) = 𝑛  

    ii.  ∑𝑛𝑖=1 ∑
𝑛
𝑗=1 𝑎𝑖𝑗 = |𝐴(Γ)| = |𝑉(Γ)| = 𝑛.  

Result 3.9.  The sum of all entries in the matrix 𝐴Γ
− is ∑𝑛𝑖=1 𝑑Γ

−(𝑣𝑖). 

Proof. The proof is left for the reader.   

Result 3.10.  The sum of all entries in the matrix 𝐴Γ
− is ∑𝑛𝑖=1 𝑑Γ

+(𝑣𝑖).  

Proof. The proof is left for the reader.   

Remark 3.3.  If 𝐴Γ
− = [𝑎𝑖𝑗]𝑛×𝑛 be an in-adjacency matrix of the digraph Γ(𝑛, 𝑘) then   

    i.  ∑𝑛𝑖=1 ∑
𝑛
𝑗=1 𝑎𝑖𝑗 = ∑

𝑛
𝑖=1 𝑑Γ

+(𝑣𝑖) = ∑
𝑛
𝑖=1 𝑑Γ

−(𝑣𝑖) = 𝑛  

    ii.  ∑𝑛𝑖=1 ∑
𝑛
𝑗=1 𝑎𝑖𝑗 = |𝐴(Γ)| = |𝑉(Γ)| = 𝑛.  

Result 3.11.  Let 𝐴Γ
+ = [𝑎𝑖𝑗]𝑛×𝑛 be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘), then the number 

of directed walks of length 𝑚 from vertex 𝑣𝑖 to vertex 𝑣𝑗(𝑖. 𝑒. 𝑣𝑖 → 𝑣𝑗  directed walk ) in Γ(𝑛, 𝑘) is the 

element in the (𝑖, 𝑗)𝑡ℎ position of the matrix (𝐴Γ
+)𝑚, where 𝑚 is a non-negative integer.  

Proof. We shall try to prove the result using mathematical induction on 𝑚. 

If 𝑚 = 0, then the number of directed walks of length 0 from vertex 𝑣𝑖 to vertex 𝑣𝑗  is 0 resulting 𝑎𝑖𝑗 =

0, for 𝑖 ≠ 𝑗. Also the number of directed walks of length 0 from a vertex 𝑣𝑖 to itself is 1 resulting 𝑎𝑖𝑗 =

1, for 𝑖 = 𝑗 which gives us the identity matrix 𝐼. So we get (𝐴Γ
+)0 = 𝐼. 

If 𝑚 = 1, then the number of directed walks of length 1 from vertex 𝑣𝑖 to vertex 𝑣𝑗  is the number of 

directed arcs from the vertex 𝑣𝑖 to vertex 𝑣𝑗  which is equal to 𝑎𝑖𝑗 of the out-adjacency matrix 𝐴Γ
+. So 

we get (𝐴Γ
+)1 = 𝐴Γ

+. 

We now assume that the result is true for 𝑚 > 1 and try to establish the result for 𝑚 + 1. Let us denote 

the (𝑖, 𝑗)𝑡ℎ element of (𝐴Γ
+)𝑚 by 𝑏𝑖𝑗 i.e. (𝐴Γ

+)𝑚 = [𝑏𝑖𝑗]𝑛×𝑛. 

As, 

 (𝐴Γ
+)𝑚+1 = (𝐴Γ

+)𝑚 ⋅ (𝐴Γ
+) 

     = [𝑏𝑖𝑗]𝑛×𝑛 ⋅ [𝑎𝑖𝑗]𝑛×𝑛 

     = [𝑐𝑖𝑗]𝑛×𝑛 



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 where, 𝑐𝑖𝑗 = ∑
𝑛
𝑘=1 𝑏𝑖𝑘𝑎𝑘𝑗. 

By assumption, 𝑏𝑖𝑘 is the number of 𝑣𝑖 → 𝑣𝑘 directed walks of length 𝑚. Also, 𝑎𝑘𝑗 = 0 or 1, so 

𝑏𝑖𝑘𝑎𝑘𝑗 = 0 or 𝑏𝑖𝑘. Then 𝑏𝑖𝑘𝑎𝑘𝑗 is exactly the number of 𝑣𝑖 → 𝑣𝑗  directed walks of length (𝑚 + 1) with 

vertex 𝑣𝑘 adjacent to vertex 𝑣𝑗 . As the sum includes this for each of the vertices, we notice that         

𝑐𝑖𝑗(= ∑
𝑛
𝑘=1 𝑏𝑖𝑘𝑎𝑘𝑗) is the number of 𝑣𝑖 → 𝑣𝑗  directed walks of length (𝑚 + 1) and hence the result 

holds for (𝐴Γ
+)𝑚+1. So by induction, the result is established.    

Result 3.12.  Let 𝐴Γ
− = [𝑎𝑖𝑗]𝑛×𝑛 be the in-adjacency matrix of the digraph Γ(𝑛, 𝑘), then the number of 

directed walks of length 𝑚 from vertex 𝑣𝑗  to vertex 𝑣𝑖 (𝑖. 𝑒. 𝑣𝑖 ← 𝑣𝑗  directed walk ) in Γ(𝑛, 𝑘) is the 

element in the (𝑖, 𝑗)𝑡ℎ position of the matrix (𝐴Γ
−)𝑚, where 𝑚 is a non-negative integer.  

Proof. It can be proven in the same way as Result 3.11, using the definition of 𝐴Γ
−.   

Result 3.13.  Let 𝐴Γ
+ = [𝑎𝑖𝑗]𝑛×𝑛 be an out-adjacency matrix of the digraph Γ(𝑛, 𝑘). Then the matrix 

𝐵Γ = [𝑏𝑖𝑗] has at least two entries which is zero, where 𝐵Γ = 𝐴Γ
+ + (𝐴Γ

+)2 + (𝐴Γ
+)3 +⋯+ (𝐴Γ

+)𝑛−1 

and 𝑛 > 1.  

Proof. By definition of Γ(𝑛, 𝑘), it is clear that the digraph Γ(𝑛, 𝑘) is disconnected for 𝑛 > 1. So there 

exists two or more than two disjoint components of Γ(𝑛, 𝑘) that have no directed arcs in between them. 

Let there be such 𝑠 number of components namely Γ1, Γ2, ⋯ , Γ𝑠. In this case, the out-adjacency matrix 

𝐴Γ
+ of Γ(𝑛, 𝑘) can be partitioned into block diagonal matrices as  

𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

 

where 𝐴Γ1
+ , 𝐴Γ2

+ , ⋯ , 𝐴Γ𝑠
+  are out-adjacency matrices of the components Γ1, Γ2, ⋯ , Γ𝑠 respectively.  

Now, let us consider the matrix, 𝐵Γ = 𝐴Γ
+ + (𝐴Γ

+)2 + (𝐴Γ
+)3 +⋯+ (𝐴Γ

+)𝑛−1. Clearly, each entry in 

(𝐴Γ
+)𝑚(1 ≤ 𝑚 ≤ 𝑛 − 1) counts the number of directed walks of length 𝑚 from vertex 𝑣𝑖 to vertex 𝑣𝑗 . 

As the digraph Γ(𝑛, 𝑘) is disconnected, so a directed walk from one component to another component 

is not possible, and hence the entry in the matrix (𝐴Γ
+)𝑚 corresponding to those directed walks will be 

zero. Thus the only non-zero entries in 𝐵Γ will come from the individual components Γ1, Γ2, ⋯ , Γ𝑠. 

Moreover, each out-adjacency matrix 𝐴Γ𝑖
+  corresponding to components Γ𝑖(1 ≤ 𝑖 ≤ 𝑠) is a non-zero 

square matrix. So, the submatrices in the diagonal blocks of 𝐵Γ will be non-zero matrices but non-

diagonal blocks will be zero because there are no arcs between the components. Hence, at least two 

entries in the matrix 𝐵Γ will be zero.   

Result 3.14.  Let 𝐴Γ
− = [𝑎𝑖𝑗]𝑛×𝑛 be the in-adjacency matrix of the digraph Γ(𝑛, 𝑘). Then the matrix 

𝐶Γ = [𝑐𝑖𝑗] has at least two entries which is zero, where 𝐶Γ = 𝐴Γ
− + (𝐴Γ

−)2 + (𝐴Γ
−)3 +⋯+ (𝐴Γ

−)𝑛−1 

and 𝑛 > 1.  

Proof. It can be proven in the same way as Result 3.13.  

 



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Lemma 3.1.  The digraph Γ(𝑛, 𝑘) has at least one vertex of in-degree 0 iff 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some 

prime 𝑝.  

Proof. Let Γ(𝑛, 𝑘) have at least one vertex of in-degree 0. To show 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝.  

Let 𝑝2 ∤ 𝑛, for any prime 𝑝. In this case, the digraph Γ1(𝑛, 𝑘) is semi-regular and so 𝑑Γ
−(𝑣) = 0 or 

𝑘𝜔(𝑛), for 𝑣 ∈ Γ1(𝑛, 𝑘), where  

 𝜔(𝑛) = {
 𝜔0(𝑛) + 1, if k2|n

 𝜔0(𝑛), if k2 ∤ n
 

 and 𝜔𝑜(𝑛) is the number of distinct primes dividing 𝑛 which are congruent to 1(𝑚𝑜𝑑 𝑘). 

As the set of residues which are co-prime to 𝑛, forms a group under multiplication modulo 𝑛 of order 

𝜙(𝑛), so the set of vertices of Γ1(𝑛, 𝑘) forms a group under multiplication modulo 𝑛 of order 𝜙(𝑛). 

Let 𝑣 ∈ Γ1(𝑛, 𝑘) such that 𝑑Γ
−(𝑣) = 𝑘𝜔(𝑛) and let H = {0 ≤ 𝑚 ≤ 𝑛 − 1  |(𝑚, 𝑛) = 1,𝑚𝑘 ≡

1(𝑚𝑜𝑑 𝑛)}. Then H is a subgroup of the group Γ1(𝑛, 𝑘) of order 𝑘𝜔(𝑛) and hence 𝑘𝜔(𝑛)|𝜙(𝑛) which 

implies 𝑘|𝜙(𝑛). 

Now, let 𝑘 ∤ 𝜙(𝑛). To show 𝑝2|𝑛, for some prime 𝑝. 

If possible, let 𝑝2 ∤ 𝑛 for any prime 𝑝, then 𝑛 is a square-free integer. Now, 𝑛 is square-free and 𝑘 ∤

𝜙(𝑛) so in this case the digraph Γ(𝑛, 𝑘) is cyclic. By definition, a digraph is cyclic if all of its 

components are cyclic. Moreover, if all the components of the digraph Γ(𝑛, 𝑘) are cycles, then the 

digraph Γ(𝑛, 𝑘) is regular and so 𝑑Γ
−(𝑣) = 1, ∀𝑣 ∈ Γ(𝑛, 𝑘), which contradicts the fact that there exists 

at least one vertex of in-degree 0. This contradiction implies that 𝑝2|𝑛, for some prime 𝑝. 

Conversely, let 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝. To show the digraph Γ(𝑛, 𝑘) has at least one vertex 

of in-degree 0.  

If 𝑘|𝜙(𝑛), then the digraph Γ1(𝑛, 𝑘) is a semi-regular digraph and hence 𝑑Γ
−(𝑣) = 0 or 𝑘𝜔(𝑛), for 𝑣 ∈

Γ1(𝑛, 𝑘). Thus, there exists at least one vertex 𝑣 such that 𝑑Γ
−(𝑣) = 0.  

If 𝑝2|𝑛, for some prime 𝑝, then some (or all) vertices of the digraph Γ2(𝑛, 𝑘) forms a rooted in-tree 

with root 0 and therefore there exists at least one leaf 𝑣 in this rooted in-tree such that 𝑑Γ
−(𝑣) = 0.   

Lemma 3.2.  The out-adjacency matrix 𝐴Γ
+ of Γ(𝑛, 𝑘) contains at least one block diagonal submatrix 

whose determinant is 𝑧𝑒𝑟𝑜 if 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝.  

Proof. Let us consider the digraph Γ(𝑛, 𝑘) with 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝. Let Γ1, Γ2, ⋯ , Γ𝑠 be 

the 𝑠 components of the digraph Γ(𝑛, 𝑘) and 𝑛 > 2. Let 𝐴Γ
+ be the out-adjacency matrix of the digraph 

Γ(𝑛, 𝑘) and  

𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

 

 where 𝐴Γ1
+ , 𝐴Γ2

+ , ⋯ , 𝐴Γ𝑠
+  are out-adjacency matrices of the components Γ1, Γ2, ⋯ , Γ𝑠 respectively. 



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By Lemma 3.1., the digraph Γ(𝑛, 𝑘) has at least one vertex of in-degree 0, so let 𝑣𝑡 be such a vertex of 

the digraph Γ(𝑛, 𝑘) such that indeg(𝑣𝑡) = 0. Therefore, each entry of the column 𝐶𝑣𝑡 (say) 

corresponding to the vertex 𝑣𝑡 in 𝐴Γ
+ will be zero. Now, some element(s) of 𝐶𝑣𝑡 is (are) also column 

element(s) of one of the block diagonal submatrix 𝐴Γ𝑖
+ (𝑠𝑎𝑦),1 ≤ 𝑖 ≤ 𝑠 and consequently one column 

of 𝐴Γ𝑖
+  is a zero column resulting 𝑑𝑒𝑡(𝐴Γ𝑖

+ ) = 0.   

Result 3.15.    If 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝 then the out-adjacency matrix 𝐴Γ
+ of Γ(𝑛, 𝑘) is a 

singular matrix.  

Proof. Let 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝. Also, let, 

𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

 

where 𝐴Γ1
+ , 𝐴Γ2

+ , ⋯ , 𝐴Γ𝑠
+  are respectively out-adjacency matrices of the components Γ1, Γ2, ⋯ , Γ𝑠 of the 

digraph Γ(𝑛, 𝑘). 

We have,  

 𝑑𝑒𝑡(𝐴Γ
+) = 𝑑𝑒𝑡(𝐴Γ1

+ ) × 𝑑𝑒𝑡(𝐴Γ2
+ ) × ⋯× 𝑑𝑒𝑡(𝐴Γ𝑠

+ ). (1) 

 By Lemma 3.2., 𝐴Γ
+ contains at least one block submatrix 𝐴Γ𝑚

+  (say), 1 ≤ 𝑚 ≤ 𝑠  such that 

𝑑𝑒𝑡(𝐴Γ𝑚
+ ) = 0 and hence from (1) we get, 𝑑𝑒𝑡(𝐴Γ

+) = 0. This shows that the matrix 𝐴Γ
+ is a singular 

matrix.   

Result 3.16.  If 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝 then the in-adjacency matrix 𝐴Γ
− of Γ(𝑛, 𝑘) is a 

singular matrix.  

Proof. Let 𝑘|𝜙(𝑛) or 𝑝2|𝑛, for some prime 𝑝.  

We have,  

 𝑑𝑒𝑡(𝐴Γ
−) = 𝑑𝑒𝑡((𝐴Γ

+)𝑡)   [By Result 3.1. ] 

                = 𝑑𝑒𝑡(𝐴Γ
+) 

                = 0                    [By Result 3.15. ] 

 This shows that the matrix 𝐴Γ
− is a singular matrix.     

4. Spectrum of the digraph 𝚪(𝒏, 𝒌) 

The characteristic polynomial of a matrix A is the polynomial 𝑑𝑒𝑡(𝐴 − 𝜆𝐼). The roots of the 

characteristic polynomial are the eigenvalues of 𝐴. A non-zero vector 𝑣 is an eigenvector of 𝐴 with 

eigenvalue  𝜆  if the equation 𝐴𝑣 = 𝜆𝑣 is satisfied. 

The eigenvalue(s) of a graph 𝐺 is (are) defined as the eigenvalue(s) of its adjacency matrix. The 

spectrum of a graph 𝐺 is the set of eigenvalues of 𝐺 together with their algebraic multiplicities. If a 



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graph 𝐺 has 𝑡 distinct eigenvalues 𝜆1 > 𝜆2 > 𝜆3 > ⋯ > 𝜆𝑡 with multiplicities 

𝑚(𝜆1),𝑚(𝜆2),𝑚(𝜆3),⋯ ,𝑚(𝜆𝑡) then the spectrum of 𝐺 is  

𝑆𝑝𝑒𝑐(𝐺)= (
𝜆1 𝜆2 𝜆3 ⋯ 𝜆𝑡
𝑚(𝜆1) 𝑚(𝜆2) 𝑚(𝜆3) ⋯ 𝑚(𝜆𝑡)

) 

Also, we have,  

 𝑑𝑒𝑡((𝐴Γ
+)𝑡 − 𝜆𝐼) = 𝑑𝑒𝑡((𝐴Γ

+)𝑡 − 𝜆𝐼𝑡), where I is an Identity matrix of order n. 

⇒ 𝑑𝑒𝑡(𝐴Γ
− − 𝜆𝐼) = 𝑑𝑒𝑡(𝐴Γ

+ − 𝜆𝐼)𝑡     [ By Result 3.1. , (𝐴Γ
+)𝑡 = 𝐴Γ

− ] 

⇒ 𝑑𝑒𝑡(𝐴Γ
− − 𝜆𝐼) = 𝑑𝑒𝑡(𝐴Γ

+ − 𝜆𝐼)       [∵ 𝑑𝑒𝑡(𝑋𝑡) = d𝑒𝑡(𝑋),where X is a square matrix. ] 

 So, the characteristic polynomial of 𝐴Γ
+ = The characteristic polynomial of 𝐴Γ

−.  

In this section, we will study some spectral properties of the digraph Γ(𝑛, 𝑘) using the out-adjacency 

matrix 𝐴Γ
+ or in-adjacency matrix 𝐴Γ

−. We define the eigenvalues of the digraph Γ(𝑛, 𝑘) as the 

eigenvalues of its out-adjacency matrix (or in-adjacency matrix) and the spectrum of the digraph 

Γ(𝑛, 𝑘) as the set of eigenvalues of Γ(𝑛, 𝑘) together with their algebraic multiplicities. 

Example 4.1.  Let us consider the digraph Γ(9, 11).   

 

Figure  2: Digraph Γ(9,11) with components Γ1, Γ2, Γ3, Γ4, Γ5. 

We have, 

𝐴Γ1
+ =

0 3 6

0 1 0 0

3 1 0 0

6 1 0 0

 
 
 
  

 ,  𝐴Γ2
+   =

 

1

1 1
 ,  𝐴Γ3

+  = 

2 5

2 0 1

5 1 0

 
 
 

 ,  𝐴Γ4
+ = 

4 7

4 0 1

7 1 0

 
 
 

 , and  𝐴Γ5
+ = 

 

8

8 1
 

Therefore, the characteristic polynomials of 𝐴Γ1
+ , 𝐴Γ2

+ , 𝐴Γ3
+ , 𝐴Γ4

+  and 𝐴Γ5
+  are 𝜆2(1 − 𝜆), (1 − 𝜆), (𝜆2 −

1), (𝜆2 − 1), and (1 − 𝜆) respectively. And, eigenvalues of 𝐴Γ1
+ ,  𝐴Γ2

+ ,  𝐴Γ3
+ ,  𝐴Γ4

+ , and 𝐴Γ5
+  are 

0,0,1;  1; −1,1; −1,1 and 1 respectively.  



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Also, 𝐴Γ
+ =  

[
 
 
 
 
 
𝐴Γ1
+ 0 0 0 0

0 𝐴Γ2
+ 0 0 0

0 0 𝐴Γ3
+ 0 0

0 0 0 𝐴Γ4
+ 0

0 0 0 0 𝐴Γ5
+
]
 
 
 
 
 

  =   

0 3 6 1 2 5 4 7 8

0 1 0 0 0 0 0 0 0 0

3 1 0 0 0 0 0 0 0 0

6 1 0 0 0 0 0 0 0 0

1 0 0 0 1 0 0 0 0 0

2 0 0 0 0 0 1 0 0 0

5 0 0 0 0 1 0 0 0 0

4 0 0 0 0 0 0 0 1 0

7 0 0 0 0 0 0 1 0 0

8 0 0 0 0 0 0 0 0 1

 
 
 
 
 
 
 
 
 
 
 
 
 
 

 

Therefore, the characteristic polynomial of 𝐴Γ
+ is                                                                      

 −𝜆9 + 3𝜆8 − 𝜆7 − 5𝜆6 + 5𝜆5 + 𝜆4 − 3𝜆3 + 𝜆2 = −𝜆2(𝜆 − 1)5(𝜆 + 1)2  

and, eigenvalues of 𝐴Γ
+ are 0, 0, −1,−1, 1, 1, 1, 1, 1. So, the Spectrum of Γ(9, 11) w.r.t. the adjacency 

matrix 𝐴Γ
+ is 𝑆𝑝𝑒𝑐(Γ(9, 11)) = (

−1 0 1
   2 2 5

 ).  

Moreover, the Characteristic polynomial of 𝐴Γ
− = The Characteristic polynomial of 𝐴Γ

+.                             

So, the characteristic polynomial of 𝐴Γ
− is  

 −𝜆9 + 3𝜆8 − 𝜆7 − 5𝜆6 + 5𝜆5 + 𝜆4 − 3𝜆3 + 𝜆2 = −𝜆2(𝜆 − 1)5(𝜆 + 1)2  

and, eigenvalues of 𝐴Γ
− are 0, 0, −1,−1, 1, 1, 1, 1, 1. So, the Spectrum of Γ(9, 11) w.r.t. the adjacency 

matrix 𝐴Γ
− is 𝑆𝑝𝑒𝑐(Γ(9, 11)) = (

−1 0 1
   2 2 5

 ).  

Result 4.1.  The digraph Γ(𝑛, 𝑘) has 𝑛 eigenvalues.   

Proof. Let us consider the digraph Γ(𝑛, 𝑘). Clearly |𝑉(Γ)| = 𝑛. 

The characteristic polynomial of the digraph Γ(𝑛, 𝑘) is given as 𝑃Γ(𝜆) = |𝐴Γ
+ − 𝜆𝐼𝑛|, which is a 

polynomial of degree 𝑛 in 𝜆. By the Fundamental theorem of algebra, we know that every polynomial 

of degree 𝑛 possesses precisely 𝑛 roots, taking into account their multiplicities within the complex 

number field. Hence, 𝑃Γ(𝜆) has 𝑛 roots. This shows that the digraph Γ(𝑛, 𝑘) has 𝑛- eigenvalues.  

Result 4.2.   If Γ1, Γ2, Γ3, ⋯ , Γ𝑠  are the 𝑠-components of the digraph Γ(𝑛, 𝑘) then  

 𝑃Γ(𝜆) = 𝑃Γ1(𝜆) ⋅ 𝑃Γ2(𝜆) ⋅ 𝑃Γ3(𝜆)⋯ ⋅ 𝑃Γ𝑠(𝜆)  

where 𝑃Γ(𝜆), 𝑃Γ1(𝜆), 𝑃Γ2(𝜆), 𝑃Γ3(𝜆),⋯ , 𝑃Γ𝑠(𝜆) are the characteristic polynomials of the digraphs 

Γ, Γ1, Γ2, Γ3,⋯ , Γ𝑠 respectively.  

Proof. Let us consider the digraph Γ(𝑛, 𝑘), where |𝑉(Γ)| = 𝑛. 

The characteristic polynomial of the digraph Γ(𝑛, 𝑘) is given as 𝑃Γ(𝜆) = |𝐴Γ
+ − 𝜆𝐼𝑛|.   

Let 𝐴Γ1
+ , 𝐴Γ2

+ , 𝐴Γ3
+ , ⋯ , 𝐴Γ𝑠

+  be the out-adjacency matrices of the component digraphs Γ1, Γ2, Γ3, ⋯ , Γ𝑠 

respectively. Also, let |𝑉(Γ𝑖(𝑛, 𝑘))| = 𝑛𝑖, 1 ≤ 𝑖 ≤ 𝑠 such that ∑𝑠𝑖=1 𝑛𝑖 = 𝑛. Then we have  



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 𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

  

 𝑎𝑛𝑑, 𝑑𝑒𝑡(𝐴Γ
+ − 𝜆𝐼𝑛) = 𝑑𝑒𝑡(𝐴Γ1

+ − 𝜆𝐼𝑛1) ⋅ 𝑑𝑒𝑡(𝐴Γ2
+ − 𝜆𝐼𝑛2) ⋅ 𝑑𝑒𝑡(𝐴Γ3

+ − 𝜆𝐼𝑛3)⋯𝑑𝑒𝑡(𝐴Γ𝑠
+ − 𝜆𝐼𝑛𝑠) 

 𝑖. 𝑒. 𝑃Γ(𝜆) = 𝑃Γ1(𝜆) ⋅ 𝑃Γ2(𝜆) ⋅ 𝑃Γ3(𝜆)⋯ ⋅ 𝑃Γ𝑠(𝜆).      

Result 4.3.    Let Γ(𝑛, 𝑘) be a digraph with 𝑠-components Γ1, Γ2, Γ3, ⋯ , Γ𝑠  then the spectrum of Γ(𝑛, 𝑘) 

is the union of the spectra of Γ1, Γ2, Γ3, ⋯ , Γ𝑠.  

Proof. To prove this, we try to show that each eigenvalue of Γ is also an eigenvalue of at least one of 

the components Γ𝑖 and conversely, each eigenvalue of Γ𝑖 is an eigenvalue of Γ ;    1 ≤ 𝑖 ≤ 𝑠 

Let 𝐴Γ
+, 𝐴Γ1

+ , 𝐴Γ2
+ , 𝐴Γ3

+ , ⋯ , 𝐴Γ𝑠
+  be the out-adjacency matrices of the digraphs Γ, Γ1, Γ2, Γ3, ⋯ , Γ𝑠 

respectively. Then we have  

𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

 

Let 𝜆 be an eigenvalue of the digraph Γ and let 𝑣 be the corresponding eigenvector, then  

     𝐴Γ
+ ⋅ 𝑣 = 𝜆 ⋅ 𝑣  

⇒

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

⋅ [

𝑣1
𝑣2
⋮
𝑣𝑠

] = 𝜆 ⋅ [

𝑣1
𝑣2
⋮
𝑣𝑠

], where 𝑣 = [

𝑣1
𝑣2
⋮
𝑣𝑠

] is an eigenvector of Γ.  

⇒ 𝐴Γ1
+ ⋅ 𝑣1 = 𝜆 ⋅ 𝑣1,        𝐴Γ2

+ ⋅ 𝑣2 = 𝜆 ⋅ 𝑣2, ⋯ , 𝐴Γ𝑠
+ ⋅ 𝑣𝑠 = 𝜆 ⋅ 𝑣𝑠  

⇒ 𝐴Γ𝑖
+ ⋅ 𝑣𝑖 = 𝜆 ⋅ 𝑣𝑖  ;     𝑖 = 1, 2,⋯ , 𝑠.  

This shows that 𝜆 is an eigenvalue of the component digraphs Γ𝑖 with eigenvalue 𝑣𝑖. Since 𝜆 is an 

eigenvalue of at least one of the components Γ𝑖, it is included in the spectrum of Γ. 

Conversely, let 𝜆 be an eigenvalue of a component Γ𝑖, then there exists a non-zero vector 𝑣𝑖 such that 

𝐴Γ𝑖
+ ⋅ 𝑣𝑖 = 𝜆 ⋅ 𝑣𝑖  ;     1 ≤ 𝑖 ≤ 𝑠  

⇒

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

⋅

[
 
 
 
 
 
 
0
⋮
0
𝑣𝑖
0
⋮
0 ]
 
 
 
 
 
 

= 𝜆 ⋅

[
 
 
 
 
 
 
0
⋮
0
𝑣𝑖
0
⋮
0 ]
 
 
 
 
 
 

 

                                  ⇒ 𝐴Γ
+ ⋅ 𝑣/ = 𝜆 ⋅ 𝑣/ ; where 𝑣/ = [0 ⋯ 0 𝑣𝑖 0 ⋯ 0]𝑡.  



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This shows that 𝜆 is an eigenvalue of the digraph Γ. Thus we have shown that every eigenvalue of Γ is 

also an eigenvalue of at least one of the components Γ𝑖 and conversely, every eigenvalue of Γ𝑖 is an 

eigenvalue of Γ. This proves that the spectrum of Γ is the union of the spectra of Γ𝑖.   

Result 4.4.  Let Γ(𝑛, 𝑘) be a digraph with 𝑠-components Γ1, Γ2, Γ3, ⋯ , Γ𝑠. Then 1 is an eigenvalue of 

each of the out-adjacency matrix 𝐴Γ𝑖
+ (1 ≤ 𝑖 ≤ 𝑠) with algebraic multiplicity one.  

Proof. Let 𝐴Γ𝑖
+  be the out-adjacency matrix of the component digraph Γ𝑖 with 𝑛𝑖 vertices, where 𝑛𝑖 ≤

𝑛 and 1 ≤ 𝑖 ≤ 𝑠. Clearly 𝐴Γ𝑖
+  is an 𝑛𝑖 × 𝑛𝑖 matrix. As the out-degree of each vertex in Γ(𝑛, 𝑘) is 1, so 

the out-degree of each vertex in Γ𝑖 is also 1 and hence 1 appears exactly once in each row of 𝐴Γ𝑖
+  with 

other entries as 0. We now consider the matrix 𝐴Γ𝑖
+ − 𝜆𝐼𝑛𝑖 and we apply the column operation 𝐶1 →

𝐶1 + 𝐶2 +⋯+ 𝐶𝑛𝑖 in the matrix 𝐴Γ𝑖
+ − 𝜆𝐼𝑛𝑖 , then it can be easily seen that each element of 𝐶1 is        

(1 − 𝜆) and hence (1 − 𝜆) will be a factor of  𝑑𝑒𝑡(𝐴Γ𝑖
+ − 𝜆𝐼𝑛𝑖). This shows that 1 is an eigenvalue of  

𝐴Γ𝑖
+ (1 ≤ 𝑖 ≤ 𝑠). 

Next, to show that the algebraic multiplicity of 1 is one. If possible, let the algebraic multiplicity of 1 

be greater than one. Then there exists at least two linearly independent vectors 𝑢 and 𝑣 with eigenvalue 

1 such that 𝐴Γ𝑖
+ ⋅ 𝑢 = 1 ⋅ 𝑢 and 𝐴Γ𝑖

+ ⋅ 𝑣 = 1 ⋅ 𝑣 which is possible if 𝑢 and 𝑣 are scalar multiples of each 

other and in this case, 𝑢 and 𝑣 are linearly dependent, which is a contradiction. Hence, the algebraic 

multiplicity of 1 is one.       

Result 4.5.   The algebraic multiplicity of 1 as an eigenvalue of 𝐴Γ
+ is the number of components of 

the digraph Γ(𝑛, 𝑘).  

Proof. Let 𝐴Γ
+ be the out-adjacency matrix of the digraph Γ(𝑛, 𝑘) where |𝑉(Γ(𝑛, 𝑘))| = 𝑛. Suppose 

Γ(𝑛, 𝑘) has 𝑠-components Γ1, Γ2, ⋯ , Γ𝑠 with their out-adjacency matrices 𝐴Γ1
+ , 𝐴Γ2

+ , 𝐴Γ3
+ , ⋯ , 𝐴Γ𝑠

+  

respectively. Also, let |𝑉(Γ𝑖(𝑛, 𝑘))| = 𝑛𝑖, 1 ≤ 𝑖 ≤ 𝑠 such that ∑𝑠𝑖=1 𝑛𝑖 = 𝑛. Then we have  

 𝐴Γ
+ = 

[
 
 
 
 
𝐴Γ1
+ 0 0 ⋯ 0

0 𝐴Γ2
+ 0 ⋯ 0

⋮ ⋮ ⋮ ⋱ ⋮
0 0 0 ⋯ 𝐴Γ𝑠

+
]
 
 
 
 

  

By Result 4.4., each block matrices 𝐴Γ1
+ , 𝐴Γ2

+ , ⋯ , 𝐴Γ𝑠
+  has eigenvalue 1 with algebraic multiplicity 1. 

Also, we have  

 𝑑𝑒𝑡(𝐴Γ
+ − 𝜆𝐼𝑛) = 𝑑𝑒𝑡(𝐴Γ1

+ − 𝜆𝐼𝑛1) ⋅ 𝑑𝑒𝑡(𝐴Γ2
+ − 𝜆𝐼𝑛2) ⋅ 𝑑𝑒𝑡(𝐴Γ3

+ − 𝜆𝐼𝑛3)⋯𝑑𝑒𝑡(𝐴Γ𝑠
+ − 𝜆𝐼𝑛𝑠)  

So, the algebraic multiplicity of 1 for the out-adjacency matrix 𝐴Γ
+ is the sum of the algebraic 

multiplicities of 1 for each 𝐴Γ1
+ , 𝐴Γ2

+ , ⋯ , 𝐴Γ𝑠
+ . Then this sum is 1 + 1 + 1 +⋯+ 1⏟            

𝑠−𝑡𝑒𝑟𝑚𝑠

= 𝑠 (as Γ(𝑛, 𝑘) has 

𝑠 -components). This shows that the algebraic multiplicity of 1 as an eigenvalue of 𝐴Γ
+ is 𝑠, which is 

the number of components of the digraph Γ(𝑛, 𝑘).   

  

 



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5. Conclusion 

we introduced the adjacency matrix of the power digraph Γ(𝑛, 𝑘), defining the out-adjacency matrix 

(𝐴Γ
+) and the in-adjacency matrix (𝐴Γ

−). We demonstrated that these matrices are singular if certain 

conditions are met and discussed the spectral properties of Γ(𝑛, 𝑘). Additionally, we proved that the 

algebraic multiplicity of 1 as an eigenvalue of (𝐴Γ
+) corresponds to the number of components in the 

digraph. 

Conflicts of Interest The authors declare no conflict of interest. 

References 

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