Communications on Applied Nonlinear Analysis 

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On Sombor Energy of Graphs with self-loops 

S. H. Pathan
1 

and S. C. Patekar
2 

1,2 Department of Mathematics, Savitribai Phule Pune University, Pune (India) 1samrinpathan508@gmail.com, 
2shri82patekar@gmail.com 

Article History: 

Received: 14-11-2024 

Revised:26-12-2024 

Accepted:10-01-2025 

Abstract:  

The goal of this paper is to broaden the concept of Sombor Energy from a simple graph 

to one containing self-loops. Let G be a simple nth-order graph, and 𝐺𝑠 be the graph 

generated by adding 𝜎 self-loops to G. Sombor matrix of 𝐺𝑠 is defined as  

 𝐴𝑆𝑂(𝐺𝑠) = (𝑎𝑖𝑗) =

{
 
 

 
 √𝑑𝑖

2 + 𝑑𝑗
2;         𝑖𝑓𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

√2𝑑𝑖;                                                   𝑖𝑓 𝑖 = 𝑗
0;               𝑖𝑓𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡.

  

If 𝜆1(𝐺𝑠), 𝜆2(𝐺𝑠), . . . , 𝜆𝑛(𝐺𝑠) are eigenvalues of 𝐴𝑆𝑂(𝐺𝑠), then Sombor Energy of 𝐺𝑠 is 

defined as 

 𝐸𝑆𝑂(𝐺𝑠) = ∑
𝑛
𝑖=1 |𝜆𝑖(𝐺𝑠) −

√2∑
𝜎
𝑗=1𝑑𝑗

𝑛
|  

 where 𝑑𝑗 is the degree of vertex 𝑣𝑗 with self loop 

Keywords: Eigenvalue, Sombor Energy, Self-loops. 

1. Introduction 

Let 𝐺 be a simple graph with vertex set 𝑉(𝐺) and edge set 𝐸(𝐺), |𝑉(𝐺)| = 𝑛. If the vertices 

𝑢, 𝑣 ∈ 𝑉(𝐺) are adjacent, then edge with end points are 𝑢 and 𝑣 is denoted by 𝑢𝑣. The neighbor 

of 𝑢 is denoted by 𝑁(𝑢) that is the set of vertices adjacent to 𝑢, |𝑁(𝑢)| is total number of 

adjacent vertices to 𝑢 and is called degree of 𝑢 and denoted by 𝑑𝐺(𝑢). 

Let 𝐴(𝐺) be the adjacency matrix of a simple graph 𝐺 with vertices 𝑣1, 𝑣2, . . . , 𝑣𝑛, elements of 

adjacency matrix are defined by  

                                        𝐴(𝐺) = (𝑎𝑖𝑗) = {
1;      𝑖𝑓 𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡        

0;     𝑖𝑓 𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗  𝑎𝑟𝑒 𝑛𝑜𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡.
  

 and let 𝜆1, 𝜆2, . . . , 𝜆𝑛 be the eigenvalues of matrix 𝐴(𝐺). 

The energy of simple graph, introduced by I. Gutman [3], is defined as  

 𝜀(𝐺) = ∑𝑛𝑖=1 |𝜆𝑖|.  

 Let S be a subset of V(G). The number of element of S will be denoted by 𝜎. Let 𝐺𝑆 is graph 

with 𝜎 self-loops. Because graphs containing self-loops are useful in chemistry, 

[heteroatoms,heteroconjugated,chemistry,molecules].  

In 2021, Gutman defined the adjacency matrix 𝐴(𝐺𝑆) [9] of the graph 𝐺𝑆. 

mailto:samrinpathan508@gmail.com
mailto:shri82patekar@gmail.com


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𝐴(𝐺𝑆) = (𝑎𝑖𝑗) =  {

  1;                      𝑖𝑓 𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗 𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

1;                                                   𝑖𝑓𝑣𝑖 ∈ 𝑆
    0;               𝑖𝑓 𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡.

 

 

Definition 1. [9] Let 𝜆1, 𝜆2, . . . , 𝜆𝑛 be the eigenvalues of matrix 𝐴(𝐺𝑆) such that ∑𝑛𝑖=1 𝜆𝑖 = 𝜎 

then energy of graph with 𝜎 self loop is given by 

 𝐸(𝐺𝑆) = ∑𝑛𝑖=1 |𝜆𝑖 −
𝜎

𝑛
|. 

 

Definition 2. [7] Let 𝐺 be a graph. If 𝑢, 𝑣 ∈ 𝑉(𝐺) and 𝑢𝑣 ∈ 𝐸(𝐺), then Sombor index of graph 

𝐺 is defined by 

 

 𝑆𝑂(𝐺) = ∑𝑢𝑣∈𝐸(𝐺) √𝑑𝐺(𝑢)2 + 𝑑𝐺(𝑣)2 

  

Definition 3. [8] The Sombor matrix of G is defined by  

 𝑆(𝐺) = (𝑠𝑖𝑗)𝑛×𝑛 = {
√𝑑𝑢2 + 𝑑𝑣2 ; if u and vare adjacent

0 ; otherwise
  

 We denote the eigenvalues of S(G) by 𝜇𝑖′𝑠 such that 𝜇1 ≥ 𝜇2 ≥. . . ≥ 𝜇𝑛. The set of all 

eigenvalues of S(G) is called Sombor spectrum and 𝜇1 is the Sombor spectral radius of G. The 

Sombor energy [8] is defined by  

 𝐸𝑆𝑂(𝐺) = ∑
𝑛
𝑖=1 |𝜇𝑖|. 

The sum of squares of eigenvalues of S(G) satisfies following equation is given by [2]  

 2𝐹 = 𝜇1
2 + 𝜇2

2 + 𝜇3
2+. . . +𝜇𝑛

2 (1) 

 where  

 𝐹 = 𝐹(𝐺) = ∑𝑛𝑖=1 𝑑𝑣𝑖
3 = ∑𝑣𝑖∼𝑣𝑗 (𝑑𝑖

2 + 𝑑𝑗
2) 

is forgotten topological index of G [1].  

 Definition 4.  Let S be a subset of V(G). The number of element of S will be denoted by 𝜎. The 

Sombor matrix of graph G with 𝜎 self loop is defined by  

 𝐴𝑆𝑂(𝐺𝑠) = (𝑎𝑖𝑗) =

{
 
 

 
 √𝑑𝑖

2 + 𝑑𝑗
2        𝑖𝑓𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

√2𝑑𝑖                                                      𝑖𝑓 𝑖 = 𝑗
0              𝑖𝑓𝑣𝑖 𝑎𝑛𝑑 𝑣𝑗𝑎𝑟𝑒 𝑛𝑜𝑡 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

 



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If 𝜆1(𝐺𝑠), 𝜆2(𝐺𝑠), . . .  𝜆𝑛(𝐺𝑠) are eigenvalues of 𝐴𝑆𝑂(𝐺𝑠), then the Sombor energy of 𝐺𝑠 (which 

is analogous to the energy of any matrix with a non-zero diagonal [Laplace energy, Laplace 

energy and radiac energy, New spectral]) must be defined as 

 

 𝐸𝑆𝑂(𝐺𝑠) = ∑
𝑛
𝑖=1 |𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1𝑑𝑗

𝑛
| 

where 𝑑𝑗 is the degree of vertex 𝑣𝑖 with self loop. 

 

2. Main Results 

Proposition 1.  Let G be graph with n vertices and S be any subset of V(G) with 𝜎 elements. 

If 𝜎 = 0, then 𝐸𝑆𝑂(𝐺𝑆) = 𝐸𝑆𝑂(𝐺).  

Proof. It is trivially obvious, since for 𝜎 = 0, the graphs 𝐺𝑆 and G coincide then Sombor matrix 

of 𝐺𝑆 and 𝐺 are same and hence Sombor energy are also same.  

 

 Let 𝑆(𝐺) be Sombor matrix of G and 𝐷(𝐺𝑆) is the diagonal matrix with diagonal entries degree 

of vertex having self loop then Sombor matrix of 𝐺𝑆 is equal to  

 𝐴𝑆𝑂(𝐺𝑆) = 𝑆(𝐺) + √2𝐷(𝐺𝑆) (2) 

 

Proposition 2.  Let G be a graph with n vertices and m edges. If S⊂V with 𝜎 elements, then 

eigenvalues 𝜆1(𝐺𝑠), 𝜆2(𝐺𝑠), . . ., 𝜆𝑛(𝐺𝑠) of 𝐴𝑆𝑂(𝐺𝑆) satisfy,   

    1. ∑𝑛𝑖=1 𝜆𝑖
2(𝐺𝑠) = 2𝐹 + 4∑𝑣𝑖∈𝑆 𝑑𝑖

2  

    2.  ∑𝑛𝑖=1 [𝜆𝑖(𝐺𝑠) −
√2∑𝜎𝑗=1𝑑𝑗

𝑛
]2 = 2𝐹 + 4∑𝑣𝑖∈𝑆 𝑑𝑖

2 −
2(∑𝑣𝑖∈𝑆𝑑𝑖)

2

𝑛
.   

Proof.      1.  From equation (2) 𝐴𝑆𝑂(𝐺𝑆) = 𝑆(𝐺) + √2𝐷(𝐺𝑆) 

Therefore, 

∑
𝑛

𝑖=1
𝜆𝑖
2(𝐺𝑠) =∑

𝑛

𝑖=1
[(𝑆(𝐺) + √2𝐷(𝐺𝑆))

2]𝑖𝑖 

      = ∑
𝑛

𝑖=1
[(𝑆(𝐺)2)𝑖𝑖 + 2√2[𝑆(𝐺)𝐷(𝐺𝑆)]𝑖𝑖 + 2[(𝐷(𝐺𝑆))

2]𝑖𝑖] 

From equation (1), 2𝐹 = ∑𝑛𝑖=1 (𝑆(𝐺)
2)𝑖𝑖 where 𝐹 = ∑𝑣𝑖∼𝑣𝑗 (𝑑𝑖

2 + 𝑑𝑗
2) 

As the diagonal entries of diagonal matrix are nonzero for 𝜎 self loop and diagonal entries of 

Sombor matrix is zero. Hence, diagonal entries of 𝑆(𝐺)𝐷(𝐺𝑆) are zero.  

Therefore, 



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 ∑𝑛𝑖=1 [𝑆(𝐺)𝐷(𝐺𝑆)]𝑖𝑖 = 0 

Also, 

 

 ∑𝑛𝑖=1 [𝐷(𝐺𝑆)
2] = 2∑𝑣𝑖∈𝑆 𝑑𝑖

2 

                           ∑
𝑛

𝑖=1
𝜆𝑖
2(𝐺𝑠)

=∑
𝑛

𝑖=1
(𝑆(𝐺)2)𝑖𝑖 + 2√2∑

𝑛

𝑖=1
[𝑆(𝐺)𝐷(𝐺𝑆)]𝑖𝑖 + 2∑

𝑛

𝑖=1
(𝐷(𝐺𝑆)

2)𝑖𝑖 

= 2𝐹 + 2(2∑
𝑣𝑖∈𝑆

𝑑𝑖
2) 

                                                                                = 2𝐹 + 4∑𝑣𝑖∈𝑆 𝑑𝑖
2  

    2.  ∑𝑛𝑖=1 [𝜆𝑖(𝐺𝑠) −
√2∑𝜎𝑗=1𝑑𝑗

𝑛
]2 = ∑𝑛𝑖=1 [𝜆𝑖(𝐺𝑆)

2 − 2√2𝜆𝑖(𝐺𝑠)
√2∑𝜎𝑗=1𝑑𝑗

𝑛
+ 2

(∑𝜎𝑗=1𝑑𝑗)
2

𝑛2
] 

=∑
𝑛

𝑖=1
[𝜆𝑖
2(𝐺𝑆) −

2√2

𝑛
∑

𝑣𝑖∈𝑆
𝑑𝑖∑

𝑛

𝑖=1
𝜆𝑖(𝐺𝑆) + 2𝑛

(∑𝜎𝑗=1 𝑑𝑗)
2

𝑛2
] 

As ∑𝑛𝑖=1 𝜆𝑖(𝐺𝑆) = √2∑𝑣𝑖∈𝑆 𝑑𝑖, 

= 2𝐹 + 4∑
𝑣𝑖∈𝑆

𝑑𝑖
2 −

2√2

𝑛
∑

𝑣𝑖∈𝑆
𝑑𝑖√2∑

𝑣𝑖∈𝑆
𝑑𝑖 + 2

(∑𝜎𝑗=1 𝑑𝑗)
2

𝑛
 

= 2𝐹 + 4∑
𝑣𝑖∈𝑆

𝑑𝑖
2 −

4

𝑛
(∑

𝑣𝑖∈𝑆
𝑑𝑖)

2 + 2
(∑𝜎𝑗=1 𝑑𝑗)

2

𝑛
 

                                               = 2𝐹 + 4∑𝑣𝑖∈𝑆 𝑑𝑖
2 −

2

𝑛
(∑𝑣𝑖∈𝑆 𝑑𝑖)

2 (𝑠𝑖𝑛𝑐𝑒 ∑𝑣𝑖∈𝑆 𝑑𝑖 =

∑𝜎𝑗=1 𝑑𝑗). 

Lemma 1.  Let 𝐺 = 𝐾𝑛 be a complete graph with n vertices, If 𝐺𝑆 is a graph obtained from G 

by adding n-1 self loops, then Sombor eigenvalues of 𝐺𝑆 are 

𝜆1 =
(𝑛2−1)√2+√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
, 𝜆2 =

(𝑛2−1)√2−√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
 , and 𝜆𝑖 = 0  𝑓𝑜𝑟 𝑖 ≥

3. 

Proof. Let J be the 𝑛 × 𝑛 matrix with all entries one and 0 is the matrix with all zero entries. 

Sombor matrix of 𝐺𝑆 is  

 

 𝐴𝑆𝑂(𝐺𝑆) = [
(𝑛 + 1)√2𝐽𝑛−1×𝑛−1 √2(𝑛2 + 1)𝐽𝑛−1×1

√2(𝑛2 + 1)𝐽1×𝑛−1 0
]

𝑛×𝑛

 

Then, 



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 𝑑𝑒𝑡(𝜆𝐼 − 𝐴𝑆𝑂(𝐺𝑆)) = 𝑑𝑒𝑡 [
𝜆𝐼𝑛−1×𝑛−1 − (𝑛 + 1)√2𝐽𝑛−1×𝑛−1 −√2(𝑛2 + 1)𝐽𝑛−1×1

−√2(𝑛2 + 1)𝐽1×𝑛−1 𝜆𝐼1×1
]

𝑛×𝑛

 

 

If M is non-singular square matrix  

 

 𝑑𝑒𝑡 [
𝑀 𝑁
𝑃 𝑄

] = 𝑑𝑒𝑡(𝑀)𝑑𝑒𝑡(𝑄 − 𝑃𝑀−1𝑁) 

Here 𝑀 = 𝜆𝐼𝑛−1×𝑛−1 − (𝑛 + 1)√2𝐽𝑛−1×𝑛−1, 𝑁 = −√2(𝑛2 + 1)𝐽𝑛−1×1, 𝑃 =

−√2(𝑛2 + 1)𝐽1×𝑛−1 and 𝑄 = 𝜆𝐼1×1, 

then 𝑑𝑒𝑡(𝑀) = 𝜆𝑛−2(𝜆 − (𝑛2 − 1)√2) 

𝑀−1 =
1

𝜆𝑛−2(𝜆 − (𝑛2 − 1)√2)
(𝜆𝑛−3(𝜆 − (𝑛2 − 1)√2𝐼𝑛−1×𝑛−1) + 𝜆

𝑛−3(𝑛

+ 1)√2𝐽𝑛−1×𝑛−1) 

𝑃𝑀−1 = −√2(𝑛2 + 1)𝐽1×𝑛−1 ×
1

𝜆𝑛−2(𝜆 − (𝑛2 − 1)√2)
(𝜆𝑛−3(𝜆 − (𝑛2 − 1)√2𝐼𝑛−1×𝑛−1)

+ 𝜆𝑛−3(𝑛 + 1)√2𝐽𝑛−1×𝑛−1) 

    = −√2(𝑛2 + 1)𝐽1×𝑛−1 × [
1

𝜆
𝐼𝑛−1×𝑛−1 +

(𝑛 + 1)√2

𝜆(𝜆 − (𝑛2 − 1)√2)
𝐽𝑛−1×𝑛−1] 

    =
−√2(𝑛2 + 1)

𝜆
𝐽1×𝑛−1 −

√2(𝑛2 + 1)(𝑛 + 1)√2

𝜆(𝜆 − (𝑛2 − 1)√2)
× (𝑛 − 1)𝐽1×𝑛−1 

    =
−√2(𝑛2 + 1)

𝜆
𝐽1×𝑛−1 −

2(𝑛2 − 1)√(𝑛2 + 1)

𝜆(𝜆 − (𝑛2 − 1)√2)
× 𝐽1×𝑛−1 

    = [
−√2(𝑛2 + 1)

𝜆
−
2(𝑛2 − 1)√(𝑛2 + 1)

𝜆(𝜆 − (𝑛2 − 1)√2)
]𝐽1×𝑛−1 

    = [
−(𝜆 − (𝑛2 − 1)√2)√2(𝑛2 + 1) − 2(𝑛2 − 1)√(𝑛2 + 1)

𝜆(𝜆 − (𝑛2 − 1)√2)
]𝐽1×𝑛−1 

    = [
−√2(𝑛2 + 1)𝜆

𝜆(𝜆 − (𝑛2 − 1)√2)
]𝐽1×𝑛−1 

    = [
−√2(𝑛2 + 1)

(𝜆 − (𝑛2 − 1)√2)
]𝐽1×𝑛−1 



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𝑃𝑀−1𝑁 = [
−√2(𝑛2 + 1)

(𝜆 − (𝑛2 − 1)√2)
]𝐽1×𝑛−1 × −√2(𝑛2 + 1)𝐽𝑛−1×1 

    =
−√2(𝑛2 + 1) × −√2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
𝐽1×1 

    =
2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
𝐽1×1 

𝑄 − 𝑃𝑀−1𝑁 = 𝜆𝐼1×1 −
2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
𝐽1×1 

Therefore , 

𝑑𝑒𝑡(𝑄 − 𝑃𝑀−1𝑁) = 𝜆 −
2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
 

𝑑𝑒𝑡(𝜆𝐼 − 𝐴𝑆𝑂(𝐺𝑆)) = 𝑑𝑒𝑡(𝑀)𝑑𝑒𝑡(𝑄 − 𝑃𝑀
−1𝑁) 

    = 𝜆𝑛−2(𝜆 − (𝑛2 − 1)√2) × [𝜆 −
2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
] 

    = 𝜆𝑛−2(𝜆 − (𝑛2 − 1)√2)(𝜆 −
2(𝑛2 + 1)(𝑛 − 1)

(𝜆 − (𝑛2 − 1)√2)
) 

    = 𝜆𝑛−2(𝜆2 − (𝑛2 − 1)√2𝜆 − 2(𝑛2 + 1)(𝑛 − 1)) 

    = 𝜆𝑛−2(𝜆 −
(𝑛2 − 1)√2 − √2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2
)(𝜆

−
(𝑛2 − 1)√2 + √2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2
) 

Hence, eigenvalues of 𝐺𝑆 are 

𝜆1 =
(𝑛2−1)√2+√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
, 𝜆2 =

(𝑛2−1)√2−√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
, and 𝜆𝑖 = 0  𝑓𝑜𝑟  𝑖 ≥

3. 

Theorem 3.  Let 𝐺 = 𝐾𝑛 be a complete graph with n vertices. If 𝐺𝑆 is a graph obtained from 

𝐺 by adding 𝑛 − 1 self loops, then Sombor energy of 𝐺𝑆 is 𝐸𝑆𝑂(𝐺𝑆) = (𝑛 − 2)√2
𝑛2−1

𝑛
+

√2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6 .  

 Proof. By definition Sombor energy of matrix with self loop is , 

𝐸𝑆𝑂(𝐺𝑠) =∑
𝑛

𝑖=1
|𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1 𝑑𝑗

𝑛
| 

By lemma 1, for complete graph with n-1 selfloop 



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∑𝜎𝑗=1 𝑑𝑗 = 𝑛2 − 1 and its eigenvalues are 𝜆1 =
(𝑛2−1)√2+√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
,  𝜆2 =

(𝑛2−1)√2−√2𝑛4+8𝑛3−12𝑛2+8𝑛−6

2
, and 𝜆𝑖 = 0  𝑓𝑜𝑟 𝑖 ≥ 3. 

𝐸𝑆𝑂(𝐺𝑠) = |𝜆1 −
(𝑛2 − 1)√2

𝑛
| + |𝜆2 −

(𝑛2 − 1)√2

𝑛
| +∑

𝑛

𝑖=3
| −

(𝑛2 − 1)√2

𝑛
| 

                                 

= |
(𝑛2 − 1)√2 + √2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2
−
(𝑛2 − 1)√2

𝑛
|

+ |
(𝑛2 − 1)√2 − √2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2
−
(𝑛2 − 1)√2

𝑛
| + (𝑛

− 2)
(𝑛2 − 1)√2

𝑛
 

                         =
(𝑛2 − 1)√2(𝑛 − 2)

2𝑛
+
√2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2

+ [−
(𝑛2 − 1)√2(𝑛 − 2)

2𝑛
+
√2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6

2
] + (𝑛

− 2)
(𝑛2 − 1)√2

𝑛
 

                                       = √2𝑛4 + 8𝑛3 − 12𝑛2 + 8𝑛 − 6 + (𝑛 − 2)
(𝑛2−1)√2

𝑛
. 

 

 Example 1. Consider 𝐺 = 𝐾3 is complete graph with 3 vertices and 𝐺𝑆 is the graph obtained 

by adding 2 loops to graph 𝐺 = 𝐾3.  

 

 

  

 The Sombor Energy of graph 𝐺𝑆 is given by  

 𝐸𝑆𝑂(𝐺𝑆) = ∑
3
𝑖=1 |𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1𝑑𝑗

3
| 

Sombor matrix of 𝐺𝑆 is 𝐴𝑆𝑂(𝐺𝑆) =

[
 
 
 4√2 4√2 2√5

4√2 4√2 2√5

2√5 2√5 0
]
 
 
 

 



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Here 𝜆1 = 10√2, 𝜆2 = −2√2, 𝜆3 = 0 

𝐸𝑆𝑂(𝐺𝑆) = |10√2 −
8√2

3
| + | − 2√2 −

8√2

3
| + |0 −

8√2

3
| = 20.741798915 

In above theorem putting n=3 we get, 

𝐸𝑆𝑂(𝐺𝑆) = (3 − 2)√2
32 − 1

3
+ √2(3)4 + 8(3)3 − 12(3)2 + 8(3) − 6 

    = √2
8

3
+ √288 = √2

8

3
+ 12√2 = 20.741798915.  

  

Lemma 2.  Let 𝐺 = 𝐾𝑛 be a complete graph with n vertices. If 𝐺𝑆 is a graph obtained from 𝐺 

by adding 𝜎 self loops, then Sombor eigenvalues of Sombor matrix 𝐺𝑆 are 0 with multiplicity 

𝜎 − 1, −(𝑛 − 1)√2 with multiplicity 𝑛 − 𝜎 − 1,  

√2((𝑛−1)2+2𝜎)+√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1, and 

√2((𝑛−1)2+2𝜎)−√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1.  

 Proof. Let J be the 𝑛 × 𝑛 matrix with all entries one, I be 𝑛 × 𝑛 identity matrix. 

Sombor matrix of complete graph 𝐺𝑆 with 𝜎 self loop is  

 𝐴𝑆𝑂(𝐺𝑆) = [
(𝑛 + 1)√2𝐽𝜎×𝜎 √2(𝑛2 + 1)𝐽𝜎×𝑛−𝜎

√2(𝑛2 + 1)𝐽𝑛−𝜎×𝜎 (𝑛 − 1)√2(𝐽 − 𝐼)𝑛−𝜎×𝑛−𝜎
]

𝑛×𝑛

 

Then, 

det(𝜆𝐼 − 𝐴𝑆𝑂(𝐺𝑆))

= 𝑑𝑒𝑡 |
𝜆𝐼𝜎×𝜎 − (𝑛 + 1)√2𝐽𝜎×𝜎 −√2(𝑛2 + 1)𝐽𝜎×𝑛−𝜎

−√2(𝑛2 + 1)𝐽𝑛−𝜎×𝜎 𝜆𝐼𝑛−𝜎×𝑛−𝜎 − (𝑛 − 1)√2(𝐽 − 𝐼)𝑛−𝜎×𝑛−𝜎
|

𝑛×𝑛

 

 det(𝜆𝐼 − 𝐴𝑆𝑂(𝐺𝑆)) =

|
𝜆𝐼𝜎×𝜎 − (𝑛 + 1)√2𝐽𝜎×𝜎 −√2(𝑛2 + 1)𝐽𝜎×𝑛−𝜎

−√2(𝑛2 + 1)𝐽𝑛−𝜎×𝜎 (𝜆 + (𝑛 − 1)√2)𝐼𝑛−𝜎×𝑛−𝜎 − (𝑛 − 1)√2𝐽𝑛−𝜎×𝑛−𝜎
|

𝑛×𝑛

 

If M is non-singular square matrix,  

 𝑑𝑒𝑡 [
𝑀 𝑁
𝑃 𝑄

] = 𝑑𝑒𝑡(𝑀)𝑑𝑒𝑡(𝑄 − 𝑃𝑀−1𝑁) 

here, 𝑀 = 𝜆𝐼𝜎×𝜎 − (𝑛 + 1)√2𝐽𝜎×𝜎 , 𝑁 = −√2(𝑛2 + 1)𝐽𝜎×𝑛−𝜎, 𝑃 = −√2(𝑛2 + 1)𝐽𝑛−𝜎×𝜎 and  

𝑄 = (𝜆 + (𝑛 − 1)√2)𝐼𝑛−𝜎×𝑛−𝜎 − (𝑛 − 1)√2𝐽𝑛−𝜎×𝑛−𝜎, 

Then, 𝑑𝑒𝑡(𝑀) = 𝜆𝜎−1(𝜆 − (𝑛 + 1)√2𝜎). Therefore 



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𝑀−1 =
1

𝜆𝜎−1(𝜆 − (𝑛 + 1)√2𝜎)
(𝜆𝜎−2(𝜆 − (𝑛 + 1)√2𝜎𝐼𝜎×𝜎) + 𝜆

𝜎−2(𝑛 + 1)√2𝐽𝜎×𝜎) 

𝑃𝑀−1 = −
√2(𝑛2+1)

(𝜆−(𝑛+1)√2𝜎)
𝐽𝑛−𝜎×𝜎 and 𝑃𝑀−1𝑁 =

2(𝑛2+1)𝜎

(𝜆−(𝑛+1)√2𝜎)
𝐽𝑛−𝜎×𝑛−𝜎. So 

𝑄 − 𝑃𝑀−1𝑁 = (𝜆 + (𝑛 − 1)√2)𝐼𝑛−𝜎×𝑛−𝜎 − (𝑛 − 1)√2𝐽𝑛−𝜎×𝑛−𝜎

−
2(𝑛2 + 1)𝜎

(𝜆 − (𝑛 + 1)√2𝜎)
𝐽𝑛−𝜎×𝑛−𝜎 

The eigenvalues of (𝜆 + (𝑛 − 1)√2)𝐼𝑛−𝜎×𝑛−𝜎 are (𝜆 + (𝑛 − 1)√2) with n-𝜎 multiplicity and 

eigenvalues of (𝑛 − 1)√2𝐽𝑛−𝜎×𝑛−𝜎 +
2(𝑛2+1)𝜎

(𝜆−(𝑛+1)√2𝜎)
𝐽𝑛−𝜎×𝑛−𝜎 are 0 with multiplicity 𝑛 − 𝜎 − 1 

and 
(𝜆(𝑛−1)√2+4𝜎)(𝑛−𝜎)

𝜆−(𝑛+1)√2𝜎)
 with multiplicity 1. 

Hence, the eigenvalues of 𝑄 − 𝑃𝑀−1𝑁 are (𝜆 + (𝑛 − 1)√2) −
(𝜆(𝑛−1)√2+4𝜎)(𝑛−𝜎)

𝜆−(𝑛+1)√2𝜎)
 with 

multiplicity 1 and (𝜆 + (𝑛 − 1)√2) with multiplicity 𝑛 − 𝜎 − 1.  

Then  

𝑑𝑒𝑡(𝑄 − 𝑃𝑀−1𝑁)

= (𝜆 + (𝑛

− 1)√2)𝑛−𝜎−1(
𝜆2 − √2𝜆((𝑛 − 1)2 + 2𝜎) − 𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

𝜆 − (𝑛 + 1)√2𝜎)
) 

𝑑𝑒𝑡(𝑀)𝑑𝑒𝑡(𝑄 − 𝑃𝑀−1𝑁)

= 𝜆𝜎−1(𝜆 + (𝑛 − 1)√2)𝑛−𝜎−1(𝜆2 − √2𝜆((𝑛 − 1)2 + 2𝜎) − 𝜎(2(𝑛2 − 1)

+ 4(𝑛 − 𝜎))) 

Hence, eigenvalues of 𝐴𝑆𝑂(𝐺𝑆) are 0 with multiplicity 𝜎 − 1, −(𝑛 − 1)√2 with multiplicity 

𝑛 − 𝜎 − 1, 
√2((𝑛−1)2+2𝜎)+√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1, and  

         
√2((𝑛−1)2+2𝜎)−√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1.  

  

Theorem 4.  Let 𝐺 = 𝐾𝑛 be a complete graph with n vertices. If 𝐺𝑆 is a graph obtained from 

𝐺 by adding 𝜎 self loops, then Sombor energy of 𝐺𝑆 is 𝐸𝑆𝑂(𝐺𝑆) =

√2
(𝑛+1)(𝜎−1)𝜎+(𝑛−𝜎−1)(𝑛(𝑛−1)+(𝑛+1)𝜎)

𝑛
+

√2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎)).   

Proof. By definition, the Sombor energy of matrix with 𝜎 self loop is , 

𝐸𝑆𝑂(𝐺𝑠) =∑
𝑛

𝑖=1
|𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1 𝑑𝑗

𝑛
| 



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By lemma 2, for complete graph with 𝜎 selfloop has ∑𝜎𝑗=1 𝑑𝑗 = (𝑛 + 1)𝜎 and eigenvalues are  

0 with multiplicity 𝜎 − 1, −(𝑛 − 1)√2 with multiplicity 𝑛 − 𝜎 − 1 , 

√2((𝑛−1)2+2𝜎)+√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1, 

√2((𝑛−1)2+2𝜎)−√2((𝑛−1)2+2𝜎)2+4𝜎(2(𝑛2−1)+4(𝑛−𝜎))

2
 with multiplicity 1. 

Then Somber energy becomes 

𝐸𝑆𝑂(𝐺𝑠) =∑
𝜎−1

𝑖=1
|0 −

√2(𝑛 + 1)𝜎

𝑛
| +∑

𝑛−𝜎−1

𝑖=1
| − (𝑛 − 1)√2 −

√2(𝑛 + 1)𝜎

𝑛
|

+ |
√2((𝑛 − 1)2 + 2𝜎) + √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2

−
√2(𝑛 + 1))𝜎

𝑛
|

+ |
√2((𝑛 − 1)2 + 2𝜎) − √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2

−
√2(𝑛 + 1))𝜎

𝑛
| 

= (𝜎 − 1)
√2(𝑛 + 1)𝜎

𝑛
+ (𝑛 − 𝜎 − 1)((𝑛 − 1)√2 +

√2(𝑛 + 1)𝜎

𝑛
)

+
√2((𝑛 − 1)2 + 2𝜎) + √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2

−
√2(𝑛 + 1))𝜎

𝑛

−
√2((𝑛 − 1)2 + 2𝜎) − √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2

+
√2(𝑛 + 1))𝜎

𝑛
 

= (𝜎 − 1)
√2(𝑛 + 1)𝜎

𝑛
+ (𝑛 − 𝜎 − 1)((𝑛 − 1)√2 +

√2(𝑛 + 1)𝜎

𝑛
)

+
√2((𝑛 − 1)2 + 2𝜎) + √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2

−
√2((𝑛 − 1)2 + 2𝜎) − √2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎))

2
 

= √2
(𝑛+1)(𝜎−1)𝜎+(𝑛−𝜎−1)(𝑛(𝑛−1)+(𝑛+1)𝜎)

𝑛
+

√2((𝑛 − 1)2 + 2𝜎)2 + 4𝜎(2(𝑛2 − 1) + 4(𝑛 − 𝜎)) . 

  



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Example 2. Consider 𝐺 = 𝐾3 is complete graph with 3 vertices and 𝐺𝑆 is the graph obtained 

by adding 1 loops to graph 𝐺 = 𝐾3.  

 

   

 Sombor matrix of 𝐺𝑆 is  

 𝐴𝑆𝑂(𝐺𝑆) =

[
 
 
 4√2 2√5 2√5

2√5 0 2√2

2√5 2√2 0
]
 
 
 

 

Here𝜆1 = 10.7234, 𝜆2 = −2.82843, 𝜆3 = −2.2381 The Sombor Energy of graph 𝐺𝑆 is given 

by  

 𝐸𝑆𝑂(𝐺𝑆) = ∑
3
𝑖=1 |𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1𝑑𝑗

3
| 

    = |10.7234 −
4√2

3
| + | − 2.82843 −

4√2

3
| + | − 2.2381 −

4√2

3
|  

    = 17.675548083 

In above theorem putting 𝜎 = 1 and n=3 we get, 

𝐸𝑆𝑂(𝐺𝑆) = √2
(3 + 1)(1 − 1) + (3 − 1 − 1)(3(3 − 1) + (3 + 1)1)

3

+ √2((3 − 1)2 + 2)2 + 4(2(32 − 1) + 4(3 − 1)) = 10
√2

3
+ √72 + 96 

                                         = 17.675526605.  

 

 

Theorem 5.  Let G be the complete graph of order n and 𝐺𝑙 be the graph obtained from G by 

adding a loop on each vertex of G then Sombor Energy 𝐸𝑆𝑂(𝐺 ⋃ 𝐺𝑙) =
2𝑛

𝑛−1
𝐸𝑆𝑂(𝐺)  

  

Proof. Let 𝐻𝑛 = (𝐺⋃ 𝐺𝑙). The graph 𝐻𝑛 contains 2n vertices and n loops. The Sombor 

matrix of 𝐻𝑛 is given by:  

 𝐴𝑆𝑂(𝐻𝑛) = [(𝑛 − 1)√2(𝐽 − 𝐼)𝑛×𝑛 [0]𝑛×𝑛 𝑛 × 𝑛|(𝑛 + 1)√2(𝐽)𝑛×𝑛] 

The characteristics polynomial of above matrix is given by:  



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 𝜙(𝐻𝑛: 𝑥) = [𝑥𝐼 − (𝑛 − 1)√2(𝐽 − 𝐼)𝑛×𝑛 [0]𝑛×𝑛 𝑛 × 𝑛|𝑥𝐼 − (𝑛 + 1)√2(𝐽)𝑛×𝑛] 

If 𝜆1, 𝜆2, . . . , 𝜆𝑛 are eigenvalues of 𝐴𝑆𝑂(𝐻𝑛), then 

𝜆1 = (𝑛 − 1)
2√2, 𝜆𝑖 = −(𝑛 − 1)√2    𝑓𝑜𝑟    𝑖 = 2,3, . . . , 𝑛, 𝜆𝑛+1 = 𝑛(𝑛 + 1)√2, 𝜆𝑗

= 0    𝑓𝑜𝑟    𝑗 = 𝑛 + 2, 𝑛 + 3, . . . ,2𝑛 

Sombor Energy of 𝐻𝑛 is given by ,  

 𝐸𝑆𝑂(𝐺𝑠) = ∑
2𝑛
𝑖=1 |𝜆𝑖(𝐺𝑠) −

√2∑𝜎𝑗=1𝑑𝑗

𝑛
| 

 

Here ∑𝜎𝑗=1 𝑑𝑗 = 𝑛(𝑛 + 1) 

Hence, 

𝐸𝑆𝑂(𝐺𝑠) =∑
2𝑛

𝑖=1
|𝜆𝑖(𝐺𝑠) −

√2𝑛(𝑛 + 1)

2𝑛
| 

    = |(𝑛 − 1)2√2 −
√2(𝑛 + 1)

2
| + (𝑛 − 1)| − (𝑛 − 1)√2 −

√2(𝑛 + 1)

2
| + |𝑛(𝑛 + 1)√2

−
√2(𝑛 + 1)

2
| + (𝑛 − 1)|0 −

√2(𝑛 + 1)

2
| 

    = [2(𝑛 − 1)2 − (𝑛 + 1)]
√2

2
+ (𝑛 − 1)(2(𝑛 − 1) + (𝑛 + 1))

√2

2
+ [2𝑛(𝑛 + 1) − (𝑛

+ 1)]
√2

2
+ (𝑛 − 1)(𝑛 + 1)

√2

2
 

    =
1

√2
[2𝑛2 − 4𝑛 + 2 − 𝑛 − 1 + (𝑛 − 1)(2𝑛 − 2 + 𝑛 + 1) + (2𝑛2 + 2𝑛 − 𝑛 − 1) + 𝑛2

− 1] 

    =
1

√2
(8𝑛2 − 8𝑛) 

    = 4√2𝑛(𝑛 − 1) 

Therefore, 

 

 𝐸𝑆𝑂(𝐺𝑠) = 4√2𝑛(𝑛 − 1) (3) 

 The Sombor matrix of graph 𝐺 = 𝐾𝑛 is , 

 

 𝐴𝑆𝑂(𝐺) = [(𝑛 − 1)√2(𝐽 − 𝐼)𝑛×𝑛] 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

296 
https://internationalpubls.com 

Eigenvalues of 𝐴𝑆𝑂(𝐺) are(𝑛 − 1)2√2 with multiplicity 1 and −(𝑛 − 1)√2 with 𝑛 − 1 

mutilplicity. 

Hence Sombor Energy of 𝐺 is given by, 

𝐸𝑆𝑂(𝐺) =∑
𝑛

𝑖=1
|𝜆𝑖| 

    = (𝑛 − 1)2√2 + (𝑛 − 1)(𝑛 − 1)√2 

    = 2(𝑛 − 1)2√2 

Therefore,  

 𝐸𝑆𝑂(𝐺) = 2(𝑛 − 1)
2√2 (4) 

 Hence, by equation (3) and (4),  

 

 𝐸𝑆𝑂(𝐻𝑛) =
2𝑛

(𝑛−1)
𝐸𝑆𝑂(𝐺) 

 

  

Example 3. Consider the graph 𝐻3 = 𝐾3⋃ 𝐾3
𝑙 and 𝐺 = 𝐾3. The graph 𝐻3 contains 6 vertices 

and three loops. It is known that 𝐸𝑆𝑂(𝐺) = 8√2  

 

 

 

 

 The Sombor matrix of 𝐻3 is  

 𝐴𝑆𝑂(𝐻3) =

[
 
 
 
 
 
 
 0 2√2 2√2 0 0 0

2√2 0 2√2 0 0 0

2√2 2√2 0 0 0 0

0 0 0 4√2 4√2 4√2

0 0 0 4√2 4√2 4√2

0 0 0 4√2 4√2 4√2
]
 
 
 
 
 
 
 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 9s (2025) 

 

297 
https://internationalpubls.com 

The eigen values of 𝐻3 are 4√2, 12√2, (−2√2) with multiplicity 2,

and 0 with multiplicity 2. 

Hence, 

𝐸𝑆𝑂(𝐻3) = |4√2 − 2√2| + 12√2 − 2√2| + 2| − 2√2 − 2√2| + 2|0 − 2√2| = 24√2. 

Therefore, 𝐸𝑆𝑂(𝐻3) =
6

2
𝐸𝑆𝑂(𝐺). 

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