Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 2 (2025) 

 

64 https://internationalpubls.com 

Bounds on Dominating Energy of Graph 

 

Shivakumar Swamy C S1, Ashwini G 2, Ramesha M S 3, Nanjundaswamy N 4 

1,2,3 Department of Mathematics, Government College for Women (Autonomous), Mandya-571401, INDIA. 

4 Department of Mathematics, Government First Grade College for Women, Byrapura, T.Narasipura taluk, Mysore 

District-571124, INDIA. 

 

Article History: 

Received: 23-07-2024 

Revised: 01-09-2024 

Accepted: 15-09-2024 

Abstract:  

The Minimum Dominating Energy (MDE) of a graph𝔄, denoted by 𝐸𝑀 𝐷 (𝔄), is nothing but 

the sum of absolute values of all minimum dominating eigenvalues of 𝔄. In this study, few 

upper and lower constraints on the minimum dominating energy are obtained. 

Keywords: Minimum dominating matrix, minimum dominating eigenvalues, minimum 

dominating energy. 

2000 AMS Subject Classification: 05𝐶50. 

 

1. Introduction 

Let 𝔈 be the edge set and 𝔙, set of vertices of a simple graph 𝔄. Let |𝔈| = 𝑚 and |𝔙| = 𝑛.  

ℇ(𝔄): =  ∑|𝒱𝑘|

𝑛

𝑘=1

 

 where 𝒱𝑘, 𝑘 = 1, 2, 3, … , 𝑛 are the eigenvalues (characteristic roots) of the Adjacency matrix (𝐴𝑀) 

of a graph 𝔄, Ivan Gutman [12] conducted this study on 𝔄 for the first time in 1978, and named it as 

Energy of a Graph  𝔄, from then numerous research has been conducted on 𝐴𝑀, with inspiration 

drawn by this, different matrix types for a graph 𝔄 [18, 17, 13, 2] are defined and studied. For basic 

mathematical properties of the theory of graph energy including its upper and lower bounds one can 

see [ 20, 21, 22]. Erich Huckle [3], employed the energy of graphs technique in the early 1930s to 

develop approximations solutions for a family of organic molecules known as conjugated hydro 

carbons. 

Let 𝒟 ⊆ 𝔙 (𝔄), if every vertex of 𝔙 − 𝒟 is adjacent to some vertex in 𝒟, then 𝒟 is referred to as a 

dominating set of 𝔄. A minimum dominating set (MDS) 𝒟 of 𝔄 is a dominating set of 𝔄 with 

minimum cardinality. 

Let 𝒟 be a MDS of 𝔄. The following kind of matrix, known as the minimum dominating matrix 

(MDM) of 𝔄, introduced by M.R.Rajesh Kanna et.al. in [22]: The 𝑛 × 𝑛 matrix 𝑀𝒟(𝔄) = [𝑑𝑘 𝑗], is 

the MDM of 𝔄, whose 𝑘𝑗 − 𝑡ℎ element is given by  

𝑑𝑘𝑗 = {

1,                 𝑖𝑓 𝑘 = 𝑗  𝑎𝑛𝑑 𝑣𝑗   𝑖𝑛 𝒟;

 1,       𝑖𝑓 𝑣𝑘 𝑎𝑛𝑑 𝑣𝑗   𝑎𝑟𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡;

 0,                                    𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

 

 



Communications on Applied Nonlinear Analysis 

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Γ(𝔄 ∶  𝜒) = 𝑑𝑒𝑡(𝜒𝐼 − 𝑀𝒟(𝔄)) is the characteristic polynomial of 𝑀𝒟(𝔄). The minimum dominating 

eigenvalues of 𝔄 are the eigenvalues 𝜒1, 𝜒2,   .  .  . , 𝜒𝑛of 𝑀𝒟(𝔄). The matrix 𝑀𝒟(𝔄) is real as well as 

symmetric. The real numbers that make up the eigenvalues of 𝑀𝒟(𝔄) are arranged to be: 𝜒1 ≥ 𝜒2 ≥

  .  .  . ≥  𝜒𝑛. The formula  

Ε𝑀𝒟(𝔄) =  ∑|𝜒𝑘|

𝑛

𝑘=1

 

defines 𝔄′𝑠 minimum dominating energy (MDE). 

Note that 𝑀𝒟(𝔄) has trace =Domination number = d, and ∑ 𝜒𝑘
2 = 2|𝔈| + |𝒟| = 2𝑚 + |𝑑|𝑛

𝑘=1  

We derive some upper and lower bounds for the MDE, Ε𝑀𝒟(𝔄), in this study. 

2. Upper Bounds for MDE 

Throughout this series 𝔄 denotes a simple graph. This section is aimed to discuss upper bounds for 

MDE of 𝔄. 

Theorem 2.1 Let 𝔄 be graph of order 𝑛 and size 𝑚. Then 

Ε𝑀𝒟(𝔄) ≤ √(
1

2
(𝑛2 + |𝑑|2) + 2𝑚(𝑚 + |𝑑|)) 

Proof: We recall the following well-known inequality from [11]:  

(∑ 𝑝𝑘𝑒𝑘
2

𝑛

𝑘=1

) (∑ 𝑞𝑘𝑓𝑘
2

𝑛

𝑘=1

) + (∑ 𝑝𝑘𝑔𝑘
2

𝑛

𝑘=1

) (∑ 𝑞𝑘ℎ𝑘
2

𝑛

𝑘=1

) ≥ 2 (∑ 𝑝𝑘𝑒𝑘𝑔𝑘

𝑛

𝑘=1

) (∑ 𝑞𝑘𝑓𝑘ℎ𝑘

𝑛

𝑘=1

)           ( 2.1) 

            

where 𝑒𝑘, 𝑓𝑘, 𝑔𝑘 𝑎𝑛𝑑 ℎ𝑘denote sequence of real numbers; 𝑝𝑘  𝑎𝑛𝑑 𝑞𝑘 denote non-negative numbers 

for 1 ≤ 𝑘 ≤ 𝑛. 

For 𝑝𝑘  =  𝑞𝑘 = 𝑒𝑘 = 𝑓𝑘 = 1  and 𝑔𝑘 =  ℎ𝑘 = |𝜒𝑘|, 1 ≤ 𝑘 ≤ 𝑛, the inequality (2.1) reduces to  

(∑ 1

𝑛

𝑘=1

) (∑ 1

𝑛

𝑘=1

) + (∑|𝜒𝑘|2

𝑛

𝑘=1

) (∑|𝜒𝑘|2

𝑛

𝑘=1

) ≥ 2 (∑|𝜒𝑘|

𝑛

𝑘=1

) (∑|𝜒𝑘|

𝑛

𝑘=1

). 

Using,  

∑|𝜒𝑘|2

𝑛

𝑘=1

= ∑ 𝜒𝑘
2

𝑛

𝑘=1

= 2𝑚 + |𝑑| 

in the above inequality, we deduce that  

𝑛. 𝑛 + (|𝑑| + 2𝑚)(|𝑑| + 2𝑚) ≥ 2. Ε𝑀𝒟(𝔄).Ε𝑀𝒟(𝔄) 

which gives  

2. Ε𝑀𝒟(𝔄)2 ≤ 𝑛2 + (|𝑑| + 2𝑚)2. 



Communications on Applied Nonlinear Analysis 

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Hence, 

Ε𝑀𝒟(𝔄) ≤ √(
1

2
(𝑛2 + |𝑑|2) + 2𝑚(𝑚 + |𝑑|)). 

Theorem 2.2 Let 𝔄 be graph of order 𝑛 and size 𝑚. Then 

Ε𝑀𝒟(𝔄) ≤ 𝑚 +
1

2
(𝑛 + |𝑑|). 

Proof: We recall the following well-known inequality from [11]:  

(∑ 𝑝𝑘𝑒𝑘
2

𝑛

𝑘=1

) (∑ 𝑞𝑘𝑓𝑘
2

𝑛

𝑘=1

) + (∑ 𝑝𝑘𝑔𝑘
2

𝑛

𝑘=1

) (∑ 𝑞𝑘ℎ𝑘
2

𝑛

𝑘=1

) ≥ 2 (∑ 𝑝𝑘𝑒𝑘𝑔𝑘

𝑛

𝑘=1

) (∑ 𝑞𝑘𝑓𝑘ℎ𝑘

𝑛

𝑘=1

)            (2.2) 

where 𝑒𝑘, 𝑓𝑘, 𝑔𝑘 𝑎𝑛𝑑 ℎ𝑘denote sequence of real numbers; 𝑝𝑘  𝑎𝑛𝑑 𝑞𝑘 denote non-negative numbers 

for 1 ≤ 𝑘 ≤ 𝑛. 

For 𝑝𝑘  =  𝑞𝑘 = 𝑒𝑘 = 𝑓𝑘 = ℎ𝑘 = 1  and 𝑔𝑘 = |𝜒𝑘|, 1 ≤ 𝑘 ≤ 𝑛, the inequality (2.2) yields 

(∑ 1

𝑛

𝑘=1

) (∑ 1

𝑛

𝑘=1

) + (∑|𝜒𝑘|2

𝑛

𝑘=1

) (∑ 1

𝑛

𝑘=1

) ≥ 2 (∑|𝜒𝑘|

𝑛

𝑘=1

) (∑ 1

𝑛

𝑘=1

). 

That is, 

𝑛2 + (∑|𝜒𝑘|2

𝑛

𝑘=1

) 𝑛 ≥ 2𝑛. (∑|𝜒𝑘|

𝑛

𝑘=1

) 

Which gives, 

𝑛 + 2𝑚 + |𝑑| ≥ 2Ε𝑀𝒟(𝔄). 

Hence,  

Ε𝑀𝒟(𝔄) ≤ 𝑚 +
1

2
(𝑛 + |𝑑|). 

3. Lower Bounds for MDE 

Throughout this section 𝔄 denotes a simple graph. This section is aimed to discuss lower bounds for 

MDE of 𝔄. 

Theorem 3.1 Let 𝔄 be a bipartite graph of order 𝑛 ≥ 2 and size 𝑚 with spectral radius 𝜒1. Then 
|𝑑|+2𝑚

𝜒1
≤ Ε𝑀𝒟(𝔄). 

Proof:Let 𝑒𝑘, 𝑓𝑘  be non-negative decreasing sequences where 𝑒𝑘, 𝑓𝑘 ≠ 0, and 𝑗𝑘 be a non-negative 

sequence for 1 ≤ 𝑘 ≤ 𝑛. Then we have the following inequality [11]: 

(∑ 𝑗𝑘𝑒𝑘
2

𝑛

𝑘=1

) (∑ 𝑗𝑘𝑓𝑘
2

𝑛

𝑘=1

) ≤ max {𝑓1 ∑ 𝑗𝑘𝑒𝑘
2

𝑛

𝑘=1

,   𝑒1 ∑ 𝑗𝑘𝑓𝑘
2

𝑛

𝑘=1

} (∑ 𝑗𝑘𝑒𝑘𝑓𝑘

𝑛

𝑘=1

)                     (3.1) 

For 𝑒𝑘 = 𝑓𝑘 = |𝜒𝑘|  and 𝑗𝑘 = 1, 1 ≤ 𝑘 ≤ 𝑛, the inequality (3.1) gives,  



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(∑ 1. |𝜒𝑘|2

𝑛

𝑘=1

) (∑ 1. |𝜒𝑘|2

𝑛

𝑘=1

) ≤ max {𝜒1 ∑|𝜒𝑘|

𝑛

𝑘=1

, 𝜒1 ∑|𝜒𝑘|

𝑛

𝑘=1

} (∑|𝜒𝑘|2

𝑛

𝑘=1

) 

That is, 

∑|𝜒𝑘|2

𝑛

𝑘=1

≤ 𝜒1 ∑|𝜒𝑘|

𝑛

𝑘=1

 

Which implies 

𝜒1Ε𝑀𝒟(𝔄) ≥ ∑|𝜒𝑘|2

𝑛

𝑘=1

 

Hence, 

|𝑑| + 2𝑚

𝜒1
≤ Ε𝑀𝒟(𝔄) 

Lemma 1 [22] Let 𝑛 be a positive integer. If 𝑙1, 𝑙2, … , 𝑙𝑛 are non-negative numbers with          𝑙1 ≥

 𝑙2 ≥ ⋯ ≥ 𝑙𝑛, then 

(𝑙1 + 𝑙2 +  … + 𝑙𝑛)(𝑙1 + 𝑙𝑛) ≥ 𝑙1
2 + ⋯ + 𝑙𝑛

2 + 𝑛𝑙1𝑙𝑛                                             (3.2) 

Further, equality holds in (3.2) if and only if for some 𝑟, 1 ≤ 𝑟 ≤ 𝑛, 𝑙1 = ∙∙∙ =  𝑙𝑟 and 

𝑙𝑟+1 = ∙∙∙ =  𝑙𝑛. 

Theorem 3.2 Let 𝔄 be a graph with order 𝑛 ≥ 2 and size 𝑚 ≥ 1. Assume that 𝜒1, … , 𝜒𝑛  are all 

eigenvalues of 𝔄, such that |𝜒𝑛| ≥ ⋯ ≥ |𝜒1| ≥ 0, then 

Ε𝑀𝒟(𝔄) ≥
2√(2𝑚 + |𝑑|)𝑛√|𝜒1𝜒𝑛|

|𝜒1| + |𝜒𝑛|
 

Proof: Since there is at least one edge in the graph 𝔄, it follows that 𝔄 has at least one eigenvalue 

different from zero. Applying Lemma 1, we get 

(|𝜒1| + ⋯ + |𝜒𝑛|)(|𝜒1| + |𝜒𝑛|) ≥ |𝜒1|2 + ⋯ +|𝜒𝑛|2 + 𝑛|𝜒1||𝜒𝑛|                                 (3.3) 

and equality holds in (3.3) if and only if |𝜒1| = ⋯ = |𝜒𝑟|  and |𝜒𝑟+1| = ⋯ = |𝜒𝑛| for some𝑟 ∈ 1,∙∙∙

, 𝑛 since |𝜒1|2 + ⋯ + |𝜒𝑛|2 = 2𝑚 + |𝑑|. By equation (3.3) we get, 

Ε𝑀𝒟(𝔄)(|𝜒1| + |𝜒𝑛|) ≥ 2𝑚 + |𝑑| + 𝑛|𝜒1||𝜒𝑛| 

Ε𝑀𝒟(𝔄) ≥
2𝑚 + |𝑑| + 𝑛|𝜒1||𝜒𝑛|

|𝜒1| + |𝜒𝑛|
                                                                                   (3.4) 

and the equality holds if and only if |𝜒1| = ⋯ = |𝜒𝑟|  and |𝜒𝑟+1| = ⋯ = |𝜒𝑛| for some                      

𝑟 ∈ 1, ⋯ , 𝑛. We all know that for every real number 𝑎 ≥ 0 and 𝑏 ≥ 0,  

𝑎 + 𝑏 ≥ 2√𝑎𝑏 

and equality holds if and only if 𝑎 = 𝑏. Using this fact in (3.4), we get 



Communications on Applied Nonlinear Analysis 

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Ε𝑀𝒟(𝔄) ≥
(2𝑚 + |𝑑|) + 𝑛|𝜒1||𝜒𝑛|

|𝜒1| + |𝜒𝑛|
 

                 ≥
2√(2𝑚 + |𝑑|)𝑛|𝜒1𝜒𝑛|

|𝜒1| + |𝜒𝑛|
 

                    =
2√(2𝑚 + |𝑑|)𝑛√|𝜒1𝜒𝑛|

|𝜒1| + |𝜒𝑛|
 

This proves the result. 

Acknowledgement: The authors are thankful to Prof.Chandrashekara Adiga for his encouragement 

and suggestions. 

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