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If $G$ is a graph, then the graph energy of $G$ denoted by $E(G)$ is defined as the sum of absolute values of eigenvalues of the adjacency matrix of $G$. It is known that $E(G)\geq E(G-v)$, where $ v$ is a vertex in $G$. But it is not known that how the energy of a graph and its subgraphs are related if the number of vertices are same in both the graphs.

I am trying to formulate increasing or decreasing pattern of energy of a graph and its subsequent subgraphs on the same set of vertices. For instance, I am taking a complete graph $G=K_4$ with four vertices. Then I have the following observations.

(1) $E(G)=6$

(2) $E(G_1)=5.1231056256=1+\sqrt{17}$, where $G_1$ is a subgraph obtained by deleting one edge from $G$.

Similarly,

(3) $E(G_2)=4=2\sqrt{4}$ (obtained by deleting two edges from $G$)

(4) $E(G_3)=4.472135955=2\sqrt{5}$ (obtained by deleting three edges from $G$)

(5) $E(G_4)=2\sqrt{2}$ (obtained by deleting four edges from $G$)

(6) $G_5=2=2\sqrt{1}$ (obtained by deleting five edges from $G$).

I would really appreciate if someone could help me arriving at a generalizing formula of the above pattern in particular. That is, what is the general pattern in which the energy is increasing or decreasing in the above example?

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  • $\begingroup$ People interested in this question may also be interested by your post here: math.stackexchange.com/questions/4002762/… $\endgroup$ Mar 5, 2021 at 20:49
  • $\begingroup$ I do not understand your second sentence: it claims that the energy decreases when we remove an edge (which seems wrong from your $K_4$ example) but then the sentence talks about vertices. Did I miss something? $\endgroup$ Mar 5, 2021 at 21:34
  • $\begingroup$ @MatthieuLatapy I have mentioned that the energy increases or drecreases. Well, the set of vertices remains same in all the graphs. $\endgroup$
    – gete
    Mar 6, 2021 at 2:25
  • $\begingroup$ @gete you have an error in the question, you say that $E(G) \geqslant E(G-e)$ where $e$ is an edge, which is not true. $\endgroup$ Mar 6, 2021 at 3:10
  • $\begingroup$ @GordonRoyle Thank you for pointing the error. It is corrected now. $\endgroup$
    – gete
    Mar 6, 2021 at 5:17

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