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4A Ky Fan theorem's application in the theory of graph energy
Zabihullah Zahir Parwan University, Afghanistan Farzana Mohammadi Balkh University, Afghanistan
Abstract: The energy of a graph G is equal to the total of its absolute eigenvalues, which is equal to the sum of its singular adjacency matrix values. Let x'y and z be matrices such that x + y = z. The Ky-Fan theorem proves an inequality between the sum of the singular values of z and the sum of the singular values of x andy. Several new inequalities as well as fresh proofs of several previously known inequalities are produced when this theorem is applied to the notion of graph energy.
Keywords: graph energy, invariants energy, Laplacian energy, Ky Fan theorem and inequality
Introduction
Simple graphs are the focus of the essay. Let G E ) be such a graph with a vertex.
Set V = V(G) and the edge setE = E(G). When the order and size ofG are n andm, respectively, i.e,V = n and E = m, we say that G is a (n'm ) - graph.
Let A = A(G^ represent the (0,1) - adjacency matrix of G. The spectrum of the graph G t2] is made up of its eigenvalues .
If V gV(G^ , then denotes the vertex's degree with 1 = 1,2, -n . D = D(G)
represents the (^1) -entry of the diagonal matrix of order n.
Then
L = L (G) = D (G)- A (G) (l)
The Laplacian matrix of G is designated as (1).
The Laplacian spectrum of the graph G is formed by its eigenvalues ^^n [7,8,17]
We will also require a second Laplacian-type matrix in the subsequent discussion, which is defined as
L = L (G) := D(G) + A{G) (2)
The definition of the graph G's energy is
E = E (G) :=£|4 |.
i=1
when contrasted with its Laplacian energy
LE = LE (G)
i=1
2m
A--
n
(3)
The graph invariants energy and Laplacian energy are currently the subject of extensive research. A common A current bibliography, which only includes papers published in 2001 and later, can be found online at http://www.sgt.pep.ufrj.br and currently includes about 150 references. For more information on graph energy theory See the review [9] and the book [12]. The most recent papers [14, 21-22].
Provide an overview of the fundamentals of Laplacian energy.
Let In be the order n unit matrix. It will be important to note for the considerations that follow that, instead, via Eq. (3), the Laplacian energy can also be expressed as
n
LE (G ) = £\y\ ( 4)
i=i
Where '1 =12".'n are the eigenvalues of the matrix L (G) (2m / n) In.
I. The Ky-Fan theorem
Assume that M is a square matrix of real and symmetric order n. Consider the
singular values of Si (M) ,where 1 =12'. .'n as well as its eigenvalues, For is then
s'(M) = X(m) for i =12'..'n The energy of the graph G is known by Nikiforov [18] to
be the sum of the singular values of its adjacency matrix A (G). The following theorem, which Fan first established [5], suggests that this observation is extremely significant for the theory of graph energy.
Theorem 1 [5]. Assume that x + y = z and Let y and z be square matrices of
n
ES (x)+£S (7)>£S (Z) ordern. Then i=1 i=1 i=1
Equality holds if and only if there exists an orthogonal matrix p where and
are both positive semi-definite.
You can read more about the Ky Fan theorem in [3, 4] and the references cited
there.
II. Some basic applications of the Ky Fan theorem
G G C
Since Theorem 1 states that for graphs y andGz whose adjacency matrices
meet the requirement A (Gx) + A (Gy) = A (Gz).
E (Gx) + E (Gy )> E (Gz).
The following corollaries list some particular examples of this inequality:
n
n
Corollary 2. Assume that G is an order n graph, and that G stands for it's complement. Then
E (G) + E (G )> 2 n-1
If and only if G = Kn or G = Kn, equality is maintained.
Proof. By noting that A (G) + A (g) A (Kn) and E(Kn ) = 2 (n -1) the inequality follows. Assume that the eigenvalues of G are for the purpose of establishing the requirements for equality.
4 (G )>4 (G )>4(G)
Then
E (G) + E (G) > 24 (G) + 24 (G).
2m (G) 2m (G)_ 4
> 2-^ +
n n n
fn\
v 2 y
= 2n - 2
G must be regular, E(G) (G), and E (G) (G) equality is only possible if
all three conditions are met. If so, it must be ^2 (G )" 0 and ^(G ^" 0 since the energy of a graph is equal to twice the sum of its positive eigenvalues. To put it another way,
equality is only achieved if and only if both G and G have one and only one positive eigenvalue. Then, we must separately take into account each of the following three scenarios:
(i) G has a negative eigenvalue, meaning thatG = Kn.
(ii)G has no positive eigenvalue. i.e, G = K" indicating thatG = Kn.
(iii) There is just one positive eigenvalue shared by both G and G.
Smith's theorem [19] states that both G and G would then be full multipartite graphs. As opposed to their complements, which are disconnected, complete multipartite graphs are connected, making this impossible.
Corollary 3: LetB be the bipartite complement of B, which is a bipartite graph
•1 ,A • E (B) + E (B )> 2jabn„ witha +b vertices. v 7 v ' follows.
fí ~ fi T> _
If and only if B ~ Kab or B = Ka+b ,equality is maintained.
Proof. The inequality is shown by noting that A(B) + A(B^ A(Ka~b^ and
E (Ka.b) - 2^ab. The case of equality is handled in a manner that is comparable to the proof of Corollary 2.
Corollary 4a: Assume thatG - e is the subgraph that results from eliminating the edge e from graph G. Then
E(G)< E(G-e) + 2 (5)
The outcome mentioned in this corollary 4a was already disclosed in [4]. The equality in (5) was demonstrated in [4] to hold only when e is an isolated edge in G.
We obtain by repeatedly applying (5) to all edges of a (n'm)-graph
E (G )< 2m (6)
with equality, which is also a previously established upper bound [1], if and only if G has m isolated edges andn -2m isolated vertices. We obtain the following by
(n m)
repeatedly apply ing equation (5) to all edges of a ' ' -graph, excluding those that end
at a maximum-degree vertex:
Corollary 4b: If A is the highest degree at a vertex in a (n'm) - graphG, then E (G)< 2m - 2 (A-VA) ( 7 )
If and only if G is a union of the star 5a+1' m -A, m isolated edges, and n - 2 m + A+1 isolated vertices, then equality in (7) holds.
Evidently, the bound (7) is superior to (6). It appears to be the first time it has been mentioned here. But if m ^ 2n, (7) is weaker than some other established upper
bounds [9], such as McClleland's J2™.
Similar to this, we also obtain the following energy upper bounds:
Corollary 4c. IfG is a connected (n 'm) -graph and T is its spanning tree, then
E (G)< 2(m - n +1) + E (T) If G t, the inequality is rigid.
Corollary 4d. If G is a Hamiltonian (n 'm)-graph, then E (G)< 2 (m n +1) + E (Cn)
C G — C
where Cn stands for the n -vertex cycle. If G — Cn, then the inequality is strict.
Corollary 4e. E(G)<2(m-d) + E(P+i),where P is the k -vertex path, if d is the
diameter of a connected graph G. The inequality is strict if We mention this in passing.
4cos V n
G = P
d+1
E (C. ) =
sin V
' n 4
sin V
n
2
sin V
if n = 0 (mod 4) if n = 2 ( mod 4) if n = 1(mod 2)
2n
And
E ( P ) =
2
sin
K
- 2 if n = 0 ( mod 2 )
2cos
2 ( n +1)
K
2 ( n +1)
sin
K
- 2 if n = 1( mod 2)
2 (n +1)
III. Relating Laplacian energy and energy
We start by pointing out yet another straightforward Ky-Fan theorem consequence. Letd represent the graph's average vertex degree(n 'm) -graph of G. Of
, 2m d = —
course, n .
Corollary 5. For a (n'm)-graph ofG, where the vertex degrees are dl'd2'''dn ,and the average vertex degree is a .
n _
LE (G)< E (G) = ^ d - d
i=1
Proof. Rewrite Eq.(1) as f L - ] = (- ,) + i . - ]
D - ^
n
has eigenvalues
Consider (4) as well as the fact that the diagonal matrix 0f di - d, i -1,2, ..., « when applying Theorem 1.
It was hypothesized in [10] that the Laplacian energy is always greater than or equal to the regular graph energy.
LE (G)> E (G) (8)
Counter examples were ultimately employed to refute the validity of the conjecture [16,20]. Now, however, we demonstrate that this hypothesis was not entirely off the mark.
Theorem 6. If the graphG is bipartite, then relation (8) is true.
Proof. Eqs.(1) and (2) are subtracted, and the result is
LL - L - 2 A
Which can be written as
2 m
2 m
L--/ - L--/ = 2A
n
n
(9)
It is generally understood that the matrices L andL have equal spectra in the case of bipartite graphs (see, for example, [8]). In context with this, we briefly address
I<l -TI.)=i* (L -2m^n j=§* -
L - ^m/n
n
= LE
<
and inequality (8) is obtained by using the Ky Fan theorem on equation (9). We can arrive at a slightly more powerful result using similar reasoning. The result of adding (1) and (2) is in passing, we state
L + L = 2D From which
2m
2m
2m
LL-—L + L-—L = 2 D-—I,
n
n
n
Then by theoreml.
LE >±\dt - d
i=1
The above inequality, along with the conclusions of Corollary 5 and Theorem 6, result in:
Theorem 7. For a bipartite (m) - graphG with vertex degrees dl'd2'"'dn and
an
d = 2m/
average vertex degree of /n
max
£h
- d
<
EL(G)< E(G) + J|d -d
IV. A difference in the intensity of the coalescence of two graphs.
Two graphs with disjoint vertex sets are G and H. Allow u = V(G^ and v = V(H^ , by figuring out the vertices"and v, create the graphG°Husing copies of GandH.
Consequently V iGoHV(G) + V)\ 1 The graph G°H is referred to as the G and H coalescence with respect to u and v.
Theorem 8: Assume that G, Hand^°^are the afore mentioned graphs. Then E(G°H)<E(G) + E(H) (10)
If and only if either u is an isolated vertex of G orv is an isolated vertex of H, or both, equality is achieved.
Proof. When the vertices of the graphs G and H are appropriately labeled, the adjacency matrix of G and H takes the form
= B + C
where x is the column vector corresponding to the vertex, and R A (G u^ and
S = A (H - v)
y is the column vector corresponding to the vertex v inH, and u inG.
Then Theorem 1 now follows directly from
Relation (10).
" R 0 " " R x 0" "0 0 0"
A = A(G°H) = xT 0 yT = xT 0 0 + 0 0 yT
0 y s 0 0 0 0 y s
Since either x or y is a zero vector in the equality case, it is simple to verify that the condition is sufficient.
The equality in (10) implies the equality in the singular value inequality for the necessity section. There is an orthogonal matrix p in the equality case of Theorem 1 such that both PB andPC are positive semi-definite. Let P now divide in the way
described below using the matrix A(G°H)
" P11 P12 P13
P = P21 P 1 22 P 1 23
_ P31 P 1 32 P 1 33
where I stands for an identity matrix that is the proper size. Recall that both
PT P + pt p + Pt p = i
1 11 111 + 1 211 21 + 1 311 31 1
pTp + pTp + pTp = 1
p3 p3 + 1 231 23 + P33P33 1
PTP + PTP + PTP = 0
p1 P13 + 1 211 23 + P31 P33 0
(11) (12) (13)
Where 1 denotes an identity matrix of appropriate size. Note that both
PB =
And
PC =
puR+P2 y
Pnx 0
P21R + P22 xT P21 x 0
P31R + P32 xT
P3lx 0
0 P3y P2/ + P13S"
0 P2,y P22yT + P23S 0 P33y P32yT + P33S
symmetry dictates thatPi x 0 and Pl3y 0 because and are positive semi-definite. At this point, by multiplying equation (13) by x from the left and y from the right, we get
(P2x)T (P23y) = xTPPlP13y + xTP2TP23y + xTP3Ti P33y = 0 Hence, one of the two scalars (p2x) and (P23y) must be zero. Case 1. (P"x f = 0
P x
The positive semi-definite matrix PB is diagonal entry P21 x should be noted. Due to
P x P x = 0
the fact that P21 x belongs in the entire column, which is zero, and so Plx 0.
Finally, xTx = xTp1pix+xTpiP2ix + xTpTiP3ix = 0 + 0+0 = °>x = 0. This is due to Eq. (11), which states that x = 0. This indicates that G is isolated vertex of u.
Case 2. (^ ) = 0
The positive semi-definite matrix PC is diagonal entry P23y should be noted. Due to the fact that P;!i x belongs in the entire column, which is zero, and so Pi3 y = 0. Finally,
/y = yrPilPuy + + /p33p33y = 0+o + 0 = 0, /X y = 0. This is due to Eq. (12), which
states that y = 0. This indicates that H is isolated vertex of v.
The idea of hypo energetic graphs was just recently developed [11, 13]. E (G) < n indicates that a graphG of order n is hypo energetic. For the purposes of the discussion
at hand, we designate a graph as strongly hypo energetic if E (G )< n - 1. (Remember that the n-vertex star for n > 5 is strongly hypo energetic.) Theorem 8 then yields the next:
Corolarlly 9a: Assume that the graphs f H and G ° H are as in Theorem 8. If G and H are both strongly hypo energetic, then GoH is also strongly hypo energetic.
Proof. From(10) and the fact thatE(G)<V(G)- 1and E(H)<V(H)-1 follows: E (G°H )< V (G)| + V (H )|- 2 = V (G°H )|-1
Corolarlly 9b: If G andH are both strongly hypo energetic (or vice versa), then G o H are both hypo energetic. Conclusion
Graph energy and Laplacian graph energy, which are the sums of the absolute price, the set of special prices, the adjacency matrix, and the Laplacian matrix. Regular r-graphs and single-round linked graphs with no hanging vertices are equal, but in all other Laplacian graphs, the energy of the graph is larger than or equal to the energy of the graph.
The energy of the graph is always smaller than its two sides if the number of
vertices in the graph G (n'm) is higher than two sides, and this property does not occur in the Laplacian energy graph.
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