On (Signless) Laplacian eigenvalues of graphs · On (Signless) Laplacian eigenvalues of graphs...

On (Signless) Laplacian eigenvalues of graphs

Kinkar Chandra Das

Department of Mathematics,Sungkyunkwan University, Rep. of Korea

Spectra of graphs and applications 2016,in honor of Prof. Dragos Cvetkovic for his 75th birthday,Serbian Academy of Sciences and Arts, Belgrade, Serbia.

May 20, 2016 (Joint work with M. Liu)

( )

Adjacency and Signless Laplacian Matrix

Conjecture 1 [1,2]:

Let G be a connected graph of order n > 3. Then

q1 − 2λ1 ≤ n− 2√n− 1

with equality holding if and only if G ∼= K1, n−1.


Conjecture 1 [1,2]:


q1 − 2λ1 ≤ n− 2√n− 1

with equality holding if and only if G ∼= K1, n−1.

Conjecture 2 [1,2]:

Let G be a connected graph n > 3,

1−√n − 1 ≤ q2 − λ1 ≤ n − 2−

√2 (n − 2)

with equality holding iff G ∼= K1, n−1 (lower) and G ∼= K2, n−2 (upper).

[1] M. Aouchiche, P. Hansen, A survey of automated conjectures...., Linear Algebra Appl. 432 (2010) 2293-2322.

[2] D.Cvetkovic, P. Rowlinson, S. K. Simic, Eigenvalue bounds for the signless Laplacian, Publ. Inst. Math.

(Beogr ) (N S ) 81 (95) (2007) 11–27

Relation between q1 and λ1

Theorem [3]:


q2 − λ1 ≥ 1−√n − 1

with equality holding if and only if G ∼= K1, n−1 or G ∼= K5.

[3] K. C. Das, Proof of conjecture involving the second largest signless Laplacian eigenvalue and the index of

graphs, Linear Algebra Appl. 435 (2011) 2420–2424.


Partial Proof of Conjecture 1:

Let x = (x1, x2, . . . , xn)T be an unit eigenvector corresponding eigenvalue

q1 of Q(G ). Then

q1 =∑

vi vj∈E(G)

(xi + xj)2




q1 of Q(G ). Then

q1 =∑

vi vj∈E(G)

(xi + xj)2 =

∑vi vj∈E(G)

(xi − xj)2 + 4

∑vi vj∈E(G)

xixj .




q1 of Q(G ). Then

q1 =∑

vi vj∈E(G)

(xi + xj)2 =

∑vi vj∈E(G)

(xi − xj)2 + 4

∑vi vj∈E(G)

xixj .

Rayleigh-Ritz theorem,

λ1 ≥ 2∑

vi vj∈E(G)

xixj .




q1 of Q(G ). Then

q1 =∑

vi vj∈E(G)

(xi + xj)2 =

∑vi vj∈E(G)

(xi − xj)2 + 4

∑vi vj∈E(G)

xixj .


λ1 ≥ 2∑

vi vj∈E(G)

xixj .

Hence

q1 − 2λ1 ≤n∑

i=1

di x2i − λ1




q1 of Q(G ). Then

q1 =∑

vi vj∈E(G)

(xi + xj)2 =

∑vi vj∈E(G)

(xi − xj)2 + 4

∑vi vj∈E(G)

xixj .


λ1 ≥ 2∑

vi vj∈E(G)

xixj .

Hence

q1 − 2λ1 ≤n∑

i=1

di x2i − λ1 ≤ Δ− λ1.



If Δ ≤ n − 2√n − 1, then the Conjecture 1 holds.



If Δ ≤ n − 2√n − 1, then the Conjecture 1 holds.Otherwise,

Δ > n − 2√n − 1.




Δ > n − 2√n − 1.

If m ≥ n2

[Δ− n + 2

√n − 1

], then also Conjecture 1 holds.




Δ > n − 2√n − 1.

If m ≥ n2

[Δ− n + 2

√n − 1


Still we have to prove our Conjecture 1 for Δ > n − 2√n − 1 and

m < n2

[Δ− n+ 2

√n− 1

].




Δ > n − 2√n − 1.

If m ≥ n2

[Δ− n + 2

√n − 1



m < n2

[Δ− n+ 2

√n− 1

].

q1 − 2λ1 ≤∑

vi vj∈E(G)

(xi − xj)2




Δ > n − 2√n − 1.

If m ≥ n2

[Δ− n + 2

√n − 1



m < n2

[Δ− n+ 2

√n− 1

].

q1 − 2λ1 ≤∑

vi vj∈E(G)

(xi − xj)2 ≤ m (xmax − xmin)

2 ≤ n

2

[Δ− n + 2

√n − 1

]

× (xmax − xmin)2 .

Conjecture 1 for tree

Proof of Conjecture 1 for tree:

Let mi be the average degree of the adjacent vertices of vertex vi .



Let mi be the average degree of the adjacent vertices of vertex vi .We have

q1 ≤ max1≤i≤n

(di +mi )




q1 ≤ max1≤i≤n

(di +mi ) = dk +mk , (say).




q1 ≤ max1≤i≤n


Moreover, for Tree

λ1 ≥ max1≤i≤n

√di +mi − 1




q1 ≤ max1≤i≤n


Moreover, for Tree


√di +mi − 1 ≥

√dk +mk − 1.




q1 ≤ max1≤i≤n


Moreover, for Tree


√di +mi − 1 ≥

√dk +mk − 1.

Then

q1 − 2λ1 ≤ (√

dk +mk − 1− 1)2




q1 ≤ max1≤i≤n


Moreover, for Tree


√di +mi − 1 ≥

√dk +mk − 1.

Then

q1 − 2λ1 ≤ (√

dk +mk − 1− 1)2 ≤ (√n− 1− 1)2 = n − 2

√n − 1.

Algebraic Connectivity

Conjecture 3 [4,5,6]:

a(G )/δ(G ) is minimum for graph composed of 2 triangles linked with apath.

[4] M. Aouchiche, Comparaison Automatisee d’Invariants en Theorie des Graphes, Ph.D. Thesis, Ecole

Polytechnique de Montreal, February 2006.

[5] M. Aouchiche, G. Caporossi, P. Hansen, Variable neighborhood search for extremal graphs. 20. Automated

comparison of graph invariants, MATCH Commun. Math. Comput. Chem. 58 (2007) 365–384.

[6] M. Aouchiche, P. Hansen, A survey of automated conjectures in spectral graph theory, Linear Algebra Appl. 432

(2010) 2293–2322.

Notation:

A path with n vertices is denoted by Pn.

Notation:


The near path Qn is the tree on n vertices obtained from a path Pn−1 :v1v2 · · · vn−2vn−1 by attaching a new pendant edge vn−2vn at vn−2.

Notation:



Let Wn be the tree on n vertices obtained from a path Pn−2 :v2v3 · · · vn−2vn−1 by attaching a new pendant edge vn−2vn at vn−2 andanother new pendant edge v1v3 at v3, respectively.

Notation:




Let Q ′n = Qn + vn−1vn, W

′n = Wn + v1v2 and W ′′

n = Wn + v1v2 + vn−1vn.

Notation:




Let Q ′n = Qn + vn−1vn, W

′n = Wn + v1v2 and W ′′

n = Wn + v1v2 + vn−1vn.

Let Zn be the tree on n vertices obtained from a path Pn−1 :v1v2 · · · vn−2vn−1 by attaching a new pendant edge vn−3vn at vn−3.

Inequality

Lemma 1:

For x > 0,

x > sin x > x − x3

6.

Inequality

Lemma 1:

For x > 0,

x > sin x > x − x3

6.

Lemma 2:

For any positive integer n > 3,

sin( π

2 n

)>

1√2sin

(π

2 (n − 1)

).

Inequality

Proof:

For x > 0,

sin( π

2n

)>

π

2n− π3

48 n3>

π

2√2 (n − 1)

>1√2sin

(π

2 (n − 1)

).

Laplacian Eigenvalues

Lemma 3:

The Laplacian eigenvalues of path Pn are

2 + 2 cos(π i

n

), i = 1, 2, . . . , n − 1 and 0.


Definition:

We define

Ad(G ) =a(G )

δ(G ).


Definition:

We define

Ad(G ) =a(G )

δ(G ).

Conjecture 3:

Let G be a connected graph of order n > 3 and minimum degree δ. Then

Ad(G ) ≥ Ad(W ′′n ) (1)

with equality holding if and only if G ∼= W ′′n .


Theorem 1:

Let G be a connected graph of order n > 3 and minimum degree δ. If δ ≤ 2or δ ≥ n/2, then

Ad(G ) ≥ Ad(W ′′n ) (2)



Theorem 1:

Let G be a connected graph of order n > 3 and minimum degree δ. If δ ≤ 2or δ ≥ n/2, then

Ad(G ) ≥ Ad(W ′′n ) (2)


Proof:

For 4 ≤ n ≤ 8, one can easily check the result by Sage. Otherwise, n ≥ 9.

Lemma 4 [7]:

Let G be a connected graph of order n ≥ 9 andG /∈ {Pn, Qn, Q

′n, Wn, W

′n, W

′′n }. Then we have

a(Pn) < a(Qn) = a(Q ′n) < a(Wn) = a(W ′

n) = a(W ′′n ) < a(G ), a(Wn) < a(Zn

Lemma 4 [7]:

Let G be a connected graph of order n ≥ 9 andG /∈ {Pn, Qn, Q

′n, Wn, W

′n, W

′′n }. Then we have

a(Pn) < a(Qn) = a(Q ′n) < a(Wn) = a(W ′

n) = a(W ′′n ) < a(G ), a(Wn) < a(Zn

Lemma 5 [8]:

If v is a pendent vertex, then a(G ) ≤ a(G − v).

[7] J.-Y. Shao, J.-M. Guo, H.-Y. Shan, The ordering of trees and connected graphs by algebraic connectivity, Linear

Algebra Appl. 428 (2008) 1421–1438.

[8] J. X. Li, J.-M. Guo, W. C. Shiu, The Smallest Values of Algebraic Connectivity for Trees, Acta Mathematica

Sinica, English Series 28 (10) (2012) 2021–2032.

Proof:

Case (i): δ = 1.

If G /∈ {Pn, Qn, Q′n, Wn, W

′n}, then by Lemma 4, we get

Ad(G ) = a(G )

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n )

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have

Ad(Pn) < Ad(Qn)

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have

Ad(Pn) < Ad(Qn) = Ad(Q ′n)

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have

Ad(Pn) < Ad(Qn) = Ad(Q ′n) < Ad(Wn)

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have

Ad(Pn) < Ad(Qn) = Ad(Q ′n) < Ad(Wn) = Ad(W ′

n).

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have


n).

Now we have to show that

Ad(Pn) > Ad(W ′′n )

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have


n).


Ad(Pn) > Ad(W ′′n ), i.e., a(Wn) = a(W ′′

n )

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have


n).



n ) < 2 a(Pn)

Proof:

Case (i): δ = 1.



Ad(G ) = a(G ) > a(W ′′n ) >

a(W ′′n )

2= Ad(a(W ′′

n )).

Moreover, we have


n).



n ) < 2 a(Pn) = 2(2− 2 cos

(πn

))

Proof:

Case (i): δ = 1.

By Lemma 1.2, we have

2 sin2( π

2n

)> sin2

(π

2(n − 1)

)

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0

Using the above result with some Lemmas 1.3, 1.4 and 1.5, we have

a(Wn) < a(Zn)

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0


a(Wn) < a(Zn) ≤ a(Pn−1)

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0


a(Wn) < a(Zn) ≤ a(Pn−1) = 2− 2 cos

(π

n − 1

)

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0


a(Wn) < a(Zn) ≤ a(Pn−1) = 2− 2 cos

(π

n − 1

)< 2

(2− 2 cos

(πn

))

Proof:

Case (i): δ = 1.


2 sin2( π

2n

)> sin2

(π

2(n − 1)

), i.e., 1− 2 cos

(πn

)+ cos

(π

n − 1

)> 0


a(Wn) < a(Zn) ≤ a(Pn−1) = 2− 2 cos

(π

n − 1

)< 2

(2− 2 cos

(πn

))

= 2 a(Pn).

Proof:

Case (ii): δ = 2.

Then by Lemma 1.4,

Ad(G ) =a(G )

2≥ a(W ′′

n )

2= Ad(W ′′

n )


Proof:

Case (iii): δ ≥ n/2.

By Case (i) and Lemma 1.1, we have

Ad(W ′′n ) =

a(W ′′n )

2

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)<

π2

n2.

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)<

π2

n2.

Since n ≥ 5, one can easily see that

n

(1− π2

n2

)> n − 2 .

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)<

π2

n2.


n

(1− π2

n2

)> n − 2 .

Thus we have

δ

(1− π2

n2

)≥ n

2

(1− π2

n2

)

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)<

π2

n2.


n

(1− π2

n2

)> n − 2 .

Thus we have

δ

(1− π2

n2

)≥ n

2

(1− π2

n2

)>

n − 2

2

Proof:



Ad(W ′′n ) =

a(W ′′n )

2< 2− 2 cos

(πn

)= 4 sin2

( π

2n

)<

π2

n2.


n

(1− π2

n2

)> n − 2 .

Thus we have

δ

(1− π2

n2

)≥ n

2

(1− π2

n2

)>

n − 2

2, i.e.,

π2

n2< 1− n − 2

2 δ.

Case (iii): δ ≥ n/2

Lemma 1.6 [9]:

Let G be a connected graph of order n, minimum degree δ(G )and algebraic connectivity a(G ). Then

a(G )− δ(G ) ≥

⎧⎪⎨⎪⎩

−n − 8 +√n2 + 8n − 16

4if n is even

−n − 3

2if n is odd

Moreover, the equality holds if and only if G ∼= Kn/2, n/2\{e}when n is even (e is any edge), and G ∼= K(n−1)/2, (n−1)/2 ∨ K1

when n is odd.

[9] K. C. Das, Proof of conjectures involving algebraic connectivity of graphs, Linear Algebra Appl. 438 (2013)

3291–3302.

Proof:


From Lemma 1.6, we have

−n− 2

2≤ −n− 8 +

√n2 + 8n − 16

4≤ −n− 3

2≤ a(G )− δ.

Proof:



−n− 2

2≤ −n− 8 +

√n2 + 8n − 16

4≤ −n− 3

2≤ a(G )− δ.

Therefore

Ad(G ) =a(G )

δ

Proof:



−n− 2

2≤ −n− 8 +

√n2 + 8n − 16

4≤ −n− 3

2≤ a(G )− δ.

Therefore

Ad(G ) =a(G )

δ≥ 1− n − 2

2 δ

Proof:



−n− 2

2≤ −n− 8 +

√n2 + 8n − 16

4≤ −n− 3

2≤ a(G )− δ.

Therefore

Ad(G ) =a(G )

δ≥ 1− n − 2

2 δ>

π2

n2

Proof:



−n− 2

2≤ −n− 8 +

√n2 + 8n − 16

4≤ −n− 3

2≤ a(G )− δ.

Therefore

Ad(G ) =a(G )

δ≥ 1− n − 2

2 δ>

π2

n2> Ad(W ′′

n ).

Proof:

Remark:

For 3 ≤ δ < n/2, Conjecture 3 is still open.

THANK YOU for attention.

On (Signless) Laplacian eigenvalues of graphs · On (Signless) Laplacian eigenvalues of graphs...

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