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2019/2/23, 中研院, The largest real eigenvalues of nonnegative matrices  by applying Kelmans transformations, (2019 Taipei International Workshop on  Combinatorics and Graph Theory, 2019/2/20-23)

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(1)

The largest real eigenvalues of nonnegative matrices

by applying Kelmans transformations

Chih-wen Weng (joint work with Louis Kao)

Department of Applied Mathematics National Chiao Tung University

14:25-15:10, February 23, 2019

(2)

Undirected graphs

G:

r r r r

r r

u v

a b c d V(G) ={u, v, a, b, c, d}

E(G) ={ua, ub, uc, uv, vc, vd}

N(u) ={a, b, c, v}

N(v) ={u, c, d}

N[v] ={u, c, d, v}

(3)

Adjacency matrices

G : r r r

1 2 3 −→ A(G) =

 0 1 0 1 0 1 0 1 0

(4)

Kelmans transformations for undirected graphs

The graph G(a, b)is obtained from G with the same vertex set and replacing edge ac by bc for each c∈ N(a) − N[b].

r r r r

r r

a b

G : −→ G(a, b) :

r r r r

r r

a b

b a







0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 1

1 1 1 0 1 0







−→







0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0

1 1 1 1 0 1

0 0 1 0 1 0







(unchanged)

(5)

A.K. Kelmans, On graphs with randomly deleted edges, Acta Math. Acad.

Sci. Hung. 37 (1981) 77-88.

(6)

Remark 1: G(a, b) is isomorphic to G(b, a)

r r r r

r r

a b

G : −→ G(a, b) :

r r r r

r r

a b

r r r r

r r

a b

G : −→ G(b, a) :

r r r r

r r

a b

(7)

Complement graph

r r r r

r r

a b

G : −→ G :

r r r r

r r

a b

b a







0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 0







−→







0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 1 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0







(8)

Remark 2: G(b, a) = G(a, b)

b a







0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 0







−→ G(a, b) =







0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 1 1 1 0 1

0 0 1 0 1 0







b a







0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 1 0







−→ G(b, a) =







0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 0

1 1 0 1 0 0







=G(a, b)

(9)

Eigenvalues and eigenvectors

Let A be a real square matrix, u is a column vector, λ is a complex number. The number λ is an eigenvalueof A with associatedeigenvector u if

Au = λu.

(10)

A few facts from linear algebra

1 An n× n real symmetric matrix has n real eigenvalues and n corresponding orthonormal eigenvectors.

2 A nonnegative matrix M has a largest real eigenvalue ρ(M), called the spectral radius of M.

3 The eigenvalue ρ(M) is associated with a nonnegative eigenvector of length 1, called Perron vector.

4 All other eigenvalues λ of M satisfy |λ| ≤ ρ(M).

5 Thespectral radius ρ(G) of a graph G is the largest real eigenvalue of its adjacency matrix A = A(G).

(11)

Example

G : r r r

a b c A =

 0 1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1 0

2

ñ2 2

 = ±√ 2

2

±2√ 2

 ,

0 1 0 1 0 1 0 1 0

 1 0

−1

 = 0.

Note that 1

2 2(

2, 2,√

2) is the Perron vector of A and the three eigenvectors

1 2

2( 2, 2,√

2), 1 2

2(

2,−2,√ 2), 1

2(1, 0,−1)

are orthonormal and ρ(G) =√ 2.

(12)

Rayleigh quotient for symmetric matrices

Lemma

Let A = (aij) be an n× n symmetric matrix. Then

ρ(A) = max

x̸=0

xtAx xtx . Proof.

Let λ1 ≥ λ2 ≥ · · · ≥ λn be eigenvalues of A with corresponding orthonormal eigenvectors x1, x2, . . . , xn. Write x =n

i=1cixi, where

n

i=1c2i = 1. Then the maximum of xtAx =

n i=1

c2iλi

is λ1 and is when x =±x1.

(13)

Non-symmetric matrices

x:xmaxtx=1xt (0 1

0 0 )

x = max

x=(a,b) a2+b2=1

ab = 1

2 ̸= 0 = ρ (0 1

0 0 )

.

(14)

Lemma

Let x = (xa) be the Perron vector of A(G). Then xb≥ xa ⇒ ρ(G) ≤ ρ(G(a, b)).

Proof.

ρ(G(a, b)) = max

∥y∥2=1

ytA(G(a, b))y

≥ xtA(G(a, b))x =

cd∈E(G(a,b))

xcxd

= ∑

cd∈E(G)

xcxd+ ∑

e∈N(a)−N[b]

xbxe− xaxe

≥ xtAx = ρ(G).

(15)

An old result for undirected graphs

Theorem

ρ(G)≤ ρ(G(a, b)) Proof.

Let x = (xa) be the Perron vector of A(G). Then

xb≥ xa ρ(G)≤ ρ(G(a, b)), xa≥ xb ρ(G)≤ ρ(G(b, a)), G(a, b)) ∼= G(b, a) ρ(G(a, b)) = ρ(G(b, a)).

(16)

Key fact in the above proof

G(a, b) ∼= G(b, a)

(17)

Corollary

ρ(G)≤ ρ(G(a, b)) Proof.

ρ(G)≤ ρ(G(b, a)) = ρ(G(a, b))

The above results ρ(G)≤ ρ(G(a, b)) and ρ(G) ≤ ρ(G(a, b)) are from [Péer Csikvári, On a conjecture of V. Nikiforov, Discrete Mathematics, 309(13), 2009, 4522-4526]

(18)

A counterexample for digraph transformation

A = a b



0 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0



 → A =



0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 0



unchanged

ρ(A)≈ 2.234 > 2.148 ≈ ρ(A)

(19)

We will define Kelmans transformations for nonnegative matrices.

Let C denote a nonnegative square matrix of order n. For a subset S of [n] :={1, 2, . . . , n} let C[S] denote the principal submatrix of C restricted to S.

Fix 1≤ a, b ≤ n.

(20)

Kelmans transformation of C from b to a (w.r.t. t

i

, s

j

)

C =





c d e c14 c15 f g h c24 c25 i j k c34 c35 c41 c42 c43 m c51 c52 c53 n o





 a b

≥ 0

→ C =





c d e c14+ t1 c15− t1

f g h c24+ t2 c25− t2

i j k c34+ t3 c35− t3

c41+ s1 c42+ s2 c43+ s3 m c51− s1 c52− s2 c53− s3 n o





≥ 0,

where cia+ ti≥ cib− ti and caj+ sj ≥ cbj− sj.

(21)

Example

For a = 4, b = 5,t1= 1, t2 = t3 = 0, s1= s3= 0,s2 = 1,

C =





c d e 0 1 f g h 1 0 i j k 1 1 1 0 1 m 0 1 1 n o





→ C =





c d e 1 0 f g h 1 0 i j k 1 1 1 1 1 m 0 0 1 n o





.

For a = 5, b = 4,t2= 1, t1 = t3 = 0, s2= s3= 0,s1 = 1,

C =





c d e 0 1 f g h 1 0 i j k 1 1 1 0 1 m 0 1 1 n o





→ C =





c d e 0 1 f g h 0 1 i j k 1 1 0 0 1 m 1 1 1 n o





(= C′′ later).

(22)

C

and C

′′

Throughout letC denote the Kelmans transformation of C from b to a with respect to ti and sj, and letC′′= (c′′ij) is the Kelmans transformation of C from a to bwith respect to cia− cib+ ti and caj− cbj+ sj.

(23)

Symmetric subgraph

A matrix C is symmetricon{a, b} of if the principal submatrix C[a, b] is symmetric with constant diagonals.

Example The matrix

C =





1 1 0 0 1 0 1 0 1 0 0 1 1 1 1 1 0 1 s t 0 1 1 t s





is symmetric on {4, 5}.

(24)

Remark: C

is similar to C

′′

If C is symmetric on {a, b}, then C is similar to C′′.

C =





c d e c14+ t1 c15− t1

f g h c24+ t2 c25− t2

i j k c34+ t3 c35− t3

c41+ s1 c42+ s2 c43+ s3 s t c51− s1 c52− s2 c53− s3 t s





C′′=





c d e c15− t1 c14+ t1

f g h c25− t2 c24+ t2 i j k c35− t3 c34+ t3

c51− s1 c52− s2 c53− s3 s t c41+ s1 c42+ s2 c43+ s3 t s





(25)

Main result

Theorem

If a nonnegative matrix C is symmetric on {a, b}, then ρ(C) ≤ ρ(C).

For the remaining of the talk, we will prove the above theorem.

(26)

Strongly connected

A digraph is strongly connectedif for any two vertices x, y there exists a directed path from x to y.

r r

r

1 2

3 - R 

V(G) ={1, 2, 3}

E(G) ={13, 12, 23, 32}

A =

0 1 1 0 0 1 0 1 0

The above digraph is notstrongly connected because there is no directed path from 2 to 1.

(27)

Digraph associated with a matrix

For an n× n matrix M, we define a digraph G(M) with vertex set {1, 2, . . . , n} and edge set

E ={xy | Mxy̸= 0}.

G(M) is called the digraph associated with M.

M =

0 0 −2

3 0 2

0 −1 0

−→

r r

r

1 2

3

 R 

(28)

Irreducible matrix

A matrix is irreducibleif its associated digraph is strongly connected.

The matrix

A =

0 1 1 0 0 1 0 1 0

is reduciblesince its associated digraph G(A) is not strongly connected.

r r

r

1 2

3 - R 

V(G(A)) ={1, 2, 3}

E(G(A)) ={13, 12, 23, 32}

(29)

A fact from linear algebra

A nonnegative irreducible matrix has a positivePerron vector.

(30)

The irreducible matrix A

ϵ

For a binary matrix A = (aij) and 0 < ϵ < 1, define the matrix Aϵ = A + ϵJ, where J is the matrix with all 1’s.

A =

0 1 1 0 0 1 0 1 0

 ⇒ Aϵ =

ϵ 1 + ϵ 1 + ϵ ϵ ϵ 1 + ϵ ϵ 1 + ϵ ϵ

(31)

The idea

Our proof is motivated by the following exercise.

Exercise

Suppose 0≤C≤ C. Show ρ(C)≤ ρ(C).

Proof.

Let ut≥ 0 be left Perron vector of C for ρ(C) and vϵ> 0 be right Perron vector of C′ϵ for ρ(C′ϵ). Then

ρ(C)utvϵ= utCvϵ≤utC′ϵvϵ= ρ(C′ϵ)utvϵ, which implies ρ(C)≤ ρ(C′ϵ) since utvϵ > 0. Then by continuity

ρ(C)≤ lim

ϵ→0+ρ(C′ϵ) = ρ(C).

(32)

The left Perron vector w

t

for ρ(C)

Let wt= (wi) denote the left Perron vector for ρ(C) of C.

We will show that

wa≥ wb ⇒ ρ(C) ≤ ρ(C),

wb≥ wa ⇒ ρ(C) ≤ ρ(C′′) = ρ(C).

By symmetric, we might assume wa≥ wb and only prove ρ(C)≤ ρ(C).

(33)

The matrix Q = I

n

− E

ba

Set vt= wtQ, where Q = In− Eba and Eba denote the binary matrix of order n with a unique 1 in the position ba.

vt=(w1, w2, w3, w4− w5, w5)

=(w1, w2, w3, w4, w5)





1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 −1 1





 a b

=wtQ

By the assumption wa≥ wb, vt≥ 0

(34)

v

t

Q

−1

C = w

t

C = ρ(C)w

t

= ρ(C)v

t

Q

−1

vt= wtQ and wtC = ρ(C)wt

⇒ vtQ−1C = ρ(C)vtQ−1

Q−1 =





1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1





 a b

(35)

The matrix Q

−1

C(Q

−1

)

t

Q−1C(Q−1)t

=







c11 · · · c1 n−1 c1n−1+ c1n

... ... ...

cn−11 · · · cn−1n−1 cn−1n−1+ cn−1n

cn−1 1+ cn1 · · · cn−1 n−1+ cnn−1 cn−1n−1+ cnn−1

+cn−1 n+ cnn





 a b

(36)

Q

−1

C(Q

−1

)

t

≤ Q

−1

C

′ϵ

(Q

−1

)

t

C =



0 0 1 1 1 0 0 1 1 1 0 1 1 1 1 0



 a b

→ C =



0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 0



⇒ Q−1C(Q−1)t=



0 0 1 2 1 0 0 1 1 1 0 1 2 2 1 2



 ≤



0 0 1 2 1 0 1 1 1 1 0 1 2 2 1 2



 = Q−1C(Q−1)t

<Q−1C′ϵ(Q−1)t

(37)

The right eigenvector vector u

ϵ

for ρ(Q

t

C

′ϵ

(Q

t

)

−1

)

Let uϵ̸< 0 denote the right eigenvector vector for ρ(QtC′ϵ(Qt)−1).

(C′ϵ is positive, but QtC′ϵ(Qt)−1 could have some negative entries) Then

(i) ρ(C′ϵ) = ρ(QtC′ϵ(Qt)−1);

(ii) C′ϵ(Q−1)tuϵ = C′ϵ(Qt)−1uϵ = ρ(C′ϵ)(Qt)−1uϵ= ρ(C′ϵ)(Q−1)tuϵ; (iii) (Q−1)tuϵ> 0is an eigenvector for ρ(C′ϵ).

(38)

Right multiplication of u

ϵ

Q−1C(Q−1)tuϵ≤Q−1C′ϵ(Q−1)tuϵ= ρ(C′ϵ)Q−1(Q−1)tuϵ.

(39)

Left multiplication of v

t

ρ(C)vtQ−1(Q−1)tuϵ= vtQ−1C(Q−1)tuϵ ≤ vtρ(C′ϵ)Q−1(Q−1)tuϵ.

(40)

Deleting the positive term v

t

Q

−1

(Q

−1

)

t

u

ϵ

ρ(C)vtQ−1(Q−1)tuϵ≤ ρ(C′ϵ)vtQ−1(Q−1)tuϵ

⇒ ρ(C) ≤ ρ(C).

The proof is completed.

(41)

Thank you for your attention.

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