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
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}
Adjacency matrices
G : r r r
1 2 3 −→ A(G) =
0 1 0 1 0 1 0 1 0
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)
A.K. Kelmans, On graphs with randomly deleted edges, Acta Math. Acad.
Sci. Hung. 37 (1981) 77-88.
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
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
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)
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.
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).
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.
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.
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 )
.
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).
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)).
Key fact in the above proof
G(a, b) ∼= G(b, a)
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]
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′)
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.
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 ℓ 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 ℓ m c51− s1 c52− s2 c53− s3 n o
≥ 0,
where cia+ ti≥ cib− ti and caj+ sj ≥ cbj− sj.
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 ℓ 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 ℓ 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 ℓ 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 ℓ m 1 1 1 n o
(= C′′ later).
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.
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}.
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
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.
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.
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
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}
A fact from linear algebra
A nonnegative irreducible matrix has a positivePerron vector.
The irreducible matrix A
ϵFor 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
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′).
The left Perron vector w
tfor 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′).
The matrix Q = I
n− E
baSet 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
v
tQ
−1C = w
tC = ρ(C)w
t= ρ(C)v
tQ
−1vt= 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
The matrix Q
−1C(Q
−1)
tQ−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
Q
−1C(Q
−1)
t≤ Q
−1C
′ϵ(Q
−1)
tC =
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
The right eigenvector vector u
ϵfor ρ(Q
tC
′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).
Right multiplication of u
ϵQ−1C(Q−1)tuϵ≤Q−1C′ϵ(Q−1)tuϵ= ρ(C′ϵ)Q−1(Q−1)tuϵ.
Left multiplication of v
tρ(C)vtQ−1(Q−1)tuϵ= vtQ−1C(Q−1)tuϵ ≤ vtρ(C′ϵ)Q−1(Q−1)tuϵ.
Deleting the positive term v
tQ
−1(Q
−1)
tu
ϵρ(C)vtQ−1(Q−1)tuϵ≤ ρ(C′ϵ)vtQ−1(Q−1)tuϵ
⇒ ρ(C) ≤ ρ(C′).
The proof is completed.