The maximum spectral radius of k -uniform hypergraphs with r pendent - - PowerPoint PPT Presentation

The maximum spectral radius of k-uniform hypergraphs with r pendent vertices

Jianbin Zhang

(Joint with Jianping Li)

South China Normal University

July 7, 2018

Let G be a hypergraph with vertex set V (G) and edge set E(G), where E(G) is a set whose elements are subsets of V (G). If each edge of G contains exactly k distinct vertices, then G is called a k-uniform hypergraph. An alternating sequence of vertices and edges is called a path if the vertices and edges are distinct, and a cycle if the first and last vertices are the same, the other vertices and all edges are distinct. If there exists a path between any two vertices of G, then G is called connected.

The adjacency tensor of a k-uniform hypergraph G on n vertices, denoted by A(G) = (ai1...ik), is an order k dimension n symmetric tensor, where ai1...ik =

• 1

(k−1)!,

if {i1, . . . , ik} ∈ E(G), 0,

• therwise.

For a complex λ, if there exists a nonzero vector x = (x1, . . . , xn)T such that A(G)x = λxk−1, where xk−1 = (xk−1

1

, . . . , xk−1

n

)T and A(G)x is a n-dimensional vector whose i-th component is (A(G)x)i =

n

• i2,...,ik=1

aii2...ikxi2 . . . xik, then we call λ an eigenvalue

• f G.

If λ and x are real, we call λ an H-eigenvalue of G. Moreover, we call λ an H+-eigenvalue of G if x ∈ Rn

+ and

H++-eigenvalue of G if x ∈ Rn

++, where Rn + is the set of

nonnegative vectors of dimension n, and Rn

++ is the set of positive

vectors of dimension n.

The spectral radius of A(G) is defined as ρ(A(G)) = max{|λ| | λ ∈ spec(A(G))}, where spec(A(G)) is the set of all eigenvalues of A(G).

A tensor T of order k and dimension n is called reducible, if there exists a nonempty proper index subset I ⊂ [n] such that Ti1i2...ik = 0 for any i1 ∈ I and any i2, . . . , ik ∈ I. If T is not reducible, then T is called irreducible. Chang,Pearson and Zhang [3] proved that if T is irreducible, then ρ(T ) is the unique H++-eigenvalue of T , with the unique eigenvector x ∈ R++, up to a positive scaling coefficient.

Yang and Yang [18] proved that if T is a nonnegative tensor of

• rder k and dimension n, then ρ(T ) is an H+-eigenvalue of T .

A tensor T of order k and dimension n is called weakly reducible, if there exists a nonempty proper index subset I ⊂ [n] such that Ti1i2...ik = 0 for any i1 ∈ I and at least one of the i2, . . . , ik ∈ I. If T is not weakly reducible, then T is called weakly irreducible. Friedland, Gaubert and Han [6] proved that if T weakly irreducible, then ρ(T ) is the unique H++-eigenvalue of T , with the unique eigenvector x ∈ Rn

++, up to a positive scaling coefficient.

It is proved that a k-uniform hypergraph G is connected if and

• nly if its adjacency tensor A(G) is weakly irreducible [6]. Thus for

a k-uniform connected hypergraph G, its adjacency tensor A(G) has a unique positive eigenvector x, called the principal eigenvector

• f G, with xk = 1, corresponding to ρ(G).

Lemma

[14] Let T be a symmetric nonnegative tensor of order k and dimension n. Then ρ(T ) = max

xk=1{xT (T x)|x is a nonnegative n-dimension vector}

Furthermore, x is an eigenvector of T corresponding to ρ(T ) if and only if it is an optimal solution of the maximization problem of above equation.

By Lemma 1 we know that for a k-uniform connected hypergraph, there is a unique positive vector with xk = 1 such that A(G)x = ρ(G)xk−1 and ρ(G) = xT(A(G)x) =

n

• i1,i2,...,ik=1

ai1i2...ikxi1xi2 . . . xik = k

• e={i1,i2,...,ik}∈E

xi1xi2 . . . xik = k

• e∈E(G)

ωx(e), (1) where ωx(e) = xi1xi2 . . . xik for e = {i1, i2, . . . , ik}. By Equation 1 we have

Lemma

[4, 11]Let G be a connected k-uniform hypergraph and H is a sub-hypergraph of G. Then ρ(H) ≤ ρ(G) with equality if and only if H ∼ = G.

Lemma

Let l ≥ 1, x be the principal eigenvector of a connected k-uniform hypergraph G. Let G ′ be obtained from G by deleting e1, e2, . . . , el and adding e′

1, e′ 2, . . . , e′ l, where ei ∈ E(G), Vi ⊂ ei, U ⊂ ei, e′ i = {ei\Vi} ∪ U

and |Vi| = |U| for i = 1, . . . , l. If G ′ has no multiple edges and Πu∈Uxu ≥ max

1≤i≤l Πv∈Vixv, then

ρ(G ′) > ρ(G).

Proof.

By Lemma 1 we have ρ(G ′) ≥ xT A(G ′)x = k

• e∈E(G ′)

ωx(e) = k  

e∈E(G)

ωx(e) +

l

• i=1
• ωx(e′

i) − ωx(ei)

• 

 = k  

e∈E(G)

ωx(e) +

l

• i=1

ωx(e′

i)

Πu∈Uxu (Πu∈Uxu − Πv∈Vi xv)   ≥ k

• e∈E(G)

ωx(e) = ρ(G).

Lemma

[13]Let x be the principal eigenvector of a connected k-uniform hypergraph G, vi, vj ∈ V (G). If vi ∈ e implies vj ∈ e for e ∈ E(G), then xvj ≥ xvi. Furthermore, if there exists an edge which contains vj but not vi, then xvj > xvi.

The complete k-uniform hypergraph on n ≥ k ≥ 2 vertices, denoted by K k

n , is a hypergraph which has all k-subsets of V as

• edges. Let n − r ≥ k + 1, An,r be the k-uniform hypergraph
• btained from the complete hypergraph K k

n−r by adding r pendent

vertices and r edges, each new edge contains exactly a new pendent vertex and the same k − 1 distinct vertices of V (K k

n−r) .

Clearly, An,0 ∼ = K k

n .

Theorem

Let G be a connected k-uniform hypergraph on n vertices with exactly r pendent vertices. If n − r ≥ k + 1, then ρ(G) ≤ ρ(An,r) with equality if and only if G ∼ = An,r.

Proof: Let G be a k-uniform hypergraph with maximum spectral radius among all connected k-uniform hypergraph on n vertices with exactly r pendent vertices. Let V ∗ be the set of pendent vertices in G. Then by Lemma 2, it is easy to obtain that G[V (G) − V ∗] is a complete hypergraph. Let E ∗ = {e1, e2, . . . , es} be the set of edges in G, each of which contains at least a pendent vertex, Vi = ei ∩ V ∗ be the pendent vertex set of ei, then s ≤ r and V ∗ = V1 ∪ V2 ∪ · · · ∪ Vs. Suppose that 1 ≤ |V1| ≤ |V2| ≤ . . . ≤ |Vs|. Let N(Vi) = ei \ Vi. Then

Claim 1. N(V1) ⊇ N(V2) ⊇ . . . ⊇ N(Vs). Let x be the principal eigenvector of G and i, j ∈ [s]. If there exist vertices u and v such that u ∈ N(Vi), v ∈ N(Vj) and u ∈ N(Vj), v ∈ N(Vi) and xu ≥ xv, then by Lemma 3 ρ(G1) > ρ(G), where G1 is obtained from G \ ej by adding the edge {ej \ v} ∪ {u}, a contradiction. Thus N(Vi) ⊆ N(Vj) or N(Vj) ⊆ N(Vi) and we complete the proof of Claim 1.

Let V 0 = V (G) \ {V ∗ ∪ N(V1)}, N∗(Vi) = N(Vi) \ N(Vi+1) for i = 1, 2, . . . , s − 1 and N∗(Vs) = N(Vs). Note the fact that |V 0| + |N(V1)| = n − r ≥ k + 1 and |V1| + |N(V1)| = k, then |V 0| > |V1|. By Lemma 4 we get that xu = xv if {u, v} ⊆ Vi(i = 1, . . . , s), V 0

• r N∗(Vj)(j = 1, 2, . . . , s). So we can suppose that xw = xi if

w ∈ Vi(1 ≤ i ≤ s), xw = x0 if w ∈ V 0 and xw = x∗

i if

w ∈ N∗(Vi)(i = 1, 2, . . . , s).

Claim 2. If N∗(Vi) is not empty for some i ∈ {1, 2, . . . , s − 1}, then xi+1 > xi ∗ > xi and x∗

s > xs.

Let u ∈ Vi, v ∈ N∗(Vi). Then xu = xi and xv = x∗

i . Note that

u, v ∈ ei, u ∈ V ∗ and v ∈ V (G) \ V ∗, that is, u ∈ ei implies v ∈ ei, and d(v) ≥ 2, by Lemma 4 we get xi ∗ > xi. Now we assume that x∗

i ≥ xi+1. Since N∗(Vi) is not empty,

|N(Vi) \ N(Vi+1)| > 0 and |Vi+1| > |Vi|. Let V 1

i+1, V 2 i+1 be the

two partition of Vi+1 such that |V 1

i+1| = |Vi|.

Let G2 be obtained from G by deleting ei+1 and adding the edges e∗

i+1 = V 1 i+1 ∪ N(Vi) and e0, where e0 is the edge containing V 2 i+1

and any k − |V 2

i+1| vertices of V (G) − V ∗.

Clearly,

ω(e∗

i+1)

ω(ei+1) = ( x∗

i

xi+1)|N(Vi )\N(Vi+1)| ≥ 1.

G2 is a connected hypergraph with exactly r pendent vertices, and ρ(G2) ≥ kΣe∈E(G2)ωx(e) = kΣe∈E(G)\ei+1ωx(e) + kω(e∗

i+1) +

kω(e0) ≥ kΣe∈E(G)ωx(e) + kω(e0) = ρ(G) + kω(e0) > ρ(G), a contradiction.

Claim 3. |V1| = |V2| = . . . = |Vs|. Assume that |Vs| > |V1|. By Claims 1 and 2 we can get that xs > x1 and xs > x∗

i for any i < s with |N∗(Vi)| = 0. By Claim 1

we know that N(V1) ⊇ N(Vs) and |e1\N∗(Vs)| = k − |N∗(Vs)| = |Vs|. Then there is a bijective σ from e1\N∗(Vs) to Vs. Let x∗ be obtained from x by exchanging the weight of e1 \ N∗(Vs) and Vs, that is, x∗

u = xσ(u), x∗ v = xσ−1(v)

for u ∈ e1 \ N∗(Vs), v ∈ Vs, and x∗

w = xw for

w ∈ V (G) \ {{e1 \ N∗(Vs)} ∪ Vs}. Clearly, x∗k = xk = 1, ωx∗(e1) + ωx∗(es) = ωx(e1) + ωx(es), and ωx∗(e) ≥ ωx(e) for any e ∈ E(G) \ {e1, es}.

Let E 0(G) = {e|e ∈ E(G), e ⊆ V (G) \ V ∗, e ∩ {e1 \ N∗(Vs)} = ∅} . Clearly, e1, es ∈ E 0(G). Note that n − r ≥ k + 1. Then |E 0(G)| = 0 and ωx∗(e) > ωx(e) for any e ∈ E 0(G). We get ρ(G) = k

e∈E(G) ωx(e)

≤ k

e∈E(G)\E 0(G) ωx∗(e) + k e∈E 0(G) ωx(e)

< k

e∈E(G)\E 0(G) ωx∗(e) + k e∈E 0(G) ωx∗(e)

= k

e∈E(G) ωx∗(e)

≤ ρ(G), a contradiction.

By Claim 3 we can get that N(V1) = N(V2) = . . . = N(Vs) and xu = x∗

s for any u ∈ N(V1). It is easy to prove that x1 = . . . = xs.

By Claim 2 we get that x1 < x∗

s . Note that n − r ≥ k + 1 and

|V 0| > |V1|. For any v ∈ V 0, ρ(G)xk−1

1

= x|V1|−1

1

(x∗

s )k−|V1|,

ρ(G)xk−1 =

• v∈e

ωx(e) x0

> x|V1|−1 (x∗

s )k−|V1|,

and x0 > x1.

If |V1| ≥ 2, let G3 be obtained from G by deleting e1 and adding |V1| edges, each contains exactly a vertex of V1, any |V1| − 1 vertices of V 0 and N(V1). It is easy to see that ρ(G3) ≥ k

e∈E(G3) ωx(e)

> k

e∈E(G) ωx(e) + (|V1| − 1)x1x|V1|−1

(x∗

s )k−|V1|

> ρ(G), a contradiction. Thus |V1| = . . . = |Vs| = 1 and s = r. By Claims 1 and 3 we get that G ∼ = An,r.

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