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 ) = ( a i 1 ... i k ), is an order k dimension n symmetric tensor, where � 1 if { i 1 , . . . , i k } ∈ E ( G ) , ( k − 1)! , a i 1 ... i k = 0 , otherwise .

For a complex λ , if there exists a nonzero vector x = ( x 1 , . . . , x n ) T such that A ( G ) x = λ x k − 1 , where x k − 1 = ( x k − 1 ) T and , . . . , x k − 1 1 n A ( G ) x is a n -dimensional vector whose i -th component is n ( A ( G ) x ) i = � a ii 2 ... i k x i 2 . . . x i k , then we call λ an eigenvalue i 2 ,..., i k =1 of 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 ∈ R n + and H ++ -eigenvalue of G if x ∈ R n ++ , where R n + is the set of nonnegative vectors of dimension n , and R n ++ 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 T i 1 i 2 ... i k = 0 for any i 1 ∈ I and any i 2 , . . . , i k �∈ 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 order 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 T i 1 i 2 ... i k = 0 for any i 1 ∈ I and at least one of the i 2 , . . . , i k �∈ 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 ∈ R n ++ , up to a positive scaling coefficient.

It is proved that a k -uniform hypergraph G is connected if and only 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 of G , with � x � k = 1, corresponding to ρ ( G ).

Lemma [14] Let T be a symmetric nonnegative tensor of order k and dimension n. Then � x � k =1 { x T ( T x ) | x is a nonnegative n-dimension vector } ρ ( T ) = max 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 � x � k = 1 such that A ( G ) x = ρ ( G ) x k − 1 and n x T ( A ( G ) x ) = � ρ ( G ) = a i 1 i 2 ... i k x i 1 x i 2 . . . x i k i 1 , i 2 ,..., i k =1 � = k x i 1 x i 2 . . . x i k (1) e = { i 1 , i 2 ,..., i k }∈ E � = k ω x ( e ) , e ∈ E ( G ) where ω x ( e ) = x i 1 x i 2 . . . x i k for e = { i 1 , i 2 , . . . , i k } . 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 e 1 , e 2 , . . . , e l and adding e ′ 1 , e ′ 2 , . . . , e ′ l , where e i ∈ E ( G ) , V i ⊂ e i , U �⊂ e i , e ′ i = { e i \ V i } ∪ U and | V i | = | U | for i = 1 , . . . , l. If G ′ has no multiple edges and Π u ∈ U x u ≥ max 1 ≤ i ≤ l Π v ∈ V i x v , then ρ ( G ′ ) > ρ ( G ) .

Proof. By Lemma 1 we have x T A ( G ′ ) x = k ρ ( G ′ ) � ≥ ω x ( e ) e ∈ E ( G ′ )   l  � � ω x ( e ′ � � i ) − ω x ( e i ) = k ω x ( e ) +  i =1 e ∈ E ( G )   l ω x ( e ′ i )  � � = k ω x ( e ) + (Π u ∈ U x u − Π v ∈ V i x v )  Π u ∈ U x u i =1 e ∈ E ( G ) � ≥ k ω x ( e ) e ∈ E ( G ) = ρ ( G ) .

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

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, A n , r be the k -uniform hypergraph obtained 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, A n , 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 ( A n , r ) with equality if and only if G ∼ = A n , 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 ∗ = { e 1 , e 2 , . . . , e s } be the set of edges in G , each of which contains at least a pendent vertex, V i = e i ∩ V ∗ be the pendent vertex set of e i , then s ≤ r and V ∗ = V 1 ∪ V 2 ∪ · · · ∪ V s . Suppose that 1 ≤ | V 1 | ≤ | V 2 | ≤ . . . ≤ | V s | . Let N ( V i ) = e i \ V i . Then

Claim 1. N ( V 1 ) ⊇ N ( V 2 ) ⊇ . . . ⊇ N ( V s ). Let x be the principal eigenvector of G and i , j ∈ [ s ]. If there exist vertices u and v such that u ∈ N ( V i ) , v ∈ N ( V j ) and u �∈ N ( V j ) , v �∈ N ( V i ) and x u ≥ x v , then by Lemma 3 ρ ( G 1 ) > ρ ( G ), where G 1 is obtained from G \ e j by adding the edge { e j \ v } ∪ { u } , a contradiction. Thus N ( V i ) ⊆ N ( V j ) or N ( V j ) ⊆ N ( V i ) and we complete the proof of Claim 1.

Let V 0 = V ( G ) \ { V ∗ ∪ N ( V 1 ) } , N ∗ ( V i ) = N ( V i ) \ N ( V i +1 ) for i = 1 , 2 , . . . , s − 1 and N ∗ ( V s ) = N ( V s ). Note the fact that | V 0 | + | N ( V 1 ) | = n − r ≥ k + 1 and | V 1 | + | N ( V 1 ) | = k , then | V 0 | > | V 1 | . By Lemma 4 we get that x u = x v if { u , v } ⊆ V i ( i = 1 , . . . , s ), V 0 or N ∗ ( V j )( j = 1 , 2 , . . . , s ). So we can suppose that x w = x i if w ∈ V i (1 ≤ i ≤ s ), x w = x 0 if w ∈ V 0 and x w = x ∗ i if w ∈ N ∗ ( V i )( i = 1 , 2 , . . . , s ).

Claim 2. If N ∗ ( V i ) is not empty for some i ∈ { 1 , 2 , . . . , s − 1 } , then x i +1 > x i ∗ > x i and x ∗ s > x s . Let u ∈ V i , v ∈ N ∗ ( V i ). Then x u = x i and x v = x ∗ i . Note that u , v ∈ e i , u ∈ V ∗ and v ∈ V ( G ) \ V ∗ , that is, u ∈ e i implies v ∈ e i , and d ( v ) ≥ 2, by Lemma 4 we get x i ∗ > x i . Now we assume that x ∗ i ≥ x i +1 . Since N ∗ ( V i ) is not empty, | N ( V i ) \ N ( V i +1 ) | > 0 and | V i +1 | > | V i | . Let V 1 i +1 , V 2 i +1 be the two partition of V i +1 such that | V 1 i +1 | = | V i | .

Let G 2 be obtained from G by deleting e i +1 and adding the edges i +1 ∪ N ( V i ) and e 0 , where e 0 is the edge containing V 2 e ∗ i +1 = V 1 i +1 and any k − | V 2 i +1 | vertices of V ( G ) − V ∗ . ω ( e ∗ i +1 ) ω ( e i +1 ) = ( x ∗ x i +1 ) | N ( V i ) \ N ( V i +1 ) | ≥ 1. Clearly, i G 2 is a connected hypergraph with exactly r pendent vertices, and ρ ( G 2 ) ≥ k Σ e ∈ E ( G 2 ) ω x ( e ) = k Σ e ∈ E ( G ) \ e i +1 ω x ( e ) + k ω ( e ∗ i +1 ) + k ω ( e 0 ) ≥ k Σ e ∈ E ( G ) ω x ( e ) + k ω ( e 0 ) = ρ ( G ) + k ω ( e 0 ) > ρ ( G ), a contradiction.

Claim 3. | V 1 | = | V 2 | = . . . = | V s | . Assume that | V s | > | V 1 | . By Claims 1 and 2 we can get that x s > x 1 and x s > x ∗ i for any i < s with | N ∗ ( V i ) | � = 0. By Claim 1 we know that N ( V 1 ) ⊇ N ( V s ) and | e 1 \ N ∗ ( V s ) | = k − | N ∗ ( V s ) | = | V s | . Then there is a bijective σ from e 1 \ N ∗ ( V s ) to V s . Let x ∗ be obtained from x by exchanging the weight of e 1 \ N ∗ ( V s ) and V s , that is, x ∗ u = x σ ( u ) , x ∗ v = x σ − 1 ( v ) for u ∈ e 1 \ N ∗ ( V s ) , v ∈ V s , and x ∗ w = x w for w ∈ V ( G ) \ {{ e 1 \ N ∗ ( V s ) } ∪ V s } . Clearly, � x ∗ � k = � x � k = 1, ω x ∗ ( e 1 ) + ω x ∗ ( e s ) = ω x ( e 1 ) + ω x ( e s ), and ω x ∗ ( e ) ≥ ω x ( e ) for any e ∈ E ( G ) \ { e 1 , e s } .