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New upper bounds on the spectral radius of trees with the given number of vertices and maximum degree
Linear Algebra and its Applications 439 (2013) 2527–2541
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Linear Algebra and its Applications www.elsevier.com/locate/laa
New upper bounds on the spectral radius of trees with the given number of vertices and maximum degree ✩ Haizhou Song ∗ , Qiufen Wang, Lulu Tian College of Mathematical Sciences, Huaqiao University, Quanzhou, Fujian 362021, PR China
a r t i c l e
i n f o
Article history: Received 9 May 2012 Accepted 9 July 2013 Available online 19 August 2013 Submitted by B.L. Shader MSC: 05C05 15A48 Keywords: Tree Adjacency matrix Spectral radius Maximum degree Upper bound
a b s t r a c t This paper studies the problem of estimating the spectral radius of trees with the given number of vertices and maximum degree. We obtain the new upper bounds on the spectral radius of the trees, and the results are the best upper bounds expressed by the number of vertices and maximum degree, at present. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality 2 holds if and only if T is an almost completely full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; (3) if n > 2 + 1, then
√
ρ ( T ) < 2 − 1 cos 2kπ+1 , where k =
log−1 ( (−2)(n−1) + 1) + 1.
© 2013 Elsevier Inc. All rights reserved.
1. Introduction Let T = ( V , E ) be a tree on n vertices with maximum degree . Its adjacency matrix is defined to be the n × n matrix A ( T ) = (ai j ), where ai j = 1 if v i is adjacent to v j ; and ai j = 0, otherwise. The ✩ This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural Science Foundation of Fujian Province (Z0511028). Corresponding author. E-mail address: [email protected] (H. Song).
*
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.07.026
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characteristic polynomial of T is det(xI − A ( T )) and is denoted by Φ( T , x). Since A ( T ) is a symmetric matrix, each of its eigenvalues is real. We assume, without loss of generality, that they are ordered in nonincreasing order, i.e., ρ1 ( T ) ρ2 ( T ) · · · ρn ( T ). We call them the eigenvalues of T . In particular, the largest eigenvalue ρ1 ( T ) is called the spectral radius of T , denoted by ρ ( T ). Since T is a connected graph, A ( T ) is irreducible, the spectral radius is simple and there is a unique positive unit eigenvector by the Perron–Frobenius Theorem, e.g. [1]. We shall refer to such an eigenvector as the Perron vector of T . Let d v be the degree of v ( v ∈ V ), = max{d v : v ∈ V }. Let N T ( v ) denote the set of vertices adjacent to v in T . A pendant vertex of T is a vertex of degree 1. Denote by ρ ( A ) the largest eigenvalue of the matrix A. The distance dist( v , u ) from a vertex v to a vertex u is the length of the shortest path joining v and u. We recall that the eccentricity e u of a vertex u is the largest distance from u to any other vertex of the graph. We recall some results on the upper bounds of the spectral radius of trees. In [2] Stevanovic´ proves that
√
ρ (T ) < 2 − 1
(1.1)
where T is a tree with maximum degree . In [3] Oscar Rojo gives an improved upper bound on the spectral radius of a tree. Let T be a tree with maximum degree , u be a vertex of T such that du = . Let k = e u + 1, where e u is the eccentricity of u. For j = 1, 2, . . . , k − 1, let δ j = max{d v : dist( v , u ) = j }. Then
ρ ( T ) < max
max { δ j − 1 +
2 j k−2
√ δ j −1 − 1 }, δ1 − 1 + .
(1.2)
In [4] Oscar Rojo gives another upper bound on the spectral radius of a tree. Let T be a tree with maximum degree and such that there exist two adjacent vertices u and v with du = d v = . Let T be the forest obtained from T by deleting the edge uv. Thus T is the union of two disjoint trees T u = ( V u , E u ) and T v = ( V v , E v ). Let ku = e u + 1, k v = e v + 1, where e u and e v are the eccentricities of u and v with respect to the trees T u and T v respectively. Now, we define k = max{ku , k v }, γ j (u ) = max{dx : x ∈ V u , dist(x, u ) = j }, 1 j k − 2, γ j ( v ) = max{d y : y ∈ V v , dist( y , v ) = j }, 1 j k − 2, and γ j = max{γ j (u ), γ j ( v )}, 1 j k − 2, where γ j (u ) = 0 for j > ku − 1 or γ j ( v ) = 0 for j > k v − 1. Then
ρ ( T ) < max
max {
2 j k−2
√
γ j − 1 + γ j−1 − 1 }, γ1 − 1 + − 1 .
(1.3)
This paper studies the problem of estimating the spectral radius of trees with the given number of vertices and maximum degree. When = n − 1 or = 2, the tree T is a star or a path, respectively, and it is easy to obtain the spectral radius of the trees. Thus in this paper we only study for 3 n − 2. We obtain the new upper bounds on the spectral radius of trees, as the following two theorems. The results are the best upper bounds expressed by the number of vertices and maximum degree, at present. Theorem 3.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; √ (−2)(n−1) (3) if n > 2 + 1, then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( + 1) + 1. Theorem 3.2. Let T = ( V , E ) be a tree on n vertices with maximum degree , where = 3 and n + 2. Let t be the cardinality of the vertices of degree 3. We prove that
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Fig. 1. Completely full-degree tree.
Fig. 2. Almost completely full-degree tree.
√
(1) if t = 1, then ρ ( T ) < 4.5; √ (2) if t = 2, then ρ ( T ) < 5; (3) if t = 3, then ρ ( T ) <
√
4 3 ; 3
√
(4) if t = 4, then ρ ( T ) < 2 + 13; √ 2 (5) if t 5, then ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ + 1. 3 2. Some basic concepts and lemmas First, we introduce some needed definitions and lemmas for getting the main results of this article. Definition 2.1. Let T be a rooted tree on n vertices with maximum degree , r be the root node of T . If u is a vertex of T and dist(r , u ) = k − 1 (1 k), then we call u is in the kth level, note that the root vertex is in the first level. If v is a vertex of T and dist(r , v ) = maxu ∈ V ( T ) dist(r , u ), then we call v is in the last level. Definition 2.2. Let T be a rooted tree on n vertices with maximum degree . If in the tree T , the degrees of all the nodes in each level except for the last level are . Then we call T is a completely full-degree tree. Definition 2.3. Let T be a rooted tree on n vertices with maximum degree . If T is a completely full-degree tree; or if in the tree T , the number of nodes in the last level has not reached the maximal value but only lacks some right nodes (nodes on the right-hand side), and the number of nodes in each level except for the last level has reached the maximal value. Then we call T is an almost completely full-degree tree. For example, Fig. 1 shows a completely full-degree tree with level 3 and degree 4, and Fig. 2 shows an almost completely full-degree tree with level 4 and degree 3. Lemma 2.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , and let T ∗ be an almost completely full-degree tree on n vertices with maximum degree corresponding to T . Then ρ ( T ) ρ ( T ∗ ), and equality holds if and only if T = T ∗ . Proof. The proof is similar to Theorem 2.1 in [5].
2
Lemma 2.2. Let T ∗ = ( V ∗ , E ∗ ) be an almost completely full-degree tree on n vertices with maximum degree . Let T ∗∗ be a tree defined as follows:
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(1) if T ∗ is a completely full-degree tree, then T ∗ = T ∗∗ ; (2) if T ∗ is not completely full-degree tree, then T ∗∗ is the completely full-degree tree obtained from T ∗ by supplying all the lacking vertices. Then ρ ( T ∗ ) ρ ( T ∗∗ ).
2
Proof. The proof is trivial.
Lemma 2.3. (See [6].) Let T = ( V , E ) be a completely full-degree tree on n vertices with maximum degree , and whose number of levels is k. Let dk− j +1 be the degree of vertices in the level j (1 j k), and ρ ( T ) be the spectral radius of T . If R k is a symmetric tridiagonal matrix of order k × k, where
⎡
√
d2 − 1
0
··· .. . .. . .. .
0
···
0
⎤
.. ⎥ √ ⎢√ ⎢ d2 − 1 0 d3 − 1 . ⎥ ⎢ ⎥ .. ⎥ ⎢ √ .. .. ⎢ ⎥ . . 0 d − 1 . 3 ⎥ , Rk = ⎢ ⎢ ⎥ .. . . . . ⎢ . . dk−1 − 1 0 ⎥ . ⎢ ⎥ ⎢ .. √ ⎥ .. ⎣ . . dk−1 − 1 dk ⎦ √0 0 ··· ··· 0 dk 0 k×k then ρ ( R k ) = ρ ( T ). Lemma 2.4. (See [7].) If b > 0, then the spectral radius of the m × m symmetric tridiagonal matrix
⎤
⎡
0 b ⎢b 0
⎢ ⎢ ⎢ D m (b) = ⎢ ⎢ ⎢ ⎣
b
⎥ ⎥ ⎥ ⎥ ⎥ .. ⎥ . b ⎥ b 0 b⎦ ..
b
..
.
.
b is ρ ( D m (b)) = 2b cos
b
m×m
π
. 2m+1
Lemma 2.5. (See [1].) Let A and B be both the nonnegative irreducible real symmetric matrices, and A B, A = B. Then ρ ( A ) < ρ ( B ).
/ V . Denote by ρ ( T ) the spectral radius Lemma 2.6. (See [8].) Let T = ( V , E ) be a tree, u ∈ V , du = 1, and v ∈ of T . If T 1 = T + uv, then ρ ( T ) < ρ ( T 1 ). Lemma 2.7. (See [9].) Let T = ( V , E ) be a tree, u ∈ V , v ∈ V , du 3 and d v 3. Let w 0 w 1 w 2 · · · w k+1 be the only path from u to v, where w 0 = u , w k+1 = v , k 1, and d w i = 2 (1 i k). If T 1 = T − w i −1 w i − w i w i +1 + w i −1 w i +1 (1 i k), then ρ ( T ) ρ ( T 1 ). Lemma 2.8. (See [10,11].) Suppose that u is a vertex of a tree T . For positive integers k and l, let T k,l be the tree obtained from T by adding two new paths P k and P l of lengths k and l at u. If k l 1, then ρ ( T k+1,l−1 ) < ρ ( T k,l ). Lemma 2.9. (See [10,12].) Suppose that u and v are two adjacent vertices of the tree T , d(u ) > 1, d( v ) > 1. For nonnegative integers k and l, let T k1,l be the tree obtained from T by adding two new paths P k and P l of
lengths k and l at u and v, respectively. If k l 1, then ρ ( T k1+1,l−1 ) < ρ ( T k1,l ).
Lemma 2.10. (See [9,13].) Suppose that T ∗ is a tree and b is a pendant vertex of T ∗ . Let a be the vertex which is adjacent to b, that is, N T ∗ (b) = a. Let Φ( T ∗ , x), Φ( T ∗ − b, x), and Φ( T ∗ − b − a, x) be the characteristic polynomials of T ∗ , T ∗ − b, and T ∗ − b − a, respectively, then Φ( T ∗ , x) = xΦ( T ∗ − b, x) − Φ( T ∗ − b − a, x).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 3. A tree T 1 .
Fig. 4. The star K 1,s .
3. Main results and proofs For the tree neither a path nor a star, we obtain a new upper bound on the spectral radius of the tree. As the following theorem. Theorem 3.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; √ (−2)(n−1) (3) if n > 2 + 1, then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( + 1) + 1. Proof. (1) If + 1 < n 2. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 1 ), where T 1 is an almost completely full-degree tree on n vertices with maximum degree (Fig. 3). Let r be the root node, then dr = . Let b1 , b2 , . . . , bd be the pendent vertices of T 1 , and b i ∈ / N T 1 (r ) (1 i d). Then d + + 1 = n. Denote by Φ( K 1,s , x) the characteristic polynomial of the star K 1,s (Fig. 4). By Lemma 2.10, we have
Φ( K 1,s , x) = xΦ( K 1,s−1 , x) − xs−1
= x xΦ( K 1,s−2 , x) − xs−2 − xs−1 = ···
= xs−1 x2 − s . Then, by applying Lemma 2.10 repeatedly, we have
Φ( T 1 , x) = xΦ( T 1 − bd , x) − Φ( T 1 − bd − a, x) = xΦ( T 1 − bd , x) − xd−1 Φ( K 1,−1 , x)
= xΦ( T 1 − bd , x) − xd−1 x−2 x2 − ( − 1)
= x xΦ( T 1 − bd − bd−1 , x) − Φ( T 1 − bd − bd−1 − a, x) − xd+−3 x2 − ( − 1)
= x xΦ( T 1 − bd − bd−1 , x) − xd+−4 x2 − ( − 1) − xd+−3 x2 − ( − 1)
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= x2 Φ( T 1 − bd − bd−1 , x) − 2xd+−3 x2 − ( − 1) = ···
= xd−1 Φ( T 1 − bd − bd−1 − · · · − b2 , x) − (d − 1)xd+−3 x2 − ( − 1)
= xd−1 xΦ( K 1, , x) − Φ( K 1,−1 , x) − (d − 1)xd+−3 x2 − ( − 1)
= xd x−1 x2 − − xd−1 x−2 x2 − ( − 1) − (d − 1)xd+−3 x2 − ( − 1)
= xd+−1 x2 − − dxd+−3 x2 − ( − 1) . Let Φ( T 1 , x) = 0. Then xd+−3 [x4 − (d + )x2 + d( − 1)] = 0. It is easy to prove that
ρ ( T ) > 1, so ρ ( T ) is the largest root of the equation x4 − (d + )x2 + d( − 1) = 0. Thus, x2 =
(d+)+ (d+)2 −4d(−1) 2
.
Since d = n − 1 − , we have x2 =
n−1+ (n−2)2 +2n−3 . 2
So
ρ (T 1 ) =
n−1+ (n−2)2 +2n−3 . 2
n−1+ (n−2)2 +2n−3
Therefore, ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels. (2) If 2 < n 2 + 1. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 2 ), where T 2 is an almost completely full-degree tree on n vertices with maximum degree . Let T 3 be a completely full-degree tree corresponding to T 2 with maximum degree and such that the number of levels in T 3 is 3 (may be T 2 = T 3 ). Then by Lemma 2.1 and Lemma 2.2, we have ρ ( T ) ρ ( T 2 ) ρ ( T 3 ). By Lemma 2.3, we have ρ ( R 3 ) = ρ ( T 3 ), where
⎡ R3 = ⎣
√
√ −1
0
−1 0
√
0
√
0
√
⎤
⎦.
0
√
√
Then ρ ( R 3 ) = 2 − √ 1, that is, ρ ( T 3 ) = 2 − 1. Thus, ρ ( T ) 2 − 1. Therefore, ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels. (3) If n > 2 + 1. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 4 ), where T 4 is an almost completely full-degree tree on n vertices with maximum degree , and the number of levels in T 4 is k. Let T 5 be a completely full-degree tree corresponding to T 4 with maximum degree and the level of T 5 is also k (may be T 4 = T 5 ). By Lemma 2.2, we have ρ ( T 4 ) ρ ( T 5 ). If T 4 is a completely full-degree tree, then
n = 1 + + ( − 1) + ( − 1)2 + · · · + ( − 1)k−2 . So we have
k = log−1
( − 2)(n − 1) + 1 + 1.
If T 4 is not a completely full-degree tree, then
1 + 1 + ( − 1) + · · · + ( − 1)k−3 < n < 1 + 1 + ( − 1) + · · · + ( − 1)k−2 . So we have
log−1
( − 2)(n − 1) ( − 2)(n − 1) + 1 + 1 < k < log−1 + 1 + 2.
Therefore, the level of T 4 is k = log−1 (
(−2)(n−1)
+ 1) + 1.
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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ρ ( B k ) = ρ ( T 5 ), where √ ⎤ 0 −1 0 ··· ··· 0 . √ √ . ⎢ .. .. ⎥ ⎥ ⎢ −1 0 −1 ⎢ .. ⎥ √ . . . ⎥ ⎢ . . . . . . 0 −1 . ⎥ ⎢ Bk = ⎢ ⎥ . . √ .. .. .. ⎥ ⎢ .. . . . −1 0 ⎥ ⎢ ⎥ ⎢ . √ √ .. .. ⎣ . −1 ⎦ √0 0 ··· ··· 0 0 k×k
By Lemma 2.3, we have
⎡
Let
⎡
√ −1
0
··· .. . .. . .. .
···
0
⎤
.. ⎥ √ ⎢√ ⎢ −1 −1 . ⎥ ⎢ .. ⎥ √ .. .. ⎥ ⎢ . . 0 −1 . ⎥ ⎢ Fk = ⎢ ⎥ , . √ . . ⎥ ⎢ .. .. .. −1 0 ⎥ ⎢ ⎥ ⎢ . √ √ . .. .. ⎣ −1 ⎦ √ 0 ··· ··· 0 k×k then F k = B k + diag{, , . . . , }. So ρ ( F k ) = ρ ( B k ) + . However, the matrix F k has the LL T -decomposition F k = LL T , where
⎡√ −1 1 √ ⎢ ⎢ 0 −1 ⎢ ⎢ ⎢ 0 0 L=⎢ .. .. ⎢ . ⎢ . ⎢ .. ⎣ . 0 ···
⎤ ··· ··· 0 .. ⎥ .. . 1 . ⎥ .. ⎥ .. .. .. ⎥ . . . . ⎥ ⎥ . .. .. ⎥ . . 1 0 ⎥ ⎥ √ .. . 0 − 1 √1 ⎦ ··· 0 0 k×k 0
Let A k = L T L, then
⎤ ⎡ − 1 √ − 1 0 ··· ··· 0 . √ √ . ⎥ ⎢ .. .. −1 ⎥ ⎢ −1 ⎥ ⎢ . √ .. .. .. ⎥ ⎢ . . . . 0 −1 . ⎥ ⎢ Ak = ⎢ ⎥ . .. √ .. .. .. ⎥ ⎢ . . . ⎥ ⎢ −1 0 . ⎥ ⎢ .. √ √ .. ⎦ ⎣ . . −1 √ −1 0 ··· ··· 0 − 1 + 1 k×k
Since L is invertible, we have Let
⎡
ρ ( LL T ) = ρ ( L T L ). So ρ ( F k ) = ρ ( Ak ).
√ −1
0
··· .. . .. . .. .
···
0
⎤
.. √ ⎥ ⎢√ −1 . ⎥ ⎢ −1 ⎥ ⎢ . √ . . ⎥ ⎢ . . . . . 0 −1 . ⎥ ⎢ Dk = ⎢ ⎥ , . √ .. .. ⎥ ⎢ . . . ⎥ ⎢ −1 0 . ⎥ ⎢ .. √ √ .. ⎦ ⎣ . . −1 √ − 1 √ 0 ··· ··· 0 − 1 + − 1 k×k then by Lemma 2.5, we have
ρ ( A k ) < ρ ( D k ).
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 5. A tree for t = 1.
However, D k = diag{, , . . . , } + H k , where
⎡
√ −1
⎤ 0 ··· 0 . √ √ . ⎢ ⎥ .. .. 0 −1 ⎢ −1 ⎥ ⎢ ⎥ √ . . ⎢ ⎥ . . . Hk = ⎢ . . 0 −1 0 ⎥ ⎢ ⎥ .. √ .. .. ⎣ ⎦ . . . 0 − 1 √ √ 0 ··· 0 −1 − 1 k×k 0
Then by Lemma 2.4, we have So
√
√
ρ ( H k ) = 2 − 1 cos 2kπ+1 . √
ρ ( Ak ) < ρ ( D k ) = + 2 − 1 cos 2kπ+1 . Thus, ρ ( B k ) = ρ ( F k ) − = ρ ( Ak ) − <
2 − 1 cos 2kπ+1 . Therefore,
√
ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( (−2)(n−1) + 1) + 1. 2
Corollary 3.1. √ Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3. If n + 1, (−2)(n−1) + 1) + 1. then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( Proof. The proof is similar to (3) in Theorem 3.1.
2
In the following, we give the upper bounds on the spectral radius of trees when = 3. Since it is easy to get the upper bounds on the spectral radius of trees when = 3 and n is small, we only study the upper bounds on the spectral radius of trees for = 3 and n is large. Theorem 3.2. Let T = ( V , E ) be a tree on n vertices with maximum degree , where = 3 and n + 2. Let t be the cardinality of the vertices of degree 3. Following results are established.
√
(1) if t = 1, then ρ ( T ) < √4.5; (2) if t = 2, then ρ ( T ) < √5; (3) if t = 3, then ρ ( T ) <
4 3
;
3 √ (4) if t = 4, then ρ ( T ) < √ 2 + 13; (5) if t 5, then ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2
n+2 + 1. 3
Proof. Let T n3 be the set of the trees on n vertices with maximum degree 3. By Corollary 3.1, we √ 2 + 1. Obviously, Theorem 3.2 holds obtain that for ∀ T ∈ T n3 , ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ 3 when ρ ( T ) 2. In the following, we assume that ρ ( T ) > 2. (1) If t = 1, then by Lemma 2.6, Lemma 2.7 and Lemma 2.8, for ∀ T ∈ T n3 , there is a nature number M 1 such that ρ ( T ) ρ ( T m ) for m M 1 and T m is a tree as shown in Fig. 5. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 1 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α1 , then ρm < α1 . Let x be the Perron vector of T m , and x v i be a componentof x corresponds to the vertex v i (1 i n). Then A ( T m )x = ρ ( T m )x. Therefore, we have ρm x v i = v j ∈ N T ( v i ) x v j . m
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 6. A tree for t = 2.
It is easy to prove that xai = xbi = xc i (i = 1, 2, . . . , m). Let xi = xai (i = 1, 2, . . . , m), xm+1 = xr . Then
ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are ρm + ρm2 − 4 ρm − ρm2 − 4 λ1 (m) = , λ2 (m) = . 2
2
So the general solution is
xk = C 1 λk1 (m) + C 2 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 1 , C 2 are constants, λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m). Since ρm xm+1 = 3xm , we have ρm = x3xm . Then m+1
α1 = lim
m→+∞
3xm xm+1
For limm→+∞ λ2 (m) =
√
m C 1 λm 1 (m) + C 2 λ2 (m)
= 3 lim
m→+∞
α1 − α12 −4 2
+1 +1 C 1 λm (m) + C 2 λm (m) 1 2
, we have
α1 = 3
α1 − α12 −4 2
= 3 lim λ2 (m).
. Then
m→+∞
√
α1 = 4.5. So for ∀ T ∈ T n3 , ρ ( T )
ρ ( T m ) < 4.5.
√ √ However, for t = 1 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ √ ously, 2 2 cos 2kπ+1 > 4.5.
√
Therefore, if t = 1, then ρ ( T ) < 4.5. (2) If t = 2, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 2 such that ρ ( T ) ρ ( T m ) for m M 2 and T m is a tree as shown in Fig. 6. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 2 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let lim m→+∞ ρm = α2 , then ρm < α2 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi m
(i = 1, 2, . . . , m), and xs = xt . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are
λ1 (m) =
ρm + ρm2 − 4 2
,
λ2 (m) =
ρm − ρm2 − 4 2
.
So the general solution is
xk = C 1 λk1 (m) + C 2 λk2 (m) where C 1 , C 2 are constants,
(k = 1, 2, . . . , m + 1),
λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m).
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 7. A tree for t = 3.
Since
ρm xm+1 = 2xm + xm+1 , we have ρm =
α2 = lim
m→+∞
2xm xm+1
For limm→+∞ λ2 (m) =
√
+ 1 = 2 lim
m→+∞
α2 − α22 −4 2
ρ ( T ) ρ ( T m ) < 5.
2xm xm+1
+ 1. Then
m C 1 λm 1 (m) + C 2 λ2 (m)
+1 +1 C 1 λm (m) + C 2 λm (m) 1 2
, we have
α2 = 2
α2 − α22 −4
+ 1 = 2 lim λ2 (m) + 1. m→+∞
+ 1. Then α2 =
2
√
5. So for ∀ T ∈ T n3 ,
√ √ However, for t = 2 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ √ ously, 2 2 cos 2kπ+1 > 5.
√
Therefore, if t = 2, then ρ ( T ) < 5. (3) If t = 3, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 3 such that ρ ( T ) ρ ( T m ) for m M 3 and T m is a tree as shown in Fig. 7. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 3 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let lim m→+∞ ρm = α3 , then ρm < α3 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi m
(i = 1, 2, . . . , m), and xs = xt . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are
λ1 (m) =
ρm + ρm2 − 4 2
λ2 (m) =
,
ρm − ρm2 − 4 2
.
So the general solution is
xk = C 3 λk1 (m) + C 4 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 3 , C 4 are constants, λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m). Let y i = xe i (i = 1, 2, . . . , m), ym+1 = xr , y 0 = 0. Then ρm yk = yk+1 + yk−1 , k = 1, 2, . . . , m. So the general solution is
yk = C 3 λk1 (m) + C 4 λk2 (m) (k = 1, 2, . . . , m + 1), where C 3 , C 4 are constants; λ1 (m), λ2 (m) are the same as the above. For ρm xm+1 = 2xm + ym+1 and ρm ym+1 = 2xm+1 + ym , we have
⎧ 2xm y m +1 ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
2xm+1 ym ⎪ ⎪ ⎩ ρm = + . y m +1
y m +1
It has been proved that limm→+∞ ρm exists, and limm→+∞ α3 − α32 −4 2
exists. So limm→+∞
ym+1 xm+1
exists. Let limm→+∞
xm xm+1
ym+1 xm+1
= limm→+∞
= β1 , then
ym ym+1
= limm→+∞ λ2 (m)=
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 8. Four 3-degree vertices.
Fig. 9. A tree for t = 4.
⎧ ⎪ ⎪ α − α32 − 4 3 ⎪ ⎪ ⎨ α3 = 2 + β1 , 2 ⎪ α3 − α32 − 4 ⎪ 2 ⎪ ⎪ ⎩ α3 = + . β1 2 So
α3 =
√
4 3 . 3
Then for ∀ T ∈ T n3 ,
ρ (T ) ρ (T m ) <
√
4 3 . 3
√ √ However, for t = 3 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ ously, 2 2 cos 2kπ+1 >
√
4 3 . 3
√
Therefore, if t = 3, then ρ ( T ) < 4 3 3 . (4) If t = 4, then according to the connected relationships of the four 3-degree vertices, we discuss the following two cases. Case 1. The connected relationships of the four 3-degree vertices as shown in Fig. 8. By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 4 such that ρ ( T ) ρ ( T m ) for m M 4 and T m is a tree as shown in Fig. 9. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 4 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α4 , then ρm < α4 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi = m
xe i = x f i (i = 1, 2, . . . , m), xs = xt = xr . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Then
xk = C 5 λk1 (m) + C 6 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 5 , C 6 are constants; λ1 (m), λ2 (m) are the same as the above.
ρm xk = xk+1 +
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 10. Four 3-degree vertices.
Fig. 11. A tree for t = 4.
For
ρm xm+1 = 2xm + x w and ρm x w = 3xm+1 , then ⎧ 2xm xw ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
3x ⎪ ⎪ ⎩ ρm = m+1 . xw
It has been proved that limm→+∞ ρm exists, and limm→+∞ α4 − α42 −4 2
⎧ ⎪ ⎪ ⎪ ⎨
exists. So limm→+∞
exists. Let limm→+∞
xw xm+1
= limm→+∞
ym ym+1
= limm→+∞ λ2 (m)=
= β2 , then
α4 = 2
α4 − α42 − 4
⎪ 3 ⎪ ⎪ ⎩ α4 = . β2 So
xw xm+1
xm xm+1
√
2
+ β2 ,
√
α4 = 2 + 13. Then for ∀ T ∈ T n3 , ρ ( T ) < 2 + 13.
Case 2. The connected relationships of the four 3-degree vertices as shown in Fig. 10. By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 5 such that ρ ( T ) ρ ( T m ) for m M 5 and T m is a tree as shown in Fig. 11. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 5 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α5 , then ρm < α5 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi , m
xe i = x f i (i = 1, 2, . . . , m), xs = xr , xu = x v . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xr . Then xk+1 + xk−1 , k = 2, 3, . . . , m.
ρm xk =
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Let x0 = 0, then we have
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ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Then
xk = C 5 λk1 (m) + C 6 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 5 , C 6 are constants; λ1 (m), λ2 (m) are the same as the above. Let y i = xe i (i = 1, 2, . . . , m), ym+1 = xu , y 0 = 0. Then ρm yk = yk+1 + yk−1 , k = 1, 2, . . . , m. So the general solution is
yk = C 7 λk1 (m) + C 8 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 7 , C 8 are constants; λ1 (m), λ2 (m) are the same as the above. For ρm xm+1 = 2xm + ym+1 and ρm ym+1 = ym+1 + xm+1 + ym , then
⎧ 2xm y m +1 ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
y m +1
y m +1
xm+1 ym ⎪ ⎪ + + 1. ⎩ ρm = It has been proved that limm→+∞ ρm exists, and limm→+∞ α5 − α52 −4 2
exists. So limm→+∞
⎧ ⎪ α5 − ⎪ ⎪ ⎪ ⎨ α5 = 2
α52 − 4
2
ym+1 xm+1
exists. Let limm→+∞
xm xm+1
y m +1 xm+1
= limm→+∞
ym y m +1
= limm→+∞ λ2 (m) =
= β3 , then
+ β3 ,
⎪ ⎪ α5 − α52 − 4 ⎪ ⎪ ⎩ α5 = 1 + + 1. β3 2 So
α5 ≈2.3569. Then for ∀ T ∈ T n3 , ρ ( T ) < 2.357.
√
√
For 2 + 13 > 2.357, by the two cases, we know if t = 4, then for ∀ T ∈ T n3 , ρ ( T ) < 2 + 13. √ √ However, for t = 4 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ ously, 2 2 cos 2kπ+1 >
2+
√
13.
√
Therefore, if t = 4, then ρ ( T ) < 2 + 13. √ 2 (5) If t 5, then by Corollary 3.1, we have ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ + 1. 3
2
It is easy to prove that, when = 3, the upper bounds in Theorem 3.2 are better than the upper bounds in Theorem 3.1, respectively. 4. The comparison of the results In the following, we compare our upper bounds on the spectral radius of trees in Theorem 3.1 with the previous results (1.1), (1.2), and (1.3). 4.1. Comparing with (1.1) We know that (1.1) gives a result
√
ρ ( T ) < 2 − 1.
(1) When + 1 < n 2. In Theorem 3.1, we get Since 3, we have 92 − 28 + 13 = 9( −
ρ (T ) 14 2 ) 9
−
n−1+ (n−2)2 +2n−3 . 2 79 > 0. Then we get 9
52 + 12n > 0. Thus, 602 − 112 +52 − 12n + 12n > 0. So we can get 8( − 1) − (n − 1). Therefore,
n−1+ (n−2)2 +2n−3 2
√
362 − 112 +
(n − 2)2 + 2n − 3 <
< 2 − 1. Thus, our result is better than (1.1). √
(2) When 2 < n 2 √ + 1. In Theorem √ 3.1, we get ρ ( T ) 2 − 1. It is easy to prove that 2 − 1 < 2 − 1 for 3. Thus, our result is better than (1.1).
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 12. A tree with 7 vertices.
Fig. 13. A tree with 6 vertices.
(3) When n > 2 + 1. In Theorem 3.1, we get
√
√
√
ρ ( T ) < 2 − 1 cos 2kπ+1 (k 4).
Because 2 − 1 cos 2kπ+1 < 2 − 1, our result is better than (1.1). Therefore, our upper bounds in Theorem 3.1 are better than (1.1). 4.2. Comparing with (1.2) We know that (1.2) gives a result
ρ ( T ) < max
max { δ j − 1 +
2 j k−2
√ δ j −1 − 1 }, δ1 − 1 + .
(1) When + 1 < n 2. Let T 1 be an almost completely full-degree tree as shown in Fig. 12,
then we can get our upper bound is
ρ (T 1 ) =
However, the upperbound in (1.2) is 2 +
√
√
√
7−1+ (7−8)2 +14−3 2
=
3+
√
3.
2.
Because 2 + 2 > 3 + 3, our result is better than (1.2). (2) When 2 < n 2 + 1. Let T be a tree with maximum degree , and , v be two adjacent √ let u√ vertices √ √ of T such√that du =√d v = . Then the upper bound in (1.2) is + − 1. Obviously, + − 1 > 2 − 1 > 2 − 1. Thus, our result is better than (1.2). (3) When n > 2 + 1. Let T be a tree with maximum degree , and√let u ,√ v be two adjacent = d = . Then the upper bound in (1.2) is + − 1. Obviously, vertices of T such that d u v √ √ √ √ + − 1 > 2 − 1 > 2 − 1 cos 2kπ+1 . Thus, our result is better than (1.2). Therefore, our upper bounds in Theorem 3.1 are better than (1.2) in some cases. 4.3. Comparing with (1.3) We know that (1.3) gives a result
ρ ( T ) < max
max {
2 j k−2
√
γ j − 1 + γ j−1 − 1 }, γ1 − 1 + − 1 .
(1) When + 1 < n 2. Let T 2 be an almost completely full-degree tree as shown in Fig. 13, √ 6−1+ 12−3 then we can get our upper bound is ρ ( T 2 ) = = 2. 2
√
However, √ the upper bound in (1.3) is 2 2. Because 2 2 > 2, our result is better than (1.3). (2) When 2 < n 2 + 1. Let T be a tree with maximum degree , and let u , v , w be the vertices u = d v = d√ w = and uv ∈ E ( T ), u w ∈ E ( T ). Then the upper bound in (1.3) √ of T such that d√ is 2 − 1. Obviously, 2 − 1 > 2 − 1. Thus, our result is better than (1.3).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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(3) When n > 2 + 1. Let T be a tree with maximum degree , and let u , v , w be the vertices of √ T such that √ du = d v = d √ w = and uv ∈ E ( T ), u w ∈ E ( T ). Then the upper bound in (1.3) is 2 − 1. Obviously, 2 − 1 > 2 − 1 cos 2kπ+1 . Thus, our result is better than (1.3). Therefore, our upper bounds in Theorem 3.1 are better than (1.3) in some cases. In conclusion, comparing our results with the previous results, we prove that our upper bounds are better than (1.1). For (1.2) and (1.3), in order to get the upper bound on the spectral radius of a tree T , we require to know the degrees of all the vertices of T . But in Theorem 3.1, we only require to know the number of vertices and maximum degree, and we prove that our upper bounds are better than (1.2) and (1.3) in some cases. Acknowledgements The authors would like to thank the anonymous referees for valuable suggestions and corrections, which have improved the original manuscript. References [1] A. Berman, R.J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New York, 1979. ´ Bounding the largest eigenvalue of trees in terms of the largest vertex degree, Linear Algebra Appl. 360 [2] D. Stevanovic, (2003) 35–42. [3] R. Oscar, Improved bounds for the largest eigenvalue of trees, Linear Algebra Appl. 404 (2005) 297–304. [4] R. Oscar, The spectra of some trees and bounds for the largest eigenvalue of any tree, Linear Algebra Appl. 414 (2006) 199–217. [5] A.M. Yu, L. Mei, Laplacian spectral radius of trees with given maximum degree, Linear Algebra Appl. 429 (2008) 1962–1969. [6] R. Oscar, S. Ricardo, The spectra of the adjacency matrix and Laplacian matrix for some balanced trees, Linear Algebra Appl. 403 (2005) 97–117. [7] S. Kouachi, Eigenvalues and eigenvectors of tridiagonal matrices, Electron. J. Linear Algebra 15 (2006) 115–133. [8] K.F. Fang, On the relationship between edge moving transformation and spectral radii in graphs, J. Huzhou Teach. Coll. 29 (2) (2007) 10–11. [9] B.F. Wu, X.Y. Yuan, E.L. Xiao, On the spectral radii of trees, J. East China Norm. Univ. Natur. Sci. 3 (2004) 22–27. [10] Q. Li, K.Q. Feng, On the largest eigenvalues of graphs, Acta Math. Appl. Sin. 2 (1979) 167–175 (in Chinese). ´ P. Rowlinson, S. Simic, Eigenspaces of Graphs, Cambridge University Press, Cambridge, 1997. [11] D. Cvetkovic, [12] J.M. Guo, J.Y. Shao, On the spectral radius of trees with fixed diameter, Linear Algebra Appl. 413 (1) (2006) 131–147. ´ M. Doob, H. Sachs, Spectra of Graphs-Theory and Applications, third ed., Johann Ambrosius Barth Verlag, [13] D. Cvetkovic, 1995.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
New upper bounds on the spectral radius of trees with the given number of vertices and maximum degree ✩ Haizhou Song ∗ , Qiufen Wang, Lulu Tian College of Mathematical Sciences, Huaqiao University, Quanzhou, Fujian 362021, PR China
a r t i c l e
i n f o
Article history: Received 9 May 2012 Accepted 9 July 2013 Available online 19 August 2013 Submitted by B.L. Shader MSC: 05C05 15A48 Keywords: Tree Adjacency matrix Spectral radius Maximum degree Upper bound
a b s t r a c t This paper studies the problem of estimating the spectral radius of trees with the given number of vertices and maximum degree. We obtain the new upper bounds on the spectral radius of the trees, and the results are the best upper bounds expressed by the number of vertices and maximum degree, at present. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality 2 holds if and only if T is an almost completely full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; (3) if n > 2 + 1, then
√
ρ ( T ) < 2 − 1 cos 2kπ+1 , where k =
log−1 ( (−2)(n−1) + 1) + 1.
© 2013 Elsevier Inc. All rights reserved.
1. Introduction Let T = ( V , E ) be a tree on n vertices with maximum degree . Its adjacency matrix is defined to be the n × n matrix A ( T ) = (ai j ), where ai j = 1 if v i is adjacent to v j ; and ai j = 0, otherwise. The ✩ This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural Science Foundation of Fujian Province (Z0511028). Corresponding author. E-mail address: [email protected] (H. Song).
*
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.07.026
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
characteristic polynomial of T is det(xI − A ( T )) and is denoted by Φ( T , x). Since A ( T ) is a symmetric matrix, each of its eigenvalues is real. We assume, without loss of generality, that they are ordered in nonincreasing order, i.e., ρ1 ( T ) ρ2 ( T ) · · · ρn ( T ). We call them the eigenvalues of T . In particular, the largest eigenvalue ρ1 ( T ) is called the spectral radius of T , denoted by ρ ( T ). Since T is a connected graph, A ( T ) is irreducible, the spectral radius is simple and there is a unique positive unit eigenvector by the Perron–Frobenius Theorem, e.g. [1]. We shall refer to such an eigenvector as the Perron vector of T . Let d v be the degree of v ( v ∈ V ), = max{d v : v ∈ V }. Let N T ( v ) denote the set of vertices adjacent to v in T . A pendant vertex of T is a vertex of degree 1. Denote by ρ ( A ) the largest eigenvalue of the matrix A. The distance dist( v , u ) from a vertex v to a vertex u is the length of the shortest path joining v and u. We recall that the eccentricity e u of a vertex u is the largest distance from u to any other vertex of the graph. We recall some results on the upper bounds of the spectral radius of trees. In [2] Stevanovic´ proves that
√
ρ (T ) < 2 − 1
(1.1)
where T is a tree with maximum degree . In [3] Oscar Rojo gives an improved upper bound on the spectral radius of a tree. Let T be a tree with maximum degree , u be a vertex of T such that du = . Let k = e u + 1, where e u is the eccentricity of u. For j = 1, 2, . . . , k − 1, let δ j = max{d v : dist( v , u ) = j }. Then
ρ ( T ) < max
max { δ j − 1 +
2 j k−2
√ δ j −1 − 1 }, δ1 − 1 + .
(1.2)
In [4] Oscar Rojo gives another upper bound on the spectral radius of a tree. Let T be a tree with maximum degree and such that there exist two adjacent vertices u and v with du = d v = . Let T be the forest obtained from T by deleting the edge uv. Thus T is the union of two disjoint trees T u = ( V u , E u ) and T v = ( V v , E v ). Let ku = e u + 1, k v = e v + 1, where e u and e v are the eccentricities of u and v with respect to the trees T u and T v respectively. Now, we define k = max{ku , k v }, γ j (u ) = max{dx : x ∈ V u , dist(x, u ) = j }, 1 j k − 2, γ j ( v ) = max{d y : y ∈ V v , dist( y , v ) = j }, 1 j k − 2, and γ j = max{γ j (u ), γ j ( v )}, 1 j k − 2, where γ j (u ) = 0 for j > ku − 1 or γ j ( v ) = 0 for j > k v − 1. Then
ρ ( T ) < max
max {
2 j k−2
√
γ j − 1 + γ j−1 − 1 }, γ1 − 1 + − 1 .
(1.3)
This paper studies the problem of estimating the spectral radius of trees with the given number of vertices and maximum degree. When = n − 1 or = 2, the tree T is a star or a path, respectively, and it is easy to obtain the spectral radius of the trees. Thus in this paper we only study for 3 n − 2. We obtain the new upper bounds on the spectral radius of trees, as the following two theorems. The results are the best upper bounds expressed by the number of vertices and maximum degree, at present. Theorem 3.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; √ (−2)(n−1) (3) if n > 2 + 1, then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( + 1) + 1. Theorem 3.2. Let T = ( V , E ) be a tree on n vertices with maximum degree , where = 3 and n + 2. Let t be the cardinality of the vertices of degree 3. We prove that
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 1. Completely full-degree tree.
Fig. 2. Almost completely full-degree tree.
√
(1) if t = 1, then ρ ( T ) < 4.5; √ (2) if t = 2, then ρ ( T ) < 5; (3) if t = 3, then ρ ( T ) <
√
4 3 ; 3
√
(4) if t = 4, then ρ ( T ) < 2 + 13; √ 2 (5) if t 5, then ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ + 1. 3 2. Some basic concepts and lemmas First, we introduce some needed definitions and lemmas for getting the main results of this article. Definition 2.1. Let T be a rooted tree on n vertices with maximum degree , r be the root node of T . If u is a vertex of T and dist(r , u ) = k − 1 (1 k), then we call u is in the kth level, note that the root vertex is in the first level. If v is a vertex of T and dist(r , v ) = maxu ∈ V ( T ) dist(r , u ), then we call v is in the last level. Definition 2.2. Let T be a rooted tree on n vertices with maximum degree . If in the tree T , the degrees of all the nodes in each level except for the last level are . Then we call T is a completely full-degree tree. Definition 2.3. Let T be a rooted tree on n vertices with maximum degree . If T is a completely full-degree tree; or if in the tree T , the number of nodes in the last level has not reached the maximal value but only lacks some right nodes (nodes on the right-hand side), and the number of nodes in each level except for the last level has reached the maximal value. Then we call T is an almost completely full-degree tree. For example, Fig. 1 shows a completely full-degree tree with level 3 and degree 4, and Fig. 2 shows an almost completely full-degree tree with level 4 and degree 3. Lemma 2.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , and let T ∗ be an almost completely full-degree tree on n vertices with maximum degree corresponding to T . Then ρ ( T ) ρ ( T ∗ ), and equality holds if and only if T = T ∗ . Proof. The proof is similar to Theorem 2.1 in [5].
2
Lemma 2.2. Let T ∗ = ( V ∗ , E ∗ ) be an almost completely full-degree tree on n vertices with maximum degree . Let T ∗∗ be a tree defined as follows:
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(1) if T ∗ is a completely full-degree tree, then T ∗ = T ∗∗ ; (2) if T ∗ is not completely full-degree tree, then T ∗∗ is the completely full-degree tree obtained from T ∗ by supplying all the lacking vertices. Then ρ ( T ∗ ) ρ ( T ∗∗ ).
2
Proof. The proof is trivial.
Lemma 2.3. (See [6].) Let T = ( V , E ) be a completely full-degree tree on n vertices with maximum degree , and whose number of levels is k. Let dk− j +1 be the degree of vertices in the level j (1 j k), and ρ ( T ) be the spectral radius of T . If R k is a symmetric tridiagonal matrix of order k × k, where
⎡
√
d2 − 1
0
··· .. . .. . .. .
0
···
0
⎤
.. ⎥ √ ⎢√ ⎢ d2 − 1 0 d3 − 1 . ⎥ ⎢ ⎥ .. ⎥ ⎢ √ .. .. ⎢ ⎥ . . 0 d − 1 . 3 ⎥ , Rk = ⎢ ⎢ ⎥ .. . . . . ⎢ . . dk−1 − 1 0 ⎥ . ⎢ ⎥ ⎢ .. √ ⎥ .. ⎣ . . dk−1 − 1 dk ⎦ √0 0 ··· ··· 0 dk 0 k×k then ρ ( R k ) = ρ ( T ). Lemma 2.4. (See [7].) If b > 0, then the spectral radius of the m × m symmetric tridiagonal matrix
⎤
⎡
0 b ⎢b 0
⎢ ⎢ ⎢ D m (b) = ⎢ ⎢ ⎢ ⎣
b
⎥ ⎥ ⎥ ⎥ ⎥ .. ⎥ . b ⎥ b 0 b⎦ ..
b
..
.
.
b is ρ ( D m (b)) = 2b cos
b
m×m
π
. 2m+1
Lemma 2.5. (See [1].) Let A and B be both the nonnegative irreducible real symmetric matrices, and A B, A = B. Then ρ ( A ) < ρ ( B ).
/ V . Denote by ρ ( T ) the spectral radius Lemma 2.6. (See [8].) Let T = ( V , E ) be a tree, u ∈ V , du = 1, and v ∈ of T . If T 1 = T + uv, then ρ ( T ) < ρ ( T 1 ). Lemma 2.7. (See [9].) Let T = ( V , E ) be a tree, u ∈ V , v ∈ V , du 3 and d v 3. Let w 0 w 1 w 2 · · · w k+1 be the only path from u to v, where w 0 = u , w k+1 = v , k 1, and d w i = 2 (1 i k). If T 1 = T − w i −1 w i − w i w i +1 + w i −1 w i +1 (1 i k), then ρ ( T ) ρ ( T 1 ). Lemma 2.8. (See [10,11].) Suppose that u is a vertex of a tree T . For positive integers k and l, let T k,l be the tree obtained from T by adding two new paths P k and P l of lengths k and l at u. If k l 1, then ρ ( T k+1,l−1 ) < ρ ( T k,l ). Lemma 2.9. (See [10,12].) Suppose that u and v are two adjacent vertices of the tree T , d(u ) > 1, d( v ) > 1. For nonnegative integers k and l, let T k1,l be the tree obtained from T by adding two new paths P k and P l of
lengths k and l at u and v, respectively. If k l 1, then ρ ( T k1+1,l−1 ) < ρ ( T k1,l ).
Lemma 2.10. (See [9,13].) Suppose that T ∗ is a tree and b is a pendant vertex of T ∗ . Let a be the vertex which is adjacent to b, that is, N T ∗ (b) = a. Let Φ( T ∗ , x), Φ( T ∗ − b, x), and Φ( T ∗ − b − a, x) be the characteristic polynomials of T ∗ , T ∗ − b, and T ∗ − b − a, respectively, then Φ( T ∗ , x) = xΦ( T ∗ − b, x) − Φ( T ∗ − b − a, x).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 3. A tree T 1 .
Fig. 4. The star K 1,s .
3. Main results and proofs For the tree neither a path nor a star, we obtain a new upper bound on the spectral radius of the tree. As the following theorem. Theorem 3.1. Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3 n − 2. Denote by ρ ( T ) the spectral radius of T . We prove that
n−1+ (n−2)2 +2n−3
(1) if n 2, then ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels; √ (2) if 2 < n 2 + 1, then ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels; √ (−2)(n−1) (3) if n > 2 + 1, then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( + 1) + 1. Proof. (1) If + 1 < n 2. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 1 ), where T 1 is an almost completely full-degree tree on n vertices with maximum degree (Fig. 3). Let r be the root node, then dr = . Let b1 , b2 , . . . , bd be the pendent vertices of T 1 , and b i ∈ / N T 1 (r ) (1 i d). Then d + + 1 = n. Denote by Φ( K 1,s , x) the characteristic polynomial of the star K 1,s (Fig. 4). By Lemma 2.10, we have
Φ( K 1,s , x) = xΦ( K 1,s−1 , x) − xs−1
= x xΦ( K 1,s−2 , x) − xs−2 − xs−1 = ···
= xs−1 x2 − s . Then, by applying Lemma 2.10 repeatedly, we have
Φ( T 1 , x) = xΦ( T 1 − bd , x) − Φ( T 1 − bd − a, x) = xΦ( T 1 − bd , x) − xd−1 Φ( K 1,−1 , x)
= xΦ( T 1 − bd , x) − xd−1 x−2 x2 − ( − 1)
= x xΦ( T 1 − bd − bd−1 , x) − Φ( T 1 − bd − bd−1 − a, x) − xd+−3 x2 − ( − 1)
= x xΦ( T 1 − bd − bd−1 , x) − xd+−4 x2 − ( − 1) − xd+−3 x2 − ( − 1)
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= x2 Φ( T 1 − bd − bd−1 , x) − 2xd+−3 x2 − ( − 1) = ···
= xd−1 Φ( T 1 − bd − bd−1 − · · · − b2 , x) − (d − 1)xd+−3 x2 − ( − 1)
= xd−1 xΦ( K 1, , x) − Φ( K 1,−1 , x) − (d − 1)xd+−3 x2 − ( − 1)
= xd x−1 x2 − − xd−1 x−2 x2 − ( − 1) − (d − 1)xd+−3 x2 − ( − 1)
= xd+−1 x2 − − dxd+−3 x2 − ( − 1) . Let Φ( T 1 , x) = 0. Then xd+−3 [x4 − (d + )x2 + d( − 1)] = 0. It is easy to prove that
ρ ( T ) > 1, so ρ ( T ) is the largest root of the equation x4 − (d + )x2 + d( − 1) = 0. Thus, x2 =
(d+)+ (d+)2 −4d(−1) 2
.
Since d = n − 1 − , we have x2 =
n−1+ (n−2)2 +2n−3 . 2
So
ρ (T 1 ) =
n−1+ (n−2)2 +2n−3 . 2
n−1+ (n−2)2 +2n−3
Therefore, ρ ( T ) , and equality holds if and only if T is an almost completely 2 full-degree tree of 3 levels. (2) If 2 < n 2 + 1. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 2 ), where T 2 is an almost completely full-degree tree on n vertices with maximum degree . Let T 3 be a completely full-degree tree corresponding to T 2 with maximum degree and such that the number of levels in T 3 is 3 (may be T 2 = T 3 ). Then by Lemma 2.1 and Lemma 2.2, we have ρ ( T ) ρ ( T 2 ) ρ ( T 3 ). By Lemma 2.3, we have ρ ( R 3 ) = ρ ( T 3 ), where
⎡ R3 = ⎣
√
√ −1
0
−1 0
√
0
√
0
√
⎤
⎦.
0
√
√
Then ρ ( R 3 ) = 2 − √ 1, that is, ρ ( T 3 ) = 2 − 1. Thus, ρ ( T ) 2 − 1. Therefore, ρ ( T ) 2 − 1, and equality holds if and only if T is a completely full-degree tree of 3 levels. (3) If n > 2 + 1. Then by Lemma 2.1, we have ρ ( T ) ρ ( T 4 ), where T 4 is an almost completely full-degree tree on n vertices with maximum degree , and the number of levels in T 4 is k. Let T 5 be a completely full-degree tree corresponding to T 4 with maximum degree and the level of T 5 is also k (may be T 4 = T 5 ). By Lemma 2.2, we have ρ ( T 4 ) ρ ( T 5 ). If T 4 is a completely full-degree tree, then
n = 1 + + ( − 1) + ( − 1)2 + · · · + ( − 1)k−2 . So we have
k = log−1
( − 2)(n − 1) + 1 + 1.
If T 4 is not a completely full-degree tree, then
1 + 1 + ( − 1) + · · · + ( − 1)k−3 < n < 1 + 1 + ( − 1) + · · · + ( − 1)k−2 . So we have
log−1
( − 2)(n − 1) ( − 2)(n − 1) + 1 + 1 < k < log−1 + 1 + 2.
Therefore, the level of T 4 is k = log−1 (
(−2)(n−1)
+ 1) + 1.
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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ρ ( B k ) = ρ ( T 5 ), where √ ⎤ 0 −1 0 ··· ··· 0 . √ √ . ⎢ .. .. ⎥ ⎥ ⎢ −1 0 −1 ⎢ .. ⎥ √ . . . ⎥ ⎢ . . . . . . 0 −1 . ⎥ ⎢ Bk = ⎢ ⎥ . . √ .. .. .. ⎥ ⎢ .. . . . −1 0 ⎥ ⎢ ⎥ ⎢ . √ √ .. .. ⎣ . −1 ⎦ √0 0 ··· ··· 0 0 k×k
By Lemma 2.3, we have
⎡
Let
⎡
√ −1
0
··· .. . .. . .. .
···
0
⎤
.. ⎥ √ ⎢√ ⎢ −1 −1 . ⎥ ⎢ .. ⎥ √ .. .. ⎥ ⎢ . . 0 −1 . ⎥ ⎢ Fk = ⎢ ⎥ , . √ . . ⎥ ⎢ .. .. .. −1 0 ⎥ ⎢ ⎥ ⎢ . √ √ . .. .. ⎣ −1 ⎦ √ 0 ··· ··· 0 k×k then F k = B k + diag{, , . . . , }. So ρ ( F k ) = ρ ( B k ) + . However, the matrix F k has the LL T -decomposition F k = LL T , where
⎡√ −1 1 √ ⎢ ⎢ 0 −1 ⎢ ⎢ ⎢ 0 0 L=⎢ .. .. ⎢ . ⎢ . ⎢ .. ⎣ . 0 ···
⎤ ··· ··· 0 .. ⎥ .. . 1 . ⎥ .. ⎥ .. .. .. ⎥ . . . . ⎥ ⎥ . .. .. ⎥ . . 1 0 ⎥ ⎥ √ .. . 0 − 1 √1 ⎦ ··· 0 0 k×k 0
Let A k = L T L, then
⎤ ⎡ − 1 √ − 1 0 ··· ··· 0 . √ √ . ⎥ ⎢ .. .. −1 ⎥ ⎢ −1 ⎥ ⎢ . √ .. .. .. ⎥ ⎢ . . . . 0 −1 . ⎥ ⎢ Ak = ⎢ ⎥ . .. √ .. .. .. ⎥ ⎢ . . . ⎥ ⎢ −1 0 . ⎥ ⎢ .. √ √ .. ⎦ ⎣ . . −1 √ −1 0 ··· ··· 0 − 1 + 1 k×k
Since L is invertible, we have Let
⎡
ρ ( LL T ) = ρ ( L T L ). So ρ ( F k ) = ρ ( Ak ).
√ −1
0
··· .. . .. . .. .
···
0
⎤
.. √ ⎥ ⎢√ −1 . ⎥ ⎢ −1 ⎥ ⎢ . √ . . ⎥ ⎢ . . . . . 0 −1 . ⎥ ⎢ Dk = ⎢ ⎥ , . √ .. .. ⎥ ⎢ . . . ⎥ ⎢ −1 0 . ⎥ ⎢ .. √ √ .. ⎦ ⎣ . . −1 √ − 1 √ 0 ··· ··· 0 − 1 + − 1 k×k then by Lemma 2.5, we have
ρ ( A k ) < ρ ( D k ).
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Fig. 5. A tree for t = 1.
However, D k = diag{, , . . . , } + H k , where
⎡
√ −1
⎤ 0 ··· 0 . √ √ . ⎢ ⎥ .. .. 0 −1 ⎢ −1 ⎥ ⎢ ⎥ √ . . ⎢ ⎥ . . . Hk = ⎢ . . 0 −1 0 ⎥ ⎢ ⎥ .. √ .. .. ⎣ ⎦ . . . 0 − 1 √ √ 0 ··· 0 −1 − 1 k×k 0
Then by Lemma 2.4, we have So
√
√
ρ ( H k ) = 2 − 1 cos 2kπ+1 . √
ρ ( Ak ) < ρ ( D k ) = + 2 − 1 cos 2kπ+1 . Thus, ρ ( B k ) = ρ ( F k ) − = ρ ( Ak ) − <
2 − 1 cos 2kπ+1 . Therefore,
√
ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( (−2)(n−1) + 1) + 1. 2
Corollary 3.1. √ Let T = ( V , E ) be a tree on n vertices with maximum degree , where 3. If n + 1, (−2)(n−1) + 1) + 1. then ρ ( T ) < 2 − 1 cos 2kπ+1 , where k = log−1 ( Proof. The proof is similar to (3) in Theorem 3.1.
2
In the following, we give the upper bounds on the spectral radius of trees when = 3. Since it is easy to get the upper bounds on the spectral radius of trees when = 3 and n is small, we only study the upper bounds on the spectral radius of trees for = 3 and n is large. Theorem 3.2. Let T = ( V , E ) be a tree on n vertices with maximum degree , where = 3 and n + 2. Let t be the cardinality of the vertices of degree 3. Following results are established.
√
(1) if t = 1, then ρ ( T ) < √4.5; (2) if t = 2, then ρ ( T ) < √5; (3) if t = 3, then ρ ( T ) <
4 3
;
3 √ (4) if t = 4, then ρ ( T ) < √ 2 + 13; (5) if t 5, then ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2
n+2 + 1. 3
Proof. Let T n3 be the set of the trees on n vertices with maximum degree 3. By Corollary 3.1, we √ 2 + 1. Obviously, Theorem 3.2 holds obtain that for ∀ T ∈ T n3 , ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ 3 when ρ ( T ) 2. In the following, we assume that ρ ( T ) > 2. (1) If t = 1, then by Lemma 2.6, Lemma 2.7 and Lemma 2.8, for ∀ T ∈ T n3 , there is a nature number M 1 such that ρ ( T ) ρ ( T m ) for m M 1 and T m is a tree as shown in Fig. 5. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 1 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α1 , then ρm < α1 . Let x be the Perron vector of T m , and x v i be a componentof x corresponds to the vertex v i (1 i n). Then A ( T m )x = ρ ( T m )x. Therefore, we have ρm x v i = v j ∈ N T ( v i ) x v j . m
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 6. A tree for t = 2.
It is easy to prove that xai = xbi = xc i (i = 1, 2, . . . , m). Let xi = xai (i = 1, 2, . . . , m), xm+1 = xr . Then
ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are ρm + ρm2 − 4 ρm − ρm2 − 4 λ1 (m) = , λ2 (m) = . 2
2
So the general solution is
xk = C 1 λk1 (m) + C 2 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 1 , C 2 are constants, λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m). Since ρm xm+1 = 3xm , we have ρm = x3xm . Then m+1
α1 = lim
m→+∞
3xm xm+1
For limm→+∞ λ2 (m) =
√
m C 1 λm 1 (m) + C 2 λ2 (m)
= 3 lim
m→+∞
α1 − α12 −4 2
+1 +1 C 1 λm (m) + C 2 λm (m) 1 2
, we have
α1 = 3
α1 − α12 −4 2
= 3 lim λ2 (m).
. Then
m→+∞
√
α1 = 4.5. So for ∀ T ∈ T n3 , ρ ( T )
ρ ( T m ) < 4.5.
√ √ However, for t = 1 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ √ ously, 2 2 cos 2kπ+1 > 4.5.
√
Therefore, if t = 1, then ρ ( T ) < 4.5. (2) If t = 2, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 2 such that ρ ( T ) ρ ( T m ) for m M 2 and T m is a tree as shown in Fig. 6. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 2 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let lim m→+∞ ρm = α2 , then ρm < α2 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi m
(i = 1, 2, . . . , m), and xs = xt . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are
λ1 (m) =
ρm + ρm2 − 4 2
,
λ2 (m) =
ρm − ρm2 − 4 2
.
So the general solution is
xk = C 1 λk1 (m) + C 2 λk2 (m) where C 1 , C 2 are constants,
(k = 1, 2, . . . , m + 1),
λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m).
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Fig. 7. A tree for t = 3.
Since
ρm xm+1 = 2xm + xm+1 , we have ρm =
α2 = lim
m→+∞
2xm xm+1
For limm→+∞ λ2 (m) =
√
+ 1 = 2 lim
m→+∞
α2 − α22 −4 2
ρ ( T ) ρ ( T m ) < 5.
2xm xm+1
+ 1. Then
m C 1 λm 1 (m) + C 2 λ2 (m)
+1 +1 C 1 λm (m) + C 2 λm (m) 1 2
, we have
α2 = 2
α2 − α22 −4
+ 1 = 2 lim λ2 (m) + 1. m→+∞
+ 1. Then α2 =
2
√
5. So for ∀ T ∈ T n3 ,
√ √ However, for t = 2 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ √ ously, 2 2 cos 2kπ+1 > 5.
√
Therefore, if t = 2, then ρ ( T ) < 5. (3) If t = 3, then by Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 3 such that ρ ( T ) ρ ( T m ) for m M 3 and T m is a tree as shown in Fig. 7. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 3 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let lim m→+∞ ρm = α3 , then ρm < α3 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi m
(i = 1, 2, . . . , m), and xs = xt . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then ρm xk = xk+1 + xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Its characteristic equation is λ2 − ρm λ + 1 = 0. Then the characteristic roots are
λ1 (m) =
ρm + ρm2 − 4 2
λ2 (m) =
,
ρm − ρm2 − 4 2
.
So the general solution is
xk = C 3 λk1 (m) + C 4 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 3 , C 4 are constants, λ1 (m)λ2 (m) = 1, and λ1 (m) > λ2 (m). Let y i = xe i (i = 1, 2, . . . , m), ym+1 = xr , y 0 = 0. Then ρm yk = yk+1 + yk−1 , k = 1, 2, . . . , m. So the general solution is
yk = C 3 λk1 (m) + C 4 λk2 (m) (k = 1, 2, . . . , m + 1), where C 3 , C 4 are constants; λ1 (m), λ2 (m) are the same as the above. For ρm xm+1 = 2xm + ym+1 and ρm ym+1 = 2xm+1 + ym , we have
⎧ 2xm y m +1 ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
2xm+1 ym ⎪ ⎪ ⎩ ρm = + . y m +1
y m +1
It has been proved that limm→+∞ ρm exists, and limm→+∞ α3 − α32 −4 2
exists. So limm→+∞
ym+1 xm+1
exists. Let limm→+∞
xm xm+1
ym+1 xm+1
= limm→+∞
= β1 , then
ym ym+1
= limm→+∞ λ2 (m)=
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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Fig. 8. Four 3-degree vertices.
Fig. 9. A tree for t = 4.
⎧ ⎪ ⎪ α − α32 − 4 3 ⎪ ⎪ ⎨ α3 = 2 + β1 , 2 ⎪ α3 − α32 − 4 ⎪ 2 ⎪ ⎪ ⎩ α3 = + . β1 2 So
α3 =
√
4 3 . 3
Then for ∀ T ∈ T n3 ,
ρ (T ) ρ (T m ) <
√
4 3 . 3
√ √ However, for t = 3 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ ously, 2 2 cos 2kπ+1 >
√
4 3 . 3
√
Therefore, if t = 3, then ρ ( T ) < 4 3 3 . (4) If t = 4, then according to the connected relationships of the four 3-degree vertices, we discuss the following two cases. Case 1. The connected relationships of the four 3-degree vertices as shown in Fig. 8. By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 4 such that ρ ( T ) ρ ( T m ) for m M 4 and T m is a tree as shown in Fig. 9. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 4 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α4 , then ρm < α4 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi = m
xe i = x f i (i = 1, 2, . . . , m), xs = xt = xr . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xs . Then xk−1 , k = 2, 3, . . . , m. Let x0 = 0, then we have ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Then
xk = C 5 λk1 (m) + C 6 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 5 , C 6 are constants; λ1 (m), λ2 (m) are the same as the above.
ρm xk = xk+1 +
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 10. Four 3-degree vertices.
Fig. 11. A tree for t = 4.
For
ρm xm+1 = 2xm + x w and ρm x w = 3xm+1 , then ⎧ 2xm xw ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
3x ⎪ ⎪ ⎩ ρm = m+1 . xw
It has been proved that limm→+∞ ρm exists, and limm→+∞ α4 − α42 −4 2
⎧ ⎪ ⎪ ⎪ ⎨
exists. So limm→+∞
exists. Let limm→+∞
xw xm+1
= limm→+∞
ym ym+1
= limm→+∞ λ2 (m)=
= β2 , then
α4 = 2
α4 − α42 − 4
⎪ 3 ⎪ ⎪ ⎩ α4 = . β2 So
xw xm+1
xm xm+1
√
2
+ β2 ,
√
α4 = 2 + 13. Then for ∀ T ∈ T n3 , ρ ( T ) < 2 + 13.
Case 2. The connected relationships of the four 3-degree vertices as shown in Fig. 10. By Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9, for ∀ T ∈ T n3 , there is a nature number M 5 such that ρ ( T ) ρ ( T m ) for m M 5 and T m is a tree as shown in Fig. 11. Denote ρ ( T m ) by ρm . It is easy to prove that {ρm } (m M 5 ) is strictly monotone increasing and has a upper bound. So limm→+∞ ρm exists. Let limm→+∞ ρm = α5 , then ρm < α5 . Similar to (1), we have ρm x v i = v j ∈ N T ( v i ) x v j . It is easy to prove that xai = xb i = xc i = xdi , m
xe i = x f i (i = 1, 2, . . . , m), xs = xr , xu = x v . Let xi = xai (i = 1, 2, . . . , m), xm+1 = xr . Then xk+1 + xk−1 , k = 2, 3, . . . , m.
ρm xk =
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Let x0 = 0, then we have
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ρm xk = xk+1 + xk−1 , k = 1, 2, . . . , m. Then
xk = C 5 λk1 (m) + C 6 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 5 , C 6 are constants; λ1 (m), λ2 (m) are the same as the above. Let y i = xe i (i = 1, 2, . . . , m), ym+1 = xu , y 0 = 0. Then ρm yk = yk+1 + yk−1 , k = 1, 2, . . . , m. So the general solution is
yk = C 7 λk1 (m) + C 8 λk2 (m)
(k = 1, 2, . . . , m + 1),
where C 7 , C 8 are constants; λ1 (m), λ2 (m) are the same as the above. For ρm xm+1 = 2xm + ym+1 and ρm ym+1 = ym+1 + xm+1 + ym , then
⎧ 2xm y m +1 ⎪ ⎪ + , ⎨ ρm = xm+1
xm+1
y m +1
y m +1
xm+1 ym ⎪ ⎪ + + 1. ⎩ ρm = It has been proved that limm→+∞ ρm exists, and limm→+∞ α5 − α52 −4 2
exists. So limm→+∞
⎧ ⎪ α5 − ⎪ ⎪ ⎪ ⎨ α5 = 2
α52 − 4
2
ym+1 xm+1
exists. Let limm→+∞
xm xm+1
y m +1 xm+1
= limm→+∞
ym y m +1
= limm→+∞ λ2 (m) =
= β3 , then
+ β3 ,
⎪ ⎪ α5 − α52 − 4 ⎪ ⎪ ⎩ α5 = 1 + + 1. β3 2 So
α5 ≈2.3569. Then for ∀ T ∈ T n3 , ρ ( T ) < 2.357.
√
√
For 2 + 13 > 2.357, by the two cases, we know if t = 4, then for ∀ T ∈ T n3 , ρ ( T ) < 2 + 13. √ √ However, for t = 4 and n + 2, we have k 3. Then 2 2 cos 2kπ+1 2 2 cos π7 ≈ 2.5483. Obvi-
√ ously, 2 2 cos 2kπ+1 >
2+
√
13.
√
Therefore, if t = 4, then ρ ( T ) < 2 + 13. √ 2 (5) If t 5, then by Corollary 3.1, we have ρ ( T ) < 2 2 cos 2kπ+1 , where k = log2 n+ + 1. 3
2
It is easy to prove that, when = 3, the upper bounds in Theorem 3.2 are better than the upper bounds in Theorem 3.1, respectively. 4. The comparison of the results In the following, we compare our upper bounds on the spectral radius of trees in Theorem 3.1 with the previous results (1.1), (1.2), and (1.3). 4.1. Comparing with (1.1) We know that (1.1) gives a result
√
ρ ( T ) < 2 − 1.
(1) When + 1 < n 2. In Theorem 3.1, we get Since 3, we have 92 − 28 + 13 = 9( −
ρ (T ) 14 2 ) 9
−
n−1+ (n−2)2 +2n−3 . 2 79 > 0. Then we get 9
52 + 12n > 0. Thus, 602 − 112 +52 − 12n + 12n > 0. So we can get 8( − 1) − (n − 1). Therefore,
n−1+ (n−2)2 +2n−3 2
√
362 − 112 +
(n − 2)2 + 2n − 3 <
< 2 − 1. Thus, our result is better than (1.1). √
(2) When 2 < n 2 √ + 1. In Theorem √ 3.1, we get ρ ( T ) 2 − 1. It is easy to prove that 2 − 1 < 2 − 1 for 3. Thus, our result is better than (1.1).
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H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
Fig. 12. A tree with 7 vertices.
Fig. 13. A tree with 6 vertices.
(3) When n > 2 + 1. In Theorem 3.1, we get
√
√
√
ρ ( T ) < 2 − 1 cos 2kπ+1 (k 4).
Because 2 − 1 cos 2kπ+1 < 2 − 1, our result is better than (1.1). Therefore, our upper bounds in Theorem 3.1 are better than (1.1). 4.2. Comparing with (1.2) We know that (1.2) gives a result
ρ ( T ) < max
max { δ j − 1 +
2 j k−2
√ δ j −1 − 1 }, δ1 − 1 + .
(1) When + 1 < n 2. Let T 1 be an almost completely full-degree tree as shown in Fig. 12,
then we can get our upper bound is
ρ (T 1 ) =
However, the upperbound in (1.2) is 2 +
√
√
√
7−1+ (7−8)2 +14−3 2
=
3+
√
3.
2.
Because 2 + 2 > 3 + 3, our result is better than (1.2). (2) When 2 < n 2 + 1. Let T be a tree with maximum degree , and , v be two adjacent √ let u√ vertices √ √ of T such√that du =√d v = . Then the upper bound in (1.2) is + − 1. Obviously, + − 1 > 2 − 1 > 2 − 1. Thus, our result is better than (1.2). (3) When n > 2 + 1. Let T be a tree with maximum degree , and√let u ,√ v be two adjacent = d = . Then the upper bound in (1.2) is + − 1. Obviously, vertices of T such that d u v √ √ √ √ + − 1 > 2 − 1 > 2 − 1 cos 2kπ+1 . Thus, our result is better than (1.2). Therefore, our upper bounds in Theorem 3.1 are better than (1.2) in some cases. 4.3. Comparing with (1.3) We know that (1.3) gives a result
ρ ( T ) < max
max {
2 j k−2
√
γ j − 1 + γ j−1 − 1 }, γ1 − 1 + − 1 .
(1) When + 1 < n 2. Let T 2 be an almost completely full-degree tree as shown in Fig. 13, √ 6−1+ 12−3 then we can get our upper bound is ρ ( T 2 ) = = 2. 2
√
However, √ the upper bound in (1.3) is 2 2. Because 2 2 > 2, our result is better than (1.3). (2) When 2 < n 2 + 1. Let T be a tree with maximum degree , and let u , v , w be the vertices u = d v = d√ w = and uv ∈ E ( T ), u w ∈ E ( T ). Then the upper bound in (1.3) √ of T such that d√ is 2 − 1. Obviously, 2 − 1 > 2 − 1. Thus, our result is better than (1.3).
H. Song et al. / Linear Algebra and its Applications 439 (2013) 2527–2541
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(3) When n > 2 + 1. Let T be a tree with maximum degree , and let u , v , w be the vertices of √ T such that √ du = d v = d √ w = and uv ∈ E ( T ), u w ∈ E ( T ). Then the upper bound in (1.3) is 2 − 1. Obviously, 2 − 1 > 2 − 1 cos 2kπ+1 . Thus, our result is better than (1.3). Therefore, our upper bounds in Theorem 3.1 are better than (1.3) in some cases. In conclusion, comparing our results with the previous results, we prove that our upper bounds are better than (1.1). For (1.2) and (1.3), in order to get the upper bound on the spectral radius of a tree T , we require to know the degrees of all the vertices of T . But in Theorem 3.1, we only require to know the number of vertices and maximum degree, and we prove that our upper bounds are better than (1.2) and (1.3) in some cases. Acknowledgements The authors would like to thank the anonymous referees for valuable suggestions and corrections, which have improved the original manuscript. References [1] A. Berman, R.J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New York, 1979. ´ Bounding the largest eigenvalue of trees in terms of the largest vertex degree, Linear Algebra Appl. 360 [2] D. Stevanovic, (2003) 35–42. [3] R. Oscar, Improved bounds for the largest eigenvalue of trees, Linear Algebra Appl. 404 (2005) 297–304. [4] R. Oscar, The spectra of some trees and bounds for the largest eigenvalue of any tree, Linear Algebra Appl. 414 (2006) 199–217. [5] A.M. Yu, L. Mei, Laplacian spectral radius of trees with given maximum degree, Linear Algebra Appl. 429 (2008) 1962–1969. [6] R. Oscar, S. Ricardo, The spectra of the adjacency matrix and Laplacian matrix for some balanced trees, Linear Algebra Appl. 403 (2005) 97–117. [7] S. Kouachi, Eigenvalues and eigenvectors of tridiagonal matrices, Electron. J. Linear Algebra 15 (2006) 115–133. [8] K.F. Fang, On the relationship between edge moving transformation and spectral radii in graphs, J. Huzhou Teach. Coll. 29 (2) (2007) 10–11. [9] B.F. Wu, X.Y. Yuan, E.L. Xiao, On the spectral radii of trees, J. East China Norm. Univ. Natur. Sci. 3 (2004) 22–27. [10] Q. Li, K.Q. Feng, On the largest eigenvalues of graphs, Acta Math. Appl. Sin. 2 (1979) 167–175 (in Chinese). ´ P. Rowlinson, S. Simic, Eigenspaces of Graphs, Cambridge University Press, Cambridge, 1997. [11] D. Cvetkovic, [12] J.M. Guo, J.Y. Shao, On the spectral radius of trees with fixed diameter, Linear Algebra Appl. 413 (1) (2006) 131–147. ´ M. Doob, H. Sachs, Spectra of Graphs-Theory and Applications, third ed., Johann Ambrosius Barth Verlag, [13] D. Cvetkovic, 1995.