the acyclic matrices with a p set of maximum size 2015 linear algebra and its applications

8/17/2019 The Acyclic Matrices With a P Set of Maximum Size 2015 Linear Algebra and Its Applications

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Linear Algebra and its Applications 468 (2015) 27–37

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Linear Algebra and its Applications

www.elsevier.com/locate/laa

The acyclic matrices with a P-set of maximumsize

Zhibin Du a, Carlos M. da Fonseca b,∗

a School of Mathematics and Information Sciences, Zhaoqing University, Zhaoqing 526061, Chinab Department of Mathematics, Kuwait University, Safat 13060, Kuwait

a r t i c l e i n f o a b s t r a c t

Article history:

Received 16 June 2013

Accepted 22 October 2013

Available online 23 December 2013

Submitted by P. Psarrakos

MSC:15A18

15A48

05C50

Keywords:

Trees

Acyclic matrices

Multiplicity of eigenvalues

P-set

P-vertices

Let m A (0) denote the nullity of a given matrix A of order n. Set

A(α) for the principal submatrix of A obtained after deleting therows and columns indexed by the nonempty subset α of {1, . . . ,n}.When m A(α)(0) = m A (0) + |α|, we call α a P-set of A. In this paper,we classify all of the trees T for which there exists a matrix A

whose graph is T and containing a P-set of maximum size. Our

characterization does not depend on whether the acyclic matrices

are singular or nonsingular.

© 2013 Elsevier Inc. All rights reserved.

1. Introduction

Let A = (ai j ) be an n × n real symmetric matrix. The graph of A, denoted by G( A), is defined tobe the graph whose vertex set is {1, . . . ,n} and edge set is {i j | i = j and ai j = 0}. In this sense, G( A)does not depend on the main diagonal entries of A. Here we focus our study on the set

S (G) =

A ∈ Rn×n A is symmetric and G ( A) = G,

where G is a given tree on n vertices. Any matrix A ∈ S (G) is called acyclic .

* Corresponding author.E-mail addresses: [email protected] (Z. Du), [email protected] (C.M. da Fonseca).

0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved.

http://dx.doi.org/10.1016/j.laa.2013.10.042

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28 Z. Du, C.M. da Fonseca / Linear Algebra and its Applications 468 (2015) 27–37

The principal submatrix of A obtained after deleting the rows and columns indexed by α ⊂{1, . . . , n} is denoted by A(α). If we consider the principal submatrix of A consisting of both rowsand columns indexed by α, then we will write A[α]. However, we reserve the notation A(G) for thestandard adjacency (0, 1)-matrix of the graph G .

Given a tree T with vertices 1, . . . , n, for any nonempty subset α of {1, . . . , n}, we denote byT (α) the forest obtained from T by deleting the vertices in α and the incident edges, and T [α] thesubgraph (forest) of T induced by the vertices in α. Clearly, if A ∈ S (T ), then A(α) ∈ S (T (α)) and A[α] ∈ S (T [α]), for any nonempty subset α of {1, . . . , n}.

Denoting by m A (λ) the (algebraic) multiplicity of an eigenvalue λ of A, we assume that m A (λ) = 0if λ is not an eigenvalue of A. As a consequence of the Cauchy Interlacing Theorem [9, Theorem

4.3.17], we have

m A (λ) − 1 m A(i)(λ)m A (λ) + 1. (1.1)

In the case of equality in the right-hand side of (1.1), with λ = 0, the index i is known as a P-vertexof A [10,11]. Notice that m A (0) corresponds to the nullity of A.

One may also derive from the right-hand side of (1.1) that

m A(α)(0) m A (0) + |α|, (1.2)

for any nonempty subset α of {1, . . . , n}. When we have equality in (1.2), we call α a P-set of A[10]. Therefore, α is a P-set of A if and only if the nullity of A(α) is |α| more than that of A. Themaximum size of a P-set of A is denoted by P s( A) [12]. Interestingly, from the Cauchy Interlacing

Theorem, every subset of a P-set of A is also a P-set of A. In particular, each vertex in a P-set of A is

a P-vertex of A [12, Proposition 5] (see also [10] for proofs).

It is also worth mentioning that, when α is a P-set of A, for each proper subset β ⊂ α (which is aP-set of A), α \ β is a P-set of A(β) .

Motivated by [11], in 2009, Kim and Shader [12] started a thorough study of the P-vertices and

P-sets of nonsingular matrices whose graph is a path or a star. Several interesting questions wereproposed there and all answered in [1–6] with some significant results in the more recent references.

Some considerations on P-sets were also made by Kim and Shader regarding slightly more general

singular acyclic matrices. Namely, they constructed the nonsingular tridiagonal matrices of order n

with P-sets of cardinality n/2, where x stands for the largest integer not greater than x. Theseauthors verified this (tight) upper bound for nonsingular acyclic matrices and for singular acyclic

matrices of odd order. But, in fact, this is the maximum size of any P-set of any symmetric matrix of

order n . Our result is motivated by [12, p. 402] which is apparently proved for all symmetric matrices.

It is stated by its importance in this paper and the proof is included for completeness.

Proposition 1.1. For any symmetric matrix A of order n, we have

P s( A)

n

2

.

Proof. Suppose that α is a P-set of A, i.e., m A(α)(0) = m A (0) + |α|. On one hand,

m A(α)(0) = m A (0) + |α| |α|. (1.3)

On the other hand, noting that the matrix A(α) is of order n − |α|, we have

m A(α)(0) n − |α|. (1.4)

Now it follows that |α| n − |α|, i.e.,

|α|

n

2

.

The result is now clear.


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Z. Du, C.M. da Fonseca / Linear Algebra and its Applications 468 (2015) 27–37 29

Finally, it is worth mentioning that for any tree T , there exists a nonsingular matrix A ∈ S (T )without P-vertex, so P s( A) = 0. Moreover, the only tree T for which each nonsingular matrix A ∈ S (T )satisfies P s( A) 1 is the star [1].

The thrust of this paper is to completely characterize the trees where the bound of Proposition 1.1

is attained. Our classification only depends on the parity of the order of the matrices. Before ourmain results, we will establish several technical lemmas which may be of independent interest. We

will also recall some previous results.

2. Preliminary results

Let us denote by P n the path on n vertices.

Our first result observes that the only tree of order n for which there exists a singular matrix of

nullity n − 1 is the path P 2 .

Lemma 2.1. Suppose that T is a tree on n 2 vertices, and A ∈ S (T ). If m A (0) = n − 1, then T ∼= P 2 .

Proof. Observe that m A (0) = n − 1 1 implies that the spectrum of A consists of two distinct eigen-values, one is zero, and the other is nonzero. Thus 0 is either the largest or the smallest eigenvalue

of A.

It is well known that whether the largest or the smallest eigenvalue of an acyclic matrix is simple

(see, e.g., [10, Corollary 7]). So m A (0) = n − 1 = 1, i.e., n = 2. Accordingly, T ∼= P 2 .

The next theorem asserts an important property for the nullity of an acyclic matrix with a P-set

of maximum size: it is at most 1. We recall that a trivial graph contains only one vertex and, conse-

quently, no edges. If, in addition, its nullity is 1, then this will mean that the weight of such vertex

is 0.

Theorem 2.2. Let T be a tree on n 2 vertices, and A ∈ S (T ). If α is a P-set of A with |α| = n2

, then one of the following conditions must be satisfied:

(i) n is even, m A (0) = 0, and each component of T (α) is trivial and of nullity 1.(ii) n is odd, m A (0) = 1, and each component of T (α) is trivial and of nullity 1.

(iii) n is odd, m A (0) = 0, and each component of T (α) is trivial and, with one exception, of nullity 1.(iv) n is odd, m A (0) = 0, T (α) contains exactly one edge, and each component of T (α) is of nullity 1.

Proof. Let T̄ = T (α) and ¯ A = A (α) ∈ S (T̄ ).

Case 1. n is even, i.e., |α| = n2 and, consequently, ¯ A is of order

n2 .

From the inequalities (1.3) and (1.4), we have

n

2 m A (0) +

n

2 = m ¯ A (0) n −

n

2 =

n

2.

Thus m A (0) = 0 and m ¯ A (0) = n2 . Since

¯ A is symmetric, the rank of ¯ A is 0, i.e., ¯ A is the zero matrix.

Accordingly, T̄ is an edgeless graph. So (i) follows.

Case 2. n is odd, i.e., |α| = n−12

and, consequently, ¯ A is of order n+12

.

From the inequalities (1.3) and (1.4), we have

n − 1

2 m A (0) +

n − 1

2 = m ¯ A (0) n −

n − 1

2 =

n + 1

2 .

Thus m A (0) = 0 or 1.


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Proof. Let us set β = {u, v, w} ∩ α and = |β|. Suppose that (i) does not hold, i.e., = 1.If = 0, then u, v, w are in the same component of T (α), which is a contradiction to the hypoth-

esis that each component of T (α) contains at most one edge. Thus = 0.If {u, v} ⊆ α, then {u, v} would be a P-set of A, which implies that u is a P-vertex of A(v), however,

which is impossible whether A[u] = 0 or not, since u is an isolated vertex of T (v). This implies that = 3, and if = 2, then β = {u, w} or β = { v, w}.

Let ᾱ = α ∩ V (T̄ ), i.e., ᾱ = α \ {u, v}, and let ¯ A = A[T̄ ].Clearly, |ᾱ| = |α| − 1, whether β = {u, w} or β = { v, w}. Now, since ᾱ ⊆ α, we have ᾱ is a P-set of

A, i.e., m A(ᾱ)(0) = m A (0) + |ᾱ|.Suppose that β = {u, w}. Then u is a P-vertex of A(w), equivalently, u is a P-vertex of A[u, v].

Thus A[v] = 0 and m A[u,v](0) = 0. On one hand, since u is a P-vertex of A, from Lemma 2.4(i), wehave m A (0) = m ¯ A (0). On the other hand, taking into account that w ∈ ᾱ, we have

m A(ᾱ)(0) = m A[u,v](0) + m ¯ A(ᾱ)(0) = m ¯ A(ᾱ)(0).

So we get that

m ¯ A(ᾱ)(0) = m A(ᾱ)(0) = m A (0) + |ᾱ| = m ¯ A (0) + |ᾱ|,

i.e., ᾱ is a P-set of ¯ A.If β = {v, w}, then we can deduce that ᾱ is a P-set of ¯ A similarly.In particular, if T has n vertices, and P s( A) =

n2 , then, from Proposition 1.1, P s(

¯ A) = n−22 .

Lemma 3.2. Suppose that T is a tree with a pendant P 3 = uvw , and A ∈ S (T ). Let T̄ be the tree obtained from T by deleting the vertices u, v. Moreover, let us assume that α is a P-set of A such that each component of T (α) is trivial and of nullity 1. Then |α ∩ V (T̄ )| = |α| − 1, and α ∩ V (T̄ ) is a P-set of A[T̄ ]. In particular, if T has n vertices, and P s( A) =

n2 , then P s( A[T̄ ]) =

n−22 .

Proof. Let β and be defined as previously in the proof of Lemma 3.1.

If = 1, then from Lemma 3.1, the result follows.Clearly, the hypothesis that each component of T (α) is trivial implies that if = 1, then β = {v}.Suppose in the following that β = { v}. Then the hypothesis that each component of T (α) is of

nullity 1 implies that A[u] = 0.Let us set ᾱ = α ∩ V (T̄ ), i.e., ᾱ = α \ {u, v}, and let ¯ A = A [T̄ ]. Clearly, |ᾱ| = |α| − 1.Since ᾱ ⊆ α , we have ᾱ is a P-set of A, i.e., m A(ᾱ)(0) = m A (0) + |ᾱ|. Recall that A[u] = 0 and,

from Lemma 2.4(ii), we can get m A (0) = m ¯ A (0) and m A(ᾱ)(0) = m ¯ A(ᾱ)(0). Now we have m ¯ A(ᾱ)(0) =

m ¯ A (0) + |ᾱ|, i.e., ᾱ is a P-set of ¯ A.

In particular, if T has n vertices, and P s( A) = n

2

, then, from Proposition 1.1, P s( ¯ A) = n−2

2

.

We now recall a key definition in [4]:

Definition 3.2. If x, y are two terminal vertices sharing a common vertex z in a tree T , then the path

of T induced by the vertices x, y, z is said to be an outer P 3 of T . In this case, we write P 3 = xz y.

Observe that we are identifying P 3 = xzy with the star S 3 on 3 vertices where z is the centralvertex.

Lemma 3.3. Suppose that T is a tree with an outer P 3 = xzy, and A ∈ S (T ). If α is a P-set of A such that each

component of T (α) contains at most one edge, then exactly one of x, y, z is in α .

Proof. Let us set γ = { x, y, z } ∩ α and = |γ |.If = 0, then x, y, z are in the same component of T (α), which is a contradiction to the hypothesis

that each component of T (α) contains at most one edge. Thus = 0.


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Observe that x is an isolated vertex of T ( z ), thus whether A[ x] = 0 or not, x would not be aP-vertex of A( z ), hence { x, z } would not be a P-set of A. Similarly, { y, z } would not be a P-set of A.

So it follows that = 3, and if = 2, then γ = { x, y}.Suppose that γ = { x, y}. Then { x, y} would be a P-set of A, i.e., m A( x, y)(0) = m A (0) + 2. So

m A( x, y, z )(0) m A( x, y)(0) − 1 = m A (0) + 1. On the other hand, since x ∈ α, x is a P-vertex of A, fromLemma 2.4(i), we have m A( x, z )(0) = m A (0). Furthermore, note that y is an isolated vertex of T ( x, z ),thus m A( x, y, z )(0) = m A( x, z )(0) or m A( x, z )(0) − 1, i.e., m A( x, y, z )(0) = m A (0) or m A (0) − 1, either of which

is a contradiction to m A( x, y, z )(0) m A (0) + 1. Thus = 2.Combining the above arguments, we have = 1.

Lemma 3.4. Suppose that T is a tree with an outer P 3 = xzy, and A ∈ S (T ). If α is a P-set of A such that eachcomponent of T (α) is trivial and of nullity 1, then A is singular.

Proof. Defining γ and as previously in the proof of Lemma 3.3.From Lemma 3.3, we have = 1. The assumption that each component of T (α) is trivial implies

that γ = { z }.If γ = { z }, then the hypothesis that each component of T (α) is of nullity 1 implies that A[ x] =

A[ y] = 0. So A is singular.

We will call each component obtained from the deletion of a given vertex v of a tree T as a branch

o f T a t v .

Lemma 3.5. Suppose that T is a tree with an outer P 3 = xzy, and A ∈ S (T ). Denote by z 1, . . . , z r the neigh-bors of z in T different from x, y, and T i , for 1 i r, the branch of T at z containing z i . In addition, let α bea P-set of A. If z /∈ α, m A (0) = 0, and each component of T (α) contains at most one edge and of nullity 1, thenα ∩ V (T i ) is a P-set of A [T i ], for 1 i r.

Proof. We only prove the result for i = 1, i.e., α ∩ V (T 1) is a P-set of A[T 1], since other cases aresimilar.

Let αi = α ∩ V (T i ) and A i = A[T i ], for 1 i r .From Lemma 3.3, exactly one of x, y, z is in α. Assume that x ∈ α and y /∈ α, since z /∈ α.Note that x ∈ α implies that x is a P-vertex of A, from Lemma 2.4(i), m A( x, z )(0) = m A (0) = 0. So

m A1 (0) = 0, because T 1 is a component of T ( x, z ).Observe that yz is an edge of T (α), so the hypothesis that each component of T (α) contains at

most one edge implies that { z 1, . . . , z r } ⊆ α. Hence { z 1, . . . , z r } is a P-set of A, i.e.,

m A( z 1,..., z r )

(0) = m A

(0) + r . (3.1)

Since { x, z 1, . . . , z r } ⊆ α, x is a P-vertex of A( z 1, . . . , z r ), equivalently, x is a P-vertex of A[ x, y, z ],i.e., m A[ y, z ](0) = m A[ x, y, z ](0) + 1. Moreover, the hypothesis that each component of T (α) is of nullity1 implies that m A[ y, z ](0) = 1, and thus A[ x, y, z ] is nonsingular.

Note that α1 ⊆ α is a P-set of A. As a consequence, α1 \ { z 1} is a P-set of A( z 1), equivalently,α1 \ { z 1} is a P-set of A1( z 1). If z 1 is a P-vertex of A1 , then clearly α1 would be a P-set of A1 . Thuswe are left to show that z 1 is a P-vertex of A1, i.e., m A1( z 1)(0) = m A1 (0) + 1 = 1.

Clearly,

A( z 1, . . . , z r ) = A[ x, y, z ] ⊕ A1( z 1) ⊕ · · · ⊕ A r ( z r ), (3.2)

where ⊕ represents the direct sum of matrices.From (3.1) and (3.2), we have

r k=1

m Ak ( z k )(0) = m A (0) + r = r . (3.3)


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Suppose that there exists an index k, for 1 k r , such that m Ak( z k )(0) 2. Note that z k is a

P-vertex of A, i.e., m A( z k )(0) = m A (0) + 1 = 1. On the other hand, observe that T k( z k) is the vertex-disjoint union of all the components of T ( z k), except the one containing z , and thus,

1 = m A( z k )(0) m Ak ( z k )(0) 2,which is a contradiction.

So we have m Ak ( z k)(0) 1, for 1 k r . Furthermore, together with (3.3), we can state that

m Ak( z k)(0) = 1, for 1 k r . In particular, m A1( z 1)(0) = 1.

4. The characterization for acyclic matrices of even order

Attaching the path P 2 to a vertex of a tree T is called adding a pendant P 2 to T . The operation of

adding pendant P 2 to a tree was crucial in our previous works. Interestingly, it also plays a funda-

mental role in this paper.

Lemma 4.1. Let T be a tree. Suppose that T̂ is a tree obtained from T by adding a pendant P 2 = u v, wherev is of degree 2. If α is a P-set of A(T ), then α ∪ {v} is a P-set of A(T̂ ). In particular, if T has n vertices, andP s( A(T )) =

n2

, then P s( A(T̂ )) = n+2

2 .

Proof. Set A = A(T ) and ˆ A = A(T̂ ).Since α is a P-set of A, we have m A(α)(0) = m A (0) + |α|.

Note that ˆ A[u] = 0 and, from Lemma 2.4(ii), m A (0) = m ˆ A (0).

Observe also that T is a component of T̂ (v), we have ˆ A(α ∪ {v}) = A(α) ⊕ ( 0 ). So we can get theequalities

m ˆ A(α∪{v})(0) = m A(α)(0) + 1

= m A (0) + |α| + 1

= m ˆ A (0) +α ∪ {v},

i.e., α ∪ {v} is a P-set of ˆ A.In particular, if T has n vertices, and P s( A) =

n2 , then, from Proposition 1.1, P s(

ˆ A) = n+22 .

Contrary to the operation of adding pendant P 2 , we can define the operation of deleting pen-

dant P 2 . Given a tree T , deleting a path P 2 , with degrees 1 and 2 in T , is called deleting a pendant P 2of T .

It is easily verified that if there is no outer P 3 in a tree T on n 3 vertices, then some deletingpendant P 2 operation can be applied to T .

We are ready to present our first main characterization.

Theorem 4.2. Let T be a tree on n 2 vertices, where n is even. The following two conditions are equivalent :

(a) There exists a matrix A ∈ S (T ) such that P s( A) = n2 .

(b) T is a tree obtained from P 2 by sequentially adding pendant P 2’s.

Proof. (a) ⇒ (b) Suppose that A is a matrix in S (T ) such that P s( A) = n2 .

Let us start from T , deleting pendant P 2’s repeatedly, until there exists an outer P 3 or a path P 2

left, the resulting graph is denoted by T ∗.If T ∗ ∼= P 2, then (b) follows clearly.Suppose that there exists an outer P 3 in T

∗ . By Theorem 2.2(i) and Lemma 3.2, we reach that

m A[T ∗](0) = 0, and there is a P-set α∗ of A[T ∗] such that each component of T ∗(α∗) is trivial and of

nullity 1. However, from Lemma 3.4, we can deduce that A[T ∗] is singular, which is a contradiction.


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(b) ⇒ (a) Suppose that T is a tree obtained from P 2 by sequentially adding pendant P 2 ’s. Clearly,P s( A(P 2)) = 1. By Lemma 4.1, we have

P s A(T )= P s A(P 2)+ n − 2

2

= n

2

.

So (a) follows.

The example of a path of even order is perhaps the most simple application of Theorem 4.2.

Therefore, the Jacobi matrix of even order

0 1

1. . .

. . .. . .

. . . 1

1 0

in [12, Theorem 4(a)] is an immediate consequence of our result.

5. The odd case

In order to establish our classification for the odd case, we need several technical lemmas.

Lemma 5.1. (See [5, Corollary 3.5].) If T is a tree obtained from P 2 by sequentially adding pendant P 2’s, then

for any vertex v in T , there exists an edge e incident with v such that T is a tree obtained from P 2 = e bysequentially adding pendant P 2’s.

Lemma 5.2. Suppose that T is a tree on n 3 vertices, where n is odd, containing an outer P 3 . If there exists

a matrix A ∈ S (T ) such that P s( A) = n−1

2 , then T is a tree obtained from P 3 by sequentially adding pendant

P 2 ’s.

Proof. Let P 3 = xz y be an outer P 3 in T , where x, y are both terminal vertices. Denote by z 1, . . . , z r the neighbors of z in T different from x, y , and T i , for 1 i r , the branch of T at z containing z i .

Suppose that A is a matrix in S (T ) such that P s( A) = n−1

2 , and α is a P-set of A with |α| = n−12 .

From Theorem 2.2(ii)–(iv), we know that each component of T (α) contains at most one edge.Furthermore, from Lemma 3.3, exactly one of x, y, z is in α.

Let n i be the number of vertices in T i and set ñi = |α ∩ V (T i )|, for 1 i r . Clearly, n1 + · · · + nr =n − 3, and ñ1 + · · · + ñr = |α| − 1 =

n−32 .

First suppose that z ∈ α . Note that (α ∩ V (T i)) ∪ { z } is a P-set of A, equivalently, α ∩ V (T

i) is a

P-set of A( z ), for 1 i r . Since T i is a component of T ( z ), we have α ∩ V (T i ) is a P-set of A[T i ], for1 i r .

Suppose now that z /∈ α , and assume that x ∈ α and y /∈ α . Then yz is an edge of T (α). ByTheorem 2.2(iv), m A (0) = 0 and each component of T (α) contains at most one edge and of nullity 1.Moreover, by Lemma 3.5, α ∩ V (T i ) is a P-set of A[T i ], for 1 i r .

Therefore, whether z ∈ α or not, α ∩ V (T i ) is always a P-set of A[T i ], for 1 i r . Consequently,by Proposition 1.1,

ñi P s A[T i]

ni

2

ni

2 ,

for 1 i r , and thus

n − 3

2 = ñ1 + · · · + ñr

n1

2 + · · · +

nr

2 =

n − 3

2 .

So, in fact, we have ñi = ni

2 , furthermore, P s( A[T i ]) = ni

2 , for 1 i r .


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Clearly, A[T i ] ∈ S (T i ), thus by Theorem 4.2 (a) ⇒ (b), T i is a tree obtained from P 2 by sequentiallyadding pendant P 2’s, for 1 i r . Furthermore, it follows from Lemma 5.1 that there exists an edge

ei in T i incident with z i such that T i is a tree obtained from P 2 = e i by sequentially adding pendantP 2 ’s, for 1 i r .

Finally we can obtain T from P 3 = xz y by sequentially adding pendant P 2’s as follows:

() Firstly, we begin with the path P 3 = xz y, sequentially adding pendant P 2 ’s, each time it wouldgive rise to two new edges: zz i and ei , for 1 i r , which finally would result in the tree

obtained by attaching r paths P 2 to the center z of the path P 3 = xz y.() Secondly, we begin with the resulting graph in (), sequentially adding pendant P 2 ’s, just as

the procedure of sequentially adding pendant P 2 ’s from ei to T i , for all 1 i r , which finally

would result in the tree T .

Then the result follows.

Lemma 5.3. Suppose that T is a tree on n 3 vertices, where n is odd, containing a pendant P 3 = uv w. Let T̄ be the tree obtained from T by deleting the vertices u, v. Suppose that A ∈ S (T ), and α is a P-set of A with|α| = n−12 . Let β = {u, v, w} ∩ α.

(i) If β = {u}, then P s( A[T̄ ]) = n−3

2 .

(ii) If β = {w}, then T is a tree obtained from P 3 by sequentially adding pendant P 2 ’s.(iii) If β = {v} and A[u] = 0, then P s( A[T̄ ]) =

n−32

.

(iv) If β = {v} and A[u] = 0, then T is a tree obtained from P 3 by sequentially adding pendant P 2 ’s.

Proof. Let ᾱ = α ∩ V (T̄ ), i.e., ᾱ = α \ {u, v}, and let ¯ A = A[T̄ ]. Denote by w1, . . . , w r (all) the neigh-bors of w in T different from v , and T i , for 1 i r , the branch of T at w containing w i . Finally, set

αi = α ∩ V (T i ) and Ai = A[T i ], for 1 i r .(i) Suppose that β = {u}. Then v w is an edge of T (α), and from Theorem 2.2(iv), { w1, . . . , wr } ⊆ α.Since {u, w 1, . . . , wr } ⊆ α , thus u is a P-vertex of A(w1, . . . , w r ), equivalently, u is a P-vertex of

A[u, v, w]. Now by Lemma 2.4(i), we can get that m A[u,v,w](0) = m A[w](0).Note that {w1, . . . , w r } ⊆ ᾱ. Clearly,

A(ᾱ) = A[u, v, w] ⊕ A 1(α1) ⊕ · · · ⊕ A r (αr )

and

¯ A(ᾱ) = A[w] ⊕ A 1(α1) ⊕ · · · ⊕ A r (αr ).

So m A(ᾱ)(

0) =

m¯ A(ᾱ)(

0)

.

On the other hand, since u is a P-vertex of A, by Lemma 2.4(i), m A (0) = m ¯ A (0).We remark that ᾱ ⊆ α is a P-set of A, i.e., m A(ᾱ)(0) = m A (0) + |ᾱ|, so m ¯ A(ᾱ)(0) = m ¯ A (0) + |ᾱ|, i.e.,

ᾱ is a P-set of ¯ A. Clearly, |ᾱ| = |α| − 1 = n−32 . Thus, by Proposition 1.1, P s(¯ A) = n−32 .

(ii) Suppose that β = {w}. Note that αi ∪ {w}, for 1 i r , is a P-set of A, which implies that αiis a P-set of A(w). Since T i is a component of T (w), we have αi is a P-set of Ai , for 1 i r . Now,similarly to the proof of Lemma 5.2, we can deduce that T is a tree obtained from P 3 by sequentially

adding pendant P 2’s.

(iii) Suppose that β = {v} and A[u] = 0. Note that ᾱ ⊆ α is a P-set of A, i.e., m A(ᾱ)(0) =m A (0) + |ᾱ|. Since A[u] = 0, from Lemma 2.4(ii), we can get m A (0) = m ¯ A (0) and m A(ᾱ)(0) = m ¯ A(ᾱ)(0).

Now we have m ¯ A(ᾱ)(0) = m ¯ A (0) + |ᾱ|, i.e., ᾱ is a P-set of ¯ A. Clearly, |ᾱ| = |α| − 1 = n−32 and, from

Proposition 1.1, P s( ¯ A) = n−32 .(iv) Suppose that β = {v} and A[u] = 0. Taking into account that u is an isolated vertex of T (α)

and A[u] = 0, we have, from Theorem 2.2(iii), m A (0) = 0, A[w] = 0, and {w1, . . . , wr } ⊆ α.Since v is a P-vertex of A, we have m A(v)(0) = m A (0) + 1 = 1. From the fact that u is an isolated

vertex of T (v) and A[u] = 0, it follows that m A(u,v)(0) = m A(v)(0) = 1. Additionally, from the fact


10/11


that v is a P-vertex of A and m A(u,v)(0) = m A (0), we can deduce that m A(v,w)(0) = m A (0) = 0, fromLemma 2.3. Thus Ai is nonsingular, for 1 i r .

Now, since {v, w 1, . . . , wr } is a P-set of A, thus v is a P-vertex of A(w1, . . . , w r ), equivalently, v isa P-vertex of A[u, v, w ], i.e., m A[u,w](0) = m A[u,v,w](0) + 1. Recall that A[u] = 0 and A[w] = 0. Thus

m A[u,w](0) = 1 and so A[u, v, w ] is nonsingular.Similarly to the proof of Lemma 3.5, we have αi is a P-set of Ai , for 1 i r . Moreover, analo-gously to the proof of Lemma 5.2, we can deduce that T is a tree obtained from P 3 by sequentially

adding pendant P 2 ’s.

We are ready now to provide our second main classification.

Theorem 5.4. Let T be a tree on n 3 vertices, where n is odd. The following two conditions are equivalent :

(a) There exists a matrix A ∈ S (T ) such that P s( A) = n−1

2 .

(b) T is a tree obtained from P 3 by sequentially adding pendant P 2’s.

Proof. (a) ⇒ (b) Suppose that A is a matrix in S (T ) such that P s( A) = n−1

2 .

Let us start from T , deleting pendant P 2’s repeatedly, until there exists an outer P 3 . This must

occur as n is odd. The deletion process is listed as follows:

T = T 1 → ·· · → T k,

where T i+1 is the tree obtained from T i by deleting a pendant P 2 , for 1 i k − 1.Let Ai = A[T i ] and ni be the order of Ai , for 1 i k. Clearly, A i ∈ S (T i ).By Proposition 1.1, P s( Ai )

ni −12

, for 1 i k.

Case 1. P s( Ak) = nk−1

2 .

Observe that there exists an outer P 3 in T k , from Lemma 5.2, T k is a tree obtained from P 3 by

sequentially adding pendant P 2’s, so T is a tree obtained from P 3 by sequentially adding pendant

P 2 ’s.

Case 2. P s( Ak) < nk −1

2 .

Recall that P s( A) = n−1

2 , i.e., P s( A1) = n1−1

2 , thus there exists an index, say j, such that P s( A j ) =n j −1

2 and P s( A j+1) < n j+1−1

2 .

Let α∗ be a P-set of A j with |α∗| =

n j −1

2 . From Theorem 2.2(ii)–(iv), each component of T j (α∗)

contains at most one edge.Suppose that T j+1 is the tree obtained from T j by deleting the pendant P 2 = u v , where u is a

terminal vertex, and w is the neighbor of v in T j different from u.

Note that P s( A j+1) < n j+1−1

2 , then by Lemma 3.1, exactly one of u, v, w is in α∗. Moreover, from

Lemma 5.3, T j is a tree obtained from P 3 by sequentially adding pendant P 2’s. So T is a tree obtained

from P 3 by sequentially adding pendant P 2 ’s.

In conclusion, (a) ⇒ (b).(b) ⇒ (a) Suppose that T is a tree obtained from P 3 by sequentially adding pendant P 2 ’s. Clearly,

P s( A(P 3)) = 1. By Lemma 4.1, we have

P s A(T )= P s A(P 3)+ n − 3

2

= n − 1

2

.

So (a) follows.

As in the even case, the path of odd order reveals to be the most simple example for the odd case.

In fact, the matrices


11/11


0 1 01 0 1

0 1 0

and

0 1 01 0 1

0 1 1

are, respectively, singular and nonsingular. Therefore, both Jacobi matrices of odd order

0 1

1. . .

. . .

. . . . . . 1

1 0

and

0 1

1. . .

. . .

. . . . . . 1

1 0 1

1 1

attain the bound of Proposition 1.1. The same is true if we append a path of even order to the central

vertex of P 3 which is the special type of the so-called comet considered by Kim and Shader in [12]:

0 11

. . . . . .

. . . . . . 1

1 0 1

1 0 1 1

1 0 0

1 0 0

and

0 11

. . . . . .

. . . . . . 1

1 0 1

1 0 1 1

1 0 0

1 0 1

.

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187 (2) (2008) 251–261.[8] C.D. Godsil, Algebraic matching theory, Electron. J. Combin. 2 (1995) #R8.

[9] R.A. Horn, C.R. Johnson, Matrix Analysis, second edition, Cambridge University Press, New York, 2013.

[10] C.R. Johnson, A. Leal Duarte, C.M. Saiago, The Parter–Wiener theorem: refinement and generalization, SIAM J. Matrix Anal.

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