The maximum multiplicity of the largest k -th eigenvalue in a matrix whose graph is acyclic or unicyclic

Discrete Mathematics 342 (2019) 2924–2950

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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc

The maximum multiplicity of the largest k-th eigenvalue in a matrix whose graph is acyclic or unicyclic ∗

Mohammad Adm a,b , , Shaun M. Fallat c a

Department of Applied Mathematics and Physics, Palestine Polytechnic University, Hebron, Palestine Department of Mathematics and Statistics, University of Konstanz, Konstanz, Germany c Department of Mathematics and Statistics, University of Regina, Regina, Canada b

article

info

Article history: Received 20 February 2018 Accepted 17 June 2019 Available online 4 July 2019 Keywords: Graphs Symmetric matrices Maximum nullity Partial inertia Trees Unicyclic graph

a b s t r a c t Given a graph G we are interested in studying the symmetric matrices associated to G with a fixed number of negative eigenvalues. For this class of matrices we focus on the maximum possible nullity. For trees this parameter has already been studied and plenty of applications are known. In this work we derive a formula for the maximum nullity and completely describe its behavior as a function of the number of negative eigenvalues. In addition, we also carefully describe the matrices associated with trees that attain this maximum nullity. The analysis is then extended to the more general class of unicyclic graphs. Further our work is applied to re-describing all possible partial inertias associated with trees, and is employed to study an instance of the inverse eigenvalue problem for certain trees. Crown Copyright © 2019 Published by Elsevier B.V. All rights reserved.

1. Introduction Suppose G = (V , E) is a simple graph with vertex set V = V (G) = {1, 2, . . . , n} and edge set E = E(G). To a graph G we associate the collection of real n × n symmetric matrices defined by S (G) = {A : A = AT , for i ̸ = j, aij ̸ = 0 ⇔ {i, j} ∈ E }.

The main diagonal entries of any A in S (G) are free to be chosen. The class of matrices S (G) has been well studied recently (see [10,11] and the references therein), and there has been significant development of the parameters M(G) (maximum multiplicity or nullity over S (G)), mr(G) (minimum rank over S (G)), and their positive semidefinite versions; see, also, the works [7,10,11]. In addition, research interest has grown in connecting these parameters to various combinatorial properties of graphs. For example, the inverse eigenvalue problem for graphs (see [14]) continues to receive considerable and deserved attention, as it remains one of the most interesting unresolved issues in this subject area. The collection of positive semidefinite matrices in S (G) can be viewed as restricting the number of negative eigenvalues on the matrices in S (G) (in this case zero). More recently, there has been interest in studying the matrices in S (G) with a fixed number, say q, of negative eigenvalues. To this end, suppose G is a graph on n vertices. Then for q = 0, 1, 2, . . . , n, we let Sq (G) denote the subset of matrices in S (G) with exactly q negative eigenvalues. Thus S0 (G) consists of all the positive semidefinite matrices in S (G), and ∪q Sq (G) = S (G). For such a fixed q, we denote the maximum nullity over Sq (G) as Mq (G). It follows that Mq (G) ≤ M(G) for any graph G. ∗ Corresponding author at: Department of Applied Mathematics and Physics, Palestine Polytechnic University, Hebron, Palestine. E-mail address: [email protected] (M. Adm). https://doi.org/10.1016/j.disc.2019.06.030 0012-365X/Crown Copyright © 2019 Published by Elsevier B.V. All rights reserved.

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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In connection with the classes S0 (G) and S (G), a combinatorial parameter called zero forcing number has been developed, originally motivated for bounding the maximum nullity. Suppose G = (V , E) is a given graph. A subset Z ⊆ V defines an initial set of black vertices (while all vertices not in Z are white). The color change rule is a rule by which we can change the color of a white vertex w to black if w is the unique white neighbor of a black vertex u; in this case we say u forces w . A zero forcing set for G is a subset of vertices Z such that if initially the vertices in Z are colored black and the remaining vertices are colored white, applying the color change rule until no more changes are possible leads to all vertices being black. The zero forcing number, Z (G), is the minimum size over all zero forcing sets Z ⊆ V . For any graph G, M(G) ≤ Z (G) ([1, Proposition 2.4]). Zero forcing for positive semidefinite matrices has also been studied. For a given graph G, let B be the set consisting of all the black vertices. Let W1 , . . . , Wk be the sets of vertices of the k components of G after removing the vertices B (note that it is possible that k = 1). Let w ∈ Wi . If u ∈ B and w is the only white neighbor of u in the subgraph induced by Wi ∪ B, then change the color of w to black. The positive semidefinite zero forcing number of a graph G, denoted by Z+ (G), is then the size of the smallest such zero forcing set. As noted above, the class Sq (G) has been considered, see, for example, the work in [8]. In this paper, they defined the q-zero forcing number of a graph, and denote it by Zq (G). Even though the Zq forcing scheme is slightly different from the two described above, it can still be used to bound the nullity of matrices describing G from above. Namely, if we let A be in Sq (G), then null(A) ≤ Mq (G) ≤ Zq (G), where null(A) denotes the nullity of A. It was noted in [8] that Z0 (G) = Z+ (G) and that Zn (G) = Z (G), where n is the number of vertices of G. For an m × n matrix A (or A ∈ Rm,n ), α ⊆ {1, 2, . . . , m}, and β ⊆ {1, 2, . . . , n}, the submatrix of A lying in rows indexed by α and the columns indexed by β will be denoted by A[α|β]. Similarly, A(α|β ) is the submatrix obtained from A by deleting the rows indexed by α and columns indexed by β . If A is square and α = β , then the principal submatrix A[α|α] is abbreviated to A[α], and the complementary principal submatrix to A(α ). For brevity, if α = {i}, then A(α ) is denoted by A(i), or Aii . In the special case when α or β is a contiguous index set, say for example, α = {2, 3, 4, 5, 6}, we abbreviate the notation A[{2, 3, 4, 5, 6}|β] by A[2, 3, 4, 5, 6|β]. By convention, A[∅] is defined to have both rank and nullity equal to 0, and we say A[∅] is a positive definite matrix. Finally, we let Eii denote the standard elementary basis matrix with a unique 1 in position (i, i) and zeros elsewhere. If G = (V , E) is a graph, and U ⊆ V , then we let H = G[U ] denote the subgraph induced by U. In this case, we may denote the vertex set of H by V (H) and its edge set by E(H). Further, if H is an induced subgraph of G, then for A ∈ S (G), we let A[H ] denote the principal submatrix of A indexed by the vertices in H, namely V (H). Further, if two vertices u and v are adjacent, we write u ∼ v , and for V ′ ⊆ V , we let N(V ′ ) = v ∈ V \ V ′ | v ∼ u in G for some u ∈ V ′ .

{

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Our interest here is to study the optimal matrices associated with the maximum nullity over the subclass Sq (G). When q = 0, there has been a tremendous amount of work done concerning the maximum positive semidefinite nullity of graphs. For example, it is well known that M0 (T ) = 1 for any tree T (see, for example, [7]). In addition, there is a rather useful result in [6] that we will make use of repeatedly, especially when considering matrices in S1 (T ) for a tree T . We state this lemma now for reference. Lemma 1.1 ([6, Lemma 3.2]). If A ∈ S0 (G) and M0 (G) = null(A), then every row (column) in A is linearly dependent on the other rows (columns). For the matrix A, σ (A) denotes the spectrum of A and for λ ∈ σ (A), multA (λ) denotes the multiplicity of the eigenvalue λ. Thus M(G) = max{multA (λ) : λ ∈ σ (A) and G(A) = G}. Two n × n real matrices A and B are said to be congruent if there exists an invertible n × n matrix C such that B = CAC T . An important result concerning congruence of matrices and the preservation of inertia is known as Sylvester’s law of inertia (see also [15]). This theorem states that two symmetric matrices are congruent if and only if they have the same inertia, i.e., the number of positive, negative, and zero eigenvalues (counting multiplicities). We denote by Pn and Cn the path on n vertices and the cycle on n vertices, respectively. For a given graph G, P(G) is the path cover number, namely, the minimum number of vertex disjoint paths, occurring as induced subgraphs of G, that cover all the vertices of G, and Γ (G) is the maximum of p − q such that the deletion of q vertices from G leaves p paths. For trees it is well-known that M(T ) = P(T ) = Γ (T ) (see [16]). For the graph G, we let e(G) = |E |, and let deg(v ) be the number of edges incident with the vertex v (called the degree of v ). We define, ∆(G) = maxv deg(v ). Further, we let MDk (G) denote the maximum disconnection number, or the largest number of components that result from deleting a collection of k vertices from G. In particular, for a tree T , we have MD1 (T ) = ∆(T ). The following parameter will be used throughout our paper. For a given graph G, q0 (G) := min q | (MDq (G) − q) attains its maximum value .

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In the case G is a tree or a forest, the definition of q0 (G) simplifies to q0 (G) := min q | MDq (G) − q = P(G) .

{

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By Observation 1.2, MDq (G) − q < P(G) for all q < q0 (G), and q0 (Pn ) = 0 and q0 (K1,n−1 ) = 1. Moreover, by repeatedly applying the fact P(G − v ) ≤ P(G) + 1 the following observation follows Observation 1.2.

For any graph G on n vertices,

MDq (G) − q ≤ P(G), q = 0, 1, . . . , n. The following lemma is a basic lemma concerning the maximum disconnection number of a forest. Lemma 1.3. Let F be a forest and let v, u ∈ V (F ) be such that deg(v ) = 1 and v ∼ u. Then for k = 1, 2, . . . , q0 (F ) we have MDk−1 (F − v ) + 1 ≤ MDk (F ), MDk−1 (F − {v, u}) + 1 ≤ MDk (F ). Proof. Let Vk−1 be a set of k − 1 vertices whose deletion gives MDk−1 (F − v ) (MDk−1 (F − {v, u})) components and let w ∈ V (F − Vk−1 ) be chosen such that degF −Vk−1 (w) ≥ 2 (such w exists since k ≤ q0 (F )). Then deleting Vk−1 ∪{w} (Vk−1 ∪{u}) will give at least MDk−1 (F − v ) + 1 (MDk−1 (F − {v, u}) + 1) components of F . Hence MDk−1 (F − v ) + 1 ≤ MDk (F ) and MDk−1 (F − {v, u}) + 1 ≤ MDk (F ). □ Given a graph G = (V , E) and subsets V ′ ⊂ V , E ′ ⊂ E, we let G − V ′ and G − E ′ denote the graphs obtained by removing the vertices V ′ or the edges E ′ , respectively. For brevity, if V ′ = {v}, then we let G − v be the graph G − {v}, and let G − e denote the graph obtained from G by removing the edge e. A simple consequence of the Cauchy interlacing inequalities is that for any symmetric matrix B, the multiplicity of any eigenvalue can change by at most one upon removal of a common row and column. For trees, the work of Parter [21] and Wiener [22] demonstrates that for multiplicity greater than one there must always be a vertex whose deletion raises the multiplicity; such a vertex is often called a Parter–Wiener (PW ) vertex (see also [9] for work on an inverse eigenvalue problem for trees). For reference we state this important result now. Theorem 1.4 ([17,21,22] Parter–Wiener Theorem). Let T be a tree and A ∈ S (T ). If λ ∈ σ (A) with multiplicity k ≥ 2, then there exists a vertex v in V (T ) such that λ ∈ σ (A(v )) with multiplicity k + 1. Moreover, v may be chosen so that deg(v ) ≥ 3 and there are at least three components T1 , T2 , and T3 of T − v with λ ∈ σ (A[Ti ]), i = 1, 2, 3. Our paper is organized as follows. In the next section we study the structure of matrices in S1 (T ) that attain M1 (T ). This is followed by work on computing the parameter Mq (T ) and studying the behavior of Mq (T ), the partial inertia of T , and some structure properties of the matrices in Sq (T ) that have nullity Mq (T ). Section 4 extends some of the previous analysis to investigations of Mq (G) beginning by studying cycles and generalized partial n-suns. The study for general unicyclic graphs is discussed in Section 5. Finally, in Section 6, we apply some of our techniques to revisit some known facts and examples concerning the spectra associated with trees. 2. Characterization of matrices in S1 (T ) with optimal nullity In this section, we concentrate on matrices in S1 (T ) for a given tree T , and derive a number of structural results concerning, in some sense, where the negative eigenvalue is ‘‘located" in the matrix. We begin with a known result that follows from the work in [5]. We offer an alternative proof of the fact below, which includes illustrations of the techniques that will be used throughout our work. Lemma 2.1.

Let T be a tree on n ≥ 3 vertices with maximum degree ∆. Then M1 (T ) = ∆ − 1.

Proof. First we prove that ∆ − 1 is an upper bound for M1 (T ). The proof uses induction on n. For n = 3, then T = P3 with ∆ = 2 and the result holds. Suppose it is true for all trees on m vertices with m < n. We now show it is true for all trees T on n vertices. Suppose, on the contrary, that there exists a tree T on n vertices such that M1 (T ) ≥ ∆. Without loss of generality, we may assume that there exists a matrix A ∈ S1 (T ) such that null(A) = ∆. Let v ∈ V (T ) be such that s := deg(v ) ≥ 2. Then, T − v = ∪si=1 Ti and, by the Cauchy interlacing inequalities, all of A[Ti ], i = 1, . . . , s must be positive semidefinite except possibly at most one since A ∈ S1 (T ). Without loss of generality, assume that A[T1 ] is positive semidefinite and by permutation similarity, we may assume that A has the following form:

⎡

A11 A = ⎣aT12 0

a12 avv a23

⎤

0 aT23 ⎦ , A33

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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where A11 = A[T1 ] or is similar to A[T1 ], with |V (T1 )| = k, and a12 ∈ Rk is a vector whose only nonzero entry is akv . We distinguish the following two cases: Case 1. A11 is nonsingular. It is easy to show that A and A′ are congruent, where A′ is given as follows:

⎡

⎤

0 a′vv a23

A11 A′ = ⎣ 0 0

0 aT23 ⎦ . A33

Hence null(A) = null(A′ ) = null(B), where

[ B=

a′vv a23

aT23 A33

]

∈ S1 (T ′ ),

and T ′ = T − V (T1 ). By the induction hypothesis, null(B) ≤ M1 (T ′ ) ≤ ∆(T ′ ) − 1 ≤ ∆ − 1, which is a contradiction. Case 2. A11 is singular. Since A11 ∈ S0 (T1 ) we have null(A11 ) = 1 = M0 (T1 ). By Lemma 1.1, A and A′′ are congruent, where A′′ is given as follows: 0 0 av k 0

A11 (k) ⎢ 0 ′′ A =⎣ 0 0

⎡

0 akv avv a23

0 0 ⎥ ⎦. aT23 A33

⎤

Now, by congruence, we use akv (av k ) to eliminate all the nonzero entries in a23 (aT23 ). Hence A′′ is congruent to

⎡

0 0 akv 0

A11 (k) ⎢ 0 C =⎣ 0 0

0 akv avv 0

⎤

0 0 ⎥ , 0 ⎦ A33

which implies that A33 is positive semidefinite with deg(v ) − 1 components (direct-sum of positive semidefinite submatrices corresponding to trees). It is easy to note that null(A) = null(C ) = null(A33 ) ≤ deg(v ) − 1 ≤ ∆ − 1, which is a contradiction. Hence M1 (T ) ≤ ∆ − 1. Now we prove that this upper bound is always attained. Let v ∈ V (T ) be such that deg(v ) = ∆. Then T − v has ∆ components, say T1 , T2 , . . . , T∆ . For each i = 1, . . . , ∆, let Ai ∈ S0 (Ti ) with nullity 1 and define B := A2 ⊕ A3 ⊕ · · · ⊕ A∆ and A:

⎡

A1 A = ⎣aT12 0

a12 avv a23

⎤

0 aT23 ⎦ , B

where a12 ∈ Rk is a vector whose only nonzero entry is akv and a23 is a vector whose only nonzero entries are those corresponding to the edges between vertex v and its neighbors. Hence A ∈ S (T ). As in Case 2. it is easy to show that A is congruent to

⎡

A1 (k) ⎢ 0 ′ B =⎣ 0 0

0 0 akv 0

0 akv avv 0

⎤

0 0⎥ . 0⎦ B

Since B′ has exactly one negative eigenvalue and nullity ∆ − 1 we conclude that A ∈ S1 (T ) with nullity ∆ − 1. Hence M1 (T ) = ∆ − 1. □ The next few results are concerned with the location of the unique negative eigenvalue in a certain sense. A common theme that appears is the connection to positive semidefinite matrices relative to a tree. Lemma 2.2.

Let T be a tree and A ∈ S1 (T ). Suppose there exists v ∈ V (T ) such that

A(v ) = A1 ⊕ · · · ⊕ Ak , k ≥ 2, and there exists i ∈ {1, . . . , k} with Ai being positive semidefinite and null(Ai ) = 1. Then A(v ) is positive semidefinite.

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Proof. Without loss of generality, let i = 1 and suppose that A has the following form:

⎡

av,v

av,vi

...

0

1

⎢ av,v i ⎢ ⎢ 01 ⎢ ⎢ . ⎢ .. ⎢ ⎢ 0 ⎢ ⎢ av,vi ⎢ 2 ⎢ 0 ⎢ ⎢ .. ⎢ . ⎢ ⎢ 0 ⎢ ⎢ . ⎢ .. ⎢ ⎢ av,v i ⎢ ⎢ 0k ⎢ ⎢ . ⎣ ..

0

av,vi

2

...

0

...

0

av,vi

k

...

0

A1

0

...

0

0

A2

...

0

.. .

.. .

..

.. .

0

0

...

.

Ak

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(1)

0

} { with N(v ) = vi1 , . . . , vik and where A1 is positive semidefinite with nullity 1. By Lemma 1.1, A is congruent to B, where B is given as follows:

⎡ a v,v ⎢ av,vi1 ⎢ ⎢ 0 ⎢ . ⎢ . ⎢ . ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ B = ⎢ .. ⎢ . ⎢ 0 ⎢ ⎢ . ⎢ . ⎢ . ⎢ ⎢ 0 ⎢ . ⎣ . . 0

av,i1 0 0

.. .

0 0

... ...

0 0

0 0

... ...

0 0

... ...

0 0

... ...

A1 (1)

0

...

0

0

A2

...

0

.. .

.. .

..

.. .

0

0

...

0 0

.. .

0

.. .

.

0

.. .

Ak

0 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0

Hence A1 (1), A2 , . . . , Ak are all positive semidefinite. Therefore, A(v ) is positive semidefinite. □ The next result is a basic, yet useful, property of acyclic positive semidefinite matrices with maximum nullity. Lemma 2.3. definite.

Let T be a tree on n vertices and A ∈ S0 (T ) with nullity 1. Then any proper principal submatrix of A is positive

Proof. Assume that A[α] is singular. Since A is positive semidefinite, it follows that any principal submatrix that includes A[α] must also be singular. So, A(k) is singular for some k ∈ {1, . . . , n}. Since null(A) = M0 (T ) = 1, it follows from Lemma 1.1, that the rank of A equals the rank of A(k). But the rank of A is n − 1 which implies A(k) is invertible. Thus we have reached a contradiction. □ Theorem 2.4. Let T be a tree with ∆ ≥ 3 and let A ∈ S1 (T ) satisfy null(A) = M1 (T ). Then there exists a unique vertex v ∈ V (T ) such that A(v ) is positive semidefinite and null(A(v )) = M1 (T ) + 1. Proof. By Theorem 1.4, there exists v such that deg(v ) ≥ 3 and null(A(v )) = null(A) + 1 = ∆ since, by Lemma 2.1, null(A) = ∆ − 1 ≥ 2. Assume that T − v = ∪ki=1 Ti , where k = deg(v ). Hence A(v ) = A[T1 ] ⊕ A[T2 ] ⊕ · · · ⊕ A[Tk ] and by Theorem 1.4 and interlacing, since A ∈ S1 (T ), there exists i0 ∈ {1, . . . , k} such that A[Ti0 ] is positive semidefinite with nullity 1. Hence by Lemma 2.2, A(v ) is positive semidefinite. Since for each i = 1, . . . , k, Ti is a tree and A[Ti ] is positive semidefinite, we conclude that null(A[Ti ]) ≤ 1. Therefore, null(A(v )) ≤ k. Hence ∆ = null(A(v )) ≤ k ≤ ∆. Thus k = ∆ and for each i = 1, . . . , k, A[Ti ] is a positive semidefinite matrix with nullity 1. Suppose there exists another vertex

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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u such that A(u) is positive semidefinite and null(A(u)) = null(A) + 1. Assume that T − u = ∪ki=1 Ti′ . Let Tk and Tk′′ be the components that contain u and v , respectively. Since A(v ) is positive semidefinite and k = ∆ it follows that for each i = 1, . . . , k, A[Ti ] is a positive semidefinite matrix with nullity 1, so by Lemma 2.3, we conclude that A[Ti′′ ] are positive definite for all i′ = 1, . . . , k′ − 1. Therefore u cannot be a Parter–Wiener vertex. □ Corollary 2.5. Let T be a tree and let A ∈ S1 (T ) with null(A) = M1 (T ) and ∆ ≥ 3. Then A has at most one negative entry on its main diagonal. Let T be a tree with ∆ ≥ 3, and let A ∈ S1 (T ) with null(A) = M1 (T ). If aii < 0 for some i ∈ {1, . . . , n}, then

Theorem 2.6. deg (vi ) = ∆.

Proof. Suppose on the contrary that there exists a matrix A ∈ S1 (T ) such that null(A) = M1 (T ) with aii < 0 and deg (vi ) = k < ∆. Without loss of generality, we assume that i = 1 and N(v1 ) = {v2 , . . . , vk+1 }. Hence A has the following form:

⎡

a12

a11 a12 a23

...

a23

⎢ ⎢ ⎢ ⎢ .. ⎢ ⎢ . ⎢ A = ⎢ a1,k+1 ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ .. ⎣ .

a1,k+1

0

...

0

A22

A23

AT23

A33

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 where A22 is a k × k diagonal matrix since T is a tree. It follows that A is congruent to

⎡

a11 ⎢ 0 ⎢ ⎢ 0

⎢ ⎢ ⎢ ⎢ B=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

.. .

0 0 0

.. .

0

0

...

0

0

0

...

B22

A23

AT23

A33

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 where B22 ∈ S (Kk ) as A22 is a k × k diagonal matrix. Hence B ∈ S (G), where G is the graph obtained from T − v1 by adding all possible edges between the vertices v2 , . . . , vk+1 . Since A ∈ S1 (T ) and a11 < 0, we have that B(1) is positive semidefinite and null(A) = null(B) = null(B(1)). If k = 1, then B(1) ∈ S0 (T − v1 ) and v1 is a leaf. Thus Z+ (T − v1 ) = 1 which implies that null(B(1)) ≤ 1. Since null(A) = null(B(1)) we have null(A) ≤ 1, which contradicts the condition null(A) = M1 (T ) = ∆ − 1 ≥ 2. So assume k ≥ 2 and let T − v1 = ∪ki=+21 Tvi , where vi ∈ V (Tvi ), i = 2, . . . , k + 1. Initially color the vertices {v2 , . . . , vk } black in the graph G − v1 . Then for some s ∈ {2, . . . , k}, vs forces vk+1 in the Z+ -game. Then each vi will force all the vertices in the tree Tvi , i = 2, . . . , k + 1. Hence Z+ (G − v1 ) ≤ k − 1. Therefore, M1 (T ) = ∆ − 1 = null(A) = null(B(1)) ≤ Z+ (G − v1 ) ≤ k − 1 < ∆ − 1, which is a contradiction. Hence deg(v1 ) = ∆.

□

Proposition 2.7. Let A ∈ S1 (Pn ) be such that null(A) = M1 (Pn ) = 1. Then A has at most two negative entries on its main diagonal. Moreover, if aii < 0 and ai+1,i+1 < 0 for some i ∈ {1, . . . , n − 1}, then det A[1, . . . , i] det A[i + 1, . . . , n] > 0. Proof. Since A ∈ S1 (Pn ), by the Cauchy interlacing inequalities A has at most two negative entries on its main diagonal and they correspond to adjacent vertices. Suppose that aii < 0 and ai+1,i+1 < 0. By the Cauchy interlacing inequalities, A[1, . . . , i − 1] and A[i + 2, . . . , n] are positive semidefinite. If A[1, . . . , i − 1] (or A[i + 2, . . . , n]) had nullity 1, then applying Lemma 2.2 to A ∈ S1 (Pn ), it follows that A(i) (or A(i + 1)) is positive semidefinite. However, this is not possible since ai+1,i+1

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in A(i) (or ai,i in A(i + 1)) is negative. Hence A[1, . . . , i − 1] and A[i + 2, . . . , n] are invertible and A is congruent to 0

⎡

0

.. .

⎢ A[1, . . . , i − 1] ⎢ ⎢ ⎢ ⎢ 0 ... 0 ⎢ B=⎢ ⎢ 0 ... 0 ⎢ ⎢ ⎢ ⎣ 0

⎤

.. .

0 a′ii

0

ai+1,i 0

0 ai+1,i+1 a′i+1,i+1 0

0

0

.. .

.. .

⎥ ⎥ ⎥ ⎥ ⎥ 0 ... 0 ⎥ ⎥, 0 ... 0 ⎥ ⎥ ⎥ ⎥ A[i + 2, . . . , n] ⎦

where A[1, . . . , i − 1] and A[i + 2, . . . , n] are positive definite. Therefore

([ null

a′ii

ai,i+1

ai,i+1 a′i+1,i+1

])

= 1,

which implies that a′ii a′i+1,i+1 > 0. By the definition of B, it is easy to verify that det A[1, . . . , i] = a′ii det A[1, . . . , i − 1] det A[i + 1, . . . , n] = a′i+1,i+1 det A[i + 2, . . . , n], which completes the proof. □ Example 2.8. Here we include an example of a matrix in S1 (P4 ) that attains M1 (P4 ) = 1 and has exactly two negative main diagonal entries.

⎡

1 ⎢1 A=⎣ 0 0

1 −2 3 0

⎤

0 3 −2 1

0 0⎥ . 1⎦ 1

The remainder of this section is devoted to further structural-type results regarding optimal matrices in S1 (T ). Theorem 2.9. Let T be a tree and let A ∈ S1 (T ) satisfy null(A) = M1 (T ) and avv = 0. If deg(v ) = 1 and v is adjacent to w , then deg(w ) = ∆ and each direct summand of A(v, w ) has nullity 1. Proof. Suppose that N(w ) = {v1 , . . . , vk } ∪ {v}. Then A can be written as

⎡

0 ⎢ avw

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ 0 ⎢ ⎢ ⎢ ⎢ ⎣

avw aww aw,v1 aw,v2

aw,v1

aw,v2

0 . . . aw,vk

.. .

⎤ 0

A(v, w )

aw,vk 0

.. .

0 Using avw as a pivot, it follows that A is congruent to

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ B=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0 avw 0 0

avw aww 0 0

0 0

0 0

0

0

.. . .. .

.. . .. .

0 0

0 0

... ...

0 0

0 0

A(v, w )

... ...

0 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

...

0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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Hence null(A) = null(A(v, w )), where A(v, w ) is positive semidefinite. Since each direct summand of A(v, w ) has nullity at most one,

∆ − 1 = M1 (T ) = null(A) = null(A(v, w )) ≤ k. Hence deg(w ) = k + 1 ≥ ∆. Therefore, deg(w ) = ∆ and null(A(v, w )) = ∆ − 1 and so each direct summand of A(v, w ) has nullity 1. □ Theorem 2.10. Let T be a tree and A ∈ S1 (T ) satisfy null(A) = M1 (T ) and let v ∈ V (T ) be such that deg(v ) = k, where 2 ≤ k < ∆. If T1 , T2 , . . . , Tk are the components of T − v , then A[Ti ] are positive definite for all i = 1, . . . , k except possibly for at most one of such A[Ti ]. Proof. By the generalized Cauchy interlacing inequalities, at most one of A[Ti ], i = 1, . . . , k is indefinite and the remaining blocks are positive semidefinite since A ∈ S1 (T ). Suppose that there exists i0 such that A[Ti0 ] is an ℓ×ℓ positive semidefinite with nullity 1. Then by Lemma 1.1 A is congruent to 0

⎡ ⎢ A[Ti ](ℓ) ⎢ 0 ⎢ ⎢ ⎢ ⎢ 0 ... 0 B=⎢ ⎢ 0 ... 0 ⎢ ⎢ ⎢ ⎣ 0

0

.. .

.. .

0 0 ai0 ,v 0

0 ai0 ,v avv 0

0

0

.. .

⎤ 0

⎥ ⎥ ⎥ ⎥ ⎥ 0 ... 0 ⎥ ⎥, 0 ... 0 ⎥ ⎥ ⎥ ⎥ A[(T − v ) − V (Ti0 )] ⎦

.. .

where A[(T − v ) − V (Ti0 )] is positive semidefinite with nullity at most k − 1. However, this leads to a contradiction since k − 1 < ∆ − 1. □ Theorem 2.11. Let T be a tree, A ∈ S1 (T ) satisfy null(A) = M1 (T ), and avv = 0 for some v ∈ V (T ) and let N(v ) = k {w1 , . . . , wk }, k ≥ 2 and T − wj = ∪s=j 1 Tws j , where kj = deg(wj ) and Tw1 j are components that have v as one of their s vertices, j = 1, . . . , k. Then A[Twj ] is positive definite, j = 1, . . . , k, s = 2, . . . , kj . Moreover, if k ≥ 3, then deg(v ) = ∆ and w

w

det A[Tv i ] = 0, where Tv i is the branch of T − v that contains wi , i = 1, . . . , k.

Proof. Let w ∈ N(v ) be such that N(w ) ≥ 2. Then observe that A can be written as

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0

∗

∗ ∗ .. .

∗

...

∗

A[Tw1 − v]

.. .

0

0

...

0

.. .

0

0

0 aww 0

0

...

0

0

0

0

∗

0

...

...

0

∗

0

...

0

0

...

0

0

...

0

...

...

.. .

∗

0

...

A[Tw2 ]

0

...

0

0

A[Tw3 ]

...

0

.. .

.. .

..

.. .

0

0

...

0 0 0

.. .

.. .

0

.. . ∗ 0

0

.. .

.

0

0

∗

0 0

.. .

0

∗

0 0 0

.. .

0

.. .

∗ avw 0 0

avw 0 0

A[Twkw ].

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0

Since A ∈ S1 (T ), all A[Tws ], for s = 2, . . . , kw are positive semidefinite. If one of A[Tws ], s = 2, . . . , kw has nullity 1, then by Lemma 2.2, A(w ) is positive semidefinite, which is a contradiction since A[T − w] is not positive semidefinite.

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Assume now that k ≥ 3. Since for each s = 2, . . . , kj , A[Tws j ] is positive definite, A is congruent to 0

⎡

⎢ avw1 ⎢ a ⎢ vw2 ⎢ . ⎢ .. ⎢ ⎢ a ⎢ vwk ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ .. ⎢ . ⎢ ⎢ 0 ⎢ B=⎢ 0 ⎢ ⎢ 0 ⎢ . ⎢ . ⎢ . ⎢ ⎢ 0 ⎢ . ⎢ . ⎢ . ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ . ⎣ . .

avw1

∗ 0

... ... ... .. .

avw2 0

.. .

∗ .. .

0

0

avwk 0 0

0

0

...

0

0

0

...

0

...

0

0

...

0

0

...

0

0

A1

0

...

0

0

0

A2

...

0

.. .

.. .

.. .

..

.

.. .

0

0

0

...

Al

...

.. . ∗

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 where B[1, . . . , k + 1] ∈ S (K1,k ) and Ai are positive definite for all i = 1, . . . , l. Hence null(A) = null(B) = null(B[1, . . . , k + 1]) ≤ k − 1, w

which is a contradiction if k < ∆. Therefore k = ∆. Moreover, bii = 0 for all i = 2, . . . , k + 1 and therefore det A[Tv i ] = 0, i = 1, . . . , k. □ 3. Mq of trees In this section, we generalize some of our ideas and techniques to studying the maximum nullity of matrices in Sq (T ). We recall a useful term concerning a special vertex in a tree. A vertex v is called an appropriate vertex if its deletion from G has at least two components that are paths, in which both such paths are joined at the end to v . For any tree T with at least one vertex of degree three or more, it is known that T contains an appropriate vertex (see [20]). Appropriate vertices in trees are directly connected to the notion of pendant generalized stars in trees (see [10, Lemma 2.4]), and such vertices are known to have the property that P(T − v ) = P(T ) + 1 (see [4, Prop. 4.2]), for any tree T with appropriate vertex v . We begin with a theorem, which was proved in [5] as a by-product of their work on the inverse partial inertia problem for trees. Theorem 3.1 ([5]). Let T be a tree on n vertices. Then for q = 0, 1, . . . , q0 (T ) the following holds: Mq (T ) = MDq (T ) − q. We now begin our independent basic analysis on comparing Mq (T ) and MDq (T ). Lemma 3.2.

Let T be a tree on n vertices. Then for q = 0, 1, . . . , q0 (T ) we have

MDq (T ) ≥ 2q + 1. Proof. We proceed by induction on n. For n ≤ 4 and for q = 0, the result is obvious. In particular, this result holds if T is a path. Moreover, if T is a generalized star that is not a path, then q0 (T ) = 1, and the result holds. Assume the result holds for all trees on m vertices where m < n. Let T be a tree on n vertices, 1 ≤ q ≤ q0 (T ), and Uq be a set of vertices in V (T ) such that T − Uq has MDq (T ) components. Applying a basic induction argument, we may assume that there exists a vertex u of T with ℓ ≥ 3 such that T − u = ∪ℓi=1 Ti , where Ti is a path for each i = 1, . . . , ℓ − 1 and each of these paths is joined at the end to u. In particular, u is an appropriate vertex. If q − 1 ≤ q0 (Tℓ ), then by the induction hypothesis MDq−1 (Tℓ ) ≥ 2(q − 1) + 1 = 2q − 1.

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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Hence deleting u together with a set of (q − 1) vertices that produces MDq−1 (Tℓ ) from T we get at least (2q − 1 + ℓ − 1) components. From which it follows MDq (T ) ≥ 2q + (ℓ − 2) ≥ 2q + 1. If q − 1 > q0 (Tℓ ), then q0 (Tℓ ) + 1 < q ≤ q0 (T ). Since u is an appropriate vertex, it follows that P(T ) = P(T − u) − 1 = P(Tℓ ) + ℓ − 2. Using this we have MDq0 (T ) (T ) − q0 (T ) = P(T ) = MDq0 (Tℓ ) (Tℓ ) − q0 (Tℓ ) + ℓ − 2. After rearranging terms, q0 (T ) = q0 (Tℓ ) + MDq0 (T ) (T ) − MDq0 (Tℓ ) (Tℓ ) + 2 − ℓ.

(2)

Given the hypothesis on the vertex u, we claim that q0 (T ) ≤ q0 (Tℓ ) + 1. To see this we consider two possible cases: (1) u ∈ Uq0 (T ) , in which case we have MDq0 (T ) (T ) − MDq0 (Tℓ ) (Tℓ ) = ℓ − 1 which and (2) imply that q0 (T ) = q0 (Tℓ ) + 1; or (2) u ̸ ∈ Uq0 (T ) , which implies that deg(u) = ℓ = 3 and the vertex in Tℓ that is adjacent to u, call it v , must be in Uq0 (T ) . If v is in a set Uq0 (Tℓ ) for the component Tℓ , then it follows easily that q0 (T ) = q0 (Tℓ ). Otherwise q0 (T ) = q0 (Tℓ ) + 1. Hence q0 (Tℓ ) + 1 < q0 (T ) ≤ q0 (Tℓ ) + 1 which is a contradiction. □ By Theorem 3.1 and Lemma 3.2 we have the following corollary. Let T be a tree on n vertices. Then for q = 0, 1, . . . , q0 (T ) we have

Corollary 3.3.

Mq (T ) ≥ q + 1. Lemma 3.4. Let T be a tree on n vertices, U = u1 , u2 , . . . , uq be a set of q ≤ q0 (T ) vertices whose deletion results in k := MDq (T ) components, T − U = ∪ki=1 Ti , and N(U) ∩ V (Ti ) = Vi , i = 1, . . . , k. Then T can be written as

{

}

T = (∪ki=1 Ti ) ∪ H ∪ (∪ki=1 Gi ),

(3)

where H is the induced subgraph on U and Gi is the induced subgraph of T on the vertices (U ∩ N(Vi )) ∪ Vi , i = 1, . . . , k, such that |E(G1 )| = 1 and |E(Gi − {u1 , . . . , ui−1 })| = 1, for i = 2, . . . , q. Proof. The existence of (3) without restrictions on the graph Gi , i = 1, . . . , k is obvious. In the following we show that such a decomposition exists with |E(G1 )| = 1 and |E(Gi − {u1 , . . . , ui−1 })| = 1, i = 2, . . . , q. Let Gi be the induced subgraph of T on the vertices (U ∩ N(Vi )) ∪ Vi , i = 1, . . . , k. Since T is connected and T − U = ∪ki=1 Ti we have |E(Gi )| ≥ 1. Suppose that for all i = 1, . . . , k we have |E(Gi )| ≥ 2. Then T has at least (n − q) − k + 2k = n + (k − q) edges. By Lemma 3.2, k − q ≥ 1, which implies that T has at least n + 1 edges that is not possible since T is a tree. Therefore, there exists i0 such that |E(Gi0 )| = 1. Choose i0 to be 1 and u1 the unique vertex u ∈ U that is adjacent to a vertex in T1 . Delete u1 together with T1 and all components Tj such that Tj coincides with one of the components of T − u1 , j ∈ {2, . . . , k}. The remainder of the argument follows from an application of the same argument to each component in the resulting forest until we obtain (3) with |E(Gi − {u1 , . . . , ui−1 })| = 1, i = 2, . . . , q. □ The following theorem is a corollary to Theorem 3.1. We include it here with a proof since the argument presented provides a method for constructing a matrix whose graph is a tree with exactly q negative eigenvalues. Let T be a tree on n vertices. Then for q = 0, 1, . . . , q0 (T ) the following holds:

Theorem 3.5.

Mq (T ) ≥ MDq (T ) − q. Proof. Let U = u1 , . . . , uq be a set of q vertices whose deletion results in k := MDq (T ) components, T − U = ∪ki=1 Ti , and N(U) ∩ V (Ti ) = Vi , i = 1, . . . , k. By Lemma 3.4, T can be written as

{

}

T = (∪ki=1 Ti ) ∪ H ∪ (∪ki=1 Gi ),

(4)

where H is the induced subgraph on U and Gi is the induced subgraph of T on the vertices (U ∩ N(Vi )) ∪ Vi , i = 1, . . . , k, such that |E(G1 )| = 1 and |E(Gi − {u1 , . . . , ui−1 })| = 1, i = 2, . . . , q. For each i = 1, . . . , k, let Ai ∈ S0 (Ti ) satisfy null(Ai ) = 1. Without loss of generality, we may assume that ui is adjacent to the first vertex in Ti , i = 1, . . . , q. Define A as follows:

⎡

B BT1 BT2 BT3

B1 A1 0 0

B2 0 A2 0

B3 0 0 A3

BTk

0

0

0

⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ .. ⎣ .

.. .

.. .

.. .

... ... ... ... .. . ...

Bk 0 0 0

⎤

⎥ ⎥ ⎥ ⎥ ⎥, ⎥ .. ⎥ . ⎦

Ak

(5)

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where B ∈ S (H) and, up to permutation similarity, we assume that Bi [i, . . . , q|V (Ti )] has exactly one nonzero entry in the position (Bi )i1 , i = 1, . . . , q. Hence A ∈ S (T ). In the following we show that A has exactly q negative eigenvalues and has nullity equal to k − q. By Lemma 1.1, each row and column of Ai can be written as a linear combination of the remaining rows and columns. First we apply Lemma 1.1 to A1 to eliminate its entries in its first row and column. Hence in the resulting matrix, the only nonzero entry in its (q + 1)th column (row) is in position (1, q + 1) ((q + 1, 1)), which is (B1 )11 . Now use (B1 )11 to eliminate all the entries in the first row except a11 . Hence after permutation similarities, A is congruent to

⎡

a11 ⎢ (B1 )11

(B1 )11 0

⎢ ⎢ ⎢ ⎢ ⎢ C1 = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0

0

0

0

...

0 0

A1 (1) 0

0 B(1)

0 B2 (1|∅)

0 B3 (1|∅)

0

0

BT2 (1|∅)

A2

0

0

0

0

A3

.. .

.. .

BT3 (1|∅)

.. .

.. .

.. .

... ... ... ... .. .

0

0

BTk (1|∅)

0

0

...

⎤ 0

⎥ ⎥

0 ⎥ ⎥ Bk (1|∅) ⎥ 0 0

.. .

⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Ak

Repeat this process sequentially for j = 2, . . . , q to deduce that A is congruent to q

q

(⊕i=1 B′i ) ⊕ (⊕i=1 Ai (1)) ⊕ (⊕ki=q+1 Ai ),

(6)

where B′i :=

[

]

aii (Bi )i1

(Bi )i1 , 0

i = 1, . . . , q.

It is easy to observe that each B′i has exactly one positive and one negative eigenvalue. Hence by (6) A has exactly q negative eigenvalues and has nullity equal to k − q. Therefore, Mq (T ) ≥ MDq (T ) − q. □ We note here a consequence, Mq0 (T ) (T ) = M(T ), which may be useful in its own right. It follows that the MDq0 (T ) (T ) components created upon deletion of the q0 (T ) vertices must be paths since if there were a high degree vertex after deleting q0 (T ) vertices, then deleting it with the q0 (T ) vertices we would obtain MDq0 (T )+1 (T ) − (q0 (T ) + 1) > MDq0 (T ) (T ) − q0 (T ). Using Theorem 3.5 and M(T ) = P(T ) (see [16]) we have M(T ) ≥ Mq0 (T ) ≥ MDq0 (T ) − q0 (T ) = P(T ) = M(T ). Remark 3.6. Let T be a tree on n vertices and A ∈ S0 (T ) with nullity 1. Then by Lemma 2.3, for any ϵ > 0 and any i ∈ {1, . . . , n}, A + ϵ Eii is positive definite and A − ϵ Eii ∈ S1 (T ) For a given n × n symmetric matrix A, we define the partial inertia of A to be the pair (q, p), where q is the number of negative eigenvalues of A and p is the number of positive eigenvalues of A. In the paper [5] a number of important results were established concerning the partial inertia of graphs and more precisely trees, including the named Northeast Lemma (see [5, Lemma 1.1]). This lemma can be stated as follows: Suppose G is a graph and A ∈ S (G) with partial inertia (q, p). Then for every q′ ≥ q and p′ ≥ p with q′ + p′ ≤ n, there exists a matrix B ∈ S (G) with partial inertia (q′ , p′ ). In what follows we describe the same result for trees, but provide a different verification based on the results developed in this paper. We refer to this result immediately following as the Right Triangle Lemma. Lemma 3.7 (Right Triangle Lemma). Let T be a tree on n vertices and q ∈ {1, . . . , q0 (T )}. Then any pair from the following set is the partial inertia of a matrix in S (T ) (r , s) | r ≥ q, s ≥ n − q − Mq (T ), r + s ≤ n .

{

}

Proof. Let A ∈ Sq (T ) be as in the proof of Theorem 3.5 with nullity Mq (T ) = MDq (T ) − q. Then A is congruent to q

q

MDq (T )

C = (⊕i=1 Ai (ui )) ⊕ (⊕i=1 B′i ) ⊕ (⊕i=q+1 Ai ), where u1 , . . . , uq is a set of vertices whose deletion gives MDq (T ) components, Ai (ui ) are positive definite matrices and B′i are 2 × 2 matrices with exactly one negative eigenvalue and one positive eigenvalue, i = 1, . . . , q, and Ai are positive semidefinite with nullity 1, i = q + 1, . . . , MDq (T ). By Remark 3.6, adding (subtracting) a positive real number from any main-diagonal entry of Ai , i = q + 1, . . . , MDq (T ), results in a matrix that has all of its eigenvalues positive (with one negative eigenvalue and the remaining positive). Therefore, for any (r , s) such that r ≥ q, s ≥ n − q − Mq (T ), and r + s ≤ n define the matrix A′ from A by modifying (adding or subtracting a positive number to a main-diagonal entry) the matrices Ai , i = q + 1, . . . , MDq (T ) as follows:

{

}

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

2935

• use r − q of them to produce r − q more negative eigenvalues; • use s − (n − q − Mq (T )) of them to produce s − (n − q − Mq (T )) more positive eigenvalues. By using congruence, as in the proof of Theorem 3.5, it follows that A′ has partial inertia (r , s). □ A consequence of the Right Triangle Lemma, is that we know exactly where to add (subtract) a nonzero real number in order to generate any pair in the right triangle whose right angle is located at the point (q, n−q−Mq (T )) and its hypotenuse is the line segment r + s = n, r ≥ q. Comparing this with the North East Lemma from [5], there is a requirement to add (subtract) nonzero real numbers to the main-diagonal entries randomly and at each step check the rank of the resulting matrix, which is rather costly and consequently inefficient. Lemma 3.8. Let n ≥ 2. Then for any q ∈ {0, 1, . . . , n − 1} there exists a matrix A ∈ Sq (Pn ) with nullity 1 such that the last row (column) is a linear combination of the remaining rows (columns). Proof. For q = 0, the result follows by the fact that M0 (Pn ) = 1 and Lemma 1.1. In the following we assume that q > 1. Let n = 2. Then define A ∈ S1 (P2 ) as follows:

[ −1 A= −1

] −1 . −1

Thus A has exactly one negative eigenvalue and nullity equal to 1. Now let n ≥ 3 and for any q ∈ {1, . . . , n − 1} there exists B ∈ Sq (Pn−1 ) such that null(B) = null(B(n − 1)) = 0 (see, e.g., [13] or [12]). Since B and B(n − 1) are nonsingular, then the nonzero vector [0 0 · · · 0 d]T ∈ Rn−1 can be written as a linear combination of the rows of B with the nonzero coefficient of the last row of B, say cn−1 . Define A ∈ S (Pn ) as follows:

aij =

⎧ bij ⎪ ⎪ ⎨ d

⎪ c d ⎪ ⎩ n−1 0

if i, j ≤ n − 1, if i = n, j = n − 1, if i = j = n, if i = n, j ≤ n − 2.

Hence A ∈ Sq (Pn ) with nullity 1. □ Before we come to one of our main results concerning the behavior of the maximum nullity, Mq , we have one more inequality to settle. Lemma 3.9.

Let T be a tree. Then

Mq0 (T ) (T ) > Mq0 (T )−1 (T ). Proof. By the definition of q0 (T ), MDq0 (T ) (T ) − q0 (T ) > MDq0 (T )−1 (T ) − (q0 (T ) − 1). Hence by Theorem 3.1, Mq0 (T ) (T ) > Mq0 (T )−1 (T ). □ Theorem 3.10. Let T be a tree on n vertices with ∆(T ) ≥ 3. Set p0 := MDq0 (T ) (T ) and let Pl1 , Pl2 , . . . , Plp0 be the paths that ∑p0 result from the deletion of q0 (T ) vertices and let ℓ := i=1 li − p0 = n − (q0 (T ) + p0 ). Then 1. M0 (T ) < M1 (T ) ≤ · · · ≤ Mq0 (T )−1 (T ) < Mq0 (T ) (T ) = M(T ); 2. Mn−k (T ) = k − ξ ,

k = 1, 2, . . . , n − q0 (T ) − ℓ − 1(= p0 − 1),

where

{ } ξ = min η | η ∈ {0, 1, . . . , q0 (T )} and η + Mη (T ) ≥ k { } = min η | η ∈ {0, 1, . . . , q0 (T )} and MDη (T ) ≥ k ; 3. Mn−1 (T ) ≤ Mn−2 (T ) ≤ · · · ≤ Mq0 (T )+ℓ+1 (T ) ≤ Mq0 (T )+ℓ (T ); 4. Mq0 (T ) (T ) = Mq0 (T )+1 (T ) = · · · = Mq0 (T )+ℓ (T ) = Γ (T ) = P(T ).

(7)

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Proof. Statement 1 follows from [5, Cor. 3.8]. We now turn to proving Statement 2. Let A ∈ Sξ (T ) be such that null(A) = Mξ (T ). By the definition it follows that ξ < k and by Lemma 3.7 there exists a matrix B ∈ Sξ (T ) such that null(B) = Mξ (T ) − (Mξ (T ) + ξ − k) = k − ξ and B has (n − ξ − Mξ (T )) + (Mξ (T ) + ξ − k) = n − k positive eigenvalues. Hence −B ∈ Sn−k (T ) and null(−B) = k − ξ . Therefore, k − ξ ≤ Mn−k (T ). In the following we show that Mn−k (T ) ≤ k − ξ . Assume on the contrary that Mn−k (T ) > k − ξ and let C ∈ Sn−k (T ) with nullity Mn−k (T ). Then Mn−k (T ) ≥ k − (ξ − 1) and −C ∈ Sk−Mn−k (T ) (T ) with nullity Mn−k (T ). Hence Mn−k (T ) ≤ Mk−Mn−k (T ) (T ) ≤ Mξ −1 (T ), where the last inequality follows by using Statement 1, since k − Mn−k (T ) ≤ ξ − 1 and ξ ≤ q0 (T ). Therefore, k − (ξ − 1) ≤ Mξ −1 (T ) and so k ≤ Mξ −1 (T ) + (ξ − 1), which contradicts the definition of ξ . Hence Mn−k (T ) = k − ξ . For Statement 3, let k ∈ {1, . . . , n − q0 (T ) − ℓ − 1}, we show that Mn−k (T ) ≤ Mn−(k+1) (T ). By Statement 2 we have Mn−k (T ) = k − ξ ,

Mn−(k+1) (T ) = k + 1 − ξ ′ .

Therefore, Mn−k (T ) ≤ Mn−(k+1) (T ) if and only if ξ ′ − ξ ≤ 1. From (7), ξ ′ + Mξ ′ (T ) ≥ k + 1 > k. Hence ξ ′ ≥ ξ which implies that ξ = ξ ′ or ξ ′ > ξ . The latter occurs when ξ + Mξ (T ) = k and ξ ′ + Mξ ′ (T ) ≥ k + 1. Since ξ < ξ ′ ≤ q0 (T ) we have by Statement 1. (ξ + 1) + Mξ +1 (T ) ≥ ξ + Mξ (T ) + 1 = k + 1 which implies by (7) that ξ ′ ≤ ξ + 1. Therefore, in both cases we have ξ ′ − ξ ≤ 1. For Statement 4, and for all j ∈ {q0 (T ), q0 (T ) + 1, . . . , q0 (T ) + ℓ} we have Mj (T ) ≤ Γ (T ) = Mq0 (T ). In order to show that Mj (T ) ≥ Γ (T ) we proceed parallel to the proof of Theorem 3.5 and replace the matrices Ai in that proof with matrices of the type that are defined in Lemma 3.8 to deduce Mj (T ) ≥ Γ (T ). Therefore, Mj (T ) = Γ (T ). □ A symmetric matrix A is said to have balanced inertia if the partial inertia of A, (q, p), satisfies |q − p| ≤ 1. Similarly, a graph G is said to be inertia-balanced if there exists a matrix A ∈ S (G) such that rank(A) = mr(G) and A has balanced inertia (see [3]). Let T be a tree on n vertices with the usual definition of q0 (T ) and let p0 and ℓ be defined as in Theorem 3.10 and let

{ℓ

if ℓ is even,

2

s :=

ℓ

2

±

1 2

if ℓ is odd.

Let A ∈ Sq0 (T )+s (T ) be such that null(A) = Mq0 (T )+s (T ) = Mq0 (T ) (T ) = M(T ) = P(T ) = MDq0 (T ) (T ) − q0 (T ) = p0 − q0 (T ). Hence the partial inertia of A is (q0 (T ) + s, n − (q0 (T ) + s + p0 − q0 (T ))) = (q0 (T ) + s, n − (p0 + q0 (T )) − (s − q0 (T )))

= (q0 (T ) + s, ℓ − s + q0 (T )). Thus,

|q0 (T ) + s − ℓ + s − q0 (T )| = |2s − ℓ| =

{

0

if ℓ is even,

±1

if ℓ is odd.

Hence all trees are inertia-balanced graphs (see [3]). Lemma 3.11 ([18, Lemma 2.3]). Let T be a tree. Consider a nonempty subset U = {v1 , . . . , vk } of V (T ) and let T ′ be the subgraph of T induced by U. Assume T ′ has e(T ′ ) edges. Then the forest T − U has 1+

k ∑

deg(vi ) − k − e(T ′ )

i=1

components. The next few results are devoted to studying the maximum disconnection number. Corollary 3.12.

Let T be a tree. Then for k = 1, 2, . . . , q0 (T ), we have

MDk (T ) = 1 − k + max

{ k ∑

} deg(vi ) − e(T ) , ′

i=1

where the maximum runs over all subsets of k vertices and where T ′ = T [{v1 , . . . , vk }].

(8)

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Fig. 1. T .

Theorem 3.13. Let T be a tree on n vertices with ∆(T ) ≥ 3 such that deg(vi ) = di , i = 1, . . . , n and d1 ≥ d2 ≥ · · · ≥ dk ≥ 3 > dk+1 ≥ dk+2 ≥ · · · ≥ dn . Let eq = min number of edges of T [ u1 , . . . , uq ] | deg(ui ) = deg(vi ), i = 1, . . . , q .

{

{

}

}

Then 1+q+

q ∑

q ∑

i=1

i=1

(di − 2) − e ≤ MDq (T ) ≤ 1 + q +

(di − 2), q = 1, . . . , q0 ,

(9)

where e = number of edges in the induced subgraph on

{ } v1 , . . . , v q .

Equality occurs in the right-hand side { inequality if}and only if eq = 0. Equality occurs in the left-hand side inequality if and only if there exists a set of {vertices u1 , u2}, . . . , uq such that deg(ui ) ≥ deg(vi ) for all i = 1, . . . , q and the maximum in (8) is attained for the vertices u1 , u2 , . . . , uq . Proof. Let {w1 , . . . , wq }be the set of vertices whose deletion will produce MDq (T ) components and let T ′ be the subgraph induced by w1 , . . . , wq . Then, by Corollary 3.12, we have for q = 1, 2, . . . , q0 (T ),

{

}

MDq (T ) = 1 − q +

q ∑

deg(wi ) − e(T ′ ) ≤ 1 +

i=1

Hence Mq (T ) ≤ 1 +

∑q

i=1

q ∑

di − q.

(10)

i=1

di − q − q = 1 +

∑q

i=1 (di

− 2). By Corollary 3.12 it follows that

q

MDq (T ) ≥ 1 − q +

∑

deg(vi ) − e(T ′ ),

i=1

where T is the induced subgraph on v1 , . . . , vq . Hence the left-hand side inequality holds. {Equality holds } on the right-hand side if and only if equality holds in (10). Equivalently, there exists a set of vertices u , . . . , u such that 1 q { } deg(ui ) = deg(vi ) and e(T ′ ) = 0, where T ′ is the subgraph induced by u1 , . . . , uq , i.e., if and only if eq = 0. Equality of ∑q ∑q the left-hand side holds if and only if MDq (T ) ={1 + q + } i=1 (di − 2) − e = 1 − q + i=1 di − e, where e is the number of edges in the induced subgraph on the vertices v1 , . . . , vq . Hence there must exist a set of such vertices for which the maximum is attained in (8). □ ′

{

}

The inequalities in the above theorem are used to estimate MDq (T ) and they may become strict as the following example demonstrates. Let T be the tree depicted in Fig. 1. Then it is easy to note that q0 (T ) = 3 and that MD3 (T ) = 7, where as the left hand side and the right hand side of (9) are 6 and 8, respectively. Theorem 3.14. Let T be a tree, and let A ∈ Sq0 (T ) with null(A) = Mq0 (T ). Then there exists a unique subset of vertices Vq0 (T ) of size q0 (T ) such that T − Vq0 (T ) has MDq0 (T ) (T ) components, A(Vq0 (T ) ) is positive semidefinite, and null(A(Vq0 (T ) )) = Mq0 (T ) (T ) + q0 (T ) = MDq0 (T ) (T ). Proof. We proceed by induction on q0 (T ). If q0 (T ) = 0, T is a path and the result is obvious. The result follows for all trees T with q0 (T ) = 1 by using Theorem 2.4. Assume the result is true for all trees T with q0 (T ) < q0 . We now show

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it is true for all trees with q0 (T ) = q0 . Let T be a tree with q0 (T ) = q0 and A ∈ Sq0 (T ) with nullity Mq0 (T ) and let v be a Parter–Wiener vertex and T − v = ∪si=1 Ti . Then deg(v ) =∑s ≥ 3, null(A(v )) = Mq0 (T ) + 1, and A[Ti ] ∈ Sqi (Ti ), qi ≥ 0, s i = 1, . . . , s. By the Cauchy interlacing inequalities we have i=1 qi = q0 − 1 since A ∈ Sq0 (T ) and null(A(v )) = Mq0 (T ) + 1. For the remainder of the proof, we distinguish between the following cases: Case 1. There exists i0 ∈ {1, . . . , s} such that A[Ti0 ] ∈ Sqi (Ti0 ), null(A[Ti0 ]) < Mqi (Ti0 ) or qi0 < q0 (Ti0 ). 0 0 For simplicity, assume i0 = 1. In either case we may modify A to derive a B ∈ S (T ) as follows: B[Tj ] := A[Tj ], j = 2, . . . , s. Replace A[T1 ] by B[T1 ], where in the first case B[T1 ] ∈ Sq1 (T1 ) with nullity Mq1 (T1 ) and in the second case B[T1 ] ∈ Sq0 (T1 ) (T1 ) with nullity Mq0 (T1 ) (T1 ). In either case null(B[T1 ]) > null(A[T1 ]) and null(B(v )) > null(A(v )) = Mq0 (T ) (T ) + 1 = M(T ) + 1. Therefore, null(B(v )) ≥ M(T ) + 2, which implies that null(B) ≥ M(T ) + 1, which is a contradiction. Thus for all i ∈ {1, . . . , s}, A[Ti ] ∈ Sqi (Ti ), null(A[Ti ]) = Mqi (Ti ) and qi ≥ q0 (Ti ). Case 2. There exists i0 ∈ {1, . . . , s} such that A[Ti0 ] ∈ Sqi (Ti0 ), null(A[Ti0 ]) = Mqi (Ti0 ), and qi0 = q0 (Ti0 ). 0 0 For simplicity, assume i0 = 1. By the induction hypothesis, there exists Vq0 (T1 ) ⊂ V (T1 ) of size q0 (T1 ) such that A[T1 ](Vq0 (T1 ) ) is positive semidefinite and null(A[T1 ](Vq0 (T1 ) )) = Mq0 (T1 ) (T1 ) + q0 (T1 ) = MDq0 (T1 ) (T1 ). If q0 (T1 ) > 0, then ′ choose a vertex v ′ ∈ Vq0 (T1 ) such that T − v ′ = ∪si=1 Ti′ with V (T1′ ) ⊂ V (T1 ), and for any vertex w ∈ V (T1′ ) and any ′ x ∈ V (T − V (T1 )), the path from w to x contains the vertex v ′ , and A[T1′ ] is k × k positive semidefinite with nullity 1. Note that such A[T1′ ] exists since T is a tree and A[T1 ](Vq0 (T1 ) ) is positive semidefinite with nullity MDq0 (T1 ) (T1 ). Otherwise if q0 (T1 ) = 0, then set v ′ := v and A[T1′ ] := A[T1 ]. By permutation similarity, we may assume that A has the following form:

⎡

av ′ ,v ′ ⎢ av′ ,w1 ⎢ ⎢ 0

av ′ ,w1

⎢ .. ⎢ ⎢ . ⎢ ⎢ 0 ⎢ a′ ⎢ v ,w2 ⎢ 0 ⎢ ⎢ .. ⎢ . ⎢ ⎢ 0 ⎢ ⎢ .. ⎢ . ⎢ ⎢ a′ ⎢ v ,ws′ ⎢ 0 ⎢ ⎢ .. ⎣ .

...

0

0

av ′ ,w2

...

0

...

0

av ′ ,ws′

...

0

A[T1′ ]

0

...

0

0

A[T2′ ]

...

0

.. .

.. .

..

.. .

0

0

...

.

A[Ts′′ ]

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

0 with N(v ′ ) = {w1 , . . . , ws′ } and wj ∈ V (Tj′ ), j = 1, . . . , s′ . Since null(A[T1′ ]) = 1 = M0 (T1′ ), by Lemma 1.1 the first row and column of A[T1′ ] can be written as a linear combination of the remaining rows and columns of A[T1′ ]. Hence A is congruent to C , where C is given as follows: av ′ ,v ′ ⎢ av′ ,w1 ⎢ ⎢ 0

⎡

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ C =⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

.. .

0 0

.. .

0

.. .

0

.. .

0

av ′ ,w1 0 0

.. .

... ...

0 0

0 0

0 0

... ...

0 0

... ...

0 0

... ...

A[T1′ ](w1 )

0

...

0

0

A[T2′ ]

...

0

.. .

.. .

..

.. .

0

0

...

0 0

.. .

0

.. .

.

0

.. .

0

Hence C can be written as ′

C = B1 ⊕ A[T1′ ](w1 ) ⊕ (⊕si=2 A[Ti′ ]),

A[Ts′′ ]

0 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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where B1 is the 2 × 2 leading submatrix of C which has exactly one negative and one positive eigenvalue, A[T1′ ](w1 )

∑s′

is positive definite, and A[Ti′ ] ∈ Sq′ (Ti′ ), q′i ≥ 0 and

i=2

i

q′i = q0 − 1. Moreover,

∑s′

∑s′

i=2

null(A[Ti′ ]) = null(A) = Mq0 (T )

′ ′ since congruence preserves the inertia. Hence null(A(v ′ )) = i=1 null(A[Ti ]) = Mq0 (T ) + 1. Therefore, v is a Parter– Wiener vertex. Because Case 1 cannot occur, q′i ≥ q0 (Ti′ ) and null(A[Ti′ ]) = Mq′ (Ti′ ), i = 2, . . . , s′ . Assume, without loss of i generality, that q′2 > q0 (T2′ ). Then define q′′2 := q′2 − 1 ≥ q0 (T2′ ). Modify A to get B ∈ Sq0 −1 (T ) as follows: B[Tj′ ] := A[Tj′ ], ′ ′ ′ ′ ′ j = 1, 3, . . . , s . Modify A[T2 ] to get B[T2 ] such that B[T2 ] ∈ Sq′′ (T2 ) with nullity Mq′′ (T2′ ). Since q0 (T2′ ) ≤ q′′2 < q′2 , we 2 2 have, by Theorem 3.10, that null(A[T2′ ]) = Mq′ (T2′ ) ≤ Mq′′ (T2′ ) = null(B[T2′ ]). Again, by using Lemma 1.1, we can show that 2 2 B ∈ Sq0 −1 (T ) and null(B) ≥ null(A). Therefore,

Mq0 −1 (T ) ≥ null(B) ≥ null(A) = Mq0 (T ), which is a contradiction since Mq0 −1 (T ) < Mq0 (T ). Thus for all i ∈ 2, . . . , s′ , q′i = q0 (Ti′ ) and null(A[Ti′ ]) = Mq′ (Ti′ ). Hence i by the induction hypothesis we can delete q′i vertices from V (Ti ), for i = 2, 3, . . . , s′ together with v ′ to produce a positive semidefinite matrix.

{

}

Case 3. For all i ∈ {1, . . . , s}, A[Ti ] ∈ Sqi (Ti ), null(A[Ti ]) = Mqi (Ti ) and qi > q0 (Ti ). Since qi > q0 (Ti ) for all such i it follows that q0 (T ) − 1 =

s ∑

qi ≥

i=1

s ∑

s ∑

i=1

i=1

(q0 (Ti ) + 1) = s +

q0 (Ti ).

(11)

Suppose there that v ∈ Uq0 (T ) and T − Uq0 (T ) has MDq0 (T ) (T ) components, then ∑sexists a subset of vertices Uq0 (T )∑such s q0 (T ) = 1 + i=1 q0 (Ti ). Otherwise, q0 (T ) ≤ s + i=1 q0 (Ti ) since in this case all of the neighbors of v except at most two must be in Uq0 (T ) . Hence in either case, by (11) and s ≥ 3 we have q0 (T ) ≤ s +

s ∑

q0 (Ti ) ≤ q0 (T ) − 1,

i=1

and so we reach a contradiction. Suppose there exist two distinct subsets V ′ and V ′′ with |V ′ | = |V ′′ | = q0 (T ) such that both A(V ′ ) and A(V ′′ ) are positive semidefinite with nullities equal to MDq0 (T ) (T ). Then one of the components of T − V ′ , say T1 will strictly contain a component of T − V ′′ , say T2 , or vice-versa. To see this, set U := V ′ ∩ V ′′ and T − U := ∪si=1 Ti (if U = ∅, then set ′ T1 = T in the following). The case in which there exists i0 ∈ {1, . . . , s{} such } } that Ti0 is a path and V (Ti0 ) ∩ V ̸={∅ or V (Ti0 ) ∩ V ′′ ̸ = ∅ cannot hold. If such a vertex existed, then T − (V ′ \ u′ ) for some u′ ∈ V (Ti0 ) ∩ V ′ (T − (V ′′ \ u′′ ) ′′ ′′ for some u ∈ V (Ti0 ) ∩ V ) has MDq0 (T ) (T ) − 1 components so that MDq0 (T )−1 (T ) ≥ MDq0 (T ) (T ) − 1, which contradicts ′ ′′ Mq0 (T ) (T ) > Mq0 (T )−1 (T ). Therefore, assume } satisfies V (T1 ) ∩ V ̸= ∅ and V (T1 ) ∩ V ̸= ∅. } that T1 is a tree {(not a path) that { Now, define V (T1 ) ∩ V ′ := v1 , . . . , vs1 and V (T1 ) ∩ V ′′ := w1 , . . . , ws2 . Then at least one of the vi′ s, say v1 , has degree at least three in T1 in which all of the components of T1 − v1 are paths except possibly at most one (that is, T1 has an appropriate vertex). Hence at least one of the wi′ s, say w1 , is adjacent to v1 . Thus if w1 is one of the vertices of the paths of T − v1 , then we are done. Otherwise, deg(v1 ) = 3 in T1 and w1 is one of the vertices of the non-path component, and hence one of the path components of T1 − w1 contains v1 as one of its vertices and our claim is proved. Furthermore, it follows that both A[T1 ] and A[T2 ] are positive semidefinite matrices with (maximum) nullity equal to one since T1 and T2 are trees, null(A(V ′ )) = null(A(V ′′ )) = MDq0 (T ) (T ), and each of T − V ′ and T − V ′′ has MDq0 (T ) (T ) components. In this case A[T2 ] is a principal submatrix of A[T1 ], and so by Lemma 2.3, we conclude that A[T2 ] is positive definite. This contradiction demonstrates the uniqueness property and completes the proof. □ By example, we will demonstrate that the above theorem cannot be extended in general to values of q with 1 < q

< q0 (T ).

Example 3.15. Let T be the tree depicted in Fig. 1 and let A be given as follows.

⎡ −1 ⎢1 ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢ A=⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎣0 0

1 0 1 1 0 0 0 0 0 0 0

0 1 1 0 0 0 0 0 0 0 0

0 1 0 1 1 0 1 0 0 1 0

0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 4 2 0 0 0 0

0 0 0 1 0 2 3 2 0 0 0

0 0 0 0 0 0 2 2 0 0 0

0 0 0 0 0 0 0 0 4 2 0

0 0 0 1 0 0 0 0 2 3 2

⎤

0 0⎥ ⎥ 0⎥ 0⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥. 0⎥ ⎥ 0⎥ ⎥ 0⎥ 2⎦ 2

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Then it is easy to observe that A ∈ S2 (T ) with maximum nullity M2 (T ) = 3 and with q0 (T ) = 3. However, for any v1 , v2 ∈ V (T ), A(v1 , v2 ) is not positive semidefinite. Therefore Theorem 3.14 may not hold in general for any such q with 1 < q < q0 (T ). 4. Cycles and generalized partial n-Sun Let Cn be an n-cycle and let U ⊆ V (Cn ). The graph G (H) obtained from Cn by appending a path (leaf) to each vertex in U is called a generalized partial n-sun (partial n-sun). If U = V (Cn ), then G (H) is called the generalized n-sun (n-sun). We need the notion of a generalized n-sun as a building block for determining Mq for unicyclic graphs. In this section we begin our study of optimal matrices in Sq (G) for unicyclic graphs by studying such matrices restricted to generalized partial n-sun’s. Lemma 4.1. Let Cn be an n-cycle and let A ∈ S (Cn ) be such that null(A) = 2. Then every row (column) in A is linearly dependent on the other rows (columns). Proof. Let A ∈ S (Cn ) be such that null(A) = 2 and without loss of generality we want to show that row n can be written as a linear combination of the other rows. Since A(n) ∈ S (Pn−1 ) we have that null(A(n)) ≤ 1. If null(A(n)) = 0, then null(A) ≤ 1, which is a contradiction. Hence null(A(n)) = 1. If A[n|1, . . . , n − 1] is not in the row space of A(n), then null(A) = 0. Hence A[n|1, . . . , n − 1] is in the row space of A(n). If row n is not in the row space of A[1, . . . , n − 1|1, . . . , n], then null(A) = 1, a contradiction. □ Let Cn be an n-cycle. Then Mq (Cn ) = 2, for q = 0, 1, . . . , n − 2 and Mn−1 (Cn ) = 1.

Lemma 4.2.

Proof. By Lemma 3.8, there exists B ∈ Sq (Pn−1 ) such that null(B) = 1 and null(B(n − 1)) = 0, q = 0, 1, . . . , n − 2. Since null(B(n − 1)) = 0, we can choose an1 and an,n−1 such that the vector [an1 0 . . . 0 an,n−1 ]T is in the row space of B[1, . . . , n − 2|1, . . . , n − 1]. Define A as follows: an1 0

⎡ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ ⎣

.. .

B

an1

0

...

0

0 an,n−1 ann

an,n−1

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

where ann is chosen such that row n of A is in the row space of A[1, . . . , n − 2|1, . . . , n]. Hence A is congruent to 0 0 ⎥

⎡

⎤

⎢ ⎢ ⎢ ′ A =⎢ ⎢ ⎢ ⎣

.. ⎥ ⎥ . ⎥. ⎥ 0 ⎥ 0 ⎦

B

0

0

...

0

0

0

Therefore, A ∈ Sq (Cn ) and null(A) = 2. The case q = n − 1 follows immediately from the fact that M0 (Cn ) = 2. □ Lemma 4.3.

Let G be a generalized partial n-sun on N vertices with q0 (G) ≥ 2. Then for 2 ≤ q ≤ q0 (G),

MDq (G) = 2q. Proof. For q = 2, 3, . . . , q0 (G), let Uq be a set of vertices such that G − Uq has MDq (G) components. It is easy to observe that each vertex of Uq must have degree 3 in G. Let U be a largest set of vertices such that all of its vertices are of degree 3 in G and no two are adjacent and |U | = k. Hence G − U has 2k components. In the following we show that k = q0 (G). If k > q0 (G), then removing any q0 (G) vertices will produce at most 2q0 (G) components. Hence MDq0 (G) − q0 (G) ≤ 2q0 (G) − q0 (G) = q0 (G) < k = 2k − k, which contradicts the definition of q0 (G). Therefore, k ≤ q0 (G). Assume that k < q0 (G). Thus for any set of vertices Uq0 (G) such that G − Uq0 (G) has MDq0 (G) (G) components, there are two vertices with degree 3 that are adjacent in G, say u1 and u2 and all the vertices in G − Uq0 (G) have degree at most 2 in G − Uq0 (G) . Hence G − (Uq0 (G) \ {u1 }) has MDq0 (G) (G) − 1 components. If each vertex of Uq0 (G) \ {u1 } has degree at most 2 in G − (Uq0 (G) \ {u1 }), then MDq0 (G)−1 (G) − (q0 (G) − 1) ≥ MDq0 (G) (G) − 1 − (q0 (G) − 1) = MDq0 (G) (G) − q0 (G) contradicting that q0 (G) is the smallest positive integer q such that MDq (G) − q attains its maximum. If there is a vertex in G − (Uq0 (G) \ {u1 }) with degree 3, then deleting this vertex together

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2941

with the vertices in Uq0 (G) \ {u1 } will result in MDq0 (G) (G) + 1 components, which is a contradiction. Hence k = q0 (G). Finally, the best strategy for producing MDq (G) components is to remove the vertices of degree 3 in which no two are adjacent. Thus removing such q vertices produces q + q = 2q components. □ Lemma 4.4 ([5, Proposition 4.3]). If v is a pendant vertex of the graph G and (i, j) is the partial inertia of a matrix in S (G − v ), then both (i + 1, j) as well as (i, j + 1) are the partial inertia of matrices in S (G). Theorem 4.5. Let G be a generalized partial n-sun on N vertices. Then M0 (G) = M1 (G) = MN −3 (G) = MN −2 (G) = 2, MN −1 (G) = 1, and for q = 2, 3, . . . , N − 4: 1. If q0 (G) = 0, then Mq (G) = 2. 2. If q0 (G) ≥ 2, then for q = 2, 3, . . . , q0 (G),

⎧ ⎨MDq (G) − q

for q = q0 (G) + 1, . . . , q0 (G) + l, ⎩ N −q−ξ for q = q0 (G) + l + 1, q0 (G) + l + 2, . . . , N − 4, { } ξ = min η | η ∈ {0, 1, . . . , q0 (G)} and η + Mη (G) ≥ N − q

Mq (G) =

Mq0 (G) (G)

and l = N − q0 (G) − MDq0 (G) (G). Proof. By Lemmas 4.2 and 4.4 we have Mq (G) ≥ 2, q = 0, 1, . . . , N − 2. It is easy to determine that Z0 (G) = Z1 (G) = 2. Thus Mq (G) = 2, q = 0, 1. By the facts MN −2 (G) ≤ 2 and M0 (G) = 2 we have MN −2 (G) = 2. For MN −3 (G) = 2, we argue by MN −3 (G) ≤ 3 and M0 (G) = M1 (G) = 2. If q0 (G) = 0, then G is a cycle, or G has exactly one vertex on the cycle of degree 3, or exactly two vertices on the cycle of degree 3 and these two vertices are adjacent, or G is a generalized 3-sun. In any of these cases Z (G) = 2 and so Mq (G) Mq (G) = 2, q = 2, 3, . . . , N − 4. { ≤ 2. Therefore, } For 2 ≤ q ≤ q0 (G). Let Uq = u1 , . . . , uq be a set of vertices whose deletion gives k := MDq (G) components. By Lemma 4.3, Uq consists of non-adjacent degree 3 vertices on the cycle. By the special structure of G, G can be written as follows: q

q

G = (∪i=1 Pji ) ∪ (∪ki=q+1 Ti ) ∪ (∪i=1 Gi ) ∪ H ,

(12)

where Pji is the path connected to ui and Gi = K2 whose one end is ui and other end is on Pji , i = 1, . . . , q, q (G − Uq ) − V (∪i=1 Pji ) = ∪ki=q+1 Ti , and H is a union of P3 paths induced by ui and its neighbors on the cycle. Now following parallel to the proof of Theorem 3.5 we obtain Mq (G) ≥ MDq (G) − q. Next we show that Mq (G) ≤ MDq (G) − q, q = 2, . . . , q0 (G). Let A ∈ Sq (G) be such that null A = Mq (G), 2 ≤ q ≤ q0 (G), Pi1 , Pi2 , . . . , Pis be the paths connected to a vertex on the cycle, and let Ait be κit ×κit submatrices of A such that Ait ∈ S (Pit ), t = 1, . . . , s. We distinguish the following two cases: Case 1. All Ait , t = 1, . . . , s are nonsingular. Hence A is congruent to Ai1 ⊕ Ai2 ⊕ · · · ⊕ Ais ⊕ C , where C ∈ S (Cn ) and Mq (G) = null(A) = null(C ) ≤ 2 < MDq (G) − q, if 2 < q ≤ q0 (G). In this case we reach a contradiction. If q = 2, then by Lemma 4.3 we have null(C ) = Mq (G) ≤ 2 = 4 − 2 = MD2 (G) − 2. Case 2. There exists h ∈ {1, . . . , s} such that Aih is singular. Without loss of generality assume that h = 1 and Ai1 ∈ Sζ (Pi1 ) for some ζ ≥ 0. By permutation similarity, we may assume that A has the following form:

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤

0

.. .

Ai1

...

0

0

av,u

0 av,u au , u au,w1 au,w2 0

.. .

0

0

au,w1

au,w2

0

C

...

0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

where v is the vertex at Pi1 that is connected to the cycle and N(u) = {w1 , w2 , v}. Since Ai1 is singular irreducible symmetric tridiagonal matrix we have the last row and column of Ai1 can be written as a linear combination of its remaining rows and columns. Therefore A is congruent to 0

⎡ ⎢ Ai1 (v ) ⎢ ⎢ ⎢ ⎢ ⎢ 0 ... 0 ⎢ 0 ... 0 ⎢ ⎢ ⎢ ⎢ ⎣ 0

.. .

0

.. .

0 0 av,u 0

0 av,u au , u 0

0

0

.. .

.. .

⎤ 0 ⎥ ⎥

⎥ ⎥ ⎥ 0 ⎥ ⎥, 0 ⎥ ⎥ ⎥ ⎥ C ⎦

where C ∈ Sq−ζ −1 (G′ ) and G′ is the tree obtained from G by deleting Pi1 and the vertex on the cycle that is adjacent to a vertex on Pi1 . Hence Mq (G) = null(A) = null(C ) ≤ Mq−ζ −1 (G′ ). It is easy to note that q0 (G′ ) = q0 (G) − 1. Hence q − ζ − 1 ≤ q0 (G′ ). By the special structure of G′ we have MDq−ζ −1 (G′ ) = 2(q − ζ − 1) + 1. So by Theorem 3.1 and Lemma 4.3 we have Mq (G) ≤ Mq−ζ −1 (G′ )

= MDq−ζ −1 (G′ ) − (q − ζ − 1) = 2(q − ζ − 1) + 1 − (q − ζ − 1) = q−ζ ≤ q = MDq (G) − q. Therefore, Mq (G) = MDq (G) − q, q = 2, . . . , q0 (G). Next we show that M(G) ≤ MDq0 (G) (G) − q0 (G). Let A ∈ S (G) be such that null(A) = M(G). Let Pi1 , Pi2 , . . . , Pis and Ai1 , Ai2 , . . . , Ais be as above. If all of Ait , t = 1, . . . , s are nonsingular, then as above we arrive at M(G) = null(A) ≤ 2 ≤ MDq0 (G) (G) − q0 (G). If there exists h ∈ {1, . . . , s} such that Aih is singular, then without loss of generality assume that h = 1 and proceeding as above we obtain that A is congruent to Ai1 (v ) ⊕ B1 ⊕ B2 , where B1 ∈ S1 (P2 ) is nonsingular, B2 ∈ S (G′ ), and G′ is the tree obtained from G by deleting Pi1 and the vertex on the cycle that is adjacent to a vertex on Pi1 . Hence M(G) = null(A) = null(B2 ) ≤ M(G′ ). As above q0 (G′ ) = q0 (G) − 1. Hence by Theorem 3.1, Theorem 3.10, and Lemma 4.3 M(G′ ) = Mq0 (G′ ) (G′ ) = MDq0 (G′ ) (G′ ) − q0 (G′ ) = 2q0 (G′ ) + 1 − q0 (G′ ) = q0 (G′ ) + 1 = q0 (G) − 1 + 1 = q0 (G) = 2q0 (G) − q0 (G) = MDq0 (G) (G) − q0 (G) = Mq0 (G) (G). Hence M(G) ≤ Mq0 (G) (G) and by definition we have { M(G) ≤ Mq0 (G) } (G). For q ∈ {q0 (G) + 1, . . . , q0 (G) + l}. Let Uq0 (G) = u1 , . . . , uq0 (G) be a set of vertices whose deletion gives k := MDq0 (G) (G) components. By Lemma 4.3, Uq0 (G) consists of non-adjacent degree 3 vertices on the cycle. Hence G can be written as follows: q (G)

q (G)

G = (∪i=0 1 Pji ) ∪ (∪ki=q0 (G)+1 Pji ) ∪ (∪i=0 1 Gi ) ∪ H ,

(13)

where Pji is the path connected to ui and Gi = K2 whose one end is ui and other end is on Pji , i = 1, . . . , q0 (G), q (G) (G − Uq0 (G) ) − V (∪i=0 1 Pji ) = ∪ki=q (G)+1 Pji , and H is a union of paths on three vertices induced by ui and its neighbors 0 on the cycle. Let A ∈ S (G) whose submatrices corresponding to Pji , i = 1, . . . , k are of the type defined in Lemma 3.8 with total q − q0 (G) negative eigenvalues. By proceeding as above we have A ∈ Sq (G) with null(A) = Mq0 (G) (G). Therefore, Mq (G) = Mq0 (G) (G), q = q0 (G) + 1, . . . , q0 (G) + l. For q ∈ {q0 (G) + l + 1, q0 (G) + l + 2, . . . , N − 4}, the proof follows along the same lines as the proof of Statement 2 in Theorem 3.10. □ As we have seen in the case of trees, Mq (T ) depends on the combinatorial parameter MDq (T ). Similarly, for generalized partial n-suns G, we have shown that the maximum nullity parameter Mq (G) depends only on the combinatorial parameter q0 (G).

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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By using Theorem 4.5 and Lemma 4.3 we have the following corollaries. Corollary 4.6.

Let G be a generalized partial n-sun on N vertices and let l = N − q0 (G) − MDq0 (G) (G). Then

1. If q0 (G) = 0, then Mq (G) = 2, q = 0, 1, . . . , N − 2. 2. If q0 (G) ≥ 2, then

Mq (G) =

⎧ ⎨2

if q = 0, 1, N − 3, N − 2,

q

if q = 2, 3, . . . , q0 (G),

⎩

q0 (G)

if q = q0 (G) + 1, q0 (G) + 2, . . . , q0 (G) + l,

MN −2q (G) = MN −2q−1 (G) = Mq (G) = q, q = 2, 3, . . . , q0 (G) − 1. Remark 4.7. We observe that in the above two results l = N − q0 (G) − MDq0 (G) (G)

= N − (q0 (G) + MDq0 (G) (G)) = N − 3q0 (G). Corollary 4.8. Let G be an n-sun with n ≥ 4, let k := ⌊ 2n ⌋, and let

δ=

0

if n is even,

1

otherwise.

{

Then Mq (G) =

⎧ ⎨2 ⎩

if q = 0, 1, 2n − 3, 2n − 2,

q

if q = 2, 3, . . . , k,

k

if q = k + 1, k + 2, . . . , 2k + 2δ, and

M2(n−q) (G) = M2(n−q)−1 (G) = Mq (G) = q, q = 2, 3, . . . , k − 1. Proof. The proof follows by using Corollary 4.6 and the fact that q0 (G) = k and l = k + 2δ .

□

In the same way as in the tree case, we can easily show that any generalized partial n-suns are inertia-balanced graphs. The next result describes, in a way, a Parter–Wiener-type behavior on the nullity of matrices associated with generalized partial n-suns. The Parter–Wiener theory is very important for studying the inverse eigenvalue problems for trees. Regarding the next statement though we caution the reader to note that it may not hold when null(A) < 3. For example, simply consider an n-cycle. Theorem 4.9. Let G be a generalized partial n-sun on N vertices and let A ∈ S (G) be such that null(A) = k ≥ 3. Then there exists v ∈ V (G) with null(A(v )) = k + 1. Proof. Let t be the number of the paths that are connected to the cycle. By Corollary 4.6, t ≥ 3 as null(A) ≥ 3. Let Ai be the principal submatrices of A which correspond to the paths that are connected to the cycle, i = 1, . . . , t. If null(Ai ) = 0 for all i = 1, . . . , t, then A is congruent to A1 ⊕ A2 ⊕ · · · ⊕ At ⊕ B, where B ∈ S (Cn ). Hence k = null(A) = null(B) ≤ 2, which is a contradiction. Therefore, there exists i0 ∈ {1, . . . , t } such that null(Ai0 ) = 1. Hence the last row and column of Ai0 can be written as a linear combination of the remaining rows and columns of Ai0 or Ai0 is the 1 × 1 zero matrix. In either case, A is congruent to C0 ⊕ C1 ⊕ C2 , where C0 is a nonsingular or an empty matrix, C1 is a 2 × 2 matrix with exactly one negative and one positive eigenvalue, and C2 ∈ S (T ) with T being the tree that is obtained from G by deleting the vertices of the path that is corresponding to Ai0 together with the vertex on the cycle that is connected to this path. It is easy to observe that C2 is a principal submatrix of A. Hence null(C2 ) = null(A) = k ≥ 3. By Theorem 1.4 there exists v ∈ V (T ) such that null(C2 ) = k + 1. Since congruence preserves nullity, we can reverse the congruence in the previous step and apply them to A(v ) and obtain null(A(v )) = k + 1. □

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M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

5. Unicyclic graphs We now turn our attention to unicyclic graphs (namely those graphs that contain a unique cycle), and determine a formula for Mq (G) for a unicyclic graph G. Theorem 5.1. Let G be a unicyclic graph on n vertices, C := {v1 , . . . , vk } ⊂ V (G) be the set of vertices that induce the unique cycle, and for q = 0, 1, . . . , n − 1, let MD′q (G) := MDq (G) + δq , where

δq =

⎧ ⎨1

if there exists Uq = u1 , . . . , uq

⎩

otherwise.

{

}

such that G − Uq has MDq (G) components and

one of them is a unicyclic graph,

0

• If there exists a set Uj0 of j0 vertices whose deletion produces MD′j0 (G) components, one of them, say H, is a generalized partial k-sun and all of the remaining components are paths, and for any q > j0 , G − Uq is a forest, where Uq is any set of q vertices whose deletion produces MD′q (G) components, then for q = 1, 2, . . . , j0 ,

MD′q (G) − q

{ Mq (G) =

for q = j0 + 1, . . . , j0 + q0 (H),

Mj0 (G) − 2 + Mq−j0 (H)

(14)

• otherwise Mq (G) = MD′q (G) − q for q = 1, 2, . . . , q0 (G).

(15)

Proof. For q = 1, 2, . . . , j0 (or q0 (G)), the decomposition given in Lemma 3.4 holds with one of the components, say Ti0 with i0 > q, being a unicyclic graph if δq = 1 or all the components of G − Uq are trees if δq = 0. Therefore, proceeding parallel to the proof of Theorem 3.5, and choosing Ai0 ∈ S0 (Ti0 ) with nullity equal to 2 if δq = 1, we obtain Mq (G) ≥ MD′q (G) − q, q = 1, . . . , j0 (or q0 (G)). Suppose now that H is a generalized partial k-sun. Then G − Uj0 is a union of paths and H. The decomposition that is given in Lemma 3.4 holds for G − Uj0 with one of the components being H. Proceeding in a manner parallel to Theorem 3.5 and choosing all of the matrices Ai to be positive semidefinite with nullity one while the matrix that is corresponding to H is in Sq−j0 (H) with nullity Mq−j0 (H). Hence Mq (G) ≥ Mj0 (G) − 2 + Mq−j0 (H), q = j0 + 1, . . . , j0 + q0 (H). In the following we show that the quantities in (14) and (15) are upper bounds for Mq (G). The proof uses induction on |G| = n. There is nothing to prove for n ≤ 4. Suppose it is true for all unicyclic graphs on fewer than n vertices. Let v, w ∈ V (G) be such that deg(v ) = 1 and v ∼ w and let A ∈ Sq (G) be such that null(A) = Mq (G). If for all such v , w ∈ C , then we follow as in the proof of Theorem 4.5. In the following we assume that there exists a v such that deg(v ) = 1, v ∼ w , and w ∈ / C . Set G′ := G − v and G′′ = G − {v, w}. Then A may be written as follows:

⎡

av,v A = ⎣ av,w 0

⎤

0T ⎦. aT23 A(v, w )

av,w aw,w a23

(16)

Hence A is congruent to

[ B=

av,v 0

0T A′

]

,

or

⎡

0

C = ⎣ av,w 0

⎤

0T 0T ⎦ . A′′

av,w aw,w 0

The matrices B and C satisfy A′ ∈ Sq (G′ ) if avv > 0, A′ ∈ Sq−1 (G′ ) if avv < 0, and A′′ ∈ Sq−1 (G′′ ) if avv = 0. In the following we distinguish between these cases. Case 1. avv > 0. By the induction hypothesis we have Mq (G) = null(A) = null(A′ ) =

{ ′

Mq (G ) ≤

for q = 1, . . . , j′0 (or q0 (G′ )),

MD′q (G′ ) − q ′

′

Mj′ (G ) − 2 + Mq−j′ (H ) 0

0

for q = j′0 + 1, . . . , j′0 + q0 (H ′ ),

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

2945

where j′0 is defined as before relative to the graph G′ . By the facts MD′q (G′ ) ≤ MD′q (G) and using Corollary 4.6, we have (14) and (15) are upper bounds for Mq (G). Case 2. avv < 0. By the induction hypothesis we have Mq (G) = null(A) = null(A′ ) =

{

for q − 1 = 0, 1, . . . , j′0 (or q0 (G′ )),

MD′q−1 (G′ ) − (q − 1)

′

Mq−1 (G ) ≤

′

′

Mj′ (G ) − 2 + Mq−1−j′ (H ) 0

0

for q − 1 = j′0 + 1, . . . , j′0 + q0 (H ′ ).

As in Case 1. the quantities in (14) and (15) are upper bounds for Mq (G). Case 3. avv = 0. If deg(w ) = 2, then G′′ is a connected unicyclic graph. Therefore, by the induction hypothesis, MD′q−1 (G′′ ) + 1 ≤ MD′q (G) if q ≤ j0 (or q0 (G)), and applying Corollary 4.6 we are done. Otherwise ℓ := deg(w ) − 1 ≥ 2. In this case A′′ is a direct-sum of ℓ matrices. Suppose that A′′ = B1 ⊕ · · · ⊕ Bℓ ,

∑ℓ

where Bi ∈ Ssi (Gi ), G′′ = ∪ℓi=1 Gi , and i=1 si = q − 1. Moreover, all Gi are trees except one, say Gℓ which must be a unicyclic ∑ graph since w ∈ / C . Since Mq (G) = null(A) = null(A′′ ) = ℓi=1 null(Bi ), we conclude that null(Bi ) = Msi (Gi ), i = 1, . . . , ℓ. In the following we assume that there exists a set Uj0 of j0 vertices whose deletion produces MD′j (G) components, one of 0 them, say H, is a generalized partial k-sun and all of the remaining components are paths, and for any q > j0 , G − Uq is a ′ forest, where Uq is any set of q vertices whose deletion produces MDq (G) components. The other case follows in a similar manner. Case 3.1. There exists a set Uh of h vertices such that MD′q (G) attains its value and w ∈ Uh , where h = q if q ≤ j0 or h = j0 if q > j0 . Case 3.1.1. q ≤ j0 . We may assume that si ≤ q0 (Gi ), i = 1, . . . , ℓ − 1 and sℓ ≤ j′0 . Otherwise by Theorem 3.10 and Corollary 4.6, we can design a matrix C ′′ ∈ Sq−1 (G′′ ) such that C ′′ := C1 ⊕ · · · ⊕ Cℓ , where Ci ∈ Ssi (Gi ) and si ≤ q0 (Gi ), i = 1, . . . , ℓ − 1 and sℓ ≤ j′0 . Therefore, by the induction hypothesis, w ∈ Uq , and q ≤ j0 we have Mq (G) = null(A) =

ℓ−1 ∑

Msi (Gi ) + Msℓ (Gℓ )

i=1

≤

ℓ−1 ∑

(MDsi (Gi ) − si ) + (MD′sℓ (Gℓ ) − sℓ )

i=1

=

ℓ−1 ∑

MDsi (Gi ) + MD′sℓ (Gℓ ) −

i=1

ℓ ∑

si

i=1

≤ MD′q (G) − 1 − (q − 1) = MD′q (G) − q, since removing si vertices from Gi together with w results in at least

∑ℓ−1 i=1

MDsi (Gi ) + MDsℓ (Gℓ ) − δ + 1 components.

Case 3.1.2. q > j0 . Let q = j0 + η for some η > 0. Again we may assume that si = q0 (Gi ), i = 1, . . . , ℓ − 1, and sℓ = j′0 + η, η > 0, where ∑ℓ−1 ′ j0 = i=1 q0 (Gi ) + j0 . Hence q−1=

ℓ ∑

si =

( ℓ−1 ∑

i=1

) ′

q0 (Gi ) + j0

+ η − 1 = j0 + η − 1.

i=1

By the induction hypothesis and w ∈ Uq we obtain Mq (G) = null(A) =

ℓ ∑

Msi (Gi )

i=1

=

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + Mj′ +η (Gℓ ) 0

i=1

≤

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + (Mj′ (Gℓ ) − 2 + Mη (H)) 0

i=1

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M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

=

( ℓ−1 ∑

) Mq0 (Gi ) (Gi ) + Mj′ (Gℓ ) 0

− 2 + Mη (H)

i=1

≤

ℓ−1 ∑

MDq0 (Gi ) (Gi ) + MD′j0 (Gℓ ) −

i=1

ℓ−1 ∑

q0 (Gi ) − j′0 − 2 + Mη (H)

i=1

≤ MD′j0 (G) − j0 − 2 + Mq−j0 (H) ≤ Mj0 (G) − 2 + Mq−j0 (H). Case 3.2. For q ≤ j0 , any set Uq of q vertices such that MD′q (G) attains its value, w ∈ / Uq . Consider all tuples (q1 , . . . , qℓ ) such that ℓ ∑

qi = q − 1

and

ℓ ∑

i=1

Mqi (Gi )

is maximum.

i=1

Since q ≤ j0 and by Theorem 3.10, we may assume that qi ≤ q0 (Gi ), i = 1, . . . , ℓ − 1, where qℓ ≤ j′0 . Since Mq (G) = ∑ℓ null(A) = i=1 Msi (Gi ) we have that (s1 , . . . , sℓ ) is one of the above tuples. By Theorem 3.1 we have Msi (Gi ) = MDs1 (Gi ) − si , i = 1, . . . , ℓ − 1, and by induction Msℓ (Gℓ ) = MD′sℓ (Gℓ ) − sℓ . Since w ∈ / Uq for any Uq it follows that MD′q (G) >

ℓ−1 ∑

MDsi (Gi ) + MD′sℓ (Gℓ ) + 1.

i=1

Hence Mq (G) = null(A) =

ℓ ∑

Msi (Gi )

i=1

≤

ℓ−1 ∑

(MDsi (Gi ) − si ) + MD′sℓ (Gℓ ) − sℓ

i=1

=

ℓ−1 ∑

MDsi (Gi ) −

i=1

=

ℓ−1 ∑

ℓ ∑

si + MD′sℓ (Gℓ )

i=1

MDsi (Gi ) + MD′sℓ (Gℓ ) + 1 − q

i=1

< MD′q (G) − q ≤ Mq (G), which is a contradiction.

/ Uj0 . Case 3.3. For q > j0 , any set Uj0 of j0 vertices such that MD′j (G) attains its value, w ∈ 0 {v, } Let q = j0 +η for some η > 0 and N( w ) = w , . . . , w . In this case we may assume that si = q0 (Gi ), i = 1, . . . , ℓ− 1 1 ℓ ∑ℓ−1 ′ and sℓ = j′0 + η − 1, where j0 = i=1 q0 (Gi ) + j0 . The following three cases then arise. Case 3.3.1. For any Uj0 , wi ∈ Uj0 , i = 1, . . . , ℓ. By the induction hypothesis, Theorem 3.1 and Corollary 4.6 we have Mq (G) = null(A) =

ℓ ∑

Msi (Gi )

i=1

=

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + Mj′ +η−1 (Gℓ ) 0

i=1

≤

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + Mj′ (Gℓ ) − 2 + Mη−1 (H) (note that Mη−1 (H) = 2 if η = 1) 0

i=1

≤

ℓ−1 ∑ i=1

=

ℓ−1 ∑

MDq0 (Gi ) (Gi ) −

ℓ−1 ∑

q0 (Gi ) + MD′j′ (Gℓ ) − j′0 − 2 + Mq−j0 −1 (H) 0

i=1

MDq0 (Gi ) (Gi ) + MD′j′ (Gℓ ) − j0 − 2 + Mq−j0 −1 (H) 0

i=1

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

2947

Fig. 2. G1 .

< MD′j0 (Gi ) (Gi ) − j0 − 2 + Mq−j0 −1 (H) ≤ Mj0 (G) − 2 + Mq−j0 (H) ≤ Mq (G), which is a contradiction. Case 3.3.2. For any Uj0 , wi ∈ Uj0 , i = 1, . . . , ℓ − 1. Mq (G) = null(A) =

ℓ ∑

Msi (Gi )

i=1

=

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + Mj′ +η−1 (Gℓ ) 0

i=1

≤

ℓ−1 ∑

Mq0 (Gi ) (Gi ) + Mj′ (Gℓ ) − 2 + Mη−1 (H ′ ) 0

i=1

′

≤ MDj0 (G) − j0 − 2 + Mη−1 (H ′ ) ≤ Mj0 (G) − 2 + Mη (H) = Mj0 (G) − 2 + Mq−j0 (H), where H ′ is either H or the unique unicyclic component of H − w . Case 3.3.3. For any Uj0 , wi ∈ Uj0 , i = 2, . . . , ℓ. This case follows by using similar arguments as in Case 3.3.2.

□

For the special case when q = 1, we have the following result. Corollary 5.2. Let G be a unicyclic graph and let {v1 , . . . , vk } ⊂ V (G) be the set of vertices that induce the cycle and Tvi be the acyclic component of G \ ({v1 , . . . , vk } \ {vi }) that contains vi , i = 1, . . . , k. Then

{ M1 (G) =

if G is a generalized partial k-sun,

2 max ∆(Tvi ) | i = 1, . . . , k

{

}

otherwise.

For graph G1 , see Fig. 2, by removing v12 we have MD1 (G1 ) = 4 and one of them is a unicyclic graph which implies that MD′1 (G1 ) = 5. By removing v4 and v12 we have MD′2 (G1 ) = 7. By removing v1 , v9 , and v12 we have MD′3 (G1 ) = MD3 (G1 ) = 9. Finally, there is no such j0 as in Theorem 5.1 and q0 (G1 ) = 3. For graph G2 , see Fig. 3, we have j0 (G2 ) = 2 and after removing vertices v12 and v3 we have a generalized partial 5-sun, say H, with q0 (H) = 2. Hence by Theorem 5.1 we have q Mq (G1 )

0 2

1 4

2 5

3 6

q Mq (G2 )

0 2

1 4

2 5

3 5

and 4 . 5

As we observed in Theorem 3.10, Corollary 4.6, and Theorem 5.1, we suspect that the following conjecture holds.

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Fig. 3. G2 .

Fig. 4. T1 .

Conjecture 5.3. Let G be a connected graph on n vertices. Then there exist ηG , γG ∈ {1, . . . , n − 1}, ηG + γG ≤ n − 1 such that M0 (G) ≤ M1 (G) ≤ · · · ≤ MηG (G) = M(G) = MηG +1 (G) = · · · = MηG +γG (G), MηG +γG (G) ≥ MηG +γG +1 (G) ≥ · · · ≥ Mn−2 (G) ≥ Mn−1 (G). 6. Application to an inverse eigenvalue problem for graphs: More on counterexamples In [2], two conjectures were resolved, both in the negative. Here we re-establish this resolution using the techniques developed in this paper, which may be viewed as following from simpler arguments. It will be shown that the minimum number of distinct eigenvalues of the tree T1 in Fig. 4 is at least 8. It is easy to note that q0 (T1 ) = 3. Hence we apply Theorem 3.1 to compute Mq (T1 ) for q = 0, 1, 2, 3. By Theorem 3.10, (4), Mq (T1 ) = 4, q = 4, . . . , 9 since ℓ = 6. For q = 10, . . . , 15, we use Theorem 3.10, (2), to determine Mq (T1 ). First we define k := 16 − q and depending on Ms (T1 ), s = 0, 1, 2, 3 to find ξ such that (7) holds. Hence we have q Mq (T1 )

0 1

1 2

2 3

3 4

4 4

5 4

6 4

7 4

8 4

9 4

10 3

11 3

12 2

13 2

14 1

15 . 1

Suppose there exists a matrix in S (T1 ) with ordered multiplicity list (m1 , m2 , . . . , mκ ) such that κ ≤ 7. Since the number of vertices is 16 and M(T ) = P(T ) = 4, we have κ ≥ 4. We know that m1 = mκ = 1 since M0 (T1 ) = 1. Hence

M. Adm and S.M. Fallat / Discrete Mathematics 342 (2019) 2924–2950

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Fig. 5. T2 .

κ ≥ 5. Further, we have that m2 ≤ 2 since M1 (T1 ) = 2. By repeated application of Theorem 1.4, it follows that if λ is an eigenvalue of multiplicity 4 for such a matrix, then λ must appear in the (1,1) entry. Thus, there exists at most one i ∈ {3, . . . , κ − 1}, such that mi = 4. If m2 = 1, then mκ−1 ≥ 3 since m3 + · · · + mκ−1 = 13. If mκ−1 = 3, then M12 (T1 ) ≥ 3 because m1 + · · · + mκ−2 = 12, which cannot occur. On the other hand if mκ−1 = 4, then M11 (T1 ) ≥ 4, which also cannot occur. Thus m2 = 2. So, by symmetry, we have that mκ−1 = 2. If κ ∈ {5, 6}, then 16 = m1 + m2 + · · · + mκ−1 + mκ ≤ 1 + 2 + 7 + 2 + 1 = 13 which is a contradiction. In the following we assume that κ = 7. Since m3 + m4 + m5 = 10, and at most one summand is 4, we are left with three possible ordered multiplicity lists: (1, 2, 4, 3, 3, 2, 1), (1, 2, 3, 3, 4, 2, 1), (1, 2, 3, 4, 3, 2, 1). Suppose there exists A ∈ S3 (T1 ) that has multiplicity list (1, 2, 4, 3, 3, 2, 1) and null(A) = 4. By Theorem 3.14, A(4, 9, 14) is positive semidefinite with nullity 7 since the only set of vertices that produces MDq0 (T1 ) (T1 ) is {4, 9, 14}. Therefore, all of the following matrices are positive semidefinite with nullity 1: A1 := A[2, 3], A2 := A[5, 6], A3 = A[7, 8], A4 := A[10, 11], A5 := A[12, 13], A6 := A[15, 16], and A7 := A[1]. Since T1 − v1 is a union of three paths it follows that v1 cannot be a Parter–Wiener vertex for λ4 and λ5 . By symmetry we may assume that v4 is a Parter–Wiener vertex for λ4 and that v9 is a Parter–Wiener vertex for λ5 . Hence applying Theorem 1.4 to λ4 and to λ5 , we have λ4 ∈ σ (A1 ) ∩ σ (A2 ), λ4 ∈ σ (A[1, 7, . . . , 16]) with multiplicity 2, λ5 ∈ σ (A3 ) ∩ σ (A4 ), and λ5 ∈ σ (A[1, . . . , 6, 12, . . . , 16]) with multiplicity 2. Hence {0, λ4 } = σ (A1 ) = σ (A2 ) and {0, λ5 } = σ (A3 ) = σ (A4 ). Again after application of Theorem 1.4 to λ4 in A[1, 7, . . . , 16] we have λ4 ∈ σ (A5 ) ∩ σ (A6 ), which implies that {0, λ4 } = σ (A5 ) = σ (A6 ). Furthermore, by Theorem 1.4, λ5 ∈ σ (A5 ) ∩ σ (A6 ) and so {0, λ5 } = σ (A5 ) = σ (A6 ). Hence λ4 = λ5 which is impossible. If there were a matrix B ∈ S9 (T1 ) with multiplicity list (1, 2, 3, 3, 4, 2, 1) and null(B) = 4, then −B ∈ S3 (T1 ) with multiplicity list (1, 2, 4, 3, 3, 2, 1) and nullity 4. From the above analysis this is impossible. Proceeding as above we can also exclude the ordered multiplicity list (1, 2, 3, 4, 3, 2, 1). Indeed, let C ∈ S6 (T1 ) with ordered multiplicity list (1, 2, 3, 4, 3, 2, 1) and nullity 4. We can easily conclude that all of the submatrices of C that correspond to the vertices

{2, 3} , {5, 6} , {7, 8} , {10, 11} , {12, 13} {15, 16} , {1} have nullity one. Now applying Theorem 1.4 as above to λ3 and λ5 we reach a contradiction. Therefore, any matrix in S (T1 ) has at least 8 distinct eigenvalues. In [19], an example was presented to show that the ordered multiplicity list (1, 2, 4, 2, 1) is realized by a matrix in S (T2 ), where T2 is given in Fig. 5, only if a certain algebraic relation holds among the eigenvalues. Here we provide an alternative proof based on the techniques within this work. Let A ∈ S3 (T2 ) with nullity 4. By Theorem 3.14, A has the following form:

⎡

a11 ⎢a12 ⎢a ⎢ 13 ⎢a ⎢ 14 ⎢0 A=⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎣0 0

a12 0 0 0 0 0 0 0 0 0

a13 0 0 0 0 0 0 0 0 0

a14 0 0 0 a45 0 0 a48 0 0

0 0 0 a45 a55 a56 a57 0 0 0

0 0 0 0 a56 0 0 0 0 0

0 0 0 0 a57 0 0 0 0 0

0 0 0 a48 0 0 0 a88 a89 a8,10

0 0 0 0 0 0 0 a89 0 0

0 0 0 0 0 0 0

⎤

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ a8,10 ⎥ 0 ⎦ 0

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The only possible Parter–Wiener vertex for λ2 and λ4 is v4 . Therefore,

λ2 , λ4 ∈ σ (A1 ) ∩ σ (A2 ) ∩ σ (A3 ), where A1 := A[1, 2, 3], A2 := A[5, 6, 7], A3 := A[8, 9, 10]. From the structure of Ai , i = 1, 2, 3, we have that each Ai has one positive, one negative, and one zero eigenvalue. Since λ2 , λ4 are the nonzero eigenvalues of A1 , A2 , and A3 ,

λ2 + λ4 = tr(A1 ) = tr(A2 ) = tr(A3 ) = a11 = a55 = a88 . Whence λ2 + λ4 = a11 = a55 = a88 . We know that tr(A) = λ1 + 2λ2 + 2λ4 + λ5 . Therefore,

λ1 + λ5 = λ2 + λ4 . Declaration of competing interest The authors declared that they had no conflicts of interest with respect to their authorship or the publication of this article. Acknowledgments The work of M. Adm leading to this publication was supported by the German Academic Exchange Service (DAAD) with funds from the German Federal Ministry of Education and Research (BMBF) and the People Programme (Marie Curie Actions) of the European Union’s Seventh Frame-work Programme (FP7/2007–2013) under REA Grant Agreement No. 605728 (P.R.I.M.E. — Postdoctoral Researchers International Mobility Experience) during his delegation to the University of Regina and work at University of Konstanz under monitoring of Prof. Dr. Shaun Fallat and Prof. Dr. Jürgen Garloff (University of Konstanz) and revised during his work at Palestine Polytechnic University. Fallat’s research was supported in part by an NSERC Discovery Research Grant, Application No. RGPIN-2014-06036. 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