Generating I -eigenvalue free threshold graphs

A graph is said to be I -eigenvalue free if it has no eigenvalues in the interval I with respect to the adjacency matrix A . In this paper we present two algorithms for generating I -eigenvalue free threshold graphs


Introduction
In 1972 A. J. Hoffman [7] presented a remarkable result about the density of eigenvalues on the real line. Let A be the set of all symmetric matrices of all orders, every entry of which is a non-negative integer and R = {ρ : ρ = ρ(A) for some A ∈ A} where ρ(A) is the largest eigenvalue of A. He determined all limit points of R 2 + √ 5. Then, in 1989, Shearer [12] extended this result. He showed that every point 2 + √ 5 is a limit point of R. Since then, several interesting results have been obtained on this topic as you can see in [13], where it is proved that every real number is a limit point of eigenvalues of graphs.
Given that, it seems counterintuitive that some classes of graphs have no eigenvalues in a prescribed, arbitrary size interval. Nevertheless, Jacobs et al. [8] showed that threshold graphs are (−1, 0)-eigenvalue free. And, in [1], Aguilar et al. proved that anti-regular graphs, a subclass of threshold graphs, have no eigenvalues in the interval except the trivial eigenvalues −1 and 0. This result was extended by Ghorbani in [4] by showing that threshold graphs have no eigenvalues in Ω except the trivial ones.
Searching for graphs that have no eigenvalues in some interval, in this paper, we provide two algorithms for generating infinite families of I-eigenvalue free threshold graphs. More specifically, given a positive real number N and a natural number r one algorithm constructs a threshold graph G with associated cotree T G (a 1 , a 2 , . . . , a r ) of depth r, which will be the initial threshold graph for generating infinite families of (0, N ]-eigenvalue free threshold graphs. And, the other one provides a similar result for the interval [M, −1) where M is a negative real number. We show that our approach also can be used to prove that threshold graphs are −1− -eigenvalue free which shows that we have a generalization of the above mentioned results.
Here is an outline of the paper. In Section 2 we describe the cotrees associated to threshold graphs, and in Section 3 we recall an algorithm for locating eigenvalues in a cograph G developed in [9] and present some of its applications that include determining the inertia of G just using its associated cotree. In Section 4 we explain our strategy based on the inertia of the graph. The algorithms for generating infinite families of Ieigenvalue free threshold graphs are given in Section 5 and 6. In Section 7 we exhibit such families. Our method is used to prove that threshold graphs are −1− -eigenvalue free in Section 8. Finally, in Section 9 we discuss about some open problems.

Threshold representation
Let G = (V, E) be an undirected graph with vertex set V and edge set E. If |V | = n, then its adjacency matrix A(G) = [a ij ] is the n × n matrix of zeros and ones such that a ij = 1 if and only if v i is adjacent to v j (that is, there is an edge between v i and v j ). For v ∈ V , N (v) denotes the open neighborhood of v, that is {w|{v, w} ∈ E} and N [v] = N (v) ∪ {v} the closed neighborhood. A value λ is an eigenvalue if det(λI n − A) = 0, and since A is real symmetric its eigenvalues are real. In this paper, a graph's eigenvalues are the eigenvalues of its adjacency matrix.
The class of threshold graphs was introduced by Chvátal and Hammer [3] and Hender-son and Zalestein [6] in 1977 and their numerous applications go from computer science to psychology [10]. A threshold graph can be constructed through an iterative process which starts with an isolated vertex, and at each step either a new isolated vertex is added or a dominating vertex is added, i.e., a vertex adjacent to all previous vertices is added. More specifically, we may represent a threshold graph on n vertices using a binary sequence (b 1 , b 2 , . . . , b n ) as follows: b i = 0 if vertex v i is added as an isolated vertex, and b i = 1 if v i is added as a dominating vertex. In Figure 1 we show the threshold graph G with binary sequence b = (1, 1, 1, 1, 0, 0, 0, 1, 1). It can also be seen as consecutive blocks of 0's and 1's, for instance b = (1, 1, 1, 1, 0, 0, 0, 1, 1) = 1 4 0 3 1 2 . Notice that we order the vertices of G in the same way they are given in their sequence. Threshold graphs are a subclass of cographs and each cograph can be represented by a cotree, for more details see [2,9]. It is interesting that for threshold graphs, the cotree is a caterpillar, as shown in [11].
In this note, we focus on representing the threshold graph using its cotree (caterpillar), that we describe below. A cotree T G of a threshold graph is a rooted path in which any interior vertex w is either of union ∪ type (corresponding to a block of 0's) or join ⊗ type (corresponding to a block of 1's). The terminal vertices (leaves) are typeless and represent the vertices of the threshold. Since we work only with connected thresholds graphs in this paper, our cotree is basically defined by placing a ⊗ node at the trees's root. And then, placing ∪ on interior nodes with odd depth, and placing ⊗ on interior nodes with even depth.
Notice that, if a cotree T G associated to a threshold graph has an even depth then its final interior node is a ∪ type, and if its depth is odd then it is a ⊗ type as in Figure 2.
The cotree denoted by T G (a 1 , a 2 , . . . , a r ) with a i 1 for 1 i r − 1, a r 2 and r odd is depicted in Figure 2. Notice that following our notation each interior node at depth i has a i terminal vertices (leaves). The binary representation corresponding to the cotree in Figure 2 is given by b = 1 ar , b a r−1 2 , . . . , b a 2 r−1 , 1 a 1 . For instance, the threshold with binary representation b = (1, 1, 1, 1, 0, 0, 0, 1, 1) = 1 4 0 3 1 2 has the cotree representation T G (2, 3, 4) depicted in Figure  3. We would like to point out that in this paper we only deal with connected threshold graphs. It means that in the binary representation we always have b n = 1 and in the cotree representation our first node always is a ⊗.
Finally, we define siblings vertices given its important role in the structure of cographs and in the following algorithm. Two vertices u and v are duplicates if N (u) = N (v) and . Therefore, we call u and v siblings if they are either duplicates or coduplicates.

Algorithm Diagonalize
The algorithm presented in this section was developed in [9]. Basically, it constructs a diagonal matrix D congruent to A + xI, where A is the adjacency matrix of a cograph G.
We would like to point out that the Diagonalize(T G , x) works bottom up since the cotree is represented in the same way. In this note we just work with threshold graphs so the cotrees we use are depicted in Figure 2. Therefore, throughout the text we represent the steps bottom up by steps from left to right.
In this article given a graph G and a scalar a ∈ R we define the triple (a + , a 0 , a − ), where a + denotes the number of eigenvalues of G that are greater than a, a 0 the multiplicity of a and a − the number of eigenvalues of G that are less than a. Therefore, the inertia of a graph G, using our notation, is the triple (0 + , 0 0 , 0 − ). The following results presented in this section are from [9] and will be used throughout the note. The next one shows that the Algorithm 1 computes the triple (a + , a 0 , a − ). ii: The number of eigenvalues of G that are less than a is exactly a − .
iii: The multiplicity of a is a 0 .
the electronic journal of combinatorics 30(2) (2023), #P2.30 The following three results show that we can obtain information about the localization of certain eigenvalues of the cograph G just by analysing its associated cotree T G . Theorem 2. Let G be a cograph with cotree T G having ⊗-nodes {w 1 , . . . , w j }, and assume each w i has t i children in T G . Then Theorem 3. Let G be a cograph with cotree T G having ∪-nodes {w 1 , . . . , w m }, where w i has t i > 0 terminal children. If G has j 0 isolated vertices, then The next two lemmas provide an alternative initialization of the algorithm i.e., first we perform assignments to the leaves with identical value of the cotree and then we move on with the specialized cotree.
Lemma 6. If v 1 , . . . , v m have parent w = ∪, each with diagonal value y = 0, then the algorithm performs m − 1 iterations of subcase 2a, assigning during iteration j: One advantage of the algorithm is that it is applied directly in the cotree so the diagonal values processed during the iterations will be labeled in the vertices. As pointed out before, the algorithm progresses bottom up but in this article we work only with threshold graphs so we will represent these iterations from left to right. The next example elucidates these observations. Example 7. We will apply Diagonalize to the cograph G with cotree T G (2, 3, 4) and x = −1. Since we process the algorithm directly in the cotree, the diagonal values d i 's will appear at the terminal vertices v i 's during the execution. Each leaf is initialized with the value d i = −1 as in Figure 4. Since multiple leaves of the same parent (⊗ or ∪) have the same diagonal value y = −1 at the initialization, we can begin performing assignments to the terminal children (leaves) of the cotree using Lemmas 5 and 6 as follows.   Notice that the vertices associated to permanent values are removed and we proceed with the cotree with the remaining vertices and its values as depicted in Figure 6 left. The last vertex with value 1 2 is relocated to the next level as in Figure 6 right. Figure 6: Applying subcase 2a.
Select the sibling pair {v k , v l } labeled with d k = 1 2 and d l = −1 3 at the ∪-node at depth 2 and initialize α ← d k = 1 2 and β ← d l = −1 3 as illustrated in Figure 6 right. Then subcase 2a is executed and the assignments d k ← 1 6 and d l ← −1 are made as shown in Figure 7 left. Then, the remaining vertex is relocated to the next level as in Figure 7 right. Now, as α = −1 and β = 0, subcase 1a is executed at the last iteration and the assignments d k ← −3 and d l ← 1 3 are made as shown in Figure 8. The algorithm stops, and the final diagonal is formed by the permanent values and the last remaining value as shown in Figure 9. Therefore, according to Theorem 1 1 + = 2, 1 − = 7 and 1 0 = 0.

Our strategy
We would like to begin this section with two observations. First, suppose we have a threshold graph G such that 0 + = N + for a positive real number N , that is, the number of eigenvalues of G that are greater than 0 is equal to the number of eigenvalues of G that are greater than N . Therefore, 0 + = N + implies that G is (0, N ]-eigenvalue free. Second, suppose we have a threshold graph G such that M + = (−1) 0 + 0 0 + 0 + for a negative real number M < −1. Then, the number of eigenvalues of G that are greater than M is equal to the number of eigenvalues of G that are greater than or equal to −1. That is, Next, we explain our strategy for generating a threshold graph G that is (0, N ]eigenvalue free for a real number N > 0. Our method consists in constructing a threshold graph G with cotree T G (a 1 , a 2 , . . . , a r ), by choosing each a i 1 for 1 i r − 1 and a r 2, and showing that this threshold graph is (0, N ]-eigenvalue free using the fact that N + = 0 + . Next we compute 0 + as a function of the number of crosses and unions in T G for the cases r odd and even.
Let G be a threshold graph with cotree T G (a 1 , a 2 , . . . , a r ) with a r 2 and r odd. When we apply the algorithm Diag(T G , 0) we are able to compute the inertia (0 and | ∪ | denotes, respectively, the number of crosses and unions in G w.r.t. to the cotree representation. Hence, by Theorem 2 we obtain the following And, by Theorem 3 If r is even, then r = | ⊗ | + | ∪ | with | ⊗ | = | ∪ |. Therefore, as before, by Theorem 2 and by Theorem 3 Therefore, as aforementioned, our approach is to construct a threshold graph such that N + = | ∪ | + 1 if r is odd and N + = | ∪ | if r is even. Now, we present our strategy for generating a threshold graph G that is [M, −1)eigenvalue free for a real number M < −1. As before, our method consists in constructing a threshold graph G with cotree T G (a 1 , a 2 , . . . , a r ), by choosing each a i 1 for 1 i r − 1 and a r 2, and showing that this threshold graph is [M, −1)-eigenvalue free using the fact that M + = (−1) 0 + 0 0 + 0 + . Next we compute (−1) 0 + 0 0 + 0 + for the cases r odd and even.
Let G be a threshold graph with cotree representation T G (a 1 , a 2 , . . . , a r ) with a r 2 and r odd.
In this case, our approach is to construct an threshold graph such that M + = (n − r) + | ∪ | + 1 if r is odd and M + = (n − r) + | ∪ | if r is even.

Diag (TG,-N )
Instead of directly applying the algorithm Diag(T G (a 1 , a 2 , . . . , a r ), −N ) for N > 0, we initialize the process using Lemmas 5 and 6 in the leaves of the cotree with identical value as described in Example 7.
First, Lemma 5 is applied at each ⊗-node at depth i of the cotree. We perform a i − 1 iterations that leave negative permanent assignments j+1 j (−N − 1) for j = 1 . . . a i − 1 by (1) and by (2) a remaining value at the last iteration, as represented in Figure 10.
And, following Lemma 6, for each ∪-node at depth i of the cotree we perform a i − 1 iterations that leave negative permanent assignments j+1 j (−N ) for j = 1 . . . a i − 1 by (3) and by (4) a remaining value d l ← −N a i at the last iteration, as depicted in Figure 11. Figure 11: Union.
We would like to emphasize that, as described in Example 7, the vertices associated to permanent values are removed from the cotree. Therefore, after the specialization aforementioned in the terminal children of the cotree with identical value, we have r i=1 (a i − 1) = n − r negative permanent assignments. And, after that we will continue processing the specialized cotree in Figure 12 using Algorithm Diagonalize from left to right. Our strategy is to show that each ∪-node assigns a positive permanent value and each ⊗-node a negative permanent value with a positive remaining value. Consequently, it will imply that N + = | ∪ | + 1 if r is odd and N + = | ∪ | if r is even. It is important to point out that throughout this paper we execute Diagonalize in two main steps. In the first one we process the leaves with identical value and obtain what we have called specialized cotree as illustrated in Figure 12. Then, we proceed with the algorithm on the specialized cotree as in Example 7.

Consider the sequence of remaining values
at the specialized cotree in Figure 12 and the functions If we choose the initial value s r+1 = 1 − N +1 ar and for each new value we pick g for a union and f for a join, then we obtain recursively s r = g(s r+1 , −N a r−1 ), s r−1 = f (s r , 1 − N +1 a r−2 ) and so on, until we reach s 1 . These are the assignments we obtain by applying the subcase 1a and subcase 2a from Diag(T G , −N ), which is performed from left to right. That is why we produce s i in the inverse order, from s r+1 to s 1 .
In the next two lemmas we show that by choosing the number of terminal children we can control the sign of the permanent and remaining assignments. Proof. Since α = s i+1 > N a i and β = − N a i , we can execute subcase 2a because α + β > 0. It returns a positive permanent assignment and a negative remaining value Proof. Since α = s i+1 < 0 and β = 1 − N +1 a i then we can execute subcase 1a because It returns a negative permanent assignment and a positive remaining value , as our hypothesis holds.
Next, we will show that the two lemmas above used together produce an algorithm to construct threshold graphs that are (0, N ]-eigenvalue free.
Initial step: Considering r odd, after the specialization, the leaf at the ⊗-node of depth r has the assignment if and only if a r > N + 1. Then, using a r > N + 1 and Lemmas 8 and 9 we can construct a threshold graph of depth r that is (0, N ]-eigenvalue free. And for r even, after the specialization at the node of depth r we have a leaf with assignment for a r 2. Then, using a r 2 and lemmas 8 and 9 we can construct a threshold graph of depth r that is (0, N ]-eigenvalue free. The next two results sum up the above observations. We would like to point out that in the next theorem we begin with a generic cotree T G (a 1 , a 2 , . . . , a r ) and, at each step, we choose specific values for a i , 1 i r. Proof. Considering the case r odd we have to prove that N + = 0 + = | ∪ | + 1. We start the Diag(T G , −N ) using Lemmas 5 and 6 to process the terminal children with identical value. As afore-explained this initial process produces (n−r) negative permanent diagonal assignments, adding (n − r) to N − . Then we proceed the Algorithm Diagonalize on the specialized cotree in Figure 12. The last ⊗-node at depth r has a leaf with assignment s r = 1 − N +1 ar > 0 which will be relocated to the next level at the ∪-node as shown in Figure 15. Figure 15: Specialized cotree relocated.
Using Lemma 8 at the ∪-node at depth r − 1 we obtain a positive permanent diagonal value p r−1 > 0 and a remaining assignment s r−1 < 0. The leaf having the value s r−1 is then relocated to the next ⊗-node at level r − 2 as depicted in Figure 16. At the ⊗-node at depth r − 2 we apply Lemma 9 and it creates a negative permanent diagonal value p r−2 < 0 and a remaining value s r−2 > 0 which will be relocated to the next level r − 3. Continuing this process, the ⊗-node at level 1 will have two leafs with assignments s 2 and 1 − N +1 a 1 as illustrated in Figure 17 left. Once we apply Lemma 9 we process the two remaining vertices whose diagonal values will be p 1 < 0 and s 1 > 0. The algorithm stops, and then each ∪-node has produced a positive permanent diagonal assignment, adding | ∪ | to N + . And each ⊗-node, except the one at depth r, has produced a negative permanent diagonal value, adding | ⊗ | − 1 to N − . And the final permanent diagonal value is s 1 > 0, adding +1 to N + . Therefore N − = (n − r) + | ⊗ | − 1, N 0 = 0 and N + = | ∪ | + 1. Theorem 11. Let N > 0 be a fixed number and r an even number, if we choose natural numbers a r 2, a r−1 > N +1 , a r−2 > N s r−1 , etc., then T G (a 1 , a 2 , . . . , a r ) is (0, N ]eigenvalue free. Proof. Considering the case r even we have to prove that N + = 0 + = | ∪ |. We start the Diag(T G , −N ) using Lemmas 5 and 6 to process the terminal children with identical value. As afore-explained this initial process produces (n−r) negative permanent diagonal assignments, adding (n − r) to N − . Then we proceed the Algorithm Diagonalize on the specialized cotree in Figure 18 left. The last ∪-node at depth r has a leaf with assignment s r = − N ar > 0 which will be relocated to the next level at the ⊗-node as shown in Figure  18 right. Figure 18: Specialized cotree relocated.
Using Lemma 9 at the ⊗-node at depth r − 1 we obtain a negative permanent diagonal value p r−1 > 0 and a remaining assignment s r−1 > 0. The leaf having the value s r−1 is then relocated to the next ∪-node at level r − 2 as depicted in Figure 19. At the ∪-node at depth r − 2 we apply Lemma 8 and it creates a positive permanent diagonal value p r−2 < 0 and a remaining value s r−2 < 0 which will be relocated to the next level r − 3. Figure 19: Specialized cotree relocated.
Continuing this process, the ⊗-node at level 1 will have two leafs with assignments s 2 and 1 − N +1 a 1 as illustrated in Figure 17 left. Once we apply Lemma 9 we process the two remaining vertices whose diagonal values will be p 1 < 0 and s 1 > 0.
The algorithm stops, and then each ∪-node, except the one at level r, has produced a positive permanent diagonal assignment, adding | ∪ | − 1 to N + . And each ⊗-node has produced a negative permanent diagonal value, adding | ⊗ | to N − . And the final permanent diagonal value is s 1 > 0, adding +1 to N + . Therefore N − = (n − r) + | ⊗ |, N 0 = 0 and N + = | ∪ |.
Remark 12. The values a 1 , a 2 , . . . , a r are natural numbers so a r−1 > N sr for r odd (case r even is similar), it means that a r−1 N sr if N sr is not an integer number, otherwise a r−1 N sr + 1. To avoid this confusion we can always choose a r 1 + N + 1 , , etc., to construct T G (a 1 , a 2 , . . . , a r ). Now we can define an algorithm to produce threshold graphs T G (a 1 , a 2 , . . . , a r ) satisfying the Theorem 10 or 11: if r is odd then Choose a r 1 + N + 1 Notice that the initial threshold is obtained by making the smallest choice for each a i at RFI (N, r).
Next, we denote the first positive eigenvalue of a graph G by a + sign in the exponent, as θ + (G).

Diag(TG,-M )
We start the algorithm Diag(T G , −M ), for a fixed number M < −1, processing the leaves with identical value using Lemmas 5 and 6 as follows.
Lemma 9 is applied at each ⊗-node at depth i. We perform a i − 1 iterations that leave positive permanent diagonal values j+1 j (−M − 1) for j = 1 . . . a i − 1 by (1). And, by (2), at the last iteration, a remaining assignment Figure 20. Therefore, after the specialization in the leaves with identical value we already have r i=1 (a i − 1) = n − r positive permanent diagonal values. Hence, following our strategy we will work with the cotree represented in Figure 22 and we will show that each ∪-node gives a positive permanent diagonal value and each ⊗-node returns a negative permanent diagonal value with a positive remaining assignment. It will imply that M + = (n − r) + (| ∪ | + 1) if r is odd and M + = (n − r) + | ∪ | if r is even. Figure 22: Specialized cotree.
Next, we prove that each ⊗-node leaves a permanent negative value and a remaining value positive. Whereas, each ∪-node leaves a permanent positive value. And, these results will prove that we obtain | ∪ | positive permanent values. In the last iteration, at the final ⊗-node, we will have a positive remaining diagonal assignment. It will imply that M + = (n − r) + (| ∪ | + 1) if r is odd and M + = (n − r) + | ∪ | if r is even.
Consider the sequence of remaining assignments at the specialized cotree in Figure 22. If we choose the initial value s r+1 = 1 − M +1 ar and for each iteration we choose the function (6) for a union and (5) for a join, then we obtain recursively s r = g s r+1 , −M a r−1 , s r−1 = f s r , 1 − N +1 a r−2 and so on, until we reach s 1 . These are the assignments we obtain by applying the subcase 1a or subcase 2a from Diag(T G , −M ), which is performed from left to right. That is why we produce s i in the inverse order, from s r+1 to s 1 .
In the next two lemmas we show that by choosing the number of terminal children we can control the sign of the permanent and remaining assignments.
Lemma 16. Suppose we have a | ∪ |-node with leaves having values s i+1 > 1 and −M a i with a i > −M + M s i+1 as illustrated in Figure 23. Then we obtain a positive permanent value p i and a remaining value 0 < s i < 1. Proof. Since α = s i+1 > 1 and β = − M a i then α + β > 1 and we can use subcase 2a. It returns a permanent diagonal assignment and a remaining one Now, we want to show that s i < 1, so if and only if as our hypothesis holds.
Lemma 17. Suppose we have a ⊗-node with leaves having assignments 0 < s i+1 < 1 Figure 24. Then we obtain a negative permanent value p i and a positive remaining value s i > 1. Proof. Since α = s i+1 and β = 1 − M +1 a i < 0 if and only if a i > − M +1 1−s i+1 . Therefore, we can use subcase 1a and the permanent diagonal assignment will be And the remaining one which is satisfied if and only if s i+1 < 1.
As in Section 5, the two lemmas above used together produce an algorithm to construct threshold graphs that are [M, −1)-eigenvalue free.
Initial step: Considering r odd, after the specialization of the leaves with identical value, the ⊗-node at depth r has a leaf with assignment if and only if −(M +1) ar > 0, which is satisfied for all a r 2. Then, using a r 2 and Lemmas 16 and 17 we can construct a threshold graph of depth r that is [M, −1)-eigenvalue free. And, for r even, after the specialization, the ∪-node at depth r has a leaf with assignment s r > 0 is trivially satisfied and s r < 1 iff a r > −M . Then, using a r > −M and Lemmas 16 and 17 we can construct a threshold graph of depth r that is [M, −1)-eigenvalue free.
Remark 18. In the left case we have a r free if r is odd and in the right case we have a r free if r is even.
The next two theorems sum up the above results and their proofs are similar to the proofs of Theorems 10 and 11. Now we can define an algorithm to produce threshold graphs T G (a 1 , a 2 , . . . , a r ) satisfying the Theorem 19 or 20: Algorithm 3 The Left Free Interval Algorithm: LFI(N, r) Input: a negative real number M < −1 and a positive integer r Output: cotree T G (a 1 , a 2 , . . . , a r ) if r is odd then Choose a r 2 As before, the initial threshold is obtained by making the smallest choice for each a i in LFI(M, r).
Next, we denote the first negative eigenvalue of a graph G smaller than −1 by a − sign in the exponent, as θ − (G).  (1, 1, 1, 1, 1, 1, 2) is the cotree associated to an anti-chain graph of order n = 8. However, as we exemplify below, this correspondence between N and M = −N − 1 does not hold for any threshold.

Infinite families
In this section, we show that the initial threshold graph generated by Algorithms 2 or 3 is an initial threshold graph for a family of threshold graphs that have the same I-eigenvalue free property.
We recall below the known interlacing property that can be found in [5].
Consider the cotree T H (a 1 , . . . , a r ) of depth r associated to the threshold graph H Suppose that M < θ − (H) and θ + (H) > N . Using the notation above we have the following: Hence, θ i+p+q+1 = θ − < M and θ i = θ + > N . Notice that, when we add a vertex v to the threshold H then we have two possibilities: 1. we add a vertex v to a ∪-node at depth l that has a l terminal vertices in H. Then G will have a ∪-node at depth l that has a l + 1 terminal vertices; 2. we add a vertex v to a ⊗-node at depth l that has a l terminal vertices in H. Then G will have a ⊗-node at depth l that has a l + 1 terminal vertices.
Using Theorems 3 and 4 we have the following: Now, we apply the interlacing Theorem 22.
Considering the two possible cases we conclude the following.
And, it implies that The above computations motivate the following definition.
Definition 23. Let T G (a 1 , . . . , a r ) with a i 1 for 1 i r − 1 and a r 2, and T G (b 1 , b 2 , . . . , b r ) be two threshold graphs. We say that G G if a i b i for i = 1, 2, . . . , r i.e., if G is generated from G by adding any amount of leaves to ∪-nodes or ⊗-nodes (of course, G ≺ G if G G and G = G ). Notice that G and G have the same depth.
Then, we have proved the next result.

Revisiting Ghorbani's work
In this section we use our approach to revisit the conjecture proposed by Aguilar et al. in [1] proved by Ghorbani in [4].
In both cases we will always have a i = 1 as the initial choice. Given the decreasing convergence towards to the fixed points µ and µ we deduce that our claim is true for a arbitrary large r, concluding our proof. ]-eigenvalue free.
Proof. Our hypothesis implies that G G and, by Corollary 24 and Theorem 25, we conclude that there is no threshold graph with eigenvalues in the interval (0, −1+ √ 2 2 ].
The proof of the next two results are similar to the ones above. = −1.207106781 and r any odd natural number (the even case is identical) then the algorithm LFI(M, r) generates the initial threshold graph with cotree T G (1, . . . , 1, 2) (with r − 1 leafs 1's) that is (M, −1]-eigenvalue free.

Future work
As a sequel of this work, we would like to investigate additional properties of the initial threshold graphs generated by Algorithms RFI(N, r) and LFI(N, r), such as the existence of minimal ones, w.r.t. the I-eigenvalue free property.
By Corollary 24, the relation G G preserves the I-eigenvalue free property. In this way, given for instance I = (0, N ] and the initial threshold graph T G (a 1 , . . . , a r ), obtained from RFI(N, r), we already know that, if G G then G is also (0, N ]-eigenvalue free. However, if G G the situation is not clear. By a simple counting procedure we can see that there are exactly K = a 1 · . . . · a r of such graphs. For a fairly small number N = √ 5 = 2.26+ the initial threshold graph, obtained from RFI( √ 5, 7), is T G (4, 10, 4, 55, 3, 12, 4), producing K = 4 · 10 · 4 · 55 · 3 · 12 · 4 = 1, 267, 200 threshold graphs smaller than itself to compare. We can perform an exhaustive search for a G G such that G is also (0, N ]-eigenvalue free, by directly computing the spectrum of each graph. This task could be extremely hard from a computational point of view but we can make it easy by using the Diagonalize algorithm (it took about 5 minutes for N = √ 5 using a naive implementation). Comparing the number of positive outputs for Diag T G , −1+ √ 2 2 (which has no zero outputs, because −1+ √ 2 2 is not eigenvalue for any threshold graph) and Diag(T G , N ) (which should not have zero outputs if N < θ + (G )) we can see that G is also (0, N ]-eigenvalue free if, and only if, this number of positive outputs remains unchanged. Remembering that, after the specialization, n − r outputs will be negative, we only need compute r values in Diag T G , −1+ √ 2 2 and Diag(T G , N ). We made this computation for several values N > 0 never founding any G ≺ G such that G is also (0, N ]-eigenvalue free. Despite the efficiency of our method, it still prohibitive, for N = 3.5 we must to perform about 22 million tests. In other words, we conjecture that the initial graph is minimal in the sense that any smaller threshold graph is not (0, N ]-eigenvalue free. Conjecture 29. Let N > 0 and r 2 be fixed numbers defining the interval I = (0, N ], and T G (a 1 , . . . , a r ), the initial threshold graph obtained from RFI(N, r), then G is minimal, i.e., ifG G is such thatG is also I-eigenvalue free thenG = G.