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An extension of the star complement technique for regular graphs
Accepted Manuscript AN EXTENSION OF THE STAR COMPLEMENT TECHNIQUE FOR REGULAR GRAPHS
Peter Rowlinson
PII: DOI: Reference:
S0024-3795(18)30397-5 https://doi.org/10.1016/j.laa.2018.08.018 LAA 14697
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
26 March 2018 7 August 2018
Please cite this article in press as: P. Rowlinson, AN EXTENSION OF THE STAR COMPLEMENT TECHNIQUE FOR REGULAR GRAPHS, Linear Algebra Appl. (2018), https://doi.org/10.1016/j.laa.2018.08.018
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AN EXTENSION OF THE STAR COMPLEMENT TECHNIQUE FOR REGULAR GRAPHS
Peter Rowlinson1 Mathematics and Statistics Group Division of Computing Science and Mathematics University of Stirling Scotland FK9 4LA United Kingdom
Abstract We extend the means by which a regular graph can sometimes be identified from a star complement. In two applications we determine (i) the regular graphs with a path as a star complement for the eigenvalue 1, (ii) the extremal regular graphs which have another type of tree as a star complement for an eigenvalue = −1, 0.
AMS Classification: 05C50 Keywords: eigenvalue, regular graph, star complement, tree. 1
Tel.:+44 1786 467468; email: [email protected]
1
Introduction
Let G be a connected regular graph of order n with μ (= −1, 0) as an eigenvalue of co-multiplicity t. Thus the corresponding eigenspace E(μ) of a (0, 1)adjacency matrix A of G has dimension n − t. From [2, Theorem 3.1], we know that if t > 2 then n ≤ 12 (t − 1)(t + 2), with equality if and only if G is an extremal strongly regular graph as defined by Seidel [11]. (Only five extremal strongly regular graphs are known; see [6, Section 3.6].) If G is r-regular and r > 2 then n ≤ 12 (r + 1)t, with equality if and only if r = 3, μ = 1 and G is the Petersen graph [8]. These results are proved using star complements, defined as follows for any graph G and any eigenvalue μ. If μ has multiplicity k there exists a k-subset X of the vertex-set V (G) such that G − X does not have μ as an eigenvalue. We say that X is a star set for μ, and that G − X is a star complement for μ. (The terminology reflects a geometrical definition in terms of eutactic stars; see [6, Chapter 5].) If μ = −1, 0 then there are only finitely many graphs with a star complement of order t because the order of G is bounded above by 12 t(t + 1) when t > 2 and by 4 when t ≤ 2 [2, Theorem 2.3]. The star complement technique [6, p.145] is a procedure for constructing all graphs with a prescribed star complement H for a prescribed eigenvalue μ. The first step requires knowledge of (μI − C)−1 , where C is the adjacency matrix of H. Here we introduce a method applicable to regular graphs that can sometime be used to deal with star complements for which (μI−C)−1 is either not known explicitly or not amenable to subsequent calculations. The method is akin to the construction of a μ-eigenvector (x1 , x2 , . . . , xn ) of G using the well-known relation μxj − Σi∼j xi = 0 (i, j ∈ V (G)),
(1)
where we take V (G) = {1, 2, . . . , n} and write i ∼ j to mean that vertices i, j are adjacent. (For an illustration, see Lemma 2.8 below.) Analogously we find successive solutions of the equation μyj − Σi∼j yi = 1 (i, j ∈ V (H)),
(2)
which is derived in Section 2. We give two applications of the method. In Section 3 we determine the regular graphs with a path Pt as a star complement for the eigenvalue 1; it turns out that apart from cycles, only two graphs arise. The graphs with Pt as a star complement for −2 are found by different means in [1] and [3]. In Section 4 we take G to be an r-regular graph in which a star complement H is the tree T (m1 , m2 , . . . , mr ) illustrated in Fig.1. Here 0 ≤ mi ≤ r − 2 (i = 1, 2, . . . , r), and so H has exactly one vertex of degree r. We bound n in terms of r and t when r > 2 and μ = −1, 0, and then determine the corresponding extremal graphs that arise. Explicitly, we find that (i) if r = 3 then n ≤ 2t, with equality if and only if μ = 1, H = T (1, 0, 0) and G is the Petersen graph, (ii) if r > 3 then n ≤ 12 rt + 1, with equality if and only if μ = 1, H = T (0, 0, 0, 0, 0) and G is the complement of the Clebsch graph, i.e. the strongly regular graph Cl5 with parameters 16,5,0,2. A tree is chosen for H because, under suitable conditions, an extremal r-regular graph necessarily has a tree as a star complement (see Proposition 2.7 below). The particular choice of T (m1 , m2 , . . . , mr ) makes for a generalization of [10, Theorem 2.2], which 1
describes the regular graphs that have a star as a star complement. It also reflects the fact that any triangle-free r-regular graph generally has numerous trees of this type as induced subgraphs. Note however that our arguments are not confined to triangle-free graphs. v0
v1
s A A A s s ... . . . As
sPP P
PP
v2
s A
s
m1
...
PP
...
PP
...
A
A s . . . As
PP
vr . .P . PH s
A H A HH HH A .H . .Hs s As . . .
m2
mr
Fig.1: The graph T (m1 , m2 , . . . , mr ), where 0 ≤ mi ≤ r − 2 (i = 1, 2, . . . , r).
2
Preliminaries
The fundamental result on star sets is the following, where we write GX for the subgraph of G induced by the subset X of V (G). Theorem 2.1. (See [6, Theorem 5.1.7].) Let X be a set of kvertices in the AX B graph G and suppose that G has adjacency matrix , where AX is B C the adjacency matrix of GX . (i) Then X is a star set for μ in G if and only if μ is not an eigenvalue of C and μI − AX = B (μI − C)−1 B. (3)
(ii) If X is a star set for μ then E(μ) consists of the vectors
x (μI − C)−1 Bx
(x ∈ IRk ). Let H = G − X, where X is a star set for μ, and for v ∈ V (G) let ΔH (v) = {u ∈ V (H) : u ∼ v}. In the notation of Theorem 2.1, C is the adjacency matrix of the star complement H, while the columns bu (u ∈ X) of B are the characteristic vectors of the H-neighbourhoods ΔH (u) (u ∈ X). We write x, y for x (μI − C)−1 y (x, y ∈ IRt ), where t = n − k. Eq. (2) shows that bu , bv =
⎧ ⎪ ⎨
μ if u = v −1 if u ∼ v ⎪ ⎩ 0 otherwise.
Thus if μ = −1, 0 then bi = 0 for all i ∈ X and bi = bj whenever i = j. In other words: Lemma 2.2. If X is a star set for μ, and μ ∈ {−1, 0}, then the neighbourhoods ΔH (u) (u ∈ X) are non-empty and distinct. 2
If X is a star set for μ in G and h of the vertices in X are deleted from G then (by interlacing) the multiplicity of μ is reduced by h. Hence we have: Lemma 2.3. If X is a star set for μ in G and if U is a proper subset of X then X \ U is a star set for μ in G − U . We write j for an all-1 vector, its length determined by context. Recall that μ is a main eigenvalue of G if E(μ) is not orthogonal to j, and that in an rregular graph, every eigenvalue other than r is non-main. The next observation follows from Theorem 2.1(ii). Lemma 2.4. (See [6, Proposition 5.2.4].) If X is a star set for the non-main eigenvalue μ then bu , j = −1 for all u ∈ X. We shall exploit the fact that each bu (u ∈ X) is a linear combination of the vectors f1 , f2 , . . . , ft which constitute the standard orthonormal basis of IRt . Let yi = fi , j (i = 1, 2, . . . , t). Then (μI − C)y = j, where y = (y1 , y2 , . . . , yt ) , and we obtain Eq.(2). Lemma 2.5. (Cf. [10, p.2240].) If G is r-regular of order n and μ < r then j, j = n/(μ − r). Proof. With the notation of Theorem 2.1, let S = (B | C − μI). Then Sj = (r − μ)j and S (μI − C)−1 S = μI − A. Now we have j (μI − A)j = jS (μI − C)Sj = (r − μ)2 j (μI − C)−1 j. Hence (μ − r)n = (μ − r)2 j, j, and the result follows. 2 Lemma 2.6. Let G be an r-regular graph with μ as an integer eigenvalue of co-multiplicity t. If r > 3 and n ≥ 12 rt + 1 then μ ∈ {−1, 0, 1}.
Proof. When r > 3 and n ≥ 12 rt + 1, the multiplicity k of μ exceeds 12 n. Hence μ is an integer, for otherwise μ has an algebraic conjugate which is a second eigenvalue of multiplicity k. We suppose that μ = 0 and that the eigenvalues = μ are μ1 , . . . , μt , with mean d. Then kμ + td = tr(A) = 0 and kμ2 + Σti=1 μ2i = tr(A2 ) = nr. Accordingly we have Σti=1 (μi − d)2 = Σti=1 μ2i − td2 = (k + t)r − μ2 (k + t(
k2 )). t2
It follows that k ≤ rt/μ2 ≤ rt. Now k ≥ 12 (r − 2)t + 1 and so rt/μ2 > 12 (r − 2)t. 2 Hence μ2 < 2 + 4/(r − 2) ≤ 4 and the result follows. We write E(G) for the edge-set of G, and for subsets U, V of V (G) we write E(U, V ) for the set of edges between U and V . When H = G − X it is convenient to write X for V (H). Recall that if G is connected then G has a connected star complement for each eigenvalue [6, Theorem 5.1.6]. Proposition 2.7 Let G be an r-regular graph with μ (= −1, 0) as an eigenvalue of co-multiplicity t. If H (= G − X) is a connected star complement for μ such that |ΔH (u)| > 1 for all u ∈ X then n ≤ 12 rt + 1. Moreover if n = 12 rt + 1 then (i) H is a tree, (ii) |ΔH (u)| = 2 for all u ∈ X, (iii) μ = 1 when r > 3. Proof. Since H is connected, it has at least t − 1 edges and so we have 2k ≤ Σu∈X |ΔH (u)| = |E(X, X)| = rt−2|E(H)| ≤ (r −2)t+2. Hence k ≤ 12 (r −2)t+1, equivalently n ≤ 12 rt + 1. If equality holds then |ΔH (u)| = 2 for all u ∈ X, while H is a tree because |E(H)| = t − 1. Moreover if n = 12 rt + 1 and r > 3 then k > 12 n and so μ = 1 by Lemma 2.6. 2 We shall also require: 3
Lemma 2.8. A graph of the form illustrated in Fig.2 has μ as an eigenvalue. Proof. Figure 2 shows entries of a 1-eigenvector (x1 , x2 , . . . , xn ) , constructed by choosing xj = 1 for an endvertex j, then using Eq.(1) repeatedly and taking xi = 0 for all i ∈ V (G ). 2
s 1 μ s s 1 @ . @ .. '$
@s 1 s 0 A s −1 A &% .. . A As s −μ HH −1 Hs −1
G
⎫ ⎪ ⎬ ⎪ ⎭ ⎫ ⎪ ⎬ ⎪ ⎭
μ2
μ2
Fig.2: A graph with μ as an eigenvalue (μ2 ∈ IN ).
3
A path as a star complement
Let G be a graph of order n with Pt as a star complement H = G − X for an eigenvalue μ = −1, 0 of multiplicity k. Lemma 2.2 shows that G is connected. We assume throughout that t > 1, for otherwise G = K2 by the same Lemma. If μ ≤ −2 then by interlacing μ is the least eigenvalue of G. The graphs which arise when μ = −2 and t = 7, 8 were determined in [1] by traditional means. The maximal graphs which arise when μ = −2 (without restriction on t) were determined in [3] by exploiting the forbidden subgraph technique in graphs with least eigenvalue −2 (cf.[5, Theorem 2.3.18]). Here we determine the regular graphs with Pt as a star complement for the eigenvalue 1: in contrast to both [1] and [3], we avoid the need to describe all possible H-neighbourhoods ΔH (u) (u ∈ X). Accordingly suppose first that G is r-regular. Then r ≥ 2, and if r = 2 then μ ≤ 2 since G is a cycle. By interlacing, we also have μ < 2 when r > 2. Thus μ = r if and only if μ = 2, in which case G = Cn and k = 1. Recall that for μ < r we have j = (μI − C)y, where y = (y1 , . . . , yt ) and yh = fh , j (h = 1 . . . t). We assign the natural ordering to the vertices of Pt , and define y0 = 0, yt+1 = 0. Then 1 = μyh − yh+1 − yh−1 (h = 1, . . . , t). If α = 1/(μ − 2) and zh = yh − α (h = 0, 1, . . . , t + 1) then zh+1 = μzh − zh−1 (h = 1, . . . , t), z0 = −α = zt+1 .
(4)
Lemma 3.1. If μ = ±2 then the sequence y1 , y2 , . . . , yt is palindromic. Proof. If μ > −2 then we may write μ = 2 cos θ (0 < θ < π). In this case the solution to Eq.(4) is zh = Qeihθ + Re−ihθ , where Qei(t+1)θ + Re−i(t+1)θ = −α = Q + R. Note that ei(t+1)θ = ±1 for otherwise θ = mπ/(t + 1) for some integer m such that 0 < m < t + 1, and then μ is an eigenvalue of Pt . It follows that Q = R = −α/(ei(t+1)θ + 1) = 0 and Q/R = e−i(t+1)θ . Therefore
4
Qeihθ + Re−ihθ = Re−(t+1−h)θ + Qe(t+1−h)θ . Thus the sequence z1 , z2 , . . . , zt , and hence also y1 , y2 , . . . , yt , is palindromic. If μ < −2 then we may write μ = −2 cosh θ (θ > 0). In this case the solution to Eq.(4) is zh = Q(−e)hθ + R(−e)−hθ , where Q(−e)(t+1)θ + R(−e)−(t+1)θ = −α = Q + R. Since (−e)(t+1)θ = ±1, we have Q/R = (−e)−(t+1)θ , and the result follows as before. 2 Lemma 3.2. (i) k = (2 − r)
t
h=1 yh − y1 − yt ,
(ii) t = (μ − 2)
t
h=1 yh + y1 + yt .
Proof. From Lemma 2.4 we have −k = u∈X bu , j. On the other hand, counting in two ways the edges between X and X, we have u∈X bu , j = (r − 2) th=1 yh + y1 + yt , and this proves (i). Now (ii) follows since t = n − k and n = (μ − r) th=1 yh by Lemma 2.5. 2 Now we suppose that μ = 1 (which is necessarily the case if r > 2 and μ is a rational greater than −2). If y1 = c then repeated use of Eq.(2) yields y2 = c − 1, y3 = −2, y4 = −c − 2, y5 = −c − 1 and y6 = 0. It follows that if t = 6m + q, where m is a non-negative integer and 0 ≤ q < 6, then the subsequence y1 , y2 , . . . , y6 is repeated exactly m times. Note that t = 6m + 2 by Lemma 3.1 because y6m+2 = c − 1 = y1 . Also, t = 6m + 5 because 1 not an eigenvalue of Pt . If t = 6m then c = 0 by Lemma 3.1, and so y3 = −2, while yh ≤ 0 (h = 1, . . . , t). It follows from Lemma 2.4 that the third vertex of H has no neighbour in X. Hence r = 2, and if u is the neighbour in X of an endvertex of H then bu , j = 0, a contradiction. If t = 6m + 1 then th=1 yh = −6m + c, and so t = 6m + c by Lemma 3.2. Hence c = 1 and (y1 , y2 , y3 , y4 , y5 , y6 ) = (1, 0, −2, −3, −2, 0). It follows that |ΔH (u)| > 1 for all u ∈ X , and so |E(X, X)| ≥ 2k. By Lemma 3.2, we have t = − th=1 yh + 2 and k = (r − 2)(t − 2) − 2, and so (r − 2)t + 2 = |E(X, X)| ≥ 2k = (2r − 4)(t − 2) − 4.
(5)
6 ≤ 10. Hence t = 7 and r = 3 or 4. We deduce that r > 2 and t ≤ 4 + r−2 If r = 4 then we have equality throughout in Eq.(5), and so |ΔH (u)| = 2 for all u ∈ X; but if u is a neighbour of the fourth vertex of H then bu , j = −1, a contradiction. If r = 3 then k = 3 and n = 10. All the cubic graphs of order 10, along with their spectra, can be found in [4, pp.292-4]. Only two of these graphs, namely those labelled 3.12 and 3.16, have 1 as an eigenvalue of multiplicity 3. We denote these graphs by G3.12 and G3.16 . Each as P7 as a star complement for 1. (To verify this, it helps to note that X is an independent set, because |E(X, X)| = 9.) If t = 6m + 3 then c = −2 by Lemma 3.1, and so (y1 , y2 , y3 , y4 , y5 , y6 ) = (−2, −3, −2, 0, 1, 0). Hence th=1 yh = −6m − 7 and |ΔH (u)| > 1 for all u ∈ X. By Lemma 3.2, k = (r − 2)(t + 4) + 4 and we have
(r − 2)t + 2 = |E(X, X)| ≥ 2k = (2r − 4)(t + 4) + 8. Then (r − 2)t ≤ 10 − 8r, a contradiction. If t = 6m + 4 then c = −1 by Lemma 3.1, and so (y1 , y2 , y3 , y4 , y5 , y6 ) = (−1, −2, −2, −1, 0, 0). Hence the second vertex of Pt has no neighbour u ∈ X, for otherwise bu , j = −1. Therefore r = 2, and each endvertex of H has a single neighbour in X. These neighbours are distinct by Lemma 2.4, and it follows that G is a cycle of length t + 2 = 6(m + 1). Such cycles do indeed have Pt as a star complement for 1, and so we can summarize our results as follows. 5
Theorem 3.3. The regular graphs with a non-trivial path as a star complement for the eigenvalue 1 are the cycles of length a multiple of 6, together with the graphs G3.12 and G3.16 .
4
Another type of tree as a star complement
In this section we consider r-regular graphs with T (m1 , m2 , . . . , mr ) (r > 2) as a star complement for an arbitrary eigenvalue μ = −1, 0. To ease notation we take V (H) = {1, . . . , t}, so that fj , j is associated with vertex j of H. Proposition 4.1 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement H = G − X for the eigenvalue μ, where μ = −1, 0 and r > 2. Let v0 be the vertex of degree r in H. If X contains a vertex u such that ΔH (u) = {v} with v ∼ v0 then either (a) r = 1 + μ + μ2 and n = 1 + (1 + μ + μ2 )2 , or (b) n = (μ − r)(μr − μ3 − 2μ2 − 3μ − 2). Proof. Note that by Lemma 2.2, G is connected because H is connected. Hence the eigenvalue r is simple [6, Theorem 1.3.6]. Then μ = r, and so μ is a non-main eigenvalue. Let v1 , v2 , . . . , vr be the neighbours of v0 . Figure 3 shows the graph H with each vertex j labelled with the value of fj , j as follows. We define c1 , c2 , . . . , cr by fvi , j = μci − 1 (i = 1, 2, . . . , r). Then, by Eq.(2), fj , j = ci for any endvertex j of H adjacent to vi . If fv0 , j = y then by Eq.(2) applied to vi (i = 0) we have y = (μ2 − mi )ci − μ − 1 (i = 1, 2, . . . , r).
(6)
y
s PPP
PP
μc1 − 1
s A A A s s ... . . . As
c1
c1
c1
s
c2
PP
μc2 − 1 s ... ...
A
PP
...
A
A s . . . As
c2
c2
μcr − 1 s . .P . PH
PP
A H A HH HH A .H . .Hs s As . . .
cr
cr
cr
Fig.3: Values for fj , j (j ∈ X) in T (m1 , m2 , . . . , mr ). Without loss of generality, v ∼ v1 and then c1 = 0 by Lemma 2.4. It follows from Eq.(6) that (μ2 − mi )ci = 0 (i = 1, 2, . . . , r). By Lemma 2.8 at most one of the mi is equal to μ2 , and so without loss of generality either (a) c1 = · · · = cr = 0 or (b) c1 = · · · = cr−1 = 0 = cr . In case (a), we have y = −μ − 1 by Eq.(6). Applying Eq.(2) to v0 we obtain 1 = μ(−μ − 1) − r(−1), whence r = 1 + μ + μ2 . Now j, j = Σtj=1 fj , j = −1 − μ − r, and by Lemma 2.4 we have n = (r − μ)(1 + μ + r) = 1 + (1 + μ + μ2 )2 . In case (b), Eq.(2) yields 1 = μ(−μ − 1) − (r − 1)(−1) − (μcr − 1), whence μcr = r − 1 − μ − μ2 . Here mr = μ2 and j, j = −1 − μ + (r − 1)(−1) + (μcr − 1) + mr cr = μr − μ3 − 2μ2 − 3μ − 2. By Lemma 2.5, n = (μ − r)(μr − μ3 − 2μ2 − 3μ − 2). 6
2
The Petersen graph provides an example of Proposition 2.1(a) with H = T (1, 0, 0). We note in passing that we may weaken the condition ΔH (u) = {v} by replacing it with the requirement that ΔH (u) = {v} ∪˙ V , where Σj∈V fj , j = 0. In these circumstances, the Hoffman-Singleton graph is an example in which the harmonic tree T (2, 2, 2, 2, 2, 2, 2) is a star complement for 2, c1 = · · · = c7 = 0 and |ΔH (u)| = 4 for all u ∈ X (see [9, Eq.(4) and Lemma 2.1]). Lemma 4.2 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement H = G − X of order t for the eigenvalue μ, where μ = −1, 0 and r > 3. If X contains a vertex u such that |ΔH (u)| = 1 then n ≤ 12 rt. Proof. We define v0 , v1 , . . . , vr as in Proposition 4.1, and as usual we let k = n − t. Suppose by way of contradiction that n ≥ 12 rt + 1; then μ = 1 by Lemma 2.6. There are two cases to consider: (a) X has a vertex u such that ΔH (u) = {v} with v ∼ v0 , (b) X has a vertex u such that ΔH (u) = {v} with v ∼ v0 .
Case (a). By Proposition 4.1, n = (1−r)(r −8) and so 4 ≤ r ≤ 6. If r = 6 then n = 10 and t ≤ 3; and if r = 5 then n = 12 and t ≤ 4. In both cases, we have a contradiction to the inequality n ≤ 12 (t − 1)(t + 2). If r = 4 then n = 12 and t ≤ 5. If t ≤ 4 then the same contradiction arises, while if t = 5 then H = K1,4 . By [10, Theorem 2.2], no regular graph has K1,4 as a star complement, and so we conclude that always n ≤ 12 rt in case (a). −1
s PPP
0 1 s s A A A s s ... . . . As 1
0
m1
0
PP
x2
s
s A
A
A s . . . As
x 2 x2
...
x2
m2
PP
...
PP
...
PP
xr . .P . PH s
A H A HH HH A .H . .Hs s As . . .
xr
xr
xr
mr
Fig.4: Entries in a 1-eigenvector of H + u (case (b) of Lemma 4.2).
Case (b). We make use of Eq.(1) in H +u and Eq.(2) in H. By Lemma 2.3 and Theorem 2.1(ii) H + u has a 1-eigenvector v whose u-th entry is 1. It follows from Eq.(1) that without loss of generality the entries of v have the form shown in Fig.4. Here xi = −1 + mi xi (i = 2, . . . , r) and −1 = 0 + x2 + · · · + xr . Hence mi = 1 (i = 2, . . . , r) and r 1 = −1. (7) m i−1 i=2 We deduce that at least one of m2 , . . . , mr is zero, say m2 = 0. Since bu , j = −1, Eq.(2) ensures that the values of fj , j (j ∈ V (H)) take the form shown in Fig.5 (with the ci defined as in Proposition 4.1). Here y = (1 − mi )ci − 2 (i = 1, 2, . . . , r), and so (1 − mi )ci = c2 (i = 1, 2, . . . , r).
7
Now, using Eq.(7), we have 1=y+2−
r
(ci − 1) = c2 + r − 1 −
i=2
r i=2
c2 = r − 1, 1 − mi
whence r = 2, contrary to assumption. Hence always n ≤ 12 rt in case (b).
2
y
−2
s A A A s s ... . . . As
−1
−1
s P PP PP
c2 − 1 s ...
PP
...
PP
...
cr − 1 s . .P . PH
PP
A H A HH HH A .H . .Hs s As . . .
−1
cr
cr
m1
cr
mr
Fig.5: Values for fj , j (j ∈ X in case (b) of Lemma 4.2.
Theorem 4.3 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement of order t for the eigenvalue μ, where μ = −1, 0 and r > 2. (i) If r = 3 then n ≤ 2t, with equality if and only if μ = 1 and G is the Petersen graph, (ii) If r > 3 then n ≤ 12 rt + 1, with equality if and only if μ = 1, r = 5 and G is the complement of the Clebsch graph. Proof. As usual, we let k = n − t and H = G − X. We define v0 , v1 , . . . , vr as in Proposition 4.1. As before, μ = r. (i) We know from [7, Theorem 1.1] that for a cubic graph we have n ≤ 2t, with equality if and only if G is the Petersen graph; and we have observed that this graph has T (1, 0, 0) as a star complement for the eigenvalue 1. (ii) By Lemma 4.2, we may assume that |ΔH (u)| > 1 for all u ∈ X. By Proposition 2.7 we have n ≤ 12 rt + 1, and it remains to deal with the case n = 12 rt + 1. Here μ = 1 and |ΔH (u)| = 2 for all u ∈ X. For i = 1, 2, . . . , r let ui be a vertex in X adjacent to vi , say ΔH (ui ) = {vi , wi }. There are three possible types for such a neighbourhood according as (a) wi ∼ vi , (b) wi ∼ v0 , (c) wi ∼ vi and wi ∼ v0 . In case (a) we have fu1 , j = 2c1 − 1 and so c1 = 0 by Lemma 2.4. As in Proposition 4.1, at most one of the ci is non-zero. If say cr = 0 then mr = 1 and fwr , j ∈ {−1, 0, cr }. Since cr = 0 and −1 = bur , j = cr − 1 + fwr , j we have cr = 1 and j, j = −2 + (r − 1)(−1) + 1 = −r. By Lemma 2.5, we have 1 2 2 rt + 1 = r(r − 1) and so rt = 2r − 2r − 2. Then r divides 2, a contradiction. Therefore c1 = c2 = · · · = cr = 0, and by Eq.(2) we have 1 = −2 − r(−1), whence r = 3, also a contradiction. In case (b), we may take w1 = v2 without loss of generality. Then the entries of a 1-eigenvector of H + u1 have the form shown in Fig.6. We have a = m1 a + 1 + x and 1 − a = m2 (1 − a) + 1 + x, whence (m1 + m2 − 2)a = m2 − 1. We have m1 + m2 − 2 = 0 for otherwise m1 = m2 = 1, contradicting Lemma 2.8. 8
Hence a=
m2 − 1 , m1 + m2 − 2
s P PP
HH
H HH H
s A A
s
s ...
a
a
1 − m1 m2 = 0. m1 + m2 − 2
x
1
s H
a
x=
1−a
HHs . . . A A A s . . . As
A
s . . . As a 1−a 1−a
PP
m1
PP
...
PP
...
PP
xr . .P . PH s
A H A HH H A HH . . .Hs s As . . .
1−a
xr
m2
xr
xr
mr
Fig.6: Entries in a 1-eigenvector of H + u1 in case (b) of Theorem 4.3(ii). Now x = 1 + Σri=3 xr and xi = mi xi + x (i = 3, . . . , r), whence 1 − mi = 0 (i = 3, . . . , r) and r i=3
1 1 (m1 − 1)(m2 − 1) =1− =2− . 1 − mi x m1 m2 − 1
(8)
By Eq.(2) we have (in the notation of Fig.3) y = (1 − mi )ci − 2 (i = 1, . . . , r) and y = 1 + Σri=1 (ci − 1). Since (c1 − 1) + (c2 − 1) = −1 we have y =2−r+
r
ci = 2 − r + (y + 2)
i=3
r
1 . 1 − mi 1=3
(9)
Since y + 2 = (1−m1 )c1 = (1−m2 )c2 and c1 + c2 = 1, we have (1−m1 )(1−m2 ) = (1−m1 + 1−m2 )(y + 2). From Equations (8) and (9) we find that r−4=−
(m1 − 1)(m2 − 1) . m1 m2 − 1
Since r > 3 either m1 m2 = 0 or r = 4, (m1 − 1)(m2 − 1) = 0. Without loss of generality, either m2 = 0 or r = 4, m1 = 1. In the former case, r = 5 − m1 and so always r ∈ {4, 5}. Consider first the case r = 4. Here m1 = 1 and either m2 = 0 or m2 = 2 by 1 1 Lemma 2.8. Eq.(8) yields 1−m + 1−m = 2, whence m3 = m4 = 0. Suppose 3 4 that m2 = 0; then H = T (1, 0, 0, 0), t = 6 and k = 7. If u is any vertex in X with ΔH (u ) = {vi , vj } (0 < i < j) then the above argument shows that i = 1 because {mi , mj } = {0, 1}. Therefore at least four of the vertices in X have as a neighbour the endvertex v of H adjacent to v1 . Hence deg(v ) > 4, a contradiction. Next suppose that m2 = 2; then H = T (1, 2, 0, 0), t = 8 and k = 9. Here there is only one vertex in X with ΔH (u ) = {vi , vj } (0 < i < j), and since there are no H-neighbourhoods of type (a), there are only three possible H-neighbourhoods of the form {vi , wi } (wi ∼ v0 ). It follows that there are at most seven possible H-neighbourhoods, conradicting k = 9. Now consider the case r = 5. Here m1 = 0, m2 = 0 and Eq.(8) yields 1 1 1 1−m3 + 1−m4 + 1−m5 = 3, whence m3 = m4 = m5 = 0. Thus H = K1,5 and G = Cl5 by [10, Theorem 2.2]. 9
1
x
s PPP
s T
T a −a T s ... s T A A A T A T A A . . . As TTs s . . . As s s ...
a
a
a
1−a −a
PP
−a
PP
...
PP
...
PP
xr s . .P . PH
A HH H A H A H As . . . .H . .Hs s
xr
xr
xr
Fig.7: Entries in a 1-eigenvector of H + u in case (c) of Theorem 4.3(ii). It follows that if G = Cl5 then each of the neighbourhoods ΔH (ui ) (i = 1, 2, . . . , r) is of type (c). In particular, the vertices u1 , u2 , . . . , ur are distinct. Without loss of generality we may take w1 ∼ v2 . Then the entries of a 1eigenvector of H + u have the form shown in Fig.7. We have a = m1 a + 1 + x and −a = 1 − a + (m2 − 1)(−a) + x, whence (m1 + m2 )a = 2a. If a = 0 then 1 = 1. Then −1 = x = Σri=3 xi and xi = mi xi −1 (i = 3, . . . , r), whence ri=3 1−m i at least one of m3 , . . . , mr must be zero, say m3 = 0. With the notation of Fig.3 we have y = (1−mi )ci −2 (i = 1, 2, . . . , r) and so y = c3 −2, (1−mi )ci = c3 (i = 3, . . . , r). By Eq.(2), we have y = 1 + ri=1 (ci − 1) = 1 − r + c1 + c2 + ri=3 ci . By Lemma 2.4 we have −1 = (c1 − 1) + c2 , whence c1 + c2 = 0, and r =1+
r i=3
ci − y = 1 + c3
r i=3
1 − (c3 − 2) = 3, 1 − mi
a contradiction. Hence a = 0 and so m1 + m2 = 2. It follows that m1 = 0 and m2 = 2. If we now repeat the argument with u2 in place of u1 we find that m2 = 0, a final contradiction. 2
References [1] F. K. Bell, Line graphs of bipartite graphs with Hamiltonian paths, J. Graph Theory 43 (2003), 137-149. [2] F. K. Bell, P. Rowlinson, On the multiplicities of graph eigenvalues, Bull. London Math. Soc. 35 (2003), 401-408. [3] F. K. Bell, S. K. Simi´c. On graphs whose star complement for −2 is a path or a cycle, Linear Algebra Appl. 377 (2004), 249-265. [4] D. Cvetkovi´c, M. Doob, H. Sachs, Spectra of Graphs, 3rd edition, Johann Ambrosius Barth Verlag (Heidelberg), 1995. [5] D. Cvetkovi´c, P. Rowlinson, S. K. Simi´c, Spectral Generalizations of Line Graphs, Cambridge University Press (Cambridge), 2007. [6] D. Cvetkovi´c, P. Rowlinson, S. K. Simi´c, An Introduction to the Theory of Graph Spectra, Cambridge University Press (Cambridge), 2010.
10
[7] P. Rowlinson, Eigenvalue multiplicity in cubic graphs, Linear Algebra Appl. 444 (2014), 211-218. [8] P. Rowlinson, Eigenvalue multiplicity in regular graphs, Discrete Appl. Math., to appear. [9] P. Rowlinson, I. Sciriha, Some properties of the Hoffman-Singleton graph, Appl. Anal. Discrete Math. 1 (2007), 438-445 [10] P. Rowlinson, B. Tayfeh-Rezaie, Star complements in regular graphs: old and new results, Linear Algebra Appl. 432 (2010), 2230-2242. [11] J. J. Seidel, Strongly regular graphs, in: Surveys in Combinatorics (Ed. B. Bollob´as), Cambridge University Press (Cambridge), 1979, pp.167-180.
11
Peter Rowlinson
PII: DOI: Reference:
S0024-3795(18)30397-5 https://doi.org/10.1016/j.laa.2018.08.018 LAA 14697
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
26 March 2018 7 August 2018
Please cite this article in press as: P. Rowlinson, AN EXTENSION OF THE STAR COMPLEMENT TECHNIQUE FOR REGULAR GRAPHS, Linear Algebra Appl. (2018), https://doi.org/10.1016/j.laa.2018.08.018
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AN EXTENSION OF THE STAR COMPLEMENT TECHNIQUE FOR REGULAR GRAPHS
Peter Rowlinson1 Mathematics and Statistics Group Division of Computing Science and Mathematics University of Stirling Scotland FK9 4LA United Kingdom
Abstract We extend the means by which a regular graph can sometimes be identified from a star complement. In two applications we determine (i) the regular graphs with a path as a star complement for the eigenvalue 1, (ii) the extremal regular graphs which have another type of tree as a star complement for an eigenvalue = −1, 0.
AMS Classification: 05C50 Keywords: eigenvalue, regular graph, star complement, tree. 1
Tel.:+44 1786 467468; email: [email protected]
1
Introduction
Let G be a connected regular graph of order n with μ (= −1, 0) as an eigenvalue of co-multiplicity t. Thus the corresponding eigenspace E(μ) of a (0, 1)adjacency matrix A of G has dimension n − t. From [2, Theorem 3.1], we know that if t > 2 then n ≤ 12 (t − 1)(t + 2), with equality if and only if G is an extremal strongly regular graph as defined by Seidel [11]. (Only five extremal strongly regular graphs are known; see [6, Section 3.6].) If G is r-regular and r > 2 then n ≤ 12 (r + 1)t, with equality if and only if r = 3, μ = 1 and G is the Petersen graph [8]. These results are proved using star complements, defined as follows for any graph G and any eigenvalue μ. If μ has multiplicity k there exists a k-subset X of the vertex-set V (G) such that G − X does not have μ as an eigenvalue. We say that X is a star set for μ, and that G − X is a star complement for μ. (The terminology reflects a geometrical definition in terms of eutactic stars; see [6, Chapter 5].) If μ = −1, 0 then there are only finitely many graphs with a star complement of order t because the order of G is bounded above by 12 t(t + 1) when t > 2 and by 4 when t ≤ 2 [2, Theorem 2.3]. The star complement technique [6, p.145] is a procedure for constructing all graphs with a prescribed star complement H for a prescribed eigenvalue μ. The first step requires knowledge of (μI − C)−1 , where C is the adjacency matrix of H. Here we introduce a method applicable to regular graphs that can sometime be used to deal with star complements for which (μI−C)−1 is either not known explicitly or not amenable to subsequent calculations. The method is akin to the construction of a μ-eigenvector (x1 , x2 , . . . , xn ) of G using the well-known relation μxj − Σi∼j xi = 0 (i, j ∈ V (G)),
(1)
where we take V (G) = {1, 2, . . . , n} and write i ∼ j to mean that vertices i, j are adjacent. (For an illustration, see Lemma 2.8 below.) Analogously we find successive solutions of the equation μyj − Σi∼j yi = 1 (i, j ∈ V (H)),
(2)
which is derived in Section 2. We give two applications of the method. In Section 3 we determine the regular graphs with a path Pt as a star complement for the eigenvalue 1; it turns out that apart from cycles, only two graphs arise. The graphs with Pt as a star complement for −2 are found by different means in [1] and [3]. In Section 4 we take G to be an r-regular graph in which a star complement H is the tree T (m1 , m2 , . . . , mr ) illustrated in Fig.1. Here 0 ≤ mi ≤ r − 2 (i = 1, 2, . . . , r), and so H has exactly one vertex of degree r. We bound n in terms of r and t when r > 2 and μ = −1, 0, and then determine the corresponding extremal graphs that arise. Explicitly, we find that (i) if r = 3 then n ≤ 2t, with equality if and only if μ = 1, H = T (1, 0, 0) and G is the Petersen graph, (ii) if r > 3 then n ≤ 12 rt + 1, with equality if and only if μ = 1, H = T (0, 0, 0, 0, 0) and G is the complement of the Clebsch graph, i.e. the strongly regular graph Cl5 with parameters 16,5,0,2. A tree is chosen for H because, under suitable conditions, an extremal r-regular graph necessarily has a tree as a star complement (see Proposition 2.7 below). The particular choice of T (m1 , m2 , . . . , mr ) makes for a generalization of [10, Theorem 2.2], which 1
describes the regular graphs that have a star as a star complement. It also reflects the fact that any triangle-free r-regular graph generally has numerous trees of this type as induced subgraphs. Note however that our arguments are not confined to triangle-free graphs. v0
v1
s A A A s s ... . . . As
sPP P
PP
v2
s A
s
m1
...
PP
...
PP
...
A
A s . . . As
PP
vr . .P . PH s
A H A HH HH A .H . .Hs s As . . .
m2
mr
Fig.1: The graph T (m1 , m2 , . . . , mr ), where 0 ≤ mi ≤ r − 2 (i = 1, 2, . . . , r).
2
Preliminaries
The fundamental result on star sets is the following, where we write GX for the subgraph of G induced by the subset X of V (G). Theorem 2.1. (See [6, Theorem 5.1.7].) Let X be a set of kvertices in the AX B graph G and suppose that G has adjacency matrix , where AX is B C the adjacency matrix of GX . (i) Then X is a star set for μ in G if and only if μ is not an eigenvalue of C and μI − AX = B (μI − C)−1 B. (3)
(ii) If X is a star set for μ then E(μ) consists of the vectors
x (μI − C)−1 Bx
(x ∈ IRk ). Let H = G − X, where X is a star set for μ, and for v ∈ V (G) let ΔH (v) = {u ∈ V (H) : u ∼ v}. In the notation of Theorem 2.1, C is the adjacency matrix of the star complement H, while the columns bu (u ∈ X) of B are the characteristic vectors of the H-neighbourhoods ΔH (u) (u ∈ X). We write x, y for x (μI − C)−1 y (x, y ∈ IRt ), where t = n − k. Eq. (2) shows that bu , bv =
⎧ ⎪ ⎨
μ if u = v −1 if u ∼ v ⎪ ⎩ 0 otherwise.
Thus if μ = −1, 0 then bi = 0 for all i ∈ X and bi = bj whenever i = j. In other words: Lemma 2.2. If X is a star set for μ, and μ ∈ {−1, 0}, then the neighbourhoods ΔH (u) (u ∈ X) are non-empty and distinct. 2
If X is a star set for μ in G and h of the vertices in X are deleted from G then (by interlacing) the multiplicity of μ is reduced by h. Hence we have: Lemma 2.3. If X is a star set for μ in G and if U is a proper subset of X then X \ U is a star set for μ in G − U . We write j for an all-1 vector, its length determined by context. Recall that μ is a main eigenvalue of G if E(μ) is not orthogonal to j, and that in an rregular graph, every eigenvalue other than r is non-main. The next observation follows from Theorem 2.1(ii). Lemma 2.4. (See [6, Proposition 5.2.4].) If X is a star set for the non-main eigenvalue μ then bu , j = −1 for all u ∈ X. We shall exploit the fact that each bu (u ∈ X) is a linear combination of the vectors f1 , f2 , . . . , ft which constitute the standard orthonormal basis of IRt . Let yi = fi , j (i = 1, 2, . . . , t). Then (μI − C)y = j, where y = (y1 , y2 , . . . , yt ) , and we obtain Eq.(2). Lemma 2.5. (Cf. [10, p.2240].) If G is r-regular of order n and μ < r then j, j = n/(μ − r). Proof. With the notation of Theorem 2.1, let S = (B | C − μI). Then Sj = (r − μ)j and S (μI − C)−1 S = μI − A. Now we have j (μI − A)j = jS (μI − C)Sj = (r − μ)2 j (μI − C)−1 j. Hence (μ − r)n = (μ − r)2 j, j, and the result follows. 2 Lemma 2.6. Let G be an r-regular graph with μ as an integer eigenvalue of co-multiplicity t. If r > 3 and n ≥ 12 rt + 1 then μ ∈ {−1, 0, 1}.
Proof. When r > 3 and n ≥ 12 rt + 1, the multiplicity k of μ exceeds 12 n. Hence μ is an integer, for otherwise μ has an algebraic conjugate which is a second eigenvalue of multiplicity k. We suppose that μ = 0 and that the eigenvalues = μ are μ1 , . . . , μt , with mean d. Then kμ + td = tr(A) = 0 and kμ2 + Σti=1 μ2i = tr(A2 ) = nr. Accordingly we have Σti=1 (μi − d)2 = Σti=1 μ2i − td2 = (k + t)r − μ2 (k + t(
k2 )). t2
It follows that k ≤ rt/μ2 ≤ rt. Now k ≥ 12 (r − 2)t + 1 and so rt/μ2 > 12 (r − 2)t. 2 Hence μ2 < 2 + 4/(r − 2) ≤ 4 and the result follows. We write E(G) for the edge-set of G, and for subsets U, V of V (G) we write E(U, V ) for the set of edges between U and V . When H = G − X it is convenient to write X for V (H). Recall that if G is connected then G has a connected star complement for each eigenvalue [6, Theorem 5.1.6]. Proposition 2.7 Let G be an r-regular graph with μ (= −1, 0) as an eigenvalue of co-multiplicity t. If H (= G − X) is a connected star complement for μ such that |ΔH (u)| > 1 for all u ∈ X then n ≤ 12 rt + 1. Moreover if n = 12 rt + 1 then (i) H is a tree, (ii) |ΔH (u)| = 2 for all u ∈ X, (iii) μ = 1 when r > 3. Proof. Since H is connected, it has at least t − 1 edges and so we have 2k ≤ Σu∈X |ΔH (u)| = |E(X, X)| = rt−2|E(H)| ≤ (r −2)t+2. Hence k ≤ 12 (r −2)t+1, equivalently n ≤ 12 rt + 1. If equality holds then |ΔH (u)| = 2 for all u ∈ X, while H is a tree because |E(H)| = t − 1. Moreover if n = 12 rt + 1 and r > 3 then k > 12 n and so μ = 1 by Lemma 2.6. 2 We shall also require: 3
Lemma 2.8. A graph of the form illustrated in Fig.2 has μ as an eigenvalue. Proof. Figure 2 shows entries of a 1-eigenvector (x1 , x2 , . . . , xn ) , constructed by choosing xj = 1 for an endvertex j, then using Eq.(1) repeatedly and taking xi = 0 for all i ∈ V (G ). 2
s 1 μ s s 1 @ . @ .. '$
@s 1 s 0 A s −1 A &% .. . A As s −μ HH −1 Hs −1
G
⎫ ⎪ ⎬ ⎪ ⎭ ⎫ ⎪ ⎬ ⎪ ⎭
μ2
μ2
Fig.2: A graph with μ as an eigenvalue (μ2 ∈ IN ).
3
A path as a star complement
Let G be a graph of order n with Pt as a star complement H = G − X for an eigenvalue μ = −1, 0 of multiplicity k. Lemma 2.2 shows that G is connected. We assume throughout that t > 1, for otherwise G = K2 by the same Lemma. If μ ≤ −2 then by interlacing μ is the least eigenvalue of G. The graphs which arise when μ = −2 and t = 7, 8 were determined in [1] by traditional means. The maximal graphs which arise when μ = −2 (without restriction on t) were determined in [3] by exploiting the forbidden subgraph technique in graphs with least eigenvalue −2 (cf.[5, Theorem 2.3.18]). Here we determine the regular graphs with Pt as a star complement for the eigenvalue 1: in contrast to both [1] and [3], we avoid the need to describe all possible H-neighbourhoods ΔH (u) (u ∈ X). Accordingly suppose first that G is r-regular. Then r ≥ 2, and if r = 2 then μ ≤ 2 since G is a cycle. By interlacing, we also have μ < 2 when r > 2. Thus μ = r if and only if μ = 2, in which case G = Cn and k = 1. Recall that for μ < r we have j = (μI − C)y, where y = (y1 , . . . , yt ) and yh = fh , j (h = 1 . . . t). We assign the natural ordering to the vertices of Pt , and define y0 = 0, yt+1 = 0. Then 1 = μyh − yh+1 − yh−1 (h = 1, . . . , t). If α = 1/(μ − 2) and zh = yh − α (h = 0, 1, . . . , t + 1) then zh+1 = μzh − zh−1 (h = 1, . . . , t), z0 = −α = zt+1 .
(4)
Lemma 3.1. If μ = ±2 then the sequence y1 , y2 , . . . , yt is palindromic. Proof. If μ > −2 then we may write μ = 2 cos θ (0 < θ < π). In this case the solution to Eq.(4) is zh = Qeihθ + Re−ihθ , where Qei(t+1)θ + Re−i(t+1)θ = −α = Q + R. Note that ei(t+1)θ = ±1 for otherwise θ = mπ/(t + 1) for some integer m such that 0 < m < t + 1, and then μ is an eigenvalue of Pt . It follows that Q = R = −α/(ei(t+1)θ + 1) = 0 and Q/R = e−i(t+1)θ . Therefore
4
Qeihθ + Re−ihθ = Re−(t+1−h)θ + Qe(t+1−h)θ . Thus the sequence z1 , z2 , . . . , zt , and hence also y1 , y2 , . . . , yt , is palindromic. If μ < −2 then we may write μ = −2 cosh θ (θ > 0). In this case the solution to Eq.(4) is zh = Q(−e)hθ + R(−e)−hθ , where Q(−e)(t+1)θ + R(−e)−(t+1)θ = −α = Q + R. Since (−e)(t+1)θ = ±1, we have Q/R = (−e)−(t+1)θ , and the result follows as before. 2 Lemma 3.2. (i) k = (2 − r)
t
h=1 yh − y1 − yt ,
(ii) t = (μ − 2)
t
h=1 yh + y1 + yt .
Proof. From Lemma 2.4 we have −k = u∈X bu , j. On the other hand, counting in two ways the edges between X and X, we have u∈X bu , j = (r − 2) th=1 yh + y1 + yt , and this proves (i). Now (ii) follows since t = n − k and n = (μ − r) th=1 yh by Lemma 2.5. 2 Now we suppose that μ = 1 (which is necessarily the case if r > 2 and μ is a rational greater than −2). If y1 = c then repeated use of Eq.(2) yields y2 = c − 1, y3 = −2, y4 = −c − 2, y5 = −c − 1 and y6 = 0. It follows that if t = 6m + q, where m is a non-negative integer and 0 ≤ q < 6, then the subsequence y1 , y2 , . . . , y6 is repeated exactly m times. Note that t = 6m + 2 by Lemma 3.1 because y6m+2 = c − 1 = y1 . Also, t = 6m + 5 because 1 not an eigenvalue of Pt . If t = 6m then c = 0 by Lemma 3.1, and so y3 = −2, while yh ≤ 0 (h = 1, . . . , t). It follows from Lemma 2.4 that the third vertex of H has no neighbour in X. Hence r = 2, and if u is the neighbour in X of an endvertex of H then bu , j = 0, a contradiction. If t = 6m + 1 then th=1 yh = −6m + c, and so t = 6m + c by Lemma 3.2. Hence c = 1 and (y1 , y2 , y3 , y4 , y5 , y6 ) = (1, 0, −2, −3, −2, 0). It follows that |ΔH (u)| > 1 for all u ∈ X , and so |E(X, X)| ≥ 2k. By Lemma 3.2, we have t = − th=1 yh + 2 and k = (r − 2)(t − 2) − 2, and so (r − 2)t + 2 = |E(X, X)| ≥ 2k = (2r − 4)(t − 2) − 4.
(5)
6 ≤ 10. Hence t = 7 and r = 3 or 4. We deduce that r > 2 and t ≤ 4 + r−2 If r = 4 then we have equality throughout in Eq.(5), and so |ΔH (u)| = 2 for all u ∈ X; but if u is a neighbour of the fourth vertex of H then bu , j = −1, a contradiction. If r = 3 then k = 3 and n = 10. All the cubic graphs of order 10, along with their spectra, can be found in [4, pp.292-4]. Only two of these graphs, namely those labelled 3.12 and 3.16, have 1 as an eigenvalue of multiplicity 3. We denote these graphs by G3.12 and G3.16 . Each as P7 as a star complement for 1. (To verify this, it helps to note that X is an independent set, because |E(X, X)| = 9.) If t = 6m + 3 then c = −2 by Lemma 3.1, and so (y1 , y2 , y3 , y4 , y5 , y6 ) = (−2, −3, −2, 0, 1, 0). Hence th=1 yh = −6m − 7 and |ΔH (u)| > 1 for all u ∈ X. By Lemma 3.2, k = (r − 2)(t + 4) + 4 and we have
(r − 2)t + 2 = |E(X, X)| ≥ 2k = (2r − 4)(t + 4) + 8. Then (r − 2)t ≤ 10 − 8r, a contradiction. If t = 6m + 4 then c = −1 by Lemma 3.1, and so (y1 , y2 , y3 , y4 , y5 , y6 ) = (−1, −2, −2, −1, 0, 0). Hence the second vertex of Pt has no neighbour u ∈ X, for otherwise bu , j = −1. Therefore r = 2, and each endvertex of H has a single neighbour in X. These neighbours are distinct by Lemma 2.4, and it follows that G is a cycle of length t + 2 = 6(m + 1). Such cycles do indeed have Pt as a star complement for 1, and so we can summarize our results as follows. 5
Theorem 3.3. The regular graphs with a non-trivial path as a star complement for the eigenvalue 1 are the cycles of length a multiple of 6, together with the graphs G3.12 and G3.16 .
4
Another type of tree as a star complement
In this section we consider r-regular graphs with T (m1 , m2 , . . . , mr ) (r > 2) as a star complement for an arbitrary eigenvalue μ = −1, 0. To ease notation we take V (H) = {1, . . . , t}, so that fj , j is associated with vertex j of H. Proposition 4.1 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement H = G − X for the eigenvalue μ, where μ = −1, 0 and r > 2. Let v0 be the vertex of degree r in H. If X contains a vertex u such that ΔH (u) = {v} with v ∼ v0 then either (a) r = 1 + μ + μ2 and n = 1 + (1 + μ + μ2 )2 , or (b) n = (μ − r)(μr − μ3 − 2μ2 − 3μ − 2). Proof. Note that by Lemma 2.2, G is connected because H is connected. Hence the eigenvalue r is simple [6, Theorem 1.3.6]. Then μ = r, and so μ is a non-main eigenvalue. Let v1 , v2 , . . . , vr be the neighbours of v0 . Figure 3 shows the graph H with each vertex j labelled with the value of fj , j as follows. We define c1 , c2 , . . . , cr by fvi , j = μci − 1 (i = 1, 2, . . . , r). Then, by Eq.(2), fj , j = ci for any endvertex j of H adjacent to vi . If fv0 , j = y then by Eq.(2) applied to vi (i = 0) we have y = (μ2 − mi )ci − μ − 1 (i = 1, 2, . . . , r).
(6)
y
s PPP
PP
μc1 − 1
s A A A s s ... . . . As
c1
c1
c1
s
c2
PP
μc2 − 1 s ... ...
A
PP
...
A
A s . . . As
c2
c2
μcr − 1 s . .P . PH
PP
A H A HH HH A .H . .Hs s As . . .
cr
cr
cr
Fig.3: Values for fj , j (j ∈ X) in T (m1 , m2 , . . . , mr ). Without loss of generality, v ∼ v1 and then c1 = 0 by Lemma 2.4. It follows from Eq.(6) that (μ2 − mi )ci = 0 (i = 1, 2, . . . , r). By Lemma 2.8 at most one of the mi is equal to μ2 , and so without loss of generality either (a) c1 = · · · = cr = 0 or (b) c1 = · · · = cr−1 = 0 = cr . In case (a), we have y = −μ − 1 by Eq.(6). Applying Eq.(2) to v0 we obtain 1 = μ(−μ − 1) − r(−1), whence r = 1 + μ + μ2 . Now j, j = Σtj=1 fj , j = −1 − μ − r, and by Lemma 2.4 we have n = (r − μ)(1 + μ + r) = 1 + (1 + μ + μ2 )2 . In case (b), Eq.(2) yields 1 = μ(−μ − 1) − (r − 1)(−1) − (μcr − 1), whence μcr = r − 1 − μ − μ2 . Here mr = μ2 and j, j = −1 − μ + (r − 1)(−1) + (μcr − 1) + mr cr = μr − μ3 − 2μ2 − 3μ − 2. By Lemma 2.5, n = (μ − r)(μr − μ3 − 2μ2 − 3μ − 2). 6
2
The Petersen graph provides an example of Proposition 2.1(a) with H = T (1, 0, 0). We note in passing that we may weaken the condition ΔH (u) = {v} by replacing it with the requirement that ΔH (u) = {v} ∪˙ V , where Σj∈V fj , j = 0. In these circumstances, the Hoffman-Singleton graph is an example in which the harmonic tree T (2, 2, 2, 2, 2, 2, 2) is a star complement for 2, c1 = · · · = c7 = 0 and |ΔH (u)| = 4 for all u ∈ X (see [9, Eq.(4) and Lemma 2.1]). Lemma 4.2 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement H = G − X of order t for the eigenvalue μ, where μ = −1, 0 and r > 3. If X contains a vertex u such that |ΔH (u)| = 1 then n ≤ 12 rt. Proof. We define v0 , v1 , . . . , vr as in Proposition 4.1, and as usual we let k = n − t. Suppose by way of contradiction that n ≥ 12 rt + 1; then μ = 1 by Lemma 2.6. There are two cases to consider: (a) X has a vertex u such that ΔH (u) = {v} with v ∼ v0 , (b) X has a vertex u such that ΔH (u) = {v} with v ∼ v0 .
Case (a). By Proposition 4.1, n = (1−r)(r −8) and so 4 ≤ r ≤ 6. If r = 6 then n = 10 and t ≤ 3; and if r = 5 then n = 12 and t ≤ 4. In both cases, we have a contradiction to the inequality n ≤ 12 (t − 1)(t + 2). If r = 4 then n = 12 and t ≤ 5. If t ≤ 4 then the same contradiction arises, while if t = 5 then H = K1,4 . By [10, Theorem 2.2], no regular graph has K1,4 as a star complement, and so we conclude that always n ≤ 12 rt in case (a). −1
s PPP
0 1 s s A A A s s ... . . . As 1
0
m1
0
PP
x2
s
s A
A
A s . . . As
x 2 x2
...
x2
m2
PP
...
PP
...
PP
xr . .P . PH s
A H A HH HH A .H . .Hs s As . . .
xr
xr
xr
mr
Fig.4: Entries in a 1-eigenvector of H + u (case (b) of Lemma 4.2).
Case (b). We make use of Eq.(1) in H +u and Eq.(2) in H. By Lemma 2.3 and Theorem 2.1(ii) H + u has a 1-eigenvector v whose u-th entry is 1. It follows from Eq.(1) that without loss of generality the entries of v have the form shown in Fig.4. Here xi = −1 + mi xi (i = 2, . . . , r) and −1 = 0 + x2 + · · · + xr . Hence mi = 1 (i = 2, . . . , r) and r 1 = −1. (7) m i−1 i=2 We deduce that at least one of m2 , . . . , mr is zero, say m2 = 0. Since bu , j = −1, Eq.(2) ensures that the values of fj , j (j ∈ V (H)) take the form shown in Fig.5 (with the ci defined as in Proposition 4.1). Here y = (1 − mi )ci − 2 (i = 1, 2, . . . , r), and so (1 − mi )ci = c2 (i = 1, 2, . . . , r).
7
Now, using Eq.(7), we have 1=y+2−
r
(ci − 1) = c2 + r − 1 −
i=2
r i=2
c2 = r − 1, 1 − mi
whence r = 2, contrary to assumption. Hence always n ≤ 12 rt in case (b).
2
y
−2
s A A A s s ... . . . As
−1
−1
s P PP PP
c2 − 1 s ...
PP
...
PP
...
cr − 1 s . .P . PH
PP
A H A HH HH A .H . .Hs s As . . .
−1
cr
cr
m1
cr
mr
Fig.5: Values for fj , j (j ∈ X in case (b) of Lemma 4.2.
Theorem 4.3 Let G be an r-regular graph of order n with T (m1 , m2 , . . . , mr ) as a star complement of order t for the eigenvalue μ, where μ = −1, 0 and r > 2. (i) If r = 3 then n ≤ 2t, with equality if and only if μ = 1 and G is the Petersen graph, (ii) If r > 3 then n ≤ 12 rt + 1, with equality if and only if μ = 1, r = 5 and G is the complement of the Clebsch graph. Proof. As usual, we let k = n − t and H = G − X. We define v0 , v1 , . . . , vr as in Proposition 4.1. As before, μ = r. (i) We know from [7, Theorem 1.1] that for a cubic graph we have n ≤ 2t, with equality if and only if G is the Petersen graph; and we have observed that this graph has T (1, 0, 0) as a star complement for the eigenvalue 1. (ii) By Lemma 4.2, we may assume that |ΔH (u)| > 1 for all u ∈ X. By Proposition 2.7 we have n ≤ 12 rt + 1, and it remains to deal with the case n = 12 rt + 1. Here μ = 1 and |ΔH (u)| = 2 for all u ∈ X. For i = 1, 2, . . . , r let ui be a vertex in X adjacent to vi , say ΔH (ui ) = {vi , wi }. There are three possible types for such a neighbourhood according as (a) wi ∼ vi , (b) wi ∼ v0 , (c) wi ∼ vi and wi ∼ v0 . In case (a) we have fu1 , j = 2c1 − 1 and so c1 = 0 by Lemma 2.4. As in Proposition 4.1, at most one of the ci is non-zero. If say cr = 0 then mr = 1 and fwr , j ∈ {−1, 0, cr }. Since cr = 0 and −1 = bur , j = cr − 1 + fwr , j we have cr = 1 and j, j = −2 + (r − 1)(−1) + 1 = −r. By Lemma 2.5, we have 1 2 2 rt + 1 = r(r − 1) and so rt = 2r − 2r − 2. Then r divides 2, a contradiction. Therefore c1 = c2 = · · · = cr = 0, and by Eq.(2) we have 1 = −2 − r(−1), whence r = 3, also a contradiction. In case (b), we may take w1 = v2 without loss of generality. Then the entries of a 1-eigenvector of H + u1 have the form shown in Fig.6. We have a = m1 a + 1 + x and 1 − a = m2 (1 − a) + 1 + x, whence (m1 + m2 − 2)a = m2 − 1. We have m1 + m2 − 2 = 0 for otherwise m1 = m2 = 1, contradicting Lemma 2.8. 8
Hence a=
m2 − 1 , m1 + m2 − 2
s P PP
HH
H HH H
s A A
s
s ...
a
a
1 − m1 m2 = 0. m1 + m2 − 2
x
1
s H
a
x=
1−a
HHs . . . A A A s . . . As
A
s . . . As a 1−a 1−a
PP
m1
PP
...
PP
...
PP
xr . .P . PH s
A H A HH H A HH . . .Hs s As . . .
1−a
xr
m2
xr
xr
mr
Fig.6: Entries in a 1-eigenvector of H + u1 in case (b) of Theorem 4.3(ii). Now x = 1 + Σri=3 xr and xi = mi xi + x (i = 3, . . . , r), whence 1 − mi = 0 (i = 3, . . . , r) and r i=3
1 1 (m1 − 1)(m2 − 1) =1− =2− . 1 − mi x m1 m2 − 1
(8)
By Eq.(2) we have (in the notation of Fig.3) y = (1 − mi )ci − 2 (i = 1, . . . , r) and y = 1 + Σri=1 (ci − 1). Since (c1 − 1) + (c2 − 1) = −1 we have y =2−r+
r
ci = 2 − r + (y + 2)
i=3
r
1 . 1 − mi 1=3
(9)
Since y + 2 = (1−m1 )c1 = (1−m2 )c2 and c1 + c2 = 1, we have (1−m1 )(1−m2 ) = (1−m1 + 1−m2 )(y + 2). From Equations (8) and (9) we find that r−4=−
(m1 − 1)(m2 − 1) . m1 m2 − 1
Since r > 3 either m1 m2 = 0 or r = 4, (m1 − 1)(m2 − 1) = 0. Without loss of generality, either m2 = 0 or r = 4, m1 = 1. In the former case, r = 5 − m1 and so always r ∈ {4, 5}. Consider first the case r = 4. Here m1 = 1 and either m2 = 0 or m2 = 2 by 1 1 Lemma 2.8. Eq.(8) yields 1−m + 1−m = 2, whence m3 = m4 = 0. Suppose 3 4 that m2 = 0; then H = T (1, 0, 0, 0), t = 6 and k = 7. If u is any vertex in X with ΔH (u ) = {vi , vj } (0 < i < j) then the above argument shows that i = 1 because {mi , mj } = {0, 1}. Therefore at least four of the vertices in X have as a neighbour the endvertex v of H adjacent to v1 . Hence deg(v ) > 4, a contradiction. Next suppose that m2 = 2; then H = T (1, 2, 0, 0), t = 8 and k = 9. Here there is only one vertex in X with ΔH (u ) = {vi , vj } (0 < i < j), and since there are no H-neighbourhoods of type (a), there are only three possible H-neighbourhoods of the form {vi , wi } (wi ∼ v0 ). It follows that there are at most seven possible H-neighbourhoods, conradicting k = 9. Now consider the case r = 5. Here m1 = 0, m2 = 0 and Eq.(8) yields 1 1 1 1−m3 + 1−m4 + 1−m5 = 3, whence m3 = m4 = m5 = 0. Thus H = K1,5 and G = Cl5 by [10, Theorem 2.2]. 9
1
x
s PPP
s T
T a −a T s ... s T A A A T A T A A . . . As TTs s . . . As s s ...
a
a
a
1−a −a
PP
−a
PP
...
PP
...
PP
xr s . .P . PH
A HH H A H A H As . . . .H . .Hs s
xr
xr
xr
Fig.7: Entries in a 1-eigenvector of H + u in case (c) of Theorem 4.3(ii). It follows that if G = Cl5 then each of the neighbourhoods ΔH (ui ) (i = 1, 2, . . . , r) is of type (c). In particular, the vertices u1 , u2 , . . . , ur are distinct. Without loss of generality we may take w1 ∼ v2 . Then the entries of a 1eigenvector of H + u have the form shown in Fig.7. We have a = m1 a + 1 + x and −a = 1 − a + (m2 − 1)(−a) + x, whence (m1 + m2 )a = 2a. If a = 0 then 1 = 1. Then −1 = x = Σri=3 xi and xi = mi xi −1 (i = 3, . . . , r), whence ri=3 1−m i at least one of m3 , . . . , mr must be zero, say m3 = 0. With the notation of Fig.3 we have y = (1−mi )ci −2 (i = 1, 2, . . . , r) and so y = c3 −2, (1−mi )ci = c3 (i = 3, . . . , r). By Eq.(2), we have y = 1 + ri=1 (ci − 1) = 1 − r + c1 + c2 + ri=3 ci . By Lemma 2.4 we have −1 = (c1 − 1) + c2 , whence c1 + c2 = 0, and r =1+
r i=3
ci − y = 1 + c3
r i=3
1 − (c3 − 2) = 3, 1 − mi
a contradiction. Hence a = 0 and so m1 + m2 = 2. It follows that m1 = 0 and m2 = 2. If we now repeat the argument with u2 in place of u1 we find that m2 = 0, a final contradiction. 2
References [1] F. K. Bell, Line graphs of bipartite graphs with Hamiltonian paths, J. Graph Theory 43 (2003), 137-149. [2] F. K. Bell, P. Rowlinson, On the multiplicities of graph eigenvalues, Bull. London Math. Soc. 35 (2003), 401-408. [3] F. K. Bell, S. K. Simi´c. On graphs whose star complement for −2 is a path or a cycle, Linear Algebra Appl. 377 (2004), 249-265. [4] D. Cvetkovi´c, M. Doob, H. Sachs, Spectra of Graphs, 3rd edition, Johann Ambrosius Barth Verlag (Heidelberg), 1995. [5] D. Cvetkovi´c, P. Rowlinson, S. K. Simi´c, Spectral Generalizations of Line Graphs, Cambridge University Press (Cambridge), 2007. [6] D. Cvetkovi´c, P. Rowlinson, S. K. Simi´c, An Introduction to the Theory of Graph Spectra, Cambridge University Press (Cambridge), 2010.
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[7] P. Rowlinson, Eigenvalue multiplicity in cubic graphs, Linear Algebra Appl. 444 (2014), 211-218. [8] P. Rowlinson, Eigenvalue multiplicity in regular graphs, Discrete Appl. Math., to appear. [9] P. Rowlinson, I. Sciriha, Some properties of the Hoffman-Singleton graph, Appl. Anal. Discrete Math. 1 (2007), 438-445 [10] P. Rowlinson, B. Tayfeh-Rezaie, Star complements in regular graphs: old and new results, Linear Algebra Appl. 432 (2010), 2230-2242. [11] J. J. Seidel, Strongly regular graphs, in: Surveys in Combinatorics (Ed. B. Bollob´as), Cambridge University Press (Cambridge), 1979, pp.167-180.
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