INVESTIGATIONS INTO THE RANKS OF REGULAR GRAPHS
by
CHARLES R GARNER, JR
DISSERTATION
submitted in the fulfilment of the requirements for the degree
PHILOSOPHIAE DOCTOR
in MATHEMATICS
in the FACULTY OF SCIENCE
at the
RAND AFRIKAANS UNIVERSITY
PROMOTER: DR E JONCK
MARCH 2004 INVESTIGATIONS INTO THE RANKS OF REGULAR GRAPHS
CHARLES R GARNER, JR.
I hearby declare that the dissertation submitted for the Philosophiae Doctor degree to the Rand Afrikaans University, apart from the help recognised, is my own work and has not been formerly submitted to another university for a degree.
Charles R Garner Jr ACKNOWLEDGEMENTS
I wish to express my heartfelt gratitude and appreciation to my supportive family: my wife and my two wonderful sons.
I wish to thank my parents for their support throughout my life in any endeavour I chose to pursue.
I wish to thank Dr. Gayla S. Domke for her amazing support and encouragement as I worked on my master's thesis and this doctoral dissertation. Her support has been instrumental in my pursuit of both my master's and doctoral degrees.
I wish to thank Dr. George J. Davis for his incredible acceptance of me as a research colleague and his willingness to also support my pursuit of a doctoral degree.
I wish to express thanks also to Dr. Elizabeth Jonck for her encouragement, support, and advice throughout this journey towards a doctoral degree. My goal could not have been achieved without her.
Finally, I wish to thank the external examiners. They provided useful references and suggestions that helped to make this work better.
Contents 0.1 Summary 1 0.2 Opsomming 3
1 Introduction 5
2 Preliminary Results 7
2.1 Results from Matrix Theory 7
2.2 Ranks of Regular Graphs 9
3 Strongly Regular Graphs 14 3.1 Rank of Strongly Regular Graphs 14 3.2 Examples of Strongly Regular Graphs 17
4 Ranks of Graph Products of Regular Graphs 23 4.1 Cartesian Products 24 4.2 Complete Products 35
5 Ranks of Line Graphs 39
5.1 Line Graphs of Regular Graphs 40 5.2 Line Graphs of Strongly Regular Graphs 50 5.3 Line Graphs of Regular Graph Products 53
6 Ranks of Complements of Regular Graphs 60 6.1 Complements of Regular Graphs 61 6.2 Complements of Strongly Regular Graphs 63
6.3 Complements of Regular Graph Products 67 7 Ranks of Regular Graphs Under Other Unary Operations 71 7.1 The Subdivision 75 7.2 The Connected Cycle 82 7.3 The Complete Subdivision 86 7.4 The Total Graph 90
8 Ranks of Graphs Involving Paths 98 8.1 Paths and Cartesian Products Involving Paths 98 8.2 The Line Graph, the Complement, and Other Unary Operations on Paths 99
9 Conclusion 106
References 107 1
0.1 Summary
Subjects: Graph Theory, Spectral Theory, Eigenvalues
In this thesis we investigate the ranks of many classes of regular graphs. We also discuss the structure of strongly regular graphs and graphs under certain unary and binary operations.
In Chapter 1 we discuss the rationale behind investigating ranks of graphs.
In Chapter 2 we give relevant definitions and results from matrix theory. We also summarise existing results concerning ranks of regular graphs.
In Chapter 3 we present the structural properties of strongly regular graphs and the relationship these properties have to the spectrum of a strongly regular graph. We then determine the rank of a general strongly regular graph, followed by the ranks of specific types of strongly regular graphs.
In Chapter 4 we investigate the structure of two types of binary graph products: the
Cartesian product and the complete product. We discuss the transformation the spectra of two regular graphs undergo to form the spectra of the product. We then determine the ranks of these products on regular graphs by examining their spectra.
In Chapter 5 we begin the investigation of the ranks of graphs under unary operations.
This chapter examines the transformation of the spectrum of a regular graph under the 2 line graph operation. The ranks of the line graphs of many regular graphs are then determined.
In Chapter 6 we continue with the unary operations by discussing the ranks of the complements of regular graphs. Here, we again examine the transformation of the spectrum to determine the rank.
In Chapter 7 we conclude the exploration of unary operations by investigating the subdivision graph, the connected cycle graph, the connected subdivision graph, and the total graph. Here, the bipartite property plays a major role in the transformation of the spectrum of a regular graph under these operations; thus, we include relevant results concerning the rank of regular bipartite and regular semi-bipartite graphs.
In Chapter 8 we investigate the rank of a special non-regular graph, the path. We determine ranks of graph products involving paths and unary operations on paths by examining both structural properties and spectrum transformations.
Chapter 9 concludes the dissertation by summarising the previous work and pointing the way towards future directions for research. 3
0.2 Opsomming
Onderwerpe: Grafiekteorie, Spektraalteorie, Eiewaardes
In hierdie proefskrif ondersoek ons die rang van baie klasse van reguliere grafieke. Ons bespreek die struktuur van sterk reguliere grafieke en grafieke onder sekere unere en binere bewerkings. In Hoofstuk 1 bespreek ons die sin van die bestudering van die rang van grafieke. In Hoofstuk 2 word relevante definisies en resultate van matriksteorie bespreek. Ons som ook bestaande resultate in verband met die rang van reguliere grafieke op. In Hoofstuk 3 gee ons die strukturele eienskappe van sterk reguliere grafieke weer, asook die verwantskap wat hierdie eienskappe met die spektrum van 'n sterk reguliere grafiek het. Daarna bepaal ons die rang van 'n sterk reguliere grafiek in die algemeen, gevolg deur die rang van spesifieke tipes sterk reguliere grafieke. In Hoofstuk 4 ondersoek ons die stuktuur van twee tipes binere grafiekprodukte: die Cartesiese produk en die volledige produk. Ons bespreek die transformasie wat die spektrums van twee reguliere grafieke ondergaan om die spektrum van die produk te vorm. Daarna bepaal ons die rang van hierdie produkte van reguliere grafieke deur die spektrum daarvan te ondersoek. 4
In Hoofstuk 5 begin ons met die ondersoek van die rang van grafieke onder unere bewerkings. Hierdie hoofstuk ondersoek die transformasie van die spektrum van 'n reguliere grafiek onder die lyngrafiekoperasie. Die rang van die lyngrafiek van baie reguliere grafieke word dan bepaal. In Hoofstuk 6 word unere bewerkings weer gebruik. Die rang van die komplemente van grafieke geniet aandag. Ook hier word die transformasie van die spektrum ondersoek om die rang te bepaal. In Hoofstuk 7 voltooi ons die ondersoek van unere bewerkings deur die subverdelingsgrafiek, die samehangende subverdelingsgrafiek en die totale grafiek te beskou. Hier speel die tweeledigheidseienskap 'n groot rol in die transformasie van die spektrum van 'n reguliere grafiek onder genoemde bewerkings; daarom sluit ons relevante resultate omtrent die rang van reguliere tweeledige en reguliere semi-tweeledige grafieke in. In Hoofstuk 8 ondersoek ons die rang van 'n spesiale nie-reguliere grafiek, naamlik die pad. Ons bepaal die rang van grafiekprodukte wat paaie betrek en unere bewerkings op paaie deur beide strukturele eienskappe en spektrumtransformasies te bestudeer. Hoofstuk 9 sluit die proefskrif of deur 'n opsomming te maak van voorafgaande werk en die pad na toekomstige navorsing aan te dui. 5
1
Introduction
The present work concerns properties of graphs. An undirected graph G = (V, E) with no loops or multiple edges is a finite set V = {1, 2, ..., n} of elements called vertices together with a set E consisting of two-element subsets of V, called edges. Much recent research has focused on relating properties of the adjacency matrix of a graph to the underlying graph. An adjacency matrix of a graph is a matrix describing which vertices are adjacent. Two vertices are adjacent if they are connected by an edge; i. e., if j} E E for vertices i and j. The adjacency matrix, A(G), of a graph G on n vertices is an n x n matrix consisting of entries ay = k, where k is the number of edges connecting vertices i and j. An example of a graph and its adjacency matrix is shown in Figure 1.
Since we only consider graphs with no loops or multiple edges in this thesis, all entries in the adjacency matrix are either 0 or 1.
One such property of interest is the rank of the adjacency matrix, also called the rank of the graph. The rank of a graph G, denoted rank(G), has been found to be the upper bound for numerous graph parameters. This importance of the rank, due to applications in physics, chemistry and combinatorics, has spurred work in the determination of the rank for many types of graphs. Previous work includes ranks of trees, grid graphs
[BDM], and circulants [DD, DDG1]; ranks of graphs after vertex addition [BBD2]; and, ranks of graphs after edge insertion or deletion [DavG]. 6
0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 0 2 0 0 0 2 0_
Figure 1. A graph and its adjacency matrix
In this thesis, the ranks of many types of regular and strongly regular graphs are determined. Also determined are ranks of regular graphs under unary operations: the line graph, the complement, the subdivision graph, the connected cycle, the complete subdivision graph, and the total graph. The binary operations considered are the Cartesian product and the complete product. The ranks of the Cartesian product of regular graphs have been investigated previously in [BBD1]; here, we summarise and extend those results to include more regular graphs. We also examine a special non- regular graph, the path. Ranks of paths and products of graphs involving paths are presented as well. 7
2
Preliminary Results
We begin the treatment of rank with some basic results from matrix theory. In particular, we present those results that are significant for the adjacency matrix of a graph. Then we present well-known results concerning the ranks of simple regular graphs. Under consideration in this section are two familiar graphs: the cycle, Cn, and the complete graph, Kn . Results concerning three- and four-circulants follow, as well as ranks of
Kneser graphs.
2.1
Results from Matrix Theory
Let G be an undirected graph with no loops or multiple edges on n vertices. Then the adjacency matrix A of G is an n x n matrix whose entries ay are 1 if vertex i is adjacent to vertex j, and 0 if they are not adjacent. A is symmetric since au = al, for every i j.
Each row of A is an n-tuple of real numbers and so can be considered a vector in 111 1 ".
The n vectors corresponding to the rows of A are called row vectors of A, and the subspace spanned by the row vectors of A is the row space of A. The nullspace is the set of all solutions to the homogeneous system Ax = 0, i.e., N(A) = e Rn I Ax = 0}. The 8 following result concerns the dimension of the row space, called the rank, and the dimension of the nullspace, called the nullity.
Theorem 1 [LT] If A is an n x n matrix, then the rank of A plus the nullity of A is equal to n.
A scalar A is an eigenvalue of an n x n matrix A if there exists a nonzero vector x e R'7 such that Ax = Ax, or equivalently, (A — A1)x = 0, where I is the identity matrix.
The set of solutions to (A — Al).x = 0 is N(A — )l). Thus, the dimension of the space of solutions is the nullity of A — Al. Now, if an eigenvalue is zero, then the nullspace is the set of solutions to Ax = 0, so that the nullity of A gives the number of eigenvalues equal to zero. As a consequence of the above theorem, the rank of a matrix is equal to the difference of the dimension of the matrix and the number of eigenvalues equal to zero.
Since A is symmetric, we state the following well-known result.
Theorem 2 [LT] Let A = [ay] be an n x n symmetric real-valued matrix. Then the n eigenvalues, A, are real, and Eaii = EA1. i=1 i=1
In our case, not only are the adjacency matrices symmetric and real-valued, each diagonal element is zero. Thus, the sum of the eigenvalues is zero.
If G is a graph on n vertices and if we know the characteristic polynomial PG(A) of the adjacency matrix of G (or, more explicitly, if we know the elements in the set of eigenvalues of the adjacency matrix, called the spectrum, sp[A(G)], of the adjacency matrix), then the rank is easily determined. We use this idea throughout this paper to 9 determine the ranks of graphs. Note that the eigenvalues (spectrum) of the adjacency matrix of G are also referred to simply as the eigenvalues (spectrum) of G.
The graphs we are investigating are all regular; that is, the number of edges incident with each vertex of G (called the degree) is the same for every vertex. The last matrix- theoretic result we state here concerns the adjacency matrix of a regular graph.
Theorem 3 [Cve2] Let Ai > A2 > • > An be the eigenvalues of the graph G. The graph G is regular of degree r if and only if EA = nAi. Moreover, the number of 1=1 connected components of G is equal to the multiplicity of Ai = r.
Thus, in summary, if G is connected and regular, and has no loops or multiple edges, the largest eigenvalue is equal to its degree; the multiplicity of the largest eigenvalue is 1; and the sum of the eigenvalues is zero.
2.2
Ranks of Regular Graphs
The spectra of Cn and Kn are well-known and can be found in [CDS], [Chu], and [GR].
For completeness, we state them here. The superscript in parentheses denotes the algebraic multiplicity of the eigenvalue; if the multiplicity is omitted, it is understood to be 1.
Fact: sp(Cn)= {2 cos(27ri/n): i = 0, 1, n — 1} and sp(Kn) = — n — 11.
The following theorem is a direct consequence of the fact above. 10
Theorem 4 Let n > 2. Then rank(Kn) = n.
As an example of proof techniques that will be employed in later sections, we establish the rank of C,,.
n— 2 n.=-Omod 4 Theorem 5 Let n > 2. Then rank(G) = { n otherwise.
Proof C„ will have a zero eigenvalue only when cos (27ri/n) = 0 for i = 0, 1, n — 1.
Hence, there is a zero eigenvalue only when 2iri/n E {7/2, 37r/2}. Thus, either
271-i/n a- 7r/2 mod n or 27ri/n 37/2 mod n. Simplifying, this implies either i EF, n/4 mod n or i 3n/4 mod n. Thus, n must be a multiple of 4 for there to be zero eigenvalues in the adjacency matrix of C„.
Now assume n 0 mod 4, so n = 4k, for some k E Z+ . Then substituting 4k for n in the above congruencies yields i k mod 4k or i -=- 3k mod 4k. Therefore, there are exactly two zero eigenvalues, and the rank of C„ is n — 2. 0
Next, we turn our attention to circulant graphs; so-called because the adjacency matrices are circulant matrices. We examine two types of circulants: three- and four- circulant graphs. We begin with 3-circulant graphs. A 3-circulant is defined by a jump set S = {a, n — a} with three elements; in other words, vertex i is adjacent to vertex j if and only if (1i — ji mod n) E S. We denote a 3-circulant on n vertices with jump set S = {a, 5, n — a} by 3C„(a). Note that a three-circulant is regular of degree 3 and that n must be even.
Four-circulant graphs are defined similarly. The jump set S of a four-circulant on n vertices is S = {a, b, n — b, n — a} and the graph is denoted by 4C,(a, b). Here, n may be even or odd and the degree is 4. 11
Figure 2. The 4-circulant graph 4C8(1, 3)
By a result of Broere (see [Bro]), we have that the greatest common divisor of n and the elements of the jump set determines the number of connected components of a circulant graph. Thus, 3Cn(a) has d connected components if and only if gcd(a, n) = d; and 4Cn(a, b) has d connected components if and only if gcd(a, b, n) = d. We use this fact, along with the following theorem that gives a formula for the eigenvalues of a circulant matrix, to find the ranks.
Theorem 6 [DavP] If A is an n x n circulant matrix with first row [ci, c2, c n], then the eigenvalues ofA are given by )gyp = Eciu(i-1)P, p = 0, 1, n — 1, where w = e2'iln. i=1
We now quote the theorems regarding the ranks of three- and four-circulants. The proofs may be found in the relevant references. 12
Theorem 7 [DD] Let G be the 3-circulant on n vertices, n even, formed by
S = {a, 2 , n — a} where d = gcd(a, n). Then
n — d n/d odd and n 0 mod 3 n — 2d n/d 0 mod 12 rank(G) = n — 4d n/d a- 6 mod 12 n otherwise.
Theorem 8 [DDG1] Let n > 5 and G be the 4-circulant on n vertices formed by
S = fa, b, n — b, n — al. Then rank(G) = n — d i — d2 d3, where
{ gcd(b — a, n) if gcd(b — a, n) divides n/2 = 0 otherwise
gcd(b + a, n) if gcd(b + a, n) divides n/2 d2 = 0 otherwise and n n n(d2 — gcd(di, d2) if di, d2 are nonzero and gcd(— — ) divides d3 = d2 2did2 0 otherwise
We close this section by presenting the ranks of Kneser graphs, Kw:k. The graph Kw:k has as its vertices the subsets of size k from a set of size w with two subsets (vertices) adjacent if and only if their intersection is empty. The eigenvalues of Kneser graphs have already been determined, as the following theorem attests.
Theorem 9 [GR] The eigenvalues of the Kneser graph Kw:k are given by the numbers (w — k — i) (-1)1 , k — i
Corollary 1 Let w > 2, k > 1. Then rank(Kwk) = 13
Proof Since all the eigenvalues are obviously nonzero, the rank of every Kneser graph is equal to the number of vertices, (I). ❑
Figure 3. The Petersen Graph, Ks:2
It is interesting to note that the Kneser graph Kw..1 is isomorphic to the complete graph on (7) = w vertices, Kw . Also noteworthy is that the Kneser graph K5:2 is isomorphic to the Petersen graph. The results above give the distinct eigenvalues of the Petersen graph as 3, 1, and —2; and the rank of the Petersen graph is ( 52 ) = 10. In a later section, we shall encounter an isomorphism class for Kw:2. 14
3
Strongly Regular Graphs
In this section, we present basic results concerning the rank of strongly regular graphs and look at some specific examples of ranks of strongly regular graphs. A regular graph G of degree r that is not empty and not complete is called strongly regular if every pair of adjacent vertices has exactly u common neighbours and every pair of non-adjacent vertices has exactly v common neighbours. The numbers r, u, and v are the parameters of the graph. (Sometimes included as a parameter is the number of vertices n.) A familiar example of a strongly regular graph is the Petersen graph. Other examples of strongly regular graphs are presented in this section.
3.1
Rank of Strongly Regular Graphs
Strongly regular graphs have a structure that is easily exploited in rank investigations.
Due to this unique structure, there exists a close relationship between the parameters of a strongly regular graph and its eigenvalues. We use this relationship to develop conditions under which a strongly regular graph will have zero eigenvalues, and thus, characterise the rank. 15
Theorem 10 [SB] A regular connected graph G of degree r is strongly regular if and only if it has exactly three distinct eigenvalues Al = r > A2 > A3. If G is strongly regular with parameters u and v, then
u = r + A2A3 + A2 ± A3 and v = r + A2A3 or, in other words,
A2,3 = 1 [u — v + V(u — v) 2 — 4(v — r),.
The parameters n, r, u and v not only determine each eigenvalue, but also the multiplicities of each eigenvalue. Since G is connected, the multiplicity of Al is 1 by
Theorem 3. The multiplicities m2 and m 3 of A2 and A3 must sum to n — 1. By Theorem
2, the sum of the eigenvalues must be zero, and by Theorem 3, the sum of the squares of the eigenvalues must be equal to nA 1 = nr. Thus, m2 and m3 must satisfy the following equations.
1 + m2 + m3 = n r + m2A 2 + m3A3 = 0 r2 + m2 A3 + m3 A3 = nr
Solving these equations results in the following theorem.
Theorem 11 [CC], [Hig] If a strongly regular graph G with parameters r, u, v exists, then r = 2v, u = v — 1, m2 = m3 = r, and G has 4v + 1 vertices, or (u — v) 2 — 4(v — r) = s 2 for some positive integer s, and the expression r m = 2vs[(r— — 1 + v — u)(s + v — u) — 2v1
is a positive integer and is the multiplicity of A2. 16
A fascinating result of the above theorems is that strongly regular graphs are characterised by their spectra. Thus any regular graph with three distinct eigenvalues must be strongly regular and must meet either condition 1 or condition 2 above.
We are interested in determining the conditions that give zero eigenvalues for a strongly regular graph; conditions which will then determine the rank. The following theorem achieves this purpose.
Theorem 12 Let G be a strongly regular graph on n vertices of degree r with parameters (2r — u)(r — u — 1) { if v = r u and v. Then rank(G) = n - u n otherwise.
Proof Let G be a strongly regular graph on n vertices with parameters r, u, and v. Since
= r 0, we set the expression for A2 and A 3 given in Theorem 10 equal to zero and solve. This gives (v — u)2 = (u — v)2 — 4(v — r), which implies v = r. Thus G will have zero eigenvalues only if v = r. Note that v = r does not satisfy condition 1 in Theorem
11; thus, we use the multiplicity expression in condition 2. Since v = r, we get that s = lu — ri; substituting, we have
m = 1 [(2r — 1 — r1 r — u) — 2r]. 21u — r1 I u —r Now, if u — r > 0, then m reduces to , which is negative. Since the multiplicity u — r cannot be negative, it must be that u — r < 0. This implies that u — v < 0, so that, since
Al > A2 > A3, it is A2 that will be the zero eigenvalue. Therefore, m2 reduces to 1 1 (2r u)(r u 1). Hence, the rank of G is n (2r — u)(r — u — 1) only if r — u r — u v = r. ❑ 17
3.2
Examples of Strongly Regular Graphs
As examples of the preceding theorems, we present five classes of graphs that are strongly regular. The first is the complete equimultipartite graph Kn/p. Next is the cocktail-party graph on 2k vertices, C P(k). Finally, we present the Paley graphs, P (q), and particular classes of Johnson graphs, J(w, 2, 1), and Kneser graphs, Kw:2. We begin with Kn/p•
This graph is the complete multipartite graph 1(„1 , n2, ..., np , where ni = n2 = • - • = ni, = n I p and n is the number of vertices. We call this graph the complete equimultipartite graph and denote it by Knip.
From [CDS] we have that sp(Kni) = { 0(n-P), n — 11 , — -r1(P 1) , for n > 3, p > 2. P P Since there are three distinct eigenvalues, by Theorem 10 the graph is strongly regular. n 2n n The parameters of Kq p can then be determined: r = n — — , u = n — — , and v = n — — .
Since there are n vertices and n — p zero eigenvalues, we have the following theorem.
Theorem 13 Let n > p > 2. Then rank(Knip) = p. 18
wWW0Irmr- 1141%■4I4oir 471-1404/04t4if _....-do- .46441100,,
Figure 4. The complete equimultipartite graph K1213
The so-called cocktail party graph, CP(k), is the complete graph with a perfect matching removed; in other words, it is isomorphic to a particular class of complete equimultipartite graphs: K2kik• It is also the only graph on n = 2k vertices that is regular of degree n — 2. The spectra of CP(k) includes 2k — 2; k — 1 eigenvalues equal to —2; and k zero eigenvalues. Again, there are three distinct eigenvalues, and by Theorem 10, we can determine the parameters of CP(k): r = 2k — 2, u = 2k — 4, and v = 2k — 2. The next result easily follows.
Theorem 14 Let k > 2. Then rank[CP(k)] = k 19
Figure 5. The cocktail-party graph CP(3)
Notice that both Kqp and C P(k) satisfy condition 2 in Theorem 11 and that both have zero eigenvalues. Thus, having r = v is expected by Theorem 12.
Next, we discuss Paley graphs. Let q > 1 be a prime power such that q -.. 1 mod 4.
Then the Paley graph P(q) consists of vertices 0, 1, 2, ..., q — 1 such that two vertices i and j are adjacent if and only if i — j is a nonzero quadratic residue modulo q. Paley r = q — 1 q — 5 graphs are known to be strongly regular with parameters u = and 2 ' 4 ' — v = q l (see [GR] and [Chu]). 4
Figure 6. The Paley graph P(9) 20
Theorem 15 Rank[P(q)] = q.
Proof Let q > 1 be a prime power such that q 1 mod 4. By a direct application of
Theorem 12, P(q) will have a rank less than q if and only if r = v. For P(q), this implies 1, q ; 1 = -- which gives the solution q = 1. Since q cannot be 1, rank[P(q)] will 4 never be less than q. ❑
Next we turn our attention to a subset of the class of Johnson graphs: J(w, 2, 1), for w > 4. The Johnson graph J(w, k, i) is defined to have as its vertices the subsets of size k from a set of size w with two subsets (vertices) adjacent if and only if their intersection contains i elements. Thus, J(w, 2, 1) has as its vertices the subsets of size 2 from a set of size w with two subsets (vertices) adjacent if and only if their intersection contains 1 element. Clearly, this graph has striking similarity to the Kneser graphs from Section 2.2.
Although not, in general, strongly regular, the Kneser graphs Kw:k may also be denoted by
J(w, k, 0).
Figure 7. The Johnson graph J(6, 2, 1) 21
Theorem 16 The Johnson graph of the form J(w, 2, 1), for w > 4, is strongly regular with parameters r = 2(w — 2), u = w — 2, and v = 4, and its rank is given by the 3 if w = 4 expression rank[J(w, 2, 1)] = { lw(w — 1) otherwise.
Proof Let set Il = {a 1 , a2, ..., aw l be of size w, and let i, j, k, 1 E 1, 2, ..., w. Then the two vertices axis, and a,ak in J(w, 2, 1) are adjacent. Their common neighbours are the remaining vertices that include the w — 3 possible pairs with a, along with vertex ajak.
Thus adjacent vertices have u = w — 2 common neighbours. The two vertices a,aj and akal are obviously non-adjacent and their only common neighbours are a,ak, ajak, and alai. Thus non-adjacent vertices have v = 4 common neighbours. Hence, J(w, 2, 1) is strongly regular. Since J(w, 2, 1) is strongly regular with parameters u = w — 2 and v = 4, we apply Theorem 12. Note that J(w, 2, 1) has (Z) = lw(w — 1) vertices and is regular of degree r = (2\ (w-2\ kl/k2-1 ) = 2(w — 2). There will be rank de ficiency if and only if r = v; implying deficient rank if and only if w = 4. Thus, if w = 4, there are 6 vertices, r = v = 4 and u = 2, and the expression in Theorem 12 is rank[J(w, 2, 1)] = 3. 0
As stated above, the Kneser graphs Kw:k are not generally strongly regular. However, an important subset of those graphs, Kw:2, are strongly regular. We prove the rank of these Kneser graphs and determine their parameters in the next theorem.
Theorem 17 Let w > 5. The Kneser graph Kw:2 is strongly regular with parameters r = 1(w — 2)(w — 3), u = 1(w — 4)(w — 5), and v = 2(w — 3)(w — 4), and rank(Kw:2) = 1w(w — 1). 22
Proof From Theorem 9 we have that the eigenvalues of Kw:2 are 1 A = r = (-1)°(w 2 ) = — (w — 2)(w — 3), 2 2 - 2 - A2 = (-1)1(w 1 = ( w 7 3 ) = 3 — w, and 2 - 1 A3 = (-1)2 (w 2 2 2 ( Iv —0 4 ) Since Kw :2 is regular and has three distinct eigenvalues, the graph is strongly regular. By Theorem 10, we compute the remaining two parameters u and v: 1 u = r + A2A3 + A2 + A3 = 2(w — 2)(w — 3) + 3 — w + 3 — w + 1 1 1 1 = (w — 2)(w — 3) + 7 — 2w = — (w2 — 9w + 20) = — (w 4)(w — 5) 2 2 1 1 v = r + A2A 3 = 2 (w — 2)(w — 3) + 3 — w = 2— (w2 — 7w + 12) 1 = — (w — 3)(w — 4) 2 Clearly, r v. Then by Theorem 12, the rank of Kw:2 is full for w > 5; that is, the rank is the number of vertices ( 12'') = lw(w — 1). ❑ 23
4
Ranks of Graph Products of Regular Graphs
A graph product is a binary operation on two graphs. The most common graph products are the Cartesian product, denoted ( x ), and the complete product, denoted ( o ). Starting from simple graphs, many other graphs can be represented by using the operations x and o ; conversely, many complicated or unfamiliar graphs may be "broken down" into the product of simpler graphs. For example, from [DD], we have that the circulant graph
3Cn(a), where d = n/gcd(a, n) is odd, is isomorphic to n I d copies of Cnid x P2. The Cartesian product of two graphs G and H is the graph whose vertex set is the Cartesian product of the vertex sets of G and H; i.e., V(G x H) = V(G) x V(H). As for the edge set, consider the set X = { fu, v11, { u, v2} I u E V(G) and {vi, v2} E AM} and the set 3) = { { fui , vl, {u2, v} I {ul , u2} E E(G) and v E V(H)}. Then the edge set of G x H is defined by E(G x H) = X U y. The complete product of two graphs G and H is the graph whose vertex set is defined by V(G o H) = V(G) U V(H). The edge set of G o H is the union of the two edge sets of G and H, along with all edges between every vertex of G and every vertex of H. In other words, £(G o H) = E(G) U £(H) U {{u, v} luEV(G) and v e V(H)}. The complete product is sometimes called the join of two graphs. Note that both the Cartesian product and the complete product are commutative and associative operations. 24
Figure 8. C3 x C4 Figure 9. C3 0 C4
4.1
Ranks of Cartesian Products
There have been many investigations into the ranks of Cartesian products of regular graphs; most notably in [BBD1] and [BDM]. We begin by presenting results concerning the spectra of the Cartesian product of two graphs, then we summarise the existing rank results. We also establish ranks of the Cartesian product of some classes of strongly regular graphs. Finally, we prove the rank of a special type of Cartesian product graph: the hypercube.
The theorem below determines the spectra of the adjacency matrix of a Cartesian product of graphs. 25
Theorem 18 [Cvel] Let G and H be graphs with m and n vertices, respectively, and let the distinct eigenvalues of G be Ari ), for i = 1, 2, ..., a < m, and the distinct eigenvalues of H be p,j(q-'), for j = 1, 2, ..., b < n. Then the eigenvalues of G x H are
(Ai ± pf )(PRJ), for every 1 < i < a and 1 < j < b.
Throughout this section, we refer to "combinations" of eigenvalues. This is not the standard mathematical combination; rather, it is a shorthand description for the way in which the eigenvalues of G x H are formed from the eigenvalues of G and H (as stated in the above theorem).
With this result at hand, Bevis, Domke and Miller and Bevis, Blount, Davis, Domke, Lalani, and Miller have proved many results for the ranks of Cartesian products of cycles and complete graphs, and the ranks of Cartesian product of cycles and complete graphs with paths. A path on two vertices is regular; the results concerning paths on more than two vertices are deferred to Section 8.1. The results are summarised in the following theorem.
Theorem 19 [BBD1], [BDM] The following statements hold. f 2m — 4 m .- 0 mod 6 Let m > 3. Then rank(Cm x P2) = 1 2m otherwise.
Let m > 3. Then rank(Km x P2) = m + 1.
Let m, n > 3 and d = gcd(m, n). Then mn m, n both odd rank(Cm x Cn) = mn{ + 1 — 2d m, n have opposite parity mn + 2 — 2d m, n both even.
Let m, n > 3. Then rank(Km x Kn) = mn.
Let m > 4 and n > 3. Then
26
{mn - 2m + 2 n 0 mod 6 rank(Km x C,,) = mn otherwise.
Part (5) of the above theorem neglects K3 since K3 is isomorphic to C3; thus
rank(K3 x Cn) = rank(C3 x Ca). Next, we examine the rank of the Cartesian product of CP(k) with Kn, Cn , P2, and
CP(j).
Theorem 20 Let k > 2 and n > 4. Then rank[CP(k) x Kn] = 2nk.
Proof The spectrum of CP(k) x Kn consists of combinations of the eigenvalues of CP(k)
and Kn: {2k+ n - 3, 2k - 3 (n-1), n - 3(k-1), n - 1(k) , i(kn—k) , _3([k-1][n-1])) Thus, we
see that there will not be any zero eigenvalues, since k must be an integer greater than 1
and n must be an integer greater than 3. D
Note that CP(k) x Kn is always full rank for k > 2 and n > 4, even though CP(k)
itself is never full rank for k> 2.
Theorem 21 Let k > 2 and n > 3. Then 4n - 2 k = 2 and n 2 mod 4 4n - 4 k = 2 and n 0 mod 4 rank[CP(k) x Cn ] = 2nk - k - 1 k > 2 and n -=- 0 mod 4 2nk - k + 1 otherwise.
Proof As with the above theorem, the spectrum consists of all possible combinations of
eigenvalues of the two graphs. To determine conditions under which zero eigenvalues exist, we set each eigenvalue equal to zero and solve. This results in the following three
equations, where i = 1, 2, ..., n. 27
2k — 2 + 2 cos(27ri/n) = 0
—2 + 2 cos(27ri/n) = 0 2 cos(27ri/n) = 0 Simplifying the first equation, we get cos(27i/n) = 1 — k, which implies there is only a solution if k = 2. When k = 2, the equation becomes cos(27ri/n) = —1. This results in a condition dependent on n: there are solutions only if i n/2 mod n. Thus the first equation results in one zero eigenvalue if n is even and k = 2. The second equation reduces to cos(27ri/n) = 1, which implies i 0 mod n will give a zero eigenvalue. So there will be one zero eigenvalue for every n, independent of k. Thus, since there are k — 1 eigenvalues of 0 from CP(k) that will combine with the one zero eigenvalue, there are a total of k — 1 zero eigenvalues, regardless of n and k. The third equation is exactly the equation describing eigenvalues of C n . From the proof of Theorem 4, we see that there will be two zero eigenvalues if n 0 mod 4, regardless of the value of k. Hence, putting the various cases together and subtracting the proper number of zero eigenvalues from the number of vertices, 2nk, the result is established. ❑
Theorem 22 Let k > 2. Then rank[CP(k) x P2] = 4k
Proof The spectrum of CP(k) x P2 consists of combinations of the eigenvalues of CP(k) and the eigenvalues of P2, which are ±1: 12k — 1, 2k — 3, 1 (k), _ 1 (2k-1) , _ 3(k-1)) Thus, we see that there will not be any zero eigenvalues, since k must be an integer greater than 1. ❑ 28
10 j = k = 2 5j + 1 j > 2 and k = 2 Theorem 23 Let j, k > 2. Then rank[CP(j) x CP(k)] = 5k +1 j = 2 and k > 2 3jk otherwise.
Proof Putting all combinations of eigenvalues together, we obtain for CP(j) x CP(k) the following eigenvalues and their multiplicities: 2j + 2k — 4, 2k — 4U -1) , 2j — 4(" ),
2k — 20, 2j _ 2(k) , 00k) , _2(2ik-r-k) , and _4(1-11[k-1]) It is now easy to see that there will only be additional zero eigenvalues ifj = 2 or k = 2. ❑
The interesting result of the above theorem is that the Cartesian product of a strongly regular graph with another strongly regular graph is not strongly regular. This motivates the following theorem.
Theorem 24 The Cartesian product of two strongly regular graphs is not strongly regular.
Proof Let G and H be strongly regular graphs and let Ai = rc > A2 > A3 be the distinct eigenvalues of G, and to = rH > /12 > /13 be the distinct eigenvalues of H. We only need to show that there cannot be exactly three different eigenvalues in sp(G x
Combining eigenvalues of G and H, we get that sp(G x H) consists of the nine numbers
rG + rH, rG + 112, rc + 113
A2 ± rH, A2 + /12, A2 ± 1-13,
A3 ± rH, A3 ± /12, A3 ±
Clearly, we have that rG + rH > rG + µ2> TG + /13 > A2 + 113 > A3 + P3.
Therefore, G x H has at least five distinct eigenvalues. ❑ 29
We now examine the ranks of the Cartesian product of two complete equimultipartite graphs, of two Paley graphs, of two Johnson graphs, and of two Kneser graphs, Kw:2.
Theorem 25 Let m, n, > 3, p, q ? 2. Then 4m — 6 n = m and p = q = 2 mp + nq — pq — p +1 m(q — 1)/q = n/p rank(K„,k x K q p.) = mp + nq — pq — q +1 n(p — 1)1p = m 1 q mp + nq — pq otherwise.
Proof The spectrum of Icniq x K,h, consists of the sum of all possible distinct
combinations of eigenvalues of Kmiq and K h,: { m — m/q + n — n/p, m — mlq(n-P) , m — m/q — n1 p (P-1) , n — n 1 p (m-q) , 0(["][n-14) —n1 p ([1"-q][P-1]), —m 1 q + n — n1p (q-1) ,
—ml e n-PP-11) , —m 1 q — n1 p ([P-1][q-1]) }.
Clearly, there are always (m — q)(n — p) zero eigenvalues. More zero eigenvalues
could only be given by the two expressions m — m/q — n/p and — m/q + n — n/p.
Setting m — m/q — n/p = 0 implies that m(q — 1)lq = n/p; thus, there are p — 1 zero
eigenvalues if m/q divides n/p, in addition to the (m — q)(n — p) already present.
Similarly, there are q — 1 additional zero eigenvalues if n(p — 1)/p = m/q. Finally, if
m(q — 1)/q = n/p and n(p — 1)/p = m/q, then (p — 1)(q — 1) = 1, whose only integral solution is p = q = 2; but this implies n = m. Since Km/q x Kqp has mn vertices, the results follow. 0
Theorem 26 Let p, q be prime powers greater than 1 such that p, q a 1 mod 4. Then pq — (q — 1)12 p = 2 + .\/4 1)/2 q = 2 + VTi rank[P(p) x P(0] = pq — (p — pq — (p — 1)(q — 1)14 p = (2 + V4) 2 or q = (2 + 05)2 pq otherwise. 30
Proof Once again, the spectrum of P(p) x P(q) consists of the sum of all possible distinct combinations of eigenvalues of P(p) and P(q). First we determine the eigenvalues of the Paley graph P(q). q — 1 5 q — 1 The parameters of P(q) are r = u = q — and v = . Thus, by Theorem 2 4 4
10, the eigenvalues are (q — 1)/2 and (-1 ± \74)/2. From Theorem 11, P(q) satisfies condition 1, which gives r as the multiplicity of both (-1 — /)/2 and (-1 + fq)/2.
Thus, the spectra of P(p) x P(q) is fl(p + q — 2),
— 2 — Vin( ) , \rp + q — 2)(E-21), 1(-2 + 05+ \g)( P-1)'lq-I)), 1 (-2 + .VT3 — \iq)- (P 1)1q I)), + q 2)M, — \rp .1.4)(P 1)4(q I)) ,
" )}. - P- T- 1 (- 2 - foC
It is only possible for four of the eigenvalues to be zero. l(p — 2 — Vq) = 0 if and
only if p = 2 + \Ai; in which case there will be (q — 1)/2 zero eigenvalues. Similarly,
1(-2 + VT, — Vq) = 0 if and only ifp = (2 + \A4)2, resulting in (p — 1)(q — 1)/4 zero eigenvalues.
The eigenvalue 1(- 1/19 + q — 2) = 0 if and only if q = 2 + VT); in which case there
will be (p — 1)/2 zero eigenvalues. Similarly, 1(-2 — + = 0 if and only if
q = (2 + 05)2 , resulting in (p — 1)(q — 1)/4 zero eigenvalues. Finally, is it possible for p = 2 + Ai = 2 + _VP = q? Yes, but only if p = q = 4.
However, this is not viable since both p and q must be congruent to 1 modulo 4. Noting that P(p) x P(q) has pq vertices gives the result. ❑
31
Theorem 27 Let w, x > 4. Then {27 w = x = 4 135 w = x = 6 rank[J(w, 2, 1) x J(x, 2, 1)] = 5x2 w = 6, x 0 6 5w2 x = 6, w 0 6 iwx(w — 1)(x — 1) otherwise.
Proof The spectrum of J(w, 2, 1) x J(x, 2, 1) consists of the sum of all possible distinct combinations of eigenvalues of J(w, 2, 1) and J(x, 2, 1). First, however, we must find the eigenvalues and their multiplicities. It suffices to determine the eigenvalues of J(w, 2, 1). From Theorem 16 we have the parameters of J(w, 2, 1) as r = 2(w — 2), u = w — 2, and v = 4. The multiplicity of A = r = 2(w — 2) is clearly 1. So we use the expression for the other eigenvalues from Theorem 10: A2,3 = 1 {}1, — 2 — 4 ± -V(w — 6) 2 — 4(4 — 2w + 4)1
= 2[w-6+\/w2 -4w+4]
= 1 [w — 6 ± (w — 2)].
Thus, A2 = W — 4 and A3 = —2. Next, since J(w, 2, 1) meets condition 2 in Theorem 11, we use the multiplicity expression given there to determine the multiplicity of each of {2(w — 2), w — 4, —2}. Upon calculating s, we have s = w — 2. Then the multiplicity of A2 = w — 4 is 2w — 4 m = [(2w — 4 — 1 + 4 — w + 2)(w — 2 + 4 — w + 2) — 8] 8w— 16 w — 2 = [(w + 1)(4) — 8] 4(w — 2) 1 —(4w — 4) = w — 1. 4 Then the multiplicity of A3 = —2 is z w(w — 1) — w = i w(w — 3). Now the eigenvalues and multiplicities of J(w, 2, 1) x J(x, 2, 1) are
32
2w + 2x — 8, 2w + x — 2w — 6(x(x-3)/2), w + 2x —
w x — 8((w-1)(x-1)) , w — 6(.*-3)(w-0/2) 2x — 6(w(w-3)/2)
x _ 6(}v(w-3)(x--1)/2), and —4(wx(w -3)(x-3)/4).
Setting each of these expressions equal to zero indicates that only three eigenvalues
could give zeros in the spectrum of the Cartesian product. The eigenvalue w + x — 8
gives 9 zero eigenvalues if w = x = 4; w — 6 gives ix(x — 3) zero eigenvalues if w = 6;
x — 6 gives lw(w — 3) zero eigenvalues if x = 6; and if x = w = 6, there are 90 zero
eigenvalues.
Since the number of vertices in J(w, 2, 1) x J(x, 2, 1) is lwx(w — 1)(x — 1), the result
follows. 0
Theorem 28 Let w, x > 5. Then
4 wx(w — 1)(x — 1) — lx(x — 3) x= 2(w2 — 5w + 12)
rank(Kw,2 x Kx;2) = lwx(w/ — 1)(x — 1) — w(w — 3) w = 1(x2 — 5x + 12) 1 wx(w — 1)(x — 1) otherwise.
Proof The spectrum of Kw:2 X Kx:2 consists of the sum of all possible distinct
combinations of eigenvalues of Kw:2 and Kx:2. First, however, we must find the
multiplicities of each eigenvalue. The eigenvalues were determined in Section 3.2 and it
suffices to find the multiplicities of the eigenvalues of Kw:2.
Kw:2 meets condition 2 in Theorem 11. Thus, upon calculating s, we have s = w — 2.
Then we substitute s along with the parameters r = 1(w — 2)(w — 3),
u = i (w — 4)(w — 5), and v = .1(w — 3)(w — 4) into the multiplicity expression. 33
m [(r — 1 + v — u)(s + v — u) — 2v] 2vs 1 r((w ixw _ 4)) 2(w 3) (w 3)(w 4) ] 2(w - 4) IA 2 ) (w - 3)(w - 4) w(w - 3) = (w + 1 1) = 2(w - 4) 2 Hence, the multiplicity of Al = r is one, the multiplicity of A2 = 1 is w - 1, and the multiplicity of A3 = 3 - w is w(w - 3). Then the eigenvalues and multiplicities of
Kw:2 X Kx:2 are
[(w - 2)(w - 3) + - 2)(x - 3)], 1(w - 2)(w - 3) + 3 - x(x(x -3)/2),
2(w - 2)(w - 3) + 3 - w + 2(x - 2)(x - 3)(w(w -3)/2),
6 - w - X(wx(w-3)(x-3)/4), 4 - Iv(w(w-3)(x-I)/2) 9
1 + 2(x - 2)(x - 3) (w-I), 4 - x(x(x-3)(w-0/2) 9 2“111- 1)(x- 1)). Setting each of these expressions equal to zero indicates that only two eigenvalues could possibly give zeros: 2(w - 2)(w - 3) + 3 - x could give lx(x - 3) zero eigenvalues if x = 1(w2 - 5w + 12), and 3 - w + 1(x - 2)(x - 3) could give lw(w - 3) zero eigenvalues if w = 1(x2 - 5x + 12).
Could a pair of integers w > 5 and x > 5 satisfy both conditions for zero eigenvalues?
The answer is no: substituting one condition in the other and solving gives only w = x = 4, or w = x = 3.
Since the number of vertices of Kw:2 x Kx: 2 is 4wx(w - 1)(x - 1), the result holds. ❑
We may also combine a Kneser graph and a Johnson graph.
Theorem 29 Let w > 5 and x > 4. Then 1.14,2(w2 _ 2w ± 5) x = w + 1 rank[Kw:2 x J(x, 2, 1)] = {-&w(w 3 - w2 - 9w + 25) x = 1(w + 1) lwx(w - 1)(x - 1) otherwise. 34
Proof The spectrum of Kw,2 x J(x, 2, 1) consists of all combinations of eigenvalues from each graph: { 1(w2 — 5w + 4x — 2), 1(w2 — 5w + 2x — 2), 1(w2 — 5w + 2), 2x — w — 1, x — w — 1, 1 — w, 2x — 3, x — 3, —11. Setting each of these equal to zero indicates that it is only possible for two of these to give zero eigenvalues. If x = 1(w + 1), then
2x — w — 1 gives lw(w — 3) zero eigenvalues and if x = w + 1, then x — w — 1 gives
1w2(w 3) zero eigenvalues. Noting that K w:2 x J(x, 2, 1) has wx(w — 1)(x — 1) vertices gives the result. ❑
Some graphs are defined in terms of the Cartesian product. The n - dimensional hypercube is one such graph. Denoted Qn, it is defined recursively by the equations
Qi = K2 and Q,±1 = Qn X K2, for n > 1.
The eigenvalues of this graph are well-known (see [BBD1], [CDS], [Chu], and [GR]) and are stated in the proof of the next theorem.
Figure 10. The 3-dimensional hypercube 35
= 2" — ( nn12 ) n even Theorem 30 [BBD1] Rank(Qn) 2" n odd.
Proof The eigenvalues of Qn are the numbers n — 2k, each with multiplicity ( in, for k = 0, 1, n. Thus, for n — 2k to equal zero, it must be that k = n/2. This will only happen if n is even, and the result follows. ❑
4.2
Ranks of Complete Products
We begin with a result that describes the characteristic polynomial of the complete product of regular graphs.
Theorem 31 [FG] Let G and H be regular graphs of degrees rG and rH on nG and nil vertices, respectively. If PG(A) and P H(A) denote the characteristic polynomials of G and H, then the characteristic polynomial of G o H is
„ P G(A)P H(A)KA — r0)(A — rH) - Pco)mA = nGn Hi (A — r G)(A — r If)
By inspection of the characteristic polynomial, we may determine the spectra, and thus, the rank of a complete product.
Theorem 32 Let G and H be two regular connected graphs. Then rank(G o H) = rank(G) + rank(H). 36
Proof Since both graphs are regular and connected, by Theorem 31, A — rG is a factor of
PG(A), A — rH is a factor of PH(A), and both factors appear exactly once in the respective characteristic polynomials. Thus, by Theorem 31, sp(G o H) includes all numbers from sp(G) and sp(H), with the exception of rG and rH. The spectrum of GoH also includes both solutions to (A — rG)(A — rH) — nGnH = 0; which, by the quadratic formula, are
A — {rG + rH — rH)2 + 4nGnH — 4rGrHi• (4.1) 2
Note that nGnH — rGrH > 0. Thus, the square root is never equal to rG + rH and so
(4.1) never gives a zero eigenvalue. Hence, the zero eigenvalues of G o H include all zero eigenvalues from G and H (with the same multiplicities). Therefore, rank(G o H) = rank(G) rank(H). ❑
Now we use this theorem to specifically determine the ranks of some complete products of regular and strongly regular graphs.
Corollary 2 Let k, m, n > 2. Then the following statements hold.
1m + n — 4 m, n .- 0 mod 4 Rank(Cm o Cn) = m + n — 2 m -- 0 mod 4 or n 0 mod 4, but not both m + n otherwise.
Rank(Km o KO= m + n. {m+n-2 ma- Omod4 Rank(Cm * Kn) = m + n otherwise. 2 n 0 mod 4 Rank[CP(k)o Cn] = r k+ n— l k + n otherwise. Rank[CP(k) o K„] = k + n. 37
Proof By Theorem 5, rank(Cm) = m — 2 if only if m 0 mod 4, and rank(Cm) = m otherwise. Also, the rank a complete graph on n vertices is n, and the rank of a cocktail- party graph CP(k) is k. Hence, the results follow easily from Theorem 32. ❑
Next, we examine the rank of the complete product of some strongly regular graphs.
Corollary 3 The following statements hold.
Let j, k > 2. Then rank[CP(j) o CP(k)] = j + k
Let p, q be prime powers greater than 1 such that p, q a- 1 mod 4. Then rank[P(p) o P(q)} = p + q.
Let m, n > 3, p, q > 2. Then rank(Kmiq * KIM) = P + q.
Let w, x > 5. Then rank(Kw:2 0 Kx:2) = l[w(w — 1) + x(x — 1)].
Let w > 5 and x > 4. Then 3 + w(w — 1) x = 4 rank[Kw:2* J(x, 2, 1 )] = [w(w — 1) + x(x — 1)] otherwise.
Let w, x > 4. Then 6 w , x = 4 3 ± lx(x — 1) w = 4 and x > 4 rank[J(w, 2, 1) c, Ax, 2, 1)] = 3 { 4_ 114)(w — 1) w > 4 and x = 4 1 [w(w — 1) + x(x — 1)] otherwise.
Proof The rank of a cocktail-party graph on 2k vertices is k, the rank of a Paley graph
P(q) is q, the rank of a complete equimultipartite graph Kq p is p, and the rank of the
Kneser graph Kiv:2 is 2 w(w — 1). Thus by Theorem 32, the first four statements follow. By Theorem 16, the rank of a Johnson graph J(w, 2, 1) is 3 if w = 4 and is equal to the number of vertices, z w(w — 1), otherwise. Again, by Theorem 32, the final two results follow. ❑ 38
As with Cartesian products, some graphs are defined by a complete product. The complete bipartite graph Kmn is the complete product of a graph consisting of m isolated vertices, denoted Nm , and another graph of n isolated vertices, N.
Figure 11. The complete bipartite graph K4,5
Theorem 33 Let m, n > 2. Then rank(Km,n) = 2.
Proof The spectrum of Np for any p clearly consists of p zeros. Thus, PNm (A) = A' and
PN„(A) = A". By Theorem 31, the characteristic polynomial of Nm .o. Nn, or Km,", is Am+n-2(A2 _ mn). Thus there are m + n — 2 zero eigenvalues, and the rank is 2. ❑
Since Cartesian products of regular graphs are regular, this naturally leads one to ask if the same can be said of complete products: is the complete product of two regular graphs regular? The answer is a qualified "no." The complete product of regular graphs
G and H is only regular if G and H have the same degree and the same number of vertices. 39
5
Ranks of Line Graphs
In this section, we present results concerning the ranks of line graphs of regular graphs.
The line graph of a graph G = (V, E) is the graph L(G) whose vertex set is E and edge set is V. For a regular graph G on n vertices, degree r, and e = Inr edges, L(G) then has e vertices, degree 2r — 2, and e(r — 1) edges. The following theorem relates the eigenvalues of a regular graph to those of its line graph.
Theorem 34 [Sac2] Let G be a regular graph of degree r on n vertices and e = edges with characteristic polynomial PG(A). Then the characteristic polynomial of L(G) is PL( G)(A) = + 2)e- P G(A — r + 2).
Having the characteristic polynomial of the line graph makes finding the eigenvalues, and thus the rank, a straightforward procedure. The following alternative statement of the above theorem is sometimes more useful in directly computing eigenvalues.
Theorem 35 [DDG2] Let G be a regular graph of degree r on n vertices and e = -}nr edges with eigenvalues AI, A2, ..., An. Then the eigenvalues of L(G) are e — n values of —2 and A, r — 2 for i = 1, 2, ..., n. 40
Figure 12. K5 and its line graph, L(K5)
We begin investigating the ranks of line graphs with regular graphs: cycles, complete graphs, 3- and 4-circulants, Kneser graphs, and complete bipartite graphs. Then we move on to strongly regular graphs, including the cocktail-party graphs, Paley graphs, and complete equimultipartite graphs. We conclude this section with line graphs of some Cartesian and complete products of graphs.
5.1
Line Graphs of Regular Graphs
We begin with a trivial line graph—that of a cycle. Since the line graph of a cycle is a cycle of the same order, we have the following proposition.
Proposition 1 For n > 3, rank[L(G)] = rank(G).
For the rank of the line graph of the complete graph we apply Theorem 34.
41
n = 2 Theorem 36 [DDG2] For n > 2, rank[L(Kn)] = {03 n = 4 in(n — 1) otherwise.
Proof The characteristic polynomial of Kn is PK„(A) = (A — n 1)(A + 1)n± 1 and K,7 is
regular of degree n — 1. Thus
PL(K„)(A) = (A + 2)e-nPK,,(A — [n — + 2) = (A + 2)"(A — 2n + 4)(A — n 4)n-1 implying that there are e — n eigenvalues equal to —2, n — 1 eigenvalues equal to n — 4, ❑ and one eigenvalue equal to 2n — 4. Since Kn has i n(n — 1) edges, the rank follows.
Note that L(Kn) has three distinct eigenvalues and so L(Kn) is strongly regular with
parameters r = 2n — 4, u = n — 2, and v = 4. These parameters are similar to those of
J(n, 2, 1). This motivates the following theorem.
Theorem 37 The Johnson graph J(n, 2, 1) is isomorphic to L(K n).
Proof The vertices of L(Kn) are the edges of Kn, which are 2-subsets of vertices. Two 2- subsets are adjacent if and only if they intersect in 1 element; in other words, if they have
one vertex in common. However, this is the definition of J(n, 2, 1). ❑
Now we determine the rank of the line graph of a 3-circulant graph.
Theorem 38 [DDG2] Let n be even and n .> 4. Then, if3C n(a) is connected, {3n/2 n/2 odd rank[L(3Cn(a))] = 3n/2 1 nI2 a- 0 mod 4 3n/2 3 n/2 2 mod 4. 42
Proof 3Cn(a) has 3n/2 edges and is regular of degree 3. By Theorem 34,
PL(3c„(c0)(A) = (A + 2)ni2P3c„(a)(A — 1). Thus, the line graph of 3Cn(a) will have a zero eigenvalue if and only if 3 Cn(a) has an eigenvalue of —1. From Theorem 6, if Ap is an eigenvalue of 3Cn(a), p = 0, 1, 2, ..., n — 1, then waP UPPI2 (.4)(n-a)p — 1. Since w = cos(2ir/n) + i sin(2ir/n), an eigenvalue of —1 implies that 2 cos(2irap/n) + cos(irp) = —1 and sin(mr) = 0. The second relationship is an identity since n is even, and the first relationship naturally splits into the two cases of p odd and p even. For p odd, 2 cos(2irap/n) = 0 if and only if ap is an odd multiple of n/4. Forp even, 2 cos(2irap/n) = —2 if and only if ap is an odd multiple of n/2. Now we have three cases. Case I. Suppose n/2 is odd. If p is odd, then ap n/4 mod n/2 has no solution since n/4 is not an integer. Similarly, ifp is even, ap -=- n/2 mod n has no solutions since ap is even and n/2 is odd. Thus L(3Cn(a)) has no zero eigenvalues, and has rank e = 3n/2. Case IL Suppose n/2 0 mod 4. If p is odd, ap n/4 mod n/2 has no solution since gcd(a, n/2) = 1 and n/4 is even. If p is even, ap n/2 mod n has exactly one solution, again since gcd(a, n/2) = 1. In fact, this solution is p = n/2. Thus rank[L(3C„(a))] = e — 1 = 3n/2 — 1. Case III. Suppose n/2 2 mod 4, so that n = 8k + 4 for some integer k. If p is odd, ap n/4 mod n/2 implies ap 2k + 1 mod 4k + 2. This equation has exactly two solutions, p = 2k + 1 mod 6k + 3, orp = n/4 and 3n/4. This is because gcd(a, n/2) = gcd(a, 4k + 2) = 1. Similarly, ifp is even, ap n/2 mod n implies ap = 4k + 2 mod 8k + 4. This equation has exactly one solution, and it is p = 4k + 2 = n/2, since a and 4k + 2 have no common factors. Thus, rank[L(3C,i(a))] = e — 3 = 3n/2 — 3. ❑ 43
The above theorem gives the rank of the line graph of any connected 3-circulant. If the circulant is not connected, it splits into isomorphic connected components. The following result describes the rank of a disconnected 3-circulant.
Theorem 39 [DDG2] Let n be even and n > 4. Then, if gcd(a, n/2, n) = d,
{n3n/2 I2d odd
rank[L(3C,(a))] = 3n/2 — d n I2d -=- 0 mod 4 3n/2 — 3d n I2d _= 2 mod 4.
Proof By a result of Broere (see [Bro]), we have that the greatest common divisor of n and the elements of the jump set determines the number of connected components of a circulant graph. Thus, from [DD], 3Cn(a) has d isomorphic connected components if and only if gcd(a, n/2) = d. If d = 1, the results follow from the above theorem. If d > 1, then 3Cn(a) has d isomorphic connected components each isomorphic to 3Cnid(a I d). If n I2d is odd, each component has a line graph of full rank, and rank[L(3C„(a))] = e = 3n/2. If n1261:_=_- 0 mod 4, then each component has a line graph deficient by 1, and
rank[L(3C„(a))] = e — d = 3n/2 — d. If n I2d 2 mod 4, then each component has a
line graph rank deficient by 3, and rank[L(3C„(a))] = e — 3d = 3n/2 — 3d. 0
Now we turn our attention to four-circulant graphs. The procedure is very much the same as that of three-circulants: first we determine the rank of the line graph of connected four-circulants, then proceed to the disconnected case. However, the additional parameter b in 4C „(a, b) creates numerous complications. We begin with determining under what
44
conditions the line graph will have a zero eigenvalue. To that end, we first quote the following theorem of Conway and Jones.
Theorem 40 [CJ] Suppose we have at most four distinct rational multiples of r in the interval (0, 7r/2) for which some rational linear combination of their cosines is rational but no proper subset has this property. Then the appropriate linear combination is proportional to one from the following list. cos(r 13)=1; cos 9 + cos(7r 13 — 9) + cos(n - 13 + 9) = 0, 0 < 0 < 7r/6; cos(ir 15)— cos(27r15) = 1; cos(7r17) — cos(27 17) + cos(37r17) = 1; cos(r 15)— cos(7r115) + cos(47r115) = 1; cos(27r15) + cos(27r115) — cos(77r115) = 11 ;
cos(7r17) + cos(37r17) — coseir 120+ cos(87r/21) = 2 ;
cos(7r17) — cos(27 17) + cos(27r/21) — cos(57/21) = 1; I . —cos(271 - 17) + cos(37 17) + cos(47 121) + cos(107/21) = 2' i cos(ir I15)+ cos(271- I15)+ cos(41- 115)— cos(77 115) = 2•
Theorem 41 [DDG2] An eigenvalue Ap of L(4Cn(a, b)) is zero if and only if one or more of the following holds. n n ap _,- — mod n and by -,4 mod — (5.1) 2 4 2
ap --_- -17 mod -12 and by _- - 11 mod n (5.2) 4 2 2 45
ap = +1113 mod n and by a- ±iIi mod n (5.3)
Proof From Theorem 35, L(4Cn(a, b)) has a zero eigenvalue if and only if )gyp = —2 in b) a)p = - 4C„ (a, b). Thus coaP + cubp + w("- p + w("- 2. This holds if and only if cos(27rap/n) + cos(27rbp/n) + cos[27r(n — b)p/n] + cos[27r(n — a)p I 12] = —2
sin(27rap/n) + sin(27rbp/n) + sin[27r(n — b)p I n] + sin[27r(n — a)p I 77] = 0.
This equation in sines reduces to an identity; the equation in cosines reduces to cos a + cos 0 = —1, where a = 271-ap I n and /3 = 27rbp I n. To solve this equation, we use Theorem 40. Hence, the only rational multiples of 7r with a, 0 E [0, 27r) that solve this equation are given in the following table: the last two columns of solutions are given by applying Theorem 40; the other solutions are found by inspection.
ir 3.7r 2/1- 471- a ir 2' 2 3 3 ir 3ir 27 471- 27r 47 2' 2 ' 3' 3 3' 3
Suppose a = 7r + 2k7r for k E Z. This implies 2ap = n(2k + 1). Hence,
ap =- - n/2 mod n. This also implies 0 = 2(2j + 1) for j E Z. Hence, 2bp = -3- (2j + 1) and
by n/4 mod n/2. Similarly, if 0 = ir + 2k7r, then by =- n/2 mod n and ap =- n/4 mod n/2.
Now suppose a = 3 + 2kir for k E Z. This implies 2ap = 11 (3k + 1). Hence, ap a- n/3 mod n. This also implies 0 = ±r + 2j7r where j E Z. Hence, by = nj ± n/3
and by =- ±n/3 mod n. Similary, if a = V + 2k7r, then ap a- 2n/3 mod n and by E.--_-- ±n/3 mod n. ❑ 46
Theorem 42 [DDG2] For n > 5, 2n — 4d n1 d 0 mod 12 and (3 divides d2 — d 1 or d2 — 2d1) and (4 divides 2d2 — d1 or d2 — 2d1) 2n — 2d (n I d 0 mod 3 and [3 divides d2 — d1 or d2 — 2d 1 ]) or rank[L(4C,(a, b))] = (n I d 0 mod 4 and [4 divides 2d2 —d or d2 — 2d1]) but not both 2n otherwise where d1 = gcd(a, n), d2 = gcd(b, n), and d = gcd(di , d2).
Proof Let d1 = gcd(a, n), d2 = gcd(b, n), and d = gcd(a, b, n) = gcd(di , d2). Assume d = 1. Then the 4-circulant is connected. By the above theorem, at least one of conditions (5.1) through (5.3) must be satisfied for there to be any rank deficiency. However, if n is not divisible by 4, neither (5.1) nor (5.2) are satisfied, and if n is not divisible by 3, (5.3) is not satisfied. Hence, in our search for rank deficiency of the line graph, we examine the following cases: n (1, 2, 5, 7, 10, 11) mod 12, n (4, 8) mod 12, n (3, 6, 9) mod 12, and n 0 mod 12. Case I. Let n (1, 2, 5, 7, 10, 11) mod 12. Then clearly n is not divisible by 3 or 4; so the congruencies (5.1) through (5.3) have no solutions. Hence, the rank is full. Case II. Let n (4, 8) mod 12. Then it is possible for conditions (5.1) and (5.2) to be satisfied. But this alone does not guarantee satisfied conditions: if 4 divides a, then 4 also divides d1, so d1 does not divide n/2, and (5.1) has no solutions; if 4 divides b, then 4 divides d2, so d2 does not divide n/2 and (5.2) has no solutions. Assume 4 does not divide a nor b. Then we determine the number of simultaneous solutions of (5.1) and (5.2). Notice that from (5.1) we have by = n/4 mod n/2, which is actually two congruencies: one for n/4 mod n/2 and another for 3n/4 mod n/2. Therefore, we only need to consider by n/4 mod n, since the number of solutions to 47 this is half the number of solutions to the original congruence. Hence, we simplify our examination of (5.1) and (5.2): ap n/2 mod n and by n/4 mod n (5.4)
ap n/4 mod n and by =- - n/2 mod n (5.5)
Now we determine the number of simultaneous solutions of (5.4) and (5.5). Assuming that each congruence in (5.4) has a solution, we can reduce the congruencies and apply the Generalised Chinese Remainder Theorem (see [Long], p.71). Condition
n(2d2 — (5.4) has gcd(di , d2) = 1 solution if and only if gcd(n/d i , n/d2) divides di) 4d1 d2 • Since d1 and d2 are relatively prime, gcd(n/d i , nId2)= n/lcm(di , d2) = n/di d2. Thus n 2d2 — di l (5.4) has one solution if and only if nIdid2 divides — 4 ; this implies one
dd2 solution if and only if 4 divides 2d2 — d 1 . Similarly with (5.5), we have one solution if and only if 4 divides d2 — 2d1. Since (5.4) and (5.5) could have one solution, the original congruencies (5.1) and (5.2) could have two solutions. Thus, there are two solutions if 4 divides 2d2 — d1 or d2 — 2d1. Hence, the rank is 2n — 2. Case III. Let n (3, 6, 9) mod 12. Then condition (5.3) could be satisfied. But this alone does not guarantee satisfied conditions. If 3 divides a, then 3 also divides d1 , but not d2. Thus 3 does not divide d2 — d1 nor d2 — 2d1 . If 3 divides b, then 3 divides d2, but not d1 . Thus 3 does not divide d2 — d1 nor d2 — 2d1. Hence, condition (5.3) is not satisfied. Assume 3 does not divide a not b. Thus, we seek the number of simultaneous solutions of (5.3). Assume each is solvable. Then we again reduce each congruence and apply the Generalised Chinese Remainder Theorem. Thus, the system has gcd(d i , 48
n(d2 - d2) = 1 solution if and only if gcd(n/d i , n/d2) divides 3diddi) or if gcd(n/c/i, n1 d2) 2 n(d -1 22di ) divides . Since d1 and d2 are relatively prime, gcd(n/di , n1 d2) = n/lcm(di, 3d d d2) = n/dic/2. Thus the positive part of (5.3) has one solution if and only if n I (11612 n d d n d2 2di), . divides — ( -2 - -1 ) or• if n1 did2 divides di d2 3 ththisis implies one solution d2 3 to (5.3) if and only if 3 divides d2 - di or d2 - 2d1. We can proceed similarly with the negative part of (5.3), resulting in one solution if and only if 3 divides 2(d2 - di) or
2d2 - d1 . Next, if (5.3) has a solution and if 3 divides d2 - d1 , then of course 3 divides 2(d2 - d1). Also, if (5.3) has a solution and if 3 divides d2 - 2d1, then 3 divides
2d2 - d1, since d1 and d2 are relatively prime. Thus, there are two solutions, and the rank of the line graph is 2n - 2. Case IV. Let n 0 mod 12. Then any of the conditions in the above theorem could be satisfied, depending on the divisibility criteria established in Cases II and III. If all conditions are satisfied, there are four solutions; if only two sets of conditions are satisfied, there are two solutions; and if none are satisfied, there are no solutions, and the rank is diminished accordingly.
Summarising all the results, we have the following: rank[L(4C,(a, b))] = 2n - 4 if n 0 mod 12 and (3 divides d2 - d1 or d2 - 2di) and (4 divides 2d2 - d1 or d2 - 2d1); rank[L(4G(a, b))] = 2n - 2 if (n 0 mod 3 and [3 divides d2 - d1 or d2 - 2d1]) or (n 0 mod 4 and [4 divides 2d2 - d1 or d2 - 2d1]), but not both; and, rank[L(4C,(a, b))] = 2n otherwise.
In the entire foregoing discussion, we assumed d = gcd(di , d2) = 1. If d > 1, then we have a disconnected 4-circulant with d components. Each of the components is isomorphic to 4Cnid(ald, b/d) (for details, see [DDG1]). Thus the rank of 4C,(a, b), 49 where d > 1, is equal to the product of d and rank[4Cnid(a/d, b 1 d)]. Hence, the results hold. ❑
The following theorem gives the rank of the line graphs of Kneser graphs, L(Kw:k).
Theorem 43 Let w > 2, k > 1 with w > 2k. Then
1w0 = 2k , 3 w = 4, k = 1 rank[L(Kw:01 = I tw\ (w — k otherwise. 1 k ) k )
—k) . Proof The Kneser graph Kw:k has (I) vertices and is regular of degree (w k Thus, there are I2 ( wk ) (wk k edges. The eigenvalues are given by A, = (-1)i (wki-k-i ) for i = 0, 1, k. By Theorem 35, the eigenvalues of the line graph are given by
A, + r — 2 = (-1)rk kTi ) + ( wk k ) — 2. Setting this equal to zero, we have that the only solutions to this equation are (-1)i(wk-k7i) = (w k-k) = 1 and (_ i y(w-k-kTi) = _ 1, (2k)! = 3. The first occurs only if w = 2k, in which case K2k:k is isomorphic to ( wk k) 2k! copies of K2. Since rank[L(K2)] = 0, rank[L(K2kk)] = 0. The second solutions occurs only if w = 4 and k = 1, in which case K4 : 1 is isomorphic to 1C4. Since rank[L(K4)] = 3, rank[L(K4:1 )] = 3. ❑
We close this section by presenting the rank of the line graph of the complete bipartite graph, Km,n . This graph is a bipartite graph with m mutually non-adjacent vertices of degree rm = n and n mutually non-adjacent vertices of degree rn = m. Clearly, this graph is not regular, but is, in fact, semi-regular. 50
Theorem 44 [DDG2] For m n and m, n > 2, m + 1 n = 2 rank[L(K,,,,,)] {= n + 1 m = 2 mn otherwise.
Proof Let the two sets of vertices of Km,n be 1, 2, ..., m and 1, 2, ..., n. Then the edges of Km,n; i.e., the vertices of L(Km,,,), are the mn ordered pairs (i, j) for i = 1, 2, ..., m, and
j = 1, 2, ..., n. Notice that all vertices of L(Km , n ) whose first ordinate is i, where i E {1, 2, ..., m}, are adjacent; creating n copies of K. All vertices of L(Km,n) whose second ordinate is j, where j E {1, 2, ..., are adjacent; creating m copies of Km .
However, this is the same structure as the product Km x K,i; hence L(Km,n) is isomorphic to Km x K. Therefore, rank[L(Km,n)] = rank[Km x Ku], and by Theorem 19, parts 2 and
4, the results hold. ❑
5.2
Line Graphs of Strongly Regular Graphs
Strongly regular graphs have parameters r, u, and v, as defined in Section 3. These parameters determine the eigenvalues of a strongly regular graph. By Theorem 35, the possible zero eigenvalues of the line graph are determined by the relation Ai + r — 2, where Ai is an eigenvalue of the graph for i = 1, 2, ..., n. The eigenvalues of the line graph of a strongly regular graph are then easily determined. We follow this procedure in determining the eigenvalues (and thus the rank) of the line graph of strongly regular graphs, beginning with the complete equimultipartite graph.
51
Theorem 45 Let n > p > 2 and let p divide n. Then 1 0 n = p = 2 2 n = 4, p = 2 3 n = p = 4 rank[L(Knip)] = 10 n = 6 and p = 3 n(n — —n ) otherwise. 2
{n -(1 9 11 - Proof Since sp(Knip) = 0(n-P), n — —n , - , the possible zero eigenvalues of the P P 2n n line graph are given by 2n — — — 2; n — p values of n — — — 2; and p — 1 values of P P 2n n — — — 2. We set each expression equal to zero and solve. We find that the first gives P zeros if and only if n = p = 2, in which case there is one zero eigenvalue. The second
gives zeros if and only if n = 4 and p = 2, in which case there are two zero eigenvalues.
The last expression gives solutions of n = p = 4 and n = 6, p = 3, in which case there
are p — 1 = 3 and p — 1 = 2 zero eigenvalues, respectively. Since L(Knip) has 1 n — — vertices, the rank result follows. 0
Note that if p = 2 and n = 2m for some integer m, Ico, is simply the complete
bipartite graph Km m. Thus, L(Km,m) is strongly regular with parameters r = 2m — 2,
u = m — 2, and v = 2.
Next, we focus upon the cocktail-party graph and the Paley graph.
2 k = 2 Theorem 46 [DDG2] Let k > 2. Then rank[L(CP(k))] = 1 10 k = 3 2k2 — 2k otherwise. 52
Proof The spectra of CP(k) includes 2k — 2; k — 1 numbers equal to —2; and k zero eigenvalues. By Theorem 35, the possible zero eigenvalues of L[CP(k)] includes 4k — 6; k —1 values of 2k — 6; and k values of 2k — 4. Thus, only when k= 2 or k= 3 will there be zero eigenvalues in the line graph, in both instances giving 2 zero eigenvalues. Since
L[CP(k)] has 2k2 — 2k vertices, the result holds. ❑
Theorem 47 Let q be a prime power greater than 1 such that q a- 1 mod 4. Then 14 q = 9 rank[L(P(q))] = 1 2 74- (q2 otherwise.
Proof The number of vertices of L[P(q)] is 1(q2 — q). From the proof of Theorem 26, we have that the spectra of P(q) consists of 1(q — 1) and 1(q — 1) values each of
1(-1 + V4)- and 1(-1 — 10. Thus, possible zero eigenvalues of the line graph are given by q— 3, and 1(q — 1) values each of 1(q — — 6) and 1(q + V4 — 6). Setting these equal to zero gives solutions of q = 3 and q = 9 and q = 4. Since only 9 is congruent to 1 modulo 4, there are no zero eigenvalues in the line graph of P(q) unless q = 9, in which case the rank is 1(q2 — q) — -1(q — 1) = 14. Hence, the result holds. ❑
Because of the unique relationship between the structure and spectra of strongly regular graphs, it is natural to ask if line graphs of strongly regular graphs are strongly regular. After all, by Theorem 35, there would only be four (possibly) distinct eigenvalues in the spectra of the line graph of a strongly regular graph: those three transformed by the relation A, + r — 2 and the new value of —2. The line graph would then be strongly regular if one of those transformed happens to be equal to —2, thus giving three distinct eigenvalues in the line graph. Indeed, under certain conditions this 53 does happen, but the statement and proof of this claim must be delayed until Chapter 7, when we discuss bipartite graphs in detail.
5.3
Line Graphs of Regular Graph Products
In this section, we present ranks of line graphs of Cartesian products and complete products. We begin with Cartesian products; in particular, the line graphs of a cycle, a complete graph and a cocktail party graph with a path on two vertices.
1n3n odd Theorem 48 [DDG2] For n > 3, rank[L(Cn x P2)] = 3n — 1 n - 2 mod 4
3n — 3 n m- 0 mod 4.
Proof The eigenvalues of C'n x P2 are 2 cos(27rj/n) ± 1 for j = 1, 2, ..., n. Since Cn x P2 is 3-regular with 2n vertices and 3n edges, the eigenvalues of L(Cn x P2) are n values of
—2, as well as 2 cos(27rj/n) and 2 cos(27rj/n) + 2 for j = 1, 2, ..., n. These eigenvalues will be zero only when either cos(27rj/n) = 0 or cos(27rj/n) = 1 for j = 1, 2, ..., n.
For cos(27rj/n) = 0, it must be true that 27rj In = (2k + 1)7r/2 for some k E Z. Hence, n must be divisible by 4 and j = n/4 or j = 3n/4. For cos(27rj/n) = —1, it must be true that 27rj/n = (2k + 1)7r for some k E Z. Hence, n must be even and j = n/2.
Now, if n is odd, neither of these equations will hold for any n. Hence, gCn x P2) will be full rank. If n -,7 2 mod 4, n is even but not divisible by 4. So there will only be one eigenvalue equal to zero. If n = 0 mod 4, then n is even and divisible by 4. So there will be three eigenvalues equal to zero. Therefore the result holds. 0 54
Figure 13. C5 x P2 and its line graph L(C5 x P2)
/2 n = 2 Theorem 49 [DDG2] For n > 2, rank[L(Kn x P2)] = 13 n = 4 n2 otherwise.
Proof The eigenvalues of Kr, x P2 are n — 1 values each of —2 and 0 along with n — 2 and n. Since Kr, x P2 is n-regular with 2n vertices and n2 edges, the eigenvalues of
L(Kn x P2) are n2 — 2n values of —2, n — 1 values each of n — 4 and n — 2 along with
2n — 4 and 2n — 2. These eigenvalues are never zero (and L(K, x P2) has full rank) whenever n 1, 2, or 4. When n = 2, the only nonzero eigenvalues are —2 and 2.
Hence the rank of L(K2 x P2) = 2. When n = 4, the only nonzero eigenvalues are n2 — 2n = 8 values of —2, n — 1 = 3 values of n — 2 = 2 plus 2n — 4 = 4 and
2n — 2 = 6. Hence the rank of L(K4 x P2) = 13. Therefore the result holds. ❑
1k9 = 2 Theorem 50 For k > 2, rank[L(CP(k) x P2} = 28 k = 3 4k2 — 2k otherwise. -
55
Proof. The spectrum of CP(k) x P2 is {2k - 1, 2k - 3, l (k) , _1(2k-1) , _3(k-1)) Since
CP(k) x P2 is regular of degree 2k - 1, the eigenvalues of the line graph are given by the transformation A., + 2k - 1 - 2 = Ai + 2k - 3. Thus, since L(CP(k) x P2) has
1(4k)(2k - 1) = 4k 2 - 2k vertices, the spectrum of the line graph is given by {4k - 4, 4k - 6, 2k - 2 (k), 2k - 4(2k-1), 2k - 6(k-1), -2(4k2-6k)}. These eigenvalues will only be zero if k = 2 or k = 3 (since k > 2). If k = 2, then there are three zero eigenvalues; if k = 3, there are two zero eigenvalues; otherwise, there is full rank. ❑
We now move on to the Cartesian product of two cycles, of two complete graphs, of a cycle and a complete graph, and of two cocktail party graphs.
Theorem 51 [DDG2] For m, n > 3, 2mn - 8 (m, n) (0, 0) mod 12 2mn - 6 (m, n) (0, 6) mod 12 or (m, n) =- (6, 0) mod 12 2mn - 4 (m, n) =- (0, 0) mod 3 or (m, n) =- (0, 0) mod 4 rank[L(Cm x C„)] = and (m, n) is not as above 2mn - 2 (m, n) (0, 2) mod 4 or (m, n) =- (2, 0) mod 4 and (m, n) is not as above 2mn otherwise.
Proof Assume both m, n > 3. The spectrum of L(Cm x C„) consists of mn values of -2 and mn values of A + 2, j = 1, 2, ..., mn, where Ai E Sp(Cm x Cn). However,
Aj + 2 = 2 cos(27a/m) + 2 cos(27rb/n) + 2, where a = 1, 2, ..., m and b = 1, 2, ..., n. Setting Ai + 2 = 0 gives cos(27ra/m) + cos(2irb/n) = -1, which has been solved previously in Theorem 41. Reworking that solution in this new context reveals the following conditions. 56
An eigenvalue of L(C „, x Cn) will be zero if and only if one or both of the following holds. a m/2 mod m and b n I 4 mod n (5.6)
a ± m/3 mod m and b ± n/3 mod n (5.7)
For any nonzero eigenvalues, (5.6) implies that both m and n must be even.
Moreover, one of m or n must have a power of 2 of at least 4. If it is the case that, say, m's only power of 2 is 2 (so that m = 2k, k odd) and n's is 4 or larger, then cos(2ira/m) = cos(air/k) will only be equal to —1 once (at a = k), and this value will combine with the two values of zero given by cos(27b/n) (at b = n/4 and b = 3n/4). Thus there will be only two solutions that satisfy (5.6). More precisely, there will be two zero eigenvalues if and only if m 2 mod 4 and n 0 mod 4 (or if and only if m= 0 mod 4 and n 2 mod 4).
However, if both m and n have at least an even factor of 4, then each cosine will give a value of zero and two values of —1, for a total of four solutions. Hence there will be four zero eigenvalues if and only if m 0 mod 4 and n 0 mod 4. Now (5.7) implies that both m and n are divisible by 3, so then each cosine gives two values of —1/2. Hence, there will be four zero eigenvalues if and only if m 0 mod 3 and n - 0 mod 3. Both (5.6) and (5.7) together imply six zero eigenvalues if and only if m 6 mod 12 and n 0 mod 12 (or if and only if m 0 mod 12 and n =- 6 mod 12); there will be eight zero eigenvalues if and only if m -=- 0 mod 12 and n 0 mod 12.
Thus, the result is established. ❑
57
Theorem 52 [DDG2] For m, n > 3, 14 m = n = 3 rank[L(K„, x K„)] = { 1 -imn(m + n — 2) otherwise.
Proof Assume that m, n > 3. The spectrum of Km x Kn consists of one eigenvalue of
m + n — 2, n — 1 eigenvalues of m — 2, m — 1 eigenvalues of n — 2, and (m — 1)(n — 1)
eigenvalues of —2. Since Km x K„ is regular of degree m + n — 2, L(Km x K„) has 1 -2 mn(m + n — 2) vertices. Then the spectrum of L(KL(Km x KO consists of -2 mn(m + n — 4) eigenvalues of —2 and mn values of A, + m + n — 4, where Ai is an eigenvalue of
Km x Kn for j = 1, 2, ..., mn. Substituting the above four possible values for Ai, we see
these mn eigenvalues of the line graph are 2m + 2n — 6, n —1 values of 2m + n — 6,
m — 1 values of m + 2n — 6 and the remaining (m — 1)(n — 1) values of m + n — 6.
Setting all four expression equal to zero gives that only m + n — 6 = 0 will have a
solution since both m and n are at least 3. Hence, there are exactly (m — 1)(n — 1) zero
eigenvalues if and only if m = n = 3, in which case, the rank of L(Km x KO is equal to
2 mn(m + n — 2) — (m — 1)(n — 1) = 14. Otherwise, the rank is full. ❑
Theorem 53 Let m, n > 3. Then {6m — 4 n = 3 and m :_=_- 0 mod 3 rank[L(Cm x K,,)] = 10m — 3 n= 4 and m even Imn(n + 1) otherwise.
Proof The graph Cm x K„ is regular of degree n + 1 and has Imn(n + 1) edges; thus the
line graph has -Imn(n + 1) vertices. The spectrum of Cm x K,, consists of the numbers 2
cos(27j/m) + n — 1 and n — 1 values of 2 cos(27j/m) — 1, for j = 1, 2, ..., m. By
Theorem 35, the spectrum of L(C„, x KO includes 2 cos(27j/m) + 2n — 2, n — 1 values
58
of 2 cos(27rj/m) + n — 2 and Imn(n — 1) values of —2. We set these expressions equal to zero and solve to determine the conditions under which eigenvalues will be zero. Setting 2 cos(27rj/m) + 2n — 2 = 0 gives cos(27rj/m) = —n + 1, which has no
solution for n > 3.
Setting 2 cos(27j/m) + n — 2 = 0 gives 2 cos(27j/m) = —n + 2, which has a solution if n = 3 or n = 4. Assuming n = 3, solving 2 cos(27j/m) = —1 results in j +m/3
mod m. Thus, there are 2(n — 1) = 4 zero eigenvalues. Assuming n = 4, 2 cos(271j/m)- = —2 results in j m/2 mod m. Thus there are n — 1 = 3 zero eigenvalues. Summarising, we have that if n = 3 and m 0 mod 3, we have a total of four zero eigenvalues and if n = 4 and m is even, we have three zero eigenvalues. Subtracting these from the corresponding number of vertices gives the result. ❑
Theorem 54 Let j, k > 2. Then {28 j = k = 2 rank[L(CP(j) x CP(k))] = 70 j = 2, k = 3 or j = 3, k = 2 4jk(j + k — 2) otherwise.
Proof The spectrum of CP(j) x CP(k) contains the following eigenvalues and their multiplicities: 2j + 2k — 4, 2k — 2j — 0-1) , 2k — 2(/), 2j — 2(k), o(k), _2(2jk-j-k) ,
and —4(U-I nk-I D. The Cartesian product of CPU) and CP(k) also is regular of degree 2j + 2k — 4, has 4jk vertices, and has 4jk(j + k — 2) edges; thus the line graph has 4jk(j + k — 2) vertices. By Theorem 35, the spectrum of the line graph consists of 4j + 4k — 10, 2j + 4k — 10-1), 4j + 2k — 10(k-1), 2j + 4k — 80, 4j + 2k — 8(k) , 2j + 2k — 60k), 2j + 2k — 8(2jk -j-k) , 2j + 2k — 10(U-I nk-ID, and _2(4.ficti+k-3]). 59
Since j, k > 2, it is clear that only the eigenvalues 2j + 2k — 8 and 2j + 2k — 10 can be zero. The former is zero when j = k = 2 (in which case there are four zero eigenvalues); the latter is zero when j = 2, k = 3 or j = 3, k = 2 (in which case there are
two zero eigenvalues). The result then follows. ❑
The final Cartesian product whose rank we prove is that of the line graph the n- dimensional hypercube Qn. Recall from Section 4.1 that the hypercube is defined recursively by a Cartesian product.
Theorem 55 [DDG2] Rank[L(Qn)] = n(2n-1 — 1).
Proof Since Q, is regular of degree n and has 2" vertices, L(Qn) has In(2n) = 2"-I n
vertices. The spectrum of Qn consists of the numbers n — 2k, each with multiplicity ( nk ) , for k = 0, 1, ..., n. By Theorem 35, the spectrum of L(Qn) consists of n — 2k + n — 2 = 2(n — k — 1), each with multiplicity (k) , and 2 n—I (n — 2) values of —2.
Setting 2(n — k — 1) = 0 results in one solution: k = n — 1. The multiplicity of this
eigenvalue is ( nn i ) = n. Thus the rank is n(2" -1 — 1). ❑ 60
6
Ranks of Complements of Regular Graphs
In this section we present results involving the rank of the complement of regular graphs. The complement of a simple graph G, denoted G, is the graph with same vertex set as G, but with two vertices adjacent in G if and only if they are non-adjacent in G. The characteristic polynomial of the complement of a regular graph has been determined previously, as the following theorem shows.
Theorem 56 [Sad] Let G be a regular graph of degree r on n vertices with characteristic polynomial PG(A). Then the characteristic polynomial of G is TA — n r 1) pG(_A In other words, if the spectrum of G is PUP') = 1)n A - F r 1
0 1 = r, A2, A3, ..., A n }, then sp(6) = {n — 1 — r, —A2 — 1, —A3 — 1, ..., —A n —
The above theorem indicates that an eigenvalue of G will be zero if an eigenvalue of
G is —1 (or if n = r 1; i.e., if G is a complete graph). This gives a relatively simple method of finding the rank of G by determining how many eigenvalues of G are —1. Determining whether the eigenvalues of a regular graph are —1 plays an important role in the rank of the neighbourhood matrix of a graph G = (V, E). Let N(x) = {y E V I y is adjacent to x} denote the open neighbourhood of y E V. Then Mx] = N(x) U {x} denotes the closed neighbourhood of x E V. The neighbourhood matrix N(G) of a graph 61
G is then the symmetric n x n 0-1 matrix with entry nu equal to 1 if x, E N[xj] and 0 otherwise.
Note that N(G) = A(G) + I, where A(G) is the adjacency matrix of G and I is the identity matrix. From matrix theory (see, for instance, [LT]) we have that if B is a symmetric square matrix, then A E sp(B) if and only if A + 1 E sp(B + I). Thus, sp[N(G)] = {A + 1 I A E sp[A(G)]}. Hence, if k is the number of zero eigenvalues of N(G), then k is also the multiplicity of —1 in A(G). Thus, for the rank of the neighbourhood matrix, we seek the eigenvalues of A(G) that are —1. Many of the following results for the ranks of the complements of regular graphs and of graph products are actually results first determined in the context of the rank of the neighbourhood matrix in the work of Domke and Miller [DM]. Here, we rephrase those results for ranks of graph complements. We also present new results for certain classes of graphs not found in [DM]. We begin with regular graphs, then strongly regular graphs, and finally, graph products.
6.1
Complements of Regular Graphs
We begin with a trivial case: the rank of the complement of a complete graph, rank(K,,). The complement of a complete graph on n vertices is a graph of n isolated vertices. The rank of a graph of n isolated vertices is clearly zero. Thus we have the following proposition.
Proposition 2 For n > 1, rank(1c) = 0. 62
The next result, the rank of the complement of a cycle, is quoted from Domke and
Miller.
n-2 nOmod3=- Theorem 57 [DM] For n > 3, rank(G) = { n otherwise.
Theorem 58 For n > 5, rank[3Cn(a)] = rank[L(3C,(a))] — n/2.
Proof The number of zero eigenvalues of 3C,(a) are given by the number of eigenvalues of —1 in 3C„ (a). But this is precisely the same condition used to determine the number of zero eigenvalues in L(3Cn(a)), as done in Theorem 38. Therefore, the number of zero eigenvalues is the same as that of L(3C,i(a)). D
Figure 14. The complement of the 3-dimensional hypercube, 0
2" — ( 1)12 ) n odd Theorem 59 For n > 2, rank(Qn) = (n4 2n n even. 63
Proof The eigenvalues of Qn are given by n — 2k, each with multiplicity (7), for k = 0, 1, ..., n; setting n — 2k equal to —1, we have that k = 2'1--1- . This implies that n must be odd for any eigenvalues to be —1. The multiplicity of this eigenvalue is ((n+)/2).
Hence, the result holds. ❑
6.2
Complements of Strongly Regular Graphs
Recall that a strongly regular graph has parameters r, u, and v: r is the degree; u is the number of common neighbours for any pair of adjacent vertices; and v is the number of common neighbours for any pair of non-adjacent vertices. The following result is well-known. We establish it here using an argument based on the eigenvalues of a strongly regular graph.
Theorem 60 [GR] Let G be a strongly regular graph on n vertices with parameters r, u, and v. Then G is strongly regular with parameters T = n — r — 1, ü = n — 2 — 2r + v, and V = n — 2r ± u.
Proof Let G be a strongly regular graph on n vertices with parameters r, u, and v, and eigenvalues r > A2 > A3. Then by Theorem 56, the eigenvalues of G are T. = n — r — 1, A2 = —A2 — 1, and A3 = —A3 — 1. Since G is regular and has exactly three eigenvalues, G is strongly regular.
64
Now we use Theorem 10 to find the parameters ii and V, in terms of r, u, and v, based
on the eigenvalues of G. Note that A2A3 + A2 + A3 = u — r and that A2 + A3 = u — v.
We first determine U.
Tj = 71 ± A2A3 ± A2 ± A3 = n — r — 1 ± A2A3 + A2 ± A3 ± 1 — A2 — 1 — A 3 — 1 =n—r-1-1-u—r+1+v—u— 2 = n — 2 — 2r + v. Now we determine V.
1.; = T. + A2A3 = n — r — 1 + A2A3 + A2 + A3 ± 1 =n—r—l+u—r+1 = n — 2r + u. Hence, the results hold. ❑
Theorem 61 Let G be a strongly regular graph with parameters n, r, u, and v. Then (n — v)(r — v) —G) = In r = u + 1 rank( r — v + 1 n otherwise.
Proof Let G be a strongly regular graph on n vertices with parameters r, u, and v.
Substituting the parameters of G given above into Theorem 12, the rank of G easily
follows. ❑
As a result of the above theorem, we have that the ranks of the complements of K„ gyp,
CP(k), and P(q) are all full, since none of these graphs have r = u + 1.
An interesting fact concerning the relation r = u + 1 is found in Godsil and Royle. 65
Theorem 62 [GR] Let G be a strongly regular graph with parameters n, r, u, and v. Then the following are equivalent. G is not connected
v = O.
r = u + 1.
G is isomorphic to r + 1 copies of Kr-Fi
For instance, CP(k) has parameter V = 2k — 2(2k — 2) + 2k — 4 = 0, thus CP(k) is n = 2k isomorphic to = 2k copies of K2. We also provide the + 1 2k — (2k — 2) — 1 following corollary.
Corollary 4 Let G be a disconnected strongly regular graph with parameters n, r, u, and v. Then G satisfies condition 2 from Theorem 11.
Proof Since G is disconnected, v = 0. From condition 1 in Theorem 11, this implies that r = 2.0 = 0 and that u = 0 — 1 = —1. Clearly, this is impossible. Thus, G must meet condition 2 in Theorem 11. ❑
With the aid of Theorem 62, we can more specifically restate Theorem 61 for the case when r = u + 1.
Restatement of Theorem 61 Let G be a strongly regular graph on n vertices. Then n r = u + 1 rank(G) ={ r — 1 n otherwise. 66
Next, to demonstrate the utility of the foregoing theorems, we show that L(Kn) is strongly regular and determine its rank.
Theorem 63 Let n > 4. The graph L(Kn) is strongly regular with parameters r = 1(n — 2)(n — 3), u = -1(n — 4)(n — 5), and v = 2(n — 3)(n — 4). Also, the rank of
L(K„) is full; that is, rank[L(K,)] = 2 n(n — 1).
Proof L(Kn) is regular of degree 2n — 4 and has a spectrum containing e — n eigenvalues equal to —2, n — 1 eigenvalues equal to n — 4, and one eigenvalue equal to 2n — 4, where e = n(n — 1)/2 is the number of edges of Kn. In other words, n(n — 1)/2 is the number of vertices of L(Kn).
By Theorem 56, L(Kn) has n(n — 1)/2 vertices as well, with a spectrum containing one eigenvalue of n(n — 1)12 — 1 — 2n + 4 = (n — 2)(n — 3)/2, e — n eigenvalues of 1, and n — 1 eigenvalues of —(n — 4) — 1 = 3 — n. Since L(Kn) is regular and has exactly three distinct eigenvalues, it is strongly regular with parameters computed from the expressions given in Theorem 10: r = 1(n — 2)(n — 3), u = 1(n — 4)(n — 5), and v = 2 (n — 3)(n — 4).
Clearly, since n > 4, none of the eigenvalues can be zero. Thus the rank is equal to the number of vertices, In(n — 1). ❑ 67 6.3
Complements of Regular Graph Products
First we examine the ranks of the Cartesian product of regular graphs. These results are restated from Domke and Miller.
Theorem 64 [DM] Let m, n > 3, and let p be either m or n and q the other. Also, let k = f 1 if p 0 mod 5 and q 0 mod 5 1 0 otherwise.
Then Imn — 12 — 8k p --_-E- 0 mod 12 and q 7,- 0 mod 12 mn — 8 — 8k p =- 0 mod 12 and q a- 6 mod 12 mn — 6 — 8k p- (4, 8) mod 12 and q a (0, 6) mod 12 rank(C m x CO = mn — 4 — 8k p -.:=_. (3, 9) mod 12 and q a- (0, 4, 8) mod 12 or p =- 6 mod 12 and q a. 6 mod 12 mn — 2 — 8k p --__ (2, 10) mod 12 and q -=- (0, 6) mod 12 mn — 8k otherwise.
Theorem 65 [DM] Let m, n > 2. Then rank(Km x K„) = nm.
Theorem 66 [DM] Let m, n > 3. Then mn — 2n + 2 m 0 mod 4 rank(Cm x = {inn otherwise.
Now we present ranks of complements of Cartesian products involving the cocktail- party graph.
Theorem 67 For j, k > 2, rank[CP(i) x CP(k)i = 4(j — 1)(k — 1). 68
Proof CP(j) x CP(k) has the following eigenvalues and multiplicities: 2j + 2k — 4,
2k — 4Y-1), 2j — 4(k-1) , 2k — 20) , 2j — 2(k), 0(/k), _ 2(2jk-j-k) and _4(v-ii[k-11) Setting each of these equal to —1 and solving, we see that there are no solutions since j, k > 2.
Thus, the rank is full. ❑
12nk — 4k + 2 n a- 0 mod 6 Theorem 68 For k > 2 and n > 3, rank[CP(k) x Cn ] = 2nk — 2k n = 3 mod 6 2nk otherwise.
Proof To determine conditions under which zero eigenvalues exist, we set each eigenvalue of CP(k) x Cn equal to —1 and solve. This results in the following three equations, where i = 1, 2, ..., n.
2k — 2 + 2 cos(27ri/n) = —1
—2 + 2 cos(27ri/n) = —1
2 cos(2iri/n) = —1
Simplifying the first equation, we get cos(2iri/n) = 2 — k, which implies there is no solution since k > 2.
The second equation reduces to cos(2iri/n) = 1, which implies i ± n/6 mod n will give a zero eigenvalue in the line graph. So there will be 2k — 2 zero eigenvalues in the line graph if n is divisible by 6, independent of k.
The third equation reduces to cos(2iri/n) = — 1, which implies i ± n/3 mod n will give a zero eigenvalue in the line graph. So there will be 2k zero eigenvalues in the line graph if n is divisible by 3, regardless of the value of k.
However, notice that if n is divisible by 6, then it is also divisible by 3. Hence, if n is divisible by 6, there will, in fact, be 4k — 2 zero eigenvalues in the line graph. This establishes the result. ❑ 69
Theorem 69 For k > 2 and n > 3, rank[CP(k) x k(n + 1).
Proof The spectrum of CP(k) x IC, consists of combinations of the eigenvalues of CP(k) and Kn : {2k+ n — 3, 2k— 3(n-1), n — 3 ("), n — 1(k), _1(kn-k), —3 ([1c-1][11-1])}. Setting each of these equal to —1 and solving, we find that there are no solutions (since n > 3 and k > 2). But there are clearly kn — k eigenvalues of —1 already present in the spectrum.
Thus the rank is 2nk — kn + k = kn + k. ❑
Next, we turn our focus upon the rank of complements of complete products.
Theorem 70 Let G and H both be regular graphs, each of degree r and each on n vertices. Then 0 if both G and H are complete graphs rank(G o H) = rank(G) + rank(H) otherwise.
Proof By the proof of Theorem 32, the spectrum of G o H consists of the eigenvalues of G, except r, the eigenvalues of H, except r, and the two eigenvalues i [2r + V4n2] = r n. Thus, if there are eigenvalues of —1 with multiplicity mG in the spectrum of G, or of
—1 with multiplicity mH in the spectrum of H, then there are mG + mH eigenvalues of —1 in G o H. If these are the only eigenvalues equal to —1 in Go H, then rank(G o II) = rank(G) rank(R).
However, there could be another eigenvalue of —1 in G o H. The only new eigenvalues of —1 will come from r f n. Setting r — n = —1 clearly gives the only solution: n — 1 = r, but this implies that G and H are complete graphs. Hence, either the only eigenvalues of —1 are those present in spec(G) and spec(), or G o H is a complete graph. Since the rank of the line graph of a complete graph is zero, the result follows. ❑ 70
Clearly, the above theorem implies that the rank of C, o C, is 2n — 4 if n is divisible by 3, and is 2n if not; the rank of CP(k) o CP(k) is 4k; and the rank of K, o K, is 0. 71
7
Ranks of Regular Graphs Under Other Unary Operations
Ranks of regular graphs transformed by unary operations, other than the line graph and the complement, are considered in this chapter. We begin by defining the four different unary operations. The subdivision graph S(G) of a graph G is obtained by subdividing every edge of G; that is, for every edge v,vi E E(G), we add a new vertex vk and replace the edge v,vi by the edges v,vk and vkvj. Note that the subdivision operation transforms any connected simple graph into a bipartite graph. The connected cycle graph R(G) of a graph G is obtained by adding a new vertex corresponding to every edge and adding two new edges from each new vertex to the endpoints of the corresponding edge. Note that this operation creates a triangle on every edge of G. The connected subdivision graph Q(G) of a graph G is obtained by inserting a vertex onto every edge of G and then connecting the pairs of new vertices that lie on adjacent edges of G. This operation creates a complete graph on every vertex of G. Finally, if G = (V, E) then the total graph T(G) has as its set of vertices V U E with vertices connected by an edge if and only if the corresponding elements of G are either adjacent or incident. 72
Figure 15. The unary operations on C4. Top row: C4, S(C4), and R(C4). Bottom row: Q(C4) and 7(C4).
Whether or not a graph is bipartite has great influence on the rank of a graph under the transformations S, R, and Q. Thus, it is now convenient to formally define bipartite. A graph G is bipartite if and only if the vertices of G can be partitioned into two subsets of
V(G), A and B, such that every vertex of A is only adjacent to vertices in B, and every vertex of B is only adjacent to vertices in A.
The following theorem is well-known and concerns bipartite graphs. The theorem is useful for determining the structure of some graphs transformed by the unary operations under consideration.
Theorem 71 [CRS] Let G be a connected graph. Then the following statements are equivalent. G is bipartite. If r is the largest eigenvalue of G, then —r is an eigenvalue of G. The eigenvalues of G are symmetric about zero on the real line. In other words, if
A is an eigenvalue of G, then so is —A with the same multiplicity. 73
As an application of the above theorem, we prove the following two statements.
Theorem 72 Let G K1 and H K1 be two regular connected graphs. Then the complete product of G and H is not bipartite.
Proof Let G and H be two regular connected graphs on nG and nH vertices with eigenvalues = rG > A2 > • • • > AnG and pi = rH > p2 > - - • > pn,„ respectively. It suffices to show that the eigenvalues of G o Hare not symmetric on the real line.
By Theorem 31, the spectrum of the complete product includes A2 > • > A nG and
11,2 > • • • > Any , along with the two numbers 0 = 2[rG + rH + (rG — 1'11)2 + 4ncrild and ¢ = 2 [rG + rH — V(rG — + 4nGnH]. Clearly, 0 is larger than any other eigenvalue in sp(G o H), so —0 (which is not equal to 0) is smaller than both )nG and itnif • Therefore, —0 is not included in sp(G o H), so the eigenvalues of G o H are not symmetric on the real line.
Theorem 73 If G is a strongly regular graph of degree r then the following are equivalent. u =0 and v = r. n= 2r and v = r. G is isomorphic to Kr,r. G is bipartite.
Proof Let graph G be strongly regular with parameters u = 0 and v = r. Then, clearly,
G meets condition 2 in Theorem 11. Since v = r, we have that s = r; evaluating the multiplicity expression, we get that A2 has a multiplicity of 2r — 2. By Theorem 10, A2 and A3 are given by 2[0 — r V(0 — r)2 — 4(r — r)] . This simplifies to the two values 74
A2 = 0 and A3 = —r. Since r is also an eigenvalue, we have that the three distinct eigenvalues are symmetric about zero, and, by Theorem 71, G is bipartite. Moreover, G then has 2r — 2 + 2 = 2r eigenvalues, implying there are n = 2r vertices.
Next, let G be strongly regular with parameters n = 2r and v = r. This implies that if two vertices are non-adjacent, they have r common neighbours. Thus, adjacent vertices cannot have any common neighbours, since each of the r edges emanating from each vertex connect to neighbours of non-adjacent vertices. Hence, not only does u = 0, but this exactly describes the structure of Kr,r. Now let G be isomorphic to Kr,,.. Then there are clearly n = 2r vertices and each pair of non-adjacent vertices has r common neighbours (r = v) and each pair of adjacent vertices has no common neighbours (u = 0).
Finally, let graph G be strongly regular and bipartite. By Theorem 10, a strongly regular graph G has three distinct eigenvalues Al = r > A2 > A3 . By Theorem 71, the fact that G is bipartite implies that A 2 = 0 and A3 = —r, since the eigenvalues of G must be symmetric about zero on the real line; moreover, the multiplicity of A3 is 1. Thus, the parameters of G must be u = 0 and v = r. Note that a strongly regular graph with these parameters meets condition 2 of Theorem 11; we then are able to use the multiplicity expression to determine the multiplicity of A2 = 0. Here, the quantity s = v = r and u = 0. Substituting gives m2 = 2r — 2. However, since Al and A3 each have multiplicity 1, m2 = n — 2. This implies n = 2r. ❑
As was promised in Section 5.2, we are now able to prove the following concerning line graphs of strongly regular graphs. 75
Theorem 74 The line graph of a strongly regular graph G is strongly regular if and only if G is bipartite.
Proof Let G be strongly regular of degree r with n vertices, e = 1-nr edges, and three distinct eigenvalues A = r > A2 > A3 such that L(G) is also strongly regular. Then, by
Theorem 35, L(G) has eigenvalues 2r — 2 > A2 r — 2 > A3 r — 2 along with e — n values of —2. Since we assume L(G) is strongly regular, then one of the three transformed eigenvalues must be equal to —2. To determine which of the three is transformed to —2, we solve A - — 2 = —2. The result is that —r = A„ where i must be 3 since —r would be the smallest eigenvalue. Therefore, by Theorem 71, G must be bipartite.
Now, let G be bipartite and strongly regular. Then by the above theorem, G has eigenvalues r, 0, and —r. Therefore, by Theorem 35, L(G) has eigenvalues 2r — 2, r — 2, and —2, along with e — n additional values of —2. Thus L(G) has three distinct eigenvalues, and L(G) must be strongly regular. ❑
7.1
The Subdivision
The transformation that the characteristic polynomial PG(A) of a graph G undergoes by S(G) has been determined by D. M. Cvetkovic. We quote that result as the next theorem.
Theorem 75 [Cve3] If G is a regular graph of degree r with n vertices, e = Inr edges and characteristic polynomial PG(A), then S(G) has e n vertices, 2e edges and the 76 characteristic polynomial ps(G)(A) = )1e-n pGo 2 r).
Equivalently, the spectrum of S(G) consists of e — n numbers equal to zero and 2n numbers given by
± + r (7.1) where A, is an eigenvalue of G for i = 1, 2, ..., n.
The next result is obvious; we formally prove it to show the utility afforded us by the foregoing theorems.
Proposition 3 S(G) is bipartite for any connected regular graph G.
Proof By Theorem 75, the nonzero eigenvalues of S(G) are given by A = + r, where A, E sp(G) and r is the degree of G. Hence, the eigenvalues of S(G) are symmetric about zero. Therefore, S(G) is bipartite by Theorem 71. ❑
Because we seek the rank, we determine the number of zero eigenvalues of the graph.
This is more easily done by setting the eigenvalue expression in (7.1) equal to zero and solving for A. Solutions to this equation are values of the original graph that will be transformed to zero eigenvalues of the graph under S. Noting that if a graph G is regular of degree r and bipartite then its largest and smallest eigenvalues are r and —r motivates the following theorem. 77
Theorem 76 [GDD] Let G be a connected regular graph of degree r with n vertices. Then G is bipartite if and only if rank[S(G)] = 2n — 2; G is not bipartite if and only if rank[S(G)] = 2n.
Proof Let G be a connected bipartite r-regular graph with n vertices, e edges and eigenvalues A i = r, A2, A3, ..., An . Setting (7.1) equal to zero we obtain the following equation.
(7.2)
This equation has only one solution: A, = —r. Both positive and negative roots satisfy (7.2). Hence, there are two eigenvalues equal to zero in addition to the e — n given by Ps(G) (A). Thus rank[S(G)] = e n — (e — n) — 2 = 2n — 2. Now, let G be a connected non-bipartite r-regular graph with n vertices, m edges and eigenvalues A = r, A2, A3 , ..., An. Then, since G is not bipartite, —r cannot be in the spectrum of G by Theorem 71. Thus there are no solutions to (7.2) which implies that there are only m — n zero eigenvalues in S(G). Hence, rank[S(G)] is not 2n — 2, but is, in fact, 2n. ❑
Now we may utilise the above theorems to determine the rank of the subdivision of many regular graphs.
Theorem 77 [GDD] The following statements hold 22nn — 2 n even For n > 2, rank[S(Cn)] = n odd For n > 3, rank[S(Kn)] = 2n. For n > 2, rank[S(Q,7)] = 2(2" — 1).
78
For k > 3, rank[S(CP(k))] = 4k For q > 1, where q is a prime power congruent to 1 mod4, rank[S(P(q))] = 2q.
For w > 5, rank[S(Kw:2)] = w(4) — 1 ). For n, p > 3, rank[S(Knip)] = 2n. For n > 2, rank[S(Kn,n)] = 4n — 2. 4n — 2 n even For n > 3, rank[S(C„ x P2)] = {4n n odd 6 n = 2 For n > 2, rank[S(Kn x P2)] = 4n otherwise. 8k k = 2 For k > 2, rank[S(CP(k) x P2)] = { 8k otherwise. 2 m, n both even For m, n > 2, rank[S(n, x C„)] = { 2mn — 2mn otherwise. 14 j = k = 2 For n, k > 2, rank[S(CP(i) x CP(k))] = {4jk j, k > 2. For m, n > 3, rank[S(K,n x Kn)] = rank[S(C„, x = 2mn.
Proof By Theorem 76, it suffices to determine under which conditions each of the graphs will be bipartite. A cycle on n vertices is bipartite if and only if n is even. A complete graph on n > 3 vertices is never bipartite. An n-dimensional hypercube is bipartite for all n > 2. Since sp[CP(k)] = {2k — 2, 0(k), —2(k-1 1 is only symmetric about zero on the real line for k = 2, which indicates that it is only bipartite for k = 2 by Theorem 71, we have that CP(k) is not bipartite for k > 3. 5) P(q) is strongly regular with parameters r = q 2 1 u = q and v= q 4 1 4 5 • By Theorem 73, P(q) is bipartite if u = 0 and r = v. Clearly, for P(q), r v. Thus, the Paley graphs are never bipartite. 79
Let w > 5, since K4 :2 is disconnected. Thus, Kw:2 is strongly regular with
r = (w2 — 2)(w — 3) -21 (W — 3)(w — 4) = v, so K,„:2 is never bipartite. Clearly, for n, p > 3, Kqp is not bipartite.
Clearly, Knm is bipartite for all n > 2. C, x P2 is bipartite if and only C, is bipartite. If the eigenvalues of x P2 are symmetric about zero, then Kn x P2 is bipartite.
The spectrum of K, x P2 is { n, n — 2, 00' 1), —20-1) 1. Clearly, only n = 2 will give a symmetric set of eigenvalues. If the eigenvalues of CP(k) x P2 are symmetric about zero, then CP(k) x P2 is
bipartite. Since sp[CP(k) x P2] = {2k — 1, 2k — 3, (k) 1 (2k— 1), _3(k-1) j only k = 2 will give a symmetric set of eigenvalues. Since a graph is bipartite if and only if it contains no odd cycles, C m x Cn is bipartite if and only if both m and n are even and m, n > 2. Since CP(k) is strongly regular with parameters r = v = 2k — 2 and u = 2k — 4, it is clear by Theorem 73 that u = 0 only if k = 2. Thus cp(j) x CP(k) is only bipartite ifj = k = 2. Since Kn is not bipartite for n > 3, the Cartesian product of two complete graphs
or of a complete graph and a cycle is not bipartite. The rank results then follow. ❑ 80
Figure 16. The graph S(C5 x P2).
Theorem 78 [GDD] For even n > 4 and gcd(a, n) = 1, 2n — 2 n/2 is odd rank[S(3Cn(a))] = {2n n/2 is even
Proof In light of Theorem 76, there will be rank deficiency if and only if 3Cn(a) is bipartite, and 3C,(a) is bipartite if and only if —r = —3 E sp[3C,(a)]. Thus, we examine conditions under which an eigenvalue of 3C,(a) is —3.
wap (n12)p + w(n—a)p = —3 By Theorem 6, Ap = + w if and only if 2 cos(27rap/n) + cos(7rp) = —3.
From this, two cases arise: p even and p odd. Ifp is even, the equation above reduces to 2 cos(27rap/n) = —4, which has no solution. Thus, p must be odd; so the equation reduces to 2 cos(27rap/n) = —2, which implies ap -.-. n/2 mod n. 81
Since gcd(a, n) = 1 and n is even, a is odd. Thus both a and p are odd, forcing n/2 to be odd for there to be a solution to the congruence; that solution being p = n/2. Hence,
3Cn(a) is bipartite for n/2 odd, and the result holds. ❑
Theorem 79 [GDD] Let d1 = gcd(a, n) and d2 = gcd(b, n). Then for n > 5 and gcd(a, b, n) = 1,
2 both n and d2 — d 1 are even rank[S(4C,(a, b))] = 2n — 2n otherwise.
Proof Proceeding in a similar fashion to the above theorem, we find conditions under which an eigenvalue of 4Cn(a, b) will be —4. By Theorem 6, we have that Ap w ap wbp w(n—p w(n—a)p = —4 if and only if 2 cos(27rap/n) + 2 cos(2irbp/n) = —4.
Clearly, this can only happen when each cosine term is equal to —1. Hence, it must be that ap n/2 mod n and by n/2 mod n. Note that this implies that n must be even.
Let d1 = gcd(a, n) and d2 = gcd(b, n). By the Generalised Chinese Remainder Theorem (see [Long], p.71), a simultaneous solution for the two congruencies exists if and only if
gcd(n/di, d2) divides n(d2 — di)/2d1d2. But gcd(d i , d2) = 1 since the graph is
connected; and so gcd(n/di, n/d2) = n/lcm(dj, d2) = n/di d2 divides n(d2 — d1)12d1d2. This implies that (d2 — d1)/2 must be an integer. Therefore, 4C,(a, b) is bipartite if and only if n is even and 2 divides d2 — d1. Hence, the result holds. ❑
Finally, we establish the rank of the subdivision of a regular complete product. 82
Theorem 80 Let G and H be regular connected graphs, both on n vertices and both of degree r. Then G o H is regular and rank[S(G o H)] = 4n.
Proof Let G and H be regular connected graphs, both on n vertices and both of degree r.
Then G o H has 2n vertices and is regular of degree r + n since each vertex of degree r is connected to n additional vertices. By Theorem 72, G o H is not bipartite. Therefore, rank[S(G o II)] = 4n. D
7.2
The Connected Cycle
The transformation of the characteristic polynomial of a graph under the operation R has been determined by D. M. Cvetkovic. That result is the following theorem.
Theorem 81 [Cve3] If G is a regular graph of degree r with n vertices, e edges and characteristic polynomial PG(A), then R(G) has e + n vertices, 3e edges and the characteristic polynomial 2 _ r p R(G)(A) Ae-no lrp G ( A ±
Equivalently, the spectrum of R(G) consists of e — n numbers equal to zero and 2n numbers given by the expression 1 1 (A, f -\/A + 4(r + A,)) (7.3) where Ai is an eigenvalue of G for i = 1, 2, ..., n. 83
Because we seek the rank, we determine the number of zero eigenvalues of the graph.
This is more easily done by setting the eigenvalue expression in (7.3) equal to zero and solving for Ai. Solutions to this equation are values of the original graph that will be transformed to zero eigenvalues of the graph under R. Noting that if a graph G is regular of degree r and bipartite then its largest and smallest eigenvalues are r and —r motivates the following theorem.
Theorem 82 [GDD] Let G be a connected regular graph of degree r with n vertices. Then G is bipartite if and only if rank[R(G)] = 2n — 1; G is not bipartite if and only if rank[R(G)] = 2n.
Proof Let G be a connected bipartite r-regular graph with n vertices, e edges and eigenvalues Ai = r, A2, A3, ..., An . Setting the expression (7.3) for the eigenvalues of
R(G) equal to zero we obtain the following equation.
A, ± VA? + 4(r + A,) = 0 (7.4)
This equation has only one solution: A, = —r. So there is only one additional zero eigenvalue of R(G). There are already e — n eigenvalues of zero given by PR(G)(A).
Hence, rank[R(G)] = e n — (e — n) — 1 = 2n — 1.
Now, let G be a connected non-bipartite r-regular graph with n vertices, m edges and eigenvalues Ai = r, A2, A3, ..., An. Then, since G is not bipartite, —r is not in sp(G) by
Theorem 71. Thus, there are no solutions to (7.4), and rank[R(G)] = 2n. ❑ 84
Now we use the above theorems to determine the rank of the connected cycle of many regular graphs.
Theorem 83 [GDD] The following statements hold. For n > 2, rank[R(C,)] = 2n — 1 n even {2n n odd. For n > 3, rank[R(Ku)] = 2n. For n > 2, rank[R(Qn)] = 2n+1 — 1. For k > 3, rank[R(CP(k))] = 4k For q > 1, where q is a prime power congruent to 1 mod 4, rank[R(P(q))] = 2q. For w > 5, rank[R(Kw:2)] = w(w — 1). For n, p > 3, rank[R(Knip)] = 2n. For n > 2, rank[R(Kmn)] = 4n — 1. 4n — 1 n even 3, For n > rank[R(C, x P2)] = {4n n odd. 7 n = 2
For n > 2, rank[R(K, x ] P2) = { 4n otherwise. 2 2, rank[R(CP(k) x P2)] = 15 k = For k > { 8k otherwise. )] = 2mn — 1 m, n both even For m, n > 2, rank[R(Cm x Ca 2mn otherwise. 14 j = k = 2 For n, k > 2, rank[R(CP(j) x CP(k))] = {4jk j, k > 2. For m, n > 3, rank[R(m x Ku)] = rank[R(m x Ku)] = 2mn.
Proof By Theorem 82, we determine under which conditions each of the graphs will be bipartite. These conditions are identical to those given in the proof of Theorem 77; hence, the results hold. 0 85
Figure 17. The graph R(C5 x P2).
To establish the ranks of R(3Cn(a)) and R(4C,,(a, b)), we must determine conditions under which these graphs will be bipartite. As with the above theorem, these conditions have been given in Section 7.1. We therefore state the result without proof.
Theorem 84 [GDD] For even n > 4 and gcd(a, n) = 1,
2n — 1 nI2 is odd rank[R(3Cn(aDi = {2n n/2 is even.
Let di = gcd(a, n) and d2 = gcd(b, n). Then for n > 5 and gcd(a, b, n) = 1,
2n — 1 both n and d2 — di are even rank[R(4C„(a, b))] = 2n otherwise.
Also similar is the proof of the rank of the connected cycle of a complete product; we therefore state the result. 86
Theorem 85 Let G and H be regular connected graphs, both on n vertices and both of degree r. Then G o H is regular and rank[R(G o = 4n.
7.3
The Complete Subdivision
As with operations S and R, the transformation of the characteristic polynomial under Q has been determined.
Theorem 86 [Cve3] If G is a regular graph of degree r with n vertices, e edges and characteristic polynomial PG(A), then Q(G) has e n vertices, e(r + 1) edges and the characteristic polynomial
( A2 — (r — 2)A — r). PQ(G)(A) = (A + 2)e-n(A + 1rPG + 1
Equivalently, the spectrum of Q(G) consists of e — n numbers equal to —1 and 2n numbers given by
1 I (A,+r-24)q+2rA,±r2 +4) (7.5) where A, is an eigenvalue of G for i = 1, 2, ..., n.
Because we seek the rank, we determine the number of zero eigenvalues of the graph.
This is more easily done by setting the eigenvalue expression in (7.5) equal to zero and solving for A. Solutions to this equation are values of the original graph that will be transformed to zero eigenvalues of the graph under Q. As with operations S and R, if a 87 graph G is regular of degree r and bipartite then its largest and smallest eigenvalues are r and —r motivates the following theorem.
Theorem 87 [GDD] Let G be a connected regular graph of degree r with n vertices. Then G is bipartite if and only if rank[Q(G)] = e + n — 1; G is not bipartite if and only if rank[Q(G)] = e + n.
Proof Let G be a connected bipartite r-regular graph with n vertices, e edges and eigenvalues A i = r, A2, A3, ..., An. Setting (7.5) equal to zero we obtain the following equation.
A,+r-2±VA+2rA,-Fr2 +4=0 (7.6)
This equation has only one solution: A, = —r. Only the negative root satisfies (7.6). Thus, there is only one zero eigenvalue and so rank[Q(G)] = e + n — 1. Now, let G be a connected non-bipartite r-regular graph with n vertices, e edges and eigenvalues Al = r, A2, A3, ..., A. Then, since G is not bipartite, —r is not in sp(G) by Theorem 71. Thus there are no solutions to (7.6). Thus, rank[Q(G)] = e + n. ❑
As with the operations S and R, we use the above theorems to determine the rank of the connected cycle of many regular graphs.
Theorem 88 [GDD] The following statements hold. 2n — 1 n even 1) For n > 2, rank[gCn)] = 2n n odd. { For n > 3, rank[Q(Kn)] = ln(n + 1). For n > 2, rank[Q(QA = 2n-1 (n + 2) — 1. 88
For k > 3, rank[Q(CP(k))] = 2k2. q(q + 3) For a prime power q > 1 congruent to 1 mod 4, rank[Q(P(q))] = 4
For w > 5, rank[Q(Kw:2)] = iw(w — 1)(w2 — 5w 10). For n, p > 3, rank[Q(Kh,)] = + 2) —
For n > 2, rank[Q(Kn,n)] = n2 + 2n — 1. 5n — 1 n even For n > 3, rank[Q(Cn x 132)] ={ 5n n odd. 7 n = 2 For n > 2, rank[Q(Kn x P2)] = {n2 + 2n otherwise. 19 k = For k > 2, rank[Q(CP(k) x P2)] = { 2k(2k + 1) otherwise.e 3 mm n — 1 m, n both even For m, n > 2, rank[Q(,„ X Ca)] = { otherwise. 47 j = k = 2 For n, k > 2, rank[Q(CP(j) x CP(k))] = 4ilo +1 1) j, k > 2. For m, n > 3, rank[Q(Km x Kn)] = lmn(m + n). For m, n > 3, rank[Q(,„ x K,,)] = Imn(n + 3).
Proof By Theorem 87, we determine under which conditions each of the graphs will be bipartite. These conditions are identical to those given in the proof of Theorem 77; hence, we need only determine the quantity e + n, then subtract 1 if the graph is bipartite. A cycle has n edges and n vertices; e + n = 2n.
A complete graph on n vertices has 2 n(n — 1) edges, so e + n = in(n + 1). An n-dimensional hypercube has 1(2"n) = 2n-1 n edges, so e + n = (n + 2).
CP(k) has 2k2 — 2k edges and 2k vertices; e + n = 2k2 . P(q) has lq(q — 1)/2 edges and q vertices; e + n = q(q ± 3) .
Kw:2 has g w(w — 1)(w — 2)(w — 3) edges and -1w(w — 1) vertices; e + n = w(w — 1)(w 2 — 5w + 10). 89
n Knh, has .1 n(n — n I p) edges and n vertices. Thus e + n = 2p—[p(n — 2) — n].
K„,„ has n2 edges and 2n vertices, giving e + n = n2 + 2n. Cn x P2 has 3n edges and 2n vertices; e + n = 5n. IC,., x P2 has n2edges and 2n vertices, so e + n = n2 + 2n. CP(k) x P2 has 2k(2k — 1) edges and 4k vertices; hence, e + n = 2k(2k + 1).
Cm x C, has 2mn edges and mn vertices, so that e + n = 3mn. CP(j) x CP(k) has 4jk(k +j — 2) edges and 4jk vertices; e + n = 4jk(k + j — 1). Km x K, 7 has 1 mn(m + n — 2) edges and mn vertices; e + n = -Imn(m + n). Cm x Kn has 1 mn(n + 1) edges and mn vertices. Hence, e + n = 2 mn(n + 3). The rank results then follow. ❑
Figure 18. The graph Q(C5 x P2). 90
To establish the ranks of Q(3C„(a)) and Q(4Cn(a, b)), we must determine conditions under which these graphs will be bipartite. As with the above theorem, these conditions have been given in Section 7.1. Noting that 3C n(a) has 3n/2 edges and n vertices and that 4C,(a, b) has 2n edges and n vertices, we have the following theorem.
Theorem 89 [GDD] For even n > 4 and gcd(a, n) = 1,
5n/2 — 1 nI2 is odd rank[Q(3Cn(aDi = { 5n/2 n/2 is even.
Let d1 = gcd(a, n) and d2 = gcd(b, n). Then for n > 5 and gcd(a, b, n) = 1, 1 both n and d2 — d1 are even rank[Q(4Cn(a, b))] = 3n — 3n otherwise.
The complete product of two graphs, with each graph on n vertices and of degree r, has n(r + n) edges. Hence, we have the following theorem.
Theorem 90 Let G and H be regular connected graphs, both on n vertices and both of degree r. Then G o H is regular and rank[Q(G o II)] = n(r + n +2).
7.4
The Total Graph
The operation T transforms the eigenvalues of a graph. This transformation has been investigated previously; we state the result as the next theorem. 91
Theorem 91 [Cve4] If G is a regular graph of degree r with n vertices and e edges, then T(G) has e + n vertices, r(e + n) edges and e — n eigenvalues equal to —2 and 2n eigenvalues given by the expression
(2A, + r — 2 ± -V4A, + r2 + 4) (7.7) where A is an eigenvalue of G for i = 1, 2, ..., n.
For the total graph, examining whether a regular graph G is bipartite will not aid in ascertaining the rank of T(G). Thus, we use a more useful restatement of the above theorem as the following corollary to aid in our development of ranks of T(G).
Corollary 5 If G is a regular graph of degree r with n vertices and e edges, then T(G) will have zero eigenvalues if and only if the numbers given by
2 (—r + 3 ± r2 — 2r + 9) (7.8) are included in the spectrum of G.
Proof Setting (7.7) equal to zero and solving for ), gives the result. ❑
We now may determine ranks of the total graph of many regular graphs. We begin with cycles, and proceed to complete graphs and hypercubes. Then we present results for the total graph of strongly regular graphs.
2n — 3 n_Omod 3 Theorem 92 [GDD] For n > 3, rank[T(G)] = {2n — 1 otherwise. 92
Proof Since r = 2, (7.8) gives that the numbers 2 and -1 should be included in sp(Cn) for there to be any zero eigenvalues. Since the spectrum of G includes the numbers = 2 cos(27rj/n) for j = 1, 2, ..., n, we set Ai equal to 2 and then to -1 and solve. 2 cos(27rj/n) = 2 27rj/n = 7r(2k), k E Z j = 1 (2k), k E j 0 mod n Thus, 2 is included in sp(G) for every n and will give one zero eigenvalue of T(G). This should be obvious, since 2 = r = A i . 2 cos(27rj/n) = -1 27rj/n = 23 r(3k ± 1), k E Z
j = (3k ± 1), k E Z
j ±-r-1- mod n 3 Hence, -1 (2) is included in the spectrum of C, only for n divisible by 3, and so will give two zero eigenvalues of T(G). The result follows since T(G) has 2n vertices. ❑
3 n = 3 Theorem 93 [GDD] For n> 2, rank[T(Kn)] = {112(n + 1) otherwise.
Proof For the rank of T(K,,) we examine (7.8) upon substitution of r = n - 1: 1 - (-n + 4 ± P12 - 4n + 12) 2 Since sp(Kn) = In - 1, -1 (n-I) }, we set the above expression equal to each of n - 1 and -1 to examine which values of n give eigenvalues that will be transformed to zero in the total graph. Solving the expression equal to n - 1 gives n = 3, and solving equal to
-1 gives n = 3. Hence, n = 3 gives the only instance of rank deficiency, in which case 93 sp(K3) = { 2, -1 (2) } so that all three eigenvalues become zeros of T(K„). Since T(K„) has In(n + 1) vertices, the result follows. ❑
7 n = 2 Theorem 94 [GU)] For n > 2, rank[T(QA = {2n-1(n + 2) otherwise.
Proof The spectrum of Q, consists of the numbers n - 2k, each with multiplicity ( nk ), for k = 0, 1, n. Substituting r = n in (7.8) gives i (-n + 3 ± V n2 - 2n + 9) . We set this expression equal to n - 2k to determine conditions for zero eigenvalues in T(Q„). -n + 3 ± V 112 - 2n + 9 = 2n - 4k ± 112 - 2n + 9 = 3n - 4k - 3 n2 - 2n + 9 = 9n2 + 16k2 - 24nk - 18n + 24k + 9 0 = n2 - n(3k + 2) + 2k2 + 3k 1 n= -2 (3k + 2 ± Vk2 + 4)
The only integral value of k that gives an integral value of n is k = 0, which implies that n = 2. Thus, if n = 2, the eigenvalue corresponding to k = 0 is n, with multiplicity (o) = 1, and is transformed to a zero eigenvalue of T(Q,). Noting that T(Q,) has 2n-1 (n + 2) vertices gives the result. ❑
Now we turn to the total graph of a strongly regular graph. As stated previously, it is irrelevant whether or not the graph is bipartite; however, the structure of strongly regular graphs leads to another condition for rank deficiency in the total graph.
Theorem 95 Let G be a strongly regular graph with parameters n, r, u, and v. Then n u = 1, r = 2 rank[T(G)] = { 2n - 1 u 1, r = 2 1 n(r + 2) otherwise. 94
Proof Let G be a strongly regular graph with parameters n, r, u, and v. Then T(G) has zero eigenvalues if and only if the numbers given by (7.8) are included in the spectrum of
G. However, by Theorem 10, the eigenvalues apart form r are given by the expression [u — v /(u — v) 2 — 4(v — r)]. Thus it must be that, if there are to be any zero eigenvalues in T(G), the values given by (7.8) equal those given by the latter expression. So, by Theorem 11, we have two cases.
Case I: G meets condition 1 in Theorem 11. Then r = 2v and u = v — 1. Substituting these values, we have that the strongly regular eigenvalue expression reduces to z [-1 ± + 4v] and that (7.8) reduces to 2 [-2v + 3 ± V4v2 — 4v + . Thus, —2v + 3 ± /4v2 — 4v + 9 = —1 ± +4v 4v2 — 16v + 16 = 4v2 + 10 ± 2 V16v3 — 12v2 + 32v + 9 —8v + 3 =±\/16v3 — 12v2 + 32v + 9 64v2 — 48v + 9 = 16v3 — 12v2 + 32v + 9 16v3 — 76v2 + 80v = 0, whose only integral solution is v = 0, which is impossible, since this would imply r = 0 and u = —1. Thus G cannot meet condition 1 for T(G) to have zero eigenvalues.
Case II: G meets condition 2 in Theorem 11. Then /(u — v) 2 — 4(v — r) is an integer, which implies )'2 and A3 in sp(G) are of the form i k for some integer k. Hence, the values given by (7.8) must be half-integers if there are any zero eigenvalues in T(G). However, Vr2 — 2r + 9 is an integer only if r = 2; this implies that 1[1 + 3] = 2 and 1- [1 — 3] = —1 must be in sp(G).
Clearly, 2 is included in sp(G) for every strongly regular graph with r = 2, which would give one zero eigenvalue in T(G) for each connected component of G. If the value —1 is in sp(G), then —1 = i [u — v f /(u — v) 2 — 4(v — 0], for some parameters r, u, and v. Solving, we have that this implies r = u + 1, and by Theorem 62, 95
G is isomorphic to copies of Kr± i . Therefore T(G) is isomorphic to copies of r + 1 r + 1 3n T(Kr+i); and, since rank[T(K3)] = 3, rank[T(G)] = But here, r = 2; thus, r + 1 . 3n 3n = — =n. r + 1 3 Since the spectrum of K3 includes —1 as well as 2, there are actually two conditions (for u = 1 and r = 2, and another for u 1 and r = 2), that give the result. 0
We use the foregoing result to greatly simplify our rank investigations for the total graph of strongly regular graphs. There will only be rank deficiency when r = 2, so we simply set any expression for the parameter r equal to 2 and solve. The proof of the next theorem demonstrates this procedure.
Theorem 96 The following statements hold. k = 2 For k > 2, rank[T(CP(k))] = {7 2k2 otherwise. Let q > 1 be a prime power such that q 1 mod 4. Then 9 q = 5 rank[T(P())] = { ,, ig(q + 3) otherwise. Let w > 5. Then rank[T(Kw..2)] = -kw(w - 1)(w2 - 5w + 10).
Let w > 4. Then rank[T(J(w, 2, 1))] = lw(w — 1)2 . 3 n = p = 3 Let n > 3 and p > 3. Then rank[T(Knip )] = 1 r2p i- [p(n + 2) — n] otherwise. u-
Proof We use Theorem 95 to find the rank of the total graphs of these strongly regular graphs. 96
For rank[T(CP(k))], we solve r = 2k — 2 = 2 to get k = 2. Since T[CP(k)] has 2k2
vertices, the result follows. P(q) has r = 1(q — 1) = 2 only if q = 5. The result follows since T[P(q)] has
41 q(q + 3) vertices.
Since r = Ov — 2)(w — 3) = 2 has no integer solution, the rank of T(K w:2) is equal to the number of vertices: i w(w — 1)(w2 — 5w + 10). Here, r = 2(w — 2) = 2 only if w = 4. The number of vertices is lw(w — 1) 2 .
For Kulp, r = n — n/p = 2 only if n = p = 3. But K313 is isomorphic to K3, whose rank of its total graph is 3. Otherwise, the rank is equal to the number of n vertices, — [p(n + 2) — n]. ❑ 2p
Next, we determine the rank of the total graph of some products of regular graphs.
Theorem 97 [GDD] The following statements hold. 7 n = 2 For n > 2, rank[T(Kn x P2)] = {n2 + 2n otherwise. For k > 2, rank[T(CP(k) x P2)] = 8k For m, n > 3, rank[T(Km x Ku)] = Imn(m + n). Let j, k > 2. Then rank[T(CP(j) x CP(k))] = 4jk(j + k — 1).
Proof First, we claim that T(G) will only have zero eigenvalues if and only if G is isomorphic to C4, in which case there is one zero eigenvalue. Let G be a connected regular graph of degree r > 2 and let the eigenvalues of G be integers. If T(G) has zero eigenvalues, then the radical Vr 2 — 2r + 9 from (7.8) must be 97 integral. This only occurs when r= 2; the only graph with r= 2 and all integral eigenvalues is C4. Conversely, if G is isomorphic to C4, T(C4) has one zero eigenvalue by the proof of Theorem 92. This completes the proof of the claim.
We use this result to easily determine ranks: Kn x P2 is isomorphic to C4 only for n= 2; however, CP(k) x P2, for k > 2, Km x Kn, for m, n > 3, and CPU) x CP(k), for j, k > 2 are never isomorphic to C4. Noting the number of vertices of the total graphs completes the proof ❑
Figure 19. The graph T(C5 x P2). 98
8
Ranks of Graphs Involving Paths
The path on n vertices, Pn, is a cycle on n vertices with an edge removed. Clearly, paths are not regular, but their structure is so unique that they lend themselves (somewhat) easily to rank investigations. We include, in this chapter, results concerning the ranks of paths, graph products involving paths, line graphs of paths, complements of paths, and the unary operations on paths.
8.1
Paths and Cartesian Products Involving Paths
The eigenvalues of a path on n vertices are well-known. From [CDS] we have that the eigenvalues are given by the numbers 2 cos(7ri/[n + 1]) for i = 1, 2, ..., n. The rank of Pn is also a well-known result and can be found in various sources.
n even Proposition 4 [BBD1], [Chu], [GR] Rank(Pn) = n {n-1 n odd.
Now we quote results concerning the rank of the following Cartesian products: a path with a path, a path with a cycle, and a path with a complete graph.
Theorem 98 [BDM] For n, m > 2, rank(P,, x PO = mn + 1 — gcd(m + 1, n + 1). 99
Theorem 99 [BDM] Let n > 3 and g = gcd(m + 1, n). Then mn+l—g nodd rank(P„, x Cn) = mn + 2 — g g even, n/g odd mn + 2 — 2g n/g even.
Theorem 100 [BBD1] Let m > 3 and n > 2. Then ) { mn — m + 1 n =- 2 mod 3 rank(K„, x Pn mn otherwise.
8.2
The Line Graph, the Complement, and Other Unary Operations on Paths
The path Pn is a well-known non-regular graph. Due to the unique structure of this graph, it is a straightforward procedure to establish basic results for Pn under all basic unary operations.
Theorem 101 For n > 2, rank[L(Pn)] = rank(Pn-1)-
Proof There are n — 1 edges in Pn which become the n — 1 vertices of L(Pn). Thus the ranks of L(Pn) and Pn_ i are the same since the graphs are isomorphic. ❑
In order to prove the following result, we use neighbourhood weights. If each vertex of a graph G is assigned some number ai, then consider the vector w = (a 1 , a2, an)T - Now, if A is the adjacency matrix of G, then Aw is a vector containing the sum of the
100
weights of the neighbours of each vertex in the graph. In particular, Aw = 0 if and only if
w is a weighting of G such that all the neighbourhood weights are zero.
in-1 n 2 mod 3 Theorem 102 For n > 4, rank(Pn) = 1 n otherwise.
Proof Suppose Pn has n vertices v1 , v2 , ..., vn, weighted al , a2, ..., an, respectively. Then
the sum of the neighbourhood weights of each vertex must be zero. So we investigate
some of the first few neighbourhood weights. n- For vi , + • • + an = 0, implying an = -Ea,. r=3 n-i For V2, a4 + a5 + + an = 0, implying a3 = 0 and an = -Eai. i=4 n-i n-i For v3, 0 = a5 + a6 + - - • + an = ai + Eai - Eai = ai - a4; thus a4 = ai i=5 i=4 n-i n-i For v4, 0 = al + a2 + a6 + a7 + • • + an = a 1 + a2 + > ai - Eai = i=6 i=4 al + a2 - a4 - a5 = a2 - a5 ; thus a2 = a5 . n-i n-i
For v5, 0 = ai + a2 + a3 + a7 + as + • • + an = ai + a2 + a3 +Eai = i=7 i=4 a1 + a2 - a4 - a5 — a6 = -a6; therefore a6 = 0.
From this, an obvious pattern emerges: for k > 2 and j < 3k, we have that l j -- 1 mod 3 aj =/a a2 j 2 mod 3 0 j _- 0 mod 3. 3k-2 n-1 n-1 Now, for V3k, 0 = Eai + E a; — Eai = al + a2 + a3 - a3k-1 - a3k - a3k+1 = i=1 i=3k+2 1=4
a1 - a3k+1 . Hence, a3k+1 =
101
3k-1 n-1 n- I For V3k+1, 0 = Eai E ai — Ea, = a1 + a2 + a3 — a3k a3k+1 a3k+2 = i=1 i=3k+3 i=4
a2 — a3k+2. Hence, a3k+2 = a2. 3k n-1 n-1 For V3k+2, 0 = Ea; + E ai —Eai = a1 + a2 + a3 — a3k4.1 — a3k+2 — a3k+3 = i=1 i=3k+4 i=4
a3k.o• Hence, a3k+3 = 0. n-3 n-2 al n 0 mod 3 For vn_i, 0 = Eai and for vn, 0 = Ea1. Hence, an-2 = a2 n 1 mod 3 Also, i=1 i=1 0 n 2 mod 3. n-3 since every ai = al or a2, and 0 = Eai, then a2 = cal , for some c E R. Therefore, for 1=1 n 2 mod 3, we have one degree of freedom and rank(P n) = n — 1; otherwise, the rank is full. ❑
We use the following theorem to determine the rank of Q(Pn).
Theorem 103 [Cve3] If G is a graph with n vertices, e edges and whose line graph has characteristic polynomial PL(G)(A), then Q(G) has characteristic polynomial — 2 \ PQ(G)(A) = A'(A + l)QPL(G) + 1 j .
Theorem 104 [GDD] For n > 2, rank(S(Pn)) = rank(Q(Pn)) = 2n — 2.
Proof Since S(Pn) is another path on m + n = (n — 1) + n = 2n — 1 vertices, the rank(S(Pn)) = rank(P2n_1). It is well-known that rank(Pn) = n — 1 if n is odd. Hence, because 2n — 1 is odd for all n, rank(P2n_i) = rank(S(Pn)) = (2n — 1) — 1 = 2n — 2. For Q(Pn), note that L(Pn) is isomorphic to Pn_ i . Now, from the above theorem, simple algebra reveals that the eigenvalues of Pn_ i are transformed by the operation of 102 connected subdivision into numbers given by the expression 2 (A, + 4A1 + 8), where A, E spec(P„_i) for i = 1, 2, ..., n — 1. Setting this expression equal to zero determines that only an eigenvalue of —2 will be transformed into an eigenvalue of zero in Q(P,,_1). Now, eigenvalues of P„_1 are given by 2 cos(7riAn — 1 + 1]) = 2 cos(7ri/n) for i = 1,2, ..., n — 1. Thus, we determine which eigenvalues are equal to —2. But cos(iri/n) = —1 has no solution since i n. So —2 0 spec(Pn_ i ) which implies additional zero eigenvalues, other than the n — m = n — (n — 1) = 1 given by the characteristic polynomial, do not exist. The result follows. ❑
To treat the rank of R(P n), we require some results from matrix theory and the block matrix representation of the adjacency matrix of R(G) for some graph G. Recall that for every graph G on n vertices, where each vertex has degree d1 for i = 1, 2, ..., n, we have the degree matrix D = diag[di , d2, ..., do]. The following theorem relates the adjacency matrix and the degree matrix.
Theorem 105 [Cve3] If A is the n x n adjacency matrix of graph G and D is the degree matrix of G, then the adjacency matrix of R(G) is of the block form
OmR A R1 where RR T = A + D and Om denotes the m x m zero matrix. 103
Next, the following two lemmas are required.
Lemma 1 [LT] If A is a nonsingular matrix and D is square, then
A B DI= IAIID - CA -1 BI. C
Lemma 2 [LT] If a square matrix A of order n is tridiagonal—that is, of the form
al b1 0 0 c1 a2 b2 0 C2 a3 A n - 0 an-1 bn-1
0 . - 0 Cn-1 an
—then IA n I = an lAn- 1 I - bn-1Cn-1IAn- , where A k_ i is Ak with the kth row and kth column deleted.
Theorem 106 [GDD] For n > 2, rank(R(P n)) = 2n - 1.
Proof By Theorem 105, the adjacency matrix of R(Pn) is [ 0n-i . Since this R A (Pn ) operation adds n - 1 vertices to the path on n vertices, the zero matrix is of dimension n - 1. Hence, the characteristic polynomial of the adjacency matrix of R(P n), denoted
A[R(P n)], is
104
I AI2,1 — A[R(Pn)]i Aln_ i — RT = —R n n) Ain-11 A — A(P n) — )R T A
= An-1 1 Ain — A(Pn ) — —RRT Awn) — RRT
= A -1 IA2 — AAP n) — A(P n) — D(P ri)1 = A-1 1 A2in — (A + 1)A(Pn) — D(13 01 A2 — 1 —A — 1 0 0 —A — 1 A2 — 2 —A — 1 1 0 —A — 1 A2 — 2 A
A2 - 2 —A — 1 0 0 —A — 1 A2 — 1 where all entries are zero except on the sub- and super-diagonal (where entries are all
—A — 1) and the main diagonal. We denote this tridiagonal matrix Bn . — (A + 1)2 By Lemma 1, IBnI = (A2 — 1Bn_21. Note that Bn_i and Bn_2 are tridiagonal; moreover, all entries are identical to Br, except for the last diagonal entry
which is A2 — 2. To simplify the following argument, we denote Bn_1 by AN. Thus we
have IBn I = (A2 — 1)IANI — (A + 1 )2 1AN-1 The determinant of Bn results in a characteristic polynomial of degree 2n; denote it by p(A). Upon multiplication of p(A) by 1/A, the result must be a polynomial of degree 2n — 1. Therefore, there will be a constant term in the polynomial p(A)/A if and only if
the A term ofp(A) is nonzero. Thus we examine the A term of IBnI. This term is obtained by the multiplication of the A terms of both IAN I and IAN_II by —1 and the constant term of IAN_I I by —2A. We show that none of these terms is nonzero; in fact, we show by 105 induction that the constant term of IAN I is (-1)N, the A term of IANI is (-1)N-I N(N— 1),
and that the A term of I/3,1 is given by 2(-1)n -I (n — 1). We begin with IAN'. Note that IA3I = (A2 — 2)IA2I — (A + 1)2 11111 = (A2 — 2)[(A2 — 2)(A2 — 1) — (A + 1)2] — (A + 1)2(A2 — 1) = A6 — 7A4 — 4A3 + 9A2 + 6A — 1. Assume that the constant term of Oki is (-1) k and the A term of lAk I is (— 1)k-1 k(k _ 1) for all k < N. Let N = k + 1. Then the constant term of lAk +11 is given by —2(-1)k — (-1)k-I = (-1)k+I and the A term is given by
—2(-1)k-I k(k — 1) — [(-1) k-2 (k — 1)(k — 2) + 2(-1)k-I ] = (-1)kk(k + 1).
Now we examine 114 Note that IB31 = (A2 — 1)IA2I — (A + 1 )2 IA1l = (A2 — 1)[(A2 — 2)(A2 — 1) — (A + 1)2] — (A + 1)2(A2 — 1) = A6 — 6A4 — 4A3 + 5A2 + 4A. Assume the A term of 'Bic ' is 2(-1)k-I (k — 1) for all k < n. Let n = k + 1. Then the A term of IBk±i I is given by —(-1)k-l k(k — 1) — [(-1)k-2(k — 1)(k — 2) + 2(-1)k- I ] = 2(-1)kk.
Hence, p(A) has a nonzero A term, which implies p(A)/A has a nonzero constant term. Since the constant term is the product of the roots (the eigenvalues) of the characteristic polynomial of R(Pn), and the constant term is nonzero, there cannot be any zero eigenvalues of R(P 0. Thus, its rank is 2n — 1. ❑ 106
9
Conclusion
The goal of this research is to determine which structural properties of a graph are related to the rank of its adjacency matrix. This investigation into regular graphs has proved to be an interesting and fruitful exercise. Many different branches of mathematics were used to determine ranks of regular graphs, including basic trigonometric identities to more advanced number theory. We have exploited some structural properties of graphs to produce rank results: the parameters of strongly regular graphs are determined by their spectra and the bipartite property has an important role in the rank of regular graph under three of the four unary operations specified. The next possible step includes determining ranks of binary operations on two graphs, such as the complete product, the strong product, more Cartesian products, the bipartite product of a graph with itself, and the composition of two graphs. More results involving specific graphs, such as the generalised Petersen graphs, cages and Cayley graphs, are also possible studies. 107
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