The Minimal Rank Problem for Finite Fields. Wayne Barrett (BYU) Hein van der Holst (Eindhoven) Raphael Loewy (Technion) Don March (BYU)

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1 The Minimal Rank Problem for Finite Fields Wayne Barrett (BYU) Hein van der Holst (Eindhoven) Raphael Loewy (Technion) Don March (BYU) 12th ILAS Meeting - Regina June 26-29, 2005

2 References [BvdHL1] Barrett, van der Holst, Loewy, Graphs whose Minimal Rank is Two, Electronic Journal of Linear Algebra, volume 11 (2004), pp [BvdHL2] Barrett, van der Holst, Loewy, Graphs whose Minimal Rank is Two: The Finite Fields Case, Electronic Journal of Linear Algebra, volume 14 (2005), pp [BM] Barrett, March, The Minimal Rank Problem over a Finite Field with a Prime Number of Elements, in preparation 0

3 Setup F - a field G = (V, E) - a graph V = {1, 2,..., n} S(F, G) - set of all symmetric n n matrices A with graph G. This means: For i j a ij 0 ij E no condition on the diagonal entries 1

4 Example: 1 2 paw 3 4 S(F, paw) = a w x 0 w b y 0 x y c z 0 0 z d a, b, c, d, w, x, y F, wxyz 0 The zeros correspond to the missing edges 14, 24. 2

5 Problem Given a field F and a graph G, find mr(f, G) = min{rank A A S(F, G)} Idea of the Question: How much can you tell about a matrix if you only know where the zeros are? Related to other questions: maximum multiplicity of eigenvalues degeneracies in chemical bonding theory 3

6 The answer can depend on the field: Let G be the graph clique sum of K 4 and K 3 on K 2 two missing edges: 15, 25 4

7 F = R: A = = A S(R, G) rank A = 2 But if char F = 2, the 3, 4 element is = 0, not 2 and this A / S(F, G) 5

8 F = F 2 : Any A S(F 2, G) has the form d d d d d 5 Know all off-diagonal entries now. A[ ] = d 5 has determinant 1 so rank A 3, and mr(f 2, G) 3. 6

9 Theorems (Barrett, van der Holst & Loewy) charf 2 F infinite: mr(f, G) 2 G c = (K s1 K s2 K p1,q 1 K pk,q k ) K r F finite with p t elements: mr(f, G) 2 Either or G c = (K s1 K s2 K p1,q 1 K pk,q k ) K r, k (p t 1)/2, G c = (K p1,q 1 K pk,q k ) K r, k (p t + 1)/2 7

10 Theorems charf = 2 F infinite: mr(f, G) 2 Either or G c = (K s1 K s2 K sk ) K r G c = (K s1 K p1,q 1 K p2,q 2 K pk,q k ) K r F finite with 2 t elements: mr(f, G) 2 Either or G c = (K s1 K s2 K sk ) K r, k 2 t + 1, G c = (K s1 K p1,q 1 K p2,q 2 K pk,q k ) K r, k 2 t 1 8

11 Corollary: mr(f 2, G) 2 Either G c = (K p K q K r ) K t or G c = (K r K p,q ) K t Recall the graph It s complement is P 3 2K 1 Is it one of the forms above? 9

12 No dominating vertex in P 3 2K 1 Is P 3 2K 1 of the form K p K q K r or K r K p,q? Not K p K q K r because of P 3. Not K r K p,q because P 3 K 1 is not induced in K p,q But P 3 2K 1 is induced in the union of two bipartite graphs or in the union of two complete graphs and a bipartite graph = its complement is a rank 2 graph for any other field. 10

13 The methods in the papers [BvdHL1], [BvdH2] do not extend in any straightforward way to the problem of characterizing graphs with mr(f, G) k for k 3. However, it is possible to obtain results of this sort for finite fields using a quite different method which makes explicit use of the finiteness of F. The method is straightforward but the complexity grows exponentially with the rank k. 11

14 We had Corollary: mr(f 2, G) 2 Either G c = (K p K q K r ) K t or G c = (K r K p,q ) K t Taking complements mr(f 2, G) 2 Either G = K p,q,r K c t or G = [(K p K q ) K c r] K c t 12

15 Assume that A S n (F 2 ) with rank A 2. rank A < 2 = A is the zero matrix, or is permutation similar to J k O n k So assume rank A = 2. Then A = U t BU, B 2 2, symmetric, invertible U 2 n 13

16 Either B is congruent to I 2 or B = So WLOG either A = U t U or A = U t U U has at most 4 distinct columns: 1 0, 0 1, 1 1, 0 0 We may assume that identical columns are contiguous Write U = [ E 1 E 2 J O ] 14

17 Either A = E1 t E2 t J t O t [ E1 E 2 J 0 ] = J p 0 J p,r 0 0 J q J q,r 0 J r,p J r,q 0 r t or A = E1 t E2 t J t O t 0 1 [ E E 2 J 0 ] = 0 p J p,q J p,r 0 J q,p 0 q J q,r 0 J r,p J r,q 0 r t The graph of the first matrix is [(K p K q ) K c r] K c t The graph of the second is K p,q,r K c t 15

18 Every block in the above matrices is either a 0 matrix or a J matrix. So it would suffice to use U = and compute A = U t U = or A = U t 0 1 U = , Then if we interpret each diagonal 1 as a clique and each diagonal 0 as an independent set, we again obtain the graphs in Corollary 1. 16

19 True also when rank A > 2. Gives a straightforward process for constructing the graphs of all rank 3, rank 4, etc. matrices. Complexity grows exponentially with rank. Definition Graph X: Marked graph X B : Substitution graph: 17

20 Example What graphs are substitution graphs of? Example: Answer K p,q,r K c t 18

21 Let G 8 be the marked graph G 8 is a planar graph as can be seen from the alternative representation Theorem: mr(f 2, G) 3 G is a substitution graph of G8. 19

22 Proof: A = U t BU, B 3 3, symmetric, invertible, U = , Consider possible forms for B. substitution graph. Interpret answer as a For each prime number p and each positive integer k, there is a family of marked graphs F p,k, such that mr(f p, G) k G is a substitution graph of a graph in F p,k. 20

23 Open Questions: 1. How to recognize a substitution graph of G 8 or X p,k 2. What are the forbidden subgraphs for G 8 or X p,k? 3. Does every substitution graph of G 8 have minimum rank 3 over R? 4. Find a graph G that has minimum rank 3 over R that is not a substitution graph of G 8. 21

24 5. Let G be any graph, let F fin be any finite field, and F be any infinite field such char F fin = char F If mr(f fin, G) k, is mr(f, G) k? 6. Have substitution graphs occurred in any other settings in combinatorial matrix theory?