Minimum rank of powers of cycles and trees

PREPRINT

國立臺灣大學 數學系 預印本 Department of Mathematics, National Taiwan University

www.math.ntu.edu.tw/ ~ mathlib/preprint/2013- 04.pdf

Minimum rank of powers of cycles and trees

Gerard Jennhwa Chang and Jephian Chin-Hung Lin

April 7, 2013

Minimum rank of powers of cycles and trees

Gerard Jennhwa Chang

and Jephian Chin-Hung Lin

1Department of Mathematics, National Taiwan University, Taipei 10617, Taiwan

2Taida Institute for Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan

3National Center for Theoretical Sciences, Taipei Office, Taiwan

April 7, 2013

Abstract

The minimum rank mr(G) (respectively, maximum nullity M(G)) of a graph G with n vertices is the minimum rank (respectively, maximum nullity) of an n× n real symmetric matrix A with its off-diagonal entry A_ij ≠ 0 whenever ij is an edge of G.

There was an incomplete proof in a previous paper by Nazari and Radpoor [7] that M(Cn^r) = 2r for the r-th power Cn^r of the n-cycle C_n when r ≤ ⁿ₂. In this paper, we give a complete proof for this result. We also determine M(T²) for the square T² of a tree T .

Keywords. Rank, nullity, power, cycle, path, zero forcing number, path cover number.

∗E-mail: gjchang@math.ntu.edu.tw. This research is partially supported by the National Science Council of the Republic of China under grant NSC101-2115-M-002-005-MY3.

†E-mail: jlch3554@hotmail.com.

1 Introduction

Graphs and symmetric matrices are in intimate relation. For an n×n real symmetric matrix A, it is natural to consider the corresponding graph G= G(A) with

vertex set V(G) = {1, 2, . . . , n} and edge set E(G) = {ij∶ i ≠ j, Aij ≠ 0},

where A_ij is the ij-entry of A. Conversely, for a graph G with n vertices, there is a class of n× n real symmetric matrices whose corresponding graph is G. Denote this class as

S(G) = {A ∈ Mn×n(R)∶ A = A^⊺, G(A) = G},

whereMn×n(R) is the set of all n × n matrices over the field of real numbers. The minimum rank of a graph G is

mr(G) = min{rank(A)∶ A ∈ S(G)};

and the maximum nullity of G is

M(G) = max{null(A)∶ A ∈ S(G)}.

It is easy to see that

mr(G) + M(G) = ∣V (G)∣.

So a result in mr(G) can be presented as a result in M(G) and vice versa. In this paper we very often write results in terms of M(G) rather than mr(G).

For a positive integer r, the r-th power of a graph G is the graph G^r whose vertex set is V(G) and two distinct vertices i and j are adjacent in G^r if their distance in G is at most r. The maximum nullity of the path P_n of n vertices is 1. de Alba et al. [2] proved that M(Pn^r) = min{r, n − 1}. It is also known that the maximum nullity of cycle Cn of n vertices is 2 for n ≥ 3. Nazari and Radpoor [7] proved that M(Cn^r) = 2r for r ≤ ⁿ₂ by using the delta Conjecture that δ(G) ≤ M(G) for any graph G, which was posted in [4] but remains unsolved. In Section 3, we prove this result without using the delta Conjecture. In Section 4, we determine the maximum nullity of the square of a tree.

2 Notation and terminology

For a positive integer n, the set {1, 2, . . . , n} is denoted by [n]. The support supp(v) of a vector v∈ Rⁿ is the index set of nonzero entries of v.

A zero forcing set of a graph G is a subset F ⊆ V (G) which can force all vertices black at the end of repeatedly applying the following color changing rule:

• initially, all vertices in F are black and all other vertices are white;

• if a black vertex x has exactly one white neighbor y, then y is changed to be black.

The zero forcing number Z(G) is the minimum size of a zero forcing set of G. In the above rule, we write x → y to refer that a black vertex x forces its only white neighbor y to be black. A chronological list is a chronological record {xi → yi}^si=1, where x_i → yi is the color changing at iteration i. A zero forcing process ζ refers to a zero forcing set together with the corresponding chronological list. For more detail on the parameter Z, see [3]. The following inequality from [1] is particularly useful in this paper: for any graph G,

M(G) ≤ Z(G). (1)

A path cover of a graph G is a collection P of disjoint induced paths that cover all vertices of G. The path cover number p(G) of G is the minimum size of a path cover of G.

It is known that M(G) = p(T ) for a tree T [6] and M(G) ≤ p(G) for an outerplanar graph G [8]. For a positive integer r, the r-th weight of a path cover P is

w_r(P) = ∑

π∈P

Z(π^r);

and the r-th path cover number of G is

p_r(G) = min{wr(P)∶ P is a path cover of G}.

Since Z(π) = 1 for any path π, it is the case that p1(G) = p(G).

The clique cover number cc(G) of a graph G is the minimum number of (not necessarily disjoint) cliques to cover E(G). It is known that cc(G) ≥ mr(G) for all G, even we replace the fieldR by any other infinite field, see [5]. The star-clique cover C of a graph G is a set of stars and cliques that cover all edges of G. The weight of C is w(C) = 2p + q when C consists of p stars and q cliques. The star-clique number of G is

scc(G) = min{w(C)∶ C is a star-clique cover of G}.

By the facts that rank(A + B) ≤ rank(A) + rank(B) and that mr(Ka,b) = 2 for a + b ≥ 3, it follows that scc(G) ≥ mr(G) for any graph G. The dual star-clique cover number is defined as scc(G) = ∣V (G)∣ − scc(G). Then, for any graph G,

scc(G) ≤ M(G) ≤ Z(G). (2)

Section 4 shows that scc(T²) = M(T²) = Z(T²) = p2(T ) for any tree T .

3 Powers of cycles

Recall that Nazari and Radpoor [7] proved that M(Cn^r) = 2r for r ≤ ⁿ₂ by using the delta Conjecture that δ(G) ≤ M(G) for any graph G, which was posted in [4] but remains unsolved.

The purpose of this section is to give a proof of this result without using the delta Conjecture.

The following lemma in [4] is useful.

Lemma 1. For positive integers k≤ n, there is a k ×n real matrix C whose k ×k submatrices are nonsingular. Also, S is the support of a non-zero vector v with Cv = 0 if and only if

∣S∣ ≥ k + 1.

Theorem 2. If n≥ 3, then M(Cn^r) = Z(Cn^r) = min{2r, n − 1}.

Proof. If 2r≥ n − 1, then Cn^r = Kn and so M(Cn^r) = Z(Cn^r) = n − 1.

We now consider the case of 2r≤ n − 2. Since each set of 2r consecutive vertices of the cycle form a zero forcing set, M(Cn^r) ≤ Z(Cn^r) ≤ 2r. Next, we shall prove that M(Cn^r) ≥ 2r by constructing a symmetric matrix A with G(A) = Cn^r and rank(A) ≤ n − 2r.

For k ∈ [n − r], let Ik be the subset {k, k + 1, . . . , k + r} of [n − r], where the addition is taken module n− r, that is, k + i is k + i − (n − r) if k + i > n − r. By Lemma 1, we may choose an r× (n − r) real matrix C whose r × r submatrices are nonsingular; also a vector v_k ∈ R^n−r such that Cv_k= 0 with supp(vk) = Ik for each k ∈ [n − r]. Next, choose appropriate coefficients a_k such that A= ∑ⁿi=1^−2ra_iv_iv^⊺_i has the property thatG(A) = Pn^r−r. This is possible because we only have to worry about that some nonzero entries vanish under the process of summation. However, there are only finitely many these conditions and we have infinitely many choices for the coefficients. Furthermore, we can choose an−2r as small as we want.

Let B be the (n − r) × r matrix whose i-th column is vn−2r+i. Since all v_i are in the null space of C, the space spanned by{v1, v₂, . . . , v_n−r} has dimension at most n − 2r. Also, rank(A) ≥ n − 2r, since mr(Pn^r−r) = n − 2r. Hence, there is a matrix X such that AX = B. As we may choose an−2r as small as we want, X^⊺AX can be chosen to contain no zero entries.

Then

D= (A B

B^⊺ X^⊺AX) = ( A B X^⊺A X^⊺B)

has rank n− 2r and G(D) = Cn^r. These prove that M(Cn^r) ≥ 2r and so M(Cn^r) = Z(Cn^r) = 2r as desired.

4 Squares of trees

The purpose of this section is to determine the maximum nullity of the square of a tree.

Besides a formula in terms of the zero forcing number, a procedure to compute it is also given.

Theorem 3. If T is a tree, then scc(T²) = M(T²) = Z(T²) = p2(T ).

The theorem follows from (2) that scc(T²) ≤ M(T²) ≤ Z(T²) and the following two lemmas. The first lemma proves that p₂(T ) is an upper bound of Z(T²) and the second lemma proves that p₂(T ) is a lower bound of scc(T²).

Lemma 4. If T is a tree, then Z(T²) ≤ p2(T ).

Proof. Choose a path cover P of T with w2(P) = p2(T ). A vertex is special if it is of degree 2 and is adjacent to a leaf. We conclude the lemma by proving Claim 1 using an induction on the number n of vertices of T .

Claim 1. ∣F ∣ ≤ w2(P) for some zero forcing set F of T² with a zero forcing process ζ for which each forcing x→ y has the property that dT(x, y) = 2 whenever y is not special in T .

The claim is clear for n = 1. Now assume that n ≥ 2. For the case when T is a star, T² is K_n. Choose F as a set of all vertices except a leaf of T . The claim follows from that P consists of paths with minimum total weight n − 1. For the case when T is the n-path v1, v2, . . . , vn with n ≥ 3, choose F = {v1, v2} which is a zero forcing set of T² using the chronological list {vi→ vi+2}ⁿ⁻²_i=1. Then Z(T²) ≤ 2. As Z(T²) ≥ δ(T²) = 2. In fact Z(T²) = 2.

The claim then follows.

Now we consider the case when T is neither a star nor a path. In this case, there is always a path π∶ v1, v₂, . . . , v_r, . . . , v_s inP such that v1 is a leaf in T and v_r is the only vertex of π adjacent to an unique vertex v∉ π in T . Then, T1∶= T − π is a tree of at least 3 vertices and P1 ∶= P − {π} is a path cover of T1. We claim that v is not a special vertex in T₁ when s= 2. Suppose to the contrary that s = 2 but v is of degree 2 and is adjacent to a leaf u in T₁. Suppose u is a vertex of a path π₁∈ P.

Case 1. π1 = u. Suppose v is in path π2 ∈ P. In this case, we may replace P by the path cover (P − {π2, π}) ∪ {π2+ π} of weight no more than P and replace π by u.

Case 2. π₁= uv. In this case, we may replace P by the path cover (P−{π1, π})∪{π1+π}

of weight no more than P and replace π by π1+ π.

Case 3. π₁ has at leas 3 vertices. In this case, we may replace P by the path cover (P − {π1, π}) ∪ {u, (π1− u) + π} of weight no more than P and replace π by u.

So, we may assume that either s ≠ 2 or v is not special. By the induction hypothesis,

∣F1∣ ≤ w2(P1) for some zero forcing set F1 of T₁² with a zero forcing process ζ1 for which each forcing x→ y has the property that dT1(x, y) = 2 whenever y is not special in T1, in particular when y is v for the case of s= 2.

Let F = F1∪ {v1} when s ≤ 2 and let F = F1∪ {v1, v₂} when s ≥ 3. We shall check that F is a zero forcing set of T² by constructing a zero forcing process corresponding to F as follows. First, if s≥ 3, then do forcing vi → vi+2 for 1≤ i ≤ r − 2. By now, vr is black unless s= 2. Next, do all forcing x → y of ζ1 until v^∗→ v. Notice that either vr is black or else s= 2 and so d_T1(v^∗, v) = 2. In either case, all the forcing of ζ1 mentioned above do not infect the vertices in π. Then do the forcing v₁ → v2 when s= 2 or the forcing vr−1 → vr+1 when s≥ 3.

Finally, do the remaining forcing of ζ₁, follow by the remaining forcing v_i→ vi+2 alone π for r≤ i ≤ s − 2. These give a zero forcing processing corresponding to F with the property that dT(x, y) = 2 whenever y is not special in T . This completes the proof of Claim 1.

For a vertex v in T , we use N(v) and N²(v) to denote the set of neighbors of v in T and T² respectively. We use κ_v to denote the clique induced by N(v) ∪ {v} in T², and use σ_v to denote the star in T² whose center is v and whose set of leaves is N²(v).

In a tree T , a pendent path is a maximal induced path that contains a leaf but no vertex of degree more than 2. A pendent branch consists of vertex v with degree k+ 1 ≥ 3 and k pendent paths each has an end vertex adjacent to v. For the case when T is not a path, a pendent branch can be obtained from a breadth first search. Equivalently, consider T rooted at a chosen vertex r and choose a vertex v of degree k+ 1 ≥ 3 farest from r. Then v has k children and all proper descendants of v form k pendent paths of the pendent branch.

Lemma 5. If T is a tree, then p₂(T ) ≤ scc(T²).

Proof. We shall prove the lemma by induction on the number n of vertices of T . For the case when T is a path π∶ v1, v2, . . . , vn, the lemma follows from considering the path cover {π} and the star-clique cover {κvi∶ 2 ≤ i ≤ max{2, n − 1}}. Now assume that T is not a path and so n≥ 4. Choose a pendent branch at a vertex v of degree k + 1 ≥ 3 with pendent paths α_i∶ v1ⁱ, v₂ⁱ, . . . , v_sⁱ_i for 1≤ i ≤ k. We consider three cases.

Case 1. One of the following conditions holds: (i) there is some s_i ≥ 4, (ii) there is some s_i = 2, (iii) k = 2 and there is some si = 3 with the other sj = 1. Let T1 be the tree obtained from T by deleting the leaf v_sⁱ

i. By the induction hypothesis, p₂(T1) ≤ scc(T1²).

Choose a path cover P1 of T₁ with w₂(P1) = p2(T1) and a star-clique cover C1 of T₁² with w(C1) = scc(T₁²). Let π be the path in P1 which contains the leaf v_sⁱ

i−1 in T₁. If ∣V (π)∣ = 2, then we change P1 and π according to three subcases:

(i) The other end vertex v_sⁱ

i−2of π is adjacent to v_sⁱ

i−3which is an end vertex of a path π₁ inP1. Since Z((π1+π)²) ≤ 2 ≤ Z(π1²)+Z(π²), we may replace P1 by(P1−{π1, π})∪{π1+π}

and replace π by π₁+ π.

(ii) The other end vertex v of π is adjacent to a neighbor of v which is an end vertex of a path π₁ inP1. Replace P1 by (P1− {π1, π}) ∪ {π1+ π} and replace π by π1+ π.

(iii) The other end vertex vⁱ₁ of π is adjacent to v. Let v is in a path π₁ in P1. If v is an end vertex of π₁, then replace P1 by(P1− {π1, π}) ∪ {π1+ π} and replace π by π1+ π. If v is not an end vertex of π₁, then the leaf v₁^j is an end vertex of π₁. In this case, replace P1

by(P1− {π1, π}) ∪ {v^j1,(π1− v1^j) + π} and replace π by (π1− v1^j) + π.

By now we may assume that∣V (π)∣ ≠ 2. Then P ∶= (P1−{π})∪{π +vⁱsi} is a path cover of T with w₂(P) = w2(P1), since Z((π + vsⁱi)²) = Z(π²) = 1 if ∣V (π)∣ = 1 and Z((π + vⁱsi)²) = Z(π²) = 2 if ∣V (π)∣ ≥ 3. Also C ∶= C1∪{κvⁱ_si−1} is a star-clique cover of C with w(C1)+1 = w(C).

Consequently,

p₂(T ) ≤ w2(P) = w2(P1) ≤ scc(T1²) = ∣V (T1)∣ − w(C1) = ∣V (T )∣ − w(C) ≤ scc(T²).

Case 2. k = 2 and s1 = s2 = 3. Let T1 be the tree obtained from T by deleting α₁, v and α2. By the induction hypothesis, p2(T1) ≤ scc(T₁²). Choose a path cover P1 of T1

with w₂(P1) = p2(T1) and a star-clique cover C1 of T₁² with w(C1) = scc(T1²). Then P ∶=

P1∪{α1+v +α2} is a path cover of T with w2(P) = w2(P1)+2. Also C ∶= C1∪{σv, κ_v, κ_v¹

2, κ_v²

2} is a star-clique cover of T² with w(C1) + 5 = w(C). Consequently,

p₂(T ) ≤ w2(P) = w2(P1) + 2 ≤ scc(T1²) + 2 = ∣V (T1)∣ − w(C1) + 2 = ∣V (T )∣ − w(C) ≤ scc(T²).

Case 3. One of the following conditions holds: (i) k = 2 with s1 = s2 = 1, (ii) k ≥ 3 with some s_i say s₁ = 1, (iii) k ≥ 3 with all si = 3. Let T1 be the tree obtained from T by deleting α₁. By the induction hypothesis, p₂(T1) ≤ scc(T₁²). Choose a path cover P1 of T₁ with w₂(P1) = p2(T1) and a star-clique cover C1 of T₁² with w(C1) = scc(T₁²). We have the following facts.

(a) We may assume that if σx ∈ C1, then x has degree at leat 3 in T1. For otherwise we may replace σ_x by κ_y for all y ∈ NT1(x) to get a star-clique cover of weight no more than C1.

(b) We may assume that if x has at least two neighbors y₁ and y₂ of degree 1 or 2 in T₁, then κ_x∈ C1. As the edge y₁y₂ can be covered only by κ_x, σ_y1 or σ_y2, this follows from (a).

(c) Under condition (i), we may assume that κ_v ∈ C1. This follows from the facts that the edge vv²₁ can only be covered by κ_v or κ_v²

1 and that κ_v covers more edges than κ_v²

1. (d) Under condition (iii), we may assume that σ_v ∈ C1. For otherwise κ_vⁱ

1 and κ_vⁱ

2 are inC1

for 2≤ i ≤ k and so we may replace κvⁱ₁ for 2≤ i ≤ k by σv to get a star-clique cover of weight no more than C1.

Now P ∶= P1 ∪ {α1} is a path cover of T with w2(P) = w2(P1) + Z(α²₁). According to (a), (b), and (c), κ_v ∈ C1 in any case. According to (d), σ_v ∈ C1 under condition (iii).

Under condition (i) or (ii), C ∶= (C1− {κv in T₁}) ∪ {κv in T}) is a clique cover of T with w(C1) = w(C). Under condition (iii), C ∶= (C1− {κv and σ_v in T₁}) ∪ {κv and σ_v in T, κ_v¹

2}) is a clique cover of T with w(C1) + 1 = w(C). In any case, w(C1) + ∣V (α1)∣ − Z(α₁²) = w(C).

Hence,

p2(T ) ≤ w2(P) = w2(P1) + Z(α²1) = p2(T1) + Z(α²1) ≤ scc(T1²) + Z(α²1)

= ∣V (T1)∣ − w(C1) + Z(α²1) = ∣V (T )∣ − w(C) = scc(T²).

The proof of the theorem in fact provides an algorithm for computing M(T²). We summary it as follows.

Corollary 6. If T is a path, then M(T²) = 1 when ∣V (T )∣ ≤ 2 and M(T ) = 2 when ∣V (T )∣ ≥ 3.

If T is a tree containing a pendent branchB which has p pendent paths with at most 2 vertices and q paths of at least 3 vertices, T₁ is obtained from T by deleting B and T2 is obtained from T by replacing B with a path of two vertices, then M(T ) = M(T1) + p + 2q − 2 if q ≥ 2 and M(T ) = M(T2) + p + q − 1 if q ≤ 1.

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