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Linear Algebra and its Applications 438 (2013) 4041–4061

Contents lists available at SciVerse ScienceDirect

Linear Algebra and its Applications

journal homepage: www.elsevier .com/locate/ laa

Bounds for minimum semidefinite rank from

superpositions and cutsets

Jonathan Beagleya, Lon H. Mitchell b, Sivaram K. Narayanc,∗, Eileen Radzwionc,

Sara P. Rimerd, Rachael Tomasinoc, Jennifer Wolfe c, Andrew M. Zimmere

aDepartment of Mathematical Sciences, George Mason University, Fairfax, VA 22030, United States

bDepartment of Mathematics, Virginia Commonwealth University, Richmond, VA 23284, United States

cDepartment of Mathematics, Central Michigan University, Mount Pleasant, MI 48859, United States

dDepartment of Civil and Environmental Engineering, University of Michigan, Ann Arbor, MI 48109, United States

eDepartment of Mathematics, University of Michigan, Ann Arbor, MI 48109, United States

A R T I C L E I N F O A B S T R A C T

Article history:

Received 10 June 2011

Accepted 10 July 2012

Available online 11 october 2012

Submitted by W. Barrett

Keywords:

Minimum semidefinite rank

Vector representation of a graph

Superposition of graphs

Cutsets

The real (complex) minimum semidefinite rank of a graph is the

minimum rank among all real symmetric (complex Hermitian) pos-

itive semidefinite matrices that are naturally associated via their

zero-nonzero pattern to the given graph. In this paper we give an

upper bound on the minimum semidefinite rank of a graph when

the graph is modified from the superposition of two graphs by can-

celing some number of edges.We also provide a lower bound for the

minimumsemidefinite rank of a graph determined by a given cutset.

When the complement of the cutset is a star forest these lower and

upper bounds coincide andwe can compute theminimum semidef-

inite rank in terms of smaller graphs. This result encompasses the

previously known case inwhich the cut set has order two or smaller.

Next we provide results for when the cut set has order three. Using

these resultsweprovide an examplewhere thepositive semidefinite

zero forcing number is strictly greater than the maximum positive

semidefinite nullity.

© 2012 Published by Elsevier Inc.

1. Introduction

A graph G consists of a set of vertices V and a set of edges E, where the elements of E are unordered

pairs of vertices. The order ofG, denoted |G|, is the cardinality ofV . A graph is simple if it has nomultiple

edges or loops.

∗ Corresponding author.

E-mail address: [email protected] (S.K. Narayan).

0024-3795/$ - see front matter © 2012 Published by Elsevier Inc.

http://dx.doi.org/10.1016/j.laa.2012.07.012


4042 J. Beagley et al. / Linear Algebra and its Applications 438 (2013) 4041–4061

The entries of an n-by-n Hermitian matrix A = (aij) over the complex numbers C naturally corre-

spond to a simple undirected graph G(A) with vertex set {1, . . . , n} and edge set {{i, j} : aij �= 0, 1 �i < j � n}. Observe that the diagonal entries of A have no effect on the structure of G(A). Within this

correspondence a number of natural extremal problems arise; one of recent interest is the minimum

rank problem, which seeks to determine the smallest possible rank of any real symmetric matrix with

a given graph (see the 2007 survey by Fallat and Hogben [5] for many references). Define

S(G) = {A ∈ Mn(R) : At = A, G(A) = G}.The minimum rank of G is defined to be

mr(G) = min{rank(A) : A ∈ S(G)}.In this paper, we consider the related problem of determining the minimum rank among positive

semidefinite (psd) matrices with a given graph (see the 2011 survey by Fallat and Hogben [6] for many

references). For K ∈ {C, R} definePK(G) = {A ∈ Mn(K) : A psd, G(A) = G}.

The minimum semidefinite rank of G is then defined to be

mrK+(G) = min{rank(A) : A ∈ PK(G)}.The real and complex parameters are actually different (see for instance [1]).

1.1. Preliminaries

Given a list of n column vectors in Kd, �X = (�x1, . . . , �xn), let X be the matrix [�x1 . . . �xn]. Then X∗Xis a psd matrix called the Gram matrix of �X with regard to the Euclidean inner product. Its associated

graph G has n vertices v1, . . . , vn corresponding to the vectors �x1, . . . , �xn and edges corresponding to

nonzero inner products among those vectors. Since X∗X ∈ PK(G) for the graph G, we say �X is a vector

representation of G in Kd. By rank �X , we mean the dimension of the span of the vectors in �X , which is

equal to rank X∗X . At times we will denote the Gram matrix X∗X by Gram(�x1, . . . , �xn).This type of vector representation is a particular instance of themore general vector representations

studied by Parson and Pisanski [10].

As every positive semidefinitematrix is aGrammatrix and conversely everyGrammatrix is positive

semidefinite, mrK+(G) � d if and only if there is a vector representation of G inKd. Wewill oftenwrite

�v to mean a vector representing a vertex v.

As our only concern is undirected graphs, an edge {v, u} will often be written as vu. The subgraph

G[R] of G induced by R ⊂ V(G) is the subgraph of G with vertex set R and edge set consisting of those

edges of G where both vertices are elements of R. For a vertex w of a graph G, let N(w) denote the set

of all vertices adjacent to w in G, called the neighborhood of w in G. By the closed neighborhood of w,

denoted N[w], we mean {w} ∪ N(w).Let Kn denote the complete graph on n vertices, Cn the cycle on n vertices, and Pn the path on n

vertices. If H and K are graphs, let H � K denote the disjoint union of H and K . For n � 3 it is well

known that mrK+(Kn) = 1, mrK+(Cn) = n − 2, and mrK+(Pn) = n − 1. If G is a graph let

nG = G � · · · � G︸︷︷︸n copies

.

1.2. Summary of paper

Let G1 and G2 be graphs with the same vertex set and define a new graph G as follows:

V(G) = V(G1) = V(G2) and E(G) = E(G1) ∪ E(G2). Then if A1 ∈ PK(G1) and A2 ∈ PK(G2) we have

A1 + εA2 ∈ PK(G) for all but finitely many ε > 0. This implies that mrK+(G) � mrK+(G1) +mrK+(G2).


J. Beagley et al. / Linear Algebra and its Applications 438 (2013) 4041–4061 4043

Fig. 1. A graph G with blocks K3, K4, K2, and C6.

Generalizing this construction, we say a graph G is a superposition of two graphs G1 and G2 if G

is obtained by identifying G1 and G2 at a set of vertices, keeping all the edges that are present

in either G1 or G2. The above argument shows that if G is the superposition of G1 and G2 then

mrK+(G) � mrK+(G1) + mrK+(G2).

In Section 2, we generalize this result to show that mrK+(G) � mrK+(G1) + mrK+(G2) when G is

modified from a superposition of G1 and G2 by the removal of the edges of a star forest common to both

G1 and G2.

For a connected graph G = (V, E), a vertex set S ⊂ V is called a cut set of G if G[V − S] is a

disconnected graph. When S = {v}, the vertex v is called a cut vertex. A maximal connected induced

subgraph without a cut vertex is called a block; see Fig. 1. Booth et al. [3, Theorem 3.4] and van der

Holst [12, Theorem2.4] have independently showed that theminimumsemidefinite rank is completely

determined by the blocks of G. More specifically, if G1, . . . , Gk are the blocks of G then mrK+(G) =∑ki=1 mrK+(Gi). Van der Holst [12, Corollary 2.9] also showed that if G has a cut set of size two, then

one can compute the minimum semidefinite rank of G in terms of the minimum semidefinite rank of

strictly smaller graphs.

In this paper we generalize these cutset results. In Section 3, we show that if S is a cut set and

G[V −S] = H1 �H2 then there exists an easily computable finite setD (G, S;H1,H2) of pairs of graphs(G1, G2) with |Gi| = |Hi| + |S| such that

mrK+(D (G, S;H1,H2)) � mrK+(G), (1)

where

mrK+(D (G, S;H1,H2)) = min{mrK+(G1) + mrK+(G2) : (G1, G2) ∈ D (G, S;H1,H2)}.We also show that if S induces a star forest in G then the inequality in Eq. (1) becomes equality. As

any vertex set of size two or fewer induces a star forest, we obtain the results of Booth et al. and

van der Holst as special cases. The primary tool used to prove these results is orthogonal projec-

tions.

A basic example of a cutset of a graph G is the neighborhood N(v) of a vertex v in G. In this case,

there are several results relating mrK+(G) to the minimum semidefinite rank of graphs with vertex set

V(G) − {v} (see for instance work of Hackney et al. [8, Proposition 4.2] or van der Holst [11, Lemmas

3.6 and 3.7]). In Section 4, we compare our results to these previous results.

We next extend our work to cutsets of size 3. Notice that if S is a cutset of size 3, then the possible

isomorphism classes of G[S] are K3, K1 � K2, 3K1, P3. In particular, if G[S] �= 3K1 we may apply the

results in Section 3 to computemrK+(G) from theminimumsemidefinite rank of strictly smaller graphs.

In Section 5, we show that when G[S] = 3K1 it is possible to compute mrR+(G) in terms of constrained

rank parameters of strictly smaller graphs. In Example 5.8 we use our cutset results to construct a

graph G such that

|G| − Z+(G) = 6 < 7 = mrR+(G)

where Z+(G) is the positive semidefinite zero forcing number of G.



2. Superpositions and edge cancellations

Theorem 2.1. Let G be modified from a superposition of G1 and G2 by the removal of the edges of a star

forest common to G1 and G2. Then mrK+(G) � mrK+(G1) + mrK+(G2).

Booth et al. [3, Proposition4.1] provedTheorem2.1 in the special case of the removal of a single edge.

As adding isolated vertices to a graph does not change the minimum semidefinite rank, Theorem 2.1

follows immediately from the next lemma.

Lemma 2.2. Let A, B ∈ Mn(K) be Hermitian matrices (not necessarily psd) corresponding to

graphs G1, G2 respectively. Assume that V = V(G1) = V(G2) and G is a graph on vertex set V modi-

fied from the superposition of G1 and G2 by the removal of the edges of a star forest common to both graphs.

With these assumptions there exists a diagonal matrix D ∈ Mn(K) such that A+D∗BD corresponds to the

graph G.

Proof. This is essentially an argument used byHackney et al. [8, Proposition 4.2].We only consider the

case inwhich the star forest removed consists of a single star. The general case can be proved similarly.

Let A = (aij) and B = (bij). Let 1 be the index corresponding to the star center and let the indices

2, . . . , k correspond to the pendant vertices of the star. Let n = |G| and define D = D(α, β) to be the

n-by-n diagonal matrix with the entries

α, − a1,2

αb1,2, . . . , − a1,k

αb1,k, β, . . . , β

on the diagonal.

For any α, β ∈ R the matrix A + D∗BD corresponds to a subgraph of G, however specific choices

of α, β may result in the matrix A+ D∗BD corresponding to a proper subgraph of G. As the next claim

shows this rarely occurs.

Claim. For almost all (α, β) ∈ R2 the matrix A + D∗BD corresponds to the graph G.

First pick α ∈ R, α �= 0 such that for 2 � i, j � k and {i, j} ∈ E(G) the entry

(A + D∗BD)i,j = ai,j +(

a1,i

αb1,i

)bi,j

(a1,j

αb1,j

)

is nonzero. Notice that if 2 � i, j � k and {i, j} ∈ E(G) then (A + D∗BD)i,j is a nonzero polynomial in

α−1, hence α can be all but finitely many real numbers. Next pick β ∈ R such that

(A + D∗BD)i,j

is nonzero whenever {i, j} ∈ E(G). Since α is already fixed, if {i, j} ∈ E(G) then (A + D∗BD)i,j is a

nonzero polynomial in β , hence β can be all but finitely many real numbers. With the choices above

the matrix A + D∗BD corresponds to G. As only finitely many values were avoided at each step the

claim follows. �

Remark2.3. Thematrixconstructionused inLemma2.2givesaminimumrankresult (seeCorollary2.4

below) aswell as aminimum semidefinite rank result; however it is useful to think of the construction

also in terms of vector representations. In the case in which A and B are psd, we may find vectors

�v1, . . . , �vn and �w1, . . . , �wn such that A = Gram(�v1, . . . , �vn) and B = Gram(�w1, . . . , �wn). Thenif D is a diagonal matrix with real numbers d1, . . . , dn down the diagonal, we have A + DBD =Gram(�v1 ⊕ d1 �w1, . . . , �vn ⊕ dn �wn). We will frequently use this direct sum construction throughout

the remainder of the paper.



Fig. 2. The graphs G1, G2, and G of Example 2.7.

Corollary 2.4. Let G be modified from a superposition of G1 and G2 by the removal of the edges of a star

forest common to G1 and G2. Then mr(G) � mr(G1) + mr(G2).

Example 2.5. Theorem 2.1 allows one to calculate tight upper bounds in a variety of cases:

(1) Let Cn be the cycle on n vertices. Then Cn is a superposition of Cn−1 and C3 = K3 modified by

the cancellation of a shared edge. It follows by induction that

mrK+(Cn) � mrK+(Cn−1) + mrK+(C3) � ((n − 1) − 2

) + 1 = n − 2.

This inequality is actually an equality.

(2) If G is a star forest, then mrK+(G) � 2. One can see this by observing that G is the superposition

of Kn and Kn modified by the cancellation of the edges of a shared star forest.

(3) If MLn is the Möbius ladder on 2n vertices, then Theorem 2.1 was used in [9] to show that

mrK+(MLn) = 2n − 4 when n � 5.

Remark 2.6. We now discuss possible extensions of Theorem 2.1.

(1) Let Pn be the path on n � 4 vertices, then it is well known that mrK+(Pn) = 3 (see for instance

Booth et al. [3, Proposition 5.8] or the AIM paper [7, Theorem 3.16]). As Pn is a superposition of

Kn and Kn modified by the cancellation of a shared copy of Pn, we see that Theorem 2.1 cannot

be extended to general trees or forests.

(2) The argument above also shows that Theorem 2.1 cannot be extended to the cancellation of a

shared copy of a graphHwhenevermrK+(H) � 3. A result of Barrett et al. [2, Corollary 1] implies

the following: if mrK+(H) � 2 then H is either a star forest, H contains a complete graph on 3

vertices, or H contains a cycle on four vertices.

In the next example we show that Theorem 2.1 cannot be extended to cycles on four vertices. In

Proposition 5.3 we will consider the case in which H = K3.

Example 2.7. Let C4 be the cycle on four vertices and consider the graphs G1, G2, and G in Fig. 2. Then

G is modified from a superposition of G1 and G2 by the removal of the edges of a copy of C4 common

to G1 and G2. In Appendix B we will show that mrK+(G) = 6 and mrK+(G1) = 4. Then

mrK+(G) = 6 > 5 = mrK+(G1) + mrK+(G2).

3. Cutset decompositions

If G[V − S] = H1 � H2 with |H1| , |H2| > 0 we say a pair (G1, G2) is a decomposition of G at S with

respect to H1 and H2 when



(1) G is the superposition ofG1 andG2 identified at the vertices of S andmodified by the cancellation

of some number of edges common to both graphs.

(2) Under the identification above G1 − S = H1 and G2 − S = H2.

Now define D (G, S;H1,H2) to be the set of all decompositions of G at the cutset S with respect to H1

and H2. The minimum semidefinite rank of the set D (G, S;H1,H2) can naturally be defined as:

mrK+(D (G, S;H1,H2)) = min{mrK+(G1) + mrK+(G2) : (G1, G2) ∈ D (G, S;H1,H2)}.Remark 3.1. Notice that G[V − S] could have many components, leading to many choices for H1 and

H2. In Section 3.1, we will see that these choices matter.

The motivation for these definitions is contained in the proof of the next lemma.

Lemma 3.2. For a graph G = (V, E) with cut set S such that G[V − S] = H1 � H2,

mrK+(D (G, S;H1,H2)) � mrK+(G).

Proof. Let G = (V, E), G[V − S] = H1 � H2, and mrK+(G) = m. Let

�V = (�ui)Li=1 ∪ (�si)Ki=1 ∪ (�wi)Mi=1

be a vector representation ofG inKm where the �ui correspond to the vertices ofH1, the�si correspond to

theverticesofS, and the �wi correspond to theverticesofH2.Define thesubspaceU = span({�ui}Li=1)and

letP : Km → U be theorthogonal projectionontoU. Then for 1 � i � M, �wi ∈ U⊥ = (I−P) Km. Now

let G1 be the graph corresponding to the vectors (�ui)Li=1 ∪ (P�si)Ki=1 and G2 be the graph corresponding

to the vectors ((I − P)�si)Ki=1 ∪ (�wi)Mi=1. Note that

mrK+(G1) + mrK+(G2) � dim(U) + dim(U⊥) = mrK+(G).

By construction, the vertices G1 − S are represented by the �ui and hence G1 − S = H1. For a similar

reason G2 − S = H2. Further

〈�si,�sj〉 = 〈P�si, P�sj〉 + 〈(I − P)�si, (I − P)�sj〉,so if si, sj ∈ S are adjacent in G then they must be adjacent in at least one of G1 or G2. If si, sj are not

adjacent in G then

〈P�si, P�sj〉 = −〈(I − P)�si, (I − P)�sj〉and {si, sj} is an edge in G1 if and only if {si, sj} is an edge in G2. This implies that G is the superposition

of G1 and G2 identified at the vertices of S and modified by the cancellation of some number of edges

common to both graphs.

In conclusion, (G1, G2) ∈ D (G, S;H1,H2) and

mrK+(D (G, S;H1,H2)) � mrK+(G1) + mrK+(G2) � mrK+(G). �

Theorem 3.3. Let G be a connected graph with cut set S such that G[V − S] = H1 � H2. If G[S] is a star

forest then

mrK+(G) = mrK+(D (G, S;H1,H2)) = min{mrK+(G1)+mrK+(G2) : (G1, G2) ∈ D (G, S;H1,H2)}.

Proof. Lemma 3.2 implies that mrK+(D (G, S;H1,H2)) � mrK+(G). To show the reverse inequality let

(G1, G2) ∈ D (G, S;H1,H2). Then G is the superposition of G1 and G2 identified at S and modified by



the cancellation of some number of edges common to both graphs. But the edges canceled must be

in G[S] which is a star forest, therefore Theorem 2.1 implies that mrK+(G) � mrK+(G1) + mrK+(G2).

Minimizing over all such G1 and G2 gives mrK+(G) � mrK+(D (G, S;H1,H2)). �

Recall that amaximal connected induced subgraphwithout a cut-vertex is called a block. In the case

in which S = {v} is a single vertex, the set D (G, S;H1,H2) is a single pair and Theorem 3.3 recovers

the cut vertex result of Booth et al. [3, Theorem 3.4] and van der Holst [12, Corollary 2.5]. Repeated

applications of Theorem 3.3 to cut vertices yields the following corollary.

Corollary 3.4. Let G be a connected graph with blocks G1, . . . , Gm. Then mrK+(G) = ∑mi=1 mrK+(Gi).

In the case in which S consists of two vertices, D (G, S;H1,H2) is also small and easy to compute.

Example 3.5. Consider the graph G with cut set S = {1, 2}:

If H1 = G[{3, 4, 5}] and G[V − S] = H1 � H2, we then see that

It iswell known thatmrK+(H) � |H|−1with equality if and only ifH is a tree. FurthermrK+(Cn) = n−2

where Cn is the cycle on n vertices. This implies that mrK+(G) = min{8, 4 + mrK+(G′)} where G′ is thegraph

Now S = {1, 2} is a cut set of G′ and

Again using the results for trees and cycles, we see that mrK+(G′) = 3. This implies that

mrK+(G) = 7.

In the above example, the minimum semidefinite rank can be computed using other methods in

the literature (for instance van der Holst’s classification of graphs with mrK+(G) � |G| − 2 [11]). In the

next example we consider a graph for which this is not true.



Example 3.6. Let G be the graph in Fig. 3. Using results in the literature one can show

|G| − 5 = 6 � mrK+(G) � 8 = |G| − 3.

Here the lower bound comes from the fact that the positive semidefinite zero forcing number is five.

The upper bound comes fromvanderHolst’s classification of graphswithmrK+(G) � |G|−2 [11],more

specifically van der Holst’s classification implies that any connected graph G with mrK+(G) � |G| − 2

must have a vertex with degree less than or equal to two (see Remark 3.7). However G has a cutset

S = {s1, s2, s3} and G[S] = P3 so we may apply Theorem 3.3 to compute mrK+(G) in terms of smaller

graphs. In Appendix B we compute the minimum semidefinite rank of these smaller graphs and show

that mrK+(G) = 6.

Remark 3.7. Van der Holst’s full classification of connected graphs with mrK+(G) � |G| − 2 is not

needed in Example 3.6. In fact it is enough to know that any graph with mrK+(G) � |G| − 2 does not

have a K4-minor (see for instance [11, Proposition 3.2]). Then a famous theorem of Dirac [4] implies

that G has a vertex of degree two or less.

As the next example shows, mrK+(D (G, S;H1,H2)) can be a poor lower bound when G[S] has fewedges.

Example 3.8. The difference mrK+(G) − mrK+(D (G, S;H1,H2)) can be arbitrarily large. For example,

let G be the graph in Fig. 4. Then mrK+(G) = 6. One way to see this is to apply Corollary 3.4 and

use the fact that mrK+(K2) = 1 and mrK+(C6) = 4. Now let S be the cycle in G and let G1 and G2

the graphs in Fig. 4. Then (G1, G2) ∈ D (G, S;H1,H2) and mrK+(G1) = mrK+(G2) = 2. To show

mrK+(Gi) = 2 again use Corollary 3.4 and the fact that mrK+(K�) = 1. These observations show

that mrK+(G) − mrK+(D (G, S;H1,H2)) � 2. By replacing the six-cycle in G with an n-cycle, we have

mrK+(G) − mrK+(D (G, S;H1,H2)) � n − 4.

Fig. 3. The graph in Example 3.6.

Fig. 4. The graphs G, G1, and G2 of Example 3.8.



Fig. 5. The graph G in Section 6.

3.1. The dependence on H1 and H2

In this sectionwe consider an example G = (V, E)with a cut set S such that G[V − S] = H1 �H2 =H′1 � H′

2 and

mrK+(D (G, S;H1,H2)) < mrK+(D(G, S;H′

1,H′2

)).

This shows that the parameter mrK+(D (G, S;H1,H2)) depends on the choice of H1 and H2.

Let G be the graph in Fig. 5 and S = {v2, v4, v5}. Then G[V − S] = H1 � H2 = H′1 � H′

2 where

H1 = G[v1], H2 = G[v3, v6, v7, v8, v9], H′1 = G[v1, v3], and H′

2 = G[v6, v7, v8, v9].Lemma 3.9. mrK+(G) = 6.

Proof. This can be computed in many ways. For instance Theorem 3.3 can be applied repeatedly to

cutsets of size 2. �

Lemma 3.10. mrK+(D (G, S;H1,H2)) � 5.

Proof. Let (G1, G2) ∈ D (G, S;H1,H2) be the pair with G1[S] = G2[S] = K3. Then G1 = K4 and

mrK+(G1) = 1. By repeatedly applying Theorem 3.3 to cutsets of size 2 and using the fact that

mrK+(K�) = 1, one can see that mrK+(G2) = 4. Then

mrK+(D (G, S;H1,H2)) � mrK+(G1) + mrK+(G2) = 5. �

Remark 3.11. Using Proposition 4.6 the proof above shows that mrK+(D (G, S;H1,H2)) = 5.

Lemma 3.12. mrK+(D (G, S;H′

1,H′2

)) = 6.

Proof. The possible graphs forG′1 andG′

2 are given in Fig. 6. AsG[S] = 3K1, (G′1, G

′2) ∈ D (

G, S;H′1,H

′2

)implies that G′

1[S] = G′2[S]. A simple argument using vector representations leads to the following

Fig. 6. Possible graphs for G′1 and G′

2 in Section 6. Here dashed edges may or may not be present.



mrK+(G′1) =

⎧⎨⎩ 3 if G′1[S] = 3K1

2 otherwise

Further, if G′2[S] contains an edge then mrK+(G′

2) > 3 and if G′2[S] = 3K1 then mrK+(G′

2) = 3. This

shows that mrK+(D (G, S;H′

1,H′2

)) = 6. �

4. Orthogonal vertex removal

Abasic example of a cutset in a graphG is the neighborhoodN(v) of a vertex v inG. Results similar to

Theorem3.3 have already appeared in the literature in this special case via the notion of the orthogonal

removal of a vertex (see for instance work of Hackney et al. [8, Proposition 4.2] or van der Holst [11,

Lemma 3.6, Lemma 3.7]). These prior results involve multigraphs in a key way and we will show that

these multigraphs are closely related to our sets D (G, S;H1,H2) when S = N(v).

4.1. Multigraphs

A graph G is amultigraph if G is an undirected graph that has no loops but may have multiple edges

between vertices. If G is a multigraph then PK(G) denotes the set of all psd matrices inMn(K) where

• aij �= 0 if vertex i and vertex j are connected by exactly one edge in G.• aij = 0 if vertex i and vertex j are not adjacent and i �= j.

No restriction is placedonaij if vertex i andvertex j are connectedbymore thanoneedge. Theminimum

semidefinite rank of a multigraph G is defined as:

mrK+(G) = min{rank A : A ∈ PK(G)}.As in the case of simple graphs, one can consider vector representations for multigraphs. Then with

the natural definition, a multigraph G has mrK+(G) � d if and only if there is a vector representation

of G in Kd.

If G is a multigraph and H is a simple graph constructed from G by either removing all edges

or removing all but one edge from each pair of vertices i and j connected by multiple edges, then

PK(H) ⊂ PK(G) and mrK+(G) � mrK+(H). Further we have the following observation which reduces

the multigraph case to the simple graph case.

Observation 4.1. If G is a multigraph, then

mrK+(G) = min{mrK+(H) : H simple and PK(H) ⊂ PK(G)}.4.2. Orthogonal vertex removal

If v ∈ V(G), the orthogonal vertex removal of v from G, denoted G� v, is a multigraphmodified from

G[V(G) − {v}] by adding P(u,w) additional edges between each pair u,w ∈ N(v), where P(u,w) is

the product of the number of edges from v to u and from v to w.

The definition of orthogonal removal has the following motivation: Let �V = (�v1, . . . , �vn) be a

vector representation of a graph G. Suppose �vn is nonzero and let U = span{�vn} and P : Km → U⊥ be

the orthogonal projection onto U⊥. Construct the vector representation �V � �vn from �V by removing

�vn and for j < n replacing each �vj with the vector

P�vj = �vj − 〈�vj, �vn〉〈�vn, �vn〉�vn.



As �vn �= 0, rank �V � �vn = rank �V − 1 and �V � �vn is actually a vector representation of G � vn. This

proves the following observation.

Observation 4.2. LetG be a graph and v ∈ V(G) such that there existsw ∈ N(v)with v andw adjacent

by a single edge. Then

mrK+(G � v) + 1 � mrK+(G).

Remark 4.3. The existence of such aw forces v to be represented by a nonzero vector in every vector

representation of G. If no such w exists there is no guarantee that there exists a vector representation

of G with minimal rank and such that v is represented by a nonzero vector.

Under certain conditions the inequality in Observation 4.2 is actually an equality. For example we

have the following theorem.

Theorem 4.4 [8, Proposition 4.2]. Let G be a connected simple graph and v ∈ V(G) such that G[N(v)] isa star forest. Then mrK+(G � v) + 1 = mrK+(G).

4.3. Comparing G � v to D (G,N(v); G − N[v], v)We now relate orthogonal removal to our sets D (G, S;H1,H2) in the case in which S = N(v). The

following observation is immediate from the definitions.

Observation 4.5. Let G be a simple graph. If v ∈ G, � = |N(v)|, and H is a simple graph then the

following are equivalent:

(1) (H, K�+1) ∈ D (G,N(v); G − N[v], v).(2) PK(H) ⊂ PK(G � v).

Motivated by Observation 4.1, one can think of the multigraph G � v as a set [G � v] consisting of

all simple graphs H with PK(H) ⊂ PK(G � v). Then using Observation 4.5, we see that this set is also

given by the following:

[G � v] = {H : (H, K�+1) ∈ D (G,N(v); G − N[v], v)}.This implies that

mrK+(G � v) + 1 = min{mrK+(H) + mrK+(K�+1) : (H, K�+1) ∈ D (G,N(v); G − N[v], v)}and that

mrK+(D (G,N(v); G − N[v], v)) � mrK+(G � v) + 1.

The next proposition shows that we actually have equality above and so the pairs

(G1, G2) ∈ D (G,N(v); G − N[v], v)) with G2 �= K�+1 can be ignored when computing

mrK+(D (G,N(v); G − N[v], v)).Proposition 4.6. If G is a connected simple graph and v ∈ V(G) then

mrK+(D (G,N(v); G − N[v], v)) = mrK+(G � v) + 1.

Proof. By the above discussion it is enough to prove the “�” direction. Let (G1, G2) ∈D (G,N(v); G − N[v], v) be such that



mrK+(G1) + mrK+(G2) = mrK+(D (G,N(v); G − N[v], v)).Let � = |N(v)|. If G2 = K�+1 then Observation 4.5 implies that PK(G1) ⊂ PK(G � v) and thus

mrK+(G1) + mrK+(G2) � mrK+(G � v) + 1.

So suppose G2 �= K�+1. Then, as G2 is connected, mrK+(G2) � 2. Now let G′1 be the superposition of

G1 with K� identified at the vertices of N(v). Then G′1 is a simple graph,

mrK+(G′1) � mrK+(G1) + 1,

and PK(G′1) ⊂ PK(G � v). Hence

mrK+(G1) + mrK+(G2) � mrK+(G1) + 2 � mrK+(G′1) + 1 � mrK+(G � v) + 1. �

5. Cutsets of size 3

In this section we only consider simple graphs. If S is a cutset of G and |S| � 2 then G[S] is always

a star forest so we are able to compute mrK+(G) in terms of the rank of smaller graphs. When |S| > 2

we have no guarantee that mrK+(D (G, S;H1,H2)) = mrK+(G); however in the case in which |S| = 3

we will see a similar, but more complicated result is possible. In this section we specialize to the real

case. We will begin by generalizing Theorem 2.1, but first we need some notation.

Let G be a simple graph with S = {vα, vβ, vγ } ⊂ V(G) and G[vα, vβ, vγ ] = K3. Define

PR(G; S; −) = {A = (aij) ∈ PR(G) : aαβaβγ aγα < 0}and

PR(G; S; +) = {A = (aij) ∈ PR(G) : aαβaβγ aγα > 0}.The K3-constrained minimum semidefinite rank parameters of G are defined to be

mrR+(G; S; −) = min{rank A : A ∈ PR(G; S; −)}and

mrR+(G; S; +) = min{rank A : A ∈ PR(G; S; +)}.

Example5.1. LetK3 be the complete graphon three verticeswithV(K3) = S. ThenmrR+(K3; S; +) = 1

and mrR+(K3; S; −) = 2.

Observation 5.2. Let G be a simple graph with S = {vα, vβ, vγ } ⊂ V(G) and G[vα, vβ, vγ ] = K3.

Then

mrR+(G) = min{mrR+(G; S; −),mrR+(G; S; +)}� max{mrR+(G; S; +),mrR+(G; S; −)} � mrR+(G) + 1.

Proof. We will only prove the last inequality. By relabeling, suppose S = {v1, v2, v3} and �V =(�v1, . . . , �vn) is a vector representation of G. Then by picking α1, α2 ∈ R correctly, the vectors

�w1 = �v1 ⊕ (α1), �w2 = �v2 ⊕ (α2), �w3 = �v3 ⊕ (0), . . . , �wn = �vn ⊕ (0)

represent G and 〈�w1, �w2〉〈�w2, �w3〉〈�w3, �w1〉 has the desired sign. �



The motivation for these parameters is contained in the following proposition.

Proposition 5.3. Let G be a simple graph modified from a superposition of simple graphs G1 and G2 by the

removal of the edges of a copy of K3 common to G1 andG2. If this copy of K3 is induced by S = {s1, s2, s3} ⊂V(G1) ∩ V(G2) then

mrR+(G) � min{mrR+(G1; S; −) + mrR+(G2; S; +),mrR+(G1; S; +) + mrR+(G2; S; −)}.

Remark 5.4. With the notation of Proposition 5.3, Example 5.8 implies that mrK+(G) will in general

not be bounded by

min{mrR+(G1; S; +) + mrR+(G2; S; +),mrR+(G1; S; −) + mrR+(G2; S; −)}.Remark 5.5. Wewill not provide the details but the argument below also proves the following: let G

be modified from a superposition of G1 and G2 by the removal of the edges of a copy of K3 common

to G1 and G2. Then mr(G) � mr(G1) + mr(G2). The reason we can ignore signs is that the set S(G) isclosed under taking additive inverses: if A ∈ S(G) then −A ∈ S(G). The same is not true for PR(G).

Proof. As adding isolated vertices does not change the constrained minimum semidefinite rank, we

may assume that G1 and G2 have the same vertex set. We will first show that

mrR+(G) � mrR+(G1; S; −) + mrR+(G2; S; +).

Let �V = (�v1, . . . , �vn) and �W = (�w1, . . . , �wn) be vector representations of G1 and G2 in Rm1 and Rm2

respectively with (�vi)3i=1 and (�wi)3i=1 corresponding to the vertices of S. Suppose

〈�v1, �v2〉〈�v2, �v3〉〈�v3, �v1〉 < 0

and

〈�w1, �w2〉〈�w2, �w3〉〈�w3, �w1〉 > 0.

We claim that there is a vector representation of G of the form

(�v1 ⊕ ε1 �w1, . . . , �vn ⊕ εn �wn)

for some ε1, ε2, . . . , εn ∈ R. If we set

ε1 =√√√√− 〈�v1, �v2〉

〈�w1, �w2〉〈�w2, �w3〉〈�v2, �v3〉

〈�v3, �v1〉〈�w3, �w1〉 , ε2 = − 〈�v1, �v2〉

〈�w1, �w2〉ε−11 , and ε3 = − 〈�v3, �v1〉

〈�w3, �w1〉ε−11

then for all distinct i, j ∈ {1, 2, 3} we have

〈�vi ⊕ εi �wi, �vj ⊕ εj �wj〉 = 0.

Now choose ε4, . . . , εn such that no additional edge cancellation occurs. Thus

mrK+(G) � m1 + m2

and minimizing over all such vector representations of G1 and G2 shows that

mrR+(G) � mrR+(G1; S; −) + mrR+(G2; S; +).

The other bound follows from relabeling G1 and G2. �



We can now prove the following result.

Theorem 5.6. Let G be a connected simple graph with cut set S = {s1, s2, s3} such that G[S] = 3K1 and

G[V − S] = H1 � H2. Then

mrR+(G) = min{mrR+(Dstar (G, S;H1,H2)),mrR+(GK1 ; S; −) + mrR+(GK

2 ; S; +),

mrR+(GK1 ; S; +) + mrR+(GK

2 ; S; −)}where the set Dstar (G, S;H1,H2) consists of all decompositions (G1, G2) ∈ D (G, S;H1,H2) such that

G1[S] and G2[S] are not isomorphic to K3 and GK1 and GK

2 are the (unique) graphs with (GK1 , GK

2 ) ∈D (G, S;H1,H2) and GK

1 [S] = GK2 [S] = K3.

Proof. By Theorem 2.1 and Proposition 5.3 we have the “�” direction.

Following the notation of Lemma 3.2, let mrR+(G) = m and

�V = (�ui)Li=1 ∪ (�si)3i=1 ∪ (�wi)Mi=1

be a vector representation of G in Rm with the �ui corresponding to the vertices of H1 and the �wi

corresponding to the vertices of H2. Again let P : Rm → U be the orthogonal projection onto the

subspace U = span{�ui}Li=1. Let G1 be the graph corresponding to the vectors (�ui)Li=1 ∪ (P�si)3i=1 and G2

be the graph corresponding to the vectors ((I − P)�si)3i=1 ∪ (�wi)Mi=1. Then as in Lemma 3.2,

mrR+(G1) + mrR+(G2) � dim(U) + dim(U⊥) = mrR+(G)

and (G1, G2) ∈ D (G, S;H1,H2). As G[S] = 3K1, we have

G1[S] = G2[S] ∈ {K3, P3, K2 � K1, 3K1}.Case 1. Assume G1[S], G2[S] ∈ {P3, K2 � K1, 3K1}. Then (G1, G2) ∈ Dstar (G, S;H1,H2) and

mrR+(G) � mrR+(G1) + mrR+(G2) � mrR+(Dstar (G,H; S1, S2)).Case 2. Assume that G1[S] = G2[S] = K3. Then by definition (G1, G2) = (GK

1 , GK2 ) and for 1 � i, j � 3

the inner products

〈P�si, P�sj〉 and 〈(I − P)�si, (I − P)�sj〉are nonzero. Further as G[S] = 3K1,

0 = 〈�si,�sj〉 = 〈P�si, P�sj〉 + 〈(I − P)�si, (I − P)�sj〉and the two inner products in the expression above have opposite signs. Thus the products

〈P�s1, P�s2〉〈P�s2, P�s3〉〈P�s3, P�s1〉and

〈(I − P)�s1, (I − P)�s2〉〈(I − P)�s2, (I − P)�s3〉〈(I − P)�s3, (I − P)�s1〉have opposite signs. This implies that

mrR+(G) � dim(U) + dim(U⊥)

� min{mrR+(GK1 ; S; −) + mrR+(GK

2 ; S; +),mrR+(GK1 ; S; +) + mrR+(GK

2 ; S; −)}. �



We also generalize Theorem 4.4.

Proposition 5.7. Let G be a simple graph. Suppose v is a vertex in G with N(v) = {s1, s2, s3} and

G[N(v)] = 3K1. Then

mrR+(G) = mrR+(G � v;NG(v); −) + 1 � mrR+(G � v) + 2.

Proof. The inequality follows from Observation 5.2. One can deduce the equality from Theorem 5.6.

However, we will provide a direct argument. As G[N(v)] = 3K1, G � v is a simple graph. Also G is

modified from a superposition of G � v and K4 by the removal of the edges of a copy of K3 common to

both. Further mrR+(K4;NG(v); +) = 1 and so

mrR+(G) � mrR+(G � v;NG(v); −) + mrR+(K4;NG(v); +)

= mrR+(G � v;NG(v); −) + 1.

For the reverse inequality, start with a vector representation �V of G in Rm. Let �v be the vector repre-

senting v. Let U = span{�v} and let P : Rm → U⊥ be the orthogonal projection onto the orthogonal

complement of U. Let �V � �v be the list of vectors constructed from �V be removing �v and replacing

every other vector �wwith P �w. As in Section 4, �V ��v is a vector representation of G� v. We now claim

that the Gram matrix corresponding to �V � �v is in the set

PR(G � v;NG(v); −).

To see this, first suppose that�s1,�s2,�s3 represent s1, s2, s3. Next let αi ∈ R be such that�si − P�si = αi�v.Notice that

αi ‖�v‖2 = 〈�si, �v〉 �= 0

since �V is a vector representation of G. Then for i �= j

0 = 〈P�si, P�sj〉 + 〈(I − P)�si, (I − P)�sj〉 = 〈P�si, P�sj〉 + αiαj ‖�v‖2 .

This implies that

〈P �s1, P�s2〉〈P �s2, P�s3〉〈P �s3, P�s1〉 = −(α1α2α3)2 ‖�v‖6 < 0

and that Gram matrix corresponding to �V � �v is in the set

PR(G � v;NG(v); −).

Hence

mrR+(G � v;NG(v); −) � rank �V � �v = rank �V − 1.

Nowminimizing over all vector representations of G yields the desired inequality. �

Example 5.8. Let G be the graph in Fig. 7. Using results in the literature one can show

|G| − 5 = 6 � mrK+(G) � 8 = |G| − 3.

Here the lower bound comes from the fact that the positive semidefinite zero forcing number is five.

The upper bound comes fromvanderHolst’s classification of graphswithmrK+(G) � |G|−2 [11],more

specifically van der Holst’s classification implies that any graph with mrK+(G) � |G| − 2 must have a

vertex with degree less than or equal to two (see Remark 3.7). However G has a cutset S = {s1, s2, s3}



Fig. 7. The graph in Example 5.8.

and G[S] = 3K1 so we may apply Theorem 5.6 to compute mrR+(G) in terms of rank parameters

of smaller graphs. In Appendix B, we compute these parameters and show that mrR+(G) = 7. In

Appendix A, we will prove that the positive semidefinite zero forcing number is five, thus providing

an example with

|G| − Z+(G) = 6 < 7 = mrR+(G).

It is expected that the inequality |G| − Z+(G) � mrR+(G) is strict for many graphs, but there are few

examples known.

Acknowledgements

We would like to thank the referees for their comments. This research was supported by NSF-

REU grant #05-52594 and NSF-LURE grant #06-36528 and was done when J. Beagley and A. Zimmer

participated in an NSF-REU program and E. Radzwion, S. Rimer, R. Tomasino, and J. Wolfe participated

in an NSF-LURE program at Central Michigan University during the summer of 2007.

AppendixA. Positive semidefinite zero forcing number calculations

Let Z+(G) denote the positive semidefinite zero forcing number of a simple graph G. This is a combi-

natorial graph parameter introduced by Barioli et al. [1] such that

|G| − Z+(G) � mrK+(G).

Delaying definitions for the moment, the purpose of this section is to prove the following theorem.

Theorem A.1. Let G be the graph in Fig. 7. If Z+(G) is the positive semidefinite zero forcing number of G

then Z+(G) = 5.

The positive semidefinite zero forcing number has a “dual” interpretation called the OS-number of

a graph. In the proof of Theorem A.1 wewill use this interpretation. This parameter was introduced by

Hackey et al. and predates the positive semidefinite zero forcing number.

Definition A.2 [8, Definition 3.1]. Let G be a simple graph. A subset O of V(G) is called an OS-set if it

can be ordered v1, . . . , vm such that for each 1 � � � m there exists aw� ∈ V(G) \ {v1, . . . , v�} suchthat the only simple path from v� to w� in the induced graph G[v1, . . . , v�,w�] is the edge v�w�. The

OS-number of a graph G, denoted by OS(G), is the size of the largest OS-set.



Fig. 8. The Möbius ladder,ML8, presented in two ways.

In the definition above, w1, . . . ,wm are called OS-neighbors of the OS-set v1, . . . , vm. In general

there will be many choices for OS-neighbors. Hackney et al. then proved the following.

Theorem A.3 [8, Proposition 3.3]. If G is a simple graph, then OS(G) � mrK+(G).

The OS-number is dual to the positive semidefinite zero forcing number in the following sense.

Theorem A.4 [1, Theorem 3.6]. Let G be a simple graph, then |G| − Z+(G) = OS(G).

Hackney et al. conjectured that OS(G) = mrK+(G) for all graphs, but a counterexample on eight

verticeswas discovered byMitchell et al. [9]. Their counterexample is the so-calledMöbius ladder given

in Fig. 8 and denoted byML8. Mitchell et al. proved the following.

Proposition A.5 [9, Theorem 3.1]. Let ML8 be the graph in Fig. 8, then

OS(ML8) = 4 < 5 = mrK+(ML8).

Mitchell et al. constructed additional graphs for which OS(G) < mrK+(G) by considering vertex

sums and graph joins of the Möbius ladder. To prove Theorem A.1 we will follow essentially the same

argument used in [9] to establish that OS(ML8) = 4. In particular, we will use the following.

Theorem A.6 [9, Proposition 2.18]. Let G be a connected simple graph, then there exists an OS-Set O of

maximal size such that G[O] is connected.The next two observations follow directly from the definition of an OS-set.

Observation A.7. If O is an OS-set of a graph G and O′ ⊂ O then O′ is an OS-set of G.

Observation A.8. Suppose G is a simple graph. Let O = {v1, . . . , vk} be an OS-set of G with OS-

neighbors w1, . . . ,wk . If G[O] is connected, then N[wk] ∩ O = {vk}.We next explicitly state the converse of Observation A.8, but we will need a little bit of notation:

for a graph G and a subset U ⊂ V(G) the closed neighborhood of U, denoted by N[U] is all vertices inG either in U or adjacent to at least one vertex in U.



Observation A.9. Suppose G is a simple graph and U ⊂ V(G). If G[U] is connected and every vertex

w ∈ N[U] \ U is adjacent to at least two vertices in U, then U is not an OS-set.

We next apply the previous three observations to the graph in Fig. 7.

Observation A.10. Suppose G is the graph in Fig. 7. If O is an OS-set of G, then O does not contain all

of {b1, b2, b3, b4}.Proof. By Observation A.9, the set {b1, b2, b3, b4} is not an OS-set. Then by Observation A.7,O cannot

contain {b1, b2, b3, b4}. �

We are now ready to prove the theorem.

Proof of Theorem A.1. By Theorem A.4 it is enough to show that OS(G) = 6. Notice that a1, a2, a3,b1, b2, b3 is an OS-set, hence OS(G) � 6. Suppose for a contradiction that OS(G) � 7 and O ={v1, . . . , v7} is an OS-set of G with OS-neighbors w1, . . . ,w7. By Theorem A.6 we may suppose G[O]is connected and thus N[w7] ∩ O = {v7}. Exploiting the symmetries of G it is enough to consider the

cases in which w7 ∈ {s1, s2, a1, a3}.

Case 1: w7 = s1. As N[w7] ∩ {v1, . . . , v6} = ∅ we must have

{v1, . . . , v6} = {a2, a4, s2, s3, b2, b4}.By Observation A.9 this is not an OS-set. This contradicts the fact that O is an OS-set.

Case 2: w7 = s2. As N[w7] ∩ {v1, . . . , v6} = ∅ we must have

{v1, . . . , v6} = {a1, a2, s1, s3, b1, b2}.By Observation A.9 this is not an OS-set. This contradicts the fact that O is an OS-set.

Case 3: w7 = a1. As N[w7] ∩ {v1, . . . , v6} = ∅ we must have

{v1, . . . , v6} ⊂ {a4, s2, s3, b1, b2, b3, b4}.ByObservationA.10, at least oneofb1, b2, b3, b4 is not inO. But for 1 � i � 4ObservationA.9

implies that the set

{a4, s2, s3, b1, b2, b3, b4} \ {bi}is not an OS-set. This contradicts the fact that O is an OS-set.

Case 4: w7 = a3. As N[w7] ∩ {v1, . . . , v6} = ∅ we must have

{v1, . . . , v6} ⊂ {a1, a4, s3, b1, b2, b3, b4}.By Observation A.10, at least one of b1, b2, b3, b4 is not in O. For i �= 3 Observation A.9

implies that the set

{a1, a4, s3, b1, b2, b3, b4} \ {bi}is not an OS-set. So we must have

{v1, . . . , v6} = {a1, a4, s3, b1, b2, b4}.Then G[v1, . . . , v6] is connected and so

|N(w6) ∩ {v1, . . . , v6}| = |{v6}| = 1.



This can only happen if w6 = b3 and v6 = b2. Then Observation A.9 implies that

{v1, . . . , v5} = {a1, a4, s3, b1, b4}is not an OS-set. This contradicts the fact that O is an OS-set. �

AppendixB. Some rank computations

B.1. Computations for Example 2.7

Lemma B.1. Let G be as in Example 2.7. Then mrK+(G) = 6.

Proof. This follows from van der Holst’s [11] characterization of graphs with mrK+(G) � |G| − 2 or

repeated applications of Theorem 4.4. �

Lemma B.2. Let G1 be as in Example 2.7. Then mrK+(G1) = 4.

Proof. As {v1, v2, v3, v4} is an OS-set of G1, mrK+(G1) � 4. Further

⎛⎝A I

I A−1

⎞⎠ ∈ PK(G1) if A =

⎛⎜⎜⎜⎜⎜⎜⎝5 1 0 1

1 5 1 0

0 1 5 1

1 0 1 5

⎞⎟⎟⎟⎟⎟⎟⎠implying that mrK+(G1) � 4. �


Thegraphsweneed to consider canbe constructed from thegraph in Fig. 9 by adding edges between

the vertices labeled s1, s2, s3.

Lemma B.3. Let H be the graph in Fig. 9. Then

mrK+(H) = 4,

mrK+(H ∪ {s1s2}) = mrK+(H ∪ {s2s3}) = 4.

Fig. 9. The graph in Appendix B presented two ways.



mrK+(H ∪ {s1s3}) = 4

mrK+(H ∪ {s1s2, s2s3}) = 4

mrK+(H ∪ {s1s2, s1s3}) = mrK+(H ∪ {s2s3, s1s3}) = 4

mrK+(H ∪ {s1s2, s1s3, s2s3}) = 3

Proof. In each case the minimum semidefinite rank can be computed using repeated applications of

Theorem 4.4. �

Lemma B.4. Let G be the graph in Example 3.6. Then mrK+(G) = 6.

Proof. By Theorem 3.3,

mrK+(G) = min{mrK+(G1) + mrK+(G2); (G1, G2) ∈ D (G, S;H1,H2)}.If (G1, G2) is such a pair, then G1 and G2 appear in the list of graphs in Lemma B.3. In particular, the

best we can do is for G1 = G2 = H ∪ {s1s2, s1s3, s2s3}. �


Using Lemma B.3 the only graph we need to consider can be constructed from the graph in Fig. 9

by adding the edges {s1s2, s1s3, s2s3}.Lemma B.5. Let H be the graph in Fig. 9. If G = H ∪ {s1s2, s1s3, s2s3} and S = {s1, s2, s3}, thenmrR+(G; S; −) = 4.

Proof. Let ML8 be the graph in Fig. 8. By Proposition A.5, mrK+(ML8) = 5. Let v be a vertex in ML8.

Then, as G[N(v)] = 3K1, Proposition 5.7 implies that

5 = mrR+(ML8) = mrR+(ML8 � v;N(v); −) + 1.

The secondpresentation in Figs. 8 and9 implies the existenceof a graph isomorphismφ : G → ML8�v

such that S ⊂ V(G) is mapped to N(v) ⊂ V(ML8). Then

4 = mrR+(ML8 � v;N(v); −) = mrR+(G; S; −). �

Lemma B.6. Let G be the graph in Example 5.8, then mrR+(G) = 7.

Proof. Let S = {s1, s2, s3}. Then by Theorem 5.6,

mrR+(G) = min{mrR+(Dstar (G, S;H1,H2)),

mrR+(GK1 ; S; −) + mrR+(GK

2 ; S; +), mrR+(GK1 ; S; +) + mrR+(GK

2 ; S; −)}.Suppose (G1, G2) ∈ Dstar (G, S;H1,H2) then by Lemma B.3, mrR+(G1) = mrR+(G2) = 4 and so

8 = mrR+(Dstar (G, S;H1,H2)).

Now if H is the graph in Fig. 9 then GK1 = GK

2 = H ∪ {s1s2, s1s3, s2s3}. Then by Lemma B.5,

mrR+(GK1 ; S; −) = 4 and by Lemma B.3, mrR+(GK

1 ) = 3. Thus mrR+(GK1 ; S; +) = 3 and



7 = mrR+(GK1 ; S; −) + mrR+(GK

2 ; S; +) = mrR+(GK1 ; S; +) + mrR+(GK

2 ; S; −).

In conclusion, mrR+(G) = 7. �

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