The inverse inertia problem for graphs
aa r X i v : . [ m a t h . C O ] N ov The Inverse Inertia Problem for Graphs
Wayne Barrett ∗† , H. Tracy Hall ‡† , Raphael Loewy §
17 November 2007
Abstract
Let G be an undirected graph on n vertices and let S ( G ) be theset of all real symmetric n × n matrices whose nonzero off-diagonalentries occur in exactly the positions corresponding to the edges of G .The inverse inertia problem for G asks which inertias can be attainedby a matrix in S ( G ). We give a complete answer to this questionfor trees in terms of a new family of graph parameters, the maximaldisconnection numbers of a graph. We also give a formula for theinertia set of a graph with a cut vertex in terms of inertia sets ofproper subgraphs. Finally, we give an example of a graph that is notinertia-balanced, and investigate restrictions on the inertia set of anygraph. Keywords: combinatorial matrix theory, graph, Hermitian, inertia, minimumrank, symmetric, tree
AMS classification: ∗ Wayne Barrett conducted this research as a Fulbright grantee during Winter semester2007 at the Technion. The support of the US – Israel Educational Foundation and of theDepartment of Mathematics of the Technion is gratefully acknowledged. † Department of Mathematics, Brigham Young University, Provo, Utah 84602, UnitedStates ‡ Research conducted while a guest of the Department of Mathematics at the Technion,whose hospitality is appreciated. § Department of Mathematics, Technion – Israel Institute of Technology, Haifa 32000,Israel
Introduction
Given a simple undirected graph G = ( V, E ) with vertex set V = { , . . . , n } ,let S ( G ) be the set of all real symmetric n × n matrices A = (cid:2) a ij (cid:3) such thatfor i = j , a ij = 0 if and only if ij ∈ E . There is no condition on the diagonalentries of A .The set H ( G ) is defined in the same way over Hermitian n × n matri-ces, and every problem we consider comes in two flavors: the real version,involving S ( G ), and the complex version, involving H ( G ). There are knownexamples where a question of the sort we examine here has a different answerwhen considered over Hermitian matrices rather than over real symmetricmatrices [BvdHL1], [Hall], but for each question that is completely resolvedin the present paper, the answer over H ( G ) proves to be the same as thatobtained over S ( G ).The inverse eigenvalue problem for graphs asks: Given a graph G on n vertices and prescribed real numbers λ , λ , . . . , λ n , is there some A ∈ S ( G )(or A ∈ H ( G ), alternatively) such that the eigenvalues of A are exactlythe numbers prescribed? In general, this is a very difficult problem. Somecontributions to its solution appear in [DJ], [JD2], [JS2], [JDS].A more modest goal is to determine the maximum multiplicity M( G )of an eigenvalue of a matrix in S ( G ). This is easily seen to be equiva-lent to determining the minimum rank mr( G ) of a matrix in S ( G ) sincemr( G ) + M( G ) = n . This problem has been intensively studied. Some of themajor contributions appear in the papers [F], [N], [JD1], [Hs], [JS1], [vdH],[BFH1], [BvdHL1], [BvdHL2], [BFH2], [BFH3], [BF], [BGL], [JLS], [Hall]. Avariant of this problem is the study of mr + ( G ), the minimum rank of all pos-itive semidefinite A ∈ S ( G ). The Hermitian maximum multiplicity hM( G ),Hermitian minimum rank hmr( G ), and Hermitian positive semidefinite rankhmr + ( G ) are defined analagously.A problem whose level of difficulty lies between the inverse eigenvalueproblem and the minimum rank problem for graphs is the inverse inertia1roblem, which we now explain. Definition 1.1.
Given a Hermitian n × n matrix A , the inertia of A is thetriple (cid:0) π ( A ) , ν ( A ) , δ ( A ) (cid:1) , where π ( A ) is the number of positive eigenvalues of A , ν ( A ) is the numberof negative eigenvalues of A , and δ ( A ) is the multiplicity of the eigenvalue 0of A . Then π ( A ) + ν ( A ) + δ ( A ) = n and π ( A ) + ν ( A ) = rank A .If the order of A is also known then the third number of the triple issuperfluous. The following definition discards δ ( A ). Definition 1.2.
Given a Hermitian matrix A , the partial inertia of A is theordered pair (cid:0) π ( A ) , ν ( A ) (cid:1) . We denote the partial inertia of A by pin( A ).We are interested in the following problem: Question 1 (Inverse Inertia Problem) . Given a graph G on n vertices, forwhich ordered pairs ( r, s ) of nonnegative integers with r + s ≤ n is there amatrix A ∈ S ( G ) such that pin( A ) = ( r, s )?The Hermitian Inverse Inertia Problem is the same question with H ( G ) inthe place of S ( G ). It is well known [JD2, p. 8] that in the case of a tree T mostquestions over H ( T ) are equivalent to questions over S ( T ), and in particularif F is a forest and A ∈ H ( F ), then there exists a diagonal matrix D withdiagonal entries from the unit circle such that DAD − = DAD ∗ ∈ S ( F ).In those sections concerned with the inverse inertia problem for trees andforests we thus assume without loss of generality that every matrix in H ( F )is in fact in S ( F ).In this paper we give a complete solution to the inverse inertia problemfor trees and forests. The statement of our solution is a converse to an easierpair of lemmas that apply not just to forests but to any graph.2 emma 1.1 (Northeast Lemma) . Let G be a graph and suppose that A ∈H ( G ) with pin( A ) = ( π, ν ) . Then for every pair of integers r ≥ π and s ≥ ν satisfying r + s ≤ n , there exists a matrix B ∈ H ( G ) with pin( B ) = ( r, s ) . Ifin addition A is real, then B can be taken to be real. In other words, thinking of partial inertias or Hermitian partial inertias aspoints in the Cartesian plane, the existence of a partial inertia ( π, ν ) impliesthe existence of every partial inertia ( r, s ) anywhere “northeast” of ( π, ν ),as long as r + s does not exceed n . We prove this lemma in Section 2 byperturbing the diagonal entries of A .To state the second lemma we need to introduce an indexed family ofgraph parameters. Definition 1.3.
Let G be a graph with n vertices. For any k ∈ { , . . . , n } we define MD k ( G ), the maximal disconnection of G by k vertices, as themaximum, over all induced subgraphs F of G having n − k vertices, of thenumber of components of F .For example, MD ( G ) is the number of components of G , and if T is a treethen MD ( T ) is the maximum vertex degree of T . Since an induced subgraphcannot have more components than vertices, we always have k +MD k ( G ) ≤ n . Remark.
As far as we can determine, MD k ( G ) is not a known family ofgraph parameters. It is, however, related to the toughness t of a graph,which can be defined [C] as t ( G ) = min (cid:26) k MD k ( G ) : MD k ( G ) ≥ (cid:27) . For a recent survey of results related to toughness of graphs, see [BBS]. Thereis also some relation between MD k ( G ) and vertex connectivity: a graph G on n vertices is k -connected, k < n , if and only if MD j ( G ) = 1 whenever0 ≤ j < k . 3 emma 1.2 (Stars and Stripes Lemma) . Let G be a graph with n vertices,let k ∈ { , . . . , n } be such that MD k ( G ) ≥ k , and choose any pair of integers r and s such that r ≥ k , s ≥ k , and r + s = n − MD k ( G ) + k . Then thereexists a matrix A ∈ S ( G ) such that pin( A ) = ( r, s ) . This lemma is proved in Section 2; the idea of the proof is that eachpartial inertia in the diagonal “stripe” from ( n − MD k ( G ) , k ) northwest to( k, n − MD k ( G )) can be obtained by combining the adjacency matrices of“stars” at each of the k disconnection vertices together with, for each of theremaining components, a matrix of co-rank 1 and otherwise arbitrary inertia.These two lemmas provide a partial solution to the Inverse Inertia Prob-lem for any graph. Our main result for trees and forests is that for suchgraphs, and exactly such graphs, the partial solution is complete. Definition 1.4.
Let G be a graph on n vertices. Then ( r, s ) is an elementaryinertia of G if for some integer k in the range 0 ≤ k ≤ n we have k ≤ r , k ≤ s , and n − MD k ( G ) + k ≤ r + s ≤ n .The elementary inertias of a graph G are exactly those partial inertiasthat can be obtained from G by first applying the Stars and Stripes Lemmaand then applying the Northeast Lemma. The partial solution given by theselemmas is the following: if ( r, s ) is an elementary inertia of a graph G , thenthere exists a matrix A ∈ S ( G ) with pin( A ) = ( r, s ). This is proved asObservation 2.4 in Section 2. Theorem 1.1.
The Stars and Stripes Lemma and the Northeast Lemmacharacterize the partial inertias of exactly forests, as follows:1. Let F be a forest, and let A ∈ S ( F ) with pin( A ) = ( r, s ) . Then ( r, s ) is an elementary inertia of F .2. Conversely, let G be a graph and suppose that for every A ∈ S ( G ) , pin( A ) is an elementary inertia of G . Then G is a forest.
4f course Claim 1 also applies for A ∈ H ( F ), since for F a forest anymatrix in H ( F ) is diagonally congruent to a matrix in S ( F ) having the samepartial inertia. Claim 2 of Theorem 1.1 is a corollary to known results, herecalled Theorem 2.4. We prove Claim 1 of Theorem 1.1 at the end of Section 5.In Section 4 we show that determining the set of possible inertias of anygraph with a cut vertex can be reduced to the problem of determining thepossible inertias of graphs on a smaller number of vertices. The formulawe obtain is a generalization of the known formula for the minimum rankof a graph with a cut vertex. In Section 5 we describe elementary inertiasin terms of certain edge-colorings of subgraphs, and we show that the samecut-vertex formula proven in Section 4 for inertias also holds when appliedto the (usually smaller) set of elementary inertias. Applying these parallelformulas inductively to trees and forests then gives us a proof of Claim 1 ofTheorem 1.1. In Section 6 we outline an effective procedure for calculatingthe set of partial inertias of any tree, using the results of Section 3 to justifysome simplications, and we calculate a few examples. In Section 7 we againconsider more general graphs, and demonstrate both an infinite family offorbidden inertia patterns, and the first example of a graph that is not inertia-balanced. The concept of an inertia-balanced graph was introduced in [BF],and determining whether a graph is inertia-balanced is a special case of theinverse inertia problem. Definition 1.5.
A Hermitian matrix A is inertia-balanced if | π ( A ) − ν ( A ) | ≤ . A graph G is inertia-balanced if there is an inertia-balanced A ∈ S ( G ) withrank A = mr( G ). A graph G is Hermitian inertia-balanced if there is aninertia-balanced A ∈ H ( G ) with rank A = hmr( G ). Remark.
Our formulation, unlike the definition in [BF], is symmetric inallowing ν ( A ) = π ( A ) + 1. This doubles the set of inertia-balanced matrices5f odd rank, but the two definitions are equivalent when applied to graphssince A ∈ S ( G ) if and only if − A ∈ S ( G ).Barioli and Fallat [BF] proved that every tree is inertia-balanced. Theo-rem 1.1, once proved, will imply a slightly stronger result. The intuition forexpecting a graph to be inertia-balanced comes from many small examples inwhich achieving an eigenvalue of high multiplicity appears to become increas-ingly difficult as the imbalance increases between the number of eigenvaluesthat are higher and the number that are lower than the target multiple eigen-value. The behavior observed in these small examples can be stated formallyin terms of the following definitions. Definition 1.6.
A set S of ordered pairs of integers is called symmetric ifwhenever ( r, s ) ∈ S , then ( s, r ) ∈ S . A symmetric nonempty set S of orderedpairs of nonnegative integers is called a stripe if there is some integer m suchthat r + s = m for every ( r, s ) ∈ S , and we specify the particular constantsum by saying that S is a stripe of rank m . A stripe S is convex if theprojection { r : ( r, s ) ∈ S } is a set of consecutive integers. Example 1.1.
The set { (2 , , (2 , , (3 , , (3 , , (4 , } is symmetric, the set { (6 , , (3 , , (0 , } is a stripe, and the stripe { (4 , , (3 , , (2 , } is convex. Observation 1.3.
Given a graph G of order n and an integer m in the range mr( G ) ≤ m ≤ n , the set { pin( A ) : A ∈ S ( G ) and rank A = m } is a stripe of rank m . The same is true for A ∈ H ( G ) with m in the range hmr( G ) ≤ m ≤ n .Proof. Symmetry comes from the fact that − A ∈ S ( G ) if and only if A ∈S ( G ), and similarly for H ( G ). The sets are nonempty by the definitions ofmr( G ) and hmr( G ) and the Northeast Lemma.6 efinition 1.7. A graph G is inertia-convex on stripes or Hermitian inertia-convex on stripes if each of the stripes defined in Observation 1.3 (with A ∈ S ( G ) or A ∈ H ( G ), respectively) is convex.In other words, a graph is inertia-convex on stripes if each stripe of pos-sible partial inertias does not contain a gap. Corollary 1.2 (Corollary to Theorem 1.1) . Every forest is inertia-convexon stripes.Proof.
Let F be a forest. By Theorem 1.1, each of the stripes defined inObservation 1.3 is the set of elementary inertias of some fixed rank m . Foreach fixed k with MD k ( F ) ≥ k we obtain a set of elementary inertias whichis a union of convex stripes. It follows that for any fixed m , the set ofelementary inertias of rank m is the union of convex stripes of rank m as k varies over all allowed integers. Since a union of convex stripes of the samerank is a single convex stripe, each of the stripes defined in Observation 1.3is convex.It has been an outstanding question if there is any graph that is notinertia-balanced. At the AIM Workshop in Palo Alto in October 2006, theprevailing opinion was that such a graph does not exist [BHS].In Section 7 we give an example of a graph that is not inertia-balanced.First we show that every graph satisfies a condition that is much weaker thaninertia-balanced (except in the case of minimum semidefinite rank 2). Thecounterexample graph and new condition together allow us to completely de-termine which sets can occur as the complement of the set of possible partialinertias of a graph G with mr + ( G ) ≤
3. The possible excluded partial inertiasets giving minimum semidefinite rank 4 or greater remain unclassified.For the most part our notation for graphs follows Diestel [D]. We makeuse specifically of the following notation throughout: • All graphs are simple, and a graph is formally an ordered pair G =( V, E ) where V is a finite set and E consists of pairs from V . When7eferring to an individual edge, we abbreviate { u, v } to uv or vu . Thevertex set of a graph G is also referred to as V ( G ), and the edge set as E ( G ). • For S ⊆ V , G [ S ] is the subgraph of G induced by S and G − S is theinduced subgraph on V ( G ) \ S . We write G − F rather than G − V ( F )and G − v rather than G − { v } . • The number of vertices of a graph G is denoted | G | . • K n is the complete graph on n vertices. • S n = ( { , , , . . . , n } , { , , . . . , n } ) is called the star graph on n vertices. This is the same as the complete bipartite graph K ,n − . • P n is the path on n vertices. Paths are described explicitly by concate-nating the names of the vertices in order; for example, uvw denotes thegraph ( { u, v, w } , { uv, vw } ). • If v is a vertex of G , d ( v ) is the degree of v . • ∆( G ) = max { d ( v ) : v ∈ V ( G ) } .We conclude the introduction with some elementary facts about inertia,and include short proofs to keep the paper self-contained. Proposition 1.4.
Let A be a Hermitian n × n matrix and let B be a principalsubmatrix of A of size ( n − × ( n − . Then π ( A ) − ≤ π ( B ) ≤ π ( A ) and ν ( A ) − ≤ ν ( B ) ≤ ν ( A ) . Proof.
By the interlacing inequalities [B] λ ≥ µ ≥ λ ≥ µ ≥ · · · ≥ λ n − ≥ µ n − ≥ λ n , where λ , . . . , λ n are the eigenvalues of A and µ , . . . , µ n − are the eigenvaluesof B , arranged in decreasing order. If µ ≤ , π ( B ) = 0 ≤ π ( A ). Otherwise,8et m ∈ { , . . . , n − } be the largest integer with µ m >
0. Then λ m > π ( B ) = m ≤ π ( A ).If λ ≤ , π ( B ) = 0 > π ( A ) −
1. Otherwise, let ℓ ∈ { , . . . , n } be thelargest integer with λ ℓ >
0. Then µ ℓ − ≥ λ ℓ > π ( B ) ≥ ℓ − π ( A ) − ν ( A ) − ≤ ν ( B ) ≤ ν ( A ). Proposition 1.5 (Subadditivity) . Let A , B be Hermitian n × n matricesand let C = A + B . Then π ( C ) ≤ π ( A ) + π ( B ) and ν ( C ) ≤ ν ( A ) + ν ( B ) . Proof. If π ( A ) + π ( B ) ≥ n , the first inequality is true, so assume that π ( A ) + π ( B ) < n . Let i = π ( A ) and j = π ( B ). Then λ i +1 ( A ) ≤ λ j +1 ( B ) ≤ λ i + j +1 ( C ) = λ i +1+ j +1 − ( C ) ≤ λ i +1 ( A ) + λ j +1 ( B ) ≤ . Therefore π ( C ) ≤ i + j = π ( A ) + π ( B ).Similarly, ν ( C ) ≤ ν ( A ) + ν ( B ). Proposition 1.6.
Let A be a Hermitian n × n matrix and let cxx ∗ be aHermitian rank 1 matrix (so c is real-valued). Then π ( A + cxx ∗ ) ≤ (cid:26) π ( A ) + 1 if c > π ( A ) if c < and ν ( A + cxx ∗ ) ≤ (cid:26) ν ( A ) + 1 if c < ν ( A ) if c > Proof.
Let c >
0. Then π ( cxx ∗ ) = 1, ν ( cxx ∗ ) = 0. By Proposition 1.5, π ( A + cxx ∗ ) ≤ π ( A ) + π ( cxx ∗ ) = π ( A ) + 1 ,ν ( A + cxx ∗ ) ≤ ν ( A ) + ν ( cxx ∗ ) = ν ( A ) . The argument is similar if c <
0. 9
The inertia set of a graph
Definition 2.1.
Let N be the set of nonnegative integers, and let N = N × N .We define the following sets: N ≤ k = (cid:8) ( r, s ) ∈ N : r + s ≤ k (cid:9) , N ≥ k = (cid:8) ( r, s ) ∈ N : r + s ≥ k (cid:9) , N i,j ] = N ≥ i ∩ N ≤ j , N i = N i,i ] (the complete stripe of rank i ) . We note that a stripe of rank i is a nonempty symmetric subset of N i . Definition 2.2.
Given a graph G , we define I ( G ) = { ( r, s ) : pin( A ) = ( r, s ) for some A ∈ S ( G ) } , and h I ( G ) = { ( r, s ) : pin( A ) = ( r, s ) for some A ∈ H ( G ) } . We call I ( G ) the inertia set of G and h I ( G ) the Hermitian inertia set of G .Now suppose ( r, s ) ∈ I ( G ) and let A ∈ S ( G ) with pin( A ) = ( r, s ). Since r + s = rank A , we have mr( G ) ≤ r + s ≤ | G | . We record this as Observation 2.1.
Given a graph G on n vertices, I ( G ) ⊆ N G ) ,n ] and h I ( G ) ⊆ N G ) ,n ] . The fact that every real symmetric matrix is also Hermitian immediatelygives us:
Observation 2.2.
For any graph G , I ( G ) ⊆ h I ( G ) and hmr( G ) ≤ mr( G ) . The Northeast Lemma, as stated in the Introduction, substantially short-ens the calculation of the inertia set of a graph.10 roof of Northeast Lemma.
Let G be a graph and suppose that ( π, ν ) ∈ h I ( G ), and let ( r, s ) ∈ N ≤ n be given with r ≥ π and s ≥ ν . We wishto show that ( r, s ) ∈ h I ( G ). If in addition ( π, ν ) ∈ I ( G ), we must show that( r, s ) ∈ I ( G ).Let A ∈ H ( G ) with pin( A ) = ( π, ν ). If π + ν = n there is nothing toprove, so assume π + ν < n . It suffices to prove that there exists a B ∈ H ( G )with pin( B ) = ( π + 1 , ν ), because then an analogous argument can be givento prove that there is a C ∈ H ( G ) with pin( C ) = ( π, ν + 1) and these twofacts may be applied successively to reach ( r, s ). We also need to ensure thatwhen A is real symmetric B is also real symmetric. Choose ε > A + εI is invertible and ν ( A + εI ) = ν . Then π ( A + εI ) = n − ν . Let A = A and then perturb the diagonal entries in order: for any i ∈ { , . . . , n } let A i = A i − + εe i e ∗ i , so that A n = A + εI . Then A i ∈ H ( G ) for i = 0 , . . . , n and by Propositions 1.5 and 1.6, π ( A i − ) ≤ π ( A i ) ≤ π ( A i − ) + 1for i = 1 , . . . , n . It follows that every integer in { π, π + 1 , . . . , n − ν } is equalto π ( A i ) for some i ∈ { , . . . , n } . Since ν = ν ( A + εI ) ≤ ν ( A n − ) ≤ · · · ≤ ν ( A ) ≤ ν ( A ) ≤ ν ( A ) = ν by Proposition 1.6, ν ( A i ) = ν for i = 0 , , . . . , n . Then for some i we havepin( A i ) = ( π +1 , ν ), and we can take B = A i . As desired, B is real symmetricif A is real symmetric, which completes the S ( G ) version of the NortheastLemma as well as the H ( G ) version: Within either one of the two inertia sets I ( G ) or h I ( G ), the existence of a partial inertia ( π, ν ) implies the existenceof every partial inertia ( r, s ) within the triangle r ≥ π, s ≥ ν, r + s ≤ n, or in other words every partial inertia to the “northeast” of ( π, ν ).11 efinition 2.3. If a graph G on n vertices satisfies I ( G ) = N G ) ,n ] we saythat G is inertially arbitrary . If a graph G on n vertices satisfies h I ( G ) = N G ) ,n ] we say that G is Hermitian inertially arbitrary . Example 2.1.
The complete graph K n , n ≥
2. Since ± J n (the all onesmatrix) ∈ S ( K n ), (1 , , (0 , ∈ I ( K n ). By the Northeast Lemma N ,n ] ⊆I ( K n ). Since I ( K n ) ⊆ N K n ) ,n ] = N ,n ] by Observation 2.1, K n is inertiallyarbitrary. Example 2.2.
The path P n , n ≥
2. A consequence of a well-known resultof Fiedler [F] is that for a graph G on n vertices, mr( G ) = n − G = P n . It follows from a Theorem in [Hald] that there is an A ∈ S ( P n )with eigenvalues 1 , , , . . . , n . Then for k = 1 , . . . , n , pin( A − kI ) = ( n − k, k − I ( P n ) = N n − ,n ] = N P n ) ,n ] , so P n isalso inertially arbitrary.The partial inertia set for a graph on n vertices can never be smaller thanthe partial inertia set for P n . Proposition 2.3. If G is any graph on n vertices, N n − ,n ] ⊆ I ( G ) .Proof. Let r, s ∈ { , , . . . , n } with r + s = n . Let D = diag( r, r − , . . . , , , − , − , . . . , − s )and let A G be the adjacency matrix of G . By Gershgorin’s theorem, B = D + n A G ∈ S ( G ) has eigenvalues λ > λ > · · · > λ r > > λ r +1 > · · · > λ n ,so pin( B ) = ( r, s ). Furthermore for r < n , B − λ r +1 I n ∈ S ( G ) has partialinertia ( r, s − N n − ,n ] ⊆ I ( G ).The fact that inertia sets are additive on disconnected unions of graphs(Observation 4.1) gives us an immediate corollary. Corollary 2.1. If G is any graph on n vertices and G has ℓ components, N n − ℓ,n ] ⊆ I ( G ) . n − ℓ playsa role in the proof of our second lemma from the Introduction. Proof of the Stars and Stripes Lemma.
Let G be a graph with n vertices, andlet S ⊆ V ( G ) be such that | S | = k and G − S has MD k ( G ) components, withMD k ( G ) ≥ k . Also, let ( r, s ) be any pair of integers such that k ≤ r , k ≤ s ,and r + s = n − MD k ( G ) + k .Without loss of generality label the vertices of G so that S = { , . . . , k } ,and for each vertex 1 ≤ v ≤ k let A v be the n × n adjacency matrix ofthe subgraph of G that retains all vertices of G , but only those edges thatinclude the vertex v . If v is isolated in G then pin( A v ) = (0 , A v ) = (1 , G − S is a graph with n − k vertices and MD k ( G ) components, so byCorollary 2.1 there exists a matrix B ∈ S ( G − S ) with pin( B ) = ( r − k, s − k ).Let C be the direct sum of the k × k zero matrix with B , so that the rowsand columns of C are indexed by the full set V ( G ), as is the case with thematrices A , . . . , A k . Let M = A + A + · · · + A k + C . Then M ∈ S ( G ), andby subadditivity of partial inertias (Proposition 1.5) we also have π ( M ) ≤ r − k + k = r and ν ( M ) ≤ s − k + k = s . Since r + s ≤ n we can apply theNortheast Lemma to conclude that ( r, s ) ∈ I ( G ).As mentioned in the introduction, the partial inertias which can be de-duced from Lemmas 1.1 and 1.2 are precisely the elementary inertias. Definition 2.4.
Let G be a graph on n vertices. Then the set of elementaryinertias of G , E ( G ), is given by E ( G ) = { ( r, s ) ∈ N : ( r, s ) is an elementary inertia of G } . We may also think of E ( G ) as follows: For each integer k , 0 , ≤ k ≤ n , let T k = { ( x, y ) ∈ R : k ≤ x, k ≤ y, n − MD k ( G ) + k ≤ x + y ≤ n } , T = n S k =0 T k . Each nonempty T k is a possibly degenerate trapezoid,and E ( G ) = N ≤ n ∩ T. Observation 2.4.
For any graph G , we have E ( G ) ⊆ I ( G ) .Proof. Let G be a graph on n vertices, and suppose ( r, s ) ∈ E ( G ). Then forsome integer k we have k ≤ r, k ≤ s, and n − MD k ( G ) + k ≤ r + s ≤ n. (Note that this implies MD k ( G ) ≥ k .) Recall that k + MD k ( G ) ≤ n , so k + k ≤ n − MD k ( G ) + k ≤ r + s. It follows that there is an ordered pair of integers ( x, y ) satisfying k ≤ x ≤ r, k ≤ y ≤ s, and x + y = n − MD k ( G ) + k. The Stars and Stripes Lemma gives us ( x, y ) ∈ I ( G ), after which the North-east Lemma gives us ( r, s ) ∈ I ( G ) since r + s ≤ n . Remark.
Given a graph F on m vertices there is a smallest integer a suchthat N a ⊆ I ( F ). If F is inertia-convex on stripes then a is the same asmr + ( F ), and if F is inertially arbitrary then a is the same as mr( F ). Supposethat F is G − S as in the definition of MD k ( G ), with | S | = k and m = n − k .Then some trapezoid of elementary inertias of G comes from the easy estimatethat the maximum co-rank of arbitrary inertia for F , i.e. m − a , is at least thenumber of components ℓ of F (Corollary 2.1). Suppose we had an improvedlower bound Ξ( F ) for this co-rank, a graph parameter that always satisfies ℓ ≤ Ξ( F ) ≤ m − a . (The improvement ℓ ≤ Ξ( F ) will be guaranteed, forexample, if Ξ is additive on the components of F and is at least 1 on eachcomponent.) We could then define a family of graph parameters analogous toMD k ( G ) by defining MΞ k ( G ) to be the maximum, over all subsets S ⊆ V ( G )14f size | S | = k , of Ξ( G − S ). Replacing MD k ( G ) by MΞ k ( G ) would then givea stronger version of the Stars and Stripes Lemma, and an expanded set ofnot-as-elementary inertias.For any graph G , the Stars and Stripes Lemma gives us a bound on themaximum eigenvalue multiplicity M( G ). Corollary 2.2.
Let G be a graph on n vertices. Then for any ≤ k ≤ n , M( G ) ≥ MD k ( G ) − k . When this bound is attained, it is attained in particular on a set thatincludes the center of the stripe N G ) . Corollary 2.3.
Let G be a graph. If MD k ( G ) − k = M( G ) for some k , then G is inertia-balanced. Example 2.3.
The n -sun H n is defined as the graph on 2 n vertices obtainedby attaching a pendant vertex to each vertex of an n -cycle [BFH1]. We haveMD ( H n ) = 1 and MD k ( H n ) = 2 k for 1 ≤ k ≤ ⌊ n ⌋ . It follows that, inaddition to (2 n − ,
0) and (0 , n − I ( H n ) contains every integer point( r, s ) within the trapezoid r + s ≤ n, n ≤ r + 2 s, n ≤ r + s, n ≤ r + 2 s. Since for n > H n ) = 2 n − ⌊ n ⌋ [BFH1], this shows thatthe n -sun is inertia-balanced for n > Observation 2.5.
The inertia set of a graph restricted to an axis gives I ( G ) ∩ ( N × { } ) = { ( k,
0) : k ∈ N , mr + ( G ) ≤ k ≤ n } , I ( G ) ∩ ( { } × N ) = { (0 , k ) : k ∈ N , mr + ( G ) ≤ k ≤ n } , and similarly for h I ( G ) and hmr + ( G ) .
15n other words, solving the inverse inertia problem for a graph G on the x -axis (or y -axis) is equivalent to solving the minimum semidefinite rankproblem for G . One well-known result about minimum semidefinite rank is: Theorem 2.4 (hmr + [vdH], mr + [BFH3]) . Given a graph G on n vertices, hmr + ( G ) = n − if and only if G is a tree, and mr + ( G ) = n − if and onlyif G is a tree. As noted in Example 2.2, if G is not P n then mr( G ) = n −
1, and thereforemr( G ) < n −
1. It follows that (cid:8) P n (cid:9) ∞ n =1 are the only inertially arbitrary trees.If G is not connected then any matrix in S ( G ) is a direct sum of smallermatrices, which shows that mr + is additive on the components of a graph. Observation 2.6.
Let G be a graph on n vertices and let ℓ be the number ofcomponents of G . Then mr + ( G ) = n − ℓ if and only if G is a forest. This gives us a statement that implies the second claim of Theorem 1.1.
Corollary 2.5.
Let G be a graph. Then (mr + ( G ) , is an elementary inertiaof G if and only if G is a forest.Proof. Let G be a graph on n vertices and let ℓ be the number of componentsof G . Since MD ( G ) = ℓ , ( i,
0) is an elementary inertia of G exactly forthose integers i in the range n − ℓ ≤ i ≤ n . In particular, (mr + ( G ) ,
0) is anelementary inertia if and only if mr + ( G ) = n − ℓ . By Observation 2.6, thisis true if and only if G is a forest.Although the Stars and Stripes Lemma only gives the correct value ofmr + ( G ) when G is a forest, we have already seen that it can give the correctvalues of mr( G ) and M( G ) for some graphs containing a cycle. Question 2.
What is the class of graphs for whichM( G ) = max ≤ k ≤ n { MD k ( G ) − k } ?Theorem 1.1 implies that this class includes all forests, and Example 2.3shows that the class includes the n -sun graphs H n for n > xample 2.4. G = S n , n ≥
4. Let A be the adjacency matrix of S n . Thenpin( A ) = (1 , S n ) = 2 and mr + ( S n ) = n −
1, by the NortheastLemma we have I ( S n ) = (cid:8) ( n − , , ( n, , (0 , n − , (0 , n ) (cid:9) ∪ (cid:8) ( r, s ) : r ≥ , s ≥ , r + s ≤ n (cid:9) . It follows that S n is not inertially arbitrary.As has already been noted, if A ∈ H ( G ) with pin( A ) = ( r, s ), then − A ∈ H ( G ) with pin( A ) = ( s, r ), and if A is real then − A is real. Aconsequence of Observation 1.3 is Observation 2.7 (Symmetry property) . The sets I ( G ) and h I ( G ) are sym-metric about the line y = x . The purpose of this section is to define some basic parameters associatedwith a tree, establish their fundamental properties, and relate them to themaximal disconnection numbers MD k ( G ). In Section 6 we will use theseresults to simplify the application of Theorem 1.1.In [JD1], Johnson and Duarte computed the minimum rank of all matricesin S ( T ), where T is an arbitrary tree. One of the graph parameters used bythem, the path cover number of T , is also needed in our work. It is definedas follows. Definition 3.1.
Let T be a tree.(a) A path cover of T is a collection of vertex disjoint paths, occurring asinduced subgraphs of T , that covers all the vertices of T .17b) The path cover number of T , P ( T ), is the minimum number of pathsoccurring in a path cover of T .(c) A path tree P is a path cover of T consisting of P ( T ) disjoint paths,say Q , Q , . . . , Q P ( T ) . An extra edge is an edge of T that is incident tovertices on two distinct Q ’s. Clearly there are exactly P ( T ) − Theorem 3.1.
For any tree T on n vertices, mr( T ) + P ( T ) = n. As indicated, P ( T ) will also be used in our work. This is not surprising,because inertia is a refinement of rank. Our use of P ( T ) will be made precisenow. First, we need another definition. Definition 3.2.
Let G = ( V, E ) be a graph, and let S ⊆ V . Let E G ( S ) = { xy ∈ E : x ∈ S or y ∈ S } , that is, E G ( S ) consists of all edges of G that are incident to at least onevertex in S .We define now an integer-valued mapping f on the set of all subsets of V by: f G ( S ) = | E G ( S ) | − | S | + 1 . Observation 3.1.
For any graph G , f G ( ∅ ) = 1 . Observation 3.2.
Let T be a tree on n vertices, and choose an integer k inthe range ≤ k ≤ n . Then • For every S ⊆ V ( T ) with | S | = k , f T ( S ) ≤ MD k ( T ) − k, and • For some S ⊆ V ( T ) with | S | = k , f T ( S ) = MD k ( T ) − k. roof. In any forest, the number of components plus the number of edgesequals the number of vertices. Let S ⊆ V with | S | = k . The forest T − S has n − − E T ( S ) edges and n − k vertices, so it has E T ( S ) − k + 1 components.By definition of MD k ( T ), E T ( S ) − k + 1 ≤ MD k ( T ), or equivalently, f T ( S ) ≤ MD k ( T ) − k . Since E T ( S ) − k + 1 = MD k ( T ) for some S with | S | = k , thesecond statement follows.Our first theorem in this section is the following: Theorem 3.2.
Let T be a tree with | T | = n and let P ( T ) denote its pathcover number. Then P ( T ) = max S ⊆ V { f T ( S ) } = max ≤ k ≤ n { MD k ( T ) − k } . The second equality is a direct consequence of Observation 3.2. The firstequality will be proved by induction on | T | , but first we prove it directly forseveral special cases. These special cases will also be used in the proof ofTheorem 3.2. Observation 3.3.
Theorem 3.2 holds for any path.Proof.
Let P n denote the path on n vertices, n ≥
1. The degree of any vertexin P n is at most two, so for any ∅ 6 = S ⊆ V ( P n ), f P n ( S ) ≤ | S | − | S | + 1 = 1.The result follows by Observation 3.1. Corollary 3.3.
Theorem 3.2 holds for any tree T with | T | ≤ . Observation 3.4.
Theorem 3.2 holds for S n , for any n ≥ .Proof. Label the pendant vertices of S n by 2 , , . . . , n and the vertex of degree n − P ( S n ) = n −
2. For S = { } , we have f S n ( S ) = n − − n −
2, while for any S ⊆ V ( S n ) it is straightforwardto see that f S n ( S ) ≤ n − Lemma 3.5.
Let T be a tree, and let ∅ 6 = S ⊆ V ( T ) . Then f T ( S ) ≤ P ( T ) . roof. The proof is by induction on | T | . Corollary 3.3 covers the base of theinduction, so we proceed to the general induction step.Let P be any path tree of T , consisting of paths Q , Q , . . . , Q P ( T ) . Thereare P ( T ) − Q is a pendant path in P (so exactly one extra edge emanates from it), and wedenote by v the vertex of Q that is incident to an extra edge.We can also assume without loss of generality that no vertex of S hasdegree 1 or 2, since deleting such a vertex cannot increase the value of thefunction f T ( S ) that we are trying to bound from above. Case . v is an end vertex of Q . In this case S ⊆ P ( T ) S i =2 V ( Q i ). Also, P ( T − Q ) = P ( T ) −
1. Applying the induction hypothesis, we get f T ( S ) ≤ f T − Q ( S ) + 1 ≤ P ( T − Q ) + 1 = P ( T ) . Case . v is an internal vertex of Q . Suppose first that one of the two endvertices of Q (call it z ) is at distance (in Q ) of at least two from v . Then P ( T − z ) = P ( T ) and S ⊆ V ( T − z ). Moreover, by the induction hypothesis, f T ( S ) = f T − z ( S ) ≤ P ( T − z ) = P ( T ) . Hence we may assume that Q has the form yvz : ♠ y ♠ v ♠ z We have P ( T − Q ) = P ( T ) −
1. If v / ∈ S then, by induction, f T ( S ) ≤ f T − Q ( S ) + 1 ≤ P ( T − Q ) + 1 = P ( T ) . If v ∈ S , then f T ( S ) = f T − Q ( S \{ v } ) + 3 − ≤ P ( T − Q ) + 1 = P ( T ) . bservation 3.6. Let T be a tree that is not a star and for which ∆( T ) ≥ .Then there exists v ∈ V that has a unique non-pendant neighbor and at leastone pendant neighbor.Proof. Let r be a vertex of degree ∆( T ). Let Q be a path starting at r ,and of maximum length. Denote by u the terminal vertex of Q , by v thepredecessor of u in Q (note that v = r ), and by w the predecessor of v in Q (it is possible that w = r ). Then u is a pendant neighbor of v , and w is theunique non-pendant neighbor of v . Remark.
A similar result appears as Lemma 13 in [S].
Proof of Theorem 3.2.
As previously mentioned, the second equality comesfrom Observation 3.2. The proof of the first equality is by induction on | T | .The base of the induction is ensured by Corollary 3.3. The theorem holds forany path and any star, by Observations 3.3 and 3.4. Hence we may assumethat T is not a star and ∆( T ) ≥
3. Let v be as in Observation 3.6, and let u , u , . . . , u m ( m ≥
1) be its pendant neighbors.
Case . m = 1.In this case, P ( T − u ) = P ( T ). By the induction hypothesis, there exists S ⊆ V ( T − u ) such that f T − u ( S ) = P ( T − u ) = P ( T ). Hence, f T ( S ) ≥ f T − u ( S ) = P ( T ) , so f T ( S ) = P ( T ) by Lemma 3.5. Case . m = 2.In this case it is straightforward to see that P ( T − { u , u , v } ) = P ( T ) − S ⊆ V ( T − { u , u , v } ) such that f T −{ u ,u ,v } ( S ) = P ( T ) −
1. Hence, for S = S ∪ { v } we have f T ( S ) = f T −{ u ,u ,v } ( S ) + 3 − P ( T ) . Case . m ≥ P ( T − u m ) = P ( T ) −
1. By21he induction hypothesis, there exists S ⊆ V ( T − u m ) such that f T − u m ( S ) = P ( T − u m ) = P ( T ) −
1. If v ∈ S then f T ( S ) = f T − u m ( S ) + 1 = P ( T ), sowe may assume that v / ∈ S . We claim that u , u , . . . , u m − / ∈ S . Supposeotherwise that u ∈ S . Then f T − u m ( S \{ u } ) = f T − u m ( S ) + 2 − > P ( T − u m ) , contradicting Lemma 3.5. Let S = S ∪ { v } . Then f T ( S ) ≥ f T ( S ) + m − ≥ f T − u m ( S ) + 1 = P ( T ) , so f T ( S ) = P ( T ) by Lemma 3.5.This completes the proof that P ( T ) = max S ⊆ V { f T ( S ) } = max ≤ k ≤ n { MD k ( T ) − k } . We pause to note a similar result to Theorem 3.2. Given a tree T , John-son and Duarte [JD1] ascertained that P ( T ) is the maximum of p − q suchthat there exist q vertices whose deletion leaves p components each of whichis a path (possibly including singleton paths). It is obvious that this max-imum is at most max ≤ k ≤ n { MD k ( T ) − k } since any components are allowed indetermining MD k ( T ), and the converse is also true: if any component of theremaining forest is not a path, then deleting a vertex of degree greater than2 increases the value of MD k ( T ) − k . The observation of Johnson and Duartecan thus be seen as a corollary of Theorem 3.2. We will see the usefulnessof allowing non-path components in Section 6, where we show that the lowervalues of MD k ( T ) provide an exact description of part of the boundary of I ( T ). Definition 3.3.
Let T be a tree.(a) A set S ⊆ V ( T ) is said to be optimal if f T ( S ) = P ( T ).22b) Let c ( T ) = min {| S | : S is optimal } .(c) We say S is minimal optimal if S is optimal and | S | = c ( T ). Observation 3.7.
For a tree T , c ( T ) = min ≤ k ≤ n { k : n − MD k ( T ) + k = mr( T ) } . Proof.
Theorem 3.1, Theorem 3.2, Definition 3.3, and Observation 3.2.
Observation 3.8.
For a tree T , c ( T ) ≤ (cid:22) mr( T )2 (cid:23) . Proof.
Let h = c ( T ) so n − MD h ( T )+ h = mr( T ). Recall that k +MD k ( T ) ≤ n for any integer k , 0 ≤ k ≤ n , so in particular h ≤ n − MD h ( T ) and therefore2 h ≤ mr( T ). Observation 3.9.
Let T be a tree and let S ⊆ V ( T ) be minimal optimal.Then d ( v ) ≥ for every v ∈ S . Example 3.1.
We calculate c ( T ) for paths and stars. • T = P n : Then f P n ( ∅ ) = 1 = P ( P n ) so c ( P n ) = 0. • T = S n , n ≥
4: Let v be the degree n − S n . Then f S n ( { v } ) = n − − P ( S n ) > f S n ( ∅ ). So c ( S n ) = 1. Proposition 3.10.
Let T be a tree and let v ∈ V ( T ) be adjacent to m ≥ pendant vertices u , u , . . . , u m , and at most one non-pendant vertex w . Thenthere is a path tree P of T in which u vu ∈ P .Proof. The claim is obvious if T is a star, | T | ≥
3, so assume this is not thecase. Then v is adjacent to exactly one non-pendant vertex.Let P be a path tree of T . Then at least m − u , u , . . . , u m give single-vertex paths in P . Let Q be a path in P containing v . Then Q v , say u . Then Q = u vv v . . . v k .Note that k ≥
1, as P is a path tree. If v = u , then Q = u vu . Otherwise, u is a single-vertex path in P . We can form a new path tree P by replacingthe path u vv v . . . v k and the singleton path u of P by the pair of paths u vu and v v . . . v k . Proposition 3.11.
Let T be a tree and let v ∈ V ( T ) be adjacent to m ≥ pendant vertices u , u , . . . , u m , and at most one non-pendant vertex w . Let T = T − { u , u , . . . , u m , v } . Then P ( T ) = P ( T ) + m − , and c ( T ) ≤ c ( T ) + 1 . If m ≥ , c ( T ) = c ( T ) + 1 . Proof.
The proposition clearly holds if T is a star, so we may assume that thisis not the case. Let P be a path tree for T . Then P ∪ u vu ∪ u ∪ · · · ∪ u m is a path cover for T , so P ( T ) ≤ P ( T ) + m − P be a path tree for T containing the path u vu (see Proposi-tion 3.10). Then R = { u vu , u , . . . , u m } ⊆ P , and P\R is a path cover for T . Therefore, P ( T ) ≤ |P\R| = |P| − ( m −
1) = P ( T ) − ( m − . Hence P ( T ) = P ( T ) + m − S be a minimal optimal set for T , so | S | = c ( T ). This impliesthat | E T ( S ) | − | S | + 1 = P ( T ). Let S v = S ∪ { v } . Since T is not a star v has a unique non-pendant neighbor w . The vertices w, u , u , . . . , u m areadjacent to v , so | E T ( S v ) | = | E T ( S ) | + m + 1. Then f T ( S v ) = | E T ( S v ) | − | S v | + 1 = | E T ( S ) | + m + 1 − | S | − P ( T ) + m − P ( T ) , S v is an optimal set for T . It follows that c ( T ) ≤ | S v | = | S | + 1 = c ( T ) + 1 . Now assume that m ≥ S is a minimal optimal set for T . By Ob-servation 3.9, none of the vertices u , u , . . . , u m is in S . If v / ∈ S , Lemma 3.5implies P ( T ) ≥ | E T ( S ∪ { v } ) | − | S ∪ { v }| + 1 ≥ | E T ( S ) | + 3 − | S | − P ( T ) + 1 , a contradiction. Therefore v ∈ S .Let S ′ = S \{ v } . Then | E T ( S ′ ) | ≥ | E T ( S ) | − ( m + 1), so f T ( S ′ ) ≥ | E T ( S ) | − ( m + 1) − | S ′ | + 1= | E T ( S ) | − ( m + 1) − | S | + 2 + 1= P ( T ) − ( m −
1) = P ( T ) . It follows from Lemma 3.5 that S ′ is optimal for T , implying c ( T ) ≤ | S ′ | = | S | − c ( T ) − . Hence c ( T ) = c ( T ) + 1.Proposition 3.11 gives us a simple algorithm to calculate P ( T ) and thusthe minimum rank of a tree. We will use the fact that if u is a pendant vertexwhose neighbor v has degree 2, then any path in a minimal path cover thatincludes the vertex v will also include the vertex u , and P ( T ) = P ( T − u ). Observation 3.12.
Let T be a tree. Then P ( T ) may be calculated as follows:1. Set G to T and set p to .2. If G has a pendant vertex u whose neighbor v has degree , then replace G by G − u and repeat step 2. . If G consists of a single edge or single vertex, then P ( T ) = p + 1 . If G is a star on m + 1 vertices, then P ( T ) = p + m − .4. In all other cases (by Observation 3.6) there will be some v ∈ V ( G ) thatis adjacent to m ≥ pendant vertices u , u , . . . , u m and exactly onenon-pendant vertex w . Replace G by G − { u , u , . . . , u m , v } , replace p by p + m − , and return to step 2. The calculation of c ( T ) is not quite as straightforward as that of P ( T ),although we can show one special case in which it is additive on subgraphs.For this we need the following definition. Definition 3.4.
Let F and G be graphs on at least two vertices, each with avertex labeled v . Then F ⊕ v G is the graph on | F | + | G | − v in F with the vertex v in G .The vertex v in Definition 3.4 is commonly referred to as a cut vertex ofthe graph F ⊕ v G . The next result determines c ( T ) when d ( v ) = 2. Theorem 3.4.
Let T and T be trees each with a pendant vertex labeled v .Let T = T ⊕ v T . Then c ( T ) = c ( T ) + c ( T ) .Proof. Let R , R be minimal optimal sets for T , T , respectively. Then f T i ( R i ) = | E T i ( R i ) | − | R i | + 1 = P ( T i ) , i = 1 , . Since P ( T ) ≤ P ( T ) + P ( T ) −
1, and v / ∈ R , R by Observation 3.9, byLemma 3.5 P ( T ) ≥ f T ( R ∪ R ) = | E T ( R ∪ R ) | − | R ∪ R | + 1= X i =1 (cid:0) | E T i ( R i ) | − | R i | + 1 (cid:1) − P ( T ) + P ( T ) − ≥ P ( T ) . R ∪ R is an optimal set for T by Lemma 3.5 and c ( T ) ≤| R ∪ R | = | R | + | R | = c ( T ) + c ( T ). We also see that P ( T ) = P ( T ) + P ( T ) − S is a minimal optimal set for T . By Observation 3.9, v / ∈ S .Let S i = S ∩ V ( T i ), i = 1 ,
2. Since v / ∈ S , S ∩ S = ∅ . Now P ( T ) = f T ( S ) = | E T ( S ) | − | S | + 1 ,P ( T ) ≥ f T ( S ) = | E T ( S ) | − | S | + 1 , and P ( T ) ≥ f T ( S ) = | E T ( S ) | − | S | + 1 . Then 1 = P ( T ) + P ( T ) − P ( T ) ≥ X i =1 ( | E T i ( S i ) | − | S i | + 1) − ( | E T ( S ) | − | S | + 1) = 1 , so we must have P ( T i ) = f T i ( S i ) , i = 1 , , and c ( T ) + c ( T ) ≤ | S | + | S | = | S | = c ( T ) . Corollary 3.5.
Let p be a pendant vertex in a tree T and suppose the neighborof p has degree . Then c ( T ) = c ( T − p ) . Corollary 3.6.
If a tree T has exactly one vertex of degree d > , then c ( T ) = 1 .Proof. It is straightforward to see, by repeated application of Corollary 3.5,that c ( T ) = c ( S d +1 ) = 1. 27 efinition 3.5. Let T be a tree and let k be an integer such that 0 ≤ k ≤ c ( T ). Then r k ( T ) = max {| E T ( S ) | : S ⊆ V ( T ) , | S | = k } . Observation 3.13.
For a tree T , r k ( T ) = MD k ( T ) + k − . The next theorem will play an important role in simplifying the compu-tation of I ( T ). Theorem 3.7.
Let T be a tree with c ( T ) ≥ . Then r k ( T ) − r k − ( T ) ≥ if k = 1 or k = c ( T ) , if < k < c ( T ) . Proof.
Since c ( T ) ≥ T is not a path. Therefore ∆( T ) ≥
3, implying r ( T ) − r ( T ) = ∆( T ) − ≥
3. If k = c ( T ), r k ( T ) − k + 1 = P ( T ) , while r k − ( T ) − k −
1) + 1 < P ( T ) . Then r k ( T ) − r k − ( T ) − ≥ . Thus, the stronger conclusion in the special cases k = 1 and k = c ( T ) hasbeen established. We proceed by induction on | T | . Since T cannot be a path,the base of the induction is | T | = 4, and the only relevant tree T with | T | = 4is S . Since c ( S ) = 1 the theorem holds in this case.Consider now the general induction step. Let T be a tree on n vertices,and let k ∈ { , . . . , c ( T ) − } . Note that if c ( T ) ≤ T is not a star. We have to show that r k ( T ) − r k − ( T ) ≥ v ∈ V that is adjacent to a unique non-pendant vertex w , and to pendant vertices u , u , . . . , u m , where m ≥ ase . m ≥ T = T − { u , u , . . . , u m , v } . Then c ( T ) ≥ c ( T ) − T (which requires c ( T ) ≥
1) and that k ≤ c ( T ).Now choose Q ⊆ V ( T ) with | Q | ≤ k − | E T ( Q ) | ≥ r k − ( T ). Thischoice is possible (with equality) by the definition of r k − . We can assumewithout loss of generality that Q contains none of the vertices { u , . . . , u m } asfollows: If v ∈ Q we delete all u i ’s that belong to Q , possibly decreasing | Q | without changing | E T ( Q ) | . If at least one of u , u , . . . , u m , say u , belongsto Q but v / ∈ Q , we replace u by v in Q and delete from Q all remaining u i ,possibly decreasing | Q | and possibly increasing | E T ( Q ) | . We give the name ℓ to | Q | , so ℓ ≤ k − Subcase . Suppose that v / ∈ Q . Let R = Q ∪ { v } . Then | R | = ℓ + 1 ≤ k ,and E T ( R ) ⊇ E T ( Q ) ∪ (cid:8) vu , vu , . . . , vu m (cid:9) . Hence r k ( T ) ≥ r ℓ +1 ( T ) ≥ | E T ( R ) | ≥ | E T ( Q ) | + m ≥ r k − ( T ) + m ≥ r k − ( T ) + 2 . Subcase . Suppose that v ∈ Q . Let Q ′ = Q \{ v } . Then | Q ′ | = ℓ − ≤ k − r ℓ − ( T ) ≥ | E T ( Q ′ ) | . By the induction hypothesis, r k − ( T ) − r ℓ − ( T ) ≥ r k − ( T ) − r k − ( T ) ≥ . Choose R ⊆ V ( T ) with | R | = k − r k − ( T ) = | E T ( R ) | . Let R v = R ∪ { v } . Then | R v | = k , and E T ( R v ) = E T ( R ) ∪ (cid:8) vu , vu , . . . , vu m , vw (cid:9) . Also, E T ( Q ) = E T ( Q ′ ) ∪ (cid:8) vu , vu , . . . , vu m , vw (cid:9) , so | E T ( R v ) | = | E T ( R ) | + m + 1 , | E T ( Q ) | = | E T ( Q ′ ) | + m + 1 . r k ( T ) ≥ | E T ( R v ) | = | E T ( R ) | + m + 1 = r k − ( T ) + m + 1 ≥ r ℓ − ( T ) + 2 + m + 1 ≥ | E T ( Q ′ ) | + m + 1 + 2 = | E T ( Q ) | + 2 ≥ r k − ( T ) + 2 . Case . m = 1.Let T = T − u . By Corollary 3.5, we have c ( T ) = c ( T ), and clearly P ( T ) = P ( T ). Since k ≤ c ( T ) − k ≤ c ( T ) − Q ⊆ V ( T ) with | Q | ≤ k − | E T ( Q ) | ≥ r k − ( T ), and can assume without loss of generality that u / ∈ Q . Subcase . Suppose that v / ∈ Q . Then | E T ( Q ) | = | E T ( Q ) | , and applyingthe induction hypothesis, we have r k ( T ) ≥ r k ( T ) ≥ r k − ( T ) + 2 ≥ | E T ( Q ) | + 2 = | E T ( Q ) | + 2 ≥ r k − ( T ) + 2 . Subcase . Suppose that v ∈ Q . Let Q ′ = Q \{ v } . Then | Q ′ | ≤ k − E T ( Q ′ ) = E T ( Q ′ ). Hence | E T ( Q ′ ) | = | E T ( Q ′ ) | ≥ | E T ( Q ) | − ≥ r k − ( T ) − . Applying the induction hypothesis to T , r k − ( T ) ≥ r k − ( T ) + 2 ≥ | E T ( Q ′ ) | + 2 ≥ r k − ( T ) . Applying the induction hypothesis again to T , r k ( T ) ≥ r k ( T ) ≥ r k − ( T ) + 2 ≥ r k − ( T ) + 2 . Corollary 3.8.
Let T be a tree. Then for ≤ j ≤ k ≤ c ( T ) , MD k ( T ) ≥ MD j ( T ) + ( k − j ) . roof. It suffices to prove the case k − j = 1. Here we have 1 ≤ k ≤ c ( T ),and Theorem 3.7 gives us r k ( T ) ≥ r k − ( T ) + 2. Making the substitution r k ( T ) = MD k ( T )+ k − Proposition 3.14.
Let T be a tree on n ≥ vertices. Then c ( T ) ≤ n − .Proof. We prove the proposition by induction on n . The cases n = 3 , T = S n , as c ( S n ) = 1, so assume T is not a star. By Observation 3.6, thereexists a vertex v that is adjacent to exactly one non-pendant vertex w andto pendant vertices u , u , . . . , u m , where m ≥ Case . m = 1.It follows from Corollary 3.5 and the induction hypothesis that c ( T ) = c ( T − u ) ≤ n − − < n − . Case . m ≥ T = T − { u , u , . . . , u m , v } . Then | T | ≤ n −
3, and by Proposition 3.11 c ( T ) ≤ c ( T ) + 1. Then, by induction hypothesis, c ( T ) − ≤ c ( T ) ≤ | T | − ≤ n −
43 = n − − . We conclude this section with a partial result toward the first claim ofTheorem 1.1.
Definition 3.6.
For a tree T we define L T , the minimum-rank stripe of T ,as the set L T = { ( r, s ) ∈ N T ) : r ≥ c ( T ) , s ≥ c ( T ) } . For the moment the name “minimum-rank stripe” is not entirely justified,since it suggests that L T = I ( T ) ∩ N T ) . In Section 6 we will show thatthis is the case, but we can already show one direction of containment.31 heorem 3.9. For any tree T , L T ⊆ I ( T ) .Proof. Let k = c ( T ). Given any ( r, s ) ∈ L T , we have r ≥ k , s ≥ k , and r + s = mr( T ) = n − MD k ( T ) + k by Observation 3.7. Then by the Starsand Stripes Lemma we have ( r, s ) ∈ I ( T ). Corollary 3.10.
Theorem 1.1 gives the correct value of mr( T ) for T a tree. In this section we interrupt our discussion of inertia sets of trees in order toderive basic formulae about the inertia set of any graph with a cut vertex.We obtain formulae for inertia sets that are the analogue of Theorem 16in [Hs] and Theorem 2.3 in [BFH1] for minimum rank.
Definition 4.1. If Q , R are subsets of N , then Q + R = { ( a + c, b + d ) : ( a, b ) ∈ Q and ( c, d ) ∈ R } . Addition of 3 or more sets is defined similarly.
Definition 4.2. If Q is a subset of N and n is a positive integer, we let (cid:2) Q (cid:3) n = Q ∩ N ≤ n . We first consider the case of disconnected graphs. Since the inertia of adirect sum of matrices is the sum of the inertias of the summands, we have:
Observation 4.1.
Let G = k S i =1 G i . Then I ( G ) = I ( G ) + I ( G ) + · · · + I ( G k ) , and similarly for h I ( G ) .
32e now determine the inertia set of a graph with a cut vertex—see Def-inition 3.4. We first recall the following useful result [Hs], [BFH1], whichreduces the minimum rank problem for graphs to the case of 2-connectedgraphs.
Theorem 4.1 (Hsieh; Barioli, Fallat, Hogben) . With F , G and F ⊕ v G as inDefinition 3.4, we have mr( F ⊕ v G ) = min (cid:8) mr( F ) + mr( G ) , mr( F − v ) + mr( G − v ) + 2 (cid:9) . Our next result generalizes this to inertia sets.
Theorem 4.2.
Let F and G be graphs on at least two vertices with a commonvertex v and let n = | F | + | G | − . Then I ( F ⊕ v G ) = (cid:2) I ( F ) + I ( G ) (cid:3) n ∪ (cid:2) I ( F − v ) + I ( G − v ) + { (1 , } (cid:3) n and similarly for h I ( F ⊕ v G ) .Proof. We prove the complex Hermitian version of the theorem; the proofof the real symmetric version is the same but with the assumption that allmatrices and vectors are real.Let v be the last vertex of F and the first vertex of G . Reverse containment : I. Let ( r, s ) ∈ (cid:2) h I ( F ) + h I ( G ) (cid:3) n . Then r + s ≤ n and there exist ( i, j ) ∈ h I ( F ) and ( k, ℓ ) ∈ h I ( G ) such that i + k = r , j + ℓ = s . Let M = (cid:20) A bb ∗ c (cid:21) ∈ H ( F ) and N = (cid:20) c d ∗ d E (cid:21) ∈ H ( G )with pin( M ) = ( i, j ) and pin( N ) = ( k, ℓ ), and let c M = A b b ∗ c
00 0 0 and b N = c d ∗ d E
33e matrices of order n . Thenpin( c M ) = pin( M ) = ( i, j ) , pin( b N ) = pin( N ) = ( k, ℓ ) , and c M + b N ∈ H ( F ⊕ v G ). By the subadditivity of partial inertias (Proposi-tion 1.5), π ( c M + b N ) ≤ π ( c M ) + π ( b N ) = i + k = r, and ν ( c M + b N ) ≤ ν ( c M ) + ν ( b N ) = j + ℓ = s. Since ( π ( c M + b N ) , ν ( c M + b N )) ∈ h I ( F ⊕ v G ) by definition, and r + s ≤ n ,( r, s ) ∈ h I ( F ⊕ v G ) by the Northeast Lemma (Lemma 1.1). Thus, we have (cid:2) h I ( F ) + h I ( G ) (cid:3) n ⊆ h I ( F ⊕ v G ). II.
Now let ( r, s ) ∈ (cid:2) h I ( F − v ) + h I ( G − v ) + { (1 , } (cid:3) n . Then r + s ≤ n andthere exist ( i, j ) ∈ h I ( F − v ) and ( k, ℓ ) ∈ h I ( G − v ) with ( i, j )+( k, ℓ )+(1 ,
1) =( r, s ). Let A ∈ H ( F − v ) with pin( A ) = ( i, j ) and let E ∈ H ( G − v ) withpin( E ) = ( k, ℓ ). Choose b , c , d such that M = A b b ∗ c d ∗ d E ∈ H ( F ⊕ v G ) . By Proposition 1.4, π ( M ) ≤ π (cid:18)(cid:20) A E (cid:21)(cid:19) + 1 = π ( A ) + π ( E ) + 1 = i + k + 1 = r, and, similarly, ν ( M ) ≤ j + ℓ + 1 = s. Since (cid:0) π ( M ) , ν ( M ) (cid:1) ∈ h I ( F ⊕ v G ), and r + s ≤ n , by the Northeast Lemma,( r, s ) ∈ h I ( F ⊕ v G ).So we have (cid:2) h I ( F − v ) + h I ( G − v ) + (cid:8) (1 , (cid:9)(cid:3) n ⊆ h I ( F ⊕ v G ) . orward containment :Now let ( i, j ) ∈ h I ( F ⊕ v G ). By Observation 2.1, i + j ≤ n . Let M = A b b ∗ c d ∗ d E ∈ H ( F ⊕ v G ) . with pin( M ) = ( i, j ). Thenrank A + rank E ≤ rank (cid:20) A b d E (cid:21) ≤ rank M ≤ rank A + rank E + 2 . If the first and third inequalities are strict, thenrank A + rank E + 1 = rank (cid:20) A b d E (cid:21) = rank M. The first equality implies that either b / ∈ Col( A ) or else d / ∈ Col( E ), whilethe second equality implies that b ∈ Col( A ) and d ∈ Col( E ). So this casedoes not occur and eitherrank A + rank E = rank (cid:20) A b d E (cid:21) or else rank M = rank A + rank E + 2 . I. rank A + rank E = rank (cid:20) A b d E (cid:21) .Then b ∈ Col( A ) and d ∈ Col( E ). So b = Au , d = Ev for some u ∈ C | F |− , v ∈ C | G |− .Define b A = (cid:20) A Auu ∗ A u ∗ Au (cid:21) ∈ H ( F ); b E = (cid:20) v ∗ Ev v ∗ EEv E (cid:21) ∈ H ( G ) . Then b A is congruent to A ⊕ [ 0 ] × ; b E is congruent to E ⊕ [ 0 ] × . Hence π ( b A ) = π ( A ); ν ( b A ) = ν ( A ); π ( b E ) = π ( E ); ν ( b E ) = ν ( E ) . b A = rank A ≤ | F | − , (1)rank b E = rank E ≤ | G | − . (2)By Proposition 1.4, ∃ a, b ∈ { , } such that i = π ( M ) = π ( A ) + π ( E ) + a = π ( b A ) + π ( b E ) + a, (3) j = ν ( M ) = ν ( A ) + ν ( E ) + b = ν ( b A ) + ν ( b E ) + b. (4)It follows from (1) and (2) that π ( b A ) + a + ν ( b A ) = rank b A + a ≤ | F | − a ≤ | F | ,π ( b E ) + ν ( b E ) + b = rank b E + b ≤ | G | − b ≤ | G | . Hence, by the Northeast Lemma, (cid:0) π ( b A ) + a, ν ( b A ) (cid:1) = (cid:0) π ( A ) + a, ν ( A ) (cid:1) ∈ h I ( F ) , (cid:0) π ( b E ) , ν ( b E ) + b (cid:1) = (cid:0) π ( E ) , ν ( E ) + b (cid:1) ∈ h I ( G ) , and since these two vectors add up to ( i, j ), by (3) and (4), we conclude that( i, j ) ∈ h I ( F ) + h I ( G ).Since i + j ≤ n we get ( i, j ) ∈ (cid:2) h I ( F ) + h I ( G ) (cid:3) n . So in this case,h I ( F ⊕ v G ) ⊆ (cid:2) h I ( F ) + h I ( G ) (cid:3) n . II. rank M = rank A + rank E + 2.By Proposition 1.4, we have i + j ≤ π (cid:18)(cid:20) A E (cid:21)(cid:19) + 1 + ν (cid:18)(cid:20) A E (cid:21)(cid:19) + 1= π ( A ) + π ( E ) + 1 + ν ( A ) + ν ( E ) + 1= π ( A ) + ν ( A ) + π ( E ) + ν ( E ) + 2= rank( A ) + rank( E ) + 2 = rank M = i + j. It follows that i = π ( A ) + π ( E ) + 1 and j = ν ( A ) + ν ( E ) + 1 and ( i, j ) = (cid:0) π ( A ) , ν ( A ) (cid:1) + (cid:0) π ( E ) , ν ( E ) (cid:1) +(1 , i, j ) ∈ h I ( F − v )+h I ( G − ) + { (1 , } , and since i + j ≤ n , ( i, j ) ∈ (cid:2) h I ( F − v ) + h I ( G − v ) + { (1 , } (cid:3) n .So in this case,h I ( F ⊕ v G ) ⊆ (cid:2) h I ( F − v ) + h I ( G − v ) + { (1 , } (cid:3) n . This completes the proof of the forward containment.It is straightforward to show that Theorem 4.1 is a corollary of Theo-rem 4.2. The proof is not illuminating, so we do not include it.
Example 4.1.
Let F = S and G = P with v a pendant vertex in S andthe degree 2 vertex in P . Then T = F ⊕ v G is the graph below. v From Examples 2.4 and 2.2 we have I ( S ): ✻ ✲ ss ss ss s ss s I ( P ): ✻ ✲ ss ss ss s I ( S − v ) = I ( P ): ✻ ✲ ss ss ss s I ( P − v ) = I (2 K ): ✻ ✲ ss ss s s It follows that 37 I ( S ) + I ( P )] is ✻ ✲ ss ss ss s ss s ss s ss s and[ I ( P ) + I (2 K ) + { (1 , } ] is ✻ ✲ ss ss s ss s ss s s Then I ( T ) = I ( F ⊕ v G ) = [ I ( S ) + I ( P )] ∪ [ I ( P ) + I (2 K ) + { (1 , } ] is: ✻ ✲ ss ss ss s ss s ss s ss s Since | T | = 6 and P ( T ) = 2, mr( T ) = 4 by Theorem 3.1. Since | E T ( { v } ) | =3, | E T ( { v } ) | − |{ v }| + 1 = 2 = P ( T ), so { v } is a minimal optimal set for T . We observe that in this case L T = I ( T ) ∩ N T ) .We pause to develop some additional fundamental properties of inertia38ets before generalizing Theorem 4.2. The next result generalizes the fact [N]that mr( G − v ) ≤ mr( G ) ≤ mr( G − v ) + 2. Proposition 4.2.
Let G be any graph on n vertices and let v be any vertexof G . Then we have:(a) (cid:2) I ( G ) (cid:3) n − ⊆ I ( G − v ) .(b) I ( G ) ⊇ (cid:2) I ( G − v ) (cid:3) n − + { (1 , } .The same inclusions hold in the Hermitian case.Proof. Let ( r, s ) ∈ (cid:2) I ( G ) (cid:3) n − . Then r + s ≤ n −
1. Let A ∈ S ( G ) withpin( A ) = ( r, s ), and let B be the principal submatrix of A obtained bydeleting the row and column v . Then B ∈ S ( G − v ) and by the interlacinginequalities pin (cid:0) B (cid:1) is one of ( r, s ), ( r − , s ), ( r, s − r − , s − I ( G − v ) so by the Northeast Lemma, ( r, s ) ∈ I ( G − v ).This proves (a).Now let ( r, s ) ∈ (cid:2) I ( G − v ) (cid:3) n − so r + s ≤ n −
2. Choose A ∈ S ( G ) in sucha way that the principal submatrix B obtained by deleting row and column v satisfies pin (cid:0) B (cid:1) = ( r, s ). Then by the interlacing inequalities, pin( A ) isone of ( r, s ), ( r + 1 , s ), ( r, s + 1), or ( r + 1 , s + 1). Since r + 1 + s + 1 ≤ n ,( r + 1 , s + 1) ∈ I ( G ) by the Northeast Lemma applied to G . This completesthe proof of (b).The proof of the Hermitian case is the same, but with Hermitian notation. Proposition 4.3. If v is a pendant vertex of the graph G and ( i, j ) ∈ I ( G − v ) , then ( i + 1 , j ) ∈ I ( G ) and ( i, j + 1) ∈ I ( G ) , and similarly for h I ( G − v ) and h I ( G ) .Proof. As usual, the proofs of the real symmetric and Hermitian versions donot differ materially. Let v be the first vertex of G and let its neighbor u be39he second. Let A ∈ S ( G − v ) with pin( A ) = ( i, j ). Then M = (cid:20) J
00 0 (cid:21) + (cid:20) A (cid:21) ∈ S ( G )and rank M = 1 + rank A . By Proposition 1.6 π ( M ) ≤ π ( A ) + 1 and ν ( M ) ≤ ν ( A ) . Then rank M = π ( M ) + ν ( M ) ≤ π ( A ) + 1 + ν ( A ) = rank A + 1 = rank M and( i + 1 , j ) = (cid:0) π ( A ) + 1 , ν ( A ) (cid:1) = (cid:0) π ( M ) , ν ( M ) (cid:1) ∈ I ( G ). Similarly, ( i, j + 1) ∈I ( G ).The following corollary of Theorem 4.1 is very useful in simplifying thecalculation of the minimum rank of a graph. Proposition 4.4 ([S, Lemma 38]) . If the degree of v is in F ⊕ v G , then mr( F ⊕ v G ) = mr( F ) + mr( G ) . The following result generalizes this fact to inertia sets.
Proposition 4.5.
If the degree of v is in F ⊕ v G , and n = | F | + | G | − ,then I ( F ⊕ v G ) = (cid:2) I ( F ) + I ( G ) (cid:3) n , and similarly for h I ( F ⊕ v G ) .Proof. By Theorem 4.2 it suffices to show that (cid:2) I ( F − v ) + I ( G − v ) + { (1 , } (cid:3) n ⊆ (cid:2) I ( F ) + I ( G ) (cid:3) n . Let ( r, s ) ∈ (cid:2) I ( F − v ) + I ( G − v ) + { (1 , } (cid:3) n . Then r + s ≤ n and ( r, s ) =( i, j ) + ( k, ℓ ) + (1 ,
1) with ( i, j ) ∈ I ( F − v ) and ( k, ℓ ) ∈ I ( G − v ). Since v is pendant in both F and G , by Proposition 4.3, ( i + 1 , j ) ∈ I ( F ) and( k, ℓ + 1) ∈ I ( G ), so ( r, s ) = ( i + 1 + k, j + ℓ + 1) ∈ I ( F ) + I ( G ). Since r + s ≤ n , ( r, s ) ∈ (cid:2) I ( F ) + I ( G ) (cid:3) n .Replacing I by h I uniformly proves the Hermitian case.40 xample 4.2. Let F = G = S and let v be a pendant vertex in each of F and G so that T = F ⊕ v G is the graph below. v By Proposition 4.5, I ( T ) = (cid:2) I ( F ) + I ( G ) (cid:3) = (cid:2) I ( S ) + I ( S ) (cid:3) . Knowingthat the inertia set I ( S ) is ✻ ✲ ss ss ss s ss s allows us to calculate I ( T ), as depicted below. ✻ ✲ ss ss ss s ss s ss s s ss s ss s Here, mr( T ) = 4 is attained only at the partial inertia (2 , c ( T ) = 2, so that L T = { (2 , } .We close this section with the generalization of Theorem 4.2. We firstextend Definition 3.4. 41 efinition 4.3. Let G , G , . . . , G k , k ≥
2, be graphs on at least two verticeswith a common vertex v and let G = k L i =1 G i be the graph on n = k P i =1 | G i | − ( k −
1) vertices obtained by identifying the vertex v in each of the G i . Wecall G the vertex sum of the graphs G , G , . . . , G k at v . Theorem 4.3.
Let G be a graph on n ≥ vertices and let v be a cut vertexof G . Write G = k L i =1 G i , k ≥ , the vertex sum of G , G , . . . , G k at v . Then I ( G ) = (cid:2) I ( G ) + I ( G ) + · · · + I ( G k ) (cid:3) n (5) ∪ (cid:2) I ( G − v ) + I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n , and similarly for h I ( G ) .Proof. The idea of the proof is the same as in the proof of Theorem 4.2,which is showing that each side of equation (5) is contained in the other.Since each of the theorems cited applies equally well to h I as to I , the sameproof demonstrates both cases. Forward containment :We prove that I ( G ) ⊆ (cid:2) I ( G ) + I ( G ) + · · · + I ( G k ) (cid:3) n (6) ∪ (cid:2) I ( G − v ) + I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n by induction on k . For k = 2 this follows from Theorem 4.2. Assume (6)holds for all integers j with 2 ≤ j < k . Let G ′ = k − L i =1 G i , the vertex sum of G , . . . , G k − at v and let n ′ = | G ′ | . Then by Theorem 4.2, I ( G ) = I ( G ′ ⊕ v G k ) ⊆ (cid:2) I ( G ′ ) + I ( G k ) (cid:3) n ∪ (cid:2) I ( G ′ − v ) + I ( G k − v ) + { (1 , } (cid:3) n . But I ( G ′ − v ) = I k − [ i =1 ( G i − v ) ! = I ( G − v ) + · · · + I ( G k − − v )42y Observation 4.1. Applying the induction hypothesis to I ( G ′ ) we have I ( G ) ⊆ hn(cid:2) I ( G ) + · · · + I ( G k − ) (cid:3) n ′ ∪ (cid:2) I ( G − v ) + · · · + I ( G k − − v ) + { (1 , } (cid:3) n ′ o + I ( G k ) i n ∪ (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n = h(cid:16)(cid:2) I ( G ) + · · · + I ( G k − ) (cid:3) n ′ + I ( G k ) (cid:17) ∪ (cid:16)(cid:2) I ( G − v ) + · · · + I ( G k − − v ) + { (1 , } (cid:3) n ′ + I ( G k ) (cid:17)i n ∪ (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n = h(cid:2) I ( G ) + · · · + I ( G k − ) (cid:3) n ′ + I ( G k ) i n ∪ h(cid:2) I ( G − v ) + · · · + I ( G k − − v ) + { (1 , } (cid:3) n ′ + I ( G k ) i n ∪ (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n ⊆ (cid:2) I ( G ) + · · · + I ( G k ) (cid:3) n ∪ h(cid:16)(cid:2) I ( G − v ) + · · · + I ( G k − − v ) (cid:3) n ′ − + { (1 , } (cid:17) + I ( G k ) i n ∪ (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n . Let Q = (cid:2) I ( G ) + · · · + I ( G k ) (cid:3) n ,Q = (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n ,Q = h(cid:2) I ( G − v ) + · · · + I ( G k − − v ) (cid:3) n ′ − + { (1 , } + I ( G k ) i n . We show that Q ⊆ Q . Suppose that ( r, s ) ∈ Q . Then( r, s ) = ( i , j ) + ( i , j ) + · · · + ( i k − , j k − ) + (1 ,
1) + ( i, j )with ( i t , j t ) ∈ I ( G t − v ), t = 1 , . . . , k −
1, ( i, j ) ∈ I ( G k ), k − X t =1 ( i t + j t ) ≤ n ′ − , and r + s ≤ n. i + j < | G k | , by Proposition 4.2(a), ( i, j ) ∈ I ( G k − v ) and then ( r, s ) ∈ Q .So suppose that i + j = | G k | . At least one of i , j is greater than 0. Withoutloss of generality, assume i >
0. By Proposition 2.3, ( i − , j ) ∈ I ( G k ), and byProposition 4.2(a), ( i − , j ) ∈ I ( G k − v ). Since n ′ = | G ′ | = k − P t =1 | G t | − ( k − n ′ − k − X t =1 ( | G t | − − k − X t =1 | G t − v | ! − . Therefore, k − P t =1 ( i t + j t ) < k − P t =1 | G t − v | . Without loss of generality, assume i + j < | G − v | . By the Northeast Lemma ( i + 1 , j ) ∈ I ( G − v ). Since r + s ≤ n , and ( r, s ) = ( i +1 , j )+( i , j )+ · · · +( i k − , j k − )+( i − , j )+(1 , r, s ) ∈ Q . This completes the proof that Q ⊆ Q . Therefore I ( G ) ⊆ Q ∪ Q , which is (6). Reverse containment :A proof by induction is not straightforward. However, one can show thetwo containments (cid:2) I ( G ) + · · · + I ( G k ) (cid:3) n ⊆ I ( G ), and (cid:2) I ( G − v ) + · · · + I ( G k − v ) + { (1 , } (cid:3) n ⊆ I ( G ), by simply imitating each step in the proof ofTheorem 4.2. As there are no new ideas in the proof, we omit it. The results of the previous section give us a way to inductively calculatethe inertia set of any graph once we know the inertia sets of 2-connectedgraphs. In this section we prove that the same inductive formula holds whencalculating the set of elementary inertias. Claim 1 of Theorem 1.1 will thenfollow because a forest is a graph with no 2-connected subgraph on 3 or morevertices.It is convenient to describe the elementary inertias of a graph G in termsof bicolored edge-colorings of certain subgraphs of G .44 efinition 5.1. Let G be a graph on n vertices, let S be a subset of V ( G ),and let X and Y be disjoint subsets of E ( G − S ). The ordered triple ( S, X, Y )is called a bicolored span of G if ( V \ S, X ∪ Y ) is a spanning forest of G − S .(A spanning forest of a graph consists of a spanning tree for each connectedcomponent.) If ( S, X, Y ) is a bicolored span of G , we say that the orderedpair ( | S | + | X | , | S | + | Y | ) is a color vector of G . The set of color vectors of G is denoted C ( G ).The color vector counts how many edges of the spanning forest have beenmarked with either the first color or the second color, but it also countsthe set S of excluded vertices twice, as though each such vertex were markedsimultaneously with both colors. Because every spanning forest has the samenumber of edges, the quantity | X | + | Y | depends only on S , and for a givensize | S | = k , | X | + | Y | is minimized if G − S has MD k ( G ) components. If G is a graph on n vertices, ℓ = MD ( G ) is the number of components of G ,and ( ∅ , X, Y ) is a bicolored span of G , then | X | + | Y | + ℓ = n . Observation 5.1. If G is a graph on n vertices and ℓ = MD ( G ) , then N n − ℓ ⊆ C ( G ) . Definition 5.2. If Q is a subset of N , we define the northeast expansion of Q as Q ր = Q + N . For example, the Northeast Lemma is equivalent to the statement that,for G a graph on n vertices, (cid:2) I ( G ) ր (cid:3) n ⊆ I ( G ). The prevalence of northeastexpansions in this section leads us to define the following equivalence relation: Definition 5.3.
Given two sets
P, Q ⊆ N , we say that P is northeastequivalent to Q , written as P ∼ Q , if P ր = Q ր . Definition 5.4.
Let G be a graph on n vertices, and let ( x, y ) be an orderedpair of integers. We say that ( x, y ) is a northeast color vector of G if x + y ≤ n and if x ≥ x and y ≥ y for some color vector ( x , y ) of G .45ote that the set of all northeast color vectors of G is (cid:2) C ( G ) ր (cid:3) n . Theterm northeast color vector is actually a synonym for elementary inertia, aswe now demonstrate. Proposition 5.2.
Let G be a graph on n vertices. Then E ( G ) = (cid:2) C ( G ) ր (cid:3) n .Proof. We show both inclusions.
Forward inclusion.
Let ( r, s ) be an elementary inertia of G . Then there exista nonnegative integer k and an ordered pair of integers ( r , s ) such that k ≤ r ≤ r, k ≤ s ≤ s, and r + s = n − MD k ( G ) + k. Let S be chosen such that | S | = k and G − S has MD k ( G ) components,and let F be a spanning forest of G − S , so that F has n − k vertices and r + s − k edges. We partition the edges of F into two sets X and Y with r − k and s − k edges respectively. It follows that ( k + | X | , k + | Y | ) = ( r , s )is a color vector of G . Since r + s ≤ n , ( r, s ) belongs to the set (cid:2) C ( G ) ր (cid:3) n ofnortheast color vectors of G . Reverse inclusion.
Let ( x, y ) be a northeast color vector of G , and let( S, X, Y ) be a bicolored span of G such that x = | S | + | X | ≤ x and y = | S | + | Y | ≤ y . Letting k = | S | , we can assume without loss of generalitythat S is chosen among all sets of size k in such a way as to minimize | X | + | Y | ,or in other words that G − S has MD k ( G ) components. Under this assumptionwe have | X | + | Y | +MD k ( G ) = n − k , so n − MD k ( G )+ k = x + y ≤ x + y ≤ n .We further have k ≤ x ≤ x and k ≤ y ≤ y , so ( x, y ) is an elementary inertiaof G .We now state some set-theoretic results that allow us to simplify certainexpressions involving Q ր and (cid:2) Q (cid:3) n . Observation 5.3.
For Q ⊆ N and nonnegative integers m ≤ n , we have1. h(cid:2) Q (cid:3) n i m = h(cid:2) Q (cid:3) m i n = (cid:2) Q (cid:3) m . . (cid:2)(cid:2) Q ր (cid:3) n ր (cid:3) m = (cid:2) Q ր (cid:3) m .3. (cid:2) Q ր (cid:3) m ∼ (cid:2) Q (cid:3) m .4. If P is a stripe of rank m , then (cid:2) Q + P (cid:3) n = (cid:2) Q (cid:3) n − m + P .5. N m ⊆ Q implies Q ∼ (cid:2) Q (cid:3) m .6. Q ∼ (cid:2) Q (cid:3) m implies Q ∼ (cid:2) Q (cid:3) n .Proof. These are all straightforward consequences of the definitions.
Proposition 5.4.
Let ℓ , m , and n be nonnegative integers with ≤ ℓ ≤ n and ≤ m ≤ n , suppose that Q ⊆ N satisfies Q ∼ (cid:2) Q (cid:3) n − ℓ , and let P = (cid:2) Q ր (cid:3) n . Then1. P ∼ (cid:2) P (cid:3) n − ℓ ,2. (cid:2) P ր (cid:3) m = (cid:2) P (cid:3) m , and3. (cid:2) P ր (cid:3) n = P .Proof. We have P = (cid:2) Q ր (cid:3) n ∼ (cid:2) Q (cid:3) n ∼ Q ∼ (cid:2) Q (cid:3) n − ℓ ∼ (cid:2) Q ր (cid:3) n − ℓ = h(cid:2) Q ր (cid:3) n i n − ℓ = (cid:2) P (cid:3) n − ℓ , (cid:2) P ր (cid:3) m = (cid:2)(cid:2) Q ր (cid:3) n ր (cid:3) m = (cid:2) Q ր (cid:3) m = (cid:2)(cid:2) Q ր (cid:3) n (cid:3) m = (cid:2) P (cid:3) m , and (cid:2) P ր (cid:3) n = (cid:2)(cid:2) Q ր (cid:3) n ր (cid:3) n = (cid:2) Q ր (cid:3) n = P. We can apply this proposition immediately. First note that Observa-tions 5.1 and 5.3 (5) give us
Observation 5.5.
Let G be a graph on n vertices with ℓ components. Then C ( G ) ∼ (cid:2) C ( G ) (cid:3) n − ℓ . C ( G ) for Q . Observation 5.6 (Northeast equivalence I) . Let G be a graph on n verticeswith ℓ components, and let m be an integer in the range ≤ m ≤ n . Then1. E ( G ) ∼ (cid:2) E ( G ) (cid:3) n − ℓ ,2. (cid:2) E ( G ) ր (cid:3) m = (cid:2) E ( G ) (cid:3) m , and3. (cid:2) E ( G ) ր (cid:3) n = E ( G ) . Observation 5.6 (3) can be viewed as a Northeast Lemma for elementaryinertias.
Lemma 5.7.
Let Q , . . . , Q k be subsets of N , and suppose that for somecollection n , . . . , n k of nonnegative integers we have Q i ∼ (cid:2) Q i (cid:3) n i for i =1 , . . . , k . Let Q = Q + · · · + Q k and let n = n + · · · + n k . Then (cid:2) Q ր (cid:3) n = (cid:2) Q ր + · · · + Q k ր (cid:3) n = (cid:2) Q ր (cid:3) n + · · · + (cid:2) Q k ր (cid:3) n k . Proof.
The first equality comes from the observation that N + N = N .For the second equality, the reverse inclusion is easy to check. Supposethen that we are given( x, y ) ∈ (cid:2) Q ր + Q ր + · · · + Q k ր (cid:3) n , so there exist k ordered pairs of integers ( x i , y i ) ∈ Q i ր with x = k P i =1 x i , y = k P i =1 y i , and x + y ≤ n . For any such collection { ( x i , y i ) } , we can definetwo quantities, a surplus s = m X i =1 max( x i + y i − n i , deficit d = m X i =1 max( n i − x i − y i , , so that x + y − s + d = n and hence s ≤ d . If s = 0, then in every case wehave ( x i , y i ) ∈ (cid:2) Q i ր (cid:3) n i , so( x, y ) ∈ (cid:2) Q ր (cid:3) n + (cid:2) Q ր (cid:3) n + · · · + (cid:2) Q k ր (cid:3) n k and we are done. But we can assume s = 0 without loss of generality forthe following reason: If s >
0, then d > i and j in the range 1 ≤ i, j ≤ k we have x i + y i > n i and x j + y j < n j . Since Q i ր = (cid:2) Q i (cid:3) n i ր , we can replace ( x i , y i ) by either ( x i − , y i ) or ( x i , y i − Q i ր , and simultaneously replace ( x j , y j ) withrespectively either ( x j + 1 , y j ) ∈ Q j ր or ( x j , y j + 1) ∈ Q j ր . This reducesboth the value of s and the value of d , so we can assume without loss ofgenerality that s = 0, giving the desired result.The following proposition is an immediate corollary. Proposition 5.8.
Given Q ⊆ N and nonnegative integers m ≤ n , supposethat Q ∼ (cid:2) Q (cid:3) m . Then (cid:2) Q ր (cid:3) n = (cid:2) Q ր (cid:3) m + n − m X i =1 E ( K ) . Proof.
Apply Lemma 5.7 with k = n − m + 1, Q = Q , n = m , and for i > Q i = { (0 , } and n i = 1. We have abbreviated (cid:2) { (0 , } ր (cid:3) by theequivalent expression E ( K ).With the necessary set-theoretic tools in place, we can proceed to demon-strate some properties of E ( G ), starting with the fact that it is additive onthe connected components of G . Proposition 5.9 (Additivity on components) . Let G = k S i =1 G i . Then E ( G ) = E ( G ) + E ( G ) + · · · + E ( G k ) . roof. We first observe that for any bicolored span (
S, X, Y ) of G , each entryof the triple is a disjoint union of corresponding entries from bicolored spansof the components G i , so C ( G ) = C ( G ) + C ( G ) + · · · + C ( G k ) . Now let n = | G | and for each integer i in the range 1 ≤ i ≤ k , let n i = | G i | .From Observations 5.5 and 5.3 (6) we can conclude that C ( G i ) ∼ (cid:2) C ( G i ) (cid:3) n i .Since n = n + n + · · · + n k , we can apply Lemma 5.7 to obtain (cid:2) C ( G ) ր (cid:3) n = (cid:2) C ( G ) ր (cid:3) n + (cid:2) C ( G ) ր (cid:3) n + · · · + (cid:2) C ( G k ) ր (cid:3) n k , which by Proposition 5.2 is equivalent to the desired conclusion.Before stating and proving the cut vertex formula for elementary inertiasets, it will be useful to split the set E ( G ) into two specialized sets dependingon a choice of vertex v , and establish some of the properties of these sets. Definition 5.5.
Let G be a graph and let v be a vertex of G . • If (
S, X, Y ) is a bicolored span of G and v ∈ S , then we say that theordered pair ( | S | + | X | , | S | + | Y | ) is a v -deleting color vector of G . Theset of v -deleting color vectors of G is denoted C − v ( G ). • If (
S, X, Y ) is a bicolored span of G and v S , then we say that theordered pair ( | S | + | X | , | S | + | Y | ) is a v -keeping color vector of G . Theset of v -keeping color vectors of G is denoted C + v ( G ). Definition 5.6.
Let G be a graph on n vertices including v . We define theset of v -deleting elementary inertias of G as E − v ( G ) = (cid:2) C − v ( G ) ր (cid:3) n and the set of v -keeping elementary inertias of G as E + v ( G ) = (cid:2) C + v ( G ) ր (cid:3) n . Proposition 5.10 (Splitting at v ) . Let G be a graph with v ∈ V ( G ) . Then E ( G ) = E − v ( G ) ∪ E + v ( G ) . There are equivalent, simpler expressions for the set of v -deleting colorvectors and v -deleting elementary inertias of G . Proposition 5.11 (The v -deleting formula) . Let G be a graph on n ≥ vertices with v ∈ V ( G ) . Then C − v ( G ) = C ( G − v ) + { (1 , } and E − v ( G ) = (cid:2) E ( G − v ) (cid:3) n − + { (1 , } = (cid:2) E ( G − v ) + { (1 , } (cid:3) n . Proof.
The triple (
S, X, Y ) is a bicolored span of G with v ∈ S if and onlyif the triple ( S − { v } , X, Y ) is a bicolored span of G − v . It follows that the v -deleting color vectors ( r, s ) in C − v ( G ) are exactly the vectors (1 + x, y )where ( x, y ) is a color vector of G − v . This gives us our first conclusion C − v ( G ) = C ( G − v ) + { (1 , } . With the first conclusion as our starting point, we now have E − v ( G ) = (cid:2) C − v ( G ) ր (cid:3) n = (cid:2) C ( G − v ) ր + { (1 , } (cid:3) n . Since { (1 , } is a stripe of rank 2, by Observation 5.3 this simplifies to E − v ( G ) = (cid:2) C ( G − v ) ր (cid:3) n − + { (1 , } = h(cid:2) C ( G − v ) ր (cid:3) n − i n − + { (1 , } = (cid:2) E ( G − v ) (cid:3) n − + { (1 , } = (cid:2) E ( G − v ) + { (1 , } (cid:3) n which completes the proof. 51 bservation 5.12. Let G be a graph whose n vertices include v , and let ℓ be the number of components of G − v . Then C − v ( G ) ∼ (cid:2) C − v ( G ) (cid:3) n +1 − ℓ .Proof. By Observation 5.5, C ( G − v ) ∼ (cid:2) C ( G − v ) (cid:3) n − − ℓ . Proposition 5.11and Observation 5.3 (4) then give us C − v ( G ) ∼ (cid:2) C − v ( G ) (cid:3) n +1 − ℓ .Substituting Q = C − v ( G ) into Proposition 5.4 now gives us a result about v -deleting elementary inertias. Observation 5.13 (Northeast equivalence II) . Let G be a graph whose n vertices include v , let ℓ be the number of components of G − v , and let m bean integer in the range ≤ m ≤ n . Then1. E − v ( G ) ∼ (cid:2) E − v ( G ) (cid:3) n +1 − ℓ ,2. (cid:2) E − v ( G ) ր (cid:3) m = (cid:2) E − v ( G ) (cid:3) m , and3. (cid:2) E − v ( G ) ր (cid:3) n = E − v ( G ) . Similar results hold for the v -keeping color vectors and v -keeping elemen-tary inertias: Observation 5.14.
Let G be a graph whose n vertices include v , and let ℓ = MD ( G ) . Then C + v ( G ) ∼ (cid:2) C + v ( G ) (cid:3) n − ℓ .Proof. It suffices to consider bicolored spans of the form ( ∅ , X, Y ), which ofcourse satisfy v
6∈ ∅ . The set of v -keeping color vectors arising from suchbicolored spans is exactly N n − ℓ , from which the desired result follows byObservation 5.3 (5).Proposition 5.4 now gives us: Observation 5.15 (Northeast equivalence III) . Let G be a graph whose n vertices include v , let ℓ = MD ( G ) , and let m be an integer in the range ≤ m ≤ n . Then1. E + v ( G ) ∼ (cid:2) E + v ( G ) (cid:3) n − ℓ , . (cid:2) E + v ( G ) ր (cid:3) m = (cid:2) E + v ( G ) (cid:3) m , and3. (cid:2) E + v ( G ) ր (cid:3) n = E + v ( G ) . It is possible to restrict the set of allowable bicolored spans that define C + v ( G ) and still obtain the full set of v -keeping color vectors of G . Proposition 5.16.
Let G be a graph with vertex v , and let E ′ = E ( G − v ) .Suppose that ( x, y ) belongs to C + v ( G ) . Then there exists a bicolored span ( S, X, Y ) of G with v S such that ( x, y ) = ( | S | + | X | , | S | + | Y | ) and suchthat ( S, X ∩ E ′ , Y ∩ E ′ ) is a bicolored span of G − v .Proof. By the definition of C + v ( G ), there exists a bicolored span ( S, X, Y ) of G with v S such that ( x, y ) = ( | S | + | X | , | S | + | Y | ). The vertex v thusbelongs to some component G i of G − S , and those edges in X and Y whichare part of G i give a spanning tree T i of G i . There is no loss of generality ifwe assume that T i is constructed as follows: First, a spanning tree is obtainedfor each component of G i − v . Each subtree is then connected to v by wayof a single edge, so that the degree of v in T i is equal to MD ( G i − v ). Withthis assumption, ( S, X ∩ E ′ , Y ∩ E ′ ) is a bicolored span of G − v .The next key ingredient is a consequence of Propositions 5.11 and 5.16. Proposition 5.17 (Domination by G − v ) . Let G be a graph on n vertices,one of which is v . Then for ǫ ∈ {− , + } we have (cid:2) E ǫv ( G ) (cid:3) n − ⊆ E ( G − v ) . Given Proposition 5.10, Proposition 5.17 is equivalent to an inclusionon elementary inertia sets which has already been proven for inertia sets asProposition 4.2 (a):
Proposition 5.18.
For any graph G and any vertex v ∈ V ( G ) , (cid:2) E ( G ) (cid:3) n − ⊆ E ( G − v ) .
53e need one more result before stating and proving the cut vertex formulafor elementary inertias.
Proposition 5.19 (The v -keeping cut vertex formula) . Let G = k L i =1 G i be agraph on n vertices which is a vertex sum of graphs G , G , . . . , G k at v , for k ≥ . Then E + v ( G ) = (cid:2) E + v ( G ) + E + v ( G ) + · · · + E + v ( G k ) (cid:3) n . Proof.
Let G , v , n , and G , . . . , G k be as in the statement of the proposition.We first establish a related identity, C + v ( G ) = C + v ( G ) + C + v ( G ) + · · · + C + v ( G k ) . This holds because1. The sets V ( G i ) − { v } are disjoint, and their union is V ( G ) − { v } , sosubsets S ⊆ V ( G ) with v S are in bijective correspondence withcollections of subsets S i ⊆ V ( G i ) none of which contain v .2. For any such set S partitioned as a union of S i , G − S is a vertex sumat v of the graphs G i − S i , and so the set E ( G − S ) is a disjoint unionof E ( G i − S i ).3. A subgraph F of the vertex sum G − S is a spanning forest of G − S ifand only if F is a vertex sum of graphs F i each of which is a spanningforest of G i − S i .For each graph G i , let n i = | G i | , so that ( n −
1) = k P i =1 ( n i − G i contains the vertex v , MD ( G i ) ≥
1. Observations 5.14 and 5.3 (6)then give us C + v ( G i ) ∼ (cid:2) C + v ( G i ) (cid:3) n i − . Thus by Lemma 5.7 we have (cid:2) C + v ( G ) ր (cid:3) n − = (cid:2) C + v ( G ) ր (cid:3) n − + · · · + (cid:2) C + v ( G k ) ր (cid:3) n k − .
54e also have C + v ( G ) ∼ (cid:2) C + v ( G ) (cid:3) n − , so by Proposition 5.8 we can add k copiesof E ( K ) to both sides to obtain (cid:2) C + v ( G ) ր (cid:3) n − k = (cid:2) C + v ( G ) ր (cid:3) n + · · · + (cid:2) C + v ( G k ) ր (cid:3) n k which gives the desired formula by Observation 5.3 (1) and Definition 5.6.The proof of the cut vertex formula depends on the following propertiesof E ( G ), E + v ( G ), and E − v ( G ): • Northeast equivalence I and III (Observations 5.6 and 5.15), • Additivity on components (Proposition 5.9), • Splitting at v (Proposition 5.10), • The v -deleting formula (Proposition 5.11), • Domination by G − v (Proposition 5.17), and • The v -keeping cut vertex formula (Proposition 5.19). Theorem 5.1 (The cut vertex formula for elementary inertias) . Let G be agraph on n ≥ vertices and let v be a cut vertex of G . Write G = k L i =1 G i , k ≥ , the vertex sum of G , G , . . . , G k at v . Then E ( G ) = (cid:2) E ( G ) + E ( G ) + · · · + E ( G k ) (cid:3) n ∪ (cid:2) E ( G − v ) + E ( G − v ) + · · · + E ( G k − v ) + { (1 , } (cid:3) n . Proof.
We manipulate both sides to obtain the same set.Define two sets Q − = (cid:2) E ( G − v ) + · · · + E ( G k − v ) + { (1 , } (cid:3) n and Q + = (cid:2) E + v ( G ) + · · · + E + v ( G k ) (cid:3) n .
55y Propositions 5.11 and 5.9, E − v ( G ) = Q − and by Proposition 5.19, E + v ( G ) = Q + , so by Proposition 5.10, E ( G ) = Q − ∪ Q + .The right hand side isRHS = (cid:2) E ( G ) + · · · + E ( G k ) (cid:3) n ∪ Q − . For each i = 1 , . . . , k , let n i = | G i | , so that E ( G i ) ∼ (cid:2) E ( G i ) (cid:3) n i − (Observa-tions 5.6 (1) and 5.3 (6), since in each case ℓ ≥ (cid:2) E ( G ) + · · · + E ( G k ) (cid:3) n = h(cid:2) E ( G ) ր (cid:3) n + · · · + (cid:2) E ( G k ) ր (cid:3) n k i n = h(cid:2) E ( G ) ր + · · · + E ( G k ) ր (cid:3) n − k i n = (cid:2) { (0 , } ր + E ( G ) ր + · · · + E ( G k ) ր (cid:3) n = E ( K ) + (cid:2) E ( G ) ր (cid:3) n − + · · · + (cid:2) E ( G k ) ր (cid:3) n k − which by Observation 5.6 (2) gives usRHS = Q − ∪ (cid:16) E ( K ) + (cid:2) E ( G ) (cid:3) n − + · · · + (cid:2) E ( G k ) (cid:3) n k − (cid:17) . By applying Proposition 5.10 to each term (cid:2) E ( G i ) (cid:3) n i − we obtainRHS = Q − ∪ (cid:16) E ( K ) + k X i =1 (cid:0)(cid:2) E − v ( G i ) (cid:3) n i − ∪ (cid:2) E + v ( G i ) (cid:3) n i − (cid:1)(cid:17) . For any α = ( ǫ , . . . , ǫ m ) ∈ {− , + } k we will define E αv = k X i =1 (cid:2) E ǫ i v ( G i ) (cid:3) n i − . This gives us RHS = [ α ∈{− , + } k Q − ∪ (cid:0) E ( K ) + E αv (cid:1) .
56e divide the 2 k choices for α into two cases: either ǫ j is “ − ” for some j ∈ { , . . . , k } , or ǫ i is “+” for all i . In the first case, by Proposition 5.11 wehave E ( K ) + E αv = E ( K ) + (cid:2) E ǫ v ( G ) (cid:3) n − + · · ·· · · + (cid:2) E ( G j − v ) + { (1 , } (cid:3) n j − + · · ·· · · + (cid:2) E ǫ k v ( G k ) (cid:3) n k − . We wish to show that this is a subset of Q − . For every i besides j , we have (cid:2) E ǫ i v ( G i ) (cid:3) n i − ⊆ E ( G i − v ) by Proposition 5.17. The remaining terms weregroup as E ( K ) + (cid:2) E ( G j − v ) + { (1 , } (cid:3) n j − = E ( K ) + (cid:2) E ( G j − v ) (cid:3) n j − + { (1 , }⊆ E ( K ) + (cid:2) E ( G j − v ) (cid:3) n j − + { (1 , } . Observation 5.6 and Proposition 5.8 give us E ( K ) + (cid:2) E ( G j − v ) (cid:3) n j − = E ( G j − v ) . We have thus shown that E ( K ) + E αv ⊆ E ( G − v ) + · · · + E ( G k − v ) + { (1 , } , and since E ( K ) + E αv = (cid:2) E ( K ) + E αv (cid:3) n , this gives us Q − ∪ (cid:0) E ( K ) + E αv (cid:1) = Q − in the case where α has at least onesign ǫ j = “ − ”.This leaves the case where α has all signs ǫ j = “+”. By Observa-tions 5.15 (1) and 5.3 (6), E + v ( G i ) ∼ (cid:2) E + v ( G i ) (cid:3) n i − . Starting with Observa-tion 5.15 (2), applying Lemma 5.7 both backwards and forwards, and finally57sing Observation 5.15 (3), we have E ( K ) + E αv = E ( K ) + (cid:2) E + v ( G ) ր (cid:3) n − + · · · + (cid:2) E + v ( G k ) ր (cid:3) n k − = (cid:2) { (0 , } ր + E + v ( G ) ր + · · · + E + v ( G k ) ր (cid:3) n = h(cid:2) E + v ( G ) ր + · · · + E + v ( G k ) ր (cid:3) n − k i n = h(cid:2) E + v ( G ) ր (cid:3) n + · · · + (cid:2) E + v ( G k ) ր (cid:3) n k i n = Q + . The entire union thus collapses to RHS = Q − ∪ Q + , and the left and righthand expressions are equal. Remark.
We can generalize the splitting of E ( G ) into E + v ( G ) and E − v ( G )for non-elementary inertias: Given a graph G with vertex v and A ∈ H ( G ),order the vertices of G such that v = 1 and decompose A as A = (cid:20) a b ∗ b B (cid:21) . If b is in the column space of B , then say that pin( A ) ∈ h I + v ( G ), and defineh I − v ( G ) as (cid:2) h I ( G − v ) + { (1 , } (cid:3) n . Define I + v ( G ) and I − v ( G ) analogously.Under these definitions we can uniformly replace E with h I or I in Observa-tions 5.6, and 5.15 and in each of Propositions 5.9, 5.10, 5.11, 5.17, and 5.19,and we claim that in every case the result still holds. We will not provethese statements, as we already have a proof of Theorem 4.3, but given thoseobservations and propositions, the proof of Theorem 5.1 demonstrates thesame cut vertex formula for inertia sets.We now state and prove the main result of the section. Theorem 5.2.
For any tree T , I ( T ) = E ( T ) .Proof. Let n = | T | .If n = 1, T = K and I ( T ) = N , . Since ( ∅ , ∅ , ∅ ) is a bicolored span of K ,58he origin (0 ,
0) is a color vector of T and E ( T ) = N , also. If n = 2, then T = K , and I ( K ) = N , = E ( K ).Proceeding by induction, assume that I ( T ) = E ( T ) for all trees T onfewer than n vertices and let T be a tree on n vertices, n ≥
3. Let v be a cutvertex of T of degree k ≥
2. Write T = k L i =1 T i , the vertex sum of T , . . . , T k at v . By Theorem 4.3, I ( T ) = (cid:2) I ( T ) + · · · + I ( T k ) (cid:3) n ∪ (cid:2) I ( T − v ) + · · · + I ( T k − v ) + { (1 , } (cid:3) n and by Theorem 5.1, E ( T ) = (cid:2) E ( T ) + · · · + E ( T k ) (cid:3) n ∪ (cid:2) E ( T − v ) + · · · + E ( T k − v ) + { (1 , } (cid:3) n . Corresponding terms on the right hand side of these last two equations areequal by the induction hypothesis, so I ( T ) = E ( T ). Corollary 5.3.
For any forest F , I ( F ) = E ( F ) .Proof. By Theorem 5.2, I ( T ) = E ( T ) for every component T of F , and byadditivity on components for both I ( G ) (Observation 4.1) and E ( G ) (Propo-sition 5.9), I ( F ) = E ( F ).Claim 1 of Theorem 1.1, which says I ( F ) ⊆ E ( F ) for any forest F , hasnow been verified.We restate Theorem 1.1 compactly as Theorem 5.4.
Let G be a graph. Then I ( G ) = E ( G ) if and only if G is aforest. Graphical determination of the inertia setof a tree
Tabulating the full inertia set of a tree T on n vertices by means of Theo-rem 1.1 appears, potentially, to require a lot of calculation: Every integer k in the range 0 ≤ k ≤ n with MD k ( T ) ≥ k gives a trapezoid (possiblydegenerate) of elementary inertias, and the full elementary inertia set is theunion of those trapezoids. (One could also construct every possible bicoloredspan of the tree, which is even more cumbersome.) In fact the calculation isquite straightforward once we have the first few values of MD k ( T ). In thissection we present the necessary simplifications and perform the calculationfor a few examples. Definition 6.1.
For any graph G on n vertices and k ∈ { , . . . , n } , let π k ( G ) = min { r : ( r, k ) ∈ I ( G ) } ,ν k ( G ) = min { s : ( k, s ) ∈ I ( G ) } . Since π k ( G ) = ν k ( G ) for each k , we will deal exclusively with π k ( G ).The main simplification toward calculating the inertia set of a tree is thefollowing: Theorem 6.1.
Let T be a tree on n vertices and let k ∈ { , , . . . , c ( T ) } .Then π k ( T ) = n − MD k ( T ) . (7) Proof.
For k in the given range, we can apply Corollary 3.8 with j = 0 toobtain MD k ( T ) ≥ MD ( T ) + k and in particular MD k ( T ) ≥ k . We can thusapply the Stars and Stripes Lemma to obtain ( n − MD k ( T ) , k ) ∈ I ( T ) andhence π k ( T ) ≤ n − MD k ( T ). It remains to prove, for k ≤ c ( T ), n − MD k ( T ) ≤ π k ( T ) . r, s ) ∈ I ( T ) with s = k and r Let T be a tree. Then L T = I ( T ) ∩ N T ) . In other words, every partial inertia of minimum rank is in the minimum-rank stripe already defined. Proof. We already have L T ⊆ I ( T ) by Theorem 3.9, and L T ⊆ N T ) bydefinition. To show equality, it suffices by symmetry (Observation 2.7) toshow that for k < c ( T ), k + π k ( T ) > mr( T ). Let c = c ( T ). If c = 0, we aredone. By Observation 3.7, n − MD c ( T ) + c = mr( T ) but n − MD k ( T ) + k > mr( T ) for k < c . It follows by Theorem 6.1 that k + π k ( T ) > mr( T ) for k < c , which completes the proof.Theorem 4.2 already gives a method for determining the inertia set I ( T )for any tree T , but with Theorem 1.1 and the simplifications above there isa much easier method, which we summarize in the following steps:1. Use the algorithm of Observation 3.12 to find P ( T ).2. Since T is connected, MD ( T ) = 1. If T is a path then c ( T ) = 0;otherwise c ( T ) ≥ ( T ) = ∆( T ). Continue to calculate highervalues of MD k ( T ) until MD k ( T ) − k = P ( T ), at which point k = c ( T ).61. The defining southwest corners of I ( T ) are ( n − MD k ( T ) , k ) and itsreflection ( k, n − MD k ( T )), for 0 ≤ k < c ( T ), together with the stripe L T of partial inertias from ( n − P ( T ) − c ( T ) , c ( T )) to ( c ( T ) , n − P ( T ) − c ( T )).4. Every other point of I ( T ) is a result of the Northeast Lemma appliedto the defining southwest corners.We give three examples. Example 6.1. Let T be the tree in Example 4.1, whose inertia set we havealready calculated.Here P ( T ) = 2 and mr( T ) = 4. We haveMD ( T ) = 3 , MD ( T ) − , so c ( T ) = 1, and from π ( T ) = 5 we go immediately to L T , starting at height1, which is the convex stripe of three partial inertias from (3 , 1) to (1 , ✻ ✲ ss ss ss s ss s ss s ss s xample 6.2. Let T be the treewhose horizontal paths realize the path cover number P ( T ) = 3, so mr( T ) =6. Taking any vertex of degree 3 we haveMD ( T ) = 3 , MD ( T ) − , and taking the non-adjacent pair of degree-3 vertices we haveMD ( T ) = 5 , MD ( T ) − , so c ( T ) = 2. Starting as always from π ( T ) = n − π ( T ) = 9 − L T from (4 , 2) to (2 , I ( T ) is: ✻ ✲ ss ss ss s ss s ss s s ss s s ss s s ss s ss s The examples we have shown so far appear to exhibit some sort of con-vexity. For F a forest we do at least have convexity of I ( F ) on stripes of63xed rank, as stated in Corollary 1.2. Based on small examples one maybe led to believe that, in addition, π k ( T ) is a convex function in the range0 ≤ k ≤ c ( T ), or in other words that π k ( T ) − π k +1 ( T ) ≤ π k − ( T ) − π k ( T ) for 0 < k < c ( T ) . However, this is not always the case, as seen in the following example: Example 6.3. Given S with v a pendant vertex, let T be the tree con-structed as a vertex sum of four copies of the marked S : T = M i =1 S = . Here P ( T ) = 5 and mr( T ) = 8. To find MD ( T ) we always take a vertex ofmaximum degree; hereMD ( T ) = 4 , MD ( T ) − . For MD ( T ) we can either add the center of a branch or leave out the degree-4vertex and take two centers of branches; either choice gives usMD ( T ) = 5 , MD ( T ) − . At k = 3 something odd happens: to remove 3 vertices and maximize thenumber of remaining components, we must not include the single vertex ofmaximum degree. Taking the centers of three branches, we obtainMD ( T ) = 7 , MD ( T ) − , and finally taking all four vertices of degree 3 gives usMD ( T ) = 9 , MD ( T ) − P ( T ) , c ( T ) = 4. The sequence π k ( T ) thus starts (12 , , , , S n and Example 4.2, we here have a tree where the minimum-rank stripe L T is a singleton, in this case the point (4 , I ( T ) is: ✻ ✲ ss ss ss s ss s s ss s s ss s s ss s s s ss s s s ss s s s s ss s s s ss s s ss s s ss s While π k ( T ) is not a convex function over the range 0 ≤ k ≤ c ( T ) in thelast example, the calculated set I ( T ) does at least contain all of the latticepoints in its own convex hull. To expect this convexity to hold for every treewould be overly optimistic, however: if we carry out the same calculation forthe larger tree L i =1 S (on 16 vertices instead of 13) we find that the points(11 , 1) and (5 , 5) both belong to the inertia set, but their midpoint (8 , 3) doesnot. Question 3. What is the computational complexity of determining the par-tial inertia set of a tree? The examples above pose no difficulty, but they doshow that the greedy algorithm for MD k ( T ) fails even for T a tree. Comput-ing all n values of MD k ( G ) for a general graph G is NP-hard because it can65e used to calculate the independence number: k + MD k ( G ) = n if and onlyif there is an independent set of size n − k .In the next section we will consider more general graphs, rather thanrestricting to trees and forests, and we will see that even convexity of partialinertias within a single stripe can fail in the broader setting. In this section we investigate, over the set of all graphs, what partial inertiasets—or more specifically, what complements of partial inertia sets—can oc-cur. Once a graph G is allowed to have cycles, we can no longer assume thath I ( G ) = I ( G ) by diagonal congruence. It happens, however, that each ofthe results in this section is the same in the complex Hermitian case as in thereal symmetric case. For the two versions of each question we will thereforedemonstrate whichever is the more difficult of the two, proving theorems overthe complex numbers but providing counterexamples over the reals.It is convenient at this point to introduce a way of representing the com-plements of partial inertia sets. Definition 7.1. A partition is a finite (weakly) decreasing sequence of pos-itive integers. The first integer in the sequence is called the width of thepartition, and the number of terms in the sequence is called the height of thepartition.It is traditional to depict partitions with box diagrams. In order to agreewith our diagrams of partial inertia sets, we choose the convention of puttingthe longest row of boxes on the bottom of the stack; for example, the decreas-ing sequence (5 , , 1) is shown as the partition . Given a box diagram ofheight h and width w , we index the rows by 0 , , . . . , h − , , . . . , w − efinition 7.2. Given a partition π = ( π , π , . . . , π k − ), let ℓ = π , and for i ∈ { , , . . . , ℓ − } let π ∗ i = |{ j : π j ≥ i + 1 }| , i.e. the number of boxes incolumn i of the box diagram of π . Then π ∗ = ( π ∗ , π ∗ , . . . , π ∗ ℓ − ) is called theconjugate partition of π . A partition π is symmetric if π = π ∗ .For example, we have (5 , , ∗ = (3 , , , , 1) and (3 , , ∗ = (3 , , I ( G ), for a graph G on n vertices, interms of its complement N ≤ n \ I ( G ). Definition 6.1 gives us a natural wayto describe the shape of this complement as a partition. We first extend tothe Hermitian case (distinguishing from the real symmetric case as usual byprepending an ‘h’). Definition 7.3. For any graph G on n vertices and k ∈ { , . . . , n } , leth π k ( G ) = min { i : ( i, k ) ∈ h I ( G ) } . The partition corresponding to a partial inertia set is a list of as many ofthe values of π i ( G ) as are positive. Definition 7.4. Given a graph G , let k = π ( G ) and let h = h π ( G ). Thenthe inertial partition of G , denoted π ( G ), is the partition( π ( G ) , π ( G ) , . . . , π k − ( G )) . The Hermitian inertial partition of G , denoted h π ( G ), is the partition(h π ( G ) , h π ( G ) , . . . , h π h − ( G )) . It would perhaps be more accurate to call these the partial inertia com-plement partition and Hermitian partial inertia complement partition, butwe opt for the abbreviated names. 67he Northeast Lemma ensures that the inertial partition and Hermitianinertial partition of a graph are in fact partitions, and by Observation 2.7the partitions π ( G ) and h π ( G ) are always symmetric. This symmetry is thereason why k = π ( G ) is the correct point of truncation: π k − ( G ) > 0, but π k ( G ) = 0. Remark. If one starts with the entire first quadrant of the plane R andthen removes everything “northeast” of any point belonging to I ( G ), theremaining “southwest complement” has the same shape as the box diagramof π ( G ). The same applies of course to h I ( G ) and h π ( G ).The partial inertia sets I ( G ) and h I ( G ) can be reconstructed from thepartitions π ( G ) and h π ( G ), respectively, if the number of vertices of G isalso known. The addition of an isolated vertex to a graph G does not change π ( G ).We begin to investigate the following problem: Question 4 (Inertial Partition Classification Problem) . For which symmet-ric partitions π does there exist a graph G for which π ( G ) = π ?Rather than examining all possible partial inertias for a particular graph,we are now examining what restrictions on partial inertias (or rather excludedpartial inertias) may hold over the class of all graphs.The Hermitian Inertial Partition Classification Problem is the same ques-tion with h π ( G ) in the place of π ( G ). While it is known that there are graphs G for which I ( G ) is a strict subset of h I ( G ), it is not known whether there arepartitions π that are inertial partitions but not Hermitian inertial partitions,or vice versa.At the moment we are only able to give a complete answer to the InertialPartition Classification Problem for symmetric partitions of height no greaterthan 3. We first list examples for a few symmetric partitions that are easilyobtained. Of course, adding an isolated vertex to any example gives another68xample for the same partition. For simplicity we will identify the partition π ( G ) with its box diagram. • For height 0, π ( G ) is the empty partition if G has no edges. • For height 1, π ( K n ) = for any n > • For height 2, π ( P ) = . • For height 3, π ( S ) = and π ( P ) = .The partitions already listed cover every possible case, up to height 3,of an inertia-balanced graph, and leave three non-inertia-balanced partitionsunaccounted for: , , and . The following theorem eliminates cases and , as well as every largersquare partition. Theorem 7.1. Let G be a graph and let M ∈ H ( G ) be a Hermitian matrixwith partial inertia ( k, , k > . Then there exists a matrix M ′ ∈ H ( G ) with partial inertia ( r, s ) satisfying r < k and s < k . Furthermore, if M isreal then M ′ can be taken to be real. Corollary 7.2 (No Square Partitions) . For any k > , the square partition π = ( k, k, . . . , k ) of height k and width k is not the inertial partition of anygraph G , and is not the Hermitian inertial partition of any graph G .Proof of Theorem 7.1. Let G be a graph on n vertices and suppose that M ∈ H ( G ) is a Hermitian matrix with partial inertia ( k, M = (cid:2) m ij (cid:3) is thus positive semidefinite of rank k , and can be factored as A ∗ A for some k × n complex matrix A = (cid:2) a ij (cid:3) . If M is real symmetric, then A can be taken to be real.We wish to construct a matrix M ′ ∈ H ( G ) with strictly fewer than k positive eigenvalues and also strictly fewer than k negative eigenvalues. By69roposition 1.5, we will have accomplished our purpose if we can find ( k − × n matrices B = (cid:2) b ij (cid:3) and C = (cid:2) c ij (cid:3) such that B ∗ B − C ∗ C ∈ H ( G ), withthe requirement that B and C be real if M is real.We need to impose some mild general-position requirements on the firsttwo rows of the matrix A , which we accomplish by replacing A by U A , where U is a unitary matrix and where U is real (and hence orthogonal) in thecase that A is real. This is a permissible substitution because ( U A ) ∗ U A = A ∗ U ∗ U A = A ∗ A = M .The first general-position requirement is that, for integers 1 ≤ j ≤ n , a j = 0 and a j = 0 unless column j of A is the zero column. The sec-ond requirement, which we will justify more carefully, is that the set of ratios { a j /a j } be disjoint from the set of conjugate reciprocals { a i /a i } , or equiv-alently a i a j = a i a j for any 1 ≤ i, j ≤ n where neither i nor j corresponds to a zero column.Now we prove the existence of a unitary matrix U with the desired prop-erties. To do so, we temporarily reserve the symbol i ∈ C to represent asolution to i + 1 = 0. For the duration of this argument, j will representany integer 1 ≤ j ≤ n such that column j of A is not the zero column.Let x j represent the vector (cid:20) a j a j (cid:21) . If C ∗ represents the set of nonzerocomplex numbers, then our first general position assumption already guar-antees x j ∈ ( C ∗ ) . We now define three functions z, z, w : ( C ∗ ) → C ∗ by z (cid:18)(cid:20) pq (cid:21)(cid:19) = p/q, z (cid:18)(cid:20) pq (cid:21)(cid:19) = p/q, and w (cid:18)(cid:20) pq (cid:21)(cid:19) = q/p. Our task is to find a unitary matrix U , orthogonal in the case that A isreal, such that the sets { z ( U x j ) } and { w ( U x j ) } are disjoint. In this casewe can achieve the desired general position of A by replacing it with U A ,where U = U ⊕ I k − . 70ow consider the unitary matrices R θ = (cid:20) e iθ/ e − iθ/ (cid:21) and Q = 1 √ (cid:20) ii (cid:21) . These matrices transform complex ratios as follows: z ( R θ x ) = e iθ z ( x ) , z ( Qx ) = z ( x ) + iiz ( x ) + 1 . We have Qx j ∈ ( C ∗ ) as long as z ( x j ) 6∈ { i, − i } and in particular as long as z ( x j ) is not pure imaginary, which is automatically true in the case A is real.In the case where A is not real, we can assume without loss of generalitythat no z ( x j ) is pure imaginary after uniformly multiplying on the left by anappropriate choice of R θ .Given x and y in ( C ∗ ) such that neither z ( x ) nor z ( y ) is pure imaginary, z ( x ) = w ( y ) if and only if z ( Qx ) = z ( Qy ). We have reduced the problem tothat of finding a unitary matrix U , orthogonal in the case A is real, such thatthe sets { z ( QU x j ) } and { z ( QU x j ) } are disjoint. In fact we will establishthe stronger condition that the two finite subsets of the unit circle (cid:26) z ( QU x j ) | z ( QU x j ) | (cid:27) and (cid:26) z ( QU x j ) | z ( QU x j ) | (cid:27) are disjoint. Let U = Q ∗ R θ Q = (cid:20) cos θ/ − sin θ/ θ/ θ/ (cid:21) . Then U is orthogonal, and z ( QU x j ) | z ( QU x j ) | = e iθ z ( Qx j ) | z ( Qx j ) | . Our general-position requirement for A thus reduces to the following fact:Given a finite subset P of the unit circle, there is some θ such that e iθ P is disjoint from its set of conjugates e − iθ P and from the set { i, − i } . To beconcrete, if ǫ is the minimum nonzero angle between any element of P and71ny element of P or { i, − i } , θ = ǫ/ A .We now construct the matrices B and C . Each column j of the matrices B and C for 1 ≤ j ≤ n is as follows: • b j = a j . • b ij = a j a ( i +1) j for i ∈ { , . . . , k − } . • c j = a j . • c ij = a j a ( i +1) j for i ∈ { , . . . , k − } .Now consider an arbitrary entry m ′ ij of the matrix M ′ = B ∗ B − C ∗ C ;this takes the form m ′ ij = a i a j + a i a j a i a j + · · · + a i a j a ki a kj − a i a j − a i a j a i a j − · · · − a i a j a ki a kj , which factors as m ′ ij = ( a i a j − a i a j )( a i a j + a i a j + a i a j + · · · + a ki a kj )= ( a i a j − a i a j ) m ij . In case either column i or column j of A is the zero column, we have m ′ ij =0 = m ij , and in all other cases we have, by the generic requirement a i a j = a i a j , that m ′ ij = 0 if and only if m ij = 0. It follows that M ′ is a matrix in H ( G ),and by construction M ′ has at most k − k − M is real symmetric then so is M ′ . 72e have determined which inertial partitions occur for all partitions upto height 3 except for one: the partition . Perhaps surprisingly, thereis indeed a graph, on 12 vertices, that achieves this non-inertia-balancedpartition in both the real symmetric and Hermitian cases. Theorem 7.3. There exists a graph G on vertices such that π ( G ) =h π ( G ) = , the partition (3 , , . The counterexample graph G will be defined directly in terms of a realsymmetric matrix with partial inertia (3 , , 1) isnot in h I ( G ).Consider a cube centered at the origin of R , and choose a representativevector for each line that passes through an opposite pair of faces, edges, orcorners of the cube. x yz 12 3 4567 89 10These 13 vectors give us the columns of a matrix M = x y z − − − − − − − − − , { x, y, z, , . . . , } . The matrix M T M is real symmetric and positive semidefinite of rank 3, and thus haspartial inertia (3 , G as the graph on 13 vertices (labeled bythe same 13 symbols) for which M T M ∈ S ( G ); distinct vertices i and j of G are adjacent if and only if columns i and j of M are not orthogonal.We note in passing that the subgraph of G induced by vertices labeled 1–10is the line graph of K , or the complement of the Petersen Graph. We nowdefine the graph G (as promised in Theorem 7.3) as the induced subgraphof G obtained by deleting the vertex labeled 10.Before proving Theorem 7.3, we prove a lemma about a smaller graph G that is obtained from G by deleting the vertices labeled 6 and 9 (whileretaining the labels of the other vertices). The vertices of G are thus labeled { x, y, z, , , , , , , } (notice that this set skips index 6). Lemma 7.1. Let A = (cid:2) a ij (cid:3) be a Hermitian matrix in H ( G ) of rank no morethan . Then the first two diagonal entries a xx and a yy are both nonzero andhave the same sign.Proof. Let d x , d y , and d z be the first three diagonal entries of A : d x = a xx , d y = a yy , d z = a zz . We show first that all three of these entries are nonzero. For this purpose itsuffices to consider only the first six rows and columns of A , correspondingto the graph G = z x y sometimes called the supertriangle graph. The automorphism group of G realizes any permutation of the vertices { x, y, z } (as do the automorphismgroups of G and of G , but not that of G ).74he principal submatrix of A on rows and columns { x, y, z, , , } , like A itself, has rank at most 3. Suppose that we had d y = 0 while d z = 0.Then the 4 × { x, y, z, } and columns { y, z, , } wouldbe combinatorially nonsingular (that is, permutation equivalent to an upper-triangular matrix with nonzero entries on the diagonal), contradicting thatrank( A ) ≤ 3. By the symmetries of G , we could have chosen any pairinstead of d y = 0 , d z = 0, and thus if any one of the three quantities d x , d y ,or d z is equal to zero, then all three must be. But if all three of the firstdiagonal entries were zero, then the 4 × { x, y, , } would be combinatorially nonsingular. It follows that inthe 10 × 10 rank-3 matrix A , the first three diagonal entries d x , d y , and d z are all nonzero.Considering once more the full matrix A , let β = a x /a z and γ = a y /a z ,so the first three rows of A can be written: d x βa z a x a x a x a x d y γa z a y a y a y a y d z a z a z a z a z a z a z . Since d x , d y , and d z are nonzero and A has rank at most 3, every otherrow of A can be obtained from these first three rows by taking a linearcombination, and the coefficients of the linear combination are determinedby entries in the first three columns. Every entry of A is thus determined bythe variables appearing in the 3 × 10 matrix above. For any i and j in theset { , , , , , , } , we have a ij = a xi d x a xj + a yi d y a yj + a zi d z a zj . In those cases where i = j and ij is not an edge of G , the entry a ij = 0gives an equation on the entries of the first three rows. Using several suchequations, we deduce that d x d y > 0, as follows:1. The entries a = 0 and a = 0 give us the pair of equations d x a z = − d z βa x and d y a z = − d z γa y . 75. Combining the equations from a = 0 and a = 0, we have a x a y = a y a x . 3. Combining the equations from a = 0 and a = 0, we have a x = βa z . 4. Combining the equations from a = 0 and a = 0, we have a y = γa z . Multiplying the first pair of equations and then substituting in each of theremaining equations in order, then canceling the nonzero term a z a z , wearrive finally at d x d y = d z ββ γγ, a positive quantity. This proves that, in any Hermitian matrix A ∈ H ( G )of rank no more than 3, the first two diagonal entries are nonzero and havethe same sign. Proof of Theorem 7.3. We review the definition of the graph G that willprovide the claimed example: starting from the diagram of the cube witha labeled vector for every pair of faces, edges, or corners, we omit the vec-tor 10 and connect pairs of vertices from the set { x, y, z, , , , , , , , , } whenever their corresponding vectors are not orthogonal.Letting M be the submatrix of M obtained by deleting the last column(labeled 10), we have M T M ∈ S ( G ) and thus (3 , ∈ I ( G ) and (3 , ∈ h I ( G ). It follows from Theorem 7.1 that the point (2 , 2) also belongs to I ( G ) and h I ( G ). To show that π ( G ) = h π ( G ) = , it suffices bythe Northeast Lemma and symmetry to show that (2 , h I ( G ).Let A be any matrix of rank 3 in H ( G ), and let d x , d y , and d z be thefirst three diagonal entries of A . Omitting rows and columns 6 and 9 gives us76 matrix of rank no more than 3 in H ( G ), and so Lemma 7.1 tells us that d x and d y are nonzero and have the same sign. However, the automorphismgroup of G inherits all the symmetries of a cube with one marked corner,and thus anything true of the pair of vertices { x, y } is also true of the pair { y, z } , so d y and d z are also nonzero and have the same sign. More explicitly,using the symmetry of a counterclockwise rotation of the cube around thecorner marked 10, we delete rows and columns 4 and 7 (instead of 6 and 9)and reorder the remaining rows and columns as ( y, z, x, , , , , , , 9) toyield a different matrix belonging to H ( G ), and invoke Lemma 7.1 againto obtain d y d z > 0, showing that the three diagonal entries d x , d y , and d z allhave the same sign.The principal submatrix of A on rows and columns { x, y, z } has eitherthree positive eigenvalues or three negative eigenvalues, and so by interlacingthe partial inertia of A must be either (3 , 0) or (0 , G achieves the inertial partition and Hermitian inertialpartition (3 , , π ( G ) = h π ( G ) = ,but we are interested in the smallest possible graph that is not inertia-balanced. The following proposition justifies our claim that G is at leastlocally optimal. Theorem 7.4. Every proper induced subgraph of G is either isomorphic to G or is inertia-balanced and Hermitian inertia-balanced.Proof. By Theorem 7.1 every graph G with mr( G ) < G with hmr( G ) < F of G other than G , (2 , ∈ I ( F ) unless | F | < G is defined by orthogonality relations between the columns77f the matrix M = − − − − − − − − − , corresponding to various axes of symmetry of a cube. In other words, M T I M ∈ S ( G ) (where the identity matrix I imposes the standardpositive definite inner product on R ) and so (3 , ∈ I ( G ).The automorphism group of G has three orbits, corresponding to thefaces ( x , y , and z ), edges (1, 2, 3, 4, 5, and 6), and corners (7, 8, 9, and 10)of the cube. The deletion of any corner yields G (perhaps with a differentlabeling) and there is, up to isomorphism, only one way to delete two corners.Every proper induced subgraph of G other than G is thus isomorphic toan induced subgraph of G − x , of G − 3, or of G − { , } . Letting G beeach of these three graphs in turn, we exhibit for each a diagonal matrix D with pin( D ) = (2 , 1) and a real matrix M such that M T DM ∈ S ( G ). G = G − x, (2 , ∈ I ( G ): D = − , M = . − − . − − − − . G = G − , (2 , ∈ I ( G ): D = − , M = . − . . G = G − { , } , (2 , ∈ I ( G ): D = − , M = − . . . . . . M and D , the matrix M T DM has partial inertia (2 , S ( G ) for the desired subgraph G . If F is a proper inducedsubgraph of G other than G , then F is an induced subgraph of one ofthese three graphs. Part (a) of Proposition 4.2 allows us to delete verticesfrom any one of the three graphs and keep the partial inertia (2 , 1) as longas at least 3 vertices remain, which gives us (2 , ∈ I ( F ) unless | F | < Question 5. Is G the unique graph on fewer than 13 vertices that is notinertia-balanced?Theorems 7.1 and 7.3 only permit us to answer the Inertial Partition Clas-sification Problem and Hermitian Inertial Partition Classification Problem upto height 3. We have shown examples of constructing a graph whose mini-mum rank realization with a particular partial inertia is sufficiently “rigid”to prevent intermediate partial inertias of the same rank between the matrixand its negative. If the rank is allowed to increase, though, it is much lessclear what restrictions can be made. The next difficult question appears tobe whether (4 , , , 3) = is an inertial partition. Question 6. Let G be a graph and let M be a matrix in S ( G ) with pin( M ) =(4 , M ′ ∈ S ( G ) with pin( M ′ ) = (3 , , 2) is of higher rank than partial inertia(4 , R could not be duplicated in R with an indefinite inner product of signature(3 , M ′ directlyfrom M using any sort of continuous map such as that employed in the proofof Theorem 7.1. 79 eferences [B] W. Barrett, Hermitian and positive definite matrices, in Handbookof Linear Algebra , edited by L. Hogben, R. Brualdi, A. Greenbaum,R. Mathias, CRC Press, Boca Raton, 2006.[BBS] D. Bauer, J. Broersma and E. 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