Singular value inequality and graph energy change

Transcription

1 Electronic Journal of Linear Algebra Volume 16 Article Singular value inequality and graph energy change Jane Day so@mathsjsuedu Wasin So Follow this and additional works at: Recommended Citation Day, Jane and So, Wasin (007), "Singular value inequality and graph energy change", Electronic Journal of Linear Algebra, Volume 16 DOI: his Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact scholcom@uwyoedu

2 Volume 16, pp 91-99, September 007 SINGULAR VALUE INEQUALIY AND GRAPH ENERGY CHANGE JANE DAY AND WASIN SO Abstract he energy of a graph is the sum of the singular values of its adjacency matrix A classic inequality for singular values of a matrixsum, including its equality case, is used to study how the energy of a graph changes when edges are removed One sharp bound and one bound that is never sharp, for the change in graph energy when the edges of a nonsingular induced subgraph are removed, are established A graph is nonsingular if its adjacency matrixis nonsingular Key words Singular value inequality, Graph energy AMS subject classifications 15A45, 05C50 1 Singular value inequality for matrix sum Let X be an n n complex matrix and denote its singular values by s 1 (X) s (X) s n (X) 0 If X has real eigenvalues only, denote its eigenvalues by λ 1 (X) λ (X) λ n (X) Define X XX which is positive semi-definite, and note that λ i ( X ) s i (X) for all i We w rite X 0tomeanX is positive semi-definite We are interested in the following singular value inequality for a matrix sum: (11) s i (A + B) s i (A)+ s i (B) and its equality case his inequality is well-known, and there are at least 4 different proofs in the literature We briefly reviewthem now he inequality (11) was first proved by Fan 4 using the variational principle: } s i (X) max{ u i UXu i : U is a unitary matrix, where {u i :1 i n} is a fixed orthonormal basis his proof also appears in Gohberg and Krein 6, and Horn and Johnson 7 No equality case was discussed in these references A different proof found in Bhatia applied a related eigenvalue inequality for 0 X a sum of Hermitian matrices to Jordan-Wielandt matrices of the form, X 0 whose eigenvalues are {±s i (X) :1 i n} Again, no equality case was discussed Another proof was provided by hompson in 10 He used polar decomposition and employed the inequalities due to Fan and Hoffman 5 (see heorem 11 below) to establish the matrix-valued triangle inequality A + B U A U + V B V Received by the editors 3 May 007 Accepted for publication 6 September 007 Handling Editor: Raphael Loewy Department of Mathematics, San Jose State University, San Jose, CA (day@mathsjsuedu, so@mathsjsuedu) 91

3 Volume 16, pp 91-99, September J Day and W So and its equality case was characterized in a later paper 11 Inequality (11) and its equality case followeasily Still another proof was given by Cheng, Horn and Li in 3 hey used a result of hompson 13 on the relationship between diagonal elements and singular values of a matrix hey also characterized the equality case in the same paper For the sake of completeness, we give the details of a proof of (11) and its equality case in heorem 14 Our proof is a variation of the one given by hompson heorem 11 For any n n complex matrix A, λ i ( A + A ) s i (A) for all i Proof Denote λ i λ i ( A+A )ands i s i (A) Let v 1,,v n be orthonormal eigenvectors of A+A such that 1 (A + A )v i λ i v i,andw 1,,w n be orthonormal eigenvectors of AA such that AA w i s i w i Now for a fixed 1 i n, consider the subspace S 1 span{v 1,,v i } and S span{w i,,w n } By a dimension argument, the intersection S 1 S contains at least one unit vector x hen Consequently, x A + A x λ i and x AA x s i λ i x A + A x Re x A x x A x A x x AA x s i Remark 1 he inequality in heorem 11 was first proved by Fan and Hoffman 5 Our proof is taken from 1 he equality case was discussed by So and hompson 9 We include a different proof of the equality case in Corollary 13 Corollary 13 For any n n complex matrix A, ( ) A + A tr tr A Equality holds if and only if λ i ( A+A )s i (A) for all i if and only if A 0 Proof It follows from heorem 11 that ( ) A + A tr λ i ( A + A ) s i (A) tr A ( ) Nowif A 0thenλ i ( A+A A+A )λ i (A) s i (A) for all i, andsotr tr A Conversely, if the equality holds then n λ i( A+A ) n s i(a), and so λ i ( A+A )s i (A) for all i because of heorem 11 Consequently, λ i ( A+A )s i (A) ( for all i, andsotr A+A ) tr (AA ) Nowwe have tr(a A )(A A) tr(aa + A A A (A ) )tr(4aa (A + A ) )0,

4 Volume 16, pp 91-99, September 007 Singular Value Inequality and Graph Energy Change 93 it follows that A A henais positive semi-definite because it is Hermitian and its eigenvalues are all non-negative heorem 14 Let A and B be two n n complex matrices hen s i (A + B) s i (A)+ s i (B) Moreover equality holds if and only if there exists a unitary matrix P such that PA and PB are both positive semi-definite Proof By polar decomposition, there exists a unitary matrix P such that P (A + B) 0 hen s i (A + B) s i (P (A + B)) trp (A + B) ( PA+ PB +(PA) +(PB) ) tr ( ) ( ) PA+(PA) PB +(PB) tr + tr tr PA + tr PB tr A + tr B s i (A)+ s i (B) Nowequality holds if and only if tr( PA+(PA) )tr PA and tr( PB+(PB) )tr PB if and only if both PA and PB are positive semi-definite, by Corollary 13 Remark 15 If both A and B are real matrices then the unitary matrix P in the equality case of heorem 14 can be taken to be real orthogonal Graph energy change due to edge deletion Let G be a simple graph ie, a graph without loops and multiple edges Also let V (G) ande(g) denote the vertex set and edge set of G respectively, while denote A(G) the adjacency matrix of G he spectrum of G is defined as Sp(G) {λ i (A(G)) : 1 i n} where n is the number of vertices of G he energy of a graph G is defined as E(G) n λ i(a(g)) 8 Since A(G) is a symmetric matrix, E(G) is indeed the sum of all singular values of A(G), ie, E(G) n s i(a(g)) 14 We are interested in howthe energy of a graph changes when edges are deleted from a graph Let us begin with a few examples Example 1 Consider a graph H on 6 vertices with an adjacency matrix A(H)

5 Volume 16, pp 91-99, September J Day and W So We have Sp(H) {1+ 3,, 0, 1 3,, } and E(H) (1+ + 3) 895 Example Let H 1 be a graph obtained from H by deleting the edge {, 3} hen Sp(H 1 ) {4383, 11386, 06180, 080, 16180, 17566} and E(H 1 ) > E(H) Example 3 Let H be a graph obtained from H by deleting the edge {1, } hen Sp(H ) {5395, 1085, 0611, 05406, 1061, 1364} and E(H ) 7766 < E(H) Example 4 Let H 3 be a graph obtained from H by deleting the edge {, 5} hen Sp(H 3 ){1+, 3, 1, 1, 1, 3} and E(H 3 )(1+ + 3) E(H) hese examples showthat the energy of a graph may increase, decrease, or remain the same when an edge is deleted heorem 6 gives bounds on the amount of change when edges are deleted, and characterizes the situation when the bounds are sharp A graph is called nonsingular if its adjacency matrix is nonsingular Let H be an induced subgraph of a graph G, which means that H contains all edges in G joining two vertices of H Let G H denote the graph obtained from G by deleting all vertices of H and all edges incident with H LetG E(H) denote the graph obtained from G by deleting all edges of H, but keeping all vertices of H If G 1 and G are two graphs without common vertices, let G 1 G denote the graph with vertex set V (G 1 ) V (G )andedgesete(g 1 ) E(G ) Hence A(G 1 G )A(G 1 ) A(G ) We need the next lemma, which appears as an exercise in 7, section 71, ex Lemma 5 If A a ij is a positive semi-definite matrix and a ii 0for some i, thena ji a ij 0for all j Our main result is heorem 6 Let H be an induced subgraph of a graph G hen E(G) E(H) E(G E(H)) E(G) +E(H) Moreover, (i) if H is nonsingular then the left equality holds if and only if G H (G H) (ii) the right equality holds if and only if E(H) Proof Notethat (1) A(H) X A(G) A(H) X + where X represents edges connecting H and G H Indeed, A(G E(H)) 0 X By heorem 14 to (1), we have E(G) E(H)+E(G E(H)), which gives the left inequality On the other hand, A(H) 0 () A(G E(H)) A(G)+ 0 0 Again, by heorem 14 to (), we have E(G E(H)) E(G)+E(H), which gives the right inequality

6 Volume 16, pp 91-99, September 007 Singular Value Inequality and Graph Energy Change 95 (i) Assume that A(H) is a nonsingular matrix For the sufficiency part, let G H (G H) then A(H) 0 A(G) 0 A(G H) which gives E(G) E(H) + E(G H) On the other hand, if the left inequality becomes equality then, by applying heorem 14 to (1), there exists an orthogonal P11 P matrix P 1 A(H) 0 0 X such that both P and P P 1 P 0 0 are positive semi-definite he symmetry of P11 P 1 A(H) 0 P 1 P 0 0 P11 A(H) 0 P 1 A(H) 0 gives P 1 A(H) 0andsoP 1 0 because of the nonsingularity of A(H) Since P is an orthogonal matrix, it follows that P 1 0andsoP 11 is nonsingular herefore 0 X P X P 0 P 0 P 11 X P X P A(G H) Since this matrix is positive semi-definite with a zero diagonal block, by Lemma 5, P 11 X 0 Hence X 0 because of the nonsingularity of P 11 Finally X 0 implies that G H (G H) (ii) For the sufficiency part, let E(H) then G E(H) G and E(H) 0 Hence E(G E(H)) E(G) E(G)+E(H) For the necessity part, assume that the right inequality becomes an equality, ie, E(G E(H)) E(G)+E(H) By applying heorem 14 to (), there exists an orthogonal matrix Q11 Q Q 1 A(H) 0 A(H) X such that both Q and Q Q 1 Q 0 0 are positive semi-definite We want to show that E(H) Suppose not, ie, A(H) is nonzero Case 1 Suppose that A(H) is nonsingular As in the proof of (i), it follows that Q 1 0 because A(H) is nonsingular, and then Q 1 0 because Q is orthogonal Hence Q is block-diagonal herefore Q 11 A(H) and Q 11 A(H) areboth positive semi-definite, so Q 11 A(H) must be 0; but that implies A(H) 0sinceQ 11 is orthogonal his leads to a contradiction because A(H) is nonsingular Case Suppose that A(H) is singular (but nonzero) Since it is symmetric and nonzero, there exist orthogonal R 1 and nonsingular symmetric A 1 such that

7 Volume 16, pp 91-99, September J Day and W So A1 0 A(H) R 1 R R1 0 LetA(H) be(n p) (n p) and define R 0 I p A(H) 0 A(H) X Since both Q and Q are positive semi-definite, 0 0 it follows that both R A1 0 QR and R 0 0 A1 X1 QR X 1 Y 1 are also positive semi-definite his cannot happen, by Case 1, because A 1 is nonsingular and R QR is orthogonal Corollary 7 Let e be an edge of a graph G hen the subgraph with the edge set {e} is induced and nonsingular, hence E(G) E(G {e}) E(G)+ Moreover, (i) the left equality holds if and only if e is an isolated edge of G, (ii) the right equality never holds Remark 8 Corollary 7 answers two open questions on graph energy raised in the AIM workshop 1 Question 1: If e is an edge of a connected graph G such that E(G) E(G {e})+, then is it true that G K? Answer: YES By (i) of Corollary 7, e is an isolated edge Since G is connected, G K Question : Are there any graphs G such that E(G {e}) E(G)+? Answer: NO By (ii) of Corollary 7 3 More examples In this section, we give examples to illuminate the significance of the condition on H and the tightness of the inequalities in heorem 6 In particular, Examples 31 and 34 showthat when H is singular, the equality E(G) E(H) E(G E(H)) in heorem 6 may or may not be true Also, we know from heorem 6 (ii) that E(G H) < E(G)+E(H) for any graph G and any induced subgraph H which has at least one edge Example 35 shows that this gap can be arbitrarily small Example 31 his example shows that heorem 6 (i) is not true for some singular graphs he graph H K K 1 is singular, where K i denotes a complete graph on i vertices Let G K K,soH is an induced subgraph of G such that E(G) E(H) E(G E(H)) But G H (G H) Lemma 3 If u is a nonzero real vector such that uv is symmetric then v λu for some λ Proof By symmetry, uv (uv ) vu Hence u(v u)v(u u), it follows that v v u u u u

8 Volume 16, pp 91-99, September 007 Singular Value Inequality and Graph Energy Change 97 Lemma 33 Let K and P 11 be a 3 3 real matrix such that a 1 1 a 1 P 11 K is positive semi-definite and P11P 11 K K henp a 1 a 1 for some a 1 Proof Write P 11 a 1 b 1 c 1 a b c a 3 b 3 c 3 Since P 11 K b 1 a 1 + c 1 b 1 b a + c b b 3 a 3 + c 3 b 3 positive semi-definite, it follows that a 1 + c 1 b a 3 + c 3, b 1 b 3, b 1 0, a + c 0, and b 1 (a + c ) b (a 1 + c 1 ) 0 Denote a a 1 a,b b 1 b, and a 3 b 3 c c 1 c c 3 Since a b a (a + c) a b b b b (a + c) b b c b c (a + c) c b P11P 11 K K is, it follows that a b c b 0,b b 1,a (a + c) c (a + c) 1, and so b (a + c) 0, a a c c Also, note that P11 P 11K K does not have any zero column and so does P 11 K, hence a + c > 0 hese facts, with some algebraic manipulations, enable one to deduce that b 0,b 1 b 3 1, c 1 a 1, c a, c 3 a 3 a 1, and finally a 1 Example 34 his example shows that heorem 6 (i) is true for some singular graphs he path graph on 3 vertices P 3 is singular If P 3 is an induced subgraph in agraphg, ande(g) E(P 3 )E(G E(P 3 )) then G P 3 (G P 3 ) Proof Let K A(P 3 ) and Y A(G P 3 ) hen A(G) K X where X represents the edges between G P X Y 3 and P 3 Wewanttoshow that X 0 from the hypothesis that E(G) E(P 3 )+E(G E(P 3 )) By heorem P11 P 14, there exists an orthogonal matrix P 1 K 0 such that both P P 1 P X and P are positive semi-definite he symmetry of X Y P11 P 1 K 0 P 1 P 0 0 P11 K 0 P 1 K 0 gives P 1 K 0, and so P 1 βk where β is an unknown vector and k he orthogonality of P gives P P I, and it follows that (31) (3) P11 P 11 + P1 P 1 I P1P 11 + PP 1 0

9 Volume 16, pp 91-99, September J Day and W So Right-multiplying equation (31) by K, wehavep11 P 11K K ogether with the fact that P 11 K is positive semi-definite, Lemma 33 gives 1 a 1 a P 11 a 1 1 a 1 Right-multiplying equation (3) by K, wehavep1 P 11K 0,andsoP 1 kα where α is an unknown vector Since P11 P 1 0 X P1 X P 11 X + P 1 Y P 1 P X Y P X P 1 X + P Y is positive semi-definite, so is P 1 X (kα )X k(x α) By Lemma 3, X α µk, andsop 1 X µkk which has a zero (,) entry Consequently, by Lemma 5, the second rowof P 11 X + P 1 Y must be zero too because it is part of a positive semi-definite matrix Let X ( ) x 1 x x 3,then 1 x 1 + x 3 0 SinceX is non-negative, x 1 x 3 0andsoP X 0 P x 0 By symmetry, the first and third rows of P 11 X + P 1 Y must be zero, ie, 1 x +(Y α) 0 1 x (Y α) 0 and hence x 0 Finally, we have X 0 Example 35 his example shows that the gap E(G) + E(H) E(G E(H)) in the right inequality of heorem 6 can be arbitrarily small if the set E(H) isnot empty Consider the family of complete regular bipartite graphs K n,n with energy E(K n,n )n and E(K n,n {e}) n +n 3wheree is any edge in K n,n Hence E(K n,n )+ E(K n,n {e}) which approaches 0 as n goes to infinity 8 n +n +1+ n +n 3, Acknowledgment he research in this paper was initiated in the workshop Spectra of Families of Matrices described by Graphs, Digraphs, and Sign Patterns held at the American Institute of Mathematics, October 3-7, 006 Both authors thank AIM and NSF for their support Example 35 is provided in a private communication by Professor Sebastian Cioaba hanks also are due to Steve Butler and Dragan Stevanovic for their helpful comments

10 Volume 16, pp 91-99, September 007 Singular Value Inequality and Graph Energy Change 99 REFERENCES 1 AIM Workshop on spectra of families of matrices described by graphs, digraphs, and sign patterns: Open Questions, December 7, 006 R Bhatia Matrix Analysis Springer, New York, C M Cheng, R A Horn, and C K Li Inequalities and equalities for the Cartesian decomposition of complexmatrices Linear Algebra Appl, 341:19-37, 00 4 K Fan Maximum properties and inequalities for the eigenvalues of completely continuous operators Proc Nat Acad Sci, USA, 37: , K Fan and A J Hoffman Some metric inequalities in the space of matrices Proc Amer Math Soc, 6: , I Gohberg and M Krein Introduction to the theory of linear nonselfadjoint operators ranslations of Mathematical Monographs, vol 18, American Mathematical Society, Providence, RI, R Horn and C Johnson Matrix Analysis Cambridge University Press, Cambridge, I Gutman he energy of a graph Ber Math-Statist Sekt Forsch Graz, 103:1, W So and R C hompson Singular values of matrixexponentials Linear Multilinear Algebra, 47:49 55, R C hompson Convexand concave functions of singular values of matrixsums Pacific J Math, 66:85 90, R C hompson he case of equality in the matrix-valued triangle inequality Pacific J Math, 8:79 80, R C hompson Dissipative matrices and related results Linear Algebra Appl, 11: , R C hompson Singular values, diagonal elements, and convexity SIAM J Appl Math, 31:39 63, V Nikiforov he energy of graphs and matrices J Math Anal Appl, 36: , 007