Eigenvectors and Reconstruction

Transcription

1 Eigenvectors and Reconstruction Hongyu He Department of Mathematics Louisiana State University, Baton Rouge, USA Submitted: Jul 6, 2006; Accepted: Jun 14, 2007; Published: Jul 5, 2007 Mathematics Subject Classification: 05C88 Abstract In this paper, we study the simple eigenvectors of two hypomorphic matrices using linear algebra We also give new proofs of results of Godsil and McKay 1 Introduction We start by fixing some notations ( [HE1]) Let A be a n n real symmetric matrix Let A i be the matrix obtaining by deleting the i-th row and i-th column of A We say that two symmetric matrices A and B are hypomorphic if, for each i, B i can be obtained by simultaneously permuting the rows and columns of A i Let Σ be the set of permutations We write B = Σ(A) If M is a symmetric real matrix, then the eigenvalues of M are real We write eigen(m) = (λ 1 (M) λ 2 (M) λ n (M)) If α is an eigenvalue of M, we denote the corresponding eigenspace by eigen α (M) Let 1 be the n-dimensional vector (1, 1,, 1) Put J = 1 t 1 In [HE1], we proved the following theorem Theorem 1 ( [HE1]) Let B and A be two real n n symmetric matrices Let Σ be a hypomorphism such that B = Σ(A) Let t be a real number Then there exists an open interval T such that for t T we have 1 λ n (A + tj) = λ n (B + tj); 2 eigen λn (A + tj) and eigen λn (B + tj) are both one dimensional; I would like to thank the referee for his valuable comments the electronic journal of combinatorics 14 (2007), #N14 1

2 3 eigen λn (A + tj) = eigen λn (B + tj) As proved in [HE1], our result implies Tutte s theorem which says that eigen(a + tj) = eigen(b + tj) So det(a + tj λi) = det(b + tj λi) In this paper, we shall study the eigenvectors of A and B Most of the results in this paper are not new Our approach is new We apply Theorem 1 to derive several wellknown results We first prove that the squares of the entries of simple unit eigenvectors of A can be reconstructed as functions of eigen(a) and eigen(a i ) This yields a proof of a Theorem of Godsil-McKay We then study how the eigenvectors of A change after a perturbation of rank 1 symmetric matrices Combined with Theorem 1, we prove another result of Godsil-McKay which states that the simple eigenvectors that are perpendicular to 1 are reconstructible We further show that the orthogonal projection of 1 onto higher dimensional eigenspaces is reconstructible Our investigation indicates that the following conjecture could be true Conjecture 1 Let A be a real n n symmetric matrix Then there exists a subgroup G(A) O(n) such that a real symmetric matrix B satisfies the properties that eigen(b) = eigen(a) and eigen(b i ) = eigen(a i ) for each i if and only if B = UAU t for some U G(A) This conjecture is clearly true if rank(a) = 1 For rank(a) = 1, the group G(A) can be chosen as Z n 2, all in the form of diagonal matrices In some other cases, G(A) can be a subgroup of the permutation group S n 2 Reconstruction of Square Functions Theorem 2 Let A be a n n real symmetric matrix Let (λ 1 λ 2 λ n ) be the eigenvalues of A Suppose λ i is a simple eigenvalue of A Let = (p 1,i, p 2,i,, p n,i ) t be a unit vector in eigen λi (A) Then for every m, p 2 m,i can be expressed as a function of eigen(a) and eigen(a m ) Proof: Let λ i be a simple eigenvalue of A Let = (p 1,i, p 2,i,, p n,i ) t be a unit vector in eigen λi (A) There exists an orthogonal matrix P such that P = (p 1, p 2,, p n ) and A = P DP t where λ λ 2 0 D = 0 0 λ n Then A λ i I = P DP t λ i I = P (D λ i I)P t = j i (λ j λ i )p j p t j the electronic journal of combinatorics 14 (2007), #N14 2

3 which equals λ 1 λ i 0 0 p 1,1 p 1,i p 1,n p 2,1 p 2,i p 2,n 0 λ i λ i 0 p n,1 p n,i p n,n 0 0 λ n λ i p 1,1 p 2,1 p n,1 p 1,i p 2,i p n,i p 1,n p 2,n p n,n Deleting the m-th row and m-th column, we obtain p 1,1 p 1,i p 1,n λ 1 λ i 0 0 p m,1 p m,i p m,n 0 λ i λ i 0 p n,1 p n,i p n,n 0 0 λ n λ i p 1,1 p m,1 p n,1 p 1,i p m,i p n,i p 1,n p m,n p n,n This is A m λ i I n 1 Notice that P is orthogonal Let P m,i be the matrix obtained by deleting the m-th row and i-th column Then det Pm,i 2 = p2 m,i where p m,i is the (m, i)-th entry of P Taking the determinant, we have det(a m λ i I n 1 ) = p 2 m,i (λ j λ i ) j i It follows that QED p 2 m,i = n 1 j=1 (λ j(a m ) λ i ) j i (λ j λ i ) Corollary 1 Let A and B be two n n real symmetric matrices Suppose that eigen(a) = eigen(b) and eigen(a i ) = eigen(b i ) Let λ i be a simple eigenvalue of A and B Let the electronic journal of combinatorics 14 (2007), #N14 3

4 = (p 1,i, p 2,i,, p n,i ) t be a unit vector in eigen λi (A) and = (q 1,i, q 2,i,, q n,i ) t be a unit vector in eigen λi (B) Then p 2 j,i = q2 j,i j [1, n] Corollary 2 (Godsil-McKay, see Theorem 32, [GM]) Let A and B be two n n real symmetric matrices Suppose that A and B are hypomorphic Let λ i be a simple eigenvalue of A and B Let = (p 1,i, p 2,i,, p n,i ) t be a unit vector in eigen λi (A) and = (q 1,i, q 2,i,, q n,i ) t be a unit vector in eigen λi (B) Then p 2 j,i = q2 j,i j [1, n] 3 Eigenvalues and Eigenvectors under the perturbation of a rank one symmetric matrix Let A be a n n real symmetric matrix Let x be a n-dimensional row column vector Let M = xx t Now consider A + tm We have A + tm = P DP t + tm = P (D + tp t MP )P t = P (D + tp t xx t P )P t Let P t x = q So = (, x) for each i [1, n] Then Put D(t) = D + tqq t A + tm = P (D + tqq t )P t Lemma 1 det(d + tqq t λi) = det(a λi)(1 + i tq 2 i λ i λ ) Proof: det(d λi + tqq t ) can be written as a sum of products of λ i λ and q j For each S a subset of [1, n], combine the terms containing only (λ i λ) Since the rank of qq t is one, only for S = n, n 1, the coefficients may be nonzero We obtain det(d + tqq t λi) = n (λ i λ) + i=1 n i=1 tq 2 i (λ i λ) j i The Lemma follows Put P t (λ) = 1 + i tq 2 i λ i λ Lemma 2 Fix t < 0 Suppose that λ 1, λ 2,, λ n are distinct and 0 for every i Then P t (λ) has exactly n roots (µ 1, µ 2,, µ n ) satisfying an interlacing relation: λ 1 > µ 1 > λ 2 > µ 2 > > µ n 1 > λ n > µ n the electronic journal of combinatorics 14 (2007), #N14 4

5 tq 2 i Proof: Clearly, dpt(λ) = dλ i (λ i < 0 So P λ) 2 t (λ) is always decreasing On the interval (, λ n ), lim λ P t (λ) = 1 and lim λ λ n P t (λ) = So P t (λ) has a unique root µ n (, λ n ) Similar statement holds for each (λ i 1, λ i ) On (λ 1, ), lim λ P t (λ) = 1 and lim λ λ + P t(λ) = So P 1 t (λ) does not have any roots in (λ 1, ) QED Theorem 3 Fix t < 0 and x R n Let M = xx t Let l be the number of distinct eigenvalues satisfying (x, eigen λ (A)) 0 Choose an orthonormal basis of each eigenspace of A so that one of the eigenvectors is a multiple of the orthogonal projection of x onto the eigenspace if this projection is nonzero Denote this basis by { } and let P = (p 1, p 2,, p n ) Let S = {i 1 > i 2 > > i l } such that (x, ) 0 for every i S and (x, ) = 0 for every i / S Then there exists (µ 1,, µ l ) such that λ i1 > µ 1 > λ i2 > µ 2 > > λ il > µ l and eigen(a + tm) = {λ i (A) i / S} {µ 1, µ 2, µ l } Furthermore, eigen µj (A + tm) contains Here the index set {i 1, i 2,, i l } may not be unique statement holds for t > 0 with I shall also point out a similar µ 1 > λ i1 > µ 2 > λ i2 > > µ l > λ il Proof: Recall that = (, x) Since (x, eigen λij (A)) 0, j 0 For i / S, = 0 Notice l tqi 2 P t (λ) = 1 + j λ ij λ Applying Lemma 2 to S, we obtain the roots of P t (λ), {µ 1, µ 2,, µ l }, satisfying j=1 λ i1 > µ 1 > λ i2 > µ 2 > > λ il > µ l It follows that the roots of det(a + tm λi) = P t (λ) n i=1 (λ i λ) can be obtained from eigen(a) be changing {λ i1 > λ i2 > > λ il } to {µ 1, µ 2, µ l } Therefore, eigen(a + tm) = {λ i (A) i / S} {µ 1, µ 2, µ l } the electronic journal of combinatorics 14 (2007), #N14 5

6 Fix a µ j Let {e i } be the standard basis for R n Notice that (A + tm) =P (D + tqq t )P t =P (D + tqq t ) e i λ i µ j λ i =P e i + t =P ( =P =µ j λ i e i µ j e i q 1 q n ) e i q 2 i (1) tq 2 i λ i µ j Notice that here we use the fact that P t (µ j ) = + 1 = 0 We have obtained that (A + tm) q λ i S λ i µ j = µ i j λ i µ j Therefore, QED eigen µj (A + tm) 4 Reconstruction of Simple Eigenvectors not perpendicular to 1 Now let M = J = 11 t Theorem 3 applies to A + tj and B + tj Theorem 4 (Godsil-McKay, [GM]) Let B and A be two real n n symmetric matrices Let Σ be a hypomorphism such that B = Σ(A) Let S [1, n], A = P DP t and B = UDU t be as in Theorem 3 For i S, we have = or = In particular, if λ i is a simple eigenvalue of A and (eigen λi (A), 1) 0, then eigen λi (A) = eigen λi (B) Proof: By Tutte s theorem, eigen(a) = eigen(b) Let A = P DP t and B = UDU t Since det(a + tj λi) = det(b + tj λi), by Lemma 1, det(a λi)(1 + i t(1, ) 2 λ i λ ) = det(b λi)(1 + i t(1, ) 2 λ i λ ) the electronic journal of combinatorics 14 (2007), #N14 6

7 It follows that for every λ i, λ j =λ i (1, p j ) 2 = λ j =λ i (1, u j ) 2 Consequently, the l for A is the same as the l for B Let S be as in Theorem 3 for both A and B Without loss of generality, suppose that A = P DP t and B = UDU t as in Theorem 3 In particular, for every i [1, n], we have (, 1) 2 = (, 1) 2 (2) Let T be as in the proof of Theorem 1 in [HE1] for A and B Without loss of generality, suppose T = (t 1, t 2 ) R Let t T and let µ l (t) be the µ l in Theorem 3 for A and B Notice that the lowest eigenvectors of A + tj and B + tj are in R +n (see Lemma 1, Theorem 7 and Proof of Theorem 2 in [HE1]) So they are not perpendicular to 1 By Theorem 3, µ l (t) = λ n (A + tj) = λ n (B + tj) By Theorem 1, (,1) eigen µ1 (t)(a + tj) = eigen µl (t)(b + tj) = R (,1) So is parallel to λ i µ l (t) Since {p λ i µ l (t) i} and { } are orthonormal, by Equation 2, (, 1) λ i µ l (t) 2 = (, 1) λ i µ l (t) 2 It follows that for every t T, (, 1) λ i µ l (t) = ± (, 1) λ i µ l (t) Recall that 1 = qi 2 t i Notice that the function ρ qi 2 λ i µ l (t) i λ i is a continuous ρ and one-to-one mapping from (, λ n ) onto (0, ) There exists a nonempty interval T 0 (, λ n ) such that if ρ T 0, then qi 2 i ( 1 λ i ρ t 1, 1 t 2 ) So every ρ T 0 is a µ l (t) for some t (t 1, t 2 ) It follow that for every ρ T 0, (, 1) λ i ρ = ± (, 1) λ i ρ Notice that both vectors are nonzero and depend continuously on ρ Either, or, (, 1) λ i ρ = (, 1) λ i ρ (, 1) λ i ρ = (, 1) λ i ρ (ρ T 0 ); (ρ T 0 ); Notice that the functions {ρ 1 λ ij ρ } i j S are linearly independent For every i S, we have (, 1) = ± (, 1) Because and are both unit vectors, = ± In particular, for every simple λ i with (, 1) 0 we have eigen λi (A) = eigen λi (B) QED the electronic journal of combinatorics 14 (2007), #N14 7

8 Corollary 3 Let B and A be two real n n symmetric matrices Suppose that B = Σ(A) for a hypomorphism Σ Let λ i be an eigenvalue of A such that (eigen λi (A), 1) 0 Then the orthogonal projection of 1 onto eigen λi (A) equals the orthogonal projection of 1 onto eigen λi (B) Proof: Notice that the projections are (, 1) and (, 1) Whether = or =, we always have (, 1) = (, 1) QED Conjecture 2 Let A and B be two hypomorphic matrices Let λ i be a simple eigenvalue of A Then there exists a permutation matrix τ such that τeigen λi (A) = eigen λi (B) This conjecture is apparently true if eigen λi (A) is not perpendicular to 1 References [Tutte] W T Tutte, All the King s Horses (A Guide to Reconstruction), Graph Theory and Related Topics, Academic Press, 1979, (15-33) [GM] C D Godsil and B D McKay, Spectral Conditions for the Reconstructiblity of a graph, J Combin Theory Ser B , No 3, ( ) [HE1] H He, Reconstruction and Higher Dimensional Geometry, Journal of Combinatorial Theory, Series B 97, No 3 ( ) [Ko] W Kocay, Some New Methods in Reconstruction Theory, Combinatorial mathematics, IX (Brisbane, 1981), LNM 952, (89-114) the electronic journal of combinatorics 14 (2007), #N14 8