Squared distance matrix of a tree R.B.Bapat Indian Statistical - PowerPoint PPT Presentation

Squared distance matrix of a tree R.B.Bapat Indian Statistical Institute New Delhi, India

Matrices associated with a graph Adjacency matrix A Incidence matrix Q Laplacian matrix L = QQ ′ Distance matrix D

◦ 1 ◦ 5 ❇ ❇ ❇ ❇ ❇ ❇ ❇ ❇ ◦ 2 ◦ 4 ◦ 6 ◦ 7 ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ 3 The distance matrix of the tree is given by  0 1 2 2 3 3 4  1 0 1 1 2 2 3     2 1 0 2 3 3 4     D = 2 1 2 0 1 1 2     3 2 3 1 0 2 3     3 2 3 1 2 0 1   4 3 4 2 3 1 0

Collaborators in the area of distance matrices: Lal, Pati, Kirkland, Neumann, Balaji, Chebotarev, Sivasubramanian, Gupta, Rekhi, Kurata

Theorem (Graham and Pollak, 1971) Let T be a tree with n vertices and let D be the distance matrix of T . Then detD = ( − 1) n − 1 ( n − 1)2 n − 2 .

Weighted version Theorem Let T be a tree with n vertices. Let the edges of the tree carry weights α 1 , . . . , α n − 1 . Let D be the (weighted) distance matrix of T . Then � n − 1 � n − 1 detD = ( − 1) n − 1 2 n − 2 � � α i α i . i =1 i =1

Matrix weights Theorem Let T be a tree with n vertices. Let the edges of the tree carry weights A 1 , . . . , A n − 1 , which are s × s matrices. Let D be the (weighted) distance matrix of T . Then the determinant of D equals � n − 1 � n − 1 � � ( − 1) ( n − 1) s 2 ( n − 2) s det � � A i det A i . i =1 i =1

Example A B C ◦ 1 ◦ 2 ◦ 3 ◦ 4 Let A , B , C be s × s matrices. Then the determinant of   0 A A + B A + B + C A 0 B B + C     A + B B 0 C   A + B + C B + C C 0 is ( − 1) s 2 2 s det ( A + B + C ) det ( ABC )

Properties of the Laplacian L   d 1 0 · · · 0 0 d 2 · · · 0   L = QQ ′ =  − A  . . .  ... . . .   . . .  0 0 · · · d n L is symmetric, with zero row and column sums. L is positive semidefinite. If G is connected then rank of L is n − 1 .

Notation and some identities T : tree with n vertices { 1 , 2 , . . . , n } , d i : degree of i , τ i = 2 − d i     d 1 τ 1 d 2 τ 2     δ =  , τ = 2 1 − δ =  .   .  . .     . .    d n τ n Q ′ DQ = − 2 I , LDL = − 2 L , D τ = ( n − 1) 1

Inverse of D Theorem (Graham and Lov´ asz, 1978) Let T be a tree with n vertices. Let D be the distance matrix and L the Laplacian of T . Then D − 1 = − 1 1 2( n − 1) ττ ′ 2 L +

Moore-Penrose inverse of the Laplacian If the Laplacian L has the spectral decomposition L = λ 1 x 1 x ′ 1 + · · · + λ n − 1 x n − 1 x ′ n − 1 , then L + = 1 1 x 1 x ′ x n − 1 x ′ 1 + · · · + n − 1 . λ 1 λ n − 1

Distance in trees and Laplacian minors Let T be a tree with Laplacian L . Then (i) d ( i , j ) = det( L ( i , j | i , j )) (ii) Any principal minor of L = a constant × (sum of the cofactors in the complementary principal submatrix of D ) (iii) d ( i , j ) = ℓ + ii + ℓ + jj − 2 ℓ + ij (iv) 1 ′ D 1 = 2(Wiener index of T ) = 2 n (trace L + )

Inverse of the resistance matrix Let G be a connected graph with Laplacian L . The resistance distance between vertices i , j is given by r ( i , j ) = det( L ( i , j | i , j )) det( L ( i | i )) Theorem Let G be a connected graph with resistance matrix R and Laplacian matrix L . Then R − 1 = − 1 2 L + a rank one matrix

A generalized Laplacian We will refer to an n × n matrix as a Laplacian if it is symmetric, has row and column sums zero, and has rank n − 1 . We seek formulae of the form (distance matrix) − 1 = − 1 2 (Laplacian) + a rank one matrix

Distance matrix − → Laplacian Let D be a symmetric, nonsingular matrix such that 1 ′ D − 1 1 � = 0 . Then L = 2 D − 1 11 ′ D − 1 − 2 D − 1 1 ′ D − 1 1 is a Laplacian. Hence we have D − 1 = − 1 2 L + a rank one matrix

Laplacian − → Distance matrix Let L be a Laplacian and define D = [ d ( i , j )] by d ( i , j ) = det( L ( i , j | i , j )) . det( L ( i | i )) Then D is nonsingular and D − 1 = − 1 2 L + a rank one matrix.

Inverse of a principal submatrix of the distance matrix Theorem Let T be a tree with distance matrix D and Laplacian L . Let � D 11 � L 11 � � D 12 L 12 D = , L = D 21 D 22 L 21 L 22 be a compatible partitioning. Then 11 = − 1 D − 1 2( L 11 − L 12 L − 1 22 L 21 ) + a rank one matrix . The result can be used to give a formula for the “terminal Wiener index” of a tree.

Schur complement in a rank one update Lemma Let A be an n × n matrix. Then a Schur complement in ( A + a rank one matrix) equals (a Schur complement in A ) + a rank one matrix . The Lemma can be applied to D − 1 = − 1 1 2( n − 1) ττ ′ 2 L + to get the previous result.

Euclidean distance matrix Let p 1 , . . . , p n be points in R k . The matrix with ( i , j )-entry equal || p i − p j || 2 is called a Euclidean distance matrix. If the points are all on a sphere then the matrix is a spherical Euclidean distance matrix.

Inverse of a Euclidean distance matrix Let D be a nonsingular, spherical Euclidean distance matrix with d ( i , j ) = || p i − p j || 2 . Then 1 ′ D − 1 1 � = 0 and D − 1 = − 1 2 L + a rank one matrix , where L is a Laplacian such that p 1 , . . . , p n are the columns of 1 ( L + ) 2 .

Remark Distance matrix of a tree is a spherical Euclidean distance matrix. In particular, it has exactly one positive eigenvalue.

q -analogue Let T be a tree with n vertices and let q be an indeterminate. If i , j are vertices of the tree T with d ( i , j ) = t , define the q -distance between i and j to be 1 + q + q 2 + · · · + q t − 1 . (We set d ( i , i ) = 0 . ) Let D q be the q -distance matrix of T .

Example ◦ 1 ❇ ❇ ❇ ❇ ❇ ❇ ❇ ❇ ◦ 2 ◦ 4 ◦ 5 ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ 3 1 + q + q 2  0 1 1 + q 1 + q  1 0 1 1 1 + q    1 + q + q 2  D q = 1 + q 1 0 1 + q     1 + q 1 1 + q 0 1   1 + q + q 2 1 + q + q 2 1 + q 1 0

Determinant of the q -distance matrix of a tree Theorem Let T be a tree with n vertices and let D q be the q-distance matrix of T . Then detD q = ( − 1) n − 1 ( n − 1)(1 + q ) n − 2 .

q -Laplacian Let L q = I − qA + q 2 ( Diag ( δ ) − I ) . If q � = − 1 , then 1 D − 1 = − 1 + q L q + a rank one matrix q The determinant of L q is related to the Ihara zeta function. A. Terras, Zeta functions of graphs, A stroll through the garden, Cambridge Univ Press, 2011.

Exponential distance matrix For a tree with n vertices, define E q = [ q d ( i , j ) ] D q and E q are closely related: (1 − q ) D q = J − E q . Theorem : (i) det E q = (1 − q 2 ) n − 1 . (ii) If q � = ± 1 , then E − 1 1 = 1 − q 2 L q . q

Squared distance matrix of a tree Let T be a tree with n vertices and let D be the distance matrix of T . The squared distance matrix ∆ of T is defined as ∆ = D ◦ D = [ d ( i , j ) 2 ] .

Consider the tree: ◦ 1 ◦ 5 ❇ ❇ ❇ ❇ ❇ ❇ ❇ ❇ ◦ 2 ◦ 4 ◦ 6 ◦ 7 ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ ⑤ 3  0 1 4 4 9 9 16  1 0 1 1 4 4 9     4 1 0 4 9 9 16     ∆ = 4 1 4 0 1 1 4     9 4 9 1 0 4 9     9 4 9 1 4 0 1   16 9 16 4 9 1 0

Determinant of the squared distance matrix Theorem Let T be a tree with n vertices and let ∆ be the squared distance matrix of T . Let d 1 , d 2 , . . . , d n be the degree sequence of T . Let τ i = 2 − d i , i = 1 , . . . , n . Then   n n   det ∆ = ( − 1) n 4 n − 2 � � �  (2 n − 1) τ i − 2 τ j  . i =1 i =1 j � = i

Theorem Let T be a tree and let ∆ be the squared distance matrix of T . Then ∆ is nonsingular if and only if T has at most one vertex of degree 2 .

Edge orientation matrix Let T be a tree with V ( T ) = { 1 , . . . , n } . We assign an orientation to each edge of T . Definition The edge orientation matrix of T is the ( n − 1) × ( n − 1) matrix H defined as follows. The rows and the columns of H are indexed by the edges of T . The ( e , f )-element of H is 1( − 1) if the corresponding edges of T are similarly (oppositely) oriented. The diagonal elements of H are set to be 1 .

� � � � � � � ◦ 2 ◦ 5 ◦ 8 i iv vii iii � ◦ 4 vi � ◦ 7 ◦ 1 ◦ 10 ix v ii viii ◦ 3 ◦ 6 ◦ 9  1 1 − 1 1 1 − 1 1 − 1 1  1 1 1 − 1 − 1 1 − 1 1 − 1     − 1 1 1 − 1 − 1 1 − 1 1 − 1     1 − 1 − 1 1 − 1 1 − 1 1 − 1     H = 1 − 1 − 1 − 1 1 1 − 1 1 − 1     − 1 1 1 1 1 1 − 1 1 − 1     1 − 1 − 1 − 1 − 1 − 1 1 1 − 1     − 1 1 1 1 1 1 1 1 1   1 − 1 − 1 − 1 − 1 − 1 − 1 1 1

Determinant of the edge orientation matrix Theorem Let T be a directed tree with n vertices and let H be the edge orientation matrix of T . Then n det H = 2 n − 2 � τ i . i =1 Corollary H is nonsingular if and only if no vertex in T has degree 2 .