Generalized Inverses in the Study of Graphs

Graphs, Matrices and Generalized Inverse

Dr. Manjunatha Prasad Karantha [email protected], [email protected] Professor of Mathematics Department of Statistics, PSPH, MAHE, Manipal Coordinator Center for Advanced Research in Applied Mathematics and Statistics (CARAMS), MAHE, Manipal

December 10-21, 2018 Contents

1 Introduction

2 Some Properties of Incidence Matrices

3 Minors of incidence matrix

4 Moore–Penrose Inverse of Incidence Matrix

5 Properties of 0-1 Incidence Matrix

6 Generalized inverse of Laplacian Matrix

7 More results on generalized inverses and graphs

8 References

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 2 / 50 Introduction Notations ℝ : set of real numbers ℝ푚×푛 : set of real matrices of size 푚 × 푛 ℝ푛 : set of real column vectors of order 푛 ퟏ : (1, … , 1)푇 {ퟏ} : span of ퟏ For 퐴 ∈ ℝ푚×푛, 퐴푇 : transpose of 퐴 퐴∗ : conjucate transpose of 퐴 (퐴) : column space of 퐴 (퐴) : (퐴푇 ) rank(퐴) : rank of the matrix 퐴 퐴푖 : 푖th row of 퐴 퐴푗 : 푗th column of 퐴 For any two subspaces 푆, 푇 of ℝ푛, 푆 ⊕ 푇 : 푆 + 푇 with 푆 ∩ 푇 = {0} 푆 ⟂ 푇 : 푆 and 푇 are orthogonal to each other 푆 ⦹ 푇 : 푆 + 푇 with 푆 ⟂ 푇

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 3 / 50 Introduction Preliminaries

Given an 푚 × 푛 matrix 퐴 over a real ﬁeld, consider the following matrix equations, called Penrose Equations;

(1) 퐴푋퐴 = 퐴 (2) 푋퐴푋 = 푋 (3) (퐴푋)푇 = 퐴푋 (4) (푋퐴)푇 = 푋퐴.

A matrix 퐺 satisfying the condition(s) ∙ (1) → generalized inverse of 퐴, denoted by 퐴− ∙ (2) → outer inverse of 퐴, denoted by 퐴= ∙ (1) − (4) → Moore-Penrose inverse of 퐴, denoted by 퐴+

{퐴−} = {퐺 + (퐼 − 퐺퐴)푈 + 푉 (퐼 − 퐴퐺) ∶ 퐺 is any g-inverse of 퐴, 푈 and 푉 are arbitrary matrices} (1)

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 4 / 50 Introduction Preliminaries

Deﬁnition (Space Equivalence) For any two matrices 퐴 and 퐵, ∙ 퐴 ≃ 퐵 if (퐴) = (퐵) 퐶푆 ∙ 퐴 ≃ 퐵 if (퐴) = (퐵) 푅푆 ∙ 퐴 ≃ 퐵 if (퐴) = (퐵) and (퐴) = (퐵) 푆푃

Lemma For any matrix 퐴, 퐴+ ≃ 퐴푇 . 푆푃

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 5 / 50 Introduction Preliminaries

Lemma For given non-null matrices 퐴, 퐵 and 퐶 in ℝ푚×푛, the following are equivalent. (i) 퐵퐴−퐶 is invariant under the choices of 퐴− (ii) (퐶) ⊆ (퐴) and (퐵) ⊆ (퐴).

Sketch of the proof. (푖) ⟹ (푖푖). Let 퐺 ∈ {퐴−}. For 퐴− = 퐺 + (퐼 − 퐺퐴)푈 + 푉 (퐼 − 퐴퐺), where 푈, 푉 are arbitrary, 퐵퐴−퐶 = 퐵퐺퐶 implies that 퐵(퐼 − 퐺퐴)푈퐶 = 0 = 퐵푉 (퐼 − 퐴퐺)퐶. Hence, 퐵(퐼 − 퐺퐴) = 0 and (퐼 − 퐴퐺)퐶 = 0, proving (푖푖). (푖푖) ⟹ (푖). If 퐶 = 퐴푋 and 퐵 = 푌 퐴 for some 푋 and 푌 , then

퐵퐴−퐶 = 푌 퐴퐴−퐴푋 = 푌 퐴푋,

independent of 퐴−. □ K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 6 / 50 Introduction Some matrices associated with graphs

Incidence Matrix

Adjacency Graph Distance Matrix Matrix

Laplacian Matrix

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 7 / 50 Introduction Incidence Matrix

퐺 푉 퐺 , , 푛 퐸 퐺 푒 , , 푒 Let be a graph with ( ) = {1 … } and ( ) = { 1 … 푚}. For any directed graph 퐺, the incidence matrix, denoted by 푄(퐺), is the 푛 × 푚 matrix deﬁned as

⎧ 푖 푒 ⎪ 0 vertex and edge 푗 are not incident (푄(퐺))푖푗 = ⎨ 1 푒푗 originates at 푖 . 1 ⎪ ⎩−1 푒푗 terminates at 푖 푒 푒 푒 1 2 3 The incidence matrix of the graph in Figure 2 4 1.1 is, 3 푒 푒 푒 푒 푒 푒 푒 푒 4 푒 6 1 2 3 4 5 6 5 1⎛ −1 1 −1 0 0 0 ⎞ 2⎜ 1 0 0 −1 0 0 ⎟ 5 푄 = 3⎜ 0 −1 0 0 1 0 ⎟ ⎜ ⎟ Figure 1.1: Directed Graph 4⎜ 0 0 1 0 0 −1 ⎟ 5⎝ 0 0 0 1 −1 1 ⎠

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 8 / 50 Introduction Some important facts about 푄(퐺)

For any graph 퐺, 푄 퐺 ∙ Always the column sum of ( ) is zero. 푒 푒 푒 푒 Therefore, the rows are dependent. 1 2 3 4 1⎛ −1 1 −1 0 ⎞ ∙ Rank of 푄(퐺) ≤ 푛 − 1. 2⎜ 1 0 0 −1 ⎟ ⎜ ⎟ ∙ Consider any left null vector 푥 of 푄(퐺). Then, 3⎜ 0 −1 0 0 ⎟ 4⎜ 0 0 1 0 ⎟ 푇 ⎝ ⎠ 푥 푄(퐺) = 0 ⟹ 푥푖 − 푥푗 = 0 if 푖 ∼ 푗. 5 0 0 0 1

⟹ 푥푖 = 푥푗∀푖, 푗, if 퐺 is connected

Lemma If 퐺 is a connected graph on 푛 vertices, then rank 푄(퐺) = 푛 − 1.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 9 / 50 Introduction Rank of 푄(퐺)

Lemma For any directed graph 퐺 with 푘 components, the rank of 푄(퐺) = 푛 − 푘.

Sketch of the proof. 퐺 , , 퐺 퐺 Let 1 … 푘 be the connected components of . Then, after a relabeling of vertices(rows) and edges (columns) if necessary, we have

푄 퐺 ⎡ ( 1) 0 … 0 ⎤ ⎢ 0 푄(퐺 ) … 0 ⎥ 푄(퐺) = ⎢ 2 ⎥ . ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎣ 0 0 … 푄(퐺푘)⎦

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 10 / 50 Introduction Rank of 푄(퐺) (contd.)

Since 퐺푖 is connected, rank 푄(퐺푖) is 푛푖 − 1, where 푛푖 is the number of vertices in 퐺푖, 푖 = 1, … , 푘. It follows that

푄 퐺 푄 퐺 ⋯ 푄 퐺 rank ( ) = rank ( 1) + + rank ( 푘) 푛 ⋯ 푛 = ( 1 − 1) + + ( 푘 − 1) 푛 ⋯ 푛 = 1 + + 푘 = 푛 − 푘

This completes the proof. □

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 11 / 50 Introduction Adjacency Matrix

퐺 푉 퐺 , , 푛 퐸 퐺 푒 , , 푒 Let be a graph with ( ) = {1 … } and ( ) = { 1 … 푚}. The adjacency matrix of 퐺, denoted by 퐴(퐺), is the 푛 × 푛 matrix deﬁned as follows. { 1 if vertex 푖 and vertex 푗 are adjacent (퐴(퐺))푖푗 = . 0 otherwise

1 푒 푒 푒 1 2 3 4 5 1 2 3 1⎛ 0 1 1 1 0 ⎞ 2 4 ⎜ ⎟ 3 2⎜ 1 0 0 0 1 ⎟ 퐴(퐺) = 3⎜ 1 0 0 0 1 ⎟ 푒 푒 푒 4 5 6 4⎜ 1 0 0 0 1 ⎟ ⎜ ⎟ 5⎝ 0 1 1 1 0 ⎠ 5

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 12 / 50 Introduction 0–1 Incidence Matrix

퐺 푉 퐺 , , 푛 퐸 퐺 푒 , , 푒 Let be a graph with ( ) = {1 … } and ( ) = { 1 … 푚}. The (vertex-edge) incidence matrix of 퐺, which we denote by 푀(퐺), or simply by 푀, is the 푛 × 푚 matrix deﬁned as follows. { 0 vertex 푖 and edge 푒푗 are not incident (푀(퐺))푖푗 = 1 otherwise.

1 The 0–1 incidence matrix of the graph in Figure 1.2 is, 푒 푒 푒 1 2 3 푒 푒 푒 푒 푒 푒 1 2 3 4 5 6 2 4 3 1⎛ 1 1 1 0 0 0 ⎞ 2⎜ 1 0 0 1 0 0 ⎟ 푒 푒 푒 푀 퐺 ⎜ ⎟ 4 5 6 ( ) = 3⎜ 0 1 0 0 1 0 ⎟ 4⎜ 0 0 1 0 0 1 ⎟ 5⎝ 0 0 0 1 1 1 ⎠ 5

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs Figure 1.2: Directed06 December, Graph 2018 13 / 50 Introduction Degree Matrix

The degree matrix of a graph is a diagonal matrix with the vertex degrees on its diag- 1 onal. 푒 푒 푒 1 2 3 1 2 3 4 5 2 4 1⎛ 3 0 0 0 0 ⎞ 3 2⎜ 0 2 0 0 0 ⎟ ⎜ ⎟ 푒 푒 퐷푒푔 퐺 3 0 0 2 0 0 4 푒 6 ( ) = ⎜ ⎟ 5 4⎜ 0 0 0 2 0 ⎟ ⎝ ⎠ 5 0 0 0 0 3 5

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 14 / 50 Introduction Laplacian Matrix

The laplacian matrix of a given graph 퐺 is, denoted by 퐿(퐺), is given by

⎧ 푖 ≠ 푗, 푖 푗 ⎪ −1 vertex and vertex are adjacent (퐿(퐺))푖푗 = ⎨ 0 푖 ≠ 푗 vertex 푖 and vertex 푗 are not adjacent . ⎪ ⎩ 푑푖 i=j

1 2 3 4 5 6 1⎛ 3 −1 0 −1 −1 0 ⎞ ⎜ ⎟ ∙1 ∙2 ∙3 2⎜ −1 3 −1 0 −1 0 ⎟ 3 0 −1 2 0 −1 0 퐿(퐺) = ⎜ ⎟. 4⎜ −1 0 0 2 −1 0 ⎟ ⎜ ⎟ ∙4 ∙5 ∙6 5⎜ −1 −1 −1 −1 5 −1 ⎟ 6⎝ 0 0 0 0 −1 1 ⎠

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 15 / 50 Introduction Laplacian Matrix (contd.)

Facts about 퐿(퐺) ∙ 퐿(퐺) = 푄(퐺)푄(퐺)푇 , considering any arbitrary directions for the edges ∙ 퐿(퐺) = 퐷푒푔(퐺) − 퐴(퐺) ∙ rank(퐿(퐺)) = rank(푄(퐺)) = 푛 − 푘, where 푘 is the number of component of the graph 퐺

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 16 / 50 Introduction Some important results

Results For any connected (directed) graph 퐺 on 푛 vertices, ♣ (푄(퐺)) ⦹ {ퟏ} = ℝ푛 ♣ (퐿(퐺)) = (퐿(퐺)) and (퐿(퐺)) ⦹ {ퟏ} = ℝ푛 퐺 퐺 푛 For any graphs 1 and 2 on vertices, ♣ 퐿(퐺 ) ≃ 퐿(퐺 ) 1 푆푃 2

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 17 / 50 Introduction Distance Matrix

∙1 ∙2 ∙3 Distance 푑(푖, 푗) between the vertices 푖 and 푗 of a connected graph 퐺 is the length of a shortest path from 푖 to 푗. ∙4 ∙5 ∙6

푑(1, 5) = 1. Distance Matrix 1 2 3 4 5 6 1⎛ 0 1 2 1 1 2 ⎞ The distance matrix of a connected 2⎜ 1 0 1 2 1 2 ⎟ graph, denoted by 퐷(퐺), is given by, ⎜ ⎟ 3⎜ 2 1 0 2 1 2 ⎟ 4⎜ 1 2 2 0 1 2 ⎟ (퐷(퐺))푖푗 = 푑(푖, 푗). ⎜ ⎟ 5⎜ 1 1 1 1 0 1 ⎟ 6⎝ 2 2 2 2 1 0 ⎠

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 18 / 50 Some Properties of Incidence Matrices Some Properties of Incidence Matrices Some Properties of Incidence Matrices

Lemma 퐺 푛 푗 , , 푗 푄 퐺 Let be a graph on vertices. Columns 1 … 푘 of ( ) are linearly independent if and only if the corresponding edges of 퐺 induce an acyclic graph.

A matrix is said to be totally unimodular if the determinant of any square submatrix of the matrix is either 0 or ,1. It is easily proved by induction on the order of the submatrix that 푄(퐺) is totally unimodular as seen in the next result. Lemma Let 퐺 be a graph with incidence matrix 푄(퐺). Then 푄(퐺) is totally unimodular. Sketch of the proof.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 20 / 50 Some Properties of Incidence Matrices Some Properties of Incidence Matrices (contd.)

Consider the statement that any 푘 × 푘 submatrix of 푄(퐺) has determinant 0 or ,1. We prove the statement by induction on 푘. Clearly the statement holds for 푘 = 1, since each entry of 푄(퐺) is either 0 or ,1. Assume the statement to be true for 푘 − 1 and consider a 푘 × 푘 submatrix 퐵 of 푄(퐺). If each column of 퐵 has a 1 and a −1, then 푑푒푡퐵 = 0. Also, if 퐵 has a zero column, then 푑푒푡퐵 = 0. Now, suppose 퐵 has a column with only one nonzero entry, which must be ,1. Expand the determinant of 퐵 along that column and use induction assumption to conclude that det B must be 0 or ,1. □ Lemma Let 퐺 be a tree on 푛 vertices. Then any submatrix of 푄(퐺) of order 푛 − 1 is nonsingular.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 21 / 50 Some Properties of Incidence Matrices Some Properties of Incidence Matrices (contd.)

Incidence Vector of a Path Let 퐺 be a graph with the vertex set 푉 퐺 , , , 푛 퐸 퐺 푒 , , 푒 푃 ( ) = {1 2 … } and the edge set ( ) = { 1 … 푚}. Given a path in 퐺, the incidence vector of 푃 is an 푚 × 1 vector deﬁned as follows. The entries of the vector are indexed by 퐸(퐺). If 푒푖 ∈ 퐸(퐺) then the 푖th element of the vector is 0 if the path does not contain 푒푖. If the path contains 푒푖 then the entry is 1 or −1, according as the direction of the path agrees or disagrees, respectively, with 푒푖. Let 퐺 be a tree with the vertex set {1, 2, … , 푛}. We identify a vertex, say 푛, as the root. The path matrix 푃푛 of 퐺 (with reference to the root 푛) is deﬁned as follows. The 푗th column of 푃푛 is the incidence vector of the (unique) path from vertex 푗 to 푛, 푗 = 1, … , 푛 − 1.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 22 / 50 Some Properties of Incidence Matrices Some Properties of Incidence Matrices (contd.)

Theorem Let 퐺 be a tree with the vertex set {1, 2, … , 푛}. Let 푄 be the incidence matrix of 퐺 and let 푄푛 be the reduced incidence matrix obtained by deleting row 푛 of −1 푄. Then 푄푛 = 푃푛.

Deﬁnition (Unimodular matrix) A square integer matrix is called unimodular if its determinant is ,1.

Theorem Let 퐴 be an 푛 × 푛 integer matrix. Then 퐴 is nonsingular and admits an integer inverse if and only if 퐴 is unimodular.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 23 / 50 Some Properties of Incidence Matrices Some Properties of Incidence Matrices (contd.)

Theorem (Smith Normal Theorem)

Let 퐴 be an 푚 × 푛 integer matrix. Then there exist unimodular matrices 푆 and 푇 of order 푚 × 푚 and 푛 × 푛, respectively, such that 푧 ⋯ ⋯ ⎡ 1 0 0 ⎤ ⎢ 푧 ⋯ ⋯ ⎥ 0 2 0 ⎢ ⋮ ⋱ ⋮ ⎥ ⎢ ⎥ 푆퐴푇 = ⎢ 0 ⋯ 푧푟 ⋯ 0 ⎥ ⎢ 0 ⋯ 0 0 ⎥ ⎢ ⎥ ⎢ ⋮ ⋱ ⋮ ⋮ ⎥ ⎣ 0 ⋯ 0 0 ⎦ 푧 , , 푧 퐴 where 1 … 푟 are positive integers (called the invariant factors of ) such 푧 푧 푖 , , , 푟 푧 ⋯ 푧 푑 푑 that 푖 divides 푖+1, = 1 2 … − 1. Furthermore, 1 푖 = 푖, where 푖 is the greatest common divisor of all 푖 × 푖 minors of 퐴, 푖 = 1, … , min{푚, 푛}. K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 24 / 50 Theorem Let 퐺 be a graph with vertex set 푉 (퐺) = {1, 2, … , 푛} and edge set 푒 , , 푒 푒 , , 푒 퐺 { 1 … 푚}. Suppose the edges 1 … 푛−1 form a spanning tree of . Let 푄 푄 , , 푛 1 be the submatrix of formed by the rows 1 … − 1 and the columns 푒 , , 푒 푞 푚 푛 . 푄 1 … 푛−1. Let = − + 1 Partition as follows: [ ] 푄 푄 푁 푄 1 1 . = 푇 푄 푇 푄 푁 −1 1 −1 1 Set [ ] 푄−1 0 퐵 = 1 , 0 0 [ ] [ ] 푄−1 퐼 푁 0 푛−1 − 푆 = 1푇 , 푇 = , 1 1 0 퐼푞 [ ] 푄 [ ] 퐹 1 , 퐻 퐼 푁 . = 푇 푄 = 푛−1 −1 1 Then the following assertions hold: (푖) 퐵 is an integer reﬂexive g-inverse of 푄. 푖푖 푆 푇 ( ) and [are unimodular] matrices. 퐼 0 (푖푖푖) 푆푄푇 = 푛−1 is the Smith normal form of 푄. 0 0 (푖푣) 푄 = 퐹 퐻 is an integer rank factorization of 푄. Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix

Lemma If 퐴 is an 푚 × 푛 matrix, then for an 푛 × 1 vector 푥, 퐴푥 = 0 if and only if 푥푇 퐴+ = 0.

Lemma 퐺 퐼 푄푄+ 1 퐽 If is connected, then − = 푛 .

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 26 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.)

Construction of Moore-Penrose Inverse : Let 퐺 be a graph with 푉 퐺 , , , 푛 퐸 퐺 푒 , , 푒 푒 , , 푒 ( ) = {1 2 … } and ( ) = { 1 … 푚}. Suppose the edges 1 … 푛−1 form a spanning tree of 퐺. Partition 푄 as follows: [ ] 푄 = 푈 푉

where 푈 is 푛 × (푛 − 1) and 푉[ is 푛 ×] (푚 − 푛 + 1). Note that, 푉 = 푈퐷 for some 퐷, 푈 has left inverse and 푈 퐼 퐷 is a rank factorization of 푄 for some 퐷. Since 푟푎푛푘(푄) = 푟푎푛푘(푈) = 푛 − 1, [ ] [ ] 퐼 푋 푄+ 퐼 퐷퐷푇 −1 푈 푇 푈 −1푈 푇 = 퐷푇 ( + ) ( ) = 퐷푇 푋

for 푋 = (퐼 + 퐷퐷푇 )−1(푈 푇 푈)−1푈 푇 .

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 27 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.)

Let 푀 = 푄푄+. Then [ ] [ ] 푋 [ ] 푀 푄푄+ 푈 퐼 퐷 푈 퐼 퐷퐷푇 푋. = = 퐷푇 푋 = +

Hence, [ ] 퐼 1퐽 푈 퐼 퐷퐷푇 푋. − 푛 = + Hence, Thus, for any 푖, 푗,

푇 푈푖(퐼 + 퐷퐷 )푋푗 = 푀(푖, 푗),

where 푈푖 is 푈 with row 푖 deleted, and 푋푗 is 푋 with column 푗 deleted.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 28 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.)

푇 By Lemma 8, 푈푖 is nonsingular. Also, 퐷퐷 is positive semideﬁnite and thus 푇 푇 퐼 + 퐷퐷 is nonsingular. Therefore, 푈푖(퐼 + 퐷퐷 ) is nonsingular and

푇 −1 푋푗 = (푈푖(퐼 + 퐷퐷 )) 1푀(푖, 푗).

Once 푋푗 is determined, the 푗th column of 푋 is obtained using the fact that 푄+1 = 0. Then 푌 is determined, since 푌 = 퐷푇 푋.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 29 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.)

We illustrate the above method of calculating 푄+ by an example. Consider the graph

푒 푒 4 3

푒 1 5 3

푒 푒 1 2

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 30 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.)

with the incidence matrix

⎡ 1 0 0 1 0 ⎤ ⎢−1 1 0 0 1 ⎥ ⎢ ⎥ ⎢ 0 −1 −1 0 0 ⎥ ⎣ 0 0 1 −1 −1⎦ 푒 , 푒 , 푒 푄 푈푉 푈 Fix the spanning tree formed by { 1 2 3}. Then = where is formed by the ﬁrst three columns of 푄. Observe that 푉 = 푈퐷, where

⎡ 1 0 ⎤ 퐷 ⎢ ⎥ . = ⎢ 1 1 ⎥ ⎣−1 −1⎦

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 31 / 50 Moore–Penrose Inverse of Incidence Matrix Moore–Penrose Inverse of Incidence Matrix (contd.) [ ] 푋 푖 푗 푄+ Set = = 4. Then = 푌 where

⎡3 −2 −1⎤ 푇 1 푋4 = (푈 (퐼 + 퐷퐷 ))−1푀(4, 4) = ⎢1 2 −3⎥ . 4 8 ⎢ ⎥ ⎣1 0 −3⎦

The last column of 푋 is found using the fact that the row sums of 푋 are zero. Then 푌 = 퐷푇 푋. After these calculations we see that

⎡3 −2 −1 0 ⎤ ⎢ ⎥ [ ] 1 2 −3 0 푋 1 ⎢ ⎥ 푄+ = = ⎢1 0 −3 2 ⎥ . 푌 8 ⎢3 0 −1 −2⎥ ⎢ ⎥ ⎣0 2 0 −2⎦

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 32 / 50 Properties of 0-1 Incidence Matrix Properties of 0-1 Incidence Matrix Properties of 0-1 Incidence Matrix

Lemma

Let 퐶푛 be the cycle on the vertices {1, … , 푛}, 푛 ≥ 3, and let 푀 be its incidence matrix. Then 푑푒푡 푀 equals 0 if 푛 is even and ,2 if 푛 is odd.

Lemma Let 퐺 be a connected graph with 푛 vertices and let 푀 be the incidence matrix of 퐺. Then the rank of 푀 is 푛 − 1 if 퐺 is bipartite and 푛 otherwise.

Sketch of the proof. Suppose 퐺 is not bipartite graph, 퐺 has a cycle of odd length. To prove 푇 푟푎푛푘(푀) = 푛, consider 푥 푀 = 0. Then 푥푖 + 푥푗 = 0 whenever the vertices 푖 and 푗 are adjacent. Since 퐺 is connected it follows that |푥푖| = 훼, 푖 = 1, … , 푛, for some constant 훼. Suppose 퐺 has an odd cycle formed by the vertices

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 34 / 50 Properties of 0-1 Incidence Matrix Properties of 0-1 Incidence Matrix (contd.)

푖 , … , 푖푘. Suppose 푥푖 = 훼, then 푥푖 = −훼, 푥푖 = 훼, … 푥푖 = 훼, which implies 1 1 2 3 푘 that 푥푖 = 푥푖 = −훼. That is 훼 = −훼, implies that 훼 = 0. Hence, 푥 = 0 푘+1 1 which proves rank of 푀 is 푛. Now suppose 퐺 has no odd cycle, that is, 퐺 is bipartite. Let 푉 (퐺) = 푋 ∪ 푌 be a bipartition. Orient each edge of 퐺 as by considering all the edges orginates from 푌 and let 푄 be the corresponding {0, 1, −1}–incidence matrix. We know that 푟푎푛푘(푄) = 푛 − 1. Also we know that 푄 = 퐷푀 where 퐷 is 푛 × 푛 diagonal matrix whose 푖th entry is −1 if 푖 ∈ 푌 , otherwise 1. Clearly 퐷 is invertible and therefore,

푛 − 1 = 푟푎푛푘 푄 = 푟푎푛푘 퐷푀 = 푟푎푛푘 푀.

Hence, the proof. □

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 35 / 50 Generalized inverse of Laplacian Matrix Generalized inverse of Laplacian Matrix Generalized inverse of Laplacian Matrix

Lemma For a tree 푇 on 푛 vertices, let 퐷 be the distance matrix, 퐿 the Laplacian matrix and 휏 is a column vector with 휏푖 = 2 − 푑푖, where 푑푖 is the degree of the 푖th vertex. Then, 퐿퐷 + 2퐼 = 휏ퟏ푇 . Sketch of the proof. Note that, 푑 푎 푎 푑 , 푑 , 푛 ⎡ 1 − 12 … − 1푛⎤ ⎡ 0 (1 2) … (1 )⎤ ⎢−푎 푑 … −푎 ⎥ ⎢푑(2, 1) 0 … 푑(2, 푛)⎥ 퐿 = ⎢ 21 2 2푛⎥ and 퐷 = ⎢ ⎥. ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎢ ⋮ ⋮ ⋱ ⋮ ⎥ ⎣ 푎 푎 푑 ⎦ ⎣푑 푛, 푑 푛, ⎦ − 푛1 − 푛2 … 푛 ( 1) ( 2) … 0

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 37 / 50 Generalized inverse of Laplacian Matrix Generalized inverse of Laplacian Matrix (contd.)

To ﬁnd, (퐿퐷)푖푗, without loss of generality, assume that 푑푖 = 푘 and {1, 2, … , 푘} are adjacent with 푖. For 푗 = 푖,

푖 (퐿퐷)푖푖 = 퐿푖퐷 푎 푑 푖, 푎 푑 푖, ⋯ 푎 푑 푖, 푘 = − 푖1 ( 1) − 푖2 ( 2) − − 푖푘 ( ) ∙푗 = −푘 = −푑푖 = 휏푖 − 2.

For 푗 ≠ 푖, ∙1 ∙2 ∙푘 (퐿퐷) = 퐿 퐷푗 푖푗 푖 ∑ 푙 푑 푖, 푗 푎 푑 푝, 푗 = 푖푖 ( ) − 푖푝 ( ) 푖 푝≠푖 ∙ = 푘푑(푖, 푗) − (푘(푑(푖, 푗) + 1) − 2)

= 2 − 푘 = 2 − 푑푖 = 휏푖

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 38 / 50 Generalized inverse of Laplacian Matrix Generalized inverse of Laplacian Matrix (contd.)

Therefore, 퐿퐷 = 휏ퟏ푇 − 2퐼. In other words, 퐿퐷 + 2퐼 = 휏ퟏ푇 . □

Theorem For any tree 푇 and its Laplacian matrix 퐿, a generalized inverse of 퐿 is given by − 1 퐷, where 퐷 is the distance matrix of 푇 . That is, 2 ( ) 1 퐿 − 퐷 퐿 = 퐿. 2 Sketch of the proof. Proof follows from the above lemma and the fact that (퐿) = {ퟏ}⟂. □

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 39 / 50 Generalized inverse of Laplacian Matrix For 푖 ≠ 푗, deﬁne For any square matrix 푋, ⎛ 0 ⎞ ⎜ ⋮ ⎟ 푒푇 푋푒 = 푥 + 푥 − 푥 − 푥 . ⎜ ⎟ 푖푗 푖푗 푖푖 푗푗 푖푗 푗푖 ⎜ 0 ⎟ ⎜ 1 ⎟ 푖 ⎜ ⎟ ⎜ 0 ⎟ Observation 푒푖푗 = ⎜ ⋮ ⎟ . For any connected graph 퐺 on 푛 ⎜ 0 ⎟ ⎜ ⎟ vertices, ⎜−1⎟ 푗 ⎜ 0 ⎟ 푒푖푗 ∈ (푄(퐺)) = (퐿(퐺)) for all 푖 ≠ 푗 ⎜ ⋮ ⎟ ⎜ ⎟ 푒푇 ퟏ ퟏ ⟂  푄 퐺 ⎝ 0 ⎠ as 푖푗 = 0 and { } = ( ( )).

The following is immediate from the observation. Lemma For any connected graph 퐺 and its Laplacian matrix 퐿,

푇 − − 푒푖푗퐿 푒푖푗 is invariant under the choice of 퐿 . K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 40 / 50 Generalized inverse of Laplacian Matrix Weiner Index of a graph

Deﬁnition For any graph 퐺, Weiner index of 퐺, denoted by 푊 (퐺), is deﬁned by ∑ 푊 (퐺) = 푑(푖, 푗). (2) 푖<푗

Theorem If 푇 is a tree with Laplacian matrix 퐿 then

∑푛−1 푊 푇 푛 1 , 휆 ≥ 휆 ≥ 휆 > 휆 퐿. ( ) = 휆 where 1 2 … 푛−1 푛 = 0 are eigenvalues of 푖=1 푖 (3)

Sketch of the proof.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 41 / 50 Generalized inverse of Laplacian Matrix Weiner Index of a graph (contd.)

휆 ≥ 휆 ≥ 휆 > 휆 퐿 Note that, if 1 2 … 푛−1 푛 = 0 are eigenvalues of , then 1 , 1 1 퐿+ 휆 휆 … 휆 are the eigenvalues of and 1 2 푛−1

∑푛−1 퐿+ 1 . trace( ) = 휆 푖=1 푖 Now for any 푖, 푗, 푖 ≠ 푗, ( ) ( ) 푇 1 1 푒 − 퐷 푒푖푗 = − 푑(푖, 푖) + 푑(푗, 푗) − 푑(푖, 푗) − 푑(푗, 푖) = 푑(푖, 푗). 푖푗 2 2

푇 − − Since 푒푖푗퐿 푒푖푗 is invariance under choice of 퐿 ,

푇 − 푇 + 푑(푖, 푗) = 푒푖푗퐿 푒푖푗 = 푒푖푗퐿 푒푖푗 + + + + = 푙푖푖 + 푙푗푗 − 2푙푖푗 ( Since 퐿 is symmetric)

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 42 / 50 Generalized inverse of Laplacian Matrix Weiner Index of a graph (contd.)

∑ ∑ ∑ ∑ + + + 푑(푖, 푗) = 푙푖푖 + 푙푗푗 − 2 푙푖푗 푖,푗 푖,푗 푖,푗 푖,푗 = 2푛 trace(퐿+)(since 퐿+ퟏ = 0) ∑푛−1 푛 1 = 2 휆 푖=1 푖 Therefore, ∑ ∑푛−1 푊 푇 푑 푖, 푗 푛 1 푛 퐿+ . ( ) = ( ) = 휆 = trace( ) 푖<푗 푖=1 푖 □

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 43 / 50 More results on generalized inverses and graphs More results on generalized inverses and graphs Absorption Law

Let 푎, 푏 be two elements in an associative ring 푅 with outer inverses 푎= and 푏= respectively. Absorption Law (Bapat, Jain, Karantha, and Raj[1]) Absorption law (Extended absorption law) for the elements 푎 and 푏 in an associative ring 푅 with reference to the outer inverses 푎= and 푏= is

푎=(푎 + 푏)푏= = 푎= + 푏=.

Lemma (Bapat, Jain, Karantha, and Raj[1])) 푎=(푎 + 푏)푏= = 푎= + 푏= if and only if 푎=푅 ⊇ 푏=푅 and 푅푎= ⊆ 푏=푅.

If 퐴 and 퐵 are two matrices,

퐴=(퐴+퐵)퐵= = 퐴= +퐵= if and only if (퐴=) = (퐵=) and (퐴=) = (퐵=).

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 45 / 50 More results on generalized inverses and graphs Absorption Law for incidence matrices

Theorem 퐺 퐺 푛 푚 Let 1 and 1 be any two connected graphs on vertices, and with edges. 푄 푄 퐺 푄 푄 퐺 . Let 1 = ( 1) and 2 = ( 2) Then the following are equivalent 푄+ 푄 푄 푄+ 푄+ 푄+ (i) Absorption law ‘ 1 ( 1 + 2) 2 = 1 + 2 ’ holds  푄+  푄+ (ii) ( 1 ) = ( 2 )  푄  푄 (iii) ( 1) = ( 2) Sketch of the proof. Since for any connected graph 퐺, (푄(퐺)) ⦹ {ퟏ} = ℝ푛,

 푄  푄 ퟏ ⟂. ( 1) = ( 2) = { }

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 46 / 50 More results on generalized inverses and graphs Absorption Law for incidence matrices (contd.)

Therefore, from the necessary suﬃcient condition for the absorption law to be satisﬁed, (푖) holds if and only if

 푄 ⊆  푄 . ( 1) ( 2)  푄  푄 푖 ⟺ 푖푖푖 Equality holds from the fact that ( 1) = ( 2). Hence, ( ) ( ). (푖푖) ⟺ (푖푖푖) follows from the fact that 퐴 ≃ 퐴+. □ 푆푃

Corollary 퐺 퐺 푛 Let 1 and 2 are two trees on vertices. Then absorption law given in the above lemma always holds.

Sketch of the proof.  푄  푄 ℝ푛−1 □ Proof follows from the previous Theorem as ( 1) = ( 2) = .

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 47 / 50 More results on generalized inverses and graphs Absorption Law for incidence matrices (contd.)

Corollary 퐺 퐺 푛 퐿 퐿 Let 1 and 2 be any two connected graphs on vertices, and let 1 and 2 be its laplacian matrices respectively. Then the absorption law

퐿+ 퐿 퐿 퐿+ 퐿+ 퐿+ 1 ( 1 + 2) 2 = 1 + 2

Sketch of the proof. Proof follows from the above Theorem as (퐿(퐺)) = (퐿(퐺)) = {ퟏ}⟂, for any connected graph 퐺. □

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 48 / 50 References References

✓ Ravindra B. Bapat, Surender Kumar Jain, K. Manjunatha Prasad Karantha, and M. David Raj. “Outer inverses: Characterization and applications”. In: Linear Algebra Appl. 528 (Sept. 2017), pp. 171–184. ✓ Adi Ben-Israel and T. N. E. Greville. Generalized inverses: Theory and applications. Second. Springer-Verlag, Berlin, 2002. ✓ C. R. Rao and S. K. Mitra. Generalized inverses of matrices and applications. Wiley, New York, 1971. ✓ Ravindra B Bapat. Graphs and matrices. Springer, 2010.

K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 49 / 50 Thank You