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Graphs, Matrices and Generalized Inverse

Dr. Manjunatha Prasad [email protected], [email protected]

Professor of MathematicsDepartment of Statistics, PSPH, MAHE, Manipal

CoordinatorCenter 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

Notationsℝ : set of real numbersℝ𝑚×𝑛 : set of real matrices of size 𝑚 × 𝑛ℝ𝑛 : set of real column vectors of order 𝑛𝟏 : (1,… , 1)𝑇{𝟏} : span of 𝟏

For 𝐴 ∈ ℝ𝑚×𝑛,𝐴𝑇 : transpose of 𝐴𝐴∗ : conjucate transpose of 𝐴(𝐴) : column space of 𝐴(𝐴) : (𝐴𝑇 )rank(𝐴) : rank of the matrix 𝐴𝐴𝑖 : 𝑖th row of 𝐴𝐴𝑗 : 𝑗th column of 𝐴

For any two subspaces 𝑆, 𝑇 of ℝ𝑛,𝑆 ⊕ 𝑇 : 𝑆 + 𝑇 with 𝑆 ∩ 𝑇 = {0}𝑆 ⟂ 𝑇 : 𝑆 and 𝑇 are orthogonal to each other𝑆 ⦹ 𝑇 : 𝑆 + 𝑇 with 𝑆 ⟂ 𝑇


Introduction

Preliminaries

Given an 𝑚 × 𝑛 matrix 𝐴 over a real field, consider the following matrixequations, called Penrose Equations;

(1) 𝐴𝑋𝐴 = 𝐴 (2) 𝑋𝐴𝑋 = 𝑋(3) (𝐴𝑋)𝑇 = 𝐴𝑋 (4) (𝑋𝐴)𝑇 = 𝑋𝐴.

A matrix 𝐺 satisfying the condition(s)∙ (1) → generalized inverse of 𝐴, denoted by 𝐴−

∙ (2) → outer inverse of 𝐴, denoted by 𝐴=

∙ (1) − (4) → Moore-Penrose inverse of 𝐴, denoted by 𝐴+

{𝐴−} = {𝐺 + (𝐼 − 𝐺𝐴)𝑈 + 𝑉 (𝐼 − 𝐴𝐺) ∶ 𝐺 is any g-inverse of 𝐴,𝑈 and 𝑉 are arbitrary matrices} (1)


Introduction

Preliminaries

Definition (Space Equivalence)For any two matrices 𝐴 and 𝐵,

∙ 𝐴 ≃𝐶𝑆

𝐵 if (𝐴) = (𝐵)∙ 𝐴 ≃

𝑅𝑆𝐵 if (𝐴) = (𝐵)

∙ 𝐴 ≃𝑆𝑃

𝐵 if (𝐴) = (𝐵) and (𝐴) = (𝐵)

LemmaFor any matrix 𝐴, 𝐴+ ≃

𝑆𝑃𝐴𝑇 .


Introduction

Preliminaries

LemmaFor given non-null matrices 𝐴, 𝐵 and 𝐶 in ℝ𝑚×𝑛, the following areequivalent.

(i) 𝐵𝐴−𝐶 is invariant under the choices of 𝐴−

(ii) (𝐶) ⊆ (𝐴) and (𝐵) ⊆ (𝐴).

Sketch of the proof.(𝑖) ⟹ (𝑖𝑖). Let 𝐺 ∈ {𝐴−}. For 𝐴− = 𝐺 + (𝐼 −𝐺𝐴)𝑈 + 𝑉 (𝐼 − 𝐴𝐺), where𝑈, 𝑉 are arbitrary, 𝐵𝐴−𝐶 = 𝐵𝐺𝐶 implies that𝐵(𝐼 − 𝐺𝐴)𝑈𝐶 = 0 = 𝐵𝑉 (𝐼 − 𝐴𝐺)𝐶 . Hence, 𝐵(𝐼 − 𝐺𝐴) = 0 and(𝐼 − 𝐴𝐺)𝐶 = 0, proving (𝑖𝑖).(𝑖𝑖) ⟹ (𝑖). If 𝐶 = 𝐴𝑋 and 𝐵 = 𝑌 𝐴 for some 𝑋 and 𝑌 , then

𝐵𝐴−𝐶 = 𝑌 𝐴𝐴−𝐴𝑋 = 𝑌 𝐴𝑋,

independent of 𝐴−. □K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 6 / 50

Introduction

Some matrices associated with graphs

Graph

IncidenceMatrix

AdjacencyMatrix

LaplacianMatrix

DistanceMatrix


Introduction

Incidence Matrix

Let 𝐺 be a graph with 𝑉 (𝐺) = {1,… , 𝑛} and 𝐸(𝐺) = {𝑒1,… , 𝑒𝑚}.For any directed graph 𝐺, the incidence matrix, denoted by 𝑄(𝐺), is the 𝑛 × 𝑚matrix defined as

(𝑄(𝐺))𝑖𝑗 =

⎧⎪⎨⎪⎩0 vertex 𝑖 and edge 𝑒𝑗 are not incident1 𝑒𝑗 originates at 𝑖−1 𝑒𝑗 terminates at 𝑖

.

The incidence matrix of the graph in Figure1.1 is,

𝑄 =

⎛⎜⎜⎜⎜⎝

𝑒1 𝑒2 𝑒3 𝑒4 𝑒5 𝑒61 −1 1 −1 0 0 02 1 0 0 −1 0 03 0 −1 0 0 1 04 0 0 1 0 0 −15 0 0 0 1 −1 1

⎞⎟⎟⎟⎟⎠

1

2 3 4

5

𝑒2𝑒1

𝑒5

𝑒3

𝑒4 𝑒6

Figure 1.1: Directed Graph


Introduction

Some important facts about 𝑄(𝐺)

For any graph 𝐺,∙ Always the column sum of 𝑄(𝐺) is zero.

Therefore, the rows are dependent.∙ Rank of 𝑄(𝐺) ≤ 𝑛 − 1.∙ Consider any left null vector 𝑥 of 𝑄(𝐺). Then,

𝑥𝑇𝑄(𝐺) = 0 ⟹ 𝑥𝑖 − 𝑥𝑗 = 0 if 𝑖 ∼ 𝑗.⟹ 𝑥𝑖 = 𝑥𝑗∀𝑖, 𝑗, if 𝐺 is connected

⎛⎜⎜⎜⎜⎝

𝑒1 𝑒2 𝑒3 𝑒41 −1 1 −1 02 1 0 0 −13 0 −1 0 04 0 0 1 05 0 0 0 1

⎞⎟⎟⎟⎟⎠Lemma

If 𝐺 is a connected graph on 𝑛 vertices, then rank 𝑄(𝐺) = 𝑛 − 1.


Introduction

Rank of 𝑄(𝐺)

LemmaFor any directed graph 𝐺 with 𝑘 components, the rank of 𝑄(𝐺) = 𝑛 − 𝑘.

Sketch of the proof.Let 𝐺1,… , 𝐺𝑘 be the connected components of 𝐺. Then, after a relabeling ofvertices(rows) and edges (columns) if necessary, we have

𝑄(𝐺) =

⎡⎢⎢⎢⎣𝑄(𝐺1) 0 … 0

0 𝑄(𝐺2) … 0⋮ ⋮ ⋱ ⋮0 0 … 𝑄(𝐺𝑘)

⎤⎥⎥⎥⎦ .


Introduction

Rank of 𝑄(𝐺) (contd.)

Since 𝐺𝑖 is connected, rank 𝑄(𝐺𝑖) is 𝑛𝑖 − 1, where 𝑛𝑖 is the number of verticesin 𝐺𝑖, 𝑖 = 1,… , 𝑘. It follows that

rank 𝑄(𝐺) = rank 𝑄(𝐺1) +⋯ + rank 𝑄(𝐺𝑘)= (𝑛1 − 1) +⋯ + (𝑛𝑘 − 1)= 𝑛1 +⋯ + 𝑛𝑘= 𝑛 − 𝑘

This completes the proof. □


Introduction

Adjacency Matrix

Let 𝐺 be a graph with 𝑉 (𝐺) = {1,… , 𝑛} and 𝐸(𝐺) = {𝑒1,… , 𝑒𝑚}. Theadjacency matrix of 𝐺, denoted by 𝐴(𝐺), is the 𝑛 × 𝑛 matrix defined asfollows.

(𝐴(𝐺))𝑖𝑗 =

{1 if vertex 𝑖 and vertex 𝑗 are adjacent0 otherwise

.

1

2 3 4

5

𝑒2𝑒1

𝑒5

𝑒3

𝑒4 𝑒6𝐴(𝐺) =

⎛⎜⎜⎜⎜⎜⎝

1 2 3 4 51 0 1 1 1 02 1 0 0 0 13 1 0 0 0 14 1 0 0 0 15 0 1 1 1 0

⎞⎟⎟⎟⎟⎟⎠K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 12 / 50

Introduction

0–1 Incidence Matrix

Let 𝐺 be a graph with 𝑉 (𝐺) = {1,… , 𝑛} and 𝐸(𝐺) = {𝑒1,… , 𝑒𝑚}. The(vertex-edge) incidence matrix of 𝐺, which we denote by 𝑀(𝐺), or simply by𝑀 , is the 𝑛 × 𝑚 matrix defined as follows.

(𝑀(𝐺))𝑖𝑗 =

{0 vertex 𝑖 and edge 𝑒𝑗 are not incident1 otherwise.

The 0–1 incidence matrix of the graph inFigure 1.2 is,

𝑀(𝐺) =

⎛⎜⎜⎜⎜⎝

𝑒1 𝑒2 𝑒3 𝑒4 𝑒5 𝑒61 1 1 1 0 0 02 1 0 0 1 0 03 0 1 0 0 1 04 0 0 1 0 0 15 0 0 0 1 1 1

⎞⎟⎟⎟⎟⎠

1

2 3 4

5

𝑒2𝑒1

𝑒5

𝑒3

𝑒4 𝑒6

Figure 1.2: Directed GraphK. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 13 / 50

Introduction

Degree Matrix

The degree matrix of a graph is a diagonalmatrix with the vertex degrees on its diag-onal.

𝐷𝑒𝑔(𝐺) =

⎛⎜⎜⎜⎜⎝

1 2 3 4 51 3 0 0 0 02 0 2 0 0 03 0 0 2 0 04 0 0 0 2 05 0 0 0 0 3

⎞⎟⎟⎟⎟⎠

1

2 3 4

5

𝑒2𝑒1

𝑒5

𝑒3

𝑒4 𝑒6


Introduction

Laplacian Matrix

The laplacian matrix of a given graph 𝐺 is, denoted by 𝐿(𝐺), is given by

(𝐿(𝐺))𝑖𝑗 =

⎧⎪⎨⎪⎩−1 𝑖 ≠ 𝑗, vertex 𝑖 and vertex 𝑗 are adjacent0 𝑖 ≠ 𝑗 vertex 𝑖 and vertex 𝑗 are not adjacent𝑑𝑖 i=j

.

𝐿(𝐺) =

⎛⎜⎜⎜⎜⎜⎜⎝

1 2 3 4 5 61 3 −1 0 −1 −1 02 −1 3 −1 0 −1 03 0 −1 2 0 −1 04 −1 0 0 2 −1 05 −1 −1 −1 −1 5 −16 0 0 0 0 −1 1

⎞⎟⎟⎟⎟⎟⎟⎠.

∙1 ∙2 ∙3

∙4 ∙5 ∙6


Introduction

Laplacian Matrix (contd.)

Facts about 𝐿(𝐺)

∙ 𝐿(𝐺) = 𝑄(𝐺)𝑄(𝐺)𝑇 , considering any arbitrary directions for the edges

∙ 𝐿(𝐺) = 𝐷𝑒𝑔(𝐺) − 𝐴(𝐺)

∙ rank(𝐿(𝐺)) = rank(𝑄(𝐺)) = 𝑛 − 𝑘, where 𝑘 is the number of component of thegraph 𝐺


Introduction

Some important results

ResultsFor any connected (directed) graph 𝐺 on 𝑛 vertices,♣ (𝑄(𝐺))⦹ {𝟏} = ℝ𝑛

♣ (𝐿(𝐺)) = (𝐿(𝐺)) and (𝐿(𝐺))⦹ {𝟏} = ℝ𝑛

For any graphs 𝐺1 and 𝐺2 on 𝑛 vertices,♣ 𝐿(𝐺1) ≃

𝑆𝑃𝐿(𝐺2)


Introduction

Distance Matrix

Distance 𝑑(𝑖, 𝑗) between the vertices 𝑖and 𝑗 of a connected graph 𝐺 is thelength of a shortest path from 𝑖 to 𝑗.

∙1 ∙2 ∙3

∙4 ∙5 ∙6

𝑑(1, 5) = 1.

Distance Matrix

The distance matrix of a connectedgraph, denoted by 𝐷(𝐺), is given by,

(𝐷(𝐺))𝑖𝑗 = 𝑑(𝑖, 𝑗).

⎛⎜⎜⎜⎜⎜⎜⎝

1 2 3 4 5 61 0 1 2 1 1 22 1 0 1 2 1 23 2 1 0 2 1 24 1 2 2 0 1 25 1 1 1 1 0 16 2 2 2 2 1 0

⎞⎟⎟⎟⎟⎟⎟⎠K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 18 / 50

Some Properties of Incidence Matrices



LemmaLet 𝐺 be a graph on 𝑛 vertices. Columns 𝑗1,… , 𝑗𝑘 of 𝑄(𝐺) are linearlyindependent if and only if the corresponding edges of 𝐺 induce an acyclicgraph.

A matrix is said to be totally unimodular if the determinant of any squaresubmatrix of the matrix is either 0 or ±1. It is easily proved by induction onthe order of the submatrix that 𝑄(𝐺) is totally unimodular as seen in the nextresult.

Lemma

Let 𝐺 be a graph with incidence matrix 𝑄(𝐺). Then 𝑄(𝐺) is totallyunimodular.

Sketch of the proof.



Some Properties of Incidence Matrices(contd.)

Consider the statement that any 𝑘 × 𝑘 submatrix of 𝑄(𝐺) has determinant 0 or±1. We prove the statement by induction on 𝑘. Clearly the statement holds for𝑘 = 1, since each entry of 𝑄(𝐺) is either 0 or ±1. Assume the statement to betrue for 𝑘 − 1 and consider a 𝑘 × 𝑘 submatrix 𝐵 of 𝑄(𝐺). If each column of 𝐵has a 1 and a −1, then 𝑑𝑒𝑡𝐵 = 0. Also, if 𝐵 has a zero column, then𝑑𝑒𝑡𝐵 = 0. Now, suppose 𝐵 has a column with only one nonzero entry, whichmust be ±1. Expand the determinant of 𝐵 along that column and useinduction assumption to conclude that det B must be 0 or ±1. □

Lemma

Let 𝐺 be a tree on 𝑛 vertices. Then any submatrix of 𝑄(𝐺) of order 𝑛 − 1 isnonsingular.




Incidence Vector of a Path Let 𝐺 be a graph with the vertex set𝑉 (𝐺) = {1, 2,… , 𝑛} and the edge set 𝐸(𝐺) = {𝑒1,… , 𝑒𝑚}. Given a path 𝑃 in𝐺, the incidence vector of 𝑃 is an 𝑚 × 1 vector defined as follows. The entriesof the vector are indexed by 𝐸(𝐺). If 𝑒𝑖 ∈ 𝐸(𝐺) then the 𝑖th element of thevector is 0 if the path does not contain 𝑒𝑖. If the path contains 𝑒𝑖 then the entryis 1 or −1, according as the direction of the path agrees or disagrees,respectively, with 𝑒𝑖.Let 𝐺 be a tree with the vertex set {1, 2,… , 𝑛}. We identify a vertex, say 𝑛, asthe root. The path matrix 𝑃𝑛 of 𝐺 (with reference to the root 𝑛) is defined asfollows. The 𝑗th column of 𝑃𝑛 is the incidence vector of the (unique) pathfrom vertex 𝑗 to 𝑛, 𝑗 = 1,… , 𝑛 − 1.




Theorem

Let 𝐺 be a tree with the vertex set {1, 2,… , 𝑛}. Let 𝑄 be the incidence matrixof 𝐺 and let 𝑄𝑛 be the reduced incidence matrix obtained by deleting row 𝑛 of𝑄. Then 𝑄−1

𝑛 = 𝑃𝑛.

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

Theorem

Let 𝐴 be an 𝑛 × 𝑛 integer matrix. Then 𝐴 is nonsingular and admits an integerinverse if and only if 𝐴 is unimodular.




Theorem (Smith Normal Theorem)

Let 𝐴 be an 𝑚 × 𝑛 integer matrix. Then there exist unimodular matrices 𝑆 and𝑇 of order 𝑚 × 𝑚 and 𝑛 × 𝑛, respectively, such that

𝑆𝐴𝑇 =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

𝑧1 0 ⋯ ⋯ 00 𝑧2 ⋯ ⋯ 0⋮ ⋱ ⋮0 ⋯ 𝑧𝑟 ⋯ 00 ⋯ 0 0⋮ ⋱ ⋮ ⋮0 ⋯ 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎦where 𝑧1,… , 𝑧𝑟 are positive integers (called the invariant factors of 𝐴) suchthat 𝑧𝑖 divides 𝑧𝑖+1, 𝑖 = 1, 2,… , 𝑟 − 1. Furthermore, 𝑧1⋯ 𝑧𝑖 = 𝑑𝑖, where 𝑑𝑖 isthe greatest common divisor of all 𝑖 × 𝑖 minors of 𝐴, 𝑖 = 1,… ,min{𝑚, 𝑛}.

Theorem

Let 𝐺 be a graph with vertex set 𝑉 (𝐺) = {1, 2,… , 𝑛} and edge set{𝑒1,… , 𝑒𝑚}. Suppose the edges 𝑒1,… , 𝑒𝑛−1 form a spanning tree of 𝐺. Let𝑄1 be the submatrix of 𝑄 formed by the rows 1,… , 𝑛 − 1 and the columns𝑒1,… , 𝑒𝑛−1. Let 𝑞 = 𝑚 − 𝑛 + 1. Partition 𝑄 as follows:

𝑄 =[

𝑄1 𝑄1𝑁−1𝑇𝑄1 −1𝑇𝑄1𝑁

].

Set𝐵 =

[𝑄−1

1 00 0

],

𝑆 =[𝑄−1

1 01𝑇 1

], 𝑇 =

[𝐼𝑛−1 −𝑁0 𝐼𝑞

],

𝐹 =[

𝑄1−1𝑇𝑄1

], 𝐻 =

[𝐼𝑛−1 𝑁

].

Then the following assertions hold:

(𝑖) 𝐵 is an integer reflexive g-inverse of 𝑄.(𝑖𝑖) 𝑆 and 𝑇 are unimodular matrices.

(𝑖𝑖𝑖) 𝑆𝑄𝑇 =[𝐼𝑛−1 00 0

]is the Smith normal form of 𝑄.

(𝑖𝑣) 𝑄 = 𝐹𝐻 is an integer rank factorization of 𝑄.


Moore–Penrose Inverse of IncidenceMatrix

Moore–Penrose Inverse of Incidence Matrix

Moore–Penrose Inverse of IncidenceMatrix

Lemma

If 𝐴 is an 𝑚 × 𝑛 matrix, then for an 𝑛 × 1 vector 𝑥, 𝐴𝑥 = 0 if and only if𝑥𝑇𝐴+ = 0.

Lemma

If 𝐺 is connected, then 𝐼 −𝑄𝑄+ = 1𝑛𝐽 .



Moore–Penrose Inverse of IncidenceMatrix (contd.)

Construction of Moore-Penrose Inverse : Let 𝐺 be a graph with𝑉 (𝐺) = {1, 2,… , 𝑛} and 𝐸(𝐺) = {𝑒1,… , 𝑒𝑚}. Suppose the edges 𝑒1,… , 𝑒𝑛−1form a spanning tree of 𝐺. Partition 𝑄 as follows:

𝑄 =[𝑈 𝑉

]where 𝑈 is 𝑛 × (𝑛 − 1) and 𝑉 is 𝑛 × (𝑚 − 𝑛 + 1). Note that, 𝑉 = 𝑈𝐷 for some𝐷, 𝑈 has left inverse and 𝑈

[𝐼 𝐷

]is a rank factorization of 𝑄 for some 𝐷.

Since 𝑟𝑎𝑛𝑘(𝑄) = 𝑟𝑎𝑛𝑘(𝑈 ) = 𝑛 − 1,

𝑄+ =[𝐼𝐷𝑇

](𝐼 +𝐷𝐷𝑇 )−1(𝑈𝑇𝑈 )−1𝑈𝑇 =

[𝑋

𝐷𝑇𝑋

]for 𝑋 = (𝐼 +𝐷𝐷𝑇 )−1(𝑈𝑇𝑈 )−1𝑈𝑇 .




Let 𝑀 = 𝑄𝑄+. Then

𝑀 = 𝑄𝑄+ = 𝑈[𝐼 𝐷

] [ 𝑋𝐷𝑇𝑋

]= 𝑈

[𝐼 +𝐷𝐷𝑇 ]𝑋.

Hence,𝐼 − 1

𝑛𝐽 = 𝑈

[𝐼 +𝐷𝐷𝑇 ]𝑋.

Hence, Thus, for any 𝑖, 𝑗,

𝑈𝑖(𝐼 +𝐷𝐷𝑇 )𝑋𝑗 = 𝑀(𝑖, 𝑗),

where 𝑈𝑖 is 𝑈 with row 𝑖 deleted, and 𝑋𝑗 is 𝑋 with column 𝑗 deleted.




By Lemma 8, 𝑈𝑖 is nonsingular. Also, 𝐷𝐷𝑇 is positive semidefinite and thus𝐼 +𝐷𝐷𝑇 is nonsingular. Therefore, 𝑈𝑖(𝐼 +𝐷𝐷𝑇 ) is nonsingular and

𝑋𝑗 = (𝑈𝑖(𝐼 +𝐷𝐷𝑇 ))−11𝑀(𝑖, 𝑗).

Once 𝑋𝑗 is determined, the 𝑗th column of 𝑋 is obtained using the fact that𝑄+1 = 0. Then 𝑌 is determined, since 𝑌 = 𝐷𝑇𝑋.




We illustrate the above method of calculating 𝑄+ by an example. Consider thegraph

1

2

3

4

𝑒1

𝑒4

𝑒2

𝑒5

𝑒3




with the incidence matrix

⎡⎢⎢⎢⎣1 0 0 1 0−1 1 0 0 10 −1 −1 0 00 0 1 −1 −1

⎤⎥⎥⎥⎦Fix the spanning tree formed by {𝑒1, 𝑒2, 𝑒3}. Then 𝑄 = 𝑈𝑉 where 𝑈 isformed by the first three columns of 𝑄. Observe that 𝑉 = 𝑈𝐷, where

𝐷 =⎡⎢⎢⎣1 01 1−1 −1

⎤⎥⎥⎦ .K. Manjunatha Prasad (MAHE, Manipal) G-inverses & Graphs 06 December, 2018 31 / 50



Set 𝑖 = 𝑗 = 4. Then 𝑄+ =[𝑋𝑌

]where

𝑋4 = (𝑈4(𝐼 +𝐷𝐷𝑇 ))−1𝑀(4, 4) = 18

⎡⎢⎢⎣3 −2 −11 2 −31 0 −3

⎤⎥⎥⎦ .The last column of 𝑋 is found using the fact that the row sums of 𝑋 are zero.Then 𝑌 = 𝐷𝑇𝑋. After these calculations we see that

𝑄+ =[𝑋𝑌

]= 1

8

⎡⎢⎢⎢⎢⎢⎣

3 −2 −1 01 2 −3 01 0 −3 23 0 −1 −20 2 0 −2

⎤⎥⎥⎥⎥⎥⎦.


Properties of 0-1 Incidence Matrix



Lemma

Let 𝐶𝑛 be the cycle on the vertices {1,… , 𝑛}, 𝑛 ≥ 3, and let 𝑀 be itsincidence matrix. Then 𝑑𝑒𝑡 𝑀 equals 0 if 𝑛 is even and ±2 if 𝑛 is odd.

Lemma

Let 𝐺 be a connected graph with 𝑛 vertices and let 𝑀 be the incidence matrixof 𝐺. Then the rank of 𝑀 is 𝑛 − 1 if 𝐺 is bipartite and 𝑛 otherwise.

Sketch of the proof.Suppose 𝐺 is not bipartite graph, 𝐺 has a cycle of odd length. To prove𝑟𝑎𝑛𝑘(𝑀) = 𝑛, consider 𝑥𝑇𝑀 = 0. Then 𝑥𝑖 + 𝑥𝑗 = 0 whenever the vertices 𝑖and 𝑗 are adjacent. Since 𝐺 is connected it follows that |𝑥𝑖| = 𝛼, 𝑖 = 1,… , 𝑛,for some constant 𝛼. Suppose 𝐺 has an odd cycle formed by the vertices



Properties of 0-1 Incidence Matrix (contd.)

𝑖1,… , 𝑖𝑘. Suppose 𝑥𝑖1 = 𝛼, then 𝑥𝑖2 = −𝛼, 𝑥𝑖3 = 𝛼,… 𝑥𝑖𝑘 = 𝛼, which impliesthat 𝑥𝑖𝑘+1 = 𝑥𝑖1 = −𝛼. That is 𝛼 = −𝛼, implies that 𝛼 = 0. Hence, 𝑥 = 0which proves rank of 𝑀 is 𝑛.Now suppose 𝐺 has no odd cycle, that is, 𝐺 is bipartite. Let 𝑉 (𝐺) = 𝑋 ∪ 𝑌 bea bipartition. Orient each edge of 𝐺 as by considering all the edges orginatesfrom 𝑌 and let 𝑄 be the corresponding {0, 1,−1}–incidence matrix. We knowthat 𝑟𝑎𝑛𝑘(𝑄) = 𝑛− 1. Also we know that 𝑄 = 𝐷𝑀 where 𝐷 is 𝑛 × 𝑛 diagonalmatrix whose 𝑖th entry is −1 if 𝑖 ∈ 𝑌 , otherwise 1.Clearly 𝐷 is invertible and therefore,

𝑛 − 1 = 𝑟𝑎𝑛𝑘 𝑄 = 𝑟𝑎𝑛𝑘 𝐷𝑀 = 𝑟𝑎𝑛𝑘 𝑀.

Hence, the proof.□


Generalized inverse of LaplacianMatrix

Generalized inverse of Laplacian Matrix


LemmaFor a tree 𝑇 on 𝑛 vertices, let 𝐷 be the distance matrix, 𝐿 the Laplacianmatrix and 𝜏 is a column vector with 𝜏𝑖 = 2 − 𝑑𝑖, where 𝑑𝑖 is the degree of the𝑖th vertex. Then,

𝐿𝐷 + 2𝐼 = 𝜏𝟏𝑇 .

Sketch of the proof.Note that,

𝐿 =

⎡⎢⎢⎢⎣𝑑1 −𝑎12 … −𝑎1𝑛

−𝑎21 𝑑2 … −𝑎2𝑛⋮ ⋮ ⋱ ⋮

−𝑎𝑛1 −𝑎𝑛2 … 𝑑𝑛

⎤⎥⎥⎥⎦ and 𝐷 =

⎡⎢⎢⎢⎣0 𝑑(1, 2) … 𝑑(1, 𝑛)

𝑑(2, 1) 0 … 𝑑(2, 𝑛)⋮ ⋮ ⋱ ⋮

𝑑(𝑛, 1) 𝑑(𝑛, 2) … 0

⎤⎥⎥⎥⎦.



Generalized inverse of Laplacian Matrix(contd.)To find, (𝐿𝐷)𝑖𝑗 , without loss of generality, assume that 𝑑𝑖 = 𝑘 and{1, 2,… , 𝑘} are adjacent with 𝑖.For 𝑗 = 𝑖,

(𝐿𝐷)𝑖𝑖 = 𝐿𝑖𝐷𝑖

= −𝑎𝑖1𝑑(𝑖, 1) − 𝑎𝑖2𝑑(𝑖, 2) −⋯ − 𝑎𝑖𝑘𝑑(𝑖, 𝑘)= −𝑘 = −𝑑𝑖 = 𝜏𝑖 − 2.

For 𝑗 ≠ 𝑖,

(𝐿𝐷)𝑖𝑗 = 𝐿𝑖𝐷𝑗

= 𝑙𝑖𝑖𝑑(𝑖, 𝑗) −∑𝑝≠𝑖

𝑎𝑖𝑝𝑑(𝑝, 𝑗)

= 𝑘𝑑(𝑖, 𝑗) − (𝑘(𝑑(𝑖, 𝑗) + 1) − 2)= 2 − 𝑘 = 2 − 𝑑𝑖 = 𝜏𝑖

∙𝑖

∙1 ∙2 ∙𝑘

∙𝑗



Generalized inverse of Laplacian Matrix(contd.)

Therefore,𝐿𝐷 = 𝜏𝟏𝑇 − 2𝐼.

In other words, 𝐿𝐷 + 2𝐼 = 𝜏𝟏𝑇 .□

TheoremFor any tree 𝑇 and its Laplacian matrix 𝐿, a generalized inverse of 𝐿 is givenby −1

2𝐷, where 𝐷 is the distance matrix of 𝑇 . That is,

𝐿(−12𝐷)𝐿 = 𝐿.

Sketch of the proof.Proof follows from the above lemma and the fact that (𝐿) = {𝟏}⟂. □



For 𝑖 ≠ 𝑗, define

𝑒𝑖𝑗 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0⋮010⋮0−10⋮0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

𝑖

𝑗

For any square matrix 𝑋,

𝑒𝑇𝑖𝑗𝑋𝑒𝑖𝑗 = 𝑥𝑖𝑖 + 𝑥𝑗𝑗 − 𝑥𝑖𝑗 − 𝑥𝑗𝑖.

ObservationFor any connected graph 𝐺 on 𝑛vertices,

𝑒𝑖𝑗 ∈ (𝑄(𝐺)) = (𝐿(𝐺)) for all 𝑖 ≠ 𝑗

as 𝑒𝑇𝑖𝑗𝟏 = 0 and {𝟏}⟂ = (𝑄(𝐺)).

The following is immediate from the observation.

LemmaFor any connected graph 𝐺 and its Laplacian matrix 𝐿,

𝑒𝑇𝑖𝑗𝐿−𝑒𝑖𝑗 is invariant under the choice of 𝐿−.



Weiner Index of a graph

DefinitionFor any graph 𝐺, Weiner index of 𝐺, denoted by 𝑊 (𝐺), is defined by

𝑊 (𝐺) =∑𝑖<𝑗

𝑑(𝑖, 𝑗). (2)

TheoremIf 𝑇 is a tree with Laplacian matrix 𝐿 then

𝑊 (𝑇 ) = 𝑛𝑛−1∑𝑖=1

1𝜆𝑖, where 𝜆1 ≥ 𝜆2 ≥ … 𝜆𝑛−1 > 𝜆𝑛 = 0 are eigenvalues of 𝐿.

(3)

Sketch of the proof.



Weiner Index of a graph (contd.)

Note that, if 𝜆1 ≥ 𝜆2 ≥ … 𝜆𝑛−1 > 𝜆𝑛 = 0 are eigenvalues of 𝐿, then1𝜆1, 1𝜆2… 1

𝜆𝑛−1are the eigenvalues of 𝐿+ and

trace(𝐿+) =𝑛−1∑𝑖=1

1𝜆𝑖.

Now for any 𝑖, 𝑗, 𝑖 ≠ 𝑗,

𝑒𝑇𝑖𝑗(−12𝐷)𝑒𝑖𝑗 = −1

2(𝑑(𝑖, 𝑖) + 𝑑(𝑗, 𝑗) − 𝑑(𝑖, 𝑗) − 𝑑(𝑗, 𝑖)

)= 𝑑(𝑖, 𝑗).

Since 𝑒𝑇𝑖𝑗𝐿−𝑒𝑖𝑗 is invariance under choice of 𝐿−,

𝑑(𝑖, 𝑗) = 𝑒𝑇𝑖𝑗𝐿−𝑒𝑖𝑗 = 𝑒𝑇𝑖𝑗𝐿

+𝑒𝑖𝑗= 𝑙+𝑖𝑖 + 𝑙+𝑗𝑗 − 2𝑙+𝑖𝑗 ( Since 𝐿+ is symmetric)



Weiner Index of a graph (contd.)

∑𝑖,𝑗

𝑑(𝑖, 𝑗) =∑𝑖,𝑗

𝑙+𝑖𝑖 +∑𝑖,𝑗

𝑙+𝑗𝑗 − 2∑𝑖,𝑗

𝑙+𝑖𝑗

= 2𝑛 trace(𝐿+) (since 𝐿+𝟏 = 0)

= 2𝑛𝑛−1∑𝑖=1

1𝜆𝑖

Therefore,

𝑊 (𝑇 ) =∑𝑖<𝑗

𝑑(𝑖, 𝑗) = 𝑛𝑛−1∑𝑖=1

1𝜆𝑖

= 𝑛 trace(𝐿+).

□


More results on generalized inversesand graphs

More results on generalized inverses and graphs

Absorption Law

Let 𝑎, 𝑏 be two elements in an associative ring 𝑅 with outer inverses 𝑎= and 𝑏=respectively.

Absorption Law (Bapat, Jain, Karantha, and Raj[1])Absorption law (Extended absorption law) for the elements 𝑎 and 𝑏 in anassociative ring 𝑅 with reference to the outer inverses 𝑎= and 𝑏= is

𝑎=(𝑎 + 𝑏)𝑏= = 𝑎= + 𝑏=.

Lemma (Bapat, Jain, Karantha, and Raj[1]))𝑎=(𝑎 + 𝑏)𝑏= = 𝑎= + 𝑏= if and only if 𝑎=𝑅 ⊇ 𝑏=𝑅 and 𝑅𝑎= ⊆ 𝑏=𝑅.

If 𝐴 and 𝐵 are two matrices,

𝐴=(𝐴+𝐵)𝐵= = 𝐴=+𝐵= if and only if (𝐴=) = (𝐵=) and (𝐴=) = (𝐵=).



Absorption Law for incidence matrices

TheoremLet 𝐺1 and 𝐺1 be any two connected graphs on 𝑛 vertices, and with 𝑚 edges.Let 𝑄1 = 𝑄(𝐺1) and 𝑄2 = 𝑄(𝐺2). Then the following are equivalent

(i) Absorption law ‘𝑄+1 (𝑄1 +𝑄2)𝑄+

2 = 𝑄+1 +𝑄+

2 ’ holds(ii) (𝑄+

1 ) = (𝑄+2 )

(iii) (𝑄1) = (𝑄2)

Sketch of the proof.Since for any connected graph 𝐺, (𝑄(𝐺))⦹ {𝟏} = ℝ𝑛,

(𝑄1) = (𝑄2) = {𝟏}⟂.



Absorption Law for incidence matrices(contd.)

Therefore, from the necessary sufficient condition for the absorption law to besatisfied, (𝑖) holds if and only if

(𝑄1) ⊆ (𝑄2).

Equality holds from the fact that (𝑄1) = (𝑄2). Hence, (𝑖) ⟺ (𝑖𝑖𝑖).(𝑖𝑖) ⟺ (𝑖𝑖𝑖) follows from the fact that 𝐴 ≃

𝑆𝑃𝐴+. □

CorollaryLet 𝐺1 and 𝐺2 are two trees on 𝑛 vertices. Then absorption law given in theabove lemma always holds.

Sketch of the proof.Proof follows from the previous Theorem as (𝑄1) = (𝑄2) = ℝ𝑛−1. □



Absorption Law for incidence matrices(contd.)

CorollaryLet 𝐺1 and 𝐺2 be any two connected graphs on 𝑛 vertices, and let 𝐿1 and 𝐿2be its laplacian matrices respectively. Then the absorption law

𝐿+1 (𝐿1 + 𝐿2)𝐿+

2 = 𝐿+1 + 𝐿+

2

Sketch of the proof.Proof follows from the above Theorem as (𝐿(𝐺)) = (𝐿(𝐺)) = {𝟏}⟂, forany connected graph 𝐺. □


References

References

✓ Ravindra B. Bapat, Surender Kumar Jain, K. Manjunatha Prasad Karantha, andM. David Raj. “Outer inverses: Characterization and applications”. In: LinearAlgebra Appl. 528 (Sept. 2017), pp. 171–184.

✓ Adi Ben-Israel and T. N. E. Greville. Generalized inverses: Theory andapplications. 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.


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