Graphs are propular to visualize a problem . Matrix representation is use to convert the graph in a form that used by the computer . This will help to get the efficent solution also provide a lots of mathematical equation .

1. Created by ,
1. Pathik De
2. Subhajit Pal
3. Biswanath Nag
4. Rounak Biswas

2. A seminar performed by RPSBR , conducted by Sanchita
ma’am .

3.  Graph is a set of some dots (known as Vertex ) and some line
which connect those dots (knows as Edge)
 The basic need of graph a better understanding , broader
meaning and easy to grab complexity
 So we use Graphs to visually illustrate relationships in the data
 This need further complicated
 Graph stands as backbone of many real world problem
 Such as network flow problem

4. Situation comes where we need a better way out to convert
the graphical notation into digital or algorithmic format
This need creates the area of this topic . { MATRIX
REPRESENTAION OF GRAPH } .
The mechanism of this representation should be effective ,
efficient and easy to use
Because we are all face a problem of limited resources

5. The general idea is that the visual graph can’t be used in a
system for work
The increasing no of complexity can not anymore operable
with pen and paper manner
The demand of graph in various sector
Various operation performed on the graph
Want a efficient output quickly

6. As matrix is widely used in mechanical manipulations
Also it has efficient to define operation with various
operators
Various previously solved properties also helps a lot
Most important , Matrix is convenient for defining graph
So this chapter is all about
 Formation of matrix
 Different kind of observation on that matrix

7. Some general properties should be kept in mind .
Followings are the terms that we will carry out in the
whole presentation
So its good to remember those things :
1) Incidence
2) Adjacent
3) Degree
4) Walk ,Path , Circuit
5) Fundamental Circuit
6) Cut set
7) Spanning tree
8) Rank and Nullity
9) Ring Sum operation
10) Isomorphism

8.  Each edge has two end points / vertex (Self loop
contains one).
 For ex , E1 as two end points {A,B} . There A and
B are called INCIDENT with respect to E1 edge
.
A B
C D
E1
E2
E3
E4
E5
E6
 Two end points of an edge are called ADJACENT to each other
 For ex , node A and D are said to adjacent
 The total count of incident , is known as DEGREE of a graph (Self loop
counts 2 )
 For ex , Degree(A) = 3 and Degree(B)=4
 IN A GRAPH , WE USE :
 V- VERTEX SET ; E – EDGE
SET

9. While traversing a graph , walk is
An alternating sequence of vertex and edge ;
 Edge repetition is not allowed
For ex , A-E2-C-E4-D-E5-A-E1-B
Path is a type of open walk
 Vertex repetition is not allowed
For ex , A-E2-C-E4-D
Circuit is a kind of closed walk
 Node repetition is allowed for only start node
For ex , A-E2-C-E4-D-E5-A
A B
C D
E1
E2
E3
E4
E5
E6

10.  In a connected graph G , Spanning tree is a subgraph
 Which contains all vertices of G
 Gives a minimal edge connected subgraph
 Edges of the spanning tree , called Branch
 Others are known as Chord .
 For ex , G” is a spanning tree .
 Brunch = n-1 = 4-1=3
 Chord = e-n+1= 6-4+1 = 3
 Chords equipped a interesting property
A B
C D
E1
E2
E3
E4
E5
E6
A B
C D
E3
E4
E1
G
G”

11. With a given Spanning tree and its chord .
 A circuit formed ,introducing every single chord .
 Those circuits are known as Fundamental circuits .
 For ex , Spanning tree of the graph is -
E4
E5
E6
E3
E2
E1
E7
E1
G
G”
 Now G” is spanning tree of G
 In G :– v = 5 , e =7
 In G”: – v = 5 ,
 Brunch = 4 & Chord = 3
 Chords are { E3 , E6 , E2 }

12. E4
E5
E6
E3
E2
E1
E7
E1
Fundamental Circuits :
E3
E2
E6
Original
Graph
Spannin
g tree

13.  Those two term comes from general
observations
 The no of component ‘k’ never cross ‘n’
 So , n ≥ 𝑘
 Rank = n-k
 Another , The no of edge ‘e’ is never
less than ‘n’
 So e ≥ n = e ≥ n - k
 Here , Nullity = e – n + k
 For ex , for the above graph ‘G’
 Rank = 4-2=2
 Nullity = 4-4+2 = 2
 Think , For connected Graph
 Rank = Branch and Nullity = Chord
 So , Total edge = Rank + Nullity
A B
C D
E1
E2
E3
E4
G
E
F

14. It’s a set which contains edge , such that
Whose removal make the graph disconnected
Also point that the set must be minimal
So, no further sub-set obtain from the cut set
In the ex , set {E1 , E5 , E6 , E3} is a cut set ,
Observed that no proper subset is Cut set .
A
D
E
C
B
E1
E2
E3
E4
E5
E6
E7
E8

15.  Drawing of the graph is not fix
 So , there are multiple view of a single graph
 Need rules for detect two same graph
 If match found , two are called Isomorphic
 Two graphs are isomorphism , if
 No of edge in both graph same
 No of vertex same
 Adjacency relation holds
 For ex , Following two are isomorphic in nature .

16.  A graph operation act as follows :
 Let G1 G2 be two graph , with {V1,E1} , {V2,E2} respectively
 Then G3 = Ring_Sum(G1,G2) ,As
 V3 = V1 ∪ V2
 E3 = Either in E1 or in E2 , but not in both
 For ex ,
A
B
A
E
D
C
B
D F
C
E1 E2
E3 E4
E5
E2E1
A
B C
D
E
FG1 G2
G3
m

18.  With a given graph G , Adjacency or connection matrix is a n by n
symmetric binary matrix .
 Consider graph has no parallel edge and self loop , because those
are not carry redundant information .
 Then each element of the graph define as ,
 Xi,j = 1 if ith and jth vertex are end point of an edge .
 Else = 0

19. A
D
E
C
B
0 1 0 1 1
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
1 1 0 1 0
A
B
C
D
E
A B C D
E

20.  All entity in principal diagonal is 0 , iff the graph has no self
loop
A
D
E
C
B
0 1 0 1 1
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
1 1 0 1 0
A
B
C
D
E
A B C D
E

21.  A graph with no parallel edge and self loop
 In its adjacent matrix , degree of a vertex define by no of 1 in row or
column .
A
D
E
C
B
0 1 0 1 1
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
1 1 0 1 0
A
B
C
D
E
A B C D
E
Degree of A - 3

22.  Permutation of any two row and equivalent column gives us Two
isomorphic graph
0 1 1 0
1 0 1 0
1 1 0 1
0 0 1 0
A B C D
A
B
C
D
0 0 1 1
1 0 1 0
1 1 0 1
0 0 1 0
A B C D
A
B
C
D
A
DC
B

23. 0 0 1 1
1 0 1 0
1 1 0 1
0 0 1 0
A B C D
A
B
C
D
0 0 1 1
0 0 1 0
1 1 0 1
1 0 1 0
A B C D
A
B
C
D
A
DC
B

24. Both are
Isomorphic

25.  Let G be a disconnected graph
 g1 and g2 be two component
 The adjacency matrix A partitioned as follows :
X(G) =
X(g1)
X(g2)0
0
 Where X(g1) and X(g2) are adjacent matrix of component g1 and g2
respectively

26. A
G
FE
CD
B
0 1 0 1 0 0 0
1 0 1 0 0 0 0
0 1 0 1 0 0 0
1 0 1 0 0 0 0
0 0 0 0 0 1 1
0 0 0 0 1 0 1
0 0 0 0 1 0 1
A
B
C
D
E
F
G
A B C D E F G

27. &
Change in character : (Part 2 …)

28. Let G be a graph with n vertices and e edges. And there is no self-loop.
The incidence matrix A (G) is an v × e matrix= [a i j ]
Whose n rows correspond to the v vertices and e columns correspond to the e
edges
The Matrix element , a i j =1 ,if edge e j is incident on vertices vi
=0 , otherwise
Incidence Matrix

29. v1
v3
v5v2
v6
v4
e1
e2
e6
e4
e3
e5
G
e1 e2 e3 e4 e5 e6
V1
V2
V3
V4
V5
V6
1 0 0 1 0
0
0 0 0 0 0 0
1 1 0 0 1 1
0 1 1 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
A(G)

30. OBSERVATIONS :-
1. Since every edge is incident on exactly two vertices, each column of A has
exactly Two 1’s.
v1
v3
v5
v2
v6
v4
e1
e2
e6
e4
e3
e5
G
e1 e2 e3 e4 e5 e6
V1
V2
V3
V4
V5
V6
1 0 0 1 0
0
0 0 0 0 0 0
1 1 0 0 1 1
0 1 1 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
A(G)

31. 2. The number of 1’s in each row equals the degree of the
corresponding vertex
v1
v3
v5
v2
v6
v4
e1
e2
e6
e4
e3
e5
G
e1 e2 e3 e4 e5 e6
V1
V2
V3
V4
V5
V6
1 0 0 1 0
0
0 0 0 0 0 0
1 1 0 0 1 1
0 1 1 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
A(G)
2

32. 3. A row with all zeros represents an isolated vertex.
v1
v3
v5v2
v6
v4
e1
e2
e6
e4
e3
e5
G
e1 e2 e3 e4 e5 e6
1 0 0 1 0
0
0 0 0 0 0 0
1 1 0 0 1 1
0 1 1 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
V1
V2
V3
V4
V5
V6

33. 4. Parallel edges in a graph produce identical columns in its incidence matrix.
v1
v3
v5v2
v6
v4
e1
e2
e6
e4
e3
G
e1 e2 e3 e4 e5 e6
1 0 0 1 0
0
0 0 0 0 0 0
1 1 0 0 1 1
0 1 1 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
A(G)
e5

34. 5. If a graph is disconnected and consists of two components
g1and g2 the incidence matrix A(G) of graph G can be written
in a block diagonal form as
A(G)
A(g1)
0
0
A(g2)
Where A(g1) and A(g2) are the incidence matrices of components g1 and g2. This
observation results the fact that no edge in g1 incident on vertices of g2 ,and vice versa
Obviously this remarks is also true for a disconnected graph with any number of
components.
Take an example to understand this property

35. g1
g2
v1
v3
v2
e3
e2
e1
e4
V4 V5
e1 e2 e3 e4 e5
V1
V2
V3
V4
V5
1 0 1 0 0
1 1 0 0 0
0 1 1 0 0
0 0 0 1 1
0 0 0 1 1
A(G)
A(g1)
0
0
A(g2)
G
No edge in g1 incident on vertices
of g2 and vice versa .

36. 6.Permutation of any two rows or columns in an incidence matrix simply
corresponds to relabelling the vertices and edges of the same graph
v1
v3
v2
e3
e2
e1
e1 e2 e3
V1
V2
V3
1 0
1
1 1 0
0 1
1
e1 e2 e3
V3
V2
V1
0 1 1
1 1 0
1 0 1

37. Path Matrix
A path matrix is defined for a specific pair of vertices in a graph ,say (x , y)
And is written as P(x ,y).
And The rows in P(x,y) correspond the different paths between vertices x & y and
columns correspond to the edges in G.
The path matrix for (x,y) vertices is P(x,y)=[pij ]
Pij = 1 if jth edge lies in ith path
= 0 , otherwise.

38. v1
v3
v2
v4
e1
e2
e4
e3
G = e5
e1 e2 e3 e4 e5
1 1 1 0 0
0
0 0 0 0
1
0 0 1 1
0
P(v1 ,v3 ) =
v1 v2
e1
v3
e2
v1
v3
e5
v1
v3v4
e4
e3
1
2
3
2
3

39. OBSERVATIONS
1. A column of all zeros corresponds to an edge that does not lie in any path between
x and y.
v1
v3
v2
v4
e1
e2
e4
e3
e5
v5 e1 e2 e3 e4 e5
e6
1 1 1 0 0 0
0
0 0 0 0 1
0
0 0 1 1 0
0
P(v1 ,v3 )
2
3
e6

40. 2. A column of all ones corresponds to an edge that lies in every path between x
and y.
v1
v3
v2
v4
e1
e2
e4
e3
e5
v5
e1 e2 e3 e4 e5 e6
1 1 1 0 0 0
1
0 0 0 0 1
1
0 0 1 1 0
1
P(v5 ,v3 )
2
3
e6

41. 3. There is no row with all zeros.
v1
v3
v2
v4
e1
e2
e4
e3
e5
v5
e1 e2 e3 e4 e5 e6
1 1 1 0 0 0
1
0 0 0 0 1
1
0 0 1 1 0
1
P(v5 ,v3 )
2
3
e6
There is no edge in that path. But it
is not possible to make path without
any edge

42. 4. The ring sum of any two rows in P(x, y) corresponds to a circuit
v1
v3
v2
v4
e1
e2
e4
e3
e5
v5
e1 e2 e3 e4 e5 e6
1 1 1 0 0 0
1
0 0 0 0 1
1
0 0 1 1 0
1
P(v5 ,v3 )
2
3
e6
1 3
v1
v3
v2
v4
e1
e2
e4
e3
e1
1
e2
1
e3
1
e4
1
e5
0
e6
0

43. Journey of life is a Circuit , two path to archive the
Goal . Choose wisely . ( Part 3 begins …)

44. Like adjacency or incident matrix , it is an another way
of representing a graph.
It is a circuit x edge matrix .From this matrix we can
identify which edge is belonged to which circuit of a graph
Formal defination:- Let q be the total number of circuits of graph
G(V,E) and the number of edges in G is e . Then the circuit matrix
of the graph G , B(G) =[bi j ] is a q x e matrix defined as
follows
bi j =1 , if ith circuit includes jth edge , and
=0 ,otherwise

45. e1
e2
e3
e4
e5
e6
circuit matrix
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0
1 1 0 1 1
1
e1
e2
e3 e3
e4
e5
e6
e1
e2
e4
e5
e6
G(V,E) (e1 e2 e3 )
( e3 e4 e5 e6 )
(e1 e2 e4 e5 e6 )
B(G) (Circuit matrix of G )

46. OBSERVATIONS
Circuit matrix
1. A column of all zeros corresponds to a non-circuit edge (i.e., a n edge that
does not belong to any circuit).
e1
e2
e3
e4
e6
G(V,E)
e4 e6e1 e2 e3 e7e5
(e1 e2 e3 )
0 0 1 1 1 1
0
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0
0
1 1 0 1 1 1
0
B(G) (Circuit matrix of G )
e5
e7

47. circuit
matrix
2.Unlike the incidence matrix, a circuit matrix is capable of representing a
self-loop-the corresponding row will have a single 1.
e1
e2
e3
e4
e5
e6
G 1(V,E)
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0
1 1 0 1 1
1
B(G)
e1
e2
e3
e4
e5
e6
G(V,E)
e7
e4 e6e1 e2 e3 e7
e5
(e1 e2 e3 )
0 0 1 1 1 1
0
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0
0
1 1 0 1 1 1
0
B(G1 )
(e7 ) 0 0 0 0 0 0
1

48. circuit
matrix
4. The number of 1's in a row is equal to the number of edges in the
corresponding circuit.
e1
e2
e3
e4
e5
e6
G(V,E)
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1
1
( e3 e4 e5 e6 )
1 1 1 0 0 0
1
B(G) (Circuit matrix of G )
( e1e2 e4 e5
e6) 0 11 11

49. circuit
matrix
4. If graph G is separable (or disconnected) and consists of two blocks (or
components) g1 and g2 , then circuit matrix B(G) can be written in a block-
diagonal form as
where B(g1) and B(g2) are the circuit matrices of g1 and g2.
This observation results from the fact that the circuits in g1 have no edges belonging to g2
and vice versa.
Take an example to understand this property in better way

50. e1
e2
e3
e4
e5
e6
g1
e7
e8
e9
e10
e11
g2
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1 0 0 0 0
0
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0 0 0 0 0
0
1 1 0 1 1 1 0 0 0 0
0
e7 e8 e9 e10 e11
(e7 e8 e9 e10 )
( e8 e9 e11 )
(e7 e10 e11 )
0 0 0 0 0 0 1 1 1 1
0
0 0 0 0 0 0 0 1 1 0
1
0 0 0 0 0 0 1 0 0 1
1
B(G)
G

51. e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1 0 0 0 0
0
( e3 e4 e5 e6 )
1 1 0 1 1 1 0 0 0 0
0
e7 e8 e9 e10 e11
( e8 e9 e11 )
(e7 e10 e11 )
0 0 0 0 0 0 1 1 1 1
0
0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 1
1
1 1 1 0 0 0 0 0 0 0
0
(e1 e2 e4 e5 e6 )
0 1 1 0
1
B(g1)
0
0
B(g2)
The circuits in g1 have no edges belonging to g2
and vice versa.
(e7 e8 e9 e10 )

52. 5. Permutation of any rows or columns in circuit matrix simply corresponds to
relabeling the circuits and edges
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 1 1 0 0 0
1 1 0 1 1
1
e4 e6e1 e5 e3 e2
(e1 e2 e3 )
0 1 1 1 0 1
(e1 e2 e4 e5 e6 )
( e3 e4 e5 e6 )
1 0 1 0 1 0
1 1 0 1 1
1

53. circuit matrix
Relationship between Incident matrix (A) and circuit
matrix (B) of a self-loop free graph G
● Let A and B be ,respectively ,the circuit matrix and the incident matrix (of a self-loop free
graph) whose columns are arranged using same order of edges. Then every row of A orthogonal
to every row of B ; That is
A . BT = B . AT = 0 (mod 2)
e1
e2
e3
e4
e5
e6
G(V,E)
e1 e2 e3 e4 e5 e6
V1
V2
V3
V4
V5
1 1 0 0 0 0
0 1 1 1 0 0
0 0 0 1 1 0
0 0 0 0 1 1
1 0 1 0 O
1
V1
V2
V3
V4 V5
e4 e6e1 e2 e3 e5
(e1 e2 e3 )
0 0 1 1 1 1( e3 e4 e5 e6 )
1 1 1 0 0 0
1 1 0 1 1 1
B(G) (Circuit matrix of G )
(e1 e2 e4 e5 e6 )
A(G)( Incidence matrix of G)

54. 1 1 0 0 0 0
0 1 1 1 0 0
0 0 0 1 1 0
0 0 0 0 1 1
1 0 1 0 O
1
A
.
1 0 1
1 0 1
0 1 1
1 1 0
1 1 0
1 1 0
2 0 2
2 2
2
2 2 0
2 2
0
2 2 2
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
(mod 2)
BT

55. Now, if we know only the fundamental circuits of a graph ,then we
easily find out all other non –fundamental circuits by applying ring
sums( i.e. linear combinations) of those fundamental circuits .
Rank of a matrix:- Rank of a matrix in defined as the order of largest
square matrix whose determinate is not zero
Fundamental circuit matrix (Bf):- A sub-matrix (of a circuit matrix ,B) in which all rows
correspond to a set of all fundamental circuits.
● It is a (e-n+k)(fundamental circuit or chord) by e matrix as total chord is (e-n+k) and each
chord make a fundamental circuit.
●As in matrix B, permutations of rows (and/or of columns) of Bf do not affect

56. circuit matrix
Cont…
G
e1
e2
e3
e4
e5
e6
e
7

57. e1
e3e5
e6
e
7
e2
e4
e4
e6e1 e2 e3 e7e5
v1
v3
v4
v2
v5
(e1 e2 e4 e7 )
(e3 e4 e7 )
(e5 e6 e7)
1 1 0 1 0 0
1
0 0 1 1 0 0
1
0 0 0 0 1 1
1
circuit matrix
Rank of a matrix is defined as the order of largest square sub-
matrix whose determinate is not zero.

58. circuit matrix
Rearrange rows and columns of Bf:-
●Arrange the columns in Bf such that all the (e -n + 1) chords correspond to
the first
(e - n + 1) columns.
●Furthermore, Rearrange the rows such that the first row corresponds to
the fundamental circuit made by the chord in the first column, the second
row to the fundamental circuit made by the second, and so on.

59. e1
e3e5
e6
e
7
e2
e4
e4
e6e1 e2 e3 e7
e5
v1
v3
v4 v2
v5
(e1 e2 e4 e7 )
(e3 e4 e7 )
(e5 e6 e7 )
1 1 0 1 0 0
1
0 0 1 1 0 0
1
0 0 0 0 1 1
1
e1 e5e2 e3 e6 e7
e4
(e3 e4 e7 )
(e5 e6 e7 )
1 0 0 1 1 0
1
0 1 0 0 1 0
1
0 0 1 0 0 1
1
(e1 e2 e4 e7 )
[ Bf ]
circuit matrix

60. circuit matrix
We can write matrix Bf (arranged) can be written as, Bf = [Iµ Bt ]
● Iµ is an identity matrix of order µ = e - n +1 ,
● Bt is the remaining µ by (n - I) sub-matrix, corresponding to the
branches of the spanning tree.
e1 e5e2 e3 e6 e7e4
(e3 e4 e7 )
(e5 e6 e7 ) 0 0 1 0 0 1
1
0 1 0 0 1 0
1
1 0 0 1 1 0
1
(e1 e2 e4 e7 )
Bf =[ Iµ Bt ]
Here Iµ is the largest square(3 x 3) sub-matrix .
Iµ is an identity matrix whose determinate is always 1 t.e. non-zero
Rank of Bf is 3
● the rank of Bf= µ = e - n + I.
● Since Bf is a sub-matrix of the circuit matrix
B, then , rank of Circuit matrix(B) ≥ e - n + I.

61. Calculate the cut set of friends . Place them in your
heart and others ; leave in past |( Chapter 4 begins …
)

62.  Let G be a graph with m edges and q cut-sets. The cut-set matrix C =
[ci j]q×m of G is a (0,1)-matrix with
1 , if ith cut set contains jth edge
0 , else
Ci,j=

63. a b c d
f e c7
c5
e1
e2
c6
c1
e8
c8
e6
e5
e7 e3
e4
c4 c3
c2
C(G1)=
c1
c2
c3
c4
c5
c6
c7
c8
e1 e2 e3 e4 e5 e6 e7 e8
0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 0 0 0 1 1 1 0
0 0 1 0 0 1 1 0
0 0 0 1 0 1 0
00 0 1 1 0 0 1 0
0 0 0 1 1 0 1 0

64.  The permutation of rows or columns corresponds simply renaming of
the cut-sets and edges respectively.
c1
c2
c3
c4
c5
c6
c7
c8
e1 e2 e3 e4 e5 e6 e7 e8
0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 0 0 0 1 1 1 0
0 0 1 0 0 1 1 0
0 0 0 1 0 1 0
00 0 1 1 0 0 1 0
0 0 0 1 1 0 1 0
C(G1)=
a b c d
f e c7
c5
e1
e2
c6
c1
e8
c8
e6
e5
e7 e3
e4
c4 c3
c2

65.  Column with all zeros corresponds to an edge forming a self-loop.
a b c d
f e c7
c5
e1
e2
c6
c1
e8
c8
e6
e5
e7 e3
e4
c2e9
C(G1)=
c1
c2
c3
c4
c5
c6
c7
c8
e1 e2 e3 e4 e5 e6 e7 e8
0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 0 0 0 1 1 1 0
0 0 1 0 0 1 1 0
0 0 0 1 0 1 0
00 0 1 1 0 0 1 0
0 0 0 1 1 0 1 0
e9
0
0
0
0
0
0
0
0

66.  Parallel edges form identical columns in the cut-set matrix.
a b c d
f e c7
c5
e1
e2
c6
c1
e8
c8
e6
e5
e7 e3
e4
c4 c3
C(G1)=
c1
c2
c3
c4
c5
c6
c7
c8
e1 e2 e3 e4 e5 e6 e7 e8
0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 0 0 0 1 1 1 0
0 0 1 0 0 1 1 0
0 0 0 1 0 1 0
00 0 1 1 0 0 1 0
0 0 0 1 1 0 1 0

67.  In non separable graph
 Since every set of edges incident on a vertex is a cut-set
 Therefore every row of incidence matrix A(G) is included as a row in
the cut-set matrix C(G).
 So C(G) contains A(G)

68.  For non separable graph
 The incidence matrix of each block is contained in the cut-set
matrix.
 Ex. In the graph the incidence matrix of the block {e3,e4,e5,e6,e7} is
4X5 submatrix of C, left after deleting row c1,c2,c5,c8 and columns
e1,e2,e8.
a b c d
f e c7
c5
e1
e2
c6
c1
e8
c8
e6
e5
e7 e3
e4
c4 c3
c2
C(G1)=
c1
c2
c3
c4
c5
c6
c7
c8
e1 e2 e3 e4 e5 e6 e7 e8
0 0 0 0 0 0 0 1
1 1 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 0 0 0 1 1 1 0
0 0 1 0 0 1 1 0
0 0 0 1 0 1 0
00 0 1 1 0 0 1 0
0 0 0 1 1 0 1 0

69.  Since the number of edges common to a cut set and circuit is always
even ,each row in C is orthogonal to every row is B, provided the
edge in both B and C are arranged in same order. In other words
B.CT=C.BT=0 (mod 2)

70.  From the last observation we can conclude that -
 The rank C(G)>= rank A(G)
 Therefore for connected graph of n vertices
 Rank C(G) >= n−1. ------------------(1)

71. So, applying Sylvester’s theorem to equation (1) we get
rank B+ rank C <=e
Now for a connected graph
Rank B=e-n+1 .
Rank C <= n−1. ----------------------------(2)
 Combining equation (1) and (2)
 We get rank C=n-1
Sylvester Theorem
If A and B are matrices of order k×m and n×p respectively,
then nullity AB ≤ nullity A+ nullity B

72.  The rank of the cut set matrix is equal to the rank of the incidence
matrix A(G), which equals the rank of graph G.
 Soln..As if circuit matrix, the cut set matrix generally has many
redundant(or linearly dependent) rows, therefore we can define Cf
(fundamental cut set matrix).
 So Cf is n-1 by e sub matrix of C such that each row correspondent to
the set of fundamental cut set of some spanning tree.
so Cf = [Cc|In-1 ]

73.  Bf=[Iμ| Bf]
 Cf=[Cc|In-1]
Where t denotes the sub matrix corresponding to the branches of a
spanning tree.
And subscript c denote the sub matrix corresponding to the chords.
b c d a e f g h
1 0 0 1 0 0 0 0
0 1 0 0 1 0 0 0
0 0 1 0 0 1 0 0
0 1 1 0 0 0 1 0
0 0 0 0 0 0 0 1
Cf=
h e
a
b
f g
d
c

74.  Similarly we can partition the fundamental incidence matrix Af into two
sub matrices.
 Af=[Ac|At] where At consist of n-1 columns of branches and Ac with rest of
the sub matrix i.e. e-n+1 number of chords.
 Since the columns in Af and Bf are arranged in the same order, the
equation
 ABT =BAT = 0(mod 2) gives Af BT
f equivalent to 0(mod 2),
Iµ
 OR AC ⁞ At … is equivalent to 0(mod 2)
BT
f
 Ac +AtBT
f is equivalent to 0(mod 2).

75.  Since At is non singular, A−1
t exists. Now, multiplying both sides of
equation by A−1
t , we get
 A−1
t Ac +A−1
t AtBT
t is equivalent to 0(mod 2),
 or A−1
t Ac +BT
t is equivalent to 0(mod 2).
 Therefore, A−1
t Ac = −BT
t .
 Since in mod 2 arithmetic −1 = 1,
 BT
t = A−1
t Ac.
Af
0 0 1 0 0 1 0 0
0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 1
1 1 0 1 1 0 0 0
0 1 1 0 0 0 1 0
Ac
At
1 0 0 0 0
0 1 0 1 0
0 0 1 1 0Bt
1 0 0
0 1 0
0 0 1
0 1 1
0 0 0
At
-1Ac= Hence At
-1Ac=Bt
T

76.  Now as the columns in Bf and Cf are arranged in the same order,
therefore
 (in mod 2 arithmetic) Cf . Bt
f is equivalent to 0(mod 2) in mod 2
arithmetic gives
 Cf .BT
f= 0. (as −1 = 1 in mod 2 arithmetic).
 Hence, Cc = A−1
t Ac

77.  We make the following observations from the above relations.
 1. If A or Af is given, we can construct Bf and Cf starting from an arbitrary
spanning tree and its sub matrix At in Af .
 2. If either Bf or Cf is given, we can construct the other. Therefore, since
Bf determines a graph within 2-isomorphism, so does Cf .
 3. If either Bf and Cf is given, then Af in general cannot be determined
completely.

78. Everything comes to earth with a lifespan . Choice is your’s ,
live it or waste it . If you waste , Please don’t blame God in
the END Game .

79. We are describe the following topic :
1. Need of a graph
2. Need of computer processing
3. Presentation of a graph with various properties
4. Observe their characteristics

80. Our topic is collected from the book
 “ GRAPH THEORY – BY NARSINGH DEO ”
“Matrix representation of Graph” – chapter 7 .

81. Beside of that book majority of our idea and concept comes
through Internet , links are as follows :
 math.com
 youtube.com
 wikipedia.org
 cs.cmu.edu
 nptel.ac.in
 quora.com
 cs.xu.edu

82.  "We can always find something to be thankful for, and there may be
reasons why we ought to be thankful for even those dispensations
which appear dark and frowning." - Albert Barnes

83. শুভ
শারদীযা