# Addition & Product of 2 Graphs Rank and Nullity of a Graph

**Introduction :**

A Graph G consists of vertices & edges. The edges are lines or arcs that connect any two nodes in the graph, and the nodes are also known as vertices.

A simple Graph G = (V, E) consists of :

- V : Finite set of vertices
- E : Set of edges

Example –

In the diagram below, Graph G consists of :

V = { A, B, C, D}

E = { {A, C}, {C, B}, {A, D}}

**Addition Of 2 Graphs :**

If we have 2 graphs, G_{1} & G_{2} such that their vertices intersection is null ( V(G_{1})∩ V(G_{2}) = ∅ ) , then the sum :

G_{1} + G_{2} is defined as the graph whose vertex set V(G_{1}+G_{2}) is V(G_{1}) + V(G_{2}) and the edge set consists of these edges, which are inG_{1} & in G_{2} & the edges contained by joining each vertex of G_{1} to each vertex of G_{2}.

Example : The addition of 2 graphs shown G_{1} & G_{2} are :

Here : V(G_{1})∩ V(G_{2}) = ∅

The already contained edges in G_{1} are, E(G_{1}) : {{A, B}} and the vertices are : V(G_{1}) = {A, B}

The already contained edges in G_{2} are, E(G_{2}) : {{A’, B’} ,{B’,C’} }and the vertices are : V(G_{2}) = {A’, B’, C’}

So the graph , G_{1} + G_{2} will have

(i) vertices as : V(G_{1}+G_{2} ) = V(G1) + V(G_{2}) = { A, B. A’, B’, C’}

(ii) and E(G_{1} + G_{2} ) = E(G_{1}) + E(G_{2}) + edges contained by joining each vertex of G_{1} to each vertex of G_{2} =

{ {A, B}, {A’, B’} ,{B’,C’} , {A,A’} , {A, B’}, {A, C’}, {B,A’} , {B, B’}, {B, C’}}

**Product of 2 Graphs :**

We define the product of 2 graphs, G1*G2, as (G_{1}*G_{2})(V, E) such that :

(i) Vertices : V(G_{1}*G_{2} ) = Cartesian product of V(G_{1}) & V(G_{2}) = V(G_{1}) * V(G_{2}) and

(ii) Edges : Considering any 2 vertices in the set of vertices of the graph G_{1}*G_{2} (i.e, the cartesian product of V(G_{1}) & V(G_{2}) ) , say A & V (Note : A & V are vertex that is a pair of 2 elements) such that : A = (a1, a2) & V = (v1,v2) , then {A, V} is an edge in graph G_{1}*G_{2} if one of the following is satisfied –

(i) a1 = v1 ( the first element of the pair is same) and a2 is adjacent to v2

(ii) a2 = v2 ( the second element of the pair is same) and a1 is adjacent to v1

Example : Consider 2 Graphs, G_{1} & G_{2} such that :

V(G_{1}) = {A, B}

E(G_{1}) = { {A, B} }

V(G_{2}) = {A’, B’, C’}

E(G_{2}) ={ {A’, B’} ,{B’,C’} }

Then Graph G_{1}*G_{2} will have :

(i) Vertices : V(G_{1}*G_{2} ) = cartesian product of V(G_{1}) & V(G_{2}) = V(G_{1}) * V(G_{2}) = { (A, A’) , (A, B’), (A, C’), (B, A’) , (B, B’), (B, C’) }

(ii) Edges : We need to check for every pair of vertex in G_{1} * G_{2} that it can form an edge / not . As we know, that if we have 2 vertices A & V in (G_{1} * G_{2}) such that : A = (a1, a2) & V = (v1,v2) , then {A, V} is an edge in graph G_{1}*G_{2} if one of the following is satisfied –

(i) a1 = v1 (first element of pair is same) AND a2 is adjacent to v2

(ii) a2 = v2 (second element of pair is same) AND a1 is adjacent to v1

We find that :

1. { (A, A’) , (A, B’) } is an edge because the first element of the pair is same (A = A) and A’ is adjacent to B’ in G_{2}

2. { (A, C’) , (A, B’) } is an edge because the first element of the pair is same (A = A) and C’ is adjacent to B’ in G_{2}

3. { (B, A’) , (B, B’) } is an edge because the first element of the pair is same (B = B) and A’ is adjacent to B’ in G_{2}

4. { (B, C’) , (B, B’) } is an edge because the first element of the pair is same (A = A) and C’ is adjacent to B’ in G_{2}

5. { (A, A’) , (B, A’) } is an edge because the second element of the pair is same (A’ = A’) and A is adjacent to B in G_{1}

6. { (A, B’) , (B, B’) } is an edge because the second element of the pair is same (B’ = B’) and A is adjacent to B in G_{1}

7. { (A, C’) , (B, C’) } is an edge because the second element of the pair is same (C’ = C’) and A is adjacent to B in G_{1}

So E(G_{1}*G_{2}) = { { (A, A’) , (A, B’) } , { (A, C’) , (A, B’) } , { (B, A’) , (B, B’) } , { (B, C’) , (B, B’) } , { (A, A’) , (B, A’) } , { (A, B’) , (B, B’) } , { (A, C’) , (B, C’) } }

In this way we can find the addition & product of any 2 graphs.

**Rank & Nullity of a Graph :**

Let G(V,E) be a graph with n vertices & m edges and K Components.

i.e.; |G(V)| = n & |G(E)| = m we define the rank P(G) & nullity μ(G) as follows :

If G Is not connected : P(G) = Rank of G = n - k μ(G) = Nullity of G = m - n + k If G Is connected : P(G) = Rank of G = n - 1 μ(G) = Nullity of G = m - n + 1

Example 1 :The Graph shown below is connected :

|G(V)| = n = 4 &

|G(E)| = m = 3

P(G) = Rank of G = n – 1 = 4 -1 = 3

μ(G) = Nullity of G = m – n + 1 = 3 – 4 + 1 = 0

Example 2 : The Graph shown below is not connected :

|G(V)| = n = 6 &

|G(E)| = m = 4 &

No of components = k = 2

P(G) = Rank of G = n – k = 6 – 2 = 4

μ(G) = Nullity of G = m – n + k = 4 – 6 + 2 = 0

## Please

Loginto comment...