𝕮-inverse of graphs and mixed graphs

Definition 1

Let G ξ be a mixed graph, then

1. A mixed subgraph X ξ of G ξ is called elementary subgraph if each component of X is either edge or cycle.

2. The rank and co-rank of an elementary mixed subgraph are defined by r ( X ξ ) = n − c and s ( X ξ ) = m − r ( X ξ ) , respectively, where n = ∣ V ( X ) ∣ , c is the number of components of X , and m = ∣ E ( X ) ∣ .

3. If G has a unique perfect matching, then a path P ξ of G ξ is called m m -alternating path if the edges of P start and end with matching edges and its edges alternate between matching and unmatching edges.

4. If P : v i 1 v i 2 v i 3 … v i k is a path in G ξ , then the weight of P in the complex unit gain graph G ξ , n ϕ is defined by

   H ϕ ( P ) = ∏ j = 1 k − 1 h i j i j + 1 .

Theorem 1

Let G ξ , n ϕ be a complex unit gain graph and H ϕ be its ϕ -Hermitian adjacency matrix. Then,

where the sum ranges over spanning elementary subgraphs of G, the product is being taken over all cycles C in X and C → is any mixed closed walk traversing C.

Theorem 2

Let G ξ , n ϕ be a bipartite mixed graph with unique perfect matching ℳ , H ϕ be its ϕ -Hermitian adjacency matrix, and

Then,

Definition 2

Let G be a unicyclic bipartite graph with unique perfect matching. Then, a matching edge is called a peg if it is incident to exactly one vertex of the cycle of G .

Theorem 3

Let G be a unicycle bipartite graph with unique perfect matching. If G has more than two pegs, and H ϕ is the ϕ -Hermitian adjacency matrix of the complex unit gain graph G ξ , n ϕ , then

Theorem 4

Let G be a bipartite unicyclic graph having a unique perfect matching. Let the cycle be of length 2 m , and suppose that there are 2 k pegs. Then, G is invertible if and only if one of the following holds:

1. k ≥ 2 and m − k is even,

2. k = 1 , m is even, and the vertices on the cycle incident with pegs are adjacent.

Definition 3

Let G ξ be a mixed graph with α -Hermitian adjacency matrix H α . Then, G ξ is called C -invertible if there is ± α r diagonal matrix D and unit complex number γ , such that D H α − 1 D * is γ -Hermitian adjacency matrix of some mixed graph ( I G ) η .

Theorem 5

Let G be a graph, A be its adjacency matrix, D be ± α r diagonal matrix, where α is a primitive n th root of unity and G ξ , n ϕ is the complex unit gain graph corresponding to the matrix D A D * . Then, the weight of any cycle in G ξ , n ϕ is 1.

Proof

Suppose that C : v i 1 v i 2 v i 3 … v i k v i 1 is a cycle G ξ , n ϕ and H ϕ = D A D * , where D = diag ( d 1 , d 2 , … , d n ) . Then,

Corollary 1

Let G ∈ U , ( I G ) η be the mixed graph corresponding to the ( − 1 ) -Hermitian adjacency matrix A − 1 . Then, A − 1 is not signable similar to 0 − 1 -matrix if and only if ( I G ) η contains a cycle of weight − 1 with respect to the ( − 1 ) -Hermitian adjacency matrix H − 1 .

Example 1

In Figure 1, the graph G ∈ U with m = 3 and k = 2 . Note that, in the mixed graph ( I G ) η , the cycles C ( 1 ) : 0 , 9 , 7 , 5 , 0 and C ( 2 ) : 0 , 9 , 7 , 5 , 2 , 3 , 0 have the weight h − 1 ( C ( 1 ) ) = h − 1 ( C ( 2 ) ) = − 1 . While the weight of the cycle C ( 3 ) : 5 , 2 , 3 , 0 , 5 has the weights h − 1 ( C ( 3 ) ) = 1 , respectively. Therefore, there is no ± 1 diagonal matrix D such that D A − 1 D is ( 0 − 1 ) -matrix (Figure 2).

Theorem 6

Let T be a tree and C ′ be a cycle of ( I T ) η . Then, there are two vertices u and v of T such that V ( C ′ ) ⊂ V ( P u v ) , where P u v is the path from u to v in T .

Proof

Suppose not, then there are vertices r , s 1 , s 2 , and s 3 such that s 1 , s 2 , s 3 are vertices of the cycle C ′ and for i ≠ j , V ( P r s i ) ∩ V ( P r s j ) = { r } , where P r s i and P r s j are paths in T . Choose such s j to be of maximum distance from r . Then, there are two vertices u 1 r s 1 and u 2 r s 2 of V ( P r s 1 ) and V ( P r s 2 ) , respectively, such that u 1 r s 1 and u 1 r s 2 are adjacent in C ′ , two vertices u 2 r s 2 and u 1 r s 3 of V ( P r s 2 ) and V ( P r s 3 ) , respectively, such that u 2 r s 2 and u 1 r s 3 are adjacent in C ′ , and two vertices u 2 r s 3 and u 3 r s 1 of V ( P r s 3 ) and V ( P r s 1 ) , respectively, such that u 2 r s 3 and u 3 r s 1 are adjacent in C ′ . This contradicts the fact that r is matched by only one matching edge (Figure 3).□

Theorem 7

Let T be a tree, A be the adjacency matrix of T, and ( I T ) η be the mixed graph corresponding to ( − 1 ) -Hermitian adjacency matrix A − 1 . Then, the weight of any cycle C ′ in ( I T ) η is h − 1 ( C ) = 1 .

Proof

Let C ′ be a cycle in ( I T ) η . As T is a tree, the edge v i v j belongs to C ′ if and only if there exists an mm-alternating path between v i and v j in T . Consequently, there exists an mm-alternating path between any two consecutive vertices of C ′ in T . Utilizing Theorem 2, the weight of cycle C ′ in I T η equals ( − 1 ) a , where a = ∑ u v ∈ E ( C ′ ) n u v , and n u v denotes the number of non-matching edges between u and v in T . Given that T is a tree and mm-alternating paths start and end at the same vertex u of T , each non-matching edge between the vertices of C ′ in T is counted an even number of times, as per Theorem 6. This implies that a is even.□

Theorem 8

Let T be a tree graph with unique perfect matching and ( I T ) η be the mixed graph corresponding to ( − 1 ) -Hermitian adjacency matrix A − 1 . Then, there is ℑ : E ( I T ) → V ( I T ) × V ( I T ) ∪ E ( I T ) such that:

1. If ( A − 1 ) u v = 1 , then ℑ ( u v ) = u v .

2. For any cycle C in ( I T ) ℑ , C is self-balanced.

Proof

Fix a vertex v of T , and define ξ v : E ( T ) → V ( T ) × V ( T ) ∪ E ( T ) as follows: For every leaf u in T , starting from v , orient the non-matching edges of the path between v and u to be opposite to each other and fix the matching edges as digons (Figure 4). Now, for any unit complex number α , suppose that H α is the α -Hermitian adjacency matrix of T ξ . Then, using Lemma 3.2 in [1], there is a diagonal matrix D with α power entries, such that D A D * = H α . Using Theorem 2, one can easily see that the entries of matrix H α − 1 are from the set { 0 , ± 1 , ± α , ± α ¯ } . Furthermore, since every tree graph with perfect matching is invertible [18] and using Theorem 3, there is a ± 1 - diagonal matrix S with the property S H α − 1 S is Hermitian adjacency matrix of a mixed graph I T η for some orientation function η . Finally, using Theorem 5, ( I T ) η is α -balanced. This is true for any value of α , and thus every cycle of ( I T ) η is self-balanced.□

Theorem 9

Let G ∈ U . Suppose that { u i u i ′ } i = 1 2 k are the pegs of the G -cycle C, where u i ∈ V ( C ) . Then,

1. For any cycle C ′ of ( I G ) η with the property { u i ′ } i = 1 2 k ⊈ V ( C ′ ) , C ′ has the weight h − 1 ( C ′ ) = 1 .

2. The cycle C ′ induced by the vertices { u i ′ } i = 1 2 k in ( I G ) η has the weight h − 1 ( C ′ ) = ( − 1 ) m − k .

Proof

(1) Suppose that C ′ is a cycle in ( I G ) η and u r ′ is not a vertex of C ′ . Observing that u r u r ′ is a peg of the cycle C in G , u r is incident to two non-matching edges of C say u r t and u r s . Obviously, T = G − u r t is a tree with perfect matching. Furthermore, since v i v j is an edge of C ′ if and only if there is an mm-alternating path between v i and v j in G , we have that C ′ is a cycle in ( I T ) η . Using Theorem 7 we obtain h − 1 ( C ′ ) = 1 .

(2) Obviously, { u i } i = 1 2 k induces a cycle C ′ in ( I G ) η . Also, h − 1 ( u i u i + 1 ) = ( − 1 ) a i , where a i is the number of non-matching edges in the mm-alternating paths between u i and u ( i + 1 ) ( m o d ( 2 k ) ) . Therefore,

where c is the number of non-matching edges in the cycle C of G . Observing that C has 2 k pegs, the number of vertices incident to a matching edge e ∈ E ( C ) is 2 m − 2 k . Therefore, the number of non-matching edges of C is 2 m − 2 m − 2 k 2 = m + k .□

Corollary 2

If G ∈ U odd and A its adjacency matrix, then there is no ± 1 diagonal matrix D with the property D A − 1 D are { 0 , 1 } -matrix.

Corollary 3

Let G ∈ U odd and suppose that { u i u i ′ } i = 1 2 k are the pegs of the G -cycle C, where u i ∈ V ( C ) . Then, a cycle C ′ in ( I G ) η has the weight h − 1 ( C ′ ) = − 1 if and only if { u i ′ } i = 1 2 k ⊂ V ( C ′ ) .

Proof

Let C ′ be a cycle in ( I G ) η with the property { u i ′ } i = 1 2 k ⊂ V ( C ′ ) . Then, observing that { u i ′ } i = 1 2 k induces a cordless cycles C m in ( I G ) η , for each i , { u i ′ u ( i + 1 ) ( m o d ( 2 k ) ) ′ } is either a chord in C ′ or an edge in C ′ . Therefore, C ′ ∪ { C m } can be partitioned into cycles { C s ′ } s = 1 j each of which contains exactly one edge of C m (Figure 5). Now, suppose that C r ′ : { u i ′ u ( i + 1 ) ( m o d ( 2 k ) ) ′ v 1 v 2 … u i ′ } is one of these cycles, then

Using Theorem 9, h − 1 ( C r ′ ) = 1 . Therefore, h − 1 ( u i ′ u ( i + 1 ) ( m o d ( 2 k ) ) ′ ) h − 1 ( v 1 v 2 … u i ′ ) = 1 , and thus h − 1 ( C ′ ) = h ( C m ) = 1 . Finally, using Theorem 9, we obtain the result.□

Theorem 10

Let G ∈ U odd with 2 k pegs. Then, G is C -invertible.