Consider a k -regular undirected network (i.e., a network in which every vertex

Question

Consider a k-regular undirected network (i.e., a network in which every vertex has de- gree k).

1- Show that the vector 1 = (1, 1, 1, . . .) is an eigenvector of the adjacency matrix with eigenvalue k.

2- By making use of the fact that eigenvectors are orthogonal (or otherwise), show that there is no other eigenvector that has all elements positive. The Perron–Frobenius theorem says that the eigenvector with all elements positive has the largest eigen- value, and hence the eigenvector 1 gives, by definition, the eigenvector centrality of our k-regular network and the centralities are the same for every vertex.

3- Find the Katz centralities of all vertices in a k-regular network.

4- You should have found that , as with eigenvector centrality, the Katz centralities of all vertices in the network are the same. Name a centrality measure that could give different centralities for different vertices in a regular network.

Explanation / Answer

1)Consider G is a K-regular undirected network.

The adjacency matrix A of G will contain k time in a row.

Therefore, By matrix multiplication Ax=(ki,k, . . . ,k)T

=>Ax=KX

K is eigen value with 1=(1,1,….,1) The eigen vector

2)

we know eigen values are orthogonal. For any other eigen vector (a1,a2,……a0)

(1,1….1.). (a1,a2,……a0)=0

=> a1+a2+……+a0=0

=> atleast one of the ai must be negative.

Now, by perron-Feobenius theorem k is the largest eigenvalue of adjacency matrix A.

3) Katz centrality of G is given by vector

v=(IA)11,

where >0>0 is a free parameter.

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