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Eigenvector centrality assigns nonnegative vertex scores proportional to the sum of neighbouring scores: $v_i={\lambda }_{\max }^{-1}{\sum }_j{A}_{ij}{v}_j$. It uses a nonnegative eigenvector of the adjacency matrix for its largest positive eigenvalue. State the normalization, such as Euclidean norm one. For a three-leaf star graph, hub and leaf scores have ratio $\sqrt{3}$.

Factor the characteristic polynomial as ${\lambda }^2({\lambda }^2-3)$. The eigenvalues, with algebraic multiplicity, are thereforeThe adjacency matrix describes a star graph with node 1 as the hub. For the positive leading eigenvalue $\lambda =\sqrt{3}$, write the hub component as $h$. The leaf equations $h=\lambda v_i$, for $i=2,3,4$, give equal leaf components $\ell =h/\sqrt{3}$. The hub equation is $3\ell =\lambda h$, which is consistent precisely because ${\lambda }^2=3$. Thus an eigenvector is $(\sqrt{3},1,1,1{)}^{\mathsf{T}}$.

Normalizing its Euclidean norm to one gives the eigenvector centralityNumerically this is $(0.707107,0.408248,0.408248,0.408248{)}^{\mathsf{T}}$. It satisfies both $Av=\sqrt{3}v$ and $v^{\mathsf{T}}v=1$. If “normalized” instead means maximum component one, use $(1,1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$; if it means components summing to one, divide $(\sqrt{3},1,1,1)$ by $\sqrt{3}+3$. The hub-to-leaf ratio $\sqrt{3}$ is invariant under these conventions. The negative leading-magnitude eigenvalue also has an eigenvector, but its mixed signs do not define nonnegative eigenvector centrality.