= Solution
Factor the <characteristic polynomial> as $\lambda^2(\lambda^2-3)$. The <eigenvalues>, with algebraic multiplicity, are therefore
$$
\boxed{\sqrt3,\ -\sqrt3,\ 0,\ 0}.
$$
The <adjacency matrix> describes a <star graph> with node 1 as the hub. For the positive leading <eigenvalue> $\lambda=\sqrt3$, write the hub component as $h$. The leaf equations $h=\lambda v_i$, for $i=2,3,4$, give equal leaf components $\ell=h/\sqrt3$. The hub equation is $3\ell=\lambda h$, which is consistent precisely because $\lambda^2=3$. Thus an <eigenvector> is $(\sqrt3,1,1,1)^\mathsf T$.
Normalizing its <Euclidean norm> to one gives the <eigenvector centrality>
$$
\boxed{v=\frac1{\sqrt6}(\sqrt3,1,1,1)^\mathsf T=\left(\frac1{\sqrt2},\frac1{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right)^\mathsf T}.
$$
Numerically this is $(0.707107,0.408248,0.408248,0.408248)^\mathsf T$. It satisfies both $Av=\sqrt3v$ and $v^\mathsf Tv=1$. If “normalized” instead means maximum component one, use $(1,1/\sqrt3,1/\sqrt3,1/\sqrt3)$; if it means components summing to one, divide $(\sqrt3,1,1,1)$ by $\sqrt3+3$. The hub-to-leaf ratio $\sqrt3$ is invariant under these conventions. The negative leading-magnitude <eigenvalue> also has an <eigenvector>, but its mixed signs do not define nonnegative <eigenvector centrality>.
Back to article page
