Regular Graphs

Abstract

Recall that a k-regular graph is one in which every vertex has degree k. Thus, every row sum (and hence every column sum) of its adjacency matrix A is k. We have seen (see Exercise 4.5.1) that k is an eigenvalue of A. Moreover, it is easy to see that all the eigenvalues λ satisfy |λ| ≤ k. Indeed, let v = (x₁, …,x _n )^t be an eigenvector with eigenvalue λ. Then

$$\lambda v = Av$$

implies that

$$\lambda {x_i} = \sum\limits_{\left( {i,j} \right)\, \in \,E} {{x_j}}$$

. Without loss of generality, we may suppose |x₁| = max _i |x _i |. Then,

$$ \left| \lambda \right|\left| {{{x}_{1}}} \right|\leq k\left| {{{x}_{1}}} \right|, $$

, from which we infer |λ| ≤ k. A similar argument shows that if X is connected, then the multiplicity of λ₀ = k is one. In fact, the same argument shows that the multiplicity of λ₀ = k is the number of connected components of X. To see this, let v = (x₁, …, x _n )^t be an eigenvector corresponding to the eigenvalue k and without loss of generality, suppose |x₁| is maximal as before. We may also suppose x₁ > 0. Then,

$$k{x_1} = \sum\limits_{\left( {1,j} \right) \in \,E} {{x_j} \leqslant k{x_1}}$$

which means that there is no cancelation in the sum and all the x _j ‘s are equal to x₁.