# Heuristic understanding of the eigenvalue equality between two Neumann quantum graph

Lemma: Let $$G$$ be a quantum graph (not necessarily connected) with two vertices $$v_{1}$$ and $$v_{2}$$ with the Neumann conditions imposed on it. Modifying the graph G by merging the two vertices into one to obtain the graph $$G'$$, then, $$\lambda_{n}(G) \leq \lambda_{n}(G') \leq \lambda_{n+1}(G)$$.

It is mentioned that an equality between an eigenvalue of $$G$$ and an eigenvalue of $$G'$$ exists if the eigenspace of $$G$$ contains an eigenfunction whose values at $$v_{1}$$, $$v_{2}$$ are equal.

In an attempt to show the above, here's my effort:

I take the author to mean that the eigenfunction of a graph $$G$$ to be the set of all eigenfunction in the graph. Let $$Eig(G)$$ contain the eigenfunction $$f$$. By hypothesis, suppose that $$f(v_{1}) = f(v_{2}) = \alpha \in \mathbb{F}$$. Since $$f$$ is an eigenfunction, $$\exists A \in \mathbb{R}^{n \times n}, \lambda \in \mathbb{F} : Af = \lambda f$$. Following the consequence of our hypothesis, $$A \alpha = \lambda \alpha$$.

Edit:

$$Af(v_{1}) = \lambda_{1}f(v_{1})$$ and $$Af(v_{2}) = \lambda_{2}f(v_{2})$$ implies $$\alpha = \lambda_{1} = \lambda_{2}$$

Is this correct? Otherwise, any hints are appreciated.